Inheritance: AQA A Level Biology Exam Questions by Topic

AQAInheritanceExam Questions
Written by Tom Whitburn

Inheritance (AQA 3.7.1) opens topic 7 and it's A-level only, so it's Year 13 content. You need to define genotype and phenotype, use dominant, recessive, codominant, homozygous and heterozygous accurately, and draw fully labelled genetic diagrams for monohybrid and dihybrid crosses, including sex linkage, autosomal linkage, multiple alleles and epistasis. You also need the chi-squared test to compare observed and expected ratios.

The examiners' reports find the same slips every year. Students leave the environment out of phenotype, drop the sex from the phenotypes in a sex-linked cross, and keep offspring that die as embryos in the ratio. Linkage is the hardest area: only 1% got full marks explaining the fruit fly results, because answers didn't link the missing phenotypes to a lack of crossing over or to the gametes made. Only 44% named the chi-squared test and gave the right degrees of freedom.

Start with the definitions and pedigree questions, then work through the crosses and finish with probability and chi-squared. Draw each genetic diagram in full, with gametes, before you open the mark scheme. If you'd like help with this topic, my AQA masterclasses run online on Tuesdays at 5.15pm for Year 12 and on Wednesdays or Thursdays at 6.30pm for Year 13.

What you need from GCSE and Year 12

Each question on this page assumes you're secure on these GCSE and Year 12 ideas. If any of them feels shaky, revise it before you start.

  • GCSE: a Punnett square predicts the offspring of a monohybrid cross, and a dominant allele is expressed when only one copy is present.
  • GCSE: human females are XX and males are XY, so a father passes his X chromosome to every daughter and his Y chromosome to every son.
  • Year 12 (AQA 3.4.1): a gene occupies a fixed position, its locus, on a chromosome, and alleles are different forms of the same gene.
  • Year 12 (AQA 3.4.3): meiosis makes haploid gametes, and independent segregation and crossing over produce new combinations of alleles.

Key terms

These words come up in the questions and mark schemes. Examiners expect you to use them exactly.

  • Genotype: the genetic constitution of an organism: the alleles it has for a gene.
  • Phenotype: the expression of the genotype and its interaction with the environment.
  • Allele: a different form of a gene, found at the same locus on a chromosome.
  • Homozygous and heterozygous: two identical alleles at a locus, or two different alleles at a locus.
  • Dominant and recessive: a dominant allele is always expressed when present; a recessive allele is only expressed when no dominant allele is present.
  • Codominant: both alleles are expressed in the heterozygote, as with the IA and IB blood group alleles.
  • Multiple alleles: a gene with more than two alleles, such as IA, IB and IO for the ABO blood groups.
  • Sex-linked: a gene carried on the X chromosome, so males have only one allele of it.
  • Autosomal linkage: two genes on the same autosome, inherited together unless crossing over separates their alleles.
  • Epistasis: one gene affecting, or masking, the expression of another gene.
  • Chi-squared test: a statistical test of whether observed numbers differ significantly from expected numbers; degrees of freedom are the number of categories minus one.

What these questions cover (AQA 3.7.1)

The questions test these statements from the AQA AS and A-level Biology specification (7401/7402):

  • Genotype: The genotype is the genetic constitution of an organism. (no written question asks for the definition of genotype, but you write genotypes in the ABO, pedigree and genetic diagram questions)
  • Phenotype: The phenotype is the expression of this genetic constitution and its interaction with the environment.
  • Alleles: There may be many alleles of a single gene. Alleles may be dominant, recessive or codominant. (the ABO blood group question covers multiple alleles, and the flower colour and cat fur crosses cover codominance)
  • Homozygous and heterozygous: In a diploid organism, the alleles at a specific locus may be either homozygous or heterozygous.
  • The use of fully labelled genetic diagrams to interpret, or predict, the results of: monohybrid and dihybrid crosses involving dominant, recessive and codominant alleles
  • The use of fully labelled genetic diagrams to interpret, or predict, the results of: crosses involving sex-linkage, autosomal linkage, multiple alleles and epistasis.
  • The chi-squared test: Use of the chi-squared (χ²) test to compare the goodness of fit of observed phenotypic ratios with expected ratios. (the 2n gametes question asks for expected numbers under a null hypothesis, the first step of a chi-squared test)

How to use this page

Write your answer on paper before you open the mark scheme. Allow about 1.3 minutes per mark: the 18 questions are worth 40 marks, so about 52 minutes in total. Then mark your answer strictly against the mark scheme and read the examiners' tip underneath it. If you'd rather print the questions, download the PDF pack.

Genotype, phenotype and alleles (3.7.1)

Question 1

A level Paper 2 · June 2025 · Q05.1 · 2 marks · Easy

Explain what is meant by the term phenotype.

Question text: explain what is meant by the term phenotype, 2 marks.
Show the mark scheme Mark scheme: expression, appearance or characteristic due to the genetic constitution, genotype or alleles; and due to interaction with the environment. Accept a named characteristic; ignore chromosomes.

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From the examiners' report: About 95% got a mark but only 60% got both. Give the genetic part (expression of the genotype) and the environment; 'the appearance of a characteristic' alone scored nothing.

Question 2

A level Paper 2 · June 2024 · Q04.1 · 2 marks · Medium

Explain how to find out whether a grey fruit fly is homozygous or heterozygous.

Question text: in fruit flies the allele for grey body, G, is dominant to the allele for black body, g. Question asks how you would determine whether the genotype of a grey fly is homozygous or heterozygous for body colour, 2 marks.
Show the mark scheme Mark scheme: cross with a homozygous recessive, black or gg fly; any black offspring means it is heterozygous (Gg), all grey offspring means it is homozygous (GG). Alternatives: DNA base sequencing compared with known alleles, or separating alleles by electrophoresis or probes.

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From the examiners' report: 46% got both marks. Cross the grey fly with a black (gg) fly, not another grey one, and say what each result would tell you: any black offspring means Gg.

Question 3

A level Paper 2 · June 2022 · Q06.1 · 1 mark · Medium

Give the ABO genotypes of two people from a family tree.

Stem: ABO blood groups are controlled by three alleles, IA, IB and IO; IA and IB are codominant and IO is recessive to both; Rhesus positive (R) is dominant to Rhesus negative (r). Figure 3: a family tree in which 1 (blood group A, Rhesus positive male) and 2 (AB, Rhesus negative female) have son 3 (A); 3 and 4 (B, Rhesus positive female) have children 5 (O), 6 (A), 7 (O, Rhesus negative male) and 8 (B). Question asks for the ABO genotypes of 1 and 2, 1 mark.
Show the mark scheme Mark scheme: (1) IAIO and (2) IAIB, in either order; accept AO and AB.

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From the examiners' report: About 54% got this. Individual 1 must be IAIO, not IAIA: work back from the group O grandchildren, who need an IO allele from 3 and so from 1.

Question 4

A level Paper 2 · June 2022 · Q06.2 · 2 marks · Hard

Use a family tree to show that the Rhesus positive allele is dominant.

Figure 3: the ABO and Rhesus family tree; 3 (Rhesus positive male) and 4 (Rhesus positive female) have three Rhesus positive children and one Rhesus negative son, 7. Question asks you to explain one piece of evidence that the allele for Rhesus positive is dominant, 2 marks.
Show the mark scheme Mark scheme: Rhesus positive parents 3 and 4 produce 7, a Rhesus negative child; both 3 and 4 have the recessive allele or are heterozygous (or, if Rhesus positive were recessive, all their children would be Rhesus positive).

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From the examiners' report: 62% scored zero and about one in four got both. Use 3, 4 and 7, not 1 and 2, and say both parents must be heterozygous, not just one of them.

Monohybrid and dihybrid crosses (3.7.1)

Question 5

A level Paper 2 · November 2020 · Q06.3 · 3 marks · Easy

Complete a dihybrid cross with one dominant gene and one codominant gene.

Information from earlier in this question:

Earlier part of the same exam question, giving the information this part uses. Stem: T (tall) is dominant to t (dwarf); CR (red) and CW (white) are codominant, so CRCW plants are pink. Question asks you to complete the genetic diagram for a dwarf, pink-flowered plant crossed with a heterozygous tall, white-flowered plant, with genotypes and ratio of phenotypes, 3 marks.
Show the mark scheme Mark scheme: parents ttCRCW and TtCWCW; offspring TtCRCW, TtCWCW, ttCRCW and ttCWCW; tall pink, tall white, dwarf pink and dwarf white in a 1 : 1 : 1 : 1 ratio. Two marks are possible from a homozygous tall parent.

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From the examiners' report: About 59% got all three. The tall parent is heterozygous (Tt), as the question says, and don't write 9 : 3 : 3 : 1 out of habit when your own diagram gives 1 : 1 : 1 : 1.

Question 6

A level Paper 2 · June 2024 · Q04.4 · 3 marks · Medium

Complete a dihybrid cross in which one genotype dies as an embryo.

Question text: curly wings (R) are dominant to normal wings (r), on a different chromosome from body colour; RR flies die as embryos. A curly-winged fly homozygous for grey body is crossed with a curly-winged, black-bodied fly. Question asks you to complete the genetic diagram with the genotypes and ratio of phenotypes expected to develop into adults, 3 marks.
Show the mark scheme Mark scheme: parents RrGG and Rrgg; offspring RrGg (×2), rrGg (and RRGg); curly-winged grey and normal-winged grey in a ratio of 2 : 1.

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From the examiners' report: About 65% got two marks but only 25% got all three. Remove the RRGg embryos before you write the ratio: it's 2 : 1, not 3 : 1 or 1 : 2 : 1.

Sex linkage (3.7.1)

Question 7

A level Paper 2 · November 2021 · Q04.1 · 2 marks · Easy

Explain why male fruit flies are more likely than females to have white eyes.

Stem: male fruit flies are XY and females XX; a gene for eye colour on the X chromosome has a dominant red allele, R, and a recessive white allele, r. Question asks why male flies are more likely than females to have white eyes, 2 marks.
Show the mark scheme Mark scheme: males have one allele; females need two recessive alleles, must be homozygous recessive, or could be heterozygous (carriers). One mark only if 'gene' is used instead of allele.

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From the examiners' report: 52% got both marks. Talk about alleles, not just X and Y chromosomes: males need only one recessive allele, while females need two, or can be heterozygous carriers.

Question 8

A level Paper 2 · June 2019 · Q06.3 · 2 marks · Hard

Use a fruit fly pedigree to show the body colour gene is not on the X chromosome.

Information from earlier in this question:

Earlier part of the same exam question, giving the information this part uses. Stem: grey body (G) is dominant to black body (g); red eyes (R) are dominant to white eyes (r) and the eye colour gene is on the X chromosome. Figure 4: a four-generation pedigree of flies 1 to 13 with body colour written under each and shading for eye colour; grey male 3 and grey female 4 have a black female, 9; black female 2 and grey male 1 have a grey son, 5. Question asks you to explain one piece of evidence that the body colour gene is not on the X chromosome, 2 marks.
Show the mark scheme Mark scheme, in pairs: fly 3 (and 4) produce black female 9, but 3 would pass his dominant allele to 9; or black female 2 (and 1) produce grey male 5, but 5 would receive a recessive allele from 2.

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From the examiners' report: Only 5% got both marks. Pick a pair of flies that breaks the X-linked pattern, such as grey male 3 having black daughter 9, then explain it with alleles: 3 would have to pass his dominant allele to every daughter.

Question 9

A level Paper 2 · June 2019 · Q06.4 · 3 marks · Hard

Complete a dihybrid cross with one autosomal gene and one sex-linked gene.

Figure 4 and stem as before: grey body (G) dominant to black (g); red eyes (R) dominant to white (r) on the X chromosome. Question asks you to complete the genetic diagram for a heterozygous grey-bodied, white-eyed female crossed with a black-bodied, red-eyed male, with genotypes and ratio of phenotypes, 3 marks.
Show the mark scheme Mark scheme: parents GgXrXr and ggXRY; offspring GgXRXr, ggXRXr, GgXrY and ggXrY; grey red-eyed female, black red-eyed female, grey white-eyed male and black white-eyed male in a 1 : 1 : 1 : 1 ratio.

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From the examiners' report: About 28% got all three. Write the eye colour alleles on X chromosomes, give the male a Y, and include the sex in every offspring phenotype, since all daughters here are red-eyed and all sons white-eyed.

Epistasis (3.7.1)

Question 10

A level Paper 2 · June 2017 · Q07.2 · 1 mark · Easy

Name the type of gene interaction controlling summer squash fruit colour.

Figure 7: a pathway from white to green to yellow fruit colour. Enzyme 1, produced with genotype aa, converts white to green, and the A allele inhibits this step. Enzyme 2, produced with allele B, converts green to yellow; bb gives no functional enzyme. Question asks you to name the type of gene interaction shown, 1 mark.
Show the mark scheme Mark scheme: epistasis or epistatic (interaction); accept phonetic spellings.

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From the examiners' report: Almost 60% named epistasis. Dihybrid and codominance were the usual wrong answers: when one gene controls whether another gene's effect can show, it's epistasis.

Question 11

A level Paper 2 · June 2017 · Q07.4 · 3 marks · Easy

Complete an epistasis cross for summer squash fruit colour.

Figure 7 as before: A inhibits enzyme 1, so any A allele gives white fruit; aa with B gives yellow; aabb gives green. Question: genes A and B are not linked; complete the genetic diagram for aabb × AaBb to show all the possible genotypes and the ratio of phenotypes, 3 marks.
Show the mark scheme Mark scheme: AaBb, Aabb, aaBb, aabb; white, white, yellow, green; ratio 2 : 1 : 1. Zero marks if the genotypes are wrong.

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From the examiners' report: Over 55% got all three, but over 25% scored zero. Use Figure 7 for each genotype: both Aa offspring are white, so the ratio is 2 white : 1 yellow : 1 green, not 1 : 1 : 1 : 1 or 9 : 3 : 3 : 1.

Question 12

A level Paper 2 · June 2023 · Q04.3 · 3 marks · Medium

Complete a cross for cat fur colour with a sex-linked gene and an epistatic gene.

Information from earlier in this question:

Earlier part of the same exam question, giving the information this part uses. Stem: in cats a gene on the X chromosome controls fur colour; G (ginger) and B (black) are codominant, and heterozygous females are tortoiseshell. A second gene, not sex-linked, has a dominant allele F that stops pigment forming, giving white fur, and a recessive allele f with no effect. Question asks you to complete the genetic diagram for the cross XGXGFf × XBYff with offspring genotypes, phenotypes and ratio, 3 marks.
Show the mark scheme Mark scheme: gametes XGF, XGf, XBf and Yf; offspring XGXBFf, XGXBff, XGYFf and XGYff; white female, tortoiseshell female, white male and ginger male in a 1 : 1 : 1 : 1 ratio.

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From the examiners' report: 44% got all three and 28% got two, usually because they left out the sex or combined white males with white females to give 2 : 1 : 1. Keep males and females separate in the phenotypes and the ratio.

Autosomal linkage (3.7.1)

Question 13

A level Paper 2 · November 2021 · Q04.3 · 3 marks · Hard

Explain why a dihybrid test cross produced only two phenotypes.

Question text: in fruit flies, grey body (G) is dominant to black (g) and long wings (L) to short (l), and neither gene is on the sex chromosomes. Flies heterozygous for both genes were crossed with black, short-winged flies. Table 2: 223 grey, long-winged and 218 black, short-winged offspring. Question asks you to explain the results, 3 marks.
Show the mark scheme Mark scheme: the genes are linked (autosomal linkage, on the same chromosome); no crossing over, or the genes are close together; no Gl and gL gametes are made (only GL and gl). Ignore independent assortment.

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From the examiners' report: Only 1% got all three. Explain why you get 1 : 1, not 1 : 1 : 1 : 1: the genes are on the same chromosome, no crossing over happened, so only GL and gl gametes were made.

Question 14

A level Paper 2 · June 2025 · Q05.3 · 3 marks · Hard

Use autosomal linkage to explain the percentages of four phenotypes from a cross.

Question text: in sweet peas, purple flowers (F) are dominant to red (f) and long pollen (L) to round (l); a plant heterozygous for both genes was crossed with a red, round-pollen plant. Table 2: purple long 48.6%, red round 47.5%, purple round 2.1%, red long 1.8% of the offspring. Question asks you to use autosomal linkage to explain the results, including what it means and the relative proportions of gamete genotypes, 3 marks.
Show the mark scheme Mark scheme: genes on the same chromosome or autosome; crossing over or recombination; fewer Fl and fL gametes, or more FL and fl gametes. Accept alleles of different genes, not alleles unqualified.

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From the examiners' report: Only 10% got all three. Define linkage as different genes on the same autosome, name crossing over, and then write about the gametes, Fl and fL being rare, not just the offspring.

Probability and the chi-squared test (3.7.1)

Question 15

A level Paper 2 · June 2025 · Q04.3 · 1 mark · Hard

Calculate the chance that two carrier parents have a son with haemophilia C.

Question text: haemophilia C is an autosomal recessive condition; two parents are both heterozygous for the haemophilia C allele. Question asks you to calculate the probability of these parents producing a son who has haemophilia C, 1 mark.
Show the mark scheme Mark scheme: 0.125, 12.5% or 1/8; accept 1 in 8; reject 1 : 8 or 8 : 1, and 12.5 without %.

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From the examiners' report: Only about 30% got this. The usual answer was 0.25, the chance of an affected child; the question wants a son, so multiply by the 0.5 chance of a boy.

Question 16

A level Paper 2 · June 2025 · Q05.2 · 2 marks · Medium

Suggest two reasons why observed ratios differ from the expected ratios.

Question text: in genetic crosses, observed phenotypic ratios in the offspring are often not the same as the expected ratios. Question asks for two reasons why, without referring to sex linkage or autosomal linkage, 2 marks.
Show the mark scheme Mark scheme, any two: small sample size; crossing over; random fusion of gametes or random fertilisation; epistasis; lethal genotypes; epigenetics. Ignore mutation and independent segregation.

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From the examiners' report: Over a third got both and almost 80% got one. Mutation was the most common wrong answer: it's far too rare to change a ratio. Random fertilisation, crossing over and small samples scored.

Question 17

A level Paper 2 · June 2022 · Q06.5 · 2 marks · Easy

Name the test that compares observed and expected phenotype frequencies and give the degrees of freedom.

Information from earlier in this question:

Earlier part of the same exam question, giving the information this part uses. Stem: scientists found the frequencies of the ABO alleles and the four ABO phenotypes in a large population and used a statistical test to see if the phenotype frequencies differed significantly from those expected from the Hardy–Weinberg equation. Question asks you to name the test and state the number of degrees of freedom, 2 marks.
Show the mark scheme Mark scheme: chi-squared; 3.

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From the examiners' report: 88% named chi-squared but only 44% got both marks. Degrees of freedom are the number of categories minus one: four phenotypes give 3, not 2, 4 or 0.05.

Question 18

A level Paper 1 · June 2018 · Q06.2 · 2 marks · Medium

Calculate expected numbers of plants under a null hypothesis for a chi-squared test.

Question text: red clover plants that produce diploid (2n) gametes were selected over breeding cycles. Table 3: observed numbers of plants that did not and did produce 2n gametes were 50 and 4 in cycle 0, 14 and 42 in cycle 1, 2 and 44 in cycle 2, and 0 and 56 in cycle 3; expected numbers for cycle 0 are 50 and 4, with blanks for cycle 1. Null hypothesis: the proportion of plants that produce 2n gametes will not change from one breeding cycle to the next. Question asks you to complete the expected numbers for cycle 1, to the nearest whole number, 2 marks.
Show the mark scheme Mark scheme: 52 and 4; one mark for numbers totalling 56 (except the observed 14 and 42), or for evidence of 56 if the table is blank.

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From the examiners' report: Only 22.3% got both marks. Keep the cycle 0 proportions (50/54 and 4/54) and apply them to the new total of 56 plants in cycle 1.

Where students lose marks (from AQA examiners' reports and mark schemes)

Each point below is followed by wording that gained marks.

  • Phenotype needs the genotype and the environment: About 95% got a mark for phenotype but only 60% got both. Answers that just said 'appearance of a characteristic' scored nothing, and many left out the environment.

    ✔ Gains credit: "Expression or characteristic due to the genotype or alleles; and due to interaction with the environment."

  • Cross with the recessive and read the offspring: 46% got both marks for testing a grey fly. Crossing two grey flies didn't score, and many said the offspring would show the genotype without saying how.

    ✔ Gains credit: "Cross with a black (gg) fly; any black offspring means the grey fly is heterozygous, all grey offspring means it's homozygous."

  • Both parents carry the recessive allele: 62% failed to score on the Rhesus pedigree. Some used individuals 1, 2 and 3, and many who chose 3, 4 and 7 said only one of the parents was heterozygous.

    ✔ Gains credit: "3 and 4 (Rhesus positive) produce 7 (Rhesus negative); both 3 and 4 have the recessive allele or are heterozygous."

  • Males have one allele; females need two: 52% got both marks on white-eyed flies. Answers that talked only about X and Y chromosomes, never mentioning alleles, scored nothing, and some wrote 'an X allele' or put the allele on the Y chromosome.

    ✔ Gains credit: "Males have one allele; females need two recessive alleles or could be heterozygous (carriers)."

  • Every sex-linked phenotype needs a sex: On the cat cross 28% got two marks because they left the sex out of the phenotypes or lumped white males with white females, giving 2 : 1 : 1. The fruit fly cross lost marks the same way.

    ✔ Gains credit: "White female, tortoiseshell female, white male, ginger male, and ratio 1 : 1 : 1 : 1."

  • Offspring that die as embryos leave the ratio: Around 65% had the right genotypes for the curly-winged cross, but only 25% got all three because many kept the homozygous curly flies and gave 3 : 1 or 1 : 2 : 1.

    ✔ Gains credit: "RrGg (×2), rrGg (and RRGg); curly-winged grey and normal-winged grey in a ratio of 2 : 1."

  • Linkage means genes on one chromosome, then explain the gametes: Only 1% got all three on the 1 : 1 fruit fly results and 10% on the sweet peas. Many wrote 'alleles are linked' with no genes, and talked about offspring instead of the gametes made with or without crossing over.

    ✔ Gains credit: "Genes are on the same chromosome (autosome); crossing over; fewer Fl and fL gametes, or more FL and fl gametes."

  • Degrees of freedom are categories minus one: 88% named the chi-squared test but only 44% also gave the degrees of freedom. Common wrong answers were 2, 4, 5 and even 0.05.

    ✔ Gains credit: "Chi-squared; 3" (four phenotypes, so 4 − 1).

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Download the PDF pack (18 questions, mark schemes at the end)

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Questions, mark schemes and examiners' report extracts © AQA.

Tom Whitburn
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