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AQA A Level Biology

Topic questions Genetics and inheritance Gene mutations

Gene mutations

8 questions · 71 marks · about 1.5 hours

Gene mutations questions

AQA 7402 Gene mutations 8 questions

Paper 2 June 2025, Question 4 10 marks

Also involves Proteins

4.1

People with the inherited blood disorder haemophilia A (HA) have a reduced ability to clot blood. One variant of HA is due to an inversion mutation in the F8 gene. This mutation causes the production of a non-functional form of the protein, factor VIII. Factor VIII is not an enzyme.

Explain how an inversion mutation can cause the production of non-functional factor VIII.

[4]
Reveal mark scheme Mark scheme for question 4.1
4.2

Currently, the most effective treatment for HA is weekly or monthly injections of factor VIII. Scientists investigated the effectiveness of a gene therapy product, Roctavian, as a treatment for HA. Roctavian contains a viral vector that has the DNA coding for factor VIII.

The scientists:

• selected 112 male volunteers with severe HA, who were at least 18 years of age

• treated each volunteer with a single injection of Roctavian into their blood

• recorded the number of bleeding episodes (internal bleeding) in these volunteers over a two-year period (set A results)

• compared these data with the number of bleeding episodes per year in the volunteers before receiving Roctavian (set B results).

Table 1 shows the scientists’ results.

A value of ±2 × SD from the mean includes over 95% of the data.

Figure for question 4

Use all the information provided to evaluate the use of Roctavian as a treatment for HA.

[5]
Reveal mark scheme Mark scheme for question 4.2
4.3

Haemophilia C is an autosomal recessive condition.

Two parents are both heterozygous for the haemophilia C allele.

Calculate the probability of these parents producing a son who has haemophilia C.

[1]
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Paper 2 June 2025, Question 8 10 marks

Also involves The cell cycle

8.1

A study of cells from a human embryo found that 2.66 × 10–7 % of the genome of one cell had changed due to mutation.

You can assume that:

• the genome contains 3 × 109 nucleotides

• the mean rate of mutation is 1.14 mutations per cell cycle

• the cell cycle lasts 24 hours.

Use this information to calculate the age of the embryo in days.

Show your working.

[2]
Reveal mark scheme Mark scheme for question 8.1
8.2

The genome of the bacterium Escherichia coli is approximately 4.6 million base pairs in length and contains around 4000 genes.

The analysis of the genome of E. coli allows the proteome to be determined more easily than that of eukaryotes.

Explain why.

[2]
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8.3

Determining the proteomes of bacteria has helped provide protection for individuals and populations against some infectious diseases.

Explain how.

[2]
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8.4

Scientists investigated the relationship between one type of variable number tandem repeat (VNTR) and the risk of breast cancer occurring in women.

The scientists determined the risk of breast cancer occurring in women:

• with 2, 3 or 4 repeats of the VNTR

• without the VNTR.

They calculated the odds ratio (OR) of breast cancer occurring with 2, 3 or 4 repeats of the VNTR.

An OR is a measure of how strongly an outcome is associated with a risk factor.

The OR formula is:

risk of breast cancer occurring with the VNTR: risk of breast cancer occurring without the VNTR

Table 5 shows the scientists’ results.

Figure for question 8

A student concluded that the number of repeats of the VNTR is associated with the risk of breast cancer occurring in women.

Evaluate the student’s conclusion.

[4]
Reveal mark scheme Mark scheme for question 8.4

Paper 3 June 2024, Question 5 10 marks

Leigh syndrome (LS) is a rare, recessive, inherited condition.

LS is caused by a mutation in any one of more than 75 different genes coding for proteins involved in oxidative phosphorylation.

In 80% of people with LS, these mutations occur in nuclear DNA. In 20% of people with LS, these mutations occur in mitochondrial DNA (mtDNA).

15% of the nuclear DNA mutations that cause LS occur in the SURF1 gene. A mutated SURF1 gene codes for a shorter polypeptide than a non-mutated SURF1 gene.

5.1

Name one type of SURF1 gene mutation and explain how this mutation could lead to production of a shorter polypeptide.

[2]
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5.2

Globally, the frequency of LS is 1 in 40 000

In the Faroe Islands, which are 18 isolated islands, the frequency of LS is 1 in 1700

The population of the Faroe Islands is 49 053

Estimate the number of people in the Faroe Islands with LS caused by a mutation in the SURF1 gene.

Use information in this question and the information on page 12.

Give your answer to the nearest whole number.

Show your working.

[3]
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5.3

The frequency of LS is higher in the Faroe Islands than globally.

Suggest and explain one reason why.

[2]
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5.4

LS usually causes death within the first three years of life.

Using all the information in this question, evaluate whether all people should be genetically screened for LS.

[3]
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Paper 3 November 2021, Question 1 9 marks

In one species of squirrel, Sciurus carolinensis, fur colour is controlled by one gene, with two codominant alleles. CG represents the allele for grey fur colour, and CB represents the allele for black fur colour.

Table 1 shows the three possible phenotypes.

Figure for question 1
1.1

In a population of 34 S. carolinensis, 2 had black fur.

Use the Hardy–Weinberg equation to estimate how many squirrels in this population had brown-black fur. Show your working.

[2]
Reveal mark scheme Mark scheme for question 1.1
1.2

The actual number of squirrels in this population that had brown-black fur was 16.

Use all of the information to calculate the actual frequency of the CG allele.

Do not use the Hardy–Weinberg equation in your calculation.

Give your answer to 2 decimal places.

[1]
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1.3

S. carolinensis were first introduced to the UK from North America in the 1870s. They are now widely distributed across the UK.

S. carolinensis from both North America and the UK show exactly the same genotypic and phenotypic variation. An identical mutation causing black fur has also been found in several other species closely related to S. carolinensis.

Use this information to deduce which one of the following conclusions is most likely true.

Tick (maths symbol) one box.

Figure for question 1
[1]
Reveal mark scheme Mark scheme for question 1.3
1.4

The mutation that caused the CB allele was due to a 24 base-pair deletion from the CG allele.

The protein coded for by the CB allele is 306 amino acids long.

Calculate the percentage reduction in size of the protein coded for by the CB allele compared with the protein coded for by the CG allele.

Give your answer to 3 significant figures and show your working.

[2]
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1.5

In S. carolinensis, fur colour depends on the distribution and relative amounts of light pigments and dark pigments in the hairs of the fur. Figure 1 shows how the protein produced from the CG allele can result in the production of a light pigment or a dark pigment.

Figure for question 1

The deletion mutation in the CB allele results in the production of a receptor protein that does not have glutamic acid. The lack of glutamic acid in the receptor protein has the same effect as αMSH leaving the receptor protein.

Use Figure 1 and this information to suggest why S. carolinensis with the genotype CBCB have black fur rather than grey fur.

[3]
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Paper 3 November 2020, Question 3 8 marks

In Europe, viruses have infected a large number of frogs of different species. The viruses are closely related and all belong to the Ranavirus group.

Previously, the viruses infected only one species of frog.

3.1

Suggest and explain how the viruses became able to infect other species of frog.

[2]
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3.2

Name two techniques the scientists may have used when analysing viral DNA to determine that the viruses were closely related.

[1]
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3.3

Determining the genome of the viruses could allow scientists to develop a vaccine.

Explain how.

[2]
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3.4

Describe how the B lymphocytes of a frog would respond to vaccination against Ranavirus.

You can assume that the B lymphocytes of a frog respond in the same way as B lymphocytes of a human.

Do not include details of the cellular response in your answer.

[3]
Reveal mark scheme Mark scheme for question 3.4

Paper 2 June 2019, Question 2 9 marks

Also involves Proteins

Sickle cell disease (SCD) is a group of inherited disorders. People with SCD have sickle-shaped red blood cells. A single base substitution mutation can cause one type of SCD. This mutation causes a change in the structure of the beta polypeptide chains in haemoglobin.

2.1

Explain how a single base substitution causes a change in the structure of this polypeptide.

Do not include details of transcription and translation in your answer.

[3]
Reveal mark scheme Mark scheme for question 2.1
2.2

Haematopoietic stem cell transplantation (HSCT) is a long-term treatment for SCD. In HSCT, the patient receives stem cells from the bone marrow of a person who does not have SCD. The donor is often the patient’s brother or sister. Before the treatment starts, the patient’s faulty bone marrow cells have to be destroyed.

Use this information to explain how HSCT is an effective long-term treatment for SCD.

[3]
Reveal mark scheme Mark scheme for question 2.2
2.3

A new long-term treatment for SCD involves the use of gene therapy.

Figure 2 shows some of the stages involved in this treatment in a child with SCD.

Figure for question 2

Some scientists have concluded that this method of gene therapy will be a more effective long-term treatment for SCD than HSCT. Use all the information provided to evaluate this conclusion.

[3]
Reveal mark scheme Mark scheme for question 2.3

Paper 3 June 2019, Question 4 9 marks

Mitochondrial DNA (mtDNA) is a small circular DNA molecule located in mitochondria. It is 16 569 nucleotides long and contains 37 genes and a control region.

Sports scientists investigated whether a mutation in the control region of mtDNA in human males was related to an ability to exercise for longer.

• The males in Group T had thymine at nucleotide position 16 519

• The males in Group C had a mutation resulting in cytosine at nucleotide position 16 519

4.1

The control regions of Group T and Group C were the same length.

Name the type of gene mutation that is most likely to have occurred at nucleotide position 16 519

[1]
Reveal mark scheme Mark scheme for question 4.1
4.2

Group T and Group C completed the same 8-week training programme. The following measurements were taken at the start of the 8-week programme, and again at the end.

1. VO 2 max (a measure of maximal oxygen uptake).

2. Citrate synthase (CS) activity (CS is an enzyme involved in the Krebs cycle).

The scientists then calculated the percentage increase in each measurement in both groups.

Figure 4 and Figure 5 show their results.

Figure for question 4

A student concluded from Figure 4 and Figure 5 that training has a positive effect on VO 2 max and CS activity.

Evaluate the student’s conclusion.

[3]
Reveal mark scheme Mark scheme for question 4.2
4.3

The mitochondrial DNA (mtDNA) control region is an area of mtDNA that is non-coding. This region stimulates the synthesis of both mtDNA and mitochondrial messenger RNA.

Use this information to suggest two reasons why the mutation at nucleotide position 16 519 could lead to the differences seen in Figure 5.

[2]
Reveal mark scheme Mark scheme for question 4.3
4.4

The sports scientists investigated whether there was a correlation between the percentage change in VO 2 max and percentage change in CS activity in Group T.

Figure 6 shows their results.

Figure for question 4

‘Having thymine at nucleotide position 16 519 in Group T causes an increase in ability to exercise for longer.’

Evaluate this conclusion.

Use all the data in this question.

[3]
Reveal mark scheme Mark scheme for question 4.4

Paper 2 June 2018, Question 2 6 marks

Also involves Cell division

2.1

There are different types of gene mutation.

Put a tick (maths symbol) in the box next to the statement which describes incorrectly the effect of the mutation in an exon of a gene.

Figure for question 2
[1]
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2.2

Describe how alterations to tumour suppressor genes can lead to the development of tumours.

[3]
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2.3

A type of malignant tumour cell divides every 8 hours.

Starting with one of these cells, how many tumour cells will be present after 4 weeks? Assume none of these cells will die.

Give your answer in standard form.

[2]
Reveal mark scheme Mark scheme for question 2.3

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