Ch 7 - Problems

Problems

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1. Estimate K from the following data describing ligand binding to a protein. [Ligand] (mM (mM )

Y

0.25

0.30

0.50

0.45

0.80

0.56

1.4

0.66

2.2

0.80

3.0

0.83

4.5

0.86

6.0

0.93

Answer:

2. Which set of binding data is likely to represent cooperative ligand binding to an oligomeric protein? (a) [Ligand] [Ligand] (mM) (mM )

Y

0.1

0.3

0.2

0.5

0.4

0.7

0.7

0.9

(b) [Ligand] [Ligand] (mM) (mM )

Y

0.2

0.1

0.3

0.3

Problems


[Ligand] [Ligand] (mM) (mM )

Y

0.4

0.6

0.6

0.8

Answer:

Set b describes sigmoidal binding to an oligomeric protein and hence represents cooperative binding.

t he cell surface and 1 torr at the t he mitochondria (the organelles organelles 3. In active muscles, the pO2 may be 10 torr at the where oxidative metabolism occurs). Use Eq. 7-6 to show how myoglobin ( p50 = 2.8 torr) facilitates the diffusion of O2 through these cells. cells. Answer:

According to Eq. 7-6,

When pO2 = 10 torr,

When pO2 = 1 torr,

The difference in values is 0.78 – 0.26 = 0.52. Therefore, in active muscle muscle cells, cells, myoglobin myoglobin can transport a signifi significant cant amount a mount of O 2 by diffusion diffusion from the cell surface to the th e mitochondria. high concentrations concent rations of CO2 in the blood; there are no direct 4. In humans, the urge to breathe results from high physiological sensors of blood pO2. Skindivers Skindivers often hyperventilate (breathe rapidly rapidly and deeply for several minutes) just before making a dive in the belief that this will increase the O 2 content of their blood. (a) Does it do so? (b) Use Use your knowledge knowledge of hemoglobin hemoglobin function to evaluate whether this practice is useful. Answer:

significantly affect the O 2 concentration, since (a) Hyperventilation eliminates CO2, but it does not significantly the hemoglobin in arterial blood is already essentially saturated with oxygen.

Problems


(b) The removal of CO2 also removes protons, according to the reaction

The resulting increase in blood pH would increase the O 2 affinity of hemoglobin hemoglobin through th rough the Bohr effect. The net result would would be that less oxygen could be delivered to the tissues until until the CO 2 balance was restored. Thus, hyperventilation hyperventilation has the opposite of the intended effect e ffect (note that since hyperventilation hyperventilation suppresses the urge to breathe, breat he, doing so may cause the diver to lose consciousness due to lack of O 2 and hence drown). 5. Drinking a few drops of a commercial preparation called “vitamin O,” which consists of oxygen and

sodium chloride chloride dissolved dissolved in water, is claimed claimed to increase the concentration conc entration of oxygen in the body. (a) Use your knowledge of oxygen transport to evaluate this claim. (b) Would vitamin O be more or less effective if it were infused directly into the bloodstream? Answer: (a) Vitamin O is useless because the body's capacity to absorb oxygen is not limited by the amount of oxygen available available but by the abil a bility ity of hemoglobin hemoglobin to bind and transport O2. Furthermore, oxygen is

normally introduced into the body via the lungs, so it is unlikely that the gastrointestinal tract would have an efficient mechanism mechanism for extracting extract ing oxygen. oxygen. delivery in vertebrates requires a dedicated O 2-binding protein (hemoglobin) (b) The fact that oxygen delivery indicates that dissolved dissolved oxygen by itself cannot attain att ain the high concentrations required. Moreover, a few drops of vitamin O would make an insignificant contribution to the amount of oxygen already present in a much larger volume volume of blood. 6. Is the p50 higher or lower than normal in (a) hemoglobin Yakima and (b) hemoglobin Kansas? Explain. Answer:

(a) Lower; (b) hig h igher. her. The Asp 99 β → His mutation of hemoglobin Yakima disrupts a hydrogen bond at the α1 – β2 interface of the T state (Fig. 7-9a), causing the T R equilibrium to shift toward R state (lower p p50). The Asn 102β → Thr of hemoglobin Kansas causes the opposite shift in the T R equilibrium by abolishing an R-state hydrogen bond (Fig. 7-9 b). 7. Hemoglobin S homozygotes who are severely anemic often have elevated levels of BPG in their erythrocytes. Is this a beneficial effect? Answer:

The increased BPG helps the remaining erythrocytes deliver deliver O 2 to tissues. However, BPG stabilizes the T conformation of hemoglobin, so it promotes sickling and therefore aggravates the disease. 8. In hemoglobin Rainier, Tyr 145β is replaced by Cys, which forms a disulfide bond with another Cys residue in the same subunit. This prevents the formation of ion pairs that normally stabilize the T state. How does hemoglobin Rainier differ from normal hemoglobin with respect to (a) oxygen affinity, (b) the Bohr effect, and a nd (c) the Hill Hill coefficient? Answer:

destabilizes the T conformation of hemoglobin hemoglobin Rainier, the R (oxy) (a) Because the mutation destabilizes conformation is more stable. Therefore, the oxygen affinity of hemoglobin Rainier is greater than normal.

Problems


normally form in deoxyhemoglobin deoxyhemoglobin absorb protons. The absence of these ion (b) The ion pairs that normally pairs in hemoglobin hemoglobin Rainier Rainier decreases the t he Bohr effect (in fact, the Bohr Boh r effect in hemoglobin hemoglobin Rainier is about half that of normal hemoglobin). hemoglobin Rainier is is more stable than the T conformation, even (c) Because the R conformation of hemoglobin when the molecule molecule is not oxygenated, O2-binding cooperativity is reduced. The Hill coefficient of hemoglobin Rainier is therefore less than that for normal hemoglobin. which can remain under water without breathing for up to 1 h, drowns its air-breathing 9. The crocodile, which prey and then the n dines at its leisure. An adaptation that aids the crocodile in doing so so is that it can ca n utilize utilize virtually virtually 100% of the O2 in its blood whereas humans, for example, example, can extract only on ly ~65% of the O2 in their blood. Crocodile Hb does not bind BPG. However, crocodile deoxyHb preferentially preferentially binds . How does this help the crocodile obtain its dinner? Answer:

As the crocodile remains under water without breathing, its metabolism metabolism generates CO 2 and hence the content of its blood increases. The preferentially preferentially binds to the crocodile's deoxyhemoglobin, deoxyhemoglobin, which allosterically prompts the hemoglobin to assume the deoxy conformation and thus release its O 2. This helps helps the crocodile stay under water long enough enough to drown its prey. 10. Some primitive animals have a hemoglobin that consists of two identical subunits. (a) Sketch an oxygen-binding curve for this protein. (b) What is the likely range of the Hill coefficient for this hemoglobin? Answer: (a)

Hill coefficient most likely likely has a value between 1 (no cooperativity) and 2 (perfect (b) The Hill cooperativity between the two subunits). 11. Is myosin a fibrous protein or a globular protein? Explain. Answer:

Myosin is both fibrous and globular. Its two heads are globular, with several layers of secondary structure. Its tail, t ail, however, consists of a lengthy, fibrous coiled coil. coil. 12. A myosin head can undergo five ATP hydrolysis cycles per second, each of which moves an actin monomer by ~100 Å. How is it possible for an entire sarcomere to shorten by 1000 Å in this same period?

Problems


Answer:

Because many myosin heads bind along a thin filament filament where it overlaps a thick filament, and because be cause the myosin molecules molecules do not n ot execute execu te their th eir power strokes simultaneously, simultaneously, the thick t hick and thin t hin filaments filaments can move past each other by more than t han 100 Å in the interval between power po wer strokes of an individual myosi myosinn molecule. 13. Rigor mortis, the stiffening of muscles after death, is caused by depletion of cellular ATP. Describe the molecular basis of rigor. Answer:

In the absence a bsence of ATP, ATP, each myosin myosin head adopts a conformation co nformation that does not allow itit to release its bound actin a ctin molecule. Consequently, thick and an d thin filaments filaments form a rigid rigid cross-linked cross-linked array. 14. Explain why a microfilament is polar whereas a filament of keratin is not. Answer:

Microfilaments Microfilaments consist consist entirely of actin subunits that are assembled in a head-to-tail head-to -tail fashion fashion so that t hat the t he polarity of the subunits is preserved in the fully assembled fiber. In keratin filaments, however, successive heterodimers align in an antiparallel fashion, so that in a fully assembled intermediate filament, filament, half the molecules are oriented in one direction and half are oriented in the opposite direction (Fig. 6-16). 15. Give the approximate molecular masses of an immunoglobulin G molecule analyzed by (a) gel filtration chromatography, (b) SDS-PAGE, and (c) SDS-PAGE in the presence of 2-mercaptoethanol. Answer:

(a) 150–200 kD; (b) 150–200 kD; (c) ≈23 kD and 53–75 kD. 16. Explain why the variation in VL and VH domains of immunoglobulins is largely confined to the hypervariable loops. Answer:

The loops are on the surface of the domain, so they can tolerate to lerate more amino acid substitutions. substitutions. Amino Amino acid changes in the β sheets would be more likely to destabilize the domain. raised against a native na tive protein sometimes fai faill to bind to the corresponding denatured 17. Why do antibodies raised protein? Answer:

The antigenic site site in the t he native protein usually usually consists of several peptide segments that are a re no longer longer contiguous when the tertiary structure of the protein is disrupted. Antibodies raised against against a macromolecular macromolecular antigen usually produce an antigen–antibody antigen–antibody precipitate p recipitate 18. Antibodies when mixed with that antigen. Explain why no precipitate forms when (a) Fab fragments from those antibodies are mixed with the antigen; (b) antibodies raised against a small antigen are mixed with that small small antigen; antigen; and (c) the antibody a ntibody is in in great excess over the antigen and vice versa. Answer:

(a) Fab fragments are monovalent and therefore cannot cross-link antigens antigens to produce a precipitate. (b)

Problems


A small small antigen antigen has only one antigenic antigenic site and therefore cannot bind more than tha n one antibody a ntibody to produce a precipitate. (c) When Whe n antibody is in great excess, most antibodies that are bound to antigen bind only one per immunoglobulin molecule. When antigen is in excess, most immunoglobulins bind to two independent antigens. antigens.

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Ch 7 - Problems

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