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Biology · AQA 7402 · Paper 1 · November 2021

AQA A Level Biology Paper 1 November 2021 question paper 7402/1

2h 91 marks

Practise the past exam questions by topic

AQA 7402 · Paper 1

10 questions
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Question 1

Describe the induced-fit model of enzyme action and how an enzyme acts as a catalyst.

[3 marks]

Scientists investigated the action of the enzyme ATP synthase. They made reaction mixtures each containing:

• ATP synthase

• buffer (to control pH)

• substrates.

One of the substrates required in these reaction mixtures is inorganic phosphate (Pi).

Tick () one box to show which other substrate the scientists must add to the reaction mixtures to produce ATP.

Figure for question 1

[1 mark]

The scientists investigated the effect of concentration of inorganic phosphate (Pi) on ATP synthase activity.

After 2 minutes, they stopped each reaction and then measured the concentration of ATP.

Figure 1 shows the scientists’ results.

Figure for question 1

Suggest and explain a procedure the scientists could have used to stop each reaction.

[2 marks]

Explain the change in ATP concentration with increasing inorganic phosphate concentration.

[2 marks]

Question 2

Explain the advantage for larger animals of having a specialised system that facilitates oxygen uptake.

[2 marks]

Figure 2 shows two models of oxygen uptake found in animals.

Figure for question 2

Suggest how the environmental conditions have resulted in adaptations of systems using Model A rather than Model B.

[2 marks]

Figure 3 shows changes in concentration of oxygen in two gas exchange systems.

Figure for question 2

A student studied Figure 3 and concluded that the fish gas exchange system is more efficient than the human gas exchange system.

Use Figure 3 to justify this conclusion.

[2 marks]

Explain how the counter-current principle allows efficient oxygen uptake in the fish gas exchange system.

[2 marks]

Table 1 shows features of two mammals.

Bats are flying mammals; shrews are ground-living mammals.

Figure for question 2

Calculate how many times the lung volume per unit of body mass of the bat is greater than that of the shrew.

Give your answer to an appropriate number of significant figures.

Give one suggestion to explain this difference.

[3 marks]

Question 3

Describe how one amino acid is added to a polypeptide that is being formed at a ribosome during translation.

[3 marks]

Table 2 shows:

• mRNA codons and the amino acid coded for by each codon

• the type of bond formed by the R group of some of the amino acids.

Figure for question 3

Crystallin is a structural protein found in the human eye. An inherited disease that leads to blindness is caused by changes in properties of crystallin. The replacement of the amino acid Arg with the amino acid Gly causes these changes.

Use information in Table 2 to suggest why this amino acid replacement changes the properties of crystallin.

[2 marks]

The amino acid replacement of Arg with Gly is caused by a single base substitution mutation in the DNA. The non-mutant DNA triplet is TCC.

Complete Table 3.

Give:

• the mRNA codon complementary to the non-mutant DNA triplet

• the mutated mRNA codon that could cause the change from Arg to Gly in the crystallin protein

• the DNA triplet complementary to this mutated mRNA codon.

Figure for question 3

[2 marks]

Question 4

A student dissected an organ from a mammal to observe blood vessels.

He dissected a slice of the organ and identified two blood vessels.

Figure 4 shows a photograph of his dissection.

Figure for question 4

Figure 5 shows a drawing of the blood vessels from his dissection.

Figure for question 4

Suggest two ways the student could improve the quality of his scientific drawing of the blood vessels in this dissection.

[2 marks]

Identify the type of blood vessel labelled as X and the type of blood vessel labelled as Y in Figure 4.

Describe one feature that allowed you to identify the blood vessels.

[2 marks]

Describe two precautions the student should take when clearing away after the dissection.

[2 marks]

Question 5

Describe how a sample of chloroplasts could be isolated from leaves.

[4 marks]

Scientists grew two groups of plants:

maths symbol control plants with all the inorganic ions needed

maths symbol iron-deficient plants with all the inorganic ions needed but without iron ions.

After 1 week, the scientists measured the mass of protein and the mass of chlorophyll in the chloroplasts isolated from samples of leaves of these two groups of plants.

Table 4 shows the scientists’ results.

Figure for question 5

Some proteins found inside the chloroplast are synthesised inside the chloroplast.

Give one feature of the chloroplast that allows protein to be synthesised inside the chloroplast and describe one difference between this feature in the chloroplast and similar features in the rest of the cell.

[2 marks]

Structural difference

The ratio of protein to chlorophyll in control plants is 9:1

Use the information in Table 4 to calculate the ratio of protein to chlorophyll in iron-deficient plants.

[1 mark]

The scientists also observed the chloroplasts from the samples of leaves using an electron microscope.

Figure 6 shows a chloroplast from a control plant (image A) and a chloroplast from an iron-deficient plant (image B).

Figure 6

This source has been removed due to third-party copyright restrictions.

Use Figure 6 to suggest why iron-deficient plants have a reduced growth rate.

[3 marks]

Question 6

Figure 7 shows the mean distance between centromeres and the poles (ends) of the spindle during mitosis.

Figure for question 6

Calculate the rate of movement of the centromeres during phase E.

Give your answer in μm minute–1 and to 3 decimal places.

[2 marks]

Name the three phases of mitosis shown by C, D and E on Figure 7.

Describe the role of the spindle fibres and the behaviour of the chromosomes during each of these phases.

[5 marks]

Question 7

ADCs are molecules made of a monoclonal antibody linked to a cancer drug.

Figure 8 shows how an ADC enters and kills a tumour cell.

The process of entering the cell and the breakdown of the antibody to release the drug is very similar to phagocytosis.

Figure for question 7

Use your knowledge of phagocytosis to describe how an ADC enters and kills the tumour cell.

[3 marks]

Some of the antigens found on the surface of tumour cells are also found on the surface of healthy human cells.

Use this information to explain why treatment with an ADC often causes side effects.

[2 marks]

Scientists investigated whether one type of ADC could be used to treat human breast

cancer.

This ADC is a monoclonal antibody combined with a drug to inhibit mitosis. The monoclonal antibody binds to a protein found on human breast cancer cells.

The scientists placed small pieces of human breast cancer tissue under the skin of mice.

The scientists then randomly divided the mice into three groups. They treated the groups as follows on day 0.

Group G – control Group H – injected with monoclonal antibody only Group J – injected with ADC (monoclonal antibody + drug).

Every few days, the scientists measured the volume of the tumours formed from the human breast cancer tissue.

Figure 9 shows the scientists’ results.

Figure 9

This source has been removed due to third-party copyright restrictions.

Mice in Group H were injected with 2 mg kg–1 of monoclonal antibody. The monoclonal antibody was in a solution of concentration 500 mg dm–3

Calculate the volume of antibody solution that the scientists would have injected into a 23 g mouse. Give your answer in dm3 and in standard form.

[2 marks]

Suggest one reason why there are no data for Group G and Group H after day 8

[1 mark]

Suggest and explain two further investigations that should be done before this ADC is tested on human breast cancer patients.

[2 marks]

Question 8

Describe how a triglyceride molecule is formed.

[3 marks]

Table 5 shows some properties of four fatty acids.

Figure for question 8

Figure 10 shows diagrams of these fatty acids.

Figure for question 8

Put a tick () in one box that contains correct information about one of these fatty acids.

Figure for question 8

[1 mark]

The percentage of saturated fatty acids compared with unsaturated fatty acids found in lipid stores in seeds differs in different populations.

Scientists investigated two populations of the plant, Helianthus annuus.

The scientists grew young plants from seeds collected from each population. They placed the seeds on wet tissue paper so that the root growth was visible.

They grew seeds from each population at two temperatures:

• warm temperature of 24 °C

• cool temperature of 10 °C

After 10 days, the scientists measured the length of each root.

Table 6 shows some of the properties of the two populations and the scientists’ results.

Figure for question 8

The mean ±2 × standard deviation includes 95% of the data.

The scientists used a data logger to measure the length of the root rather than a ruler.

Suggest one reason why they used a data logger and explain why this was important in this investigation.

[1 mark]

It is known that:

• 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.

[4 marks]

Although these two populations are completely separate and show genetic variation, they are both called Helianthus annuus.

Explain why they are both given this name.

[1 mark]

Question 9

Complete Table 7 with ticks () to show which elements are found in the following biological molecules.

Figure for question 9

[2 marks]

After Watson and Crick proposed the model of DNA structure, scientists investigated

the possible mechanisms for DNA replication.

Two scientists grew a bacterial population, providing them with a nitrogen source containing only the heavy isotope of nitrogen,15 N. As soon as all the DNA in this population contained15 N, the scientists changed the nitrogen source to one containing only the lighter isotope of nitrogen,14 N. They changed the nitrogen source at 0 hours.

During the investigation, the scientists measured the size of the population of bacterial cells.

Figure 11 shows the scientists’ results.

Figure for question 9

The generation time for a population of bacteria is the time taken for all the bacteria to divide once by binary fission.

Use Figure 11 and the following equation to calculate the generation time for this population of bacteria. Give your answer in hours.

size of population at time maths symbol4 hours

Question 9, shown as an image

[2 marks]

Question 9, shown as an image

At intervals during this investigation, the scientists removed samples of the bacterial population, isolated the DNA and measured the density of the DNA.

DNA made using15 N has a higher density than DNA made using14 N.

Figure 12 shows the scientists’ results.

Figure for question 9

There are three possible models of DNA replication.

These models are shown in Figure 13.

Figure for question 9

Which of these models, P, Q or R, is supported by the results shown in Figure 12 ?

Give the letter and name of the model supported and explain why the results do not support the other models.

Explanation for first unsupported model

Explanation for second

Question 9, shown as an image

[3 marks]

Question 10

Describe the structure of DNA.

[5 marks]

Name and describe five ways substances can move across the cell-surface membrane into a cell.

[5 marks]

Figure 14 shows transmission electron micrographs of two cells, one animal cell and one prokaryotic cell.

Figure for question 10

Contrast the structure of the two cells visible in the electron micrographs shown in Figure 14.

[5 marks]

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