AQA GCSE Combined Science: Trilogy Biology Paper 2 (Higher), November 2021: Question 6

11 marks · Standard Demand difficulty · Extended Answer

Define 'heterozygous', determine the probability of having an affected male child using a pedigree tree and Punnett square, and explain the sudden occurrence of polydactyly via mutation.

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Question

Question 06 contains three parts: 06.1 asks to define the term 'heterozygous' (1 mark). 06.2 shows Figure 6, a family pedigree chart tracking a genetic disorder across three generations. Parents 1 and 2 (both unaffected) have children 5 (unaffected female), 6 (affected female), and 7 (affected male). Male 7 and unaffected female 8 (daughter of unaffected parents 3 and 4) have children 10 (affected male) and 11 (unaffected female). The student is instructed to draw a Punnett square to find the probability of person 7 and 8 having another child that is male and has the disorder, using symbols H for dominant allele and h for recessive allele (6 marks). 06.3 asks to explain how polydactyly suddenly occurred in a child whose parents do not have any alleles for polydactyly in their ordinary body cells (4 marks).
Question text

06 This question is about genetic disorders.

06.1 Some people are heterozygous for a genetic disorder.

Define the term ‘heterozygous’.

[1 mark]

06.2 Figure 6 shows the inheritance of a genetic disorder in a family.

Figure 6

Person 7 and person 8 plan to have another child.

Determine the probability that the child will be a male who has the disorder.

You should:

• draw a Punnett square diagram

• identify the genotype of person 7 and the genotype of person 8

• identify the phenotype of each offspring genotype

• use the symbols:

H = dominant allele

*22* h = recessive allele

[6 marks]

Probability of having a male child with the disorder =24

06.3 Polydactyly is a different inherited disorder.

Two parents do not have any alleles for polydactyly in their ordinary body cells.

These parents produced a child with polydactyly.

Explain how polydactyly suddenly occurred in this family.

[4 marks]

Mark scheme

Show the mark scheme Mark scheme for Question 06: 06.1 awards 1 mark for having two different alleles for a gene/trait. 06.2 awards 6 marks: 1 mark for father genotype hh, 1 mark for mother genotype Hh, 1 mark for offspring genotypes hh (x2) and Hh (x2), 1 mark for phenotypes (hh has disorder, Hh does not), 1 mark for probability of having disorder (0.5), and 1 mark for probability of male with disorder (0.25). 06.3 awards 4 marks for: caused by mutation, during meiosis (or in gamete formation), causing a change in amino acid sequence, and causing a different protein (or no protein) to be produced.

AO /

Question Answers Extra information Mark

Spec. Ref.

06.1 any one from: 1 AO1

• (having two) different alleles ignore examples such as Hh 4.6.1.4

for a gene / trait / ignore having two different

characteristic / disorder alleles unqualified

• (having) the dominant and

recessive allele for a gene /

trait / characteristic / disorder

06.2

allow hh and Hh parental

father / person 7 hh genotypes with each parent 1 AO2

unidentified or reversed for

mother / person 8 Hh 1 mark 1 AO2

(possible offspring correctly allow correctly derived offspring 1 AO2

derived) from incorrect parental

hh (× 2) genotype(s)

Hh (× 2)

(each different phenotype 1 AO2

identified)

hh = has the disorder allow from incorrectly derived

Hh = does not have the disorder offspring

if incorrectly have HH = does not

have the disorder

0.5 allow 50% or ½ or 1:1 or 1 AO3

1 out of 2 or 1 in 2

do not accept 1:2

allow probability of having

disorder correctly derived from

incorrect parental genotypes

(probability of male with allow 25% or ¼ or 1:3 or 1 AO3

disorder) 1 out of 4 or 1 in 4

0.25 do not accept 1:4

4.6.1.4

allow probability of male with 4.6.1.5

disorder correctly derived from 4.6.1.6

incorrect probability of having

the disorder

06.3 caused by mutation allow description, for example 1 AO2

change in the genetic code or 4.6.2.1

change in base sequence 4.6.1.4

4.6.1.5

4.6.1.3

during meiosis allow in (germ) cells prior to 1 4.6.1.1

meiosis 4.6.1.2

allow in (the formation of)

gametes / egg / sperm

allow during mitosis between

fertilisation and birth

causing a change in amino acid 1

sequence

causing a different (specific) causing a different (specific) 1

protein to be produced enzyme to be produced

or or

causing none of a (specific) causing none of a (specific)

protein to be produced enzyme to be produced

allow polydactyly is caused by a

dominant allele so if child has

one / the allele (with the

mutation) they will have the

disorder

if no other mark awarded allow

parents used donated egg /

sperm for 1 mark

Total 11

How to answer it

Inheritance, Pedigree Trees & Genetic Mutations

WHAT THIS QUESTION TESTS

Specification Topics: AQA GCSE Biology / Combined Science Trilogy (Topic 4.6 - Inheritance, Variation and Evolution)

  • Genetic Vocabulary: Defining heterozygous accurately using terms like "alleles" and "gene".
  • Pedigree Tree Analysis: Deducing genotypes of parents from phenotypes of family members across generations.
  • Monohybrid Crosses: Constructing a Punnett square, linking genotypes to phenotypes, and calculating combined independent probabilities.
  • Mutations & Protein Synthesis: Explaining how changes in DNA base sequences lead to altered proteins and new phenotypes.
QUESTION 06.1 • 1 MARK

Defining 'Heterozygous'

Part (a): Recall of fundamental genetic terminology

✅ Model Answer

Having two different alleles for a gene (or characteristic).

OR: Having one dominant allele and one recessive allele for a gene.

💡 Key Knowledge

  • Alleles are different versions or forms of the same gene.
  • Heterozygous: different alleles (e.g. Hh ).
  • Homozygous: two identical alleles (e.g. HH or hh ).

❌ Common Errors

  • Writing just "different genes" instead of "different alleles" (genes and alleles are NOT the same).
  • Writing only an example like Hh without an explanation — the mark scheme explicitly says: ignore examples unqualified.
  • Vague statements like "having two different types of DNA".

🧠 Exam Technique

Always state the word allele and mention for a gene / trait. For a 1-mark definition question, be precise and concise.

Mark allocation: 1 mark for stating "different alleles for a gene/trait" or "dominant and recessive allele".
QUESTION 06.2 • 6 MARKS

Pedigree Analysis, Punnett Square & Probability

Part (b): Determining parental genotypes, offspring phenotypes, and combined probability

💡 Step 1: Deducing the Pattern of Inheritance

  • Is the disorder dominant or recessive? Parents 1 and 2 do not have the disorder, but have children who do (6 and 7). This proves the disorder is caused by a recessive allele (h) because parents 1 and 2 must be unaffected carriers ( Hh ).
  • Parent 7 (Male): He has the disorder, so his genotype must be homozygous recessive: hh .
  • Parent 8 (Female): She does not have the disorder, so she must carry at least one dominant allele ( H_ ). However, look at their child (person 10): he has the disorder ( hh ). To produce an affected child, person 8 must pass on an h allele. Therefore, person 8 is heterozygous: Hh .

✅ Model Answer & Punnett Square

Parental Genotypes:

  • Father (Person 7): hh
  • Mother (Person 8): Hh
Gametes h h
H Hh Hh
h hh hh

Offspring Phenotypes:

  • hh = has the disorder
  • Hh = does not have the disorder (unaffected)

📐 Step-by-Step Probability Calculation

  1. Probability of having the disorder:
    Out of 4 outcomes, 2 are hh .
    Probability = 2 ÷ 4 = 0.5 (or 50% or 1/2)
  2. Probability of the child being male:
    Biological sex is determined independently (XX vs XY).
    Probability = 0.5 (or 50% or 1/2)
  3. Combined probability (male AND disorder):
    Probability = 0.5 × 0.5 = 0.25
    (Acceptable forms: 0.25, 25%, 1/4, 1 in 4, or 1:3 ratio)
Final answer on line: 0.25 (or 1/4 or 25%)

🧠 Exam Technique: Securing all 6 Marks

  • Mark 1: Identify Person 7 as hh
  • Mark 2: Identify Person 8 as Hh
  • Mark 3: Correct offspring genotypes inside the grid ( Hh , Hh , hh , hh )
  • Mark 4: Match both genotypes to phenotypes ( hh has disorder, Hh does not)
  • Mark 5: Probability of having disorder = 0.5
  • Mark 6: Multiplied by 0.5 for male = 0.25

❌ Common Errors

  • Forgetting the sex condition: Many students stop at 0.5 (the chance of inheriting the disorder) and forget the question asked for a male child who has the disorder.
  • Incorrect ratio format: If expressing 1 in 4 as a ratio, it is 1:3 (1 affected : 3 unaffected). Writing 1:4 as a ratio gets zero marks for probability. Decimals ( 0.25 ) or percentages ( 25% ) are much safer!
  • Not labelling which parent is which.
QUESTION 06.3 • 4 MARKS

Spontaneous Mutations & Polydactyly

Part (c): Linking mutations to changes in protein structure and phenotype

✅ Model Answer (4 Marking Points)

  1. The condition was caused by a mutation (a random change in the DNA base sequence / genetic code).
  2. The mutation occurred during meiosis (during the formation of gametes / egg or sperm cell).
  3. This caused a change in the amino acid sequence of the coded protein.
  4. This resulted in a different (or non-functional) protein / enzyme being produced (which caused extra digits to develop).

💡 The Biological Chain of Cause and Effect

  • DNA base sequence changes (mutation).
  • Occurs during gamete formation (meiosis) so it can be passed to offspring even if parents' body cells lack it.
  • Changes mRNA codon sequence, so a different sequence of amino acids is assembled at ribosomes.
  • The protein folds differently, producing an altered or novel protein/enzyme that changes phenotype.
  • Note: Polydactyly is caused by a dominant allele, so only a single mutated allele is needed for the child to display the disorder.

🧠 Exam Technique: "Explain how..."

When an exam question asks how an inherited condition suddenly appears in a child when neither parent has it in their body cells, follow the genetic hierarchy:

Mutation in gametes → Meiosis → Amino acid order → Altered protein

Each step in this four-part sequence corresponds directly to 1 mark on the AQA mark scheme.

❌ Common Errors

  • Saying the parents were "carriers": Polydactyly is dominant. If parents had the allele, they would already have extra digits! The question explicitly stated neither parent had the allele in their body cells.
  • Confusing mitosis and meiosis: The mutation must happen during the production of sex cells (meiosis) or gametes to be passed on from healthy parents.
  • Vagueness: Saying "it changed the gene" without naming mutation, or saying "it changed the cell" without mentioning amino acids and protein.

Topics

Biology · B6: Inheritance, Variation and Evolution

Question and mark scheme from the AQA GCSE Combined Science: Trilogy examination, Biology Paper 2 (Higher), November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.