AQA A-Level Biology Paper 2, June 2025: Question 9

7 marks · Medium difficulty · Short Answer

Calculate the difference between observed and expected cystic fibrosis cases in 2022, define a DNA probe, and deduce family members' genotypes from gel electrophoresis bands.

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Question

Question 9 contains Table 6 showing the number of people with cystic fibrosis and estimated UK population (in millions) for 2020 (10,837 cases, 67.06 million) and 2022 (11,148 cases, 67.51 million). Part 9.1 asks to calculate how many more people had CF than expected in 2022. Part 9.2 asks to define a DNA probe. Part 9.3 presents Figure 5, an electrophoresis gel with 4 lanes: lane 1 has two bands (one upper, one lower), lane 2 has one lower band, lane 3 has two bands (upper and lower), lane 4 has one upper band. Students must fill Table 7 identifying each lane as parent or child with their genotype, and explain the identification of the child with CF.
Question text

09 Cystic fibrosis (CF) is a recessive inherited disorder caused by a defective CFTR

gene.

Table 6 shows the number of people with CF and the estimated population of the UK

in 2020 and 2022.

Table 6

Estimated population

Year Number of people with CF 6

of the UK × 10

2020 10 837 67.06

2022 11 148 67.51

09.1 Calculate how many more people had CF than was expected in 2022.

You can assume the relative frequency of CF in a population remains constant.

Show your working.

[2 marks]

Additional number of people

CF can be caused by a three-base deletion mutation in the CFTR gene. This deletion

creates a recessive allele called F508del.

DNA probes and gel electrophoresis were used to screen two parents and their two

children for the F508del allele. One of the children was diagnosed with CF.

Figure 5 shows the results.

Figure 5

09.2 What is a DNA probe?

[2 marks]

09.3 The identity of each family member can be determined from the relative position of the

bands shown in Figure 5.

Use information in Figure 5 to:

• complete Table 7

• explain how you identified the child with CF.

You can use F to represent the dominant CFTR allele and f to represent the recessive

F508del CFTR allele.

[3 marks]

Table 7

*28* Lanes Parent or child Genotype

Explanation

Mark scheme

Show the mark scheme Mark scheme for Question 9. For 09.1, 2 marks for answer in range 238 to 240, or 1 mark for intermediate working (expected 10909 to 10910, or ratio). For 09.2, 2 marks for stating a short single strand of DNA with bases complementary to the target DNA/allele. For 09.3, 1 mark per correct column in Table 7 (Lanes 1 and 3 are parents with genotype Ff; Lane 2 is child with genotype ff; Lane 4 is child with genotype FF), and 1 mark for explaining that Lane 2 contains only the recessive allele, which travels further due to having fewer bases from the deletion mutation.

Question Marking Guidance Mark Comments

Correct answer in range 238 to 240 = 2 marks;;

OR

In working shows:

16 (ignoring preceding zeros or subsequent numbers

and position of decimal point e.g. 0.016) = 1 mark

09.1 OR (2 x

AO2)

10909 or 10910 (ignore subsequent numbers and

position of decimal point) = 1 mark

OR

1.00656 (ignore subsequent numbers and allow

correct rounding) = 1 mark;

2. Accept

1. (Short) single strand of DNA; 2

complementary base

09.2 (2 x

sequence

2. Bases complementary (with DNA allele/gene); AO1)

1. and 2. 1 mark per correct column;; 1 and 2. Accept if

columns are reversed

Lane Parent or child Genotype

2. Accept

1 Parent Ff / fF

heterozygous,

2 Child ff homozygous recessive,

3 Parent Ff / fF heterozygous and

4 Child FF homozygous dominant

3. (Lane 2 only contains) recessive/f allele which

travels further due to fewer bases 2. Accept use of

3 different letters other

09.3 OR (3 x than F and f

AO3)

(Lane 2 only contains) recessive/f allele that 3. Accept ‘triplet’ for (3)

travels further due to deletion of (3) bases; bases

3. Accept F508del or

mutation for recessive

allele

3.In this instance

accept gene for allele

How to answer it

Cystic Fibrosis: Population Proportions, DNA Probes & Gel Electrophoresis

📋 What this question tests

This question assesses your ability to calculate projected population frequencies from demographic data, recall the precise definition and structural function of a DNA probe, and analyse gel electrophoresis banding patterns to deduce family relationships, genotypes, and the physical effect of a deletion mutation on fragment mobility.

Part 09.1 • 2 Marks

Population Frequency Calculation

Calculate how many more people had CF than was expected in 2022

📐 Step-by-Step Calculation

  1. Find the relative frequency in 2020:
    Relative frequency = 10 837 / (67.06 × 10⁶) = 1.6160155 × 10⁻⁴ (or ~0.0001616)
  2. Calculate expected cases in 2022:
    Expected = (1.6160155 × 10⁻⁴) × (67.51 × 10⁶) = 10 909.69 (accept 10 909 or 10 910)
  3. Find the difference from observed cases:
    Additional people = 11 148 − 10 909.69 = 238.31

✅ Correct Answer

Any integer in the range:

238 to 240

Mark Breakdown:
• 2 marks: Correct final answer in range 238 to 240.
• 1 mark (method): Showing intermediate relative frequency (e.g. 1.6 × 10⁻⁴), expected population calculation (10 909 or 10 910), or population scale factor (67.51 / 67.06 = 1.00656).

🧠 Exam Technique: Proportional Scaling

You can solve this in one simple ratio step:

Expected in 2022 = 10 837 × (67.51 / 67.06) = 10 909.7

Always subtract the expected figure from the actual observed figure (11 148). Never leave your answer as a formula—compute the final number.

❌ Common Errors

  • Simply subtracting raw cases: 11 148 − 10 837 = 311. This fails to account for population growth from 67.06 million to 67.51 million.
  • Rounding too early: Rounding 1.616 × 10⁻⁴ to 1.6 × 10⁻⁴ introduces significant rounding error (giving 10 801 expected, missing the mark range). Keep numbers in your calculator memory!
Part 09.2 • 2 Marks

Definition of a DNA Probe

What is a DNA probe?

✅ Marking Points

  1. (Short) single strand of DNA; [1 mark]
  2. Bases complementary (with DNA allele / target gene); [1 mark]
Marking note: Also accept "complementary base sequence".

💡 Key Knowledge

A DNA probe has two essential characteristics:

  • Structure: It is a short, single-stranded length of DNA nucleotides.
  • Specificity: It has a specific sequence of bases complementary to the target mutant/normal allele, allowing it to anneal via complementary base pairing.
  • Detection: In practice, it also carries a label (radioactive ³²P or fluorescent marker), though the two primary defining features in AQA mark schemes are single-stranded and complementary bases.

❌ Common Errors

  • Forgetting "single-stranded": Saying only "a piece of DNA" or "double-stranded DNA" scores 0 for MP1. A probe cannot hybridise if it is double-stranded.
  • Vague targeting: Stating it "finds" or "attaches to" the gene without using the keyword complementary fails to secure MP2.

🧠 Exam Technique

Whenever asked for the definition of an analytical biotechnology tool (DNA probe, primer, restriction enzyme), always specify: (1) what molecule it is physically made of and (2) how it binds/interacts via base pairing or specific sites.

Part 09.3 • 3 Marks

Analysing Gel Electrophoresis & Identifying Family Members

Complete Table 7 and explain how you identified the child with CF

✅ Table 7 Completed

Lanes Parent or child Genotype
1 Parent Ff (heterozygous)
2 Child ff (homozygous recessive)
3 Parent Ff (heterozygous)
4 Child FF (homozygous dominant)
• 1 mark: Entire "Parent or child" column correct.
• 1 mark: Entire "Genotype" column correct.

✅ Explanation Point

(Lane 2 only contains) recessive / f allele which travels further due to fewer bases / deletion of (3) bases; [1 mark]

Marking notes: Accept "triplet" for 3 bases. Accept "F508del" or "mutation" for recessive allele. Accept "gene" for allele in this context.

💡 Gel Electrophoresis Logic

  • Why do fragments separate? The electric current pulls negatively charged DNA towards the positive anode. Smaller fragments encounter less resistance through the gel matrix and migrate further / faster.
  • The mutation: F508del is a 3-base deletion. The recessive allele ( f ) is shorter than the normal allele ( F ).
  • Bands: Therefore, the lower band (travelled further) represents the mutant allele f , while the upper band represents the normal allele F .
  • Family genetics: A child with recessive CF must be ff (Lane 2). Both parents must be carriers ( Ff ) to produce an affected child, matching Lanes 1 & 3. The second child is unaffected and non-carrier ( FF ), matching Lane 4.

❌ Common Errors

  • Inverting fragment speed: Thinking larger fragments move faster or further. Remember: smaller = travels further down.
  • Swapping parents and child 4: Assuming lane 4 is a parent. Since both parents must pass an f allele to the affected child, both parents must be heterozygous ( Ff ).
  • Incomplete explanation: Stating only "Lane 2 has one band" without explaining why that band represents the mutant allele (it travelled further because of the 3-base deletion / fewer bases).

Topics

Biology · Practical skills · 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.