BIOLOGY Form 4 Chapter 7

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4551/2 Chapter 7:Respiration 2014

7.1 The respiration process in energy production No (a)

Marking scheme

Marks

Aerobic respiration Anaerobic respiration

OR Process

Respiration equation

S

Glucose +oxygen

R

Glucose

Carbon dioxide +Water +2898 kJ energy

Carbon dioxide +ethanol+210 energy

Name the process R and S

(b)

R:Anerobic respiration

1

S:Aerobic respiration

1

Table 1 shows the respiration equation shown by muscle cells and yeast cells during cellular respiration Cell type

Respiration equation

(Smooth) Muscle cells

Glucose +oxygen

Yeast cells

Glucose

Carbon dioxide +Water +2898 kJ energy

Carbon dioxide +ethanol+210 energy

1 1

(a) Fill in the table by writing in muscle cells or yeast cells that matches with its respiration equation (c)

2

State where tissue V(smooth muscle cell) can be found in the body Blood vessel/alimentary canal/oeosophagus/stomach/uterus/urinary bladder/etc

(d)

2

1

1

Write the equation of process S and R Process R Glucose

lactic acids + energy

2

Process S Glucose +Oxygen

Carbon dioxide + water +2898 kJ

138

2

Reactant- 1m

1

Product -1m

1

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(e)

Explain process P and Process Q / Explain the cellular respiration process that occurs in individual P and Q Process P F1 - aerobic respiration. P1 - glucose is completely oxidized/breakdown in the presence ofoxygen P2 - releases more energy/2898 kJ of energy ( per mole of glucose) E3-Produce carbon dioxide and water Process Q F2 - Anaerobic respiration P3 - glucose is not completely oxidized// the glucose molecules breakdown partially (into lactic acid) P4 - releases less energy/150 kJ of energy 9 per mole glucose) E6-Produce lactic acid

(f)

1 1 1 1 1 1 1 1

6

Anaerobic respiration in cells Explain the condition of a person after completing a 100 meter race in 12 seconds 2

(g)

F-the person is panting /higher breathing rate

1

E1-As he is in oxygen debt//anaerobic respiration

1

E2-Much lactic acids is produced (in his muscle cells)

1

E3-Causes muscle cramp Any 2

1

4

When a person is resting, the heartbeat rate is 61 to 71 beats per minutes .During vigorous activity, the heartbeat rate increase to 120 beats per minute Explain this statement

(h)

F1 - (During the vigorous activity) the muscle cells are in state of oxygen deficiency/oxygen debt //the blood cannot supply oxygen fast enough to meet the demand for oxygen ATP

1

P1-( The increase in heated beat rate ) is to deliver more glucose to muscle cells

1

P2-To induce extra energy cellular respiration

1

P3-To remove more carbon dioxide from the muscle cells Any 2

1

4

After completing vigorous exercise, an athlete will gasp heavily Based on the above statement the condition faced by the athlete Oxygen debt (reject: anaerobic respiration is a process, not a condition) Explain why E1-Because of oxygen deficiency//lack of oxygen

1

E2-To get more oxygen immediately

1

E3-To oxidize lactic acids Any 2E

1

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(a)

(b)

(c)

(d)

(e)

Explain how the oxygen intake by the athlete returns to the normal level at the 25th minute P1-Lactic acid has been removed from the muscle

1

P2-The lactic acids has been converted to energy/convert to glucose

1

2

Explain the condition of a person after completing a 100 meter race in 12 seconds F-the person is panting /higher breathing rate

1

E1-As he is in oxygen debt//anaerobic respiration

1

E2-Much lactic axids is produced ( in his muscle cells)

1

E3-Causes muscle cramp Any 2

1

2

Explain the usage of cell W in bread making industry F1-Carbon dioxide released

1

E1-Traps in the dough

1

E2-Causes the dough to rise

1

2

Explain what happen to the yeast cells if there is too much ethanol produced P1-( too much ethanol0 causes unsuitable medium /condition //toxic/poisonous medium /condition

1

P2-For yeast cells to reproduced //yeast cell die

1

2

State the differences between the process that mention I 6(a) (i) Diagram shows respiratory organs in an insect and human (Prefer) Aerobic respiration

Anaerobic respiration

D1-Oxidation of glucose in present of oxygen/ Oxygen is required

D1-Oxidation of glucose in absent of oxygen / Oxygen is not required

1

D2-Oxidation of glucose is complete/ Complete breakdown of glucose

D2-Oxidation of glucose is not complete/ Incomplete breakdown of glucose

1

D3-Produced higher/large energy/38 ATP/2898 kJ of energy 9 per mole of glucose)

D4-Produced lower energy /2 ATP/150 kJ of energy ( per mole of glucose)

1

D4-Produced carbon dioxide and water

D4-Produced lactic acid

1

D5-Occurs in mitochondria

D5-Occur in cytoplasm

1

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(f)

Diagram shows the rate of oxygen intake before, during and after a vigorous Exercise of an athlete.

(i) Based on the graph, compare the respiration before and during the vigorous Exercise. 4

1

Before (A)

During (B)

Explanation (E)

Aerobic respiration

Anaerobic respiration

Before-Oxygen Intake is low/the same as oxygen required /enough oxygen is supplied to the cell

3

During-Oxygen required is more than oxygen intake 2

The muscle are in normal condition

The muscle are in the atate of oxygen debt

Before-Oxygen is sufficient

3

During-Oxygen is insufficient/oxygen supplied is less than oxygen supplied 3

Energy produced is more /38 ATP

Energy produced is less /2 ATP

Before-complete breakdown of glucose (produce more energy ) During-incomplete breakdown of glucose (produce less energy)

4

No/less accumulation of lactic acid in the muscle

High accumulation of lactic acids in the muscles

Before-complete incomplete break down of glucose produce carbon dioxide and water Dduring -Incomplete breakdown of glucose produce lactic acid

A+B=1m

3

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E=1m (Any 1 E)



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(g)

Explain what happens to cell w when there is no oxygen F1-Cell W undergoes anaerobic respiration

1

E1-Glucose break down (partially/incompletely)

1

E2-To produce ethanol, carbon dioxide

1

E3-Less ATP/2 ATP is produce

1

2

F1 and any of E1/E2/E3 (h)

the above process takes place in tissue P in the presence of oxygen .Name and explain the process F-Process is called aerobic respiration

1 1 1

P1-Glucose diffuses into cells P from the blood capillary P2-Cells P contain a lot of mitochondria P3-Mitochondria ( contain enzymes) for cell respiration //mitochondria carry out cell respiration P4-Oxidation of glucose (take placed in mitochondria) P5-In a series of reaction catalyzed by respiratory enzymes in mitochondria P6-1 molecule of glucose will produce 38 molecule ATP/ More ATP P7-water and carbon dioxide are released as waste material in this process (i)

1 1 1 1 1 1

8

Explain the importance of increased pulse rate during vigorous activity and why it takes several minutes for the pulse rate to return to normal after activity 6 During vigorous activity, P1 more blood is sent to the muscles P2-so that oxygen supply to the muscles is increased P3-The heart beats faster P4-to deliver more blood, hence the pulse rate increases After some time during the activity, P5-respiration takes place anaerobically P6-because the maximum rate of oxygen uptake is less than oxygen demand. P7-there is build up of lactic acid P8-After activity, a period of recovery is needed to provide the oxygen P9-so that the lactic acid can be oxidized and to provide the energy for the recovery of the muscles

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(a)

Molecule X + 2ATP Process Q

- Anaerobic respiration

Molecule X - Lactic acid P1- Inhale more oxygen by doing fast and deep breathing. P2-Excess oxygen taken in during inhalation is used to oxidize lactic acid to carbon dioxide

1 1 1 1

and water.

1

P3-This oxidation process takes place in the liver. P4-Thus the oxygen debt is the amount of oxygen needed to remove the lactic acid from the muscle cells. Lactic acid + oxygen

carbon dioxide + water + energy

1 6 1

(b)

P1-The muscle cells of the athlete undergoes anaerobic respiration to produce energy P2-During intensive physical activity / running / sprinting// when the athlete start running (t =

1 1

0), oxygen requirement increase immediately to produce large amount of energy P3-The athlete holds his breath for a short period of time // the athlete breath is shallow during running P4-The oxygen supplied by breathing between t = 0 minute to 6 minute is insufficient for cellular respiration P5-The muscle cells are now in the state of oxygen debt // undergo oxygen deficit P6-Glucose is broken down incompletely without the presence of oxygen

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P7- Small amount of energy is released to continue the activity P8-Lactic acids produced accumulate in the muscle causing the muscular pain and fatigue P9-The anaerobic respiration occurs in the cytoplasm P10- (after the activity is over), the athlete breathes faster and deeper to supply more oxygen P11-Oxygen is used to oxidize the lactic acid into carbon dioxide, water and energy // converted into glucose and stored as glycogen

1 1 1 1 1

10

7.2the respiration structure and breathing mechanism in human and animal No (a)

Marking scheme

Marks

Adaptation of the respiratory structures State two characteristic shown by the respiratory surface of animal(common characteristic) P1-the respiratory surface is moist P2-Cells lining respiratory structure are thin P3-Thr respiratory structure has a large surface area

1 1 1

3

The respiration structure and breathing mechanism insects Aspect Respiratory structure

Question & Marking Scheme

Marks

The respiration structure and breathing mechanism insects P:Air sac Q: Muscle T: Spiracle S: Trachea

R:Tracheole

Name the part labeled P ,Q ,Rand S Which organism has the respiratory structure? Insect Name the respiratory system shown in diagram 2.1 Tracheal system State the function of the following (i) Chitin support the tracheal/prevent the tracheal form collapsing (ii) Air sac Speeds up the movement of gases exchange to and form tissue during vigorous body movement

5

5

1

1

1

1

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Aspect


Marks

Explain one adaptation of the respiratory structure in diagram for efficient gaseous exchange

Structural Adaptation

P1-The large number of tracheoles provides a large surface for the diffusion of gases

1

P2-Tips of tracheoles have thin permeable wall and contain fluid in which respiratory gases can dissolved

1

P3-Terminal ends the tracheoles remains moist which allows teh gases to be dissolved

1

2

Explain how structure Q and S increase the efficiency of gaseous exchange in each organism 2 F-Consists of million alveoli in lungs and many tracheal Tubes/Tracheole/thin layer/1 cell thick

1 1

P1-To increase total surface area per volume rate for gaseous exchange F2-The inner surface of alveolus and tracheoles end consists of tissue fluid moisture P2-To provide moist surface for gas diffusion /to dissolve oxygen /gases for diffusion Any F +P Breathing mechanism

1 1

2

State how air is drawn from T to S 2 P1-By(rhythmic) movements, of the abdominal muscles P2-Decreasing of air pressure inside trachea, ( so the air is drawn in) P3-Gases diffuses into the cells(s)

1 1 1

2

Diagram 7.1 (i), (ii) and (iii) show the respiratory structure of an insect. Describe the respiratory structure and breathing mechanism of and insect R-respiratory structure R1-The tracheal system consists of network of trachea

1

R2-The trachea is lined with chitin to prevent dorm collapsing R3-Spiracles is tiny opening thet allow air to go in and out

1

R3-spiracles is tiny opening that allow air to go in and out

1

R4-The trachea branch into fine tubes celled tracheole

1

R5-The tracheole branch throughout the body and temperature and penetrate into body tissues / muscle

1



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4551/2 Chapter 7:Respiration 2014 Breathing mechanism B1-When inside inhales, the abdominal muscles relax and spiracles open

1

B2-air pressure inside the trachea decrease and air is drawn in

1

B3-When the insect exhale, the abdominal muscle contract

1

B4-So increase air pressure in side trachea and forces air out through spiracles

1

B5-Inesct inhale and exhale through rhythmic contraction and expansion of their abdominal muscles

1

B6-the body movement and contraction of abdominal muscle speed up the rate of diffusion of gases from trachea into body cells

1

8

Breathing mechanism

Explain the gases exchange between tracheol and body cell. P1-Partial pressure/concentration of oxygen in the tracheole is higher /than partial pressure/concentration of oxygen in body cell

1

P2- Oxygen diffuse from tracheole to body cell

1

P3- Partial pressure/concentration of carbon dioxide in the body cell is higher than partial pressure/concentration of carbon dioxide in tracheole . P4- Carbon dioxide diffuse from tracheole to body cell

1

1

4 1

Chitin is a polysaccharide on the outer surface of structure P. Due to the change in the environment, the insect is unable to form the polysaccharide. Explain how the absence of chitin affects inhalation and the energy production. 6 P1- The function of chitin is to prevent trachea from collapsing/sustain the air pressure P2- During inhalation high pressure air moves into the trachea. P3 -The absent of chitin will cause the trachea / P to collapse / burst /rupture. P4 -Air with oxygen cannot reach tracheal. P5-Body cell cannot get enough oxygen for cellular respiration P6-The insect does not produce enough energy and respire anaerobically. P7-Less energy produced. (Any 6)

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Aspect


Marks

Breathing mechanism

Diagram show a trachea system of and insect Based on the diagram explain the gases exchange between the tracheoles and muscle cells F-there are concentration gradient of oxygen and carbon dioxide between tracheoles & body cells

1

E1-(simple) diffusion can take place

1

E2-Oxygen concentration /partial pressure is higher in the tracheoles while the concentration of oxygen is lower in the cells

1

E3-Oxygen diffuses directly form the tracheoles onto the cells

1

E4-Carbon dioxide concentration is higher in the cells while lower in the tracheoles

1

E5-Carbon dioxide diffuses directly form the cells into the trachoeles

1

4

The respiratory structure and breathing mechanism of fish Aspect Respiratory structural

Marking scheme

Marks

The respiratory structure and breathing mechanism of fish

What is X ?/ Name the respiratory structure of the organism in diagram Gills/ gill filament

1

1

1

1

1

1

State the function of structure P P-Speed up the movement of gases to and from the insect’s tissue The efficiency of gaseous in organism Y is further enhanced by a mechanism. Name the mechanism Countercurrent exchange mechanism State two characteristic of X, which makes it a good respiratory structure for fish 2

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P1-Have lamella and filament to increase total surface area

1

P2-Numberous blood capillaries for efficient transport of respiratory gases

1

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Aspect Structural adaptation

Breathing Mechanism

Breathing mechanism

Inhalation


Marks

Explain one adaptation of the respiratory structure in diagram 1.1 (b) and diagram 1.2 (b) for efficient gaseous exchange P1-Th e have numerous thin walled lamellae to maximize the surface area for gaseous exchange

1

P2-The gills filament have numerous thin membrane and covered by net work of capillaries to transport respiratory gases

1

P3-the surface of gills Is moist which allows the gases to be dissolved

1

Based on the diagram explain how the oxygen is drawn from mouth to X(gill) P1-Mouth closes

1

P2-The floor of buccal cavity raised (water contain air flow to X)

1

2

Describe the inhalation in fish E1-th floor of cavity lowers

1

E2-At the same time, the opercular cavity enlarges and operculum closes

1

E3-This lowers the pressure in buccal cavity

1

E4-Water with dissolved oxygen is drawn into the mouth

1

4

Describe the breathing mechanisms in fish.

P1 - When the mouth opens, the floor of the buccal cavity is lowered./Increase the volume/ space of the buccal cavity P2-opercular cavity enlarges and operculum closes P3 - This lowers the pressure in buccal cavity . P4 - Water with dissolved oxygen is drawn into the mouth. Exhalation

2

P5 - When the mouth closes, the floor of buccal cavity is raised. P6 - Water flow through the lamellae and gaseous exchange between the blood capillaries and water takes place. P7 - Oxygen diffuses from the flowing water through the gill lamellae into the blood capillaries. P8 - Carbon dioxide diffuses from the blood capillaries via the gill lamellae into the flowing water. Any 4

1 1 1 1

4

1 1 1 1

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The respiratory structure and breathing mechanism of amphibians Aspect


Marks

Respiratory structural

Name structure X and Y in diagram 3.1

2

1

X: Bucco-pharyngeal

1

Y: Glottis Structural adaptation

Breathing Mechanism

2

Respiratory gases flow in and out through the lungs .Describe the characteristic of the frog’s lungs E1-Numerous inner partition to increase the surface area

1

E2-Membrane of lungs are thin and moist to facilitate the efficient diffusion of respiratory gases

1

E3-Supplied with a rich network of blood capillaries to transport respiratory gases to the body cells

1

3

Structure Y in diagram 3.1 had been injured .Describe how this condition affect the respiration of the frog E1- Glottis unable to open and close

1

E2-Air pressure is not increased /decrease in the bucco-pharyngeal cavity

1

E3-Air cannot be forced into /out the lungs

1

E4-Lung ventilation is not efficient

1

4

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The respiratory structure and breathing mechanism of humans Aspect

Marking scheme

Marks

Respiratory structure

Name the parts labeled Y Y-Alveolus

1

1

1

1

1

1

1

1

What is the function of alveoli? Place for gaseous exchange //store the oxygen gas before gaseous exchange occur State the organ in which the tissue in Diagram 4.1(alveolus) can be found Lung State the function of organ stated in Gaseous exchange//respiration Respiratory gases flow in and out through the trachea .Describe the characteristic of trachea F-Have C-shaped cartilage rings //cartilage rings

1

P1-keep the trachea open permanently

1

P2-Avoid the trachea form collapse when the out side pressure is higher than inside pressure

1

P3-oxygen can continuously flow through trachea to the alveoli/lung F-1m P-1m Explain the effects of the breathing mechanism if structure R is unable to function P1-Structure R is diaphragm.

1

P2-Less/no change in volume in the thoracic cavity/ lung

1

P3-Less/ no change in air pressure in the thoracic cavity/ lung

1

P4-Less/ no air exchange/ less/no intake of O2/ less/no CO2 expelled

1

3

1

4

Resulting difficulty in breathing in and out Structural adaptations

State the important characteristic of alveoli to ensure the function in (a) is efficient 1 P1-Have very large total surface area//

1

P2-Have moist surface all the time//

1

P3-have very thin wall/one cell thick Note ( any 1P)

1

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4551/2 Chapter 7:Respiration 2014 Describe the characteristic of the respiratory structure of human that enable gaseous exchange to be carried out efficiently

Breathing mechanism

P1-the ratio total surface area per volume (TSA/V) is high for the exchange of gases

1

P2-the cells lining the respiratory surface is a single layer of cell which is very thin to allow gases to diffuses easily

1

P3-the respiratory surface is constantly moist to allow gases to dissolved in water before diffusing in and out of the respiratory surface

1

P4-the respiratory surface is covered with a dense network capillaries to allow rapid diffusion and transport of gases

1

3

Describe how intercostals muscle and diaphragm can change the volume and pressure in the thoracic cavity during inhalation P1-External intercostals muscle contract/internal intercostals muscle relax caused the ribs cage moves out wards and upwards

1

P2-Diaphgram muscle contract , the diaphragm lower and flattenP3-The volume of thoracic Cavity increase but the pressure decrease (lower the atmospheric pressure)

1

P3-The volume of thoracic cavity increase but the pressure decrease ( lower the atmospheric pressure)

1 1

P4-Air forced into the lung//alveolus

3

Describe the breathing mechanism of human Inhalation: P1-External intercostals muscle contract//internal costal muscle relax

1

P2-ribcage move upwards and out wards

1

P3-diaphragm contracts/flattens

1

P4-Volume of thoracic cavity increase // pressure of thoracic cavity decrease

1

P5-So air ( form outside) is forced into the lungs

1

Exhalation : P1-External intercostals muscle relax//internal costal muscle contract

1

P2- ribcage move downwards and inwards

1

P3-diaphragm relax/curved upward

1

P4-Volume of thoracic cavity decrease // pressure of thoracic cavity increase

1

P5-So air ( form inside) is forced out of lungs

1

6

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Constructing a model of human lung study the breathing mechanism in humans No

Marking scheme

(a)

Marks

Rubber cork Glass tube Bell Jar Balloon Thin rubber sheet String Based on the model of the lungs in Figure 3.1, what are the equivalent structures to the glass tube and the bell jar in the human respiratory system? Glass tube: Trachea / Bell jar : Rib cage / ribs Balloon : lung Rubber sheet: diaphragm

(b)

1 1 1 1

4

1

1

The thin rubber sheet represents the diaphragm in the human respiratory system. What is the function of the thin rubber sheet in the model of the lungs? To increase / decrease the pressure / volume in the bell jar

(c)

The balloons represent the human lungs. Explain one characteristic of the balloons which is similar to the human lungs[2 marks] F- elastic E- can expand (inhalation) and contract/ decrease in size (exhalation )

(d)

1 1

2

1

1

(c) (i) The string in the model of the lungs is released.. Draw the changes to the balloons in Diagram 3.2 below.

-both balloons decrease in size (e)

(ii) Observe your drawing in (c)(i). Explain the relationship between the changes in the model of the lungs you have drawn and the real human respiratory system. P1- the string represent the diaphragm P2- when the diaphragm muscles contract, P3- the volume of the thorax increase

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P4- this will decrease the thorax pressure P5- air will be inhale (f)

1

3

1

1

The percentage of oxygen and carbon dioxide gases in inspired and expired air is determined by using the J-tube. Why is the end of the J-tube dipped in potassium hydroxide solution and then followed by potassium pyrogallol solution? 1 To prevent oxygen gas being absorbed by the potassium pyrogallol solution as it can absorb both carbon dioxide and oxygen

(g)

(ii) Table 3.3 shows the result of a study on the content of inspired and expired air. Type of gas

Inspired air / %

Expired air / %

Oxygen

21.0

16.0

Carbon dioxide

0..04

4.0

Nitrogen gas

78.0

78.0

Water vapour

Vary

Saturated

Explain why there is an increase in percentage of carbon dioxide in the expired air. P1-The concentration of carbon dioxide is higher in the cell body; released from the cellular respiration P2-Carbon dioxide diffuses into the blood to be transport to the lungs.

1 1

2

Comparison of respiratory system between human and insect No

Marking scheme

Marks

(a) S

R

P Q

State one similarity and one difference of structure P in diagram 2.1 and 2.2 Similarity: both wall of P consisting ring to strengthen it Differences: the wall of P in insect consists of chitin ring while P in human consists of cartilage ring (b)

1 1

2

Humans and cockroach have different respiratory system .Explain one difference between the respiratory system of human and a cockroach F1-Respieratory structure of cockroach consists of trachea and spiracles while the respiratory structure of human consists of a trachea and a pair of lungs

1

P1-tracheae of cockroach are branch into 2 bronchi which enter the right and left lungs Module Biology Trial Paper Collection


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P2-Thr trachea of human branched into 2 bronchi which enter the right and left lungs

1

P3-The bronchi of human branched ito smaller tubes called bronchioles which ends in a cluster of sacs called alveoli

1

3

(c)

Explain one similarity and four differences between the respiratory organs of insect and human Similarities

1

S1-Both of respiratory organs has thin wall/one cell thick E1-Incrase rate of diffusion of respiratory gaseous

1

OR

1

S2-Both of respiratory organs has respiratory surface such as alveolus in human and tracheole in an insect

1

E2-Provide a large surface area for the diffusion Differences D1-Trachea in human is supported by cartilage and traches in insect is supported by chintin

1 1

E1-To prevent them form collapsing D2-The wall of alveolus is moist surface but the tracheole has fluid

1

E2-To dissolve the respiratory gases D3-Alveolus is covered by network of blood capillaries but not for trachoele

1

E3-T provide a large surface area for rapid diffusion of gases 9 to and form the alveoli0 in human but tracheole direct contact to the tissue ( and organs)

1

D4-Haemoglobin is needed in transport of oxygen nt but in insect

1

E4-oxygen combine with heamoglobin in (erythrocyte) to form oxyhaemoglobin but not in insect

1

D5-(larger) insect have air sacs but not in human

1

E5-to speed up the movement of gases to and form the insect’s tissue

1

D6-in human air enters the lungs through the nostrils but spiracles in insects

1

1

E6-to allow gases in and out of the body any 4 pairs

10

What differences between the respiratory system of frog and fish D1-Gills is the respiration organ for fish but lung and skin ids for frog

1

D2-Gill have filament and lamella to increase the surface area, but lung of frog have numerous inner partition to increase the surface area

1

D3-Gill received oxygen directly form water , but lungs and skin of frog received oxygen form the atmosphere

1

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(d)

Describe the comparison between the respiratory system in insect and human 8 Similarities:

1

F1-The structure of tracheal system and trachea branches into small tubes E1-increase the total surface area of tracheole/alveolus so that increase the efficiency of gases exchange F2-moist surface on tracheole and alveolus

1

E2-Oxygen and carbon dioxide can be dissolve easily

1

F3-Very thin wall of tracheole and alveolus/one cell thick

1

E3-To ensure the simple diffusion can take place /Increase rate of diffusion of respiratory gaseous Insects Aspect Human F4-Consists of spiracles, Respiratory structure Consists of nose trachea, trachea and tracheoles bronchus, bronchioles ad alveolus E4-Air enters through Air enter through nose into spiracles into tracheoles lungs/alveolus F5-Tracheoles directly contact Alveolus is surrounded by a with the muscle cells large network of blood capillaries F6-Trachea is reinforced/ supported with ring of chitin E6-Prevent the trachea form collapsing due to different air pressure F7-Does not have red blood cell to transport oxygen

Trachea is reinforced/ supported with ring of cartilage P5-Prevent the trachea form collapsing Oxygen transportation

E7-Oxygen is not transported in the body F8-Oxygen diffuses directly form the respiratory tructure into the cells E8-Carbon dioxide is directly released form the cells into tracheoles

The diffusion of oxygen into the cells Product of respiration

1

Has red blood cells to transport oxygen through blood vessels Oxygen is transported by red blood cells around the body Oxygen needed to be transported into the cells and then diffuses into the cells Carbon dioxide produced diffuses into the blood capillary then transported into the lungs

1 1

1 1

1 1

1

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Comparison of respiratory system between human and fish No

Marking scheme

Marks

(a)

Explain three adaptation from structure show in diagram 2 (b)(ii) to carry out its function efficiently

(b)

(c)

(d)

P1-Thin membrane /one cell thick for easily diffusion of respiratory gases

1

P2-Moist surface for respiratory gases easily dissolve

1

P3-Numerous blood capillaries for efficient transport of respiratory gasesAny 2

1

3

Y is the respiratory surface in human, explain how gaseous exchange occurs between structures Y and blood capillary P1-t he partial pressure of oxygen in Y is higher than in blood capillaries

1

P2-Oxygen diffuses form Y into blood capillaries by simple diffusion

1

2

Humans and fish have different respiratory systems, Explain one differences between the respiratory system of human and fish 3 F1-the respiratory system of fish of gills while the respiratory system of human consists of a trachea and pair of lungs

1

P1-A fish has four pairs of gills which are covered by operculum//the surface of each gills Filament has many plate –like projections called lamella

1

P2-the trachea of human branched into 2 bronchi which enter the right and left lungs//The bronchi of human branched into smaller tubes called bronchioles which ends in a cluster of sac called alveoli

1

3

What are the differences between respiratory system of human and fish? P1-gill is the respiratory organ for fish nut is for human

1

P2-gill have filament and lamella to increase the surface area, but lung have alveoli to increase the surface area

1

P3-gill touch /surrounded by water

1

P4-Gill receives oxygen directly from water, but lung received oxygen form atmosphere via trachea , bronchus and bronchioles

1

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7.3Gaseous exchange across the respiratory surfaces and transport of gases in humans The process of gaseous exchange across the surface of the alveolus and blood capillaries and between the tissue capillaries and the body tissue cells No (a)

Marking scheme

Marks

State the importance of gaseous exchange in human P1-To get oxygen for (cellular) respiration

1

P2-To get rid of/excrete the carbon dioxide

1

2

(b)

Name gas X and Y

1

X : Oxygen Y : Carbon dioxide (c)

1

Explain the difference between the concentration of gas x and Y in blood vessel Q F1 : The concentration of gas X in blood vessel Q is lower than gas Y E1 : Oxygen has been used by the body cells /cellular respiration E2 : (Cellular respiration) produces gas Y E3 : to be sent to the lung (to be excreted)

(d)

(e)

(f)

2

1 1 1 1

2

Name blood vessel P and Q P: Pulmonary veins

1

Q:Pulmonary artery

1

2

State the function of blood vessel P and Q P: Carries deoxygenated blood to lungs

1

Q: carries oxygenated blood back to heart

1

2

Describe the role of blood vessel P in transporting oxygen form alveolus to muscle cells P1-In the blood, Oxygen form alveolus combine with respiratory pigment/haemoglobin to form oxyhaemoglobin /oxygenated blood

1

P2-Transport oxygenated blood //oxyhaemoglobin to heart

1

P3-the heart pump the oxygenated blood to muscle cells via the aorta Any2

1

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(a)

State the process by which gaseous exchange takes place across alveolus1 (Simple) diffusion (b)

1

1

1

1

1

Explain how the process occurs F-Partial pressure of oxygen /carbon dioxide in the air of the alveolus is higher than in blood capillary

(c)

1

Gaseous exchange takes place across structure Y Name structure Y Alveolus/ Alveoli

(d)

State two ways how the alveolus are adapted for efficient gaseous exchange P1-Thin wall P2-Moist P3-Rich with blood capillary

(e)

1 1 1

2

Explain how the alveolus is structured to increased the efficiency of gaseous exchange F1 : Alveolus has thin wall ( one cell thick) E1 : Gaseous can diffuse in and out through the wall more efficiently / Quick /easy gases diffusion

1

F2 : The (inner) surface of the alveolus is moist E2: Allowing oxygen to dissolve first before diffusing out F3 : A large number of alveoli /The (outer surface) of the alveolus is covered by a network of blood capillaries

1 1 1

P1-Large total surface area per volume for gaseous exchange F4-Network of blood capillaries P4-To increase the rate of gases transportation F+P=1m E3 : Increase the surface area for rapid diffusion of gaseous Notes : F1/2/3 + E 1/2/3 = 2 mark F1/2/3 = 1 mark E1/2/3 = O mark

1

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(f)

Describe the movement of respiratory gases across structure Y P1-Partial pressure of oxygen on alveolus is higher than the partial pressure of oxygen in the blood capillaries//oxygen concentration is higher in alveolus than in the blood capillaries

1

P2-Oxygen Diffuses form alveolus into the blood capillaries

1

OR P3- Partial pressure of carbon dioxide on alveolus is higher than the partial pressure of oxygen in the blood capillaries/Carbon dioxide oxygen concentration is higher in alveolus than in the blood capillaries P4- Carbon dioxide diffuses form alveolus into the blood capillaries (g)

1 1

4

Explain the role of oxygen in the muscle cells F-oxygen oxidized the glucose molecule

1

E1-Cellular respiration /aerobic respiration takes place in muscle cells

1

E2-ATP/energy released

1

E3-Produced carbon dioxide and water as by product/waste products

1

E4-energy is used for contraction and relaxation of muscle cells/movement of insect

1

No

Marking scheme

4

Marks

(a)

R P Q Based on the diagram 3.2 name X and Y

(b)

(c)

X: oxygen

1

Y: Carbon dioxide

1

2

Name structure P and Q X: Red blood cell

1

Y:Alveolus

1

2

1

1

Name the complex substances contained in X Haemoglobin

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(d)

Explain how the gaseous exchange occur across the alveolus 3 P1 : Oxygen diffuse/ moves across /through ( plasma membrane) to blood capillary P2: From higher (oxygen ) concentration ( in alveolus )to lower concentration ( in blood capillary) P3: On the other hand the partial pressure of carbon dioxide is lower in the air of the alveoli compared to the blood capillaries. P4: Carbon dioxide diffuses out of the blood capillaries into the alveoli. P5 : expelled through the nose or mouth into the atmosphere

1 1 1 1 1

3

(e)

Explain how gaseous exchange occurs during respiration in Diagram 4.1 (in human )

(f)

F1-Oxygen diffuses from alveolus into blood capillaries

1

E1-Oxygen concentration /partial pressure in alveolus is higher than in blood capillaries

1

F2-Carbon dioxide diffuses from blood capillaries to the alveolus

1

E2-Carbon dioxide concentration /partial pressure in the blood capillaries is higher than in alveolus MAX:2

1

2

Explain how the red blood cell accepts oxygen form alveolus and transfer to the cell P1-Oxgen diffuses into the blood plasma

1

P2-Combine with haemoglobin

1

(g)

2

CO2 O2

R

S

. Based on the diagram , explain the exchange of respiratory gases P1-Respiratory surfaces in human are alveoli.

1

P2-The concentration of oxygen in the alveoli is higher than its concentration in the blood

1 160

capillaries. 1 P3-Oxygen in the alveoli diffuses into the blood capillaries. Module Biology Trial Paper Collection


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P4The concentration of carbon dioxide in the blood capillaries is higher than its concentration

1

in the alveoli. P5-Carbon dioxide diffuses from the blood capillaries of the lungs into the alveoli. P6-Blood leaving the blood capillary of the lungs has higher concentration of oxygen and lower concentration of carbon dioxide

1 1

6

7.4 The Regulatory mechanism in respiration The human respiratory response and rate of respiration in different situation

Diagram 7 (ii) shoes 3 different situation of human activities Diagram 7 (ii) (a)) shows a boy watching television Diagram m 7 (ii(b)) shows a man is chased by a fierce dog Diagram 7 (ii(c)) shows a man climbing a mountain Explain the effect of the 3 different situations towards the physiological process that occur in organ X as shown in diagram 7 (ii)



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Aspect Diagram 7 (ii) (a)) shows a boy watching television

Marking scheme

Marks

F1-At rest, the respiratory rate is normal /12-20 breaths per minute

1

P1-The partial pressure of O2and CO2 are normal

1

F2-When a person is in fear, breathing rate increase

1

P2-It’s needed because the demand of a higher respiration rate in cells

1

P3-In order to oxidize more glucose

1

P4-To produce more energy

1

P5-(then), rapid muscles contraction (as a responded to the dangerous situation /running)

1

F3-( in mountain climbing) as the altitude increase, the atmospheric pressure of decrease

1

P6-Thus, partial pressure of O2becomes lower

1

P7-Causes a drop in the oxygen level in blood

1

P8-(the person will face difficulty in breathing

1

P9-So, the person will experience headache/nausea/dizziness

1

2

(Relaxing) Diagram m 7 (ii(b)) shows a man is chased by a fierce dog (In fear)

Diagram 7 (ii(c)) shows a man climbing a mountain (At high altitude)

5

5

The regulatory mechanism of carbon dioxide content in the body No

Marking scheme

Marks

(a)

30 breath per minute while the heartbeat rate increase to 120 beats per minute .Explain how the body During vigorous activities such as swimming running and aerobic the breathing rate increase to about regulates the carbon dioxide content in human body 7 P1-during vigorous exercise , the partial pressure of carbon dioxide increase //rate of cellular respiration increase

1



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P2-Thus , carbon dioxide reacts with water to form carbonic acids

1

P3-(due to high level of co2 in blood ), its results in a drop im the pH value of the blood ( and)/cerebrospinal fluid

1

P4-The drop in pH is detected by (central) Chemoreceptor’s (in the medulla oblongata P5-Send the nerve impulse to the respiratory centre / (which is in turn sends nerve impulse to) diaphragm and intercostals muscles

1

P6-Pespiratory muscle to contract and relax faster

1

P7-breathing and ventilation rates faster

(b)

1

1

P7-Breathing and ventilation rates increase

1

P8-Excess CO2is eliminated from the body

1

P9-CO2concentration /pH value so blood return to normal levels Any 7p

1

7

1 1 1 1 1 1 1

4

In an experiment, a boy takes part in an 800 meter event track. His exhaled air was obtained three times which were before running, right after he finished running and 10 minutes after running to determine the percentage of carbon dioxide. Table 3.1 shows the result of the experiment.

Percentage of carbon dioxide (%)

Before running

Right after he finishes running

After 10 minutes running

4%

7.5%

4%

Based on the table 3.1, Explain how the percentage of carbon dioxide is returned to normal after 10 munites running 4 E1 : The high concentration of carbon dioxide E2 : decreases the blood pH E3 : Detected by central chemoreceptor and/ peripheral chemoreceptor E4 : Impulses are sent to the respiratory centre E5 : (Impulses are sent to) the cardiac and respiratory muscles E6 : Increase the heart beat and breathing rate E7 : To remove excess carbon dioxide (so that the of carbon dioxideis returned tonormal) Notes : Choose any three Es

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7.5 the importance of maintaining a healthy respiratory system No (a)

(b)

Marking scheme

Marks

Explain how smoking can harm the respiratory system in human F1-Cigarette smoke contain tar

1

E1-Causes lugs cancer

1

F2-cigarette smoke contain acidic gases

1

2

Explain why does this occur? F1 : Cigarette smoke contains carbon monoxide E1 : (Carbon monoxide) has higher affinity to bind with hemoglobin compared to oxygen E2 : forms carbaminohaemoglobin E3 : Therefore, less oxygen will bind with hemoglobin to be transported in blood vessel

1 1 1 1

2

P Notes : F1 + any two Es (c)

(d)

(e)

Explain why carbon monoxide is poisonous to the body cells P1-C02 has higher affinity to bind with heamoglobin the with oxygen //CO2 reduce the ability of haemoglobin to combine with oxygen

1

P2-the body cells lack oxygen //Less oxygen is transported to the body cells

1

2

Smoker do not realize that they destroy their respiratory organ during smoking, Explain how this habit will affect the intake of oxygen efficiency E1-Carbon monoxide

1

E2-Bind with haemoglobin to form carboxyhaemoglobin

1

E2-Less oxygen combine with haemoglobin

1

E4-Tobacco tar will be deposited/logged /accumulate (inside the lungs)

1

E5Reduce diffusion of oxygen

1

E6-Haet fom the smoke m

1

E7-Dry the surface of the alveoli

1

E8-Oxygen cannot be dissolved Any 4

1

4

Explain the effects of smoking on the human respiratory system. P1-Carbon monoxide competes with oxygen to bind with haemoglobin and forms

1 1

carboxyhaemoglobin. It reduces the supply of oxygen to the cells. P2Nitrogen dioxide can dissolve in mucus to form an acidic medium which erodes lung tissue.

1

P3- BENZO-(α)-PYRENE is carcinogenic chemical that can cause cancer.

1

P4-Nicotine can stimulate the production of cancer cell in trachea and lung.

1

164

P5-Heat and dryness irritation the lungs and can lead to laryngitis 1

4



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7.6Respiration in plants The intake of oxygen by plants for respiration No (a)

Marking scheme

Marks

Like animals, plants also respire aerobically to obtain energy for metabolism . They derive most their energy from cellular respiration .during cellular respiration, the plants cells take in oxygen and release carbon dioxide Based on the above statement, describe the intake of oxygen by the plants for respiration S1-The intake occurs by diffusion mainly through stomata and lenticels S2-Stomata can be found in epidermis of leaves. the stem of herbaceous plants S3-Lenticels can be found on the stems and root of plants

1 1

Explanation P1-When stomata open, they connect the air space (within the leave) to atmosphere P2-Oxygen form the atmosphere diffuses into the air spaces P3-then dissolves in the film of water around the mesophyll cells P4-So the concentration of oxygen in the cells becomes lower than in the air spaces P5-Thus, oxygen diffuse continuously form air space to the cell P6-During daytime, carbon dioxide that is produced during respiration is used in photosynthesis P7-The excess carbon dioxide diffuses into the air spaces and then through stomata into atmosphere (b)

1

1 1 1 1 1 1

7

Diagram 6.1 shows the surface view of lower epidermis in a leaf of a plant. Diagram 6.2 shows part of cross section of a woody stem. Broken epidermis

Pore M Epidermal cell

Cork tissue

Guard cell

Pore M

Explain the gas uptake for respiration through pores M and N in the plant Through M: F- (In day time) stoma / M (in the epidermis of the leaf) open P1-Oxygen from the atmosphere diffuses (through stoma) into intercellular air spaces ll (and palisade mesophyll) P2- follow the concentration gradient Through N: P3- At the lenticels (N) oxygen from atmosphere diffuses into the air spaces between cork cells which are loosely arranged P4- then diffuses into the cells at the stem /and old roots

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Respiration and photosynthesis in plants No (a)

Marking scheme

Marks

Diagram 6.4 shows the changes in the volume of carbon dioxide absorbed or released by a plant in different light intensity

State the relationship between light intensity and rate of transpiration P1-Light increase as the rate of transpiration increase P2-The plant carries out anaerobic respiration (b)

(c)

Explain the changes in the volume of carbon dioxide absorbed or released by a plant in different light intensity P1-glucose is broken down in the absence of oxygen to release energy produces ethanol, CO2 (and energy) P2- cells in the roots of rice plants are extremely tolerant of ethanol P3-Many of the roots are very shallow P4-the roots use the oxygen which diffuses into the water surface. P5-Rice stem contain a large number of air spaces P6-(the air space) allow oxygen to penetrate through to the cells of roots ( growing in the absence of oxygen) Explain the relationship between the rate of photosynthesis and the rate of respiration in the plant at points P, Q, R and S. At P : P1-In the dark / low light (intensity), only respiration occurs P2-hence large quantity of CO2 is produced/released P3-As light (intensity) increases the quantity of CO2 / produce decreases P4because part of CO2 produced during respiration is used for photosynthesis P5-sugar used in respiration more rapidly than it is produced in photosynthesis At Q: P6- (At this point of light intensity) all the CO2 release from respiration is reused / equivalent to CO2 used up during photosynthesis // no net gain or loss in CO2 / sugar produced P7- rate of photosynthesis is equal to the rate of respiration P8-this point is called compensation point P9-net gaseous exchange is zero

1

2

1

1 1 1 1 1 6

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At R: P10- as light intensity increases, the rate of photosynthesis become faster than / exceed the rate of respiration P11-the CO2 needed is obtained from the atmosphere (at the same time) excess O2 is releases (into the atmosphere) At S: P12- is the light saturation point P13-an increase in light intensity does not increase the rate of photosynthesis // maximum rate of photosynthesis (Any 8) (d)

1 1 1 1 10

An experiment on a plant is carried out to study the rate of water loss from 0500 to 0300 the next day. Graph 6.1 shows the result of the experiment and diagram 6.2 shows the structure of a stoma and the cells found in the epidermal layer of a leaf.

Based on the graph, explain how light intensity and the structure in diagram 6.2 affect the rate of water loss 10 F1 : From 0500 to 0170, the rate of water loss increases E1: Light intensity increases E2 : stimulates photosynthesis in the guard cells./ (The guard cells) start producing glucose E3 : This makes energy available for potassium to move into guard cells E4 by active transport E5 : (The guard cells) become hypertonic (compared to the cell sap) of the epidermal cells.

1 1 1 1 1



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E6 : Water molecules from the epidermal cells diffuse into the guard cells by osmosis E7 : Causing the guard cells to bend outwards E8 : the stoma opens (to allow water to escape to the atmosphere through it) F2 : From 0170 to 0300, the rate of water loss decreases E9 : Light intensity decreases / causes the rate of photosynthesis to decrease / soon stop. E10 : The guard cells become flaccid E11 : and bend inwards E12: The stoma closes and this prevent water molecules to escape through it. Notes : (F1 + any 5 Es) + (F2 + 3 Es)

1 1 1 1 1 1 1 1

10

Comparision between photosynthesis an d respiration No (a)

Marking scheme

Marks

Explain the differences between the process in organelle P and Q Site

Organelle P / mitochondria

Organelle Q/ chloroplast

Process

Respiration

Photosynthesis

Aim /purpose

Released energy

Stores energy

Raw material

Glucose, oxygen

Water, carbon dioxide, light

Products

Energy, water , carbon dioxide

Glucose / starch water and oxygen

1 1 1 1

Energy

Not required light energy

Required in form of light 1

(b)

4

The intake of oxygen by plants for respiration

State two differences between tissues in diagram 4.1 and 4.2 Tissue in diagram 4.1

Tissue in diagram 4.2

D1-Alveolus

Leaf

D2-Carry out transpiration

Carry out photosynthesis

D3-Absent of chlorophyll

Presence of chlorophyll

1 1 1

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Extra Question Diagram 7.1 shows how the respiratory gases are transported in the human body

(i) Based on Diagram 7.1, explain how the transport of oxygen and carbon dioxide takes place in the body cells Aspect Transport of oxygen

Marking scheme P1: The blood circulatory system transport oxygen from the alveoli to the body cells. P2: Oxygen combines with the haemoglobin in the red blood cells P3: to form oxyhaemoglobin (which is unstable.) P4: Oxygen is carried (in form of oxyhaemoglobin) to the tissues (which have a low partial pressure of oxygen.) P5: The (unstable) oxyhaemoglobin breaks down into oxygen and haemoglobin again. P6: Oxygen (molecules are) transferred to the body cells

Transport of

P7: Carbon dioxide binds (itself) to the haemoglobin P8: (and is) transported in the form of carbaminohaemoglobin. P9: Carbon dioxide is (also) transported as dissolved carbon dioxide (in the blood plasma.) P10: Most of carbon dioxide is carried as bicarbonate ions (dissolved in the blood plasma.) P11: When the blood carrying carbon dioxide reaches the body cells, the carbon dioxide diffuses into the blood plasma and combines with the red blood cells. P12:Carbon dioxide reacts with water to form carbonic acid. P13:Carbonic anhydrase in the red blood cells catalyse the formation of carbonic acid. P14: The carbonic acid then dissociates into a hydrogen ions and bicarbonate ions. MAXIMUM: 6 marks

carbon dioxide



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BIOLOGY Form 4 Chapter 7

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