Biology for CAPE Unit 2 Chapter 2 Answers

2 Answers to end-of-chapter questions Multiple choice questions 1

C

[1]

2

A

[1]

3

A

[1]

4

B

[1]

5

B

[1]

6

D

[1]

7

C

[1]

8

C

[1]

9

D

[1]

10 D

[1]

Structured questions 11 a • Weigh three mung beans after removing testa • • • • • • • • •

b

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• • • • • •

Place beans in barrel of syringe Allow apparatus and beans to equilibrate for three minutes Germinating mung beans would take up oxygen and give off carbon dioxide during respiration Carbon dioxide is absorbed by soda lime, so the pressure inside of syringe would decrease This would cause the coloured water to move towards the syringe Distance moved by coloured water is directly proportional to volume of oxygen uptake Using a graph page/ruler, measure distance (d) moved by coloured water per minute for five minutes Calculate volume of O2 uptake using πr2d (r = radius of capillary tube) per minute Calculate volume of O2 uptake using πr2d (r = radius of capillary tube) in mm3 min-1 g-1 Capillary tube could be too short Variations in temperature can affect the volume/pressure in the syringe Variations in atmospheric pressure can affect the volume/pressure in the syringe Connections may not be airtight Intrinsic error due to measuring with a ruler Soda lime can become saturated with carbon dioxide Any correct answer

Biology Unit 2 for CAPE® Examinations

Any 2 points well explained [1] Max [3]

Any 1 point [1] Max [2]

Original material © Cambridge University Press 2011

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c

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d

• • • •

Replace germinating seeds with an inert material/glass beads/boiled seeds of equivalent mass to mung beans Leave for same length of time Control would compensate for pressure changes in the apparatus/changes due absorption of atmospheric carbon dioxide by soda lime Distance moved by meniscus can be added (if meniscus moved away from syringe)/or subtracted (if meniscus moved towards the syringe) from experiment results

Any point from 1st and 2nd [1] Point 3 [1] Point 4 [1]

x-axis correct (independent variable – time in seconds) and properly labelled y-axis correct (independent variable – distance moved by meniscus/mm) and properly labelled Points plotted correctly Appropriate title

Title: Graph showing the distance moved by meniscus during the uptake of oxygen by mung beans

e

-1

•

Average distance moved per min = 67.0/3 = 22.3 mm min Average volume = 3.14 × 0.12 × 22.3 = 0.70 mm3 min-1 Rate of O2 uptake per min per g = 0.70/0.5 = 1.4 mm3 min-1 g-1

•

Remove soda lime from experiment tube Replace soda lime with same mass of glass beads/inert material Reset the meniscus/push plunger of syringe down Leave seeds for same length of time and conditions/temperature as experiment tube Measure the distance moved per minute for same length of time as experiment tube Determine average distance moved per minute Use oxygen uptake data to calculate carbon dioxide release

• •

f

Each point [1] Max [4]

• • • • • •

Biology Unit 2 for CAPE® Examinations

3 points [2] 1–2 points [1] Correct answer with no working [1] Max [3]

Point 1 [1] Any 2 points [1] Max [2]

Original material © Cambridge University Press 2011

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g

i

ii

iii

Acts as a control chamber/acts as a thermobar/thermobarometer • Eliminates the effects of external temperature or pressure changes • External pressure or temperature changes act equally on both sides of the manometer and cancel them out

•

• Used to reset equipment • When opened, used to equalise atmospheric pressure • When closed, used to keep the apparatus airtight/close off apparatus to external atmospheric pressures • • •

iv 12 a

b

c

[1]

To maintain a constant temperature

[1]

Each point [1]

Actual diameter of mitochondrion = Length of A Magnification = 58000 μm = 0.71μm 82000

ii

[1]

Allows movement in the tube to be measured directly as volume. This removes the need to find out how much liquid in the tube Allows for resetting of manometer fluid Keeps apparatus airtight

I – matrix II – cristae III – outer membrane IV – intermembrane space V – inner membrane i

Any point [1]

Allows for rapid diffusion of gases and other substances • Short diffusion pathway to middle of organelle

Complete working [2] Partial [1] Correct answer with no working [1]

•

Any point [1]

i

Structure labeled I

[1]

ii

Structure labeled II

[1]

Biology Unit 2 for CAPE® Examinations

Original material © Cambridge University Press 2011

3

d

•

• • • • • • • • • e

• • •

13 a

• •

• • b

•

• • • • c

i

Folded inner membrane/cristae – increases surface area available for more stalked particles and electron carriers /more oxidative phosphorylation/ATP synthase complexes/more protons pumped across membrane Intermembrane space – allows accumulation of protons/H+ Impermeability of inner membrane to protons/H+ – maintains proton gradient/protons only go through channels in ATP stalked particles Stalked particles/ATP synthase channel for protons Linear arrangement on ETC on inner membrane – greater efficiency Membranes separate mitochondrion from cytoplasm – allows for different pH Inner membrane has attachment of stalked particle protruding into matrix – allows for passage of protons down a diffusion gradient Matrix contains enzymes for oxidation and decarboxylation Diameter does not exceed 1.0 μm to ensure short diffusion path/distance to centre: allows for rapid diffusion Any correct answer Enters: ADP/Pi/NADH/O2/pyruvate/fatty acids/amino acids Leaves: water/carbon dioxide/ATP/ NAD Any correct answer Is a series of reactions in which a 6C sugar is split into two molecules of pyruvate, a 3C acid Two molecules of ATP are used to phosphorylate glucose but four molecules of ATP are produced by substrate level phosphorylation, yielding a net synthesis of two ATP for each molecule of glucose It is the first of a series of reactions in the respiration process Location – cytoplasm Enzymes specific for a limited range of reaction Allows greater control over the breakdown pathway Allows intermediates to be available for other reactions Allows for coupling with ATP synthesis To prevent large heat losses

Any point well explained [1] Max [4]

Both points correct [1]

Term well explained [1] Location [1]

Any 2 points [2]

Between glucose and glucose-6-phosphate • Between fructose-6-phosphate and fructose-bisphosphate

•

Biology Unit 2 for CAPE® Examinations

Original material © Cambridge University Press 2011

[1] [1]

4

d

ii

It increases the activation energy of glucose thereby making the molecule unstable • It blocks the glucose from leaking out since the cell lacks transporters for glucose-6-phosphate which no longer fits the glucose carrier • This ensures the pure glucose is kept at a very low concentration inside the cells so it will always diffuse down a concentration gradient from the tissue fluid into the cell • Glucose-6-phosphate is the starting material for pentose sugars (and therefore nucleotides) and glycogen Point 1 and any other point [2]

•

When the phosphate group is added to fructose-6phosphate, fructose bisphosphate is formed. This molecule has symmetry and would allow for formation of two smaller reactive trioses This allows for another phosphate group to be added thereby increasing the activation energy of the molecule

•

e

f

g

i

On arrows from fructose-bisphosphate to triose phosphate

ii

• Opening of a stable ring structure • Lowers molecular mass • Makes available bonding groups in the form of phosphates • Production of the three carbon end product, pyruvate, which has a lower energy level

i

•

[1]

Any 2 points [2]

Produces a high energy phosphate group in an organic substrate To phosphorylate ADP to form ATP by substrate level phosphorylation

[1]

Any point [1]

i

2 ATP

[1]

ii

Substrate level/ground level phosphorylation

[1]

•

2 molecules of pyruvic acid/pyruvate 2 molecules of reduced NAD/NADH/NADH + H+ 2 ATP 2 molecules of water

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j

Any point [1]

It is at the point where NAD is converted to reduced NAD (between fructose bisphosphate and triose phosphate)

•

h

•

• •

Enters the mitochondrion to be oxidised to carbon dioxide and water if oxygen is available Is reduced to lactate if no oxygen is available

Biology Unit 2 for CAPE® Examinations

All products [2] 2–3 points [1]

Any point [1] Max [2]

Original material © Cambridge University Press 2011

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Essay questions 14 a

•

• • • •

b

•

Decarboxylation: removal of carbon dioxide Dehydrogenation: removal of hydrogen/H

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Pyruvate enters mitochondrion by active uptake/ATP used Into matrix of the mitochondrion CO2 is removed/decarboxylation from pyruvate and removal of hydrogen/dehydrogenation (oxidation) NAD reduced/reduced NAD formed 1 molecule of CO2 formed per pyruvate molecule Forms 2-carbon acetyl compound Which combines with CoA to form acetyl CoA Acetyl CoA combines with a 4C molecule/oxaloacetate to form citrate in Krebs cycle

•

c

• •

• • • • •

d

Coenzyme for dehydrogenase When reduced, carries electrons/protons/H+/hydrogen/H/2H to electron transport chain/cytochromes from glycolysis, link reaction and Krebs cycle When reoxidised, 3 ATP formed per molecule Aids in oxidation of triose phosphate to pyruvate in glycolysis Is regenerated in anaerobic respiration to allow glycolysis to continue

•

• • • •

• • • • • • •

[1] [1]

Occurs in matrix of mitochondria Allows for repeated oxidation and decarboxylation by building up the number of carbon bonds and attached H2OH groups. Series of steps/intermediate occurs Enzyme catalysed reactions Decarboxylation/removal of CO2 from 6C compound and then 5C compound Decarboxylation allows for oxidation of the group Dehydrogenation/oxidation Dehydrogenation/oxidation occurs four times (with NAD being used as the hydrogen acceptor three times and FAD used once) per turn of cycle 1 ATP produced by substrate level phosphorylation Oxaloacetate regenerated Produces 2CO2, 1 ATP, 3 NADH and 1 FADH2 per turn of cycle Any correct answer

Biology Unit 2 for CAPE® Examinations

Any 2 points [2]

7–8 points [4] 5–6 points [3] 3–4 points [2] 1–2 points [1]

Any point well explained [1] Max [7]

Original material © Cambridge University Press 2011

6

15 a

i

ii

• Found in all cells/all organisms • Easily transported because it is small and water soluble • Produced where energy is released (ADP + Pi + energy = ATP) • Breaks down to release energy where required by removal of third phosphate group by hydrolysis • Immediate source of energy • Couples energy-releasing reactions/catabolic and energy-requiring/anabolic reaction •

• • • • • • • •

Glycolysis Active transport Muscle contraction DNA replication Protein synthesis Cell division Flagella beating Endocytosis Any correct answer

Any 2 points [2]

Any 2 points [2]

b

•

• • • •

• • • • • • • • •

Occurs in ETC stage of respiration ETC located in the cristae of the mitochondria ETC made up of co-enzymes and cytochromes/electron carriers It is made up of four complexes: complex I–IV reduced NAD and reduced FAD from glycolysis, link reaction and Krebs cycle enters chain dehydrogenases/enzymes present removes H from coenzymes H split into H+ + e_ electrons flow through carriers/cytochromes release energy energy used to pump protons/H+ across membrane into intermembrane space protons accumulate in the intermembrane space proton gradient established/proton motive force/ electrochemical gradient/pH gradient proton flow back through ATP synthase/stalked particles into matrix

Biology Unit 2 for CAPE® Examinations

Original material © Cambridge University Press 2011

7

• • • •

c

•

•

Since oxygen is the final electron/proton/hydrogen acceptor in the ETC, it therefore allows the ETC to continue by maintaining a flow of electrons and protons If absent, reduced NAD and FADH2 would remain reduced Flow of electrons down the chain would stop No release of energy to create proton motive force No phosphorylation of ADP Cytochromes and hydrogen carriers would not be reoxidised/regenerated The Krebs cycle stops, no oxidised NAD and FAD for oxidation No substrate level phosphorylation

i

•

• • • • • •

16 a

formation of ATP from ADP + Pi/chemiosmotic synthesis of ATP oxygen acts as final acceptor water is formed for every reduced NAD, 3 ATP formed and for every reduced FAD, 2 ATP formed

Diagram [2] 6 well explained points [6] Each point [1]

Any well explained point [1] Max [3]

Reduced NAD and pyruvate remain in cytoplasm

[1]

Yeast: • Pyruvate converted into ethanal with removal of carbon dioxide/decarboxylation • Ethanal reduced to ethanol • Using 2H/H from reduced NAD • Using alchohol dehydrogenase • NAD reoxidised and recycled to glycolysis • Ethanol passes into the medium

2 well explained points [2]

Mammalian cells: • Pyruvate reduced to lactic acid/lactate • Using 2H/H from reduced NAD • Using lactate dehydrogenase • NAD reoxidised and recycled to glycolysis • Lactate passes into blood/liver

2 well explained points [2]

Biology Unit 2 for CAPE® Examinations

Original material © Cambridge University Press 2011

8

ii • • • • • b

Mammalian muscle cell Occurs in one step No decarboxylation/CO2 released Lactate dehydrogenase used Lactate passes into blood/liver Process reversible

• • • • •

Yeast Occurs in 2 steps Decarboxylation Alcohol dehydrogenase used Ethanol passes into medium Irreversible

Any 2 points [2]

•

Production of carbon dioxide to make dough rise

• •

Using sugar from the hydrolysis of starch in flour Production of ethanol and carbon dioxide in alcoholic drinks Using sugar/maltose, from germinating cereal grains/other

•

named sources c

i

ii

Any 2 points [2]

• During vigorous exercising, oxygen deficit occurs • The additional oxygen that must be taken into the

• • • • •

body after vigorous exercise to restore all systems to their normal states is called the oxygen debt

Well explained [2] Incomplete [1]

Lactate formed Removed to liver via blood Lactate oxidised to pyruvate Pyruvate forms glucose or glycogen Or enters Krebs cycle via link reaction

Point 1 [1] Any other point [2] Max [3]

Biology Unit 2 for CAPE® Examinations

Original material © Cambridge University Press 2011

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