AQA A-Level Biology Paper 2, June 2025: Question 8

10 marks · Medium difficulty · Extended Answer

Calculate the age of an embryo from mutation rate data, explain why prokaryote proteomes are easier to determine than eukaryote proteomes and how this aids disease prevention, and evaluate evidence linking VNTR repeats to breast cancer risk.

Practise this question

Question

Question 8 consists of four parts: Question 08.1 asks to calculate embryo age in days given that 2.66 × 10^-7% of a 3 × 10^9 nucleotide genome mutated, with a mutation rate of 1.14 per cell cycle and 24 hours per cycle (2 marks). Question 08.2 asks to explain why determining the proteome of E. coli is easier than that of eukaryotes (2 marks). Question 08.3 asks how bacterial proteomes provide protection against infectious diseases (2 marks). Question 08.4 gives Table 5 showing odds ratio of breast cancer for 2 repeats (0.55:1), 3 repeats (1.58:1), and 4 repeats (1.94:1) of a VNTR, asking to evaluate the student's conclusion that repeat number is associated with breast cancer risk (4 marks).
Question text

08.1 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 marks]

Answer days

The genome of the bacterium Escherichia coli is approximately 4.6 million base pairs

in length and contains around 4000 genes.

08.2 The analysis of the genome of E. coli allows the proteome to be determined more

easily than that of eukaryotes.

Explain why.

[2 marks]

08.3 Determining the proteomes of bacteria has helped provide protection for individuals

and populations against some infectious diseases.

Explain how.

[2 marks]

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.

Table 5

Number of VNTR repeats Odds ratio

20.55 : 1

31.58 : 1

4 25 1.94 : 1

08.4 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 marks]

Mark scheme

Show the mark scheme Mark scheme for Question 8: 08.1 awards 2 marks for 7 days (or 1 mark for intermediate working such as 700, 168, 9, 7.98 or 8). 08.2 awards 2 marks for noting prokaryotes lack introns/non-coding DNA and eukaryotes contain more regulatory genes. 08.3 awards 2 marks for identification of antigens used to develop vaccines or monoclonal antibodies. 08.4 awards up to 4 marks for evaluation points: 2 repeats decrease risk, 3 and 4 increase risk (both required for max marks), no data for 1 or >4 repeats, other factors affecting risk, different breast cancer types, or unknown sample size.

Question Marking Guidance Mark Comments

Correct answer of 7 (days) = 2 marks;;

Incorrect answer but shows 7 to incorrect

magnitude e.g. 700 (ignore preceding/subsequent

zeros and position of decimal point i.e. did not

divide by 100 or incorrect use of order of

magnitude) = 1 mark

OR

Incorrect answer of 168 (incorrectly multiplied by

24) = 1 mark

OR 2

08.1 (2 x

Incorrect answer shows 9 ignoring subsequent AO2)

numbers (multiplied rather than divided by 1.14) =

1 mark

OR

Incorrect but shows 6.93 (incorrect rounding of

7.98) = 1 mark

OR

Incorrect but shows 7.98 or 8 (did not divide by

1.14) = 1 mark;

1. (E. coli / prokaryotes/bacteria) do not have 1 and 2 Accept ‘It’

introns /non-coding (DNA) refers to E. coli

OR

(E. coli / prokaryotes/bacteria) only have exons /

coding (DNA)

OR

Eukaryotes contain introns / non-coding (DNA);

2 2. Accept prokaryotes

08.2 (2 x have no/fewer

2. Eukaryotes contain (more) regulatory AO1) regulatory genes

genes;

– A-LEVEL BIOLOGY – 7402/2 –

1. (Identification/isolation of) antigens; 3. 2. Accept descriptions

of (active) immunity

2. (Antigens/proteins used to develop)

2 2. Ignore antibiotics

vaccines/antitoxins/antiserums

08.3 (2 x

OR AO1)

(Antigens/proteins used) to develop

(monoclonal) antibodies;

Ignore length of

Mark points 1 and 2 are required for maximum

investigation and use

marks.

of a statistical test

1. Two repeats decreases (risk of) breast cancer;

2. Accept ‘more than

2. Three and four repeats increases (risk of) breast two repeats’

cancer;

3. Accept only 2,3 and

4 repeats

3. No results for more than 4 repeats

OR

Ignore increase in

VNTRs repeats

No results for 1 repeat

4 max increase risk of breast

08.4 (4 x cancer

OR

AO3)

Few repeats (studied);

4. Other factors affect (risk of) breast cancer

OR

Named factor (e.g. age) affects (risk of) breast

cancer;

5. Breast cancers not all same (type);

6. Sample size not known;

How to answer it

Genomes, Proteomes and Evaluating Risk Factors

📋 What this question tests

This question brings together mathematical problem-solving and molecular genetics across Topic 8 (Control of Gene Expression) and Topic 2 (Cells/Pathogens):

  • Multi-step quantitative skills: Calculating cell cycles from mutation percentages and orders of magnitude.
  • Genome vs. proteome sequencing: Why prokaryotic proteomes are simpler to deduce than eukaryotic proteomes (introns and regulatory DNA).
  • Medical applications of genomics: Identifying surface antigens to create vaccines and monoclonal antibodies.
  • Critical data evaluation (AO3): Interpreting odds ratio (OR) tables and evaluating associative claims concerning VNTRs and cancer risk.

Question 08.1

Embryo Age Calculation from Mutation Frequency (2 Marks)

📐 Step-by-Step Calculation

  1. Find total mutations in the genome:
    Convert percentage to a decimal/fraction by dividing by 100:
    Fraction mutated = (2.66 × 10⁻⁷) ÷ 100 = 2.66 × 10⁻⁹
    Number of mutations = (2.66 × 10⁻⁹) × (3 × 10⁹) = 7.98 mutations
  2. Calculate total cell cycles:
    Mean rate = 1.14 mutations per cycle.
    Cell cycles = 7.98 ÷ 1.14 = 7 cell cycles
  3. Convert cell cycles to days:
    Each cell cycle lasts 24 hours (= 1 day).
    Age = 7 × 1 = 7 days

✅ Correct Answer

7 (days) [2 marks]

Partial Credit (1 mark):
  • Shows an answer of 7 to an incorrect power of ten (e.g. 700 due to forgetting to divide percentage by 100).
  • Calculates 168 (incorrectly multiplied cycles by 24 instead of converting to days).
  • Shows 7.98 or 8 (calculated total mutations but forgot to divide by mutation rate 1.14).
  • Shows 9 (multiplied by 1.14 instead of dividing).

❌ Common Errors

  • Forgetting the % sign: Multiplying 2.66 × 10⁻⁷ directly by 3 × 10⁹ gives 798 mutations instead of 7.98. Remember: "%" means per hundred!
  • Unit confusion (hours vs days): Multiplying 7 cycles by 24 hours gave 168, forgetting the question explicitly asks for the answer in days.

🧠 Exam Technique

Always write every single step of your working clearly. Even if you press a wrong key on your calculator and get an incorrect final number, clear intermediate values (e.g., showing 7.98 ) secure method marks.

Question 08.2

Determining Proteomes: Bacteria vs. Eukaryotes (2 Marks)

✅ Mark Scheme Points

  • Mark 1: (E. coli / prokaryotes / bacteria) do not have introns / non-coding DNA
    OR (E. coli) only contain exons / coding DNA
    OR Eukaryotes contain introns / non-coding DNA.
  • Mark 2: Eukaryotes contain (more) regulatory genes / prokaryotes have fewer regulatory genes.

💡 Key Knowledge

  • Proteome: The full range of proteins that a cell is capable of producing.
  • In simple prokaryotes, there is almost a 1:1 relationship between DNA base sequences and mRNA/protein sequences because there is virtually no non-coding DNA.
  • In eukaryotes, vast swathes of DNA consist of non-coding introns and complex regulatory regions that determine whether and when a gene is expressed.

❌ Common Errors

  • Vague claims like "prokaryotes have less DNA" or "eukaryotes have more genes". You must refer specifically to introns / non-coding DNA or regulatory genes.
  • Confusing introns with exons. Remember: Exons are expressed.

🧠 Exam Technique

Comparative questions require explicit points about either group or both. State clearly: "Bacteria lack introns, whereas eukaryotic DNA contains non-coding introns and more regulatory genes."

Question 08.3

Bacterial Proteomes and Disease Protection (2 Marks)

✅ Mark Scheme Points

  • Mark 1: Identification / isolation of (surface) antigens;
  • Mark 2: These antigens are used to develop vaccines (or antitoxins / antiserums / monoclonal antibodies);
Examiner note: Descriptions of inducing active immunity are accepted. Ignore antibiotics (antibiotics target bacterial processes, not sequenced protein antigens).

💡 Key Knowledge

By mapping the bacterial proteome, scientists can pinpoint which specific surface proteins act as antigens. Synthesising or purifying these antigens allows production of vaccines, which stimulate memory cells and confer herd immunity on populations.

❌ Common Trap: Stating "Antibiotics"

A very common student error is suggesting proteome analysis helps develop antibiotics. Antibiotics are chemical compounds targeting cell wall synthesis or bacterial ribosomes, not antigen-specific immune products. The mark scheme explicitly states: "Ignore antibiotics".

Question 08.4

Evaluating VNTR Repeats and Breast Cancer Risk (4 Marks)

✅ Mark Scheme Points (Max 4 Marks)

*Note: Mark points 1 and 2 are essential to achieve full marks.*

  • 1. [Essential]: Two repeats decreases the risk of breast cancer (odds ratio = 0.55, which is < 1).
  • 2. [Essential]: Three and four repeats increase the risk of breast cancer (odds ratio = 1.58 and 1.94, which are > 1).
  • 3. [Limitation]: No data / results for more than 4 repeats OR no data for 1 repeat OR only a few repeat lengths studied (2, 3, 4).
  • 4. [Limitation]: Other named factors affect breast cancer risk (e.g. age, family history/genetics, lifestyle, HRT).
  • 5. [Limitation]: Breast cancers are not all the same type (different mutations/mechanisms).
  • 6. [Limitation]: Sample size is not known.

🧠 Understanding Odds Ratios (OR)

The definition was given in the question: risk with VNTR : risk without VNTR

  • OR < 1.0 (0.55 : 1): Risk with 2 repeats is lower than without it → decreased risk.
  • OR > 1.0 (1.58 : 1 and 1.94 : 1): Risk with 3 or 4 repeats is higher than without → increased risk.
Examiner Warning: Stating simply that "increasing VNTR repeats increases risk" was rejected! 2 repeats actually lowers risk compared to baseline. You must mention 2 repeats separately from 3 and 4.

❌ Common Errors

  • Generalising the trend: Writing "as repeats increase, cancer risk increases" loses marks because 2 repeats actually protects against cancer compared to having no repeats.
  • Vague evaluation: Writing "they didn't do a statistical test" or "study was too short" — the mark scheme explicitly says to ignore these. Stick to sample size, missing data points, and confounding risk factors.

💡 High-Scoring Structure for "Evaluate" Questions

Always balance your argument:

  1. Support the claim: Quote the data accurately (distinguish between OR > 1 and OR < 1).
  2. Challenge/limit the claim: Identify gaps in data (missing repeat numbers: 1, 5, 6+), confounding lifestyle/genetic variables, and unknown sample size.

Topics

Biology · Practical skills · 3.2 Cells · 3.4 Genetic information, variation and relationships between organisms · 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.