AQA A-Level Biology Paper 3, June 2025: Question 5

6 marks · Medium difficulty · Short Answer

Suggest why the Hardy–Weinberg equation cannot be used for CLP allele frequencies, describe the effect of DNA methylation on CDH1, and evaluate evidence for the 'two-hit' model.

Practise this question

Question

Question 05 focuses on cleft lip and palate (CLP) and the CDH1 gene under a 'two-hit' model involving genetic mutation and inflammation-induced methylation. A bar graph (Figure 4) displays 'Percentage CDH1 methylation' on the y-axis (0 to 80%) for two cell types: normal human cells and human cells with loss of CDH1 function. Each cell type has an untreated bar and a bar for cells treated with a drug causing inflammation. Normal untreated is ~36%, normal treated is ~58%, mutated untreated is ~40%, and mutated treated is ~72%. Sub-question 5.1 asks for two reasons the Hardy–Weinberg equation cannot estimate allele frequencies despite a live birth frequency of 1 in 700. Sub-question 5.2 asks for one effect of methylation on CDH1. Sub-question 5.3 asks to suggest and explain three reasons why the data in Figure 4 might not support the two-hit model.
Question text

05 Cleft lip and palate (CLP) is a split in the upper lip and/or the roof of the mouth.

CLP is present from birth.

CLP can be caused by various genetic and environmental factors.

CDH1 is a gene involved in lip and mouth development in a fetus before a baby

is born.

Scientists hypothesised that CLP is caused by a ‘two-hit’ model:

• a ‘genetic hit’ through loss of CDH1 function

• an ‘environmental hit’ leading to CDH1 methylation caused when inflammation

affects a fetus.

Inflammation can result from the mother having an infection or diabetes.

Scientists investigated the ‘two-hit’ model using human cells grown in vitro.

They:

• grew normal cells and cells with a loss of CDH1 function

• treated some of each type of cell with a drug to cause inflammation and left some

of each type of cell untreated

• measured the percentage CDH1 methylation of the cells.

Figure 4 shows the scientists’ results.

Figure 4

05.1 The frequency of live births with CLP is 1 in 700

*16Despite knowing this frequency, the Hardy–Weinberg equation* cannot be used to

estimate any allele frequencies for CLP.

Suggest two reasons why.

[2 marks]

05.2 Describe one effect methylation could have on CDH1

[1 mark]

05.3 Suggest and explain three reasons why the data in Figure 4 might not provide

evidence for the ‘two-hit’ model.

[3 marks]

Mark scheme

Show the mark scheme Mark scheme for Question 05. For 05.1 (2 marks max): not caused by one gene, environmental/epigenetic factors involved, mutations occur, or human populations migrate/are not isolated. For 05.2 (1 mark max): reduce/prevent transcription or expression of CDH1, prevent transcription factor or RNA polymerase binding to promoter, or reduce/prevent CDH1 protein production. For 05.3 (3 marks max): inflammation alone increases methylation in both cell types; loss of CDH1 function may not affect methylation since untreated levels are similar (36% vs 40%); drug-induced inflammation may differ from infection/diabetes; cells grown in vitro may not represent living organisms or whole fetuses; cells not allowed to develop into mouth/lip to verify CLP outcome.

Question Marking Guidance Mark Comments

1. Not caused by one gene; Ignore any references

to random mating

2. (Hardy-Weinberg) cannot predict (the effect of) 1. Ignore ideas

environmental factors relating to not

knowing about

OR dominant/

recessive

Environmental factors involved; 2. Accept

‘methylation/

3. (Various genetic factors suggests) mutations

2 max epigenetic factors’

occur;

05.1 (2 x for ‘environmental

AO2) factors’

4. Human populations migrate/emigrate/

immigrate; 4. Accept

descriptions of

human migration

4. Accept human

populations are

not isolated

4. Ignore population

size changes due

to births/deaths

Reject (methylation)

1. Reduce/prevent expression/transcription (of

causes mutation

CDH1/the gene);

1. Accept

2. Prevent transcription factor binding

reduce/prevent

production of

OR

messenger RNA/

1 max

mRNA

05.2 Prevent binding to promotor (of gene); (1 x

1. Ignore ‘switching

AO1)

off’

3. Reduce/prevent production of CDH1/protein;

2. Accept prevent

binding/activation

of RNA

polymerase

1. Inflammation (alone) may be the cause as

both (types of) treated cells have increased

methylation;

2. Accept loss of

2. Loss of CDH1/gene function might not have

CDH1/gene function

an effect as (both) untreated cells have

might not have an 15

similar (percentage) methylation;

effect as 36% and

40% are similar

3. Drug used to cause inflammation, so might

not be the same as infection/diabetes;

3. Accept ‘normal

environmental

4. Cells grown in vitro, so does not represent a

conditions’ for

living organism/human/body/fetus

‘infection/diabetes’

3 max 3. Accept ‘artificial

05.3 OR (3 x

conditions’ for ‘drug

AO3)

Cells grown in vitro, so might not respond in

4. Accept ‘in vivo’ OR ‘in

the same way as a living

utero’ OR ‘pregnancy’

organism/human/body/fetus;

for ‘living organism’

4. Accept only cells

5. Cells not (allowed to) develop into

grown, not a whole

lip/mouth/baby, so do not know any would

organism/human/body/

develop CLP

fetus, so might not

respond in the same

OR

way

Cells not (allowed to) develop into

lip/mouth/baby, so do not know what

percentage (CDH1) methylation leads to

CLP;

How to answer it

Cleft Lip & Palate: Genetics, Epigenetics & Data Evaluation

📋 What this question tests

This question assesses your ability to apply core ecological and molecular genetics principles to medical research:

  • Assumptions & limitations of the Hardy–Weinberg principle: Knowing when population genetics models break down due to multifactorial inheritance, migration, or environmental interactions.
  • Control of gene expression (Epigenetics): Understanding how DNA methylation inhibits transcription at the molecular level.
  • Critical evaluation of experimental data (AO3): Interpreting bar charts, distinguishing between correlation and causation, and evaluating the ecological validity of in vitro cell cultures versus whole organisms.

Question 05.1

Hardy–Weinberg Assumptions & Inheritance of CLP (2 marks)

✅ Acceptable Answers (Any 2 of the following)

  • Not caused by a single gene: CLP is polygenic / caused by multiple genes.
  • Environmental factors are involved: Hardy–Weinberg assumes phenotype is purely genetic and cannot account for environmental/epigenetic influences (such as maternal infection or methylation).
  • Mutations occur: The condition arises from various genetic factors/new mutations, violating the assumption of no mutation.
  • Human populations migrate: Gene flow occurs through immigration/emigration (populations are not geographically or genetically isolated).

💡 Key Knowledge: Hardy–Weinberg Conditions

The Hardy–Weinberg equation ( p² + 2pq + q² = 1 ) strictly assumes:

  • Organisms are diploid and trait is controlled by a single gene with two alleles.
  • No migration (no gene flow).
  • No mutation.
  • Random mating.
  • No natural selection.
  • Large population size.

The stem explicitly states CLP is caused by "various genetic and environmental factors", immediately invalidating conditions 1, 2, and 3.

🧠 Exam Technique: Use the Question Stem

Always inspect the introductory text when a question asks "Suggest reasons why...":

  • The text says "CLP can be caused by various genetic and environmental factors". That directly provides two marking points!
  • Do not just write generic Hardy-Weinberg assumptions without linking them to the context of CLP.

❌ Common Errors & Mark Losses

  • Mentioning random mating: Explicitly ignored by the mark scheme because humans do mate largely at random with respect to cleft palate alleles.
  • Mentioning dominant/recessive unknowns: Ignored. Even if allele dominance is unknown, you cannot use the equation if multiple genes or environmental causes exist.
  • Population size changes from births/deaths: Ignored unless framed strictly as genetic drift in very small populations.

Question 05.2

Molecular Effect of Methylation on CDH1 (1 mark)

✅ Acceptable Answers (Any 1)

  • Reduces / prevents expression or transcription of CDH1 (or prevents production of mRNA).
  • Prevents binding of transcription factors (or prevents binding to the promoter region).
  • Prevents binding / activation of RNA polymerase.
  • Reduces / prevents production of the CDH1 protein.

💡 Key Knowledge: Mechanism of Methylation

Increased methylation of DNA involves adding methyl groups ( –CH₃ ) to cytosine bases in CpG islands located near the promoter region.

  • This causes chromatin to condense (more closed/heterochromatin state).
  • Prevents transcription factors and RNA polymerase from binding to the promoter.
  • As a result, transcription is inhibited and the protein is not synthesised.

❌ Common Misconceptions

  • "Methylation causes a mutation": REJECTED. Epigenetic changes alter gene expression without altering the underlying base sequence of DNA.
  • Vague wording like "switches off the gene": IGNORED. A-Level requires precise biological terminology (e.g., prevents transcription, stops transcription factor binding).

🧠 Exam Technique

When asked for the effect of methylation or acetylation, specify the exact biological stage affected: transcription factor binding, transcription, or translation/protein yield.

Question 05.3

Critiquing the 'Two-Hit' Model Data (3 marks)

⚠️ Requirement: "Suggest AND explain" — each point must have a reason linked to data or design

✅ Mark Scheme Pairs (Need 3 distinct suggestions + explanations)

  • Point 1: Inflammation alone may cause methylation AS both normal and mutant cells show increased methylation when treated (~58% and ~72%).
  • Point 2: Loss of CDH1 function might have no effect AS both untreated cell types have very similar baseline methylation (36% vs 40%).
  • Point 3: An artificial drug was used to induce inflammation, SO it might not mimic natural conditions like maternal infection or diabetes.
  • Point 4: Cells were grown in vitro (cell culture), SO they do not represent whole living organisms/fetuses and may not respond the same way in vivo.
  • Point 5: Cells were not allowed to differentiate/develop into a lip or mouth, SO we do not know if this level of methylation actually leads to cleft lip and palate.

🧠 Top-Level Response Strategy

For high-scoring evaluation answers on 3-mark questions:

  1. Analyse the graph directly: Look at differences between bars. Untreated normal (36%) vs untreated mutant (40%) is barely different; both treated groups go up dramatically. Use the connective "AS" or "BECAUSE".
  2. Critique the experimental method: Look at how the experiment was run:
    • In vitro (isolated cells) vs in vivo (developing fetus).
    • Chemical drug used vs real maternal infection/diabetes.
  3. Check the dependent variable: They measured % methylation, NOT whether cleft palate actually formed!

📐 Data Reading from Figure 4

  • Normal cells untreated: 36% methylation
  • Normal cells + inflammation drug: 58% methylation (increase of +22%)
  • Loss of function cells untreated: 40% methylation
  • Loss of function cells + inflammation drug: 72% methylation (increase of +32%)

Notice that the "first hit" (loss of function) only changes baseline methylation by 4%, which is unlikely to be significant without error bars or statistical testing.

❌ Common Errors on AO3 Evaluation

  • Stating an observation without explaining: Writing just "cells were grown in vitro" gets 0 marks. You must add the consequence: "...so does not represent a whole fetus / may behave differently in vivo".
  • Assuming methylation equals disease: Forgetting that high methylation in a petri dish does not prove anatomical clefting of the fetal palate.
  • Ignoring the 'two-hit' hypothesis definition: The hypothesis requires both genetic loss of function AND environmental hit to cause the outcome. Untreated mutant cells had virtually no extra methylation, undermining hit #1 on its own.
Examiner Insight: Full mark scripts almost always combined one critique of the graph trends (e.g. baseline similarity between normal and mutant cells) with two critiques of experimental validity (e.g. in vitro limitations and artificial chemical induction vs actual physiological infection).

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 3, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.