AS BIOLOGY PAST PAPER QUESTIONS

Name the three phases of mitosis shown by C, D and E on the figure

above.

Describe the role of the spindle fibres and the behaviour of the

chromosomes during each of these phases.

b) 1. C = prophase and

D = metaphase and

E = anaphase;

2. (In) prophase, chromosomes condense;

Accept chromatin for ‘chromosomes’ and for

‘condense’, shorten and thicken

3. (In) prophase OR metaphase, centromeres attach to spindle fibres;

4. (In) metaphase, chromosomes/pairs of chromatids at

equator/centre of spindle/cell;

5. (In) anaphase, centromeres divide;

6. (In) anaphase, chromatids (from each pair) pulled to (opposite)

poles/ends (of cell);

Accept for ‘chromatids’, chromosomes but reject

homologous chromosomes

7. (In) prophase/metaphase/anaphase, spindle fibres shorten;

If mark point 1 is not credited = 4 max

Do not carry forward error from 1.

Accept letters for stages as indicated in 1.

Accept for ‘shorten’, contract

explain results shown in figure above

Group A – initial uptake slower because by diffusion (only);

Group A – levels off because same concentrations inside cells

and outside cells / reached equilibrium;

Group B – uptake faster because by diffusion plus active

transport;

Group B fails to level off because uptake against gradient / no

equilibrium to be reached;

Group B – rate slows because few / fewer chloride ions in

external solution / respiratory substrate used up.

A scientist treated growing tips of onion roots with a chemical that stops roots growing. After 24 hours, he prepared a stained squash of these root

tips. Figure 2 is a drawing showing the chromosomes in a single cell observed in the squash of one of these root tips in anaphase. This cell was typical of other cells in anaphase in these root tips. Use all of this information to suggest how the chemical stops the growth of roots

1. Stops anaphase / cell division / mitosis;

Accept prevents telophase / cytokinesis

2. 3. 4. (By) stopping / disrupting / spindle fibres forming / attaching / pulling;

Ignore affects anaphase

Preventing separation of (sister) chromatids;

Ignore chromosomes separate / split

Accept chromatids split

(So) no new cells added (to root tip);

The arrows in Figure 2

show the directions in which

each new DNA strand is

being produced.

(d) Use Figure 1,

Figure 2 and your knowledge

of enzyme action to explain

why the arrows point in opposite directions.

______________________________________

1. 2. 3. 4. (Figure 1 shows) DNA has antiparallel strands / described;

(Figure 1 shows) shape of the nucleotides is different /

nucleotides aligned differently;

Enzymes have active sites with specific shape;

Only substrates with complementary shape / only the 3’ end

can bind with active site of enzyme / active site of DNA

polymerase.

(c) Which of these models, P, Q or R, is supported by the results shown in

Figure 2?

Give the letter and name of the model supported and explain why the

results do not support the other models.

(b) Correct answer for 2 marks, 0.8376308/0.84/0.8 (hours);;

Accept for 1 mark,

Evidence of 4 x 106 and 3 x 109 (written in any format, for correct readings

from graph)

OR

Evidence of 9.550746785 (correct number of generations)

OR

Evidence of 1.1938443348 (correct generations/ hour)

OR

Evidence of 50.26 (correct generation time in minutes)

AQA Biology A-Level - Nucleic Acids MS PhysicsAndMathsTutor.com

Incorrect reading of graph, 3 x 106 and 2 x 109

Accept for 1 mark, calculation carried out correctly

Evidence of 9.380821784 (correct calculation of number of generations)

OR

Evidence of 1.172602723 (correct calculation of generations/ hour)

OR

Evidence of 51.16822503 (correct calculation of generation time in

minutes)

OR

Evidence of 0.8528037505 (correct calculation of generation time in hours)

Incorrect reading of graph, 106.4 and 109.3 OR 106.3 and 109.2

Accept for 1 mark, calculation carried out correctly

Evidence of 9.633591475 (correct calculation of number of generations)

OR

Evidence of 1.204198934 (correct calculation of generations/ hour)

OR

Evidence of 49.82565445 (correct calculation of generation time in

minutes)

OR

Evidence of 0.8304275742 (correct calculation of generation time in hours)

OR

Evidence of 0.83, with no other working

Accept correct rounding to any number of decimal

places

2

(c) 1. (Model) Q and

(Name) Semi-conservative (replication);

Explanation

2. (Model) P (is unsupported because)

There should be two peaks in generation 1

OR

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(Only) one peak is shown in generation 1

3. OR

There should be 3:1 (ratio) of peaks in generation 2

OR

There should not be an intermediate/15N 14N peak in generation 1/2/3

OR

The original/generation 0/15N peak should be in generation 1/2/3;

(Model) R (is unsupported because)

There should be >2 peaks in generation 2/3

OR

There should be one wide/overlapping peak in generation 3;

(c) Which of these models, P, Q or R, is supported by the results shown in

Figure 2?

Give the letter and name of the model supported and explain why the

results do not support the other models.

(b) Correct answer for 2 marks, 0.8376308/0.84/0.8 (hours);;

Accept for 1 mark,

Evidence of 4 x 106 and 3 x 109 (written in any format, for correct readings

from graph)

OR

Evidence of 9.550746785 (correct number of generations)

OR

Evidence of 1.1938443348 (correct generations/ hour)

OR

Evidence of 50.26 (correct generation time in minutes)

AQA Biology A-Level - Nucleic Acids MS PhysicsAndMathsTutor.com

Incorrect reading of graph, 3 x 106 and 2 x 109

Accept for 1 mark, calculation carried out correctly

Evidence of 9.380821784 (correct calculation of number of generations)

OR

Evidence of 1.172602723 (correct calculation of generations/ hour)

OR

Evidence of 51.16822503 (correct calculation of generation time in

minutes)

OR

Evidence of 0.8528037505 (correct calculation of generation time in hours)

Incorrect reading of graph, 106.4 and 109.3 OR 106.3 and 109.2

Accept for 1 mark, calculation carried out correctly

Evidence of 9.633591475 (correct calculation of number of generations)

OR

Evidence of 1.204198934 (correct calculation of generations/ hour)

OR

Evidence of 49.82565445 (correct calculation of generation time in

minutes)

OR

Evidence of 0.8304275742 (correct calculation of generation time in hours)

OR

Evidence of 0.83, with no other working

Accept correct rounding to any number of decimal

places

2

(c) 1. (Model) Q and

(Name) Semi-conservative (replication);

Explanation

2. (Model) P (is unsupported because)

There should be two peaks in generation 1

OR

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(Only) one peak is shown in generation 1

3. OR

There should be 3:1 (ratio) of peaks in generation 2

OR

There should not be an intermediate/15N 14N peak in generation 1/2/3

OR

The original/generation 0/15N peak should be in generation 1/2/3;

(Model) R (is unsupported because)

There should be >2 peaks in generation 2/3

OR

There should be one wide/overlapping peak in generation 3;

Give two pieces of evidence from Figure 1 that this cell was undergoing

mitosis. Explain your answers.

(a) 1. 2. 3. The (individual) chromosomes are visible because they have

condensed;

Both parts of each answer are required – evidence

and explanation.

For ‘they’ accept ‘chromosomes/chromatin/DNA’

Accept ‘tightly coiled’ or ‘short and thick’ for

condensed but do not accept ‘contracted’.

Ignore references to nucleus/nucleolus/nuclear

membrane.

(Each) chromosome is made up of two chromatids because DNA has

replicated;

Both parts of each answer are required – evidence

and explanation.

Accept ‘sister chromatids’ for ‘two chromatids’.

Ignore references to nucleus/nucleolus/nuclear

membrane.

Name structure Q in Figure 2 and suggest how it is involved in the absorption of

lipids.

1. Golgi (apparatus);

2. Modifies / processes triglycerides;

3. Combines triglycerides with proteins;

4. Packaged for release / exocytosis

OR

Forms vesicles;

The scientists expressed their results as Percentage of lipid in plasma membrane by mass. Explain how they would find these values.

1. Divide mass of each lipid by total mass of all lipids (in that type of cell); 2. Multiply answer by 100.

(d) Use the information in the table above to calculate the number of polypeptides:

6 amino acids

20 amino acids

(d) 1. (6 amino acids in length) 1; 2. (20 amino acids in length) 2; Accept for 1 mark, 55 (2 5 + 3 15) if no other mark awarded.

Which tissue sample, C or D, was taken from a cancerous tumour?

Use information in the table to explain your answer.

(iii) D (no mark)

1. 2. Lower % (of cells) in interphase / higher % (of cells) in mitosis / named

stage of mitosis;

1. Accept: ‘less’ or ‘more’ instead of ‘%’

1. Do not accept: higher % (of cells) in each / all stage(s)

(So) more cells dividing / cells are dividing quicker;

2. Accept: uncontrolled cell division

2. Do not award if Tissue C is chosen

Washing powders often contain enzymes from bacteria. These enzymes include

proteases that hydrolyse proteins in clothing stains.

Figure 1 shows the effect of temperature on a protease that could be used in

washing powder.

1. Both denatured (by high temperature);

2. Denaturation faster at 60 °C due to more

(kinetic) energy;

3. Breaks hydrogen/ionic bonds (between amino

acids/R groups);

4. Change in shape of the active site/active site no

longer complementary so fewer enzyme-

substrate complexes formed/substrate does not

fit;

Suggest why the development of a monopolar mitotic spindle would prevent

successful mitosis.

1. No separation of

chromatids/chromosomes/centromeres;

2. Chromatids/chromosomes all go to one

pole/end/sides of cell/not pulled to

opposite poles;

3. Doubles chromosome number in

cell/one daughter cell gets no

chromosomes or chromatids

1. (No, because) at 100 there are still

some (7%) cancer cells

dividing/undergoing mitosis;

2. 3. 4. 5. 6. So, cancer not destroyed/may continue

to grow/spread/form tumours;

Best concentration may be between

100 and 1000/need trials between 100

and 1000;

This research in culture, don’t know

effect of KI on people;

(Yes, because) above 100 produces

little increase in % of cells not

dividing/undergoing mitosis/at 100,

most (93%) cancer cells unable to

divide/dead;

Above 100 may be harmful (to body);

7. Higher concentrations more expensive;

8. (above 100) will have more effect on

(rapidly dividing) cancer cells;

Evaluate the use of 2,4-D as a herbicide on a wheat crop that contains wild oats as

a weed. Use all the information provided.

1. 2. 3. 4. 5. MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2017

2,4-D causes an increase in release of ions

from wild oat cells and 2,4-D does not

affect/has little effect on the release of ions

from wheat cells;

(For wheat) Difference is less than LSD/7 so

difference is not significant;

OR

(For wild oats) Difference is more than LSD/10

so difference is significant;

Loss of ions from cells (likely to) lead to

cell/plant death/damage;

OR

Disruption of cell membrane (likely to) lead to

cell/plant death/damage;

No evidence here about death of plants as a

result of this ion loss;

No evidence here of other

ecological/environmental impact;

Two solutions often used to stain tissues are haematoxylin solution and

iodine solution.

• Haematoxylin solution stains DNA a blue colour.

• Iodine solution stains starch a blue-black colour.

The scientist used haematoxylin solution and not iodine solution to stain

the lung tissue.

Suggest why.

1. This/animal/lung tissue does not contain starch;

Accept cell(s) for ‘tissue’

2. (Makes) nucleus visible;

OR

Nucleus contains DNA;

Do the data in the graph above show a linear relationship between

concentration of particulate matter and percentage of dead cells?

Use suitable calculations to justify your answer.

No)

EITHER

1. 2. 9 (percent per 5 µg cm–3);

1.42/1.8 (percent per 5 µg cm–3);

Accept any number of significant figures as long as

rounding correct, full answer for mp2 is

1.42105263.

OR

3. 4. 1.8 (percent per 1 µg cm–3);

0.28/0.36 (percent per 1 µg cm–3);

Accept any number of significant figures as long as

rounding correct, full answer for mp4 is

0.28421053.

OR

5. 6. 9% and 36/27% increase here;

(To be linear) 100 (µg cm–3) would be 180/171% (increase)

OR

(To be linear) 5 (µg cm–3) would be 1.8% (increase)

OR

% increase is x4 (0-5 µg cm–3 compared with 0-100 µg cm–3) but 5-

100 is more than x4

OR

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% increase is x3 (0-5 µg cm–3 compared with 5-100 µg cm–3) but 5-

100 is more than x3;

OR

7. (Using y = mx + c) at 5 (µg cm–3) m = 1.8;

8. (Using y = mx + c) at 100 (µg cm–3) m = 0.36;

9. At 100 (µg cm–3) y would be 186%;

10. At 5 (µg cm–3) y would be 7.8%;;

If no correct answers accept for one mark

Evidence of incorrect graph reading but division by 19

OR

Evidence of incorrect graph reading but division by 95

Accept 1 and 2 OR 3 and 4

OR 5 and 6

OR 7 and 8

OR 7 and 9

OR 8 and 10

The scientist concluded that the cells in flask G took up sodium ions by active

transport. Explain how the information given supports this conclusion.

1. 2. 3. 4. Uptake in flask G much greater than in flask F;

Showing use of ATP in flask G;

Sodium ion concentration in flask G falls to

zero;

Showing uptake against a concentration

gradient;

chromatids are (in two groups) at poles of spindle / at

ends of spindle;

Do not accept ‘ends of cell’

V-shape shows that (sister) chromatids have been pulled apart at their

centromeres / that centromeres of (sister) chromatids have been pulled

apart.

Many sports drinks contain water, sodium chloride and carbohydrates. The manufacturers of the

sports drinks claim that carbohydrates provide an energy boost. The sodium chloride is used to

increase absorption of glucose in the small intestine.

Scientists investigated the effect of a sports drink on the performance of runners in 5 km races.

They recruited 100 runners who had previously run a 5 km race in similar times. During this race,

Race 1, they had water they could drink.

The scientists divided the runners into two equal groups, P and Q. 5 km race, Race 2. During this race:

Both groups ran a second

l group P had water available

l group Q had the sports drink available.

The scientists recorded the mean time for each group to complete this race.

One of the runners concluded that the sports drink improved performance.

Do these data support his conclusion?

Yes)

1. Faster running time after sports

drink;

(No)

2. 3. 3 max Can mix and match yes or no

approach, all 5 responses are

available

1. ‘Faster running time in group

Q’ is insufficient but accept

‘faster running time in group Q

in Race 2’

3. Accept ‘no stats analysis’

4. 5. Mean times given so there will be

variation in the group;

No standard deviations to know the

spread of the data (about the

mean)/whether they overlap;

Improvement in running time only

small in both groups / both groups

improved in Race 2;

Did not drink the same volumes;

The scientist used units of µmol g−1 h−1 for the rate of oxygen uptake.

Suggest why he used µmol in these units.

(Measures) small uptake / amount / quantity /

volume / concentration / rate (of oxygen

uptake);

OR

Avoids use of powers of ten / standard form /

many decimal places;

explain how relationship between blood and water shown in micrograph is useful for fish

(b) Maintains concentration gradient (over whole length of

gill) / diffusion can occur over whole gill;

More oxygen enters blood (/ more CO2 leaves);

More (aerobic) respiration / more energy release in muscle

/ for swimming;

‘more’ needed ONCE only

the scientist decided to use the ratio of surface area to mass, rather than the ratio of

surface area to volume. He made this decision for practical reasons.

Suggest one practical advantage of measuring the masses of frog eggs, tadpoles and

adults, compared with measuring their volumes.

More accurate / less error (in measuring

mass);

OR

Causes less distress / damage to animal (to

measure mass);

OR

Easier / quicker (to find mass) because

irregular shapes;

OR

1 Ignore references to

human error

Accept converse if

reference made to volume

Reject if comparison is

made with surface area.

Fewer measurements / calculations;

(a) Describe what occurs between

(i) stage A and stage B;

nuclear membrane disappears (once only);

chromosomes become shorter / condense /coiling;

arranged on equator;

spindle formation (once only);

centromeres attach to spindle;

accept reference to centrioles (once only)

(a) Describe what occurs between

(i) stage B and stage C;

nuclear membrane disappears (once only);

chromosomes become shorter / condense /coiling;

arranged on equator;

spindle formation (once only);

centromeres attach to spindle;

accept reference to centrioles (once only)

(iii) Describe and explain the appearance of one of the chromosomes in stage B.

(original) chromosome/DNA has been replicated;

each chromosome consists of two chromatids/

chromatids attached at centromere;

(accept reference to condensed state of chromosomes

The diagram shows one method by which amino acids are absorbed from the small intestine into

the blood. They are co-transported into the epithelial cell with sodium ions (Na+) at point X on

the diagram. Normally, the concentration of sodium ions inside the epithelial cell is low.

Why does pump Y no longer work?

Lack of ATP;

Pump = active transport / requires energy / ATP provides energy /

transport is up

concentration gradient;

Describe the role played by organelle U in the formation of chylomicrons.

proteins are synthesised by U;

involvement of ribosomes;

protein isolation / transport (inside RER);

vesicle formation;

interphase

Apart from an increase in the amount of DNA, give one process which occurs during

stage X which enables nuclear division to occur.

cell replication

short duration of interphase

a. B,D
b. idea of molecules/named molecules moving = fluid

mosaic= both proteins and phospholipids

Calculations made from raw data/ raw data would have been initial and final mass

Q12.Turkey meat can dry out when it is cooked in an oven. One way to overcome this is to soak

the meat in a salt solution before cooking it. This is called brining.

A food writer organised a demonstration. He treated three similar pieces of turkey in

different ways.

• Piece A was untreated.

• Piece B was soaked overnight in a 6% solution of salt. A 6% solution of salt has a

greater solute concentration than the cells in turkey meat.

• Piece C was soaked overnight in water.

He put all three pieces in an oven at 150 °C. He left each piece until it was cooked and the

temperature in its centre was 65 °C. The writer weighed each piece at different stages in

the demonstration. The graph shows his results.

(i) Explain the advantage of using percentage change in mass in this

investigation.

(ii) The pieces of turkey meat were cooked. Explain the advantage of leaving them

in the oven until the temperature in the centre of each piece was 65 °C.

i. 1. Allows results to be compared;

2. Because initial masses may have been different;

ii. 1. Quantitative measure (of cooking);

2. Ensures all cooked to same extent as not all turkey pieces same

shape / thickness

Short diffusion pathway (to cells)

OR

It has a surface permeable (to water/ions into cells);

1. (Red blood cells) do not have a nucleus/DNA;

2. Haemoglobin;

32;

28 32 26;

Use your knowledge of breathing to explain the changes in pressure during inspiration

and expiration. The letters on the graph are to help you to refer to different parts of the

curve.

A - B: volume of lungs/ thorax increases;

due to contraction of diaphragm muscles;

contraction of (external) intercostal muscles raising ribs;

pressure falls / decreases;

B - C: atmospheric pressure / difference in pressure forces air in;

C - D: Pressure increased by elasticity of lungs;

contraction of (internal) intercostal or abdominal muscles;

D - E : air expelled, so pressure falls / due to thorax pressure.

Explain why the figures in the table are given per gram of tissue.

Allows comparison; As organs differ in size; Larger organs will need more blood;

The scientists put the same number of skin tumour cells in each dish at the start of

thisinvestigation. Explain why it was important to put the same number of cells in each

dish.

To ensure the colour is the same at the start;

The scientists concluded that bromelain did not kill cancer cells but stopped them

dividing.Does the graph support this conclusion? Explain your answer

Yes – curve on graph with bromelain present remains approximately constant / risesvery

slightly;

Would decrease if killing of cells occurred / would increase if cells still dividing;

An article in a newspaper claimed that these data show that bromelain can be used to

treatcancer.

Give three reasons why we should be careful about accepting this claim.

Use of mouse cells (rather than human);

(Carried out) in vitro / not in living organisms;

Only tested on one type of cancer;

Not possible to predict effect on humans (as no data collected);

Describe and explain what the student should have done when counting cells to make

sure that the mitotic index he obtained for this root tip was accurate.

Suggest how the environmental conditions have resulted in adaptations of systems

using Model A rather than Model B.

1. Water has low(er) oxygen partial

pressure/concentration (than air);

2. So (system on outside) gives large surface area

(in contact with water)

OR

So (system on outside) reduces diffusion

distance (between water and blood);

3. Water is dense(r) (than air);

4. (So) water supports the systems/gills;

Lactose is present in milk. It is broken down by lactase into glucose and galactose. This is

shown in the equation

Explain the results for the lactose intolerant man.

No lactase;

Therefore lactose not digested/glucose not produced;

No glucose absorbed therefore concentration in blood stays the

same/does not rise;

Suggest a difficulty of counting movements of gill covers as a method of measuring

rate of ventilation in fish.

(FIGURE 2 X AXIS: WATER TEMPERATURE)

  Fish keep moving / swimming / movement of gill covers too fast to count (at higher temperatures).

Which species, A or B, would you predict grew in a drier environment?

Explain one feature that caused you to choose this species.

  (i)      Species B (no mark)

Correctly selected feature and the explanation required for 1 mark:

1.      Smaller surface area

         so

         less evaporation / less heat absorbed;

2.      Thicker leaves

         so

         greater diffusion distance (for water);

Accept ‘thicker leaves so more water storage’.

3.      Fewer stomata / lower stomatal density

         so

         less diffusion / evaporation (of water);

4.      Smaller surface area to volume ratio

         so

less evaporation.

Species C has a high number of stomata per mm2. Despite this it loses a small amount of water.

use data to explain why

  Small leaves / surface area so (total) number of stomata is low.

Both aspects needed for mark.

Explain what the standard deviations suggest about the difference in mean total

number of dung beetles between the different types of farm.

1. No overlap in standard deviations;

2. (Difference in mean total)

significant/is not due to chance/is

real;

Explain the results shown in Figure 4.

1. (Rate of) increase in concentration

of maltose slows as

substrate/starch is used up

2. OR

High initial rate as plenty of

starch/substrate/more E-S

complexes;

No increase after 25 minutes/at

end/levels off because no

substrate/starch left;

1. With 16 or fewer than 16 (repeats the

association) is significant;

2. With 17 or more than 17 (repeats the

association) is not significant;

3. With 16 or fewer than 16 (repeats) there is

less than a 5% or less than 0.05 probability

of being due to chance

OR

With 17 or more than 17 (repeats) there is

more than a 5% or more than 0.05

probability of being due to chance

3 max

OR

Explanation of a probability value e.g. 0.30

is a 0.30 or 30% probability of being due to

chance;

4. With 16 or fewer than 16 (repeats) reject the

null hypothesis

OR

If none of the marks is

awarded allow principle

mark of (prostate) cancer

more likely with 16 or less

than 16 (repeats) or

(prostate) cancer less

likely with 17 or more than

17 (repeats)

OR

Alternative principle mark

Correctly links

significant/not significant to

correct probability

value/percentage or to

rejecting/accepting the null

hypothesis.

1. Reject ‘the results are

significant’.

1. Accept ‘difference in

results is significant’.

2, 3 and 4. Accept

reference to any number of

repeats (e.g. 18) between

7MARK

With 17 or more (repeats) accept the null

hypothesis;

17 to 20 for 17 or more

than 17 (repeats).

3. Accept equivalent

responses in terms of 95%

or 0.95 probability

correct answer for 2 marks = 7

Accept for 1 mark,

7.2 (correct answer but not rounded)

OR

Evidence of 1.6 ÷ 4.0/0.4/40% (correct dilution

factor)

OR

Evidence of 0.08 (correct amylase volume in

0.2cm3)

1. Cell ingests/engulfs the antibody/ADC

OR

Cell membrane surrounds the antibody/ADC (to take it inside the cell);

3

2. Lysosomes fuse with vesicle/phagosome

(containing ADC);

3. Lysozymes breakdown/digest the antibody/ADC

to release the drug;

0 8 . 3 The pressure of the equilibrium mixture was increased by reducing the volume of the

container at constant temperature.

Predict the effect of increasing the pressure on the equilibrium yield of hydrogen.

Explain your answer.

Predict the effect of increasing the pressure on the value of Kc

[4 marks]

Effect on equilibrium yield of hydrogen:

Explanation:

Effect on value of Kc:

M1 yield would decrease

M2 M3 equilibrium (position) moves left / shifts left / in direction of

reverse reaction

to oppose increase in pressure / to reduce pressure

fewer moles/molecules of gas on left hand side / fewer

moles/molecules of gaseous reactants

M4 no effect on Kc

1. Chromosome is formed of two chromatids;

2. 3. (Because) DNA replication (has occurred);

(Sister) chromatids held together by

centromere

A section/slice (so nucleus in another part of cell)

1

OR

(Nucleus) not stained;

As shown in Figure 1, ATP synthase has two functions.

• It catalyses the synthesis of ATP.

• It allows the movement of H+ ions.

Suggest how the shape of the ATP synthase allows it to have these two functions.

Explain your answers.

[4 marks]

Catalyses the synthesis of ATP:

Allows the movement of H+ ions:

(Catalyses the synthesis of ATP)

1. Active site complementary to ADP + Pi;

2. Enzyme-substrate complex forms;

2. ‘E-S’ alone is

insufficient

(Allows the movement of H+ ions)

3. Channel (in membrane/protein/enzyme);

4. Allows facilitated diffusion of H+

OR

(Channel) has tertiary structure specific for (only)

H+

;

Name the type of blood vessel that controls blood flow to muscles and explain how

these blood vessels change blood flow during exercise.

[3 marks]

Name of blood vessel:

Explanation:

1. Arteriole;

2. (Circular/smooth) muscle relaxes;

3. Vasodilation increases blood flow

OR

Widens/dilates (lumen of) blood vessel so

increases blood flow;

Before meiosis, a cell of a rice plant has 12 pairs of homologous chromosomes

(24 chromosomes in total).

Give the letter of the diagram from Figure 3 that correctly shows the chromosome

content of rice cells after the first meiotic division and after the second meiotic

division.

[2 marks]

After first meiotic division:

After second meiotic division:

After first meiotic division – B;

After second meiotic division – E;

1. Abdominal pumping/pressure in tubes linked to carbon

dioxide release;

MP1 relates to description of link shown in graphs

2. (Abdominal) pumping raises pressure in body;

Needs idea of causation, not just description of correlation

Air/carbon dioxide pushed out of body /air/carbon dioxide moves down pressure gradient (to atmosphere);

Reject ref to concentration gradients/diffusion

1. Mass of DNA equals 1 (indicates) (early)

interphase

OR

No cells with mass of DNA less than 1 (indicates)

mitosis only type of division occurring;

2. Mass of DNA equals 2 (indicates) mitosis

OR

Mass of DNA equals 2 (indicates) each

chromosome consists of 2 chromatids;

3. Mass of DNA between 1 and 2 (indicates) some

of the DNA has been replicated;

1. Accept G1 phase

Accept no meiosis

2. Accept G2 / any

named phase of

mitosis

3. Accept S phase

1. Random samples;

2. Large sample size;

1. Allow in context

of fish or gills

2. If a specified

number is given,

it must be 10 or

more.

2. Accept ‘many’/

‘multiple’ for large

sample but ignore

‘several

Figure 5 shows a test that has been developed to find out if a person has antibodies

to the human immunodeficiency virus (HIV) antigen.

This test only detects the presence of HIV antibodies. Give two reasons why it cannot

be used to find out if a person has AIDS.

(To diagnose AIDS, need to look for/at)

1. (AIDS-related) symptoms;

2. Number of helper T cells;

Figure 5 shows a test that has been developed to find out if a person has antibodies

to the human immunodeficiency virus (HIV) antigen

The solution will remain yellow if a person is not infected with HIV. Explain why.

1. HIV antibody is not present;

2. (So) second antibody/enzyme will not bind/is

not present;

2 1. Accept HIV

antibodies will not

bind (to antigen)

Explain how the monoclonal antibody would prevent the regulator protein from

working (lines 7−8).

(c) Any two from:

1. (Monoclonal antibody) has a specific tertiary structure / variable region /

is complementary to regulator protein

Do not award MP1 if reference to active site.

2. 3. Binds to / forms complex with (regulator protein)

“It” refers to monoclonal antibody in MP1 and MP2

(So regulator protein) would not fit / bind to the receptor / is not

complementary to receptor

3. Reject receptor on LDL

1. Aortic/semi-lunar valves is closed;

2. Because pressure in aorta higher than in

ventricle;

At Q on Figure 3 there is a small increase in pressure and in rate of blood flow in the

aorta.

Explain how this happens and its importance.

1. Elastic recoil (of the aorta wall/tissue);

2. Smooths the blood flow

OR

Maintains rate of blood flow

OR

Maintains blood pressure;

Multiple copies of the AMY1 gene is an adaptation to a high-starch diet.

Use your knowledge of protein synthesis and enzyme action to explain the

advantage of this adaptation.

1. More mRNA / more transcription;

2. More translation / enzyme;

3. So reaction faster;

Multiple copies of the AMY1 gene is an adaptation to a high-starch diet.

Suggest how this evolved through natural selection.

1. Mutation(s) produce extra copies of

(AMY1) gene;

2. Those with more copies / this

adaptation/mutation reproduce / survive

better on high starch diet;

3. And pass on multiple copies / this

adaptation/mutation (to offspring);

Explain the purpose of Tank B (on page 21).

1. As a baseline/control;

2. To show effect of no selection

OR

To show what happens in a normal

population/naturally

OR

To show effect of/ compare with tank A/tank

C;

2. Ignore reference

to type of selection

2. Accept not

removing/not

catching/not fishing

for ‘selection’

2. Accept genetic

drift for ‘no

selection’

2. Accept no

fishing/no

selection/no caught

fish for ‘normal

population’

2. Accept to

compare with other

results

Explain why all the cells of the body will have this mutation.

1. Mutation / extra chromosome in

gamete/egg/sperm (that formed zygote);

2. All cells derived (from a single cell/zygote) by

mitosis;

3. OR

All cells derived from a single cell/zygote by

mitosis;

4. Mitosis produces genetically identical cells / a

clone;

Use Figure 5 to explain how human mass at birth is affected by stabilising selection.

1. (Most likely to be) transferred to a special care

unit are those under 2800 g

OR

(Most likely to be) transferred to a special care

unit are those over 4200 g;

2. Extreme mass babies least likely to survive (to

reproduce) and so less likely to pass on their

alleles (for extreme mass at birth);

3. Extreme mass at birth decreases in frequency

(in the population)

OR

Alleles (for extreme mass at birth) decrease in

frequency (in the population);

If neither 1 or 2 awarded allow correct stated mass

less/more likely to survive for 1 mark

3 Accept converse

answers linked to

those with mass at

birth at any value

between 2800 and

4200 g.

1. For ‘2800 g’ accept

any value between

1400 g and 2800 g.

1. For ‘4200 g’ accept

any value between

4200 g and 5200 g.

1. If values for both

extremes are given,

both must be correct.

1. Reject data quoted

below 1400 g or above

5200 g.

3. Accept

‘proportion/percentage’

for ‘frequency’.

3. Do not accept

‘number’ for

‘frequency’.

1. Directional;

2. Fish with more spots are more likely to be

predated;

3. Alleles for (more) spots not passed on;

4. (So) frequency of (more) spots alleles

decreases;

4

(4 x AO2)

2. 3. and 4. Accept

converse

2. Accept killed/eaten

for ‘predated’

2. Accept more of

them killed/eaten, for

‘more likely’

3. and 4. Reject ‘gene’

once

4. Accept ‘proportion’

for frequency

Ignore ‘number of

alleles decreases’

0 7 . 1 Can you conclude that the insect pest resistant to Bt toxin found in the years 2002 to

2005 was the same insect species? Explain your answer.

[1 mark]

7.2 One farmer stated that the increase in the use of Bt crop plants had caused a mutation

in one of the insect species and that this mutation had spread to other species of

insect. Was he correct? Explain your answer.

[4 marks]

7.2 Mutations are spontaneous/random;

2. Only the rate of mutation is affected by

environment;

3. 4. Different species do not interbreed/do not

produce fertile offspring;

So mutation/gene/allele cannot be passed from

one species to another;

It is known that:

Do not write

outside the

box

• during respiration saturated fatty acids yield more energy than unsaturated fatty

acids

• saturated fatty acids have higher melting points than unsaturated fatty acids

• lipases in seeds act more rapidly on liquid substrates.

Use this information and Table 6 to show how each population is better adapted for its

natural environment when compared with the other population.

Q3.Scientists studied the rate of carbon dioxide uptake by grape plant leaves. Grape leaves

have stomata on the lower surface but no stomata on the upper surface.

The scientists recorded the carbon dioxide uptake by grape leaves with three different

treatments:

Treatment 1 − No air-sealing grease was applied to either surface of the leaf.

Treatment 2 − The lower surface of the leaf was covered in air-sealing grease that

prevents gas exchange.

Treatment 3 − Both the lower surface and the upper surface of the leaf were covered in

air–sealing grease that prevents gas exchange.

The scientists measured the rate of carbon dioxide uptake by each leaf for 60 minutes in

light and then for 20 minutes in the dark.

The scientists’ results are shown in the diagram below.

Describe the method the student would have used to obtain the results in

the graph. Start after all of the cubes of potato have been cut. Also

consider variables he should have controlled.

Species B is more active than species A. Use Figure 5 to explain how the

haemoglobin of species B allows a greater level of activity.