Biology for CAPE Unit 2 Chapter 2 Answers

Full text

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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

• 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

b

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

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

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c

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

d

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

Average distance moved per min = 67.0/3 = 22.3 mm min-1

• 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

f

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

Each point [1] Max [4] 3 points [2] 1–2 points [1] Correct answer with no working [1] Max [3] Any point from

1st and 2nd [1] Point 3 [1] Point 4 [1]

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g i

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 Any point [1]

ii

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 [1]

iii

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 [1]

iv To maintain a constant temperature [1]

12 a I – matrix

II – cristae

III – outer membrane IV – intermembrane space

V – inner membrane Each point [1]

b i

Actual diameter of mitochondrion = Length of A

Magnification

=

58000 μm = 0.71μm

82000

ii

Allows for rapid diffusion of gases and other substances

• Short diffusion pathway to middle of organelle Any point [1]

c i Structure labeled I [1] ii Structure labeled II [1]

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

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d

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

e

Enters: ADP/Pi/NADH/O2/pyruvate/fatty acids/amino

acids

• Leaves: water/carbon dioxide/ATP/ NAD

• Any correct answer Both points correct [1]

13 a

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

b

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 2 points [2]

c i

Between glucose and glucose-6-phosphate [1]

• Between fructose-6-phosphate and fructose-bisphosphate [1]

Any point well explained [1] Max [4]

Term well explained [1] Location [1]

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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]

d

When the phosphate group is added to fructose-6-

phosphate, 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 Any point [1]

e i On arrows from fructose-bisphosphate to triose phosphate [1] 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 Any 2 points [2]

f It is at the point where NAD is converted to reduced NAD

(between fructose bisphosphate and triose phosphate) [1]

g

Produces a high energy phosphate group in an organic

substrate

• To phosphorylate ADP to form ATP by substrate level

phosphorylation Any point [1]

h i 2 ATP [1]

ii Substrate level/ground level phosphorylation [1]

i

2 molecules of pyruvic acid/pyruvate

• 2 molecules of reduced NAD/NADH/NADH + H+

• 2 ATP

• 2 molecules of water

j

Enters the mitochondrion to be oxidised to carbon dioxide

and water if oxygen is available Any point [1]

• Is reduced to lactate if no oxygen is available Max [2]

All products [2] 2–3 points [1]

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Essay questions

14 a

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 Any 2 points [2]

b

Decarboxylation: removal of carbon dioxide [1]

• Dehydrogenation: removal of hydrogen/H [1]

c

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

d

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

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

Any point well explained [1] Max [7]

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15 a i

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 Any 2 points [2]

ii

Glycolysis • Active transport • Muscle contraction • DNA replication • Protein synthesis • Cell division • Flagella beating • Endocytosis

• Any correct answer 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/

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• 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

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

16 a i

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

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

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

Any well explained point [1] Max [3]

2 well explained points [2] 2 well explained points [2]

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ii

Mammalian muscle cell Yeast

• Occurs in one step

• No decarboxylation/CO2

released

• Lactate dehydrogenase used

• Lactate passes into blood/liver

• Process reversible

• Occurs in 2 steps • Decarboxylation

• Alcohol dehydrogenase used • Ethanol passes into medium • Irreversible

b

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

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

ii

Lactate formed

• Removed to liver via blood • Lactate oxidised to pyruvate

• Pyruvate forms glucose or glycogen • Or enters Krebs cycle via link reaction

Any 2 points [2]

Well explained [2] Incomplete [1]

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

Figure

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