AQA GCSE Biology Biology Paper 2 (Higher), June 2024: Question 6

11 marks · High Demand difficulty · Short Answer

Explain the inheritance of polydactyly using a family pedigree and a Punnett square, and calculate the incidence of cystic fibrosis in a population.

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Question text

06 Some human disorders are inherited.

Polydactyly is an inherited disorder.

• A person with polydactyly has extra fingers or toes.

• Polydactyly is caused by a dominant allele.

06.1 What is a dominant allele?

[1 mark]

Figure 10 shows the inheritance of polydactyly in one family.

Figure 10

In questions 06.2 and 06.3, use the following symbols:

D = allele for having polydactyly

d = allele for not having polydactyly.

*0246.*2 Person 1 is heterozygous.

Explain how Figure 10 shows that person 1 is heterozygous.

[2 marks]

06.3 Persons 6 and 7 are expecting a fourth child.

A doctor states that the probability of having a child with polydactyly is 0.5

Explain how the doctor determined this probability.

[4 marks]

You should:

• draw a Punnett square diagram

• give the genotype of person 6 and the genotype of person 7

• identify all the offspring that will have polydactyly.

06.4 Cystic fibrosis (CF) is another inherited disorder caused by a mutation.

The mutation occurs in a gene called CFTR.

For the CFTR gene, one allele in every 50 in the UK population is the cystic fibrosis

allele.

Explain why only one person in 2500 in the UK population has cystic fibrosis.

[4 marks]

Mark scheme

Show the mark scheme

Question 6

AO /

Question Answers Extra information Mark

Spec Ref.

06.1 (an allele) that is always allow always shows in the 1 AO1

expressed phenotype 4.6.1.6

or ignore stronger

(an allele) that is expressed

even when the other / recessive

allele is present

(an allele) that is expressed in

the heterozygote

(an allele) that is expressed

when only one copy is present

AO /

Spec Ref.

06.2 allow an annotated genetic AO3

diagram for up to 2 marks 4.6.1.6

4.6.1.7

(person 1) has polydactyly so allow (person 1) has polydactyly 1

must have D so must have a dominant allele

has offspring / (person) 5 who 1

does not have polydactyly so

must have d from person 1

person 5 does not have

polydactyly so must be dd and – – –

must inherit d from person 1

AO /

Spec Ref.

06.3 Female / 6’s gametes correct 1 AO2

D + d 4.6.1.6

allow 1 mark for both sets of

4.6.1.7

gametes if parents not

Male / 7’s gametes correct 1

identified

d + d

correct derivation of offspring derivation must be consistent 1

genotypes: Dd Dd dd dd with parental gametes

Dd correctly identified as mp4 only awarded if mp3 is 1

polydactyly in only half of correct

offspring

20 AO /

Spec Ref.

06.4 CF allele is recessive allow CF is recessive 1 AO1

to have CF, must have 2 CF 1 AO2

alleles

chance of having one CF allele ignore chance of having one CF 1 AO2

1 allele is one in 50

(chance of having two CF alleles ignore 50 × 50 = 2500 1 AO2

is)

4.6.1.6

11 1

× = 4.6.1.7

50 50 2500

Total Question 6 11

How to answer it

Inherited Disorders: Polydactyly & Cystic Fibrosis

📌 What This Question Tests

This question evaluates your core understanding of genetics, monohybrid inheritance, and probability calculations:

Key terminology: Defining dominant vs recessive alleles accurately without vague language.
Pedigree tree deduction: Using offspring phenotypes to prove parental genotypes.
Punnett square crosses: Constructing genetic crosses, assigning gametes, and calculating outcome probabilities.
Mathematical genetics: Linking recessive inheritance to population allele frequencies and combined probability rules.

Question 06.1 • 1 Mark

Definition of a Dominant Allele

AO1 Recall

✅ Acceptable Answers (Any 1)

An allele that is always expressed (even if only one copy is present).
An allele that is expressed even when the other / recessive allele is present.
An allele that is expressed in a heterozygote.
An allele that always shows in the phenotype.

❌ Common Errors & Trap Words

Writing that a dominant allele is "stronger" or "overpowers" the other allele (the examiner will ignore or reject this).
Saying it is the allele that is "most common" in the population (dominance does not mean commonality).

Mark Scheme Note: 1 mark for stating that it is always expressed / expressed when only one copy is present. Ignore any reference to "stronger".

Question 06.2 • 2 Marks

Deducing Genotype from a Pedigree Chart

AO3 Analysis and Deduction

✅ Model Answer

Person 1 has polydactyly, so they must have at least one dominant allele ( D ). [1 mark]

Person 1 has an unaffected child (person 5) who does not have polydactyly ( dd ). Person 5 must have inherited one recessive allele ( d ) from person 1. [1 mark]

🧠 Exam Technique: Two-Step Deduction

Step 1 (Parent): State what their own phenotype tells you about their genotype (polydactyly means they must carry D ).
Step 2 (Child): Look for an unaffected offspring ( dd ). Because one allele comes from each parent, the parent must supply one d .

Mark Breakdown:
• Mark 1: (Person 1) has polydactyly so must have D (or has a dominant allele).
• Mark 2: Has unaffected offspring (person 5) who must have received a recessive allele ( d ) from person 1.

Question 06.3 • 4 Marks

Punnett Square and Probability Determination

AO2 Application

💡 Genotypes of Parents

Person 6 (Mother): Has polydactyly, but her mother (2) was unaffected ( dd ). Therefore, person 6 is heterozygous: Dd .
Person 7 (Father): Unaffected by polydactyly, so must be homozygous recessive: dd .

✅ Punnett Square Diagram

	d (Father)	d (Father)
D (Mother)	Dd (Polydactyly)	Dd (Polydactyly)
d (Mother)	dd (Unaffected)	dd (Unaffected)

Offspring genotypes: Dd, Dd, dd, dd

📐 Identifying Phenotypes & Probability

Offspring with polydactyly: Dd (2 out of 4)
Offspring without polydactyly: dd (2 out of 4)
Probability = 2 ÷ 4 = 0.5 (or 50%)

❌ Common Errors

Assuming Person 6 is DD because three previous children are unaffected (10, 11, 12). Previous children do not alter the probability of the next child!
Forgetting to explicitly state which offspring have polydactyly ( Dd ).

Mark Scheme Breakdown:
• Mark 1: Correct female gametes ( D and d ).
• Mark 2: Correct male gametes ( d and d ).
• Mark 3: Correct derivation of offspring genotypes ( Dd , Dd , dd , dd ).
• Mark 4: Correctly identifying that Dd gives polydactyly in only half (0.5) of the offspring.

Question 06.4 • 4 Marks

Explaining Population Frequency for Cystic Fibrosis

AO1 / AO2 Application & Maths Skills

📐 Step-by-Step Mathematical Explanation

Inheritance Type: Cystic fibrosis is caused by a recessive allele.
Genotype Required: A person must inherit two copies of the faulty allele (homozygous recessive) to have the disorder.
Single Allele Probability: The chance of inheriting one CF allele from any parent is 1 in 50 (or ¹/₅₀).
Combined Probability: Inheriting two alleles requires multiplying the independent probabilities:
¹/₅₀ × ¹/₅₀ = ¹/₂₅₀₀
Therefore, only 1 in 2500 people will have both alleles and develop cystic fibrosis.

❌ Common Errors & Mark Traps

Just writing 50 × 50 = 2500: The mark scheme says "ignore 50 × 50 = 2500". You must express it as a probability: ¹/₅₀ × ¹/₅₀ = ¹/₂₅₀₀.
Failing to state it is recessive: If you do not state that cystic fibrosis is recessive, you will miss the key biological reasoning for needing two alleles.
Confusing alleles with people: 1 in 50 alleles is not the same as 1 in 50 people having the condition.

Mark Scheme Breakdown (4 Marks Total):
• Mark 1: CF allele is recessive.
• Mark 2: To have CF, an individual must have 2 CF alleles.
• Mark 3: Chance of inheriting one CF allele is ¹/₅₀.
• Mark 4: Chance of inheriting two CF alleles is ¹/₅₀ × ¹/₅₀ = ¹/₂₅₀₀.

Topics

Biology · B6: Inheritance, Variation and Evolution

Question and mark scheme from the AQA GCSE Biology examination, Biology Paper 2 (Higher), June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.