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Biotogy

Nerve & Muscle

Types Of Tissues

Types Of Tissues Let's consider a series of discussions on cellular physiology. For example, we will consider how muscle and nerve cells function. How does the chemical energy of

ATP (which was generated in glycoiysis, the Krebs cvcle, and oxidative phosphorylation) become converted into the mechanical movement of, say, muscle cells? How is it that the chemical energy of ATP is converted into an electrical signal that al1ows various nerves to communicate with those muscles? Before we can discuss the cellular mechanisms of muscles and nerves, we first need to consider some of the general characteristics of cells, tissues, and organs. The general body plan of an animal is fairly simple and can be divided into a number of systems that represent a variety of organs working in concert with one another. For example, one body system you are probably quite familiar with is the skeletal system. Another is the muscular system. Others are the circulatory, integumentary (skin), endocrine, nervous, and digestive systems, to name but a

few. The digestive system is formed by an alimentary canal (gastrointestinal "tube") that begins at the mouth and ends at the anus. This system is suspended within a body cavity re-ferred to as the coelom. The coelom is separated into a thoracic cavity (upper) and an abdominal cavity (lower). These two cavities are separated by the dome-shaped mass of skeletal muscle called the diaphragm. Within the thoracic cavity, one finds the lungs and the heart. The abdominal cavity contains the liver, stomach, and iatestines.

As we examine

l"hese

various systems, we will find different levels of

organization. There are individual cells, and then there are cells of a particular type which coalesce to form tissue. One example of a tissue is the layer of epithelial cells that line one of the principal organs of the alimentary canal, the

stomach. some of the simple epithelial cells within the siomach secrete hydrochloric acid (pH = 1) to aid in the digestion of food. Other epithelial cells of the stomach secrete mucus to help prevent that acid from digesting the lining of the stomach. Still other epithelial cells secrete enzymes. These epithelial cells are just one type of tissue that is involved in forming the stomach. The stomach is also composed of other types of tissue. For example, nervous tissue helps to innervate the stomach, connective tissue helps to hold the stomach in its proper position, and muscle tissue helps to propel food through the stomach. Thus, these four groups of primary tissue (epitheliai, connective, muscle, and nerve) have the abiiity to form the various organs of the body. An organ is n structure that is composed of two 0r ffLlre tissues that act in such a way as to perform a specific .function.

Epithelial Tissues ce11s in a little more detail. The epithelial tissue that constitutes the various organs of the body can be either simple epithelium (consisting of a single layer of cells) or stratified epithelium (consisting of two or more layers of cells). These epithelial celtrs come in a variety of shapes and sizes. For example, there are squamous (flat), cuboidal, and columnar epithelial cells (refer to Figure 1-1).

Let's examine the epithelial

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Nerve & Muscle

Types Of Tissues

On the lumenal side of the simple epithelial cells are projections cailed microvilli (singular, microviilus--see Figure 1-1 and Figure 1-2). These projections increase the total absorptive area of the cell (sometimes by as much as 25%). Sometimes you find specialized structures cailed cilia (singular, cilium) projecting outward on the apical surface of these cells. For example, in the respiratory tract these hair-like appendages move in a coordinated unidirectional wave to move foreign particles out of the mucous lining of the lungs and bronchial tubes.

Simple squamous epithelial cel1s

Simple columnar epithelial cells

\-J

Basal .--------\ lamina

Cuboidal and columnar epithelial cells

Stratified squamous epithelial cells (non-keratinized)

Figure l - l Types of epithelial cells.

Thebe cells are !_l-undea by a number of specialized junctions. For example, tight junctionS aet-fs a permeability barrier (see Figure 1-2). Not oniy do they prevent

the transport of protein molecules from the lumenal side of the cell towards the basolaterai side of the cell, but they also act to hold neighboring cells together.

Epithelial cells are also held together by structures called desmosomes (see Figure 1-2) One type of desmosome joins the epithelial cell to a structure on the basal side of the ceil called the basal lamina (or basement membrane). The basal iamina is in close contact with connective tissue that helps to anchor the cells in place.

Gap junctions provide a means for water-soluble molecules to pass from the cytoplasm of one cel1 to the cytoplasm of another cell (see Figure 1-2). These lunclions allow for equilibration within the connected epithelial celis and therefore allow those celis to function as a unit. For example, the beating cilia appear to be coordinated by waves of calcium, which flow in the plane of the juxtaposed epithelial cells.

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Biology

Nerve & Muscle

Types Of Tissues

LUMEN Apical surface

€ (=l

Tight junctions

Microvillus Tightjunctions

Gap

.lunctlon Desmosome Desmosomes Basal lateral surface

C) €f

BLOOD

Basal iamina (basement membrane)

Figure 1.2 Different components of an epithelial cell.

As we have mentioned, epithelial cells can secrete substances into a lumenal space. For example, hydrochloric acid can be secreted into the lumen of the stomach. If a cell secretes a substance into the lumen by way of a duct, it is referred to as an exocrine gland. Endocrine glands secrete substances into the blood. For example, insulin is a protein hormone secreted into the blood by clusters of specialized epithelial cells in the pancreas. Dead skin cells

&r \t-

(keratinizedl

I I noio..-ur f cells

)

Basement membrane

Collagen fibers

] conn..,,u. tissue J

t)

Figure 1.3 Stratified squamous epithelial

ce1ls.

Stratified squamous epithelium usualiy has a protective function. Your skin is composed of many layers of stratified squamous epithelial cells. The outer cells of your skin are dead, and they contain a large amount of the fibrous protein keratin (Figure 1-3). These cells are constantly being lost and replaced, as cells begin to move toward the surface from beiow.

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Biology

Nerve Er Muscle

Types Of fissues

Consider a segment of skin. This organ comprises about 15o,i, of r-our total body weight. The epidermal region contains stratified epithelial cells that act to protect the deeper layers of the skin. Belou'the epidermis is the dermis. Within the dermis are a variety of structures. Surrounding the hair follicles are erector muscles, which act to straighten the hair shaft. This causes the skrn close to the hair follicle to become depressed and gives the characteristic appearance of "goose pimples." Those erector muscles are innervated biz nerves which cause them to contract at specific times (e.g., when it is cold outside). The skin is also a highiy vascularized organ. When it is hot outside, the blood is shunted towards the surface of the skin where it can dissipate some of its heat to the outside environment. Below the dermis is the subcutaneous tissue. This is where one

finds adipose deposits.

Connective fissues This type of tissue helps to anchor and support the various structures of the body. There are a variety of types of connective tissues, a few of which are structural, blood cells, mast cells, adipose cells, and melanocytes. Many of the proteins that make up structural connective tissue are secreted by cells cailed fibroblasts. Collagen, reticulirl and elastin are structural proteins which are secreted by these cells. Collagen is a triple-stranded, insoluble, fibrous protein (see Figure 1-3) that is highly ooss-linked, a feature that makes these fibers quite strong and rather flexible. Besides having a very high tensile strength, collagen is also the most abundant protein found in mammals. Reticulin is a thin fiber found in the spleen and lymph nodes. It is not as highly coiled as collagen. Elastin is also a highly cross-linked protein found associated with organs that require some degree of elasticity (like the lungs, skin, and blood vesseis).

Another type of structural connective tissue, cartilage, is secreted by a specialized fibrobiast cell called a chondrocyte. There are different types of cartilage, but in general it is found in places where there is a certain amount of stress placed on the body. For example, cartilage can be found in the nose, on the articulating surfaces of bones (including the intervertebral discs of the vertebrai column), and in the external ear.

Bone is also a structural connective tissue. About one-third of the weight of bone comes from organic materiai such as collagen, while the remaining two-thirds is inorganic material such as calcium phosphate and calcium carbonate. The collagen found in bone matrix is secreted by specialized fibroblast cells called osteoblasts. Collagen lends flexibility to bone, while the inorganic crystals lend rigidity. Within the centrai cavity of bone, we find a spongy marrow where red blood cells and white blood cells are formed. Towards the surface of bone the

ceilular arrangement is more compact. [As a comparison, the main structural component of chitin (found in the exoskeleton of insects) consists of specially modified glucose residues linked to one another to form long polymers. Associated with these polymers is calcium carbonate (CaCO3). This combrnation adds rigidity to the exoskeleton, but offers little in the way of flexibility.l We mentioned that blood cells and mast cells are kinds of connective tissue. We discuss blood cells in a separate lecture. Mast cells can be found in the respiratory tract, as well as in the gastrointestinal tract. Mast cells can release

will

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Biology

Nerve 8r Muscle

TYpes Of Tissues

histamines in response to an allergic reaction, an infection, or even an injury. Histamine causes an increase in blood flow to the blood vessels of the affected region.

Other types of connective tissue involve adipose cells and melanocytes. Adipose cells are simply ceils that store fat whereas melanocytes are cells which store pigments.

Muscle fissues will be discussing various types of muscie in future lectures. For example, r,vhen we examine skeletal muscle we will find that it is voluntary muscle. That

We

is, we can generally control its action. Cardiac muscle and smooth muscle are examples of involuntary muscles.

Nervous fissues The nervous systems allow one to adapt rather quickly to external stimuli. For example, consider a simple reflex arc. If someone were to tap on your knee with a

rubber hammer, your lower leg would extend outward. As the hammer impinged upon the patellar tendon in your knee, an eiectrical impuise was generated and traveled via a sensory nerve to your spinal cord. That sensory neuron svnapsed with a motor neuron, which returned the impulse to the muscle that was initialiy stimulated and caused it to contract. We will come back io this exampie and examine it in a bit more detail later. First, let's consider some terrninology.

Neurotransmitters

Dendrites

are released from sYnaPtic Uulbs

--i

A Typical Neuron Ceil body

Figure I-4

1r

:nrjor components of a neuron.

associated supporting cells make up the newous system. Nerve are also called neurons, and ihey are the basic structural unit that make up --:r: r.ervous system. The major anatomical features of a neuron are the cell body

\en-e cellsand :=

-s

-:.-,-oh'ed in integration of information), the dendrites (involved in receiving and ::a:.smitting information towards the cel1 body), and the axon (involved in :,:.ducting information away from the cell body). When a neuron becomes :\.-ri€d and receives electrical information in the form of a stimulus, the celi body ::it.esses that information and transmits it down the axon in the form of a nerve rnpuise called an action potential. When that action potential reaches the end*o{ -j-.. a\on (referred to as the synaptic bulb or bouton terminal), it causes the

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Biology

Nerve & Muscle

Types OfTissues

release of a chemical substance called a neurotransmitter (see Figure 1-4). The neurotransmitter diffuses across the synaptic cleft and induces an identical action potential in an adjoining neuron, muscle cell, or gland cell. The junction between two such cells is called a synapse.

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Biology

Nerve & Muscle

Membrane Potentials

The generation of electrical signals in the nervous system is concerned with the diffusion of ions from a high to a low concentration (see Figure 1-5). in other words, charged ions diffuse down their concentration gradient. In the extracellular space of vertebrates the concentration of Nao is about 150 mM, while that of Ko is about 5 mM. The concentration of Cle is about 130 mM, while ihat of HCO3e is about 25 mM. Note that the concentrations of the cations (Nae and KCI) and the anions (Cl€ and HCO3O) balance one another. In other words, we find 155 mM of the cations and 155 mM of the anions. This represents

electroneutrality. Outside Cell

Inside Cell

K* =

120 mM

to

140

K+=5mM

mM

Na* = 10 mM to 15 mM

]..,,"",

Na* = 150 mM

Cl-=5mMto40mM

Cl- = 130 mM

HCO3- = 12 mM to 25 mM

HCO3- = 25 mM

-

Proteins

Figure I -5 I. :.::r ceiiular

l^","",

concentrations of the common ions.

Ke (about 120 mM to 140 mM) =,:. a lorv concentration of Nao (about 10 mM to 15 mM). We also find a lower : -:.::nfation of Cle (about 5 mM to 40 mN! and usually a lower concentration "

r--r'n the cel1, we find a high concentration of

: :-rCO3 3 (about 12 mM to 25 mM). There are many negativeiy charged proteins ,' :--:ir. the cell. Electroneutrality can also be found on the inside of the cell as

:

-.

the distribution of ions across the cell's membrane, you will find it - r: :.::.-mrnetri.cal. Let's consider a resting nerve (i.e., a nerve that is waiting to :: .=:;-:: an action potential) that has a permeabiiity to potassium which is much .r:-:i:r :han its permeabiiity to sodium. In other words, PKe >>> PNao (where F :=:::s to permeability). Because the concentration of Ke is higher inside the ,: --r.;r: outside the cell, potassium will diffuse down its concentration gradient .: : -.:'" e 'Jee ceil (see Figure 1-6). As the positive Ko cation leaves the cell, there : -::::>:cndrngiy less positive charge remaining inside the cell. In other words, -- = ,: s:ie oi the cell is now more negatively charged with respect to the outside . ---- ::--. This sets up a voltage that is positive on the side of the celi to which ::::::-'':. is trving to diffuse to (i.e., the outside of the celi). As that positive :-,:i- ::gins to build up, it tends to push potassium back into the cell :;::::,:.:. irke charges repel each other).

:

. - .. -ook at

I


Biology

Nerve & Fluscle

Membrane Potentials

Those tn'o forces (chemical and electrical) do not exactly match each other. It lums out that the force of diffusion is a little larger than the electrical force. This results in a little bit of leakage of Ko out of the cell, as well as a little bit of Ieakage of Nao into the cell. The Ke that leaked out of the cell has to be pumped back into the cell, and the Nao that leaked into the cell has to be pumped out of the cell. The pumping action of these two ions is provided for by the Nao/K@ ATPase pump. This pump is responsible for the generation of the asymmetrical concentration gradient of Nao and Ke across the cell's membrane. Cell membrane

Diffusion

Inside cell

Outside cell

Where P* ))) P"u

Na+

-87mV

+ 87

mV

Na*

and the anions are

impermeable

Figure l-6 Cellular gradients where PK >>> PNa.

We can calculate the membrane voltage (the potential difference) across the cell's membrane using the Nernst equation as shown in (L-1). In this equation, V is the voltage in miilivolts (1 mV = 10-3 volts), i refers to inside, o refers to outside, R is the gas constant, T is the temperature in Kelvin, Z is the ion's valance, and F is the Faraday constant. If we let the cell's membrane be permeabie to just Ko, we find that the voitage is '87 mV inside the cell with respect to outside the cell. Remember, this is if potassium is the only permeable ion. It is ihe gradient of potassium alone across the cell's membrane that is able to generate this potential.

[K*]o V'^ = 2.3 RL ,on " ZF [K*]r

(t-1)

[5 mM]" [140 mM]r

(1-2)

Vio = 60 log

Vio=-87mV

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Biology

Nerve & Muscle

Membrane Potentials

a nerve, it leaves its resting state and enters an active state in r,r'hich the cell's membrane is more permeable to sodium (Nuo) than it is to potassiurn. In other words, PNa >> PK. Since the concentration of sodium is higher outside the ce1l than inside the cel1, sodium will tend to diffuse down its concentration gradient and into the cell.

If we stimulate

Once again, we can establish a separation of charge. As the Nao ions enter the ce11 ivith their positive charges, there is that much iess positive charge outside the .e11. The outside of the cell becomes more negativeiy charged than it was before \a3 started to diffuse into the cell. Similarly, as the Nao ions diffuse into the -e11, the inside of the celi begir-rs to accumulate more positive charge. The inside :- rhe cell becomes more positively charged than it was before the Nao ions :t.:ied to diffuse in (see Figure 1-7). As the Nao ions diffuse into the cell down --:-:ir concentration gradient, a chemical and an electrical equilibrium is being

=r:a:rished.

Inside cell

Where P*, )) F

Outside cell

" Na+ Voltage

:x,qLu'e l. / : . - -r :-:- :1::. 'ihere P\a >> PK

: -:-- -:-,:-:ie:he magnitude of the potential across the ceil's membrane by -- - - : I\-:llL,
:

-

:e::eci

r

to outside the cell)

[Na*]o [Na*]i

11

i 150

mMlo

(1-4)

(1-s)

[15 n-A4]i (1-6) I -::

.

f:.: B:rielev

Review

lt


Biology

Nerve & Muscle

Membrane Potentials

suppose we were to measure the potentiar across the plasma membrane of a neuron (nerve cell). We can do this by using two electroi"es. A microelectrode is inserted into the neuron itself while another electrode is placed in the fluid that surrounds the neuron (see Figure 1-g). If we connect these two electrodes to a voltmeter that can read the potential difference between the two environments, we will find that the inside of neuron registers about -g0 mv (with respect to the extracellular space). Note that this voltage is quite close to tire voltage that we would get (-87 mv) if this plasma membrane were permeable only to ro. Ttu reason that the inside of the neuron is not -gr rny is because the plasma membrane is not exclusively permeable to K@.

Figure l-B Measuring electrical potential across a plasma membrane.

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Biology

Nerve & Muscle

Action Potentials

We can reduce the plasma membrane potential by transiently increasing the membrane's permeability to Nao. This can be accomplished by stimulating the nerve. The transient reduction in the membrane potential is referred to as a depolarization of the membrane. If this stimulus is strong enough and the initial depolarization exceeds a specific minimum value characteristic of the cell (called the threshold potential), then within milliseconds a burst of Nao ions lr'iil enter the cel1 and generate an action potential (see Figure 1-9). As the depolarization of the membrane continues, its permeability to Nao becomes much greater than its permeability to Ko (i.e., PNa >> PK). What we find is that the membrane

potential goes from a resting membrane potential of about -80 mV to

a

membrane potential that approaches that for Na@. In this example, at the height of the wave of depolarization, the membrane potential would be about +40 mV. [Why isn t the membrane potential +60 mV as would be calculated using equation (1-4)?l

+60

Na+ equilibrium potentiai

+40

(A) Stimulus

,.I

-=

.t)

=

Px >>

PNo

PNr>PK

Threshold potential

!

-z

* 't *

{BtDepolarization (C t Repolarization (D) Hyperpolarization (E)Relractory

-

8t)

Resting potential

(E)

Q7

K+ equilibrium potentiai

Tirne (msec) Figure I -9 - :-:.. .ction potential. ._"

-: :- :'.e membrane potential reaches

the equilibrium potential for Nao there is r- , ::-:::e influx of Nao ions into the celi. At this point (indicated by the peak of -,- : ,'.::,... poiential) the membrane potential, which is +40 mV inside the cell, is :: ,-.-=j r\ the concentration gradient of Nae. Recall that the concentration of ,, = - ':=r,je the cell is greater than the concentration of Nao inside the cell.

,.

::.-,-rsecond after the burst of Nao into the cell, the ion channels -::phorest that let Na@ into the celi close and become temporarily inactive. -:: -:,- \a3 channels are temporarily inactive they are said to be in a _:z':r::rin' period, which usually lasts for several milliseconds. During this time a -:*r-r n-il1 not be able to generate another action potential, because the Nao -" .::.-. ::-nnot open to allow the cel1 membrane to depoiarize.

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Biology

Nerve 6r Muscle

Action Potentials

once the Nao channers have closed, the Ke channels begin to open and Ko begins to leave the cell. This phase of the action potent"ial is referred to as repolarization. As the ceil membrane's permeability to ro increases, a massive amount of Ko flows out of the cell and the membrane potential passes its origrnal resting state of about -80 mv. This event is cailed hyperpolarization (see Filure 1-9). Eventually the ce1l wilr reach equilibrium, u.,i'o.,i" again an appropiiate stimulus will generate an action potential. The generation of an action potential is said to be an all-or-none phenomenon. What does this mean? If.we apply a stimulus to a nerve (see point (A) in Figure

1-9), the plasma membrane of the neuron

wil

become depolarizea.

ine

membrane potential will become less negatively charged during the increase in the permeability of the plasma membrane to Nab io.,.]R"*"rnber, Nao wants to flow down its concentration gradient and into the cell. As Na@ enters the cell, the membrane becomes more deporafized.. This increase in depolarization tends to induce neighboring Nao channels to open, thus retting more Nae ions into the cell' However, we know that the celr is dynamic rathe"r than static; Ko ions are leaving the celI and flowing down their concentration gradient. This action has a tendency to repolarize the prasma membrane. In other-words, the generation of an action potential depends on the ratio of membrane permeabilitlito these two ions. If there is not a sufficient depolarization (i.e., influx of Nab ions) of the membrane, then there will be no action potential. Flowever, if there is a sufficient depolarization of the membrane because there are more Nao ions entering the cellthan Ko ions leaving the ceil, then an action potential will be generated if that depolarization exceeds some threshold var'e. once an action potentiar is generated in a nerve cell it witt always haae the same magnitude. Thus, an action potential is either all-or none. rt either happens or it doeJn't.

In Figure 1-9 we have indicated how an action potentiar is generated. we mentioned that in order to generate an action potential there must be a depolarization of the membrane above some threshoid value, a reporatization of the membrane, a hyperpolarization, and a refractory p"rioa that ailows the membrane potential to return to its resting value before a new action potentiai can be generated. Now let's examine how an action potential propagates arong a

neIVe.

Consider the charge separation across the plasma membrane (see Figure 1-10) of a nerve as an action potential moves from, say, the center of the axon outwards. Suppose we have a stimulus that exceeds a given threshold value. As the action potential is generated a separation of charge will be established. Because Nae

ions are rapidly entering the cell, the inside of the cell will become more positively charged (+) with respect to the outside of the cell. Since there is a deficit of positive charge on the outside of the cell (compared to the original resting condition), we say that the outside of the cell is io." negati"vely charged (-) with respect to the inside of the cell. In other words, -or" at thai pirticuta. piace on the axon, where the action potential has been generated, there is a reversai in the membrane's polarity. This is shown in Figure"l-10a. The depolarization of_the central region of an axon spreads outward in both directions as more Nae ions enter the cell. In Figure 1-10a the arrows represent a locai flow of current between the area of the membrane that has been flur*u depolarized and the adjacent areas of the membrane which are still at their Copyright

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Biology

Nerve & Muscle

Action Potentials

resting potential. The flow of positive charge within the ceil causes the adjacent areas to become more depolarized.In other words, the polarity of the membrane

is being decreased in front of the propagating action potentials. Because the membrane potential in these adjacent areas is becoming depolarized, more Na@ channels begin to open. This lets more Nae into the ce1l, which increases the depoiarization until an action potential is generated.

+ + + + + + + '_j*t_j

+++++++

f'l = Na+ Channels

Propagation

of

(b)

+

Action Potential

++++++++

+++

Axon ++++++++

+++ Figure l-lO iction potential propagation.

that as the action potentials spread outward, they create a region which is refractory (see Figure 1-10b). Recali that the refractory period is due to nactivation of the Nao channels. If the plasma membrane has been depoiarized rom an action potential, then another action potential cannot be generated until rat membrane has once again reached its resting membrane potential.

\ote

)urrng the state of hyperpol arization, the K@ channels are wide open and K@ is :orrriig out of tne iett. Even though an action potential could be generated ,."urrng this time (of hyperpolarization), it is rather difficult unless a strong ..o.rgh stimulus is applied. Also, the action potentials in Figure 1-10b appear :ackiards compared to the action potential in Figure 1-9, because the action : otential in Figure 1-9 shows a fixed point on the axon and the membrane's .ot".,trul variation with time at that point. The action potentials in Figure 1-10b :ilo\\' a distribution of the membrane potential along an axon (not at a fixed

:ont)

at a particular instant in time.

lirere are a many kinds of nerve fibers in the body, and they do not all conduct :-::ion potentiali at the same rate. For example, large diameter neurons will Jcrl'rrght O by The BerkeleY Review

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Biology

Nerve & Muscle

Action Potentials

conduct a depolarization or action potential further and faster than a small diameter neuron. \Alhat this means is that the ability of a given neuron to conduct a current (the flow of charge represents a current) depends on the cross-sectional area of that neuron. The larger the cross-sectional area of a neuron, the more cytosol il that neuron, and the more ions there will be to conduct that current. The more ions available to conduct the current, the faster and farther that current

will spread.

The nerves which rve harre been discussing so far are referred to as unmyelinated ner-"'es. Myelinated nerves have the ability to greatly increase the rate at which action potentials are conducted. Myelin is a specialized membrane that is composed of just a few types of proteins and a large proportion of phospholipids. A specialized type of cell (called a glial cell) attaches itself to a section of an unmyelinated axon and begins to rotate itself around that axon a number of times. During this process a myelin sheath is deposited on the axon. Sometimes this myelin sheath is thicker than the axon itself. The deposition of myelin along the axon is interrupted by areas where there is t-ro -yeiitl. These areas are called the nodes of Ranvier. See Figure 1-11.

TJ:ln.'*1'l t\

+JL

Axon \_-J

Plasma membrane

Node of

(contains a high density of Na+ channels)

R.anvier

Figure l- I I Action potential propagation lrom node to node.

Later we wiii learn that the body's nervous system is separated into two divisions (based on function). The central neryous system is macie up of the brain and spinal cord whiie the peripheral nervous system represents the nerves in the periphery. The point here is that the glia1 ce11s in the centrai nervous system are called oligodendrocytes, while the gliai cel1s in the peripheral nervous system are ca1led Schwann cells. We mentioned that myelinated nerves increase the rate at which action potentiais are conducied. The myelin that surrounds the axon acts as an electrical irsulator and prevents the transfer of ions across the plasma membrane of the axon. The only region of the axon where ions can pass across the plasma membrane is at the nodes of Ranvier. These nodes contain a high density of Nao channels. As an action potential is generated, there is a flow of current through the cytopiasm of the axon (called the axoplasm) and the extracellular fluid from node to node. The

plasma .membrane at each of these nodes rs Copyright

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Biology

Nerve

6e

Muscle

Action Potentials

generate an action potential. The nerve impulse seems to "jump" from node to node along the axon. Thus, this type of impulse transmission is referred to as saltatory conduction (from the Latin saltatorius--pertaining to dancing or Ieaping). See Figure 1-11.

Nerve fibers can be grouped into different classes. The nerve fibers in some of these classes are myelinated, whiie those in other classes are unmyelinated. Some of the myelinated fibers that impinge on skeletal muscle can transmit motor impulses quite rapidly, about 60 to 120 m/sec. In contrast, some of the r:nmyelinated fibers which transmit sensory impulses of certain types of pain, do so rather slowly (relative to a myelinated fiber), about 0.6 to 2 m/sec. As you can see, myelinated nerve fibers greatly increase the rate at which information, in the form of action potentials, is conducted. So far, we have been considering the propagation of an action potential within a single nerve cell. Let's see how this information is transmitted to other cells.

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Biology

Nerve Et Muscle

, tt

|ll

The Neuromuscular Junction

ffiuscunffffi

,,!,'u;,.''

At the end of the axon is a specialized

area called the terminal bouton (or syraptic bulb, synaptic knob, or even terminai foot). This specialized end region of the axon can impinge or synapse on another axon of a different neuron, on a dendrite, or even on some type of cell body (like a muscle ce11). The space between this synaptic junction is called the synaptic cleft. The plasma membrane at the terminal bouton is calied the pre-synaptic membrane while the piasma membrane that the synapse is being made to is called the post-synaptic membrane. See Figure 1-12.

Presynaptic

Synaptic cleft

membrane

memhrane

-@\

ACh

ACh receptor ACh

+@

rcs

Postsynaptic

complex

@,

X

aAC])

\ Terminal bouton

@

2

Acetlycholinestrase

I - l2 Presynaptic and postsynaptic membranes.

Figure

'

Let's consider a neuromuscular junction. This is the synapse between the terminal bouton of an axon and a muscle fiber. Within the terminal bouton of this

junction are hundreds of thousands of synaptic vesicles that contain the excitatory neurotransmitter acetylcholine (abbreviated ACh). There may be thousands of molecules of acetylcholine in each synaptic vesicle. Acetylcholine itseif is synthesized in the cytosol of the neuron from the molecules acetyl-CoA (which caries an acetate moiety) and choline.

As the action potential reaches the terminal bouton, it triggers the openrng of calcium (Ca2o)channels" A transient influx of Ca2e into the terminai region causes the synaptic vesicles to fuse with the presynaptic membrane and release their neurotransmitters into the synaptic cieft (via exocytosis). Calcium is pumped out of the cytosol and back into the extraceilular fluid. The membranes of the synaptic vesicies that fused with the presynaptic membrane are recycled Copyright O by The Berkeley Review

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Biology

Nerve & Muscle


(via endocytosis). The released neurotransmitters diffuse ihrough the synaptic

cleft and bind to specific postsynaptic membrane receptors. Binding of acetylcholine to this postsynaptic membrane receptor conformationally changes that receptor into a channel (ionophore) that is large enough to allow cations such as Nae through. [Note that this channel is not a voltage-activated channel like the channels we examined in the axon itself. This channel is a ligandaciivated channel. In this case, the ligand (a molecule or ion bound to a protern) rs acetylcholine.] As Nas enters the postsynaptic membrane, the muscle fiber begins to depolarize and an action potential is eventualiy generated.

If acetylcholine were to remain in the synaptic cleft, then it would continue to bind to the postsynaptic membrane and action potentials would continually be qenerated. This cor,rid result in prolonged muscle spasms. However, the enzyme

acetylcholinesterase, which is bound to the surface of the postsynaptic membrane, hydrolyzes acetvlcholine to acetate and choline. These two products are then recycled as they are transported back into the presynaptic terminal .,r-here they are used in the svnthesis of acetylcholine (Figure 1-12).

fhere are many different types of neurotransmitters. For example, the amino acid *lutamate and the amino acid derivatives epinephrine, norepinephrine, iopamine, and serotonin can act as neurotransmitters. There are also drugs that ;an act at the level of these postsynaptic receptors.

EPSPs and IPSPs ',\-e said that the binding of acetyichohne to its postsynaptic membrane receptor :iicits an excitatory response that generates an action potential. These types of :oientials are referred to as excitatory postsynaptic potentials (EPSPs). There can aiso be inhibitory postsynaptic potentials (IPSPs) as we11.

Action Potentials Excitatory Post Svnaptic Potential

u

J The result of synapse with

AandB Depolarization

I

Inhibitory

L,;*:',t:il",'"::l^,

{r=u t-fi] ' fi loo''l ,"r'liiili#iiiir::i,"

Hyperpolarizarron

Figure l-15 IPSPs and IPSPS.

Suppose neuron .

A

synapses

with neuron C, as shown in Figure

1-13a.

If the

-,'naptic connection betr.t'een these tr,vo neurons is excitatory, then the same

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l9


Biology

Nerve & Muscle


number of action potentials that pass down the axon of neuron A will also pass down the axon of neuron C. In the case of an excitatory postsynaptic potential, there is an increase in the permeability of the postsynaptic membrane to Nao (i.e., neuron C will depolarize). Suppose neuron B, which is inhibitory, also synapses with neuron C, as shown in Figure 1-13b. In the case of an inhibitory postsynaptic potential, there is an increase in the postsynaptic membrane to Ko and to Cl- ions. Neuron C will hyperpolarize. Remember, becoming more hyperpolarized does not generate an action potential. In fact, it will keep neuron C quiescent. In other words, if we just excited neuron B alone, then nothing would happen to neuron C.

However, if we integrated the action potentials that neuron C receives from both neuron A and neuron B, we find that neuron C only generates 2 action potentials (Figure 1-13b). In this case, neuron B will decrease the excitability of neuron C to the stimulation it is receiving from neuron A. Consider two action potentials coming toward one another, as shown in Figure 1-14. When those two action potentiais meet what will happen? Wiil they continue past one another? Wiil their amplitudes add together? The region in front of each action potential is continually being depoiarized, while the region behind the action potential is in its refractory period for a few milliseconds and cannot immediately be depolarized. Therefore, when the two action potentials meet one another ihey will stop their propagation along the axon.

+++++++++++++++

Axon

:

+++++++++++++++ Figure l-14 Propagation of action potentials.

Copyright

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20


Biology

Nerve & Muscle

The Sarcomere

The Sarcomere: '!',-e mentioned that the synapse in Figure 1-12 involved a neuromuscular 'rnction. We have considered the neuronal portion of that junction already. Let's :.or.,- examine the muscular portion and follow the fate of the action potential that ','.-a. generated.

-- -,-ou rvere to examine a typical muscle, you would find that in most cases it is :::ached to a specific area of the skeleton by tendons at either end of the muscle' f:.= region between the tendons is referred to as the belly of the muscle. A cross-

.=:rion of the beliy of the muscle shows a great number of multinucleated n-ruscle fibers (ce11s), as shown in Figure 1-15. These muscle cells do not divide to ::r,iuce new muscle cells. However, if you were to exercise your muscles, those :=Ls rr-ou1d just get iarger.

Individual muscle cells

Actin

it !

l-llne

H-zone

Sarcomere il-g.ure

1-

l5

- :,=: eramination of skeletal muscle fibers shows striations in both the *l::-.::>e and longitudinal directions. The striations in the longitudinai -,:::

3

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Biology

Nerve

& Muscle

The Sarcomere

direction are due to myofibrils. The myofibrils contain the contractile units of the muscle. Each contractile unit (Figure 1-16), bounded by a structure referred to as the Z-line, is referred to as a sarcomere and each sarcomere contails tn'o types of contractiie proteins. The thin contractile protein is called actin and the thick contractile protein is called myosin. The myosrn thick filament comprises the Aband and the region between two A-bands, where there is just the thin actin filaments, is the l-band. The H-zone is that region in the center of the A-band and between the ends of the actin filaments. I-band A-band

A-band

Actin

I

H"od'+>8ffi=%Quyosin Z-line

H-zone

Sarcomere

Figure

H-zone

Z-line

Sarcomere

I - 16

Sarcomere details

Myosin fiiaments are arranged toward the center of the sarcomere. They give rise to the characteristic dark bands one sees when examining muscle tissue. Actin filaments are attached to the Z-lines. The actin and myosin filaments interact with each other through projections stemming from the myosin fiiaments. Those projections are sometimes referred to as myosin heads. The myosin thick filament does not have any of these head groups in its central region but rather concentrates those head groups towards its terminal regions. The actin thin filament is composed of a protein subunit called G actin ("G for giobular), which is roughly spherical in shape. The actin filaments can grow by the addition of G actin to the ends of the already existing filament. Each actin

fiiament is composed of two rows of G actin monomers wound around each other to form a helix.

Actin and Myosin Let's consider how myosin and actin interact with each other at the molecular level. When a muscle is in its relaxed state, ATP is bound to the myosin head groups. The myosin head is not bound to the actin filament, because ATP reduces myosin's affinity for actin. \,Vhen ATP is bound to the myosin head, the myosin head itself is at about a 45' angle with respect to the actin filament. Since actin is not bound to myosin, the ATP on the myosin heads is hydrolyzed to ADP and Pi (inorganic phosphate). These hydrolysis products remain on the

myosin head. More importantiy, the myosin head now undergoes a conformational change, so that it is situated at a 90' angle with respect to the actin filament. This high energy, stable, myosin'ADP-P1 head complex binds to the actin filament. As we will see, this step is dependent on Ca2e being present. Copyright O by The Berkeley Review

22


Biology

Nerve & Muscle

The Sarcomere

The interaction between the actin filaments and myosin head groups causes the reiease of Pi and then ADP from the myosin heads. This process causes a conformational change iir the myosin heads, so that they shift by about 45' in a direction that is away from the Z-lines. This step, in which the actin filaments move relative to the myosin filaments, is cailed the power stroke, and the product of this step is referred to as the rigor complex or rigor state. See Figure 1-17.

ATP binds and causes myosin head

Mvosin

This is the rigor complex

r

ADP

f-ilament

:ioVes

Actin and myosin bind

rxgure l.I/ -'

a

:

-l::

-:Lrn Cf'Cle.

:-i.:: --:.. :TL\-osirl heads have swiveled and have pulled the actin filament past : : '- : :-rr, tilament, the myosin heads remain bound to the actin filaments. This

'-"

:-:

:=:..:-:alied rigor state. In order for the myosin heads to dissociate from the

i::*- 1ir1ents, ATP needs to bind to those myosin

heads. Remember, ATP

'.:,::s:r-.-osin's affinity for actin. Without ATP, muscles remain in a state of .".-:-: -:: : gir-en period of time. This is what happens to muscles in your body .-.-- -.----. :ie (i.e., rigormortis). The myosin head groups can no longer bind

:lT :=-;use the metabolic ",i = -=-..i to function.

pathways that generate this energy-rich nucleotide

lfi'u'oponin, Tropomyosin, Et Calcium r" r: : ': .'ra::Lln€d actin, we saw that it was composed of two long rows of

*::- , ;::.:rica1 protein subunits, which polymerized together and then wound : -r :;:: :--.Ler rn a helical fashion. If we look closely at the helical structure of r--:-i .:.:ii:e tn-o grooves. Running the length of these grooves is a coiled : *, ::*- :.--=j tropomyosin, composed of two helical polymers wound about --: =: :ee Figure 1-18). When tropomyosin resides in the actin groove, it -; - : :u:ding sites for the myosin heads and prevents those head groups -: -:':.1

:o the actin filament.

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The Sarcomere

In the case of striated (skeletal) muscle, this is often referred to as the actin-based regulation of muscie contraction. Cardiac muscle and smooth muscle are both controlled by myosin-based regulation. Troponin Protien Complex

ca2+ Binding Tropom yosin

Site

o Relaxation

/t

"t llll f,l Calcium \1/

Removal

Addition ot Calcium

li

Contraction Myosin Head Binding Sites

Figure

l-lB

Sarcomere detail.

Troponin, a multi-subunit binding protein, interacts with tropomyosin, actin, and Ca2o. \A/hen Ca2o is bound to a particular subunit of the troponin complex, it causes tropomyosin to shift its position and expose the myosin head binding sites. The myosin heads then bind to the actin filaments (see Figure '1'-17) and muscle contraction follows. [In Figure L-17, Ca2@ would bind at Step 3.] lf Ca2o is not in the medium, there witl be none to bind to the troponin complex and tropomyosin wili remain in the actin gloves and cover the myosin head binding sites. This is the relaxed state.

Surrounding each myofibrii is a membranous structure,

a

modified version of the

endoplasmic reticulum called the sarcoplasmic reticulum. Calcium is sequestered within this smooth membranous structure. Also surrounding each mvofibril is an invaeination of the sarcoleuuna (i.e., the plasma membrane) called the transverse tubule (abbreviated as T-tubules). These T-tubules follow lhe Zlines of each myofibril. Selected anatomical features of the structures associated with the sarcoplasmic reticulum are shown in Figure 1-19. After an action potential crosses the iast synaptic junction on its way to a muscle cell, it passes down each T-tubule and somehow stimulates the release of Ca2e from the sarcoplasmic reticulum.' [At the present time, it is thought that the lumen of the T-tubules and the lumen of the sarcoplasmic reticulum are not

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24


Biology

Nerve

8e

Muscle

The Sarcomere

continuous.] As the Ca2s flows from a region of high concentration (the sarcoplasmic reticulum) to a region of low concentration (the cytosol), it binds to

its binding site on the troponin complex. Each sarcomere

contracts simuitaneously because of the way in which the nerve impuise is carried along the sarcolemma and into the T-tubules.

+

Sarcolemma

Myofibrils

ransverse tubule

Figure l-19 -

-

=,.:,'oplasmic reticulum.

- ,- :-

:cntraction has taken place anC the nerve impulse has ceased, the Ca2e in --= :-.:osol is pumped back into the sarcoplasmic reticulum by a Ca2o-ATPase --::-: In the sarcoplasmic reticulum Ca2o associates with a specific binding ::,:: :.. \Vhen the next action potential passes down the T-tubuies, the cycle will -:.larn and another muscie contraction will take place. :.:e a number of ways ir-r which the strength of muscle contraction can be The strength of contraction can be varied by (1) the size of the motor unit :efined in a moment), (2) the number of available motor units, and (3) the ': of acth and myosin contained with each cell.

tor unit is simply a motor neuron and the muscle fibers that it innervates. .. e already examined the muscle fibers. What is a motor neuron? Motor . :-s are nerve cel1s whose ceII bodies are located in the centrai nervous :- e.g., the spinal cord or brain stem), and whose myelinated axons :,e skeletal muscle. Reca11 that myelinated axons allow action potentials to ::-,-::-,itted rapidly to the desired effector organ (in this case, a muscle) ,, anted to precisely move a muscle (e.g., the muscles associated with your - :ngers), then you wouid need motor units which were rather small in -:r smalier the size of the motor unit, the smaller the strength of . : ; ::- rn, In contraSt the posturai muscles of the back or the legs require large - . : .-::.:ts to be effective. Not only can strength be controlled but precrsion can

. : .--led as wel1. ::. :


25


Biology

Nerve 6t Muscle

The Sarcomere

The number of motor units also controls the strength of contraction. For example,

if you wanted to pick up a light object (e.9., u pencil), you would need to employ only a few motor units. However, if you wanted to pick up heavier objects (e.g., gym weights), you would need to utilize more motor units. When you lift weights regularly, the size of each muscle cell increases, because the amount of actin and myosin in each muscle cell has increased. The more actin and myosin in a muscle cell, the greater the

strength o{ contraction that can be generated.

Even though the strength of muscle contraction can vary with the size of the motor units, the number of motor units, and the amount of actin and myosin in each muscie cell, the ultimate determinant of muscie contraction is the concentration of ATP in your muscle cells. The energy for contraction comes from the hydrolysis of ATP to ADP and Pi (inorganic phosphate). See equation (1-7\:

ATP

----a

ADP + P; + Energy

(r-7)

For muscie contraction to continue, ATP must be constantly regenerated. Recail

from basic biology that under aerobic conditions (when 02 is not limited) glucose will be oxidized to CO2 and H2O. Energy can be extracted (in the form of 36 molecules of ATP--if we use the glycerol-phosphate shuttle) during this catabolic process from glycolysis, the Krebs cycle, and oxidative phosphorylation/electron transport. See equation (1-8): Aerobic (Slow)

Glucose

CO2

+ H2O +

36 ATP

(1-8)

During anaerobic conditions (when 02 is limited), glucose is metabolized to lactate (the anion of lactic acid). Only 2 ATP molecules are extracted for the use of energy in this process. See equation (1-9). [This equation is not balanced.] Anaerobic

Glucose

(Fast)

Lactate

+ 2ATP

(1-e)

Even though the efficiency of ATP production is greater for aerobic metabolism than for anaerobic metabolism (36 ATPs compared to 2 ATPs), ATP can be produced much more quickly through anaerobic metabolism than through aerobic metabolism, owing to enzyme regulation in the glycolytic pathway.

During anaerobic metabolism, the concentration of lactic acid begins to increase. This means that the pH of the medium will decrease (i.e., become more acidic). One of the key regulatory enzymes in the glycolytic pathway cannot function weli below a certain pH value. This enzyme has some optimum pH range at whictr it functions and if the pH falls below (or above) that range, the enzyme is inJribitedQlycolysis comes to a halt and the ATP yield becomes rnsufficient to /iatry out ndlmal metabolic processes. In other words, fatigue sets in.


26


Biology

Nerve Ef Muscle

ti$,#$,ft€

N.€

Nervous System Components i,,,.'r,,".,,,,' ,..

t,

..,.,.

The nervous system in its simplest form can be found among the cnidarians. All the neurons in these creatures are linked together in a nerve net (much like a rr.eb). stimulation of the nerve net causes muscle contraction. This simple procedure is referred to as a reflex arc. In the case of the cnidarians there is no

associative activity, meaning that the transmission of the action potential is not linfluenced by other neuronal activity.

-{ssociative neurons are {ound in higher animals (Platyhelminthes and above). Here, different neurons can interact to produce a given response. when a rrrlnber of these associative neurons are grouped together, the nervous system erpands. A grouping of nerve cells is called a ganglion (or a nucleus or nuclei). Groupings of neurons in higher animals also lead to more elaborate sense organs, :jtterentiation into a central region of cells (the central nervous system, ot -Ns; =ld a peripheral region of celis (the peripheral nervous system, or pNS), various :rssociative areas, and a brain. The three basic anatomicai divisions of the vertebrate brain can be seen in Figure 1-?0- Those divisions are: (1) the forebrain (prosencephalon), (2) the midbrain :resorcephaion), and (3) the hindbr ain (rhombencephalon).

Rhombencephalon

rfignr l-2o 'Mlmcai

Cirisions of the vertebrate brain.

erufurlcr icl rmeaning "below") the brain is the spinal cord. Recall that the brain ffiie spinaf cord together are the central neraous system (CNS). From the spinal ''rlrmfi r@!d* rrrmr-es extend into the limbs and extremities (the periphery) of the body. ffime mru1:es represent lhe peripheral neroous system (PNS). If a neuron carries

into the spinal cord and brain, that neuron is said to be an afferent i{lm*.'*yt nenron- U a neuron carries the information away from the brain or mnffimnnnmrnn#.ncne

qryilltumf rmmd-

that neuron is said to be an efferent (motor) neuron.

l[ffie Sunbrain has several anatomical subdivisions, including the cerebrum, md hrpothalamus (see Figure 1-21'.a). The cerebrum is divided into @{ilfril mr,{ lc{t cerebral hemispheres, joined by the corpus callosum. The cerebral lfirnlil'ilr1iql'ruttmsres are dirtded into the frontal lobes (associated with movement and Wmrmnlilittmy ri" parietal lobes (associated with touch and stretch sensation),

ilhihr

niltntldmdtll

rfti

ldes

nirilmtwr,,os

i{0fu@fh,@

(associated

shonrr in Figure

@,

with vision), and temporal lobes (associated with

'1,-21b.

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27


Biology

Nerve Er Muscle (a)

Nervous System Components (b)

Thalamus Cerebral

Central Sulcus

Cortex Corpus

Callosum

)/s /a

."o

\ Parjetal

o .)

Pons

The Cerebrum and its Four Lobes

\ Medulla

Spinal Cord

Cerebe llum

Figure l-2 I Anatomical divisions of the vertebrate brain.

The outermost layer of the cerebrum is called the cerebral cortex. It consists of gray matter, which is simply nerve cel1 bodies and their dendrites. Beneath the gray matter is the white matter, or myelinated axons of the nerve cells. In the spinal cord, the situation is reversed: The gray matter is more centralized, while the white matter is more peripheral. The cerebral cortex has many important landmarks, one being the central sulcus, a prominent groove that separates the frontal lobes and the parietal lobes.

Anterior to this sulcus is the motor cortex, which controls the movement of individual muscles. Posterior to this sulcus is the (somatic) sensory cortex, which detects sensations in various parts of the body.

Somatic receptors send their information into the spinal cord via afferent nerve fibers, which either cross over to the opposite side of the spinal cord and then ascend to the sensory cortex in the brain, or ascend in the spinal cord and then cross over in the brainstem before reaching the sensory cortex.

flomunculus Wilder Penfield, a Canadian neurosurgeon, was able to rnap the sensory and motor areas of the brain by electrically stimuiating certain regions of the brains of

his patients during sulgery and observing their actions. Throughout this procedure his patients were conscious but were unable to feel any pain because there are no sensory receptors for pain il the cerebrum. He was able to show that the largest number of cortical neurons found in the sensory cortex register sensation in the fingers, hands, lips, and tongue. This is represented as the sensory homunculus (a schematic model of a human being mapped out on the sensory cortex) shown in Figure 1-22. Penfield was also able to show that the largest number of cortical neurons found

in the motor cortex control individual movement of the fingers, hands,

and speech. Groups of muscles are controlled by an area just anterior to the motor cortex called the premotor cortex. This is an association area. Stemming from

2a


Biotogy

Nerve & Muscle

Nervous System Components

this association area are neuronal connections with the thalamus and cerebellum. Together these structures plan specific movements that the motor cortex then executes. It is thought that cognitive functioning like speech, writing, and reading are localized in the left hemisphere of the brain, whiie intuitive functions are localized in the right hemisphere of the brain. This is not proven, only theoretical.

-l

-J

Sensory receptors on the right side of the body project to the somatosensory cortex on the left hemisphere of the brain and vice versa

Left

i

Hemisphere J ,

L ,

Tongue

Right Hemisphere

Figure l-22

l:.:

:.c,munculus.

.:.

thalamus is a relay station for much of the visual and auditory information -,\-e receive from our environment. The hypothalamus is concerned with the activities of the body. The pituitary gland is the master endocrine gland -':t:a1 ,,' -:.: body. It receives information from the hypothalamus and sends out ::::rahon to regulate different parts of the body. -*-::

'' = brainstem contains such anatomical feafures as the midbrain, cerebellum, : ::ls medulla, and the reticular formation. These areas coordinate motor and -.:=:;-- activities. The midbrain detects movement and can direct the head and : :!

:iir-drds that movement. The midbrain can also sense pleasure and pain. is responsible for the bulk of regulation and coordination of -: ..-.:-îar activity. The pons and medulla coordinate visceral activities. The -. * ----:-:: formation, which is the core of the brainstem, is essentially an activating . :::::.;esigned to alert the brain. The reticular formation also inhibits motor i.- - ::.*ior:\- impulses and can induce sleep. Below the medulla is the spinal cord. --- = :e:ebel1um

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Biology

Nerve & Muscle

Control Of Body Activity

G;dffitfdlrri drdyrfffifiiffii Recall that when we first mentioned the nervous system, we iooked at a simple reflex arc. If someone were to tap you on your knee with a rubber-headed hammer, your lower leg would extend outward. The mechanism behind this is quite simple. As the hammer impinges on the patellar tendon, sensory neurons located in the tendon of the quadriceps muscle are excited. These impuises travel along an axon that enters the spinal cord through the dorsal root gangiion and synapses with two neurons (Figure 1-23). This area, even though it is supposed to be gray matter, is shown as being clear so you can see the synapses. Biceps ('hamstring") muscie is a flexor

\*u.

I

Ganglion

Dorsal

Grav Matter

Root

Spinal Cord

Motor Nerve Polysynaptic Reflex Arc

Tibia

Interneuron

Figure l-23 The reflex arc.

One of the synapses is to a motor neuron that immediateiy leaves the spinal cord and returns to the quadriceps muscle, causing a contraction. As this muscle contracts, the leg straightens (extends) at the knee joint. Because it elicits this kind of action, the quadriceps muscle is termed an extensor muscle. The type of synapse just described (making just one synaptic connection), is referred to as a

monosynaptic reflex arc. The other synapse is made to an interneuron which, in this case, is inhibitory. This interneuron in turn synapses with a motor neuron that innervates the bicep ("hamstring") muscle in the back of the leg. When the bicep muscle contracts, the lower portion of the leg bends or flexes at the knee joint. We call this kind of muscle a flexor muscle, and this type of synapse (because it makes at least two synaptic connections) is referred to as a polysynaptic reflex arc.

If contraction of the biceps muscle is inhibited

as the quadriceps muscle

contracts, then one observes a smooth and coordinated movement at the knee joint, as the lower portion of the leg extends outward after stimulation by the tapping of the hammer on the patellar tendon.


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Biology

Nerve & Muscle


Neurovisceral Control The autonomic nervous system, which is part of the efferent division of the peripheral nervous system, can be subdivided into the sympathetic and the parasympathetic systems. Nerve fibers from the autonomic nervous system leave the spinal cord to irrrervate various glands, smooth muscle, and cardiac muscle. Let's examine these two subdivisions.

Paras;rrmpathetic Division The parasympathetic division of the autonomic nervous system has nerve fibers which leave from the sacral portion of the spinal cord and from the midbrain (mesencephalon) and medulla (part of the rhombencephalon), as shown in Figure L-24. Parasympathetic nerve impulses tend to increase the rate of digestion and lower the heart rate. The blood pressure is also iowered, and the pupils constrict (contract). In general, the parasympathetic division conserves energy and helps in the restoration of various bodily functions (".9., by aiding in

the digestion of food for later use in metabolic processes). Ganglion

Neurotransmitter is acetylcholine

Eve

Ganglion

1 Lacrimal glands

o

Postganglionic nerve fiber

\ur, \Itrl.a

lr

o (,)

I

Small intestines

Heart

Lungs and Bronchi

(tQ.i

\ \ Large intestines I

(

Sexuai organs

1

Figure l-24 Parasympathetic nerves.

The parasympathetic division has both preganglionic and postganglionic nerve fbers. The cell bodies of the preganglionic neurons are found in the sacral region Copyright

@


3l


Biology

Nerve & Muscle


of the spinai cord and in the brainstem. The ganglia of the parasympathetic division lie near or in the organs that are to be innervated. Therefore, the preganglionic nerve fibers are rather long, while the postganglionic nerve fibers are rather short. Both the preganglionic and the postganglionic nerve fibers in the parasympathetic division release acetylcholine (A C h ) as the

neurotransmitter. INerve fibers that release acetylcholine as their neurotransmitter are called cholinergic nerve fibers.]

The most prominent nerve in the parasympathetic division is the vagus nerve [also called the (tenth) X cranial nerve]. Roughly three-quarters of all the neurons in the parasympathetic division can be found in the vagus nerve. The reason for this should become obvious, if you look at Figure 1-24 The vagus nerve innervates the heart, lungs, stomach, liver (not shown), small intestine, large intestine, and kidneys (not shown), among other organs.

Sympathetic Division The sympathetic division of the autonomic nervous system has nerve fibers branching off from the thoracic and lumbar regions of the spinal cord, as shown in Figure 1-25. Sympathetic nerve fibers tend to condition the body for a "fightor-flight" response (a term first proposed by the Harvard physiologist Walter Cannon in the early 1900s). The heart rate increases, blood pressure elevates, pupils dilate (open wider), and the digestive functions decrease, all as a result of sympathetic nervous innervation. The sympathetic nerves that leave the thoracic and lumbar regions of the spinal cord first enter chains of ganglia connected by nerve fibers on either side of the spinal column. These chains of ganglia are collectively called the sympathetic trunk. As these spinal nerves, called preganglionic nerve fibers, enter the sympathetic trunk, they can either (a) pass through this collection of ganglia to synapse with other ganglia outside the sympathetic trunk, (b) pass into the sympathetic trunk and ascend or descend to synapse with ganglia at other levels,

or (c) pass into the sympathetic trunk and directly

synapse

with a given

ganglion. The nerve fibers leaving a synapse in a given ganglion are referred to as postganglionic nerve fibers. In the case of the sympathetic division the preganglionic nerve fibers tend to be short, while the postganglionic nerve fibers tend to be longer. The preganglionic fibers release acetylcholine, while the postganglionic fibers release norepinephrine as their neurottansmitters. [Nerve fibers that release norepinephrine (or epinephrine (adrenaline)) are called adrenergic nerve fibers.l One set of spinal nerves that originates in the lower thoracic region of the spinal

cord send long preganglionic nerve fibers to the adrenal medull4 a region within the adrenal glands (located on the superior surface of the kidneys). These nerve fibers synapse directly on the ganglia in the adrenal medulla. There are no postganglionic nerve fibers. When the adrenal medulla is stimulated, both norepinephrine and epinephrine are released directly into the bloodstream. Because these substances are released into the bloodstream, we can refer to them as hormones. Therefore, the adrenal medulla can be considered as an endocrine

gland. These hormones, distributed throughout the body by the circulatory system, can increase the heart rate and cause the pupils to dilate.


32


Biology

Nerve & Muscle Postganglionic nerve fiber Eye


Neurotransmitters are acetylcholine and norepinephrine

{Vi,

t( .& LJ /^

Preganglionic nerve flber

Hearr

Thoracic Lungs and Bronchi

bo

Stomach O d)

Pancreas

o,

Small (n

Kidney

intestine

Large intestine

'o k O

aa

Bladder

Figure I -25 . -:.:::irC nefves.

Somatic vs Autonomic Nervous System

-: , : :evierv the basic differences between the somatic nervous svstem and

the

-: - ,-:r-ic nervous system. In the somatic nervous system, we find that (a) once . :.=:r,'e fibers ieave the central nervous system, they do not make a synapse :-- -:,ev have reached their effector organ. When the synapse is made at the -:--: organ, (b) the neurotransmiiter that is released is acetylcholine. The :

nervous system (c) innervates skeletal muscle. Innervation of that (d) Ieads to excitation of the muscle itseif. nuscle =:-':--:-:

autonomic nervous system, we find that (a) once the nerve fibers leave the

:::- :,en-ous system, they synapse with a ganglion before they make the final =:-= rvith their effector organ. The preganglionic fibers in both the ::':-rathetic and sympathetic divisions release (b) acetylcholine as the -:,-:=.smitter. The postganglionic fibers in the parasympathetic division =:'= a:eivicholine; in the sympathetic division, norepinephrine is released. ::nomic nervous system (c) innervates glands, and smooth and cardiac

l:'e

ce1ls innervated

by the autonomic nervous system can (d) be either

or inhibitory

The Berkelev Review


Biology

Nerve & Muscle


The adrenal medulla, a specialized ganglion in the sympathetic division of the autonomic nervous system, is directly stimulated by a preganglionic fiber. This nerve fiber releases acetylcholine, which causes the ceils of the adrenal meduila to release two types of hormones into the biood. The major hormone released is epinephrine. The minor hormone released into the blood is norepinephrine. These differences are illustrated in Figure 1-26.

CNS Skeletal muscle

Somatic Nervous System

Acetvlcholine Ganglion

Giands, cardiac and smooth muscle

Postganglionic nerve fiber

l-IE t'ts

Acetylcholine

Autonomic Nervous System

.*_|or-g".e

F+

Biood

(Epineprhine) Adrenal Medulla

Ganglion

Preganglionic nerve fiber

r-)

Norepinephrine

>s t4 l5 lo l-

)l6'

Organ

Glands, cardiac and smooth muscle

t la

t: IrJ ID J5

@

o

Figure l-26 CNS and PNS review.


34


Biology

Nerve & Muscle

KeGeptof5,:,and,ri .S€n5,6ff

Receptors and Sensory Input

,.,Iffit

There are many different types of receptors receiving sensory information from

the environment and passing that information to the nervous system. For erample, there are mechanoreceptors, which are concerned with pressure, hearlng, balance and blood pressure. Nociceptors sense pain. Mechanoreceptors :espond to a change in their configuration. Some can respond to a light pressure, -,r-hile others respond to a deeper pressure. There are specialized regions on the =:des of fish called lateral line organs that respond to a change in the pressure of :re rvater. There are pressure receptors in the walls of the aorta that are able to ::nse an increase in blood pressure.

Thermoreceptors detect cold and warmth, while chemoreceptors can detect -:--te, smell, oxygen, carbon dioxide, hydrogen ions, and blood giucose ievels. -:-.e taste receptors in the tongue can distinguish between food which is bitter, .:'.r, salty, or sweet. Olfactory cells rn the nasal cavity can distinguish hundreds - ilfferent odors. Photoreceptors in the retina of the eye can respond to the ::.sence of a single photon of light. There are also receptors concerned with :,ectricity and magnetism. For example, catfish have electrical receptors that ..=.r them detect prey, and birds have magnetic receptors that help them to ---:' 1gate. - -.= sensory information that a receptor receives is specific for that particular :=:E:tor. The stimulus that is received by that receptor changes (converts or ::ansduces) the potentiai (V*) across the receptor's membrane. This change in =::'.biane potential is called a receptor potential. If this receptor potential were ', ='.:eed a specific threshold, an action potential would be generated. We also :: -,rat if the pressure on this receptor is great and the change in membrane : :=:.iial is large, then the receptor potential will exceed the threshoid level. The :::*-: is an increase in the frequency of the action potentials being generated. . . = ,:nplitude of the action potentials r,vi11 not be change, only the frequency.

Receptors and Transducers .: . :onsider the receptor potential for a special type of mechanoreceptor ca11ed :acinian corpuscle. A pacinian corpuscle has an unmyelinated nerve ending . -::. is encapsulated in layers of connective tissue. However, the afferent nerve -- :, . -=ar-es

this encapsulated ending is myelinated. Saltatory Pressure

_/\:*v + +

\----l--\------r'

'','gUre 1.27

',, - .--. -'.rrpuscle.

.ii

0 by The Berkeley Review

cc


Biology

Nerve Et Muscle


ending is depressed, a local deformation causes sodium channels to open and Nao ions to rush into the nerve ending. An action potential is not generated in the receptor region. Instead, the establishment of this receptor potential causes a localized flow of current to be propagated along the nerve fiber. If the threshold has been reached, then this local current flow will jnitiate an action potential at the first node of Ranvier, located at the outer edge of the pacinian corpuscle. The action poiential can spread down the afferent nerve and towards the central nervous system by saltatory conduction, as shown in Figure \zVhen the nerve

7-27.

Suppose we were to depress the pacinian corpuscle just a smail amount. If the threshold is not reached, an action potentiai will not be generated. This means one would not feel the pressure that is being applied. If we apply a second stimulus, we might depolarize the membrane even more. If we were to apply a third stimulus that was even stronger, we could exceed threshold and an action potentiai would be generated. If we maintain this pressure such that we are just

above threshold by a certain amount, we will continue to generate action potentials at the rate of, say,2 action potentials every 10 milliseconds. 2 action potentials per 10 miliiseconds

3 action potentials per 10 milliseconds

Threshold

o lst Weak

[fr

3rd

Time (ms)

tl

Strength

of

stimulus

Strong

4rh

Figure l-28 Acti0n potential frequencies.

What wouid happen if we apply a fourth stimulus, much stronger than the previous stimuli? The membrane would become more depolarized, and we would move further above the threshoid level. Once this happens, the frequency of action potentials generated would increase, say, to 3 action potentials every 10 milliseconds. This is shown in Figure 1-28. Note that in each case the amplitude of the action potentials remains the same. \Alhat changes is the frequency of the action potentials propagating along the axon" It is the frequency of action potentials that signals to the central nervous system the magnitude of the stimulus being received. All receptors function on this basic principle.


36


Biology

Nerve 6r Muscle

Receptors and Sensory tnPut

Adaptation

wheriyou wake up in the morning and take a cool shower, you {eel the coolness of the water on your skin. When you get dressed, you feel the clothes touching your skin. In each case, though, after a brief period of time you get used to the water or the clothes touching your body. These are examples of sensory adaptation. One type of receptor that undergoes sensory adaptation is the pr"ir.r." receptor. if we plot the frequency of action potentiais (as they travel Lack to the central ,r*tno.tt system aiong a given axon) as a function of the time of a sustained stimulus, we would find that pressule receptors adapt rapidly' F{owever, pain receptors do not adapt raprdly, as is shown-in Figure 1-29' If the pressure ,u."ptott did not adapt to the touch of the clothes that we wear, it rvould plove to be rather inconvenient. [On the other hand, the body does not adapt ai readily to the sensation of pain--a good thing, evolutionarily speaking, silce pain is a warning of potentiai damage to the body's tissue and not something to be responded tb as a matter of convenience, but as a matter of sunival.]

)\= 'r O

Time of sustained stimulus (sec)

Figure l-29 -.:.::aiion to pressure

and to pain.

Tactile Discrimination

-"-=:J that the end of a neuron can be divided into many branches. Each of these ::::rches in turn can end at a receptor (such as a pacinian corpuscle). These 1.,--ra-l branches and their corresponding receptors constitute a receptive field. there can be many -.=:ending on which area of the body one is describing, :=,::iiit'e f-ieids, some of which overlap, or there can be a few receptive fields, :::-: o j tvhich do not overlap. Let's consider a set of receptive fields that overlap : :,rier io d.etermine how iwo points in space can be distinguished from one ci- :

same touch. =er bv the

:-::l=e \\e have a set of receptive fields like the ones shown in Figure 1-30a. ,--.., ."" generally finds is thit a greater frequency of action potentiais will be :

l 0 3

e

n .e

ra'

tn

if the central region rather than seem *: = :e:rphery of a glven receptive field is stimulated. In other words, there itsperiphery. in ,, , ," receptors in the central region of a receptive field than

r.:=:.:ed fass,tming threshold has been

reached)

-ore

outd happen if receptive fields overlapped, as shown in Figure 1-30? If ri: -.r-ere to sti.mulate the receptive field represented by (b) in Figure 1-30 ,:r:t:-i.*- enough, then because of the field overlap, we would also stimulate l::-= :::e::ive fields in (a) and (c). It would feel to a subject as if three different .r::. _: e body were being stimulated when, in fact, just the receptive field in from these b -. :eing stimulated. HJw can the information being received (b) is being field only .:5s.-r-. a-ierent neurons be fine-tuned to let us know that

r,,:::

-,,.

- :' :::::

3

b1' The BerkeleY

Review

5/


Biology

Nerve Ef Muscle


stimulated, and not field (a) or field (c)? This is accomplished by a process called lateral inhibition, mediated through interneurons within the spinal cord. In Figure 1-30, the axon that leads away from field (b) has lateral connections to interneurons that inhibit the impulses being sent down the axons from fields (a) and (c). In other words, because of lateral inhibition the impulses generated in receptive fields (a) and (c) are suppressed, allowing the impulses from the receptive field in (b) reach the proper spinal tracts that ascend to the brain. To Brain

6 a

(a)

+

(b)

D

(c)

+ Lateral Inhibition (interneuron)

Figure l-3O Receptive fields.

Somatic Sensory Pathways Once the sensory afferent nerve fibers enter the spinal cord they cross to the side opposite from the side they entered, either in the spinal cord or in the brainstem.

In other words, the sensory input from the right side of the body witl

be

represented on the somatosensory cortex of the left cerebral hemisphere, and sensory input from the left side of the body will be represented on the somatosensory cortex of the right cerebral hemisphere. [In Figure 1-39 which side of the body are the receptive fields located, the right side or the left side? Will the sensory input go to the somatosensory cortex of the right or left cerebral hemisphere? How do you know?]

Where do these sensory neurons synapse as they ascend to the brain? As a general rule, there are three neurons involved in sensory pathways. In the case of pressure, we find that the first-order neurons carrying information from the receptive field(s) enters the spinal cord and synapse with second-order neurons that ascend on the opposite side of the spinal cord to the thalamus. Here, another synapse is made with third-order neurons that continue to ascend, until they reach a specific region of the somatosensory cortex of the cerebral hemisphere opposite to the side of the body in which the sensation was perceived. The cerebral cortex itself contains ceils that are organized into 6 horizontal layers. The sensations of pressure would be registered in cells arranged in columns that cross these different lavers" Copyright @ by The Berkeley Review

3a


Nerrre & Musc[e To Go 15 Passages

98 Questions

Time for All Passages Taken Together as a Practice Exam 125 Minutes

I. il. IIII. IV. V. VI. VIII. WII. IX. X. XL XIII. XIIII. XIV. XV.

Passage Titles Types ofTransport Autonomic Nervons SYstem Action Potentials Local Anesthetic The Lens, the lris, & Associated Muscles Resting Membrane Potential Nicotine Replacement RetinalProjection Axonctl Transport Huntington's Disease Photoreceptors Sound Transmission in the Ear T-r,-ptophan & Serotonin Experiment Frog Muscle Experiment Skeletal Muscle Groups

,TftE

EIEru{EMY

L,g.p-y.1.fl-1ry'

Specializing in MCAT Preparation

Questions

1-8 9-1.3

14-20

2t-26 a l /./- - JJ

34-40

4r-46

47 -52

53-58 59-64 65-71 12-19 80-85 86-92 93-98

Suggestions The passages that follow are designed to get you to think irr a conceptual manner about the processes of physiology at the organismal ievel. If you have a solid foundation in physiology, *u1y of these answers will be straightforward. If you have not had a pleasant experience with the topic, some of these answers might appear to come from the void past the Oort field of the solar system. Pick a few passage topics at random. For these initial few passages, do not worry about the time. Just focus on what is expected of you. First, read the passage. Second, look at any diagrams, charts, or graphs. Third, read each question and the accompanying answers carefully. Fourth, answer the questions the best you can. Check the soiutions and see how you did" \A4rether you got the answers right or wrong/ it is important to read the explanations and see if you understand (and agree with) whal is berng explained. Keep a record of your results.

After you feel comfortable with the format of those initial few passages, pick another block of try them. Be aware that time is going to become important. Generally, you will have about

passages and 1

minute and 15 seconds to complete a question. Be a little more creative in how you approach this next

group. If you feel comfortable with the outline presented above, fine. If not, then try different

approaches to a passage. For example, you might feei well versed enough to read the questions first and then try to answer some of them, without ever having read the passage. Maybe yor.un answer some of the questions by just looking at the diagrams, charts, or graphs that are pr"r"rrtud in a particular passage. Remember, we are not clones of one another. You need to begin to develop a format that works best for you. Keeping a record of your results may be helpful.

The last block of passages might contain topics that are unfamiiiar to you. Find a place where the level of distraction is at a minimum. Get out your watch and time yontr!ff on these parruges, either individually or as a group. It is important to have a feei for time, and how much is passrng aslou try to answer each question. Never let a question get you flustered. If you carurot figure ont *nui the answer is from information given to you in the passage, or from your own knowledge-Lase, dump it and move on to the next question. As vou do this, make a note of that pesky question ind come back to it at the end, when you have more time. When you are finished, check your answers and make sure you understand the solutions. Be inquisitive. If you do not know the answer to something, look it up. The solution tends to stay with you longer. (For example, what is the Oort field?)

The estimated score conversions for 100 questions are shown below. At best, these are rough approximations and should be used only to give one a feel for which ballpark they are sitting in.

Section I Estimated Score Conversions Scaled Score

>12

l0-

11

8-9 7 6 5

<4

Raw Score 86

- 100

19 -85

6s-78 59-64 54-s8 48-53 0-47

Biology

Types of Transport

Passage I (Questions l-8)

Passage I

The transport of substances across a cellular membrane by a carrier protein occurs either through a uniport, a symport, or an antiport mechanism.

Animal ceils have membranes containing a vast array and proteins. Some of the transmembrane proteins in these membranes act as channels that allow for the simpie diffusion (passive transport) of water-soluble molecuies or ions of low molecuiar weight between the intracellular and extracellular regions of the cell. The ionic components of these two regions differ considerably

of lipids

see Tabie

i

Organelie

o O

l).

ADP + P1 Selected Cellular Ionic Components

Ion

[Intracellular] [Extracellular]

mM mM 120 mM 4 mM

O k

tEion

(.)

mM

140

+58 mV

K+

135 mM

4

-92 mV

'o

cl-

4mM

-89 mV

rl

15

Na+

^ 1L Lil-'

2

x

10-4 mM

\'lg2*

ATp

Ca2+ ADp +

pi

Ca

o.

+129 mV

2mM

40 mM

'Eion = Equilibrium potential fbr the ion

Table

1

Figure I

Other transmembrane proteins act as carriers to r:nsport ions and molecules across the cell membrane. If :"-.: substance is too large (e.g., glucose), it will require a : =-ier protein. This type of carrier-mediated transport is

1.

: ''i.et .faciLitated diffusion, and the rate of diffusion wiil

:--r-ise until all of the carrier binding sites on the

::ir.brane transport protein

are filled.

A. B. C. D.

I- energy is required to move a substance across a r::r,:iane, the process is called active transport. Primary r- -r-.

i roDsport uses the energy

of ATP directly to move a

,rS.ir,Ce against its concentration gradient (Figure 1). .--,: --arrier molecule is the membrane-bound protein :- - F.::. Only three primary active transport systems ::'.. teen identified: one for Na@ and Ko ions, one for -

,l i

:rns. and one for Ho ions.

A11

-r

A. B. C. D.

--rtr oi ATP directly, but rather use the energy stored in gradient established across a celluiar

. :.-:inlration

- :;::::ne.

-::nsmembrane proteins that act as channels in nerve

-: ceils can be electrically or chemically activated. -"": -s--le ':'ting membrane potential (Em) in these ce1ls is ,' - - -i,j mV. The driving force that moves an ion 'lr'-:

!

*

:::

Ion

of the following will lead to inhibition of

the

intraceliuiar[ATP]. intracellularlNa@1. extracellular [Ke]. extracelluiar [Nae].

all of the following EXCEPT:

A. B. C. D.

- E* - Eion


11s

Glucose can be carried across the cell membrane by

-:-. ,:ne of these channels is the difference between ,- : :,on. as shown in the foliowing equation: Force Moving

protein HCO3Po43-

Na+/K+-ATPase EXCEPT a low:

!:--ondary active transport systems do not use the ;

Compared to the extracellular fluid, the intracellular fluid will contain higher concentrations of all of the following substances EXCEPT:

4l

faciiitateddiffusion. primary active transport. a symport mechanism.

secondary active transport.


Biology 4.

Tylres of Tfansport

Glucose, despite

its reiatively high intracellular

All of

concentration, can dissociate from its carrier protein on the cytosolic side of the membrane and into the cytoplasm. because rhe:

A.

low intracellular [Na@] allows sodium to

the following characteristics are common to

A. B. C.

be

D.

high extracellular [Nao] allows glucose to be

cytoplasmiccontinuity. rapid transmission of action potentials.

coordination

of muscle contraction in

the

heart. a high resistance electrical pathway.

released from the carrier protein.

C.

affinity of the carrier protein for intracellular sodium is high.

8.

D. affinity of

the carrier protein for glucose decreases, once sodium binds to the carrier protein on the extracellular side of the

A.

Decreased primary active transport. Decreased secondary active transport. Increased sodium concentration gradient. Increased intracellular calcium levels.

?

B.

/--

/: /,"

FI)

/

// ,,' /

SD

/ IGradient]

IGradient]

/to. / ./*

st)

-

//

// lGradientl

6.

4, B. C. D.

Which of the following graphs best represents the relationship between simple diffusion (SD) and facilitated diffusion (FD)

Which of the following will NOT occur across a cell's membrane, if the eiectron transport chain is

inhibited?

membrane.

5.

IGradient]

The acetylcholine receptor within the postsynaptic membrane of a neuromuscular junction binds two molecules of acetylcholine and allows Na@ and Ke to flow through the membrane. The acetylcholine receptor can best be described by its ability to allow:

A. B.

Na@ to enter the cell, leading to repolarization.

more Na@ to enter the cell than K@ to leave the cell.

C. D.

Ko to enter the cell than Nae to leave the cell. a change in the membrane potential to lead to more

its activation. Copyright @ by The Berkeley Review

I

gap junctions that exist between cells EXCEpT:

released from the carrier protein.

B.

Passage

42


Biology

Autonomic Nervous System

Passage II (Questions 9-15)

Passage Il

10. Which of these situations would stimulate the SNS?

I. il. ilI. W.

The autonomic nervous system (ANS) is concerned primarily with the regulation of visceral function in response to external and internal stimuli. The two divisions of the ANS are the parasympathetic nervous system (PNS) and sympathetic nervous system (SNS).

Extreme heat. Extreme cold. Sleeping.

Running a marathon on a sunny day.

A. I and II only B. I, II, and III only C. IIi and IV only D. I, II, and IV only

The SNS is associated with readying the body for stressful situations, while the PNS is involved in maintaining bodily functions at basal levels. These subgroups use a two-neuron system. A primary neuron rvill extend from the central nervous system (CNS) while a secondary neuron wiil extend fiom a ganglion to the nerve's target organ.

11. Which

of the following

symptoms would be

considered a PNS response?

The primary and secondary neurons for the PNS secrete the neurotransmitter acetylcholine (ACh). The SNS, however, uses two different neurotransmitters. ACh is secreted fiom the primary neuron, while norepinephrine ,NE) is secreled from the secondary neuron.

A. B. C. D.

Blushing Gastrointestinal immotiltiy Piloerection Penile erection

The SNS has large coliections of ganglia that form an intricate network. The most prominent of these ganglia are the paravertebral ganglia, which are adjacent to the

vertebral column. The prevertebral ganglia lie along the anterior surface of the great vessels of the abdominal :avity and on the adrenal medulla. The PNS has ganglia ihat are immediately adjacent to the organs that are being

The adrenal gland is considered an extension of the SNS. Stimuiation of the SNS will initiate the release of the hormones epinephrine and norepinephrine from the adrenai medulla. Cortisoi and aldosterone, produced in the adrenal cortex, will also be released. Which of the effects below can be attributed to the hormones released by SNS stimulation of the adrenal cortex ?

innervated.

9.

The controi of individual organs is considered an important function of the ANS. However, the PNS component is thought to be more specific in its regulation than the SNS. Which statement(s) best explain this theory

I.

I. U. III. IV.

?

A response to stressful stimuli

III. IV.

specific.

The SNS developed much later in evolution

Storage of glucose in the liver.

and thus is not as refined as the PNS.

The regulation needed to maintain the bodily systems at equilibrium and after stress must be more specialized to each organ innervated. The location of the gangiia of the PNS allows it to regulate organ function more closely.

13. The vagus (1Oth craniai) nerve is a component of the PNS. Which of the following effects is LEAST like1y to be associated with the PNS?

A. I only B. II, III, and IV only C. III and IV only D. I, III, and IV only

Copyright

Retention of ions and water in the kidney. Proteolysis

A. I and II only B. I, II, and III only C. III and IV only D. I, il, m, and IV

needs a general

stimulation of the body, thus the SNS is less

il.

Gluconeogenesis

@


A.

Constriction of the pupils.

B.

Increased heart rate. Increased gastric motility. Increased secretion of HCI in the stomach

C.

D.

43


Biology

Action Potentials

Passage III (Questions 14.2O) Many animal cells have an intracellular fluid (ICF) that is electrically negative relative to the extracellular

fluid (ECF), creating a voltage difference across the ceil's membrane called the resting membrane potential (E^). The E* in nerve cells and muscle cells is about -80 mV, while in epithelial and red blood cells it is abour -30 mV. In nerve and muscle cells the concentration gradients and the permeability of the membrane to both Na@ and Ko determine the E6. The concentrations of these ions in

the ICF and ECF are shown in Table

1.

They

Passage III

If a stimulus is strong enough so that the threshold potential of a cell is reached (about 15 mV from the E,n), an action potential (Figure 2) will be generated that will

propagate along the nerve

cell until it

reaches the

presynaptic ending. A neurotransmitter is released and diffuses across the synaptic cleft, where it binds to the postsynaptic membrane. A postsynaptic potential of equal magnitude is generated in the postsynaptic cell and the signal continues. The response of the action potentiai is all-or-none. The phases are shown in Table 2.

are

maintained by the Nao/Ke-ATPase pump, a pump that transports 3 Nao ions to the ECF for every 2 Ke ions transported to the ICF. At rest, the relative permeability of the membrane to Ko is roughly 10 times greater than it is to Nao.

Table

Ion Na+

K+

8-+l

1

flntracellular] fExtracellular] fEion mM

140 mM

+58 mV

135 mM

4mM

-92mY

15

rEion = Equilibrium potential for the ion

If the En-' transiently changes from its resting value, an electrical signal is generated in the nerve cell. There are two types of electrical signals: graded potentials and action potentials (AP). Graded potentials are signals that operate over short distances, while action potentials are signals that operate over long distances.

Time (msec)

Figure 2 Table 2

1. Resting State: The cell at its En'. 2. Threshold: Potential that generates an Ap. 3. Depolarization: Due to influx of Na+. 4. Overshoot: AP with a positive membrane potential. 5. Repolarization: Due to efflux of K+. 6. Hyperpolarization: Membrane potential is more negative than the E'n.

Along the plasma membrane of a nerve cell are voltage-sensitive Nae and Ko channels that contain gates. The Nao channel has an m gate that opens quickly and an h gate that closes slowly upon depolarization. The K @ channel has an n gate that opens slowly upon depolarization. The Nae gate can be blocked by tetrodotoxin (TTX), and the K@ gate can be blocked by tetraethylammonium (TEA). In the resting state the Nao and K@ channels have their gates arranged as shown in Figure

7. 8.

stimulus is strong enough.

14.

1:

Absolute Refractory Period: AP cannot be generated. Relative Refractory Period: AP can be generated if the

The threshold potential for the generation of

an

action potential can be found at that value of the

E1n

where the:

A. B. C. D.

influx of K@ balances the efflux of Nao. efflux of Ko is 10 times greater than the influx of Nao. efflux of K@ balances the influx of Nae. influx of Nao is 10 times greater than the elflux of Ke.

Figure

1



Biology

Action Potentials 20.

15. The threshold potential for the action potential shown in Figure

A. B. C. D.

I

is:

Passage Itr

Gamma-aminobutyric acid (GABA) is a major inhibitory neurotransmitter in the central nervous system. It is synthesized from glutamate, a major excitatory neurotransmitter in the brain.

-50 mV. -65 mV. -80 mV.

COOH

@I H.N-C-H"

-92mY.

i

Ot

.

Enzyme

H

HrNC-l

H

CH.

16.

During depolarization, the quick opening of the voltage dependent m gates results in all of the

C. D.

an

I

cH,

COOH

COOH

t-

following EXCEPT:

A. B.

I

CH,

Glutamic acid

influx of Nao ions.

t-

GABA

The enzyme that catalyzes this reaction is best

an increase in the opening of fast m gates in neighboring Na@ channels. an increase in the opening of fast n gates in neighboring K@ channels. an absolute refractory period.

described as:

A.

GABA decarboxylase.

B. C. D.

GABA carboxylase.

glutamic acid decarboxylase. glutamic acid carboxylase.

17. Which of the following voltage gated positions best represents the repolarization phase?

A. B. C.

D.

1&

m m m m

gate gate gate gate

open; & gate closed; /, gate open. closed; h gate closed; r? gate open. open; I gate open; n gate open. closed; h gate open;n gate closed.

During the last half of the relative refractory period shown in Figure 1, the membrane potential is beginning to:

d B. C. D.

repolarize. hypelpolarize. depolarize. hypopolarize.

ft- If TIX and TEA are added to a nerve cell preparation and a depolarizing stimulus is applied to tire presynaptic ending, a postsynaptic potential

will

te produced. This shows that:

-{-

Nae and K@ ions are needed for the release of the neurotransmitter to occur.

B

Nao and Ke ions are not responsible for the release of the neurotransmitter.

C. Tfi

and TEA can amplify the postsynaptic potential by allowing neurotransmitter release.

D.

an increase in ICF K@ ions and an increase in

ECF Nae ions inhibit the release of the neurotransmitter, unless a depolarizing stimulus is applied.

I tr

q'ryynght

@


45


Biology Passage

Local Anesthetic

IV (Questions 21-26)

Passage

23. Which of the following sratemenrs is

IV

LEAST

compatible with the effects of local anesthetics?

Local anesthetics provide a reversible block in the conduction of impulses in nerve fibers. A11 local

anesthetic drugs contain a lipophilic group, an intermediate chain, and a hydrophilic group. The

A.

Local anesthetics slow the rise of the action

B.

intermediate chain may be either an ester (as in procaine) or an amide (as in lidocaine). The acriviry and potency of

Local anesthetics slow the propagation of

C. Local anesthetics

nerve impulses.

any local anesthetic depends on several aspects, including its chemicai structure, and the pKu of the substance.

D.

Local anesthetics are weak bases, so they are kept in an acid solution in which they form a water-soluble salt. The base is lipid-soluble and can penetrate various tissue barriers. The concentration of base or cation in the solution depends on the pKu of the local anesthetic. A decrease in the amount of base facilitates removal of the local anesthetics, resulting in a shorter duration of action. The duration of action of a local anesthetic is also affected by the local vasodilation which most of them produce. Finally, experiments have shown that a nerve which has been recently and repetitively stimulated is more sensitive to a local anesthetic than a resting nerve.

J

24.

The pKu of iidocaine is somewhere between 1.6 and

statements is the MOST likely conclusion?

A. B. C. D.

CH.

C

H

t<

CH:

If

F'igure I

Local anesthetics penetrate the blood-brain barrier

acidosis develops after a locai anesthetic has

A.

the amount of ionized form will be decreased, and the local anesthetic will freely cross the

B.

the amount of ionized form wiil be decreased, and the local anesthetic will be trapped in the

C.

the amount of ionized form will be increased, and the local anesthetic will freely cross the tissue barrier. the amount of ionized form will be increased, and the local anesthetic will be trapped in the

with great ease, most likely because of their:

c.

large molecular size and high lipid solubility. large molecular size and low lipid solubility. small molecular size and high lipid solubility.

D.

small molecular size and low lipid solubility.

B.

The onset of lidocaine is more rapid, because more exists in its base form at body pH. The onset of lidocaine is more rapid, because more exists in its acidic form at body pH. The onset of lidocaine is slower, because more exists in its base form at body pH. The onset of lidocaine is slower, because more exists in its acid form at body pH.

penetrated a tissue barrier, it is most likely that:

Lidocaine at pH = 9.0

A.

decrease threshold for

eiectrical stimulation in nerve fibers. Local anesthetics prevent the depolarization of cell membranes.

7.8. The pKs of procaine is between 8.i and 8.6. Based on this information, which of the followins

/-\ .. 'i cHz_ j,i_l'"'qHs \ ttr y- c-

21.

potentiai in nerve fibers.

D.

tissue barrier.

tissue.

tissue.

ta

Based on the passage, which statements is most

A. B.

C.

D.

Copyright

of the following

26. The structure of

likely true?


in Figure

tr.

than orally. This is because lidocaine:

The higher the pKu of the local anesthetic, the lower the concentration of base in the tissue. The lower the pKa of the local anesthetic, the higher the concentration of acid in the tissue. The higher the Ku of the local anesthetic, the lower the concentration of base in the tissue. The lower the Ku of the local anesthetic, the lower the concentration of acid in the tissue.

@

lidocaine is given

Lidocaine is administered through injection, rather

A.

will be deactivated by the acidic conditions of

B. C.

the stomach. will not be absorbed across the intestinal wall . is biotransformed by reactions occurring in the

D.

liver. is biotransformed by reactions occurring in the spleen.

46


Biology

The Lens, the lris, & Associated Muscles

Passage V (Questions

2'7

27.

-33)

Passage V

For

a person watching a meteor shower at midnight, what are the contraction states of the circular muscle

of the iris, the radial muscle of the iris. and the

Figure 1 shows the lens, the iris, and the muscles of

ciliary muscle?

the eye.

A. \

Retina

B.

Vitreous Chamber

C.

Suspensory

Ligaments Optic Nerve

D. Figure

i

The circular muscle of the iris is contracted, the radial muscie of the iris is relaxed, and the ciliary muscle is relaxed. The circular muscle of the iris is relaxed. the radial muscle of the iris is contracted. and the ciliary muscle is contracted. The circular muscle of the iris is relaxed, the radial muscle of the iris is relaxed. and the ciliary muscle is relaxed. The circular muscle of the iris is relaxed, the radial muscle of the iris is contracted, and the ciliary muscle is relaxed.

The opening of the pupil of the eye is controlled by :'.;,o sets of muscles. The circularly arranged smooth :uscle is under parasympathetic control, and the radially .-an-sed smooth muscle is under sympathetic control. F-qure 2 shows the relation between these muscles and the of the eye. The size of the pupil reacts to the amount

28. As people age, the lens becomes less flexible and

:::il

therefore less able

: iight present.

R.,lial Smooth

\lrscle

to

change shape during

accommodation. What happens to a person's vision with these age changes? A.

Circular Smooth Muscle Fibers

Fibers

The lens is less able to focus the iight rays from near objects.

The lens is more able to fbcus the light rays Si'mpathetic

C.

Parasympathetic

Fiber

Normal

D.

29.

Bright Light

Dim Light

from near ob.jects. The lens is more able to focus the iight rays from far objects. The lens is less able to focus the iight rays from far objects.

Special solutions are used to dilate the pupil during a retinal exan. so that the retina can be viewed. The

Figure 2

soiution causes an inability to focus on near objects, reduces clear distance vision, and enlarges the pupil.

The lens of the eye is suspended within and -r:rnded by a ring of tissue called the ciliarir muscle. : -,.r:nsory ligaments connect the lens to the ciliary - '..:ie. When the ciiiary muscle is relaxed, the

What is the stimulation or inhibition required to

-:::rsory

produce these changes in the eye?

A.

ligaments pull the lens taut and flat. Since we

B.

:-:: most of our time in far-away vision (20 feet or - ',:: this is ideai. For closer work, the lens must r : . :-.modate and become thicker for focusing. In the :. -:.ss. the ciliary muscle tightens, the suspensory :::.3nls become slack, and the lens thickens

-

::

C. D.

Sympathetic nerves are inhibited, and parasympathetic nerves are stimulated.

Sympathetic nerves are stimulated, and parasympathetic nerves are inhibited. Both are stimulated.

Both are inhibited.

and

,.:res mOIe COnVeX.

- - : .:ieht €l by The Berkeley Review

47


Biology 30.

The Lens, the lris, & Associated Muscles

When light rays travel through the pupil and the lens, upon which part of the eye does the image focus?

31.

A.

Retina

B. C. D.

Cornea Optic nerve Vitreous body

One eye exercise invoives focusing on an object while it moves from arm's length to the closest point upon which you can focus. This causes the lens to undergo accommodation, that is, the lens must change shape to keep the object in focus as it moves cioser. As the object moves nearer, what changes occur in the eye?

I. il. III.

The pupil gradually contracts. The ciliary muscle gradually contracts. The suspensory ligaments gradually contract.

A. I only B. II only C. III only D. II and III only 32. The iris is pigmented epithelial tissue. The color of the eye is determined by the amount of pigment. Biue eyes have the least pigment, brown eyes have more, and black eyes have the greatest amount of pigment. What coior are the eyes of a person who has no pigment in her iris?

A. B. C. D. 33.

Black White Pink Green

Astigmatism is the condition of having a nonuniformly shaped lens or cornea. This means that parallel light rays do not focus, so that a sharp image

is not formed. How could this eye condition be corrected?

A. By providing a convex corrective lens. B. By providing a concave corrective lens. C. By providing a non-uniform conective lens. D. By providing a uniform corrective lens.

Copyright

@


4A


Biology

Resting Membrane Potential 34.

Passage VI (Questions 34-40)

This electric potential is more commonly known

The cell membrane acts as an insulator, because the:

A.

Separated electric charges of opposite sign have the capacity to do work if they are allowed to come together.

B.

as

simply the potential. A11 cells in a resting state have a difference in potential across their cell membranes, the

of three kinds of ions within

Ion

are

charged.

proteins making up the cell membrane

C. lipids making up the cell membrane

are are

charged.

D. iipids making up the cell membrane

The magnitude of the resting membrane potential is determined primarily by two factors. One is the typical intracellular and extracellular fluids (Table

proteins making up the cell membrane uncharged.

inside being negative with respect to the outside.

concentration

Passage VI

are

uncharged.

the

1):

Extracellular Intracellular (mmol/L) (mmol/L)

Nao

150

15

cle 6o

1i0

10

5

150

35.

The resting membrane potential of a typical neuron is:

A. B.

Table I The second factor is the ability of each kind of ion to penetrate the plasma membrane, an ability that fluctuates frorn one moment to the next. In Table 1. notice that the sodium and chloride ion concentrations are generally lower inside in the cell, while potassium ion concentration is lower outside the cell. The concentration differences between sodium and potassium are produced by a plasma membrane active transport system that simultaneously purnps sodium out ofthe cell and potassium into the cell.

C. D.

As an ion moves down its concentration gradient, an electric force is established that opposes the movement of that ion. The membrane potential that exists when the electric force is equal in magnitude but opposite in direction to the concentration force is the equilibrium potential for that ion. The equilibrium potential for each ion species is different in magnitude--sometimes even in direction--and can be calculated using the Nernst

36.

positive, with the excess charge representing a very large fraction of the total number of ions inside and outside the cell. positive, with the excess charge representing a very small fraction of the total number of ions inside and outside the cell. negative, with the excess charge representing a very large fraction of the total number of ions inside and outside the cell. negative, with the excess charge representing a very small fraction of the total number of ions inside and outside the cell.

Which of the following diagrams BEST represents the equiiibrium potential for potassium across a nerve cell membrane?

A.

B.

equation:

e=#,,? In a nerve cell at rest, the cell membrane is 50 to 75

times more permeable to potassium than to sodium. During the resting potential, there is a net diffusion of ions into and out of the cell. The concentration gradients, rvhich would eventually dissipate, are maintained through

the Nao/Ke pump. If the concentration gradients remain fixed and the permeabilities of ions do not change, the

B

High

resting membrane potential remains constant.

11o


49

The Berkeley Review Speciatizing in MCAT Preparation

Biology

Resting Membrane Potential

37. The resting membrane potential lies

Passage VI

above the

equilibrium potential for potassium, because a small number of:

A.

chloride ions diffuse into the cell. chloride ions diffuse out of the cell. sodium ions diffuse into the cell. sodium ions diffuse out of the cell.

B. C.

D.

38.

According to information in the passage, the concentration gradients of ions are maintained through the Nae/ I(9 pump. The ultimate source of energy for this pump is the:

A. B. C. D.

39.

formation of ATP. breakdown of ATP. oxidation of nutrients. reduction of nutrients.

The constancy of the resting membrane potential is an example of:

A. B. C. D.

40.

an equilibrium state. a dynamic state. a steady state. a static state.

The plasma membrane of many cells is permeable to

chloride ions and does not contain chloride ion

pumps. The membrane potential set up by other ions

thus acts on chloride ions. According to this information and information in the passage, the new resting membrane potential is:

A. B. C. D. Copyright

changed only slightly by the chloride ion. changed significantly by the chloride ion. changed geatly by the chloride ion. not affected at all by the chloride ion.

@


50


Biology Passage

VII

Nicotine Keplacement Therapy 42.

(Questions 41-46)

The following graph shows the results of a binding assay on the muscarinic acetylcholine receptor:

Acetylcholine is one of the body's most important neurotransmitters, responsible for the transmission of nerve impulses across synaptic junctions. There are two rnain classes of acetylcholine receptors. The nicotinic acetvlcholine receptor responds to nicotine as an agonist

and to curare as an antagonist. The

acetvlcholine receptor responds :gonist and to scopolamine

to

Er :t/ dl/ u l/ vi tt .) a li

muscarinic

muscarine as

an

f4|u

'n

as an antagonist.

=

ll

tl

€t z

\icotine is a psychoactive alkaloid extracted from ::,bacco piants. It is a toxic substance, one that places a :Iiiss on the heart and the entire cardiovascular system. 3:;ause this drug is quite addictive, several techniques :-i; been designed to cure nicotine dependence.

IMuscarine]

Scopolamine, a competitive inhibitor of muscarine, is added to the solution and the assay is repeated. Which of the following graphs BEST represents the results of the second assay?

In a novel nicotine replacement therapy, a : ,;::bination of physostigmine and scopolamine is b:Trinistered. Physostigmine is an acetylcholinesterase :,::'ritor at both muscarinic and nicotinic acetylcholine -i:E:iors. It must be noted that scopolamine is functional lr '.n'e i€ntr?l nervous system, but receptors in heart tissue ,rr:: :nsensitive to scopolamine. Figure 1 schematically !d:r,;:-i the action of these substances in the autonomic iltf,

Passage VII

B.

A.

ot al OI

o

a 9x oO

EgF Fo

I

3o 91./ | z

3o

-,ii-< S\-Stem,

/'

/

'-

IMuscarine]

i:r:.i

:npathetic division

...,--.J. -2

^_-_

/

Nicotinic

receptor

-1,

+_

/'

Muscarinic receptor

D.

C.

:

t

s_6

I /

:t dl/dl/ ol/al/

=

aEt/

t

aaLl ll eC .d llll 6u =

zvz

I

IMuscarine]

IMuscarine]

=----

/

-'-

vlll

oal cE I // 'cE II // d.l ot/o[ 6l/ail l/ =

mrr,::,eric dir ision

h

./'

\icotinic

Adrenergic receptor

ieceptor

43.

ll{i-i - -.-rr:i nen'ous system !.l = -!.9T"{::i:1j-.3 : = Norepinephrine

The nicotinic acetylcholine receptor, which is made of a pentamer of subunits, is essentially a sodium

channel. Which

of the following occurs directly

after acetylcholine binds to a nicotinic acetylcholine Figure

receptor?

1

A. An action potential is

iltlll* 'trtlriir:: :i fte lbllowing ions is responsible for the 1! ie:&:ce .-f acety'lcholine-filled vesicles from a mflbs'

B. An inhibitory

{T,artr J nen'e terminal?

@l

:1"

in

the

postsynaptic potential is

generated.

C. An excitatory

& 5i+ t. a€ . LL,_:L-

X'.urummlruqilrm

generated

postsynaptic cell.

postsynaptic potential is

generated.

D.

There is no change in the resting membrane potential of the postsynaptic cell.

lhe Berkeley Review

5l


Biology

Nicotine Keplacement Therapy

Passage VII

44. The novel nicotine treatment described in the passage theoretically should be superior to traditional nicotine addiction treatment, because it offers patients:

A. B. C. D.

45.

no net nicotine excitation. a net nicotinic and sympathetic nervous system excitation.

a net nicotinic and parasympathetic nervous system excitation. a net nicotinic and neuromuscular excitation.

Cardiovascular stress is one common result of ingesting nicotine by smoking cigarettes. This supports the idea that:

A. nicotine stimulation of the acetylcholine receptor

nicotinic

is dominant in

the

sympathetic nervous system.

B. nicotine stimulation of the nicotinic acetylcholine receptor is weaker in the C. D.

sympathetic nervous system.

nicotine acts directly on cardiac muscle to promote overstimulation and stress.

nicotine acts indirectly through the central nervous system to produce cardiac stress.

46.

As part of an experiment, a molecule is introduced at the synapses between heart muscle and nerve fibers

of both the sympathetic and parasympathetic nervous systems. The molecule moves in a retrograde fashion through both of these divisions. By the time it reaches the next synapse, the molecule will have traveled through: A. B.

a dendrite, moving a greater distance in the parasympathetic nervous system than in the sympathetic nervous system.

a dendrite, moving a shorter distance in the parasympathetic nervous system than in the sympathetic nervous system.

C. an axon,

moving a shorter distance in the sympathetic nervous system than in the

D.

parasympathetic nervous system.

an axon, moving a greater distance in the sympathetic nervous system than in the parasympathetic nervous system.

Copyright

@


52


Biology VIII

Passage

Passage VItr

Ketinal Projections 49.

(Questions 47-52)

The blind spot, located at the optic disc, is called so because:

The optic nerve is formed from the axons of all retinal ganglion cells. The optic nerves from each eye

this is the region where ganglion celis leave the retina.

roin at fue optlc chiasm and eventually enters either the .:ft or right bptic tract. The optic tract projects to three s.rbcorti&l areas. One is the lateral geniculate nucleus,

color vision is not available at this retinal C. D.

hich is responsible for processing visual information' which produces pupillar reflexes 'lne is the pritectal area, -:sed on lnformation from the retina' Finally, the .:perior colliculus uses the information from the retina to

',,.

location. this area is free of photoreceptors. light is unabie to reach this small area of the retina.

:::]erate eye movement.

\\'hen light is shone upon one eye, it

causes 50.

: , nsrriction of the pupil rn both eyes. Constriction of the : . : in n,hich the light is shone is the direct response whlle : ,-:,striction of the other is known as the consensual '.:.:,1nse. The pupillary reflexes are mediated through -.:-:a, ganglion neurons that project to the pretectal area :--:h lies interior to the superior colliculus. The cells in '-.: F:3tecta1 area project bilaterally to preganglionic ,.,.:., mpathetic n"urons in the Edinger-Westphal -- l::s. This is also known as the accessory occulomotor neurons in -:-:*s. The preganglionic parasympatheticthrough the axons send nucleus -". =:inger-Westphal . :: --::i-io1or n"tuJ to innervate the ciiiary gangiion' The - .-- '. eangiion's postganglionic neuron innervates the :: ., :,- nuscle of the pupillary sphincter.

The neurotransmitter released by the axons in the Edinger-Westphai neurons is most likely:

A. B. C. D.

epinephrine. norepinephrine.

acetylcholine. glutamine.

eye exam' the following is shone directiy into the patient's left eye, the patient exhibits a consensual but not a direct response. Which of the following is a likely

51. As part of a routine notiled: If iight is

explanation?

A. ::'

,. .:s been determined that the frequency of action :, ::rdals increases dramatically in axons once they .

B.

:..: lefi the optic nerve. The most likely

.'.:-.nation for this increase is:

C.

\. ; hi-sher density

D.

of sodium channels are found

;n the axons leaving the oPtic disc'

. [] rl

11

, l":

The optic nerve of the left eye is intact, but the efferent limb of the left eye is damaged. The optic nerve of the left eye is damaged, but the efferent ofthe left eye is intact' The optic nerve of the right eye is intact, but the efierent limb of the right eye is damaged' The optic nerve of the right eye is damaged, but the efferent limb of the right eye is intact'

: lou er density of sodium channels are found :i the axons leaving the oPtic disc.

ine a\ons are myelinated by Schwann cells' :he a.rons are myelinated by oligodendrocytes'

patient, the following is observed: Light shone into ihe right eye does not elicit a response in either pupil. Light shone directly into the left eye causes a direct and consensual responsg' It can be concluded that there is a iesion in the:

52. In another

,-.:-t optic tract can be described as an:

A. B. C. D.

left oPtic nerve. right oPtic nerve. Ieftpregangiionicparasympatheticneuron' rightpregangiionicparasympatheticneuron'

nerve axons from

from

5J

The BerkeleY Kcview Specializing in MCAT PreParation

Biology Passage

Axonal Tfansport

IX (Questions 53-58)

Movement

of

56.

substances from the soma

to

the

Passage IX

Glucose does not simply diffuse over the inner mitochondrial membrane. Glucose enters the mitochondrial matrix as:

synaptic endings of a nerve cell by simple diffusion is an inefficient process: because axons are often very long. However, components which originate in the soma must be distributed along the axon. Axonal transport, a process

A. B. C. D.

which costs metabolic energy] is a special trinsport

mechanism that accomplishes this function. Membrinebound organelles are transported rapidly by fast axonal transport, while substances dissolved in cytoplasm are moved by slow axonal transport. Microtubuies provide a pathway along which membrane-bound organelles can move. These organelles may interact with the

glucose-6-phosphate.

phosphoenolpyruvate. pyruvate. oxaloacetate.

57. Phaseolus vulgaris leucoagglutinin is taken up by neurons and is transported anterogressively. marker substance is most likely injected at the:

microtubules through a Iinkage similar to that between thick and thin filaments of skeletal muscle cells.

A.

Transport from the soma toward the axonal terminal is known as anterograde axonal transport. Retrograde axonaL transport is transport in the opposite direction. In order to trace neural pathways experimentally, marker

synapse

to trace the pathway of

axons. synapse to trace the location

ihis

neuronal

B. C.

soma

D.

dendrites. soma to trace the synaptic endings of neurons.

ofcell bodies.

to trace the pathway of

neuronal

substances can be used that are transported either

retrogressively or anterogressively.

58.

The graph shown beiow represents the response of a

postsynaptic cell to acetylcholine in the synaptic

cleft.

Each vertical line (l) represents an action

potential.

53. A neurotransmitter traveling

|||ll||t

from the soma to the

presynaptic terminal will travel:

A. B. C. D.

time [:)

anterogressively, via fast axonal transport. anterogressively, via slow axonal transport. retrogressively, via fast axonal transport. retrogressively, via slow axonal transport.

Which of the following graphs BEST represents the respo,nse following acetylcholinesterase?

the addition of the enzyme

A.

54.

Which of the following ions mosr likely triggers

ll||ll||t

an

interaction between organelles and microtubules?

A. B. C. D.

Nao

Lime

r-)

hme

L_2

6@ B.

Ca2@

Mg2e

55. Neuropeptide

packaging into vesicles all of the following EXCEPT the:

will

A. B. C.

complete translation

D.

cytoplasm. transport through the Golgi apparatus.

include

C.

signal recognition particle receptor. signal peptide.

of the peptide in


llllllrt tlme I

)

trme I

>

D.

the

s4


Biology Passage

Iluntington's Disease 61.

X (Questions 59-64)

Passage X

The following molecule is referred to as GABA. What is irs IUPAC name?

Huntington's disease is a relatively rare disease that causes progressive degeneration of the cerebral cortex and the basal ganglia of the brain. The basal ganglia are

o

o

H3N- CH2- CH2- CH2- coo

of neuron cell bodies (gray matter) located deep n,ithin the cerebrum. Within the basal ganglia, certain

masses

types of neurons are destroyed, while others remain intact.

A. B. C. D.

Huntington's disease is a genetic disolder caused by an

autosomal dominant genetic defect on the tip of chromosome 4. The symptoms of the disease are choreiform movements (rapid, uncontrolied, jerky novements), mental deterioration, and emotional iisturbances. The disease is relentless and usually leads to complete dehabilitation and death within 15 years of its rnset. The age of onset is usually from 35-50 years,

y-amino butyrate 3-amino butanoic acid 1-amino butyrate 4-aminobutanoicacid

usually after the patient has had children.

62. Which amino acid is the precursor to GABA?

The proposed mechanism of the defect lies with :ertain neurotransmitters and is twofold. First of all, there -s a decreased level of y-aminobutyric acid, an inhibitory

A. B. C. D.

:eurotransmitter, and a decreased level of the enzyme that .i'nthesizes it, glutamic acid decarboxylase. Secondly, a :eliciency of the enzyme choline acetylase leads to

jecreased levels

of the excitatory

Histidine Leucine Tyrosine Glutamate

neurotransmitter,

.;et,.lcholine. The delicate homeostatic balance between j're r\\'o neurotransmitters is disrupted. These deficiencies

:re

thought to lead to the observed characteristic lovements and mental symptoms of Huntington's :-sease.

63.

After having one child, a 4}-year old woman with a paternal family history of Huntington's disease is diagnosed as having the disease. Assuming that the

father is not affected, what is the MOST probable likelihood that the child will also develop the

59.

Based on information in the passage, Huntington's

disease?

disease is:

A. B. C. D.

\. B. C. D.

more common in men than in women. more common in women than in men. equally common in both men and women, more common in children.

64. In what structure

w"rt. Based on this passage, the basal ganglia most likely -- rntrol:

A. B. C. D.

;

on

of the neuron are neurotransmitters

stored?

A. B. C. D.

pituitary gland secretion. speech.

the parasympathetic nervous system.

voluntarymovements.

::ght O by The Berkeley Review

100% l5%o 50Vo 25Vo

5D

The The The The

synaptic cleft. vesicles ofpresynaptic neurons. receptors ofpostsynaptic neurons. vesicles of postsynaptic neurons.


Biology

Photoreceptors

XI (Questions 65-71)

Passage

Passage XI

65. Which of the following rerinal cells is

LEAST

responsible for vision in the dark?

The retina of the human eye contains two types of photoreceptors known as rods and cones, both locited at the back of the retina, Upon stimulation by a photon, the sodium channels of these photoreceptor cells close. The highest density of cones found on the retina will give the

A. B. C. D.

greatest visual acuity, or highest visual precision. In addition to the photoreceptor cells, the retina houses four types of neurons. These cells are the bipolar, ganglion, horizontal, and amacrine cells. The rods and iones synapse with bipolar cells, which then go on to synapse with ganglion cells. The axons of the ganglion cells converge and leave Lhe eyes as the optic nerve.

66.

Cone cells.

Rod cells. Ganglion cells.

Bipolar cells.

Stimulation of a photoreceptor by a single photon

will result in a:

The visual field is the view seen by the two eyes without movement of the head. As shown in Figure 1, ihe left visual hemifield projects ro rhe righr side of the brain, while the right visual hemifield projecrs to the lefr half of

4. B. 9. D.

monopolarization depoiarization of hyperpolarization micropolarization

of the photoreceptor. the photoreceptoi. of the photoreceptor. of the photoreceptor.

the brain.

67. To view an object with greatest acutiy, one will focus which of the foliowing structuies on that

Left Visual

object?

Heniifield

A. B. C. D.

Temporal

Fovea centralis Cornea Optic disc

Choroid

68. An experiment involved the removal of the lateral geniculate nucleus of the thalamus. To determine the

effects on processing of visual information, electrodes were placed in which of the following lobes ofthe brain?

A.

Parietal

B. C. D.

Frontal Temporal Occipital

69. Light entering Left Optic Tlact

Right Optic Tract

A. B. C. D.

Figure I

Copyright

@


the eye

will pass which of these cells

first?

56

Rod cells Cone cells

Bipolar cells Ganglion cells


Biology 70.

Photoreceptors

Passage Xl

The region of the retina where the axons of the ganglion cells converge and leave as the optic nerve is best described as the region:

I.

of

highest visual precision, because

of

the

dense axonal population.

II. of the retina mbst sensitive to stereovision, UI.

ofa high density ofphotoreceptors. where no vision is possible, because of a lack of photoreceptors. because

A. I only B. I and II only C. III only D. II only

71.

According to Figure 1, light originating from the left hemivisual field will strike the nasal hemiretina of the:

A.

right eye and the temporal hemiretina of the

B.

right eye and the temporal hemiretina of the

left eye. right eye. C. D.

left eye and the temporal hemiretina of the right eye. left eye and temporal hemiretinal of the left eye.

lGqryngfu O by The Berkeley Review

5t


Biology Passage

Sound Transmission In The Ear 72.

XII (Questions 72-79)

Sound is created by disturbances within a medium resulting in the production of pressure waves detected by the ear. These pressure waves consist of aiternating compressions and rarefactions of the surrounding medium. The loudness of a sound is determined by the

Passage XII

The human ear normally responds to a range of

sound covering about 120 dB. The loudest sound pressure level that can be heard over this range is:

A.

ten thousand times less than the reference prgssure.

B.

one million times less than the reference pressure.

amplitude of these pressure waves and is measured on the decibel (db) scale. The relationship is as foliows:

C.

ten thousand times greater than the reference pressure.

Sound Pressure Level (SPL) = 20 log1g

P1

/

D.

Pp

one

million times greater than

the reference

pressure.

where P1 is the test pressure and Pt is the reference pressure. P, has the value of 20 micro-Newtons/m2. The pressure waves travel through the external ear canal and will cause the tympanic membrane of the middle ear to vibrate.

73.

A set of smail bones is responsible for transmitting this vibration throughout the middle ear, and the last of

this design is that the total:

these bones (the stapes) is attached to the oval window. The oval window is the connection between the middle and internal ear, which is composed of the cochiea and the vestibular apparatus (Figure 1).

A. B. C. D.

Scala Media

6lo

c'

According to Figure 1, the tympanic membrane has an arga greater than the oval window. A result of

force acting on the oval window is increased. forcelunit area acting on the oval window is decreased.

force acting on the oval window is decreased. force/unit area acting on the oval window is increased.

Incus Stapes

Malleus

" \

Scala Vestibul

74.

Tympanic Membrane

The relationship between the speed of sound (C), wavelength (7"), and frequency (v) is:

Scala

Round Tympani indow

Auditory

C = (),Xv)

Canal Internal Ear

where C = 340 m/s. What is the period of a sound wave with a wavelength of 3.4 x t0-4 km?

Middle Ear

Figure

1

A. 1x10-6s. B. 1x10-3s. C. 1x103s. D. lx106s.

The transduction of pressure waves into electricai signais occurs in the internal ear. The vibration of the stapes produces pressure waves

in the fluid of the scala

ueJnUuti and scala tympani. This wave motion in the paralymph naturally sets up oscillations in the endolymph, located in the scala media. The transduction organ (organ of Corti) is located on the basement membrane of the scaia media. Oscillating movements of this organ excite

and inhibit sensory transduction cells' which transmit

75.

impulses to the brain.

potential that is:

Inducing a vibration in the temporal bone will cause sounds to reich the cochlea. This sound transmission is unique in that the middle ear is bypassed' While this is an inefficient method of energy transfer, such a method is clinically useful for diagnosing auditory problems'

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As a result of movement in the organ of Corti, the sensory transducing cells most likely repond with a A. B. C.

D.

58

depolarizing. hyperpolarizing. oscillatin g depolarizing-hyperpolarizin g. returning to its resting membrane potential

The BerkeleY Keview Speciatizing in MCAT PreParation

Biology

Sound Tfansmission In The Ear

76, The following picture

represents

the

78. A patient complains

basilar

of auditory problems and so a tuning fork is set into vibration and placed at the patient's ear. When the patient indicates the sound is inaudible, the fork is placed on the temporal bone. If the sound becomes audible to the patient, the

membrane uncoiied and stretched out flat.

n-

i#

l',,,,,,,1;,,t",",',111'I-$;1il;;!ff

damage can be narrowed down to the:

A.

o

external ear. middle ear. inner ear. auditory cortex.

B. C.

D.

region

Passage XII

Flexible region

79. It has been discovered

According to the resonance theory, different areas of the basilar membrane are atfected by varying sound frequencies. High fi'equency sounds resonate best in

that every hair cell responds maximally to a particular frequency. Which of the following graphs best represents this idea?

which region?

A.

Region

B. C. D.

Region 2. Region 3. Region 4.

A.

1.

-o

!

o

!

o. E

The foiiowing circuit model attempts to explain away the problem that sound arrives at the two ears at different times. Cells wiii fire with maxirnum output when bilateral inputs anive at the same time.

l/\ l/\

B. -o

o o. E

Frequency (kHz)

C.

Frequency (kHz)

D.

Input from

right ear

-o 13

.o

O

€O

o.

o.

E

E

Frequency (kHz)

Frequency (kHz)

Input from

left ear

Ii sound to the right ear were delayed relative to the -:ii. which of the following cell(s) would most likely

-{. B. C. D.

Cells 1 and

2.

CelI3. Cells 3 and 4.

Cell

-5.

:'. ::rht O by The Berkeley Review

59


Biology Passage

XIII

Tryptophan and Serotonin Experiment 81.

(Questions 80-85)

The response in brain tryptophan (Trp) concentration and serotonin synthesis after the ingestion of two meals

was examined

in

Passage XIII

Table 1 shows the study results for the animals receiving CHO first (Trial 1). Serum Trp and Serum LNAA are in pmol/i, and cortex Trp is in pmol/g.

rats. The synthesis and release of

serotonin, a neurotransmitter, by brain neurons is rapidly influenced by the local Trp concentration.

Serum Serum Cortex

Trp

Group No Food

Brain Trp concentration reflects uptake of Trp from the blood into the brain. This is accomplished by a transporter in the blood-brain barrier that is shared by the large, neutral amino acids (LNAAs): leucine, isoleucine,

LNAA

96

Trp

482

28

CHO - CHO

140*

3'76

36*

CHO - 67a Protein

1

50*

385

JO-

35*

CHO - 127a Protein

I

56*

43'/

tyrosine, phenylalanine, and tryptophan.

CHO - 2470 Protein

1

60*

6'7 5

28

Food ingestion can alter levels of serotonin in the brain by altering blood concentrations of Trp and its LNAA competitors. The following study was designed, in which 2 sequential meals were fed to rats that fasted

CHO - 40Va Protein 162* tp 0.05 versus no food

9"t6

23

=

Table

1

overnight.

Table 2 shows the study results for the animals receiving protein first (Trial 2). Serum Trp and

The experimental diets contained varying ratios of protein and carbohydrate, with fat held a[ a low and constant percentage in ail the diets. The experimental

diets were CHO (a11 carbohydrate, no protein),

Serum LNAA are pmol/g.

6Vo

protein, I27o protein,24Vo protein, and 40Vo protein. For the sake of reference, standard rat chow is about 1070 protein, and the typical human eats about 12-l5%o protein.

in pmol/l,

and cortex

Trp is in

Serum Serum Cortex

T.p

LNAA

Trp

142

51r

26

162*

49',7

33*

CHO

lb)^

533

31*

lTVa Protern

- CHO

190*

621

27*

247o Protetn

- CHo

784*

743

28

407o Protein - CHO

194*

814

25

Group

For the experimental design, the rats received either: 1 (the CHO diet followed by one of the protein diets 2 hours later) or Trial 2 (one of the protein diets followed by the CHO diet 2 hours later). The rats were sacrificed and the brain tissue examined for Trp concentration at 2

Trial

No Food

CHO - CHO 6Va

hours following the second meal.

Protein

-

*p = 0.05 versus no food

80.

In the following reaction, the molecular structures of Trp and serotonin are shown. Trp is converted into serotonin by first undergoing

Table 2

Which of the following statements is FALSE regarding the data in Table 1 and Table 2? NHt

NH"

t-

H.C_ C_

A.

I

COOH

H2C-

CH2

ffi"o .+""cd Tryptophan

Serum

LNAA concentration increases as

dietary protein content increases. B.

A

24Va protein meal at 2 hours following a CHO meal attenuates the increase in serotonin due to Trp.

Serotonin

c.

At higher levels of protein, the serum LNAAs are converted to serotonin.

A. B. C. D.

reduction followed by carboxylation. hydroxyiation foliowed by decarboxylation. amidation foilowed by hydroxylation. amination followed by decarboxylation.


D.

60

Serum Trp concentrations increase as dietary protein content increases.


Biology 82,

Tryptophan and Serotonin Experiment 85.

Which of the following statements is TRUE of the data in Table

I

and Table 2?

A.

In Trial 1, the higher protein levels (24Vo and 40Vo) caused competition between Trp and the

il.

In Trial 2, the higher protein levels (lZVo,

B. Dietary Trp alone would

In.

24Vo, and 40Vo) caused competition between Trp and the other LNAAs. In Trials 1 and 2, serum Trp increased with feeding.

C.

Dietary Trp alone would not be absorbed readily in the small intestine, so it would have no metabolic effect.

increase brain serotonin concentrations, since few LNAAs

D.

A. I only B. II only C. II and trI only D. I, II, and III

83.

What would be the metabolic effect of eating a large amount of pure Trp as a dietary supplement?

I.

other LNAAs.

Passage XItr

would be competing.

Dietary Trp alone would increase synthesis of all classes of neurotransmitters. Dietary Trp alone would cause more alert behavior, due to the production of serotonin.

Which of the following statements is TRUE of Trp?

I.

Trp is a neurotransmitter.

n. Trp is an essential amino acid. IIr. Trp is an acidic amino acid. A. B.

c. D.

84.

I only II only I and II only I, II, and III

As part of this experiment, researchers administered an inhibitor of one of the enzymes that participates in the conversion pathway of Trp to serotonin in the brain. In this part of the experiment, what parameter could be examined to estimate serotonin synthesis rate?

A.

Accumulation of an intermediate metabolite in the brain.

B.

C. D.

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Concentration of Trp in the blood. Concentration of the inhibitor in the brain. Accumulation of the inhibitor in the blood.

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6l


Biotogy Passage

Frog Muscle ExPeriment 86.

XIV (Questions 86-92)

The gastrocnemius muscle along with the sciatic nerve from a frog's leg were dissected and attached to a

device to measure action potentials and record them on chart paper. Figure 1 is a diagram of the experimental

Passage XIV

The muscle is stretched more and more as heavier weights are added to the experimental apparatus. How is the intensity of the muscle stretch indicated by the activity ofthe nerve fiber?

A.

A larger amplitude of the action

potentials

indicates greater intensity.

setup:

B.

A

larger ampiitude of the action potentials

indicates lower intensity. C.

A greater frequency of the

action potentials

indicates greater intensity. D.

A greater frequency of the action

potentials

indicates lower intensity.

87.

Figure

1

Two more trials were performed on the same muscle preparation. A 75-g and a 100-9 weight were used to stretch the muscie. The printouts of the action potentials looked exactly like the 50-g trial. What is the explanation for this? A.

The action potentials had increased to

a

maximum with the 50-g weight and increased no further, due to the refractory period of the neuron.

IO, 20, and 50 gm) were hung opposite the muscle, so that a known set of tensions were apptlea to it. The action potentials generated by each Werghts (1, 2,

5,

weight were recorded on the chart paper' This is illustrated in Figure 2. Each vertical (l) line represents an

B.

The nerve and muscle were damaged by the weights, and the nerve therefore slowed its transmission of action potentials.

C.

The speed of the chart recorder could not keep up with the increased action potentials.

action potential.

D.

The action potentials were identical, due to

a

neurotransmi tter defect.

I

I |

lsm 88.

lllll -

lll llllllll

2sm

Figure 3 indicates the output of action potentials

after a 5-mg weight was added to the muscie preparation. What is the explanation for this effect?

llllllllllllllllllllllllllllllllllll 5sm 5mg lll ll ll

llllllll lllll ll llll ll lllllllll ll llllllllllll

1ll I I

I

10 gm

tlme L-) Figure

time

A.

t+

B.

Figure 2

C. D.

The purpose of this experiment was to learn more

about the intiraction between muscle and nerve, and how muscle stretch is communicated.

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62

3

The weight was not heavy enough to exceed the action-potential threshold.

The weight was too heavy and suppressed the aciion-potential threshold. The weight must produce an action potential, so the recording equipment must be damaged' The size of the action potential was too small to read on the Printout.

The BerkeleY Review Specializing in MCAT PreParation

Biology 89.

90.

Frog Muscle Experiment

Passage XIV

Actions potentials are conducted by the nerve cell through which structure? A.

Cytoplasm

B. C. D.

Schwann cell membrane

Endoplasmic reticulum Plasma membrane

When an action potential is communicated, what is the sequence and direction of movement of sodium and potassium?

A.

B.

C.

Potassium moves from the intracellular fluid to the outside of the cell, and then sodium moves from the extracelluiar fluid into the cell. Sodium moves from the extracellular fluid into the cell, and then potassium moves from the intracellular fluid to the outside of the cell. Sodium moves from the intracellular fluid to

the outside of the cell, and then potassium moves from the extracellular fluid into the cell. D.

qI.

Potassium moves from the extracellular fluid into the cell, and then sodium moves from the intraceliular fluid to the outside of the cell.

\\-hat type of nerve function is studied using this erperimental design?

.{. B. C. D.

04" Tle i\

Efferenr function Afferent function

Sympatheticfunction Parasympatheticfunction

sodium-potassium pump, an active transport of intracellular

stem. restores the proper balance

nnd extracellular

ions. When an organism

is

:oisoned with cyanide, which halts ATP production, ;hat happens to neural action potentials?

B.

Action potentials stop immediately. Action potentials will continue, even though ATP production stops.

C.

Action potentials continue briefly but stop

D.

-\ction potentials will generate ATP during

-4.-

eventually, due to a lack of ATP. cvanide administration.

g by The Berkeley Review

O.'


Biology Passage

Skeletal Muscle Groups

XV (Questions 93-98)

il.

\4uscles of the lower extremities aid in the movement bones such as the femur, tibia, and fibula. They help irarntain an erect skeletal system and keep the body in

r-

balance by acting as an antagonist to gravity. These

diagram

Hamstring Group Made up of three muscles. Al1 originate from the ischium and run along the posterior side of the femur. Two of the muscles of the hamstring group insert onto the medial condyle of tne tibia-, wtritl

the third inserts onto the lateral condyle of the tibia and the head of the fibuia. The hamstrings

muscles provide the pumping action that helps circulate blood and lymph through the lower extremities.

A

Passage XV

are innervated by branches of the sciatic nerve.

of the major bones found in the lower i.

extremity is shown in Figure

III.

Adductor Group Several muscles comprise this group. All of these muscles originate from the pubic bone region and

o

extend to inserr along the medial and posterior inferior shaft of the femur. The adduitors are

peilor Super

innervated mainiy by the obturator nerve.

€

Port.rio,

AnteriorS pubic bone femur

ischium

93. Judging by where rhey atrach, which of the following muscle groups would play the strongest role in flexing the hip joinr (i.e., pulling the kiee upwards

medial condyle lateral condyle

towards the chest)?

fibula

Figure

A.. B. C. D.

Hamsrrrngs. Adductors. Quadriceps and hamstrings together"

1

A iist of the major muscle groups of the thigh region of the lower extremity is provided below:

94. The pumping action of muscles in the lower extremity is crucial in order for blood to properly flow back to the heart. Which of the following

Muscles of the Thigh

I.

Quadriceps.

statements must be true if blood is to be returned to the heart effectively while a person is standing?

Quadriceps Group

Made of four muscles, three of which originate from the upper, anterior portion of the femur and one that originates from the anterior inferior iiiac spine of the ilium. They all descend and insert into the broad patellar tendon which in turn crosses the

A. B. C. D.

The veins involved must have inelastic walls. The arteries involved must have elastic walls. The veins involved must have valves.

The arteries involved must contract under sympathetic control.

joint and inserts into the tibial tuberosity of the upper, anterior tibia. These muscles are

knee

innervated by branches ofthe femoral nerve.


64


Biology 95. If

Skeletal Muscle Groups

Passage XV

an individual contracts the hamstring muscles, the:

A. B. C. D.

quadriceps must contract. quadriceps must relax.

quadriceps are unaffected. obturator nerve is responsible.

96. Paralysis

of the f'emoral nerve would

MOST

STRONGLY aff'ect the ability:

A. B. C. D.

to stand on tip-toes. to extend the hip posteriorly. to kick a ball. of the adductor muscles to contract.

97. Often when a person suffers a herniated vertebral disk in the lower back, the sciatic nerve is compressed and damaged. Which of the following would most likely NOT be a symptom of such a condition?

A. B. C.

D.

Difficulty walking. Difficulty moving the lower leg posteriorly. Pain radiating down the back of the leg. Weakness in the adductor muscles.

98. Ciaudication is a medical condition in which arteries which feed the muscles of the lower extremity become partially occluded. In these patients, moderate exercise might cause all of the following in the affected muscles EXCEPT:

A. B. C. D.

ATP deprivetion. lactic acid buildup. lower-than-normal CO2 leveis. 02 deprivation.


ttb


Biology

Nerve & Muscle

Section I Answers

B is correct' HCo3e' Roughly 60va of the body by weight is due to water, and all of the warer can be divided into two compartments separated by a cellular membrane. orie compartment contains the intracellular fluid (ICF) whiie the other compartment contains the extracellular fluid (ECtr): itr" tu.g". of the two compartments js the ICF compartment, and it contains about2/3 of all the water in the body. The ECF compartment t, ;;;;;;;;; il ,;; types of fluids: interstitial fluid and biood plasma. Table 1 in the passage lists the major ions of the ICF and ECF. The first thrng to note is that the ICF has a higher concenrrarion oi Ko"than rh. ECF: i;i. utto*, us to eliminare

choice D.

The remaining three choices are not mentioned in the passage and require a little thought. phosphate (po43e) is required not only in celiular DNA and RNA, brt it is also Lquired by a vast anay oiproteins that participate in phosphorylation reactions within the cell. Simply_consider priffioryrating ADp with p; to make ATp. Therefore, we would expect the levels of phosphate to be higher in th.: ICFlhun in tt.r, ECF, and they are. phosphate leveis in the ICF are about 20 mM' while in the ECF they are about 4 *irrr. sti-inate choice C. we can use the same reasoning for the relative levels of proteins. Ail of our ceils are carrying out metabolic reactions, and to do the majority of these reactions enzymes (proteint) be employed. thl, -"un, rhat there will be more prorein inside a cell than outside a cell' Eliminate choice A. Tr:J cetts which are catabolizing rnot".ut"r-ur" proaucing co2 gas (e.g., via the Krebs cycle) carbon dioxide can combine with water to form carbonic acid, which can then dissociate to form the bicarbonate ion (HCo3e). Since Co2 is a wasre and it needs to be eliminared via the blood, the cell wants to get rid of it we find that the levels of bicarbonate are higher in the ECF (i.e., the blood plasma) than in the icF. Bicarbonate levels in the ICF are about 10 mM, while in trr" iCn irr"f-ar" about 24 mM. The correct choice is B.

D is correct' extracelluiar [Nae]. Extracellular refers to what is happening outside the cell and intracellular refers to what is happening inside the cell. The Nae/Ko-ATPase requires ATp in order to funcrion. If the levels of ATp inside the cell are low,.the ATPase activity will decrease and eventually be inhibited. The remaining ATp left within the cell will be diverted to reactions ihut more important to the cell's survival. we can eliminate choice A' A low intracellular concentration of Nao will^r" also lead tolnhibition of the ATpase, because if there is no Nao to pump to the extracellular space, then the enzyme cannot catalyze the transport of sodium out of the cell and potassium into the ceil' we can eliminate choice B. The ,a-e .earoning applies to low extracellular concentrations of Ko ' This allows us to eliminate choice c. A low extracellular concentration of Na@ will not directly affect the ATPase' because the ATPase is located within the cell. It is not found on the outside of the cell. The correct choice is D. B is correct, primary active transport. In the second paragraph of the passage, we learned that glucose can cross the cell's membrane by facilitated diffusion' At the bottom of nlgur. 1 in the passage, we see glucose and Na@ being transported across the cell's membrane in the same directioti rn. protein i. ;;G as a symport. If the carrier-were a uniport, only one molecule would.be crossing trre mlinrane."u.."i"i An antipoJ;l;"*, two molecules to cross the membrane' but each is crossing opposite to the Jther. Glucose is entering the cell along with sodium through a symport mechanism. The sodium was actively transported out of the cell. sodium can re-enter the cell down its concentration gradient, allowing glucose to b" t.unrpoit"J *;tt it by a secondary active transport process. A,primary active transport systen uses the energy of ATP directly to move a substance across a cell,s membrane. we do not observe this happening in the oiglu.ose, especiafiy since we read in the third paragraph that only "ot" three primary active transport systems have been identified: one for Nao and K@ ions, one for a;t6 ;;r,';;;; for Ho ions. The correct choice is B.

A is correct, low intracellular [Nao] allows sodium to be released from the carrier protein. A high extracelluiar Nao concentration is what allows glucose to bind to the symport on the extracelluiar side of the cell,s membrane and be transported into the cytosol. we can eliminate choice B and D. if the affinity of the carrier protein for Nao were high on the cytosolic side of the membrane, then sodium and glucose could noi be released into the cytosol. when Nae is released into the cytosol from the carrierprotein, the carrier protein's affinity for glucose is reduced" we can now eliminate choice C. The correct choice is A. A is correct' (see the graph below)' Simple diffusion (SD) involves a substance moving from a high concentration to a low concentration' Diffusion of lipid-soluble substances can occur rhrough u lipid-til;t"r,

while diffusion of

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The Berkeley Review Specializing in MCAT preparation

Biology

Nerve 6r Muscle

Section I Answers

water-soluble substances wili occur through a transmembrane channel. The larger the graclient, the more a substance will diffuse across the membrane. We would expect to see a straight line showing that as the gradient of a substance increases, the flux of the substance across the membrane increases as well-

//'/:

ti=

FD

/ IGradient]

Facilitated diffusion (FD) requires a carrier protein. These cauier proteins have a limited number of carrier sites for the substance that is to be transported across the membrane. Therefore, at a higher concentration gradient ail of these carrier sites will be filled. The carrier will then be saturated (think of Michaelis-Menten kinJtics). At this point, the rate of diffusion of the substance across the membrane will no longer increase with an increasing concentration gradient. We would expect to see a plateau in the curve as soon as all of the carrier sites become saturated. The correct choice is A.

6'

B is correct, more Na@ to enter the cell than Ko to leave the cell. The resting membrane potential of a neuromuscular cell is about -80 mV. If a positively charged Nao ion enters the cell, it will not make the inside of the cell more negative, but rather make it more positive. The resting membrane potential will be depolarized, not repolarized. Eliminate choice A. Table 1 in the passage tells us that the concentration of Ko is higher in the cell than it is outside the cell. It also tells us that the concentration of Nao is higher outside the cell than it is in the cell. Therefore, when the acetylchoiine receptors open and allow for the flow of Ko and Nao ions, we find that Nao ions enter the cell, while Ko ions leave the cell down their concentration gradients. This allows us to eliminate choice C. The question tells us that two molecules of acetylcholine will bind to the acetylcholine receptor. This is

a

chemicaliy mediated and activated response. The channel in the acetylcholine receptor is not activated (opened) by an electrical stimulus (i.e., a depolarization). If that were the case, acetylcholine would not need to bind to the receptor. We can eliminate choice D.

B. But how do we know that more Nae will enter the cell than leave the cell? The electrochemical gradient for Nae is greater than the electrochemical gradient for Ke . The electrochemical gradient that aliows an ion to pass through a membrane is simply the difference between the membrane potential (E.) and the equilibrium potentials for the ion in question (Qon). The resting membrane potential for our cell is about -80 mV. The resting membrane potential for Ko is about -92mY, whiie the resting rembrane potenlial for Na@ is about +58 mV. Clearly, the electrochemical gradient for Nao is greater than that for K3. The correct choice is B. B.v the process of elimination, we arrive at choice

Ko will

D is correct, a high-resistance electrical pathway. Gap junctions join the cytopiasm of one cell to the cytoplasm of a r:ighboring cell through intramembrane proteins. These proteins (connexons) contain 6 subunits that form a central :ore that allows molecules with molecular weight up to 1500 to pass through. Molecules iike ATP and ions like

\a3 can easily

pass through these channels. This helps to establish a cytopiasmic continuity between the cells. \\-hen an action potential reaches a gap junction, electrical coupling between the two cells occurs, and the action propagates to the next cell. This type of coupling is important in heart tissue, where entire collections of cells must contract in a coordinated fashion. Therefore, a gap junction must provide a iow-resistance (not a high-resistance) electrical pathway that enables current to pass from one cell to the next. The correct choice is D.

5

C is correct, increased sodium concentration gradient. Inhibition of the electron transport chain means that the amount of ATP being synthesized is substantially reduced. If the amount of ATP is reduced, then the amount of energy transferred from ATP to the membrane ATPases is reduced. The primary active transport systems (those e-ATPase) ali show adecrease in their activity. In secondary involving the Nae/Ke -ATPase, Ca2o-ATPase, and Heactive transport, the energy that was stored in the Nao concentration gradient across the cell's membrane (high extracellular [Nae] and low intracellular [Nao]) is used to transport molecules like glucose and amino acids into

Sl right

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67


Biology

Nerve 8f Muscle

Section I Answers

the cell and calcium out of the cell. If the synthesis of ATP decreases, the Nao concentration gradient across the cell's membrane is not as great (i.e., it decreases, not increases), and the rate of secondary active transp ort decreases . Low levels of ATP also mean that the Ca2 o-ATPase cannot adequately pump calcium out of the cell. Since calcium is also transported out of the cell (against its concentralion gradient) by a secondary active transport system that utilizes the sodium concentration gradient (which is now rather low), the levels of intracellular calcium begin to increase. Calcium is able to diffuse back into the cell though calcium channels. The correct choice is C.

D is correct, I, m, and iV only. The sympathetic system is important in getting the body ready for a perceived stressful situation. This response is a general response, affectrng ail parts of the body almost simultaneously. This is essential to the survival of the animal. You wouldn't want to have to wait for each organ or muscle to detetmine that this is a stressful situation while a lion is running at you! But the parasympathetic system has the luxury of determining specificaily for each structure it innervates whether the job is done during this time of stress and when it's time to return to basai levels. The parasympathetic system can do this type of individual monitoring, partialiy because of the proximal location of its ganglia next to the structure in question. The correct choice is D. 10.

D is correct, I, II, and IV on1y. All three situations are interpreted by the body

as demands to increase particular functions beyond their normai basal levels. In other words, they represent a stress to the body. Sweating on a wann day is a response to release heat and cool down the external body. On cold days shir,ering and piloerection are stimulated to increase heat in the body. Both of these reactions are instigated by the sympathetic system. Furtherrnore, running is a stressor on the body. The correct choice is D.

D is correct, penile erection.

11.

Penile erection can be caused by direct stimulation invoiving penile mechanoreceptors, or it can be caused by central nervous system activity stemming fiom sights and smells or even thoughts and emotions. In either case, input reaches the neurons of the penis. Parasympathetic nervous system activity is increased while sympathetic nervous system activity is decreased. When parasympathetic activity increases and sympathetic activity decreases, vascular dilation of the arterioles occurs so that the tissues of the penis become engorged with blood. As erection continues, the veins leaving the penis become compressed and little blood is allowed to leave, thus maintaining the penis in the erect state. The correct choice is D.

12.

B is correct, I, II, and III oniy. The adrenal gland sits on top of the kidney. Cortisol, a glucocorticoid produced by the adrenal cortex, controls various aspects of metabolism. An increase in piasma concentrations of cortisoi will lead to an increase in giuconeogenesis (i.e., the synthesis of glucose from precursor moiecules like iactate), a decreased uptake of glucose by celis, an increase in protein catabolism, and an increase in triacylglycerol breakdown (i.e., the release of free fatty acids). Aldosterone, a mineralocorticoid produced by the adrenal cortex, controls electrolyte balance by stimulating sodium reabsorption and potassium secretion in the kidney. Sodium reabsorption also leads to water reabsorption in the kidney. The correct choice is B.

13.

B is correct, increased heart rate. The vagus (1Oth cranial) nerve is a component of the parasympathetic nervous system. Think of the parasympathetic system as being passive. The iris muscle of the eye will contract, making the pupii smaller. The motility and secretions (e.g., HCl from the parietal (oxyntic) cells) of the stomach will increase, êspecially after a meal. An increased heart rate and contractility is due to the sympathetic nervous system (think of the.figt'rt-or-flight response). The correct choice is B.

14.

C is correct, efflux of Ke balances the influx of Nae. A small depolarization in the membrane will allow some Na@ to enter the cytoplasm of the cell. At the end of the second paragraph in the passage we see that the relative permeability of the membrane to Ko is roughly 10 times greater than it is to Nae . Since there is more Ke in the iCp tt un Xa@ ifrom Table l), we see that more Ko will flow out of the cell than Na@ into the cell. Therefore, if there is a smal1 depolarization, then the influx of NaO will be more than balanced by the efflux of Ko ' If the efflux of Ko is greater than the influx of Nae , the explosive nature of an action potential will not resuit. What this means is that the threshold potential for the generation of an action potential is that point where the influx of Nae exactly matches the eff'lux of K@. Remember, there is a higher concentration of Na@ in the ECF and a higher concentration

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Biology

Nerve 8r Muscle

Section I Answers

of Ke in the ICF. The nanosecond that there is more of an influx of Nae than there is an efflux of Ke. an action potentiai will be generated. The correct choice is C. B is correct, -65 mV. As stated in the fourth paragraph of the passage, the threshoid is about 15 mV away from the In Figure 2 of the passage, the E* is at -80 mV. The threshold is given by the dashed line labeled with the number 2. Therefore, the threshold potential must be -65 mV. The correct choice is B.

15.

E-.

C is correct, an increase in the opening of fast n gates in neighboring Ko channels. As depolarization takes piace, Nae rapidly enters the cell (Table 2inthe passage). In order for Nae to enter the cell, the f'ast acting ar gate must open to allow Nao to rush into the cell down its concentration gradient. As Nae enters the cell, it causes further depolarization and allows more m gates to open. This is the explosive nature of the action potential. The absolute refractory period begins at about the time the depolarization of the membrane takes place. Another action potential cannot be generated at this time because of the opening of the m gates from the first action potential. This allows us to eliminate choices A, B, and D. Depolarization of the membrane will lead to an eventual opening of the n gates.

16.

The problem is that the

/?

gates are slow and not fast. The correct choice is C.

A is correct, m gate open; h gate closed; n gate open. Table 2 in the passage says that during the repolarization Ko is flowing out of the cell and down its concentration gradient. If Ko is flowing out of the cell, the n gate

11.

phase

must be open. Looking through our choice of answers, we see that the iz gate is closed in choice eliminate choice D.

D.

Therefore, we

We know that during repolarization, we do not want Nae entering the cell. If Nao enters the cell as Ko leaves, we will not be abie to obtain the resting membrane potential of the cell as quickiy as we would like. Which one of the Nao gates is closed, or are they both closed? The iast paragraph of the passage teils us that during depolarization the ft gate begins to slowly close. Let's assume that the /r gate has closed as repolarization begins. If this is the case, then we can eliminatb choice C.

However, all through depolarization Nae is entering the cell, which means that the m gate is remaining open. By the time the peak of the action potential has occurred, the h gate has closed, the m gate is still open but beginning to close, and the n gate is open. The correct choice is A.

.\ is correct, repolarize. An action potential will initially depolarize the resting membrane potential (E6) and cause that potentiai to be less negative. As the Nao gates close and the K@ gates open, the membrane potentiai is being repolarized toward the E* of the cell. Because of the high permeability to Ke during repolarization, the membrane polential becomes even more negative than the E*. This is called hyperpolarization. Once the cell has completed n;"'perpolarized, it will return to the En., value of the cell. This is what is occurring during the last half of the relative r:;ractory period. Whenever a membrane potential returns to its true E,o, the process is referred to as a :erolarization. The correct choice is A.

ns_

:l'

B is correct, Nae and K@ ions are not responsible for the release of the neurotransmitter. The passage states that

TT\ blocks the Nao channels, while TEA blocks the Ko channeis. If this is the case, then the depolarization of the :::re cell membrane will not occur. However, if the presynaptic ending is stimulated with a depolarizing signal, a :,- srs\ naptic potential is produced. The last paragraph of the passage says that a neurotransmitter is released from -:-- presynaptic terminal, diffuses across the synaptic cieft, binds to the postsynaptic terminal, and produces a

:,-sr:ynaptic potential. Since we are using a stimulus (signal) of depolarization to cause the reiease of a :,=urorransmitter, it must mean that Nae and K@ are not responsible for the reiease of the neurotransmitter. If they ;:.e needed, then the stimulus that we applied would have no effect on the generation of a postsynaptic potential. '-.--':

.-an eliminate choice A.

i'

nentioned above, TTX and TEA block Nae and K@ ion channels, respectively. They do not allow 3-it.rirrsrrritter release. If they did, we would not have needed the depolarizing stimulus. We can eliminate choice {-- Because the channeis for Nae and K@ are biocked, it does not (necessarily) mean that the concentrations of ICF !.: "nd ECF Nao are going to increase. Remember, the membrane is permeable to both Nae and Ko (see the ,"-;:r:l paragraph in the passage). These ions wili tend to diffube down their concentration gradients and the -

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The Berkeley Review

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Biology

Nerve Er Muscle

Section I Answers

NaeiKe-ATPase pump will redistribute them again. Blocking a stimulus. The nerve cell

20.

will

the Nao and Ko channels is analogous to not having be at its resting state. We can elirninate choice D. The correct chJice is B.

B is correct, glutamic acid decarboxylase. The cx-carboxyl group of glutamic acid is being removed. When CO2 is removed from a molecule, the reaction is called a decarboxylation and is carried out by a decarboxylase enzyme. If C02 were added to a molecule, it would be a carboxylation reaction, which wouli be carried out bv a carboxvlase

enzyme.

@t

COOH

HrN- C-

H

H

Enzyme

I

Ot H3N - c-

H

CH,

CH"

CH,

CH,

COOH

COOH

tt-

t-

I

Glutamic acid

GABA

Since the CO2 is being removed from glutamic acid, the enzyme that catalyzes this reaction is a _ulutamrc acid decarboxvlase. Note that once GABA is fbrmed, the numbering of the carbon atoms changes. What was the o,position becomes the y-position. There is no carboxyl group at the 1-position. Instead, there"is an amino group. If we were to remove the amino group of GABA, the four carbon compound would be butyric acid. Hence tt . ni.", y-aminobutyric acid. The correct choice is B.

C is correct, small molecuiar size and high lipid solubility. The passage tells us that the base fbrm of the local anesthetic, being lipid-soluble, can penetrate various tissue barriers. The blood brain baryier is certainly one of these tissues. The reason lipid soluble substances are able to penetrate the blood brain barrier is that the leaflets of the cell membrane, including those of the capillary endothelium, are composed of lipid molecules. Lipid soluble molecules can thus diffuse through the two-dimensional liquid and through ihe membrane. Furthermorg it is logical tliat the smaller the molecule, the greater its ease in crossing a membrane. The correct choice is C. 7','

A is

correct, the higher the pKs of the local anesthetic, the lower the concentration of base in the tissue. The passage tells us that the concentration of base or cation in the solution depends on the pKa of the local anesthetic. Remember that local anesthetics are weak bases. Look at the structure of lidocaine. When it is protonated, it carries a positive charge. In other words, the cation is the acid form. Therefore, when we have a high pKu, we have a small

Ka. This, of course, means the reaction does not go very far to the right. In other words, we have a large concentl'ation of acid and a small concentration of base in the tissue. The correct choice is A. 23.

C is correct, local anesthetics decrease threshold for electrical stimulation in nerve tlbers. We are looking for

a

statement that is least compatible with the way in which local anesthetics act. Remember that local anesthctics act to produce a reversible, local nerve conduction block. Therefore, we are looking for a statement that contradicts this goal. A decrease in the threshold value indicates it is now easier to initiate a nerve action potential. Crearing conditions where action potentials are formed more easily does certainly contradict the goal of the locai anesthetic.

The correct choice is C. 2.1.

A is correct, the onset of lidocaine is more rapid, because more exists in its base fbrm at body pH. We know from the passage that the base form is lipid-soiubie, and therefore it penetrates the tissue. One can logically conclude that the local anesthetic that exists more in its base form (as a percentage) at tissue pH will have a quicker onset of action because more penetration of tissue will occur. That is why the pKos of these local anesthetiis are important. The local with the lower pKu will have a higher percentage of molecules in the base form and thus have a more rapid onset. Lidocaine has a more rapid onset because more of it exists in its base form. The correct choice is A.

25.

D is correct, the amount of ionized lbrm will be increased, and the local anesthetic will be trapped in the tissue. Again, the local anesthetic penetrates the tissue in its neutral, base form. If acidosis (lowering of the pH) occuned in lissue after penetration of the local anesthetic, this would drive the base form back into its acidic form. In other rvords, a lower percentage of the molecuie would exist in the base form. The problent with this is that the acidic

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Biology

Nerve & Muscle

Section I Answers

form is charged and thus cannot escape the tissue barrier. The trapping of the charged form of the local anesthetic can contribute to an increase in the degree of toxicity. The correct choice is D. 26-

C is correct, is biotransfbrmed by reactions occurring in the liver. Lidocaine is injected (rather than taken orally) to avoid biotransformation reactions which occur in the liver. If the drug is taken orally, it is not destroyed by the harsh conditions of the stomach. For example, the drug is not a protein, so there is no concern over the protease content of the stomach. After absorption across the intestinal wall occurs (most wili probably occur in the smal1 intestine), the drug enters into the hepatic portal system and is transported to the liver. Remember that the liver contains enzymes (we do not need to concern ourselves with the exact reactions) which are involved in modification of substances. Such modification can render the iocal anesthetic inactive. Furthermore, we wish to avoid having a local anesthetic circulating in the blood stream. The correct choice is C.

)1

D is correct, The circular muscle of the iris is relaxed, the radial muscle of the iris is contracted, and the ciliary muscle is relaxed. Although this answer seems complicated, we can figure it out. First, figure out what the iris is doing. In the dark, the pupii will be large to iet in light. The radial muscle will contract, and the circuiar muscle will relax. This imrnediately eliminates choices A and C. We must now figure out what the ciliary muscle is doing to choose between choices B and D. The ciliary muscle, as rve are told in the passage, is relaxed for vision beyond about 20 feet. Since the night sky extends beyond 20 feet, the ciliary muscle is relaxed. Eliminate choice B. The correct choice is D.

28.

A is correct, the lens is less able to focus the light rays from near objects. Remember, the "default" shape of the lens in the relaxed state is flatter, for far vision. This really does not change with age. Choices C and D are incorrect. Near vision requires a flexible lens that can plump up into a convex shape to aliow focusing on near objects. A less flexible lens is less able to focus on near objects. Choice B is incorrect. The correct choice is A

to

B is correct, sympathetic nerves are stimulated, and parasl'mpathetic nerves are inhibited. What situation leads to large pupils? The radial muscles must contract under sympathetic stimulation. The circuiar muscies must be relaxed under parasympathetic inhibition. Choice A is backwards and thus incorrect. Two different stimuli are required, so choices C and D are incorrect. The correct choice is B.

30.

A is correct, retina. Rays converge and focus on the retina. The cornea is a clear, tough membrane covering the front of the eyeball. Choice B is incorrect. The optic nerve is responsible for transmitting visual information to the brain, but it does not focus the image. Choice C is incorrect. The vitreous body is the inert stuff inside the eyeball. Choice D is inconect. The correct choice is A.

31.

B is correct, II only. As the object moves closer, the lens must become more convex. This is accomplished by contraction of the ciliary muscle. Choice II is correct. The ligaments do not contract, they merely pull the lens taut by defauit. Choice III is incorrect. The pupil responds to changes in light, not in focusing. Choice I is inconect. Since choice

1)

III is false, choice D is incorrect

as

well. The correct choice is B.

C is correct, pink. The condition of not having epitheliai pigment is call albinism. A person or animal with albinism is called an albino. Most strains of laboratory rats are albinos. They have pink eyes, due to the color of the blood vessels visible in their iris. Also, you may have seen white bunnies with pink eyes. Pigment covers up the color of the blood vessels in people and animals who don't have albinism. Black eyes corrtain the most pigment. Choice A is incorrect. Green eyes contain some pigment, as well. Choice D is incorrect. You may not know about the lab rats, but a person would not have white eyes due to the presence of blood vessels in the iris. Choice B is incorrect. The correct choice is C.

33.

C is correct, by providing a non-uniform corrective lens. Since the person has an irregularly shaped lens or cornea, the corrective lenses used should be shaped to correct these problems and allow focusing of the image on the retina.

This would require that the corrective lens have a compatible non-uniform shape. Convex lenses correct farsightedness. Choice A is incorrect. Concave lenses correct nearsightedness. Choice B is incorrect. A uniform lens would not help the person with astigmatism. Choice D is incorrect. The correct choice is C.

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Biology

Nerve & Muscle

Section I Answers

D is correct, lipids making up the cell membrane are uncharged. Cell membranes are made up primariiy of lipids. Lipids contain very few charged groups, cannot carry current, and have a high electrical resistance. Materials with high electrical resistance are known as insuiators. The lipid layers of the flasma membrane are regions of high

34.

electrical resistance separating two water compartments of low

."iistun.". The correct choice

is D.

35.

D is correct, negative, with the excess charge representing a very small fraction of the total number of ions inside and outside the cell. This problem requires outside knowledge about the resting membrane potential of a typical neuron. The typical re.stilg membrane potential lies somewhere between -40 mv to -75 mV. The minus sign indicates that the potential is negative with reference to the inside of the cell. In other words, there is an excess of negative charge inside the cell relative to the outside. This eliminates choices A and B. Now, do these excess charges represent a small or large fraction of the total ions in the cell? The excess charges are but a very small fraction of the total number of ions inside and outside the cell. The correct choice is D.

36.

B is correct. We are looking for a picture that represents the potassium equilibrium potential. The equilibrium potential will exist where the fbrce due to the concentration gradient is equal to tlie electrical force gradient. Potassium ions have a higher concentration inside the cell, elimilnating choices C and choice A and choice B, we simpry iook at the magnitude of the arrows.

D.

To discriminate between

Again' at the equilibrium potential, the two forces will be equal. Answer choice A represents the resting membrane potential. The correct choice is B. -'\t.

C is correct, sodium ions diffuse into the cell. One can arive at this answer by thinking about the membrane potential and the direction in which it is going. We are told the resting membrane potentiailies above that for the potassium equilibrium potential. This means that some positive charge is coming back into the cell, making the potential higher. of the possible answers, the o-nly_ way this is going to -hupp"n (givln the true ion concentratiois) is to have sodium coming into the cell. This is indeed the case. There is a permeability to sodium ions over the plasma membrane, and this contribution causes a rise in the membrane potential. Chloride lons have a higher conceniration outside of the cell, so if they move down their gradient, they ;i11 decrease the membrane potential. The correct

choice is C. 38.

C is correct, oxidation of nutrients. The pump that transports sodium and potassium ions does breakdown ATp to couple the energy of that reaction to the work it must perform in transporting the ions. However, the question alludes to the ultimate source of the energy. Therefore, we need to think about how we are getting the ATp that is used by this pump. This answer comes from our knowledge of metabolism and an undeistanOing that food is oxidized in the course of glycolysis and the Krebs cycle. The electrons released are eventually transpoited down the electron transport chain and are coupled to a proton gradient, which results in the fbrmation-of ATF. The correct choice is C.

39.

C is correct, a steady state. The constancy of the resting membrane potential is best described as a steady-state potential. The answer really comes down to two choices. Is it an equilibrium, or is it a steady state? With an equilibnum, no energy input is needed to maintain the state. Is this the case? No. We use the eneigy of ATp to run the Nao/Ke ATPase pump that maintains the concentration gradients of the these ions. Since there is energy used, this cannot be described as an equilibrium. When a state is constant, but energy is used to maintain that statE, this is termed a steady-state system. The correct choice is C.

40.

D is correct, the new resting membrane potential is not affectqd at all by the chloride ion. We are told that the plasma membrane of many celis are permeable to chloride ions and do not contain chloride-ion pumps. Therefore, in these cells, the membrane potential set up by other ions will act on chioride ions. The inside negativity moves

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Biology

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Section I Answers

chloride out of the cell until a concentration gradient develops. Looking at Table 1, we see the concentration of chloride is larger extracelluiarly. This concentration gradient will force ihloride ions into the cell. However, the diffusion fbrce will be exactiy equal to the electrical force pushing chloride ions out of the cell. The resuit is that the equilibrium potential for chloride ions is equal to the resting membrane potential, and the chloride ion makes no contribution to the magnitude of the membrane potential. The correct choice is D.

4I.

D is correct, CaZ@. Calcium ions are the link

between depolarization

of the presynaptic membrane and

neurotransmitter reiease. Depolarization of the terminal causes voltage-sensitive calcium channeis in the membrane to open, and calcium diffuses into the axon terminal from the extracellular fluid, including that in the synaptic cleft. The increase in the calcium level in the terminal causes vesicles filled with neurotra*nsmitter to flse with the presynaptic membrane and release their contents into the synaptic cleft. The correct choice is D. 42.

C.is correct. In the question, we are given a graph of a binding assay for the muscarinic acetylcholine receptor, where the substrate is muscarine. The question tells us that scopolamine, a competitive inhibiior, is added to the solution and the assay is run again. A competitive inhibitor will compete for the same site on the receptor as muscarine. It will not remove or incapacitate any of the receptors (this would be a non-competitive inhibitor). Since the total number of receptors remain in place, the levei of maximum binding will not &ung". This eliminates choices A and B. What will change is the apparent affinity of the receptor for muscarine, whicl can be represented by half'-maximum binding. The apparent affinity of the receptor for muicarine will decrease, because the inhibitor is competing for the same site. Therefore, we will see an increase in the half-maximum binding site of muscarine, because it now takes a higher concentration of muscarine to reach the half-maximum binding ,It". R"-"mber, the tell-tale sign of a competitive inhibitor is that if we increase the concentration of either substra"te. one of them should be able to out-compete the other and reach maximum binding. The correct choice is C.

-13.

C is correct, an excitatory post-synaptic potential is generated. We are told from the question that the nicotinic acetyicholine receptor is essentially a sodium channel. Therefore, when acetylchoiine binds to a nicotinic acetylcholine receptor, we are essentially activating a sodium channel. When a sodium channel opens, what

happens? Since sodium always has an extracellular concentration higher than its intracellular concentrati,cn, sodium comes into the cell and causes a net movement of positive ions in the cell. At this point, we can eliminate choice D. This causes a slight depolarization and is known as a excitatory post-synaptic potential (EPSP). Eliminate choice B. This is a graded potential, and it is not an all-or-nothing action potential. in order to generate an action potential, the depolarization must be beyond a threshold value set by that particular cell. Usuallyl several EPSps added together will eventually generate an action potential at an area known as the axon hillock, where a hish igh concentration of sodium channels are found. The correct choice is C.

14.

C is correct, a net nicotinic and parasympathetic nervous system excitation. We want to offer the person who is addicted to nicotine the "high," without the cardiovascuiar risks associated with nicotine. Therefore, we wiil want to

stimulate the nicotinic receptors in the central nervous system to achieve this sensation. The problem is that nicotinic receptors are located in the pre-ganglionic synapse of both the sympathetic and parasympathetic nervous systems. We might think that since both are stimulated, they simply cancel each other out. However, we know that nicotine causes stress on the heart, so the nicotinic receptor in the sympathetic nervous system must be dominant. With this in mind, let us look at the two drugs added. The first is physostigmine (eserine). This acetylcholinesrerase inhibitor will cause increased levels of acetylcholine in both nicotinic and muscarinic receptors. This will give us the nicotine "high" that we want. Yet we do not want to stimulate all the muscarinic receptors, so we add scopolamine. This acts to block all muscarinic receptors. This is what we want. We get rhe familiar feeling caused by nicotine and have blocked all muscarinic receptors. We are not done, though. We are still left to deal with the cardiovascular stress. The physostigmine will increase acetylcholine (ACh) levels at the first synapse in the sympathetic nervous system. But in the parasympathetic system, it increases ACh levels at both the nicotinic and the muscarinic receptors (remember that scopolamine does not work directly upon heart tissue). Therefore, we have two stimuiations in the parasympathetic system while we only have one in the sympathetic system. Therefore, we are left with a net nicotinic stimulation with a parasympathetic excitation. The correct choice is C. -t5.

A is correct, nicotine stimulation of the nicotinic acetylcholine receptor is dominant in the sympathetic nervous system. We would think that because the first synapse in both the sympathetic and parasympathetic are nicotinic that nicotine would stimulate both of them equally, and therefbre there would be no stress on the heart. However, we

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Biology

Nerve

& Muscle

Section I Answers

know this not to be the case, as nicotine places considerable stress on the cardiovascular system. Therefore, nicotine stimulation of the nicotinic acetylcholine receptor in the sympathetic nervous system must be dominant over the one in the parasympathetic nervous system. There is no evidence in the passage fbr the claim made in answer choice C. Cardiovascular stress is not brought about by activity in the centrai n".uou, system, but by activity in the peripheral nervous system. we can eliminate choice D. The correct choice is A. 46.

D is correct' axons' moving a greater distance in_ the sympathetic nervous system. The question tells us we have a molecule that will be moving in a retrograde (backward) fashion from the synapses io the heart between both sympathetic and parasympathetic nervous systems. The molecuie will thus ue movinj *r.ougrr axons (eiiminating choices A and B), because the axons, not the dendrites will be closest to the organ. "In other words, an electrical signal reaching the organ will travel through a nerve's axon to reach the synapsJbetw""" tr,. nerve and the organ. So the next question becomes one of distance. Remember that in the sympathetic nervous system, the preganglionic fiber is short while the postganglionic fiber is long. The molecule wili be traveling through the postganglionic fiber to reach the first synapi" (the iirst synapse is th{unction between the pre- and postganglionic fibers). Therefore, the molecule will travel a longei'distance in the sympathetic nervous system. The correct choice is D.

D is correct, the axons

47.

by oligodendrocytes. This question calls on our knowledge of the nervous system outside of what is stated in the passage. We are looking for thb most likely explanation for the increase in the frequency of the action potential' Myelinated nerves have th*e ability to increase thl frequency of action potential conduction. Therefore, we can nalrow the options down to choices Ctr D. The question tiren becomes: which celis are responsible for the myelination? In both cases, glial cells are responsible for laying down the myelin sheath. In the central nervous system, these cells are called oligodendrocyter, r"itit" in the pNS tt"y u." called Schwann cells. Since we are talking about nerves located in the CNS, the best unr*", becomes choice D. The correct choice is D.

48.

D is correct, afferent pathway, containing nerve axons from both eyes. Again, this question requires us to draw on our knowledge of the nervous system, and in particular, the Cin. must remember that the information ieading "y". towards the brain is found in the afferent nerJes. Information leading away from the CNS is found in the efferent nerves. Therefore, answer choices A and B can be eliminated. Now, the qlestion becomes whether the optic tract

are myelinated

contains information from one eye or two. The answer is two. Information from the nasal hemiretina (medial half

of the retina) of the left eye cross the optic chiasm and enters the right optic tract. The right optic tract is also made up of nerve fibers originating from the temporai hemiretina of the r'ight -eye. In that waylthe right side of the brain processes information from the ieft side of the visual world. Because the tract contains informaiion from both eyes,

D is the correct answer. The correct choice is D.

C is

49.

correct, this area_is free of photoreceptors. The question is fairly straightforward. It asks for the best explanation of the "blind spot." The biind spot is the area on the retina where the 6ptic nerve bundle leaves the eye. That is not the reason why the area is termed the blind spot. The region is blind because it has no photoreceptors. If there are no photoreceptors to be found, there can be no transduition of light into a visual image. The correct choice is C.

50.

C is correct, acetylcholine. We are told from the passage that the neurons which make up the Edinger-Westphal nucleus are parasympathetic neurons. Therefore, this question is really testing one's knowLdge of the

neurotransmitter used by parasympathetic neurons. We cannot be expected to know from the question al6ne which neurotransmitter these neurons use. However, we are supposed to be aware that neurons that are parasympathetic use the neurotransmitter acetylcholine. The correct choice is C. 51.

A is correct, the optic nerve of the left eye is intact, but the efferent limb of the left eye is damaged. The question

tells us we see a consensual but not a direct response. We can therefore conclude that ihe optic nJrve of the left eye is intact because the optic nerve of the right eye is not involved in the response. The ."rponr" involves information going down the left optic nerve to the pretectal area. From the pretectal area, n"u.ons project bilateraliy to the Edinger-Westphal nucleus. Axons from neurons in the nucleus innervate the ciliary ganglion. We see a consensual response in the right eye. We can therefore conclude that the bilateral projection and the efferent pathway to the right eve are unharmed. In addition, we can conciude that there is some problem with the efferent patirway to the Ieft eye. We are not seeing a constriction of the pupil in response to the light being shone. We aie left with the


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Section I Answers

fbllowing conclusion: The ieft optic nerve is intact, but the efferent pathway somehow damaged. The correct choice is A. it

of the left eye is somewhere

and

B is correct, right optic nerve. The light shone in the right eye elicits neither a direct or consensual response. This most likely means that there is damage to the right optic nerve. The result of the damage is that the infbrmation leading away ftom the eye is unable to elicit a reflex which results in the constriction of both eyes. This can be the only explanation for the inforn'ration given in the question, because light shone in the other eye elicits both a direct and a consensual response. The correct choice is B.

,{ is correct, anterogressiveiy, via fast axonai hansport. Recali from the passage that transport from the soma ton'ards the axon terminal is known as anterograde axonal transport. Therefore, this eliminates choices C and D. \ext. is the neurotransmitter rroving via fast or slow axonal transport? Recall that neuropeptides are packaged into membrane bound vesicles as they are formed near the soma. From the passage, we know that membrane-bound c.rganelles are transported rapidly by fast axonal transport. While the vesicle may not be technically considered an orsanelle, given the choice between membrane-bound organelle and cytoplasm, a vesicle is best aligned with the :rembrane-bound organelle system. Our best choice includes fast axonal transport. The correct choice is A. C is correct, Ca2o. We are told from the passage that organelies interact with microtubules through a linkage -'-nilar to that between thick and thin filaments of skeletal muscle cells. Recall that thick filaments refer to myosin, 'rile thin filaments refer to actin. Recall that calcium is needed 1br actin and myosin interaction, as calcium :tltoves the troponin-tropomyosin complex which covers the actin binding sites. Since the interaction of the ',-:anelles with the microtubules is simiiar to that of actin/myosin, one can best conclude that calcium triggers the ..=.,:r;rction. The correct choice is C. C is correct. complete translation of the peptide in the cytoplasrn. Recall that peptides that are produced by a ceil - .:e purpose of secretion (neulopeptides are certainly an example) are processed through the rough endoplasmic -.---:1um. The peptide being translated will have a signai peptide that attracts a cytoplasmic signai receptor particle SRP . When the SRP attaches to the signal peptide, apause in translation will occur. The translation does not .:.:r aqain until the SRP complex (SRP/signal peptide/ribosome) binds to an SRP receptor located on the rough a-. \\-hen the SRP complex binds to the SRP receptor, translation begins aiong with transport of the new protein :: .:.3 lumen of the RER. At that point, the signal peptide is cleaved off and the proteins are modified in both the : :F. ::d the lumen of the Golgi apparatus. From the trans face of the Golgi, the peptide buds off as a vesicle. --,:, l:rs to this hypothesis, there is no complete translation of the neuropeptide in the cytoplasm. The correct

:ir

,ice is C.

- is torrect. pyruvate. Glucose will go through the glycolytic pathway and the end product of glycolysis is I -- :.3. Recall that giycolysis occurs in the cytosol of cells. This is a very straightforward question asking us to ':- :.i'.::: that pyruvate is transported over both the outer and inner mitochondrial membranes to enter the matrix. T:c i,,rrrect choice is C.

- r: i,rrrect. so1.na to trace the synaptic endings of neurons. The marker substance moves anterogressiveiy. - . - . - '_l :o the passage, the marker substance will move from the soma toward the synaptic ending. If we are - :.: j.:r :r tracing a neural pathway, we should use this marker to identify synaptic endings. In other words, we , - .,':-': this marker in a given cell body, and it will trace the axon and the synaptic endings of the neuron" If : - : -: ::: narker into the soma, it will not trace the dendrites because of its direction of movement. Therefore, ; --- :.::-lnate choice C. Furthermore, we do not want to inject the marker by the synapse, because there is ' - ; -:. :r trace. The correct choice is D. I * i rrect \\'e are told from the graph in the question that action potentials arise in a postsynaptic cel1 after the - . : '-'::.. lcholine. We then add acetylcholinesterase, which breaks down ACh. We must look at the change ... : ::.:::r --ithe action potential response. Why? Recall that action potentials are all-or-nothing events, so they - -:,- :3 rn amplitude. We can easily eliminate two of the choices. We then look at the pattern, and in this , : -. .:----lencv of response. If we are going to have lower leveis of ACh due to the addition of . : - - .:!.:iase. r.ve should see a less frequent train of action potentials in the postsynaptic cell. The correct 1 -:L:f)

:;:i

Review

/b


Biology

Nerve & Muscle

Section I Answers

59.

C is correct, equally common in both men and women. The passage tells you the defect is an autosomal dominant one. This means it is not sex-iinked, so eliminate choices A and B as incorrect. We also learn iiorn the passage that the age of onset is from 35 years of age. Eliminate choice D. The correct choice is C.

60.

D is correct, voluntary movements. The hypothalamus produces releasing hormones that aff'ect the pituitary. Choice A is incorrect. The control of speech is localized in the cerebral cortex. Choice B is incorrect. The parasympathetic nervous system (PNS) is not controlled by the basal ganglia. In the second paragraph, we learn that symptoms of the disease include rapid, uncontrolled, jerky movements. The correct choice is D.

67.

D is correct, 4-aminobutanoic acid. This molecule is a 4-carbon organic acid. The skeletal structure is called butanoic acid. The carbon of the carboxyl group is designated 1. This means the amino group is attached to the 4th carbon. This gives the name 4-aminobutanoic acid. The correct choice is D.

62.

D is correct, glutamate. This information is given in the second paragraph of the passage. A key missing enzyme that leads to GABA deficiency is glutamate decarboxylase. If we correctly named the previous molecule, GABA, then we can imagine adding on a carboxyl group on the C-zl carbon to make glutamate. Choice A is incorrect: histidine is the precursor of histamine. Choice B is incorrect. Choice C is incorrect: tyrosine is the precursor of epinephrine and norepinephrine. The correct choice is D.

63.

C is correct, 507o. Since the gene is autosomal dominant and the father is not affected, he is homozygous recessive for this disease. The gene is rare, and only one of the woman's parents shows a tamilial history for the disease. The woman is probabLy heterozygous. Therefore, she either passed on the dominant, disease-causing gene or the recessive gene. The chance is 50i50, since she has 2 X-chromosomes and only one is affected. The correct choice is C.

64.

B is correct, the vesicles of presynaptic neurons. Neurotransmitters are synthesized in presynaptic neurons, stored in vesicles there, and released by the arrival of a neural impulse. The released neurotransmitter crosses the synaptic cleft and interacts with the receptors on the postsynaptic neuron's membrane. The correct choice is B.

65.

A is correct, cone cells. This answer can be arrived at using previous knowledge, and not information specifically from the passage. The fact is that cones are used for color vision, while rods are used for black-and-white vision. The question asks about seeing in the dark, which definiteiy qualifies for a lack of color. Of course. one may see sonre color in the dark and the cones would be responsible, but the question asks which cell is ntost likeb, y1s1 involved. Consider the other answers. We have already discussed that rods play a role in black-and-white vision, making their role in dark vision likely. The other cells are involved in taking information from the photoreceptors to the brain. It is stated in the passage that the photoreceptors synapse with the bipoiar cells, which then synapse with the ganglion celis. Therefore, for both color and black/white vision, these cell types will be involved in the transmission of visual information. The correct choice is A.

66.

C is corrrect, hyperpolarization of the photoreceptor. It is stated in the passage that upon stimulation of a photoreceptor cell by a photon, the sodium channels close. This implies that when there is no light stimulation, the sodium channeis are open. When sodium channels are open, the sodium ion comes in (following its gradient) and causes the cell to be depolarized. In an unstimualted cell, a very active Nae/K@ pump is constantly restoring the sodium gradient. Again, the net result of this is that a photoreceptor cell in the dark is normally depolarized. When a photon hits, the sodium channels close and the pump continues to work. Both of the these events cause a hyperpolarization of the photoreceptor cell. This is a very special case for photoreceptors, and it is important to be aware of this phenomonon. Based on this information, the other answers can easily be eliminated. The correct choice is C.

67.

area where there is the highest density of cones on the retina will give the greatest acuity, or precision. At this point, one must recall the anatomy of the eye. The area of highest cone density is the fovea centralis, and when we look at objects, we move our eyes to focus light onto this region of the retina. This answer could not be obtained from the passage, but we thought it would be nice to review some eye anatomy. Look

A is correct, fovea centralis. The

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76

The Berkeley Keview Speciatizing in MCAT Preparation

Biology

Nerve Ef Muscle

Section I Answers

at the other answers. The cornea is the transparent structure located at the front of the eye and is the structure through which light rays enter. The optic disc is where the optic nerve leaves the eye and the retinal blood vessels enter. Since there are no photoreceptors overlying this disk, it is known as the blind spot. The choroid is a pigmented layer behind the retina, which contains many of the blood vessels that nourish structures in the eye. It also absorbs light not taken up by photoreceptors. It absorbs rather than reflects, so as not to alter the photoreceptor responses to other light. The correct choice is A. 68.

D is correct, occipital. We need to recall our knowledge of the cortices of the brain. The experiment removed the thalamus. The thalamus. in particular the lateral geniculate nucleus, is a relay station for visual information. From this nucleus, information then goes on to the visual cortex, located in the occipital lobe of the brain. The occipital lobe is located above the cerebellum. Based on this, the only possibie answer choice is D. As a very brief review, one can associate the frontal lobe with association processes, the temporal lobe with hearing, and the parietal lobe with high level sensory and motor control. The correct choice is D.

69.

D is correct, ganglion cells. The answer can be obtained by either previous knowledge, or by reading the passage carefully. It is stated that both photoreceptors are located at the back ofthe retina. It clearly states that the four other types of cells (bipolar, etc.) are part of the retina. Since the photoreceptors synapse with the bipoiar cells, and the bipolar cells synapse with the ganglia, one could create a picture in their head where the ganglion cells are furthest away from the photoreceptor celis, or the first layer in the retina. Based on this thought process, light entering the eye will first pass over the ganglion cells. The correct choice is D.

70.

C is correct, III only. The place where the axons of the ganglion cells leave as the optic nerve is called the optic disc. There are no photoreceptors at this region of the retina, because the axons are leaving and blood vessels are entering. If there are no photoreceptors, then vision is not possible. This is ofcourse one's "blind spot." Consider the other possibilities. The region of highest acuity is where one finds the highest density of cones. This is certainly not the case. In the case of stereovision, it is not the structure of one eye that allows for depth perception, but the fact that we have two eyes looking at one object from different angles. This gives us stereovision. Therefore, statement II does not apply. Statement III is the only applicable statement. The correct choice is C.

7t.

C is correct, left eye and the temporal hemiretina of the right eye. If light originates in the left visual hemifield, then we known that our right side of the brain will be processing this information. In order to get to the right side of the brain, we need to have light strike the temporal hemiretina of the right eye. If you follow the optic nerve, you will see it goes to the right side of the brain. In addition, light from the left side will strike the nasal hemiretina of the left eye. The optic nerve from this part of the retina will cross over at the optic chiasm and move on to the right side of the brain. Draw some lines of light from the left visual hemifield; it may augment your understanding of how light originating from one side of the visual field becomes processed in the opposite side of your brain. The correct choice is C.

72.

D is coruect, one million times greater than the reference pressure. The question asks for the relationship between the loudest sound pressure level and the reference pressure. We know from the question that the upper available dB range is 120. In order to arrive at this figure, we must have a test pressure that is 106 times as large as the reference pressure, because 20 log 106 = 120. The conect choice is D.

73.

D is correct, force/unit area acting on the oval window is increased. We know the sound wave is conducted through the set of small bones in the middle ear. Since the passage does not mention any significant loss of force during this transmission, we can assume the force associated with the wave does not increase or decrease. Well, how about the pressure? Pressure is the force/unit area. The force does not change. However, the area becomes smaller. Therefore, the total force/unit area acting on the ovai window is increased. The correct choice is D.

74.

B is correct, 1 x 10-3 s. The question is asking us to carry out some calculations. First, let us convert the 3.4 x 10- 4 kilometers into meters. This gives us 0.34 m. Based on this value and the speed of sound, we get a frequency of 1000 Hz. Recall that the period of a wave is the inverse of its frequency. Therefore, the period of this wave is 1/1000 Hz, or 1 x l0-3 seconds. The correct choice is B.

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Biology

Nerve & Muscle

Section I Answers

C is correct, oscillating

depolarizing-hyperpolarizing. The passage tells us that oscillating movements of the organ of Corti both excite and inhibit sensory transducingiells, bbcauie the hair cells attached-to these cells are *oriing back and forth' In one direction, these hair celis cause cation channels to open. This will cause a depolarization. However, in the other direction, the hair cells cause a closure of these channels. This is associated with a hyperpolarization. Therefore, the cell undergoes an oscillating depolarizing-hyperpolarizing potential. The correct

choice is C.

A is correct, region 1. The question is asking which region of the basilar membrane would be associared with highil-equency sounds. High-frequency sounds are associaied with shorr wavelengths and wiil be more energetic than lower-frequency sound waves. We can assume that the broad flexible ,egion-of the basilar membrane will not be able to resonate with these short energetic movements. In other words. thJ broad flexibie region will not

76.

receive high frequency sound as The correct choice is A.

well. Such movements

be able to are achieved by the short, stiff region of the membrane.

D is coffect, cell 5. Let us look at the circuit model. We know that information is arriving at the left ear at a tirne before that of the right ear. we also know that these cells fire maximally when the inputi from both sides arri'e simultaneously. For example' if sound arrived at both ears at the same time, ccll : woutd fire maximally, because input would arrive at that neuron simultaneously. Based on that logic, if information is ariving at the left ear befbre the right' it will travel up the entire circuit before the information-from the right ear will arrive. Il the tu,o inputs arrive at any cell at the same time, that cell must be located in the upper region of the model. 'rherefore, cell 5 is the best answer. The correct choice is D.

77.

B is correct, middle ear. The passage tells us that conducting sound through the temporal bone is one way to bypass the middle ear' The passage also tells us that this means of transmitting sound is quite inefficient. Therefore, when the fork is placed on the patient's temporal bone and the patient hears the sound (afier they no longer heard the sound with the fork at their ear), we can narrow down the p.oblem to conduction of air in the middle ear. The correct choice is B.

79.

C is correct, -eraph C. The Y-axis is ampiitude and th_e X-axis is frequency. If cells have a characteristic fiequency. the data should reach only one point on the X-axis. Some graphs ,hor tr"o. Choices A and D can be eiiminatecl. Now that we know thai only one characteristic frequency point ihould be reached, which direction will the curve go? At a cell's best frequency, does the loudness of a sound need to be high or low to elicit a response? It is low. The amplitude of the sound does not need to be high, if we are at the ceil's characteristic frequJncy. However, if we move away fl'om that frequency in either direction, the amplitude of the sound needed to elicit a response increases. Based on this infbrmation, graph C is best. The correct choice is C.

80.

B is correct, hydroxylation foilowed by decarboxylation. An -OH (hydroxyl) group is added to lrp and a -COOH (carboxyl) group is removed to form serotonin. The terms for this are-hydroxyiation and decarboxyiation. Choices A. C, and D are incorrect. The correct choice is B.

81.

C is correct' at high LNAAs are converted to serotonin. We are looking lor the false answer. Serum LNAA concentration does increase in both trials as dietary protein content increases. Choice A is true. Serum Trp also increases with dietary protein content. Choice D is true. In Trial 1,the24Eo protein meal does attenuate the increase in serotonin due to Trp (competition from LNAAs). Choice B is true. Otheiserum LNAAs are not converted to the neurotransmittel serotonin. Choice C is False, and we are looking for the false answer. The correct choice is C.

82.

D is correct, I, II, and III. In examining the data tables, look at how the serum LNAAs increased with the higher If these were competing with Trp for the LNAA receptor, then the brain Trp level should decrease as protein levels increase. This is true for the higher protein diets. In trial 1, the levels niZ+Eo and,40Va cause this competition. InTriai2,thelevelsof l27o,24Vo,and,40Vocausethiscompetition. Iandllarecorrect. Inalltr.ials, the ievel of serum Trp increased over the fasted group. III is correct. Theiorrect choice is D. protein diets'

B is correct, Ii only. Trp is not a neurotransmitter, but it can be converted into the neurotransmitter. serotonin. I is incorrect. Trp is an essential amino acid and cannot be synthesized by the body. II is correct. As we are told in the passage, Trp is a neutral amino acid. III is inconect. The correct choice is B.

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Biology

Nerve & Muscle

Section I Answers

84.

A is correct, accumulation of an intermediate metabolite in the brain. If one of the enzymes in a pathway is blocked by an inhibitor, then some intermediate metabolite is not processed through to the final product. This means the intermediate form may remain and accumulate in the tissue. In this case, some intermediate form between tryptophan and serotonin will accumulate. We have no real interest in the concentration of the inhibitor itself either in the blood or the brain, just whether it is working or nol. Choices C and D are incorrect. The Trp in the blood would not provide any information about lhe Trp conversion pathway rate in the brain. Choice B is incorrect. The correct choice is A.

85.

B is correct, dietary Trp alone would increase brain serotonin concentrations, since few LNAAs would be cornpeting. Trp is an amino acid, and it would be easily absorbed in the srnail intestine, either alone or in combination with other foods. Choice A is incorrect. Trp would only increase synthesis of serotonin, since it is not the precursor lor other classes of neurotransmitters, Choice C is incorrect. Actually, an increase in serotonin levels leads to a feeling of relaxation, not alertness. Choice D is incorrect. The correct choice is B.

86.

C is correct, a greater frequency of the action potentials indicates greater intensity. Action potentials act in an "allor-none" fashion. There is no amplitude modulation. Choices A and B are incorrect. Identical action potentials are produced more frequentiy to indicate a stronger stimulus, such as the stretch caused by a set of increasing weights. The correct choice is C.

87.

A is correct, the action potentials had increased to a maximum with the 50-g weight and increased no further, due

to

the refractory period of the neuron. The action potentials were limited by the nonconducting refractory period of the neuron. Increasing the weight could not speed up the "recovery time." The nerve was not damaged by the weight. Choice B is incorrect. Do not assume equipment f-ailure when a biological mechanism is being studied. Choice C is incorrect. A neurotransmitter defect would probably mean no action potentials rather than slowed ones, exactl]' identical to another weight. Choice D is incorrect. The correct choice is A. 88.

5 mg (1/200) of the 1 gram weight caused only a few action potentials. Since there was not an action potential generated, this weight must have been too small to overcome the threshold for generating action potentials. The weight was not too heavy, so choice B is incorrect. The weight does not have to produce an action potentiai, if its stimulus is lower than the threshold.

A is correct, The weight was not heavy enough to exceed the action-potential threshold.

Choice C is incorrect. Alt the action potentials were identical in size, so frequency was modulated, not the amplitude. Choice D is incorrect. The correct choice is A.

89.

D is correct, plasma membrane. The cytoplasm of a neuron is a poor conductor. Choice A is incorrect. The Schwann cell acts to insulate nerve fibers and to provide gaps between the conducting regions. Choice B is incorrect. The endoplasmic reticulum does not function as a conductor. Choice C is incorrect. The plasma membrane is the site of conduction of action potentials. The correct choice is D.

90.

B is correct, sodium moves from the extracellular fluid into the cell, and then potassium moves from the intracellular fluid to the outside of the cell. This is an easy question, but all the words make it easy to make a mistake. First, eliminate wrong answers based on ion location. Sodium is the principle extraceliular ion, while potassium is the principle intracellular ion. Choices C and D are incorrect based on this fact. Then eliminate based on orAer of acdon. The sodium channeis open first, fbllowed by the potassium channels. Choice A is

"hoi."r incorrect. The correct choice

91.

is B.

B is correct, afferent function. Afferent pathways carry information from the tissues to the nervous system. Information of muscle stretch was being conveyed back toward the (no-longer-present) nervous system. Efferent function would be tested, if a nerve were stimuiated with electrodes to make a muscle twitch. Choice B is inconect. Since the nerve and muscle under study are isolated from the intact frog, there is no parasympathetic or sympathetic function. The correct choice is B.

o't

C is correct, action potentials continue briefly but stop eventually, due to a lack of ATP. The Nao/Ke pump requires ATP to perfoim active transport. The ceil has a smail reserve of ATP that would provide brief function of the pump if ATP production stopped. Choice A is incorrect. Action potentials stop when all the ATP is gone. Cholce B is incorrect. Action potentials require ATP, they do not generate it. Choice D is incorrect. The correct choice is C.

- :'pr right

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79


Biology 93.

Nerve & Muscle

Section I Answers

has at reastone muscre that t;l:'"1::*,1'11'll:1"';"*:::*lf-t:,'l:i::':*.?t1"-3i:d'i'"t'c'9:p rt attiches above.tiie hip joint u;h";";;i* j;-ff;;ffi'rr,""'""iil1'";;ifi#i:;crosses the hip j:f,-ilf.t":1v.

. dru rr il;:;i*Ti';iJ;il;TT,?ifr':i"#ff"i'lj;'lfij,ffJ',','fl rre^run or lne nip by pulling the temur and tibia upwards ::::::,,:,1, 'l.,:r."{;1;;;;"-,#,;;",;;rr"# Refer to the skeleton diagram and try ro visualize-,''i exacrly how this works. -fhe The other :"":,Td^t lli :!"tj' orher answer ili t,ii iii ri"l"J'iffi r # :ff; shorten fhev urnrrld nrrll rho f^^'. L^-t^,,^.")^ rhis at:^ is :^ hip 1,:,, fe1:i b.aclwar{. extJnsion,;;;'fi;;i";;;j;.i""#d;i*[:"i1,i1.i;Hi ',ip' "-i

ll,j^.,f1.i1:ll:ildf,rlrh:

when they contract, they pull the thigh inwards. rrris

11"^,1"::t; choice is A.

is'aaJJ.,-",'itirr""iffii::tfi: #::i

94.

c is correct' the veins involved must have valves. How can blood be returned from the lower extremity against gravity and under the low pressure that is normally found in the venous system? Muscles of the thigh and lower leg "massage" the elastic veins, pumping blood in onl direction: ,prurar. -Blood can only go in this direction because veins in the lower extremity havJ one-way valves which p."u"it uu.tlo* of blood u;uj f-- rhe heart. These are very important in maintaining proper circulation. varicose u"inr r"rullir"*â"i*ir-l"i'tr.ri Let,s consider the other answer choices' Eiiminite choices B and D because "u,""r. arteries do not return blood to the heart, veins do. Eliminate choice A because if veins had inelastic walls, they .ouldnlt b" squeezed by muscles and the process would not work. The correct choice is C.

95.

B is correct, the quadriceps must relax. The quadriceps and the hamstrings are opposi.ng muscle groups with opposite actions' The hamstrings extend the hip posterioity while the quadriceps flex the hip anteriorly. In order for one of these antagonistic muscle groups to woik, the otler must relax. tf both li-ultaneously, nothing would get accomplished because the two actions would cancel each "onr.u.i other out. Knowing-rh"r" should lead us to eliminate choices B and c. choice D can be ruled out because the obturator n"ru", u-..o.ding to the passage, innervates the adductor muscles, which aren't involved here. The correct choice is B.

96.

c is correct' to kick.1 .ba]l. The passage states that the femoral nerve innervates the quadriceps muscles. The quadriceps are responsible for flexing tnJrrip anteriorly (see prior question) and also to, ,riir.arrgthe knee. In other words, since the quadriceps crosses the knLejoint,.-ont.".iion putt, tt'r" lo*er leg and ioo, unt".iorly (a kicking action' known as extension of the knee joint). Paralysis of the fem'oral-nerve might tf,erefore afI-ect the ability to kick a ball, among other things. Let's coniider the other answer choices. choice A can be eiiminated because the quadriceps muscle doesn't cross the ankle joint and hence can't plantar flex the foot (i.e., cause a person to stand on tip-toes)' Choice B can be ruled out because extendin-g the hil pori".io.ty is the job of the hamstrings, not the quadriceps' Choice D can be eliminated because the adductor muscles aren't innervated by the 1emoral nerve; the obturator innervates them. The correct choice is C.

97.

D is correct' weakness in the adductor muscles. what would happen if an intervertebrai disk herniated (squeezed out of its sheath) and smashed the adjacent sciatic nerve? l rom the passage, we learn that the sciatic nerve innervates the hamstring muscles. These muscles are.normally responsible fbr extending the hip posteriorlv and flexing the knee (i.e., bringing the heel upwards posterioriy). These actions of the hamstring group can

be explained by looking at where they attach. The himstringi attach to the ischium posreriorly, cross the hip joint, and attach inferiorly to the f'emur' tibia, and fibula. They thirefore cross both the hip .loint uni ir," r.n"";oint. shortening of the

hamstrings results in flexion

of the knee and exte.nsio^n of the hip posterioriy. Since the hamsirings are innervated by the sciatic nerve, damage to this nerve would affect walking (i.e., hip extension), difficulty moving the leg posteriorly

(knee flexion), and sensory stimulus along the back o1 tire thigtr and lower ieg (which manifests"as painf The combined conditions are known as sciatica. Choice D is our answer because the adductor muscles are not innervated by the sciatic nerve. According to the passage, they are innervated by the obturator nerve. We would not expect them to be affected by damage to the sciatic. The correct choice is D. 98.

C is correct, lower-than-normal CO2 levels. What effect would reduced blood flow have on exercising muscle 02 (which would be in high demand in active muscle) would not reach muscle fast enough due to poor circulation. This would lead to a decline in aerobic respiration (oxidative phosphorylation, etc.) reJulting in ATP deprivation. In order to partially compensate, anaerobii respiration (glycolysis and lactic acid f'ermentation) would be stepped up, resulting in a buildup of the by-product lactic aciA ,"uere pain). Co2 levels lpJssiUly would actually rise because it could not be carried away effectively by the blood. The "urrlng coriect choice is C. cells? To start,

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Biology

Ileart & Lungs

Cardiovascular Anatomy

Within iiving creatures there are two types of circulatory systems. In an open circulatory system the circulating fluid within the body is mixed with the body fluid itself. In other words, the blood in these creatures does not circulate entirely within the confines of vessels like arteries and veins. Most mollusks and arthropods have an open circulatory system. Conversely, in a closed circulatory

system the blood that flows throughout the body is confined to vessels like arteries and veins. It does not freely mix with the fluid of the body. Annelids and mammals like ourselves have a closed circulatory system. The function of the circulatory system is to bring nutrients and oxygen to the tissues of the body while simultaneously removing waste producis from those very tissues. Because the circulatory system is continually flowing it helps to maintain body temperature. Also, the circulatory system can act as a means to transport hormones to various locations within the body. Ultimately, though, as the blood enters into the smallest vessels, the capillaries, there will be diffusion between those capillaries and the cells in the immediate environment.

General Anatomical Features Let's examine the general anatomical features of the mammalian cardiovascular system. The foundation of the circulatory system is the heart. The heart can be thought of as being divided into two halves. These two haives are often referred to as the right heart and the left heart. \44ren we look at a picture of a heart we must remember that we are viewing that heart in its anatomical position. In the anatomical position the body is standing upright, the arms to the sides, hands

with plams up, and the face positioned forward. Thus, when we look at a diagram of a heart on paper, the right side of that paper really represents the left side of the heart and vice versa. The right heart pumps blood to the lungs and back to the left heart. The left heart pumps that blood to the remaining tissues of the body and back to the right heart. The blood that is pumped from the right heart to the lungs and back to the left heart is called the pulmonary circulation while the blood that is pumped from the left heart to the rest of the tissues and back to the right heart is called the systemic circulation. Both the pulmonary and systemic circulations lie in series

with

one another.

We can pick a starting point in the circulatory system and follow a red blood cell as it rnigrates through the pulmonary and systemic circulations. Let's start with a red blood cell in the right heart. The right heart is composed of the right atrium and right ventricle. As deoxygenatedblood passes from the right atrium into the right ventricle it is pumped rnto the pulmonary artery and to the lungs where it is oxygenated. The oxygenated blood returns to the left heart by way of the (left and right) pulmonary veins. The blood enters the left atrium and passes into the left ventricle where it is pumped out the aorta and to the branching arteries, arterioles, and capillaries. It is at the level of the capiilaries that the blood exchanges nutrients and oxygen for waste products created by metabolism. Deoxygenated blood passes from the capillaries to venules and then to larger veins and eventually to the superior and inferior vena cava which enter the right Copyright

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atrium. The circulation of our red blood cell can now start again. As you are studying this circuit it is important to note that arteries carry iiood away from the heart while veins carry blood from the tissues and lungs towards the heart. This can be seen in the diagram shown in Figure 2-1. Capillaries of Head and Arms

Superior Vena

Pulmonary

Cava

Artery

O

J

\3

.c

bo

t

o o

o

q

a)

-

!

(D

l-

+ O

0a

Right

Atrium

7

tr

-'

lnferior Vena

Pulmonary

""*,

Vein Left Atrium

Cava

Left Ventricle

Capillaries of Abdomen and Legs

Figure 2- I Blood flow through the cardiovascular system.

Aorta and Arteries As shown in Figure 2-L, blood ieaves the left ventricle of the heart and exits through the aorta and passes to both the superior and inferior portions of the body. The total cross-sectional area of the aorta is about s cm2. under resting conditions the velocity of blood flow in the aorta is about 30 cm/sec. The blooJ pressure fluctuates betrveen 120 and 80 mmHg with an average being about 100 mmHg. The volume of blood found in the aorta and arteries ui u.ty given time is about'l.6oh to 20'/..

As the left ventricle contracts (systole) it propeis blood out the aorta and into the arteries with a pressure of about 120 mmHg. At sea level this force is enough to raise a column of mercury 120 mm above the ground. After contraction takes Copyright O by The Berkeley Review

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Biotogy

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irlace and the ventricles begin to relax (diastole) and is about 80 mmHg.

n the arteries


fill r /ith blood, the pressure

\Ieasurement of Blood Pressure: When you have your blood pressure taken

a

'-riessure cuff is placed around your upper arm and a stethoscope is piaced over

-re antecubital artery (the artery at the bend between your upper and lower :.:rnl. The cuff is inflated above arterial systoiic pressure. This causes the artery in , .ur arm to collapse thus stopping the flow of blood. The pressure in the cuff is :e:reased and the pressure needle monitored. As soon as the pressure in the cuff -' relou' the systolic pressure, blood will jet through the small opening in the '::er\'. The flow through the artery is turbulent because of the great pressure :ushing the blood through such a narrow opening. The "tapping" sounds that are ,-.=a:d in the stethoscope approximately correspond to the systolic pressure in the ,-.=:.rt, The cuff is continualiy deflated and the artery slowly regains its original :j-.:le. The blood has an easier time passing through the vessel and the fiow :=i:rr.s to change from turbulent to laminar (smooth). When the turbulent sounds -rr,rfopear it means that you are recording the diastolic pressure of the heart. ---..d pressure readings are given as systolic/diastolic (e.g., 720 mmHg/8O ,:::'i{g). These are the approximate readings that the baroreceptors are sensing -:: --.re\- monitor vour blood pressure. - -.- rlood that moves through the arteries is under a lot of pressure. Therefore, ,: -.e arteries have to be rather durabie. They have thick walls and are composed : smooth muscle and connective tissue that contain both collagenous and "1,::tic fibers. The elasticity of the arteries prevents the blood pressure from - :: -::ing too high when it is ejected out of the heart and it also maintains a high : r.=r- jrl pressure between the systolic and diastolic phases of the heart. This : - ,.': blood to flow to the rest of the circulatory system without a sudden loss of

- = -::nen of all the blood vessels in the body are lined with epithetial cells and -: i>e these ceils are inside the cardiovascular system they are referred to as .rl,rihelial cells. Damage to these endothelial celis b,v the pulsating arterial ::

--::s.,:ie or even by abrasive substances in the blood can lead to the disease --.,':-: as atherosclerosis. Once these ceils are damaged choiesterol can build up :-- = site of the lesion and a plaque will form. During the later stages of the - ,:: is- the arteries become "hardened" from layer upon iayer of deposit. This is -: -:-:= j to as "hardening of the arteries" or arteriosclerosis. -

:

- :'.^-::ron

of the circulatory system is controlled (in part) by the sympathetic

--r :::asvmpathetic divisions of the autonomic nervous system.

The

:-::.:.eti.c division is the more important of the two. Besides nervous control of : :- :-,.ir- there is also humoral control from the action of ions or hormones and -.- ::rtrol at the level of the individual tissues from various metabolites.

:re metabolites important? If you were to start exercising a particular ".:--, "-ou would notice that the blood flow to that muscle would increase.

:'

.,

--,rrat

u.hen muscles contract ATP is hydrolyzed io ADP and Pi, lactic acid

::',=i, and otirer such metabolites circulate in the tissue. When

.;:

these

-:;es diffuse from the tissue out to the arterioles the smooth muscle dilates -- -: :::ases the flow of blood to that particular area. The blood brings not only :-::.:s and oxygen to the working tissue, but it carries away the waste . r-::i of metaboiism as well. r



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Arterioles After the aorta branches to form a variety of arteries, the arteries themselves

branch to form arterioles. The arterioles are important because they represent the major area of resistance in the cardiovascular system. Arterioles have strong muscular wails and are regulated as mentioned above.

Capillaries Diffusion takes place at the level of the capillaries. The total cross sectional area of the capiliaries is about 3000 cm2. The velocity of blood flow has been red.uced to about 2 cm/sec. The pressure is roughly 25 mmHg and the total volume of blood is about 5o/o to7%. The wall of the capillaries are"composed of a unicellular iayer of endothelial cells. surounding tirese cells is a basement membrane. Flowever, there is no connective tissue or smooth muscle. The capillary itself is just large enough for a red blood cell to squeeze through. At the entrance to the capillary bed is a precapillary sphinct"r .o*pos*d Lf smooth muscle which helps to regulates the flow of blood to the area.

Veins once the blood flows through the capillaries and reaches the veins the blood pressure has been reduced to a value between 0 mmHg and 25 mmHg. The velocity of blood flow through the veins is about 20 cm/Jec and the total cross sectional area is roughly T.cm2. Roughly s0'/. ofthe total blood volume is present in the veins at any given time.

Figure 2-2 A valve in

a vein showing directionality.

since there is not much pressure in the veins the amount of smooth muscle and elastic tissue surrounding the veins is reduced. However, they are under control of the sympathetic nerves. Note that the rate of blood flow inihe veins is similar to that in the arteries. This is due to specialized valves that allow the blood to flow in only one direction (see Figure 2-2). As the muscles that surround the veins in your body contract they squeeze the blood back towards the heart. Since these valves are "one-way valves" blood is prevented from flowing backward. trf the valves become damaged and blood is aliowed to flow backwarls, pressure in the veins can increase. This has a tendency to cause varicose veins, which are protrusions of the dilated veins beneath the skin.

Blood FIow in the tleart We have mentioned that deoxygenated blood returns from the tissues and enters the right atrium via the superior and inferior vena cava. As the right ventricle

begins to relax blood from the right atrium is pumped into the rig:ht ventricle. The right ventricle contracts and forces blood out the pulmonary irtery and to the lungs where it is oxygenated. Blood returning to the heart enters the left atrium via the right and left pulmonary veins. As the left ventricle begins to relax Copyight O by The Berkeley Review

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blood from the left atrium is pumped into the left ventricle. Contraction forces the blood in the left ventricle out the aorta and to the tissues of the body. See Figure 2-3.

Pulmonary

Pulmonary Artery

Artery

Pulmonary Veins

Superior Vena Cava

SA

Pulmonary Valves

(semilunar)

Aortic Valves (semilunar)

AV

Kight Atrium

L. Atrioventricular Valves (mitral)

R. Atrioventricular

Left

\ alves (tricuspid)

Ventricle

Inferior Vena Cava

Chordae tendineae

Kight Ventricle

.:->

::1. Descending

Aorta

Papillary

Bundle of flis gives rise to L. E( R. Bundle Branches (frrrther divides into Purkinje Fibers)

Muscle

ri.lre 2.5 - -.. -:ndmarks of the heart. -

-:

::.e heart there are valves between the right atrium and right ventricle valve), between the right ventricle and the pulmonary .:.. rulmonary valve, or tricuspid), between the left atrium and the left -:. :,= 'ie left atrioventricular valve, or mitral), and between the left ventricle -- .: r ird (the aortic valve). Once the ventricles are filied with blood and they - =:, :ontract the valves between the atria and the ventricles close. This : i: :r.\- backflow of blood into the atria and ensures that the blood will be : :-- rhe pulmonary and systemic systems of the body. The closing of the :.ricular valves between the atria and the ventricles as the ventricies are .- : ::1\'es the characteristic "lub" sound when listening to the heart. As ,: :-:rrs out of the pulmonary artery and the aorta, the pulmonary valve - :lrlic valve close, giving the characteristic "dub" sound. The closing of -. as blood is pumped either from the atria into the ventricles or from :-=s lo the puimonary or systemic tissues prevents backflow into either

: r l-- i atriorrentricular

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the atria or the ventricles, respectivery. The valves between the atria and the ventricles themselves do not invert because of tendinous cords, called chordae tendineae, which hoid them in place.

Electrical Activity of the lleart

Located near the junction of the superior vena cava and the right atrium is a specialized region of myocardium called the sinoatrial node (sA node) or the pacemaker of the heart (Figure 2-4).

SA

AV Node

Bundle of His

Figure 2.4 The SA and AV nodes of rhe heart.

The SA node is the point of origin for the electrical impulse that propagates through the rest of the heart. Th[ electrical impulse spreads out over the atria causing them to contract and filt the ventricles. Located in the lower portion of the right atrium and near the right ventricle is the atrioventricular node (AV node). Impulses from the sA node also spread to the AV node and then from the AV node through a coliection of fibers called the bundle o{ His. Branches of the bundle of His surround the ventricles, and when this bundle receives an impuises it causes the ventricles to contract and eject blood to the pulmonary and systemic systems. Anatomicaily speaking, ventricular contraction is from the apex of the heart towards the base of the heart. It is interesting to note that if the sA node is damaged, the AV node takes over and slows the heart down to about 40 beats per minute.

Cardiovascul.rr Control The average blood pressure leaving the aorta is about 100 mmHg. This blood pressure is monitored by baroreceptors and chemoreceptors located in the carotid arteries and in the aortic arch. The baroreceptors (pressure receptors) are continually monitoring how much the aortic and carotid arteries are eipanding Copyright

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and contracting. Suppose we were to decrease the arterial pressure. There would be less stretch on the aortic and carotid arteries. The baroieceptors would sense this and send impulses to the medulla in the brain stem. The medulla responds by activating the sympathetic nerves of the autonomic nervous system. Impulses are sent via the sympathetic nerves and norepinephrine is released at the sA node to increase the rate of the heart. This action helps to increase the contraction of the heart, which in turn wiil elevate the blood pressrrre. Sympathetic nerve fibers will also stimulate the adrenal medulla to releise epinephrine (adrenaline) into the blood. This hormone acts to increase the rate of the heart, therefore increasing the contraction of the heart. Both of these actions act in a negative feedback manner to "negate" the initial loss of pressure due toi say, hemorrhaging. The sympathetic nerves will also cause constriction of the blood vessels that lead to the gastrointestinal system and the kidneys. During hemorrhaging the brain and heart receive first priority in terms of blood. The rest of the organ systems of the body see a decrease in the flow of blood until the probiem is corrected.

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Cardiac Output :iiiiiijr;iilliijiiiti::

Every time your heart beats, a certain volume of blood is pushed out into the circulatory system. The cardiac output is that amount of blood which is pumped per minute by each of the two individual ventricles of the heart. We can define the cardiac output (for either ventricle in liters/minute) as being equal to the heart rate (in beats/minute) times the stroke volume (in liters/beat). This is shown in equation (2-1). The stroke volume is simply the amount of blood ejected by each ventricle during one beat of the heart.

If the average heart rate is T2beats per minute and the average stroke volume is 70 milliliters (or 0.07liters) per beat, then the cardiac ouput would be about 5 liters per minute. This value is for the average resting adult male. As we will see later, the volume of the cardiac output can be influenced by the diameter of the blood vessels in the periphery, the amount of blood returned to the heart by the superior and inferior vena cava (i.e., return of the venous blood), and the heart rate and force of ventricular contraction.

Cardiac Output = (Stroke Volume) (Heart Rate)

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Poiseuille's Law

Poiseuille's Law Not all of the arteries and veins in the circulatory system have the same diameter. flow of biood in blood vessels, established a relationship between the radius and length of a tube, the change in pressure between the two ends of the tube, the viscosity, and the flow rate of a fluid in that tube. This relationship, known as Poiseuille's law, is given in equation (2-2). Jean Poiseuille, a French physician studying the

Flow = AP ftRr = (P, - P, 'SnL ) 7IR*

SnL

(2-2)

in this equation

(2-2), AP is the pressure drop between the two ends of the tube (i.e., P1 - Pz), R is the radius of the tube, eta (r1) is the coefficient of viscosity, L is the length of the tube, and n/8 is a proportionality constant adjusting for the cross-sectional area of the tube. What can we say about this equation?

(a)

Notice that the fiow rate is proportional to R4. This tells us that the rate of blood flow is extremely dependent on the radius of the vessel. If the radius of the vessel were reduced by a factor of 2, then the rate of blood fiow would be reduced by a factor of 16. similarly, if the radius of the vessel were increased by a factor of '1,.5, then the flow rate would increase by a factor of 5.1.

(b)

The

(c)

The flow rate is also inversely proportional to the viscosity of the solution. This tells us that a high viscosity gives a 1ow flow rate.

(d)

The value of AP is provided for by the strength of the heart's contraction. In

flow rate is also inversely proportional to the length of the vessel. In other words, the longer the vessel, the slower the rate of flow. The shorter the vessel, the faster the rate of flow.

other words, the difference in the pressure is what drives ihe blood in the cardiovascular system.

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Diffusion

Diffusion is simply the process by which molecules randomly move from one place to the next. The molecular weight of a moiecule and the temperature of the medium have a lot to do with the velocity at which a molecule moves. Smaller

molecules tend to move faster than iarger molecuies. Similarly, a higher temperature provides more energy to a system and therefore imparts more thermal motion to the various moiecules in that system. you might think that because smaller molecules move relatively fast they would have no problem traversing their environment. However, don't forget that there are millions and millions of molecules within a given system and that each of those molecules, even thought they are moving at a given velocity, collide with their neighbors. These collisions tend to alter the path o{ the molecules, thus confining them to a random walk through their medium. Consider an imaginary sphere of water with a given radius. suppose we place some dye outside this sphere of water and ask how long it will take to reach the center. The answer clearly depends on the radius of that sphere. If the radius were 7 microns, it would take the dye about 5.4 seconds to reach the center. Flowever, if the radius were L centimeter, it would take about 11,000 seconds or a little more than 3 hours to reach the center. [As a reference a red blood cell is about 7 microns in length. There are about 1,400 red blood cells end to end in 1 centimeter.l what this is telling us is that simple diffusion is a rather poor way for a molecule to trek across long distances. The Law that governs diffusion is given in equation (2-3) where J is the net flux or net rate of diffusion (in moles per unit time, usually in seconds; that is, mol/sec), D is the proportionality constant called the diffusion coefficient, A is the area of the plane of interest (in cm2), and AC/Ax is the concentration gradient across that plane (in mol/cm4, because concentration is in units of mol/cri3 and distance is in units of cm). Since the net flux always proceeds down a concentration gradient, from a high concentration to a low concentration, we need to add a minus sign in front of this equation. [The minus sign indicates the direction of the flux or diffusion.] This equation is sometimes referred to as Fick's law (after the German physiologist who postulated it in the 19th century). In equation (2-3), what are the units of the diffusion coefficient, D?

r=-(DX^)f

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(DXAXCour - Cin) Ax

(2-3)


Biology

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Osmosis

Osmosis In order to understand hydrostatic pressure we must first review osmosis. Recall that osmosis is simply the net movement of water from a region of high concentration to a region of 1ow concentration. General chemistry tells us that the concentration of pure water is 55.5 moles/liter. If we had a beaker of pure water and u,e added a solute to that beaker (say glucose), then we would decrease the concentration of the water in the beaker. Remember, when we add a solute nolecule (or molecules) we are occupying a volume of space that was once occupied by a water molecule. The more solute molecules we add to our beaker ,.f pure water, the more water molecules we will displace and the lower r.t'ill be re concentration of pure water. Jlucose, when added to a solution of pure water, does not ionize. It stays as :1ucose. Flowever, if we add a molecule of sodium chioride to a solution of pure .,-ater, it u'ill ionize into a Nae ion and a Cle ion. Because we now have added a .-.,lium chloride molecule which has dissociated into two ions in solution, we :.ar-e displaced (iowered the concentration of) the water molecules twice as much :-s rre would if we had added a glucose mo1ecu1e. \\hat this is telling us is that --:.e concentration of water in a given solution depends on the number of solute :a:ticles (e.g., glucose, Nae, or Cle) in that solution. We can define the total .-ir-rte concentration in our solution as the osmolarity, where one osmol is :-::prlv one mole of a molecule that does not ionize. If we had a 1M solution of :-r.ose, it would have a concentration of 1 osmol per liter. If we have a 1M .: -'.ihon of sodium chloride, we r,r,ould find that it wouid have a concentration of I osmols per liter (one from the Nae ion and one from the Cle ion). If we have a 1i0mM concentration of NaCl, then after ionization we would have 150 mM of :: rons and 150 mM of Cle ions or a total of 300 milliosmols (mosmols) per ,-:=:. Therefore, the osmolarity refers to the concentration of solute particles that : :.ave in our solution.

I

Ah

L

a

Hzo- >. oo lj

o

o

to toa a Oa a

'a )ao .o

Semipermeable Membrane

Permeable Membrane

F,gure 2-5 -

:

.: ::.ovement across a permeable

-::ose

rve have a U-tube

and semipermeable membrane.

with water and apermesble mernbrane as shown in

I fure 2-5a. Water is free to pass back and forth across this membrane and | : - irS€ of this the height of the water in each of the columns of the U-tube will r : --.: Sdrrl€. However, suppose we now replace the permeable membrane with a ""

''.'-'::'neable membrane and then add some protein to the right side of the U1s shown in Figure 2-5b. \44rat will happen? The level of water in the right

::

:,:::ht

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Biology

lleart & Lungs side of the tube drop. \Alhy?

Osmosis

will rise and the level of water in the left side of the tube will

In order to understand this we must examine the semipermeable membrane and the solutes which we have added to the right side of the tube in a little more detail. First, the concentration of solute on the right side of the tube is greater than the concentration of the solute on the left side of the tube. This established a solute concentration gradient in which the solutes on the right side of the U-tube want to diffuse down their concentration gradient to the solution on the left side of the U-tube. This cannot happen, because we have said that the membrane is semipermeable. In other words, the membrane will not allow these solutes to pass through only water. What about the concentration of the water? The concentration of the water on the left side of the U-tube is greater than the concentration of water on the right side of the U-tube. Again, a concentration gradient has been established that will allow water to diffuse from the left side of the U-tube to the right side of the U-tube. Since the membrane is permeable to water we find that water diffuses down its concentration gradient (from left to right). This leads to an increase in the volume in the right side of the U-tube and a decrease in the volume in the left side of the U-tube. The effect is to decrease the solute concentration in the right side of the U-tube. An equilibrium will eventualiy be established when the concentrations of both the water and the solute are equal on both sides of the semipermeable membrane. Once equilibrium is reached no more water will flow from the left side of the Utube to the right side of the U-tube. This is because the pressure has increased in the right side of the U-tube (because there is now a larger volume of solution pushing on the semipermeable membrane). The amount of pressure that stopped osmosis is referred to as the osmotic pressure (abbreviated as nosm). A direct measure of the osmotic pressure is the difference in the levels of water in the left and right sides of the U-tube. This difference, Ah, is referred to as the hydrostatic pressure (or fluid pressure) which can be abbreviated as PH2O. The osmotic pressure is proportional to the number of dissolved molecules in a solution and is represented in Figure 2-6. As we increase the concentration of the protein in solution (i.e., solute in solution) we find that the osmotic pressure increases as well. o

()O ov tr q

o [Protein]

Figure 2-6 Relationship between osmotic pressure and dissolved particles.

Now, let's return to the hydrostatic pressure generated by the heart which forces fluid out of the capillaries and into the interstitial space. Since the hydrostatic pressure in the capillaries turns out to be a bit greater than the osmotic "pulling pressure" of the soiutes in the blood, there is a net movement of fluid from the capillaries to the interstitial space and eventually into the lymphatic capillaries.


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Lymphatic System

Lymphatic System The lymphatic system lies parallel to the systemic and pulmonary circulations. Tl'Lis can be seen in Figure 2-7.The lymphatic system ioliects the excess fiuid about 4 liters per day) that leaks into the interstitial space from the capillaries and returns it by way of the vena cava back to the ciriulatory system. Lymph :-.cdes located along the lymphatic system help to filter out foreign particles that :ould potentially lead to disease. If the lymph flow through the iymphatic system -"''ere biocked, edema wouid result. This is simply an increase in the inteistitial :-';id (because it cannot be reabsorbed by the 1ymphatic system). patients who :-ar-e heart surgery usually have swollen legs and ankles. This is because the --:art cannot pump the blood out fast enough, and as a result blood within the ---:art begins to back up. This translates to a back-up in the veins and eventually - lne lvmphatic system. Since gravity pools fiuid toward the lower extremities, :

*::fla results in the legs and

ankles.

Q

t-ymphatic capillaries

Pulmonary capillaries

Artery

.,:$

\ Vein

Heart

\ Systemic

-.nph

capillaries

:de

Q

Lymphatic capillaries

i.r1Ule 2-7

:

,

*::=:lC

SVStem.

..., 3

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Blood Ctotting

B[db i]iffiurfifig Blood clotting occurs via a cascade prlcess. Thrombin, which is involved in bibod clotting, is a serine protease. We can achieve amplification of a very weak signal by a cascade process. If we consider blood clotting, we will find that there is an intrinsic route (due to contact with some abnormal surface) and an extrinsic route (due to trauma to the tissue), both shown in Figure 2-8.

In the outline shown in Figure 2-8 we wili use the Roman numerals to represent the clotting factors. The subscript "a" means that we are dealing with the actiae

form of the molecule. Again, this would be some type of conversion of a proenzyme to an enzyme. Factor IX will be converted to Factor IXa by some intrinsic factor. Factor IXa wili be the trigger which will convert Factor X to Factor Xa. From the extrinsic portion of this scheme we start with some tissue factor which will convert Factor VII to the active form, Factor VIIa. Factor VIIa can also bring about the conversion of Factor X to Factor Xa. It is Factor Xa which is involved in the conversion of prothrombin (II) to thrombin (lla). Thrombin (IIa) wili convert fibrinogen (I) to fibrin (Ia). These fibrin fibers are then crossed-linked by Factor XIIIa (which is an enzyme called transglutaminase) to form the mature cross-linked fibrin ciot. Some of the serine proteases mentioned

in Figure 2-8 are Factors VII, VIIa, X+ process there are more than 15 different factors involved, and about 8 or so of them are serine proteases.

II, and IIa. In the blood clotting

Intrinsic Pathway

Factor

(Damaged Surface)

Extrinsic Pathway

o o r-------)

vil

IX

(Trauma) n

Factor IXo

Factor

I vrrr"

X t-----1

Factor VII

n

il

1'

Factor

VII^

Factor Xu

{}v^

Prothrombin (II)

Fibrinogen (I)

1,' !--------------- tsactorX

Thrombin (IIo)

vil

Fibrin (Io)

vfl *,,,. {transgturaminase} Crossed-linked Fibrin Clot

Figure 2-B The intrinsic and extrinsic pathways.

A lot is known about the biochemistry of blood clotting. In the conversion of Factor X to Factor Xa by Factor IXa, we find that we need an additional factor called Factor VIIIa. Individuals lacking Factor VIIIa have hemophilia (a sex-linked recessive characteristic). Factor VIIIa is sometimes called the antihemophiliac JLtLtUt.


96


Biology

fleart & Lungs HO

o il

Carboxylase enzyme

lil

l\

lt H -

requiring Vitamin K

L

and HCO3

CH.

l'

CHr I

C.

Oo//

\o

.

L__r'

0,

Blood Clotting

OH

ill C_N_C tt HC

o il H2

I

p

y '

Y-Crrboxyglutamate

o=c/

resrdue

I

ô

L-.-'/

Preprothrombin lragment

C -H

C:O

o^ I

*Can

l:,/

Prothrombin fragment chelated with calcium.

Figure 2.9 Cheiating action.

vitamin K is one of the fat-soluble vitamins that is found in green leafy

''-egetables. in addition, our intestinal flora can make a form of vitamrn K. \A{rat is the importance of this vitamin? Prothrombin exists in even an earlier form :ailed preprothrombin. In preprothrombin there are certain Glu residues which are carboxylated by a carboxylase enzyme. This carboxylase enzyme has an absolute requirement for vitamin K. The carboxylase enzyme adds another carboxyl group lo the gamma position of the first ten Glu residues located in the amino terminal region of preprothrombin. Thus, in the presence of vitamin K, l{CO3 and the carboxylase enzyme, we will be able to convert preprothrombin to

prothrombin. This structure now has a great ffinity (a good chelating agent) for lir.alent ions such as Ca2o. This is shown in Figure 2-9. (Calcium ions are essential for biood clotting.)

Platelet Membrane

following injury.

r", *Ca+ a

0C)

rl

"-\

,L-u C-H

*ca* o ooo u-L

rl \

C

a=a)H

I

CH.

NH:

t-

i/\

A section of prothrombin

Aftet cur is made. thlombin is released.

_A_

o Cut

Figure 2. I O Binding of prothrombin to

a membrane.

Prothrombin is next converted to thrombin in the presence of Factor Xa. The :onversion of prothrombin to thrombin in the presence of Factor Xa can be pictured as shown in Figure 2-10. Blood platelets have phospholipid molecules in their :rernbrane. The head of the phospholipid is negatively charged. This will allow Cop;,right O by The Berkeley Review

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Blood Clotting

the y-carboxyglutamate residues of prothrombin to bind via ca2e. Factor Xa also

has y-carboxyglutamate residues on it, and can also bind to the membrane via interaction with Ca2@. The enzyme Factor Xa wiil make cuts (at Arg-X residues) in the prothrombin molecule as shown. The portion of the prothrombin molecule that is cut away is called thrombin. Thrombin will drift away and convert fibrinogen into fibrin in the vicinity of the damaged area. It is fibrin that wili

{orm the blood clot. There is also an auxiliary factor, Factor Va, which is

involved in this process. So, what you need for this clotting process to take place is the (a) platelet membrane, (b) enzyme, (c) Ca2e ions, (d) an auxiliary factor, and (e) the substrate prothrombin. Thrombin can now act as a proteoiytic enzytne that converts fibrinogen to fibrin. Fibrinogen is a large solubie protein. Its solubility is due to an excess of negatively charged amino acids (Glu, Asp, Tyr-so4), particuiarly in the central domain of the molecule. The net charge in the central domain of the fibrinogen molecule is -8, while the net charge at the terminal ends is -4. If very large molecules have a net charge of zero, they will tend to come together. However, if we have an excess of negative charges or positive charges, the molecules will repel one another.

Fibrinogen

Fibrinopeptides

hFibrin

fl ll t,

Aggregation of

ribrin monomers

(

Fibrin Clot

o

Figure 2-l I Excess charge aliows for aggregation.

Thus, one way to make things soluble is to have an excess of charge. In order to convert a soluble protein to an insoluble protein one must remove that portion of the molecule contributing all those negative (or positive) charges. When fibrinogen is converted to fibrin we find that 4 Arg-Gly bonds are broken in the central domain. This releases four peptides containing an excess of negative charge called fibr

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ep

tid e s.

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Biotogy

fleart & Lungs

Blood Clotting

Once these fibrinopeptides are released, the overall net charge in the central domain now becomes +5. The fibrin nLonlmer that is now forir.ed (due to the release of the fibrinopeptides) has the ability to interact with other fibrin nonomers through electrostatic interactions between the terminai and central domain regions of the polypeptide. This is shown in Figure 2-r2. This aggregation of fibrin monomers leads to the formation of the fibrin clot.

lhe clot that is initialiy formed is called a soft clot. The hard clot involves a :-rrther step, in which there is a cross-linking via the enzyme transglutaminase -r'r Factor XIIIa). Cross-links are formed between the individual subunits that :--ake up the aggregate. Those cross-iinks occur between Gln and Lys resid.ues as !.-o\\'n in Figure 2-12. This type of reaction is referred to as a trnnsaminntion - . :lian. ";

Fibrin-

CHz

-

CHz

C- NH2

-

NH3-

GIn

I

il XIIIa U

Fibrin-

cHz

-

CHz

-

(CHr4-Fibrin

Lys (transglutaminase)

C- N- (CHz)+-Fibrin

Cross-linked

Fibrin Clot F1Eure --tng Ieads to a hard clot.

damaged area has been repaired, a serine protease called plasmin in the fibrin ciot in order to dissolve it into smaller ::'1: rragments (to remove the clot). Tissue plasminogen actipstor (TpA) ;;-,. plasminogen into this active protease. ;'Le

'.'zes specific regions

:-;

:::rirn to vitamin K for a moment. vitamin K is one of the fat-soluble ,:=,-:.s 'hat is found in green leafy vegetables. in addition, out intestinal flora - :::-
s

-t'arious feeds, rLLUD/ some Dvu.E of vr which vvrrrLrr were wstE LduD[lB causing rrcrt[(-,rrrrcl6.tttB hemorrhaging IIt in the LIte cows. COWS.

:':: that dicoumarol is a naturaily occurring material in the rotting clover :==is that the cows consumed. Cows that ate the dicoumarol had an

'. :'":ritrombin that did not bind Ca2o. br The Berkeley Review

99

The Berkeley Review Specializing in MQAT preparation

Biotogy

tleart & Lungs

Blood Ctotting

It occurred to some chemists that they might be able to find an inhibitor similar to dicoumarol which would not be toxic to cows, chickens, or even people, but might have an effect on rats or mice. One of the things that farmers didn't like was rats and mice running around eating their chicken feed. The chemists at the Wisconsin Agricultural Research Station developed a vitamin K antagonist called warfarin. If you mixed warfarin wiih chicken feed, and that feed was eaten by the rats and mice, then eventually they wouid die of hemorrhaging. it would take much greater levels of warfarin to be toxic to humans. It turns out that warfarin is much more active in mammals than in birds, so the chickens were not affected. Warfarin is a Vitamin K antagonist that is used as rat poison. The structures of vitamin K, Warfarin, and dicoumarol are shown in Figure 2-13.

HCHT \ ttl CH2 - C= C- CH2+

e(x" \"-Yri:e

H

I

Vitamin K2

Warfarin

C=O I

CH:

Dicoumarol

CH;

Figure 2. l3 Chemical structures of vitamin K, Warfarin, and dicoumarol.


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Biotogy

Gases

Ileart

6r Lungs

Gases

:

Giucose can be oxidized to carbon dioxide and water as shown by equation (2-4). The majority of carbon dioxide is generated i'r the Krebs cycle. oxygln is used as the final electron acceptor in the electron transport chain. Water ls generated in that reaction. out of thjs metabolic pathway -" at" able to extract energy in the form of ATP and that ATP is ultimately usecl to keep the organism alive.

60z + C6H12O6 r->

6CO2 + 6I{2O + Energy

(2-47

In

:

general, respiration is the process by which oxygen is brought to the cells of the tissues and carbon dioxide is removed ur u -urt" product. As we breathe air into our lungs it first enters our system by way of the nose or mouth. The air passes form the oral cavity to the pharynx, into the larynx, and then down the trachea. At the end of the trachea the air passes into two tubuiar passageways called bronchi. One bronchus enters into eich lung and continues to divide into smaller passageways called bronchioles, ending eventually in the functional units of the lungs which are the alveoli. Each alveolus consists of a single layer of epithelial cells juxtaposed to a very thin basement membrane. Sulrou1,air1g each single alveolus is a capillary network. The epithelial cell layer of eacfi alveolus and the endothelial layer of the capillaries are separated from each other by a very narrow interstitial space (if they are separated by an interstitial space at all). This means that the air in each alveolus and the blood in the capillaries are separated by a very small distance (about 0.2 to 0.3 pm, compared to the average diameter of an erythrocyte, which is about 7 pm). Th" t*o irrrg, are composed of millions of alveoli, and if they were all laid flat on a surfac"e, their combined total surface area would be between 70 m2 to 100 -2 1*ni.h i, about the size of a tennis court). Therefore, Iarge quantities of oxygen in the alveoli of the lungs can quickly be equiiibrated wlth-the blood in the "capiliaries bec.auss of the large surface area and the thin barrier to diffusion for gas exchange.

The air that we breathe on a normal day is composed of roughly 7g"/o N2,21"/. oz' 03% Co2, and 0.7"/o H2o. [These values vary slight from textbook to textbook.] Because al1 o{ these gas molecules are (normally) quite far apart, they tend not to interfere with one another. This means that the pressure exerted bv one gas rs independeruf of the pressure exerted by ail of the other gases. hr oth& words, the sum of each of these individual gas pressures equals the total pressure of this mixture of gas. The partial pressure of a gas (e.g., pN2, po2, pco2, etc.) is therefore a measure of the concentration of a given gas (such as 02) in a mixture of gases (i.e., the air). If we add up all the partial pressures of the individual gases in the atmosphere, we will come up with the atmospheric pressure. At sea level the atmospheric pressure is 760 mmHg. partial pressure of oxygen gas (Po2) at sea level? Roughiy 21o/, of the total atmospheric pressure at sea levei is oxygen gas. Therefore, the partial

\MFrat is the

pressure of oxygen gas at sea level is about 160 mmHg (from 0.21 x 7G0 mmHg = 16o mmHg). what happens to the parti.al pressure of oxygen as we ascend to the Copyright

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Biology

Ileart & Lungs

Gases

top of a high mountain? The concentration of oxygen in the air on this mountain top will still be about 21"/o.However, as we increaJe our altitude the atmospheric pressure begins to decrease. The result is a decrease in the partial pressure of oxygen.

How does a gas behave in a iiquid? Consider an open glass of water sitting on a table as shown in Figure 2-L4. The surface of that water is constantly being bombarded with gas molecules from the air.

'.,., "'l ,',,

',9j,,

............

Watei,,

Figure 2-14 Gas movement across a water surface.

when a gas molecure such as 02 comes in contact with that water surface it can dissolve in the liquid. The number of O2molecules which dissolve in the water is directly proportional to the partiar pressure of the 02 gas. Just as 02 molecules in the air can hit the surface of the water and dissolv;; th" ltq*d so too can 02

molecules

in the liquid hit the surface of the water and

into the air.

Th.erefore, at equilibrium the number of 02 molecules "slcape dissolving the water will equal the number of 02 molecures leaving the water. In otherinwords, the partial pressure of oxygen in the gas phase (air) Is equar to the partial pressure of oxygen in the liquid phase (water), as shown in equation (2_5).

(Por)gu. = (Poz)liquia

(2-s)

W5 cl say that the pressure of the air acting on the membranes of the epithelial cells in the alveoli of the lungs is the sum puitial pressures of all the gases in the air. Thus, the total pressure for any given gas molecule is directly proportional to the concentration of that gas molecure in the air. For example, -" .u1 account for

the many factors that a{fect the rate of diffusion of a gas into a liquid by considering equation (2-3). In this case J is the flux or difiusion rate of ihe gas molecule across the membrane, D is the diffusion coefficient, A is the area of the plane the molecule is diffusing across, C is the concentration of the moiecule, and x is the distance of diffusion. Since the concentration of a gas is proportional to the pressure of the gas, we can write that as shown in equati"on (2-6).

J_ Copyright

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(DXA)(C1O, - C2o')

Ax t02

s

-

(DXA)elo, - p2o) Ax

(2-6)


Biology

tleart

6r Lungs

Gases

The respiratory passages have a unique anatomical structure. Let's consider a :ross-section of one of these passageways and work our way from the lumen rutward. The epithelial cells which line the lumen of the passageways to the end ri the bronchioles have cilia which are continualty beating towards the pharynx. Scattered among these epithelial cells are glands which secrete mucus. As the :ilia beat towards the pharynx the mucus is moved upward towards the oral :ar"ity. Any foreign matter that has become trapped in the mucus is eventually

::ansported to the oral cavity where it is swallowed in a normal reflex action. ,.-\'re of the reasons that smoking is bad for you (besides its carcinogenic nature) is ::.at it decreases the activity of the cilia and lowers the body's defenses agarnst -'-::rg infection by bacteria that can enter the respiratory tract on airborne dust ::rticles. Immediately beneath the layer of mucus is a iayer of fiuid in which the :rlia operate. Air that flows within the passageways of the respiratory tract is

'..-arrned and moistened.

-:.: upper

passageways of the respiratory tract maintain their opening by means - cartilage rings that surround most of the diameter of the passageways. By the ,-:-Le the bronchioles are reached the cartilage has disappeared. smooth muscle is -l-,ji-rd in almost all areas of the respiratory tract where there is no cartilage. For = '.anple, the walls of the bronchioles are mainly smooth muscle. The bronchioles : :re lungs are innervated by nerve fibers from parasympathetic nerves (which ::-,'ei ir-t the vagus nerve). -

{sthma is usually caused by an allergic hypersensitivity to airborne antigens ---ch have entered the respiratory tract. The direct result is to cause the mast -=---' rr-ithin the bronchioles to release a number of different substances which -,'--'e the smooth muscle surrounding bronchioles to spasm and constrict.

ltechanics of Breathing -' = thoracic cage which contains the lungs is separated from the abdomen by a :,-::i oi skeletal muscle and connective tissue called the diaphragm (see Figure --i5. The lungs themselves are encased in a pleural membrane. The visceral :-eura covers the lungs, while the parietal pleura adheres to the diaphragm and : = - l-.est wall. Between the viscerai pieurai and parietal pleura is the intrapleural .:ace rr'hich contains a watery fluid. As the muscles of the diaphragm contract -: .- rull the diaphragm itself downward. Simultaneousiy the muscles of the rib .:= ccntract and cause the rib cage to move upward and outward. Both of these -

"-::-s enlarge the area of the thoracic cage that contain the lungs.

:

-= the diaphragm is attached to the parietal pleura this pleura is also pulled ard. The watery fluid in the intrapleural space is rather indistensible and ,-= : -:Iling of the parietal pleura translates through the intrapleural fluid. It also --: 'rre visceral pleura downward as well. Enlargement of the thoracic cage .

,

",-:,'i!

:::.ls the lungs and creates a subatmospheric pressure in the alveoli. Air -,.:.=. dorvn its pressure gradient from 760 mmHg at sea level to whatever lower r::.i jl€ i.s found in the lungs at inspiration.

: :- u1e muscies of inspiration stop contracting, the elastic tissue found within '= --:.lracic cage and lungs returns to its normal length. Air within the alveoli is ":::esseci and forced out through the passageways of the respiratory tract -:-:.i expiration.

-:it

O bv The Berkelev Review

r03


Biology

tleart & Lungs If the lungs were to

Gases

become separated from the visceral pieura, they would

collapse. This is because they have no anatomical structures io maintain rigidity. one way to separate the rungs from the visceral preura is to receive a very ltrorrg blow to the chest area (as often happens during football games). To the pharynx and oral cavity

f

Larynx

E) Intrapleural

Cartilage r

space

nngs

Parietal

pleural Diaphragm

Figure 2'15 The thoracic cage.


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Biology

Ileart & Lungs

Gas Exchange

Gas Exchange It is important to remember that

molecules u'il1 diffuse from a site of high

concentration to a site of low concentration. As deoxygenated blood (a dark red color) is coming back from the tissues it enters the right atrium via the superior and inferior vena cava and then passes into the right ventricle, where it is pumped out to the lungs. The PO2 is about 40 mmHg and the PCO2 is about 46 mmHg as this deoxygenated blood enters the capillaries surrounding the alveoli of the lungs. The PO2 and PCO2 within the alveoli are about 105 mrnHg and 40 mmHg, respectiveiy. Passage of the blood through the capillaries is relatively slow and an equilibration can be reached between the gas exchange in alveoli and the capillaries. Oxygen will diffuse down its concentration gradient from the alveoli to the capillaries, and carbon dioxide will diffuse down its concentration gradient from the capillaries to the alveoli. As the oxygenated blood (a bright red color) Ieaves the capillaries of the lungs and enters the left atrium of the heart, the PO2 is about 100 mmHg and the PCO2 is about 40 mmHg. See Figure 2-16. PO2

- 160

PCO2 = 0.3

PO2,= QQ P,COz= 46

PO2 = 100 PCOz = 40 Capiliaries at level of alveoli

I H

Capillaries at level of tissues

PO2=40,-

PCOr'='46

FO2=l[g

i

: FCO2=49

PO2 < 40

Figure 2- 16 Gas exchange at the level

ofthe lungs and tissues.

Oxygenated blood will pass from the left atrium to the left ventricle of the heart and then be pumped to the tissue capillaries. At the level of the tissues the PO2 in the cells (depending on which cells you are considering) is less than 40 mmHg, while the PCO2 in those cells is greater than 46 mmHg. Oxygen will diffuse from the blood in the capillaries to the ceils while carbon dioxide wiil diffuse from the cells to the blood in the capillaries. The deoxygenated blood that leaves the tissue capillaries returns to the right atrium of the heart via the venous system. Copyright O by The Berkeley Review

l05


Biology

Ileart & Lungs

Gas Dxchange

Oxygen can travei in the blood by being dissolved in the blood itself or by being

bound to a transport protein in the red blood cells (erythrocytes) called hemoglobin (abbreviated as Hb). Since oxygen is rather insoluble in water, not much is actuaily dissolved in the blood and transported in that manner. However, since hemoglobin has such a high affinity for oxygen, more than 987o of the oxygen in contact with hemoglobin is transported by this protein.

Hemoglobin is a protein that is composed of four polypeptide subunits. When these subunits interact with each other to form the hemoglobin molecule, they give hemoglobin a quaternary structure. Located within each of the four polypeptide subunits is a heme prosthetic group that has an iron atom in the center which is in the ferrous (Fe2e, oxidation state. Since one hemoglobin molecule has four binding sites for oxygen, there is a potential for 4 02 moiecules to bind to one hemoglobin molecule. Every time hemoglobin takes up 02 from the blood more 02 can leave the gas phase in the alveoli and enter into the liquid phase in the blood. This increases oxygen's solubility in the blood.

Oxygen Saturation Curve for tlemoglobin In Figure 2-17 we have a plot of the percent saturation of hemoglobin with oxygen as a function of the partial pressure of oxygen at various places in the body. Recall that when the PO2 is about 100 mmHg, we are at the ievel of the alveoli and when the PO2 is about 40 mmHg, we are at the level of the tissues. [It is best to read this curve from right to left and from top to bottom.] Venous

Blood

pH7.4

Arterial Blood 02 released

to tissues

€80 b0

Extra 02 released

f;oo

to tissues

qi

'E

Shift due to:

40

Decrease in pH Increase in temperature Increase in 2,3-DPG

! d a

*o

o (.)

?O

Resting Muscle

Working Muscle

Figure 2- l7 Oxygen-hemogiobin dissociation curve

At the level of the capillaries (PO2 = 100 mmHg) surrounding the alveoli roughly 98'/" of the hemoglobin is saturated with oxygen. As the blood passes to the tissues (PO2 = 40 mmHg) roughly 25"/' of the oxygen that was bound to hemoglobin is given up to the tissues. This can be seen in Figure 2-17. Copyright

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l()6


Biology

Ileart & Lungs

Gas Exchange

Effects of ptl, Temperature, and 2,3,BP(i The upper oxygen dissociation curve shown in Figure 2-17 is for normal conditions in which the body temperature is about 37 "C and the blood pH is about 7.4. If we were to decrease the pH of the blood (to a pH of 7.2) or increase the temperature, then the oxygen dissociation curve would shift to the right and downward as shown in the lower curve in Figure 2-17. This happens when you exercise. The same iype of shift occurs when a by product of glycolysis, 2,3bisphosphoglycerate (abbreviated as 2,3-BPG), binds to hemogiobin. 2,3-BPG is usually slnthesized in increased amounts when the body has been deprived of oxygen for an extended period of time (e.g., when you visit high altitudes).

AII three of these interactions with hemogiobin (the lowering of the pH, the increase in temperature, and the binding of 2,3-BPG) cause hemoglobin to release more oxygen to the tissues at the same partial pressure of oxygen as in the standard case. The oxygen dissociation curve can be shifted to the le{t and upward by reversjng these interactions.

Carbon Dioxide How is CO2 carried in the blood? It can be carried by (a) dissolving in the plasma and the red blood ceiis, (b) binding to a specific site on the hemoglobin molecule, or (c) in the form of bicarbonate ions (HCO3e), About 70% of the carbon dioxide rs carried in the blood in the form of bicarbonate ions, roughly 20o/o is carried by the hemoglobin itself, and about 10"/' is dissolved in the plasma and red blood cells. See Figure 2-18.

7jVo

Plasma

Red Blood Cell

CO2,

+ H2O

anhydrase

-:-+

20Vo

H2CO3

Hb-COo,

/

^

^

H@ + HCO3e 57a CO,t

57a COo dissolved

Capillary

CO" from

tissues

Figure 2-IB ,las exchange at the level of the tissues.

the level of the tissues the PCO2 is greater than 46 mmHg. However, in the :Iood of the capillaries it is about 40 mmHg. Therefore, CO2 will diffuse down its :oncentration gradient and into the blood. Some of the carbon dioxide will jissolve in the biood plasma, some will dissolve in the red blood cell, and some ,,, il1 bind to hemoglobin. The remaining CO2 will react with water and be :onverted to carbonic acid (H2CO3) by the enzyme carbonic anhydrase. Carbonic acid will ionize to the bicarbonate ion and a proton (H@). Bicarbonate will diffuse rto the biood plasma and be carried by the circulatory system to the capillaries

-\t

-,: the lungs. See Figure 2-18.

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Biology

tleart

E(

Lungs

Gas Exchange

At the level of the capillaries in the lungs, the PCO2 in the blood is about 46 mmHg, while the qCOZ in the alveoli of the iungs is about 40 mmHg. Once again, CO2 wiil diffuse down its concentration gradient and into the alveolar space. The CO2 that is dissolved in the plasma and in the red blood cell diffuses into the alveoli as does the CO2 that was bound to the hemoglobin. Bicarbonate ion in the plasma will diffuse into the red blood cell and combine with a proton to become carbonic acid. Carbonic anhydrase will convert carbonic acid to water and carbon dioxide. The CO2 diffuses into the lungs. This is shown in Figure 219.

Red Blood Cell

CO2

+ HzO

H2CO3

<--:Ho

+ HCO3o

CO.r dissolved

To the alveoli

Figure 2-

l9

Gas exchange at the level of the alveoli.

Control of Breathing The coordinated rhythm for breathing is generated by the medulia and the pons

in the brainstem. Nerve impulses from these centers cause the respiratory muscles to contract. \Alhat is the stimulus that causes contraction? The major stimuius is a change in the concentration of CO2 and Ho that affects chemosensitive areas in the medulia. The concentration of 02 in the blood is sensed by chemoreceptors in the carotid arteries and in the arch of the aorta. These chemoreceptors transmit signals to the respiratory center in the brainstern, informing it of the levels of 02 in the blood. If you were to breathe into

a container that collected your exhaled CO2, you wouid soon find yourself breathing deeper and faster. The increase of CO2 in your blood indicates to your respiratory center that CO2 is building up rn the body. One way to eliminate CO2 wouid be to breathe faster and deeper. What wouid happen if you were to breathe pure 02? The chemorecePtors for 02 would relay a message to the respiratory centers indicating that there is too much oxygen in the blood. Impulses from the brain stern would be sent to the respiratory muscles in order to slow their action.


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Ffleart & Lungs To Go 15 Passages

100 Questions


Passage Titles

I. III. ilI. IV. V. VI. VII. VUI. IX. X. XI. XII. XIIL XIV. XV.

Mechanics of Blood Flow Enzymes in Blood Cloning Cardiac OLttpnt & Venous Return Circulatoryt Pressure, Area, Velocity, & Volume Thoracic Cavitlt Electrocardiogrant Folate Experiment Measurement of Blood Pressure Aortic Compliance Heart Muscle Action Potentials

Capillary Filtration Respiratorry Calculations

Aspirin Sickle Cell Anemict VentilationRegulatiott

Speiializing in MCAT Preparation

Questions t -7 8-14 15

11

-2r l0

29 -34

35-40

4t-41 48 -

51,

55-60 61-66 6t -72 73 -78 19-85 86-92 93 - 100

Suggestions - The passages that follow are designed to get you to think in a conceptual manner about the processes of physiology at the organismal level. If you i uv" a solid foundation in prryriotogy, *uny of these answers will be straightforward' If you have not had a pleasant experience with the topic, some of these answers might appear to come from the void past the Oort field of the solar system. . Pick a few passage topics at random. For these initial few passages, do not worry about the time. Just focus on what is expected of you. First, read the passage.'s"coila, look at uny'diugru*s, graphs' Third, read each question and the u""o*punying answers carefully. Fourth, charts, or answer the questions the best you can. Check the solutions and see how you did. \A4reth", yt.r got the answers right or.wrong/ it is important to read the explanations and see if you understand iur,a ugr"u with) what is being explained. Keep a record of your results. After you feel comfortable with the format of those initial few passages, pick another block of try them. Be aware that time is going to become importani. Gerieral$, you will have about 1 minute and 15 seconds to complete a question. Be a little more creative in how yo, upprouch this next group' If you feel comfortable with the outline presented above, fine. If riot, then try different passages and

approaches to a passage' For example, you might feel well versed enough to read the questions first and then try to answer some of them, without ever having read the passage. Maybe you can answer some of the questions by just looking at the diagrams, charts, 6r graphs ttnut ui pr"r".rt"i in a particular passage. Remember, we are not clones of one another. You need-totegin to devetop a format that works best for you. Keeping a record of your results may be helpful.

The last block of passages might contain topics that are unfamiliar to you. Find a place where the levei of distraction is at a minimum. Get out your watch and time yorrr""lf on these passages/ either individually or as a gr9uP. It is important to have a feel for time, and how much is passing as you try to answer each question. Never let a question get you flustered. If you cannot figure ort *hut the answer is from in{ormation given to you in the passage, or from your own knowledge-Euru, Jrr*p it and move on to the next question. As you do this, make-a note of that pesky question and come back to it at the end, when you have more time. When you areanishe.l, check yolr, ur,r-"rs and make sure you understand the solutions' Be inquisitive. If you do not know the answer to something, look it up. The solution tends to stay with you longer. (For example, what is the Oort field?)

The estimated score conversions for 100 questions are shown below. At best, these are rough approximations and shouid be used only to give one a feel for which ballpark they are sitting in.

Section tr Estimated Score Conversions Scaled Score

>t2 10-

11

8-9 7 6 5

<4

Raw Score

- 100 19-8s 65-78 59-64 54-58 48-53 0-41

86

Biology Passage

I (Questions

Mechanics Of Blood Flow 3.

1-7)

Flow can be used in three different contexts: (1) as flow, (2) as average velocity, and (3) as the linear velocity of a small element of fluid. The total volume of fluid

Passage I

Which of the following equations BEST represents blood flow for the entire vascular system acting as a unit?

A. QxTPR=(CVP-MAP) B. MAPxQ=(TPR-CVP) C. QxTPR=(MAP-CVP) D. CVP x MAP = (Q _ TPR)

passing a given point per time is known as the flow rate. The flow at a given point divided by the cross-sectional area is known as the average velocity of flow. The linear velocity is the distance traveled by small volume of blood per unit time. The flow rate (Q) must remain constant throughout the entire cardiovascular system.

4.

The increased viscosity of whole blood relative to piasma is best explained by the contribution of:

Blood flow through a blood vessel results from a driving force in the form of a pressure gradient. The

A. B. C. D.

petfusion pressure along the length of any blood vessel is

given by the differences between the pressure at the proximal and distal end of the vessel. The following relationship between blood flow and perfusion pressure

albumin. glucose.

white biood celis. red blood cells.

can be established:

Blood Flow

-

The structure of systemic veins differ from the

Perfusion Pressure

structure of systemic arteries in that only:

Flow Resistance

The mean arterial pressare (MAP) is a time-weighted average of the arterial pressure over the entire heart cycle.

The net flow resistance of the systemic loop can

be

deflned as the total peripheral resistance (TPR) while the entire venous pressure (entire vascular system acting as one unit) is termed the central venous pressure (CVP).

6.

Capillary walls provide the minimum possible

C.

veins, the blood in capillaries has a:

D.

1

D.

low average velocity because of a high total

veins contain valves that favor one-way blood

flow. arteries contain a layer of smooth muscle.

cross-sectional area of the veins is larger than the arteries. venous pressure does not change significantly over the heart cycle. arterial pressure does change significantly over the heart cycle. arterial pressure rises rapidly with increased

volume while the venous pressure remains within a few mmHg with the same increase in

cross-sectional area. low average velocity because of a low total cross-sectional area. high average velocity because of a high total cross-sectional area. high average velocity because of a low total cross-sectional area.

volume.

1

The diameter of a uniform blood vessel is reduced by a factor of 2. The resistance of this blood vessel increases by a factor of:

In the classical method of measuring blood pressure, pressure from a cuff is applied to the brachial artery in the arm to coliapse the artery and prevent blood flow. The pressure of the cuff in this initial step must exceed:

A.

CVP.

MAP.

A.

2.

B. C.

B. C.

4.

D.

D.

16.

Jopyright

endothelial layer.

The most likely reason that the central venous

B.

barrier to diffusion. Relative to the arteries and

C.

arteries contain a component of elastic tissue,

A.

approximately three times that of water.

B.

B. C.

pressure (CVP) is not an average value similar to the MAP is because the:

flow velocity. Plasma has a viscosity of about 3, twice that of water, while whole blood has a viscosity

A.

the inner surface of veins is lined with a single

D.

The viscosity of a fluid is a measure of the internal u'ork necessary to make a fluid flow. The viscosity of blood is determined by the composition of blood, the nature of the vessel in which it is flowing, and the mean

1.

A.

arterial diastolic pressure. arterial systolic pressure.

8.

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lll


Biology Passage

II

Enz;rmes ln Blood Clotting

(Questions 8-14)

Passage II

Hemostasis is the process by which bleeding is arrested. At the time of injury, enucleated blood celis cailed plateiets bind to the exposed subendothelium ofthe damaged blood vessel. As platelets begin to aggregate to one another, a platelet plug is formed. Vasoconstriction

Figure 1. Blood Clotting Cascade

Intrinsic Pathway Kallrkrern (___ ,] Plekallikrein J_L A V

of the blood

vessel

is stimulated by the

release of

thromboxane ,A2 and serotonin from the platelets. The purpose of the platelet plug is to provide a surface that allows for the activation of blood coagulation factors (Table 1), many of which are synthesized in the liver. Activation of these coagulation factors initiates the blood clotting cascade (Figure 1). The end result of this cascade

is to form a blood clot (thrombus) that will seal

the

damaged blood vessel.

Many of the coaguiation factors are glycoproteins that can be ciassifled as being pre-enzymatic. They exist in their inactive form as zyllogens of serine proteases,

proteolytic enzymes that have a common catalytic mechanism based on a highly reactive serine resiclue at the active site. Once activated these serine proteases

j .,.,oQ.Gu.,o.

ixtrx

enhance the coagulation cascade.

"T"'x-"'ll iL--------__l \.2

intrinsic pathway, initiated by contact of an appropriate surface, or the extrinsic pathway, initiated by tissue

The blood clotting cascade can follow either the

trauma. Both pathways converge at factor Xa, the activated form of factor X. Factor Xa initiates the finai sequence of events in the common pathway that will lead

/?\ i! P.ottrro-olA i-----a-------

vl

Factor--J--

xilr

Pibrin

Vitamin K is essential for the synthesis of many of the coagulation factors, including prothrombin. The reduced form of this cofactor, vitamin K hydroquinone, is

12

\ XIIIa

I.,-.

-

€

(hard clot)

to the formation of a clot.

Thrombin

Fibrinogen

involved in a carboxyiation reaction that converts many of the glutamate residues iocated in the N-terminal segment of these clotting factors into y-carboxyglutamate. Vitamin K hydroquinone is regenerated in a two-step reaction that

piurin

(soft ciot)

Table 1. Biood Coagulation Factors Common Name

Factor

can be inhibited by antagonists like dicoumarol and HalfJifea

warfarin.

Fibrinogen * Tissue factor (thromboplastin) Calcium Proaccelerin Proconverrint * Antihemophilic factor Christmas factortf Stuart-Prower factort*

I 90-120 II 50- 120 III N/A IV N/A V 12-24 VII 2-6 VIIi 10-12 IX 18-30 X 25-60 45-80 Plasma thromboplastin antecedent+ XI XII 40-'74 Hageman factort '70-200 XIII Fibrin-stabilizing factor Prekallikrein 48-52 Fletcher factor*

y-carboxyglutamate is an excellent chelator of Ca2+ and, following injury, acts as an anchoring mechanism between calcium-dependent coagulation factors and the phospholipid membranes of platelets^ The functionai significance of this event is that it brings specific factors together that aid in the formation of a clot, thereby accelerating clot formation many fold.

Prothrombinl

S Ca2*, Phospholipid Membrane f Dependent on Vitamin K

Defects in the blood clotting cascade may be assessed using screening tests. Partial thromboplastin time (PTT)

screens the intrinsic and common pathways while prothrombin time (PT) screens the extrinsic and common pathways. A deficiency in factor XIII cannot be detected hy using lhese two lests.

A Reported in hours

$ Undergoes activation to Serine Protease

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tt2


Biology 8.

Enz5rmes In Blood

12.

Mature red blood cells obtain their energy from:

A. B. C. D.

Clotting

electron transport.

oxidativephosphorylation.

Passage Il

Hemophilia A results from a deficiency of factor VIII while hemophilia B results from a deficiency of factor IX. Genes coding for both hemophilias reside on the X chromosome and are genetically transmitted as a sex-linked recessive trait. What is the probability that a woman whose father has hemophilia, and who marries a normal man, wili

Krebs cycle. glycoiysis.

have an affected son?

A. B.

Glutamate is converted to y-carboxyglutamate in the vitamin K dependent reaction shown below:

c. D.

HO tlt -N-Ctl

1.00

0.75 0.50 0.25

I

HCH2H

I

Glutamate

CHr

l6 coo CO;

13.

Which of the following compounds, when added stored whole blood, BEST prevents clotting?

Y

tlr-",',ll3illi"5"r

l\* {,}

u,,u',"

A.

" -)

B.

Quinone

coo

HO lll tl -N-C-

r-o

HO

I

- C-

COO

CH"

I

r

l-

CHt

ro

y-Carboxyglutamate

CH.

l6

coo

-CH\9

OOC

^HO elilo H"N-C-C-O ,t

I

CH,

HCH?H a

6

coo

COO

Glutamate

Citrate

l-e

:nzvme which BEST regenerates vitamin K

1r;.-rquinone is called

C.

a:

A

-{- decarboxylase. K --arboxylase. (--. :e,luctase. nD.

D. coo I

CH"

t-

CHt

rrotease.

IA

coo Succinate

HO @tilo H3N- C- C-

O

CH.

t'

o,cfo OOC

COO

y-Carboxyglutamate

tlilt Tr* ,:,111 coagulation I'u

factor listed below that DOES -'T :e,quire vitamin K is:

q., S{:" 0lr

LlililIl,L

*,ill

F';ror

[.

-::omboplastin. Cr:istmas Factor.

14. The most common oral anticoagulant in use at the present time is warfarin. Administration of this

F::ror X.

:i

--::e

compound begins with a daily dose of about 5 mg and adjustments are made 3 days later to maintain a PT between I.2-I.5 times the control. This is most likely attributed to a decrease in:

tollowing mechanisms will limit clot

r'Tntu: L\CEPT:

A. B. C. D.

fl. :,:'r-.::rnFttion of leafy green vegetables. il, =-r:.:,;al from circulation by the liver. d- n;'::iors of serine proteases. 0u :r:r:r:,c 1-lo$' dilution. tlMtryMpmlr"Wtn

3

1

![he Berkeley Review

I l5

prothrombin. proconvertin. Christmas factor. Stuart-Prowerfactor.


Biology Passage

III

Cardiac Output & Venous Return

(Questions 15-21)

If there is an increase in the venous pressure, then during diastole there wili be more filling of the ventricles. Thi;

The blood returning to the right heart from the venous

stretches the muscle fibers in the heart tissue and leads to a greater force of contraction during systole. This is known as Starling's law of the heart, and simply stated says that an increase in venous pressure leads to an

system must equal the blood leaving the left heart to the arteriai system over any extended period of time. In a normal individual the venous return and the cardiac output is about 5 L/min.

The vascular function curve in Figure 1 depicts

Passage III

increase in cardiac output.

a

relationship between venous pressure and venous return. As the heart contracts, blood is removed from the vessels

15. The component of the circulatory system with the LEAST total percentage of blood volume is the

of the venous system and pumped into the vessels of the arterial system.

network of:

a10 ! e6

A.

capillaries.

B. C.

arterioles. arteries.

D.

small veins and venules.

94 oL o^

_4 0 4 8

16. The pH of the blood passing through the pulmonary artery, compared to the pH of the blood passing

1216

Venous pressure (mmHg)

through the renal artery, is:

Figure I

A. B. C. D.

If

the cardiac output and the venous return is increased, the pressure in the arterial system will increase while the pressure in the venous system will decrease. This is due to the increased transfer of blood from the venous to the arterial system.

17. Which of the following curves BEST represents

If

the heart were stopped and the cardiac output was reduced to zero, the pressures in the venous and arterial systems would soon equilibrate. However, the pressure in the venous system will be at its highest value, simply because the veins now contain more blood than when the heart was contracting. This pressure is referred to as the

A.

mean circulatory pressure (MCP) and varies with

B.

a10

a10

SU g6

J

tr

g6

E4 =)

a

E4

o^

relationship between venous pressure and cardiac output.

-4 0 4 8 1216

O^ >U


410

C.

E6 o

?10

@


o

-4 0 4 8 1216 Venous pressure (mmHg)

rt4

4 2

o^

Figure 2

Copyright

o

a)


8

6

94

0 4 I 12t6

1216

10

i!

J c6

a

-4 0 4 8


D.

AA

8o -4

an

increase in only arteriolar resistance to blood flow?

sympathetic stimulation and blood volume.

The cardiac function curve in Figure 2 depicts

lower due to the loss of 02. higher due to the loss of 02. lower due to the accumulation of CO2. higher due to the accumulation of CO2.

0

-4 0 4 8

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Biology

Cardiac Output & Venous Keturn

Passage Itr

20. Sympathetic nerve fibers are supplied to all parts of the heart while parasympathetic nerve fibers are located primarily at the sinoatrial (SA) node and the atrioventricular (AV) node. The neurotransmitter

18. Changes in blood pressure or blood volume are detected by stretch receptors in the atria and arteriai baroreceptors. Which of the following pairs of hormones act directly to increase blood volume?

released by sympathetic heart nerves is:

A.

B.

C. D.

Renin and vasopressin Ang)otensln II and aldosterone

A.

Renin and angiotensin I

norep)nephrine, because it opens Na@ and Ca2@ channels in myocardial cells.

Vasopressin and aldosterone

19. Both the cardiac output (C.O.) from the heart and the venous return (V.R.) to the heart depend on venous pressure. This allows the curves in Figure I and Figure 2 of the passage to be combined into one

B.

acetylcholine, because it opens myocardial cells.

C.

norepinephrine, because it opens in myocardial cells.

D.

Ko

channels in

Ko

channels

acetylcholine, because it opens Nao and Ca2o channels in myocardial cells.

graph as shown below:

c lU

21.

:R

All of

the following compensatory mechanisms

would be expected to occur during a hemorrhage

B6 a

EXCEPT:

o+ o da

A.

E

-4 0 4 8

1216

B.


C. D.

Which of the following graphs BEST represents

increased sympathetic discharge

to

the

to

the

arterioles.

decreased parasympathetic discharge heart.

increased cardiac output. decreased arterial pressure.

activation of the sympathetic nervous system? (Solid lines represent normal curves; dashed lines represent changes.)

B.

A.

410

410 SR

E6 a

a O+ d1 E I

UU

e4 o 6a

€

-4 0 4 8 12t6

5o -4

1216

D.

C.

ai0

al0

58

I

q =o

q =O

a^ a+

oa o

c

b

2

do -4

0 4 8

1216

d0 -4 0 4 8


Copyright

0 4 8



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1216


ll5


Biology Passage

Circulatory Pressure, Area, Velocity, & Volume


Passage IV

The pulmonary veins return the oxygenated blood to the left atrium of the heart which then passes it to the left ventricle. Contraction of the left ventricle sends the blood into the aorta under high pressure.

The total blood volume in an average human being is about 5.5 L. Blood flows through the circulatory system,

which is divided into the systemic and pulmonary circulatory systems. The systemic system supplies biood to all of the tissues of the body except the iungs. The

Biood passes down the descending aorta and into the systemic arteries, arterioles, capillaries, venules, and veins before returning to the right atrium by way of the superior and inferior vena cava (the great veins).

pulmonary system supplies biood to the lungs. The passage of biood through the circulatory system can be examined in terms of differences in the pressure, cross-sectional area, velocity, and total blood volume in different segments of this closed system.

22. The flow of blood throughout the body is primarily

100

reguiated at the level of the:

480 :(

A. B. C. D.

Eoo O

!

=40 o

veins.

capillaries. arterioles. aorta.

AIU

20

40

60

80

23. Which point on the graph of pressure versus total blood volume BEST represents blood flowing

Total Biood Volume (%)

through the pulmonary artery?

5000

J

4ooo

E

O

II

A. B.

II

D.

IV

I

c.m

:ooo

1000

20

40

60

80

24. Which point on the graph of area versus total blood voiume BEST represents blood flowing through the lungs?

Total Blood Volume (7o)

A. I B. II c.m D. IV

50

tr

940 O

t0

0

20

40

60

Total Blood Volume

80

25. Which point on the graph of velocity versus total blood volume BEST represents blood flowing

100

(7o)

through the vena cava? Blood returns to the right atrium of the heart from the systemic system under low pressure and is passed to the

A. I B. II C. ilI D. IV

right ventricle. Contraction of this ventricie pushes the blood into the pulmonary arteries where it flows to the lungs.

Copyright

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Berkeley Review

I l6

The BerkeleY Review


Biology

Circulatory Pressure, Area, Velocity, & Volume

Passage fV

26. During circulation of the blood, turbulent flow could increase with a decrease in the:

A. B. C. D.

viscosity of the blood. velocity.of the blood. diameter of the blood vessels. Reynold's number.

In the circulatory system, the vascuiar distensibilitv is the LEAST for: A. B. C.

D.

systemic veins. pulmonary veins. systemic arteries. pulmonary arteries.

l8- \\hich of the following is NOT characrerisric of the s\ stemic and pulmonary capillaries of an average adult human?

{. B. C. D.

llxnrr

The diameter of a capillary is the smallest of any of the vessels. The thickness of a capillary wall is the thinnest of any of the vessels. The capillaries contain between SVo to l\Vo of the total blood volume. The capillaries contain about 20Vo of the total blood volume.

r-r 3 The

Berkeley Review

tt7


Biology

Thoracic Cavity .tr ,o

Passage V (Questions 29-34)

Passage V

100

o

bo

The thoracic cavity of the body protects the heart and lungs and is important in the mechanics of respiration. The thoracic cage is composed of the ribs, costal

cEo

o

o.^ troU

cartilages, sternum, vertebral column, primary and associated musc

u

I

o

ature.

d

The walls of the thoracic cage are constructed of the These are curved bones that articulate with the

ribs.

(n

vertebral column posteriorly and extend anteriorly and medially to articulate with the sternum. Out of the 72

9"

pairs of ribs, only the upper six to seven pairs of ribs will articulate directly to the sternum. The remaining ribs will attach by costal cartilages.

sao

20

X

20

40

60

80

r00

Partial Pressure of Oxygen

Spanning the distance between each

rib are

the

(external and internal) intercostal muscles. Attaching to the most inferior ribs and costal cartilages, and extending posterior to the vertebral coiumn, is a large, flat, domeshaped muscle called the diaphragm. These muscles,

Figure

tr

along with several of the abdominal muscles, are important in the expansion and compression of the thoracic cage, which is necessary for the function of

i

100

,.o

o bo o

c6U +6) o60

respiration.

Lining the internal wall of the thoracic cage is a sheet of tissue referred to as the parietal pleura. Lining the external surface of the lungs is a similar sheet of tissue called the visceral pleura. Between these two layers of tissue is the pleural cavity, a space containing a small amount of fluid and registering a negative atmospheric

o cc

k

:40 cd

(A

a0

pressure.

2\)

X

ra a\0

The functional unit of the iungs is referred to as the alveolus, a small sac lined with epithelial cells. Each alveolus is surrounded by a network of capillaries. The

20

40

60

80

100

Partial Pressure of Oxygen

wails of the alveoli and the capillaries are quite thin, allowing for maximal efficiency in gas exchange between

Figure 2

the two structures.

Red blood cells are the primary means of

gas

transport between the lungs and the tissues of the body. The protein hemoglobin, a component of red blood ceils, binds oxygen at the level of the alveoli and transports this gas by way of the circulatory system to the tissues of the body. Carbon dioxide, a waste product of metabolizing tissues is transported back to the lungs by the circulatory

29. As the intercostai muscles lift the thoracic cage, how will pressure in the pleural cavity change?

system.

A. B. C. D.

The transport of both gases is dependent not only on their partial pressures but also on the temperature and pH of their environment. Figure 1 and Figure 2 represent a relationship between the partial pressure of oxygen and the percent oxygen saturation of hemoglobin.


tlB

No change to pleural pressure. Pleural pressure will become more negative. Pleural pressure will become more positive. None of the above.


Biology

Thoracic Cavity

33. In the graph shown in Figure 1, curve I represents a relationship between the partial pressure of oxygen

30. A pneumothorax is the presence of gas in the pleural space of the thoracic cavity. This condition leads to a collapsed lung. In order for a collapsed lung to occur, which structure associated with the thoracic

and the percent oxygen saturation of hemoglobin at a pH of 7.40 and a temperature of 38 'C. If the pH of the blood is increased, curve II depicts the result. Which of the statements below explains this

cavity needs to be punctured? A. B. C.

D.

phenomenon?

Internal intercostal muscles Parietal pleura Ribs Heart

I. il. III.

31.

Passage V

A.

I only

B.

II only III only I and III only

c.

Several muscles are involved in respiratory function.

D.

If the external intercostal muscles are non-functional, which muscle group will take over the primary

A rise in pH requires a lower partial pressure of 02 to bind a given amount of 02. A rise in pH requires a greater partial pressure of 02 to bind a given amount of 02. A rise in pH requires a lower partial pressure of CO2 to bind a given amount of 02.

functions of normal respiration?

I. il. ilI.

Abdominal muscles Diaphragm

Internalintercostals

A. I and III only B. II only C. III only D. Normal respiration

34. In the graph shown in Figure 2, curve I represents the relationship between the partial pressure of oxygen and the percent oxygen saturation of hemoglobin at a

"C. If the temperature ofthe blood is increased, curve II depicts the result. Which of the statements below explains this phenomenon?

pH of 7.40 and a temperature of 38

cannot occur.

L

A rise in temperature requires a lower partial

of 02 to bind a given amount 02. A rise in temperature requires a greater partial pressure of 02 to bind a given amount of 02. A rise in temperature requires a lower partial

pressure

U.

32. The main gases that are exchanged in the lungs are carbon dioxide and oxygen. This exchange depends to a large extent on the partial pressures of these gases (pCO2 and pO2) in the capillaries and the alveoli. Which situation is ideal for normal gas exchange between the capillaries and the alveoli of the lungs as blood enters the lungs from the

III.

A. I only B. II only C. III only D. II and III only

pulmonary arteries?

A.

pressure of CO2 to bind a given amount of 02.

High pCO2 in the capillaries; low pO2 in the capillaries.

B.

Low pO2 in the capillaries; low pCO2 in the

C.

capillaries. High pCO2 pressure in the lungs; high pO2 in

D.

Low pO2 in the lungs; high pCO2 in the lungs'

the lungs.

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119


Biology Passage

Electrocardiogram 36.

VI (Questions 35-40)

Potentiai differences within the heart rissue

Passage VI

The phase of contraction of the ventricies is called systole. Based on the passage, to which part of,the ECG does systole best correspond?

are

conducted to the surface of the skin because the body is a good conductor of electricity. A reading of these patterns provides a graph representing the electrical activity of the heart, called an eiectrocardiogram (ECG).

A.P B. QRS C.T D. T-P phase

During each heartbeat cycle, 3 distinct wave patterns are produced: P, QRS, and T waves. The waves represent changes in potential occurring across heart tissue as the

37. Fibrillation involves random contraction of myocardial fibers at different times. It can be

result of many action potentials inside the myocardial cel1s.

monitored with an ECG. What is the effect on an individual who experiences ventricular fibrillation?

The P wave represents the spread of depoiarization through the atria. The QRS wave represents the spread of depoiarization into the ventricles. The T wave represents the repolarization of the ventricles. The graph shown in Figure 1 indicates a standard cycle of these waves.

A. B. C. D. 38.

More heartbeats than usual. Fewer heartbeats than usual.

Unaffected. Death, if a normal rhythm is not introduced.

Which group of pacemaker cells provides the trigger for depoiarization in the right atrium and initiates the P wave?

-

0.5

o

39.

200

Sinoatrial node

Atrioventricular node Bundle of His

D.

Purkinle fibers

Why is the body a good conductor of eiectricity?

A.

400

Milliseconds

Figure

A. B. C.

B.

Tissue fluids contain ions that participate in conduction.

Tissue fluids are free

of ions that hamper

conduction.

1

C. Tissue fluids contain ions that hamper D.

conduction. Tissue fluids are free of ions that participate in

conduction.

40. The ECG can also be used to count the number of

35. Which of the following

conditions could

heartbeats per minute (the cardiac rate). Which person would most likely have the slowest cardiac

be

monitored with an ECG?

rate at rest?

A. B. C. D.

Copyright

Blood pressure Cardiac output

A. B. C. D.

Arhythmias Hematocrit

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An infant A marathon runner

A weight lifter A sedentary adult


Biology Passage

VII

Folate Experiment 42. A microbiological

(Questions 4l-47)

assay using a folate-dependent strain of Lactobacillus casei is used to determine levels of folate in the plasma and in the red blood cells. What type of laboratory technique is used to

Currently, the recommended dietary allowance for the B-r'itamin fblate is 200 trtg/day fbr non-pregnant women. One group of scientists explored the adequacy of the 200 ,reidal' recommendation in the fbliowing experiment.

Erueriment

separate the plasma and the red biood cells?

A.

I

30 non-pregnant women were assigned to 3 different Jietary groups: either a diet containing 200 pg/day, 300

-tgiday, or 400 rp"glday of folate. They foilowed the :resclibed diet for 10 weeks. Levels of folate in the :iasma and in the red blood celis (RBCs) were measured "t the beginning of the experiment and at the conclusion. Results of these measurements are shown in Table

Initial

Final

Final

Plasma

RBC

Plasma

RBC

Folate

(nglml)

Folate (ng/ml)

Folate (ng/ml)

(ne/ml)

200

3.6

444

3.1

381

300

3.5

450

3.3

4t1

400

3.6

449

6.7

610

Table

B. C.

Extraction with an organic solvent Centrifugation Thin layer chromatography

D.

Dialysis

How would a microbiological assay for foiate performed using Lactobacillus casei

A.

1.

Initial

Dosage

per Day (pg)

Passage VII

?

The subject's sample is added to a known amount of bacteria, incubated for a time period, and the turbidity of the medium is measured.

The subject's sample

is

added

to a

known

amount of bacteria along with a known amount of folate, incubated for a time period, and the turbidity of the medium is measured.

Folate

The subject's sample is added to a known amount of bacteria, incubated for a time

D.

1:

Folate concentrations in plasma and RBCs before and aiter folate diets.

period, and the amount of folate produced is measured by an enzyme acliviry assay. The subject's sample is added to a known amount of bacteria, incubated for a time

period, and the number of mutated colonies is counted.

If

plasma foiate drops below 3.0 ng/ml, then a folare

:eiiciency is indicated. Furrher deficiency will deplete

44. Based

on the results in Table l, what conclusion wouid you make about the leveis of fblate these women were consuming in their diets before the

RBC stores and couid lead to anemia.

study began?

11.

of

two different body pools of folate. Which body pool does each reflect?

A. B. C. D.

B. C.

They were consuming more than 400 pglday. They were consuming between 200 and 300

D.

They were consuming less than 200 ytg/day.

pg/day.

Plasma folate represents foiate intake over the past year, while RBC folate represents fblate intake over the past day. RBC folate represents folate intake over the past year, while plasma folate represents folate intake over the past day. Plasma folate represents folate intake over the past day, while RBC folate represents folate intake over the past few months. RBC folate represents folate intake over the past day, while plasma folate represents folate intake over the past few months.


They were consuming between 300 and 400 pg/day.

In this study, the outcome variables of piasma folate and RBC folate were chosen to reflect the status

A.

45. When folate is deficient, red blood cell synthesis is dramatically affected. Folate is required for synthesis of the nucleotide thymidylate. Developing cells will enlarge too much before they are mature

and are reieased liom the bone marrow. What type

of anemia is this called?

A. B. C. D.

t2t

Microcytic anemia Megacytic anemia

Millicyric anemia Macrocytic anemia


Biology

Folate Experiment

46. What could you infer

Passage Vtr

about the folate needs in

pregnant women?

A.

Folate needs are

B. C.

still about 200

1tglday in

pregnancy. Folate needs increase during pregnancy due to synthesis of many new cells. Folate needs decrease during pregnancy due to synthesis offoiate by the placenta. Folate needs of the mother are met by liver stores of folate, since folate is a fat-soluble

D.

vitamin.


structures indicates the pyrimidine nitrogenous base thymine?

A.

B. o

HzN

NH"

.A -t r") N

I

13);

',t'

H

C.

D. o

NH,

H.

'"^

"L*) ;

Copyright

@

N

\,

cHr

.A*)


'l

122


Biotogy Passage

VIII

Measurement of Blood Pressure

(Questions 48-54)

49, If

attached

to the

cannula. Although this

a person has a large amount

arms, how

The first measurement of blood pressure, in the early 1700s by Stephen Hales, used a cannula inserted into the artery of a horse. The pressure of the blood was determined by the height it achieved in a vertical tube

Passage VItr

will this

of fat tissue on their change the blood pressure

reading?

A.. B. C. D.

is a direct

technique, we currently use an indirect technique based on arterial sounds to measure blood pressure.

The reading will be falsely high. The reading will be falsely low. The reading will be accurate. One cannot tell from this passage.

A pressure cuff is wrapped around the arm and inflated, compressing and closing the brachial artery. Using a stethoscope, the reader listens for the beginning of blood flow as the pressure is released from the cuff. 50.

The tapping sounds she hears are spurts of blood, pushing through the partially obstructed artery with each heartbeat. She also listens for the final sound and notes at what pressure the first and last sounds occur. As the artery becomes completeiy opened, no sounds are heard. The lirst sound is recorded as the systolic pressure, and the final sound is recorded as the diastolic pressure. An attached meter called a sphygmomanometer is used to

What is laminar flow?

A. A fluid moves as a group of particles, each

read the pressures.

A fluid moves as a series of individual layers,

C.

A fluid moves as single particles, each with its

D.

A fluid moves

In Table 1 are some average arterial pressures nm/Hg) at various ages for men and women.

Systolic Systolic Diastolic Men Women Men

Age 20-24

23

16

76

72

25-29

25

I'7

'78

74

30-34

26

20

'/8

75

35-39

2'7

1^

80

78

40-44

29

t21

81

80

45-49

30

3l

82

82

50-54

35

3l

83

84

55-59

38

39

84

84

60-64

r42

144

85

85

Table

51.

Diastolic Women

each of the same velocity.

own random course. as a series of individual layers, each of a different velocitv.

Gravity plays a role in reading blood pressure. For this reason, blood pressure measurements are conventionally made with the cuff approximately level to the heart. For a 46-year-old woman, to what value would the systolic pressure change if it were measured at the level of the ankle? Assume the ankle is 100 cm below the heart, and gravity increases blood pressure 0.8 mm/Hg per cm.

A. 162 B. 2t0

c. D.

211 313

1

52. +8.

with its own random course.

B.

Astronauts on the space shuttle routineiy participate in research experiments. Where should the blood

As the pressure in the cuff is released, what does the first sound, read as the systolic pressure, indicate?

pressure

A.

A.

The minimum pressure of the blood following a heartbeat.

B. C.

The beginning of laminar flow in the artery. The maximum pressure of the blood following

D.

The heartbeat itself.

cuff be placed for a reading of blood

pressure on an astronaut in space?

B. C. D.

a heartbeat.

The cuff must be located in the conventional position at the level of the heart. The cuff should be located at the ankle, as far as possible from the level of the heart. The location of the cuff does not matter.

The cuff wili not work in an anti-gravity situation.

J:pyright

@


123


Biology

Measurement of Blood Pressure

53. When during the blood pressure

measurement is the

Passage VIII

blood flow turbulent?

L il.

Between the systolic and diastolic

IfI.

measurements After the diastolic measurement

Before the systolic measurement

A. B. C. D.

I only

II only II and III only I, II, and iII

54. The puise pressure is the difference

between the diastolic pressure and the systolic pressure. It can be used to calculate the mean arterial pressure (MAP) during the cardiac cycle.

MAP = diastoiic pressure + (pulse pressure/3)

What is the MAP, on average, for a 27-yeat-old man?

A. B. c. D.

51

62 18 94


124


Biology Passage

Aortic Compliance


55.

To obtain curves shown in Figure 1, aortas were obtained at autopsy from individuals in different age groups. Successive volumes of liquid were injected into this closed elastic system and after each increment of

Compliance is best represented by which of the following relationships ?

A. B. C. D.

volume the internal pressure was measured. The graph be used to extract information about the elfects of ase

:an

on aortic compliance.

56.

tr

;

150

A. B. C. D.

100

d

o o

-)U-^

58.

0

50 100 150 Figure

1

effects of aging on the elastic modulus (Ep) is 2. The elastic modulus is defined as Ap/ "l.-, D . AP is the aortic pulse pressure, D is the mean -,:-.-: diameter during the cardiac cycle, and AD is the -:".,-::irm change in aortic diameter during the cardiac - :.: The fractional change in diameter (AD/D) of the - ,. :-.rnng the cardiac cycle is a reflection of the change .::n: during left ventricular contraction. Finaily, the "::,-, :s unable to eject its stroke volume into a rigid -:--:--.- s'\'stem as rapidly as into a more compiiant system. .'"

Curve D. Curve E.

As compliance diminishes, it is most likely that peak arterial pressure occurs progressively:

A. B. C. D.

l:e

--

Curve A. Curve B.

200

Pressure (mmHg)

.

constant over a lifetime. not related to aortic compliance. directly proportional to aortic compliance. inversely proportional to aortic compiiance.

represents the oldest age group?

C)

i

D/AD

57. According to Figure 1, which curve most likeiy

200

6-.'

E

AD/D AV/AP AP/AD

The elastic modulus is most likely:

A. B. C. D. :

Passage IX

59.

r, rn Fisure

earlier in systole. later in systole. earlier in diastole. later in diastole.

According to Figure 1, aortic compliance in the youngest individuals is:

smaller over ail pressures when compared to other age groups.

C.

D.

:r

greater over all pressures when compared to other age groups. greatest at very high and low pressures and least over the usual range of pressure variations.

least at very high and low pressures ancl greatest over the usual range of pressure variations.

60. According to Figure 1, it can most likely

be

concluded that compliance:

A.

increases

with age,

a

manifestation of

increased arteriai rigidity.

B.

increases

with age, a

decreased arterial

40 60

decreases 80

100

with age,

a

manifestation of

increased arterial rigidity.

Age (years)

D.

decreases

with age,

decreased arterial

Figure 2

- bi The Berkeley Review

manifestation of

rigidity.

t25

a

manifestation

of

rigidity.


Biology Passage

fleart Muscle Action Potentials

Passage X

Figure 3 represents the membrane potentials of a cell from the sinoatrial node, which is located in the right

X (Questions 61-66)

Contraction of cardiac muscle is triggered by depolarization of the plasma membrane. A typical ventricuiar action potential from a contractile myocardial cell is iilustrated in Figure 1. In order to bring about this action potential, there must be inherent changes in the permeabilities of ions flowing into and out of the muscle

atrium of the heart.

ce11.

61,.

In

passing from the right atrium

to the right

ventricle, blood wiil have passed through the:

(€

cq)

A. B. C. D.

F. a-

e5

-50

-o

mitral valve. semilunar valve. tricuspid valve. pulmonary valve.

2 i00

0

0.15

0.30

Time (seconds)

62. Figure

Figure

2

1

Figure 2 depicts the permeabilities of ions during the action potential in a myocardial contractile cell.

Graph C most likely depicts the permeability of which of the following ions?

illustrates the membrane permeability

changes which are depicted by the action potential.

o

A. Na@ B. 60 C. Ca2@ D. CIE

10.0

h>' o-o

2 ;i srF .ai

t.u

dLL CJ

&

63.

0.1

0

0.15

excitation is best manifested by which region of the action potential shown in Figure 3?

0.30

Time (seconds)

A. B. C. D.

Figure 2 Special cells located in the heart itself are responsible for the heart's spontaneous and rhythmic self-contraction. These specialized cells of the sinoatriai (SA) node have significant differences in their membrane potentials when compared to contractile myocardial cells described in

Figure

The capacity for spontaneous and rhythmic self-

Region A.

Region B. Region C. Region D.

64. A membrane

1.

potential plateau region, similar to the

one shown in Figure 1, is best explained by:

A. B.

qa'

C. -o

o

D.

the flow of positive ions out of the cell equaling the flow of positive ions into the ceil.

both Nao and Ko

channels becoming

inactivated.

Time (seconds)

Figure

inactivation of Nao channels. a significant increase in the permeability of the membrane to CaZ@.

3


126


Biology 65.

Ileart Muscle Action Potentials

The action potential initiated in the SA node spreads throughout the right atrium from contractile cell to contractile cell. The means by which this spread takes place is most

A. B. C. D.

66.

Passage X

likely through:

tight junctions between cells.

neurotransmitter communication between cells. gap junctions between cells. desmosomes, which hold adjacent cardiac cells together.

The following graph depicts three different action potentials from a SA nodal cell. Which of the

following statements is most likely true of the graph? 50 6

o

aaa4

n

6* C)

50

Time (seconds)

A.

Graph B is the result of acetylcholine release onto the heart muscle.

B is the result of

B.

Graph

C.

stimulation. Graph C is the result of norepinephrine release onto ths heart muscle.

D.

Graph

vagus nerve

C is the result of vagus nerve

stimulation.


127


Biotogy Passage

Capillary Filtration

XI (Questions

Passage XI

67. An increase

67-72)

intracapillary hydrostatic pressure

would:

The direction and magnitude of water movement across a capillary wall are determined by the algebraic sum of the hydrostatic pressure and osmotic pressure that exist across that membrane (Figure 1).

of

A.

increase the concentration

B.

active particles within the vessel. result from a decrease in the arterial pressure.

C.

D.

favor movement

osmotically

of fluid from

vessel to

interstitial space. increase the venous resistance.

Filtration

fii6#-"P*""'"

68.

Reabsorption

subcutaneous tissue indicate a P1 of

-I

in

the

Io -7 mmHg. If such values are accurate, which of the following could most likely be concluded?

Lumen of Capillary

lnterstitial Space

Figure

Studies using plastic capsuies implanted

A.

Pc

B.

Pc P; Pc

c.

1

D.

The hydrostatic pressure gradient across the wall of a capillary, P", depends on the arterial pressure (35 mmHg), the venous pressure (15 mmHg), and the interstitiai fluid pressure, P;, which is about 2 mmHg. Precapillary and post capillary resistances are also important. A reduction in the arterial or venous pressure reduces capillary

69.

Pi > Pc. Pi < Pc. Pc> Pi.

Pi < Pi.

A reduction

the diameter of a precapillary vessel

would:

hydrostatic pressure. The interstitial fluid pressure opposes capillary filtration, and the difference between the two provides the driving force for filtration.

A. B.

Important factors responsible for fluid retention in the capillaries are the osmotic pressure of the plasma proteins, nO, and the osmotic pressure of the interstitial proteins, n1. The osmotic pressure exerted by these proteins is also cailed the colloid osmotic pressure or the oncotic

favor movement of fluid from the vessel to the interstitial space. result in a reduction of hydrostatic pressure.

C.

be equivalent to an

increase

venous

resistance.

D.

pressure.

significantly increase the concentration of osmotically active particles in the interstitial space.

While this oncotic pressure is low compared to the totaL plasma osmotic pressure, it plays a major role in

fluid reabsorption. This is due to the fact that

the

electrolytes that are responsible for the major fraction of plasma osmotic pressure are equal in concentration on both sides of the endothelium.

70-

The protein albumin is dominant in determining plasma oncotic pressure. While aibumin is relatively

Fluid movement is most likely described by which of the following equations? Note: k is the filtration constant for the capillary membrane.

impermeable to the capillary membrane, small amounts of the protein do leak into the interstitial fluid and create a very small osmotic force.

A. B. C.

D.


r28

Fluid Fluid Fluid Fluid

movement movement movement movement

= k[(Pc +

nil-(Pi+?rp)1.

= k[(Pc + P1)-(np-n1)1. = k[(Pc Pf+(n1+rrp)1. = k[(ni

np)-(Pi+Pc)1.


Biology 71.

Capitlary Filtration

Passage

Iil

Albumin exerts a greater osmotic force than can be accounted for soleiy on the basis of the number of

molecules dissolved in plasma. Based on this information, it can BEST be described that albumin: A.

may be replaced by inert substances with no effect.

B. C. D.

72.

dissolves into more than one protein molecule in plasma. carries electrical charges at blood pH which attracts various eiectrolytes. is permeable to the capillary endotheiium.

Only a small percentage (ZVo) of the plasma flowing through the vascular system is filtered, and of this, about 857o is reabsorbed in the capillaries and the venules. The remaining 157o of fluid:

A. B. C. D.

acts to increase intracapillary hydrostatic pressure. remains in the interstitial fluid.

returns to the arterial circulation via the lymphatic system.

returns to the venous circulation via the lymphatic system.

r n

Copyright

@


129


Biology Passage

XII

Kespiratory Calculations 74. A respiration

physiologist measures a volume of air to be 1L at 0 degrees Celcius and 1 ATM. What will be the volume of this air at 25 degrees Celcius and

(Questions 73-78)

The bronchial circulation is the nutritive supply to all of the lung support structures. Such structures include airways, connective tissue, and pleura. The pulmonary circulation is the nutritive supply to the alveolar walls. The pulmonary capillaries form an extensive network within the alveolar walls. The maximum capillary volume is about 200 ml, while the normal capillary volume at rest is about 70 ml. This capillary volume can be increased by opening compressed or closed capillaries. Such opening is callled recruitment.

lATM? A. B. C.

D. /5.

02 in systemic arterial blood - l0 mmHg less than that in alveolar gas. The difference in partial pressure exists

The partial pressure of

because:

A.

room atmospheric pressure is given the value 760 mmHg. This air contains 2IVo 02,79Vo N2, and }Vo CO2. Blood

B.

that leaves the pulmonary capillaries has come into equilibrium with all of the alveolar gases. In addition, small quantities of venous biood from bronchial venules and vessels from the heart join the pulmonary venous

C. D.

outflow.

76.

the

a fraction of 02 is replaced by the partial pressure of water.

the venous blood from bronchial venules and heart vessels contaminate the pulmonary venous outflow. the partial pressure of gases dissolved in liquid is not equal to the partial pressure of the gas

Pulmonary physiologists are often interested in only anhydrous gas volumes and partial pressures. Taking

A. B. C.

D. 77.

oxygen binds to the protein hemoglobin, which is able to

hemoglobin

concentration in blood being 150 g/L ofblood.

102 115 150 170

mmHg. mmHg. mmHg. mmHg.

in the passage, the anhydrous partial pressure of oxygen in alveolar gas is:

Based on the information

A. B. C.

D.

102 115 150 170

mmHg. mmHg. mmHg. mmHg.

78. At an oxygen partial pressure

73. Besides recruitment, capillary volume can be increased by enlarging open capillaries as their internal pressure rises. Such distension is most likely caused by:

of 90 mmHg,9'77o of

hemogiobin is fully saturated with oxygen. At 90 mmHg, which of the following is the BEST calcuiation of the TOTAL concentration of oxygen (ml O2lL) in the systemic arterial blood?

increased cardiac output. increased left atrial pressure. decreased left ventricular pressure. right heart failure.


the

this into consideration, the partial pressure of

In the alveoli, some oxygen diffuses directly into the pulmonary capillary blood. In fact, 3.0 ml of 02 are dissolved in iL of blood for a pOz -- 90 mmHg. This reduces the fraction of 02 in alveolar gas to 0.143. Most

A. B. C. D.

of 02 is directiy dissolved into

oxygen in dry inspired air at the trachea is:

obligatory daily water loss from the body.

bind 1.34 ml O2lg, with the normal

a fraction

pulmonary blood flow.

phase at equilibrium.

When air is inspired, it becomes saturated with water vapor at 37 degrees Celcius in the nose, throat, and trachea. The source of the heat and water vapor are the pulmonary and bronchial blood flows. The water vapor exerts a mandatory partial pressure, with the pH2O = 47 mmHg. The total quantitiy of water in expired gas over a

for nearly one-half of

L L 1.0 L 1.09 L

0.82 0.91

is normally 5

For a given volume, the total gas pressure of all molecular species is the sum of the individual pressures. This law of partial pressures makes the assumption that gas molecules do not interact with each other. The dry

24-hour period accounts

Passage XII

A. B. C. D. 130

145 mlO2/L. 195 ml O2/L. 197 ml O2lI-.

200mlO2/I-.


Biotogy Passage

XIII

Aspirin

Passage XItr

(Questions 79-85)

Aspirin is a non-steroidal anti-inflammatory agent that works by inhibiting the initial enzyme in prostaglandin

O

//

synthesis pathway, cyclooxygenase (CO). Aspirin acetylates a serine residue in the active site of CO, inactivating it. This makes aspirin a useful drug for

Endothelial :lialCells Cells

IJO

/

>0)

<=

modifying prostaglandin synthesis.

o

*i c\

The diagram in Figure I shows the pathways of prostaglandin synthesis from the fatty acid, arachidonic acid.

123 Days Following Aspirin Dose

Leukotrienes

I I

I

Figure 2.

Lipoxygenase

Arachidon ate

l'"

looxygenase

Thromboxanes

Prostaglandins

Figure

79. Which of the following statements BEST describes 1.

what is happening in Figure 2?

A.

Low-dose aspirin treatment is often used to reduce risk

heart attack or stroke in people with a history of cardiovascular disease. Aspirin lowers the chance of blood clots traveling through the vessels and obstructing blood flow through the coronary arteries of the heart or through the carotid arteries leading to the brain. Aspirin *'orks by altering the ratio of various prostaglandins to

of

After aspirin treatment, platelets produce new CO, but endothelial cells do not.

B. After aspirin treatment,

endothelial cells

produce new CO, but platelets do not.

C. After aspirin D.

treatment, both platelets and

endothelial cells produce new CO.

After aspirin treatment, neither platelets nor endothelial cells produce new CO.

reduce clot formation.

Specifically, the important alteration is in the ratio of prostacyclin to thromboxane ,{2. Platelets, which have a

life span of four days, are important in the clotting process. Platelets secrete thromboxane A2, which

80.

promotes platelet aggregation and vasoconstriction. The endothelial cells of the blood vessels secrete prostacyclin, ir hich inhibits platelet aggregation and causes

How many degrees of unsaturation are present in the following structure, thromboxane 82?

asodilation. The balance between the two causes appropriate clotting: a strong clot that does not cover an ercessive area, and that permits unrestricted blood flow

OH

r

around the clot.

Aspirin inactivates CO permanently in both platelets and endothelial cells. Endothelial cells rapidly produce more CO enzymatically, but platelets cannot manufacture

OH

A. B. C. D.

co. The model shown in Figure 2 indicates CO levels in platelets and vascular endothelial cells after the ingestion aspirin: ".f 300 mg of Copyright

@


l5l

One degree of unsaturation.

Two degrees of unsaturation. Three degrees of unsaturation Four degrees of unsaturation.


Biology

Aspirin 84.

81.. When aspirin modifies CO, inactivating it, which

Passage XIII

structure indicates the product amino acid residue?

Low-dose aspirin therapy is used every other day to prevent heart attacks in some people with heart disease. Why does this dosing regimen help?

A.

A.

The dosing regimen keeps platelet aggregation

B.

The dosing regimen keeps platelet aggregation high.

C.

The dosing regimen keeps endothelial cells proliferation low. The dosing regimen keeps endothelial cells

HO til -NHr-C-C-O* CH" t'

B.

I

HO lil _NH,-C-C-O_ -t CH"

t'

o

S I

I

a

/l-\

o

low.

/t\

c.

D.

C

o

cHj

cHl

proliferation high.

D.

HO I lt -NHr-c-C-oCH. t' I

HO Iil -NHr-C*C-OCHt

t-

o

S

lo

o=

85.

I

p- o" I

la

o=P-o" I

OO

OO

Although the mechanisms behind the birth process are not well understood, it is clear that prostaglandins play an important role in initiating and promoting parturition. What does this indicate for late pregnancy (months 7-9)?

A.

Aspirin would probably stimulate an early

B.

Aspirin would have no effect on pregnancy. Aspirin should not be used in late pregnancy. Aspirin would probably cause birth defects.

labor. C.

D.

82. What symptoms would accompany an intake of aspirin that was too high?

I. II. III.

83.

Easy bruising. Prolonged bleeding time. Rapid clotting.

A.

only

B. C. D.

and and

II only III only II, and trI

The fatty acid molecule that is the precursor to prostaglandins and the substrate for cyclooxygenase has the common name of arachidonic acid. What is the IUPAC name?

A. B.

c. D.

Copyright

5, 8, 11, 6, 9, 12, 6, 9, 12, 5, 8, 11,

@

14 arachidonic acid. 15-arachidonic acid.

1S-eicosatetraenoic acid. 14 eicosatetraenoic acid.


132


Biology Passage

Sickle CelI Anemia and IIbF 87.


Sickle cell anemia is caused by a singie substitution of one amino acid on the B chain of the adult hemoglobin

In the blood cells of an adult heterozygote with the sickle cell trait, how are the hemoglobin types distributed?

(HbA) molecuie. This change produces the form of hemoglobin called HbS. Under conditions of low blood partial pressure of oxygen (pO), HbS comes out of solution and forms a cross-linked crystaiiine structure inside the red blood cell. This leads to the characteristic sickle shape of the red blood cells. Complications can arise because the sickied red blood cells cannot properly fold up enough to pass through small capillaries of the circulatory system. Tissues are deprived of some of their blood supply by blockages caused by sickled cells.

A. B. C.

D.

All cells contain both HbS and HbA. Some cells contain exclusively HbS, while others contain exclusively HbA. All cells contain HbA only. All cells contain HbS only.

88. Treatment with hydroxyurea in

Homozygotes are severely affected by the sickling cells, but heterozygotes rarely experience sickling under normal oxygen levels. Heterozygotes may experience problems at high altitudes (such as in a depressurized piane) or under some types of anesthesia.

homozygotes

promotes fewer crisis events. HbF is increased to about 20Vo of the cell's hemoglobin content by treatment with hydroxyurea. What beneficial role does HbF play inside the cell?

A. B.

Certain drug treatments for sickle cell anemia focus on increasing levels of fetal hemoglobin (HbF), which does not sickle. The following chart shows types of Hb present at various ages: g chain

r00

Passage XIV

HbF binds and inactivates HbS inside the cell. HbF promotes a lower oxygenation state inside the cell.

C.

HbF promotes a higher oxygenation state

D.

inside the cell. HbF binds and activates HbA inside the cell.

b\

;80 o

o-^ !. ttu !o

-o

89.

+O

{) q

Birth 3 Gestation

(months)

6

Age (months)

Sodium metabisulfate (Na2O5S2) functions as an antioxidant in many medical preparations. It is transformed to sodium persulfate (Na2OgS2) in its role as an antioxidant. How could sodium metabisulfate be used to test for sickle cell trait when combined with hemoglobin from red blood cells?

A. Sodium metabisulfate would

Figure I

B.

86. Sickle

cell anemia is usually diagnosed by 3 months of age, but not always at birth. This is most likely due to:

a decrease of HbF and an increase of HbA

B.

persistent maternal hemoglobin

C. Sodium metabisulfate would

with age.

until age

3

D.

months.

D.

Copyright

a decrease in HbF and an increase in HbS with age. persistent maternal antibodies to HbS until age 3 months.

@

hemoglobin from red blood cells, and the sickling pattern would be noted in affected individuals.

A.

C.

promote

reduction of the hemoglobin to a rusty color, which would identify affected individuals. Sodium metabisulfate would oxygenate


promote oxidation of the hemoglobin to a rusty color, which would identify affected individuals'

Sodium metabisulfate would deoxygenate hemoglobin from red blood cells, and the sickling pattern would be noted in affected individuals.

tJc


Biology 90.

Sickle Cell Anemia and HbF

What would happen as

if

an adult had too much HbF,

is the case in some of the thalassemias, in which

too much of the

A. B. C. D.

y

chain of hemoglobin is produced?

Decreased delivery of oxygen to the tissues, compared to other adults.

Increased delivery of oxygen to the tissues, compared to other adults.

The red blood cells would sickle, bur at high pO2levels. The red blood cells would sickle, but at low pO2levels.

91. Which diagnosric

technique would distinguish

heterozygotes, homozygotes with sickle cell anemia, and wild-type homozygotes?

I. U. rII.

Restriction fragment length poiymorphisms (RFLPs) on red blood cell DNA.

Gel electrophoresis of hemoglobin samples from the red blood cells. Centrifugation of hemoglobin samples from the red blood cells.

A. B. C.

D.

92.

I only II only I and II only I, II, and III

Which of the following srarements is TRUE of HbF?

I

I.

rfl.

HbF does not bind 2,3-BPG. HbF has a higher oxygen saturation at a given pO2 than HbA.

HbF contains two s and two p chains.

A. I only B. I and II only C. II and III only D. I,II, and trI


t34


Biology

Ventilation Kegulation

Passage XV (Questions 93-100)

Passage XV

Figure 2 shows the results of the same experiment with the variable being arterial pCO2. The reflex involves the peripheral and central chemoreceptors and is closely tied to changes in the proton concentration.

A group of inspiratory neurons located in the medulla is responsible for controlling ventilation. Inputs to these

neurons from both the peripheral and central

chemoreceptors are important in regulating involuntary control of ventiiation. The peripheral chemoreceptors consist of two different bodies of receptors located in the carotid artery of the neck and in the arch of the aorta. The nerve terminals are intimate with the arterial blood and respond to changes in pCO2 , pO2, and Ho. Impulses from these nerve terminals travel up afferent fibers toward the brainstem and eventually synapse with inspiratory

An increased arterial pCO2 wili cause an increase in the arterial concentration of protons. The peripheral chemoreceptors are stimulated by this iowered pH. At the same time, the elevated arterial pCO2 causes an increase in the pCO2 of extracellular brain fluid. This causes a rise in the proton concentration of the fluid, stimulating the central chemoreceptors. It is the role of the central chemoreceptor which is most important in mediating this

neurons of the breathing center. The central

ventilation reflex.

chemoreceptors, located in the medulia, monitor the brain's extracellular fluid for changes in the concentration of protons. Fibers from this receptor synapse with inspiratory neurons.

^30

93.

To

assess the effects

of a changing arterial pO2 on

the ventilation rate accurately, which of

J q?O d

A. The arterial pCO2

O

>10 o 2,

(l

the

foliowing conditions should exist?

B.

0 20 40 60 80

C. 100

D.

Arterial pO2 (mmHg)

should decrease at

a

constant rate. The arterial pCO2 should increase at a constant rate.

The arterial pCO2 should be maintained at 40 mmHg. The arterial pCO2 should be maintained at 46 mmHg.

Figure I

Figure 1 shows the relationship between arterial pO2 the ventilation rate. The graph indicates that a :'.', ering of the arterial pO2 will stimulate the rate of :nlilation. This reflex is mediated by the peripheral

.:J

94.

because the:

-:3;noreceptors which respond to a lower arteriai pOZ by :.easing their rate ofdischarge to the breathing center.

A. pCO2 is increasing. B. total amount of oxygen

16

C. D.

<1' J

=6 o

95.

40

44

48

Arterial pCO2 (mmHB)

The inspiration of carbon monoxide gas


A.

an unchanged arterial pO2.

B. C.

a

D.

will result

lowered arterial pO2.

an increased arterial pO2.

an increased percentage of

hemoglobin

saturated with 02.

Figure 2

@

transported is relatively unaffected. body responds with a decreased use of oxygen. pCO2 is decreasing.

in:

O

Copyright

According to Figure 1, a drop of 30 mmHg from the normal resting value of pO2 does NOT radically increase the ventilation rate. This occurs most likely

t35

The BerkeleY Review Speciatizing in MCAT Preparation

Biotogy

Ventilation Regulation

96. According to Figure 1 and Figure 2, changes arteriai pCO2 are:

A. B.

C.

D.

97.

Passage XV

100. During times of sleep, the body's ventiiation

in

rate decreases more than the consumption of oxygen by ceils. The result ofthis situation is:

augmented by reflexes regulating ventilation to a greater degree than are equivalent changes in arterial pO2. augmented by reflexes regulating ventilation to a lesser degree than are equivalent changes in arterial pO2.

A.

an increased arterial pCO2 and a increased arterial pO2.

B.

an increased

pCO2 and a decreased

arterial pO2. C.

resisted by reflexes regulating ventilation to a greater degree than are equivalent changes in arterial pO2. resisted by reflexes regulating ventilation to a lesser degree than are equivalent changes in arterial pO2.

a decreased arterial pCO2 and an increased arterial pO2.

D.

a decreased arterial pCO2 and a decreased arterial pO2.

Which of the following statements provides the BEST support for the type of ventilation found early on in metabolic acidosis?

A. _ B. C. D.

98.

Hypoventilation is caused by increased neural gutput from the peripheral chemoreceptors.

Hypoventiiation is caused by increased neural output from the central chemoreceptors. Hyperventilation is caused by increased neurai gutput from the peripheral chemoreceptors. Hyperventilation is caused by increasid neural output from the central chemoreceptors.

Which of the staremenrs below is FALSE regarding pCO2 during exercise?

A. B. C. D.

The alveoiar pCO2 determines arterial pCOZ.

Alveolarvenrilationincreases. Venous pCO2 increases. Arterial pCO2 increases.

99. In order

to hold their breath for a longer duration, swimmers often hyperventilate immediately before competing in a swimming event. This activity can be dangerous because even though the:

A. low pCO2 is permitting one to hold

their breath, the exercise may lower the pO2 to

levels which may induce unconsciousness.

B. Iow pCO2 is permitting one ro hold

their

breath, the exercise may raise the pO2 to levels which may induce unconsciousness.

C.

high pCO2 is permitting one ro hold their breath, the exercise may lower the pO2 to

D.

high pCO2 is permitting one ro hold their

levels which may induce unconsciousness. breath, the exercise may raise the pO2 to levels

which may induce unconsciousness. Copyright @ by The Berkeiey Review

136


Biology

1.

.,

tleart & Lungs

Section II Answers

A is correct, low average velocity because of a high total cross-sectional area. This is according to the continuity equation describing fluid flow. This equation states that since blood flow must be equal throughout an entire closei system, the cross-sectional area times the average velocity equals the cross-sectional area times the velocity. The total cross-sectional area of aii the capillaries is very large, the largest of all types of yessels seen in the cardiovascular system. Therefore, the blood flow through the capiliaries will have a low average velocity because of their high total cross-sectional area. The correct choice is A. D is correct, 16. The problem requires that we be farniliar with the relationship between resistance of a vessel and the radius. The resistance of the vessel varies inversely with the radius to the fourth power. Therefore, decreasing theradiusbyafactorof2leadstoanincreaseintheresistancebyafactorof

16. TheiorrectchoiceisD.

3.

C is correct, Q x TPR = (MAP - CVP). We are looking for an expression to represent blood flow. The general equation is given in the passage. Biood flow (Q) = Perfusion Pressure/Flow R.eslstance. We are asked about the entire vascuiar unit. The perfusion pressure is therefore going to be the mean arterial pressure minus the central venous pressure. The resistance for the entire unit is given by the total peripheral resistanie. Placing these values in the formula and multiplying both sides by TPR will give the solution. The correct choice is C.

4-

D is correct, red blood cells. One can arrive at this answer by thinking about the composition of biood. Blood is made up of plasma and cells. The question is asking about the increase in viscosity ieen in whole blood versus plasma. Knowing the composition of biood ieads one to believe that increase must come from the cellular contribution' It then becomes a matter of WBCs or RBCs. At this point, one must be aware that there are millions of red blood cells in whole blood. In other words, there are many more red cell than white cells. Therefore, the major contribution must come from the presence of red blood cells. The correct choice is D.

5.

C is correct, veins contain valves that favor one-way biood flow. We are required to draw on knowledge about the structure of blood vessels. Both arteries and veins have a layer of endothelial ceils, a layer of elastic tissue, ancl a layer of smooth muscle. Therefore, we can eliminate answer choices A, B. and D. However, only veins contain one-way valves which prevent back flow of blood. One must understand that the pressure in the vlins is not very large, and so there is not a tremendous amount of force to propei blood. When ihe blood is propelled (skeletil muscle contraction is one method of providing the force), we wish the b1oo6 to remain *ou1ng forward. The presence of these one-way valves carry out this function. The correct choice is C.

6.

B is correct, venous pressure does not change significantly over the heart cycle. The MAP is an average figure because the pressure in the arterial system changes significantly over the heart cycle. It is very high right out oltne aorta, but becomes smaller and smaller toward the capiiiaries. Therefore, we need to take un aui.ug" value. The pressure in the veins does not change very dramatically over the entire heart cycle. The total pressure difference in the peripheral veins is usuaily 5 to 10 mmHg. Therefore, we do not need to take an average value. The fact that the arterial pressure does change dramatically is true, but it is not proper rationale for trying to explain why we do not need to take an average for the veins. Furthermore the cross-sectional area and the pressure responses to increases in volume offer us no explanation as to why no average is taken. The average is not taken because the value remains relatively constant. The correct choice is B.

7.

D is correct, arterial systolic pressure. We are told in the question that we need to use the pressure in the cuff to collapse the artery totally and prevent blood flow. Therefore, the pressure in the cuff must counter the maximum pressure offered by the artery. The maximum pressure in the artery will be represented by the arterial systolic pressure. Recall that systoiic pressure is the maximum arterial blood pressure during cardiac cycle. To coliapse the artery, the pressure in the cuff must exceed this value. The correct choice is D.

D is correct, glycoiysis. It is important to know which cells have organelles and which do not. It will also be important to know certain metabolic processes and where they take place. E,ven though we have not formally discussed metaboiism yet, this question is designed to get you to think about cellular function. Mature red blood celis are enucleated eukaryotic cells. They do not have any organelles and they have only one membrane, the plasma membrane. As we will see in future discussions, the Krebs cycle, electron transport, and oxidative phosphorylation are all associated (in eukaryotic celis) with mitochondria. Even though these three systems are responsible for the majority of energy generated in metabolism, they are not found in red blood cells. However, red blood cells do require energy and they obtain that energy from glycolysis. The correct choice is D. Copyright O by The Berkeley Review

L37


Biology 9.

Ileart & Lungs

Section II Answers

C is correet, reductase. The vitamin K dependent reaction, as written from glutamate to y-carboxyglutamate, is a carboxylation and would require a carboxylase enzyme. If we wanted to go from y-carboxyglutamate to glutamate, it would be a decarboxylation reaction, involving a decarboxyiase enzyme. Proteases are proteolytic enzymes that catalyze the hydrolysis ofpeptide bonds.

HO I tl _N-C*C-N_ ttt HCH2H

Vitamin K Vitamin K Hydroquinone Quinone

HO Iil _C_C-Ntt cH, H

,ll 'l /H

I

I

coz

cH"

l'6

CH

/:

coo Glutamate

coo

O

y-Carboxyglutamate

In this reaction we are not degrading a protein. We are regenerating vitamin K hydroquinone, which is the reduced form of vitamin K. In order to go from the oxidized form (vitamin K quinone) to the reduced form, a reduction needs to occur. This reaction will be catalyzed by a reductase. The correct choice is C. 10.

B is correct, thromboplastin (Factor III or Tissue factor). This question can be answered by looking directly at Table 1 and Figure 1 in the passage. In Table I the symbol " f " denotes dependency on vitamin K. All that is needed is a correlation between the common names and the factor numbers. The correct choice is B.

11.

A is correct, consumption of leafy green vegetables. The major source of vitamin K is from one's diet, especially in leafy green vegetables. There are also intestinal bacteria that can synthesize vitamin K. Since vitamin K is b"ing made availabie to be a cofactor in the conversion of glutamate to y-carboxyglutamate, we would expect an increase in clot growth. Since blood clots can break free from blood vessels and wreak havoc within the circulatory system (by causing heart attacks and strokes), their growth must be limited. Clot growth can be limited by inhibiting the serine proteases which act at various steps in the cascade. Since blood is dynamic (constantly moving), the areas near i clot are always being diluted with fresh blood. This helps to prevent a build up of cascade intermediates involved in the clotting process. The intermediates that are removed by the flow of blood are transported to the liver where they are removed and degraded. The correct choice is A.

12.

C is correct, 0.50 (or 50Vo). Men have an X and a Y chromosome; women have two X chromosomes. We are toid in the question that the defect for both types of hemophilia (h) resides on the X chromosome and that it is recessive, The woman's father has hemophilia, which we can designated as

XhY.

Since her mother is normal (XX) the woman

must be a carrier (XXh) because she received one good X chromosome from her mother and one defective Xh chromosome from her father. This is shown in the Punnett square in Figure I below:

X

X

yh

XXh

XXh

Y

XY

XY

Figure I

ah

X

X

XXh

XX

Y

xhv

XY

Figure 2

The woman (XXh) now marries a man who is normal (XY). The mode of inheritance of their offspring is indicated in Figure 2 above. There are two sons and two daughters. However, we care about only the sons. One will be normal, while the other will express the trait. Therefore, the probability that this woman will have an affected son is 0.50 or SOVo. Even though we have not formally discussed simple Mendelian genetics yet, this question is designed to get you to think along those lines. The correct choice is C.

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138


Biology 13.

Ileart

Er

Lungs

Section II Answers

A is correct, citrate. Calcium is an important cofactor in the cascade mechanism outlined in Figure 1 in the passage. This ion bears a +2 charge and can be chelated by certain compounds that bear negative charges. As stadd in the passage the modified amino acid y-carboxyglutamate is an excellent chelator of Ca2@. When this amino acid is tied up in a protein, it presents two carboxyl groups to the calcium ion. Based on this we could assume that the more carboxyi groups on a molecule, the better the chelator. That is, the better the molecule is at sequestering calcium.

If

we follow this assumption, we can eliminate choice B first. Glutamate has two carboxylate groups. However, note the positively-charged amino group. This would tend to repel a positively-charged caliium ion. Not only that, but the positiveiy-charged amino group will be tied up with the cr-carboxyl group through electrostatic interactions. This will diminish the chelating abiiity of that molecule. We can eliminate choice D based on similar reasoning. The major difference between the last two choices is that choice C (succinate) has just two full carboxyiate g.ouft whereas choice A (citrate) has three full carboxylate groups, In this case, the more carboxyl groups, ttie Uetter ttre chelator, and the better the molecule is at preventing ciotting in stored blood. The correct choiie is A. 14.

B is correct, proconvertin (Factor VII). This question is designed to get you to think about a number of items at the same time. As stated in the passage, warfarin is an antagonist (i.e., a competitive inhibitor) of vitamin K. Now, you need to ask yourself, "Which coagulation factors in the cascade are affected by inhibition of the use of vitamin k?"

The factors which are affected (from Table 1) are those which are dependent on vitamin K. Those factors turn out to be choices A'D in the.answers (i.e., prothrombin, factor II; proconvertin, factor VII; Christmas factor, factor IX; Stuart-Prower factor, factor X). Since there is just one answer (and not four), we need to consider some more

information.

In the question it states that 3 days after oral administration, the PT (prothrombin time) is maintained at 1.2-1.5 times the controi. The PT screens the extrinsic and common pathways. Note that in the cascade shown in Figure 1, the extrinsic pathway includes factor VII and the common pathway includes factor X and factor II. The exlrinsic pathway does not screen factor IX as it is part of the intrinsic pathway. This allows us to eliminate choice C (Christmas factor, factor IX) as a possible solution. How do we distinguish between the remaining choices? Another way to look at this is to ask yourself, "Which single factor decreases so much after 3 days as to maintain a PT between 1.2-1.5 times the control?" you should now be think about the haif-life ofthese factors (see Table I in the passage).

The one factor which has the shortest hatf-life is factor VII (between 2-6 hours). Let's assume that its half-life is about 4 hours' In 3 days (72 hours) factor VII will have been reduced by 18 half-lives. The half-life offactor X is between 25-60 hours. Let's assume that its half-tife is about 43 hours. Factor X wiil have been reduced by about 2 half-lives. Finally, factor II has a half-life between 50-120 hours. Let's assume that its halflife is about 85 hours. Factor II will have been reduced by about t half-tife. [A11 values have been rounded up for simplicity.] Based on this analysis, we see that factor VII will have been decreased the most, because it has iuch a shbrt fiie span. The correct choice is B.

B is correct, arterioles. The answer to this question is not found in the passage but rather comes from our knowiedge of the flow of blood through the circuiatory system.

Roughiy 64Vo of biood in the circulatory system is contained in the veins while about l5Vo ts contained in the arteries. The veins include the large veins, small veins, and venules. The small veins and venules together contain abott 25Vo of the total blood volume, leaving about 39Vo of the total blood volume found in the large veins. The arteries include the large arteries, small arteries, and arterioles. The large arteries and small arteries together contain about l3Vo of the total blood volume, leaving about 2Vo of the total blood volume found in Ihe arterioles. About 5Vo of the blood is contained in the capillaries. The remaining 76Vo of the blood is contained in the heat and pulmonary vessels (associated with the lungs). This allows us to pick the arterioles as having the LEAST blood volume (at any given time) in the circulatory system. The correct choice is B. 16.

C is correct, lower due to the accumulation of CO2. Recall that as blood passes through the tissues it gives up 02 as a nutrient and picks up CO2 as a waste product. The blood that is leaving the pulmonary artery is coming from the right ventricle, which in turn came from the right atrium, which in turn came from the venous system.

CO2+

-

::r

nght @ by The Berkeley Review

H2O:

o H2CO3

139


Biology

Ileart & Lungs

Section II Answers

Blood in the venous system is, for all practical purposes, deoxygenated blood. The oxygen had been given up at the level of the respiring tissues. The CO2 that was picked up by the venous system is either dissolved in the blood as free CO2 (about 57o), bound to hemoglobin as the carbamate (about 5Vo), or in the form of the bicarbonate ion (about 90Vo). When the bicarbonate ion is formed (Bohr reaction), hydrogen ions are produced which tend to lower the pH of the blood. The correct choice is C.

t7.

A is correct,

(see the graph below). When there is a constriction of a blood vessel there is more resistance to blood ieads to a greater pressure drop (i.e., a greater change in pressure, AP) between the veins and arteries. The result is a lower venous pressure. Therefore, the curve would be expected to drop along the y-axis.

flow. This

The curve intersects at the same point on the x-axis because

if

the heart were to stop and the cardiac output reduced

tozero,thepressuresinthevenousandarterialsystemswouldsoonequilibrate. Inotherwords,thevenouspressure would not change. This makes choice A a good answer. l0 -1

8

6

g

4

o

2

o

0

-4 0 4 8

1216


Choice B indicates a decrease in the blood volume. When the blood volume decreases there is less blood to fill the ventricles. This leads to less of a contractile force and a lower blood pressure. Choice C indicates decreased arteriole resistance while choice D indicates and increase in blood volume. The correct choice is A. 18.

D is correct, vasopressin and aldosterone. When the blood volume is low, neural impulses form the atria and arterial baroreceptors are sent to the hypothalamus and vasopressin (antidiuretic hormone, ADH) is released into the blood. Also, if the blood volume is low, chances are that the blood osmolarity has increased. An increased osmolarity is also detected at the level of the hypothalamus and this allows for the release of ADH as well. ADH acts at the levei of the late distal tubules and the collecting ducts in the kidney and directly promotes water reabsorption (back into the blood). This increases the blood volume.

A low blood volume is usually

associated with a low blood pressure. This is detected by the juxtaglomerular

apparatus in the kidney and granular cells in that complex release renin. This enzyme converts angiotensinogen to angiotensin I, and then angiotensin I is converted to angiotensin II by the angiotensin-converting enzyme. Not only does angiotensin II cause vasoconstriction of the arterioles (to help increase blood pressure), but it also stimulates the release aldosterone from the cortex of the adrenal glands. Aldosterone stimulates the reabsorption of sodium at the level of the kidney. Water follows down its concentration gradient into the blood and leads to an increase in blood volume. The correct choice is D. 19.

C is correct, (see the graph below). The sympathetic nervous system is a component of the autonomic nervous system. Recall that the sympathetic nervous system is involved in the fight-or-flight response. Activation of the sympathetic nervous system will stimulate the heart and cause it to be a stronger pump. In other words, the activity of the heart increases. This will cause the cardiac output (C.O.) curve to shift to a higher level. We see this response in choices A and C. Z\

310

i0

l_l c.o.

58 B6 o

,t

O+ Idr E =t

(50

Copyright

@

-4 0 4 8

1216

Ll c.o.

8

a

6

o o

4

O

0

2

-4 0 4 8

1216



Choice A

Choice C


740


Biology

fleart & Lungs

Section II Answers

Activation of the sympathetic nervous system will also cause vasoconstriction of arteries, arterioles, and veins. Contraction of venous smooth muscle will lead to a decrease in the diameter and increase in the pressure within the vessel. This will allow more blood to be returned to the right heart (i.e., the right atrium and right ventricle) from the veins. The result is an increase in venous return and an increase in the mean circulatory pressure. Since cardiac output must equal venous return (see the first paragraph in the passage), we can equate the y-axis of our curve in the question not only to cardiac output but also to venous return. Therefore, we would expect the curve to move upwards along the y-axis for venous return (V.R.) and outwards along the x-axis for an increase in mean circulatory pressure (see above). The correct choice is C.

[Note: In the third paragraph of the passage it states that if the cardiac output and the venous return is increased, the pressure in the arterial system will increase while the pressure in the venous system will decrease. Be careful of what it means when it says "the pressure in the venous system will decrease." If we are at a venous return of 0 L/min, the venous pressure is about 8 mmHg (see Figure 1 in the passage, or below). If we now move to a venous return of 5 L/min, the venous pressure is about -4 mmHg. It is a decrease form the mean circulatory pressure (MCP). l0 J E

8

6

,i)

4

o o

2

0

*""3, 0.1,,,1,ii'il In question 5 we have increased the MCP by constricting vessels. An increase in cardiac output due to sympathetic stimulation will decrease the venous pressure relative to the new MCP.I 20.

because it opens Nao and Ca2o channels in myocardial cells. This question ties in a when we discussed the nervous system. Recall that the preganglionic and postganglionic we examined concept nerve fibers of the parasympathetic nervous system release the neurotransmitter acetylcholine (ACh). The preganglionic fibers of the sympathetic division release ACh while the postganglionic fibers release the neurotransmitter norepinephrine. We can now eliminate choices B and D.

A is correct, norepinephrine,

Next, we need to recall that sympathetic fibers will stimulate the heart (think of the fight-or flight response). if the heart is to be stimulated, we would expect to see rapid depolarization of the myocardial cells. Depolarization of a cell's membrane is due to an influx of Na@ into the cell. This allows a negative resting membrane potential to become more positive. An influx of both Nao and Ca2e into a cell will allow the resting membrane potential to become depotarized. Choice A looks like a good answer at the moment. What would happen if the Ko channels were to open up? Potassium would leave the cell, making the resting membrane potential more negative. The cell would have a higher threshold potential and would not be depolarized as readily. We can eliminate choice C. The correct choice is A. t1

decreased arterial pressure. The key word here is compensatory. How does the body handle a The first thing that must be done is to prevent more loss of blood. Constriction of blood vessels would hemorrhage? help, whilh means that anlncreased sympathetic discharge to the arterioles (choice A) is true. We still want the heart to pump blood. Therefore, we do not want to inhibit its action. We can remove inhibitory signals to the heart by decreasing the amount of parasympathetic discharge at the heart (choice B). After a blood loss the body still nleds to maintain a reasonable cardiaCoutput. Therefore, we would expect to see an increase in the cardiac output (choice C). What we would not expect to see is a decreased arterial pressure (choice D). Why? A decreased urt".iul pressure is the RESULT of a hemorrhage. We want the COMPENSATION, which would be to increase the arterial pressure. We can reason this out as outlined below.

D is correct,

When a hemorrhage takes place there is a blood loss and a decrease in the blood volume. The venous pressure, venous return, and-atrial prirrur" will all decrease accordingly. The volume of blood ejected by a ventricle in the heart (i.e., the stroke volume) during one heartbeat is decreased as a decreased arterial blood pressure.

141

well. This leads to a decreased cardiac output and

The BerkeleY Review

Specializing in MCAT PreParation

Biology

Heart & Lungs

Section II Answers

Located in the upper neck region are small arteries which branch off of the common carotid arteries. The common carotid arteries stem from the aortic arch of the aorta, the main vessel that leaves the left ventricie of the heart. At the fork where the small arteries leave the carotids is a region called the carotid sinus. The caroticl sinus contains nerve endings which are sensitive to the size of those smail arteries. The carotid sinus acts as a baroreceptor, sensing the pressure differences within the small arteries. There are also baroreceptors in the aortic arch.

lf

there is a decrease in arterial blood pressure, baroreceptors will sense this and compensate accordingly. As the blood pressure drops there is less expansion of^the the arterial walls and a decreased discharge of'th" n".u" endings at the baroreceptors. The result is an increased.sympathetic discharge ro the arterioles"lvlins and heart) ancl a decreased parasympathetic discharge to the heart (becaur" *" do not want to inhibit its lunction). The parasympathetic and sympathetic divisions will act to increase heart rate and therefore increase cardiac output.

Sympathetic discharge to the arterioles and veins will constrict those vessels. Constriction of the arrerioles will lead to an increase in the total peripheral resistance of that system, thereby increasing arteriai p.Lrrr.". This is the end resuit of the compensatory mechanism. Constriction of the veins will lead to a"n increasi in ttre venous pressure, venous return' and cardiac output. An increased cardiac output leads to an increase in the arterial p.".rrr". e.gain, this is the end result of the compensatory mechanism. The correct choice is f).

22.

C is correct, arterioles. Arterioles have strong muscular walls. When stimulated by the sympathetic nervous system, they constrict. Metabolic products at the level of the arterioles, such as CO2, He, and 02, regulate the degree of constriction. The correct choice is C.

23.

C is correct, III. It is important to pay attention to the labeis on the x-axis and y-axis. If we look at posilion I on the graph we note that the total blood volume is quite low, even though the presiure is qulte high. Where is blood experiencing the highest pressure? As it is being ejected from the left ientricie. The pressure is going to decrease as the blood reaches the arteries and arterioles. once the blood reaches the capillaries and the venules, the pressure of the blood will be approaching an even lower value as inrticated by position II. As the blood begins to enter rhe veins, the pressure continues to decrease. As blood enters the right atiium, it is coming frornitre vena cava. This is represented by position III. Note that once the blood leaves the vena cava and enterithe right atrium, it will then flow into the right ventricle. Contraction of the right ventricle increases the blood pressure"and thus increases the velocity of the blood leaving the right ventricle as it flows into the pulmonary arteiy. Biood from the pulmonary artery will enter the lungs where the pressure drops to a rather low value. The iorreci choice is C.

24.

IV. Consider the x-axis (total blood volume) and the y-axis (area). The total area of the arteries will be small (point I), and it will calry a small amount of blood volume. The total area of the venules and veins will be the largest (point II), and it will carry the largest blood volume. As we move from the veins ro the vena cava and to the right atrium, right ventricle, and pulmonary artery, the total blooci volume wilt again u" rn;nrrrrut while the area wiil also be minimal (point III). Ilowever, as the blood leaves the pulmonary artery and enters the capillaries of the lungs, the total area of the vessels begins to increase as does the total bloodvolume lpoint iv1. Not" that the total blood volume in the lungs will be less than the total blood volume in the veins simpiy because (a) there are 11ore D is correct,

veins in the systemic circulation than in the pulmonary circuiation, and (b) the veins are quite distensible, The correct choice is D.

1<

C is correct, III. If we look at position I on the graph, we note that the total blood volume is quite low even though the velocity is quite high. Where is blood experiencing the highest velocity? As it is being-ejected from the left ventricle. The velocity is going to decrease as the blood reaches the arteries and arterioles. dn"" the blood reaches the capillaries and the venules the velocity of the biood will be at its minimum vaiue as indicated by position II. Blood then begins to enter the veins and the velocity begins to increase again due to muscular contractio; squeezing the veins to help return the blood to the right atrium. As blood enters the right atrium it is coming from ih" uenl cava' This is represented by position III. Note that once the blood leaves the vena cava and enters ihe right atrium, it will then flow into the right ventricle. Contraction of the right ventricle increases the blood pr"rru.:" and thus increases the velocity of the blood leaving the right ventricle as it flows into the pulmonary artery. Thus position IV will be the velocity of the blood as it leaves the pulmonary artery and enrers the capillary system of the iungs. The correct choice is C.

26"

A is correct, viscosity of blood. Laminar blood flow is blood which flows at a steady rate through a blood vessel, Blood flow is said to be streamlined. Turbuient flow involves blood which is not flowing in a steaiy stream through

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142


Biology

Ileart

Er

Lungs

Section II Answers

the blood vessel. Instead, turbulent flow creates eddy currents. These currents result from the blood passing over a rough surface in the vessel or.encountering obstructions in the vessel such as a bifurcation in the path of blo'od flow (i.e., a splitting of the vessel itself). Turbulent flow, measured by the Reynold's number (Rs), tends to increase as the velocity of the blood increases and as the diameter of the vessel increases. Turbulent flow is inversely proportional to the viscosity of blood. By using the equation R" = (v)(d)/(n/p), where v velocity, d diameter of = = vessel, n = viscosity, and p = density, we find that as the Reynold's number increases, the turbulence of blood flow can increase. Usually turbulence will occur when the Reynold's number increases above 2000. The correct choice is A)1

C is correct, systemic arteries. Arterial walls are stronger than the walls of veins. Blood vessel diameter increases as the internal pressure within the blood vessel increases. This is because the vessels are distensible. Because arterial walls are much stronger than the walls of veins, the arteriai walls are about 6 to l0 times less distensible than the walls of veins. What this is saying is that for an increase in pressure there will be 6 to 10 times more blood in a vein as there will be in a comparabie artery. Even though pulmonary arteries still have stronger walls than those of the pulmonary veins, pulmonary arteries are under less pressure (about 1/6 less) than systemi arteries. It turns out that pulmonary arteries have distensibilities about 1/2 that of veins. Even though the pulmonary arteries are still distensible, they are not as distensible as the pulmonary-veins or systemic veins. The vascular distensibility would be least for the systemic arteries. The correct choice is C.

28.

C is correct, The capillaries contain between 57o to l\Vo of the total blood volume. Capillaries are about I mm in length and, on the average, most celis are about 0.01 cm away from any given capillary. This allows for a highly efficient form of diffusion of 02 from a red blood cell within a capillary io th. su.rounding cells. r00

480

The value between the ttvo dashed lines represents the

E

E60

total blood volurne within

4)

=_

40

a)

&20 0

20

40

60

80

100

Total Blood Volume (%)

In order for this diffusion to be optimal, capillaries must have not only a very small diameter (about g pm, which is

just large enough for the passage of a red blood cell), but they must aiso have a very thin wall (about 0.5 pm). The thickness of the capillary wall is just one endothelial cell layer thick. Even thoughihese values were not presented in the passage, it shouid be realized that gas exchange (for alt practical purposes) occurs ar the level of the capillary. make sense for them to have the thinnest walls and the smallest diameter, to a1low for maximal and efficient diffusion. We can eliminate choices A and B.

It would

Capillaries in the systemic system contain about 5Vo of the total blood voiume, while capillaries in the pulmonary system contain about 4Vo of the entire blood volume. This adds up to between 5% aid l07o of the iotal btoob volume' Even though this value is not mentioned in the text of the passage, it can be obtained from the first graph in Figure 1 of the passage. In Figure 1 note that the dashed lines border the value of 20Vo. At first you tiight U" inclined to pick choice D, which states that total blood volume in the capiliaries is about 20Vo. But think aboui what the x-axis is telling us. If we move out to l00vo total blood volume, does that mean that at that point in the curve a vessel contains l007o of the total blood volume? No, it does not. What the area between the borders of the dashed lines is telling us is that the total blood volume in the capiliaries is represented by what we see between those two dashed iines. In other words, the total blood volume in the bapillaries is less than lovo of the total blood volume in the system. The correct choice is c.

29- B is correct, pieural

pressure

will

become more negative. The two lungs

in the chest cavity are individually

surrounded by a thin layer of cells called pleura. The visceral pleura attaches to the lung while the parietal pleura

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Biology

Ileart

6r Lungs

Section tr Answers

attaches to the diaphragm and interior region of the thoracic cage. Belween the two pleural surfaces is a vely thin layer of fluid called the intrapleural fluid. The volume of the intrapleural fluid is constant. Pressures in the respiratory system are listed relative to atmospheric pressure, which is 760 mmHg at sea level. Between inspiration and expiration the alveolar pressure is the same as atmospheric pressure. Since pressures are measured relative to atmospheric pressure, the alveolar pressure is said to be 0 mmHg. As stated in the passage the intrapleural pressure is negative relative to atmospheric pressure (about -4 mmHg).

As an individual inspires the thoracic cavity begins to enlarge. The rib cage moves upward and outward. This is due to contraction of the external intercostal muscles. The diaphragm descends towards the abdomen. Both actions move the wall of the thoracic cage away from the srirface of the lungs. However. the parietal pleura is in contact with the intrapleural fluid which is in turn in contact with the visceral pleura. This intrapleural space increases in volume by a very smal1 amounl. As the volume of the intrapleural space increases (ever so siightly), the pressure within that space must decrease (Boyle's Law). In other words, the pressure within the intrapleural cavity will become more negative (relative to atmospheric pressure). The same reasoning applies to the iungs. As the thoracic cage moves away from the wal1s of the lungs, the volume of the iungs increases. As the volume of the lungs increases, the pressure within the lungs must decrease. In other words, the alveolar pressure becomes subatmospheric and air flows from the outside into the lungs. The correct choice is B. 30.

B is correct, parietal pieura. The parietal pleura and/or the visceral pleura are the most important layers of the thoracic wali for maintaining the pressure component of the thoracic cage. A puncture of the intercostai muscles, ribs, or heari is not going to lead to a collapsed lung because these structures are outside the pleura, The correct choice is B.

31.

B is correct, lI only (diaphragm). The diaphragm is the primary muscle involved in normal respiration. The intercostal and abdominal muscies are excessory muscles and aid in respiration during times of forced expiration and inhalation. The correct choice is B.

32-

A is correct, high pCO2 in the capillaries; low 02 in the capillaries. As blood returns to the capillaries of the lungs liom the systemic circulation via the pulmonary arteries, it will have a low pO2 (deoxygenated blood) and a high pCO2. The partial pressures of these gases in the aiveoii of the lungs will be reversed. This sets up a concentration gradient that will allow oxygen to flow liom the alveoli to the capillaries and carbon dioxide to flow from the capillaries to the alveoli. The correct choice is A.

33.

A is coruect,

34.

B is correct, a rise in temperature requires a greater partial pressure of 02 to bind a given amount of 02. The graph shows that as the temperature increases, the curve moves to the right. The shift to the right indicates that a higher partial pressure of oxygen is needed to reach half saturation of hemoglobin with oxygen. The correct choice is B.

?{

C is correct, arrhythmias. Arrhythmia means an irregular heart beat or having an irregular rhythm. Blood pressure is measured with a blood pressure cuff. Choice A is incorrect. Cardiac output is a measure of the volume of blood pumped with each heartbeat. Choice B is incorrect. Hematocrit is a measure of the ToRBCs in whole biood. It is used for checking on anemia. Choice D is incorect. The correct choice is C.

36.

B is correct, QRS. P represents the depolarization of the atria. During depolarization, they are contracting. Choice A is incorrect. QRS represents spread of depolarization through the ventricles. During this depolarization, they are contracting. Chbice B is corect. During the T wave, the ventricles are relaxed and repolarizing. Choice C is incorrect. There is no T-P phase, it is a fake answer. Choice D is incorrect. The correct choice is B.

37.

D is correct, death, if a normal rhythm is not introduced. Ventricular fibrillation means the ventricles are not pumping together. The cells are contracting in their own unique patterns, producing an unproductive twitching'

a rise in pH requires a iower partial pressure of 02 to bind a given amount of 02, The graph shows that as the pH rises, the curve shifts to the ieft. This shift indicates that less of a partial pressure of oxygen is nceded in order to reach half saturation of hemoglobin with oxygen. As the blood becomes more alkaline (due to hyperventilation), hemoglobin shows a tendency to retain more oxygen (at lower partial pressures of oxygen). The correct choice is A.

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Section II Answers

This can be fatal pretty quickly--within minutes. The individual is definitely affected, therefore choice C is inconect. If there are no heartbeats at all, choices A and B are incorrect. The correct choice is D. 38.

is correct, sinoatrial node. All the choices are involved in conduction of action poientials in the heart, but only the sinoatrial node is actually in the right atrium, as the question asks. The atrioventricular node is located on th! ,4.

septum between the atria and the ventricles. Choice B is incorrect. The bundle of His travels though the interventricular septum, and breaks into branches, forming the Purkinje fibers. Choices C and D are incorre"i. the correct choice is A. 39.

A is correct, tissue fluids contain ions that participate in conduction. Tissue fluids contain ions, such as sodium, potassium, bicarbonate, and chloride. Choices B and D are incorrect. Ions are required for conduction. That is, deionized watet'does not conduct electricity. Choice C is incorrect. The correct choice is A.

40.

B is correct, a marathon runner. Exercise conditions the heart, so that greater volumes can be pumped per beat. This means the heart rate can slow and still maintain the same cardiac output as an untrained p"rion. -Infa;B have faster heart rates than adults. Choice A is incorrect. Weight lifting does not provide aerobic conditioning to strengthen the heart. Choice C is incorrect. A sedentary adult is also less conditioned than a trained athlete. Choice D is incorrect. The correct choice is B.

41.

C is correct, plasma folate represents folate intake over the past day, while RBC folate represent folate intake over the past few months. As stated in the passage, folate is a water-soluble vitamin. It is excreted by the kidney, so it is required every day. The serum contains folate eaten in the diet on that day. Choices A and D are incorrect. RBCs live about 120 days, so their supply of folate represents the composite intake of folate over their lifetime of a few months, not a year. Choice B is incorrect. The correct choice is C.

42.

B is correct, centrifugation. Both extraction with an organic solvent and thin-layer chromatography would break the membrane of the RBC, mixing the plasma and the cellular contents. This would not be helpful. Choices A and C are incorrect. Dialysis tubing would contain the RBCs and probably the folate, too. This would not lead to a separation. Choice D is incorrect. Finally, centrifugation is the perfect way to separate the heavy cells from the plasma. Following a short spin (10 minutes at 3000 rpm) the cells would sink to the bottom of the centrifugation tube and the plasma would move to the top, allowing separation for independent analysis of the two components. The correct choice is B.

43.

A is correct, The subject's sample is added to a known amount of bacteria, incubated for a time period, and the turbidity of the medium is measured. In this assay, the growth of the folate-dependent bacteria provide the indication of the folate level present. The serum or RBC sampie is added to a known amount of the bacteria in a liquid medium, incubated, and the growth is measured by the turbidity (cloudiness) of the solution. The more growth, the more turbid the solution. No extra folate is added to confuse the results. This makes choice B incorrect. No further enzyme activity assay is required, the bacteria provide the enzymatic assay. And they do not produce folate. Choice C is incorrect on both these points. Choice D is trying to remind us of the Ames's test for mutagens. It is also incorrect. The correct answer is A.

JJ.

A is correct, they were consuming between 300 and 400 pg/day. Compare the initial levels with the final levels lor each dosage group. Although all the women started with nearly identical piasma and RBC folate levels in each group, the 200 and 300 groups's levels fell. This means they were consuming less than their usual levels of folate. This means choices C and D are incorrect. However, the 400 group had a rise in both plasma and RBC folate levels. This means the experiment provided more folate than their usual levels. Choice B is therefore incorrect. The correct answer is A.

-15.

D is correct, macrocytic anemia. The question tells you that the cells become too large before they are released from the bone marrow. The prefix "macro" refers to iarge, while "micro" refers to small. Choice A is incorrect. Choices B and C are made up words. The correct choice is D.

16.

B is correct, folate needs increase during pregnancy due to synthesis of many new cel1s. Since the experiment indicated 2A0 'p"glday was not adequate for nonpregnant women, and pregnant women are synthesizing more nucleotides for the fetus, it makes sense that folate needs are increased during pregnancy. Choice A is incorrect. The placenta does not synthesize folate. Choice C is incorrect. Foiate is a water-solubie vitamin, and is not stored in the liver. Choice D is incorrect. The correct choice is B.

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Biology 41.

fleart & Lungs

Section II Answers

D is correct, thymine' First, make a distinction.between purines and pyrimidines. purines have2rings, while pyrimidines have 1. Choices A islhe purine guanine while choice B is the purine adenine. This leaves us with the structures shown in choice c and D' Is there iny information the passage that will allow us to distinguish between these two nitrogenous bases? No. This questionis designed ; ;"i;;l; eLminare as many of the possible choices as we can. and then to make an educated guess at the correct onr*Jr. The correct choice is D.

of the blood following a hearrbear. when the first sound is heard, rhe arreriat S,::":::t"""::.:1"^:11lt^1:T_pj:-ttr'" *: p.",,,." lT:::"î:^ri::.T"::"Tl"r, "j 1;; tJi;;;;;",rr."fi;,"', ,n" JilHJ#'il"',:ff"':the artery reaches.

tr;',; ##;;;#;r-

n:*. o{ brood is interrupied compressing the arm at all.'"Tli"' Choice B is incorrect. The heartbeat ' not represented by taking blood pressure. presiure is ireasured. choice D is incorrect. The correct choice i, c.tt

:*:.".^*,:'^t:::'::1

49.

Il:

#i,ii!'J*ii!'""j;

A is correct, the reading will be falsely high. If a cushion of fat absorbs some of the pressure from the cufT, then the cuff must be inflated more to co-ptess the artery compietely. This would mean that the reading of pressure needed to close the artery is falsely high. This would leid to un influt"o blood pressur".;i;;.- Ti. correct choice is A.

Discorrect'afluidmovesasaseriesofindividual

layers,eachofadifferentvelocity.

Ifyouhavenotseenthisin physics' try to figure out the answer based on vocabuiary. A laminated countertop *"riril-or a layer of plastic or other material glued onto a different material. Laminar."r".r ,o iuf"r.. gtirninur" .hoi"" A. since the question asks about layers (laminar) eliminate choice c as well, although urorn, ao have their own ranclom activity. To get the correct answer' think of water moving in a stream. It is siowest closest to lhe bank, where there is a lot of friction with dirt and rocks' It is fastest in thi center" Each layer moves with a differ"rt choice B is incorrect. The correct choice is D. "Lro.rt). 51.

c is correct, 211' Read the tabie ofblood pressures The systolic pressure in a 46-year-old woman is 131 at the leveloftheheart. Addinthegravitvfactorli00l0.l:80)ri;t;;ler131+80=zir. riloureadthecolumnfor men' 210 is your answer and is incorrect. choice B is incorrect. ir you read the a;astotic column, 162 is your answer. choice A is incorect. choice D is just incorrect. The correct choice is c. c is correct' the location of the cuff does not matter. on Earth, gravity changes the reading fiom site to site on the body' In space, since there is no gravity, this change will not o.!ui. crroi.es A and B are incorrect. The cuff will function.f ust fine, as will the heart, in an anti-gravitf situation ctroice D is incorrect. The correct choice is

c.

B is correct, II only' Normaily, arterial blood flow is laminar and smooth. It makes no sounds or vibrations. Turbulent flow is noisy, just iike in a stream. when the artery is Jgr,tiy by ,ir" ,rrf the flow becomes turbulent and noisy' Before the systolic sound is heard, no blloa "onstri"tJ is fio*ing tkough the cuff. choice I is incorrect, After the diastolic sound, flow is silent and therefore laminar. choice IIi is inJorrect. only during the interval between the systolic and the diastolic sound can noises be detected, so that flow must be turbulent. Choice II is correct. The correct choice is B. D is correct, g4' The calcuiation uses data from the table. A 27-year-old,man, on average, would have a systolic pressure of 125 and a diastolic pressure of 78. The difference bet*Len the two readings is 125 - 7g !1. The pulse = pressure is 47. MAP = 78 + (41/3) j8 + 16 94. The correct choice = is D. =

B is correct, av/aP' we. are looking for

the- best relationship for aortic compliance. Think about what "compliance" means' compiiance is the act of conforming. In this case, it is the conforming (stretch) of the aorta in response to successive injections of liquid. In the body biood is injected into the aorta from the heart which will cause a rise in pressure' The volume of the aorta will change, depending on ttr" p."rr,r...-irr" compliance of the aorta is the ratio of the change in volume in response to,a ciiangeîn pressure" T[is relationship is denoied on the graph by the slopes of the lines, as the y-axis represents the voluire uno th" *-u"ir ."p."r"nir-p."rrur". compliance is the change in volume in response to a change in pressure. The correct choice is B.

D is correct, inverseiy proportional to aortic compliance. We can arrive at this answer by the fbllowing logic: We are told from the passage that the fractional change (AD/D) is a reflecrion or itr" ;h;";""i; volume of rhe aorra during left ventricular contraction. Therefore, one can say that (AD/D) can be repla8ed by AV. Making this Copl,right O by The Berkeley Review

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Biology

fleart

6r Lungs

Section II Answers

substitution in the equation for the elastic modulus, one sees that Ep = AP/AV. This is, of cours,e the inverse of compliance, so the elastic modulus is inversely related to the compliance. As the compliance increases, the eiastic modulus decreases. Looking at Figure 2, one sees that with age, the elastic modulus increases. This should clue one to the idea that with age, compliance decreases. Nevertheless, the relationship is one of an inverse proportion. The correct choice is D.

5/.

D is correct, Curve E. In order to answer this question correctly, one must have made the conclusion that compliance decreases with age. From Figure 2 and the ideas in the passage, we can come to this conclusion. We have already established that in Figure 1; the compliance will be represented by the slopes of the lines. Each line represents a par:ticular age group. Let us look at the normal pressure range of 80-120 mmHg. It is clear that Curve A has the highest slope, which represents the highest compliance. We are looking for the oldest age group, which should have the lowest compliance (according to our conciusion). This is represented by the curve with the lowest slope which is best represented by curve E. The correct choice is D.

58.

B is correct, later in systole. This question requires a careful reading of the passage. First, think about when the peak arterial pressure is going to occur. Is it going to occur during diastole or systole? The heart is contracting during systole, and that is when we shall see the most force and hence the peak arterial pressure. Therefbre, *" .un eliminate the answers which claim that the peak arteriai pressure will occur during diastole. Then the question becomes when will we see the peak arterial pressure. The last sentence in the passage tells us that as compliance decreases, the heart is unable to eject its stroke volume as rapidly when compared to a more compliant system. If the heart is slowed at ejecting its stroke volume. it can be logically concluded that it will take progressiveiy longer for the peak arterial pressure to occur. The correct choice is B.

59.

D is correct, least at very high and low pressures and greatest over the usual range of pressure variations. We can gatl-rer this information from looking at Figure i. We shouid already be aware that Curve A represents the youngest age group. We simply need to look at siopes. It is ciear that at certain points in the curve, the compiiance of the youngest individuals is smaller than those of older individuals (this does not invalidate our general conclusion that compliance decreases with age). On the other hand, it is clear that for the youngest individuals, their compiiance is higher for most of the range of pressures. Therefore, when compared to the rest of the age groups, one cannot say that the compliance for the youngest individuals is either greater or less than for the entire range of pressures. Oni

can then eliminate choices A and B. We then look at choices C and D. Curve A is sigmoidal. The ilopes are least at the extreme pressures, and greatest during the normal range of pressures. This translates into the compliance for the youngest individuals is least at very high and low pressures, and greatest during the normal range ofpr"ssu."s. The correct choice is D. 60,

C is correct, decreases with age, a manifestation of increased arterial rigidity. One should easily be able to eliminate choices A and B. These claim that compliance increases with age, and Figure 1 clearly does not support this clairn. Therefore, we now look for a more likely manifestation of the decreased compliance. If the aorta does not change its volume as well in response to a given change in pressure (less compliant), the arterial system is more likely to be rigid. The more rigid the aorta, the less it can conform, or comply with a given change in pressure. The increased rigidity is a result caused by progressive changes in the contents of collagen and elastin in the arterial walls. The increased rigidity with age results in a decreased compliance with age. The correct choice is C.

61.

C is correct, tricuspid valve. This question draws not on information from the passage, but from our own knowledge of the anatomy of the heart. There are vaives between the atria and the ventricles known as the atrioventricular valves. The valve between the right atrium and the right ventricles has three cusps, making it the tricuspid valve. When the right ventricles begins to contract, the valve between the two chambers closes and prohibits the back flow of blood into the right atrium. The correct choice is C.

62.

B is correct, Ko. We are looking at the permeability of potassium. As the cardiac action potential begins, the permeability of potassium decreases as potassium channels ciose. This is the key to answering this question. Decreasing the permeability of potassium while increasing the permeability of sodium will result in depolarization of the membrane. During the plateau of the action potential, the potassium permeability stays below the resting value. Ultimately, repolarization does occur when the petmeabilities of calcium, sodium, and potassium all return to their original state. As a note, Graph A represents the permeability of sodium, while Graph B represents the permeability of calcium. Because Graph C dips right as the depolarization begins, one can safely conclude that Graph C represents the permeability of potassium. The correct choice is B.

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Section II Answers

63.

A is correct' Region A' This region of the action potential is unlike a normar resting potential. Note that the resting potential of the SA node cell is not steady but instead manifests u .ro* depolarizat[". rrrlr gradual depolarization is known as a pacemaker potential. The pacemaker potential u.i"gr trr" membrane potential to threshold, at which point an action potential occurs. After the peak of irr" u.tion p"[ntial, the potentiai repolarizes, and the gradual repolarization begins over again. Thus, the capacity forupontu'n"ou, rhythmic setr excitition is best manifested by Region A of the action potential. The correct ctroice is a.

64.

c is correct' the flow of positive ions out of the cell equaling the flow of positive ions into the cell. In myocardial cells' the original membrane depolarization causes uottage-gaiea calcium c^hannels in the plasma membrane to open. This results in a flow of calcium ions down their eiecfto;rrJ-i"uig.uaient into the cell. i""uur" there is a delay in their opening, these are called slow channels- The-flow oiporiii"L calcium ions into the cell, along with some sodium also entering through the slow channels just balancesirr" Ro* ofpositive potassium charge out of the cell. This keeps the membrane depolarized at the plateau value. In other words, the flow of positive ions out of the cell equals the flow of positive ions into the cell. Do not be fooied uy ttr" other answerr. wt ii" they may be true for certain aspects of the cardiac-,action potential we are rooting io; the best ."u*n ,o the fact that the membrane potential is essentially not changing throughout trr" piit"* region. The correct "*iiuin choice is c. c is correct' gap -iunctions between cells. The wave of depolarization spreads from contractile cell to contractile cell via gap junctions' Recall that gap junctions are channeli which allow cells to share cytoplasm. In this way, the current responsible for depolarization in one cell can spread to another ceil in a u".y qui"t'*anner. There are no neurotransmitters involved, which greatly reduces the amount of time needed to cause contraction over the entire chamber. The correct choice

65.

is C.

D is correct' Graph c is the result of vagus nerve stimulation. we know from the diagram that all three graphs are pacemaker potentials' The difference between the three graphs ouuiou.rry lies in the slope at the beginning part of the potential' In Graph B, we see a large slope. In other rriords, we.each threshold very quickiy. on the other hand, Graph C reaches threshold after quite-some time. one could .onclude that Graph n *ouiJ-ôst likely represent sympathetic stimulation, while Graph c represents parasympathetic stimulation.' ttereiore c.upt C most likely represents stimulation of the vagus nerve (a major paiasympathetic iiber). The correct choice is D.

66.

67.

c is correct, favor movement of fluid from vessel toînterstitial space. one needs to make the fbllowing conclusion from the information in,the passage. There are two forces which favor movement of fluid out of the vessel. These forces rvrvuJ 4re are the urc lrrlr'luaplllary-nyorostatlc intracapillary hydrostatic pressure and the interstitial fluid oncoric nreqsrrrc ,r.hara .*- ?,-,^ r^-^^^ oncotic pressure. There are two forces which oppose movement of fluid out of the vessel. These are ,h;l;;;, protein oncotic pressure and the interstitial fluid hydrostatic pressure. with this understanding, we ,""'rhui un increase in the intracapillary hydrostatic pressure would most iikely favor movement of flui-d'from"un the vesselio the interst;t;ot ,piri. such acrion will not affect the concentration of osmotically active moleculls, and therefore we can eliminate choice A. Also, such action will not aff-ect the

venous resistance, and therefore we can eliminate choice D. From the passage we can euminate choice B because we are told a decrease in the arterial pressure r"rutt, in a decrease in th" intru.upillary pressure. The correct choice is C. 68.

69.

A is correct, P" - Pi > P.. According to the figure in

the passage, filtration is the movement of

fluid out

of the vessel' In addition, *"^ told from the passage that the foice oifiltration comes from the difference T between the P. and P1 ' The force of filtration is thus P. - P1. According to the diagram, the p" always has a positive value and is thus larger than the P1 ' If P1 is a negative number, then the force of filtration will actually become larger than the capillary hydrostatic pressure. This is indicated by the equation in choice A. None of the other equations follows this conciusion, and all of them can be eliminated. The correct choice is A. B is correct, result in a reduction of capillary hydrostatic pressure. we are reducing the diameter of a precapillary vessel' This will increase the resistance of that vessel so biood will not flow througlithat vessel as well. The result of this increased resistance is that the capillary will not receive as much blood volume. In turn, this will result in a decrease in the capillary hydrostatic (blood) pressure. A reduction in blood volume will decrease the amount of force/ unit area placed on the walls of the capillary. A reduction in P. will not favor movement of fluid from the

vessel to the interstitial space, eliminating choice A. The result of this action will not be equivalent to an increase in venous resistance, but.is_exactly the opposite. Think about it! Finally, resistance-cffies will not affect the concentration of osmotically active particles in the interstitial space, so choice D can be eliminated. The correct choice is B.

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Biology

Ileart

6r Lungs

Section II Answers

A is correct, Fluid movement = k[(Pc + It) - (Pi + np)l. This question requires us to put together knowledge gathered in the passage into an equation form. It brings us back to our discussion of forces. The forces moving water out of the vessel are the capiilary hydrostatic pressure and the interstitial oncotic pressure. Therefore, we

70.

should see these two forces added together. This eliminates choices C and D. The two opposing forces, P1 and the plasma protein oncotic pressure, should be added together because they move fluid in the same direction. This eliminates choice B and leaves us with choice A as the correct answer. We see the opposing forces subtracted from each other will result in fluid movement. If the aigebraic sum is positive, filtration will occur, and if the sum is negative, absorption will occur. The correct choice is A.

-1.

C is correct, carries electrical charges at blood pH which attracts various electrolytes. The answer can be arrived at through a process of elimination. We are told that albumin exerts a greater osmotic force than can be accounted fbr solely on the basis of the number of molecules dissolved in the plasma. Therefore, there is something special about albumin. It is not very likely that it could be replaced by an inert molecule and have no effect on the plasma protein oncotic pressure. Therefore, we can eliminate choice A. Consider choice B. How many proteins do you know of that dissolve into more than one protein? Proteins are not simple ionic salts. We are talking about a large protein. It ri i1l not dissolve into more than one molecule, making choice B an unlikely choice (be aware that this would increase its osmotic activity, if it happened). Choice D provides no explanation for the claim in the question. In iact. choice D would allow us to predict a smaller plasma protein oncotic force. Therefbre, the best answer is choice C. Nothing in the passage tells us that it does carry a charge, but that is not the point. One should be able to reason ihat this is the best answer given our knowledge. This is indeed the case. The protein carries negative charges, \'' hich attract primarily sodium ions. In addition, chloride ions attach themselves to the protein, which attract even :nore sodium ions. The additional electrolytes provide the increased osmotic strength not accounted for by the .',rncentration of plasma albumin. The correct choice is C.

D is correct, returns to the venous circulation via the lymphatic system. Since the remaining fluid is picked up by ::e lvmphatic system, we can eliminate choices A and B. The fluid will not remain in the interstitial space. If it did, 3lema would result. Furthermore, the fluid in the interstitial space is certainly not going to increase the r:racapillary hydrostatic pressure. Having decided that the lymphatic system is going to pick up the fluid, one has t: ask where is the fluid being returned to. Nothing in the passage tells us this, so one must draw on their own o;::ou ledge. The lymphatic system returns fluid into the venous system, not the arterial system. In fact, the lymph :.:rllaries drain into larger vessels that finally enter the right and left subclavian veins at theirjunctions with the ::spective internal jugular veins. The correct choice is D.

B is correct, increased left atrial pressure. We are looking for a situation which causes an increased volume of :.: rJ tn the pulmonary capillaries. The increase in volume will cause the distension of the vessels. Increased left .-:-.1 pressure indicates that blood is filiing the left atrium, but it is not being injected into the left ventricle. This :'.:-. 'oe due to left heart failure. If blood is not being ejected from the pulmonary atrium, the pressure will increase : ': llood from the pulmonary capillaries will not be able to move into the atrium. This will cause congestion and ", .-. ir.-rease the volume in the pulmonary capillaries, causing distension. All of the other answers act to keep blood : '. ::, rlom or out of the puimonary capiilaries. The correct choice is B. D is corrrect, i.09

PiVr -

Tr

L.

According to the ideal gas law, which states that PV= nRT, we find that:

PzVz

and thar:

Tz

vz=v';

H(|i)

: P- = 1 and T2lT1 (in Kelvin) > 1. The only possible answer choice is D, because we are iooking for an answer i:.:i:: than one. The answer then becomes (lL)(2981273) = 1.09 L. The correct choice is D. il- is correct. the venous blood from bronchial venules and heart vessels contaminate the pulmonary venous outflow. 02 partial pressure. From the passage, we know that the "re iooking for an explanation to the decrease in :: .:i:ra1 \'enous flow along with vessels from the heart join with pulmonary venuies. In effect, by increasing the -rrr3 eii blood without changing the amount of oxygen, we are decreasing the concentration of oxygen in blood.

',;,':

ll":r correct choice is C.

t_t'-' 'r

r

lli-:

Berkeiey Review

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Biology 76.

77.

Ileart & Lungs

Section II Answers

C is correct, 150 mmHg. The question asks for the partial pressure of oxygen in dry inspired air at the trachea. First, we need to know the fiaction of oxygen in the air. We get this from the passage, and it is 21%. Note that in the alveolus the fraction of oxygen dips to l4.3Vo, but in the trachea it still is 217o. Next, we must get the total pressure of the dry air. The total pressure of the air is 760 mmHg, but 47 mmHg is due to the vapor pressure of

i{2O. The total dry air presssure is 760 - 4l = 713 mmHg. The partial pressure of oxygen in anhydrous air becomes 0.21 x113 mmHg = 150 mmHg' The correct choice is C. A is correct, 102 mmHg. The total anhydrous pressure is 760 - 47 mmHg. The only difference is the fraction of oxygen found in the air. In alveolar gas, the value is 0.143 (from the passage). To find the partial pressure of oxygen'wesimplymultiply0.l43xTl3mmHg=l02mmHg.ThecorrectchoiceisA.

78.

Ciscorrect. l97mlO2/L. Wearelookingforthetotal concentrationof oxygeninthesystemicarterial blood. This includes oxygen found in hemoglobin and oxygen dissolved in the blood itself. Let us start with the latter. We know this vaiue straight from the passage. The value is 3 ml O2lL. Now, let us determine the oxygen content for hemoglobin. From the passage, we see that hemoglobin binds 1.34 ml 02lg. The concentration of hemoglobin is 150 gI-. We need to multiply these values. We get 1.34 ml O2lgx 150 g/L = 201 ml O2lL. The question states that only 91% of hemoglobin is saturated with oxygen, so let us do an approximate calculation. We multiply 0.97 x 200 ml O2lI-. This gives us 194 ml O2/L. Adding the 3.0 ml O2lL that are dissolved in the plasma, we obtain 197 ml O2lL. The correct choice is C.

79.

B is correct, after aspirin treatment, endothelial cells produce new CO, but platelets do not. From the passage, we learned that the turnover time of platelets is four days. Since the platelets in Figure I have the most active CO after 4 days, this means they cannot produce more when aspirin blocks CO. This eliminates choices A and C. However. endotheiial cells show a rapid rise in CO after the aspirin dose. These cells simply produce more CO enzyme since they have the machinery to do so. Choice D is incorrect. The correct choice is B.

80. 81.

D is correct, four degrees of unsaturation. Each double bond is counted is counted as one degree of unsaturation. The correct choice is D.

as one degree

of unsaturation, and each ring

A is correct, acetylated serine residue. The passage tells us that a serine is acetylated to inactivate CO. Choices B and D indicate a cysteine residue, and are incorrect. Choice C is a phosphorylated serine residue, and is incorrect. The correct choice is A.

82.

B is correct, I and II only. Too much aspirin would lead to poor clotting. This means that minor accidents could more easily lead to bruising, and bleeding time would be prolonged. Choices I and II are correct. A prolonged bleeding time means that the wound is slow to clot, not rapid. Choice III is incorrect. The correct choice is B.

83.

D is correct, 5, 8 1 1, l4-eicosatetraenoic acid. Start counting from the carboxyl end. There are twenty carbons, which requires the prefix "eicosa". There are 4 double bonds and the carboxyl group, giving the "tetraenoic acid" part. Choices A and B are incorrect. The carbons are counted from the carboxyl end, giving 5, 8, I I, and 14 as the positions for the double bonds. The correct choice is D.

84.

A is correct, the dosing regimen keeps platelet aggregation low. The passage said nothing about the prolif'eration (growth) of endothelial ceils. Eliminate choices C and D right away. To prevent heart attacks in people with nirrowed arteries fiom cardiovascular disease, it is often desirable to keep the blood from excessive clotting. This means that platelet aggregation (clumping together) should be low. Choice B is incorrect. The correct choice is A.

85.

C is colect, aspirin should not be used in late pregnancy. Since aspirin inhibits the synthesis of prostaglandins by inactivating the first enzymatic step, it could possibly interfere with the normal prostaglandin cycles involved in labor and delivery. This would probably not stimulate labor. Choices A and B are incorrect. Birth defects are initiated in the first few weeks of pregnancy. Choice D is incorrect. The correct choice is C.

86.

C is correct, a decrease in HbF and an increase in HbS with age. HbA is normal, nonsickling hemoglobin. An increase in HbA would not promote sickling of cells. Choice A is incorrect. The maternal hemoglobin is separate from the fetal hemoglobin, and they do not mix. Choice B is incorrect. There are no antibodies to HbS, or that would lead to an euen bigger obstruction problem. Choice D is incorrect. The correct choice is C.


150


Biology 87.

tleart & Lungs

Section II Answers

A is correct, all cells contain both HbS and HbA. The hemoglobin in mixed in each blood ceil. There are no exclusive HbS-containing cells or HbA-containing cells. Choice B is incorrect. In a heterozygote, both HbS ancl HbA are present. Choices C and D are incorrect. The correct choice is A.

88.

C is correct, HbF promotes a higher oxygenation state inside the cell. Adult (HbA) and I'etal (HbF) hemoglobins do not bind to each other. This is seen in question 1, in which sickle cell anemia is diagnosed only after f'etafHb levels fail. We also know that adults who have only HbA do not sickie. Choices A and D are incorrect. The cells sickle under lower oxygen conditions. This means that a beneficial action is to raise oxygen levels inside the cell, so that the HbS does not fall out of solution and cause sickling. Since HbF is beneficial, it rnust promote a higher oxygen state inside the cell. Also, we know from our reading that HbF binds oxygen more aviclly then HbA. This is how the oxygen gets transfered from the mother to the fetus. Choice B is incorrect. The correct choice is C.

89.

D is correct, sodium metabisulfate would deoxygenate hemoglobin from red blood cells and the sickling pattern would be noted in atfected individuals. Forget about the rusty color answers. We are interested in the sickling phenomenon itself. Choices A and C are incorrect. The conversion of sodium metabisuifate to sodium persulfatE involves the removal of oxygen from the other substrate. Since HbS crystaliizes and causes sickling under lowered oxygen tension, this removal of oxygen would allow cells to sickle if the individuai has the sickle cell trait. Hemoglobin would not be oxygenated under these conditions. Choice B is incorrect. The correct choice is D.

A is correct, decreased delivery of oxygen to

tl.re tissues compared to other adults. HbF binds oxygen more tightly and therefore reieases less oxygen to the tissues. This would be a problem in disease states in *t-riil Unp is really high, such as the thalassemias. Choice B is inconect. This question has nothing to do with HbS, but rather it aski about HbF. HbF, as is stated in the passage, does not sickle. Eliminate choices C and D. The correct choice is A.

91.

C is correct, I and II only. A single base pair is changed to produce the HbS product. Since the DNA diff'ers at least in this spot, specific restriction enzymes can used to cleave the DNA to produce a unique pattern of fraginents. When fragments are separated by gel electrophoresis, the pattern of homozygotes (both sickie cell afTected and *itAtype) and the heterozygotes can be easily distinguished. Choice I is correcr. The hemoglobin proteins will move differently during gel electrophoresis. The pattern of bands can also be used to distinguish the groups. Choice II is correct. Centrifugation is used to separated things of different masses or viscosities. The hemoglobin proteins are too similar to be separated this way for diagnosis. Choice III is incorrect. The correct choice is C.

q7

B is correct, I and Ii only. HbF does not bind 2,3-BPG. This is one feature that increases its oxygen saturation capacity. Choice I is correct. HbF does have a higher oxygen saturation at a given pO2 than HbA. This is a mechanism that ensures the fetus gets oxygen preferentially from the mother. Choice II is true. HbF contains two alpha and two gamma chains. Choice

III is incorrect. The correct choice is B.

C is correct, the arteriai pCO2 should be maintained at 40 mmHg. We want to assess the effects of a changing arterial pO2 on the ventilation rate. To do this, we shouid keep the pCO2 at a constant va1ue, This elin.rinates

A and B. The question becomes whether the pCO2 is maintained at 40 or at 46 mmHg. We are interested in the effects of the arteriaL pO2. where the pCO2 is 40 mmHg. Therefore, to obtain the most accurate results, we should keep constant the pCO2 at the value which can be found in the arterial circulation. The correct choice is C.

choices

B is correct, total amount of orygen transported is relatively unaffected. The question tells us that a drop of 30 mmHg from the normal resting value of arterial pO2 does not change the rate of ventiiation. Recall the oxygen dissociation curve, which tells of the percent saturation of hemoglobin with oxygen depending on the partial pressure of oxygen. Remember that the curve is sigmoidal, with the top reaching an asyn'rptotic value. One can descend fiom 100 mmHg of oxygen to 60 mmHg of oxygen and the percent of hemoglobin saturated with oxygen does not change dramatically. In other words, the amount of oxygen delivered to the body is relatively unaffected. The correct choice is B.

Qi

A is correct, an unchanged arterial pO2. Carbon monoxide has a very strong affinity for the iron on the heme of hemoglobin which is responsible for binding oxygen. For this reason, the number of hemoglobins able to transport oxygen is decreased. Does this decrease the arterial partial pressure of oxygen? No. The reason is that only the oxygen dissolved in the plasma contributes to the partial pressure. The oxygen on hemoglobin is bound and is thus not soluble in the fluid. For that reason, carbon monoxide will not affect the partial pressure of orygen. The correct choice is A.

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Biology

tleart

E(

Lungs

Section II Answers

96'

C is correct, resisted by reflexes regulating ventilation to a greater degree than are equivalent changes in arterial pO2. This question is asking us to look at the graphs and ask ourselvei the following question: Is o"ur body more sensitive to changes in the partial pressure of oxygen or to changes in carbon dioxidl? Looking at the curves, the answer is clearly changes in the partial pressure of carbon dioxide. We also know that the reflexes discussed in the passage are involved in resisting changes that.occur to the partial pressure, in hopes of restoring values to normal. For that reasonr changes in the arterial partial pressure of carbon dioxide are r.esisted by ieflexes regulating ventilation to a greater degree than are equivalent changes in arterial pO2. The correct choice is C.

97.

C is correct, hyperventilation is caused by increased neural output from the peripheral chemoreceptors. In a metabolic acidosis' we have an overproduction of protons from some metabolic ioniition, like excessive exercise. These extra protons will stimulatc both the central chemoreceptors and the peripheral chemoreceptors to increase their output, and cause a rise in the rate of ventilation. Therefore, we can elimlnatl choices A and B, which call 1or a hypoventilation' Now, while the rise in protons will stimulate both central and peripheral receptors over time, ask yourself which one,will be the primary receptor. Can protons simply .ro5 ou'". tie biood-brain barrier to have access to the central receptors? The answer is no, because they are a charged species. Therefore, we can deduce that it is the peripheral receptors which are primarily responsible for the risJin the rate of ventilation. The correct

choice is C.

98'

D is correct, arterial pCO2 increases. During exercise, it is clear that the cells will be using more oxygen and thus producing more carbon dioxide' For that reason, the levels of venous pCO2 will increase. Will this lead to a rise in the arterial partial pressure of carbon dioxide? The answer is no, because the raie of ventilation will increase. We know that we breathe "harder" when we are exercising. In other words, the excess carbon dioxide is released as a result of the higher rate of ventilation. The increase in the carbon dioxide production is equivalent to the increase in breathing rate. The result of this is that.the arterial partial pressure of carbon clioxide does not change, making D a false statement. The correct choice is D.

99'

A is correct' low pCo2 is permitting one to hold their breath, the exercise may lower the pO2 to levels which may induce unconsciousness. The swimmer's hyperventilation will result in a lowered partial pressure of carbon dioxicle, because they are eliminating a good deal of gas. During the race, that lowered level of carbon dioxicie will allow them to hold their breath. However, the exercise will lowir the ievel of oxygen. Recall the body is more sensitive to leveis of carbon dioxide relative to oxygen. For that reason, the body does" not know to breathe, but oxygen levels may be reaching dangerous lows. The level may become low enough to induce unconsciousness, as nol enough oxygen is reaching the brain. The correct choice is A.

100. B is correct,

an increased arteriai pCo2 and a decreased arterial po2. This cluestion is very straightforward. Durrng times of sleep, the body is not breathing as often. The end result of this is that we are bringing in less amount of oxygen' and eliminating lesser amounts of carbon dioxide. The question tells us that the body;s risage of oxygen per production of carbon dioxide is unchanged during this dormant period. Therefore, we should to"see an increased pCO2 and a decreased pO2. The correct choice is B. "^p""t

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ffiffiffiKwffiy Sec&Ewm

EEE

A" The Gastrointestinal T.ract

B. Gas&rep6m&es&€naaE

€ract, Se Ln

K€dsa€ys

l.

Nutrients & Digestion

The Kidneys

l.

Renal Function Renal Physiology

2. 5. Homeostatic Mechanisms Practice Passages & Answers

a^\

.-lji,ll ,a_a .,.,, )F7.1; *

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f,ffiKf,fuffiY R.n.v.I"E.kV* Specializing in &fCAT Preparation

Gastrointestinal Tract & Kidneys Top lO Section Goals Be familiar with the anatomy of the gastrointestinal tract. Know how food passes from themouth, down the esophagus, into the stomach, through the iniestinal svstem. and how it is eliminated as feces.

Understand the interactions between the various gastrointestinal secretions. The for.rr gastrointestinal peptide hormones of importance are gastrin, cholecystokinin, secretin, and

glucose-dependent insulinotropic peptide. Be aware of their functions.

@w

Be familiar with the process of digestion and the areas of nutrient absorption. Know what types of nutrients are broken down in the stomach and what types are degraded in the intestines. Understand how peristalsis is related to the movement of thyme in"the system. Be familiar with the anatomv of the kidneys. Understand how molecules are filtered at the glomerulus and which ones are t"uUtorU*a f.o,r.r tt * different parts of the nephron tubular system. " Be aware of the effect of the autonomic nervous

on the

Know how the kidney responds to sympatJrehic and parasympathetic stimu-lation. Know tire difference

between vasoconstriction and vasodilation at the jevel

"w

of the glomerular arierioles.

Understand how the renin-angiotensin-aldosterone system functions. This system stimulates the reabsorption of Nae in.the disial convoluted tubule and in the collccting duct. iJnderstand rvhy this is imptirtant to the body.

Be aware of the different types of transport systems that line the tubules. Understand the differences between passive reabsorption (no energy required) and acfive reabsorption (energy required) and generally where and what ii reabsorbed ilong'the tubules.

Understand how the kidneys respond to a decrease in arterial blood pressure. Be aware thai there is a short-term adjustment and a long-term ad justrnent for a decrease in arterial blood pressure. Understand how ihis works in terms of ihe glomerular filtration rate.

,w

Be familiar with diabetes mellitus and diabetes insipidus. key feature of diabetes insipidus is a vasopressin deficiency. Both lead to a^large urine loss,

,g?

Be familiar with some of the conditions brought about by renal failure. The two most life-threatening consequences of renal failure involve the retention of Ke, due to poor tubular secretion of Ke, ind mefabolic acidosis, due to improper tubular secretion of Ho.

Biology

Iiutrients

Gastrointestinal Tract & Kidney

Er


Digestion

-:: s examine the gastrointestinal system. Digestion is the process by which food :.:i is eaten is broken down into progressiveiy smaller particles and uitimately

::-.crbed by the intestinal tract. The gastrointestinai system includes the moutir :---j associated salivary glands, esophagus, stomach, small intestine, large -- -=strne, and certain aspects of the iiver and pancreas. In the average adult this -:-::. running from mouth to anus, is about 30 feet in length and is actually an . , :=:.sron of the external environment.

-- = ilod that we eat has a diverse composition. A typical -,-

meal might include

-;3n1qlg6ules such as starch in the form of bread or potatoes, cellulose in the ,:'f =-. a salad or other greens, protein in the form of meat or cheese, and fat in .--= :,'rm butter or ice cream. Each of these dietary categories needs to be - = :: a led before they can be absorbed inio the body.

::r::i"r and glycogen are both important

polysaccharides

in nature. starch,

. ;--rJ:eristic of plant cel1s, and glycogen, characteristic of animal cells, are - -:--.''zed within the digestive tract by enzymes called amylases.rJpon ' --:- -.'sis both polymers release the monosaccharide glucose. Cellulose is '-

'--

---

: - -r

-.

.

''

''iii-rin the cell wa11s of plants and is a polysaccharide consisting of glucose =-. rinked together. The enzyme cellulqse can hydrolyze cellutse lnto its ::.:*:nt glucose residues. See Figure 3-1. Amylase

-._:,jt L-)

Cellulase

UluCose

Cellulose An example of bacteriai symbiosis

'

- -..::oivsis of starch and celiulose.

- : -::ant to note that even though starch, glycogen, and cellulose all contajn '. ==rdues linked together to form iong polymers, we as humans cannot :' -: ' ;lose. The reason that we cannot digest cellulose is because the glucose - :: -r. ,e11ulose are linked together in a different configuration than the -- :=srdues in starch or giycogen. This special linkage can only be " -:--'-nlr.by the enzyme cellulase--an enzyme that we do not have. Why, :. -r;\-s eat grass? The only vertebrates that can utilize celluiose are the '': :rimais (1ike the cow and goat). The rumen of these vertebrates . ,': .--.-::oorganisms which manufacture and secrete the cellulase that can - --ie cellulose in the vegetation being consumed. This is an example of ,:-c ielationship between the host ruminant and the microorganisms ---:- ihe rumen. The microorganisms get a warm place to live and the .:.> energy in the form of digested celiulose. ::.entioned in previous sections, proteins are composed of amino proteases can hydrolyze proteins to their constituent :=sidues. Fat cells or adipocytes store triacylglycerols (also called a

::-.es called

The Berkeley Review

155


Biology



triglyceride but better known as just plain fat). Fats are hydrolyzed into fatty acids and glycerol by the enzyme lipase. See Figure 3-2.

proteins Protease ----)-

Amino Acids

Lipase patsr-:)

%l%:::1'

Figure 3-2 Enzymatic hydrolysis of proteins and fats.

Vitamins are needed in only small amounts in the diet yet they are essential in order to prevent certain nutritional deficiencies. [For example, the disease beriberi is due to a deficiency of vitamin 81 or thiamine.] As a group, vitamins participate in many different chemical reactions in the body. Inorganic minerals such as iron, potassium, calcium, and zinc are also important for proper development. Iron is the atom in the center of the heme group that carries oxygen in both hemoglobin and myoglobin, two proteins thal transport oxygen within the body.

All of the proteins that we are concerned with in the biological sciences are composed of a basic set of 20 different amino acids joined together in peptide linkages. Out of these 20 different amino acids we need to obtain 9 of them in our diet. The rest we can synthesize. In other words, we cannot synthesize certain amino acids from our metabolic reactions and must therefore obtain them from the food we eat. The amino acids that we cannot synthesize are referred to as being essential amino acids. Thus, the five general components to a complete diet are carbohydrates in the form of sugars like glucose, proteins, fats, vitamins,

and minerals.

The Gastrointestinal Tract The gastrointestinal tract includes the mouth, esophagus, stomach, small

intestines and large intestines. Let's consider a cross section of the small intestine and examine the tissue layers as we work outward from the lumen. The first barrier we encounter is a convoluted layer of epitherial cells (Figure 3-3a).

The convolution of this layer helps to increase the surface area of the gastrointestinal tract for absorption of nutrients. Scattered throughout this epithelial layer are ducts from external exocrine glands like the paricreas and liver (and the salivary glands in the oral cavity). ]uxtaposed to many of the

epithelial cells are endocrine cells which contain hormones that can be released

into the blood. These hormones influence other cells in the gastrointestinal system. Both the parasympathetic and sympathetic nerves (of the autonomic nervous system) innervate the gastrointestinal system. The more important nerves stem from the parasympathetic system. Recall that the parasympathetic nerves are more active during times of relaxation and digestion while the sy'rnpathetic nerves are more active during times of ,'fight or flight."

As food is passed from the mouth to the lower portions of the gastrointestinal tract the smooth musculature surrounding the epithelial cells begin to contract in peristaltic waves. These waves (peristalsis) are not controlled by conscious thought but rather by the action of the parasympathetic division and by the action of hormones. Peristaltic action is rather quick because of the elecirical

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r56


Biology -



-':r.ltnuity imposed by the gap junctions in the circular and longitudinal muscles Jre gastrointestinal tract.

-'

(a) Blood Vesseis and/or Lymphatics Epithelial Cells

Parietal Cells (HC1)

Chief Cells (Pepsinogen)

:';rgitudinal

\Iuscie Gland outside the gastrointestinal tracl

tiil

Nerves to and from the CNS

4r

il.

F':,:ure 5-3 ,

:l

-

:

:

:

--

-r:n through the small intestines a bolus of food as

--:

it passes its way

through the gastroiatestinal tract. -o\\r -- ': - ::-outh the muscles of mastication move the jiw and the food is ground ': ::r. ihe teeth. salivary amylase is secreted into ihe oral cavity and be"gins to ' =! , :ie starch and glycogen. This secretion is contiolled u-v tr,e r- : - i : " :'.:athetic nerves.

:rrd

is swallowed and passed into the pharynx access to the nasal cavity covers the opening to the larynx :=''ents food from entering into this passageway (to prote-t the airway). :re food passes into the esophagus and then down into the stomach. : t:ol\ring reflex is controlled by centers in the medulia. =

-=: -\ flap of tissue called the epiglottiJ

":

' -:

= food reaches the stomach a sphincter (circular muscle) called the regurgitation of the food . , ::-'lhe esophagus. If this sphincter were not closed off, stomach acid r - : =:-:er the esophagus and irritate the nerve endings in the smooth muscle. - ": --::ir1g sensation is referred to as heartburn. r-^

: r.-'::: esophageal sphincter contracts and prevents

-:

:f

enters irrto the stomach a little more digestive action takes place. Two

the stomach is to break the food down into smaller partiiles and to :: ibv acidic secretions). There are four major types of secretion within . .:::'.. Mucus is secreted by surface cells and acts to protect the lining of ::-::n and lubricate the food. Gastrin, located in endocrine cells in the

. ti

The Berkeley Review

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Biology

Gastrointestinal Tract & KidneY


lower portion of the stomach, is secreted in response to protein entering into the stomach. This hormone stimulates the secretion of HCI and pepsinogen. HCl, at a pH of about 1, is secreted by parietal cells. Pepsinogen is secreted by the chief cells. Parietal cells and chief cells are located in the gastric pits the line the epithelium of the stomach (see Figure 3-3b). Pepsinogen is the inactive form of the peptidase enzyme pepsin. Peptidases, like pepsin, cleave peptide bonds. Since protein is composed of amino acids and amino acids are linked via peptide bonds, pepsin hydrolyzes proteins (at specific places in their amino acid sequence). Pepsinogen is converted to pepsin by the action of HCl. However, once pepsin is formed it can autocatalytically act on pepsinogen to form more pepsin. Besides secreting HCI the parietal cells also secrete a glycoprotein called intrinsic factor. This glycoprotein is important because it complexes with vitamin B12 and

is then absorbed by the intestinai epithelial cells and transported by the bloodstream. if there is a defect in the synthesis of intrinsic factor, vitamin 812 cannot be bound and therefore cannot be absorbed by the epithelial cells in the intestine. Vitamin B12 is important in erythrocyte (red blood cell) formation.

If too much acid is secreted into the stomach, ulcers can occur in the stomach and

in the small intestine (which is directly connected to the stomach). Ulcers are simpiy erosions of the walls of these two organs, and if they are extensive enough can cause bleeding. One of the most powerful stimulants that causes HCI

to be released into the iumen of the stomach is histamine. It turns out that a compound called cimetidine (trade name is Tagamet) inhibits the binding of histamine to its receptor on the parietal cells. This reduces the amount of HCI secreted in the lumen of the stomach. Cimetidine and its analogs are quite commonly prescribed by doctors as a non-surgical treatment for ulcers. As the dissolved food (refered to as chyme) passes from the lower stomach and into the small intestine more enzymatic activity takes place. This is partly due to distension of the stomach and smail intestine and ihe generation of nerve impulses that stimulate enzymatic secretions. Roughly 90"/o of the digestion and absorption that takes place in the gastrointestinal tract occurs in the small intestine. Not only do fiuids enter the small intestine from the stomach but the pancreas and liver also add secretions as well. The smali intestine acts to neutralize the acid which has been secreted by the stomach and to further digest and absorb food particles.

Distension of the small intestine also causes the hormone cholecystokinin (abbreviated as CCK) to be released from the intestinal mucosa. CCK diffuses by the way of the bloodstream to the pancreas where it causes the pancreas to release digestive enzymes. The hormone secretin is released from the small intestine in response to the entering chyme from the stomach. Secretin is also absorbed by the blood and is transported to the pancreas where it causes the release of bicarbonate ion and other fluids. The pancreas not only has endocrine cells which secrete insulin and glucagon but it also contains secreting structures called acini that secretes a fluid into the small intestine which has a high bicarbonate content that is rather alkaline. The bicarbonate ions combine with the protons from HCI to produce carbonic acid which is then converted into carbon dioxide and water. Carbon dioxide is absorbed into the blood and transported to the lungs where it is expired' This


l5a


Biology

Gastrointestinal Tract 6r Kidney

:-echanism increases the pH in the small intestine to a more alkaline value. Figure 3-4.

\a*+HCO3-+H++Cl-

Nutrients & Digestion See

HCO3 (bicarbonare ion) H2CO3 (carbonic acid)

H2Co3

'-)

Co2

+ Hro

F-igure 3-4

I .:: :'..1 of the bicarbonate ion.

-: - lir-er has a diverse

set of functions, one of which is to synthesize a compound concentrates and stores in the galibladder. The'ma1or : i gment in bile is a compound called bilirubin, which ii a breakdown produci of - .:.::Iobin. Bile also contains bile salts which are important in the digestion r ' ::;orption of fats. when bile is rereased from the guilbludde, it passeJdown

:.--=: bile which it

r ri:i that joins with

the pancreatic duct, througlia constriction called the

*':-:.incter of oddi, and empties into the small intestine. Bile acts to emulsifv fats : -1e.rease their surface tension in order to break them up into smaller sizes) r: - -: :lso helps the epithelial cells of the small intestine absorb those fats. The ::::.:-:e of fats in the small intestine releases CCK which then acts on the

:**:,::;er

and causes contraction, and on the sphincter of oddi and causes pass into the lumen of the small intestine. [The :o1or of the gallbladder --i:is due to the various breakdown products found -::::: r- :-:: :,-...entrated biie.]

-:-r":.:,-n, so the bile can

Basolateral

', ',Na+ 1:

lL

li.t

COS€ :

\ **ouô

Glucose

tr-umen ii*rruumne

Blood

3.5

' \ : j:'. .

ir.entioned, the small intestine is where the majority of absorption of -".* r- r --;:tients takes place. Located on the apical and basolateral regions of r , ::"r.--i="". cel1s are specialized transport proteins ("carriers") r"spor,ribl" fo, r"u,' irrt:", -:l':1 of sodium, chloride, sugars, amino acids, vitamils, and other such Tr .r -i '.. For exampie, consider the absorption of glucose into the epithelial ,r-" - - :: s. .-i oniv able to enter the cell if sodium is coiransported along with it. - . r - -: .:.-:ers the cell down its concentration gradient, it "drags" glucose into

'''r -- -.:-:::-". of the cell. Because the intracellular concentration of glucose I r''' r:'::i -- - able to diffuse out the basolateral side and eventually it,to th" , ,':r --.: --:-= .oncentration of sodium increases in the cell it is aiso pumped out '

"

;L:

The Berkeiey Review

159


Biology



on the basolateral side in exchange for potassium via the sodium-potassium pump. See Figure 3-5. As the ions and nutrients are being absorbed by the epithelial celis of the small intestine, water is diffusing_ through the membrine. witer is trying to equilibrate on both sides of the membrane by osmosis. Recall that choleri toxiln causes massive loss of fluid by diarrhea. In the majority of cases this can be treated by giving the individual large quantities of glucose and saiine solutions.

How much fluid is being absorbed by the small intestine? Roughly 1.2 iiters of water per day are taken in from the foods we eat. Another 7.0 literi are secreted by the stomach, pancreas, iiver, small intestines and so on. Roughly g.1 liters of the total 8.2 liters is reabsorbed in the small intestine. similirly, out of the rgughly 800 grams of solid food consumed, roughly 570 grams are absorbed in the small intestine. The remaining fluid and solid"material i"s excreted as waste. Recall that fats are degraded to fatty acids and glycerol by the action of the enzyme lipase. These compounds can diffuse intJtire intesiinal

epithelial cells where they are resynthesized into triglycerides (fats) and aggregate into structures called chylomicrons. These aggregates are released at the basolateral membrane and into the extraceluiar spice (Figure 3-6). The chylomicrons enter into the

lyt

ph and are transported to the veins and eventually the tissues.

)r

Fatty acids

IFATS

Glycerol

@ o

To the lymph and tissues

Figure 3-6 Release of chyiomicrons.

The large intestine absorbs most of the water and ions that are left in the chyme

as it passes from the small intestine. As the sodium and chloride ions are absorbed an osmotic gradient is established that allows for the absorption of water. \44rat is not absorbed is passed out the body in the feces.

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Biology

Gastrointestinal Tract & Kidneys

Renal Function

Renal Function r.:ush1v 60% (a littie less than 2/3) of an,averageT}-kg (154-1b) man's body ' ::ght is water- Most of this fluid is located insidJ the ceiis (called intraceilular :-*:;,and about 113 of it is located in the interstitial spaces outside the cells ::-'ec the extracellular fluid). If you think about the organ systems which we -=..: pre'iously discussed, you will reaiize that watei"u."b" vs lwDL lost rrullt from the lltt

.:".y:i"_"|,!*," r\t"lby

evaporation, the gastrointestinal tract, and the lungs.

:::: can also be tost fiom;;;;;U;;il;. ne body by these four-avenues exictry

il;;;;ffi#riilliSl,

matches the water gained by the ::rrough consumption of fluids and regulation at the level of tte kidneys.

.:.-- ihat we have mentioned that osmoiarity refers to the totai solute

r:

- --:.:.fation of a solution. Remember, the highbr the osmolarity of a given ' -:-rr, the lower will be the concentration Jf water in that solution. If an : r::-.r1 can change the internai ionic concentration of its body fluids to meet : :-'--:.e surrounding environment, they are referred to as osmoc."f";;;;. :..: --.arine invertebrates are osmoconformers. However, most vertebrates do - : ::.=.qe the internal ionic concentration of ':

-

;:

their body fluids to meet that of a this are referied to as osmoregulators. and osmoregulators can be seen on the

j:.:::ig

environment. organisms like "-: " :--:::fence between osmoconformers

-

: :: -: Figure 3-7.

ri

Osmoreguiator

Environmental osmolarity lllL{rnrrc

''

5-;

iLr ..:t'' :rx-. -

'r titl"

,.,i:

ir

: :,i:';

een osmoconformers and osmoregulators.

,t-:'rr.ii rid.ev presumably arose from freshwater teleosts (fishes). The . :r - -::i1- of a fieshwater tlleost is about 300 milliosmols/liter, which is

,:r :-':-- ::.a. of the surrounding aqueous environment. This means that the

-:-'1----:- :: rfater is less in the freshwater fish than it is in the water in ' -": 1i:. :s sh-rmming. water from the environment will diffuse d.own its

,i

"

rrr

r ':::*r--::- ::adlent and into the fish. The excess water that is contrnually : . : jr - -:-- =€ fish is filtered and removed by the kidneys and excreted in tlle i' -..-t

:-" The Berkeley Review

r6l


Biology

Gastrointestinal Tfact & Kidneys

Renal Function

form of a dilute urine. Since copious amounts of urine are excreted. on an hourly basis the freshwater fish also loses important ions like sodium and chloride ioni. These ions are in rather low concentration in freshwater habitats and sd the

freshwater teleost has developed (via evolution) gills {or active transport of sodium and chloride into its circulatory system. See Figure 3-g. Water enters by osmosis

Sallt though

Active transport Copious of salt through gills urine (salt and water loss)

Figure 5-8 Salt regulation.

Generalized Kidney Function The kidneys in vertebrates like the freshwater teleost have three main functions. The first function is that of filtration. The functional unit in the kidney is the nephron which consists of a glomerulus (Latin, meaning "little ball"), Bowman's capsule, and a tubular system. The glomerulus is a collection of capillaries that receives blood from an artery terminating in the renal system. Blood is pumped into the glomerulus by the hydrostatic pressure of the heart and that pressure -The forces the blood through the capillary walls and into Bowman's capsule. cellfree ultrafiltrate found in Bowman's capsule lacks many of the plus*a proteins found in the blood. Blood plasma is simply a solution which is about 906/o water that is composed of organic (e.g., proteins, sugars, amino acids, etc.) and inorganic substances (e.g., various ions like sodium and chloride). within the blood plasma one also finds red blood ce1ls, white blood cells, and platelets. The filtrate in Bowman's capsule is essentially the plasma minus the (large molecular weight) proteins. The second function

of the kidney is reabsorption of important organic and inorganic compounds from the filtrate in Bowman's capsule. Reabsorption occurs through many of the epithelial cells which line the tubular lumen of the nephron. The cilia of these epithelial cells move in such a way as to help propel the filtrate through these renal tubes. Glucose, small proteins, amino acids, salts, bicarbonate ions, and water are reabsorbed along this tubular system and transported back into the blood. The epithelial cells also have the ability to secrete protons, potassium, urea, uric acid, and ammonia. The third function of the kidney is excretion of waste, salts, and excess water.


r62


Biology R

Gastrointestinal Tfact & Kidney

$Ey.lffiir;rrt r.

Renal Physiology

, .otl rrffi-

Both human kidneys sit on the dorsal side of the abdominal cavity and are beanshaped organs about the size of a fist. Blood from the descending aorta enters into the renal artery and eventualiy into the glomeruli of the r,uph.onr. Bioocl ieaves the kidney by way of the renal vein which itself empties into the inferior ','ena cava. Leaving each kidney is a ureter which transports the urine to the bladder. urine exits the bladder by way of the urethra. These anatomical :onsiderations are shown in Figure 3-9. Inferior Vena

,- ::

'

.

::-

Cava---

position of the kidneys.

..jnev contains more than a million nephrons. The glomeruius and : - s capsule of each nephron is located in the cortex and gives this portion

---

r-- = .-:lnel' a granular appearance. In contrast, the striated appearance of the * :': ;--a '- primarily due to a portion of the tubular system of the nephron called : , I r ! of Henle and the collecting duct. The collecting ducts, whiih collect the '-- . =:,:tv into the renal pelvis of the kidney and then into the ureter. This is . -: Figure 3-10.

= br The Berkeley Review


Biology


Renal Physiology

Renal Pelvis

Figure 5-lO Kidney anatomy.

The Nephron We mentioned that the functional unit of the kidney is the nephron. As the renal

artery branches into smaller divisions and enters the kidney it becomes the afferent arteriole. It is the afferent arteriole which enters into Bowman's capsule and forms the capillary bed called the glomerulus. An efferent arteriole leaves the giomerulus and forms a capillary network which surrounds the renal tubules. The blood that leaves this capillary network does so by a venule which later empties into the renal vein that leaves the kidney. Extending from Bowman's capsule is a long tubular structure which is divided into the proximal convoluted tubule (PCT), loop of Henle, distal convoluted tubule (DCT), and the collecting duct. Many different nephrons can attach to a single collecting duct. These structures are shorlrn in Figure 3-1.1. The cardiac output of the heart is roughly 5 liters of blood every minute. Out of this 5 liters of blood roughly 20% of it (about 1/5) passes through the kidneys each minute. Thus, the volume of filtrate passing into Bowman's capsule will be rather large. Roughly 180 liters (47 gallons) of filtrate per day passes into all Depending these capsules and of that about 1.0 t6 1.S liters enters into the ,rtit ". on how much fluid you drank, the upper end of urine production might be about 3 liters per day.

The osmolarity of the filtrate in Bowman's capsule and in the proximal convoluted tubule is about 300 milliosmols per liter. Recall that this is essentially the same osmolarity we found in the plasma. The concentration of the urine that is collected varies depending on the circumstance. If you wanted to excrete a dilute urine, then the concentration of the fluid leaving the collecting ducts

should be lower than 300 milliosmols per liter. For example, a urine concentration which is low might have an osmolarity which is 0.7 times that of


164


Biology


Renal Physiology

osmolarity of the prasma. In other word.s, the osmolarity of this ,,d1ute,, urine -tight be 200 milliosmols per liter. h contrast, a concentrated urine would have a :.::h osmolTity, :9Tuth11s in the range of 4.2 times thar oiirr" osmolarity of :-tu
Bowman's

Efferent Arteriole

crp;;;"

Glomerulus

Proximal Convoluted

Afferent Arteriole

Cortex

----ule

tr)

l

Loop of Henie

Medulla

Capillary )'ietwork

qliquunc

''

ll

}t

=)

I

:ruo![tr-]-:r

=:

iis related capillary system.

;rir:

proximal convoluted tubure. The pCT rs the obligatory section of roughly 6s./" of arl reabsorptio, ;;,;;r"iio" occurs here. "'llrLu': *'rr*r*al cells rn the pCT are quite metatohcally active and support a 'iiir""rrrlrrri :,r ::ti-e transport processes. Glucose, small moiecular weight proteins, rrrrru*iun:r ;l.. a'd vitaminsâre completery (100%) ,"uurortua in the pcr. ril*,;11111L*9111r:; lJ', of the Nae, Cle, and *ui"r are also reabsorbed in the pCT. Many rrrt' lg4s '- 'ms for reabsorption involve sodium. ro, ''s'-' grrr"o"" ur,i ,lir Lir'ulrfT r:rrs are reabsorbed "ru*pt", together as are the amino acids and sod.ium ions. ii'lr flrr,rrur]r r:r5 a:e also reabsorbed with chl0ride ions. Little ,"grrlurion occurs in lhlT ,n F

lrrt#

=e-:ecause 'n*s;;r:'11

ulllltrrrlri

,11llrll"ti.gl:

I

The Berkelev Review

165

The Berkeley Review Specializing in MCAT pre;raration

Biology


Renal Physiology

As the PCT begins to descend from the cortex into the medulla it forms the loop of Henle. The loop of Henle has a thin portion and a thick portion (see Figure 3Jt) rn9 descending thin portion of the loop of Henle is very permeable to water but only relatively permeable to ions like sodium and molecules like urea. The ascending thin portion of the loop of Henle is much more permeable to urea but much less permeable to water than the thin descending portion of the loop. As the loop of Henle ascends it becomes thicker. The epithelial cells in this region of the loop actively transport ions like sodium and potassium from the lumen of the loop into the interstitial fluid. However, this region is impermeable to urea and water. This means that as the filtrate is passing up the thick portion of the loop of Henle it is becoming more dilute beciuse of the fact that ions are being transported out into the interstitial fluid. Remember, the lower the osmolarityl the higher will be the concentration of water (i.e., a more dilute fluid). The filtrate from the thick portion of the loop of Henle passes into the distal convoluted tubule. The segment of the DCT closer to the loop of Henle is also quite impermeable to urea and water but rather permeable toions like sodium. Therefore, as the filtrate passes through this segmlnt of the DCT it will become a bit more dilute. The epithelial cells of the segment of the DCT closest to the collecting duct and that portion of the col]ecting duct located in the cortical region of the fidney are still rather impermeable to urea. However, these two areis of the nephron are quite sensitive to the hormone aldosterone (secreted by the cortex of the adrenal glands which sit on top of the kidneys) which regulates sodium absorption. An increase in the concentration of aldosterone causes sodium to be reabsorbed by the epithelial cells. When sodium ions are pumped out into the interstitial space, potassium ions are simultaneously transported into the lumen of this .eglol of the nephron. Epithelial :]lr in this tegiot of the nephron also resp6nd to

antidiuretic hormone (ADH, produced by hypothalamls and releasea uy

tt-re

posterior pituitary) which regulates zuater absorption. If. the concentration of ADH increases, then water will be reabsorbed from the epithelial cells in the coilecting duct and passed to the iaterstitial space. Removing water from the lumen of thE collecting duct acts to concentrate the urine. The celis of the collecting duct can also secrete hydrogen ions (as can the cells of pCT and DCT).

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166


Biology


tlomeostatic Mechanisms

Homeostatic Mechanisms

:

Suppose we want to maintain the levels of sodium in the body. Certain ce1ls in :re cortex of the adrenal gland are sensiiive to the leveis of soiium in the blood. ',"-lten the concentration of sodium decreases in the blood these cells release the :.'Jrmone aldosterone. Aldosterone acts at the level of the DCT and the collecting :-rcts and stimuiates the epithelial cells in those regions to reabsorb sodium. As i ::sult the blood concentration of sodium begins to rise. This acts as a signal and =eds back on the cortical cells of the adrenal gland telling it to reduce the s-cretion of aidosterone.

::ppose we want to maintain the proper levels of water in the body. If there has ':en a decrease in the plasma volume of the bodv, there will be a tendency to - r\-e a lower than normal blood pressure (detected by baroreceptors) but a - -gher than normal osmotic pressure (detected by osmoreceptors). Receptors =ceiving this information stimulate specific cells in the hypothalamus to .', nthesize and transmit ADH to the posterior pituitary where it is released into :.e blood. ADH stimulates the epithelial cells in the latter portion of the DCT '":rd the collecting ducts to reabsorb water. This leads to less water excretion and

: righer

plasma volume. If you were to drink too much riquid (e.g., pitchers of :=er), then the opposite process would happen and you would excrete copious ::-Lount of urine. It turns out that the release of ADH is inhibited bv alcohol.

Excretion of Nitrogenous Waste

.:oteins and nucleic acids are the two primary metabolic sources of nitrogenous ,','astes. There are essentially three ways to get rid of the nitrogen ihut i, ::oduced during the metabolism of these compounds. Nitrogen can be removed .. ammonia/ as urea, or as uric acid. The structures of these compounds are --- 'rr n in Figure 3-12.

I

-C

H- N'

NH: r

Ammonia toxic, soluble)

H2N-

C

-

o,c'ry NHz

Urea

(iess-toxic, soluble)

H

Uric acid (toxic, insoluble)

Figure 3-12 "'

:.iogenous waste compounds.

.--' glutamine is metabolized near the gills of fish, the nitrogen in this amino acid, :':r be converted to ammonia which can then combine with a proton and, in =''.:hange for a sodium ion, be carried away with the passing water. The ::lrnonium ion can also be excreted by mammals. one of the by-products of the :::irno acid glutamine is urea which is carried by the biood to the kidney where it --. excreted in the urine. Urea is excreted by most mammals, amphibians and reptiles and birds. The ammonia that is produced from the metabolism ' --rre -:rm the amino acids glycine, aspartate, and glutamate can be converted to uric acid which is the excretory product of birds and reptiles. - rpvright

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167


Biology

Gastrointestinal Tract & Kidney Control of pII and

Homeostatic Mechanisms

tI@

The hydrogen ion concentration of the extraceliuiar fluid is closely monitored. The normal pH of the body is 7.4. This varue can fructuate between a pH of 7.35 and.7.45.. If the pH of the body decreases to values beiow 7.3s for an extended period of time, the individual becomes lethargic and death can result. If the pH is above 7.45 for extended periods of time, the individual becomes hyper and ieath can res-ult. The regulation of brood pH is by means of a buffeiing system in which hydrogen ions are reversibly Lound to a buffer like the bicarbonate ion (HCo3e), the hemogrobin morecule, or even certain plasma proteins. For example, if there are hydrogen ions in the plasma, they can be taken up by the bicarbonate ion as shown in equation (3_1) to form carbonic acid. This is a weak acid.and decarboxylates to form carbon dioxide and water. Carbon dioxide is eliminated by respiration. In fact, the pCo2 levels in the alveoli of the lungs are quite important in regurating the prasma pH. This can be seen if you examine the (Henderson-Hasselbalch) equation shown in (3-2). The levels of bicarbonate and hydrogen ion are controlled by the kidney whereas the levels of carbon dioxide levels are controlled by respiration.

H* +

HCO3-

H2CO3

pH = pKu + tog tlqqd [Coz]

---=

CO, + HrO

(3-1)

(3-2)

Cholera Toxin S-uppose you had a severe case of diarrhea caused by the bacterial cholera toxin

(vibrio cholera) that we have previously discussed. This would result in a decrea.se in the pH and ur in"r"ur" in the pCo2. why? Recall that the gastrointestinal tract secretes a large amount of sodium bicarbonate. lvhat happ-ens if you lose a lot of sodium bicarbonate by way of the gastrolntestinal

tract? Metabolic acidosis results. Look at equation (3_1). If you decrease the concentration of HCore, thel by LeChdtelier's principll (from general chemistry) Co2 will combine with H2o to make H2co3. This wiil dissociate into HCo3, and H@. In other words, the hydrogen ion concentration has increased or, if you like, the pH has decreased. This same conclusion can be derived from

equation (3-2).

A

decre_ase in pH will stimulate the DCT to secrete hydrogen ions and reabsorb as much bicarbonate_ as possible. By a negative reeabalk loop this tends to counter the initial decrease in pH. simultaneousry, an increase in pCo2 stimulates the central chemoreceptors in the medulla'and pons to increase the contraction of the diaphragm and chest muscles in order to'increase respiration. An increase in the respiratory rate will eliminate more Co2 from the lungs.

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l6a


GastrointestinaE Tract

& Kfidneys To Go 15 Passages

100 Questions


Passage Titles

I.

il.

III.

IV. v.

VI.

vII.

VIII. IX.

X.

XI.

xu.

XIII. XIv. XV.

G as tro int e s tinal S mo o th Mus cle Protein Tuntover Livef Pancreas, & Intestines Digestion Safeguards Digestion o.f Fats D i aberi c D i ct Expe rim enr The Kidney Kidnelt & pH Dialysis & Ultrcfiltration

CaJfeine

Kidney & Calcium Lymph Renal Clearance G as t ro inte stinal J unc tion Cholera Toxin

Specialtzrng in MCAT Preparation

Questions 1-5 6-12 13 -20 1l

LI

1'7

- L I

28-33 31-40

4l-48

49-53 54-60

6t-66 6l -12 73 -19

80-86 87 -93 94 - 100

Suggestions The passages that follow are designed to get you to think in a conceptual manner about the processes of physiology at the organismal level. If you have a solid foundation in physiology, many of these answers will be straightforward. If you have not had a pleasant experience with the topic, some of these answers might appear to come from the void past the Oort field of the solar system.

Pick a few passage topics at random. For these initial few passages, do not worry about the time. Just focus on what is expected of you. First, read the passage. Second, look at any diagrams, charts, or graphs. Third, read each question and the accompanying answers carefully. Fourth, answer the questions the best you can. Check the solutions and see how you did. \Mhether you got the answers right or wrong, it is important to read the explanations and see if you understand (and agree with) what is being explained. Keep a record of your results.

After you feel comfortable with the format of those initial few passages, pick another block of try them. Be aware that time is going to become important. Generally, you will have about 1 minute and 15 seconds to complete a question. Be a little more creative in how you approach this next group. If you feel comfortable with the outline presented above, fine. If not, then try different

passages and

approaches to a passage. For example, you might feel well versed enough to read the questions first and then try to answer some of them, without ever having read the passage. Maybe you can answer some of the questions by just looking at the diagrams, charts, or graphs that are presented in a particular passage. Remember, we are not clones of one another. You need to begin to develop a format that works best for you. Keeping a record of your results may be helpful.

The last block of passages might contain topics that are unfamiliar to you. Find a place where the level of distraction is at a minimum. Get out your watch and time yourself on these passages, either individually or as a group. It is important to have a feel for time, and how much is passing as you try to answer each question. Never let a question get you flustered. If you cannot figure out what the answer is from information given to you in the passage, or from your own knowledge-base, dump it and move on to the next question. As you do this, make a note of that pesky question and come back to it at the end, when you have more time. When you are finished, check your answers and make sure you understand the solutions. Be inquisitive. If you do not know the answer to something, look it up. The solution tends to stay with you longer. (For example, what is the Oort field?)


Section III Estimated Score Conversions Scaled Score

>12 10- 11

Raw Score

- 100 79-85

86

8-9

65 -',78

7 6

59-64 54-58 48-53 0-47

5

<4

Biology Fassage

I

Gastrointestinal Smooth Muscle

(Questions 1-5)

3.

Passage I

According to the diagram shown below, the force of smooth muscle contraction is most likely dependent upon the:

The smooth muscle ceiis of the gastrointestinal (GI) :r:i are typically of the single unit type, roughly 50 to - -,, microns long and 2-5 microns wide. The action :,::-ntlals found in smooth muscle are smaller in L:-:-jtude, but longer lasting than those in neurons. The

i:-s:le cells can be arranged in sheets and can be excited :, .::ion potentials from neighboring cells. These sheets :, ::1is also exhibit an intrinsic pattern of periodic :e:'--larization known as pacemaker activity. Slow waves : :::'olarization followed by repolarization make up the rr.es:: :iectrical rhythm. In the presence of excitatory i:-:; or hormonal input, a slow wave may exceed the

Ilr-::-rld for action potential

:it:-: .-

generation and initiate

:-r contractions.

),f :,r:1it1, in the GI tract falls into three major -ri::.::ies. The first is peristalsis, which involves

.:n:r'::ated waves of contraction and relaxation. The ;:-:-: is segmentation, which involves alternating ^ rq--:;::xon and reiaxation of the circuiar muscle layer. -:e -:--d is the mixing motions of villi and microvilli r ilrrr::, r3Juce the effect of the unstirred layer of fluid that

size of the action potential.

B.

interval between successive sets of action potentials.

C. D.

i,s, !"._':-'ent to the apical surface of cells.

4.

l t : t)'pes of smooth muscle, single-unit

A.

number of action potentials within a set. time duration of individual action potentials.

Segmentation, in contrast to peristalsis, is primarily responsible for:

and multi-

*:--. are distinguishable based on their electrical .u:"'in'. Which of the following statements about ::-: :istinction is most likely true?

B.

d

movement of intestinal contents through the GI tract.

Single-unit fibers are coupled, with many

C.

ffi*

individual cells containing gap junctions. Single-unit fibers are uncoupled, with few

D.

mixing of intestinal contents, breaking larger particles into smaller ones. stimulating the secretion of digestive enzymes from the intestinal exocrine glands.

A.

-ndir idual cells containing gapjunctions.

{--. \Iulti-unit fibers are coupled, with

D

absorption of nutrients from the lumen of the

GI tract.

few individual cells containing gap junctions. \Iulti-unit fibers are uncoupled, with many

:idividual cells containing

gap junctions.

The mixing of intestinal contents most aids in digestion because:

"li lt* ::nduction veiocity of action potentials along , r:':--: rnuscle fiber is low because activation of:

A.

smaller particles move faster through the GI

B.

tract. smaller particles are able to cross the epithelial layer with increased efficiency.

C.

L fl. r;[

smaller particles carry less charge, enabling

of particles over biological membranes. breaking larger particles into smaller particles increases the surface area available to digestive

more efficient passage

-:omssium channels is slow. =rdium channels is slow. :ai:ium channels is slow. :asnesium channeis is slow.

D.

enzymes.

,,,.,,rllltfi't'lr'"nn:r

; bt The Berkeley

Review

t7l


Biology Passage

II

Protein Turnover

(Questions 6-12)

8.

Passage II

The urine carries 80 gm of protein equivalents per

day. The term protein equivalent is used

because

The diagram shown in Figure I is a summary of daily protein intake and output-turnover-in a 70 kg human. fNote: The gastrointestinal (GI) system is equivalent to

protein is degraded and another molecule is usejd

the gut.l

product?

Protein Intake (100 gm)

as

the excretory route for nitrogen. Which of the following is the chemical formula for this excretory

Body Protein

A"\ht"

(10,000 gm)

Turnover of Body Protein

White Cells (20 gm)

(250 gm)

B.

NH:

C.

H2N- C-

D.

H2N- NHz

o

il

NH2

Ahsorbed -i--;--)

lnmino Rctos ----) (16ogmt l_ilggd

I

.skin

Urinary

Fecal (10 gm protein)

(80 gm protein equivalent)

9.

Rank in increasing order the following organs or tissues according to their amount of daily protein turnover-

Figure I A. 6.

You can see from the diagram that the gut secretes more protein than the person ingests. Which of the following gut secretions contain recyclable protein?

I. il. UI.

Digestive enzymes Mucus Bicarbonate

A. B. C.

I only I and II only II and trI only

D.

I, II, and

B. C.

D.

10.

White cells, gut, liver, muscle. Gut, muscle, liver, white cells. Muscle, liver, gut, white cells. White cells, liver, muscle, gut.

Albumin is the most abundant plasma protein. What type of experiment could be used to quantify the albumin turnover rate?

III

A.

Give a known dose of albumin and sample the

blood periodically to measure albumin concentration.

7.

Given the following equation, what is the efficiency ofprotein digestion and absorption in the gut?

B.

Give isotopically labeled albumin, sample the blood periodically, and calculate the decay rate of labeled albumin.

' -

Efficiencv

c.

Protein Absorbed

Total Protein Available

Give a known dose of albumin and sample the blood periodically to measure albumin decay rate.

A.

63Vo

B. C.

77Vo

Combine a blood sample with isotopically labeled albumin in a test tube and calculate a

9OVo

decay rate.

D.

94Vo

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D.

172


Biotogy 11.

Protein Ttrrnover

Passage tr

During a fast, which tissue is the first to break down its protein to provide amino acids for other tissues?

A. B. C. D.

Smooth muscle. Cardiac muscle. Enzymes of energy metabolism. Skeletal muscle.

12. The following

experiment, a nitrogen balance study,

was performed to determine nitrogen needs in human subjects. To be in nutrient balance means that intake - output = zero for that nutrient. For nitrogen, this means that the nitrogen contained in protein in food eaten and secreted by the gut equals the amount of nitrogen excreted in urine, feces, and through skin losses.

Subjects were placed

on 3 different

diets

sequentially. The results of their nitrogen balance are noted in Table 1.

Protein Intake (gm/kg)

Nitrogen Balance (mg/kg)

0.1

-42

0.6

a1

1.5

+12 Table I

What is the amount of protein needed by these subjects to maintain a nitrogen balance?

A. B. C. D.

Less than 0.1 gm/kg. Between 0.1 and 0.6 gm/kg. Between 0.6 and 1.5 gmikg. Greater than 1.5 grn/kg.


173


Biology Passage

III

Liver, Pancreas, and Smalt Intestines 13.

(Questions 13- 20)

In controiling the processes of digestion, the pancreas and liver work in tandem to prepare food that exits the stomach for absorption in the small intestine. Secretion of

formation

of micelles, water soluble

complexes from

which lipids can be more easily absorbed by the intestinal lining. Secretions from the pancreas contain a highly

alkaline bicarbonate solution that is important

in

CCK

Bile

Salts

Secretin Pancreatic Fluids

the

neutralization of the hydrochloric acid (HCl) secreted by the parietal cells of the stomach. Secretion of HCI can be

inhibited by gastrin inhibirory pepride (GIp),

Which secretion from the small intestine would be most appropriate for processing a meal containing a high fat content?

A. B. C. D.

bile salts from the liver (sodium and potassium salts conjugated to glycine and taurine) is important in the

Passage Itr

L4, Which of the pancreatic

enzymes activated

in

the

small intestine is important for the initiation of the cascade shown in the diagram below?

an

enterogastrone released by the duodenum. HCI can leave the stomach and enter into the duodenum through the pyloric sphincter.

Proenzyme

The pancreas also secretes enzymes needed to breakdown proteins entering the duodenum. All of these

_/\---------.-._ /\

pancreatic enzymes are secreted as inactive proenzymes. Trypsin inhibitors are secreted by the same pancreatic

lrlpsrnogen

.

Enzyme

r_l

Enterooentidase

-+-+ Chymotrypsinog.n, SI5rTrvnsin ProelastaseJ}

cells to prevent activation of these enzymes. These proenzymes travel through the ductal system to the

duodenum were the enzyme enteropeptidase

(enterokinase) is secreted from the brush border of the small intestine. Enteropeptidase initiates the conversion of the pancreatic enzymes to their active state.

Procarboxypepri6ur", Trypsin--

Bile salts and exocrine secretions of the pancreas are controlled by secretin and cholecystokinin (CCK), two enterogastrones released from endocrine cells in the upper

A. B. C. D.

portion ol the small intesrine.

TrYpsin

Chymotrypsins Elastase Carboxypeptidases

Bile salts Secretin

CCK Enterokinase

Secretin is released from the S cells in the intestinal mucosa when the pH of the duodenum falls below 4.5. Secretin release is increased by gastric acid bathing the

lining of the small intestine and products of protein digestion. Secretin stimulates the ductal cells of the

15.

pancreas to release a fluid containing a high concentration

below?

ofbicarbonate ion but a low concentration ofchloride ion. Secretin, also increases the production of pancreatic enzymes, decreases acid secretion in the stomach, and may cause contraction of the pyloric sphincter.

A. B. C. D.

CCK secretion is stimulated by peptide, amino acid, and fatty acid products coming in contact with the intestinal lining. CCK stimulates the contraction of the gall bladder, a storage organ that concentrates bile salts produced in the liver. Furthermore, CCK increases the secretion of pancreatic fluids rich in enzymes and alkaline phosphates, inhibits gastric emptying, increases enteropeptidase production, and increases motility of the

@


Fat deficiency Increases in fat absorption

Protein deficiency Vitamin B12 deficiency

16. Which

hormone is important in initiating the neutralization of gastric secretions in the small

intestine?

small intestine and colon.

Copyright

Deficiency of the enzyme enteropeptidase results in a disorder displaying which of rhe symproms listed

A. B. i. D. 174

Secretin

Enterokinase Proenzymes

CCK


Biology 17.

Liver, Pancreas, and Small Intestines

The enzyme phospholipase A can remove a fatty acid residue from lecithin (a component of bile) to form lysolecithin, a compound that damages cell membranes. Activation of phospholipase A in the pancreatic ducts causes damage to the surrounding tissue and can lead to acute pancreatitis, a severe and often fatal disease. Which of the following catalytic enzyme is most likely involved with this disorder? A.

Trypsin

B. C.

Bile salts

D.

Secretin

20.

Passage III

Antacid tablets, which are mixtures of magnesium hydroxide and aluminum hydroxide, have been shown to heal duodenal ulcerations.

Mg(OH)2 + 2 HCI

+

MgCt2 + 2H2O

AI(OH)3 + 3 HCI -+ AlCt3 +3H2O Based on the diagram shown below, antacids should be taken 1 and 3 hours after a meal because they:

Endopeptidase

o

19. -\s acidic chyme

o

enters the small intestine, secretin stimulates the release of the bicarbonate ion from the pancreas. The following reaction will result in the duodenum:

E

a

HCI + NaHCO3 -+ NaCl + H2CO3

J-)

Hours After Eating a Meal

Carbonic acid (H2CO3) will immediately:

B.

A = No Antacid B = Antacid taken t hour after meal C = Antacid taken 1 and 3 hours after meal

begin to increase the acidity of the duodenum and activate CCK in order to decrease motilitv of the small intestine and increase absorption. decrease the pH of the intestinal lumen so the entering chyme can easily be hydrolyzed by

A.

the acidic medium.

C.

B.

dissociate into CO2 and H2O in order to allow

C.

the NaCl solution in the small intestine to remain neutral.

D.

dissociate into

will allow for additional release of acid to aid in food digestion. delay the appearance of acid in the stomach. combine with either magnesium hydroxide or aluminum hydroxide to form an acid which is more concentrated.

Ho

and HCO3o in order to increase the acidity of the intestinal lumen and prepare it for nutrient absorption.

D.

produce water as a by-product of the reaction and reduce the acidity of the stomach.

S:,.ere duodenal ulcers, leading to the removal of ::-: duodenum, will cause an increase in all of the ::;iou'ing EXCEPT:

{B. C.

D.

gallbladder relaxation. parietal cell activity. pancreatic ductal cell activity. passage of chyme into the small intestine.


175


Biology Passage

IV (Questions

Digestion Safeguards 21-27)

23. Which parts of the digestive mucus?

The body is abour 95-98Vo efficient in digesting

protein of animal origin. SinceZ}Vo of the human bodv ii protein oJ animal origin, we require some sophisticated safeguards to avoid digesting our own tissue.

I. il. UI.

First, mucopolysaccharides line the epithelial tissue of the digestive tract. Mucus is indigestiblJ and protects the

A. B. C.

tissues from the actions of acids and enzymes of digestion. Second, the pancreas contains prbteolytic enzymes in an inactive form (zymogens) to avoid digestion of the pancreatic cells. These zymogens are

D.

released unchanged through the pancreatic duct into the duodenum.

24,

molecule so that the active site is exposed. A further mechanism for protection of the GI iract is the rapid turnover of cells. The epithelial lining of the GI traci is replaced every 5-7 days. Finally, although the stomach

I. II. A. B. C. D.

Raw soybeans contain a protein that acts as a trypsin

inhibitor. What would be the digestive effect of

eating a large quantity

A. B. C. D.

ofraw soybeans?

6.8

0.2 9.0 2.0

enterokinase entered the

The zymogens would be activated.

Pancreatic enzymes would appear bloodstream. Insulin secretion would be impaired.

in

the

I only I and II only

II and III onty I, II, and Itr

Trypsin would be more active, but the other

Trypsin would be inactive, and the other pancreatic enzymes would function normally. Trypsin would be inactive, and the otirer

pancreatic enzymes would remain

26. What type of acid(s) do rhe parietal cells of the stomach secrete?

as

L Hydrochloric acid

zymogens.

u. Sulfuric acid

Trypsin and other pancreatic enzymes would function normally.

IU.

Carbonic acid

A.

I only I and II only II and III only I, II, and III

C.

D.

22. Chemotherapeutic agents target rapidly

growing and dividing cells. This is beneficial for attacking tumor cells, but may be detrimental to other body cells. Which cells would be most adversely affected?

Copyright

I only I and II II and III I, II, and III

enzymes would remain as zymogens.

B.

A. B. C. D.

Colon Small intestine

pancreatic duct and went to the pancreas?

of the duodenal

ilI. 21.

Mouth

25. What could happen if

acidifies the bolus of food considerably-, the pancreas

tract are lined with

What is the approximate pH of the stomach?

A. B. c. D.

Trypsin, a pancreatic protease, is activated in the

duodenum by the enzyme enterokinase. Trypsin, in turn, activates the other zymogens by cleaving a portion of the

secretes bicarbonate to neutralize the pH contents.

Passage IV

27. To what major class of molecules

neurons

cardiac muscle cells gastric epithelium skeletal muscle ceils

@


do

mucopolysaccharides belong?

t76

A.

Protein

B.

Lipid

C. D.

Carbohydrate None of the above

The Berkeley Review Speciatizing in MCAT preparation

Biology

Digestion of Fats 29.


The digestion of fats and cholesterol by the body is :rimarily carried out in the duodenum of the small

Passage V

Which of the components of fat digestion is mosl ciosely associated with principles utilized by household detergents to remove grease?

A. B. C. D.

:rtestine. Triglycerides are the most common fat products :n the diet. The initial process in the digestion of fats is .le emulsification of fat droplets by bile salts released -iom the iiver and gall bladder. Biie salt release is .irmulated by increased leveis of long-chained fatty acids --ontaining more than 10 carbon atoms) within the :uodenum. Bile salt emulsification of fat is not essential

rr digestion and absorption since as much as 60Vo of all .:iel1'cerides can be absorbed without the aid of bile salts. ,:olesterol processing, however, is dependent on the :::lons ofbile salt.

30.

,

31.

...:r'acids, and glycerol. The hydrolyzed products offat -:rllets are then packaged by bile salts into water soluble --:elles and then transported to the cellular membrane of

Packaging of fat products in the endoplasmic reticulum.

Sprue, a disorder of malabsorption in the small intestine, is characterized initially by the increased soap-like appearance of fat in the stools. Which of the following reasons couid lead to this symptom?

A. B. C. D.

Cel1s within the duodenum sensitive to fatty acids will :iease cholecystokinin (CCK), a hormone which will .--:nulate contraction of the gall bladder and increase fluid .:cretion from the pancreas. Once the fats have been -rnulsiiled, pancreatic lipase, an enzyme produced from .:: exocrine pancreas and activated by the enzyme :-. psin, will hydrolyze the fats into monoglycerides, free

Emulsification of fats. Micelleendocytosis. Exocytosis of fatty acids.

Decrease in chylomicron production. Decrease in bile salt secretion. Decrease in intestinal cell microvilli. Increase in pancreatic lipase production.

What will occur

if bile salt secretion

is decreased?

A.

Increases in cholesterol absorption in the small intestine.

B.

Complete blockage of fatty acid digestion in

The

C.

Inhibition of cholesterol absorption in the

-:.,. lomicrons are then exocytosed into the intercellular ,:":e and taken up by the lacteals ofthe villi.

D.

small intestine. Decreased levels small intestine.

.:

intestinal

microvilli. The micelles

cross the cell

--:mbrane and are incorporated into vesicular packages

--:."lomicrons)

in the endoplasmic reticulum.

the small intestine.

Chylomicrons also contain triglycerides, cholesterol, are transported by way of the .::rphatics and the capiliary system to the liver. In route , tne liver, lipoprotein lipase, an enzyme located in the -,:rllaries, strips triglycerides from the chylomicrons for :.rra-ee in adipose tissue. The remaining components of -,: chylomicrons are degraded in the liver and stored for :l:: production of steroid hormones, VLDLs, or excreted :.:r bile salts into the intestine.

-:-: phosphoiipids. They

:8. A trypsin

antagonist

32.

of CCK secretion from

the

Chylomicrons are degraded in the liver. Which of the following compounds is a component of a

chylomicron as

it

enters

into the liver to

be

degraded?

A. B. C. D.

is present during fat

metabolism. Which of the following processes will

33.

NOT occur?

Fatty acids.

Biie salts. Lipoprotein iipase. Cholesterol.

Disorders characterized by increases in fat deposition in the liver may be produced by one of the abnormalities below.

A. B. C.

Emulsification of fats. Increase in fat content in the small intestine. Hydrolyzation of fat droplets in the small

D.

intestine. Decrease in CCK production.

- rpvright @ by The Berkeley Review

A. B. C. D. t77

Decreased Decreased Decreased Decreased

leveis leveis levels levels

of pancreatic lipase. of CCK. of lipoprotein lipase. of chylomicrons.


Biology Passage

Diabetic Diet Experiment


Table

diabetes mellitus (NIDDM), not only to improve control of hyperglycemia but also to reduce the risk of coronary heart disease by improving plasma lipid levels.

The American Diabetes Association (ADA) currently recommends the following diet for patients wittr XmnU: 30Vo

of energy

as

Plasma

lipid and lipoprorein levels

during the study.

Diet control is important in non_insulin-dependent

. . .

2:

Passage VI

Measurement

Base.

(mg/dl,) Total Cholesterol

line )) \+

Triglycerides

285 58

43

135

131

32

30

VLDL Cholesterol LDL Cholesterol HDL Cholesterol

fat (with l\Vo as saturated fat)

40-50Va complex carbohydrates (starches) less than l}Vo simple carbohydrates (sugars)

t t

P < 0.02 for baseline versus P < 0.01 for

$ P < 0.005

Lastly, protein makes up l0-20Va of the diet's energy.

ADA Diet

Diet

20s

r96

218

1

HF

631 28$

134 34$

ADA and HF values

ADA versus HF values.

for ADA versus HF values.

Some studies indicate that diets like the ADA diet, that are high in carbohydrates, Iead to increased triglyceride levels, increased very low density lipoprotein tevels, and reduced high density lipoprotein i"u"tr. These would not

be beneficial in a diabeiic population concerned with

34.

maintaining optimal plasma lipid levels.

Which of the following statements are supported by

the data in Table 1?

following experiment was designed to test the

The HF diet lowered plasma glucose in these

^^Theof the ADA effect diet versus a higher rionounsaturated

subjects.

fat diet (HF) on plasma glucose and-tipid levels.

II.

IIr. Experiment

measurements.

After a baseline measurement of 1 week, l0 NIDDM patients ate the ADA diet and the HF diet for 5 weeks in random order during 2 hospital admissions. Metabolic mea-surements regarding glucose (Glc) are outlined in Table i. lNote: Hbrefers iohemogiobin.l

A. B. C.

D.

Table 1: Merabotic variables during the study.

Variable

line

ADA Diet

Plasma Glc (mg/dl-)

129

117

101*

Urinary Gtc (mg/day) Insulin Requirements

550

142

0t

84

81

70t.

11.3$

7.8

B ase.

35.

HF

i

t

P<

Diet

(G_Hb) is

beneficial or detrimental? A.

ADA and baseline. 0.0i for baseline versus HF values. $ P < 0.005 for baseline versus HF and ADA values.

B.

C.

lipoproteins are outlined in Table 2.

I and III only I and II only II and III only I, II, and III

In Table 1, glycosylared hemoglobin

8.1

Metabolic measurements on plasma lipids

baselinE

reported as an indicator of glucose metabolism in the body- Glycosylated meanJthat glucose residues are attached. G-Hb decreases on both the HF and the ADA diets during the study. Is this decrease

(unirs/day)

P < 0.02 for HF versus

than at the baseline measurement. I-ess glucose was excreted in the urine during

the HF dier compared to the

I

Glycosylated Hb (Vo)

Patients required more insulin on the HF diet

This decrease is

detrimental,

because

This decrease is

beneficial.

because

This decrease is

beneficial,

because

glycosylated hemoglobin indicates persistent levels of low plasma glucose. This decrease is detrimental . because glycosylated hemoglobin indicates persistent levels of high plasma glucose.

glycosylated hemoglobin indicates persistent levels of low plasma glucose.

and D.

glycosylated hemoglobin indicates persistent levels of high plasma glucose.


178


Biotogy

Diabetic Diet Experiment 39.

-16" The table shown below indicates the fatty acid composition of both ADA diet and HF diet in Vo energy intake/day.

Passage VI

Some patients with NIDDM maintain good glucose control using drugs that increase insulin secretion. What is the target of these drugs that affect insulin secretion?

Fatty Acid

ADA Diet

HF

2:0

0.2

0.2

4:0

0.3

0.2

6:0

6^8

7.6

6: I

u.l

0.4

8:0

2.4

2.2

8:i

t.4

29.4

8:2

'7.2

8.0

8:3

0.0

0.2

204

0.2

0.2

22:0

0.3

1.5

Total:

25.0

49.9

Table

A.

Diet

B. C.

D.

The alpha cells of the pancreas. The exocrine cells of the pancreas. The beta cells of the pancreas.

Both the exocrine and beta cells of

the

pancreas.

40.

Insulin is secreted from the pancreas into the hepatic portal vein. Which is the first major organ insulin encounters after entering this portal system?

A. B. C. D.

3

Small intestines Stomach

Brain

Liver

Which of the fatty acids was used to supplement the monounsaturated portion of the HF diet ?

A. B. C. D. 37.

Palmitoleic acid Stearic acid

Oleic acid

Palmitic acid

Which of the following oils or fats could be used to increase the amount of monounsaturated fat in a diet?

38.

A.

Olive oil

B. C.

Lard

D.

Beef ta1low

Butter

Which of the following statements is/are supported by the data in Table 2?

L IL III.

A. B. C.

D.

Copyright

Both VLDL and LDL were lower in the HF diet compared to the ADA diet.

Triglycerides were significantly lowered by the HF diet compared to baseline. HDL cholesterol was increased by the HF diet compared to the ADA diet.

I and III only I and II only II and III only I, II, and trI @


t79


Biotogy Passage

VII

The Kidney

(Questions 41-48)

The functional unit of the human kidney is the nephron

(Figure 1), and each kidney contains approximately t million of these tubular structures.

Efferent afteriole

Passage Vtr

Antidiuretic hormone (ADH) and aldosterone are both regulators of urine composition. Both hormones promote NaCl reabsorption in the ascending limb of the loop

Henle and

in the DCT and cofecting

of ducts. ADH

promotes water reabsorption in the DCTand collecting ducts while aldosterone promotes Ko secretion in the DCT.

Glomerulus

41. Constriction

A.

of the efferent arterial will lead to:

a. decrease

in glomeruiar

pressure, followed

first by a decrease and then by an increase in glomerular filtration rate.

B. al increase in glomerular

pressure, followed

first by an increase and then by a jecrease in

C.

o c

glomerular fi ltration rate. a. decrease

in glomerular

pressure, followed

first by an increase and then by a decrease in

@

o

D.

0c

glomerular fil tration rate. an increase in glomerular pressure, followed first by a decrease and then by an increase in glomerular fil tration rate.

Figure I 42.

Blood is supplied to each kidney by a renal artery. The smallest branch of this artery is the afferent arteriole, a ves.sel-influenced by sympathetic stimulation. A capillary bed (glomerulus) is fbrmed from the afferent arteriole and is contained within Bowman's capsule. The efferent arteriole that leaves the glomerulus iJ further divided into a capillary system that surrounds different portions of the

The re-absorption of sodium across the epithelial cells of the proximal convoluted tubule and into the peritubular capillary would be significantly reduced by'

A. B.

nephron.

C. D.

The plasma ultrafiltrate that passes through the glomerulus and into Bowman's capsule is essentially _

reduced secretion of ADH from the posterior pituitary. an increased rate of secretion of aldosterone from the adrenal cortex. reduced plasma glucose concentrations.

an increase

in the tubular colloid

osmotic

pressure.

protein-free and devoid of cellular structure. This filtrate

flows inro the proximal convoluted tubule (pCT), through the loop of Henle and into the distal convoluted tubule (DCT), through rhe collecting ducts and inro rhe renal p:lytt before making its way to the urerer and urinary bladder.

43.

The PCT reabsorbs essentially all of the glucose and amino acids fiitered by the glomerulus and r6ughly 70Vo

A.

of the liltered Ko, Nao, Cle, and water. Approximately 2_OVo of the Ke, Nao, Clo, and water ttrat passes from the

B.

PCT into the loop of Henle is reabsorbed it the loop. The remaining 10Vo of the Ke, Nao, and Cle is reabsorbed at the DCT. Variable amounts of water also reabsorbed at the DCT Copyright

@


Sympathetic stimuiation affects the:

C. D. lao

of the kidneys primarily

afferent arterioles and decreases urine flow. afferent arterioles and increases urine flow. efferent arterioles and decreases urine flow. efferent arterioles and increases urine flow.

the volume of the volume

of

the volume

of

the volume of

The Berketey Review Specializing in MCAT preparation

Biology 44.

The Kidney

Which of the following graphs BEST represenrs the plasma osmolarity (X-axii) of the blood and the plasma concentration of antidiuretic hormone (y_

Passage Vtr

46, All of the following true EXCEpT it:

axis)?

A.

A.

B.

B.

C. q

a

X

x

I

I

X-Axis

D.

X-Axis

47.

statements about aldosterone are

promotes sodium reabsorption

in the distal convoluted tubule. promotes the secretion of potassium into the lumen of the distal convoluted tubule. is a steroid hormone secreted by the adrenal medulla. is the major mineralocorticoid in humans.

Excessive aldosterone secretion will result in all of the following EXCEpT:

C.

D.

a

A. B. C.

q

X

X

I

D.

I

X-Axis

hyperpolarization

of nerve and

membranes. increased muscle contraction.

an excessive loss of potassium ions from the extracellular fluid. hypokalemia.

X-Axis

48, Which of the following subsrances

is filtered by

glomerulus and into Bowman's capsule?

Which of the following graphs best represents the plasma concentration of glucose 1X_aiis) and the rate ofglucose excretion in the urine (y-axis)? A.

muscle

A. B. C. D.

the

Platelets Proteins

Electrolytes Erythrocytes

B.

o

x

X

X-Axis C.

D.

x

X

I I

X-Axis

Copyright

@

by.The Berkeley Review

X-Axis

l8l


Biology Passage

VIII

Kidney & ptl

Passage VItr

(Questions 49-53) Tubular Lumen

Most metabolic reactions are highly sensitive to the pH of the fluid in which they occur. Due to this sensitivity, the hydrogen ion concentration of body fluids is closely

Interstitial Fluid

Filtered

regulated.

J II

'\l

H9o-r

,HCO:'

Buffering systems, both intracellular and extracellular,

act to minimize changes in the hydrogen ion concentration. However, these buffering systems do not have the ability to eliminate or retain hydrogen ions from the body. This role is left for the kidneys, which accomplish this regulation in two ways. The first is by altering the secretion of hydrogen ions and the second is

H,COT

-t"

HrCOr

1.^

II'O + CO2

Ir.20 + Q,O,

by altering the reabsorption of bicarbonate. Hydrogen ion excretion and bicarbonate reabsorption are both achieved by the secretion of hydrogen ions. CA = Carbonic Anhvdrase

Tutrular

lnterstitial Fluid

Lumen

NHr

Figure 2

In the body, the events shown in Figure I and Figure 2 occur together, producing a situation where a high rate of hydrogen ion secretion achieves complete reabsorption of filtered bicarbonate.

HrO + COo

t.o-

=(l 49. Urine

Which of the following situations will NOT produce a gain of hydrogen ions?

CA = Carbonic Anhydrase

Figure I

According to Figure i, the hydrogen ion to be secreted into the lumen is generated in the tubular cell rather than coming from the blood. Once in the lumen, the hydrogen combines with a buffer and is secreted in that form. In Figure 1, the bicarbonate is transported to the interstitial

A.

An increase in the concentration of CO2.

B.

Production

C. D.

exerclse. Excessive vomiting. Loss of bicarbonate due to diarrhea.

of lactic acid during

severe

fluid. 50.

The hydrogen ion in the lumen can also bind with filtered bicarbonate to form CO2 and H2O according to Figure 2. This is in essence bicarbonate reabsorption. The net result is that filtered bicarbonate disappears while bicarbonate (the result of intracellular events) appears in the blood. This situation is equivalent to one where filtered bicarbonate is reabsorbed back into the blood.

Copyright

@


ta2

The events shown in Figure 1 and Figure 2 MOST likely illustrate renal compensation for: blood hydrogen ion concentration.

A. B.

a decreased

C.

a decreased

D.

an increased sodium ion concentration.

an increased blood hydrogen ion concentration.

sodium ion concentration.


Biology

Kidney & pll

Passage VItr

51. The transport

of a hydrogen ion from a tubular cell to the tubular lumen shown in Figure I represents:

A. B. C. D.

52.

simple diffusion. facilitateddiffusion. primary active transport. secondary active transport.

Respiratory alkalosis is best characterized by

A.

an increase in the hydrogen ion concentration, resulting from an increase in the level of CO2. a decrease in the hydrogen ion concentration, resulting from a decrease in the level of CO2. an increase in the hydrogen ion concentration, resulting from a decrease in the level of CO2. a decrease in the hydrogen ion concentration, resulting from an increase in the level of CO2.

B. C. D.

53. During CO2 A. B. C. D.

Copyright

:

metabolic acidosis, the partial pressure of

can be expected to:

increase, due to increased ventilation. increase, due to decreased ventilation. decrease, due to increased ventilation. decrease, due to decreased ventilation.

@


ra5


Biology Passage

Dialysis and UltrafiItration

Passage IX

55. Which structure of


the

ultrafiltration?

Dialysis is a process used to separate molecules in solution based on their size. Semipermeable membranes that have pores smaller than the macromolecule of interest are used to contain one or more large molecules. Dialysis tubing is filled with the solution of interest, and the ends of tubing are tied. The package is immersed in a large quantity of solvent. Diffusion of molecules smaller than the pores of the tubing allows exchange of solvents, salts, and small metabolites between the solution in the tubing and the surrounding solution. This can be used to remove excess salts, for example, from a protein to be purified.

A. B. C. D.

kidney

performs

Peritubuiarcapillaries Loop of Henle

Afferent arterioles Glomerularcapiliaries

56. A student is purifying an enzyme in the laboratory by overnight dialysis. The student tests for

Ultrafiltration is a related technique that is used to concentrate macromolecules. Pressure is used to force the

solution through the semipermeable membrane. The solvent and small molecules pass through the membrane, while a more concentrated solution of macromolecules is

successful dialysis using an enzyme activity assay. What would rhe srudenr find if the dialysis tubing was punctured with a pin before the experiment?

constantly in the removal of waste products from the blood. The entire blood supply of 5-6 liters is filtered

A. B.

left behind. Normal kidneys perform ultrafiltration

approximately every 45 minutes.

C.

Diseased kidneys require exterior dialysis support to prevent dangerous increases in the plasma solute load,

D.

including, salts, electrolytes, and urea. Uremia is the condition of having an increased urea plasma concentration.

No enzyme activity in the bag. Increased enzyme activity in the bag compared to the activity measured before the experiment. There is no effect caused by the pinhole, and the enzyme activity in the bag is the same as before. Slightly decreased enzyme acrivity in the bag compared to the activity measured before the experiment.

Dialysis support is conventionally performed in one of

two ways. The first involves passing the blood through

a

machine containing semipermeable membranes that allow waste products to diffuse out. An in vivo technique uses the patient's peritoneal membranes (in the abdominal

cavity) to form an intraabdominal dialysis system. A sterile fluid is instilled through an abdominal catheter, allowed to remain from 15 minutes to t hour, and drained through a second abdominal catheter. This process is repeated many times. Both methods are time-consuming, but they allow the necessary removal of waste products

57. Which of the following hormones

would most likely be supplernented during chronic kidney disease?

A. B. C. D.

from the blood.

54.

Erythropoetin Antidiuretic hormone (ADH) Epinephrine and norepinephrine Insulin

What complications could result from the peritoneal dialysis procedure?

I. II. nI.

Increased risk ofabdominal cavity infections. Changes in electrolyte balance. Kidney damage by ultrafiltration.

A. B. C. D.


Copyright

58.

A. B. C. D"

II and III only I, II, and III

@


What would be

ta4

a

physiological effect of uremia?

Increased blood pH Edema Increased urination

Alkalosis


Biology

Dialysis and Ultrafiltration

Passage IX

59. When a dialysis machine is used to clean blood, what precautions are taken to avoid loss of glucose and salt?

A.

Use

of a dialysis

membrane that retains

glucose and salts. B.

Use a dialysis solution that is isotonic to the

C.

blood in glucose and salts. The patients drinks an electrolyte and glucose drink during dialysis.

D.

Nothing can be done to avoid loss, so the patient has an IV of saline and glucose at the same time.

60. If a person has 5 liters of blood and it is passed completely through the kidneys every 45 minutes, what is the glomerular filtration rate?

A. B. C. D.

Copyright

225 mUmln

111ml/min 66 ml/min 9 ml/min

@


la5


Biology Passage

Caffeine

Passage X


X (Questions 61-66)

hormones would most

increase plasma FFA concentration

Caffeine is a commonly used beverage in the U.S. The average individual consumes 2-3 cups per day, with

A. B. C. D.

some people drinking 10-15 cups. Coffee contains about 150 mg caffeine/cup for drip-style makers. Caffeine is the

physiologically active agent in coffee. The following experiments were carried out to examine the effects of

?

CCK Thyroid hormone Epinephrine

Insulin

caffeine in coffee on free fatty acid (FFA) metabolism: 63.

All

subjects followed on overnight fast. At 8:00 AM, each subject drank one of the following beverages:

Experiment

In Experiment 4, how did the sucrose lead to lower FFA concentrations?

A.

l:

B.

500 mL hot water containing 5 g instant coffee (250 mg caffeine) and 3 tablets of sodium saccharin (15 mg).

C.

Experiment 2:

D.

500 mL hot water containing 5 g decaffeinated coffee (20 mg caffeine) and 3 tablets of sodium saccharin

Sucrose was digested to glucose and fructose; fructose mediated the effects on FFAs. Sucrose stimulated glucagon, which decreased release offatty acids from adipocytes.

Sucrose acted as a competitive inhibitor of caffeine. Sucrose stimulated insulin, which decreased release offatty acids from adipocytes.

(15 mg). 64. The action of caffeine is mediated through increased cAMP levels in target cells. What is the enzyme that is activated by second messengers to make cAMP?

Experiment 3:

500 mL hot water containing 3 tablets of sodium saccharin (15 mg)

A. B. C. D.

Experiment 4: 500 mL hot water containing 5 g instant coffee (250 mg caffeine) and 25 g sucrose

Blood samples for all experiments were taken before the start of the experiment and at 1, 2, and 3 hours following ingestion of the test beverage. The following table follows the FFA concentrations throughout the

65.

study:

+93*

C.

There are slight decreases

-59

+11

D.

-e0t

+1201

3 hours

82t

+245{

+384+

16x

+18*

0

19

0

-e6I

I

0

2

0

--)

4

+1

* Not signihcant compared to Experiment Signihcant compared to Experiment

Table 1: Mean Changes in Plasma FFAs (ttEQ/L)


mechanisms is the most important in raising plasma FFA concentrations? Lipolysis

B. C. D.

Lipogenesis Increased fat oxidation lncreased dietary fat

Copyright

@


in FFA concentration due to the 20 mg of caffeine present. The effects are the same as drinking water.

3

1

A.

with 25 g sucrose.

66. What was the point of Experiment 3?

3

{ Signihcant compared to Experiment 2 and Experiment

f

A. B.

2 hours

0 hours

From Experiment 2, what would you conclude about the effects of the caffeine present in decaffeinated coffee on plasma FFA concentration?

There are slight increases in FFA concentration due to the 20 mg of caffeine present The effects are the same as drinking coffee

t hour

Expt.

Epinephrine G protein Adenylate cyclase Proteinphosphorylase

I. il. III.

To determine the effects of hot water To determine the effects of sodium saccharin To establish control data

A.

I only II only I and II only I, II, and Itr

B. C.

P.

ta6


Biology Passage

Kidney & Calcium


68.

Kidney function can be charactefized by three

:rLrcesses. The first involves movement of an essentialiy

:

Passage XI

According to the diagrams in Figure l, all of the following statements concerning substances X, y. andZ are false EXCEpT:

liein free plasma from renai glomerular capillaries into

!

r'.iman's capsule. This is referred to as giomerular . -i::tion. Tubular reabsorption refers to the movement of : -::lances from the renal tubular lumen to the peritubular

Afferent Arteriole

//

-.;:-iary' plasma. Movement in the opposite direction is -:r:d tubular secretion.

Efferent

Arteriole

./

'.

The kidney, among its many

functions, is -, -',6fl with reguiating levels ofother calcium ion. The , :,.,-e11u1ar calcium concentration is normally held ' : :.r'. 3ly constant. However, a low calcium ion - .::lrration increases the excitability of nerve and

Peritubuiar

Capillary

- ---..' .'ell membranes.

Urine

.-.1;rum homeostasis is achieved through the nins of three major effector sites. These sites are : l:e\'. bone, and gastrointestinal (GI) tract. Almost " ' : - i total-body calcium is contained in bone, which is . :.::: * ork of organic molecules upon which calcium - -. : -- : ^ :..

-

'

):.-rie crvstals are deposited. Bone is a dvnamic

tissue " -. ran either remove or deposit calcium into the , ..-:--ular fluid. In the Gi tract, control of active --.-:: :t svstems that move calcium into the blood can : - .: larqe fluctuations in the amount of calcium

.:::J. The kidney modulates calcium levels by :, :::'on of both calcium and phosphate ions. " :- '--t:3e effector sites are under control of parathyroid -- -:. uhose plasma concentration fluctuates directly 1 ,: '_sr ro changes in plasma calcium levels.

"" :

Figure

A. B. C. D.

i

Z is filtered and partially reabsorbed. Y is totally reabsorbed. X is filtered and secreted, but not reabsorbed.

X, Y, and Z are all secreted.

- :he following statements about glomerular ::::t t,l are true EXCEPT: -rlid

pressure

in Bowman's capsule

69.

opposes

-.-rration.

rsnotic fbrce due to protein in plasma favors

In response to a decrease in plasma calcium leveis, one would most iikely expect to see: A.

__-rat10n.

increased plasma parathyroid levels, with increased urinary calcium excretion.

_:-omerular capillary blood pressure favors :-."ration. - -,:ces favoring filtration are larger than those

increased plasma parathyroid levels, with increased removal of bone. decreased plasma parathyroid levels, with decreased urinary calcium excretion. decreased plasma parathyroid levels, with decreased removai of bone.

::osing filtration.

ta7


Biology 70,

Kidney & Calcium

According to information in the passage, it can

Passage XI

be

concluded that:

A. B.

the concentration of extracellular calcium in no way effects neuromuscular excitability.

persons with low extracellular calcium

concentrations may suffer total skeletal muscle spasms.

C.

persons with increased extracellular calcium concentrations may suffer from frequent muscle spasms.

D.

the effect of extracellular

calcium

concentration on membranes is distinct from its role as a mediator of muscle contraction.

71.

A decrease in calcium will most likely

A. B.

cause:

an increase in the extracellular concentration of phosphate. no change in the extracellular concentration phosphate.

C. a

decrease

of

in the urinary excretion of

phosphate.

D. a reduction in the tubular reabsorption

of

phosphate.

72.

Parathyroid hormone is secreted by the:

A. B. C. D.

Copyright

anterior pituitary.

posteriorpituitary. parathyroid gland.

thyroid gland.

@


ta8


Biology Passage

XII

Lymph 75.

(Questions 73-79)

The lymphatic system has 3 functions: (1) its .''mphocytes provide immunological defenses, (2) it 'iansports absorbed fat from the small intestine to the :.ood, and (3) it transports interstitial fluid back to the :1ood.

-...

Passage XII

Fat-soluble vitamins are carried from the cells of the small intestine into the blood system via the lymph. They are packaged into chylomicrons and with fatty acids. Which of the following vitamins are fatsoiuble vitamins?

I.

il.

nI.

Both lymph nodes and lymphocytes of the immune stem play important roles in immunological defense.

j -' mph nodes contain phagocytic cells through which -'.mph is filtered before returning to the blood supply. l'. mph nodes also contains germinal centers that produce

B.

-'.mphocytes, which are present in the lymph and in the

D.

A. C.

:-:sma.

76. The lymphatic system also carries dietary fat from the :rall intestine to the blood supply. Absorbed fat, r:luding long chain fatty acids and cholesterol, along .'" ith fat-soluble vitamins, are packaged into chylomicrons

:t, the cells of the small intestinal

::r'lomicrons transport fat from the intestine to

I only II only II and III only I, II, and III

What is the function of the lymph node?

A. B. C.

mucosa. These

Vitamin A Vitamin K Vitamin E

D.

To destroy microorganisms. To filter chylomicrons out of the iymph. To produce chylomicrons. To add intestinal nutrients to the lymph.

the

blood circulation. Other nutrients from digestive =:neral :iocesses in the small intestine travel first to the liver via

What force moves the lymph inside the lymphatic vessels?

:,e hepatic portal vein.

q

The liquid part of lymph is an ultrafiltrate of plasma, : -'led interstitial fluid. This is the fluid that is squeezed :,, of capillaries under pressure. Interstitiai fluid enters :: ivmph capillaries through its endotheliai cells which

:;\'e porous junctions. This fluid, along with

Blood pressure.

vessels.

organs also produces another ultrafiltrate?

What is the consequence of the chylomicron system of fatty acid packaging?

A. B.

A. B. C. D.

Muscle and adipose tissue have first access to dietary fatty acids from chylomicrons. Liver tissue has first access to dietary fatty acids from chylomicrons.

79.

C. Microorganisms are packaged into D. -{.

III.

The term ultrafiltrate means fluid formed by the hydrostatic pressure of the blood acting against a semi-permeable membrane. Interstitial fluid is one example of an ultrafiltrate. Which of the foliowing

r" the right and left subclavian veins.

-3.

Squeezing by the muscles. Contraction of fhe muscular layer of the lymph

A. I only B. I and II only C. II and III only D. I, II, and III

::', iomicrons, microorganisms, and lymphocytes, is then ,z.led lymph. From the lymph capillaries, the lymph r-rves into the lymphatic vesseis, which have a 3-layer . -iucture simiiar to veins. The lymphatic system -,:imately returns the fluid to the cardiovascular system .

I. II.

Liver Kidney Pancreas

Salivary gland

Which of the following statements are TRUE of both lymph and plasma?

chylomicrons.

Interstitial fluid carries chylomicrons.

I. [I. [II.

Both contain chylomicrons after a meal. Both contain lymphocytes. Both contain erythrocytes.

A.


What is the major component of lymph?

A. B. C.

D.

Fat-soluble vitamins Chylomicrons Water Triglycerides

-rrpyright O by The Berkeley Review

B. C. D.

la9

II I,

and III only and III

II,


Biology XIII

Passage

Kenal Clearance 81.

(Questions 80-86)

The concept of renal clearance, based on the principle

of mass balance, is a means to quantify the excretory function of the kidney. The relationship of clearance (C) to the concentration of a substance in the urine ([ ]r), the urine flow rate (Fu), and the plasma concentration of a

c=([]u)x

What is the volume of urine that could be collected in 30 seconds if the [ ]u of substance Y is 50 mg/ml, the [ ]O of substance Y is 1 mg/ml, and the clearance for substance Y is 50 ml/min? A. B. C. D.

substance ([ Jp) is

Passage XIII

0.00025 L 0.0005 L 0.001 L 0.0015 L

(Fu)/ []p

The glomerular filtration rate (GFR) is the volume per unit time that enters the kidney from the plasma. Under certain conditions, the amount of a substance filtered is equal to the amount excreted in the urine. To calculate the amount filtered, one can multiply the GFR by the 82.

Isubstance]0.

Which of the following equations represents the glomerular filtration rate (GFR)?

Similar to the plasma, not all of a substance traveling through the renal artery is filtered at the glomerulus. The fraction of plasma filtered can be calculated as the ratio of GFR to renal plasma flow (RPF). The fraction normally takes on values

of

75Vo

A. B. C.

to 20Vo.

D.

= [substance]p x []u x Fu x (Fu) = ([ substance Jp) x ([ Ju) GFR = (l Ju x Fu) / ([ subsrance ]o) (GFR) x ([ ]u) x ([ Jp) x (Fu) = GFR

(GFR)

1

The renal handling of metabolites may be instrumental

in diagnosing certain clinical disorders such as diabetes. When determining the amount of glucose reabsorbed in the nephron, one must subtract the amount excreted from the amount filtered at the glomerulus. The amount of glucose reabsorbed by epithelial cells of the tubule can be thought of as having a maximum transport rate, known as the tubular transport maximum. The [plasma glucose] will ultimately determine whether glucose is excreted or is 1007o reabsorbed. The level of plasma glucose which first gives rise to glucose in the urine is termed the plasma

83.

According to the passage, the amount of a substance filtered is equal to the amount excreted in the urine

only under certain conditions. Which of the following conditions CANNOT exist for this statement to be TRUE?

threshold.

A.

80.

Mass balance describes the principle that the mass of a substance entering the kidneys must equal the mass of that substance leaving the kidneys. Using the following key, and information from the passage, which relationship BEST describes this principle?

B. C.

D.

The substance has no barrier to filtration at the glomerulus. The substance must not be reabsorbed and must be secreted by the nephron. The substance must not be metabolized or produced by the kidney.

The rate

of glomerular filtration

cannot be

altered by the substance. Key

[ ]A = concentration [ ]V = concentration

of substance in renal artery of substance in renal vein

PFo = plasma flow rate in renal artery 84.

PFy = plasma flow rate in renal vein

A. B. c. D. Copyright

[]e x PFy = ([ ]u x Fu) + ([ ]y x PF4) ([]n x PFu) + ([]u x Fo) = []v x PFa ([]e x PFa) + ([]ux Fu) - []y xPFy

[]r

@

x PF4

=

([ ]u

x F") +


([ ]u

x

Glucose is reabsorbed through the epithelial cells the:

A. B. C.

PFv)

D.

19()

of

proximal convoluted tubule. loop of Henle. distal convoluted tubule. collecting duct.


Biology $5.

Kenal Clearance

Passage XIII

According to the graph beiow, the plasma threshold ls:

a^ ;E

FE"

Ho vro

E< o

J

0246810 Plasma Glucose (mg/ml)

-{. B. C. D.

3.5 mg / ml. 7.6 mg / ml.

380 mg / ml. 400 mg / ml.

The following graph illustrates the influence of size and charge on a substance's filterability at the glomerulus. Which of the following statements is most likely true based on this graph?

.:.

0.8

-o

b ii 9

0.6

0.4

6 O &

0.2

Polyanionic dextran

18 22 26 30 34 38

42

Effective Molecular Radius (A)

The cationic species has the filterability,

highest

because cationic atoms are smaller

than anionic atoms.

The cationic species has the filterability,

highest

because of the presence of anionic

glycoproteins on the surface of all glomerular

filtration components. C.

The anionic species has the

highest filterability, because cationic atoms are larger than anionic atoms.

The anionic species has the highest filterability, because of the presence of cationic glycoproteins on the surface of all glomerular filtration components.


191


Biology

Gastrointestinal Junction

89. It is known that a hypertonic solution in


Passage

Passage XIV

duodenum decreases the rate This mechanism prevents a:

The pyloric sphincter separates the terminal portion of the stomach (gastric antrum) fiom the beginning segment of the small intestine (duodenum). This sphincter, which consists of a ring of connective tissue preceded by two distinct rings of circular smooth muscle, carries out two important functions. First, it heips to regulate the rate of gastric emptying into the duodenum. The transfer of the stomach's contents to the duodenum is dependent on the duodenum's ability to process chyme. Second, it prevents the reflux of small intestinal contents into the antrum. These two functions are controlled by both hormones and neurai input. Norepinephrine and cholecystokinin (CCK), for example, both cause a constriction of this sphincter.

90.

A.

rise in the blood pH.

B. C. D.

decrease in the blood pH.

the

of gastric emptying.

rise in the biood volume. decrease in the blood volume.

100 mM HCI was injected into the duodenum of a dog while the contractiie activities of both the antrum (AC) and duodenum (DC) were measured. Which of the following graphs BESTS represents the expected results of the experiment?

It

has been discovered that several distinct mechanisms

can lead to an alteration in the rate of gastric emptying. A lowering of the pH in the duodenum has been shown to slow gastric emptying. This decrease in pH has been shown to not only affect motility, but also increase the release of bicarbonate ion into the small intestine via the hormone secretin. In addition to pH, the presence of fat

A.

DC

results in a retarded gastric emptying. This response is believed to be carried out mainly by the hormone CCK. The hormone gastric inhibitory peptide (GIP) is also believed to be a component of this mechanism, as GIP has been shown to decrease the rate of gastric emptying.

Starts

Ends

lllllililtililtililtI

of amino acids and peptides in the stomach.

|ilrililtlliltIlII

Based on information in the passage, which of the following statements is TRUE?

A. A low rate of gastric emptying

t HCL Infusion

B.

Finally, the secretion of gastrin lowers the rate of gastric emptying. Gastrin is secreted in response to the presence

87.

t HCL Infusion

t

i

HCL Infusion

HCL Infusion

Starts

Ends

C.

may lead to

duodenal ulcers.

B. A high rate of gastric emptying may lead to

IIIII

stomach ulcers.

C.

Regurgitation of duodenal contents may lead

D.

to duodenal uicers. Regurgitation of duodenal contents may lead to stomach ulcers.

ililililililil1ililil

IIrI

AC I

ilililililt I I I I I I I I I ilillllllll

t

t

HCL Infusion

HCL Infusion

Starts

Ends

DC

D.

88. Norepinephrine is released in response

to sympathetic stimuiation and acts to increase pyloric sphincter contraction. Norepinephrine is an example

of

ilil1|iltilllrililrI

a:

A.

mineralocorticoid.

B. C. D.

catecholamine. steroid. peptide.

Copyright

@


t 192

i

HCL Infusion

HCL Infusion

Starts

Ends

The Berkeley Revier'r' Specializing in MCAT Preparation

Biology 91. As

stated

Gastrointestinal Junction in the passage, GIp is

Passage XIV

involved in

decreasing the rate of gastric emptying in response to fats in the small intestine. This hormone is mosr

likely to increase the secretion following hormones?

A. B. C. D.

qrL

of which of

the

Glucagon

Bile Insulin Cortisol

Cholecystokinin is known to stimulate contraction of the antrum and promote constriction of the pyloric sphincter. The overall effect of CCK is to: -{.

increase the rate

of gastric

emptying, and

contribute to the mixing of stomach contents.

B.

C. D.

increase the rate of gastric emptying and inhibit the mixing of stomach contents. decrease the rate of gastric emptying and contribute to the mixing of stomach contents. decrease the rate of gastric emptying and inhibit the mixing of stomach contenis.

\\-hich of the following statements is FALSE regarding parietal cell secretion of protons and -nancreatic release of bicarbonate ion?

-\. The pH of the stomach decreases. B. The pH ofthe blood increases. C. The acid entering the smali intestine D.

is

neutralized.

The bicarbonate ion is derived from carbon dioxide.

-

,r:;ight

@


193


Biology Passage

Cholera Toxin 96.


Cholera is a potentially lethal disease caused by the bacterium Vibrio cholerae. Individuals afflicted with cholera suffer from severe dehydration due to diarrhea,

Cholera toxin affects a second-messenger signal cascade. The toxin would NOT affect the actions of:

A. B. C. D.

which left untreated can be fatal. V. choLerae bacteria replicate in the small intestine of the host after infection (usually from contaminated drinking water), producing a protein known as cholera

97.

Passage XV

epinephrine. gastrin.

norepinephrine. testosterone.

Why is rehydration therapy alone not sufficient cure severe cholera?

toxin (CT). CT then binds to the surface of intestinal epithelial cells, and through a complex series of events

A.

The body cannot establish normal blood pressure after severe dehydration.

irreversibly alters membrane-associated G-proteins such

that they are constitutively active, allowing for the constant overproduction of the second messenger cyclicAMP. This triggers a sodium and chloride efflux that is responsible for the water loss associated with diarrhea.

B.

Pores

in the plasma membrane will

allow

water to eventually leak back into the intestinal

C.

Rehydration therapy for cholera sufferers takes advantage of an inwardly-pumping sodium/glucose symport present in the apical membranes of intestinal cells.

D.

Iumen.

Chloride ion must be ingested to replace that which is secreted.

The underlying bacterial infection must

be

treated so that dehydrating diarrhea ceases. 98.

CT is frequently used in research to help elucidate the functions of G-protein-related signal transductional

mechanisms. Recently, lines of mice have been engineered which have a transgene encoding the CT

Humans are one of the few mammals affected by cholera. Unaffected mammals :

A. B. C. D.

protein linked to a promoter which allows protein expression only in cells of the pituitary gland which produce growth hormone. These mice exhibit gigantism, an effect of excessive growth hormone.

99.

Iack G-protein mediated signaling. always produce large amounts of cyclic-AMP. iack membrane receptors for cholera toxin. normally have low cyclic-AMP leveis.

Assuming they could be applied specifically to the

pituitary, which of the lbllowing would be an effective inhibitor of the gigantism seen in the transgenic mice described above?

94.

A. Salt loss triggers the opening of B. C. D.

95.

I. il. III.

Cholera toxin causes intestinal cells to secrete large amounts of Nao/Cle into the intestinal lumen. How does this cause dehydration?

Excessivecyclic-AMP

A. I only B. II only C. III only D. II and III only

water-

permeable pores in the plasma membrane.

Salt loss creates an osmotic gradient, which water follows out of the cell. Salt loss triggers the active transport of water out of the cel1. The distal convoluted tubules of the kidney reabsorb more water.

100.

A recent hypothesis

suggests that individuals who are heterozygous for the recessive gene that causes cystic fibrosis (CF) may exhibit some resistance to

cholera (i.e., the CF gene confers some selective advantage). Which of the following statements provides evidence in support ofthis theory?

What should be given to help rehydrate a cholera

A.

patient?

A. B. C. D.

An enzyme which degrades cyclic-AMP Excessive normai G-proteins

The frequency of the CF gene is

1ow

populations chronically exposed to choiera. B.

Distilled water.

The frequency of the CF gene is high populations chronically exposed to cholera.

Water and Na@ ions. Water and glucose. Water, glucose, and Nae ions.

C.

The frequency of the CF gene is high

D.

Individuals with cholera are

popuiations rarely exposed to cholera.

often

heterozygous for the CF gene.


194


Biology 1.

i l

i

Section III Answers

A is correct, single-unit fibers are coupled, with many individual celis containing gap junctions. This questioir draws on our knowledge of smooth muscle types. Even if we did not know the ansivJr ;grrt rro11 the question, the answer can be arrived at through some common sense. If something is acting as a singlJunit, then it is titety'tnat the muscle fibers which make up the unit will be coupled to each other. T=hereforelr" eliminate choice B. Furthermore, in order for the muscle fibers to be electrically coupled to each other, many"un individual cells should have gap junctions. Renember, gap junctions offer connections between two different c"llulu. cytoplasms, which will allow current to move from cell to cell in a rapid fashion. The correct choice is A. C is correct, calcium channels is slow. As in cardiac muscle, an inward calcium current is an important component of the action potential in smooth muscle. We are told that the conduction velocity is low aiong smooth muscle fiber. It is logical to assume that the conduction velocity of action potentials aiong Ct smooth-muscle fibers is slow because activation of the calcium channels is slow. This question was just designed to make one think about the roie of calcium in the conduction velocity of an action potential of smooth muscle. the correct choice is C.

:

',1.

Gastrointestinal & Renal

C is correct' number of action potentials within a set. We are looking at the diagrams to get an understanding of the relationship between force and action potentials. In the graphs *e r"e three acion poten;ials giving rise to i larger force of contraction when compared to two action potentials within the set. Therefore, we can say that the sizJof the force depends on the_ number of action potentials within a burst. There is simply no evidence or support from the graph to consider the other_answers as viable possibilities. We cannot compareâmplitudes as the aciitn potentials are all of the same size. Thus, the force of contraction depends on the number of action potentials. The correct choice is C.

C is correct, mixing of intestinal contents, breaking larger particles into smaller ones. As stated in the passage, segmentation allows alternating contraction and relaxation within the circuiar muscle. This type of movement-is similar to opening and closing your hand. If you had a tomato in your hand, it would soon turn to mush. The same happens to food within the intestines. It is continually being broken down into smaller and smaller pieces. There is no indication that absorption of nutrients occurs because of segmentation, nor is there any direit evidence that segmentation causes movement of the intestinal contents through the GI tract. Even tirough peristalsis and segmentation may stimulate (indirectly) the secretion of digestive enzymes, this is not the best ansr"er choice. The conect choice is C.

D is correct, breaking larger particles into smaller particles increases the surface area available to digestive enzymes. We are told from the question that miring aids in digestion. We know that mixing will break-larger particles into smalier ones. This aids in digestion because the surface area exposed to digestive inzymes will hive increased. These enzymes wili break down complex molecules into those molecules that ian be absorbed across the

layer and eventually be used by the body. While the other answers may seem attractive, one has to ask themselves, rvhat would this do to aid in digestion. Moving through the gut faster does not seem to aid in digestion, thereby eliminating choice A. Furthermore, there is simply no evidence from the passage that smaller moleiules will cany less charge or cross over the epithelial layer of the lumen any easier. Absorption will most likely be occurring through channels in the cell layer. Therefore, we can say that increasing the surface area availabie to digestivE enzymes will aid in digestion. The correct choice is D.

A is correct, I only. Enzymes are proteins, and their amino acids, can themseives be digested into amino acids and taken back up to feed the body. Choice I is correct. Mucus, a protective coating on the GI tract, is made of polysaccharides. Mucus is not digestible. How could it protect tissue if it were digested by our enzymes? Choice II is incorrect. Bicarbonate is not protein. Choice

III

is incorrect. The correct choice is A.

D is correct,947o. If we merely looked at food intake and fecal output, we would have calculated 90/100 =90Vo. This did not consider secretion of proteins from the gut that were recycled. Choice C is incorrect. If we did not include food intake, we would have calculated 63Va. Choice A is incorrect. Choice B is also incorrect. The total protein available was 100 gm from the diet and 70 gm from gut secretions to make a total of 170 gm ofprotein available. 10 gm was excreted in feces, so 160 gm was digested and absorbed. 160/l-10 =94Vo. Notice that the skin is not included in this calculation The correct choice is D.

C is conect, urea. Choice A is the chemical structure for uric Choice D is hydrazine. The correct choice is C.

: .:ight O by The Berkeley Review

195

acid. Choice B is ammonia. Choice C is urea. The Berkeley Review Specializing in MCAT Preparation

Biology

Gastrointestinal & Kenal

D is correct, white cells, liver, muscle, gut. From Figure

9.

day: The correct choice is D.

1,

Section III Answers

look at the amount of protein each tissue turns over each

10.

B is correct, give isotopically labeled albumin, sample the blood periodically, and calculate the decay rate oflabeled albumin. The concentration of albumin will not give information about the turnover rate. Choice A is incorrect. An isotopic label will allow quantification of the decay rate of the label is the blood as the albumin is recycled. A nonlabeled albumin will not help. Choice C is incorrect. Since a turnover rate in the body involves 111uny diff"r"nt interactions, studying this is a test-tube would not work. Choice D is incorrect. The correct choice is B.

11.

D is correct, skeletai muscle. The first tissue to break down would be the most expendable one. Smooth muscle is the muscle of the diaphragm and many other internal organs. This is not expendable. Choice A is incorrect. Cardiac muscle is heart tissue. It is not the first to be broken down either. Choice B is incorrect. Enzymes of energy metabolism are required even during a fast, especially the enzymes controlling gluconeogenesis. ihey are not used first, either. Choice C is incorrect. Skeletal muscle serves as a reserve of amino acids foi use during a fast. It is the first tissue to break down. D is correct.

12.

C is correct, between 0.6 and 1.5 gn/kg. The nitrogen balance witl be zero mg/kg at the appropriate protein intake. The chart moves from negative to positive between 0.6 and 1.5 gm/kg. This means zero iJcontainediomewhere in that intake range. The correct choice is C.

13.

A is correct, CCK. CCK from the small intestine is stimulated by free fatty acids entering the duodenum. CCK secretion will stimulate bile salt and pancreatic juice secretion. Secretin will be stimulated by gastric acids and high

protein content in the small intestine. The correct choice is A. 14.

D is correct, enterokinase. Enterokinase (enteropeptidase) is the enzyme located within the mucosal layer of the small intestine. It catalyzes the conversion of trypsinogen to trypsin. Trypsin will catalyze the remaining conversions. Secretin and CCK are both hormones that will stimulate release of these proenzymes from the

pancreas, while bile salts, which are not enzymes, are produced in the

liver. The correct choice is D.

15.

C is correct, protein deficiency. The lack of enteropeptidase will result in decreases in the breakdown of protein products entering the small intestine. Breakdown of these proteins is essential for absorption by the small iniestine. Enteropeptidase does not affect fat or B12 metabolism. The correct choice is c.

1,6.

A is correct, secretin. Secretin is stimulated by gastric acids entering the small intestine. It will stimulate secretion of bicarbonate from the pancreas to neutralize tf,e acidity of the gaitric juices. Enterokinase and proenzymes are

enzymes that do not affect gastric acids. CCK will stimulate the secretion of bile salts needed for the emuliification of fats in the small intestine. The correct choice is A.

17.

A is correct, trypsin.

Acute pancreatitis is produced by abnormal levels of trypsin, which is produced as

a

proenzyme (trypsinogen) in the pancreas. Trypsin is a powerful enzyme that is not normally active in the pancreas. Activation of this enzyme within the pancreas will cause the breakdown/digestion of the pancreas itself. One type of damage to the pancreas involves lysolecithin (as outlined in the question). Secretin is a hormone in the small intestine. Bile salts are not found in the pancreas. Endopeptidase is a fictitious enzyme. It sounds like enteropeptidase (also known as enterokinase), which is the enzyme responsible for converting trypsinogen to trypsin. The correct choice is A. 18.

C is correct, dissociate into CO2 and H2O in order to allow the NaCl solution in the small intestine to remain neutral. The whole point behind the function of the small intestine is to neutralize the acidic chyme that enters from the stomach. The bicarbonate ion (HCO:e) that is released from the pancreas neutralizes the HCI from the stomach by forming the neutral salt, NaCl. If we want to form more of this neutral salt, then we must pull the reaction given in the question to the right (via LeChAtelier's principle). We can do that by removing carbonic acid (H2CO3). How? By allowing it to dissociate to CO2 and H2O. The CO2 is absorbed into the blood and carried back to the lungs where it is blown off as a gas. The H2O becomes part of the intestinal fluid. One reason that we do not want to increase the acidity of the intestinal lumen is because peptic ulcers can result. The correct choice is C.

19.

C is correct, pancreatic ductal cells. Severe duodenal ulcers, leading to the removal of the duodenum, will result in the removal of cell-types associated with the upper small intestine. Two important cell types are those that secrete CCK and secretin. CCK stimulates gallbladder contraction. The gallbladder stores and concentrates bile. If the

Copyright

@


t96


Biotogy


Section III Answers

cells that secrete CCK are removed, then contraction of the gallbladder will not increase. The gallbladder tends to stay in the relaxed state for longer periods of time. Secretion of HCl from the parietal cells in the stomach is inhibited by GIP, an enterogasterone released by cells in the duodenum. Removal of GlP-secreting cells will remove the inhibition on the parietal cells and lead to an increase in parietal cell secretion (of HCl). Both CCK and secretin act to inhibit gastric emptying into the smail intestine. Removal of the cells that secrete these two hormones will result in an increase in the passage of chyme into the small intestine. Secretin also acts to increase the production of bicarbonate form the pancreatic ductal cells. Removal of the cells which secrete secretin will lead to a DECREASE in the production of pancreatic bicarbonate and a DECREASE in pancreatic ductal ceil activity. The correct choice is C. 20.

B is correct, delay the appearance of acid in the stomach. The name antacid tells us what it does. The antacid tablets act to neutralize the increase of gastrointestinal acid. This is exactly what the two equations in the question are addressing. We do not see acid being formed in either equation. Instead, we see the neutralization of HCl. We can eliminate choices A and C. Even though it is true that water is a by-product of the reaction with the antacid and HCl, it is not addressing the question as to why antacids should be taken 1 to 3 hours after a meal. The deiay in taking antacids after a meal is simply to delay the appearance of acid in the stomach. This is seen from the graph in the question. If the antacid is taken t hours after a meal, the appearance of acid in the stomach is delayed for about 2 hours. We can eliminate choice D. The correct choice is B.

2t.

C is correct, trypsin would be inactive, and the other pancreatic enzymes would remain as zymogens. This is to test our understanding of the passage. Trypsin is required to activate all other secreted pancreatic digestive enzymes. Raw soybeans contain a trypsin inhibitor, so trypsin would be inactivated. Choices A and D are incorrect. Since the other enzymes require cleaving by trypsin to achieve activity, they will remain inactive as zymogens. Choice B is incorrect. The correct choice is C.

))

C is correct, gastric epithelium. The cells that are most rapidly turning over of those listed are in the gastric epithelium. Neurons and muscle cells turnover minimally, if at all, in adults. The correct choice is C.

23.

D is correct, I, II, and

III.

Since the passage tells you mucus lines the digestive tract, we could just pick choice D mucus is present in the mouth to protect the tissue and to lubricate food. Choice I is conect. and be done. However,

Mucus is present in the colon. Choice II is correct. Mucus is very important in protecting the small intestine. particularly the duodenum, from auto-digestion. Choice III is correct. The correct choice is D. 24.

D is correct, 2.0. The stomach is acidic. choice C, a basic pH, is incorrect. A pH of 6.8 is almost neutral, so choice A is incorrect. A pH of 0.2 is 10 times more acidic than a pH of 2.0, and is much too acidic. A pH of 2.0 is corect. The correct choice is D.

)<

D is correct, I, II, and III. Enterokinase activates trypsin. If enterokinase entered the pancreas via the pancreatic duct, then trypsin would be activated, and it would activate the other zymogens. They would all go to work on the pancreatic tissue, causing damage so that pancreatic enzymes leak into the blood. Extensive damage could damage the islet cells that secrete insulin, leading to impaired insulin secretion. Actually, a person would probably be very sick or have some sort of surgery before the damage reached this level. It is important that we think of both the exocrine and the endocrine parts ofthe pancreas. The correct choice is D.

26.

A is correct, I only. The stomach's parietal cells

are the acid-secreting

cells. HCl, hydrochloric acid, is the acid in

stomach contents. The correct choice is A. 'r1

C is correct, carbohydrate. The give-away is in the name polysaccharide. This term refers to sugar units, so it is carbohydrate derivative. The correct choice is C.

28.

C is correct, hydrolyzation of fat droplets in the small intestine. The hydrolyzation of fat droplets is dependent on the enzyme pancreatic lipase. This enzyme is activated by the pancreatic enzyme trypsin. Emulsification of fats is dependlnt on bile salts iecreted by the gall bladder. Stimulation of bile secretion is induced by increase in fat content in the small intestine and CCK stimulation. The correct choice is C.

29.

A is correct, emulsification of fats.

a

The removal of household grease does not involve micelle endocytosis, exocytosis of fatty acids, or packing of fat products in the endopiasmic reticulum. The correet choice is A.


t97


Biotogy

Gastrointestinal 6t Kenal

Section III Answers

30.

C is correct, decrease in intestinal cell microvilli. Because of the soap-like appearance of the stools, bile salt secretion is probably normal. Therefore, the problem most likely lies in the absorption of the fat droplets. Decreases in the surface area of the small intestine due to decreases in the microvilli will initiate this respbnse. Increases in pancreatic lipase alone will not cause this problem because any increase in free fatty acids and micelles will usually be absorbed by the intestines. Chylomicron production is not involved with this process. The correct choice is C.

31.

C is correct, inhibition of cholesterol absorption in the small intestine. Cholesterol is dependent on bile salts for absorption in the small intestine. Fatty acids, even though aided by bile salts, are still able to be absorbed. With the increase in fatty acid and cholesterol content in the small intestine, CCK secretion will be elevated. The correct choice is C.

32.

D is correct, cholesterol. By the time chylomicrons reach the liver, lipoprotein lipase wiil have cleaved all the fatty acids from the molecule. Lipoprotein iipase is found in the arterial system. The bile salts remaining in the circulatory system are not attached to the chylomicrons. The correct choice is D.

33.

C is correct, decreased levels of lipoprotein lipase. If levels of lipoprotein lipase are decreased, chylomicron deposition of fats in the liver will increase. Lipoprotein lipase, while in the peripheral circulation, will cleave triglycerides from the chylomicrons for deposition in fat stores in the periphery. Without lipase the liver will cleave and store these same fats. Decreased levels of CCK and pancreatic lipase will only allow fatty acids to be lost in the stoois due to malabsorption. The correct choice is C.

34.

A is correct, I and III only. Table I indicates a drop in the plasma glucose levels from the baseline to the HF group. Choice I is correct. Insuiin requirements were lower in the HF diet phase compared to baseline. Choice Ii is incorrect. Less glucose was excreted in the urine in the HF group compared to baseline. Choice III is correct. The correct choice is A.

35.

D is correct, this decrease is beneficial

because glycosylated hemoglobin indicates persistent leveis of high plasma glucose. When blood glucose is elevated persistently, then plasma proteins become glycosylated (glucosi residues are attached to the proteins). A decrease in glycosylated hemoglobin means that, on the averagg blood glucose levels were lower during the past few months. (The life span for a RBC is about 120 days). This would be a beneficial change. Choices A and B are incorrect. Glycosylation occurs when plasma glucose is high. Choice C is incorrect. The correct choice is D.

36.

C is correct, oleic acid. The notation for the fatty acids is (# of carbons: # of double bonds). In the HF diet, the amount of 18:1 is dramatically increased. In the passage, you are told that the HF diet is specifically higher in monounsaturated fatty acids. Monounsaturated means having one double bond. Palmitic acid (16:0) nor stearic acid (18:0) have double bonds. Choices B and D are incorrect. Palmitoleic acid is 16:1. and oleic acid is 18:1. Choice A is incorrect. The correct choice is C.

37.

A is correct, olive oil. Even if you have no clue about nutrition,

38.

C is correct, II and III only. Although VLDL was lower in Table 2, LDL was nor. Choice I is incorrect. Triglycerides are lower on the HF diet. (Use the significance symbols to help interpret the data. Researchers must apply statistical methods to see if a finding is real or not.) Choice II is correct. HDL did increase on the HF diet. Choice III is correct. The correct choice is C.

39.

C is correct, the beta cells ofthe pancreas. The alpha cells ofthe pancreas secrete glucagon. Choice A is incorrect. The exocrine portion of the pancreas secretes digestive enzymes, while the endocrine portion secretes hormones. Choice B is incorrect, as is choice D. The correct choice is C.

40.

D is correct, the liver. The word hepatic refers to the liver. The hepatic portal vein drains the blood supply of most of the organs of digestion and passes it to the liver first for processing. Insulin released from the pancreas first reaches the liver. Choices A, B, and C are incorrect. The correct choice is D.

use your test-taking skills to eliminate wrong answers or see patterns. One of these things is not like the others. Olive oil is from piants and the others are from animal fats. Anyway, oiive oil is a good source of monounsaturated fatty acids (oleic acid to be specific). The other fats are more saturated (they are soiid at room temperature). The correct choice is A.


l9a


Biology

Gastrointestinal & Renat

Section III Answers

41.

B is correct, an increase in glomerular pressure, fbllowed first by an increase and then by a decrease in glomerular fiitration rate. Ifthe efferent arteriole is constricted, blood cannot flow pass the point ofconstriction. This ieads to an increase in the glomeruiar pressure, and allows us to immediately pick choice B or D. Constriction of the efferent arteriole also means that the blood flow in the glomerulus will decrease. and more plasma will begin to filter out into Bowman's capsule. With this information we can pick choice B. The more the plasma filtered out into Bowman's capsule, the more the solute concentration begins to increase in the glomerulus. Eventually this will decrease the glomerular filtration rate. The filtrate simply wili not be able to flow out of the glomeruius and into the capsule because of the change in the concentration gradient. The correct choice is B.

42.

C is correct, reduced plasma glucose concentrations. Recall that in the proximal convoluted tubule the transport of sodium across the apical membrane (from the lumen of the proximal convoluted tubule to the cytoplasm of the lumenal epithelial cell) is mediated by proteins called symports (a cotransporter). For example, thesesymports allow for the passage of both sodium and glucose together into the cell. Another example is the symport that aliows for the passage of both sodium and various amino acids into the ceil. If the concentration of glucose is decreased in the filtrate, the amount of sodium that can be reabsorbed at the level of the proximal convoluted tubule is reduced. The correct choice is C.

43.

A is correct, afferent arterioles and decreases the volume of urine flow. Sympathetic innervation of the kidneys primarily affects the afferent arterioles. This innervation constricts the afferent aiterioles and reduces the glomeruiar filtration rate. The urine output will be dramatically reduced. The correct choice is A.

14.

A is correct,

(see the diagram below). Notice that along the X-axis the plasma osmolarity is steadiiy increasing. At some particular point the solute concentration in the plasma will be so great that osmorecepto.s, located ii the supraoptic nuclei of the anterior hypothalamus quickly respond to a change in the osmolarity of the extracellular fluid (especially to the sodium ion).

x

X-Axis Since the osmolarity of the plasma is becoming higher (hyper-osmotic), water wiil flow down its concentration gradient (osmosis) and out of the osmoreceptors and cause them to shrink. As the receptors shrink their rate of discharge increases. If the rate of discharge increases, the posterior pituitary gland releases more ADH into the plasma. This is exactly what we see as the plasma ADH concentration along the Y-axis begins to increase. If there is more ADH in the system, then there is more ADH to act at the baso-lateral membrane of the late distal tubules, collecting tubules, and collecting ducts. In other words, the cells in those iocations are more permeabie to water. Water is reabsorhed into the blood and the urine becomes more concentrated. Notice that as ADH is being released, the osmolarity does not change that much. The correct choice is A. -15-

D is correct, (see the diagram below). Again, along the X-axis the plasma concentration of glucose begins to increase. As soon as the concentration of glucose reaches a particular point where it can no longer be reabsorbed from the lumen of the proximal convoluted tubules, it is passed through the lumen of the kidney tubules in an ever increasing rate.

't

X-Axis Copyright @ by The Berkeley Review

r99


Biology

Gastrointestinal

E( Renal

Section III Answers

This is shown along the Y-axis where we observe a continual increase in the rate of excretion of glucose in the urine of the kidney. we do not see this type of curve in the other tt r"e unrr"rs. The correct choice is D. c is correct' is a steroid hormone secreted by the adrenal medulla. Aidosterone is a cholesterol-derived steroid which is synthesized in the zona glomerulosa (ihe outermot, ."gi"rj.r rhe adrenal aioorterone is referred to as a mineralo-corticoid because of its affect on electrolyt", "o.t"*. ,u-"h as Na@ K*.-i;i;'rro*ron" has a half-life of about 30 minutes' Aldosterone promotes Nae ""d at the late to u g."u, extent distal tubular region, the ,reabsorption cortical collecting tubules, and the collecting ducts. Aidosteroneilso controls the secretion of K@ ions into the lumen of the distal convoluted tubule. The coirect choice is C.

46.

47.

B is correct' increased muscle contraction. Excessive aldosterone secretion results in an excessive loss of Ke from the extracellular fluid and into the urine. This lowers rhe prasma K* .o";;;;;;ii"",'"'.*oirion referred ro as hypokalemia' Loss of K@ ions from a cell tends to decrease the K@ concentration in that cell, making the resting membrane potential for K@ more negative that it would have been had aldosterone Ko rushes out of the nerve cell, that cell tends to. hyperpolarize. Hyperpolarization leveis been normal. Recali that as leads to a decrease in the muscle contraction, not an increase. The correct choice is B.

48.

electrolytes. If we were to centrifuge down a sample of blood, we would find ,,formed elements,,such as red blood cells (erythrocytes), white biood cjts^6e*ocy,.rl, pi.r"r"o, and proreins (high molecular weight) at the bottom of the testtube' In the plasma we would find such ruurtunr", as sugars, amino aclds, and the electrolytes, a few' Electrolytes are rather small molecules rhat can u"

C is correct,

:i:rxT"rrt3:

"utlry

c

49.

is correct' excessive vomiting' we are looking for a situation which.wiil not result in the gain of hydrogen ions. or net 6odilv c"t" hydrogen ions is gasrroinresrinal secrerions :"."1.," 3L:,*'-T"'?1:T leaving the bodv. Recall tha_t vomiting has irs origins'in"the j:tiil:;;ffi::il1i::t::t".1: r.-"rrr, vomitus leaving the body will result in a net loss, not a net gain of hydrogen ions. The correct choice is c. B is correct, a increased blood hydrogen ion concentration. It is clear from figures one and two that two events are occurring' one event is the secretion of hydrogen ions from the iunutu, celli of the tiJ*y in,o the renal tubule. These hydrogen ions, after reacting with a buffe'ring system,-will r"uu" trr" body in irr" *i""] The body is removing hydrogen ions' This tells us thaithe hydrogen ion-concentration must havl been elevated, and the kidney is compensating for this elevation. second, Figure 2 is indicating th. ."ubrorption of bicarbonate. one must understand that reabsorption of a bicarbonate ion is eqriivalent ro rhe l"r;;i; ion. Therefore,

;;-i";;i

*:jj,ryfl$

50.

acting as a compensatory mechanism to reiieve an elevated 51.

filtered by the glomerulus. The correct

,'ri";:#r"*,

il;rogin

hyd-;;

rhis reabsorprion is also

Ln concentration. The correct choice is B.

D is correct' secondary active transport. This question-draws on your previous knowledge of transport systems. As one can see' a hydrogen ion is transported out of the tubular cell white a sodium ion is nioved tiom the renai tubule lumen to the cell' Sodium moves inlo the cell because it is falling down its concentration gradient. Even though this is not stated in the passage, one should know that the sodium ion"concentration is always higher extracellularly when compared to intracellularly, The reason sodium can continue to fall down its concentratlon"!.uoi"nt

is because of a sodium/potassium ATPase pump which uses the. energy of ATF to p"-p sodium out of the cell. One can ref'er to sodium moving down its gradient as primary active trinsport. To trispo.t the hydrogen ion across the membrane, one can harness the energy found in the sodium-concentration gradient (really a form ofpotential energy). since the is cotransported along with the sodium ion, it is &;; ;d rerm iecondary aitive transport. The correct

:il:i:ff;:" <)

B is correct, a decrease in the hydrogen ion concentration, with a decrease in the level of Co2. rtshould be clear that an alkalosis is a situation where the pH is above that of neutral, and there i this case, we have u" u#;#i, the equation that is shown in Fj ,: produce a siruation

l:I

":g::

ffi'; ffi,iffJ:l?";:":l?:;il"I.jli,:l'.':ffi;:**:i::?;,t l{f:"j:ij"Jl:,ll *h;,;;fi;i,la-"irii"".i##il;;;;',H?;H:,'ff:i'":l?:il,i:1ff;

i

I:::":'iT^':^1i:':?.?1.1ll":g7 i;,;;r;;.u"il';'^#;tii;;,i'i#""'ol,'*""lffi#JJu5;',:'fffi:',,J,:,'r:l The result of thatshift is-that there is a decrease in the amount

53.

v L

.v

vv'LLPwt

of hydrogen ions. A reduction in the level of lll"llt_t: ions clearly results hydrogen in alkalosis. The correct choice is B. c is correct, decrease, due to increased ventilation. we have a situation here where there is simply too much hydrogen ion' we want to restore the original levei of hydrogen ionAgain, refer to the equation shown in Figure 1l

Copyright

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200


Biology


Section III Answers

One way we can reduce the hydrogen ion concentration is to lower the ievel of carbon dioxide. This should sound very familiar, like the previous question. How do we lower the partial pressure of carbon dioxi
B is correct, I and II only. Since the solution is introduced into the abdominal cavity through catheters, there is always a possibility of infection. Choice I is correct. Depending on the person's state of hydration, dialysis can leacl to electrolyte imbalance. Choice II is correct. Since there is no pressure involved in this technique, there is no ultrafiltration occurring. Choice III is incorrect. The correct choice is B.

D is correct, glomeruiar capiliaries. The glomerular capillaries are fenestrated, that is, they contain large pores. Although large proteins cannot pass, small molecules and water are free to pass. The pressuie of the bloo? acts to force the molecules through the semipermeable capillaries. Choice D is correct. Thi ultrafiltrate coliects in the glomerular capsule. The afferent arterioles are not permeable to these molecules. Choice C is incorrect. The peritubular capillaries are involved in resorption. Choice A is incorrect. The loop of Henle is involved in concentrating the urine. Choice B is incorect. The correct choice is D. :!6.

A is correct, no enzyme activity in the bag. Although

enzymes are large compared to atoms, they arc small compared to a pinhole. This is why you test your dialysis tubing for holes before you begin enzyme purification, Since there is no way the amount of enzyme can increase, choice B is incorrect. Since theie is an efl'ect caused by the pinhole, choice C is incorrect. Choice D is incouect since the enzyme would all (or almost all) move out during an overnight dialysis. The correct choice is A.

A is correct, erythropoetin. The kidney produces erythropoetin in response to low blood volume, This hormone signals the bone marrow to make more red blood cells. If a kidney were diseased, this hormone may not be produced. Choice A is correct. Choices B, C and D are incorrect because the kidney does not make these hormones. ADH is released from the posterior pituitary where it is stored after its manufaiture in the hypothalamus. The adrenal giand produces epinephrine and norepinephrine. Insulin is produced by the beta cells oft-he pancrcas. The

correct choice is A. 58.

B is correct, edema. You are given 2 choices that give the same answer, choices A and D, Elirrinate them. If uremia is present, the kidney is probably diseased, so urination is probably decreased. Choice C is incorrect. Edema means an accumulation of fluid in the intracellular spaces, leading to tissue sweiling. Uremia indicates kidney function is declining. This would lead to an increased solute load in the blood and retention of fluid. The fluid increase would mainiy move to the extraceilular spaces, since cells have only a little room for expansion. The correct choice is B.

59.

B is correct' use of a dialysis solution that is isotonic to the blood in glucose and salts. The pores of dialysis membrane are not small enough to retain the small molecules of glucose and salts. Also, if glucose an
60.

B is correct, I 11 ml/min. 5 liters = 5000

6X.

A is correct' iipolysis. Lipolysis is the mechanism of breakdown of stored triglyceride to release FFA and glycerol into the blood. This would raise FFA concentrations. Lipogenesis is the synthesis of fat. This wouid not raise FFA levels. Choice B is incorrect. Increased fat oxidation would decrease plasma FFA levels. Choice C is incorrect, Increasing dietary fat would increase chylomicrons, but not FFAs. Choice D is incorrect. The correct choice is A.

62.

C is correct, epinephrine. CCK is a hormone secreted from the duodenum in response to dietary fat. It causes gall bladder contraction and release of pancreatic enzymes. Choice A is incorrect. Thyroid hormone does not increasc


mi.

5000 m1/45 min = I I

20t

I ml/min.

The correct choice is B.


Biology


Section III Answers

plasma FFA concentration. Choice B is incorrect. Insulin decreases lipoiysis, and increases fat storage. Choice D is incorrect. Finally, epinephrine promotes lipolysis to increase FFA concentration in the blood in response to stresses. Think of the fight or flight mechanism. The muscles will need fuel and the elevated FFAs will provide that. The

correct choice is C. 63.

D is correct, sucrose stimulated insulin, which decreased release of FFAs from adipocytes. Insulin turns down lipolysis. If you have just eaten, and insulin is secreted, your body does not need the extra fuel that circuiating FFAs provide. You have switched from withdrawal from stores to depositing into your stores. Sucrose does not stimulate glucagon, so choice B is incorrect. Fructose does not mediate the effect on FFAs, so choice A is incorrect. Sucrose is not an inhibitor ofthe action ofcaffeine, so choice C is incorrect. The correct choice is D.

64.

C is correct, adenylate cyclase. The wrong answers in this question are all part of the cAMP activation pathway. Be careful ! A hormone such as epinephrine stimulates a receptor on the cell. A second messenger, the G protein, communicates the message internally to the enzyme adenylate cyclase. Adenylate cyclase produces cAMP. cAMP activates protein phosphorylases. The question is for the enzyme, so adenylate cyclase is your answer. The correct choice is C.

65.

D is correct, the effects are the same as drinking water. The answer to this question in the table. The ,,, f, and t indicate if differences between groups are significant. This is a very common way for data to be summarized.

Experiment 2 has the symbol "*", which indicates that the results were not significantly different from Experiment 3, water. Although the numbers look like a slight increase, the x indicates there is actually no difference. The correct choice is D. 66.

D is correct, I, II, and III only. This is an experiment to determine the control data for plasma FFA concentrations over time. To simulate the real experiment as much as possible, hot water and sodium saccharin were both included. This means I, II, and III are all valid choices. The correct choice is D.

67.

B is correct, osmotic force due to protein in plasma favors filtration. This is a false statement, Filtration is the movement of substances from the glomerular capillaries into Bowman's capsule. The protein that is unable to be filtered creates an osmotic force which pulls the movement of fluid away from Bowman's capsule and into the glomerular capillaries. Therefore, the protein in the plasma creates an osmotic force which opposes, not favors, filtration. We are looking for a false statement, and choice B certainly fits the description. The correct choice is B.

68.

C is correct, X is filtered and secreted, but not reabsorbed. This question is requiring one to interpret Figure 1. It is clear that substance X leaves the glomerular capillaries and enters the renal tubule at Bowman's capsule. This classifies as filtration. Also, it is apparent that substance X leaves the peritubular capillaries and enters the renal tubule after Bowman's capsule. This classifies as secretion. The only event that does not occur is movement of substance X back into the peritubular capillaries. In other words, there is no reabsorption. Therefore, X is filtered and secreted, but not reabsorbed is a true statement. The correct choice is C.

69.

B is correct, increased levels of parathyroid hormone, with increased removai of bone, This question is re quiring one to think about the role of parathyroid hormone. We have a decrease in the level of calcium, and so we will therefore want to act to increase the levels of calcium. This results in an increase in the levels of parathyroid hormone. Why? One of the functions of the parathyroid hormone is to remove bone. Removal of bone releases calcium and phosphate from the organic molecules. The release of calcium acts to counteract the original decrease. Therefore, we will see an increase in the level of parathyroid hormone, and an increase in the removal of bone. All of the other choices are either fa1se, or lead to a flrther decrease in calcium levels. The correct choice is B.

70.

D is correct, the effect of the extracellular calcium concentration on membrane is distinct from its role as a mediator of muscle contraction. We are told from the passage that a low extracellular calcium concentration leads to an increase in nerve and muscle cell excitability. In fact, a person with a iow calcium concentration may suffer from muscle spasms. We often think of calcium as being involved with the troponin-tropomyosin complex, where calcium is needed to initiate contraction. In that situation, a low calcium concentration would lead to decreased excitability or contraction. We are told that in this case, a low extracellular concentration leads to increased nerve and muscle excitability. Based on this apparent contradiction, we can conclude that the effect of calcium concentration on the membrane is distinct from its role as a mediator of contraction. In fact, these effects reflect calcium's abiiity to bind to piasma membrane proteins that function as ion channels. The binding alters the state of the these channels. The correct choice is D.

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202

The Berkel€y Kevi€w Specializing in MCAT Preparation

Biology D is correct,

Gastrointestinal & Renal a reduction

Section III Answers

in the tubuiar reabsorption of phosphate. This question is not easy. The following logic

can be used to arrive at the answer: We have a low calcium concentration. We will want to iounteract that decreise t'ith an increase. We will see an increase in the removal of bone due to increased levels of parathyroid hormone. T1-ris not only reieases calcium, but from the passage, one should be able to conclucle that it releases phosphate ion as

n'e11. Now, if the extracellular concentration of phosphate were to increase as a result of this^release, further novement of calcium from bone would be hampered because a high extracellular concentration of phosphate would rind with calcium and cause the salt to be re-deposited on bone and other tissues. The point is that we want to get rid of the phosphate ion. One way to do this is to decrease the amount reabsorbed in the kidney. In that wiy, phosphate ion is released with the urine and calcium levels can return to normal. Again, we will most likely see'a :e

duction in the tubular reabsorption of phosphate ions. The conect choice is D.

C is correct, the parathyroid gland. This question is simply asking us to recall the function ofour endocrine glands. Parathyroid hormone, a protein hormone, is reieased by the parathyroid gland. Whiie the parathyroid gland is :mbedded in the thyroid gland, the giands are distinct. The correct choice is c.

A is correct' muscie and adipose tissue have first access to dietary fatty acids from chylomicrons. The chylomicron packaging system avoids direct delivery of fatty acids to the iiver. This is stated in the text of the passage. Choice B i: incorrect. Although microorganisms may be present in the lymph, they are not packaged intoihe chylomicrons. Choice C is incorrect. Lymph, not interstitial fluid, canies chylomicrons. Choice D-is inconect. The coirect choice

is A.

C is correct, water. 'Water is the main component of our body fluids. Since lymph is a derivative of interstitial fluid, -: is mostly water. Fat soluble vitamins are usually present in very small amounir (pg), triglycericles may be p."r"ni i larger amounts (grams), but water is the most abundant. There must be more wateithanihylomicrons so that they ::main soluble. Although chylomicrons carry fat, neither they nor the fat they carry is thl major component oi . mph. Choices A , B, and D are incorrect. The correct choice is C. D is correct, I, II, and III. AIl three vitamins are fat-soluble vitamins. The correct choice is D.

-\ is correct, to destroy microorganisms. Chylomicrons are procluced and adde
:raeocytic cells,

as

mentioned in the passage. The correct choice is A.

-\ is correct, I only. In the passage, the structure of

the lymph vessel is described as similar to the vein. The vein l --es not have a muscular layer like the artery. Choice II is incorrect. Blood in the vein is moved by squeezing by the .,;eletal muscles. Choice I is correct. Blood pressure leads to the creation of interstitial fluid. it does not moue . nph. Choice III is incorrect. The correct choice is A.

B is correct, kidney. Urine is formed as an ultrafiltrate of blood in the glomeruiar capsule of the kidney. Choice B is - rffect. Neither the liver, the pancreas, nor the salivary gland produces an ultrafiltrate. Choices A. C, and D are .:.Jonect. The correct choice is B.

B is correct, I and II only. After a meal, chylomicrons first enter the lymphatic vessels, then the blood vessels. - roice I is correct. Lymph does not contain erythrocytes (red blood cells). Choice III is incorrect. Lymphocytes are .:: second most abundant type of white blood cells in the plasma. They are aiso he lymph. Choice II is , -.irect. The correct choice is B.

]l X PFa = (l lu X FU) + (t lV X PFy). The principle of mass balance is the idea of whar enrers the jney must leave the kidney. That which enters the kidney wili be from only the plasma of the renal artery. This ::n be quantified as [ ]e X PF4. There are two ways substances can leave the kidney. One is through the urine,

D is correct, [ ).-

---:ntifiedasFuX[]u. Theotheristhroughtherenalvein,quantifieciasPFuX[]u.Therefore,massbalanceis .:st represented as []a X PFa = ( []u X Fu) + (l lv X PFv). The correct choice is D.



Biology


Section III Answers

81.

B is correct, 0.0005 L. We are interested in knowing the volume of urine collected in 0.5 minutes. We need to know the rate of urine flow to figure out the volume. The rate of urine flow can be calculated using the relationship established in the passage. In other words, Fu = (C) X (t lp)/t lu. Putting in the values, we get Fu = 50 ml/min X 1 mg/ml divided by 50 mg/ml, which turns out to be 1 ml/min. We are asked for 0.5 minutes, so we have 0.5 ml as our volume. 0.5 ml X I L/1000 ml = 0.0005 L. The correct choice is B.

82.

C is correct, GFR = [ ]u X Fu/[ substance]p. From the passage, we know that under certain conditions, the amount of a substance filtered is equal to the amount excreted in the urine. The amount of a substance excreted in the urine will be [ ]u X Fu. To calculate the amount filtered, we can multipiy the GtrR by the concentration of the substance in theplasma. Equating thetwo, wehave GFRX IJp = 1]u XFu. Therefore, theGFR = [ ]u XFu = [1p. The


B is correct, the substance must not be reabsorbed, and must be secreted by the nephron. From the passage, we know that only under certain conditions can the amount filtered at the glomerulus equal the amount excreted in the urine. It should make logical sense that one of those conditions should be that nothing is added or taken away from the filtrate as it passes through the nephron. Statement B claims that the substance must be secreted into the nephron. If this was the case, the amount excreted will be more than likely greater than the amount filtered. Therefore, condition B cannot exist if our claim is to be valid. The correct choice is B.

84.

A is correct, Proximal Convoluted Tubule. This question is simply calling on our previous knowledge of kidney function. Recall that most reabsorption occurs in the proximal convoluted tubule. This is the first part of the nephron. The substances reabsorbed are returned to the venous system through the peritubular capillaries. The reabsorption of glucose takes place via a secondary active transport, and under normal conditions, l00Vo of glucose is reabsorbed. The correct choice is A.

85.

A is correct, 3.5 mg/ml. The plasma threshold is defined as the level of plasma glucose which first gives rise to glucose in the urine. Looking at the curve, we are interested in the line labeled excretion. When this line first becomes positive, we need to look at the value of the blood glucose which gave rise to this first appearance of glucose in the urine. This line becomes positive between the values of 2 and 4. Looking carefully, the value of the blood glucose is around 3.5 mg/ml. The correct choice is A.

86.

B is correct, the cationic species has the highest filterability because of the presence of anionic glycoproteins on the surface of all glomerular filtration components. Let us read the curve given in the question. Draw an imaginary vertical line in the graph. With this line, we can say that if a molecule has the same radius, the cationic species will have the higher filterability. Therefore, any choice that uses the argument that cationic or anionic species have different radii is invalid. In other words, we can compare a positive and negative molecule that has the same radii, and know from the graph that the positive cationic species will have the greater filterability. That means there must be some other reason for the enhanced filterability. The most likely reason is the presence of anionic glycoproteins on the surface of all filtration components. The opposite signs will create an electrical attraction that can lead to an enhanced filterability. The correct choice is B.

87.

D is correct, regurgitation of duodenal contents may lead to stomach ulcers. We know from the passage that the pyloric sphincter carries out two essential functions. One of these functions is to prevent the reflux of small intestinal contents into the antrum. This is the case because the lining of the stomach is not protected from the bile that is secreted into the small intestine. The reflux of this bile containing content may contribute to the development of stomach ulcers. So again, while the stomach is protected from the acid it secretes, it is vulnerable to the bile secreted by the small intestine. The opposite holds true for the small intestine. The small intestine is protected from the bile, but is vulnerable to the acid from the stomach. Therefore, regurgitation of duodenal contents may lead to stomach ulcers. The correct choice is D.

88.

B is correct, catecholamine. This question cannot be answered from any of the information in the passage, but must be answered solely on our own knowledge of hormones and hormone classification. Norepinephrine, along with epinephrine and dopamine, are catecholamines which fall under the major category of amine hormones. The correct choice is B.

89.

D is correct, a decrease in blood volume. If a hypertonic solution is introduced into the lumen of the small intestine, this creates a situation where water will enter into the lumen from the vascular system. In other words, the water

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Section III Answers

from the vascular system will cross the intestinal wail and diffuse to where there is a higher concentration of solute. This will certainly impair the vascular system. In particular, it will cause a decrease in btood volume which

lead to a dangerous drop in arterial blood pressure. The correct choice is D.

could

90.

A is correct. We are told from the passage that a decrease in the pH of the duodenum will cause a decrease in the rate of gastric emptying. One way to achieve a decrease in gastric emptying would be to decrease the contractility of the antrum and to increase the contractiiity of the duodenum. In this way, we could achieve our goal of a decreised flow of material from the antrum into the duodenum. Looking at Graph A, the injection of HCI into the duodenum certainly does cause a decrease in the contractility rate of the antrum. In addition, the injection of the HCl certainly does cause a significant increase in the rate of contraction for the duodenum. Therefbre, Graph A BEST representl our expected results. The correct choice is A.

91.

C is correct, insulin' We are iooking for a hormone that is most likely to be released in response to a rise in GIp. Since we are looking for a hormone, bile can automaticaily be eliminated. We know that GIi, is itself released as a result of fat in the small intestine. The fat will soon find its way into the vascular system by first going through the lymphatic system. The fat will want to enter into the adipose tissue. Insulin is the hormon" i"rponrlbl"*;or taking fat from the vascuiar system and facilitating its uptake into adipose tissue. Think in very general terms. We are in an absorbing state for the body, as food is being absorbed from the small intestine. Foi this reason, insulin is the hormone that is necessary' and GIP is stimuiating the release of insulin in anticipation of the arrivai of f'at into the vascuiar systen. The conect choice is C.

92.

C is correct, decrease the rate of gastric emptying and contribute to the mixing of stomach contents. We know from the passage that CCK does retard the rate of gastric erhptying. For that reason ilone, r" can eliminate choices A and

B. We now must concentrate our efforts into understanding whether the actions of CCK contribute or inhibit the mixing of stomach contents. The question tells us that CCK causes a constriction of the pyloric sphincter. However, it also causes an increase in antrum contractility. Think about this. If the antrum has an in"r"ur"d contractility and the contents of the stomach cannot enter into the small intestine, they will simply be pushed back into the bohy of the stomach. The contents of the stomach have no where to go but baikwards. Therefore, there will be an increase in the mixing of stomach contents and an overall decrease in the rate of gastric (remember, the pyloric "-ptying sphincter is closed). The correct choice is C. 93.

B is correct, the pH of the blood increases. We are looking for afalse statement surrounding the notion of parietal and pancreatic cell secretion. We know that a parietal cell secretes protons into the lumen of tlhe stornach. Firit, how is this proton obtained. The parietal cell converts a carbon dioxide molecule into a bicarbonate ion and proton. a The cell secretes the proton into the lumen. What does it do with the bicarbonate ion? It transports the ion into the b1oo4. In that way, the parietai cell itself does not become basic. However, that does present i problem for the pH of the blood, right? Well, not exactly. A pancreatic cell does just the opposite. The ceil secretes a bicarbonate ion into the lumen of the small intestine and is left with a proton. The protonls released into the blood. If the secretions of the

parietal cell and pancreatic cells are equivalent, the pH of the blood does not change throughout the digestive

process. The correct choice is B.

94.

B is coffect, salt loss creates an osmotic gradient, which water follows out of the cell. The epithelial cells lining the intestine form a selectively permeable boundary between the blood and the lumen, o. .uuiiy, of the intestin"i. In cholera, these epithelial cells secrete sodium and chlorine ions into the lumen. This changes the relative concentrations of solutes, meaning that there is a lower concentration of solutes inside the celt in comparison to outside the cell. Water diffuses across membranes by osmosis, foliowing a gradient from iow dissolved solutes to high dissolved solutes (i.e., high water concentrations to low water concentraiions; the concentration of water is iess in a solution with a lot of ions in it.). Therefore, water leaves the epithelial cells and enters the lumen of the intestine, following the sodium and chloride ions. This causes diarrhea. Choice A is wrong because the membrane is already highiy permeable to water, and extra pores won't make a difference. Choice C is wrong because water cannot be actively transported. Choice D involves reabsorption of water and is completely incorrect. The correct choice is B.

qi

D is correct, Water, glucose, and Na@ ions. From the passage, we learn that rehydration therapy involves an inwardly-pumping glucose/sodium symport. A symport is a protein which transports two ions at a time across a membrane, in this case, into the cell. If a cholera patient is given both glucose and Nao with water, the glucose/sodium symport will transport both of these ions into the intestinai cells, increasing the solute concentration inside. This will cause water to follow osmotically and reenter the blood (via rhe epithelial cells). A11 of the other

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answer choices are wrong because none balance. The correct choice is D.

Section III Answers

of them allows solute to enter the cell, favorably changing the osmotic

96.

D is correct, testosterone. This probiem requires that we know that mechanisms of the hormones that are mentioned. Epinephrine' gastrin, and norepinephrine are peptide hormones which operat" uy ul"oing tJ receptors on the outer surface of ceil membranes'-This then triggerJ a signai cascade involving ,".oni-..r"rr"n-g"r, ut" cyclic-AMp. As such' the cholera toxin would affect their actions by. increasing cyclic-AMp levels. Testosi".on", on the other hand, is a steroid hormone. Steroid hormones can pass through t# piur*u membranes of cells, directly (or through a bound protein complex) {fectr-ng DNA tranicription. Steroid'hormones therefore do not use second-messenger systems. The correct choice is D.

97.

D is correct' the underlying bacterial infection must be treated so that dehydrating diarrhea ceases. Rehydration therapy is just that, a therapy, not a cure. It does not address the underlying piolr"*,"*hi.h i, ,t . uu.r".ial infection that produces the cholera.toxin in the first place. In severe infections, anfmicrobial'agents must be administered to destroy the bacteria. otherwise, dehydrition witl simpiy ,"o""ri after each rehydration artempt due to the persistence of the cholera toxin which causes salt secretion and water loss. choice a it rvronl because the body can reestablish a normal blood volume and pressure. choice B is inco'ecl ti6;" il;;", af1'ect the exisring osmotic gradient (see Question 1). choicJ C is not right because more chloride wiliju;i m;ie the osmotic gradienr steeper and more water will leave the body. The correct choice is D.

c

98.

is correct, lack membrane receptors for cholera toxin. The questionîs basically asking why

some mammals (a class of animal closely related to humans, relatively speaking) u."nit uff".t"d adversely by cholera in the way that humans are' Choices A and.B are wrong because mammals, like humans, use G-protein -"diut"o signaling (in fact, most studies are done in.mice).and generally only_have high cellular cyclic-AMp levels when cells are stimulated (i'e'' by hormones)' choice D is wrong beciuse if cyctic-itrlp tevets are normally low, then cholera toxin would increase them' choice c is correct beciuse the cholera toxin binds to specific ..""pto., on tn" ,r.tu.e of intestinal cells' If these receptors are absent, as in many mammal species, the toxin can,t attach and exert its effects. The correct choice is C.

A is correct, I only' The transgenic mice described in the passage are engineered to have the gene which encodes the cholera toxin' This gene is attached to a promoter (l ,"q"u"n"* thit allows DNA to be transcribed) which is ?NA only active in the growth-hormone producing cells of the, pituiiary gland. Therefore, these mice produce cholera toxin specifically in their pituitary cells' causing a sustained in"r"ur" ii cyclic-AMp levels in these cells. These mice subsequently exhibit gigantism, which results-from,excesr g-*1tr-i'o.*one. Therefore, we can assume that the

99.

increased cyclic-AMP levels are somehow signal-ing these.cell"s to proouce more growth hormone. The question asks what would inhibit this effect. Excessive noimal d-proteins wouldn'r F"-gr" tr,'ry *oriJri-ply be alrered by the cholera toxin and would become constantly active. bxcessive would add tothe gigantism because it seems to be high levels of cyclic-AMP which is.causing it in the"y"ii"-evtr iirst place. An enzyme which degrades cyclic-AMp would reduce the effect. The correct choice is A. 100.

populations

chronically exposed 3":,,'"'"||e|:,,i1"^T::,X:1"."^l:?:.91c^:i:,1'-llil.in 'r Lrrv YuvJLrv' DLcLeD rudl LIE cysuc ltDrosls gene' wnen present

ro cholera. rhe theory

ln one copy (i.e., heterozygously) may

confer a selective advantage because it provides a degree-ofresistanci to cholera. This is a similar case ro sickte-cell anemia hererozvgore., a selective advantage on y" woutd expecr ro see a.higi frequencv of tr'i" g"* because people with the gene would be more hkJly to survive epidemlcs and would evoiutionarily be more ,,fit,, than people without the gene. All of the other answers are wrong because they do not provide .uiJ.n.. that supports this theory, even though they may not contradict it. The correcichoice is B.

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Berkeley Review

206


ffig#HCIgry SeeEioxr

EV

Reproduction

&

Eleve[oprment

A. B.

Keproduction

l. 2.

Maie Reproduction Female Reproduction

Development l. DevelopmentalStaees 2. Dei'elopmental Mechanisms 5. Human Embryo Development

Practice Passages & Ansrt'ers

'Tfte

NKKETEY R.n.v.I.E.l[It Speciahztng in MCAT Preparation

Top lO Section Goals

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Be familiar w'ith lhe male reproductive anatomy.

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Know how s uced Be familiar with the actions of LH and FSH on developirn *.*. $now the different cell tvpes involved in sperm production. M;k; ;;;;;;'".ffi?.r",1"a both mitosis and meioiis.

sv

Be familiar u'ith the female re uctive ana KnowthegeneralstepsofhowanoVumdevelops'wi+* of cells wit-hin the ovdry and their relarionship toin* J*rli"p*g -gg

Know how in egg production. Make sure you understand botlifritSri" ir",Jrr-,"ioris.

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Be able to and contrast male and female ction. Understand whv the maie Leydig cells are analoqous to the female theca cells and why the maie 5ertoli cells are analogous to the female follicle ceils.

Be

familiar with thq different hormonal feedback mechanisms at the brain.

Understand urhat ha

once fertilization of the takes place. familiar wjth the process of fertilization, where ii is most likely to occur, and where the fertilized egg is most likely to impiant in the uterus. Be

Understand the hormonal relationships associated nith fetal development.

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Biology

MaI,€i,lffi HIU.A.U

Reproduction & Development

Male Reproduction

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General Reproductive Endocrinotogy

During the endocrinology of reproduction a hormone called gonadotropin releasing hormone (GnRH) is released from the hypothalamus. GnRH signals the anteriol pituitary to release the gonadotropins luteinizing hormonelLH) ar-rd follicle stimulating hormone (FSH). Both LH and FSH travel to the gonads. In the maie the gonads are the testes while in the female they are the ovaries. Each of these gonadotropins elicits a number of different responses at their target iissues. There are two general types of responses that we will be considering. The :irst is the production of steroid hormones. The second is the production of germ :eils. In the male the germ cells are cailed spermatozoa while in the femalelhey are the ova.

Male Keproductive Anatomy The organs which carry out the reproductive functions in the male are the testes,

:pididymis, vas deferens, ejaculatory ducts, seminal vesicles, prostate, iulbourethrai glands, and penis. These structures are shown in Figure 4-1. Ureter

--* I

)

Prostate

\

gland

Urethra Seminal vesicle

Ejaculatory duct

Corpus sponglosum

Vas deferens

Bulbourethral

Scrotum

Testis

gland

Epididymis

Figure 4- I : lale reproductive anatomy. .-11 of these structures are important in the synthesis and delivery of sperm to the :emale vagina. The actual production o{ sperm is referred to as spermatogenesis. lecause the formation of sperm requires a temperature lower than that of body :emperature, they are produced in the testes which iie in the scrotum outside of -ile body cavity. Within each testis are a series of convoluted tubules called the seminiferous tubules. It is within these tubules that we find the spermatogenic

di

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Male Reproduction

cells. Not oniy are sperm produced within the testes but the hormone

testosterone is synthesized within specialized interstitial cells called Leydig celts

that lie outside the seminiferous tubules. A third cell type found within the area of the seminiferous tubule is the Sertoli cell. As we will see, these cells act to promote spermatogenesis and they also produce the protein hormone inhibin.

Sperm Production The target tissue for hormones like FSH and

LH are the gonads. In the male the gonads are the testes. Within the testes are a series of convoluted tubules referred to as the seminiferous tubules. There are severai different cell types that interact within the seminiferous tubules to allow for the development of spermatozoa. Between adjacent seminiferous tubules are the Leydig cells (also called the interstitial cells). If we were to take a cross-section of a seminiferous tubule, we would note a lumenal space in the center. Adjacent to the lumenal space are fairly large cells called Sertoli cells. In direct contact with the Sertoli cells are a series of spermatogenic cells. As you proceed from the basement membrane of the seminiferous tubules toward the lumen, the cells become more and more differentiated. In other words, those cells are becoming more and more developed. An important aspect of this development is that the Sertoli cells are in constant contact with the spermatogenic cells. You could view this as the Sertoli cells regulating the development of the spermatogenic cells. Let's consider the development of the spermatogenic cells. If we consider a portion of the seminiferous tubule, we will see a basement membrane separating the Sertoli cells from the interstitial cells. The Sertoli cells are in contact with one another and essentially form a ring around the inside of the basement membrane. Different developmental events take place on different sides of the Sertoli cells. For example, spermatogenic cells closer to the basement membrane turn out to divide in a completely different way than they do further away from the basement membrane. The spermatogenic cells that are closer to the basement membrane are called spermatogonia. They have 46 chromosomes. The way that spermatogonia divide is by mitosis. The resulting "daughter" cells are exactly the same as the parental cell. In the area ciosest to the basement membrane the spermatogonia are undergoing constant division. Recall that the

first thing that happens in mitosis is duplication of the chromosomes. These chromosomes eventually migrate towards the cell's equator and the cell undergoes cytokinesis to produce two daughter cells which are exactly like the parental cells. The spermatogonia that undergo mitosis by the basement membrane are referred to as primary spermatocytes. As the primarv spermatocytes form they begin to squeeze by the juxtaposed Sertoli cells and move towards the lumen of the seminiferous tubule. In the area near the lumen a completely different set of events takes place. The primary spermatocytes undergo meiosis and will reduce their genetic complement to 23 chromosomes. Meiosis is simply one duplication with two divisions. Recall that during the meiosis the first thing that happens is duplication of the chromosomes. The chromosomes line up at the cell's equator and the cell undergoes cytokinesis. The cells that are formed after the first meiotic division are called secondary spermatocytes. At this point a second meiotic division takes place. Each secondary spermatocyte will again divide to give two daughter cells. The daughter cells of the second meiotic division are referred to as spermatids and they have half the chromosomal complement (i.e., 23 chromosomes) as the

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Biology

Keproduction & Development

Male Reproduction

original spermatogonia. Therefore, every primary spermatocyte that undergoes mitosis and meiosis ends up producing four spermatids. The very next event that takes piace is a transformation event. Each of the spermatids is transformed rnto a spermatozoa. The transformation process has an interesting characteristic to it. When the spermatids are initially formed, they are all connected to each other by cytoplasmic bridges. During the transformation process the spermatozoa essentially bud out from the spermatids.

At the very tip of a spermatozoan there is a structure called the acrosome which contains a lot of digestive enzymes. These digestive enzymes help the spermatozoa gain access to the interior of the egg once fertilization has taken place. Inside the head of the spermatozoa is a nucleus which contains DNA. in the midsection are mitochondria, providing energy for the whipping movement of the tail. This enables the sperm to swim towards their destination (the egg).

Hormonal Control of Sperm Production It is important to remember that the spermatogenic ce1ls are in constant contact with the Sertoli cells. The Leydig ceils have a rather special function. They convert cholesterol to testosterone (which we can abbreviate as T). Testosterone has several roles. Testosterone can diffuse out of the Leydig ceils and move to other target tissues of the body. It can aiso leave the Leydig celis and diffuse into the Sertoli cells. Within the Sertoli cell testosterone binds to a specific receptor and is converted to a compound called dihydrotestosterone (which we can abbreviate as dHT). This complex diffuses into the nucleus of the Sertoli celi and instructs the DNA to synthesize RNA. It is the products of the RNA synthesis that affect the spermatogenic cells. See Figure 4-2. Recall that we mentioned that the hypothalamus releases GnRH which then acts

on the anterior pituitary and causes it to release FSH and LH. These two gonadotropins irave a direct effect on the Sertoli cells and Leydig cells. LH binds to a specific receptor on the membrane of the Leydig cell. The secondary messenger produced from this binding interaction increases the conversion of cholesterol into testosterone. FSH has a different function. FSH will bind to a surface receptor on the Sertoli cell and induce a secondary response within the

Sertoli cell. This secondary response helps convert testosterone into dihydrotestosterone. It also induces the synthesis of the receptors. [A thought question. If you had to make a male contraceptive, what events would you disrupt in this pathway? You would disrupt the production of dihydrotestosterone or FSH. One of the problems with producing a male contraceptive is that it is rather difficult to stop the synthesis of FSH.I

As with any hormonal system there is always the consideration of a feedback mechanism. Lr males there is a constant production of LH and FSH. The reason is due to a feedback mechanism. One of the consequences of testosterone diffusing out of the Leydig celis is that it has a negative feedback at the anterior pituitary and at the hypothalamus. Testosterone is preventing the synthesis of GnRH and LH. The Sertoli cells are involved rn a different feedback mechanism. The Sertoli cells also secrete a compound called inhibin. Inhibin acts as a negative modulator of the anterior pituitary. The reason that the gonadotropin levels are relatively constant is that you have a constant synthesis of the spermatogenic cells. If testosterone levels are too high, the feedback mechanism will decrease the levels of LH. If the levels of dihydrotestosterone are too high, there is an increase in inhibin svnthesis which results in a decrease in the levels of FSH. Copyright

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Male Reproduction

Hypothalamus I GnRH

(-)

<:.::-.-_

I I I I I I I I

Testosterone

U)

o t

dHT

O

Cholesterol

oc

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\

Testosterone

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Figure 4.2 Hormonal regulation in the testes.

Secondarlr Sex Characteristics in the Male

Almost all of the secondary sex characteristics in the male are due to testosterone. some of those characteristics are development of a beard, pubic hair, deepening of the voice, texture of the skin, muscle distribution, bone development, sexual drive, and, if you are genetically predispositioned to it, baldness.

Sperm Delivery Sperm leave the seminiferous tubules and enter into the epididymis and then the vas deferens. Sperm are stored in the epididymis and vas deferens for about L4

days before ejaculation. Contraction of smooth muscle tining the wal1s of these structures ejects the sperm down the vas deferens and into the ejaculatory duct.

The seminal vesicles, prostate gland, and bulbourethral gland

secrete components that make of the remainder of the fluid that is ejaculated (now called semen). Some of the components secreted by these glands are fructose vitamins,

bicarbonate zinc, prostaglandins, and mucus. The total volume of the fluid ejaculated with the sperm is about 3 to 5 milliliters. Copyright

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Reproduction

Feffial€,]REp'fddub

E(

Development

Female Reproduction

i

Female Reproductive Anatomy The essential anatomical features of the female reproductive system includes the r-agina, uterus, Fallopian tubes, and ovaries. Once sperm have been ejaculated irto the vagina they can iive for about 48 hours. However, about a half an hour after ejaculation the leading sperm arrive at the site of fertilization which is usually the oviduct (also called the Faliopian tubes). The fertilized egg continues to move down the oviduct towards the uterus where it will implant in the uterine Lrning. About a week after ovulation the fertilized egg, no\4/ called a blastocyst,

rmplants in the iining of the uterus where it will continue to grow and deveiop r-urtil parturition (delivery). A few of the anatomical structures mentioned here are shown in Figure 4-3.

Fallopian

OvarY

Fimbriae

:,".:.

Urethra

/,t Vagina

Figure 4'3 '-

emale reproductive anatomy.

Ova Production

}e

production of the female germ cells occurs via a process called oogenesis. hr :ertain aspects this process is similar io that of spermatogenesis in males. The -::rportant sirnilarity in the female is that when cell division is complete the ovum .',-iii have half the DNA as the original parent cell. The regulation of oogenesis, :iowever, is quite different. First, there are a series of events in which the oogonia ritotically divide to form primary oocytes. The primary oocytes, which have 46 :hromosomes, are analogous to the primary spermatocytes in the male. kr the :emale ail of these divisions occur ruithin the first two to three months of fetal :.":oelopment. About the third month of gestation all the mitotic divisions cease. In rther words, al1 the oogonia that a female will produce will be produced within -;re first three months of fetai development. In fact, at birth a female will have ,'.'bout 400,000 primary oocytes in both of her ovaries. F{owevet, by the time she :.as finished her reproductive life only about 400 will have ever matured. The :emaining primary oocytes will have degenerated at various stages during the 'evelopmental process. This process of degeneration is referred to as atresia. Copyright @ by The Berkeley Review

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Female Reproduction

The primary oocytes will eventually undergo the first meiotic division, caused by a surge in the gonadotropin LH. The first meiotic division happens in

monthly cycles. Some of these primary oocytes could take as long as 50 years to have their first meiotic division. Once the first meiotic division is complete the secondary oocyte, which has 23 chromosomes, is formed. During this division one of the two daughter cells obtains all the cytoplasm while the other cell is just a small sphere of DNA called the first polar body. The secondary oocyte undergoes the second meiotic division only after fertilization hss taken place. Tine product of that division is the ovum and a second polar body. [The first polar body also divides to give two second polar bodies as well.] The ovum now has 23 chromosomes. v\4ren sperm and ovum unite in fertilization, the complement of 46 chromosomes will be restored (23 from the male and 23 from the female).

If we take

a cross-section of the ovary, we will see a series of specific cell types called primary follicles which are in different developmental states. The primary

follicle is actually a primary oocyte surrounded by a layer of foilicle celli. These follicle cells are in constant contact with the primary oocyte. Eventually one of these primary follicles will start to develop.[An exception arises with fraternal twins in which case two follicles will form and develop at the same time.] Theca

Cells

Growing Follicle

Primary

Follicle Zona Pellucida

Granulosa

Cells

Regressing Corpus Corpus

Luteum

Luteum

Figure 4-4 The ovarian cycle.

In order to mature the primary oocyte estrogen, LH, and FSH are needed. This developing time period is referred to as the follicular phase and lasts up to about the 14th day of the woman's monthly cycle. During this period ihe primary follicle gradually develops. Surrounding the primary oocyte will be a membrane called the zona pellucida. The zona pellucida is surrounded by more follicle cells called granulosa cells and finally by theca cells. See Figure 4-4. These cell types are a direct result of the estrogen, LH, and FSH that are present. The theca cells are analogous to the Leydig cells in the male while the granulosa cells are analogous to the Sertoli cells in the male. Copyright

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Female Reproduction

\ Estrogen

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the primary follicie a

fluid starts to build up forming the antrum.

: :l€rir is now primed for ovulation.

The The next event that takes place is the LH

ir:ge. This sr,rrge causes the primary oocyte to undergo the first meiotic ;irision, forming a polar body and the secondary oocyte. The LH surge also .;--
in --= orimary follicle. Once the secondary oocyte is released, ovulation has '-:rrred. Ovulation usually comes at about the 14th day in the woman's monthly "- le. As we have mentioned, the process leading up to ovulation is referred to as r--= ioliicular phase, The left over follicle (after ovulation) is transformed into a ;;rd-like structure called the corpus luteum. One of the main functions of the . -:!us luteum is to produce estrogen and progesterone. If fertilization does not . ---;r and there is no pregnancy, the corpus luteum will degenerate and the :.:1e cycle wiil start again. From the point of ovuiation, at about the 14th day, *:r-l the beginning of the menstrual fiow is the luteal phase. See Figure 4-5.

fflormonal Control of Ova Production -"":ring a woman's monthly cycle her gonadotropic hormones fluctuate rather :--arLatical1y. LH remains relatively constant until just before ovulation at which --:e there is a sudden surge. FSH will also have a surge, but not as great as the LH surge. There will be a surge of estrogen before ovulation and then a smaller ,:le after ovulation. The increase in estrogen before ovulation actually leads to ':"e increase in L H. Another hormone called progesterone will surge after -.-ulation as well. F{ow are these hormones regulated? -

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Biology

Reproduction & Development (+)

(-)

Hypothalamus

r r r

!r

Female Reproduction

(-) <... .;'.

I

GnRTI

i. . . (-l

Anterior Pituitary

,r//\

FSH

_

(+) on

LH

(-) on

FSH

\"nFSH Inhibin

I

: I I I I I f I I I

Testosterone

O

Cholesterol

()tr

b0 gl

!() o C\0

n9

rl'

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Testosterone

Ar

Theca

cell

Corpus

membrane Follicle cell

High level

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Luteum Figure 4-6 The hormones of the ovarian cvcle.

In the female the theca cells convert cholesterol into testosterone. The testosterone diffuses into the follicle cells where it is converted into estrogen. One of the effects of estrogen is to help in the development of the primary foftcle (i.e., the primary oocyte). We had mentioned that ihe primary oocyte reaches maturit.v

by undergoing reactions with estrogen, LH, and FSH. Recall that the anterior pituitary secetes LH and FSH. LH affects the theca cells while FSH affects the follicle cells. As estrogen is being synthesized, the primary oocyte is developing. At the same time we have a proliferation of the follicie cells. If we have more follicle cells, then we will be able to slmthesize more estrogen. This is where the first estrogen surge comes from. It turns out that at low concentrations of estrogen there is a negative feedback on FSH production. See Figure 4-6.

Flowever, the follicle cells are growing and at a certain point in time there is a threshold level in estrogen. This ineans that relative to the system we no longer have a low concentration of estrogen but instead have a high concentration of estrogen. At high concentrations we see that estrogen has a positive feedback on Copyright

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Keproduction

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Female Reproduction

the anterior pituitary and LH production. Thus, with a high level of estrogen production we now have the LH surge. This is import"ant. The increased concentration of estrogen is having a positive feedback and therefot" more LH production--giving tfe LH r,-rige. The reason that one sees these"u"ri"g ,rrg"! is because of the rapid growth of the rouicte cerls during deveropment or ine

primary follicle.

Immediately following

LH surge the levels of LH clrop down to very low

-the levels. \Ahat causes the drop in LH utrd th" increase in estrogen und progesterone curing.the iuteal phas.el If we no longer have a high concentration of lstrogen,

*'e will lose the positive feedback on the anterlor pituitary. Recall thai ai ovulation all the other follicle cells become transformed into the corpus luteum. ftuY, no longer follicle cells. During that transformation they transientiy 1T lose lhe ability to produce estrogen. f-hls abihty to produce estrogen is lost lusi tong enough so the LH ievels decrease. However, the revels ofioth u'i :rogesterone begin to increase. \M-rere is this coming from? "strog"r, The corpus luteum becomes an endocrine gland and begins to synthesize

estrogen and progesterone. Estrogen and progesterone, in combination, have a negative feedback on LH- and FSH ptod,rciion ir-, the anterior pituitary as we1 as :- negative feedback on the hypothilamus and the synthesis or cnnH. what is :ne result of having a negative feedback on LH and FSH? Essentially what is rappenrng is that the primary follicle is prevented from developing. As long as 'ne corpus luteum is pr9{uginq estrogen Ld progestetrone, the primary foilicie is ;nable to develop. But this is zuhat rue want to inppZn. At this poiit we do not need 'rother primary foliicle being made. lwith this information, what is the basis for

:-rnale oral contraception? Birth control pills are essentiaily a combination of :strogen and progesterone in concentrations that represent the needs of the : ''oman.] Remember, estrogen arone has a positive feedback. Estrogen and ::ogesterone in combination have a negative feedback. different set of errents takes place during pregnancy. what happens 'r when 'er)r spermatozoan comes in coniact with the orium? ih" '..e of the :'-êrmatozoan allows for the digestion of the membrane of the ".rorome secondary oocyte. the membrane of the ovum (zona pellucida) are receptor sites for ""-ithin the ::erm' These receptor sites prevent cross-species fertilizatiorr. Thu only thing that - .ters the secondary oocyte is the nucleus of the sperm. After the nucleus of the j:erm enters the secondary oocyte the zona pellucida changes and prevents any . ::'Ler spermatozoa from entering. This process js referred to as fertilization. At -- rs point the secondary oocyte undergoes the second meiotic division to form --.-: ovum and the poiar body. The nuileus of the sperm and the nucleus of the ::g fuse together to form the zygote, which now has a complemeni of 46 :,-.:omosomes.

- -e zygote rapidiy begins, to deveiop and in about 7 d,ays attaches itself (as a :-astocyst) to the uterine lining. A blastocyst is essentiaily a smali baII of cells -} a central fluid-filled cavity. \zVhen the implantation takes place the placenta, ::de up of maternal and fetai cell types, begins to form.

lr

ii

'1at

one does not want to happen during pregnancy is the development of

pt"gn*iy there aie high levels oiestrogen .---i progesterone. The reasons for maintain'ing high levJls of estrogen ind :: rgesterone are different during the first three months of pregnancy than they '-- other prirnary follicle. Throughout

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Female Reproduction

are during the last six months of pregnancy. shown in Figure 4-7 is adiagram of the hormonal considerations during the iirst three moriths of pregnaniy. The placenta itself is an endocrine gland. During the first three monthJ thu "orpr6 luteum is still a viable gland and it secretes estrogen and progesterone. The placenta synthesizes chorionic gonadotropin (CG) *ni.n stlmulites the corpus luteum to make estrogen and progest"totr". orre of the best pregnancy tests is to detect cG in the blood-stream. The only time CG is made is aurLg thl first three months of pregnancy.

Corpus

Anterior Pituitary

Luteum cclf r.,

Mammary

Placenta

Glands

Figure 4.7 The placenta and its relationships.

what are the functions of estrogen and progesterone? Estrogen and progesterone have a negative feed-back on the anterior pit.titury, tnus titttUitinglH"and rSH

production. This prevents the formation of the primary foihcles during

Chorionic Gonadotropin

pregnancy. It,also means that ovulation and menstrual cycles will be eliminateJ for the duration of the pregnancy. Another important gland during pregnancy is the mammary gland- There are a series of hormones 6at positivell.feeiback on

the mammary glands. one hormone, prolactin, from the anterior "o*". chorionic pituitary. The placenta forms another hormone called somatomammotropin (cs) which also acts on the mammary glands. They help the

o d

o

marunary glands to grow. Estrogen and progesterone have positive feedback on the mammary glands as well.

x IJ.]

D

o 7 2 3 4 5 6 7 8 9 l0 Months after start of iast menstrual period

Figure 4-8 Placental synihesis of estrogen and progesterone.

During the last six months of pregnancy there in only a slight variation on this theme. CG is no longer made. This results in a loss of feedback to the corpns luteum. After three months the corpus luteum breaks down and the supply of estrogen and progesterone from the corpus luteum comes to a halt. Howevei, at that point in time the placenta itself starts to synthesize estrogen and progesterone. In fact, the placenta makes much more estrogen and progesterone that the corpus luteum did. This can be seen in Figure 4-g.

Secondary Sex Qharacteristics in the Female

Some of the effects of the female sex steroid estrogen involve development of

female body configuration, growth of breasts, growth of external genitaLia, pattem of pubic hair, fluid retention, influence on the cardiovascular syslem, and it may be involved in inhibiting atherosclerosis.


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Developmental Stages

DE,VEl;ogffiilffilstegC$ Recall that when we discussed meiosis we mentioned that the haploid sperm and the haploid egg can unite to form a zygote. The zygote will undergo a series of mitotic ceil divisions which will form a clone of cells. A11 of those mitotically dividing cells were derived from a singie precursor cell and eventually they wiil give rise to an embryo. The embryo wiii continue to develop and lafer foims a sexually immature juvenile. The juvenile undergoes a maturation process and becomes the adult.

Thoracic Cavity

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Abdominal Caviry

t

-

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Figure 4.9 -..e body outline.

-:- our discussion on development we will focus on the z:ertebrate.In particular, ' e wiil want to examine the arrangement of the body plan in vertebrates. As we .-li see, human beings, as all other vertebrates, have essentiaily the same body :-an. For example, the skeletal system in vertebrates heips to support the body. }e aspect of the skeletal system is a dorsally located vertebral iol.tm.,. Within -:.:s vertebral column is the spinal cord. Anterior to the spinal cord is the brain.

- ie brain and the spinal cord comprise the central nervous system. Nervous :-ssue outside the central nervous system is referred to as being part of the ;eripheral neryous system. ventral to the vertebral column is the body cavity, or :telom. The coelom is divided into the upper thoracic caaity ur,d thu lower '-.bdominql caoity. The thoracic cavitv contains the heart and lungs while the -:dominai cavity contains a variety of organs like the stomach and the iatestinal --.-stem. These basic iandmarks are shown in Figure 4-9.

he body is composed of a variety of highly differentiated cell types. For =''.ample, there are stratified epithelial cells found in the epidermis. Cuboidal

It rn

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epithelium can be found in the kidney. Smooth muscle cells can be found in the walls of the intestine. Skeletal muscle cells can be found in the voluntary muscles and cardiac muscle cells can be found in the heart. This is just a sample of some of the cell types fond in the body. Cells of similar type form tissues and tissues can come together to form organs like the heart and liver. During development the cells, which initiaily all start out alike, begin to grow and differentiate. This happens even though those cells all contain the same genetic constitution. Th"y all have the same genes. Not only is there differentiationbut some of these cells undergo morphogenesls (i.e., they begin to take on different shapes). There are a number of stages that one can use to characterize development (see Table 4-1). The first deaelopmental stage involves the union of the male and

female gametes. This process is called fertilization. The gamete producing organs are the gonads. In the male these are the testes and in the female they are the oaaries. V\4ren the gametes unite a zygote is formed. The next developmental stage involves cleaaage of the zygote. During cleavage the zygote rapidly divided into many smaller cells without an overall increase in size. Gqstrulstiotr is the third developmental stage. Many embryologists consider this to be the most important part of an organism's development. During this stage the cells of the zygote move to form the three primary germ layers (ectoderm, mesoderm, and endoderm) of the organism. The next developmental stage is neuyulation. During neurulation we begin to see the formation of the nervous system. This is the first organ system to begin differentiation. Neural crest f.ormation is the fifth developmental stage. The neural crest cells help to form parts of the nervous system, skull, and sensory organs. The last stage of development is olganogenesls. It is during this stage that the different organs of the body are formed.

l. 2. 3. 4. 5. 6. Table

Fertilization Cleavage Gastrulation

Neurulation Neural Crest Formation Organogenesis 4-l

Development In The Ffog

We can uie the frog as an example of vertebrate development because of ihe large size of its eggs (about 1.5 millimeters in diameter). The life cycle of the frog is shown in Figure 4'10. As we have mentioned, males give rise to haploid sPelm (1N) while females eventually give rise to haploid eggs (1N). Initially the female frog will give rise to a primary oocyte which then undergoes the first meiotic divlsion. The secondary oocyte is formed but it is arrested at metaphase of the second meiotic division. This second meiotic division is not completed until the sperm from the male frog fertilize the secondary oocyte. Fertilization initiates a pio."rr called egg activation. This is a signal for development to begin. The lecond meiotic division in the secondary oocyte is completed and the zygote nucleus is formed. This nucleus is now diploid or 2N. The zygote undergoes a series of cleavages and an embryo is formed. Further cell division leads to the

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formation of the tadpore. The tadpole, being an aquatic organism, eventualy it can begin its ierrestriat life aJan adult frog. This is a process of metamorhosis. loses its tail and forms legs so

Male

2n

M eta morph osis

6

o

Errly t.rdpole Zr Secondary spermJtoc ytes

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5permatids

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Lgg's second meiotic division

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body

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Figure 4-lO The

life cycle of the frog.

o egg there is a large amount of yolk that resides in the zsegetal -eg."tf",ttilized gole (i.e., the lower hemisphere). This yotk will act as food for the devetolping ,animsl p.ole (i.e., the upperhemisphere) contains mainly cytoplasml n a fertilized egg an interesting phenomenon iakes place. on the side of =blyo:Jhe tir" ugg :pposite to where the sperm penetrates the egg's membrane, a structure called ie gray crescent forms. The gray crescent islocated on the dorsal aspect (the :uture back) of the animal. This is where we will eventually find the spinal cord :nd brain. The side opposite the dorsal aspect is tine oentral aspect. This is where lopyright

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the belly of the organism will form. An imaginary line which connects the two poles is called the meridian. Imaginary lines which run around the circumference of the cell are equatoriai in nature. The middle of the gray crescent defines the body midline of the future organism. By the time the gray crescent has appeared the main axes of the body of the organism has been established.

Cleavage Once the zygote is formed it undergoes a special cell division to increase its mass but not its overall size. The first ceilular division occur along the body midline which bisects the gray crescent. This division gives two cells. The next division also occurs along the meridian, but this time at right angles to the last division. We now have four cells. The third celluiar division occurs equatorially and gives

four cells in the animal pole and four cells in the vegetal pole. The individual cells involved in this growth are called blastomeres. Eventually a small, solid ball of cells will be formed called a morula. Further celluiar division results in the formation of a hollow ball of cells called the blastula. Within the blastuls is a fluid-filled cavity called the blqstocoel. These structures are outlined in Figure & 11.

Animal Pole

Vegetal Pole

dlflD#Tffiffi / \17 \y \.1.7 V Grey

Crescent Unfertilized

Cleavage Down Meridian Body Midline Cleavage (2 cells) (4 cells)

Foo "bb

Equatorial Meridian Cleavage Cleavage (8 cells) (16 cells) Blastocoel

Morula (32 cells)

Blastula

Cross Section

of Blastula

Figure 4- I I Different stages in embryonic development.

Gastrulation During gasf rulation rearrangement of cells occurs. Not far from the gray crescent an open develops in the blastula called theblastopore. Cells from the animal pole begin to migrate inwards through the dorsal lip of the biastopore. As this outer layer of cells migrates inward they form a second layer of cells immediatelrbelow that outer iayer. The blastocoel is reduced in size and is eventuallreliminated. in its place a new cavity is formed called tlte archenteron. At thts stage the embryo is referred to as a gastrula (because gastrulation has taken place). Invagination of this outer cell layer produces two cell layers. The outei cell layer is called the ectoderm. The inner cell layer is called the endoderm. A layer of mesoderm will later form between these two cell layers. These structures are shown from different viewpoints in Figure 4-12.


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The events that lead up to gastrulation


differ widely in the animal world.

Flowever, gastrulation is common to all of them. It seems that the generation of the three cell layers (ectoderm, mesoderm, and endoderm) has been conserved in throughout the evolutionary process. An animal cannot be made if gastrulation does not occur.

Blastocoel

Dorsal Lip of Blastopore


Blastopore

Archenteron

Blastocoe

Lateral Margin

Dorsal Li of Blastopore

Ectoderm

ofEndoderm

Mesoderm

Archenteron


Dorsal Lip of Blasto

Yolk Plug

Yolk Plug Remains of Blastocoel

Figure 4-12 Tire process of gastrulation.

lle ectoderm, mesoderm, and endoderm cell layers play different roles during ire developmental process. They have different deuelopmental fates. For

:rample, the ectoderm will eventually differentiate into structures like the skin, --ne lens of the eye, and the brain and nervous system. Mesoderm will tiiferentiate into structures like the notochord, heart, skeleton, muscle, the outer :overings of internal organs, and the reproductive organs. Endoderm will iiJferentiate into the inner lining of the digestive tract and the respiratory tract, and major glands of the body like the liver and pancreas. These are just a few of --he

clevelopmental fates of these cell layers.

Xeurulation During this stage of development the ectoderm, mesodetm, and endoderm begin :o form the structures that will eventually define the embryo and later the adult.

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During neurulation the formation of the notochord takes place along the body midline. This structure is derived from mesoderm. Superior to the notochord is a mass of ectodermal cells called the neural plate. The neural plate will begin to fold in on itself and form the neural grooae. As the edges of this folding fuse with one another the neural tube is formed. Within the neural tube wili form the spinal cord (encased in the spinal column) and anterior to the spinal cord will form the brain. In other words, the neural p1ate, which is composed of ectodermal cell, gives rise to the neraous system. This tissue is referred to as primordium (from the Latin primus, first, + ordior, to begin) and it is the earliest stage of development of a structure. During neurulation the embryo is sometimes called aneurula. As the neural plate begins to fold in on itself it will form the neural groove. Two views are shown in Figure 4-13. Belolv the neural groove is the notochord and on either side of the notochord is mesoderm. As the neural groove begins to form the mesoderm is split and forms a coelom (i.e., a body cavity). The lungs will eventually develop within this coelom. When the edges of the neural groove fuse together the neural tube will be formed. Within the neural tube forms the spinal cord and anterior to the spinal cord forms the brain. Neural Crest Neural Crest Cells

Neural Plate

Neural Tube

I

Notochord

Neural Groove

Neural Plate

Notocho Ectoderm

Figure 4-

Neural Tube

Mesoderm Coe

l3

Neural tube formation.

Neural Crest Formation

As the edges of the neural groove fused together and became the neural tube, specialized ectodermal cells were left in a more dorsal position on the neural tube. These specialized ce1ls are called the neural crest cells (Figure 4-1-3)' As these cells begin to move to the sides of the developing embryo, they begin to functionalize. For example, ectodermal cells in the anterior portion of the developing embryo associate with the neural crest cells and forms placodes. These structures will eventually form the sense organs located in the head. Some neural crest cells will help lorm sensory cells (e.g., olfaction and touch). Other neural crest cells will form the adrenal medulla. During times of stress the animal Copyright

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will experience the "fight or flight" syndrome. In order to respond to this stress the adrenal medulia releases adrenaline. This hormon" prepurur the animal to respond by increasing biood sugar leveis, heart rate, and blood pressure.

Organogenesis During the initial stages of organogenesis there is an interaction between

ectoderm and mesoderm. The neural fube becomes longer and thinner and has an anterior to posterior developmental gradient. Neural crest cells begin to migrate and take up positions in the vicinity of the neural tube. Mesodermal cells migrate towards the neural tube. Eventually they will form the vertebral column. The brain begins to form at the anterior portion of the neural tube. Optic vesicles begin to form. As the neural tube continues to form segments of mesodermal tissue called somites begin to appear. The somites will eventually give rise to the vertebrae, connective tissue, and the muscles of the body.

In the frog embryogenesis is complete with the appearance of the sexualiy irnmature tadpole. These cteatures have gills and live in an aquatic environment. They have no limbs. Instead, they have a tail which is used to propel them through the water. The last critical phase in the life of a tadpole L one of morphogenesis. The tadpole will change from a sexually immafure organism into

a sexually active organism, which is the frog. During this process of

morphogenesis the tadpole grows limbs, develops lungs, loses its tail and its

gi-Ils. This change is initiated by a hormone which is reieased during a specific lirrre of development. The hormone which is released is thyroxin and it is released from the thyroid gland.

\Iany organisms pass through a larval stage (sexuarly immature) like

the

:adpole. \rvhen the frog eggs hatch the tadpoles are left to fend for themselves. If Jre post-embryonic organism is left on its own, then the development is termed ndirect. IrL contrast, -direct development involves care being given to the postembryonic organism by the mother. For example, a human fetui develops within --:e female of the species. Even after birth the female (and male) care for the '.-oung until they are able to take care of themselves.

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Developmental Mechanisms

Developmental Mechanisms If the cells of the body all contain the same genetic information, how can they become so differentiated? There are two geneial classes of interaction associated with the differentiation of cells that we need to consider. There are intrscellular interactions which involve interactions between the components within the cells themselves. There are also intercellulqr interactions in which the interactions are between cells. The intracellular interactions usually result in the setting up of a

prepattern. The intercellular interactions usually undergo deaeloimintal induction.

Intracellular Interactions we have mentioned that the unfertilized egg has an animal pole, which contains the cytoplasm, and a vegetal pole, which contains the ytk. The egg is not homogeneous. However, at this point in the egg's existence a longitudiria"l axis is already present. In the frog the animal pote witt give rise to the head while the vegetal pole will give rise to the tail. This is true for the frog but it is not a truth

set in stone. Remember, there are always variations on a give"n theme.

Recall that after fertilization and induction of development the gray crescent f9r1s at a point opposite to where the sperm penetrate-d the egg. T;he formation of the gray crescent is due to an intracellular interaction. The"iormation of the

gray crescent is not a prepatterned phenomenon that took place before the sperm interacted with the egg. This is clearly shown by the fact that the gray crescent developed after the penetration of the egg by the sperm. The entry*of ihe sperm in the animal half of the cell is random. However, one entry is made and the grarv crescent forms, the dorsal midline is established. This allows us to define

directions. In other words, prior to the first cellular division the axes of the organism is established. In this case a prepattern is laid down which is adhered to during the rest of development.

llans Spemann During the 1920's Hans Spemann was able to demonstrate that the gray crescent is an important landmark in the future development of the embryo. In his experiment, spemann took a newly fertilized frog egg and tied a string around it such that the string bisected the gray crescent (see Figur" 4-l4a). He Jlowly tied the string tighter and separated the egg into two hal-ves, each with half a grav crescent. Two separate blastomeres developed and two separate embryos riere formed. since these embryos were twins, ihir proc"dure was called iwinning. This experiment demonstrated that each blaitomere was equivalent in G potential to form a complete embryo. Even though two complete embryos were formed, they were smaller than the normal frog embryot. wnyz Because when the fertilized ugg was cut in half, each blastomere receivea nil1 of the original amount of yolk. There is one other important point to make note of in this experiment. During ttle last lecture we said that the body midline bisected the gray crescent and thaithis

was what eventually formed the neural tube. In the spemann experiment the body midline was itself bisected. In other words, instead of the midhne being in the middle it was to the side of the bisected gray crescent. However, the embrlm developed normally. what this means is that the midline had to migrate to a position which was in the middle of the bisected gray crescent. Hence, Ih"r" *o Copyright @ by The Berkeley Review

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a deaelopmental


regulation inside the blastomere which allowed for the

movement of the body midline.

Spemann did the same experiment on another fertilized frog egg but this time he cleaved the egg at right angles to the cleavage in the pt"r1o.ir""^periment (see Figure 4-14b). Two blastomeres were formed. one blaitomere had the compiete gray crescent while the other blastomere did not contain any piece of the gray crescent. As development proceeded a complete embryo was from the blastomere with the complete gray crescent. The other blastomere formed a mass of cells that did not differentiate into an embryo. It did not gastrulate. Again, the embryo that was formed was smaller due to the reduction in the amount of volk received. These two experiments proved for the first time that there *u, u Jl"u, crepattern during the development of an egg.

(a)

Bisection

Bisection

.+.-

I

/'"T-,\

r )

(i\

^Gray i_ Crescent)>\itry

t-"1$' {\\J-,/

I

fr /1\

I

I

+

zr-f|f\\

/t'\

/'l\

\\.rll1-,t J4.l

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ng f^>

\ ti I \9 /

lr 11

f

Grav Crescent

gs

/^>.(

/

("\

/,{

1r

l

9

Complete Embryos will Deveiop

Undifferentiated Mass of

Complete

Cells

Embryo

Figure 4-14 :i:ns

Spemann's gray crescent experiment.

Hilde Mangold and [Ians Spemann - 7924Hilde Mangoldperformed an experiment in which she removed a section

: the dorsal lip from one species of salamander embryo (before gastrulation .:arted) and transplanted it into the belly of another species of salamander =nbryo. These two species differed in their pigmentation 1tn" donor being of ,-:hter pigmentation that the recipient). Recall that the dorsal tip is that irea '""-here cells from the animal pole begin to invaginate into the blastula. This :rocess leads to the formation of the blastopore and the development of the :ichenteron. The blastopore, and hence the dorsal lip of the blastopore, form :-cse_ to theboundary of the vegetal hemisphere and the gray crescent. [n Figure 1-14b one blastula did not receive the gray crescent. Therefore, it could not ::r'elop the dorsal lip and ectodermal cells could not invaginate. This resulted in :,e blastula never developing into an embryo.]

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The recipient salamander embryo now has two dorsal lips. \Alhen it developed it formed a second embryo (of a lighter pigmentation) at the location of the

transplant (Figure 4-15). The transplanted dorsal lip deveioped into a notochord. In turn the notochord induced the formation of a neural plate and eventually a neural tube. What this experiment is saying is that the ectoderm of the recipient salamander was induced by the dorsal lip of the donor to take on a fate that it was not normallr'

designed to fulfill. In other words, the dorsal

lip acts as an organizer

and

organizes the cells of the recipient such that a second embryo is formed. This process later became known as embr)'onic (or developmental) induction. The mesoderm is removed opposite the dorsal lip of recepient embryo.

'*>

^ f"'4*r

'T

f:

;ry

Mesoderm is removed from a donor embryo near the dorsal lip and transplanted into the recepient embryo.

a

@

\2$

Recepient Embryo

Primary Neural Fold

Recepient Embryo

Primary Neural

Donor Embryo

Secondary Neural

Development

)

Secondary

+ Development of a double embryo.

Neural Fold

Figure 4- 15 Transplant of the dorsal lip.

Hilde Mangold was a student of Hans Spemann when they did this experimm: She later died while cooking in her kitchen when her gas stove blew up. -{ number of years later Spemann got the Nobel Prize for his participation in ti':s work.

In another experiment Spemann examined a chain of successiae inductions th,n leads to the formation of the eye. In the anterior most portion of the neural tuire (the forebrain) there are two symmetrical protrusions from the developing brab* These protrusions extend to the ectoderm. Upon contact the ectoderm begins trn invaginate and pushes into the growing optic stalk. This creates lhe optic ry The optic stalk will eventually become the optic nerae. The lens (placode)of rm eye develops from ectoderm which is in contact with the edges of the optic on* As the developing lens pulls away, the remaining ectoderm develops into =u Copyright

@


224

The BerkeleY Reriem Specializing in MCAT Preparatic

Biology


cotrre&. The remainder of the optic cup develops into the the photoreceptors necessaly foi vision (Figure 4-16).


retina,which contains

Ectoderm Optic Cup

Optic Stalk

#y

Invagination

Figure 4- I6 )evelopment of the eye.

Erperrments like these indicate that there is a hierarchy of developmentar stages. )uring the 1930's this gave rise to the field of chemicai Researchers ai the time thought that induction was due to some kind"*ury"logy. of biochimi"ut

ii :rrms out that this line of thought was a waste of time and effort."whyi \{olecular biology had not yet been invented. The selfish g"r" ,,""a"d to be ug"rrt.

=rplored.


229


Biology


Human Embryo Development

,ruffiH.

t*ti;+,...'1,,,

;' ,

,1' .,,.

Recall that fertilization of the egg by the sperm usually takes place in the oviduct (Fallopian tube). Within the first 35 hours following fertilization the zygote will undergo its first mitotic division. It will continue to divide as it travels down the oviduct toward the uterus. By the fifth or sixth day the embryo will reach the uterus. At this point the embryo is called a blastocyst. The blastocyst is a hollow ball of cells with a mass of cells on one side. The surrounding cells which form the ball are called trophoblast; the others are called the inner cell mass. If the lining of the uterus is prepared to receive the embryo, then the embryo will implant. fOtherwise, the embryo is rejected and

sloughed off during the next menstrual period.] After implantation the trophoblastic cells grow into the lining of the uterus. In fact, the cells of the trophoblast grow little finger-like projections into the uterus. As the fetus gets bigger it will need to get nourishment from the mother's blood. The blood vessels from the umbilical cord will grow into the projections of trophoblast. There is a layer of cells between the trophoblast and the fetal blood

vessels called the chorion. The chorionic villi are the projections of fetal/umbilical blood vessels and the chorion that covers them. The trophoblast abuts the uterine cells of the mother, but allows arteries in the uterine lining to drain into sinuses around the chorionic villi. The chorion preserves the barrier between the mother's and fetus's blood (but diffusion of nutrients and waste products may still occur). This whole exchange apparatus is called the placenta. The inner cell mass will undergo changes similar to those in the frog such that the three basic cell layers are formed. First, the inner cell mass forms a cavity within itself. This is called the amniotic cavity and the cells which line it are the ectoderm. Below the amniotic cavity and its ectoderm, on the hollow side of the blastocyst, a layer of cells will form along the ectoderm and will evenfually cover the entire hollow blastocyst. This layer is the endoderm. So what we have at this stage is a hollow ball of cells, the blastocyst, which is surrounded by trophoblast.

Inside this ball there are two compartments; an amniotic cavity defined by ectoderm and yolk cavity defined by endoderm. It is the two-layered sheet of cells formed by the ectoderm and endoderm that will develop into the fetus. So unlike the frog which developed from the whole ball of cells, the human only develops from the sheet of cells suspended in middle of the blastocyst. The rest of the cells of the blastocyst will be important in forming the placenta and chorion which protect the fetus inside the womb. The primitive streak, equivalent to the neural plate in frogs, forms in the ectoderm above the endoderm. Cells from the primitive streak migrate down between the ectoderm and endoderm to become mesoderm. Further folding of the primitive streak gives rise to a neural groove and then a neural tube, etc. So

the formation of the primitive streak in mammals marks the beginning of gastrulation. It is quickly followed by neurulation. The fate of cells from the three basic cell layers will be the same as in frogs: endoderm will become the GI lining; mesoderm will become connective tissue, bones, muscle, blood; ectoderm will become skin and nervous tissue (from the neural grove and neural crest cells), etc.


230


Biology


Human Embryo Development

tlormones and Pregnancy

The embryo cannot implant if the uterus is not receptive, that is, if it is not quiescent. Furthermore, pregnancy will not be maintained by the uterus even after implantation unless it remains quiescent. So to maintain the uterine lining through the first part of pregnancy the trophoblastic cells secrete chorionic gonadotropin. Chorionic gonadotropin (CG) is a hormone which causes the .orpus luteum to continue to produce estrogen and progesterone. As long as the ,evels of estrogen and progesterone, particularly progesterone, are sufficient, irev keep the uterine lining quiescent and pregnancy continues. Eventually, the :iacenta is able to take over the production of estrogen and progesterone itself ::rd the corpus luteum is not needed.

rhe piacenta secretes increasing amounts of estrogen and progesterone. iorvever, the levels of estrogen increase faster than the levels in progesterone. -'-ld at some point near the end of pregnancy the ievels of progesterone plateau. ri a crucial ratio of [progesterone]:[estrogen] the uterus is no longer quiescent; it ::gins to have contractions. The smooth muscle of the uterus contracts, putting it

'-:rder tension. Also the size of the fetus at this stage can put the smooth muscle -:L the walis of the uterus under tension. when the walls of the uterus are put '::rder tension thev send nervous impulses to the hypothalamus. The -,"-pothalamus sends signals to the posterior pituitary (via nerves) to release -'rr-tocin. oxytocin is released into the blood and is a strong inducer of more :rntractions of the smooth muscle of the uterus. Oxytocin also stimulates the ::oduction/secretion of prostaglandins which further induce contractions. of these hormones and nerves form a positive feedback system. Once started :.e., contraction) each step stimulates the next and eventually stimulates the first

.i11

::ocess more (stronger contractions). Eventually the contractions force the fetus

rrough the vagina and into the hard, cold world. we call the birth process parturition.

Lactation -here are two different processes that go on in lactation. First, there must be milk :roduction, and second, there must be milk ejection. Breast milk is a fairly :rmplex fluid containing proteins, fats, vitamins, and other goodies (antibodies, =ic.). There are epithelial cells within the breast that make up the glands that ::oduce the milk. Around these epithelial cells are myoepithelial cells which can :r-)ntract around the milk glands, ejecting the milk that has been produced.

lroduction of milk is stimulated by the suckling of the infant on the mother's :ipple. The nipple sends a nervous impulse to the hypothalamus, The :..-pothalamus releases PRH, prolactin releasing hormone, which acts on the :nterior pituitary. PRH stimulates the anterior pituitary to release prolactin. lrolactin goes into the bloodstream and stimulates the epithelial cells in the ::easts which comprise the milk glands to produce more milk. l"tilk ejection uses a similar process, but different hormones. The suckling on the :reast sends a nervous signal to the hypothalamus. The hypothalamus send a :.ervous signal to the posterior pituitary. The posterior pituitary responds by :eleasing oxytocin. Oxytocin causes contraction of the myoepithelial cells in the ::east. Milk gets squeezed out.

Llopyright @ by The Berkeley Review

23t


Reproduction & Development To Go 15 Passages

102 Questions

Time for All Passages Thken Together as a Practice Exam 125 Minutes

Passage Titles

I. U. ilI. IV. V. VI. VII. VIII. IX. X. XI. XU. XIIII. XIV. XV.

Spermatogenesis, Oogenesis, & RU486 Male Contraception Fertilization Of The Ovum Pregnancy Fuel Utiliz.atiort Female Reproductive Hormones Endocrine Control Of Ovarian Cycle Gestation Female Birth Control Vaccine Estrogen Isoflavone Experiment Ovulation Vertebrate Gastrulation Oxytocin & Labor Puberry Testicular Cancer

ffi EKT{ruNY V. f .,E' Speciabzing in MCAT Preparation E'.

t

t

trli

il

Questions

1-8 9-15

t6-2t

22-29 30-35 36-43 44-49 50-56 57 -63

64-69

l0 -76 11 -83

84-90

9r -96

97 - r02

Suggestions - The passages that follow are designed to get you to think in a conceptual manner about the processes of physiology at the organismal level. If you have a solid foundation in physiology, *u1y of these answers will be straightforward. If you have not had a pleasant experience with the tJpic, some of these answers might aPpear to come from the void past the Oort field of the solar system. - Pick a few passage topics at random. For these initial few passages, do not worry about the time. Just focus on what is expected of you. First, read the passage. - Seco-nd, look at u.,y diugru-s, charts, or graphs. Third, read each question and the accompanying answers carefully. Fourth, answer the questions the best you can. Check the solutions and see how you did. \A/heth"t ybu got the answers right or.wrong/ it is important to read the explanations and see if you understand with) whaiis 1urdugr"" being explained. Keep a record of your results. After you feel comfortable with the format of those initial few passages, pick another block of Passages and try them. Be aware that time is going to become importani. Gerierally, you will have about 1

minute and 15 seconds to complete a question. Be a little more creative in how yo,, upprouch this next

group. If you feel comfortable with the outline presented above, fine. If not, then try different

approaches to a passage. For example, you might feel well versed enough to read the questions first and then try to answer some of them, without ever having read the passage. Maybe you can answer some of the questions by just looking at the diagrams, charts, or graphs that ai pt"r"r,t"a in a particular passage. Remember, we are not clones of one another. You need to-begin to devllop a format that works best for you. Keeping a record of your results may be helpful.

The last block of passages might contain topics that are unfamiliar to you. Find a place where the level of distraction is at a minimum. Get out your watch and time yo,rrr"lf on these passages, either individually or as a group. It is important to have a feel for time, and how much is passing as"you try to answer each question. Never let a question get you flustered. If you cannot figure o.rt 1"hut the answer is from information given to you in the passage, or from your own knowledge-tase, dump it and move on to_ the next question' As you do this, make a note of that pesky question Jt d .ornu bacl to it at the end, when you have more time' \Mhen you areJinished, check yow urr*"rs and make sure you understand the solutions. Be inquisitive' If you do not know the answer to something, look it up. The solution tends to stay with you longer. (For example, what is the Oort field?)


Section IV Estimated Score Conversions Scaled Score

>12 10- 11

8-9 7

6 5

<4

Raw Score

- 100 79-85 6s-78 59-64 54-58 48-53 0-47

86

Biology Passage

Spermatogenesis, Oogenesis. & RU 4g6

I (Questions 1-8)

following cell types?

A. B. C. D.

tormed is called the zygote. The formation of the male and female gametes occurs during gametogenesis, a process in which there are two successive meiotic divisions to reduce the chromosome number from a diploid state to a haploid state.

.,

During spermatogenesis the male germ cells (called sf.ermatogonia) grow and are transformed into primary

3.

:rature spenn. Spermatogenesis takes about 64 days.

Spermatogonia Oogonia Spermatids

Primary oocytes

Before fertilization the secondary oocyte is arrested in meiosis II at:

A. B. C. D.

spermatocytes. Each primary spermatocyte forms a :aploid secondary spermatocyte after the first meiotic jivision. After the second meiotic division, secondary .rermatocytes form spermatids which are then gradually :ransformed by differentiation (spermiogenesis) into

interphase. prophase. metaphase. anaphase.

Roughly 3 days after fertilization in the ampulla of the human fallopian tube, the morula:

During oogenesis the femaie germ cells (called

I. II. IfI. IV.

grow and are transformed into primary oocytes.

These oocytes begin their first meiotic division before :.:.rturition but are arrested in prophase until after puberty.

Tris reductive division is not completed until shortiy ::lore ovulatiion. As the primary oocyte grows duri unng

develops a blastocoel. contains about 16 blastomeres. passes into the uterus.

forms an outer cell layer called the tropoblast.

A. I and II only B. II and III only C. III and IV only D. I and IV only

: *berty, follicular epithelial cells begin to surround the .,rcvte, forming a primary follicle. As soon as a second .','e1 s; epithelial cells is formed the primary follicle :-romes a secondary (or mature) follicle. The second

:-:iotic division begins at ovulation but is arrested ::dway through the cycle unless the secondary oocyte is ::rilized by a sperm. Once fertilization takei place the :rrure oocyte is referred to as the ovum.

4.

As the mature ovarian fotlicle begins to swell,

a

small oval protrusion called a stigma forms on the side of the follicle juxtaposed to the peritoneal cavity. Rupture of the stigma:

The monthly reproductive cycles of the female is -;_lulated by the gonadotropins, follicle-stimulating -:,!mone (FSH) and luteinizing hormone (LH). FSH ' luces the development of an ovarian follicle. However, , ih FSH and LH are required for follicular maturation. --s the mature follicle grows, it begins to produce ::'-rogen. High estrogen levels produce a surge in the ,r:thesis ofLH and this surge triggers ovulation ofthe :-:ondary oocyte. The mature foliicle develops into a :.:ndular structure called the corpus luteum and begins to

I. [. rrl. IV.

releasesprogesterone. is due to increased foilicular pressure. releases a secondary oocyte. results from stimulation by prostaglandins.

A. I oniy B. II and III only C. IV only D. I, II, III, and IV

":rrete both progesterone and estrogen. These hormones ::Ein to prepare the endometrium of the uterus for .' :lantation of the blastocyst.

Dispermy is an abnormal fertilization process in which two sperm fertilize an ovum. The resulting

If fertiiization occurs, the corpus luteum becomes the rpus luteum of pregnancy and is maintained for about - - $'eeks until the placenta can assume production of ::rsesterone and estrogen. If fertilization does not occur, ,--,: corpus luteum begins to degenerate about 12 days .-:er ovulation and becomes the corpus luteum of :,:nstruation. -

Cellular division does NOT occur in which of the

1.

Human development begins with the union of the sperm (male gamete) and the ovum (female gamete) during fertilization. The unicellular organism which is

- r_eonia)

Passage I

embryo will:

-

- :,pvright @ by The Berkeley Review

A. B. C. D.

235

show trisomy 21. triploid and therefore exhibit aneuploidy. show monosomy.

be

will not exhibit polyploidy.


Biology 6.

Spermatogenesis, Oogenesis, & KU 48G

The molecular structure that BEST represents

8.

oxytocin is:

Passage I

The molecular structure of the synthetic steroid RU 486 is shown in Figure 1.

A. OH

f",

I

C_ CH"

t'

H

NHz

CH l

- o

I

N. HrC

B. RU 486

HOHO otiltlt

n.N*F c-f-?-c-o-cHl

i3""*'0

o

o : -/z

HO

H:,

(", c= c-

cH.,

, Figure

1

C.

The molecular structures of four naturally occurring steroids derived from cholesterol are shown below.

D.

o

s-s------l

H,N- Cys- Tyr- Ile- Gln- Asn- Cys HrN- Cil

Gly

- lru-

Pro

o Progesterone

7.

Testosterone

RU 486, used widely in France to terminate unwanted pregnancies, was originally tested as an

endometrial lining of the uterus. However, when

OH

o

antagonist for glucocorticoid receptors. During testing this compound was also shown to have a high affinity for other steroid receptors as well. Administered alone, RU 486 has about an SOVo success rate in expulsion of the embryo from the

I

C

-

cH2

a

drug which causes uterine contractions is administered about 36 to 48 hours after receiving

RU 486, the success rate increases by about 26V0. RU 486 has a high success rate because it blocks

Cortisol

Estradiol

steroid receptors from binding:

A.

glucocorticoids and allows administered

B.

progesterone to cause uterine contractions. testosterone and allows administered oxytocin

C. D.

Copyright

Based on the structures of the four steroids showu

above, which receptors

to cause uterine contractions. progesterone and allows administered prostaglandins to cause uterine contractions. estradiol and allows administered glucocorticoids to cause uterine contraction.

@


will RU 486 bind to with

the strongest affinity?

A. B. C.

D.

236

Progesterone and testosterone receptors. Estradiol and cortisol receptors. Testosterone and estradiol receptors. Cortisol and progesterone receptors.

The Berkeley Revieu Specializing in MCAT Preparation

Biology Passage

II

Male Contraception

(Questions 9-15)

12. What changes

does a vasectomy cause in the hypothalamus, particuiarly invoiving the secretion of FSH and LH?

Sterilization is a contraceptive option for people who are positive they wish to have no children. About 500,000

men in the U.S. choose to have vasectomies per year. Vasectomy is the surgical ligation of the vas defeiens, resulting in sterilization. Although sperm are produced normally in men with vasectomies, the sperm cells are not able to exit the testis. In a local phenomenon, sperm cells are phagocytosed in the testis by macrophages and other rmmune system cells in men with vasectomies. As a result

rf

Passage tr

A. B. C. D.

Both FSH and LH are increased. FSH is increased, and LH is decreased. FSH is decreased, and LH is increased. Both FSH and LH are unchanged.

this immune attack, antibodies to sperm are elevated in

nen with vasectomies.

13. What is the anatomical location of production?

9.

What is the effect of vasectomy on spenn production and secretion after several months?

A.

A.

Vas deferens

B. C. D.

Seminiferous tubules Prostate gland

sperm

Testicle interstitium

Both sperm production and sperm secretion lncrease.

B.

C. D.

Both sperm production and sperm secretion decrease.

Neither sperm production nor sperm secretion

L4,

change.

Sperm production

is unaffected and

Which of the following statements are TRUE?

sperm

The sperm count ultimately drops to zero following a vasectomy.

secretion stops.

II. The female counterpart of vasectomy

ilr. n{i. What is a potential

consequence

of reversing a

B.


C.

II

D.

I, II, and

A.

vasectomy?

A. Antibodies to sperm may reduce fertility. B. Antibodies ro sperrn may increase fertility. C. Birth defects are more likely in subsequent

1S

hysterectomy, the removal of the uterus. Vasectomy decreases testosterone levels.

and

III only

III

children.

D. Birth defects

are less likely

in

subsequenr

children.

15. In an experiment on dogs, researchers used an injection of a stable polymer to block the vas deferens. What results could be expected from this experiment?

11. If

a man were diagnosed with oligospermia, what implications would this have for his fertility status?

A. B. C. D.

A.

The dogs undergoing this treatment would remain fertile.

He could not ever father a child. He has reduced fertility status compared to the population average. He has increased fertility status compared to the population average. He has the same fertility status as the population average.

,:p1'right O by The Berkeley Review

B.

The dogs undergoing this treatment would

C.

The dogs undergoing this treatment would

develop enlarged testicles. become sterile.

D.

The dogs undergoing this treatment would develop autoimmune disease.

237


Biotogy Passage

III

Fertilization Of The Ovum

(Questions 16-21)

Passage Itr

18. During pregnancy the levels of progesterone

and

estrogen:

In fertile women, ovulation occurs approximately 12of the previous menses. The ovum must be fertilized within 24-48 hours if conception is to 16 days after the onset

A.

result.

B. C. D.

Fertilization of the ovum by the sperm usually occurs rube. The fertilized ovum, now called a blastocyst, moves down the uterine tube into the uterus. Once in contact with the endometrium, the blastocyst becomes surrounded by an outer layer of syncytiotrophoblast and an inner layer of

in the mid portion (ampulla) of the urerine

increase during the first trimester, but decrease during the remaining six months.

increase during the

first six months,

but

decrease during the last trimester.

increase steadily until parturition.

mimic the levels found during the menstrual cycle.

cytotrophoblast. The syncytiotrophoblast erodes the endometrium, and the blastocyst burrows into it, a process called implantation. After this process, the development of the placenta then proceeds.

to the passage, it can be inferred that during pregnancy leutenizing hormone and follicle

19. According

Human chorionic gonadotropin (hCG) is secreted by the placenta, and during the early stages of pregnancy, acts to maintain the corpus luteum, an endocrine organ that was maintained by luteinizing hormone (LH) during the luteal phase of the menstrual cycle. The corpus luteum, now termed the corpus luteum of pregnancy, primarily secretes estrogens and progesterone. In addition

stimulating hormone levels would:

A. B.

C.

to secreting hCG, the placenta, after about the 6th week of

D.

pregnany, takes over the function of the corpus luteum and produces sufficient estrogen and progesterone to continue suppression of gonadotropin releasing hormone (GnRH), follicle stimulating hormone (FSH), and LH.

16. Removal

remain relatively low, thereby eliminating further follicle development and ovulation. be undetectable in either fetal or maternal blood. increase in the maternal blood but decrease in the fetal blood.

progressively increase until ovulation

has

taken place.

20. Ovariectomy (removal of the ovaries) or damage to the ovaries during the last six months of fetal development would:

of hCG during the first 6 weeks of

pregnancy would result in:

A.

L

tr.

uI. A. B.

c. D.

continuation of the pregnancy. regression of the corpus luteum. termination of the pregnancy.

B. C. D.

have no effect at all on the prcgnancy. lead to termination of the pregnancy. result in a decrease in the levels of hCG. result in an increase in follicular development.

I only II only I and II only II and trI only 21. If implantation of the blastocyst does not occur in the uterine lining, then the proper hormonal levels

needed

17. The protein hormone hCG is very similar but NOT identical in structure to:


to maintain the lining are removed and

menses normally follows. Which of the following structures is sloughed off during menses? A.

Endometrium

LH

B.

Myometrium

FSH GnRH

c.

Primary oocyte Primary fbllicle

D.

Estrogen

@


23a


Biology Passage

Pregnancy Fuel Utilization

IV (Questions

23.

22-29)

Human gestation is 40 weeks. Along with the changes apparent to the eye, fuel metabolism changes considerably irom early pregnancy (0-25 weeks) to late pregnancy (25-

Passage IV

Insulin is important for energy storage during early pregnancy. Which of the following statements is FALSE regarding the effects of insulin on adipose tissue?

A. B.

weeks). In early pregnancy, the main goal is to build up maternal stores of adipose tissue. The fetus is quite small and has low energy needs during this time. The nother experiences increased insulin sensitivity which increases glucose and fatty acid uptake by adipose tissue, especially in the lower body depots. -10

Insulinincreases synthesisoftriglyceride.

Insulin increases production of o-glycerol phosphate.

C. Insuiin D.

increases uptake

of lipid

from

circulating lipoproteins.

Insulin increases synthesis of glucose.

However, in late pregnancy, the fuel needs of the fetus .rre increased. The mother becomes insulin resistant, so

ihat glucose is diverted to the fetus. The growing fetus

24.

-eeds continuously. Glucose, received through the :lacenta via facilitated diffusion, is its prefened fuel. The :nother mobilizes fatty acids from her adipose tissue to :leet her energy needs. Maternai gluconeogenic enzymes .re increased, as well, during late pregnancy.

resistance. Therefore, many obstetricians screen all their patients for abnormal glucose tolerance using an oral glucose tolerance test. An oral load of 50 grams (200 kcals) of glucose is given to a fasting person. A blood sample is taken t hour later. What changes in glucose and insulin compared to pretest values would be seen in a woman with sestational

22. When triglycerides are mobilized from adipose tissue, which of the following compounds are released into the blood?

diabetes?

A. B. C. D.

I.

HO til H-C-O-C-(CH,),,-CH,

I' tfl

H- C- O-C-

Il?

Some women experience gestational diabetes during

late pregnancy. This is partially due to insulin

Increased glucose, increased insulin. Increased glucose, decreased insulin. Decreased glucose, decreased insulin. Decreased glucose, increased insulin.

(CH2)14- CH3

I

H

- C-O-C-

(CH2)ri-

25.

CH3

Which organ produces insulin?

H

A. B. C. D.

II. H I

Pancreas

Thyroid gland

Liver Spleen

H-C_OH I

H- C_

OH

I

H- C_

OH

26.

I

H

Which of the following statements are TRUE for late pregnancy?

I.

III.

I IfI.

et

cH3- (cH2)r4- coo

Gluconeogenic enzymes provide glucose for the fetus when the mother is not eating. Fatty acids are a required fetal energy source.

Maternal fat stores are rapidly filled during late pregnancy.

A.

I only

B. C. D.

II only II and trI only I and III only

-:p1,right

@


A. I only B. I and II only C. II and III only D. I, II, and III 239


Biology 27. How would untreated gestational

Pregnancy Fuel Utilization

Passage IV

diabetes most

likely affect the size of the fetus?

A. Lower birthweight. B. Higher birthweight. C. No change in birthweight. D. During the first trimester fetal

weight is slightly above normal, but at parturition the

birthweight is much lower.

28.

Which of the following is TRUE regarding glucose transport across the placenta?

I. il. IU.

It is an energy requiring process. It occurs by facilitated difussion. It is higher during early pregnancy.

A. I only B. II only C. I and lll only D. I and IV only.

29.

What type of cells contain gluconeogenic enzymes?

A. B. C. D.

Copyright

Placental cells Neurons Hepatocytes

Intestinal mucosal cells

@


240


Biology

Female Reproductive tlormones


32.

The female reproductive system undergoes a series of regular cyclic changes termed the menstiual cycle. The

most obvious

of

these changes

is periodii

vaginal bleeding resulting from shedding of the endometrial lining of the uterus. It results primarily from the interaction of hormones derived from the hypothalamus, pituitary gland, and ovaries. In most women in the middle reproductive vears, menstrual bleeding recurs 25_35 days, with a median cycle length of 28 days. The interval from the onset of menses to ovulation is termed the follicular or proliferative phase. The time proceeding ovuiation to the

onset of menstrual bleeding is termed the luteal or secretory phase. Ovulation normally occurs at about the 1.lth day of the cycle.

Passage V

According to the passage, the preovulatory decline of FSH is due to:

A. B. C. D.

positive feedback of estradiol. negative feedback of FSH. negative feedback of estradiol. positive feedback of FSH secretion.

33. In the absence of

pregnancy, menses normally occurs. This may be due to the decline of thl hormonal requirement needed by the endometrium.

Th." primary hormones required by

In normal menstrual cycle, serum concentrations of both leutenizing hormone (LH) and follicle stimulating

endometrium, so that menses does not occur. are:

hormone (FSH) begin to increase prior to menses. FSH ;oncentrations attain maximum levels during the first half of the foilicular phase and, with the exceptlon of a brief

I.

u. Itr.

peak at midcyle, continue to fall uniil th* lowest ;on^centration in the cycle are reached during the second nalf of the luteal phase. The preovulatory deiline of FSH .s due to the increasing concentration of estradiol. LH levels increase gradually throughout the follicular phase .nd at midcyle, there is a large peak in serum :oncentration of LH. Subsequentiy, LH levels gradually jecline reaching their iowest concentration late in the

IV.

the

progesterone. estradiol. hCG. FSH and LH,

A. I and II only B. I and III only C. II and IV only D. III and IV only

.rteal phase.

34. All of

Women treated over a long period of time with relatively large doses of progesterone and estrogen

direct inhibition of progesterone by estrogen. over stimulation of FSH by estrogen.

A.

I only

B.

II only I and II only II and III only

C.

D.

A. B. C. D.

inhibition of FSH.

II. rII.

35. The Ovulation is marked by a maximal peak concentration. This is primarily due to:

A. B. C. D.

anterior pituitary. placenta. granulosa cells. adrenal cortex.

stimulus

for FSH and LH production

and

secretion is governed by the pulsatile release of gonadotropin releasing hormone (GnRH). Which one of the following structures produces GnRH?

of LH

A. B. C. D.

positive feedback by progesterone. positive feedback by estradiol. negative feedback by progesterone.

negative feedback by estradiol.

- -rp1'right @ by The Berkeley Review

will either secrere

progesterone together EXCEpT the:

do not ovulate. This is most likely due to:

I.

the following structures

estrogen alone, progesterone alone, or estrogen and

241

Anterior pituitary Posterior pituitary Hypothalamus Pineal gland


Biology Passage

Endocrine Control Of Ovarian Cycle


36.

Endocrine control of the ovarian cycle is complex, with cases of both positive and negative ieedback control.

What is meant by negative feedback?

A. -B.

The following diagram illustrates interactions between the hypothalamus, the anterior pituitary, and the ovaries. The

following abbreviations ar" used in Figure 1: GnRH (gonadotropin releasing hormone), nSH ltottictestimulating hormone), and LH 0uteinizing hormone).

Passage VI

The rate-limiting enzyme in a pathway is activared by the producr(s) of the pathway. The rate-limiting enzyme in a pathway can only be inhibired.

C.

The rate-limiting enzyme

D.

inhibited by rhe producr(s) of the pith.ay.The rate-limiting enzyme in a pathway. can

in a pathway is

only be activated.

37.

Hypothalamus

?

o

Anterior Pituitary

Decreased

(Day

inhibition

o

ts

o

onlrfr

I-H

fr

Anterior pituitary

Estradiol

l-l

\z

fr

(+) Feedback

J cnRtt

Anterior Pituitary

s s

t

t

rtU

c)

o tf

.: (D

I. II. III.

Decreased FSH and LH Increased growth of follicles

A. B. C. D.

II only II and III only I and II only I, II, and III only

inhibition of ovulation

40. Menopause is the cessation of the menstrual

cycie. New follicles cease to develop in the ovary, although the anterior pituitary continues to function normalh.. What levels of FSH and LH would be seen in the

Estradiot

Progesterone

exogenous hormones? Empty follicle will

r----\

Figure

@

and

blood of a menopausal women who takes no

A. B. C. D.

become the corpus luteum

(Dav 14i '

Copyright

a

cD

(-) Feedback'f

Ovaries

Ovuiation

LH FSH

Hypothalamus

LH Surge

\.2

Progesterone

contraceptives are taken?

J r-n J psu

n

H

GnRH

B. C. D.

progesterone. Usually, the pills are taken for 3 weeks and stopped for I week to allow menstrual flow. What happens to the ovarian cycle when oral

Ovaries

\.2

t

A.

39. Oral contraceptives contain both estradiol

Growth of Follicies

()

The corp'rs luteum produces which of the following hormones?

regresses

n \,2 p.

Uterine tube Cervix Abdominal cavity

Corpus luteum

Estradiol

Increased sensitivity of foilicles to FSH

a-)

38.

(Day 28) J Progesterone

Ovaries

t

FO

IJ J Estradioi

1)

Uterus

TD

'U

^4,

s s s

A. B. C. D.

E

negative feedback

FSH & LH

Where is the ovum released upon ovulation?

1


242

Increased FSH, decreased LH. Decreased FSH, increased LH. Decreased FSH, decreased LH. Increased FSH. increased LH.


Biology

Endocrine Control Of Ovarian Cycle

41. To what class of hormones do estradiol

Passage VI

and

progesterone belong?

A. B. C. D.

Steroid hormones.

Growth hormones. Peptide hormones.

Glucoregulatoryhormones.

42. A hysterectomy is the surgical removal of the uterus. What effect would this have on the cycle depicted in Figure 1?

A. No B. C. D. 43.

changes

in ovarian cycle, but

no

menstruation.

Anovulation due to lack of negative feedback from estradiol and progesterone. Decreased development of corpus luteum. Prolonged menstrual flow.

What is the trigger for ovulation?

A. B. C. D.

Regression of corpus luteum. Follicle sensitization to FSH.

LH

surge.

Increasedestradiol.


243


Biology Passage

VII (Questions

Gestation 44-49)

45. One of the factors involved in the differentiation of the Wolffian ducts is testosterone. The production and secretion of testosterone, during this stage of development, is under the influence of:

By the 42nd day of gestation, the embryonic gonads are distinguishable. Under the influence of the genes that code for male sex determination, the gonad will begin testicular differentiation by 43-50 days of gestation. In ihe gonad destined to be an ovary, the lack of differentiation persists. AtlT-84 days, a significant number of germ cells enter meiotic prophase to characterize the transition of oogonia into oocytes, which marks the onset of ovarian differentiation from undifferentiated gonads.

A. B. C. D.

follicle stimulating hormone (FSH). human placental lactogen (hpl). human chorionic gonadotropin (hCG).

gonadotropin releasing hormone (GnRH).

46. Testosterone is produced by which of the following structures?

In the 7th week of gestation, the embryo has both male and female primordial genital ducts. In a normal female fetus, the Mtillerian duct system develops into the uterine tubes, uterus, cervix, and upper one-third ofthe vagina. In normal male fetus, the Wolffian duct system on each side develops into the epididymis, vas deferens, seminal vesicles, and ejaculatory ducts.

In the presence of a functional testes. the Mtillerian ducts involute under the influence of "Mtillerian inhibiting factor" secreted by Sertoli cells. The differentiation of the Wolffian duct is stimulated by testosterone secretion from the testes. In the presence of an ovary or in the absence of

47.

a functional fetal testis, Mtllerian duct differentiation

occurs and the Wolffian ducts involute. Experimental evidence has accumulated demonstrating

that a small portion of the distal short arm of the

Passage Vtr

I. il. III.

Interstitial cells of Leydig.

A. B. C. D.

I only II only I and II only II and III only

Corpus luteum.

Sertoli cells.

XY embryo at gestational day 77 with anit-Mrillerian inhibiting factor antibody wouid result in development of: Treatment of an

A. B.

Miillerian duct structures and male external

C.

Wolffian duct structures and male externai

D.

Miillerian duct structbres and female external

i

Wolffian duct structures genitalia.

chromosome, termed H-Y antigen, is critical for testicular organogenesis of the bipotential gonads. The following experiments are a summary that supports this hypothesis.

genitalia.

genitalia. Experiment

I

48. Administration of anti-H-Y antigen antibody ar gestational day 80 to an XY embryo would result in:

Using the "moscona" technique, dissociated cells derived

from either mouse or rat newborn testes in culture reorganized to form seminiferous tubules. Another group

A.

rather than seminiferous tubules.

treated with anti-H-Y antibody resulted in the reorganization of cells into "follicular-like" structures

B.

undifferentiated testes. formation of "follicular-like" structures.

C.

no change.

D.

undiff'erentiated testes and formation of "follicular-like " structures.

Experiment 2

In a similar

49.

experiment, bovine or human fetal XX

of an XX embryo with H-y antigen during gestational day 90 would result in

Treatment

undifferentiated gonads formed testicular-like structures when incubated with H-Y antigen.

development of:

44. The Sertoli cells, which produce

I. II. Itr. IV.

female external genitalia. male external genitalia.

seminiferoustubules. epididymis.

A.

I only

vas deferens. corpus luteum.

B. C.

II only I and III only II and III only

"Mrillerian

inhibiting factor" are iocated in the:


@


D.

244

Wolffian duct structures. Miillerian duct'structures.


Biology Passage

VIII

Female Birth Control Vaccine

52. If the study were continued by giving a repear injection of the same concentration of vaccine at week 6, what would happen to the anti-hCG

(Questions 50-56)

A birth control vaccine for women has been researched by the World Health Organization since 1974.

antibody titer?

This vaccine promotes antibodies against the hormone

A. B. C. D.

human chorionic gonadotropin (hCG) which is produced by the embryo. hCG maintains a functional corpus luteum, allowing it to continue to produce estrogen and progesterone during pregnancy, and promotes normal

implantation of the embryo into the uterine lining.

Normal, nonpregnant levels of hCG are 0.000 nmol/L in plasma, and levels increase during the first trimester of

53.

pregnancy.

to diphtheria toxin to fbrm a

hapten-protein carrier complex. Antibodies to hCG are raised with this complex in female baboons and in women, following one injection. The vaccine was tested in 30 sterilized, human subjects, rvith the foilowing results:

54.

A.

To make a larger complex for antibodies to recognize, allowing more effective antibodv

B.

To protect LH against antibody formation r.ia

C. D.

The toxin prevents implantation of the embrvo

dose (pg)

Dal 1*

Day

7

Week 5

0.000

0.033

0.226

2

100

0.000

0.025

0.268

J

200

0.000

0.022

0.229

4

500

0.000

0.044

0.950

5

1000

0.000

0.058

0.183

55.

51.

Human chorionic villi Human chorionic gonadotropin Progesterone

B.

The presence of hCG would mimic pregnanc)

The anti-hCG antibody would also bind LH. preventing ovulation, and causing irregular menstruation-

C.

A. B. C. D.

D. The vaccine would not

Although the vaccine would raise hCG leveis to pregnancy levels, it would allow normal menstruation.

Affinity chromatography using LH. Ion exchange chromatography.

change normal

menstrual cycies.

Affinity chromatography using hCG. Bradford protein assay.

56-

Researchers estimated that a concentration

of 0.52 nmol/L were required to provide a contraceptive

Copyright

.

halting menstruation.

How could a researcher determine the concentration of an antibody in a blood sample?

A. B. C. D.

,d

Estrogen

How would this anti-hCG vaccine affect a woman's

A.

I. The hCG synthetic peptide

Group I Group 2 Group 3 Group 4

@


Which of these compounds can act as an antigen in the vaccine?

II. Diphtheria toxin IIr. LH

effect. Which group achieved this effect by Week 5?

_-s-

To provide simultaneous diphtheria immunln and birth control.

normal menstrual cycles?

* Injection

50.

cross-reactivity

At-home pregnancy tests contain monoclonal antibodies in a test kit to which a small volume of urine is added. Which of these compounds are

A. B. C. D.

(nmol/L) (nmol/L) (nmol/L)

50

lormation

present in urine only during pregnancy?

Anti-hCG antibody concentration

Vaccine

Increase and stay constant Decrease and stay constant No change in titer Increase and then decrease

Why is diphtheria toxin used?

The vaccine incorporates a synthetic peptide representing the amino acid sequence 109-145 of the Cterminal region of the B subunit of hCG. This particular region was chosen to avoid homology between hCG and luteinizing hormone (LH). The hCG peptide is conjugated

Group

Passage VItr

A.

I only

B.

II only I and II only II and III only

C.

D.

245

The Berkeley Beview Specializing in MCAT Preparation

Biology Passage

Estrogen


Passage IX

59.

Estrogen, a steroid hormone, is not limited in its role as being one of the major hormones present during a woman's reproductive cycle. Estrogen receptors can be found in liver cells, melanocytes, neurons, vascular

Throughout a women's life, estrogen is produced primarily by rhe:

A. B. C. D.

endothelial and vascular smooth muscle ceils. After menopause, a prolonged estrogen deficiency results. This

uterus. ovaries.

anrerior pituirary.

posteriorpituitary.

prolonged deficiency has been shown to promote osteoporosis and atherosclerotic disease. Studies indicate that women using estrogens after menopause have half the risk of dying of a myocardial infarction when compared to

60.

matched women not using estrogens. Furthermore,

women with established coronary artery disease show a J}Vo decrease in mortality over a l0 year period when compared to similar women who do not receive estrogen.

Increasing the diameter of a blood vessel bv a factor of 2 results in a resistance which is:

A. B. C. D.

In the liver, estrogen stimulates enzyme production that affects cholesterol catabolism. The overalllffect is to impede the development of a lipid profile consistent with atherosclerosis. In addition, estrogens have direct effects on arterial walls. Recent cell culture studies suggest that estrogen may inhibit plarelet aggregation and adhesion seen in early atherosclerosis.

1/2

of

the

original resistance.

/4 of the original resistance. 1/g of the original resistance. 1/16 ofthe original resistance. 1

61. Estrogen has been shown to induce arteriolar relaxation in arteries lacking endothelial cells. The BEST explanarion for this is that:

Endothelial derived relaxing factor (EDRF) is the most

important molecule produced by the endothelium. EDRF

inhibits myofibril contraction in smooth muscle, leading to vasodilation. Estrogen appears to stimulate EDRF. EDRF is opposite in function to endothelin, a potent vasoconstrictor released by the vascular endothelium.

A. B. C.

estrogen directly acts

D.

muscle to cause relaxation. estrogen stimulates the production of EDRF.

EDRF is nor a vasodilator. EDRF is not produced by the endothelial celis.

on vascular smooth

Estrogens have been shown to induce arteriolar relaxation

in which the endothelium has been removed. Subsequent studies have shown that estrogens block voltage-gated calcium channels.

Which of the following effects of estrogen is not consistent with the idea that estrogens benefit the circulatory system?

57. An

I.

Introduction ofestrogen leads to a rapid

Ir.

Introduction of estrogen leads to decrease;

III.

Estrogens inhibit endothelial celi expression adhesive molecules.

62.

estrogen receptor complex within a cell will

most likely:

A.

be carried through the blood in search of

a

target organ.

B.

result in the production of an intracellular

C.

create mutations within estrogen regulated

secondary messenger.

changes in

transcriptional rates.

58.

EDRF most probably inhibits myofibril contraction

63.

by,


permitting calcium entry into the muscle cell. inhibiting calcium entry into the muscle cell. permitting potassium entry into the muscle inhibiting sodium entry into the muscle cell. by The Berkeley Review

Based on information in the passage, concluded that estrogen:

A. B. C. D.

cell.

@

catabolism of LDl-cholesterol. ..,-

A. I only B. II only C. I and II only D. I and [I I only

elements of the genome.

D. bind to DNA, resulting in

decrease in vascular resistance.

246.

it

couid

:re

moderately stimulates release of endothelin. significantly stimulates release of endothelin inhibits release of endothelin. plays no role in regulating levels of endothel:l

The Berkeley Revierr Specializing in MCAT Preparatia

Biology Passage

Isoflavone Experiment 64.

X (Questions 64-69)

The rate of breast cancer is about JSVo lower in Far Eastern countries compared to Western countries. E,pidemiological studies from migrant populations suggests that an environmental rather than a genetic :rpianation is plausible. When people immigrate to other ,-ountries, their rates of breast cancer come to resemble

Isoflavones are antagonists to estrogen. What is the role of an antagonist?

A. An antagonist mimics a hormone,

interacts

with the hormone's receptor, and leads to very

B.

-he country where they are living rather than their home :ountry.

In the Far East, a significant quantity of

Passage X

similar intracellular effects.

An agonist mimics a hormone, interacts with the hormone's receptor, and leads to very similar intracellular eff'ects.

C. An antagonist mimjcs a hormone,

interacts

with the hormone's receptor, and leads to a biock of the intracelluiar effects of the

soybean

nrotein is consumed in the diet in many different forms, :rcluding beans, miso, tofu, and soy milk. Soy provides a :rch supply of isoflavones, nonsteroidal compounds with :strogenic properties. Isoflavones are structurally similar ,o estrogens. They bind to the estrogen receptor and act to

D.

hormone.

An agonist mimics a hormone, interacts with the hormone's receptor. and leads to a block of the intracellular effects of the hormone.

:artially antagonize estrogen. In

a

clinical trial, 6 women were studied during several

65.

renstrual cycles while living in a research center. After

concentration of FSH and LH at ovulation?

:ecording usual changes during severai menstrual cycles, "rey began the ciinical trial. First they followed a control :iet for 1 menstrual cycle, then they followed another 1:onth diet with similar composition of macronutrients, :\cept the protein was largely from soy isolate (60 g of a '3 g protein diet).

A. B. C.

fhe effects of a soy diet on the concentrations of LH and :SH

1^ t1 ii

as compared

to the control diet is shown in Figure

The soy diet group has suppressed levels of FSH and LH compared to the controls. The control diet group has suppressed levels of FSH and LH compared to the soy group. FSH is suppressed by rhe soy diet, bur LH is unchanged. LH is suppressed by rhe soy dier, but FSH is unchanged.

66. If the amount of soy protein could be increased

to

completely suppress LH and FSH, what would be

3ro

16

,-

D.

1.

J

>

What effect does the soy-protein diet have on the

the theoretical effect on the subject?

a I! 5

A.

The subject could become pregnant at any phase of her menstrual cycle.

-3

-2-t 0

I 234

Day of Ovulation

Day of Ovulation

Figure

1

B.

The normal menstrual cycle wouid continue

C.

Menstruation would begin sooner than

D.

predicted by the past menstrual cycie. The subject would not ovulate.

without any change.

'b,le 1 indicates the results of several blood parameters --.lt were followed during the study.

-

Plasma Hormone] Testosterone

(nmoVl)

Cholesteroi (mmol/L)

Control Diet

Soy Diet

i.25 + 0.38

\.46 + 052

4.21

+

1.08

3.86

+

67.

from the passage?

A.

1.01

B.

Table I

C.

The soy protein for this dietary intervention provided -5 mg of isoflavones. For comparison, a Japanese diet :

D.

rntains about 150-200 mg of isoflavones/day.

lopyright

@


Which of the following statements could you infer

247

The sole risk factor for breast cancer is genetic background. Japanese women and European women have comparable rates of breast cancer. Exposure to estrogen is a probable determinant of developing breast cancer. Breast cancer is not related to diet.


Biology 68.

Isoflavone Experiment

Passage X

What percentage of the protein in the soy diet was provided by the soy isolate?

A. B. C. D.

98Vo 25Vo 33Vo

617o

69. What was the effect of the soy protein diet on cholesterol levels?

A. B. C.

D.

Copyright

The The The The

@

soy diet soy diet soy diet soy diet

was was was was

hypercholesterolemic. hypocholesterolemic. isocholesterolemic. microcholesterolemic.


The BerkeleY Revieu Specializing in MCAT Preparation

Biotogy Passage

Ovulation


Passage XI

70, What structure is the source of the rise in

By studying physiological

progesterone after ovulation?

changes during the

nenstrual cycle, a woman may plan a pregnancy based on :he time interval during which she is fertile. One change lhat is frequentiy tracked by hopeful parents is the basal lody temperature, that is the temperature taken just after arr akening in the morning. The hormone progesterone "eads to a rise in basal body temperature of about 0.5oF. This rise signals ovulation has occurred. The temperature

:emains high until menstruation. By developing

A. B. C. D.

Cervix

a

:emperature log over several months, a woman may use --iris information to predict times when she is most iikely :!a

Corpus luteum Egg Corpus albicans

71.

Conceive.

Suppose a woman is planning a pregnancy and is plotting her basal body temperature daily. What

conditions should she observe to make an accurate measurement?

A.

ir.

B. C. D.

=9

Eating or drinking nothing before arising. Lying still during the measurement. Taking measurement at same time daily.

All of the above.

L

72.

-:l

Which hormonal changes precede ovulation?

I.

il.

IIr. A. I

B. C. D.

{

Rise in luteinizing hormone. Rise in progesterone. Rise in estrogen.

I only I and III only

II

and

III only

I, II, and

III

j o

73.

Contraceptive methods either inhibit ovulation, act

as barriers to prevent conception, or prevent

23 I 5'l

14

2t

28

implantation of a fertilized egg. Which of the follow methods inhibits ovulation?

5

Days of the Ovarian Cycle

Figure

A. B. C. D.

1

Condoms

RU-486 Birth control pills Diaphragm

-\nother predictor for ovulation is a changes in the :ucus produced by the cervix. Cervical mucus, under the ,.:'luence

of

estrogen, becomes more clear and watery

*rinediately prior to ovulation, to allow penetration by i;€rrn. The mucus may be placed on a microscope slide,

74. If

a clinician tells a woman planning a pregnancy that she is exhibiting ferning, what does this mean?

:::ed. and examined for a pattern the resembles a fern :-r.nt, a phenomenon called "ferning". This phenomenon .j :resent at ovulation and signals fertility.

Jrpyright

@


A. B. C. D. 249

She She She She

is pregnant. is anovulatory. is infertile. is ovulating.


Biology 75.

Ovulation

Passage

K

Which of the following hormone(s) directly triggers ovulation?

I. II. III.

Luteinizing hormone Follicle stimulating hormone Estrogen

A. I only B. I and II only C. II and III only D. I,II, and III

76,

What molecule is the precursor for the hormones estrogen and progesterone?

A. B. C. D.

Copyright

Sphingomyelin Cholesterol

Arachidonic acid Phosphatidylcholine

@


250


Biology Passage

Vertebrate Gastrulation

XII (Questions 77-83)

79. During

Gastrulation is one of the most dramatic and crucial

into the inrerior of the embryo. Which of

the

following could be responsible for this process of

cell sheet extension?

I. II. III.

bilateral symmetry. The outer layer of cells formed during gastruiation is considered ectoderm, while the middle and inner layers are considered mesoderm and endoderm, respectively. Since interactions between these three ceil layers will determine the further deveiopmental fate of the embryo, events occuring during gastruiation are vital for ihe proper development of the organism.

Cell shape changes within the cell sheet.

Mitotic division within

the cell sheet.

Unidirectional migration of every cell in the sheet.

A. B. C. D.

Gastrulation begins when cells around the embryonic :lastopore begin to invaginate, or move towards the inside rf the embryo. The site where this invagination initiates is :eferred to as the dorsal lip of the blastopore. In the first raif of this century, it was learned that the dorsal lip acts ts an organizer, inducing the tissue directly around it to be-uin invaginating while triggering tissue further away to :dopt other specific cell fates.

I only II only I and II only I, II. and III

80. In an embryo

that is about to undergo gastrulation, the outermost celis protect more interior cells from

the external environment. This is

partially

accomplished by tight junctions which form near rhe outer surface of these cells. This outer cell layer is most similar in structure and function to:

was later found that the cells ofthe dorsal lip secrete

' diffusible signalling substance that slowly degrades after ::s secretion. The resuit is a concentration gradient of the .ubstance, with higher concentrations existing closer to ,re source of the secretion. Cells at different distances ::om the dorsal lip are exposed to different concentrations i the substance, leading to the signaliing of different ::haviors. Due to its effect on morphogenesis (the :haping" of the embryo), this substance was referred to ':

invagination, sheets of cells around the

blastopore increase in length and extend themselves

stages of vertebrate embryogenesis. The embryo, at first simply a hollow bail of cells, undergoes a transformation into a multilayered structure with a central gut tube and

It

Passage XII

A. B. C. D. 81.

a mctrphogen.

epithelial cells. neuronal cells. smooth muscle cells. connective tissue cells.

The early stages of human embryogenesis look very similar to comparable stages in the formation of

other vertebrate embryos. For example, human embryos have gill slits and a tail at a certain stage in development. This observation does NOT support which of the following statements?

'i. If the mesodermal

cell layer is improperly formed during gastrulation, which of the following later structures would most likely be affected?

A, B. C. D.

A. B. C. D.

Nervous system Heart

Humans evoived from lower vertebrates. Primate embryos should not have gill slits. Vertebrates may share a common ancestor. There is little selective advantage in altering these stages of embryogenesis.

Stomach lininc

Liver

82. If the morphogen mentioned in the passage were nol slowly degraded after being secreted, which of the following would result?

-8. Which of the following could NOT act as a morphogen?

A. B. C. D.

A. A very steep concentration gradient

B. The concentration gradient of morphogen would persist indefinitely. gradient of morphogen couid never be formed. There would be no effect on gastruiation.

C. A concentration D.

- -pyright O by The Berkeley Review

of

morphogen would form.

A small, inorganic compound A steroid hormone A secreted protein A transmembrane protein

25t


Biology 83.

V

ertebrate Gastrulation

Passage )ilI

The dorsal lip from a frog embryo is removed and grafted opposite the dorsal lip of a second frog

embryo. Assuming this was done prior to gastrulation, which of the following would be like1y to result?

A. B. C. D.

Gastrulation would never occur and the embryo would die.

The grafted cells would be rejected by the immune system of the host embrYo. The embryo would develop into Siamese twin

tadpoles, the result

of two independent

invaginations during gastrulation.

The embryo would gastrulate and develop normally.

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The BerkeleY Keviw

Specializing in MCAT PreParatiu

Biology Passage

_

XtrI

Oxytocin and Labor

(Questions 84-90)

Passaqe XIII

86. Which of the following would NOT

increase

contractions in a female in labor?

Human pregnancy lasts on average 270 d,ays. During

the last month of pregnancy, irregular uterine contractioni increase in frequency. At the time of labor and delivery,

A. B-. C.

the cervix dilates and softens, and the muscular body of

the uterus contracts to push out the fetus.

Intravenous injection of oxytocin. Intravaginal administration ofprostaglandins.

Intravenous injection

D.

Oxytocin plays both a direct and an indirect role in labor. It acts directly on uterine smooth muscle cells to

make them contract. It aisqstimulates the production of prostaglandins in the endometrium of the uterus. which enhance the contractions produced by oxytocin.

87,

of

human chorionic

gonadotropin.

Oral administration of prostaglandins.

What strategy could be used to prevent a pre-tefin birth?

During early labor, the uterus contracts without significantly higher ievels of oxytocin compared to

prepregnancy. Once labor is initiated, a positive feedback

,oop is set up between the cervix and secretion of oxytocin. Signals of dilation from the cervix lead to sisnals in afferent nerves that promote increased secretion of oxytocin. In late labor, the plasma oxytocin level rises, leading to higher concentrations of oxytocin presented to

A. B.

Administration of an oxytocin secretagogue. Administration of prostaglandins such as RU_

C.

Administration of inhibitors of prostaglandin

D.

486. production.

Administration of an oxytocin agonist.

the uterus.

88. A prostaglandin is a twenty-carbon long fatty acid that contains a five-membered carbon rins. pictured below is the prostaglandin thromboxane n"2.

Oxytocin

&{.

Where is oxytocin produced?

A. B. C. D. 95.

Posterior pituitary Hypothalamus Anrerior pituirary

Which fatty acid is the precursor for prostaglandins?

A. B. C. D.

Placenta

During early labor, plasma oxytocin levels are not higher than prepregnancy levels of 25 pgiml. What mechanism(s) could explain an increase in uterine contractions without an increased plasma level of

89.

oxytocin?

I.

Oxytocin is a polypeptide hormone. What does this

cells?

III. A. B. C. D.

I only I and II only II and III only I, II, and trI

Jopyright

Arachidonic acid

imply about its method of interaction with uterine

An increase in the number of uterine receptors for oxytocin during late pregnancy. Paracrine release of oxytocin that does not communicate with the bloodstream. Increased conjugation ofoxytocin in the liver.

il.

Stearic acid

Myristic acid Linolenic acid

@


I. il. Itr.

It works through a second messenger system. It stimulates mRNA synthesis directly. It is binds to a cellular membrane receptor.

A. I only B. I and III only C. II and trI only D. I,II. and Itr 253


Biology

Oxytocin and Labor

Passage )iltr

90. Studies indicate that paraplegic

women can go into labor and deliver with few difficulties. What would this finding suggest?

A.

Delivery can occur without stretch signals

B.

from cervix. Delivery requires voluntary contractions of the

C. D.

abdominal muscles.

Prostaglandins play only a minor role in the delivery of children to paraplegic mothers. Smooth muscle contraction does not occur in these women.

Copyright

@


254

The Berkeley Specializing in MCAT

Biology Passage

Puberty


93. The development of the seminiferous

After an interval of childhood hypothalamic-pituitary-gonadal

Passage XIV

ac ri

vity

in

te

tubules is

under the influence of:

quiescence,

I. II. IrI.

niifies in the

peripubertal period leading to increased secretion of gonadal sex steroids that cause secondary sexual development, the pubertai growth spurt, and fertility. This stage of development is often termed puberty or

leutenizing hormone (LH). follicle stimularing hormone (FSH). gonadotropin releasing hormone (GnRH).

A. I only B. II only C. I and II only D. I,II, and III

adolescence.

The first sign of puberty in the female is an increase in growth. This is accompanied by breast development, a process stimulated by increase in estrogen levels. Other deveiopmental changes influence by estrogen include; enlargement of the labia minora and majora, dulling of the

vaginal mucosa, production of a clear whitish vaginal secretion prior to menarche, and changes in uterinqsize and shape. The development of pubic hair is primarily

94.

The stimulation of the inrerstitial cells of Leydig is under the direct influence of:

influenced by adrenal and ovarian androgen secretion.

A. B. C. D.

In males, the first sign of normal puberty is an increase in the size of the testes, primarily due to seminiferous tubuiar development. The stimulation of the interstitial cells of Leydig also plays a small component in the increase in testicular size. Pubic hair development is under the influence of the adrenal gland, as well as the

follicle srimulating hormone (FSH). leutenizing hormone (LH). gonadotropin releasing hormone (GnRH). Miillerian inhibiting hormone (MIH).

lestes, through secretion androgens.

The dramatic increase in overall body growth during the growth spurt at the time of puberty is under the

95. In

addition to growth hormone's direct effects on

muscle and bone tissue growth, it also acts indirectly on these tissues by stimulating the production and secretion of another hormone or factor which then

influence of complex endocrine control. In addition to sex steroids, growth hormone (GH) plays an important role in ilris phenomenon.

acts on these tissues. Which of the following hormones mediate tbe indirect effects of growt[ hormone?

91.

Breast development in females is stimulated by an increase in estrogen secretion. Which of the following hormones must increase in order for this

A. B. C. D.

phenomenon to occur? A.

Thyroid stimulating hormone (TSH)

B. C. D.

Gonadotropin releasing hormone (GnRH) Prolactin

I (IGF-I)

one of the following organs?

A. B. C. D.

reproductive system? Spermatogenesis.

Testosteronesynthesis. Secretion of androgen-binding protein by the Sertoli cells. Enhancement of sperm production by follicle stimulating hormone (FSH).


The hormone that stimulates the production and of growth hormone is derived from which

release

92. Which of the following events does NOT occur within the seminiferous tubules of the male

D.

Thyroxine Epidermal growth factor Insuiin-like growrh factor

Progesterone

96.

A. B. C.

Insuiin

255,

Anterior pituitary. Posteriorpituitary. Bone and liver. Hypothalamus.


Biology Passage

XV (Questions

Testicular Cancer 97

99.

-102)

Cancer of the testes if found in only l7o of malignant tumors of male internal organs. This cancer, which shows little basis in genetics, is ten times more likely to be found in males afflicted with cryptorchidism, a condition where the testes have failed to descend down into the scrotum during development. Only 5Vo of testicular tumors do not originate from germ cells. Because germ cells undergo spermatogenesis, it is difficult to determine the exact origin of tumor cells. There is some evidence that spermatogonia are the likely candidates giving rise to testicular cancer. However. researches have found that early carcinoma cells bear more of a similarity to fetal germ cells than spermatogonia.

Passage XV

Based on information in the passage, one could argue that germ cell tumors that come from fetal germ cells have a retarded developmental fate. Which of the following DOES NOT support rhis hypothesis?

A. B.

Evidence shows that tumor development

C.

In mouse studies, cancer genes act in fetal

D.

Infertile men run a greater risk for carcinoma

Cryptorchidism is

a

risk factor.

occurs in gestation. germ cells.

in situ.

It has been discovered, through studying teste tissue samples from infertile men, that certain morphologically distinct cells were precursors to invasive cancer. These malignant cells were termed carcinoma in situ (CIS). Most CIS cells will evolve into invasive cancer. Looking for differences between invasive and CIS cells, researches noted that CIS cells have a tetraploid complement of

100. How many chromosomes

are found

in a carcinoma

in situ cell?

A. B. c. D.

chromosomes.

The initial events transforming germ cells into tumor cells is not well understood, but has been likened to parthenogenesis in the female. Tumors arising from parthenogenesis, known as teratomas, contain multiple tissue types with little organization. It is postulated that

23

46 69 92

testicular cancers may have similar origins.

101. Which statement is TRUE regarding the

97, Which of the following

statements supports the theory that spermatogonia give rise to testicular cancer?

A.

A.

B.

B.

Sertoli cel1s give rise to tumors which differ significantly from tumors originating from

C. D.

germ cells.

Cryptorchidism is a risk factor for testicular cancer.

C. Testicular cancer is a rare cancer. D. Tumor cells do not have a

genetic

basis of testicular cancer?

Testicular cancer is an autosomal recessive trait.

Testicular cancer is found commonly withirfamilies prone to testicular cancer. Testicular cancer is rare among identical twins Testicular cancer is largely solved by surgicd resection.

haploid

chromosome number.


statements

is

TRUE

t02.

regarding testicular cancer?


Parthenogenesis development of

At least

90Vo of testicular cancers are found the vas deferens. At least 90Vo of testicular cancers are found the seminiferous tubules. At least 907o of testicular cancers are found the epididymis. At least 90Vo of testicular cancers are found the rete testes.

@


is BEST described as

th:

a:

in

A. B. C. D.

in in

tetraploid adult from a fertilized egg. tetraploid adult from an unfertilized egg. haploid adult from a fertilized egg. haploid adult from an unfertilized egg.

in

256,


Biology 1'


Section IV Answers

C is correct, spermatids. Cellular division does not occur in spermatids. During spermatogenesis male germ cells called spermatogonia are converted into mature spermatozoa. A single diploid ;;;.t""ogonia is rransformed izlif into a 1" spermatocyte which then undergo meiosis I to form two traptola 1x; z"'rp"r.atocytes. The two 2" spermatocytes undergo meiosis II to form four haploid spermatids. Each spermaiid n"*t undergoes a transformation process called spermiogenesis to form the mature spermatozoa. The spermatids do not undergo a cellular differentiation to form a spermatozoa. They simply diffeientiate. The whole ;."";;;;l Jp.r-urog.n"sis takes about

64 days.

During oogenesis female germ cells called oogonia are transformed into mature ova. This process involves the cellular divisions of meiosis I and meiosis II,F{Y in the development of the fetus oogonia undergo mitosis and are transformed into 1" oocytes' The diploid (2N) 1" oocyte enters into meiosis I and"is arrested at prophase until puberty. A short time before ovulation meiosis I is completed and a haploid z; oocyte is formed atong witfr iire 1p first polar body (which may divide into two more polar todies). At ovuiation the z' oocyteundergoes meiosis II but is arrested at metaphase. At fertilization the 2" oocyte completes meiosis II and forms the mature ovum and a second polar body' union of spermatozoa and ovum gives rise to tie zygote which is diploid. The coryect choice is c. C is correct' metaphase' See the explanation for oogenesis given above. The correct choice is C. B is correct, II and III only' Fertilization leads to the formation of a zygote.The zygote undergoes numerous mitotic divisions to increase the number of cells but not the cytoplasmic miss. wrren ilie zygote divides into two cells (about 30 hours after fertilization) those two cells u." r"i".."d to as blastomeres. Aboui 5 auy, after fertilization the dividing zygote has formed about 16 blastomeres (a specialized cell-rype). At rhis stage the?ll il;;;;;;j';; as a morula and it begins to enter into the uterus.

3.

Roughly four days after fertilization a fluid-filled cavity, called the blastocyst cavity or a blastocoei, begins to form in the central portion of the morula. As this cavity foims the cells within the *o.Llu ;;"r" rhemselves into two

parts' one group of cells forms the outer layer of cells called the tropoblast. The othei gr""pir cells forms an inner cell mass called the embryoblast. The tropoblast gives rise-to the piacenta while the Jmbryoutast gives rise to the embr.yo' By this stage in development ihe morula is referred io u, u blastocyst. Around the sixth day after fertilization the biastocyst (normally) attaches to the endometrium of the posterioi wall of the uterus. The correct choice is B. 4.

D is correct, I, II' ru, Iy. As the primary follicle.begins to develop, an antrum is formed and the primary oocyte is moved to one side. Td The formation bf the antrum signals the development of the secondary follicle. Towards the the foliicular phase the granulosa cells (i.e., rotticte cells) begin to secrete large ariounts of progesterone 919 "l ( 17u-hydroxyprogesterone) into the antrum. As the fluid in the antrum-increases tn" piir*. uuilds up ani places a stress on the stigma. Prostaglandins, which are synthesized and released from almost all cells in the body, can act to stimulate the smooth muscle of the uterus and the smooth muscle surrounding the outer layer of theca cells (called the theca externa). contraction of this muscle can cause an increase in pressurJ in theantrum and rupture the siigma. The contents of the antrum, including the secondary oocyte and the progesterone, are released into the peritoneal cavity (the space outside the ovary). The correct choice is D.

B is correct, be triploid and therefore exhibit aneuploidy. If two sperm fertilize an ovum, then each sperm will contribute a chromosomal complement of 23. Since the ovum also has a chromosomal complement of 23, this results in an embryo that is triploid and contains 69 chromosomes. If there is a deviation from the normal chromosomal complement of 46, then that deviation is referred to as aneuploidy.

Trisomy 21 involves the presence of three chromosomes instead of the usual pair of chromosomes. This type of chromosomal abnormality is referred to as Down's syndrome and usually results in g".u with 24 instead of 23 chromosomes. Monosomy involves embryos that are missing a chromosome. Celli ""iiare polyploid contain which multiples of the haploid number of chromosomes. The haploid-number of chromosomes is 46. If a cell contained 69 chromosomes, it would be triploid. A triploid cell is a cell which exhibits polyploidy. Itre correct choice is B. D is correct. oxytocin is considered tobe apeptide hormone, as are the other hormones of the hypothalamus and pituitary gland. in general, peptide hormones contain from 3 to over 200 amino acid residues. Choice A represents adrenaline (epinephrine), a water-soluble amine hormone.It stems from a class of hormones called catecholamines. In the brain catecholamines act as neurotransmitters. Choice B is the methvl ester of a

-.pyright

@


257


Biotogy


Section IV Answers

dipeptide called aspartame (trade name is NutraSweet@), an artificial sweetener which is commercially synthesized in huge quantities. This is not a hormone. Choice C is estradiol, a lipid soluble steroid hormone that is synthesized in the ovaries. The correct choice is D. 7.

C is correct, progesterone and allows administered prosta-glandins to cause uterine contractions. RU 486 resembles

both progesterone (a progestin) and cortisol (a glucocorticoid) and will bind to both of their receptors with high affinity. RU 486 does not resemble testosterone or estradiol as weil as it resembles progesterone and cortisol. Therefore, binding to the receptors of these two steroids is minimal (if at all). The reasoning and structures associated with question #7 would help you with this answer, but it is not necessary.

RU 486 is a progesterone antagonist and binds to progesterone receptors. Recall that progesterone aids in

the

conversion of the growing endometrial lining of the uterus into a tissue that can accept a developing embryo (i.e., the blastocyst). Progesterone allows the cells of the endometrium to synthesize and store glycogen, increases the secretory activity of the endometrial lining, and causes an increase in the vascularization of the tissue. Progesterone also acts to inhibit expulsion of the implanted embryo by reducing uterine contractions and constricting the opening of the cervix. If RU 486 binds to progesterone receptors, then the developing embryo will be facing a hostile environment. Furthermore, 36 to 48 hours after RU 486 has been taken by the pregnant woman, prostaglandins are administered. Recall that prostaglandins (released by most cells of the body) cause contraction of smooth muscle. especially the smooth muscle of the uterus.

Glucocorticoids, such as cortisol, have a variety of effects on metabolism. In the liver cortisol increases gluconeogenesis and glycogen synthesis. In skeletal muscle it decreases glucose uptake, increases protein degradation, and decreases protein synthesis. In adipose tissue it decreases glucose utilization and increases lipid metabolism. Glucocorticoids tend to impair wound healing. An antagonist to glucocorticoid receptors might actuall] increase wound healing. Glucocorticoids do not cause uterine contractions. The correct choice is C. 8.

D is correct, cortisol and progesterone receptors. Glucocorticoid and progestin receptors bind RU 486 with the greatest affinity. Look for similarities in the molecules. First, note the three substituents at the C-3, C-11, and C-1iarbons of RU 486 (see below). Estradiol has a hydroxyl function at the C-3 position; RU 486 does not. Estradio, does not have a substituent at the C-11 position; RU 486 has a bulky dimethylaminophenyl group. Estradiol only ha*' a hydroxyl group at the C-17 position. RU 486 has both a hydroxyl and a bulky alkyne residue. A similar analysi. can be done for testosterone. We will find that the substituents on RU 486 do not match well with those on both estradioi and testosterone. If we now compare RU 486 to both progesterone and cortisol, we will find more similarities. For example, both progesterone and cortisol have a carbonyl function at the C-3 carbon. They also have a (somewhat) bulky substituent aatheC-11 position. The correct choice is D.

9H,

H:c'

N

Z\

HrC

.,.4

'/ c=

RU 486

o

H-o

c- cH:

3n *

a pretty poor form of birtl answers. Since tir C as incorrect A and choices eliminate Immediately sperm secretion was unchanged.

D is correct, spefin production is unaffected, and sperm secretion stops. This would be control

if

will be present in the semen several months following a vasectomy. Usuall b1 the patient is checked for aspermia-before the procedure is considered a success. Sperm production is not affected is D. a vaiectomy. Choice B is incorrect. The correct choice vas deferens is cut and closed, no rper.n

10.

A is correct, antibodies to sperm may reduce fertility. Antibodies to sperm are produced.when macrophages entr rm the testis to catabolize the lxtra spenn. These would hamper fertility if they immobilized sperm. Choice B Thc weeks. 10 is about turnover cycle the sperm since incorrect. There is no effect of vasectomy on birth defects, man simply makes new spefin. Choices C and D are incorrect. The correct choice is A.

Copyright

@


258

The BerkeleY Revieu Specializing in MCAT PreParatia

Biology


Section lV Answers

11.

B is correct, he has reduced fertility status compared t9 the population average. The prefix ,'oligo,, means fe.o,. oligospermia is the condition of having few sperm. (This is compared to asplrmia, the condition of having no sperm)' Although this man could impregnate a woman, his chances are less than that of the population aveiage. However, his cl.rances are not zero. Choice A is inconect. Choices C anci D are incorrect. The correct choice is B.

12.

D is correct, both FSH and LH are unchansed. Since sperm production continues along its normal course, really the only interruption is in transport of speim outside of the bocly. This has no effect-on the hormones controlling .the sperm production. Choices A, B, and c are incorrect. The correct choice is D.

13.

B is correct, seminiferous tubules. The testis is the site of sperm production. It is compartmentalized into seminiferous tubules and the interstitium. The interstitium c-ontains androgen-se.reting Leydig

the

cells.

Spermatogenesis occurs in the seminiferous tubules, not the interstitium. Choice D ls incorrect. The vas deferens drains sperm from the testis. Choice A is inconect. The prostate gland is not invoived in spermatogenesis. Choice C is incorrect. The correct choice is B.

A is correct, I oniy. Foliowing a vasectomy, the sperm count rvill eventually drop to zero, since the sperm

14.

cannot

exit the testis' Choice T is correct. The female counie.pa.t of vasectomy is the ligarion of the fallopian rubes (a tubal ligation)' choice II is incorre_ct. Since sperm production continues normally, the"re is no."uron to suspect decreased testosterone leveis. Choice III is incorrect. The correct choice is A. 15.

c is correct, the dogs undergoing this treatment would become sterile. This treatment achieves the same etf-ect as a vasectomy. The vas def'erens is blocked, so that sperm cannot exit the testis. The effects would be the same as expected for vasectomy. The dogs would eventually have a sperm count of zero and be sterile. Choice A is incorrect. Normal sperm production continues, but the cells are destioyed by phagocytes in the testis. There would be no enlargement of the testicles. Choice B is incorrect. There is no evidenc" ttrit an autoimmune disease would develop, since the phagocytic cells are localized in their attack on the sperm cells. not the testis itseif. Choice D is incorect. The correct choice is C.

16.

D is correct, II and III only (regression ofthe corpus luteum and termination ofpregnancy). The corpus luteum is quite dependent on hCG during the first 8 weeks of pregnancy. During the luteal p'rr.r.'"t the uterine cycle the corpus luteum was maintained by LH, even though the LH ievels were relatively low. 'rhe corpus luteum is responsible fbr the secretion of large amounts of estrogen and progesterone. These two steroids. in concert. inhibir release of GnRH. The end result is that FSH ard LH levels are ref,t tow and follicular development and ovulation are suppressed. Increased levels of estrogen and progesterone lead to further endometrial development.

After i'ertilization the outer layer of cells (trophoblast) of the blastocyst begin ro secrete hcG. This pepride stimulates and the corpus luteum to continue its synthesis of estrogen and progesterone. why? Because the deveioping placenta will not be able to secrete adequate amounts of estrogen or progesterone until about the 3rd month of pregnancy. High levels of these two steroids allow the endometriaiiining io piolif'erate and implantation of the blsatocyst is maintained. In other words, pregnancy will continue. What happeis if nCC is removed? Removal of hCG during the first 6 weeks of pregnancy means that the corpus lureum is not being stimulated. The corpus luteum begins to regress and the output of estrogen and progesterone decreases. Uterine bl6od vessels begin to constrict, depriving cells of nutrients and oxygen. The uterine lining begins to siough. pregnancy is terminated."ln turn' GnRH, LH, and FSH levels begin to.is. and the cycle begins agiin. ttre correct choice is D.

1l

A is correct, LH. We can reason out this answer based on information in the first sentence of the second paragraph of the passage. We are told that during the luteal phase of the urerine cycle LH maintains the corpus luteum.

After fertilization we are told that hCG maintains the corpus luteum. We could assume that if both hormones maintain the corpus luteum' then they must have similar structures if not similar receptors. It turns out that hCG is a glycoprotein that-consists of an aipha and a beta chain. The alpha chain is aimost identical in sequence to the alpfia chains of TSH' FSH' and LH. The beta chain' which determines biologic effect, has a 6i vo hoiology with the beta chain of LH. Consequently, hCG and LH have very similar biologic u.tions. The correct choice is A.L\.

C is correct, increase steadily until parturition. Although FSH and LH levels are low during pregnancy, human chorionic gonadotropin takes over the functions ofLH to stimulate estrogen and progesterone secretion. In addition to the placenta, the fetal and maternal adrenal cortices may also produce estrogen ind progesterone. The correct

choice is C.

: . right

@


259

The Berkeley R.eview Specializing in MCAT Preparation

Biology 19.


Section IV Answers

A is comect' remain relativ-ely low, thereby eliminating further follicle development and ovulation. LH and FSH are inhibited by high levels of estrogen and progesterone during pregnancy. Therefore, they remain relatively lou,

throughout the duratlon ofpregnancy. The correct choice is AI 20.

A-is correct, have no effect at all on the pregnancy. The ovaries produce estrogen and progesterone during the first 6 weeks ofpregnancy' after which, other tissues such as the plaCenta take ovJr tnis rote.inerefore, pr"lnunry occur in the second term without the ovaries. The correct choice is A. "un

21.

A is correct, endometrium. Unless implantation.of the blastocyst occurs, the continuing secretion of progesterone and estrogen by the corpus luteum is stops. The result is that the glandular epithefal thut muiie up tt e endometrium begin to slough. Immediately beiow the endometrium is tf,e myometilu-, *rrl.fr ""tt, is the smooth muscle tissue of the uterus. A primary oocyte will become surrounded by a layer of granulosa cells to become a primarl. follicle. The primary foilicle will be-converted to a secondary oo"yt", und it i, ti'e secondaif oo"yt" that will become the ovum. If an ovum is not fertilized, it will be removed in the monthly menstrual flow. Thus, the only structure

mentioned that is sloughed during menses is the endometrium. The correct choice is A.

)t

c is correct, II and III only. Triglycerides in the adipose tissue are hydrolyzed by the enzyme triglyceride lipase. This yields 3 fatty acids and 1 glycerol molecule per iriglyceride. All iire gtycerot er"a"r ti-trr" blood. Some fatn, acids may remain in the adipocyte for re-esterification, ind the remaindJr"will escapi t" trr" ur""o.

choice is C. .,?,

il;^;i,.J";;

D is correct, insulin increases synthesis of glucose. We are looking for the FALSE choiice. We have learned frorn the passage that r'c Pabs'1Btr Llat rhe ure goal goar or of early pregnancy ls is storage ot of energy tn ii adipose adi tissue. If a person eats, insulin levels t",T: f.o1 the food will.be stored asTat. Insulin does increase iv",rt"rir of triglyceride in ill"lt^":111 1l lL" :l"..cJ adipose tissue. It activates triglyceride synthase. Choice A is correct. Glucose that is taken up by the udipo"yt. l' used-to make a glycerol phosphate, an activated glycerol backbone for triglyceride synthesis. Choice B is correcr Food lipids are packed into chylomicrons and are taken up by cells, inciualng Insulin increases rhe activity of lipoprotein lipase. Choice C is correct. Finally, tlhe false choice ls"di;";i;. O. rn" synthesis of glucose is gluconeogenesis, and that occurs in the hepatocytes. Insuiin does not increase gtu"on"og"*sis after a meal. The correct choice is D.

24.

A is correct, increased glucose, decreased insulin. Insulin resistance means that the insulin in less effective ir producing its effects compared to normal insulin sensitivity. The effects of insulin are to lower blood giucose bi activating glucose transpofiers and moving glucose into cells for processing. A f'eed-back loop is set"up so the: insuiin will decrease as glucose decreases. Since the woman with gestatiorial diabetes is quite insulin resistanl, glucose will remain higher longer after the test load. Insulin will also iemain high because the feed-back loop to rur: it down is not completed. The correct choice is A.

?(

A is correct, pancreas. Insulin is produced by the beta celis in the pancreas. The correct choice is A. A is correct' I only' Glucose is the required fetal fuel source. The mother builds fat stores during early pregnancl

26.

This makes choices B, C, and D incorrect. The correct choice is A.

)1

B is correcto higher birthweight. Since the fetal preferred source of energy is glucose and glucose is higher jr women with gestational diabetes, the fetus will be able to "overeat". fnis conAition is called macrosomy. Choice E is correct. This makes choices A and C incorrect. As for choice D, we have no way of inferring from the passage (or from our own knowledge) that during the first trimester fetal weight will be slightly above no#at. Not only tha*t, trr the second part of choice D states that the birthweight is lower when we knbw it should be higher. The correcr choice is B.

28.

B is correcto II only. Glucose transport is by facilitated diffusion, which is passive transport. It occurs throughou: pregnancy, not just following a meal. It is lower in early pregnancy, when tfie fetal needs are lower, and motf,er:-i insulin sensitive. The correct choice is B.

29.

C is correct, hepatocytes. The liver's cells are called hepatocytes. The hepatocyte is the only location oi gluconeogenic enzymes listed here. There is some evidence that the kidney uiro do gluconeogeneiis, but thar ls "un not an issue in this question.The correct choice is C.


26o


Biology


Section IV Answers

Aiscorrect,Ion1y(inhibitionofFSH)'Together,::|.-oc"n.an_d.progesteroneactatthelevelffi

30.

$:":H?n,Slli*lll,l:,nr",f*:l:ii':ry

jiT"?-",

*"ffi

prevents follicular growth and suppression of LH prevents estrogen inhibits progesterone or

rhe_anrerior pituitary. Suppression or FSH

#'il#il;iilii!'i'"X1#"iil|

"""1"ii""*ri";;:"J thif estrogen stimulaies *,r tyntrr"Jr oi psH. The correct choice is A.

B is correct' positive feedback by estradiol..Estradiol-is_normally an inhibitor of LH secretion, but at increasing rhe LH due ro rhe rise revels during thE

31.

p;;[;

;?::i'",x,:.T;ffi.:;i'flTnf:,:f,Jijftion 1)

;;;;;g""

c

is correct' negative feedback ofestradiol. Estradiol inhibits FSH secretion. The correct choice is c. A is correct' I and II only (progesterone and estradiol). Progesterone and estradiol are the main hormones that act on the endometrium' FSH, LH, ano hcG primarily act on the ovaries. The correct choice is A,

33.

A is comect, anterior pituitary' The anterior pituitary is (GH)' thyroid stimulating hormone (TSH), prolactin, responsible for the release of FSH, LH, growth hormone una'ua."no"orricorropic hormone (AC'H). All of these hotmones are protein hormones' This is information that is ,mforiu,iiro ."-"-ber. Notice that the steroids estrosen and progesterone are not produced at the level or tn" u.uit 'Rig;;'uruy this allows us to select choice A as ihe correct answer' The placenta produces both estrogen

34.

and progestlrone. dranuloru ."rrr?rlell as thecal cells) can produce estrogen' Recail thaithe granulosa cells-are ttrose cJtts wt i.r, m-"aiately surround the oocyte while the thecal cells are those ceils that suriound the granulosa."itr. iog"t;"i, trr" g.unulosa and thecal cells are called the follicle celis' The adrenal cortex helps to define the outer r"gion"oiirr" adre-nal glands. irril,irru" secetes a variety including androgens (male sex hor,non'"9 uni (femate sex hormones). The correct ff::"rttj:rmones, "rr.og"n,

c is correct' hypothalamus' we have learned that both FSH and LH are synthesized and released from the anterior pituitary' The release of these two hormones (and the other hormones oi.the hypothalamic hypophysiotropic hormones, such -"r1". pii"iiary) are regulated by as GnRH. G"RH ;;;"T-"9_fr:orn the hypothalamus once every rwo to three hours in a pulsatile fashion. In turn, the release of borh FSH and LH from rh" ;;;;i; pihirary is seen in a pulsatile fashion as well'-The poterior pituitary releases uuropr"rrrn (ADH) and oxytocin. two peptide hormones are svnthesized within the hypothaia.us, pu.rug-J^*.;"';;;;;yH;i,,":Xd These rransported ro rhe posterior pituitary for release. The pineai gland, derivid froir the roo^f of the diencephalon and posterior to the and amino acii derivative. rr," run"iion or,r'i, hormone r,!, nolu""n

:1Ti1:hT::".'::#il,Ti;tl;

entirery worked

C is correct, the rate-limiting enzyme in a pathway is inhibited by the product(s) of the pathway. Firsr of all, eliminate choices B and-D. Enzymes may be both activated or inhibited, turned up o.io*n. Nlgative controi means th'at the product of a pathway acts to decrease its own synthesis. This means the pathway is inhibited, not activated. Choice A is wrong. The correct choice is C. D is correct, abdominai cavity. The ovum is reieased from the mature follicle directly into the abdominal cavity. It is picked up and directed into the uterine tube by the fimbriated ends of the tube. Finaliy passes it into the uterus and leaves by the cervix if it is not fertilized. The correct choice is D. -18.

B is correct, progesterone. From the diagram, we can

see that the hypothalamus produces GnRH, so choice A is incorrect. We can also see that the anterior pituitary produces LH and FSH, ro cholces C and D are incorrect. This

ieaves B, progesterone, which is indeed made by the Corpus luteum. The correct choice is B. j ,t.

C is correct, I and II only. Look at the diagram. At around Day 1, the ievels of estradiol and progesterone are falling. This decreases the negative inhibition on the hypothalamus. Then the normai cycle of follicle?evelopment and ovulation follows- Therefore, if estradiol and progesierone are kept artificially high with oral contraceptives, the negative feedback control is not removed. The brief time between cyiles of pills doei not allow the normal time for all the ovulatory steps to get in sync. So there is no ovulation. Choice I is correct. There is also decreased FSH and LH because the fall in progesterone and estradiol does not occur in pill users. Choice II is correct. Choice IiI is incorrect, because we have already established that normal follicle deveiopment and ovulation does not occur. The correct choice is C.

-lht

O by The Belkeley Review

26t


Biology


Section IV Answers

40.

D is correct, increased FSH, increased LH. This is anothgr questiol regarding feedback control. During the secretory phase, high levels-o_f e_stradiol and progesterone feed back on the hfpothuiu*u, unj d""."ur" GnRH. This. in turn, decreases FSH- and LH. After menopause, levels of estradiol and"irogesteron" ui" low because normal follicular development has ceased. This means FSH and LH would remain high-aue to lack of feedback inhibition. This means choices A, B, and c are incorrect. The correct choice is D.

4'1.

A is correct, steroid hormones' The precursor of estradiol and progesterone is cholesterol, which is a steroid. This means choice A is correct. Growth hormone is a_ specific peptloe hormone. So choice B is incorrect. peptide hormones are made of chains of amino acids, so choice C is lnionect. Finally, glucoregulatory hormones control

plasma glucose levels. Estradiol and progesterone do not do this. The correct choice is A.

A is correct, no changes in ovarian cycle, but no menstruation. Removal of the uterus alone will not affect the cycle involving the ovaries, hypothalamus, and anterior pituitary. Therefore, choices B and C are incorrect. There should be no changes in the ovarian cycle. Choice D is inconecf because there can be no menstrual flow without the uterus-

42.

which is the source. The correct choice is A.

C is correct, LH surge. Although the follicle is prepared by FSH and LH, the final stimulus for ovulation is a surge of LH about 16 hours before ovulation. Choices B and b precede ovulation, but are preparatory steps, not t53 trigger' choice A is about 10 days following ovulation, so it c;uld not be a trigger. ttre correct choice

43.

is C.

44'

A is correct, seminiferous tubules. Sertoli cells are located within the seminiferous tubules. Their cell body extend" from the base of the-tub.ules to the cytoplasm. In addition to producing Miillerian intrioliing factor, they alsl provide nutrients to the developing sperm. The correct choice is A.

45.

C is correct, human chorionic gonadotropin Qcc;. The production of steroid during the early part of gestation is dependent on hcG derived from the placenta. The correct choiceis c.

46.

B is correct, II only (interstitial cells ofLeydig). The interstitial cells ofLeydig are the testosterone producing celis located in the testes, outside of the seminiferous tubules. The correct choice is B.

47.

C is correct, wolffian duct structures and male__external genitalia. According to the passage, testicular differentiation occurs at 43-50 days of gestation and MIF has alre-ady taken effect. ThJrefore, the ferus will develog,

male genitalia. The correct choice is C. 48.

C correct, no change. to experiment l, treatment of anti-H-Y antigen to Xy, newborn testes would nm. -is Âccording differentiate into seminiferous tubules. In the passage, testicular differentiatio-n occurs at 43-50 days of gestationTheoretically, anti-H-Y antigen should not have an effect at gestational day 80. The correct choice

is C.

49.

D is correct, II and III only (Mrillerian duct structures and female external genitalia). Similarly, there should not te an effect since treatmenf occurs beyond the critical period of ovarian diffeientiation (day ll-g;4 of gestation), Th correct choice is D.

50.

C is correct, affinity chromatography using hCG. Affinity chromatography consists of an inert column

whicfu

supports compounds that bind your molecule of interest- Since you want anti-hCG, you attach hCG to the supporL This will allow binding of your antibody. LH is not the hormone of interest, so choiie A is incorrect. Ion exchalge chromatography is for separating charged molecules, so choice B is incorrect. The Bradford protein assay only gir--es total protein, so choice D is incorrect. The correct choice is C. 51.

D is correct, Group 4. We can read this answer off the table. Only group 4 had antibody concentrations above the 0.52 level. Only group 4 could be considered protected against pregnancy by the anti-hCG antibody. The correct choice is D.

{t

D is correct, increase and then decrease. Compare this to what we know about being immunized. We get a tetanus series when we are young, and then we get booster shots to elevate our antibody level as it decreases over time. With a

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booster, as in this question, we seen an increase and then a decrease over time. The correct choice is D.

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Reproduction & Development,

Section IV Answers

A is correct' to make a larger compiex for antibodies to recognize, allowing more effective antibody formation. This is called a hapten-carrier complex. Small molecuies that in immunolo-gically inert alone are -bound to larger, immunologically robust carriers. This elicits the strongest antibody ."rponr" and ailows your small molecule of interest to have antibodies_produced against it. Choice B is incorrect, b"cuur" the toxin does not protect LH. Choice C is incorrect, the toxin has no effect on reproductive status. Choice D is incorect, the goui i, not to provide

simultaneous immunity, but contraception. The correct choice is A. 54.

C is correct, human chorionic gonadotropin. hCG is present only during pregnancy. Estrogen and progesterone are present throughout pregnancy and also during the normal menstrual cycte. f.nis eliminates Jhoi"", A unA n. Human chorionic villi are part of the developing tissues accompanying the fetus. The correct choice is C.

5-l)-

D is correct, the vaccine would not

change normal menstrual cycles. The vaccine causes the body to produce antibodies to hCG, so that any hCG produced by an embryo is trapped by the antibodies. hcG is critical for the progression of pregnancy, so the pregnancy is terminated at this stag-e. Choice A is inconect, the vaccine does not contain the hCG molecule, merely a piece of it, made synthetically. Choi." B is incorrect, the peptide is chosen to avoid cross-reactivity with LH. Choice C is incorrect, the hCG levels will be reduced to nonpregnant levels, 0.000 nmol/L. The correct choice is D.

56.

C is correct, I and II only. The intended antigen is the hCG synthetic peptide. Choice r is correct. However, since the diphtheria toxin is present, a small antibody response o..ur. to it is weil, so choice II is correct. LH is a hormone and is not an antigen. Choice IrI is incorrect. The correct choice is c.

57' D is correct'

bind-to DNA, resulting in changes in transcriptional rates. As stated in the passage, est1ogen is a steroid hormone. Therefore, we must think about how steroid hormones bring about changis on"a cellulir level. Since steroids are very hydrophobic in nature, they can cross the cell membra"ne and bind Io receptors within the cell's cytoplasm. These. steroid/receptor complexes have the ability to bind to DNA and alter transcriptional levels of certain proteins. This is very different from protein hormones which bind to receptors found on the cell surface. These types of interactions lead to the formation of secondary messengers that bring about change within the cell. Eliminate choice B as a possible answer. Answer choices A ind C ur"l bit non-r"niibl" and do iot fit the general scheme of how hormones bring about change. The correct choice is D.

58'

B is correct, inhibiting calcium entry into the muscle cell. Entrance of calcium into a muscle cell is what triggers contraction. We have seen this in skeletal muscle cell. The question asks what inhibits myofibril (there are many myofibrils in a muscle cell, and each myofibril contains many sarcomeres) contraction. Weli, the only real possibility should involved the calcium ion. As for sodium and potassium, r" huu" no evidence they are involved with the contraction of a myofibril. They will be involved in the nerve transmission to stimulate contiaction, but we cannot assume they take part in the actual contraction of the muscle. If we want to inhibit the myofibril contraction which will lead to vasodilation, we will want to inhibit calcium entry into the cell. The correct choice is B.

59.

B is correct, ovaries. This is based entirely on our knowledge of hormones. Estrogen is produced primarily by the ovaries, and its target tissue is the general female reproductive structures. Its principal actions stimulate deveioplent of secondary sex characteristics in females and growth of sex organs at pubeity. It also primes the uterus for pregnancy on a monthly basis. Hormone questions are not going to be tough in you know your hormones, where they come from, and what they accomplish. The correct choice is B.

60.

,1 I 6 of the original resistance. If we recali from physics or discussions of the cardiovascular system, the resistance of a tube is inversely proportional to the radius of the tube to the fourth power. If we increase the diameter by a factor of two, we are increasing the radius by a factor of two. We are decreaiing the resistance by 2 to the fourth power. Thus, the resistance is one sixteenth of what it was previously.. The correct choice is D.

61.

C is correct, estrogen directly acts on vascular smooth muscle to cause relaxation. We know from the passage that the endotheliai celis produce this EDRF which causes relaxation of the smooth muscle, and ultimately le-*ads to vessel diiation. Since this is stated in the passage, we have to take it as the truth and thus we can eiiminate choices A and B. If one removes the endothelial layer, one is removing the ability to produce EDRF. If estrogen stili acts as a dilator in this condition (no EDRF because no endothelial iayer), then estrogen must have the abilily to directiy acr on the smooth muscle to cause its relaxation and ultimately vessel dilation. Considering choice D, we run into a familiar problem. Estrogen cannot stimulate the production of something that cannot be produced without the presence of endothelial cells. Since these have been removed, estrogen must act directly on smooth muscle. The correct choice is C.

D is correct

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263


Biology 62.


Section IV Answers

B is correct, II only (introduction of estrogen leads to decreased catabolism of LDl-cholesterol.). Statement I is consistent with the idea the estrogen benefits cardiac tissue because less resistance to blood flow allows blogd to flow-easier, allowing your heart to work easier (assuming our heart was working harder than it shouid). In other words, more resistance to blood flow would force our heart to work harder, piacing stress on the vital organ. Statement III is consistent with this idea because inhibiting the endotheiiat cett exprer"ti"" adhesion molecules inhibits platelet aggregation and other adhesion that is normally seen in the early "r stages of atherosclerosis. This idea is that if less "stuff" can stick to the walls of the vessel, the less chance exists oflto"llng Utooa no* Statement

II is

inconsistent because the statement claims a decreased breakdown, or catabolism of LDL cholesterol is beneficial to the vascular system. An LDL particle is a low density lipoprorein *ilh ;;;;;orrs cholesrerol from the liver to the body cells via the blood. The cells_tak_e up rhese puil"i"t'ttrrough;Je;;;;i";;;6;;. However, high levels of cholesterol will inhibit LDL rLceptor'synthesis, anithe pirti"f"r will remain in the circulatory system. The build up of cholesterol in the blood is thought to ie a p.l-" ruur" of atherosclerosis, a condition where the arteries are no longer compliant and blood *ou", though with difficulty. This condition is clearly not beneficial to the vascular system, making statement II inconsistent. ihe correct choice is B.

63.

c is correct, inhibits release of endothelin. Throughout the entire passage, we have seen how estrogen benefits the cardiovascular system. It is stated in the passage that endothelin is a potent vasoconstrictor. In keeping with the theme of the passage' it is most likely that estrogen will act to block the activity of endothelin. Answer choices A and.B call for estrogen to.stimulate this potent vasoconstrictor. This is no evidence for such an interaction, an,l choices A and B can be eliminated. Choice D wants us to believe that estrogen will not affect endothelin. Again.

this is not keeping with the theme that estrogen benefits cardiac tissue. Therefore, estrogen

will most likely inhibir

the release of endothelin. This has been demonstrated in rabbit coronary arteries. The comect choice is C.

64.

C is correct, an antagonist mimics a hotmone, interacts with the hormone's receptor, and leads to a block of the intracellular effects of the hormone. Choices B and D are incorrect, because they refer to agonists, not antagonis.. choice B is the colrect definition of an agonist. An antagonist acts against the action of a"hormone and blocks irs effects. Choice A is incorrect. The correct choice is C.

65.

A. is correct, the soy diet group has suppressed levels of FSH and LH compared to the controls. Examination ot Figure 1 indicates that both FSH and LH levels are decreased (suppressed) in the

soy diet trial compared to the

control trial. The correct choice is A.

D

66.

is the subject would not ovulate. ovulation is triggered by a precise sequence of hormonal events. LH ii -correct, the direct stimulus for ovulation. The subject would not oiitut" if-LH were,o-"to* ffiressed (similar to the suppression of LH by the birth control pill). This.is a theoretical question only. The amount of soy protein may ne'er completely halt LH production. The focus of this question is more on ovulation than on soy prbGin. If the iubjecl did not ovulate, she couldnotbecome pregnant. Choice A is incorrect. Since she did not ovuiate, there is a change ir the normal menstrual cycle. Choice B is incorrect. One cannot predict when menstruation would occur due to the lack of influence of the corpus luteum. Choice C is incorrect. The correct choice is D.

67.

C is correct' exposure to estrogen is a probable determinant of developing breast cancer. From the passage. \r3 learned that women who moved to countries other than their native countries

lot

breast cancer at rates

n

natives of the adopted country. This implies an environmental cause rath-er than a genetic cause."ompu.uble Choice A i; incorrect. The fact of importance in this passage is women in different countries do not iontract breast cancer at the same rate. Choice B is inconect. Diet can modify estrogen levels, as seen in the experiment. Choice D is incorreci

The correct choice is C. 68.

D is correct , 6I%. This is taken directly from a,tiny-facet in the passage. The diet provided 98 grams of protein. ol which 60 grams was the soy protein. 60198 = 6lvo. The correct choice is D.

69.

B is correct, the soy diet was hypocholesterolemic. This question tests our knowledge of terminolog y. Hypermean; increased. Hypo means decreased. 1so means the same or unchanged . Cholesterol"ii, level of cholesterol ll ^"un"by the soy-protein die.,the blood. From the data table, we can see that cholesterol levels in the blood were lowered The correct choice is hypochoiesteremic. Choices A and C are inconect. Choice D is a nonsense answer and u incorrect. The correct choice is B"


26,4


tsiotogy -0.


Section IV Answers

A is correct, corpus luteum. The corpus luteum secretes estrogen and progesterone after the egg is released from follicle. The egg itself does not secrete hormones. Choice B iJinconeit. ih" .o.pu, is a dyin! corpus luteum

the

which is no longer secreting hormones. Choice C is incorrect. The cervix secretes mucus, but nbt lio.-on"s. Choice D is incorrect. The correct choice is A. -1.

D is correct, all of the above. This answer goes back to our understanding of basal metabolic rate. A basal measurement is made at rest, before getting out of bed after a night's sleep, after all your food is digested, while lying quietly. The correct choice is D. B is correct, I and IIi only. Look at Figure 1. We can see that LH and estrogen rise dramatically right before ovulation. Choice I and III are correct. However, progesterone is iow before ovulation and rises drlmatically

afterwards. Choice II is incorrect. The correct choice is B.

C is correct, birth controi pills. Both condoms and diaphragms are barrier methods of contraception. RU-4g6 prevents continued implantation of a fertilized egg. The ievels of estrogen and progesterone in birih control pills inhibit ovulation, so no egg is produced. The correct choice is C. D is correct, she is ovulating. Ferning is an indicator of ovulation. The woman is therefore not pregnant. Choice A is incorrect. She is not anovulatory (not ovulating). Choice B is incorrect. She is ovulating and

t.

therefore fertile.

Choice C is incorrgct. The correct choice is D.

A is correct, I only. Althoug! FIH and estrogen play critical roles in preparing the follicle and positive feedback on the hypothalamus, respectively, LH is the actual irigger for release of tf,e egg-from the follicle. The correct choice is A. h.

B is correct, cholesterol. The ring structure of cholesterol is the base molecule for the synthesis of sex hormones in men and women. Arachidonic acid is a long-chain fatty acid, a precursor of the class of molecules called prostagiandins. Phosphatidyl choline is a phospholipid, used in membrines. Sphingomyelin is a complex lipid found

in the brain. The correct choice is B.

B is correct, the heart'.The question is asking what structure would be affected if the embryonic mesodenn were defective. This is really a matter of memorizing which structures are derived from each of the germ layers (ectoderm' mesodetm, endoderm). Choice, A, the nervous system, is derived from ectodermal precursors, as is the skin and the lens of the eye. Choice C, the stomach lining, is derived from the endodermal germ layer, which also gives rise to other linings of the digestive and respiratory tracts, as well as differentiating into itre majbr glands of the body such as the pancreas and the liver (choice D). This leaves choice B, the he#, as the structure which is mesodermal in origin (along with the notochord, skeleton, muscle, outer coverings of internal organs, and

reproductive organs). The correct choice is B.

''!"

D is correct, a transmembrane protein. From the passage, we learn that a morphogen is defined as a diffusible "substance." Answer choices A, B, and C could all fit this broad definition. Answei choice D, however, is far from a diffusible substance. Rather, a transmembrane protein is anchored into the plasma membrane and cannot either be secreted or diffuse anywhere except within the confines of the membrane. Such a confined protein would not be very effective at signalling distant cells. The correct choice is D.

C is correct, I and II only. The question inquires about the processes which could bring about the extension, or the increase in length, of a sheet of cells. Statement I, cell shape changes, is a correct possibility. If individual cells r'vithin the cell sheet were to change their shape such that they were narrower and longer, the entire sheet would get nalrower and longer, resulting in extension. Statement II is also a valid possibility. Mitotic division of cells within the sheet would cause an increase in the total number of cells in the sheet. If these cell divisions were in strategic locations, the entire cell sheet would become longer and extension would occur. Statement III, on the other hand]is not a valid explanation for sheet elongation. The unidirectional migration of every cell in the sheet would simply result in the entire sheet of cells moving somewhere synchronously. This would not result in the elongation ind extension described in the question. The correct choice is C.

tr,

A is correct, epithelial cells. The question gives us some clues about the type of cells on the surface of the embryo. We learn that they have tight junctions, which are cell-cell adhesions which prevent molecules from slipping in :,ght

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Biology


Section IV Answers

between these cells' as well helping to establish an apical/basolateral polarity. We also learn that they serve a role of protecting inner embryonic celis from the external environment. From these clues alone, the most iikely candidate would be answer choice A, epithelial cetls. Epithelial cells can have tight junctions, often to aid them iniheir major role of protecting underlying tissues from arbitrary inward diffusion oi lumenaL, or gut, contents. We can also approach this problem via a process of elimination. Neuronal, smooth muscle, and coinective tissue ceils are all highly differentiated cell types. This question inquires about an embryo which has yet to undergo gastrulation. It is very unlikely that these advanced cell types would have differentiated at this stage. The correct choice is A. 81.

B is correct, primate embryos should not have gill slits. The question asks which statement is NOT supported by the observation that human embryos have giil slits and a tail at some time during their development. This observation does suggest that humans and the other vertebrates share a common ancestri (choice C) as well as supporting the theory that humans evolved from lower vertebrates (choice A). The fact that human pass through sLges which resemble the embryonic stages of lower vertebrates (fish and tailed mammals"rnbryo, in partiiular; sugi"sts tiat humans might have evolved from these species. Since these stages of development haue not been drasticiliy altered from fish to human, it could be said that there is little selective uduuntug" in altering these stages of embryogenesis. This supports choice D.. Since we are looking for the unsupported statement, by a piocess of elimination we are Ieft with choice B. Since primates are evolutionarily close to humans, we would that they would display similar embryonic stages. Therefore, they should have gill slits at some embryonic stage. "^p*.tIncidentaily, the conservation of -developmental embryonic stages is often referred to as "ontogeny recapitulating phylogeny,""ttat ir, stages reiterate the developmental stages of other species in the iame phylogenetic tiee. The "u.iy choice is B.

82.

C is coffect' a concentration gradient of morphogen could never be formed. First, let's think about what normallv occurs. If the morphogen is produced at a point source (i.e., the dorsal lip cells), ii is degraded ur it oifi"r", u*"i. The result is a concentration gradient of morphogen with a maximum neaithe ,our"", beciuse;";h"g";;;""i; that get further away are more likely to be degraded because they have been present tong".. If the mo:rphogen were not slowly degraded, as the question postulates, what would happen? The gradient would"never form lmatirig choice B incorrect). This is because the morphogen would diffuJe uruy ft6- the dorsal lip cells ana equ"alir" in concentration throughout the embryo. This wouid affect gastrulation because proper signalling depends on the establishment of a gradient of morphogen to which cells at different distances ,"rponJ to Oirl"r"nily; therefore choice A is incorrect. Choice D is likewise invalid because there would be no"ould gradieni, not a very steep one. The correct choice is C. "rp""iutty

83.

C is correct, the embryo would develop into two S]g.T_"tg twin tadpoles, the result of two independent invaginations during gastrulation. This question is referring to Hilde Mangold'i famous dorsal lip transplantation expJrimensFirst of all, we must keep in mind that the function of the dorsal lip cells is to induce the'invagination event thar begins gastrulation. This is accomplished via the secretion of a morphogen gradient. From this, it follows ther introducing an additional dorsal lip would induce a separate invagination iu"ni in essence causing the embryo o develop into two attached organisms (Siamese twins). Answerlhoice A is incorrect because not only would gastrulation occur, it would occur in two separate places. Answer choice D is incorrect because two dorsal lips in one embryo would lead to two invaginations. Answer choice B is a tempting one, but remember that an r

"orr""i

this early stage doesn't have a differentiated immune system

yel The coriecfchoice is C.

"*bryo

84.

B is correct, hypothalamus. Oxytocin is produced within the hypothalamus by neurons and transported to the posterior pituitary, where it is secreted. Choice A is incorrect. Choice C is incorrect. The placenta iiself does nd produce oxytocin, but the fetus does. Choice D is incorrect. The correct choice is B.

85.

B is correct, I and II only. If the uterus is responding more strongly to normal, nonpregnant levels of oxytocin, it ir reasonable to expect that the number of oxytocin receptors has increased. This w-ould mean that u s-ull dose sf oxytocin would be very effective at stimulating uterine contractions. Choice I is correct. If oxytocin levels did not increase in the blood, there is the possibility that some sort of paracrine transfer is occurring. This means th* hotmone is secreted from one cell to another without actually moving through the bloodstream. Choice II is correct Usually when things are conjugated by the liver, they are being processed for excretion. If oxytocin wexc increasingly conjugated, then there would be less present, leading to fewer contractions. Choice III ls incorrec Choice B is the correct choice. The correct choice is B.

86.

C is correct, intravenous injection of human chorionic gonadotropin. You are looking for the FALSE answef,. Injection of oxytocin (intravenously) would increase the circulating levels of oxytoiin. This would increas

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Biology

Keproduction Er Development

Section IV Answers

contractions. Choice A is true. Intravaginal administration of prostaglandins wouid increase contractions by acting on the uterus, just as if the prostaglandins were produced by the endometrium. (This is the idea behind oral or vaginal prostaglandin administration to allow abortion.) Choice B is true. Prostaglandins also work orally (see above). Choice D is true. Human chorionic gonadotropin (hCG) is produced durinf early pregnancy to make sure estrogen and progesterone levels remain high. Although it tapers ofi after about 10-weet
\

C is correct' administratio{ of inhibitors of prostaglandin production. Both oxytocin and prostaglandins must be present for labor to occur. By eliminating one or the other from the equation, on" avoid an lubo. in which "un would not the fetus is premature' A secretagogue encourages the secretion of iomething. This "uJy be favorable in a person in whom iabor threatens. Choice A is incorrect. An oxytocin agonist would act on the oxytocin receptors just as oxytocin would. Another unfavorable choice. Choice D is incorrect. RU-486 is used to promote abortion. Choice B is incorrect. Finally, inhibitors of prostaglandin production, such as indomethacin, can be used to prevent the positive interactions between prostaglandins and oxytocin. Choice C is correct. The correct choice is C.

iq.

D is correct, arachidonic acid. This is a bit of trivia, and you will either know it or not. Anyway, stearic is 1g:0, myristic is 14:0, and linolenic is 18:2. only arachidonic has the required 20 carbons 120:$. ti is the p.".u.roi

prostaglandins. Choices A, B, and C are incorrect. Choice D is correct. The correct choice is D. 59.

B is correct, I and III only. Peptide hormones do not cross the lipid bilayer directly. They usuaily interact with cellulal membrane receptors and signals are transferred to the intracellular medium by s"cond messengers. Choices I and III are coffect. If the hormon" enter the cell, then it cannot stimulate mRNA synthesis directly. Choice II is "un't incorrect. The correct choice is B.

9{J.

A is correct, delivery can occur without stretch signals from cervix. A paraplegic woman may not be able to

voluntarily contract her abdominal muscles to assist in labor. This is not a rLquiienient. Choice B is incorrect. The whole passage discusses the importance of the interactions between oxytocin and prostaglandins. They are required. Choice C is incorrect. The smooth muscle of the uterus contracts without voiuntari stimulation. Choice D is incorrect. What is missing in these women is the positive feedback (through afferent nerves; from cervical stretching that increases oxytocin secretion. choice A is correct. The correct choice is A.

-

91.

B is correct' Gonadotropin releasing hormone (GnRH). Estrogen is under primarily under the control of leutenizing hormone (LH) derived from the anterior pituitary. The production and releise of LH, in turn, is under the influencI of gonadotropin releasing hormone (GnRH). Therefore, estrogen production and secretion is indirectlv stimulated through the indirect effect of GnRH. The correct choice is B.

92.

B is correct, testosterone synthesis. The seminiferous tubules are located in the male testes. These tubules are the site of sperm production, or spermatogenesis. The important anatomical features of each seminiferous tubule are the Leydig cells, the basement membrane, and the Sertoli cells. The Leydig cells are found between the seminiferous tubuies, in the interstitial space. These cells respond to iuteinizing hormone (LH) and it helps regulate the conversion of cholesterol to testosterone. Therefore, testosterone synthesis does NOT occur within the seminiferous tubuies. Once synthesized, testosterone can cross the basement membrane and influence the Sertoli cells. The Sertoli cells secrete androgen-binding protein which binds testosterone (an androgen). This helps to concentrate testosterone within the seminiferous tubules. The Sertoli cells also respond to FSH, which helps to control spermatogenesis. The correct choice is B.

93.

D is correct, I, II, and III. The seminiferous tubules are under the influence of LH, FSH, and GnRH. GnRH, derived from the hypothalamus, stimuiates the secretion of LH and FSH from the anterior pituitary. LH stimulates the interstitial cells of Leydig to produce testosterone. FSH and testosterone, in turn, stimulate the deveiopment of the seminiferous tubules and spermatogenesis. The correct choice is D.

9{.

B is correct, LH. As mentioned in the previous answers, LH binds to receptors on the interstitial cells of Leydig and stimulates the synthesis of testosterone from cholesterol. The correct choice is B.

95.

D is correct, Insulin-iike growth factor I (IGF-I). The clue to answering this question comes from the question itself. We are looking for the secretion of a hormone or factor that can indirectly act on muscle and bone tissue. The word factor might clue us into what we are looking for. Insulin is secreted from the B-celts of the islets of Langerhans in the pancreas. Insulin has a direct effect on the cellular uptake of glucose, fatty acids, and amino acids, and their

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Biotogy


Section IV Answers

subsequent conversion into_glycogen, triglycerides, and protein. Thyroxine (or T4, tetraiodothyronine) is a hormone that has a global effect on basal metabolic rate and is critical for nbrmal growth. Thyroxine iirectly influences the secretion of growth hormone and helps it to promote protein synthesis utid bon" growth. Epidermai growth f'actor

of skin cells and promotes their growth. Insulin-like growtlifactor I (IGF-I), also called a for growth hormone. In other words, GH stimulates the production of IGF-I which in turn promotes -"diuto. things like frotein acts at the level

somatomedian, is simiiar to insulin in its structure. IGF-I is synthesized in the liver and acts as a

synthesis, cellular division, and overall growth. The correct choice is D.

D is correct, hypothalamus. Growth hormone-is.synthesized by the anterior pituitary and is regulated by growth hormone releasing hormone (GHRH) and growth.h?rTone inhibiling hormone (GHIH) from the hypothala*ui. Th. posterior pituitary releases antidiuretic hormone (ADH) and oxytocin. Bone and liver can react to growth hormone.

96.

but do not synthesize or release it. The correct choice is D.

D is correct, tumor cells do not have a hapioid chromosome number. We are looking for a statement thaf support5

97.

the idea that germ cell tumors arise from spematogonia. In the process of spermatog#esis, spermatogonia will gir.e rise to primary spermatocytes. In this process, the genetii complemint witi be reduced. In other words. spermatogonia have 46 chromosomes while primary spermatocytes have 23. Therefore, if tumor cells did have their origin in spermatogonia, they_should be diploid cells. The fact that tumor cells do not have a haploid chromosome

number supports this theory. The correct choice is D. 98.

99.

100.

B is correct, at least 90Vo of testicular cancers are found in the seminiferous tubules. The passage tells us that onh 57o of testicular cancers do not originate from germ ceils. That means that aboutg5vo do oiigina*te from germ celli.

Where are germ cells located? They are located in the seminiferous tubules of the teste. The Jorrect choice is B. D is correct, inf'ertile men run a greater risk for CIS. This answer could be anived at by a process of elimination. Cryptorchidism is a risk factor. From the passage, we know that this condition arises from a developmental error. This supports our theory. The fact that tumors have been shown to deveiop in gestation supports the theory that germ cells tumors have their origins in fetal germ cells and have a misguided aevelopment. Furthermore, the evidence thar cancer genes act in fetal germ cells supports the theory that tumors come from fetal germ cells and have a disrupte6 developmental fate. The only statement which does not support the theory is that inflrtile men run greater a risk for CIS- This really says nothing about misguided developmint and its connection to testicular cancer. The correct choice is D.

D is correct, 92. This question is very straightforward. The number of chromosomes in a human haploid cell is 2_r We are looking for a human cell that has a tetraploid number of chromosomes. Therefore,23 X 4 9l =

chromosomes. The correct choice is D. 101.

C is correct, testicular cancer is rare among twins. The passage informs us that testicular cancer has little basis 1r genetics. We are looking for a statement that provides evidence for this notion. The fact that testicular cancer is raf;: among identical twins is such a statement. If one twin developed testicular cancer, and the cancer had its basis ir genetics, it is likely the other twin develops the cancer. Since testicular cancer is rare among twins, this statemenL. supports the idea that testicular cancer has iittle basis in genetics. The correct choice is C.

102.

D is correct, haploid adult from an unfertilized egg. This question is a strict definition type question that ca: sometimes be found on the MCAT. Nothing in the passage directly tells us this definition. Often, these types o: questions pertain to hormones. In this case, a developmental term was used. The correct choice is D.

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The Berkeley Revier Specializing in MCAT Preparation

ffifi€3Hogy Section V Endocrinology

& Immunology

A. The Endocrine Systern

l. 2. 3. 4. 5.

Caiecholamine-Llormones Peptide Hormones Steroid & Thyroid Hormones RegulatoryMechanisms The Pituitary Cland

The Immune $ystem

I. 2. 3. 4. 5. '

Cell Types Antibody Structure

Antibody Action T Cells

Humoral & Cellular Immunitv ---'* J -

Practice Passages & Answers

,Ifu

n

Specidnztng in MCAT Preparation

Endocrinology & Immunotogy Top lO Section Goals CI?

Be familiar with catecholamine hormones and their actions. Catecholamines like dopamine, noreprnephrine, and epinephrine all contain a catechol ring ring with two adjacent hydroxyl groups). They act ai the'level of the cNS and pNS.

(a

benzene

Be familiar with de hormones and their actions. Peptide hormones contain two or more amino acids linked together by a peptide bond. These hormones are secreted by a wide range of structures aid have a'va'riety of actions. Be familiar the actions of the G and Understand-what happens when a hormone like epinephrine binds to a cell surface receptor. Be iamiiiar with the action of the C-protein and conversion of ATP to cAMP by adenylate iyclase.

At a fleneral level understand how phosphoryIation events activate Droteins. A, classic example involves the phosphorylation of glycogen phosphorylase by protein kinase A. rhosphorylated glycogen phosphorylase is activated and can-convert glycogen to glucose.

Be

familiar with steroid hormones and their actions.

The common plecursor to all steroid hormones is cholesterol. Steroid hormones are not stored after their synthesis, but instead are used right away. Know the major groups of steroid hormones.

Be

familiar with the mechanisms by which homeostasis is maintained.

Understand the concept o{ negarive feedback. Understand what is meant by endocrine, newoendocrine, parac.rlne, ano autocrlne -regulatlon.

Know the hormones of the pituitarv. It is.important to know.w_here the € majol hormones of the anterior and posterior pituitary are synthesized and stored. Know why and how they are released into ttie circulat'ory sydtem.

ffi

Be aware of the different celt involved with the immune whlte blood cells/ or leukocytes, are involved in the body's immune response. The 5 different types of leukocytes leukocvtes are neutrophils, neutroohils. eosinophils, eosinonhils. basophiis, basonhils mcinocytes, mrinorrrteq :nd lrimnh^"',ro. . and ly?nphocytes

gi -?

O

Understand the basic structure and frrnction of an antibody.

Be familiar with the differences between humoral and cellular immunitv. Understand the workings-of macrop_hages, MHC receptors, the different interleukins, the T cells and B cells, the different interferons, and hdw they relate io foreign antigens.

Biology

Endocrinology & lmmunology

Catecholamine Hormones

CatechdIah.b'o0ilcs.:i:....l....]..;.'.. One way that cells can communicate with one another over long distances is through extracellular products calied hormones. Hormones can generally be divided into peptide hormones (e.g., insulin), amine hormones (e.g., epinephrine or adrenaline which are classified as catecholamines), and steroids (e.g., the male and fernale sex hormones). Hormones are released bv endocrine organs into lhe blood and travel by way of the circulatory system to various target tissues. \Vhen a particular endocrine cell is stimulated to release a hormone into the ':lood the concentration of that hormone increases. Once the stimulation is ierminated the hormone is no longer released the concentration falls back to some normal resting level. Hormones generally have a very short lifetime in the r1ood. Even though the time spent in the blood may be relativeiy short, hormones can act within seconds or they can take hours and quite possibly dalrs :o act on their target tissue. By comparison the action of the nervous system is much faster. l'Vhen a hormone acts on a target cell

it can either do so at the level of the cell's membrane by binding to a specific receptor in that membrane or by passing :hrough the membrane itself and binding with a specific target protein within the

'ell's cytoplasm. After the hormone has bound to its specific receptor that :omplex undergoes a conformational change that allows for the synthesis of an ntracelluiar messenger. This second messenger passes the information mediated bv the hormone (i.e., the first messenger) to some specific reaction within the :e11. For example, the second messenger might convey instructions that allow a :articular set of reactions to release glucose into the biood or it might even act at :re level of the gene to turn off or turn on gene expression. Let's consider the action of the catecholamine epinephrine (adrenaline) on a :r-pical hepatic (liver) cell. when the body is under some type of stress iike :hysical exercise or even fright, an increased need for glucose arises. Glucose is slored in the body in the form of glycogen and through a series of reactions can :-.e mobilized and used as a source of energy during these times of stress. Cnce a stress has been perceived the nervous system responds by signaling the adrenal medulla (part of the adrenal gland which sits on top of the kidneys) to :elease epinephrine into the extracellular fluid. Epinephrine diffuses into the :1ood and is carried to the hepatic cells where it binds to a specific cell membrane :eceptor called a B-adrenergic receptor. This action causes the activation of the -nzyme adenylate cyclase (bound on the cvtoplasmic membrane surface) which increases the concentration of the second messenger cyclic adenosine monophosphate (cAMP) within the cell. The binding of epinephrine, a water solubie hormone, to the cell surface receptor

and the synthesis of cAMP within the cytosol is coupled through the action of a G protein. The name of this protein stems from the fact that it binds guanosine triphosphate (GTP) and GDP. GTP and GDP are both nucleotides just like ATP and ADP.

Initially, before epinephrine binds to its receptor, GDP is bound to a particular subunit of the G protein. However, once epinephrine binds to its target receptor Copyright O by The Berkeley Review

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Catecholamine llormones

this hormone-receptor complex interacts with the G protein and stimulates the for GTp. The G protein-GTp complex diffuses through the membrane and activates the membrane bound adenylate cyclase enzyrile. In turn, adenylate cyclase catalyzes the conversion of AT-p to .Rwtp. These events are shown in the sequences (a) through (d) in Figure 5-1. exchange of GDP

Hormone (Epinephrine)

s Extracellular Space

(a)

Mernbraneof iLiVer

C€ll

,

Cytosol

Inactive

Extracellular Space

(b)

MernbraneOf :Livei'cCitrr.

,

Cytosol

Extracellular Space

(c)

IVlemli,fmb:,o-f .

Adenylate

tivireeti..

Cyclase

Cytosol

Active

Extracellular Space

Membianeo-f

(d)

LiVer,,GCll. Cytosol

ATP

cAMP + PP;

Figure 5.1 Action of adenylate cyclase.

After

a

brief period of time the G protein hydrolyzes the bound GTp to GDp and

Pi (not shown in Figure 5-1) thus returning the G protein back to its inactive

state (Figure 5-1a). However, during this period of activation (Figure 5-1d) the G protein-GTP complex was able to activate many adenylate cyclaie enzymes thus forming many molecules of cAMp. In other words, by the time G protein is Copyright

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Catecholamine Hormones

ractivated the hormonal signal induced by epinephrine has been amplified :ranv times. Since cAMP is synthesized

within the cytosol of the liver cell it can diffuse within

rat cytoplasm and interacts with a variety of molecules. One of the molecules rat cAMP interacts with is a particular type of protein kinase called protein kinase A (abbreviated as PKA). A protein kinase is simply an enzyme that =ansfers the end phosphate (the gamma phosphate) of ATP to a specific amino aid residue of a substrate protein. cAMP will bind to PKA and stimulate it to

:hosphorylate (add a phosphate group) an enzyme called

glycogen

ghosphorylase. Once glycogen phosphorylase has become phosphorylated it is :"orv an active enzyme and will catalyze the conversion of glycogen into glucose. ItLis is shown in Figure 5-2. Glucose is released into the blood and travels --rroughout the body by way of the circulatory system. [Recall that glycogen is srmply the storage form of glucose in mammals. Also, there are other reactions ::rr-olved in this ampiification process. Only the important ones have been shown rere.]

cAMP ).PKA

I

Glycogen

Inactive

{<

phosphorylase

Glycogen

g

Glucose

Figure 5.2 -{ctivation by phosphorylation.

in the cascade of reactions shown in Figure 5-2 promotes the activation of many more molecules in the next step in the sequence. This is exactly what we saw when epinephrine bound to its receptor. This rapid amplification process allows the binding of just a few epinephrine molecules to release grams of glucose into the blood. [In Figure 5-1 and Figure 5-2 the sites of amplification are indicated by asterisks (*).1

Each enzyme

Cholera is an intestinal disorder caused by a bacterium (Vibrio cholerae). The major symptom of this disorder is diarrhea and if left untreated will result in ,revere dehydration and eventual death. This toxin binds to the active state of the G protein and prevents the GTP from being hydrolyzed to GDP and P1. This means that the adenylate cyclase enzyme is continually active and massive amounts of cAMP are synthesized. cAMP causes the intestinal cells to secrete idigestive) fluids.

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Peptide llormones

P-e$fiH The water soluble peptide hormone gastrin stimulates the secretion of HCI and pepsinogen from the_ stomach in response to stimulation from the vagus nerve and partially digested protein.

Gastrin (the first messenger) binds to a specific ceII surface receptor in the plasma membrane and activates a particular G protein (different tharrthe one in Figure ,!-3). The activated G protein interacts with the membrane enzyme phosphohlase C (abbreviated as PLC) and induces that enzyme to hydroiyze phosphatiaytinositol-4,5-bisphosphate (PiP2) to inositol-1,4,5-triphosphate (ip:J -and 1',2diacyiglycerol (DAG). Hormone

tGastrinl\ V

Extracellular Space

(a)

Plasma,Memtiiane Cytosol

(b)

Extracellular Space

(c)

Stimulates the release of Ca2+

Ca2*

Protein

Protein-P

(inactive)

(active)

Irr

\.2

HCI secretion into the lumen of the stomach

Figure 5-3 Activation of

a receptor

by a peptide hormone.

Next, the second messenger, IP3, which was released into the cytoplasm, interacts with the endoplasmic reticulum and stimulates the release of Ca2e into Copyright

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Peptide llormones

the cytoplarT by some unknown mechanism. Meanwhite DAG, diffusing through the plasma membrane, interacts with a molecule called protein kinase d and stimulates that kinase, with the help of Ca2 e just released from the endopiasmic reticulum, to phosphorylate an unknown protein which in turn causes HCI secretion into the lumen of the stomach. These mechanistic events are outlined in (a) through (c) in Figure 5-3. Insulin is a water soluble peptide hormone that binds to a specific transmembrane receptor in the cell membranes of liver, fat, and muscle cells. once insulin binds to the receptor on the cell surface the cytoplasmic portion of the receptor is converted into a tyrosine kinase that autophosphorylates the amino acid tyrosine found within the cytoplasmic portion oi the te""ptor. This acts to further enhance the activity of the tyrosine kinase. presumably the insulin receptor can also internalize and (somehow) act as a second messenger. This action, along with enhanced tyrosine kinase activity, leads to the internalization of glucose into these cells (see Figure 5-4). The actual events in the insulin signaling mechanism that leads to the uptake of glucose are somewhat obscure at the present time. [Recail that if giucose is in abundance (after a meal), it can be stored in the form of glycogen.l Hormone

(Insulini

\ Extracellular Space

Plasma Nlembrane Tyrosine kinase

activity

Internalization

----->

2nd Messenger

Cytosol

Phosphorylation

activiry

Glucose uptake

Figure 5.4 insulin signaling mechanism.

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Steroid & Thyroid Hormones

$',G-

These are also lipid solubte hormones which can pass through the cell's plasma membrane and interact with a receptor either in the cytosol or in the nucleus. Thyroid hormones, secreted from the thyroid gland, can diffuse across the plasma membrane and into the nucleus where they bind with specific receptors. The hormone receptor complex then activates transcription essential for certain metabolic processes. Thyroid hormones help to regulate growth and differentiation and they can stimulate the breakdowr-r of proteins, fats, and glucose.

Steroid hormones all originate from a single precursor molecule, cholesterol. Two important classes of steroid hormones are the androgens (male sex hormones ) and estrogens (female sex hormones). Estrogery is lipid soluble, can -hich diffuse across the membrane of a target cell and bind with a specific receptor with the cytoplasm. This steroid-receptor complex can then enter into the nucleus where it influences the transcription of nRNA for certain protein products. The general scheme for this action is shown in Figure 5-5. Hormone-Receptor Complex

Hormone

D

@ 6)

Receptor

oi: =fl OH

DNA

'RNA 0

S

mRN A

Proteins

-\-^,r-^r

Nucleus

c.)

Figure 5-5 General action of a steroid hormone.

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Endocrinology 8r Immunology

d b

R.egulatory Mechanisms

s

General Mechanisms

How is homeostasis maintained in the body? Homeostasis is simply the maintenance of stability within the body. one way that homeo.iurl, l,

maintained is by negative feedback. ln this case there is some controlled variable which is sensed by a particular sensor in the body. Through a variety of mechanisms these sensors send input to control centers which in turn ,"g.,lut" the controlied variable (see Figure 5-6). For example, if a certain produci was being produced in a large quantity before negative feedback was initiated, then after initiation of negative feedback a sma[er quantity would be produced. Negative feedback reverses the output from elfectois such as gLnds and muscles.

Controlled -.);"#" Variable

Fnt --\

Sensor

Mechanism

=>

Negative Feedback

Figure 5-6 \egative feedback.

Endocrine Regulation thg endocrine system r,ve might have a grand secreting a hormone into the ]iblood. This

hormone, as we have seen, can infruence otheicells in the bodv. For example, let's consider the regulation of blood glucose levels. \Arhy are the tlood levels of glucose so important? If glucose supplies to the brain ur"-i11terr.rpted for ar extended period of time, brain damage can resuit. Glucose turns out to be the only fuel_utlrizedby the brain. [That is, -xcept during extended fasting in which

;ase the brain can use compounds called ketone bodles.] Thus, the controiled i'ariable in this regulatory mechanism is the concentration of blood glucose.

B Ceils Blood glucose

-+

rnsurin

es5 hil:ll::,t

increases

r----\

'

Uotake of: Glucose

-

IGlucose] stooo gru.os. fl decreases I I

+

Blood

A cells

Glucagon

-)

Liver & Fat cells

-

Release of:

'-).

Glucose

Fatty Acids

Figure 5-7 :ifects of glucose on the B-cells and u,-cells of the pancreas.

,lle pancreas secretes two hormones that are important in maintaining the :roper levels of blood glucose. Both of these hormones are secreted from clJsters 3opyright

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Regulatory Mechanisms

of specialized cells called the islets of Langerhans. Insulin is secreted by the B cells (or B cells) while glucagon is secreted by the cr cells (or A cells). After a meal the levels of blood glucose begin to increase. This increase stimulates the B cells to release insulin into the blood. Insulin binds to specific receptors on liver, fat, and muscle cells and through a complicated mechanis- pro*oi", the uptake of glucose into these cells. This action tends to lower (by negative feedback) the blood glucose levels. See Figure 5-7. Conversely, if the levels of blood glucose levels begin to decrease below some normal value, then the u cells are stimulated to secrete glucagon. This hormone circulates in the blood and binds to specific receptors ot ii.ru. and fat cells. Glucagon not only stimulates the liver cells to degrade glycogen to glucose but it also stimulates the fat cells to release fatty acids. fatty iiias can be metab olized and used in the Krebs cycle where they supply energy. This helps to alleviate the need for glucose as an energy source. Through these negative feedback actions the levels of glucose will begin to increase. SeeFigure 5-7.

Neuroendocrine Kegulation

In this case the hormone is not release for an endocrine cell but rather from a nerve cell which releases its neurotransmitter in the form of a hormone into the blood. For example, the adrenal medulla can receive sensory input from a sympathetic nerve, which tells it to release epinephrine into the blood. Other

examples of neuroendocrine regulation involve the hypothalamus and the pituitary gland. The pituitary gland can be divided -into the anterior and

posterior pituitary.

Paracrine Regulation In paracrine regulation the chemical that acts as a signal is released from one cell and influences a cell immediately adjacent to it (Figure 5-g). An example of such a paracrine cell would be mast cells, which contain large amounts of histamine" The substances released by the paracrine cell are generally dumped into the extracellular space and not into the bloodstream. other examples o? a paracrine signal would be neurohormones and neurotransmitters.

ffi :;;" [I-J/A-/

Paracrine

I

b

I ^ -r\

: :-----

a Signaring

compound

O

a'.'

I a

a

Autocrine

Figure 5-B Paracrine and autocrine regulation.

Autocrine Kegulation In autocrine regulation cells can release certain chemicals which they can then respond to themselves (Figure 5-8). For example, certain cells (like tumor cellsrl can release growth factors which can then bind to specific receptors on the

membrane of that same cell. Thus, the cells that releasedthe growth hormone are

stimulated to grow. Prostaglandins, which are lipid-soluble chemicals, alsn appear to show autocrine regulation.

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Biology


The Pituitary Gland

Posterior Pituitary The posterior pituitary releases oxytocin and antidiuretic hormone (Figure 5-9). These two hormones are synthesized in specific cells of the hypothalamus. The

major effect of oxytocin is to stimulate {emale uterine contraction while antidiuretic hormone (ADH) stimulates water reabsorption in the kidneys and also helps to increase the blood volume (pressure).

Hypothalmic Secretory Neurons

v-V,

Pituitary Stalk

Artery

Capillariestr)

Anterior

fr

Pituitary

Capillaries

Posterior

Pituitary TSH, ACTH, FSH, LH GH, PRL

Oxytocin

ADH

Figure 5-9 The hormones released from the pituitary gland.

The ADH that is synthesized in the nerve cell bodies in the hypothalamus are

packaged into vesicles and transported down the axon to the terminal bouton in the posterior pituitary. A nerve impulse (which propagated down the same axon) causes the release of these hormones into a system of nearby capillaries. ADH circulates in the biood and eventually reaches the kidneys where it stimulates ivater and Na@ reabsorption.

The nerve cell bodies in the hypothalamus where ADH was synthesized can sense a change in biood volume (and a change in the concentration of Nae). If these cells sense a low biood volume, they release ADH into the blood. At the

Ievel of the kidneys ADH will stimulate the retention of as much water as In other words, ADH prevents diuresis which is the excessive loss

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The Pituitary Gland

urine. By preventing diuresis water can be reabsorbed and the blood volume increased.

Anterior Pituitary pituitary can secrete six hormones (Figure 5-39). They are thyroid stimulating hormone (TSH), adrenocorticotropic hormone (ACTH), ronicte stimulating hormone !FSH), luteinizing hormone (LH), growth hormone (GH), and proiactin (PRL). we will discuss the function of ihese hormones as we proceed in the course. The anterior

These hormones are reguiatedby a second set of hormones which are stored in hypothalamic nerves. For example,let's consider TSH. within the hypothalamus are nerve cells that contain thyrotropin releasing hormone (TRH). when this nerve is stimulated it secretes TRH into a set of capillaries which extend into the anterior pituitary. TRH stimulates the synthesis and release of TSH from the

anterior pituitary into the biood. TSH binds to specialized receptors in the thyroid gland and causes the release of thyroxine. Tiris hormone, when released from the thyroid gland, influence metabolism and growth.

If

the metabolic rate of the body is too low, TSH is released from the anterior pituitary and stimulates the release of thyroxine from the thyroid gland. once thyroxine increases metabolism to just the right level (homeostasis)1t acts back on the nerve cells in the hypothalamus and the cells in the anterior pituitary and inhibits the release of rRH and rsH, respectively. Again, this is ur,oih", example

of negative feedback.

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Biology

Gellii


Cell Types

S"'

Every day your body is subiect to countless attacks by microorganisms and viruses which can cause a variety of diseases. Bacteria can enter itttJyoru system and disrupt cellular function. One example that we have mentioned is massive diarrhea caused by the bacterium Vibrio cholerae. The bacterial toxin from this organism can cause over a liter of {luid to be lost every hour. Parasitic viruses can also reap havoc on the body. Viruses are much smallr than bacteria and contain a nucleic acid core (either RNA or DNA) surrounded by a protein coat. viruses lack their own metabolicrnachinery. once they invade ur-, orgunism they can take over the host's metabolic machinery for their own use und ge.rerate more progeny viruses which, after the infected cell lyses, can spread and infect other cells. However, there are certain conditions *il"t" the viral genome integrates rnto the chromosome of the host cel1. In this case the genome can be 'iril genome replicated when the host chromosome is replicated. If this viral removes itseif from the host genome, it can take orr"i th" metabolic machlery of the host cell and produce more progeny viruses. Cancer causing viruses can insert their genetic material in to host chromosomes as well. Viruses can cause a variety of ciseases ranging from smali pox and infiuenza to measles and the common cold.

Ii we were to examine some blood uncler the microscope, we would find two lifferent cells types-the erythrocytes (red blood cells) and the leukocytes (white rlood cells). Erythrocytes in adults are produced in the maffow of thre sternum,

:ibs, and vertebrae. Leukocytes uru ptod.r""d partially in the tissues of the lymph ,:nd partiy in bone marrow. we have already coniidered the erythrocytes'in -:-revious sections. Let's now briefly consider the leukocytes.

Sere are 6 types of leukocytes found in the blood. The three types of leukocytes rat we are interested in are the monocytes, neutrophils, andiymphocytes. "The ::.onocytes and neutrophils are considered to be phagocytes. .iheie ceils, along '"'ith mast cells and a variety of other ceil typei, participate in the immune response.

\Iast Cells: These ceils are derived from leukocytes and then migrate out into the r>uc> where wrtcre they Lrrey resloe. reside. \44ren vvnen mast CelS cells are stimulated they reiease -sues ,iiuur" histamine

' hich acts on endothelial cells and causes an increased permeability to cells like :,e neutrophils. This increased permeability ailows the neutrophils easy access to --:.e surrounding tissue in order to defend against foreign bacteria or viruses.

?hagocytes: As we have mentioned, monocytes and neutrophils are both ::-agocytes. when monocytes leave the blood through pores in the blood vessels

':-d enter into the tissues, they

can be transform"d ir-tto macrophages. eutrophils circulate in the blood. Flowever, they too can leave the blood *^:ough pores in the blood vessels and enter into the tissues. The macrophages ,:,d neutrophils are_lhe plimary ceII types that attack and destroy ior"ig' , .:teria and viruses. T"y do this by the process of phagocytosis, enguifing tie ,:eign invader by endocytosis. once inside the phagocyte iyscsomi-l ur,r1i*u, I

*

,

: hydrogen peroxide is released which can then d-egrade ihe foreign objects.

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'Endocrinology & Immunology

Cell Types

Lymphocytes: There are 2 types of l1'rnphocytes. Lymphocytes which are derived from the thymus gland are called T lymphocytes (or T cells). The other type of lymphocytes are called B lymphocytes (or B cells). Let's briefly consider each type'

T lymphocytes: These cells are responsible for cell-mediated immunity. These cells are responsible for the destruction of foreign microorganisms and other such agents harmful to the body. There are 3 types of T lymphocytes which we will be considering: (1) cytotoxic T cells (also called killer T cells), (2) helper T cells, and (3) suppressor T cells. B lymphocytes: These cells circulate in the blood and to the lymph organs like the spleen and lymph nodes and are responsible for humoral mediated immunity. [Humor is another way of saying fluid.] Upon infection the B lymphocytes can differentiate into plasma cells which have the ability to slmthesize and secrete antibodies. Antibodies are proteins that are synthesized in response to an antigen that has been introduced into the body. An antigen is simply a foreign substance (protein or polysaccharide) which has a high molecular weight that has entered the body and induces a particular immune response. Antigens interact with specific antibodies.

General Immune Kesponse Let's consider a general response to the invasion of a bacterium or even a vims. These invaders have on their surface an antigen. The cellular immune response begins with a macrophage engulfing a foreign particle and phagocytizing it, The antigenic fragments released into the cytosol of the macrophage are transported

to the macrophage's membrane where they bind to a specific surface proteiru This specific surface protein is called a major histocompatibility compler (MHC) protein (of the Class I typu). The antigen is being presented or "displayed" on the surface of the macrophage. Certain receptors on the class of T lymphocytes called cytotoxic T cells can recognize this antigen-surface protein complex on the macrophage. When the cytotoxic T cell binds to the antigen-Class I MHC protein complex of the macrophage, a growth factor called interleukin-L is released by the macrophage-

T cell itself releases a growth factor called interleukin-2 Interleukin-l, interleukin-2, and interleukins released by helper T cells (discussed below) stimulate the slmthesis of more cytotoxic T cells. These kilier T The cytotoxic

cells proliferate and bind to the invading foreign cells beadng the antigen and induce them to lyse. This type of response is referred to as cell mediated

immunity. On the surface of B lymphocytes are Class II MHC proteins and antibodiesWhen a B lymphocyte finds an antigen that has specifically bound to its antibody, it engulfs that antigen-antibody complex. After degrading the antigerr antibody complex the B lymphocyte transports a portion of that antigen to Class II MHC protein and "displays" the complex on the surface of i membrane.

Helper T cells with the right receptors are able to bind to the antigen-Class MHC protein complex. This binding stimulates the helper T cells to re interleukins (a lymphokine). This stimulates the B lymphocytes to pro and form plasma cells. The plasma cells in turn produce a vast amount Copyright

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Immunology

Cell Types

antibodies which are specific for the antigen. when these circulating antibodies bind to the antigen they act as a tag tnut rigg fnugocytes to engurf the antigen-antibody comprex and-destroy"it. "rr".riutirg This type of?elfoLr" i, referred to as humoral mediated immunity. The human immunodeficiency virus (HIV), thought to be responsible for the acquired immune deficiency syndrome (AIDS), ac"ts at the levei of the h"rp";i cells by infecting them.

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The Berkeley Review Speciatizing in MCAT prelraration

Biology


Antibody Structure

ffi Most human antibodies, or immunoglogulins (abbreviated as Ig), are composed of 4 subunits arranged in a "Y" configuration as shown in Figure 5-10. There are 2 light chains and 2 heavy chains. These subunits are joined to one another br, disulfide bonds. Within each heavy and light chain are variable domains and constant domains.

At the terminal ends of the heavy and light chains

are variabie (v) regions that can differ in amino acid sequence from immunoglobulin to immunoglobulin. The constant (C) regions of the heavy and light chains are found in the lower portion-.

of the immunoglobulin. The antigen binding site for a particular antibody i: located at the end of the variable regions of the heavy and light chains.

Two other regions of the immunoglobulin are important. They are the diversitr(D) region and the joining (j) region. The genes that code for the I, D, and \regions of the variable domain greatly increase the diversity one finds among the immunoglobulins.

V = variable D = diversity ; = joining

Antigen

binding site

.il$

Antigen

^$'

binding site

^no*7

t

Light Chai n

Heavy Chain

)

Constant Domains

Disulfide bond

B Cell Membrane Figure 5-lO Generalized antibody structure.

Classes of Antibodies We have 5 classes of immunoglobulins which differ in the composition of heavy chains. They are (in alphabetical order) IgA, IgD, IgE,IgG, and IgM. functions are:

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Biotogy IgA: IgD: IgE: IgG: IgM:


Antibody Structure

Found in milk and helps to protect nursing infants. Has an unknown function. Binds to mast cells and is invoived in the allergic reaction. The only antibody able to cross the placenta. iI is also the most abundant and is produced within days after the IgM antibody is secreted.

Produced a few days after detection of an antigen and antibody produced in response to an antigen.

it is the first

Combinatorial Diversifi cation

The light chain contains at least 300 different genes that can determine the variable region and about 4 different genes thit can determine that joining region. Therefore, there can be about (300) x (4) or -1.,200 different combinations for the light chain. The heavy chain contains about 1,000 different genes for the variable region, about 'l I different genes for the diversity region, and about 4 different genes for the joining region. Thus, we find that tirere ire about (1,000) x (rz1 * 141 or 48,000 different combinations for the heavy chain. If we associate all the different possible light chains with all the different possible heavy chains, we wili get a combinatorial diversification of about (.1,,200) * (+S,OOO) or roughly 5.g x 107 different possibie antibody combinations. In other words, the immune system_can generate an antibody for practically any antigen that invades the body. [The point behind this is the enormous antiboJy diversity that is gene-rated. Depending on how the calculation is done and what values are used the diversity can be as high as 1.1 x 1011.1


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Biology


Antibody Action

Antibodies do not destroy foreign antigens. They simply recognize and identify them. Antibodies can do this is a number of ways. One method is by directly blocking the foreign invader from gaining access to host cells. This is accomplished by the antibodies binding to the antigens of, say, a virus as shown in Figure 5-11. Antigen

Figure 5.1I Direct block.

Complement is a rather complicated system for disposing of invaders. The essence of this process is as follows: An antibody has already recognized and bound to a specific antigen on a bacterial cell that is considered an intruder. A complement protein (a plasma protein) recognizes this antigen-antibody complex and binds to the Fc domain of the antibody. After a series of reactions the complement protein is activated and triggers an immune response. Further reactions form a membrane attack complex (MAC) that inserts into the bacteriai cell's membrane and forms a channel that lets water into the cell. The bacterial cell swells with water and eventually lyses (bursts). See Figure 5-12.

Activation

Complement Protein

Complement Protein

Figure 5. l2 Complement.

Antibodies can bind to specific antigens on the surface of

a

bacterial cell and cod

the cell surface. Once the antibodies have attached to the bacterial ce* phagocytes and/or killer T cells can bind to the terminal portion of the Fc domain of the antibodies and begin to engulf the foreign invader as shor,rrr:mr Figure 5-13. Figure 5-15 Cell surface coating

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Endocrinology 6r Immunology

T Cells

The receptors for T cells are composed of two polypeptide chains, each with a constant and a variable domain. within the variable domain in each polypeptide

chain we find a variable (v) region and a joining (j) region. on one of the polypeptide chain is a diversity (D) region. see Figlre s-r+. Thls is analogous to the immunoglobulin structure in that a great amount of (antigen uiiraingy diversity can be generated from different combinations of the genes that produJe ihese polypeptide chains. Antigen binding site

v*,ur.

fv

Domain 1

il

Lr

V = variable D = diversity 3 = joining

Variable Domain

Constant Domain

T Cell Membrane

Cytoplasm

Figure 5-14 T

Ce11

receptor.

Cytotoxic T cells have cell receptors like the one shown in Figure 5-15. If a virus infects a host cell, then that virus will begin to take over tie host's metabolic machinery. As this happens some of the iiral antigens are transported to the surface of the host cell where they can complex *ith u Class I MHC protein receptor. class I MHC receptors are found on almost every one of our celli.

MHC I receptor

o-

&;

MHC I receptor antigen complex

r-)

i)*:r'1 ,Y{ \r_.

f.-dffi^

Cytotoxic T cell

Figure 5- l5 lvtotoxic T Cell interaction with MHC I receptor.

-he antigen-Class I MHC protein compiex is recognized by the cytotoxic T cell. 3inding occurs between the cytotoxic T cell and the antigen-bearing host cell as =hown in Figure 5-15. The cytotoxic T cells induce lysis in the host cell contain Jopyright

@


247


Biology


T Cells

the viral particles in order to prevent more progeny viral particles from spreading.

Immature helper T cells recognize macrophages which have presented an antigen on their Class II MHC protein receptors as shown in Figure 5-16. Binding induces the macrophage to synthesize and reiease interleukin-l which acts on the immature heiper T cell and causes it to synthesize and release interleukin-2. Interleukin-2 further stimulates the immature helper T cell to proliferate into a mature helper T ceil. The mature form of the helper T cell secretes interleukin-2 which can activate cytotoxic T cells, B cells, and more helper T cells. MHC II receptor

2a 50> dE O

2-

OI)

antigen complex

r-)

MHC tr receptor

H,r ry Vnal

HeIper T cell

/

poteins-\r1

Activates:

cytotoxic T cells, B cellsInterleukin-

helper T cells 1

Figure 5.16 Cytotoxic T Cell interaction with MHC I receptor

.r*



Biology

Endocrinology

Er

Immunology

tlumoral & Celular Immunity

ffi* Generalized Review

-

€t's summarize the basic events in humoral and cellular immunity. The :r:nbers in these steps will correspond to the numbers in Figure S-17. 7.

A virus enters the body by the blood and is engulfed by a macrophage.

on the surface of the macrophage II MHC receptors.

are Class

I MHC receptors and

Class

A Class II MHC receptor presents the virar antigen to the receptor of a

J.

helper T cell. t-i

This causes the macrophage to release interleukin_l

_i.

IL-1 stimulates the helper T celis to proliferate.

(fl-l).

The helper T cell is stimulated to release IL_2 which enhances proliferation of helper T cells.

{i.

A B cell with

a Class

cell.

rI MHC protein presents virar antigen to heiper T

IL-2 released from herper T celr stimulates B ceils to proriferate.

B cells produce plasma cells and memory B cells. Memory B cells 'remember" antigen and proliferate faster iuring a future inlasion of

!B

the same virus. !3. I-llt-

Plasma cells secrete antibodies specific for the viral antigen.

The.antibodies respond by direct brock, comprement, and cell surface

coating.

rL-2 from the helper T cells stimulates cytotoxic T cells which have bound to the class I MHC protein-antigen complex of an infected cell to lyse that infected

cell.

:ir

.

L-i.

lnterferon is secreted by the infected host cell and acts on the cytotoxic T cell to help enhance the immune response.

n4-

Cytotoxic T cell aiso make memory T cells (not shown) which proiiferate faster during a future invasion of the same virus.

ht O by The Berkeley Review

249

will

Biotogy


Humoral & Cellutar Immunity

t

# \\

)

(rr/-

\,",!liJ,"" Y /

Lf"_\ )" Y

\,v

Direct Block

6",;N

\u

Complement

v

{YP",

/\. Antibodies (10)

Cell Surface Coating

Mast

V Cells

s s

Histamine

Leaky Endothelial Cells

Figure 5.17 Review of humoral and cellular immunity.

Copyright

@



Endocrinology & Immlrlrology To Go 15 Passages

I03 Questions

Time forAll Passages Taken Together as a practice Exam 125 Minutes

Passage Titles

I. U. IIII. IV. V. VI. VIII. VUI. IX. X. XI. XII. X[I. XIV. XV.

Antibody Structure Acquirer{ Imntunint Major Histocompatibility Complex (MHC) IgA Antibody Experiment Complement System MyastheniaGravis/AutoimmuneDisease Giucotu, Glucagon, & Insulin Calcium, PTH, Calcitonin, & Calcitriol Erythroblaslosls Fetalis Septic Shock Calcitonin & Osteoporosis Vertebrate Immune System Insulin Receptor Vitamin D3, PTH, & Calcitonin Sepsis Syndrome

Questions | -7

8-14

T5 -21 11 11 LL--I

28-33 34-4A

4t-46

47 -54 55-61

62-61 68-15 76-83 84-90 91-96 97 - r03

Suggestions The passages that foliow are designed to get you to think in a conceptual manner about the processes of physiology at the organismal level. If you have a solid foundation in physiology, many of these answers will be straightforward. If you have not had a pleasant experience with the topic, some of these answers might appear to come from the void past the Oort field of the solar system.

Pick a few passage topics at random. For these initial few passages, do not worry about the time. Just focus on what is expected of you. First, read the passage. Second, Iook at any diagrams, charts, or graphs. Third, read each question and the accompanying answers carefully. Fourth, answer the questions the best you can. Check the solutions and see how you did. \ trhether you got the answers right or wrong, it is important to read the explanations and see if you understand (and agree with) what is being explained. Keep a record of your results.

After you feel comfortable with the format of those initial few passages, pick another block of them. Be aware that time is going to become important. Generally, you wiil have about

passages and try

minute and 15 seconds to complete a question. Be a little more creative in how you approach this next above, fine. If not, then try different approaches to a passage. For example, you might feel well versed enough to read the questions first and then try to answer some of them, without ever having read the passage. Maybe you can answer some of the questions by just looking at the diagrams, charts, or graphs that are presented in a particular passage. Remember, we are not clones of one another. You need to begin to develop a format that works best for you. Keeping a record of your results may be helpful. 1

group. If you feel comfortable with the outline presented

The last block of passages might contain topics that are unfamiliar to you. Find a place where the level of distraction is at a minimum. Get out your watch and time yourself on these passages, either individually or as a group. It is important to have a feel for time, and how much is passing as you try to answer each question. Never let a question get you flustered. If you cannot figure out what the answer is from information given to you in the passage, or from your own knowledge-base, dump it and move on to the next question. As you do this, make a note of that pesky question and come back to it at the end, when you have more time. When you are finished, check your answers and make sure you understand the solutions. Be inquisitive. If you do not know the answer to something, look it up. The solution tends to stay with you longer. (For example, what is the Oort field?)


Section V Estimated Score Conversions Scaled Score

>72 10-11

8-9 7

6 5

<4

Raw Score

- 100 79-85 65 -78 59-64 54-58 48-53 0-47

86

Biology Passage

I

Antibody Structure

(Questions 1-7)

Immunoglobulins fall into five classes (isotypes). The approximate percentage of total antibody in the serum immunoglobulin pool is indicated in Table 1.

In order to protect themselves from pathogens, animals have developed an array of protective mechanisms within fieir immune system. Immunity can be divided into two ivpes: innate (nonspecific) and acquired (specific).

Tabie

Antibody Classes 7o

IgG

Innate immunity affords protection against a multitude

IgA IgM

:ecognition of foreign macromolecules. The skin, mucous :rembranes, and enzymes in secretions are all examples of nonspecific immunity.

of plasma Ig

70-75 15-20 10

IgD IgE

= 0.005

In humans IgG is the only antibody that can cross the placental membrane. IgA is primarily found in external secretions such as tears, saliva, colostrum, and milk. IgM is the first antibody produced by rhe plasma cells during an tmmune response and is the predominant antibody produced by the fetus during development. IgD has an

Acquired immunity enhances the protection provided

:''' innate immunity and is the result of an initial encounter îth a foreign macromolecule (immunization). If an -nmune response is activated, the agent inducing that :.sponse (an immunogen) will initiate a series of reactions :at lead to the production of antibodies (proteins) that are

.:ecific against the inducing agent. The ability of the ,:ducing agent to combine with an antibody is calied

unknown function, but since it is found as a surface protein in many B cells, it may play a role in B cell differentiation into a more mature form like plasma cells or memory B cells. IgE is thought to confer immunity to certain parasites, such as worms. IgE, can also bind to basophils and mast cells where, in mast cells, it promotes

;.trigeniciry, and an agent with this characteristic is called a recovery from a disease the measles are examples of specific immunity. --i.e

::, antigen. Vaccinations and

-\ntibodies are globuiar glycoproteins that participate the immune response. These immunoglobulins (Ig) are

the release of histamine, a chemical that causes piripheral vasodilation and increased capillary permeability, leading to anaphylactic reactions.

::3sent in the y-globulin component of blood serum and

':: produced by plasma cells. All antibodies are related . ,: another through a common motif (Figure 1).

l:

,l!gtyp"

of different pathogens and is not dependent on specific

.:

Passage I

to

The fragment of the antibody that binds anrigen is the antigen binding domain. The binding occurs between the

paratope

of the antibody and the

epitope (antigenic

determinant) of the antigen. An antigen may possess just a single epitope on its surface, making it a unideterminant, univalent antigen. Antigens may also be unideterminant and multivalent, multideterminant and univalent, or multideterminant and multivalent. Antibody-antigen interaction can result in a variety of outcomes, including precipitation and agglutination. These complexes can be dissociited by high salt concentration or either high or low pH.

1.

nonspecific immune system?

Figure I

A. B. C. D.

Key for IgG

C V L H Fab Fc I ---!-

= = = = = = = =

Which of the fbllowing is NOT a componenr of the

Constant region

Variable region

Light chain Heavy chain

2-

antigen binding fragment

Which of the following types of immunity would result from a vaccination?

crystailizable fragment

Disulfide bonds

A.

Papain cleaves

B.

c.

Pepsin cleaves

D.

: n nght

@

Keratinized and epidermal cells. Beating of epithelial cell cilia. Placental transfer of antibody. Normal body temperature.


293

Active immunity that is naturally acquired. Passive immunity that is artificially acquired. Active immunity that is artificially acquired. Passive immunity that is naturally acquired.


Antibody Structure 3.

The binding of antibody to antigen occurs through all of the foliowing EXCEPT;

A. B. C. D.

4.

A paratope can best be described

A. B. C. D.

5.

Covalent bonding. Hydrogen bonding. Hydrophobic forces. Van der Waals forces.

7.

Lysozyme inhibits peptidoglycan synthesis in the cell wall of gram negative bacteria. Which of the following immunoglobulins would act agonistically with lysozyme?

A. B. C. D.

as

Passage I

IgG

IgA IgM IgE

that area ofthe:

antigen which reacts with the antibody. antibody designated by the Vg regions of the Fab.

antigen which exhibits a multideterminant and univalent expression. antibody which has similar dimensions as the antigenic determinant.

Papain, a protease extracted from papaya plants, cieaves primarily on the carboxyl side of lysine and arginine amino acid residues. Treatment of IgG with papain before IgG is exposed to a unideterminant, multivalent antigen will result in:

A. B. C.

cross-linking, because the antigen possesses just one type of determinant but many of them. an Fc fragment and a bivalent Fab fragment, which can participate in cross-linking. antigens that are not cross-linked, because each Fab fragment is monovalent.

D. two monovalent Fab fragments,

which can

participate in cross-linking.

6.

Treponema pallidium, an organism that causes syphilis, can cross the placental membrane and enter

into the fetal circulation. Which of the following immunoglobulins would be expected to increase in concentration in the blood serum ofthe fetus?

A. B. C. D.

Copyright

IgG

IgA IgM IgE

@



Biology Passage

II

Acquired Immunity

(Questions 8-14)

10.

The following diagram indicates the primary crosses the blood vessels easily, while pentamer, does not cross as well.

acquired actively or passively. Active immunity occurs rvhen an organism is attacked by a foreign substance or a microorganism (an antigen). Antibodies and specialized ivmphocytes are produced by this exposure that confer a 'memory" when the foreign agent attacks again at a later rime. Passive immunity occurs when an organism receives pre-formed antibodies from another organism.

3 !

o!

^.=

0 7 t421 28 35 42

Which of the following statements is FALSE?

B.

C. D.

Yes, receiving antibodies confers permanent immunity. Yes, making antibodies confers permanent immunity. No, receiving antibodies confers transient immunity. No, making antibodies confers transient immunity.

D.

-:pyright

does

this confer on the first individual?

A. B. D.

Artificially acquired acrive immunity. Artificially acquired passive immunity. Naturally acquired active immunity. Naturally acquired passive immunity.

12. The first vaccine by Edward

Jenner

in

1798

protected against smallpox. He used a related strain

measles in this situation. The newborn child quickly made antibodies to the measles virus. The cousin provided antibodies by kissing the newborn child. Paternal antibodies provided immunity to measles in this situation.


Before the primary exposure, there are no antibodies to a particular antigen. Both IgG and IgM remain at high levels following a "booster shot". IgM provides the primary response to an antigen.

individual. What type of acquired immunity

Maternal antibodies provided immunity to

@

for an adult exposure to

11. A susceptible adult individual received pre-formed antibodies isolated from the serum of in immune

newborn child is accidentally kissed by her cousin, who is contagious with measles. Why does the newborn stay healthy?

C.

"booster shot"

tetanus promotes formation of mainly IgG.

C.

B.

A

A.

q. A

A.

49

Time (days)

When a person is passively immune to a foreign agent, is this a permanent immunity?

D.

100

'-Elo

to the chicken pox virus, develop a rash, and beiome rmmune to chicken pox for the rest of his life. This :rrmunity is actively acquired through a natural means. Conversely, a new chicken pox vaccine would provide tnificially acquired active immvnity. Vaccines doiontain rricroorganisms that may be dead or weakened. The ';ubstances in vaccines stimulate an active immune :esponse. An example of passive natural immunity is the naternal provision of antibodies to the fetus via the blood md to the neonate via breastmilk. The baby has an rmmature immune system, and these antibodies function :or several months as the baby's system matures. An :rample of passive artificial immunity is receiving "atibodies from a rabbit that was immune to a spidei - enom, as a treatment for a spider bite.

C.

6

IgM,

e7 g>'

Acquired immunity may be achieved through natural or artificial means. For instance, a person may be exposed

B.

and

secondary response to an antigen. IgG, a monomer,

Acquired immunity is the resistance to disease that an organism develops during its lifetime. Immunity may be

A.

Passage II

of

microorganisms that caused cowpox as

a

vaccination against the microorganisms of smallpox. What type of immunity was conferred in this casl?

295

A. B.

Artificially acquired active immunity. Artificially acquired passive immunity.

C. D.

Naturally acquired active immunity. Naturally acquired passive immunity.


a

Biology 13.

Acquired Immunity

Passage

Blood serum is subjected to electrophoresis in order to separate the proteins. Antibodies are present in the gamma globulin fraction. The following diagram shows serum proteins following gel electrophoresis:

Direction of migration

Cathode

Anode

(-)

(+)

B Globulins

Albumin

In this diagram, which protein is the largest,

based

on its migration pattern?

A. B. C. D.

albumin

a-globulin B-globulin y-globulin

14. If

a newborn is orphaned at birth, which process would provide more antibodies for the child?

I.

Feed the chiid breastmilk from another nursing

mother.

II. Seclude the child in a sterile hospital unit. IfI. Vaccinate the child immediatelv to

all

childhood diseases.

A. B. C.

D.

Copyright

i only I and II only II and III only I, II, and III

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The Berkeley Reviev Specializing in MCAT Preparation

Biology Passage

III

Major llistocompatibility Complex (MHC)

(Questions 15-21)

Passage III

Endosomes are membrane vesicles that often contain surface proteins and their associated ligands. Endosomes are known to contain proteases, which are beiieved to degrade the invariant chain and allow the molecule to bind peptide. However, a small part of the invariant chain (known as CLIP) remains at the binding cleft until a DM

Transplantation antigen proteins expressed on the of a cell and recognized by the immune system are encoded by the major histocompatibility complex (MHC) genes. The MHC genes are of two fundamental types, class I and class II. While mouse and human populations carry more than one hundred forms of the molecules, only between three and six of each class are surface

molecule (an MHC class

Il-like molecule)

enters the

endosome and binds CLIP, actively removing it from the MHC class II molecule. At this time, the MHC class II

molecule, bound with peptide,

expressed.

is placed on the cell

surface-

Both classes of MHC molecules are involved with antigen processing, which includes the ingestion of antigens, the fragmentation of antigens into peptides, and

15.

the binding of these peptides to MHC molecules. The formation of the MHC-peptide complex is a critical event in the effective elimination of intracellular parasites. The peptides associated with MHC class I have invariably

The amino acid sequence of both the MHC class I and MHC class II molecules should show:

A. B. C. D.

been found to originate from a cell's own proteins, while the peptides found bound to MHC class II are normally located on the outer membrane.

The MHC class I protein usually binds peptides that are eight to nine amino acids long. The two ends of the peptide chain bind to discrete binding sites located in the cleft of the MHC class I molecule (Figure 1). The binding cleft of an MHC class II molecule is similar in shape to that of an MHC class I molecule, but the MHC class II molecule usually binds peptides in the middle of the cleft (Figure 2).

hydrophilicregions. hydrophobicregions. both hydrophobic and hydroapathetic regions. both hydrophilic and hydrophobic regions.

16. On the surface of a cell infected by majority of MHC class I molecules

a virus,

the

have bound

peptides that originate from:

A. the virus. B. akillerTcell. C. a B cell. D. a host cell.

MHC ciass

II

17.

molecule

A. B. C. D.

Membrane

,\

MHC class II molecules have been shown to assemble in the endoplasmic reticulum. However, immediately after their synthesis, the MHC ciass II subunits associate with a

A. B. C. D.

class II molecules and causes the MHC class II molecules to leave the Golgi complex and fuse with endosomes.

Berkeley Review

are formed by which

of the following

processes?

invariant chain inhibits peptides from binding to the MHC

@ by The

Shine-Dalgarnosequence. signal peptide sequence. pyrimidine-richsequence. purine-richsequence.

18. Endosomes

third molecule known as the invariant chain. The

Copyright

The two protein subunits that constitute the MHC class II molecule are MOST likely to contain a:

297

Megacytosis Exocytosis Endocytosis Transcytosis


Biology

Major llistocompatibitity Complex (MflC)

19. The protein DM is structurally

A. B. C. D.

20.

Passage III

similar to:

the invariant chain.

MHC class lI. CLIP.

MHC class I.

Which of the following statements is true regarding peptides bound to MHC class I and MHC class II proteins?

I. Peptides bound to MHC class I should exhibit

II.

2I.

a greater size variation. Peptides bound to MHC class a greater size variation.

II should exhibit

III.

The amino acids binding to a cleft should be conserved despite the variety of peptides the cleft can bind.

A. B. C. D.

I only II only I and III only II and III only

Which of the following is LEAST likely to be a defense mechanism used by a pathogen to deter the antigen processing system?

A. B.

Suppression

of MHC molecules early in

an

infection

Production of molecules that bind to MHC I molecules in the ER and prevent cell

class

C. D.

Copyright

surface expression

Production

of a

transcription factor that of the MHC

increases the transcription rate gene early in an infection Suppression of the DM molecule

@ by

The Berkeley Review

294


Biology

IgA Antibody Experiment


Passage

The graphs in Figure 2 represent the numb,er o: trg,\ secreting cells in mouse lung tissue I and I rl eek" following exposure to a virus carrier alone or a r-irus canier plus the IL-6 gene.

In the following experiment, researchers studied the role of the cytokine, interleukin-6, as a factor in the response of immunoglobulin A (IgA) to foreign molecules in mice. IgA is the antibody group that is released from epithelial surfaces in secretions such as saliva or breast milk. IgA often represents a first line of defense against

22.

invading pathogens.

L Mouth

II.

The IgA responses of mice unable to make IL-6 (IL-6) and control mice (IL-6+) were studied. Ovalbumin, an egg

Immunoglobulin response is depicted in Figure q)

40 30

a_)

20

O

A.

1.

50

C)

10 (-)

0

B.


C.

II

D.

I, II, and

IgM

IgA

IgG

Figure

IgE

A.

1

Experiment 2

I

IL-6- and IL-6+ mice were immunized with a virus construct carrying the mouse IL-6 gene. The virus was theorized to insert the IL-6 gene into the DNA of the host

C.

D.

cells it infected.

3

and

23. Which of Figure

Lr

Small intestine

IU. Urethra

protein, was added to the mouse intestinal mucosa.

o

Which of the following tissues are lined by epithelial tissue?

I

Experiment

namgn m

III only

III

the following statements is TRUE of

1?

IL-6- mice produce higher ievels of IgA than IL-6+ mice. IL-6- mice produce lower levels of IgA than IL-6+ mice. The presence of IL-6 did not affect the production of IgA. IL-6- mice respond strongly to ovalbumin.

100

Q9 bo ri -x

IL.6+ carrier alone

Q(

With virus

0!

LA \C)

IgA is composed of what type of molecules?

With virus carrier plus IL-6 gene

A. B. C. D.

;1 O

9n

t2l2

Amino acids Fatty acids Sphingomyelins Phospholipids

Weeks following innoculation of mice

3o-

With virus carier plus IL-6 gene

1oo

Which statement is TRUE of Figure 2?

Ug' :9q

A. The virus carrier

OO (.) e

v)s

B.

;aO

.tfF!

Qh

€o

alone transformed both

strains of mice. Restoring IL-6 in the IL-6- mice improved IgA

production.

1212

C.

IL-6- mice were hypersensitive to the virus

D.

IL-6+ mice did not respond to the virus carier.

carrier.

Weeks following innoculation of mice

Figure 2 ti

Copyright O by The Berkeley Review iti

&

299


Biology 26.

IgA Antibody Experiment

Passage IV

The researchers did not report IL-6 concentrations in the blood. Why is this the case?

A. B. C. D.

Cytokines are hormones and act while passing through the entire circulatory system. Hormone concentrations cannot be measured in blood. IL-6 never leaves the cell that produces it. Cytokines are local hormones and often act without passing through the entire circulatory system.

27.

What term refers to the medical alteration of genes to correct an inherited or acquired disease?

A. B. C. D.

Copyright

Gene therapy

Vaccination Geneticimmunization Pleiotropy

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500

The Berkeley Kevier Specializing in MCAT Preparatiot

Biology Passage

Complement System

V (Questions 28-33)

Complement was given

29.

its name because it

D. 30.

A. B. C. D.

Complement consists of about 20 interacting proteins. The components involved in reactions are known as C1C9, factor B, and factor D. The rest of the proteins are involved in the regulation of this system. These proteins circulate in the blood in an inactive form unless aitivated directly by an invading microorganism or indirectly by an immune response. The final result of activation is assemblage of the late complement components (C5-C9) into a membrane attack complex.

31.

lymph nodes. bone marrow. thymus. spleen.

The classical pathway is usually activated

A. B.

polysaccharides on a microbial envelope. There are two distinct pathways of eariy component activation. Cl, C2, and C4 belong to the classical pathway and is triggered by antibody binding. Factor B and D belong to rhe alternative pathway and are triggered by microbial polysaccharides. Both pathways wiil act on C3, a central component in the complement system.

C. D.

32.

by IgG or

antigenic determinant. variable region of the antibody. cell membrane. constant region of the antibody.

As stated in the passage, protease cleavage acts to expose

a membrane binding site on the

larger

fragment. The most likely reason for this is to:

4. B.

All early components

and C3 are proenzymes that are activated by each other through proteolytic cleavage. As each proenzyme is cleaved, it is activated to generate a serine protease which cleaves the next proenzyme in the sequence. Each activated enzyme cleaves many molecules of the next proenzyme in the chain. The cleavage normally exposes a membrane binding site on the larger fi'agment and liberates a small peptide fragment into the blood stream. The C3 molecule is eventualiy cleaved, with its larger fragment binding both the cell membrane and C5. Activation of C5 will initiate the spontaneous assembly of C5 through C9, forming the membrane attack complex.

avoid precipitation of complement proteins.

C.

have the larger fragment act as a diffusible signal in the bloodstream. confine complement activation to the cell

D.

inhibit the next reaction in the

surface where it began.

cascade

sequence.

33.

During the proteolytic complement cascade, ser,eral small biologically active fragments are generateci. One of the end results of these molecules' activir;, is an increase the permeability of local blood vessels. Which is the most likely explanation for such an increase?

A.

passage, complement may

normally help:

B.

produceantigen-antibodycomplexes. destroyantigen-antibodycompiexes.

C.

solubilizeantigen-antibodycomplexes.

D.

precipitateantigen-antibodycomplexes.


a

to the:

The early complement components are activated by

@

the

igM antibodies bound to antigens on the surface of a microorganism. The Cl complex mosr likely binds

either antibodies bound to a microorganism or by

Copyright

The production of antibodies used in

complement process would be greatly affected b;' disease of the:

skin, joints, and brain causing the destruction of tissue.

A. B. C. D.

focuses the complement system away from cel1 membranes. does not take place in the alternate pathway. provides a means of amplification, ultimatelv leading to many MACs. uses serine proteases at all serine residues in a protein.

B. C.

and attracting phagocytic cel1s to the site of infection. Individuals who are complement deficient also suffer from immune complex diseases, in which antibodyantigen complexes precipitate in small blood vessels in

28. According to the

The proteolytic cascade described in the passage:

A.

complements the action of antibodies and is the principal means by which antibodies defend vertebrates against most bacterial infections. A system of serum proteins are activated to form a membrane attack complex (MAC) which forms holes in microorganisms. Complement also amplifies the defense system by dilating the blood vessels

Passage V

501

These molecules stimulate the histamine from T lymphocytes. These molecules stimulate the histamine from basophils. These molecules stimulate the histamine from macrophage. These molecules stimulate the histamine from erythrocytes.

secretron of

secretion of secretion of secretion of


Biology

Myasthenia Gravis/Autoimmune Diseases


Passage

35.

Myasthenia gravis is a rare, chronic, neuromuscular It is characterized by skeletal muscle weakness and fatigability in response to repeated contraction. Resting partially restores muscle stre;gth. The muscies of the eyes, face, jaw, and neck are usually affected first. As

Passage VI

Shown below is a diagram of a neuromuscular junction. Which number indicates an acetylcholine receptor?

disease.

the disease progresses, weakness spreads to rhe fn severe cases, all the

extremities and the diaphragm. muscies are weakened.

Research indicates myasthenia gravis

is

an

autoimmune disorder in which antibodies are produced to the acetylcholine receptor that is present at the neuromuscular junction. Antibodies to the acetylcholine receptor have been found in 85Vo of patients with

generalized myasthenia. Although the mlchanism for antibody production is unclear, one hypothesis is that certain thymus cells that resemble musci" (myoid cells) are.damaged by a virus. The virus may have a molecular region that mimics part of the acetylcholine receptor, such as-the herpes simplex virus. A virus may damage myoid

cells so that antibodies are produced against

tt"-

By whatever mechanism, the viraf infection

antibody production.

a a

I ol

air"ttty. induces

The actual interaction between the antibody and the receptor is not fully understood. The antibody may block the receptor, it may cause faster receptor breakdown, or it may promote complement-mediated damage.

A. I B.II c.m D.IV

Autoimmune diseases as a whole are relativelv In the following table is a list of soml autoimmune diseases and the antibodies produced in the common.

36.

Which of the following is NOT an example of autoimmune disease?

disease state:

A.

Antibody Against

Disease Type I

Destroys B cells

Graves' disease

Thyroid stimuiating

Stimulares TSH

hormone receptor

receptor on thyroid

Multiple

Myelin basic protein

Disrupts

sclerosis

(hypothesized)

myelination

Glomerulonephritis

34,

Effects

B celis of pancreas

diabetes

Basement membrane

B. C. D.

37.

Copyright

B.

Destroys a variable

C. D.

of glomular capillaries number of glomeruii


Which of the following is NOT a symprom of A.

38.

Increased metabolic rate. Weight loss.

Lethargy.

Hyperactivity.

Secretion

of which of the following from

the

pancreas is halted by antibodies to the B cells?

An inhibitor of acetylcholinesterase. An immunostimulant. A paralytic agent, like curare. An inhibitor of acetylcholine synthase. @

Type II diabetes. Addison's disease. Myasthenia gravis. Graves' disease.

Graves' disease?

Which drug could be given to counteract the effects of the antibody produced in myasthenia gravis?

A. B. C. D.

an

A. B. C. D. 302

Glucagon.

Insulin. Bicarbonate. Digestive enzymes.


Biology

Myasthenia Gravis/Autoimmune Diseases

Passage VI

How can a viral infection lead to an autoimmune disease?

II.

III.

A. B. C.

D.

40.

The virus resembles a "self'molecule, leading to antibodies that cross-react with other body molecules The virus damages a cell so that unrecognized cell proteins are released, causing antibody production The virus resembles a "non-selfl' molecule, leading to antibodies that cross-react with other body molecules

I only I and II only II only II and III only

Often patients with autoimmune diseases are treated

with corticosteroids to reduce immune responses. Which organ in the body produces corticosteroids? A. B. C.

D.

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Pancreas. Adrenal gland. Pituitary gland. Thyroid gland.

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Biology Passage

Glucose, Glucagon, & Insulin

VfI (Questions 4l-46)

Passage Vtr

41. During times of stress

the importance of having

adequate levels of glucose, for utilization as energy, is dependent on hormonal secretions from the pancreas. Which choice below will BEST readv the

The regulation and metabolism of carbohvdrates in the

body are controlled by the pancreas. The endocrine component of the pancreas, the islets of Langerhans, are specifically responsible for carbohydrate control. These small clusters of cells imbedded within the exocrine portion of the pancreas contain peptides with specific

body for stressful situations?

A. B.

hormonal activity.

C.

Glucagon, secreted by the a cells (or A cells), liberates glucose from storage areas in the body, stimulates glucose production, increases lipid concentration in the blood stream by releasing free fatty acids, and increases the production of ketones. u cells are stimulated to secrete glucagon during increases in plasma amino acid levels, cortisol secretion, exercise, and sympathetic nervous system stimulation. Inhibition of glucagon is promoted by

D.

increases in plasma glucose, ketone, free fatty acids, insulin, and somatostatin levels.

Increased levels of glucagon and insulin.

Increased levels of glucagon and decreased levels of insulin. Decreased levels of glucagon and decreased levels of insulin. Decreased leveis Ievels of insulin.

42. Ketosis is

of glucagon and

increased

developed from an increase

in

the

conversion of free fatty acids to ketone bodies. These ketone bodies are an important source of energy in times of fasting. However, prolonged

Insulin is secreted by the B cells (or B cells) and functionally is important in increasing the storage of

ketosis will lead to a plasma acidosis due to liberated hydrogen ions from ketone bodies. A patient witb

glucose, fatty acids, and amino acids in the cells of target tissues. Furthermore, insulin decreases the release of glucose, mannose, amino acid, and glucagon plasma levels. Many intestinal hormones also stimulate insulin secretion. Parasympathetic stimulation of the cells will B increase insulin secretion, while sympathetic stimulation

acidosis

will develop

shortness

of

breath.

dehydration, hypervolemia, and hypotension. In severe cases the acidosis and dehydration will depress consciousness to the point of coma. Which choice below will lead to the development of ketosis

will inhibit secretion.

and acidosis?

Somatostatin is a peptide secreted by 5 cells (or D cells) in the islets. Somatostatin inhibits the production of both insulin and glucagon, and it acts as a regulator of

islet secretion.

The effects ofinsulin and glucagon target very specific

regions of the body where their cellular actions occur. Insulin's effects are generally associated with muscle and adipose tissue, leukocytes, fibroblasts, and mammary glands. Insulin does not directly affect brain and kidney tissue, intestinal mucosa, and red blood cells. Glucagon's effects are targeted mainly on the liver.

43.

A. B.

Increased levels ofglucagon and insulin. Increased levels of glucagon and decreased

C. D.

Decreased levels of glucagon. None of the above.

levels of insulin.

Symptoms reported by a patient include weakness. dizziness, confusion, and hunger. Furthermore, somr

tremors, palpitations, and nervousness are alsp reported. These last symptoms are characteristic d hypoglycemia. The development of these sympron$ is due to:


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abnormal increases in insulin secretion.

abnormal decreases in insulin secretion. abnormally high levels of plasma glucose. abnormally low levels of free fatty acids.


Biology 44.

Qlucose, Glucagon, & Insulin

Passage

YI

What pancreatic hormonal response is expected after a heavy protein intake?

A. B.

c. D.

Increased levels of insulin and glucagon. Increased levels ofinsulin and decreased levels of glucagon. Decreased levels ofinsulin and glucagon. Decreased levels of insulin and increased levels ofglucagon.

45. From the data reported on glucose utilization of tissue and responsiveness of tissue to insulin, it may

be determined that the main function of insulin secretion is:

A. B. C.

D.

46.

increasing glucose uptake in the brain. increasing glucose release from the liver. access and storage of glucose in cells of the peripheral tissues. increasing glucose loss in the kidney.

Patients diagnosed as having diabetes mellitus are said to be in "a state of starvation in the midst of plenty." This analogy refers to:

A. B. C. D.

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extracellular glucose excess. extracellularglucosedeficiency. decreased effects of insulin on intestinal mucosa uptake of glucose. deficiency of fatty acids in neural tissue.

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Biology Passage

,

VIII

Calcium, PTII, Calcitonin, Calcitriol

(Questions 47-54)

Calcitonin acts to inhibit osteoclast activity and reduce bone resorption. This hormone also inhibits talcium and

Calcium and phosphate both play an important role in

the mineralization of bone

in vertebratei. Calcium

phosphate reabsorption in the kidney and increases the excretion ofthese ions in the urine.

is

obtained from the diet and is largely absorbed at the level of the intestine. The intracellular concentration of calcium is about 10-7 mol/I- while the extraceliular concentration of calcium is about l0-3 mol/L. Blood calcium levels are primarily determined by bone metabolism and urinary

47.

excretion.

Calcium metabolism is regulated through the actions of parathyroid hormone (PTH), calcitonin, and vitamin D3 (choiecalciferol). PTH is synthesized and secreted by the parathyroid glands, usually located in the central region of the thyroid gland near the trachea. Calcitonin is synthesized and secreted from parafollicular (or C) cells located in the thyroid gland. Cholcalciferol is synthesized

Administration of PTH leads to changes in plasma calcium and phosphate concentrations. Whicli of the following graphs BEST represents these changes? A. J o

p o

0 O

must first be hydroxylated in the liver and

B. J

increased by PTH.

o tr

p

Bone is composed of an organic matrix consisting of collagen fibers and a ground substance composed of extracellular fluid and proteoglycans. The collagin fibers

o

o 6

help to give bone its great tensile strength, white tne ground substanca helps to control the deposition of

t-

0 0

Time (hours)

PTH added C. ..1

The collagen matrix and ground substance is laid down

o

o

E

by bone cells called osteoblasts. The tissue which is formed, called an osteoid, can enlrap some of the osteoblasts. Entrapped osteoblasts are called osteocytes.

Calcium

E

As the bone grows, calcium salts precipitate on the collagen fibers. Bone is also undergoing resorption by

0 fr

PTH causes calcium absorption from bone by

fc p o

Phosphate

(-)

cells called osteoclasts.

Time (hours)

PTH added

stimulating osteociastic activity and transiently inhibiting osteoblastic activity. At the level of the kidney, pTH increases calcium absorption in the distal tubules and collecting ducts and greatly decreases the reabsorption of phosphate at the proximal tubules.

D. J o

'6 p

6'

The activated form of cholecalciferol (1,25-(OH)Z-D/

o

()

has target receptors in the intestine, bone, and kidney (to

name but a few tissues). In the intestine this hormone promotes absorption of calcium and phosphate (following as the counterion), while in bone it promotes resorption of both calcium and phosphate. In the kidneys this hormone promotes the reabsorption of both calcium and phosphate so that little is excreted in the urine. by The Berkeley Review

B

U

calcium salts, like hydroxyapatite (Ca5@O4):OH). The calcium salts help to provide for the great compressional strength found in bone.

@

Time (hours)

PTH added

then hydroxylated in the kidney. The activity of the hydroxylase enzyme in the kidney is regulated and

Copyright

I

O

Q

animals from a photolytic reaction involving UV light and a sterol derivative. In order for this inactive hormone to be

it

o "o

in an inactive form in the skin of

activated,

Passage Vtr

0 G

Time (hours)

PTH added

506


Biology 48.


Passage Vtr

49. Calcium and phosphate absorption in the intestines

Which of the following structures BEST represents vitamin D3?

is stimulated by an increase in:

A.

I. II. III.

1,25(OH)2D3 Calcitonin PTH

A.

I only I and II only III only I and III only

B. C. D.

50.

B.

In order for the secretion of calcitonin to have a greater effect on the concentration ofcalcium ions in the plasma, which of the following statements must

be true regarding osteoclast activity and plasma calcium levels?

A. B. C. D.

C.

51.

Increased osteoclast activity hypercalcemic plasma. Decreased osteoclast activity hypercalcemic plasma. Increased osteoclast activity hypocalcemic plasma. Decreased osteoclast activity hypocalcemic plasma.

coupled with

a

coupled with

a

coupled with

a

coupled with

a

Hypophosphatemic rickets is an Xlinked dominant trait that leads to decreased levels of phosphate reabsorption in the kidneys. Which pedigiee sho*n below BEST represents this disease?

A.

B.

':l1IT

rr+\-u+-l fl-rl [i J-I-J-I

D.

Jopyright

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,,,

u,

C.

D.


Biology


Passage Vtr

52. Familial

hypophospharemia is BEST treated by diet modification and supplying adequate amounts of:

A. B. C. D.

53.

calcium. phosphate. calcium and phosphate. phosphate anA t,ZS1Ofq2O3.

Hypoparathyroidism is BEST characrerized by:

I. Ir. III. IV.

increased increased increased increased

A.

I only II and III only Itr only I and fV only

B. C.

D.

54-

osteoblast activity. osteoclast activity.

neural excitability. plasma calcium concentrations.

In the graph shown below, all of the following statements concerning the relationship between PTH, calcitonin, and calcium are true EXCEpT:

o o c€

o o o o

o

o o

a

:-

5

Total Ca2+ concentration in piasma

A. B. C. D.

PTH is a hypercalcemic hormone. calcitonin is a hypocalcemic hormone.

a positive linear relationship exists between calcitonin secretion and the concentration of plasma calcium.

a positive linear relationship exists between PTH secretion and the concentration of plasma calcium.

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@


The Berkeley Kevier Specializing in MCAT

\

Biology

Erythroblastosis Fetalis


Passage

Erythroblastosis fetalis (EF)

56,

is also known

Which of the following statements is TRUE?

A.

as

hemolytic disease of the newborn. Hemolysis is rupture of red blood ceils, so in hemolytic disease, anemia is

B.

.ommon! due to the lysis of red blood cells. In the EF --ondition, maternal antigens cross the placental barrier, auack proteins on the surface of the red blood cells of the r-etus, and lyse the cells. A very specific set of conditions must exist for this disease to occur.

Passage IX

C. D.

The placenta allows passage of all blood products from the mother to the fetus. The Rh factor is nor a component of the ABO blood group system. The mother makes red blood cells for the fetus

in the placenta. Terminated pregnancies have no effect on future development of EF.

The mother must be Rhesus (Rh) factor negative, the :etus musf be Rh factor positive, and the mother's immune s\ stem must be sensitized to the Rh positive antigen

tlrough previous full-term pregnancy or abortion. The Rh :actor antigen is transmitted as a dominant trait, so that rnly people who are homozygous recessive are Rh factor

57.

What preventive measure couid protect subsequent fetuses il an Rh-negative mother gave birth to an

regative.

Rh-positive fetus?

Roughly 90Vo of the cases of EF result from ':nsitivities to the D antigen on the Rh factor. When Rh :ositive blood enters the circulation of an Rh negative lother, antibody formation against D may be induced. This exposure may be during an accidental infusion, iuring pregnancy, delivery, or during a miscarriage or "nortion. During a first pregnancy, there is usually little

A. B. C. D.

:rchange of fetal and maternal blood, except near the end -

i

Give the fetus a blood transfusion with

Rh_

negative blood.

Treat mother with a set of antibodies directed against the anti-Rh antibodies. Treat fetus with antibodies against the anti-Rh antibodies.

the pregnancy or during delivery.

This time frame does not allow for antibody formation by an attack on the fetal biood cells. The :ioblem lies in subsequent pregnancies. Small amounts of rtigen, even the amount in I mL of fetal blood entering .:e mother's circulation, promote rapid increases in hei .r:i-D antibody titer. IgG is produced, and it can easily :.oss the placental barrier into the fetal blood supply. ;','en in the ideal conditions for EF, sometimes the diieaie : res not manifest, due to variable physiological .

Give the mother a blood transfusion with Rhpositive biood.

--'ilowed

58. An infant wirh severe EF has jaundice, a yellow

coloring due to excess bilirubin, a breakdown product of heme. In what tissue or organ is heme degraded into bilirubin?

A. B. C. D.

::nditions.

Spleen Bone marrow

Liver

All of the above

59. An Rh-positive mother is pregnant with an Rhnegative fetus.

55, A

Rh-negative woman and a heterozygous Rhpositive man have one child together. The woman has never been pregnant before. What is the

-

A.

likelihood that the child wiil be born with EF?

B.

A. B. C. D.

C.

rpyright

l)AVo l5Vo

D.

50Vo

jVo

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309

Will the fetus develop EF?

No, there are no maternal antigens to the Rh factor antigens. Yes, the mother can still make antigens to the Rh factor of the fetus. No, the fetal antibodies protect its red blood cells.

No, but fetal antibodies attack the maternal red blood cells.


Biology

Erythroblastosis Fetalis

Passage IX

60. What variables could affect the severity of hemolysis in an Rh-positive fetus whose mother is Rh-negative?

I. II. Iil.

Amount of blood transferred. Sensitivity of mother to D antigen. Number of pregnancies.

A. I only B. I and II only C. II and III only D. I,II, and IIi

61. Which

of the following clinical signs could

be

consistent with a diagnosis of EF in a newborn?

I. il. ilI.

High levels of bilirubin in the blood. Low levels of hemoglobin in the blood. Increased levels of erythrocytes.

A. B. C. D.

I only I and II only II and III only I, II, and III

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Biology

Septic Shock

Passage X (Questions 62-67)

64. A

defect

in which of the following cells

would

inhibit the production of soluble anribodies?

Septic shock, a disease characterized by hemodynamic derangements and multi-organ malfunction, has generally been associated with a gram-negative infection. However,

A. B. C. D.

it is becomming increasingly clear that gram positive organisms are equally responsible for sepsis. Many of these gram positive ogranisms release molecules known as superantigens. These superantigens can induce

Passage X

Mast cells

Cytotoxic T cells Plasma cells

Erythrocytes

T cell

proliferation without regard to the antigenic specificity of ihe cell.

65. In

A

common sign of septic shock is widespread rctivation of coagulation leading to widespread rntravascular clotting. Microbial products activate Factor -\I[, a molecule involved in b]ood clotting. Activation of :ris factor initiates the intrinsic coagulation pathway and .:so the bradykinin pathway. Bradykinin is a potent ',

would expect to see

A. B. C. D.

asodilator and also increases the permeability of vascular

:ndothelial cells. Cytokines, such as interleukin-1 and tumor necrosis :.;tor, activate tissue factor III. This factor is found on the - iler membrane of macrophage and endothelial cells, and ,

an experiment, a sepsis patient is treated with XII antibodies. After treatment. one

anti-factor

a:

total lack of intravascular clotting. rise in the level of intravascular clotting. rise in the patient's blood pressure. decrease in the patient's blood pressure.

66. Bradykinin

acts to increase the radius of a given blood vessel by a factor of 2. The flow of blood

::mulates the extrinsic coagulatory pathway.

through the vessel should increase by a factor of:

A. B. c. D.

r,l- A Gram-positive cell differs from a Gram-negative

2. 4. 8. 16.

cell in that a Gram-positive cell:

-{.

does not have an outer membrane on its cell

B. C.

does have an outer membrane on its cell wall. contains a thin peptidoglycan layer adjacent to the plasma membrane. contains no peptidoglycan layer adjacent to the plasma membrane.

D.

67.

wall.

The activation of bradykinin most likely results in:

A. hypotension in the patient. B. cytokine release. C. macrophageactivation. D. stimulation of the extrinsic

coagulation

pathway.

\lacrophages destroy microorganisms through:

A. B. C. D.

r'

:.qht

exocytosis, secreting toxins which eventually form membrane attack complexes. xxocytosis, surrounding the foreign particle with a lipid bilayer which protects the host organism. endocytosis, engulfing foreign particles which eventually will fuse with a lysosome. endocytosis, engulfing foreign particles which eventually will fuse with a peroxisome.

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Biology Passage

Calcitonin and Osteoporosis


71.

Calcitonin (CT) is a polypeptide hormone secreted by the parafollicular cells of the thyroid gland in mammals.

32 amino acids make up the hormone, and a disulfide

CT works as an antagonist of parathyroid hormone (PTH) In response to small increases in plasma calcium, CT is released and acts on the kidnw and bone to decrease the calcium level. In the bone *airix, osteoblasts synthesize bone, and osteoclasts catabolize bone. The

choices is a probable explanation why salmon CT is 30 times more active in humans than human CT?

I. II.

Calcitonin is used pharmacologicaily as a treatment for osteoporosis. Saimon CT is commonly used. Although salmon CT differs markedly from human CT, it is about 30 times more potent when used in non-ailergic humans to treat osteoporosis. Treatment with CT is not without its own side effects. CT treatment for osteoporosis increases

IfI. A. B. C. D.

supplementation to avoid hyperparathyroidism.

Osteoporosis is a disorder ofbone characterized by a decrease in bone quantity, most common in women

Salmon CT attaches more tightly to the CT

I only I and III only II and III only I, II, and III

73. If

a person begins a calcium supplement regimen and doubles calcium intake, whit would be the

following menopause and in elderly men and women. Which of the following conditions would lead to the GREATEST decrease in bone quantity?

response in CT secretion?

A. B. C. D.

Decreased osteoblast activity, decreased osteoclast activity.

Increased osteoblast activity, decreased

B.

Salmon CT is more resistant to degradation by human enzymes. Salmon CT attaches more strongly to the DNA of the osteoclasts. receptor.

plasma PTH and requires simultaneous calcium

A.

Methionine-threonine Methionine-methionine Cysteine-cysteine Cystine-cysrine


main eff-ects of CT are ( i ) inhibition of osteoclasts and (2) a transient increase in urinary calcium and phosphate.

58.

Which two amino acids form a disulfide bridse?

A. B. C. D.

bridge links residues 1 and 7. The entire CT molecule and the. disulfide bridge are required for full biological activity.

Passage XI

CT CT CT CT

secretion secretion secretion secretion

increases. decreases.

remains unchansed. halts completelyl

osteoclast activity. C.

Decreased osteoblast activity, increased

D.

Increased osteoblast activity, increased

osteoclast activity.

74.

osteoclast activity.

69.

What is the most abundant mineral in the human

A. Bn B. Water C. Calcium D. Zinc

75.

Ingestion to avoid allergic reaction.

Injection to avoid hydrolysis.

C.

Ingestion to avoid hydroiysis. Injection to avoid allergic reaction.

What are the symptoms of an allergic reaction it" foreign protein in the bioodstream?

Based on the passage, what is the role of pTH?

A. B. C. D.

A. I only B. I and II only C. III only D. I, II, and III

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312

polypepride, administered?

B.

I. II. IU.

PTH increases plasma calcium. PTH decreases plasma calcium. PTH increases urinary calcium. PTH increases urinary phosphate.

a

A.

D.

body?

70.

How is salmon CT,

Flushing or reddening ofthe skin. Skin welrs (hives).

Difficultybreathing.

The Berkeley Rerieu Specializing in MCAT preparatim

Biology Passage

XII

Vertebrate Immune System

(Questions 76-83)

77.

When certain types of antibodies bind to their tarset

cells, they attract a series of proteins coiiectivJly

In order to protect vertebrates from infection, the im,mune system has evolved both an antibody-based and a cell-mediated response to foreign antigens. B cells, which

known as complement. These pioteins can then form pores which allow small molecules to freely diffuse across the plasma membrane. What effect would this have on a target cell?

originate and develop in hemopoietic tissues (bone

marrow and fetal liver), are responsible for producing and secreting antibodies which bind to antigen particl6s. T celis also originate in hemopoietic tissues but later migrate to and mature in the thymus during early development. Cytotoxic T celis are mainly .esponsible for mounting a cell-mediated defense by directlyiausing the death of infected celis. While B cells can be activat; by the binding of extracellular antigen to special receptors on the plasma membrane, T cells must come in direcicontact r.r,ith

Passage XII

A.

The ceil would die due to its inability to

initiate action potentials. The celi would become hyperpolarized.

B. C.

The cell would lyse due- to an upset water

D.

The cell would shrink due to an upset water

balance. balance.

infected ceils in order to become activated.

How the immune system differentiates between self and foreign antigens has been the topic of much study. Immunologists in the first half of thii century p.opor"d two main theories: Theory

78. Which of the foliowing

I

consistent with

Vertebrates inherit genes that encode receptors ipresent on the surface of B and T cells) that are capable

of binding oniy foreign antigens. The immune

doesn't react against host tissues because lacks the receptors which bind selfantigens.

it

be

L A foreign cell line is injected into a mouse

embryo. Further injections of the cell line into

system

the adult mouse do not elicit an immune

geneiically

response.

il. Transplantation of organs

between monozygotic twins does not result in orsan

Theory 2

The.. immune system

is inherently

ilI.

capable of responding to both self and foreign antigens, but it becomes "tolerant" to self antigens during early

rejection by the immune system.

Cells transplanted between mice which

are

genetically identical (i.e., from the same inbred strain) are tolerated by the new host's immune

development. Since foreign antigens aren't preseni during embryonic stages, the immune system does not deveiop i tolerance to them.

76.

statements would BOTH Theory 1 and Theory 2?

system.

A. B. C. D.

I oniy II only II and III only I, II, and III

According to the passage, removal of the thymus from an adult human would most likely result in:

A. drastically B.

decreased antibody-mediated

lmmune fesponse. drastically decreased cell,mediated immune

79.

response.

C. drastically D.

Copyright

decreased antibody and cell_ mediated immune response. littie or no change in the effectiveness of either type of immune response.

@

by The Berkeley Revierv

Tolerance to self antigens breaks down in the human autoimmune disease myasthenia gravis, resulting in the production of 4ntibodies against the patient's skeletal muscle acetylcholine receptors. Wt ictr of the foliowing is a likely symprom of this disease?

A. B. C. D. .'t^t

Irregularities in heart contraction. Weakness and difficulty breathing. Paralysis of the gastrointestinal tract. Dementia.


Biology 80.

Vertebrate Immune System 83.

Recently activated B and T cells are examined via electron microscopy. Which of the following would be the most likely observations?

A. The B cells have a

mitochondria than the T cells.

B

B.

The

C. D.

endoplasmic reticulum than the T cells. Both the T and B cells lack nuclei. No differences between the two types of cells are revealed at the level of electron

Virally infected cells are usually killed by cytotoxic T cells. The T cells can most likely target:

A. B. C. D.

greater number of

cells have considerably more rough

Passage XII

most types of cells in the body. only cells lining the blood vessels.

epithelial cells only. other blood cells onlv.

microscopy.

81.. A sample of B cells is removed from an adult mouse. A highly radioactive antigen X is added to the B cells, killing the few that bind strongly (<0.0lVo). The remaining B cells are injected into mice whose own B cells were destroyed by irradiation. These mice can now make no antibodies to antigen X but do respond to other antigens. Which of the following can most likely be concluded?

A.

Each B cell is predetermined to bind a specific

B.

B cells can recognize new antigens and "learn" to bind to them. T cells must interfere with B cell binding. Antigen X is normally non-immunogenic.

C. D.

82.

antigen.

Secreted antibodies often act collectively to bind to

large target antigens and cross-link them into insoluble masses which are easily phagocytosed. Which of the following would NOT be consistent with this phenomenon?

A.

Antibodies each have two identical binding

B.

Antigens can be bound by only one antibody at a time.

C. D.

sites.

Antibody binding sites are connected by a flexible "hinge" region. Large antigens contain multiple antibodybinding sites.

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314

The Berkeley Revier Specializing in MCAT Preparatic

Biology XIII

Passage

Insulin Receptor

(Questions 84-90)

86.

The human insuiin receptor is an

integral

transmembrane protein located in the plasma membrane of many celis throughour the body. Studies have shown that a functional receptor consists of two alpha and two

beta subunits which are connected using disulfide bonding. The receptor consists of several domains, inciuding one which binds insulin, a transmembrane binding region, and an intracellular tyrosine kinase

Passage XItr

The response to glucagon is believed to involve the production of cAMp as a secondary messenger. Which of rhe following graphs BEST describes"the sequence of events associated with the binding of glucagon to its receptor? fNote: CA cAMp; p:p = =

protein phosphorylation; CR cellular response.l =

A. C) cg

()

phosphokinase (TPK) segment.

b3 >t

How does ligand binding create intracellular changes . which bring about the effects of the insulin hormone? There is experimental evidence which suggests that the human ,insulin receptor is a insulin activited tyrosine kinase. While many details remain unciear, the binding of

<€o

the hormone leads to both an autophosphorylation of two tyrosine residues on the TpK and u ptorptorylation of cytoplasmic constituenrs. In addition, the binding of insulin to irs receptor is believed to activate phosphohf,ase C, which converts glycan phosphatidyl inositol pfrospiate to glycan inositol phosphate (GIp) and 2_diacylgiyceiol.

o

B. a d !i

O

.: a

I !r/

GIP is known to mimic certain insuiin responses and 2-diacylglycerol, in the presence of elevated calcium ievels, is known to activate protein kinase C. protein

kinase C

is known to influence the activity of

enzymes and metabolic pathways.

Time

other

.-

C. a)

.J

84.

>.9 >t

Which of the following procedures is used on cells to isolate the entire insulin receptor for experimental

studies?

A. B. C. D.

Acidic

Time

wash.

Oxidizing wash.

-

D.

Detergent wash. Reducing wash.

C) dJ

i'H >t

85.

:o <3 o

Through genetic manipulations, the insulin binding region of the insulin receptor (IR) was fused with the kinase region of the epidermal growth factor (EGF)

o

A

receptor (also a ligand activated kinase). The

resulting hybrid receptor was transfected into a cell and the addition of insulin resulted in a functional

EGF kinase. From this experiment, concluded that IR and EGF have similar:

A. ligand binding regions. B. signaltransductionmechanisms. C. secondary messenger cascades. D. effects on carbohydrate, fat,

it

can

be

87. In a type I, insulin

A. B. C. D.

and protein

metabolism.


dependent diabetes, the individual

alflicted wirh the disease most likely

315

has:

a mutated insulin binding site. a mutated membrane

anchoring site.

a normal insulin receptor. an inactive phospholipase C.


Biology 88.

Insulin Receptor

Passage XItr

Which of the following is consistent with the actions of GIP? A. B. C. D.

Inhibition of hexokinase. Inhibition of phosphofructokinase. Stimulation of glycogen breakdown. Stimulation of pyruvate dehydrogenase.

89. In cell #1, the addition of bivalent anti-insulin receptor antibodies induces a response without the addition of insulin. In ce17 #2, addition of a fragment anti-receptor antibody (monovalent) induces no response without insulin. To cell #2, antibodies against the monovalent anti-receptor antibodies are added. This will result in:

A. B. C.

D.

no response. decreased glucose uptake . increased CO2 production. increased insulin binding.

90. To disrupt the dimer in order to study components, a researcher

A. B. C. D.

will most likely

its

use:

oxidizingconditions. reducingconditions. high speed centrifugation. high pH.


316


Biology Passage

t Vitamin D3, PTH, & Calcitonin

XIV (Questions 9l-96)

Passage )ilV

93. PTH and CT secreting cells are found in which of the following glands?

In all vertebrates the maintenance of calcium (Cu2@), phosphate (PO+3e), and magnesium (Mg2o) homeostasis is governed primarily by the vitamin D3 derivative 1,25_ dihydroxycholecalciferol (1,25-(OH)2-D3), calcitonin

A. Pituitary gland

(CT), and parathyroid hormone (pTH).

Humans obtain vitamin D3 either by ingestion or through synthetic mechanisms that are initiated by cells in the epidermis of the skin. Vitamin D3 is hydroxylated first in the liver and then is transported to the kidney where the enzyme 1-hydroxylase converts it to the active form of the hormone, 1,25-(OH)2-D3. The activity of 1o-hydroxylase durin g hypocalcemia and hypophospharemia. 1s -e$119ed 1,25-(OH)2-D3 passes inro a rarger cell ind then into the nucleus where it complexes with a receptor protein that has a DNA binding site. Gene transcripti,on can either be

B.

Adrenal gland

C.

Thyroid gland

D.

Thymus gland

94. Which of the following categories represent the

enhanced or suppressed. Enhancement leads to the synthesis of calcium binding proreins (calbindins) in intestinal villi and crypt cells. Calbindin promotes the uptake of Ca2@ from the intestinal lumen. A similar mechanism allows for the absorption of phosphate and magnesium from the intestinal lumen.

I.

Amine

u. IIr.

Polypeptide

A. B. C. D.

II only and III only II and III only III only

Steroid

I

95. Which of the following conditions besr describes patient with a deficiency in pTH secretion?

I.

PTH is released from the chief cells of the parathyroid glands while CT is released from the parafollicular cells (or C cells) of the thyroid gland. Both hormones have _senes encoded in the short arm of human chromosome l1 and are synthesized as a preprohormone from different

il. ur.

Hypocalcemia Hypercalcemia Hyperphosphatemia

A.

I only

primary RNA transcripts.

B.

PTH acts on bone and on the distal tubule of the iiidney to promote Ca2@ reabsorption and inhibits the reabsorption of PO43e in the proximal tubule of the oJdney. PTH also stimulates rhe synthesis of 1,25-(OH)2_ D3 in the kidney. CT acts ro lower both plasma calcium

D.

III only II and III only I and III only

C.

96. Which of the following curves BEST represents the relationship between blood levels of parathyroid hormone (PTH) and calcironin (CT) in rerms of total

:nd plasma phosphate levels.

plasma Ca2@ concentration levels?

A.

91. Ultraviolet light stimulates vitamin D3 production in

B.

which of the following cell types?

o

c)

ats

o

A. Hepatocytes B. C.

D.

o

Keratinocytes Leukocytes Lymphocytes

low

o a

B.

Q

Intestine Integument Kidney


D.

U

low

high

Totai Plasma [Ca2+1

317

high

Total Plasma [Ca2+1

C.

A. Liver D.

low

high

Total Plasma ICa2+1

0t The primary precursor for 1,25-(OH)2-D3 is located in which one of the following organs?

C.

three

hormones described in the passage?

low

high

Total Plasma [Ca2+1


Biology Passage

Sepsis S;rndrome

Passage XV

99. Which cell

products are used reactions to destroy pathogens?


The normal response to an infection is a complex system of activations and inactivations of the members of the immune system. Neutrophils are the first to respond to an infection by squeezing through blood vessel walls and

A.

destroying pathogens directly by the release

D.

B. C.

of toxic

free radical

Chemical attractants. Proteolytic enzymes. Cytokines. Toxic oxygen products.

oxygen products, such as superoxide, hydrogen peroxide,

and nitric oxide, and by the release of proteolytic enzymes. These white blood cells follow chemical attractants given off by the pathogen. Monocytes, another type of white blood cell, arrive next. They may be transformed into macrophages in the tissue and engulf pathogens and cellular debris. Monocytes and macrophages release cytokines, a powerful class of chemicals that modulate responses of different members of the immune system. T cells and B cells are attracted and activated by cytokines at the infection site. Another set of cytokines are released to signal "stop" when the

100. What

pathogens are neutralized. This is the normal progression ofevents when the body attacks an invading bacteria'

is the chemical formula for nitric oxide?

A. B.

NO HNO3

C.

H2N202

D.

Noz

10L. Researchers are attacking sepsis syndrorne from many perspectives. In one approach, synthetic receptors for tumor necrosis factor (TNF), a cytokine, are injected to bind to TNF and halt the

However, in the U.S. each year, about 175,000 people die from sepsis syndrome, in which the sepsis process is actually amplified rather than terminated. The cytokines for signal termination are not released correctly, leading to an imbalance with the amplifying cytokines. This attracts all the white blood cell participants to the area, which release more activating cytokines. Some of the

cytokine cascade. What assumption does this model make?

A.

TNF halts the abnormal cytokine cascade of

B.

TNF receptors enter cells by endocytosis

sepsis syndrome.

substances that are released lead to increased permeability

bind intracellular TNF.

of the blood vessels. In normal situations, this allows WBCs to more easily enter the infected tissue' In excess, this leads to blood celi leakage and damage, and positively feeds back to attract more WBCs. Ultimately, if this process is unchecked, blood vessels will deteriorate'

C.

TNF is mainly

D.

circulation. TNF moves from cell to junctions.

present

and

in the general only through gap

leading to tissue death, organ death, or even patient death.

102. What is the most abundant type of white blood cell? A. 97.

Which is the term for the process of white blood cells squeezing through blood vessels?

A. B. C. D. 98"

B. C.

Neutrophils Erythrocytes Leukocytes

D.

Macrophages

Mitosis Diapedesis

103. What is the role of cytokines

Phagocytosis

Endocytosis

I.

IL

celluiar defense?

ilI.

To activate WBCs. To deactivate WBCs. To kill bacteria directly.

A. B. C. D.

A.

I only

B. C. D.

I and II only

What purpose do proteolytic enzymes serve in Inactivate bacterial lipid signals. Digest bacterial cell surface carbohydrates. Destroy bacterial DNA. Hydrolyze bacterial Proteins.


the inflammatory

process?

5la

i and III only I, II, and III

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i:t

Section V Answers

immunity is what the individual has deveroped 1T11..1":",oecific) ,i;;";"";;;;;.;ffi;;;J'oj'*ili"'ii;'J.l:.T,gffi:li1'?:fi1

l: *:"::::j-':rii:T*lL1ft'"4:

"1':'5 i"v.. r1*"";il'"';;;;'. ff:i;'J"J#; ffi;T;:T,"""X':*f:"""t,*115":::r::*:5:f-11llfl"" "? are rined ;i,h i,ii, .,",li.ili;"iii#J:?,?::: :31,:l:::":j#i::k::Ti,:l. yyrrrvtr i::1.":!1.:,,ru." T,t:;,";#;;;5; Dvurere trruLUus allu LIte gprlnella celrs th"-,;;"",';;;;1i;i;:iffii#.i]ji: 119 :H:,,,"5;,.,?1,,.:i"*'.:Y::,::i1:,::11:'-"1f,.^,j_li_",1i lh.""pitheiiai called cilia that move particuiate matter over

:|;:'::::i:l,|::ll:le

lfga

with cliated

the epitheliar surface. In tr the respiratory tract the mucous membianes of

"pitrrerium

thai;;;;,;;ffi;ffi;;i;Ji;ilffiiilH,"fi?ffi:'r:l

$:#Jlf#?J#::il?L?":;:llS",:"f;*jT_1.^"-Tjt9:11,t";il;;;;;il;ffi;".""p:,,"#|"'Jiifj':il11i ;;;;;il#il;;f"ilT"'i'Jl'*?lli"i# ffiiTll"o:*::T:::?:1fi1::::::1. 9.Y1ry-:1q""i""'iri"'n"g"";) i*-,"" 3::J*::3T::li,:l'::i:i'.:l:'il:l:".:":t"::'o: F; il;;;;iil"'f*;ili'l]r;ff1TrTi; rrom mother to retus via the piacenta. rhe retus acquires

il:T'j;'"iffi:':J:::l,T.lT the mother. The correct choice is C. 2.

"v,iJ-l'

ffiffi;;;"#tjil ri

c is correct' active immunity which is artificially acquired. In the first paragraph of the passage we see thar acquired imrnunity is specific immunity. In the thud pu.ug.uph *e see tnit valcinations and a recovery from a disease like the measle.s are exampies of specific immunitlr. i vacclnation i, ,i_fiy un immunization against a pathogen' It is something that is irtfficiat[1' acquired. Theindividual who receives'the ua".ination then begins to active\' produce

his or her own immunological agents agains, G *i"nuu,ed (weakened; putt og"n. Even though the pathogen has been made innocuous, it still ietains'its *rig""i"iiy io 1i"., n".J,i.ry rio- a disease like the measles is an example of. active immunity wiich is. #,r1"rry antitoxin aoministration would be an ffii."d."pir"p.ri. example of passive immunity which is artificially acquired. transter ft tgG from trr" -otrr".-io the fetus through the piacental membrane would be an example of pasiive immunity whichis naturally acquired. The correct choice is c.

A is correct' covalent bonding. In the 7th paragraph of-the passage it says that antibody-antigen complexes can be dissociated by high salt concentrations or higtt or low irH. iiir -"st mean that the binding forces between "ith"r antibody and an antigen are relatively weak. All oithe bondi ana7o, uona;ng forces listed in rhe answer choices an are weak except for the covalent uona. we would not expect o .ouut"ni bond to b" u.oten b1, a change in (physiological) pH or by treatment with a high salt concentration. The correct choice is A. 4.

D is correct' antibody which has similar dimensions

as_the antigenic determinant. This question is simply asking for an understanding of a paratope. In 7th paragraph.of. tt purlug" *e find rhat th; p#t"; _the is associared with the antibody' This allows us to immediatety eiiminate choices " A anic. Look at the structure'ot,n" tgc molecule in Figure i of the passage' This antibody can bind two epitopes or tn" *-" ,pecificity (unideterminant). The portion of the antibody that binds the epitope resides near the N-ierminus of both the vu'"i,a vi1"gion, of the Fab. The epitope does just

not bind the vg regions alone. we can eliminate choice s. rh. u,r"*g" size of antigenic determinants is roughly equivalent to about 6 amino acids. Therefore, the size of the paratope musr be simiiar in dimensions' If this were not the case, then binding between irr"-"pitop" una pu.utJf" woutd be unfavorable. Remember, hydrophobic forces, van der waai forces, electrostatic {orces, and hydrogen bonding all require relatively close proximities between participating parties before they can be of any use. The correct choice is D.

C is correct, antigens that are not cross-linked, because each Fab fragment is monovaient. In order to answer this question, we must consider the structure of the IgG molecule in Figure'l of the passag". N.r" irr" there are two Fab domains' It

is a bivalent structure. As it sLnds, this antibJdy could participate in cross-linking with

unideterminant, multivalent antigen (see the diagram below).

a

However' after addition of the protea.se papain, the antibody is cleaved into an Fc segment and two Fab segments. The important point here is that the Fab iegments-are now indepenclent of each ottr"r.'irr"f .un no* independently bind to the antigen's epitope (see above). Cioss-linking wiil not iesurt. In choice-A.the first part of the answer (cross-linking) is wrong, but the second part is correct. This makes the whole answer choice incorrect. In choice B the answer ii correct, "but for treatment of IgG with pepsin, not papain. In choice D two monovalent fragments are produced, but they cannot participate in crJss-linkir! 1r" correct choice

is C.

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319


Biology

Immunology

E(

Endocrinology

Section V Answers

epitope

Before IgG is treated with papain. 6.

After IgG is treated with papain.

C is correct, IgM. In the 6th paragraph of the passage we find that IgG is the oniy antibody to cross the placental membrane. Even though this antibody enters into fetal circulation and can confer fetal immunity to certain diseases, it is IgM which is the first antibody produced during an immune response. Also, it is the predominant antibody produced by the fetus during fetal development. The correct choice is C. B is correct, IgA. In the 6th paragraph of the passage we see that IgA is primarily found in external secretions such as tears, saliva, colostrum, and milk. This question is asking us to recall a little information about lysozyme. In vertebrates, this enzyme is widely distributed in a variety of cells and secretions. The connection that needed to be made was that both lysozyme and IgA are found in bodily secretions. The actions of the other immunoglobulins are outlined in the passage. Ihe correct choice is B.

C is correct, no, receiving antibodies confers transient immunity. Passive immunity means receiving antibodies from another organism. The recipient does not make antibodies in this situation. Thus, choices B and D are incorrect. Injected foreign antibodies eventually degrade and are not replaced, since an immune response was not triggered in the recipient. The effect is not permanent. It is transient. Eliminate choice A. The correct choice is C. 9.

A is correct, maternal antibodies provided immunity to measles in this situation. The newborn has an immature immune system. She is protected by passive immunity via maternal antibodies she received in the uterus and in breastmilk, if she is fed that way. She cannot respond quickly to primary exposures of viruses until after about 1 month of age. Choice B is incorrect. Although antibodies may be present in saliva in an extremely small amount. this is not how passive immunity is acquired. Choice C is incorrect. Paternal antibodies stay with the father. Choice D is incorrect. The correct choice is A.

10.

C is correct, both IgG and IgM remain at high levels following a "booster shot". We are looking for the FALSE statement. Al1 of the questions can be answered by consulting the graph. A "booster shot" means receiving a vaccination to boost the antibody effects ofan earlier vaccination. On the second (and further) exposures, we can

see

in the figure that IgG provides the primary response. Choice A is TRUE. Before a person is exposed to an antigen. there are no antibodies for that particular antigen. Choice B is TRUE. The primary response to an antigen is a rise in IgM. This, as you can see in the figure, precedes a rise in IgG. Choice D is TRUE. Both IgG and IgM fall off in their levels following each expose. They do not remain at high levels. Choice C is FALSE. Since we are looking for the FALSE answer, choose C. The correct choice is C. 11.

12.

B is correct, artificially acquired passive immunity. Since the person receives antibodies from someone else who is immune to the particular antigen, then it must be passive. Choices A and C are incorrect. Passive natural immuniq' comes from a mother-baby relationship. This person is an adult, so it must be artificial. Choice D is incorrect. The correct choice is B.

A is correct, artificially acquired active immunity. A vaccine provides artificially acquired immunity. Choices C and D are incorrect. Active immunity is created by the immune response of the body to the antigens in the vaccine.

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Section V Answers

when a second exposure occurs, the body has antibodies and lymphocytes ready to attack. choice B is incorrect. The correct choice is A. D istorrect, y-globulin' A gei electrophoresis retards larger molecules,,while allowing smaller ones to migrate more quickly' The larger molecules are the slowest. The y-glo"bulin moved the least in a given time, so they must be the largest' For your information, the y-globulin fractio-n contains antibodies which

13.

lan be used to confer passive

immunity. The correct choice is D. -

A is correct, I only. Breastmilk' especiaily colostrum, the breastmilk of several days immediately following birth, is important for all babies. This chiid coujd benefit from the antibody prorection

14.

conferred by another mother,s breastmilk' choice I is correct. Seclusion wouid protect the baby from exposure to antigens, but it still would not make more antibodies. when the babv's immune system matures, it needs exposure t;;;i;"r, in order to make antibodies' Eiiminate choice II' I'rmediate vaccinatron would not produce considerably greater antibodies, since the baby's immune system cannot function really effectively at birth. ihoi"" ilI is incorrect. The correct choice is A.

D is correct' both hydrophilic and hydrophobic regions. The first sentence of the passage states that the proteins which are encoded from

15.

the class I and class II MHC genes are expressed on the surfice oFa cett. This should strike a nerve in your brain. In particular, it should tell.youihat proteins expressed on the surface of the membrane must have both a hydrophilic

region and a hydrophobic region rrre rrvarophilic region in this case, -urt most of the protein is sitting in the extracellular medium, which wili most like"ly be polar"*iriu"cause, due to extracellular fluid. The hydrophobic region is necessary because the protein is anchored to the cell by the plasma membrane of the cell, which is ofcourse lipid in nature. Therefore, the protein should contain.both r,yar"pfi"oi"-ino r,ya.optilic region. Knowing this information' choices A and B are rnerely incomplete. choice " c is reounaant oecaus" hydroapathetic is similar to hydrophobicity. The correct choice is D, 16.

D is correct, host cellular in origin. This question is not really a trick q.uestion, it simply requires a little thinking. There are about one half of a million to one million^class I vigc proteins. Th;r" ir ;;-;uy'you should know rhat number, but you should tearize that there are a lot of these proteins on the surface. Furthe#rore, the passage states that the peptides found with class I MHC are invariably ttre celt's orn prot"inr. when the cell is infected with virus, the virai proteins/peptides will be presented to stimuiaie an immune ,"rponr". But this does not change the fact the most class I MHC molecules will still have their own proteins being presented on the cell surface. choices

B and c do not make any sense because those cells are involved with the reffiition/destruction of cells infected with virus. There is no evidence to suspect those cells will have their peptides pre"sented. The correct choice is D.

[-

B is correct' signal peptide sequence. This question requires a solid knowledge of cellular action. It is stated in the II MHC molecules have been shown io assemble in the endiplur*i. ."ti"uium. This requires that the proteins get to the endoplasmic reticulum (ER). Proteins that ar; destined'for the ER tuu" u special sequence called the signal peptide sequence which causes the ribosome they are being translated on io uina to the ER. The subsequent translation of the protein occurs, and the entire protein then finds i-tself within ttre BR. consider the other answers' Choice A is incorrect because a Shine-Dalgarno s"quen.e is a segment of RNA found in prokaryotes that is recognized by ribosomes as a place of binding. Therefore, choice A is lncorrect. Choices C and D are incorrect passage that class

because the question asks about proteins, not

DNA

sequences. The

correct choice is B.

c is correct, endocytosis. Recall that the ingestion of material by the invagination of the plasma membrane is called endocytosis. This is exactly the process that occurs for the rvrur4rrerr formatiJn ur oi ttrese Lrl959 elt(]osomes. eriJ;;;;"r. This Inls 15 i a Very straightforward Consider the Lus uLrlEl other allswcls. answers. Choice Yqvolrvu' vurrDruer Lfrolce A ls 'Yqas question' is a non-senslcal non-sensical answer and can automatically automaticallv be eliminated' choice B is the opposite of endocytosis, and would call for the expuision of cellular material through the plasma membrane' Transcyto-sis is the transport of material by a vesicle from one side of ,h" fi";;;";"rd: ili. usually occurs in epithelial cells. The correct choice is C. B is correct, structurally similar to that of the class_Il MHC The passage states that the CLIp part of the invariant ;hain is actively removed by binding to the DM molecule. The clip used to be bound to the class II MHC molecule. Based on this information, one could logically assume that the bu and class II MHC molecule are structurally similar, if they bind the same molecule. Considering the other answers, there is no evidence for any of the :iatements' The DM molecule is binding CLIP, which does not mean they are structurally similar. it just means they r,-

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Section V Answers

form a bond. Choice D is incorrect because class I MHC is not involved with DM. Finally, DM is not likely similar to the invariant chain. Remember, CLIP is merely a part of the invariant chain, and if DM is not like CLIP, there is no reason to believe it will be like the invariant chain. The correct choice is B. 20.

D is conect, II and III only. A given class I MHC molecuie will bind to a variety of peptides. However, the cleft where the peptides bind does not change for a given class I molecule. In other words, the amino acids that make up the cleft are invariant. The peptides that bind to the cleft can only do so if certain amino acids in their chain are able to bind to the amino acids which make up the cleft. That is, amino acids at certain positions in the peptides must be highly conserved if a large variety of peptides are able to bind to a class I molecule which does not change its amino acid composition. Therefore, choice fII is correct. We find that choice II is conect based on the following logic: Iooking at Figure 1 and Figure 2, we can see that in class I molecules, the peptides are bound in the cleft by their ends. Therefore, the peptides can only be so long, and as stated in the passage, about eight or nine amino acids long. However, with class II, the binding occurs in the middle, and therefore we will see more of a variety of lengths in the peptides which bind. The peptides which bind to class II molecuies are not constricted to binding to the ends, so they do not have to be of a certain length. The correct choice is D.

21.

C is correct, production of a transcription factor which increases the transcription rate of the MHC gene early in an infection. If there was a transcription factor that promoted the transcription of the MHC gene early in infection, this would promote an immune response against the pathogen. This is clearly not a defense mechanism for the pathogen. All of the other answers result in a depression of the immune response, and therefore can be categorized as a defense mechanism used by a pathogen to disrupt the antigen processing system. The correct choice is C.

7)

D is correct, I, II, and III. Any body surface that can contact the outside environment is lined with epithelial tissue. The mouth, is, of course, lined with epithelium. Choice I is correct. The entire GI tract is essentially "outside the body." It is completely lined with epithelial tissue. Choice II is correct. The urethra also has contact with the external environment. It, too is lined with epithelium. Choice III is correct. The correct choice is D.

23.

B is correct,IL-6- mice produce lower levels of IgA than IL-6+ mice. Read the bar graph. The IL-6-mice have almost no IgA compared to the IL-6+ group. Choice A is incorrect. Choice C is incorrect. The response to ovalbumin was almost nil in the IL-6- mice. Choice D is incorrect. The correct choice is B.

24.

A is correct, amino acids. Immunoglobulins are proteins, made of amino acids. Fatty acids are part of many types oi lipid molecules. Choice B is incorrect. Sphingomyelins are special fats that are found in nervous system tissue Choice C is incorrect. Phospholipids are part of cell membranes. Choice D is incorrect. The correct choice is A.

?(

B is correct, restoring IL-6 in the IL-6- mice improved IgA production. The virus carrier alone did not change the genetic makeup of either strain of mice. Choice A is incorrect. The IL-6+ mice simply responded to a viral infection by increasing IgA slightly. Choice D is incorrect. The IL-6- mice did not respond at all to the virus carrier. Thel were hyposensitive. Choice C is incorrect. The correct choice is B.

26.

D is correct, cytokines are local hormones and often act without passing through the entire circulatory systexn Cytokines are special "local hormones" that do not always pass through the entire blood supply. Measurements maoi on blood samples may not reflect what is happening at the tissue level. Choice A is incorrect. Hormone concentrations can be measured in blood samples by radioimmunoassay. Choice B is incorrect. IL-6 can and does leave the cell that produces it, like other cytokines. How could they interact with IgA if they never left the cell an; IgA is present in secretions? Choice C is incorrect. The correct choice is D.

27.

A is correct, gene therapy. You may have read about the first gene therapy trials in humans during the last year ff' so. This was to alter genes in people with inherited, disease-causing genetic defects. Specific viruses carried the nem genetic material. Hopefully, these viruses will infect the host's DNA with a good copy of their defective gene Vaccination is just plain, old immunization with a modified form of a pathogen. Choice B is incorrect. Geneu; immunization is a fake answer, and Choice C is incorrect. Pleiotropy having a gene that affects many differen: characteristics in an individual. Choice D is incorrect. The correct choice is A.

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\'

{nsu-ers

28, c is correct, soiubilize antigen-antibody

complexes. This question jusr re:*-.;. . : The passage states that persons who are complement deficient sui-i.. 1....-r:-:--_ antibody complexes form precipitates in the blood vessels of skin. joints anci 'n.-,. : cause tissue damage. One can assume from this statement that complemenr piri s . antibody complexes because when the compiement is not present, the precipitatis torn,

29.

'

.. .,i

C is correct, provides a means of amplification, uhimately leading to many MACs. The ker ,.;. .:: ,.. .-:_: : -:: * , r the word cascade. one should be familiar with the idea of biirogical;;;J;. nu.ru.,i',-.. -.._.,;-., _-, pathway cleaves many molecuies of the next proenzyme in the chain. The activario" --' components consists of an amplifying proteolytic cascade where each molecuie "i*.-,.,.. --=.,.-. *activated uiifr.-U._*i".,_-. sequence leads to the production of nany membrane attack complexes. The correct choice is C.

.

30.

B is correct, bone marow. This answer is arrived at through one's knowledge of immune system development. The question is asking about what disease would affect the pioduction of antibodies. This would clearly be the bone marrow' The bone marrow is the site of B-cell birth and maturation. B-cells u.. r"rfonriu;;;#;[i#;;;;; antibodies' Therefore' a disease o{'the bone.marrow would greatly affect B-cell maturation and thereby affect antibody production. The thymus is the site of r-cell maturatioi, whiie the ,pt""n unJ tn" ly,npr, nodes are what are calied secondary lymphoid organs' where mature B-cells and r-ceils "hang out" to r"aci rith to."lgn antigens. The correct choice is B.

31.

D is correct, constant region of the antibody. we know that there exists interaction/activation between the c1 complement component and either IgG or tgM. rhe question.is asking what is the most likely means by which C1 is activated' Cl does not change depending on the micioorganism. In Jther words, we do not have a difrerent Cl for

different microorganisms. Howevet,

ti

do have diffeient IgGs and IgMs for different-microorganisms.

The difference lies in the variable region of the antibody. That variable ,"g;oniinos t; ,h"-;;;i;;nic dererminanr of the microorganism' The antibodies also have a constant region, which does not change (except for class switching) or depend on the particular microorganism. It most likely*tharthe cl, which remain"s.onrlul|^riu recognize and be activated by the constant region of the antibody. If this were not the case, cl would itself have to change to recognize the variable region of the antibody. The correct choice is D. 2,)

c is correct, confine complement activation to the cell surface where it began. The early components and c3 are proenzymes that are activated sequentially by limited proteolytic cleavage. vihen each tr;nryn . in the sequence is cleaved, it is activated to^generate a serin-e piotease *hich cleaues the next protein in the sequence. Many cleavages liberate a small peptide fragment and expose a membrane-binding site on the larger fragment. The larger fragment can now bind tightly to the target cell membrane and carry out"the next reaction in ihe sequence. The reaction cascade is constrained to the cell surface where it began, and where we want it to be. The correct choice is C.

33.

B is correct, these molecules stimulate the secretion of histamine from basophils. The question tells us that an increase is seen in the permeability of blood vessels. From the answers given, we know this is due to histamine secretion' This histamine that is secreted increases the permeability of local blood vessels which will allow white blood cells, antibodies. and more complement to enter tlie site of iniection. The question becomes what cell secretes histamine' The answer is that mast cills and basophils secrete histamine. Moit are familiar with mast cells, but basophils are a type of white blood cel1. They falt under the-category of granulocytes secrete histamine to help mediate inflammatory responses. The other types of cetti in the"uns*e.s u."""Jtrr"y rl*frj not involved in the secretion of histamine. The correct choice is B.

34

A is correct,

an inhibitor of acetylcholinesterase. Since the receptors are blocked or not as numerous, an increase in acetylcholine (ACh) should remove some of the effects. one way ro accomplish this is to give an inhibitor of acetylcholinesterase so the ACh is not degraded as quickly. This would give higher of ACh at the receptors' An immunostimulant would further complicate the problem by encouiaging "on""nt.utions antibody production. -or" choice B is incorrect. A paralytic agent would add io the weakness. choice cls incorrect. Decreasing the -uril" amount of ACh would be a bad idea, so choice D is incorrect. The correct choice is A.


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Biology 35.


Section V Answers

C is correct, III. The nerve is the top or presynaptic part, while the muscle is the bottom, or postsynaptic part. I indicates a vesicle containing ACh. Choice II indicated vesicle fused with the membrane and releasing

Choice

ACh into the synaptic cleft. Choice III is the acetylcholine receptor on the muscie cel1. Choice IV is a receptor on the nerve cell. The correct choice is C. 36.

A is correct, Type II diabetes. Type II diabetes is not caused by an autoimmune agent. It is mostly related to insulin resistance. Choices B, C, and D are clearly autoimmune diseases from the passage. The correct choice is A.

5t.

C is coruect, lethargy. The effect of stimulating the TSH receptor causes the thyroid to secrete more thyroid hormone. This increases metabolic rate and can iead to hyperactivity in severe cases. Weight loss is probably connected due to increased metabolic rate. The person is not lethargic, which means tired and sluggish. The correct choice is C.

38.

B is correct, insulin. The beta cells of the pancreas produce insulin. Lack of insulin is the defect in Type I diabetes. You could figure this out from the table if you forgot. The other hormones are produced by the pancreas, but not in the beta cells. The correct choice is B.

39.

C is correct, I and II only. A virus can resemble the structure of a body molecule and lead to autoimmune attack. Choice I is correct. A virus can damage a cell so that cell components that do not usually travel in the blood are placed in the blood. This could also lead to an autoimmune response. Choice II is correct. Choice III is the opposite ofchoice I and is incorrect. The correct choice is C.

40.

B is correct, adrenal gland. The adrenal cortex secretes steroid hormones. The correct choice is B.

41.

B is correct, increased levels of glucagon and decreased levels of insulin. It is important to remember that glucose is needed by the body as an energy source. Thus, in times of stress, when the body needs energy to deal with the stressor, glucose plasma levels must increase. Sympathetic stimulation of giucagon secretion will increase glycogen breakdown and increase glucose plasma levels. This will make glucose available for tissues to utilize as an eneigy source. The sympathetic system will inhibit insulin secretion. The correct choice is B.

42.

B is correct, increased levels of glucagon and decreased levels of insulin. The major function of glucagon secretion is to increase the amount of products in the plasma that can be utilized for energy production. During fasting, blood glucose levels are low, stimulating glucagon secretion. Glucagon secretion will increase the release of free fatty acids, which can be converted to ketone bodies and utilized for energy. However, if insulin levels are low, as in diabetes, ketone utilization is low and ketone products increase in the plasma beyond normal levels leading to ketosis and acidosis. The correct choice is B.

43.

A is correct, abnormal increases in insulin secretion. Hypoglycemia, by its name, means "low blood glucose." Thus, abnormal increases in insulin secretion will decrease levels ofblood glucose. Decreases in insulin secretion and high levels of plasma glucose are signs of hyperglycemia. Low levels of free fatty acids may indicate increases in insuiin secretion or decreases in glucagon secretion. The correct choice is A.

44.

A is correct, increased levels of insulin and glucagon. Increases in plasma amino acid concentrations will stimulate insulin secretion to promote protein production in cells. Furthermore, glucagon secretion from the pancreas is also increased. Glucagon secretion is needed to prevent hypoglycemia after a heavy protein meal since amino acids also stimuiate insulin secretion. The correct choice is A.

45.

C is correct, access and storage of glucose in cells of the peripheral tissue. Insulin stimulates the storage of glucose and eases the entry of glucose into tissue. The tissue affected most readily appears to be muscle and connective tissue of the periphery. The brain does not need insulin for glucose uptake, as brain tissue is not affected by insulin. Insulin stimulates glucose uptake and storage by the liver. Insulin will decrease glucose loss in the kidney by decreasing plasma glucose levels and increasing uptake of glucose by the tubular cells of the kidney. The correct choice is C.

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Section V Ansners

46.

A is correct' extracellular glucose excess. The lack ofinsulin secretion will decrease the uptake ofglucose bi..celis. Even with extracellular concentrations of glucose rising, cells cannot utilize the glucose U""uut" it cannot enter the cells. Therefore, tissues die of starvation, with high concentration of glucose*located just outside the cell. The correct choice is A.

47.

B is correct, (see the graph below). In the 5th paragraph we read the following: "pTH causes calcium absorption from bone by stimulating osteoclastic activity and transiently inhibiting osteoElastic activity. At the level of the kidney, PTH increases calcium absorption in the distal tubules and collecting ducts and greatly decreases the reabsorption of phosphate at the proximal tubules." This means that at the tirne PTH is adininisiered we would expect to see an increase in plasma calcium levels and a decrease in plasma phosphate levels. Since pTH is a

hormone and goes everywhere the blood goes, we would also expect to iee these effects happen at about the same time, which is exactly what we observe in the graph for choice B. J o

!

o

E

o E O

U

F

0 0

Time (hours)

PTH added

We-would not expect to see the levels of phosphate or calcium lag behind one another. This immediately allows us to eliminate choices C and D. We can eliminate choice A simply because we know (from the passage) that the levels of plasma calcium will increase. The correct choice is B. 48.

B is correct, (see the reaction below). The answer to this question is found in the first paragraph of the passage. We are told that vitamin D3 is also called cholecalciferol and that it was synthesized irom"a'sterol derivative in a photolytic reaction involving UV light. A sterol derivative is based on the ring structure of cholesterol. A photolytic reaction involves the breaking of bonds. Think of "lytic" as something being;'lysed" or broken. you If co'mpare'the four structures which are given as_answers, you will note that choicJs B, i, and D are rather similar in that they contain just three complete rings. Choice A contains four complete rings. One of the rings (called the',B,,ring) in choice A has been open up by the breaking of a bond. In this case that bond was broken "by itre action of UV li-ght. Since vitamin D3 is formed from a photolytic reaction, we can eliminate choice A as a possible answer. turns out

[It

that choice A is called 7-dehydrocholesterol and is the precursor to vitamin D3.l CH,

cH,

UV radiation (a photolytic reaction)

7-Dehydrocholesterol

Vitamin D3 (cholcalciferol)

We are left with choices

B, C, and D. It

was stated in the passage that vitamin D3 is the inactive form of the hormone. It was activated by first hydroxylating it in the liver and then by hydroxylating it in the kidney. In other words, the active form of vitamin D3 (called lc,25-dihydroxycholecalciferol (or 1,25-(OH)2-Df )) has two more

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Section V Answers

hydroxyl groups on it than the inactive form. Therefore, we look for that structure which has not had two hydroxylation events. Based on this analysis we can eliminate choices C and D. The correct choice is B. D is correct, I and III 9!1y In the second paragraph it was mentioned that in order to synthesize the activated form of vitamin D3 (1'25-(OH)z-Di, the hydroxyiase enzyme in the kidneys musr be activated (stimulated) by pTH.

49.

Once the activated hormone is synthesized, calcium and phosphate can be absorbed across the mucosa membrane of the intestines. The phosphate ion follows the calcium ion across the membrane so electrical neutrality is maintained. We can eliminate choice C because PTH by itself has no direct effect on either calcium or phosphaie absorption in the intestines. We can also eiiminate choice B because calcitonin does not allow for intestinal abiorption of ialcium

or phosphate either. Even though choice A is a correct answer, it is not the BEST answer. In brder to get the activated form of vitamin D, PTH is needed. Therefore, an increase in both PTH and subsequently an incrJase in 1,25-(oH)2-D3 is needed for absorption in the intestine. The correct choice is D. s0.

A is correct' increased osteoclast activity coupled with a hypercalcemic plasma. Be careful with what this question is asking you to answer. The bottom line of what is being asked is: "What does it take in order to secrete calcitonin?', Calcitonin acts to loser plasma calcium levels. In order to lower those ievels they must first be high. What leads to high calcium levels? Increased osteoclast activity (i.e., bone resorption and the ieturn of calcium-and phosphate to the blood) and a hypercalcemic plasma (i.e., a plasma in which the level of calcium is above normal). These two conditions demand a decrease_in the levels of plasma calcium. The secretion of calcitonin by the parafollicular cells of the thyroid gland reduces the concentration of calcium in the blood. Calcitonin does this by inhibiting osteoclast

activity (i'e., inhibits bone resorption) and inhibiting calcium reabsorption at the level of the kidney.

choice is A. 51.

T-he

correct

C is correct, (see the pedigree below). The characteristic to look for in an X-linked dominant trait is that alt of the daughters and none of the sons of males who carry the trait are affected. This is exactly what we see in this pedigree. Interestingly, the phosphate levels are not as low and the rickets are not u. i"u"." in an afflicted heterozygous female as in the afflicted male.

rT'oTo

+" r, tr-I '+ [I'rt i-l rrn

,,,

,,,

Autosomal Dominant

If

any son is affected or

Autosomal Recessive

if

X-linked Dominant

Autosomal Dominant

any daughter is unaffected, then the inheritance is autosomal and not X-linked.

Remember, autosomal recessive is oniy expressed in homozygotes. The correct choice is C.

D is correct, phosphate and 1,25-(OH)z-D:.Familial hypophosphatemia refers to the fact that there is inadequate phosphate in the plasma. As mentioned, this is an X-linked recessive disease and stems from the fact that reabsorption of phosphate at the levei of the kidneys is impaired. The addition of caicium will not affect the levels of phosphate and so we can eliminate choice A. Addition of phosphate alone will not make that much of a difference because it will not have an adequate means by which to be absorbed. We can eliminate choice B. We cam also eliminate choice C for the same reasoning. The addition of 1,25-(OH)2-D3 along with the phosphate insures adequate absorption by the intestines and (presumably) an increased reabsorption at the level of the kidney. The correct choice is D.

E2

C is correct, increased neural excitability. Lefs consider the list ofpossible answers. In order to do this we need to know the meaning of hypoparathyroidism. The parathyroid giands secrete PTH and PTH allows for calcium

absorption from bone by stimulating the osteoclasts and transiently inhibiting osteoblasts. PTH stimulates calcium reabsorbtion at the kidneys and indirectly allows for calcium absorption at the intestine through the action of the active form of vitamin D. A hypoparathyroid gland is one that does not secrete sufficient amounts of PTH. As a

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Section V Answers

result the levels of calcium in the body decrease below normal values. Not only is osteoclast activity decreased, but because calcium levels are low, osteoblast activity is decreased as wett. with this information we can eliminate choices A, B, and D' we are left with choice c as the correct answer. why would hypoparathyroidism lead to increased neural activity? A decrease in calcium in the extracellular fluid allows -o." ,o'airi.n to flow through the membranes of nerve cells. some nerves, like those.in the peripheral n"ruous ;t;4";,-;;;in to sponraneousiy discharge' This will eventually give rise to tetany and if not coriect immediut"ry'.ur r*Jio a"utt . The correct choice is C. 54.

D is correct,

a positive linear reiationship exists between PTH secretion and the concentration of plasma calcium. This statement is false. In order to makJit rue we would say that: "an inverse relationship exists between prH secretion and the concentration of plasma caicium. we can see this from rhe t.;ph ;;;;'b"jor. The first three choices are all true as discussed in the passage and in the previous unr*"rr. The correct choice is n.

so

tr

o

o

cd

O

o

o 5 o o

O

L-

$

Total Ca2+ concentration in piasma

she is very untikery to have antigens ro rhe Rh facror. Exposure 113::::1",1:111:jTl :hild, aborrion, ormiscarriaie e""" ti,""ei, "".::,:"";::,"",:?l:. ;J;; ffi:i,f:TT.r].1,:f*:,1:It,9.:ll":.t, rl: -?r*I musr arso r,uu" tr," +:o:li:""""1^"1fl llt^","111::llld, In this case, she is probabry not sensitized. The correit choice is D.

;;;;;:;;r;;;;ffiffii;#:

untiil;;il#;;;,Jffi;",:',1,ffi:ffi;.Y;

56.

is n1t a component of the ABo blood group sysrem. The piacenra acts as a barrier for l:^":j::"^':11" 1*11 some ova'w things r'urE;) and 4rru allows iluuws passage otners. There lhere ls is not tndiscriminate indiscrimi Passage to otheri. mixing of maternal and fetal blood. Nor does the mother make the red blood cells for the fetus. Choices A and C are incorrect. As we learned lenrnerl in in rha --,, the passage, -ooo,-o any is taken to completion or not, can trigger antibody formarion in Rhl, fetuses. Choice D is incorrect. The correct choice is B.

*l

,;#;^;f;il$;'Jili

lLllli"p; Yl*:t 57.

C is correct, treat mother with antibodies against the anti-Rh antibodies. If the antibodies can be trapped and removed by other antibodies, then the mother is.not sensitized against Rh antigen for subsequent pregnancies. Treating the fetus would not help future fetuses. choice D is inconEct. In extreme circumstances, the fetus, not the mother' receives a blood transfusion. This would not help future pregnancies, either. The correct choice is C.

58.

D is correct, all of the above. All three of these organs produce bitirubin, a break-down product of heme. The liver conjugates the bilirubin with glucuronic acid, so it is ioluble, and secretes it into the gail bladder. The correct choice is D.

59.

A is correct, no, there are no maternal antigens to the

Rh.

factor antigens. The term positive means the antigen is

present on the RBC' The mother does not have antigens to herself. ThJfetus is negativi,

roit"." are no Rh antigens on the RBCs to make antibodies against. choice B is incorrect. The newborn is not immunologically competent, able to make antibodies, until aboul l month after birth. It relies on the antibodies donateJd;" mother. Choices C and D are incorrect. The correct choice is A. 60.

D is correct, I, II, and III. Hemolysis in the Rh-positive fetus depends on the amount of antibodyproduced by its Rh-negative mother. If a large amount of blood has been transferred between the fetus and the mother, then she is more likely to produce anti-Rh factor antibodies. Choice I is correct. How her antibody proau"tion responds to a given stimulus is also a factor. choice II is correct. As the number of pregnanci"r in"i"'ur"i her exposure to Rh


327


Biology

Immunology

Er

Endocrinology

positive blood increases, and the likelihood of developing antibodies increases. Choice choice is D.

Section V Answers III is correct.

The correct

61.

B is conect, I and II only. The primary problem in EF is lysis of red blood cells. This leads to anemia and lower hemoglobin level. Choice II is correct. If RBCs are destroyed, this does not lead to an increased number. Erythrocytes are RBCs. Choice III is incorrect. Bilirubin is a breakdown product of heme. It is elevated in neonates and fetuses with EF. Choice I is correct. The correct choice is B.

62.

A is correct, does not have an outer membrane on its cell wall. This question requires one to know a little about the differences between Gram-positive and Gram-negative cells. A Gram-positive cell does not have an outer membrane on its cell wall. This information eliminates choice B. Furthermore, a Gram-positive bacterium has a very thick

peptidoglycan layer adjacent to the plasma membrane. The correct choice is A. 63.

C is corrrect, endocytosis, engulfing foreign particles which eventually will fuse with a lysosome. The question requires us to use our knowledge about how macrophages are functioning. They are phagoiytes in the body along with the neutrophils. The cells engulf microorganisms (this eliminates choices A and B) which then become phagocytic vesicles. These vesicles fuse with the cell's lysosome (eliminating choice D), an organelle with contains highly reactive molecules like superoxides. The fustion of thephagocytic vesicles with the lysosome exposes the foreign particles to these caustic molecules, acting to destroy the microorganism. The correct choice is C.

64.

C is correct, plasma cells. This question requires us to draw on our knowledge of the immune system and the cells which comprise that system. Antibodies are associated with B cells, which are lymphocytes. Antibodies can be found in two forms. One is the membrane bound form, attached to the B cell at its plasma membrane. When a B cell becomes activated, it will proliferate. Some of the new B cells will produce the antibody (it will be the same antibody) in a soluble form, enabling the antibody to circulate in the lymph and blood. B cells which make soluble antibody are called plasma cells. Therefore, a defect in plasma cells would affect production of soluble antibodies.

The correct choice is C. fl5.

C is correct, rise in the patient's blood pressure. Anti-Factor XII antibodies will take out (render non-functional) XII. This will affect the prodirction of bradykinin and the intrinsic coagulatory pathway. Since not all Factor XII is non-functional, some intrinsic pathway leading to coagulation may occur. We still have the entire extrinsic pathway. Therefore, we should not see either a total loss (choice A) or a rise (choice B) in the intravascular clotting of the sepsis patient. As stated above, we are attenuating the role of bradykinin and one of its role is to increase the permeability of the vessel. Increasing the permeability leads to a loss of fluid and hypotension. Since we are attenuating the role of bradykinin, we should therefore see a rise in the patient's blood pressure. The some of the available Factor


D is correct, 16. The flow of blood through a vessel is proportional to the radius to the fourth power. increase in the radius of a factor of two, the increase in flow is 24 = 16. The correct choice is D.

67.

A is correct, hypotension in the patient. One should realize from the passage that bradykinin acts in two ways. It increases the amount of blood flowing as a vasodilator, and it increases the permeability of the endothelial cells making up the blood vessels. The increased permeability causes a lot of the plasma in the blood to leaves the vessel

If we see an

(through osmotic forces ) and enter into the tissue. The result of this is that the patient becomes hypovolumeic which

will lead to a drop in blood pressuie. Therefore, bradykinin will lead to hypotension in the patient. The correct choice is A.

68.

C is correct, decreased osteoblast activity, increased osteoclast activity. When bone synthesis is low (decreased osteoblast activity) and bone breakdown is high (increased osteoclast activity), the amount of bone decrease is the greatest. The correct choice is C.

69.

C is correct, calcium. The question asks for a mineral. Eliminate choice B, water, because water is a compound, not a single atom. B 12 is a vitamin, not a mineral, so choice A is incorrect. Since calcium was mentioned in the passage,

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Biology

Immunolotry & Endocrinology

Section V Answers

consider it. Calcium is the major mineral making up bones. our body contains about 1 kg of calcium. Although zinc is part of many enzymes, our body contains only a I'ew grams of zini. The correct choic! is C. 70.

A is correct, PTH increases piasma calcium. The passage states the CT is an antagonist of pTH. This means they have opposite actions' CT,works to decrease plasma caicium, and PTH works to inc?ease piasma calcium. Choice B is incorrect' To increase plasma calcium, the kidney decreases urinary calcium and phosphate. Choices C and D are incorrect. The correct choice is A.

71.

C is correct, cysteine-cysteine. This is a biochemistry trivia question. Two cysteine residues link together via their sulfur atoms to form a cystine molecule that contains a disuifide bridge. Cysteine is the singie amino acid, while cystine refers to a dimer of cysteine. The correct choice is C.

72-

B is correct, I and III. Choice II is inconect. Since CT is

a polypeptide hormone, it does not enter the cell, but rather is conect. The salmon CT is less recognizable to the human enzymes' so it resists degradation more and has a longer-iasting action. Choice III is corrlct. Saimon CT has a stronger interaction with the CT receptor and promotes a longer signal on the receptor. The correct choice is ts.

acts on a receptor on the piasma membrane. Choice

I

73.

A is correct, CT secretion increases. In the beginning of a supplement program, a person would have increased calcium levels. In response to these increased levels, CT secretion wouldlnciease'to Lring the levels back down to normal. The correct choice is A.

74.

B is correct, injection to avoid hydrolysis. Since.CT is a polypeptide, it is subject to digestion (hydrolysis) in the stomach' Ingestion means to take orally, so choices A and C aie incorrecr. Th"r" is a"possibiiiiy of un aitergic reaction when proteins are injected, so choice D is incorrect. The correct choice is B.

t5.

D is correct, I, II, and III. In a systemic allergic reaction, histamine, leukotrienes, and prostagiandins are released and lead to skin reddening, hives, and constriction of the bronchioles (leads to bieathing difficulty), among other

symptoms. The correct choice is D.

76.

D is correct, Iittle or no change in the effectiveness of either type of immune response. The question asks about the consequences of removing the thymus from an adult human. From the passage w" learn that i cells are produced in the bone marrow and later mature in the thymus d.uf1s early development.the thymus is therefore crucial during

early development for the production of T cells, which are reiponsible for the cell-mediated response. If the thymu! is removed from a mature adult, however, there will be little or no impact on the immune system since most T cell formation has aiready occurred. The tempting answer is choice B, bui remember to read uil of th" answer choices before making a decision. A careful reading of the question in this case is also a must. The correct choice is D.

C is correct' the cell would lyse due to an upset water balance. When pores are formed in the membrane of the target cell, smali molecules (i.e. charged ions) are suddenly able to flow freely along their electrochemical gradients. The effect of this would be to disrupt the osmotic baiance maintained by the cell. T;his balance is necessar! because the large number of negatively charged macromolecules in the cell (proteins, nucieic acids, etc.) would normally attract a large number of cations in order to balance the charge. The entry of such a large number of cations woulb draw water into the cell with them, a catastrophe which would cause lysis, or bursting, 6f the plasma membrane. In normal cells, this is prevented because the membrane is mostly impermeable to small iharged i,ons, and the action of membrane transporters serves to maintain an appropriate osmotic balance. The action of tie complement proteins is to poke holes in the membrane, allowing ions and water to rush into the cell, causing lysis; this is an efficient way of killing invading microorganisms. Choice D is the opposite of what was just desciibed and is therefore incorrect. Choice A is also wrong; an inability to initiate action potentials is not letiral to the cell. Additionally, most cells which are attacked by the antibody/complement system are invading microorganisms which do not have anything to do with the neuronal function of conducting action potentials. Choice B is wrong because if the cell is nor-itty negativeiy charged inside, holes in the membrane would allow cations to rush in, leading to depolarization rather than hyperpolarization. The correct choice is C. 78.

C is correct, II and III only. Statement II states that organ transplants between monozygotic twins do not result in immune rejection of the transplanted organ. This implies that the recipient's immune ryrt"m regards the new organ

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Immunology 6r Endocrinology

Section V Answers

as being "self'" Since monozygotic twins are genetically identical (as opposed to dizygotic twins), this statement is consistent with Theory 1, which states that individuals lacks the genesio produce u iJtponr" against self antigBns. Since both twins have the same genetic structure, they both lack the genes necessary to mount u-n im,nun" r"rionr" against their own tissues; therefore, according to Theory I the transplanted organ is not rejected. Statement II is also consistent with Theory 2, which in essence states that the immune system "learns" not to attack its own tissues early in deveiopment. Since both twins had the exactly the same tissues (the result of their identical genes), both of their immune systems "learned" not to attack the same tissues. This also explains why the transplanted organ was not rejected" Statement III is, for all intents and purposes, an identical situation to that presented in Statement II. In this case' geneticaily identical mice from the same inbred strain are able to tolerate iranspiants from each other. An inbred strain, incidentally, is a strain of mice that has been inbred for many generations, resulting in siblings which are genetically identical to one another. Statement I states that cells from a foreign cell line are injected into*a mouse at the embryonic stage. When the mouse matures, it does not mount an immunJresponse to further injections of the s_ame cells. This supports Theory 2 inthat it provides evidence that the immune system "learns" during embryonic development not to attack cells that are present at the time (i.e., self cells). ThL immune system is*fooled into thinking the fbreign cells are self ceils. This contradicts Theory 1. According to this theory, the foreign cells would still be attacked even if they were injected at the embryo stage" The correct choice is C.

79.

B is correct, weakness and difficulty breathing. The question states that antibodies are produced against skeletal muscle acetylcholine receptors. This would most likely result in muscle weakness (dueio an inabi-lity of muscle fibers to receive signals from effector nerves). Difficulty breathing would also be likely because miny skeletal muscles (i.e., the intercostals) are involved in the respiratory process. This problem ulro be solved via a "ould process of elimination. Answer choice A can be eliminated because the heart is made up of cardiac muscle. Since the antibodies only attack skeletal muscle receptors, the heart should not be affected. Likewise, the smooth muscle of the gastrointestinal tract should not be affected, ruling out answer choice C. Choice D is unlikely to be true as well. Neurons in the brain should not be affected by skeletal muscie receptor antibodies, and therefore no dementia should occur. The correct choice is B.

80.

B is correct, the B cells have considerably more rough endoplasmic reticulum than the T cells. Since the function of activated B cells is to produce and secrete large amounts of antibody, it would be likely that under electron microscopy they would appear to have more rough endoplasmic reticulum than activated T ceils, which do not have as great of a secretory role. Rough ER is where proteins that are to be secreted are mainly produced. The "rough,, appearance of this organelle is due to the presence of ribosomes dotting the convoluted membrane of the ER. Thise ribosome produce the antibody proteins. Choice A is not necessarily a likely observation, as neither cell has an especially great need for mitochondria relative to the other. Answer choice C ii also incorrect as T and B cells both have nuclei. Do not confuse them with another type of blood cell which does lack a nucleus: the erythrocyte. Answer choice D is incorrect because electron microscopy is the most powerful means of visualizing und as such is ""llr very likely to detect differences berween B and r cells. The correct choice is B.

81.

A is correct,

each B cell is predetermined to bind a specific antigen. First, let's go over the experiment in the question. A population of B cells is removed from an adult mouse and mixed with an effectively lethal antigen X. The B cells that have the appropriate surface receptors bind the antigen are killed due to the radioactivity. Veiy few B cells are killed, therefore very few out of the total population (<0.017o) recognize antigen X specifically. Whin the surviving B cells are injected into an irradiated mouse (lacking B cells), the mouse is now capibl" of reiponding to other antigens, but not antigen X. It can be implied that the reason for this was the death of the B cblls which specifically recognized antigen X. The remaining population of B cells had no members which recognized antigen X, though they could recognize other antigens. Therefore we can conclude that each B cell is predetermined to bind a specific antigen. B cells can not recognize new antigens and "learn" to bind to them, ma*ing answer choice B incorrect. Answer choice C is wrong because the above experiment implies nothing about T cell interference with B cell binding. Answer choice D is wrong because an antigen, by its very definition (antibody generator), is immunogenic. Therefore antigen X does elicit an immune response normally. The correct choice is A.

82.

B is correct, antigens can only be bound by a single antibody at a time. The question is asking which condition would not hold true if antibodies were capable of "teaming up" and cross-linking antigens together. We can approach this problem by first theorizing how antibodies could accomplish this feat. First, ifantigens could be bound by more than one antibody, and each antibody had two antigen binding sites connected by a flexible "hinge" region,

cross-linking could easily occur. Each antibody would bind two antigen particles, which would then bind

an

additional two antibodies, which would then bind another two antigen particles, etc. The flexible hinge region would allow greater flexibility during the cross-linking process. The above description would make answer choices A, C,

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Section V Answers

and D all consistent with the cross-linking phenomenon. This leaves choice B as the answer. If antigens could only be-bound-by a single antibody at a time, ihe above scheme would not be possible and there could be no crosslinking. The correct choice is B.

A is correct, most types of cells in the body. The key concept to remember here is that viruses can inf'ect most types ofcelis in the body. Since the question states that ceis infecied with viruses are usually killed by cytotoxic T cells, it logically follows that the T cells should be able to target most cells in the body. The other

83.

specific cell types and are therefore incorrect. The correct choice is A.

unr*". choices mention

c- is conect, detergent wash. The passage informs us that the insulin receptor is an integral transmembrane protein. The receptor is thus embedded in the lipid bilayer and actually spans the entire layer flom the extracellular to the intracellular face. These proteins are tightly bound to the lipidbiiayer by hydroptrouic foices ano harsh conditions are needed to remove them' In particular, detergents must be useO. rh" ,"uron is that these molecules are

84.

amphipathic, containing both a polar and

non-polar region. The non-polar region binds to the hydrophobic regions of the membrane proteins., disruPting the lipid bilaye.. The polar region creates a water soluble micelle with the membrane protein (hydrophobic portion) in the middle. The t"""p-to. is thus isolated


and ready to study. The

B is correct, transduction mechanisms. we know from the question that we have a hybrid receptor which is made up a insulin binding region and a EGF tyrosine kinase segmint. From the passage, rL r.nor it ut the insulin receptor is believed to be a insulin activated tyrosine kinase. Recall that ttre uinainf of insulin leads to tyrosine kinase activiry' The mechanism which brings about this activity is termed the signal tinrou.iion puril*uy. with our hybrid receptor, the binding of insulin leads to kinase activity in the EGF segment. This information indicates a similar signal transduction mechanism. The correct choice is B.

of

C is correct' The question tells us

86.

that the binding of glucagon will result in the production of the secondary messenger cAMP' The values of cAMP should thus rise wittr tirne. Using this piece of information, we can eliminate choices A and B. The question then becomes one of knowing the sequence of events. To answer this, one has to use previous knowledge and/or a little common sense. The seco-ndary *"rr"ng"r, produced lead to changes in the cell. Recall that to lruiy bring about change for the cell, there must be some ctrige il pr"bi;r,-;;proreins are the key to function. one very common way to regulate proteins is through their phosph&ylation ana ineii dephosphorylation. It is the secondary messenger that promotes the phosphorylation of certain peptides, converting these proteins into their active forms. The. active forms of the pr-oteins bring about the change. From this i"f;-;ii;n; ;;-;;.;;;; sequence of graphs can be seen. The correct choice is C.

87.

C is correct, a normal insllin receptor. The question refers to a type I, insulin dependent diabetes. These individuals do not produce insulin. That is why they are dependent. This should clue us inlo the fact that their receptors more than likely have nothing wrong with them. The problem is that these individuals have no beta cells in ttre pancreas which release insulin. These individuals take insulin and their body responds appropriately. From this information, we can conclude that the receptors are more than likely normal. The correct ctroice fu C. -

88.

D is correct, stimulation of pyruvate dehydrogenase. We know from the passage that GIp mimics certain insulin responses. Therefore, we should look for an insulin response. Choice D inlicates a stimulation of pyruvate dehydrogenase, the enzyme converting pyruvate into acetyl CoA. This would indicate an acceleration of giycolysis and the Krebs cycle. This is certainly one of the effects of insulin, as the hormone promotes the use of glucose as a source of fuel. The correct choice is D.

89.

C is correct, increased CO2 production. The question tells us that the addition of bivalent anti-insulin receptor antibodies induces a response without insulin. What will antibodies cause? The key word here js bivalent. The bivalent structure will induce crossJinking, leading to a clustering of insulin receptors. This conclusion is bolstered by the next sentence of the question which states that monovalenl andbodies, *hi.h produce cross-linking, do not induce a response in the absence of insulin. However, if we add antibodies "unnot against these monovalent antibodies/fragments, we should be able to induce crossJinking and thus clustering. Therefire, we should be able to induce an insulin-like response, which involves the promotion of glucose u. u fu-"1. When glucose is oxidized, the carbons are released as carbon dioxide. The correct choice is C.

Copyright

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Biology 90.


Section V Answers

B is correct, reducing conditions. This question is answered from our understanding of the conditions which are necessary to form or break disulfide bonds. When two cysteine residues are oxidized, the bond is formed and they fotm cystine. In order to break the bond, cystine must be placed in reducing conditions. From the passage, we know

that the dimer insulin receptor is held with disulfide bonds. To isolate a component of the receptor for study, reducing conditions would be used to break these disuifide connections. The correct choice is B.

91.

B is correct, keratinocytes. Hepatocytes are cells found in the liver. The liver has numerous functions, such as the regulation of metabolism of carbohydrates, lipids, and proteins. The liver stores glycogen and is the primary site of gluconeogenesis. The liver is a major source of cholesterol in the body and is a mijor sto.uge site forlron. The liver

is also the site where many hormones are degraded and where many toxins are inaitivate d,. Leukocyte.r (white blood cells) are a celluiar constituenf of blood, aiong with erythrocytes and platelets, and they play an important role in the body's defensive system. Leukocytes come in five classes: neutrophils, eosinophils,'baiophils, monocytes, and lymphocytes. Lymphocytes can be categorized as being either B-cells, which mature in the bone marow, oi T-cells, which mature in the thymus. Lymphocytes participate in the immune response.

Keratinocytes are skin cells of the epidermis. This cell type (of all the cell types listed) is closest to the exposure of ultraviolet light and allows for the conversion of 7-dehydrocholesterol to pievitamin D3. In the skin previiamin D3 can be converted to vitamin D3, which is the precursor to the active form, 1,25-(OH2)-D3. The correct choice is B. 92.

C is correct, integument. In the passage it states that humans obtain vitamin D3 either by ingestion or through synthetic mechanisms that are initiated by cells in the epidermis of the skin. If we ingest vitamin D3, it will end up being absorbed by the cells of the intestine. However, vitamin D3 is not the primary precursor for t,is-ioH 2)-D3. in the skin (integument) the primary precursor (7-dehydrocholesterol) reacts with UV light to form previtamin b3, which in turn will form vitamin D3. The correct choice is C.

93.

C is correct, thyroid gland. In the passage it states that PTH is referred to

as parathyroid hormone. It would seem logical that PTH would therefore be derived from the parathyroid gland. There are four parathyroid glands imbedded within the thyroid gland. One pair is located in the superior poles of the bi-lobed thyroid gtanA, wnlte the orher pair is located in the inferior poles. The parafollicular cells (C cells) are located within the thyroid gland itself. The correct choice is C.

94.

C is correct, II and III only (polypeptide and steroid). In the second paragraph of the passage we are told that 1,25(OH)2-D3 passes into a target cell and then into the nucleus where it .ompie*es with a releptor protein that has a DNA binding site. This is the general mechanism for action of steroid. hormones. We can eliminate choice A as a

possible answer.

In the third paragraph we find that both PTH and CT have genes encoded in the short arm of human chromosome 1l and are synthesized as a preprohormone from different primary RNA transcripts. If they are both synthesized as a preprohormone from an RNA transcript, it must mean that amino acids are being linked together in peptide linkages at the ribosome. In other words, a polypeptide is being synthesized. The mature PTH polypeptide contiins 84 amino acids while the mature CT polypeptide contains 32 amino acids. These three hormones fail under the steroid and polypeptide classes of hormones. The correct choice is C. 95.

D is correct, I and III only (hypocalcemia and

hyperphosphatemia). What does PTH do under normal

circumstances? Based on information in the passage we find that PTH acts on bone and on the distal tubule of the kidney to promote Ca2o reabsorption and inhibits the reabsorption of PO43 e in the proximal tubule of the kidney. PTH aiso stimulates the synthesis of 1,25-(OH)2-D3 in the kidney. We saw that 1,25-(OH)2-D3 promores the uprake

of Ca2o and PO43e from the intestinal lumen. If we did not have adequate levels of PTH secretion, we would expect low levels of Ca2@ reabsorption (hypocalcemia) and a decrease in urinary POo3e excretion. A decrease in urinary PO43e excretion means that PO43e is being reabsorbed by the proximal tubules of the kidney. This will lead to high (hyperphosphatemia) plasma leveis of PO43 e. Th" correct choice is D.


332


Biotogy 96.

lmm unology & Endocrinology

B is correcr [Ca23,. :. .: goin-t to '.," .- ' nar. -r - ! 1]qaq:1L!..

reabsor:,- , \\'e se- ,:. _:.;

Section V Answers

Be careful it reading the graph. Arong the x-axis we see the totar plasma ,t_,'' ,,ues. what iswhen .-. '_- .t --=:going to happen when the plasma levels of cad ;;;;;;we are ' - :-: :' :: c:-= tn the plasma' How can we do this? we increase the levels of pTH. In the fourth '-: :::!::: -: s'ries that PTH acts on bone and on the distal tubuie of

".,-

:.,

f;;ilr:;;:trTrto

increase the levels of

PrH ,"in

t;;;,o

the kidney to p.o,not" cuze d""."ur". ir,i, i, exactly whar

o o I

low

high

Total Plasma [Ca2+1

what happens when plasma Ca2 e levels are high? The parafollicular cells begin to secrete calcitonin which acts on osteoclasts of bone tissue' As the calcium levelJare iowered 1oy a m"cirunismihat i, not .o-pi"teiy understood) the calcitonin leveis begin to drop. This is exacrly wtrat is trro*n )n'*r";;;il l;;;;il;;;:'rdJ'J;.rect choice is B.

97.

iljl,i"liTl:111"*",,,,:",1'|}?::1,:"^1::i1:::lil.ven il

the passage, rer,s use our tesr_raking skius. choice A is

il,:"_T:;liiJil:.#i::l:": l*^l^::.""::,i:*.',l:i 111,;?;;ffiH;;ii,Tio"liiil'f:'Jii,'f i:;*'J:$,:: ilyf:.?'#T.il?"".1i,T1'Jil'":u:j:""x:l'^:*:::li'lr:::Tid!11'!''i'tr:ilru:#J:i'J'll,i,?l?li,iilii, the co*ecr term for wBCs squeezing out is criaieded. i;;;;"ii ifroi"" i, n.

98'

D is correct' hydrolyze bacterial proteins' Think of what enzymes do and how they are named, A lipase hydrolyzes lipids' so choice A is incorrect' An amvtas"_o;si;.o;i;#;;;;;;ffiffiil,1"t1'j, ,o B is incorrecr. A DNA, so cho'ce c i' "rroi." or proreoryric

H:r:::."ffiXf,lf1'tt'e 99'

il;;;.;

f.olur"

D is correct' toxic oxygen products. Toxic-oxygen products are form free radicals' The pasidge explains the cJmmuni.ution toxic

enzyme hydrolyzes proreins.

rhe

because they are fiee radicals or decompose to choice C as incorrect.

,ot" or.ytoti*r, ,'o

The proteases directly. attack bacteria protein and hydrolyze i""at, buido not usl-rre. "rj-inur" iaai"ats. Eliminate choice B. Chemicai attractants signal neutrophils to move to infection area, so eliminate ciroic" a. The correct choice is D.

100'

ililffiTl"TiJ"

is nitric oxide' HNo3 is nitric acid. H2N2o2 is nitramide. No2 is nirrogen dioxide. The

101' c is correct,

TNF is mainly present in the general circulation. In the question, we are told that binding TNF halts the abnormal cytokine cascade' Therefore choice A is incorrect. usuatiy proteins ,r-rui cells by endocytosis are degraded in lysosomes'.If tle receptors w-ere taken "rrt", up in this .rrunn"., then they would be degraded and not able to interact with TNF' choice B is incbnect. rf the rxr i"cepto.s are injected in the general circularion and work, then TNF must be present in the general circulation. This means choice b l, .ort""t-uiJ-rr, contradict choice D. The correct choice is C.

lo2'

A is correct, neutrophils. By correct reading, we can eiiminate choices B and c. Erythrocytes are red blood cells, so choice B is incorrect. T-eukocvte is simplv Inother name fo. rhi,J;iJo"d;i, #i;i": c is incorrect. Neutrophils are 60-7}vo of the wBC population, *uling them the mort abundant. The correct choice is A.

103' B is correct' I

and-II only. cytokines do not

kill

bacteria directly, they serve as communicators to activate or

deactivate wBCs. This is stated in the passage. The correct choice is B.

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333


EBI{ELEY R.E.V.l.D.W'"

PERIODIC TABLE OF THE ELETTENTS

2

1

H

He 4.0

1.0

4 Be

5

6

7

8

9

10

Li

B

C

N

o

F

Ne

6.9

9.0

10.8

12.O

r4.0

16.0

19.0

20.2

3

l1

12

13

14

15

16

17

l8

Na

Mg

AI

Si

P

s

CI

Ar

23.0

24.3

27.0

28.1

31.0

32-t

35.5

39.9

19

20

21

22

ZJ

24

25

26

2'7

28

29

30

31

3Z

33

34

35

Ca

Ti

Cr

Mn

Fe

Co

Ni

Cu

Zn

Ga

Ge

As

Se

Br

Kr

39.1

40.1

Sc 45.0

v

36

K

47.9

50.9

52.0

54.9

55.8

58.9

58.7

63.s

65.4

69.7

72.6

74.9

79_O

79.9

83.8

52 Te 127.6

53

54

I

Xe

126.9

131.3

3t

38

39

40

41

42

44

46

47

48

49

50

5l

Rb

Sr

Y

Zr

Nb

Mo

43 Tc

45

Ru

Rh

Pd

Ag

Cd

In

Sn

Sb

85.5

87.6

88.9

95.9

(e8)

101.1

102.9

106.4

107.9

t12.4

114.8

118.7

127.8

77

78

79

80

81

82

83

84

85

Ir

86

Pt

Au

Irg

TI

Pb

Bi

Po

At

Rn

91.2

92.9

55 Cs

56

57.

72

t3

74

75

76

Ba

Lal

Hf

Ta

w

Re

Os

132.9

137.3

138.9

178.5

180.9

183.9

186.2

190.2

104

105

106

1W

108

58 Ce

59

60

Pr

Nd

140.1

140.9

r44.2

90

9l

92

Th

Pa

U

8-l

88

t92.2 195. I r09

r10

197.0 200.6

lll

2M.4 201.2 209.0 (20e) (210)

1t2

i?u Rf Db Sg Bh Hs Mt Uun Uuu Uub Ra (223) 226.0 227.0 (261) (262) (263) (262) (26s) (266) (269\ (272) (27',|) Fr

232.0 (231) 238.0

61

62

Pm Sm (145) 150.4 93

94

63

64

65

66

67

68

69

70

7l

Eu

Gd

Tb

Dy

Ho

Er

Tm

Yb

Lu

152.0

r5'7.3

158.9

167.3

168.9

173.0

175.0

95

96

97

101

102 No

103

r62.5 164.9 98

99

100

BK Am Np Cm cf Es Fm Pu (237) (244) (243) (247) (247) (251) (252) (2s7)


Md

Lr

(2s8) (2s9) (260)

(222)

TBR Bio1 Opt

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