AQA A-Level Biology Paper 2, June 2025: Question 4

10 marks · Medium difficulty · Extended Answer

Explain the effect of an inversion mutation on protein structure, evaluate clinical trial data for a haemophilia A gene therapy, and calculate the inheritance probability of an affected son.

Practise this question

Question

Question 04 contains three parts. Part 04.1 asks to explain how an inversion mutation can cause the production of non-functional factor VIII (which is not an enzyme) for 4 marks. Part 04.2 describes a trial of gene therapy Roctavian involving 112 adult males with severe haemophilia A, and includes Table 1 comparing mean number of bleeding episodes per year (±2 × SD): Volunteer set A (received Roctavian) is 0.8 ± 1.2, and set B (received factor VIII) is 4.8 ± 0.6. Students must evaluate the use of Roctavian as a treatment for HA for 5 marks. Part 04.3 asks for the probability of two heterozygous parents producing a son with haemophilia C, an autosomal recessive condition, for 1 mark.
Question text

04.1 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 marks]

04.2 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 the9

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.

Table 1

Mean number of bleeding episodes

Volunteer set per year (±2 × SD)

A

0.8 ± (1.2)

(Received Roctavian)

B

4.8 ± (0.6)

(Received factor VIII)

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

HA.

[5 marks]

04.3 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 mark]

Probability =

Mark scheme

Show the mark scheme Mark scheme for Question 04. 04.1 awards 4 marks for: change in sequence of DNA bases/inversion, change in amino acid sequence/primary structure, change in hydrogen/ionic/disulfide bonds, and altered tertiary structure. 04.2 awards up to 5 marks (point 1 required for maximum): (For) non-overlapping SDs indicating significant reduction, single injection needed, compares same volunteers; (Against) only males studied, possible side effects, only individuals ≥ 18 studied, only severe HA treated. 04.3 awards 1 mark for 0.125, 12.5%, or 1/8.

Question Marking Guidance Mark Comments

1. Change in sequence of DNA bases 1. Accept if answer

refers to DNA and

OR shows base sequence

inverted e.g. TATGC

DNA base sequence inverted/reversed; to CGTAT

1. Accept nucleotides

2. Change in (sequence of) amino acid(s)

for bases

OR 1. Accept changes

DNA triplet(s)

Change in primary structure; 4

04.1 (4 x 2. Reject amino acids

3. Change in hydrogen/ionic/disulfide bonds; AO2) are ‘formed’ or

‘produced’

4. Alters tertiary/3° structure (of protein);

2. Accept shorter

amino acid sequence

or terminated

4. Reject active

site/enzyme

4. Ignore quaternary

4. Ignore 3D

– A-LEVEL BIOLOGY – 7402/2 –

Mark point 1 required for maximum marks. 1. Reject ‘results are

10 significant’

Max 3 marks from mark points 4 to 7

1. Accept ‘not due to

(For) chance’ for

‘significant’

1. SDs do not overlap so significant

2. Accept fewer/less

difference/decrease/change (with Roctavian);

for ‘one’

2. Only one injection/treatment is required

Ignore ‘sample size’

‘length of

OR

investigation’ and

‘repeat studies

Weekly/monthly/regular treatments/injections not

required’

required;

Ignore reference to

3. (Investigation) compares same volunteers;

5 max statistical test

04.2 (5 x

(Against)

AO3)

4. Only males (investigated/treated)

OR

No females (investigated/treated);

5. (Possible) side effects;

6. Accept

6. At least 18 (years of age investigated/treated)

children/young for

under 18

OR

6. Reject only 18

No under 18 (years of age investigated/treated);

years of age treated

7. Only severe HA (investigated/treated);

11 Reject 1 : 8 or 8 : 1

0.125 / 12.5% / /8 ;

04.3 (1 x Accept 1 in 8

AO2) Reject 12.5 without %

How to answer it

Haemophilia A: Mutations, Gene Therapy & Genetics

What this question tests

This question integrates core molecular biology, clinical data interpretation, and classic Mendelian inheritance across three distinct areas:

  • Gene Mutations & Protein Structure: Linking a change in DNA base sequence (inversion) through transcription and translation to altered primary and tertiary structures of a non-enzymatic protein.
  • Experimental Evaluation (AO3): Interpreting standard deviations (overlap vs significance) and critically balancing pros (effectiveness, convenience) and cons (demographics, unstudied groups, side effects) of gene therapy.
  • Monohybrid Inheritance & Probability: Calculating the combined probability of an autosomal recessive condition alongside biological sex determination.
Question 04.1 • 4 Marks (AO2)

Effects of an Inversion Mutation on Factor VIII

Explaining how a base inversion leads to a non-functional protein

✅ Mark Scheme Model Answer

  1. Change in sequence of DNA bases (or DNA base sequence is inverted/reversed).
  2. Change in sequence of amino acids (or change in primary structure).
  3. Change in hydrogen, ionic, or disulfide bonds.
  4. Alters the tertiary (3°) structure of the protein.

🧠 Exam Technique & Logic Chain

Always construct a seamless 4-step chain of causation for mutation questions:

DNA Sequence (bases inverted)
  ↓ alters codons on mRNA
Primary Structure (sequence of amino acids)
  ↓ bonds form in different places
Hydrogen / Ionic / Disulfide bonds
  ↓ protein folds differently
Tertiary Structure lost/altered

❌ Critical Trap Alert (Examiner Commentary)

  • Do NOT mention active site: The question stem explicitly says "Factor VIII is not an enzyme". Saying "active site is no longer complementary" immediately loses mark point 4!
  • Do NOT say amino acids are "produced": Amino acids are already present in the cytoplasm; they are joined in a different sequence or order.
  • Do NOT just say "3D shape": AQA specifically requires the precise term tertiary structure. "3D shape" is ignored.

💡 Key Knowledge: Inversion Mutation

An inversion mutation occurs when a section of DNA bases detaches, rotates 180°, and reattaches. This changes the triplet codons transcribed into mRNA, altering the specified amino acids and shifting the R-group bonding interactions in tertiary folding.

Question 04.2 • 5 Marks (AO3)

Evaluating Roctavian Gene Therapy

Using experimental data and clinical context to evaluate effectiveness

Special Marking Rule: Mark point 1 (Standard Deviation analysis) is strictly required to score the maximum 5 marks. Furthermore, a maximum of 3 marks can come from the "Against" points (points 4–7).

✅ Accepted Evaluation Points

For (Benefits):

  • MP1 (Compulsory): Standard Deviations do not overlap (A: 0.8 ± 1.2 vs B: 4.8 ± 0.6) so there is a significant difference / decrease in bleeding episodes.
  • MP2: Only one injection/treatment is required (regular weekly/monthly injections no longer needed).
  • MP3: Investigation compares the same volunteers (reduces participant variation).

Against (Limitations - Max 3):

  • MP4: Only males investigated / no females tested.
  • MP5: Potential side effects (e.g. from viral vector/immune response).
  • MP6: Only tested on adults (at least 18 years old) / not tested on children.
  • MP7: Only tested on individuals with severe haemophilia A.

🧠 How to Calculate & Discuss Overlap

Calculate the 95% data intervals (Mean ± 2 × SD):

  • Set A (Roctavian): 0.8 + 1.2 = 2.0 (upper limit = 2.0)
  • Set B (Factor VIII): 4.8 − 0.6 = 4.2 (lower limit = 4.2)

Since the highest value of A (2.0) is far below the lowest value of B (4.2), the standard deviation ranges do not overlap. This shows the difference is significant / not due to chance.

❌ Common Pitfalls in AO3 Evaluate Questions

  • Vague significance statements: Writing "the results are significant" scores 0. You must state why: "The SDs do not overlap, so the difference is significant".
  • Misreading the age criteria: Writing "only 18-year-olds were treated" is rejected. The study investigated people aged at least 18 (adults only).
  • Generic stock criticisms: Stating "small sample size", "need a longer study", or "repeats required" received no credit. 112 volunteers over 2 years is a substantial clinical trial. Always inspect the specific demographic controls provided in the stem!
Question 04.3 • 1 Mark (AO2)

Inheritance Probability Calculation

Determining the probability of producing a son with Haemophilia C

📐 Step-by-Step Probability Calculation

  1. Identify parental genotypes:
    Haemophilia C is autosomal recessive. Let H = normal, h = haemophilia C.
    Both parents are heterozygous: Hh × Hh
  2. Determine probability of child inheriting Haemophilia C:
    Offspring ratios: 1 HH : 2 Hh : 1 hh
    Probability of affected child ( hh ) = ¹/₄ (0.25)
  3. Factor in sex determination:
    Probability of having a son = ¹/₂ (0.50)
  4. Calculate combined probability:
    P(Son with Haemophilia C) = ¹/₄ × ¹/₂ = ¹/₈ or 0.125 or 12.5%

✅ Correct Values Accepted

¹/₈  or  0.125  or  12.5%

Accept: "1 in 8"

❌ Calculation Traps

  • Missing the word "son": Stopping at 0.25 (¹/₄). Because Haemophilia C is autosomal (not sex-linked), sex inheritance (0.5) is independent and must be multiplied.
  • Writing odds/ratios: 1:8 or 8:1 are rejected. Probability must be written as a fraction, decimal, percentage, or "1 in 8".
  • Missing the percentage symbol: Writing 12.5 without the % sign is rejected.

Topics

Biology · Practical skills · 3.1 Biological molecules · 3.7 Genetics, populations, evolution and ecosystems (A-level only) · 3.8 The control of gene expression (A-level only) · Data analysis

Question and mark scheme from the AQA A-Level Biology examination, Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.