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

• 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]

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 mm}3 _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]

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]

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]

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]

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]

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/

• 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]

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]