AQA AS Level Biology Paper 1, June 2024: Question 9

10 marks · Medium difficulty · Short Answer

Analyze a passage on spontaneous DNA mutations, nucleotide changes, and the effects of ataxia telangiectasia (AT) enzyme deficiency through calculation, structural comparisons, and explanations of mutations and cell cycle control.

Practise this question

Question

An exam page containing a reading passage about spontaneous DNA mutations and four sub-questions (09.1 to 09.4) regarding DNA structure, percentage calculations in standard form, explanations of how mutations can still produce functional proteins, and why mutations occur at a higher rate in people with ataxia telangiectasia.
Question text

09 Read the following passage.

DNA is a stable molecule but, even under normal cell conditions, spontaneous

changes occur to the DNA nucleotide sequence. One example of a

spontaneous change occurs when a cytosine base in a guanine–cytosine

nucleotide pair is changed to a uracil base. This produces a guanine–uracil

nucleotide pair in the DNA molecule. Scientists estimate this type of 5

spontaneous change occurs to 100 guanine–cytosine nucleotide pairs in the

genome of healthy human cells every day.

In healthy cells, enzyme-controlled processes repair these spontaneous

changes in the DNA molecule by changing uracil bases back to cytosine

bases. If these repairs do not happen, the uracil DNA nucleotide attracts an 10

adenine DNA nucleotide when the DNA is replicated in the cell cycle. A

mutation of the original DNA has now occurred.

Healthy cells with damaged DNA produce enzyme X. This enzyme slows the

cell cycle by delaying the start of DNA replication. People with the disease

ataxia telangiectasia (AT) do not produce functional enzyme X. Mutations 15

occur at a higher rate in people with AT.

Use the information in the passage and your own knowledge to answer the

following questions.

09.1 Give one similarity in structure between a guanine–cytosine nucleotide pair and a

guanine–uracil nucleotide pair in a DNA molecule (lines 3–5).

Do not refer to guanine in your answer.

[1 mark]

09.2 The DNA in a human genome contained 3 × 109 nucleotide pairs.

Assume 40% of these nucleotide pairs are guanine–cytosine nucleotide pairs.

Use this information and lines 5–7 to calculate the percentage of

guanine–cytosine nucleotide pairs that change to guanine–uracil nucleotide pairs in

this genome every day.

Give your answer in standard form.

Show your working.

[2 marks]

Answer %

09.3 The type of mutation that occurs when ‘repairs do not happen’ (lines 10–11) may still

produce a functional protein.

Suggest and explain why.

[4 marks]

09.4 Suggest and explain why ‘mutations occur at a higher rate’ in people with AT

(lines 15–16).

[3 marks]

Mark scheme

Show the mark scheme A mark scheme table showing acceptable answers and marking guidance for questions 09.1 through 09.4, including required points for DNA structural similarities, calculation workings for standard form percentages, degenerate code explanations, and enzyme function impacts on mutation rates.

Question Marking Guidance Mark Comments

(Has) phosphate Ignore the number of

hydrogen bonds

OR

09.1 (Has) deoxyribose (1 x Accept both contain a

AO2) pyrimidine/single ring

OR (structure)

Accept ‘H bonds’

(Has) hydrogen bonds;

Correct answer of 8 × 10–6 OR 8.3 × 10–6 Accept any number of

decimal places that

= 2 marks;;

round to 8.3

Incorrect answer of

0.000 008 3 (correct answer but not in standard

form) = 1 mark

OR

8.3 × 10–8 (correct division using correct number of 2

09.2 G-C pairs, and in standard form, but not shown as (2 x

a percentage) = 1 mark AO2)

OR

Correct answer in incorrect standard form; eg 83 ×

10–7 = 1 mark

OR

1.2 × 109 (correct number of G-C pairs in the

genome in standard form) = 1 mark;

1. Substitution (mutation occurred):

2. (Only) one nucleotide/base pair is changed (in a

gene)

OR

(Only) one (DNA) triplet/codon changed; 3. Reject same amino

3. Same amino acid (coded for); acid is produced

3. Accept one amino

4. (Because) DNA/genetic code is degenerate; acid changed

4 max

09.3 (4 x 4. Accept a

AO2) description of

5. (So) tertiary structure is not changed;

20 degenerate code

6. (Change) could be in an intron; 3 and 4 can be

awarded together, e.g

7. Removed during splicing; ‘different codons/

triplets code for the

same amino acid’ =

MP3 and MP4

1. No (functional) enzyme/X;

2. Ignore ‘cell cycle

2. (So) more/faster cell cycles;

isn’t slowed down’ on

3. More(frequent) DNA replication 3 max its own

09.4 OR (3 x 3. Accept ‘faster DNA

AO2) replication’

DNA replication not delayed;

4. (So) mutations (more likely to) occur in DNA

replication;

How to answer it

DNA Structure, Mutations, and the Cell Cycle Study Guide

What this question tests

This passage-based question assesses your understanding of nucleic acid biochemistry (nucleotide structure, complementary base pairing), mathematical and standard form manipulation in a biological context, the genetic code (degeneracy and substitution mutations), and the control of the cell cycle in relation to disease (Ataxia telangiectasia).

Question 09.1

Similarity in DNA Nucleotide Structure

Give one similarity in structure between a guanine–cytosine nucleotide pair and a guanine–uracil nucleotide pair in a DNA molecule. Do not refer to guanine.

✅ Correct Answer

Any one of the following:

  • (Has) phosphate
  • (Has) deoxyribose
  • (Has) hydrogen bonds

💡 Key Knowledge

Both base pairs share a common DNA backbone structure consisting of deoxyribose sugar and phosphate groups linked by phosphodiester bonds, as well as hydrogen bonds holding the nitrogenous bases together across the double helix.

❌ Common Errors

Students often lose this mark by mentioning guanine (which is explicitly forbidden by the question stem) or incorrectly stating that both contain thymine/cytosine.

Mark: 1 mark [AO2] — Note: Hydrogen bonds count can vary, but accepting the presence of hydrogen bonds scores the mark.
Question 09.2

Calculation of Spontaneous Mutation Percentage

Calculate the percentage of guanine–cytosine nucleotide pairs that change to guanine–uracil nucleotide pairs in this genome every day. Give your answer in standard form.

📐 Step-by-Step Calculation

  1. Find total G-C pairs in the human genome:
    Total genome = 3 × 10⁹ nucleotide pairs.
    40% are G-C pairs: 0.40 × (3 × 10⁹) = 1.2 × 10⁹ G-C pairs.
  2. Determine daily changes:
    Passage states: 100 G-C pairs change every day.
  3. Calculate fraction changing:
    100 / (1.2 × 10⁹) = 8.333 × 10⁻⁸
  4. Convert to a percentage and standard form:
    Fraction × 100 = 8.333 × 10⁻⁸ × 100 = 8.333 × 10⁻⁶%
    Rounding to appropriate sig figs gives 8.3 × 10⁻⁶ (or 8 × 10⁻⁶ ).

✅ Correct Answer

8 × 10⁻⁶ OR 8.3 × 10⁻⁶ (2 marks)

❌ Common Calculation Traps

  • Forgetting to convert the final decimal fraction into a percentage (which loses 1 mark, yielding 8.3 × 10⁻⁸ ).
  • Failing to use standard form correctly (e.g., writing 0.0000083 gets 1 mark).
Mark: 2 marks [AO2]
Question 09.3

Mutation Consequences on Protein Function

Suggest and explain why the type of mutation that occurs when 'repairs do not happen' may still produce a functional protein. (4 marks max)

✅ Correct Answer / Marking Points

Any 4 of the following marking points:

  • 1. Type of mutation: Substitution (mutation occurred).
  • 2. Scale of change: (Only) one nucleotide/base pair is changed in a gene OR only one triplet/codon changed.
  • 3 & 4. Degeneracy of code: Same amino acid is coded for because the genetic code is degenerate (different codons can code for the same amino acid).
  • 5. Tertiary structure: Therefore, the primary structure and tertiary structure of the protein are not changed.
  • 6 & 7. Introns: Alternatively, the change could be located in an intron and be removed during pre-mRNA splicing.

🧠 Exam Technique & Examiner Commentary

To access all 4 marks, you must link the type of mutation (substitution affecting a single triplet) to the property of the genetic code (degenerate code resulting in the same amino acid), and finally explain the consequence on protein structure (tertiary structure unchanged).

Mark: 4 marks max [AO2]
Question 09.4

Ataxia Telangiectasia and Mutation Rates

Suggest and explain why 'mutations occur at a higher rate' in people with AT (lines 15–16). (3 marks max)

✅ Correct Answer / Marking Points

Any 3 of the following:

  • 1. Lack of enzyme: No (functional) enzyme X is produced in people with AT.
  • 2. Cell cycle impact: Cell cycle is not slowed down / results in more frequent or faster cell cycles.
  • 3. DNA replication: More frequent DNA replication occurs (or DNA replication is not delayed for repairs).
  • 4. Mutation likelihood: Therefore, mutations are more likely to occur during DNA replication.

💡 Key Knowledge

Enzyme X normally acts as a checkpoint control mechanism that delays the start of DNA replication to allow DNA repair. Without it, damaged DNA is replicated unchecked, cementing spontaneous errors into permanent mutations.

Mark: 3 marks max [AO2]

Topics

Biology · 3.1 Biological molecules · 3.4 Genetic information, variation and relationships between organisms

Question and mark scheme from the AQA AS Level Biology examination, Paper 1, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.