Respiration

HKALE 1994 Biology Paper I.......................................................................................7

HKALE 1994 Biology Paper II .....................................................................................8

HKALE 1996 Biology Paper I.......................................................................................9

HKALE 1994 Biology Paper I.....................................................................................22

HKALE 1994 Biology Paper II ...................................................................................23

HKALE 1996 Biology Paper I.....................................................................................25

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HHHKKKAAALLLEEE 111999888999 BBBiiiooolllooogggyyy PPPaaapppeeerrr IIIIII 4. (a) Define the term respiratory quotient. (2 marks)

(b) What information about the metabolism of an organism can be deduced from the

values of the respiratory quotient ? (4 marks)

(c) Give an account of the various means by which oxygen and carbon dioxide are

transported in the blood of a mammal. Indicate also their relative importance to the

total carrying capacity of blood for oxygen and carbon dioxide. (7 marks)

(d) Give an account of the factors that exist under physiological conditions to favour

(i) the release of oxygen from blood at the level of the tissue, and

(ii) the release of carbon dioxide from blood in the lungs. (7 marks)

Suggested Solution…

HHHKKKAAALLLEEE 111999999000 BBBiiiooolllooogggyyy PPPaaapppeeerrr IIIIII 1. (a) What is oxygen debt ? Explain its physiological significance. (5 marks)

(b) Explain the effect of an increase in surrounding temperature from 20� to 30� on

the respiration rate of a named terrestrial ectotherm. (4 marks)

(c) Explain why aquatic ectotherms using gills for gaseous exchange may be more

affected than terrestrial ectotherms by the same increase in surrounding

temperature. (2 marks)

(d) Describe how a change in the concentrations of carbon dioxide and oxygen in the

inspired air can affect the rate and depth of breathing in man. (4 marks)

(e) With reference to the graph below, explain the relative positions of the three

oxygen dissociation curves in relation to the size of the respective animals. (5 marks)


HHHKKKAAALLLEEE 111999999111 BBBiiiooolllooogggyyy PPPaaapppeeerrr III 4. By means of a flow diagram, briefly outline the three main stages of aerobic respiration

with a carbohydrate substrate. Annotate your diagram to show the essential features of

each of the three main stages.

(N.B. Chemical formulae of individual compounds are NOT required .) (7 marks)


HHHKKKAAALLLEEE 111999999222 BBBiiiooolllooogggyyy PPPaaapppeeerrr III 10. The ability of cells to synthesize ATP from ADP and inorganic phosphate is coupled with

the oxidation of substrates by oxygen.

(a) In an experiment to study the effect of potassium cyanide (KCN) on the respiration

of yeast cells, 4 tubes (with contents as shown in the table below) were prepared

and the corresponding rates of oxygen uptake were measured.

Volume of substances added (ml) Tube

No. Buffered yeast suspension 2M glucose solution 0.1M KCN solution

Rate of

oxygen uptake

(arbitrary units)

1 2.00 0 0 6.1

2 2.00 0.01 0 17.8

3 2.00 0.01 0.01 6.7

4 2.00 0.01 0.02 2.2

(i) Calculate the final concentrations (in moles per litre) of the following

substances in tube 3 :

(1) glucose

(2) KCN (3 marks)

(ii) Why was respiration observed in tube 1 even in the absence of added

glucose ? What was the result of adding glucose to tube 2 ? (2 marks)

(iii) What can you conclude about the effect of KCN on the respiration rate of

yeast cells ?

(2 marks)

(b) Mitochondria were isolated from the yeast cells and incubated in an isotonic buffer

containing a respiratory substrate, ADP and inorganic phosphate. The amount of

oxygen uptake and the loss of substrate and inorganic phosphate were measured.

Time(min) Oxygen uptake

(µmole)

Loss of substrate

(µmole)

Loss of inorganic

phosphate (µmole)

5 0.33 0.62 2.38

15 0.78 1.58 5.22

30 1.49 2.94 9.39

45 2.15 4.35 13.37

(i) Plot the amount of oxygen uptake, consumption of substrate and phosphate

loss against time. (4 marks)

(ii) Determine, from your graph, the quantitative relationship between

(1) oxygen uptake and the consumption of substrate.

(2) oxygen uptake and the loss of inorganic phosphate. (2 marks)

(iii) How would the relationship in (ii) differ from those when glucose is oxidized in

intact cells under aerobic conditions ? Explain your answer. (4 marks)

(iv) Explain briefly the functional significance of the cristae membranes of

mitochondria. (3 marks)


HHHKKKAAALLLEEE 111999999444 BBBiiiooolllooogggyyy PPPaaapppeeerrr III 9. With the aid of an oxygen dissociation curve, explain how the relationship between

oxygen tension and haemoglobin saturation facilitates oxygen uptake at the respiratory

surface and oxygen release in the tissues. (6 marks)


HHHKKKAAALLLEEE 111999999444 BBBiiiooolllooogggyyy PPPaaapppeeerrr IIIIII 2. (a) Describe the three stages of cellular respiration for carbohydrate metabolism. (10 marks)

(b) Compare and contrast the products of the metabolic process in (a) in the presence

and absence of free oxygen. (3 marks)

(c) What is the role of ATP ? Describe the part ATP plays in three other named

metabolic processes and cite an example for each process. (5 marks)

(d) Explain why protein is less efficient for energy production when compared with (i)

lipid ; and (ii) carbohydrate. (2 marks)


HHHKKKAAALLLEEE 111999999666 BBBiiiooolllooogggyyy PPPaaapppeeerrr III 4. The following shows the initial set-up of an experiment performed at room temperature.

Readings were taken at specific time intervals.

(a) Suggest the purpose of this experiment. (1 mark)

(b) To achieve the purpose of this experiment, three parameters need to be

considered. Using these three parameters, construct a formula for such a purpose. (2 marks)

(c) State one precaution for this experiment. (1 mark)


HHHKKKAAALLLEEE 111999999777 BBBiiiooolllooogggyyy PPPaaapppeeerrr III 11.

The figure above shows an experimental set-up used to measure the oxygen

consumption rates of animals at room temperature A mouse was placed into the

respiratory chamber A syringe was inserted into the inlet hose to inject 10 mL of air into

the respiratory chamber The clamp was then closed and the time was noted; this was

the starting time of the experiment When the fluid levels in the manometer arms were

the same, the time was again noted; this was the ending time. The experiment was then

repeated using several grasshoppers A set of the experimental data is presented in the

table below:

Mouse Grasshopper

Body weight

Starting time

Ending time

20 g

2 : 00 p.m.

2 : 09 p.m.

5 g

2 : 30 p.m.

8 : 00 p.m.

(a) At the starting time of the experiment, what happened to the fluid levels in the two

arms of the manometer ? Account for this observation. (1� marks)

(b) Account for the fluid levels in the manometer arms at the ending time of the

experiment. (3� marks)

(c) Calculate the oxygen consumption rates (mL O2 h-1g-1) of the mouse and the

grasshoppers.

(3 marks)

(d) What is the function of the water bath ? State another precaution for this

experiment. (2 marts)

(e) How would the oxygen consumption rates of the mouse and the grasshoppers

change if the temperature in the respiratory chamber dropped from 25� to 5� ?

Explain your answer.

(3 marks)

Total :13 marks


HHHKKKAAALLLEEE 111999999777 BBBiiiooolllooogggyyy PPPaaapppeeerrr IIIIII 1. (a) Explain the role of adenosine triphosphate (ATP) in cellular metabolism. (3� marks)

(b) Describe how ATP is produced in three biochemical processes in mesophyll cells.

In each case, state the site where ATP is formed. (13� marks)

(c) List two animal cell types and one plant cell type where abundant ATP is required.

How is the ATP requirement related to the functions of these cells ? (3 marks)


HHHKKKAAALLLEEE 111999999888 BBBiiiooolllooogggyyy PPPaaapppeeerrr IIIIII 5. Give a comparative account on how the chloroplast and mitochondrion process energy

and discuss the inter-relationship between these two organelles in cellular metabolism. (20 marks)


HHHKKKAAALLLEEE 111999888999 BBBiiiooolllooogggyyy PPPaaapppeeerrr IIIIII 4. (a)

consumed oxygen of volumeproduced dioxide carbon of Volume

= (R.Q.) quotient yRespirator 1

during the same period of time, �

under a steady state condition �

(2)

(b) Types of substrate being metabolized : �

R.Q. = 1 for carbohydrates; �

R.Q. = 0.7 for fats; �

R.Q. = 0.5� 0.8 for proteins �

Occurrence of anaerobic respiration when R.Q. is exceptionally high (>3), 1)

Conversion of carbohydrate to fat (R.Q. > 1) 1) any 2

Occurrence of carbon dioxide fixation (R.Q. < 0 5) 1)

(e.g. plant � for photosynthesis,

animal � for calcareous shell construction) (4)

(c) Oxygen :

(i) In combination with ferrous ions of the heme groups of

haemoglobin to form oxyhaemoglobin (95%), 1

one molecule of haemoglobin when fully oxygenated carries four

molecules of oxygen

(ii) In physical solution as dissolved gas (5%) 1

Correct order of relative importance �

(2�)

Carbon dioxide :

(i) As bicarbonate ions (85%)

� carbon dioxide diffuses into RBC and combines with water to

form � carbonic acid, catalyzed by carbonic anhydrase

present in RBC.

� haemoglobin (deoxygenated) accepts hydrogen ions from

carbonic acid, leaving behind bicarbonate ions. 2

(ii) In combination with amino groups of proteins haemoglobin in RBC

(major) and plasma proteins (minor), to form carbamino

compounds(e.g. carbamino� haemoglobin (10� 20%) 1+1(bonus)

(iii) In physical solution as dissolved gas and with a small amount as

carbonic acid (5%) 1

Correct order of relative importance �

(4�)

(d) (i) The factors that reduce the affinity of haemoglobin for oxygen are:

� Increased partial pressure of carbon dioxide (Bohr Effect)

produced from tissue respiration. 1

� decreased pH (Bohr Effect) : carbonic acid formed from

hydration of carbon dioxide and lactic acid produced from

tissue anaerobic respiration. 1

� increased temperature due to active tissue metabolism. 1

Steep gradient in the partial pressure of oxygen from blood to

tissue. 1

(4)

(ii) Oxygenation of haemoglobin releases hydrogen ions which shift

the equilibria involving bicarbonate ions; and carbamino

compounds to favour respectively the formation and unloading of

carbon dioxide (Haldane Effect). 2

Steep gradient in the partial pressure of carbon dioxide from blood

to alveolar air. 1

(3)

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HHHKKKAAALLLEEE 111999999000 BBBiiiooolllooogggyyy PPPaaapppeeerrr IIIIII 1. (a) Oxygen debt is a state of oxygen depletion after extreme physical

exertion; measured by the amount of oxygen required to restore the

system to its original state. During vigorous exercise, the metabolic rate

and hence the demand of oxygen of the active muscles increases

greatly. The oxygen delivered to the muscle is insufficient to keep pace

with the demand and the muscle cells may undergo anaerobic

respiration and produce ATP by lactic fermentation. 2

As a result, lactic acid builds up in the muscle and the oxygen deficit

resulting from this temporary employment of anaerobic pathway is to be

paid off when muscle returns to rest and adequate oxygen is available.

At this time, lactic acid is converted back into pyruvic acid. Some of the

lactic acid built up is exported to the liver and converted into glycogen. 2

Oxygen debt enables the animal to carry out vigorous exercise beyond

the capacity of aerobic respiration AND without having an unnecessary

large reserve of oxygen and blood supply. Such exercise may be of

great survival value to the animal (e.g. run away, catching preys) 1

(5)

(b) An ectotherm may not be able to regulate its body temperature which

increases with an increase in ambient temperature. 1

With a 10°C rise in body temperature, the metabolic rate (rate of

enzymic reactions) and 1

hence the respiration rate will increase / double according 1

to the Q10 rule / explanation of increasing temperature on rate of

enzymic reaction. 1

(4)

(c) Aquatic ectotherms will be more affected since an increase in water

temperature will decrease the solubility of dissolved oxygen in water,

making oxygen less available (NOTE : saturated value of dissolved

oxygen decreases from 6.19 to 5.27 ml / l in freshwater and from 5.35 to

4.5 ml/l in 30% sea water when water temperature increases from 20 to

30°C) while at the same time, the respiration rate (and hence oxygen

demand) of the animal will double / increase. (2)

(d) The rate and depth of breathing are controlled by a respiratory centre in

the medulls, 1

which is responsive directly or indirectly (through chemoreceptors : the

aortic and carotid bodies in the walls of major arteries) to changes in

CO2, H+ and O2 concentrations in blood. 1

An increase in the concentration of CO2 and H+ in the blood stimulates

the respiratory centre which in turn, increases depth of breathing (and

vice versa) 1

Lack of oxygen stimulates chemoreceptors in carotid and aortic body

and stimulates respiration. Oxygen receptors however, are sensitive

only to large changes in blood PO2 (<70 mm Hg) and is little affected by

slight changes of oxygen in blood. 1

Since oxygen in blood PCO2 and PO2 normally are proportional to one

another, breathing is generally regulated by CO2 in blood. �

(4)

(e) The smaller the animal, the farther its oxygen dissociation curve-is

shifted to the right. 1

The haemoglobin of small animals therefore unloads more of its oxygen

at any given pressure than does the haemoglobin of a larger animal. 1

Heat loss from an endotherm is proportional to its body surface area.

The smaller the animal, the larger is its S.A. / V ratio and hence its rate

of heat loss 1

In order to compensate for the higher rate of heat loss to maintain a

constant body temperature, the smaller animal needs to have a higher

metabolic rate 1

The shifting of the curve to the right allows small animals to obtain more

oxygen to sustain their higher metabolic rates at any given oxygen

partial pressure. (NOTE : partial pressure of oxygen in air and the lung

is normally higher than 100 mm Hg, and animals of all size should have

their haemoglobin fully saturated with oxygen at such partial pressure) 1

(5)

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CO2

loss of molecule of CO2 to

produce a 2C molecule

ELECTRONTRANSPORT

CHAIN (system) (�)

(C4)(Glucose)SUGARS

GLYCOLYSIS

PYRUVIC ACID

Phosphorylation of glucose

which eventually splits into

2 trioses (some ATP

production)

(progressive transfer of

electrons from one

carrier to another, with

the energy released to

form ATP.)

(as the finalelectron

acceptor)

4C cpd regenerated and

perpetuate the cycle (�)

2C compound

6C4C

reducing power

(�)

ATP

(�)

(�)

(�)

(�)

KREBS / CITRIC ACID /

TRICARBOXYLIC ACID

CYCLE (�)

(�)

(�) O2

(�)

(�)

H2O (�)

(�)

(7)

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HHHKKKAAALLLEEE 111999999222 BBBiiiooolllooogggyyy PPPaaapppeeerrr III 10. (a) (i) Either one of the following formulae can be used :

(1) M1V1 = M2V2

M1 = 1

22

V)VM(

(2) volume Finalsubstances dissolved

of Amount

= ionconcentrat Final

The final format in both cases are the same.

M 0.01 = ml 0.02) + (2ml 0.01 2M

= ionconcentrat Glucose×

/ 9.9 × 10-3M 1�

KCN concentration = 0.1M 0.01 ml(2 + 0.02) ml

= 0.0005 M×

/ 4.95 × 10-4M 1�

N.B. For correct answers (1�) each, correct equation but wrong answer (1)

each.

(ii) In tube 1, respiration observed in the absence of added substrate

is due to the presence of endogenous substrate in each yeast cell. 1

In tube 2, the presence of added substrate led to a three fold

increase in respiration as reflected by the increase in the rate of

oxygen uptake. 1

(iii) In tube 3, KCN addition resulted in an inhibition of the respiratory

process despite the presence of added substrate. 1

Double the amount of KCN in tube 4 as compared with that in tube

3 resulted in a further decrease in the respiratory process to levels

lower than in tube 1 i.e. the inhibition is concentration dependent. 1

(N.B. No marks should be given to a description of the mechanism

involved.)

(b) (i) title of graph � T

properly drawn and fully labelled x and y axes 1� A

curves correctly plotted 1� P

curves correctly keyed � K

Uptake of oxygen & disappearance of substrate & inorganic

phosphate from the medium with time

0 5 10 15 20 25 30 35 40 45

8

6

4

2

12

10

14

5

4

3

2

1

inorganic phosphate

Oxygensubstrate

Time (min.)

(ii) (1) There is approximately 2 ± 0.5 µ mole of substrate lost

for 1 µmole of oxygen uptake.

OR There is approximately 0.5 ± 0.1 µ mole of oxygen uptake

for 1 µmole of substrate lost 1

(2) There is 6 ± 0.5 µ mole of inorganic phosphate lost

for 1 µmole of oxygen uptake.

OR There is 0.17 ± 0.02 µ mole of oxygen uptake

for 1 µmole of inorganic phosphate lost. 1

Oxy

gen

up

take

OR

Lo

ss o

f su

bst

rate

( µ

mo

le)

(iii) In the experiment, only 3 ATPs were generated per molecule of the

given substrate oxidized in the mitochondria 1

(suggesting that only 2 electrons were generated in the oxidation

of 1 molecule of substrate).

In the intact cell, glucose undergoes glycolysis in the cytoplasm; 1

the products of glycolysis enter the Krebs cycle in the mitochondria.

1

Electrons generated in the processes are passed down the.

electron transport chain (in the mitochondria) to the final acceptor

oxygen. As a result, the total number of ATPs generated is much

greater (38) than that in the experiment. / more energy is

generated in intact cells 1

(4)

(iv) � infolding of the cristae membrane increases its surface area 1)

� thus providing more surface area for the embedding

enzyme systems / ATPase 1)

� these infoldings (cristae) also allow greater access to enzymes any 3

present in the matrix 1)

� mention of the positioning / sequencing of the enzyme systems to

facilitate efficient biochemical reactions 1)

(20)

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� sigmoid shape curve (�) with plateau near to 95% (�) 1

� axes named 1

� Haemoglobin on r.b.c. has high affinity for oxygen (�)

� At high oxygen tension (�), e.g. at lung surface: % saturation of

haemoglobin with O2 is near to 95%, if this facilitates uptake of O2 (�)

(oxyhaemoglobin at the respiratory surface) 2

� at the tissues, a slight decrease in O2 tension results in rapid

dissociation of oxyhaemoglobin. (Steep part of the curve). This allows

oxygen to be readily diffused to the tissues. 2

(6)

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HHHKKKAAALLLEEE 111999999444 BBBiiiooolllooogggyyy PPPaaapppeeerrr IIIIII 2. (a) Respiration involves three distinct stages : glycolysis, Krebs cycle and

oxidative phosphorylation (electron transport chain).

Glycolysis (�), it is an anaerobic process (�) that occurs in the cytoplasm. (�) 1+ �

The intermediate stages involve the phosphorylation of hexose (�), the splitting

of hexose �

phosphate into 2 triose phosphate (�) before the conversion into 2 pyruvate (�)

� + �

with the production of NADH / NADH2. and ATP (�). �

Before entering the Krebs cycle, pyruvate is both oxidized and

decarboxylated into a two-carbon acetyl (�) group. A molecule of

NADH / NADH2 is produced and the acetyl group is temporarily attached

to coenzyme A (�) (CoA). Acetyl-CoA is the starting substrate for the

Krebs cycle. 1

Upon entering the Krebs cycle (�), the acetyl group is combined with a

four-carbon compound (oxaloacetate) to form a six-carbon compound

(�) (citrate). In the course of the cycle, two of the six-carbons are

oxidized to CO2 (�) and a four-carbon compound oxaloacetate is

regenerated (�), Energy is also released to form ATP (�). NADH /

NADH2, and FADH / FADH2 are also formed (�). (decarboxylation and

dehydrogenation) This occurs in the mitochondrial matrix (�). 5

In the course of the oxidative phosphorylation / electron transport chain

(�), hydrogen / electrons are passed “down-hill” to oxygen (�), and

the energy released is used to form ATP from ADP(�). 2

(b) Similarities

Both aerobic and anaerobic respiration yield energy / ATP �

Differences

More ATP is produced in aerobic respiration. �

In anaerobic respiration (fermentation) – �

in glycolysis, pyruvate is not the end product. In many bacteria, fungi

and animal cells (�), this anaerobic process may result in the

formation of lactate (�). � +�

In yeast and most plant cells (�), pyruvate is broken down to ethanol

(�). � +�

(3)

(c) ATP is hydrolyzed to ADP and Pi, this process liberates energy for

energy consuming metabolic processes. 1

Biosynthesis (any 1 example) (�) �

� ATP provides bond energy in synthesis e.g. formation of protein

from a.a., formation of polysaccharide from monosacchoride,

formation of nucleic acids from nucleotide � +�

Muscle contraction (any 1 example) (�) �

� ATP provides energy for actin-myosin interaction 1

Active transport (any 1 example) (�) �

� energy is required for movement of substances across membranes

against a concentration gradient e.g. food absorption at the small

intestine, reabsorption of NaCl (�) at the nephron (�) for

homeostatic control of salt balance, sodium potassium pump in

maintenance of resting potential in neurones �+�

max. (5)

(d) (i) energy yield is less than lipid 1

(ii) energy generation is less efficient compared to carbohydrate

because it involves more complex metabolic pathway. E.g.

deamination of a.a.

max. (20)

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HHHKKKAAALLLEEE 111999999666 BBBiiiooolllooogggyyy PPPaaapppeeerrr III 4. (a) To find out the rate of anaerobic respiration (�) (rate of carbon dioxide

released or rate of fermentation) (�) of the yeast suspension. (1)

(b) (� mark for each parameter)

volume (amount) of gas produced (�)

mass (concentration) of yeast used (�) ¡ Ñtime taken to collect the volume (amount) of gas (�)

correct formula. (�) (2)

(c) Any one

� Glucose solution is boiled (�)and cooled (�) (just mention

cooled, no mark) �+�

� No air trap inside syringe initially (1)

� Ensure that the set-up (�) is air-tight (�) (just mention set-up,

no mark) �+�

� The yeast used is viable (1)

(no mark for constant temperature maintenance)

(if more than one precaution are given, mark the first precaution only) (1)

(Q4 = 4 marks)

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HHHKKKAAALLLEEE 111999999777 BBBiiiooolllooogggyyy PPPaaapppeeerrr III 11. (a) Start : fluid level on the left arm of manometer was higher (�) (accept

the reverse description for the right arm of the manometer) (1�)

Reason : air was injected and air pressure inside the whole system

increased (1)

(b) Reason : O2 in the chamber is used up / absorbed (�) during

respiration (�) of the experimental animals resulting in

reduction of air pressure (�), CO2 emitted / released (�)

from respiration is absorbed by soda-lime (�). (3�)

Pressure is equalized (�) when 10 mL of oxygen is used up

(�).

(c) Mouse : Rate = 15.020

10×

= 3.33 mL O2 h-1 g-1 1�

Grasshoppers : Rate = 5.55

10×

= 0.36 mL O2 h-1 g-1 1�

[formula (�), correct data (�), correct answer with unit (�)]

(3)

(d) To stabilize the temperature during the course of the experiment (1). 1

Another precaution :

Allow the animal(s) to acclimatize to the temperature before starting the

experiment(1) / To make sure the apparatus is air tight. 1

(2)

(e) Mouse : rate increases (�)

Grasshoppers : rate decreases (�) 1

Mouse : more heat is lost to the environment as temperature gradient

between the body and that of the environment increases (�),

more heat is generated (�) by increased respiratory rate to

maintain constant body temperature (�) 1�

Grasshoppers : body temperature decreases as the external temperature decreases (�) and this lowers respiratory rate (

(max. 3)

(Total : 13 marks)

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HHHKKKAAALLLEEE 111999999777 BBBiiiooolllooogggyyy PPPaaapppeeerrr IIIIII 1. (a) ATP is an immediate source of energy / energy carrier (�). In the

presence of ATPase (�) (or suitable enzymes), its terminal high energy

phosphate bond(s) (�) can be hydrolysed to liberate energy (�) used

to drive other biochemical reactions (�) / coupled to endergonic

reactions. ATP can act as source of phosphate (�) in phosphorylation

of substrate (�) in cellular metabolism. (3�)

(b) Glycolvsis (�) (N.B. accept substrate level phosphorylation) � occurs

in the matrix (�) of the cytoplasm. (�) 1�

During the conversion of trios phosphate (PGAL) (�) to pyruvate (�),

energy is liberated to form ATP (�) from ADP and Pi. 1�

(Also accept Krebs cycle (�) occurs in the mitochondrial (�) matrix

(�).

Max. 3

Oxidative phosphorviation (�) � takes place on the cristae (�) of

the mitochondria (�) 1�

In glvcolvsis (�) and the Krebs cycle (�), various intermediates are

dehydrogenated / oxidized generating NADH2, (1). In the presence of

free oxygen (�), electrons from NADH2 and FADH2 (�) will be

transferred along the electron transport chain (�). During this process,

there is a step-wise release of energy from the oxidation of electrons

and such energy is used to form ATP (1) from ADP and Pi. 4�

Photo-phosphorviation (��) light reaction � in thylakoid / grana (�)

of chloroplast (�) 1�

Chlorophyll absorbs light energy (�) this causes the electron in the

chlorophyll (�) to be excited (�) which then passes through a chain

of electron carries (�). During this process energy captured in the

electron is released in a step-wise manner (�) and used to synthesize

ATP (�) from ADP and Pi.

(13�)

(c) Example (�) mark each

(2 animal cells and 1 plant cell)

liver cell

muscle cell

Function (�) mark each

(2 on animal cells and 1 on plant cell)

ATP is required for various metabolic processes

ATP is required for muscle contraction

sperm cell

companion cell

cambium / meristematic cell

ATP is required for the motility of the sperm

ATP is required for active movement of substances

into the companion cells

ATP is required for anabolic processes that

accompanies cell division

(3)

Marker should accept alternatives which are correct. In case of doubt, please consult your C / E or A / E.

(Total : 20)

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HHHKKKAAALLLEEE 111999999888 BBBiiiooolllooogggyyy PPPaaapppeeerrr IIIIII 5. (a) Energy source :

Mitochondria Chloroplasts

� potential / chemical energy in

food (�)

� light energy (�)/ sun / photon 1

(b) Process energy

� release energy by breaking down

food (1), in Krebs cycle (�)

� capture energy by

photophosphorylation (�) to build up

food (�) in photosynthesis (�)

3

� hydrogen is removed (�) from

intermediates of Krebs cycle (�)

forming NADH(�)

� after absorption of light (�)

chlorophyll is excited (�), release

excited electron (�), to a chain of

electron carriers (�),

oxidation-reduction process of carriers

release energy (1) used to synthesize

ATP (�) from ADP (�), NADPH is also

formed (�),

6

� electron from NADH passes along

the electron transport chain (1) where

members of the chain undergo

oxidation-reduction reactions releasing

energy (1)to build up ATP (�) from ADP

(�). This is oxidative phosphorylation

(�).

� ATP (�)and NADPH (�) carrying

the energy are used in the dark reaction

to synthesize carbohydrate (1) / hexose /

starch, thus chloroplast converts light

energy to potential / chemical energy

(�) stored in food

6

Both make ATP (�) involving electron transport chain (�) 1

max. 13

(Overflow : 4)

(c) Inter-relationship between mitochondria and chloroplast

(1) The two organelles bridge the flow of biological energy (1). 1

(2) Light energy from the sun (1), together with CO2 (1) and water (1)

from products of cellular respiration in mitochordrina are processed

in the chloroplast (1) to form carbohydrate (1). 5

(3) The carbohydrate with stored energy (1) and O2 (1) formed in

chloroplast are eventually taken up and processed by the

mitochondria (1). ATP is liberated (1). 4

Deduct � mark for no comparative account, denote as C = -� Max. 7

Deduct � mark for note-form answer, denote as N = -� ( Overflow : 3 )

Question Total : 20

Overflow : 7

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Respiration

Documents

marks c

marks d

release of oxygen

oxygen release

marks e

marks iv

oxygen tension

oxygen debt