A Level Biology: Inheritance — Practice Questions (Cambridge 9700)

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs — Cambridge International holds copyright in its own papers. Use these alongside the official past papers available from your board.

Each question practises a skill tested in the June 2024 Paper 42. After each answer there is an examiner insight, a mark-scheme insight or a tip and the real question to try next.

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Questions

1. In rabbits, coat colour is controlled by a gene, C, which has four alleles: C (full colour), cᶜʰ (chinchilla), cʰ (Himalayan) and c (albino).

(a) A student writes: “Coat colour in rabbits shows multiple alleles because rabbits come in lots of different colours.” Explain what is wrong with this statement, and give a correct explanation of why coat colour in rabbits involves multiple alleles.

(b) State the maximum number of these alleles that a single rabbit can carry, and give a reason. [2]

2. In a species of garden flower, petal colour is controlled by one gene, F, with three alleles: Fᴾ, Fᵂ and Fᴳ. The genotypes and phenotypes are: FᴾFᴾ and FᴾFᴳ give purple petals; FᵂFᵂ and FᵂFᴳ give white petals; FᴾFᵂ gives purple-and-white striped petals; FᴳFᴳ gives gold petals.

(a) Identify the two codominant alleles, and list all three alleles in order of dominance, starting with the most dominant.

(b) A plant with purple petals was crossed with a plant with white petals. Some of the offspring had gold petals. Set out this genetic cross in full, giving the parental genotypes, the gametes, the offspring genotypes and phenotypes, and the ratio of offspring phenotypes. [6]

3. In fruit flies, females are XX and males are XY. Grey body (E) is dominant to ebony (very dark) body (e). Two crosses were carried out using true-breeding flies. In cross A, ebony females were mated with grey males, and all 214 offspring (109 females and 105 males) had grey bodies. In cross B, grey females were mated with ebony males, and all 196 offspring (95 females and 101 males) had grey bodies.

(a) Assuming the gene is autosomal, give the genotypes of the female and male parents in cross A.

(b) Explain how the results of the two crosses show that the gene is not on the X chromosome.

(c) The offspring of cross A were allowed to interbreed. State the ratio of phenotypes expected among the male offspring and among the female offspring, if the gene is autosomal. [4]

4. Two bean plants, both heterozygous for seed coat colour, were crossed. Black seed coat (B) is dominant to white seed coat (b). Of 400 offspring, 290 had black seed coats and 110 had white seed coats. Use the chi-squared test to decide whether these results fit the expected 3 : 1 ratio. The critical value of χ² at p = 0.05 with 1 degree of freedom is 3.84. Show your working. [4]

5. Albinism in humans is caused by a recessive allele of the TYR gene. Use albinism to show how a change in one gene can alter a protein and so change a person’s appearance (phenotype). [4]

6. A gardener grows two rows of pea plants from the same packet of seeds of a dwarf variety, all with genotype lele. One row is sprayed with active gibberellin (GA₁) every week; the other row is not sprayed. After six weeks, the sprayed plants are about as tall as a tall variety, while the unsprayed plants are still short. Explain why the unsprayed plants are short and why spraying makes the plants grow tall. [4]

7. A mutant plant makes a DELLA protein with a changed amino acid sequence, so that the protein can no longer be broken down. The plant is short, and it stays short even when it is sprayed with gibberellin. Explain why. [4]

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Answers

1. (a) The student describes phenotypes (colours), not alleles. Coat colour involves multiple alleles because one gene (C) has more than two alleles — here, four alleles [1].

(b) Two alleles, because a rabbit is diploid: it has two copies of each chromosome, so two copies of the gene (one inherited from each parent) [1].

Examiner insight (Cambridge 9700 June 2024 examiner report, Paper 42, Question 3(a)): Answers needed to say how many alleles the gene has, or that one gene has more than two alleles. Simply repeating the phrase “multiple alleles” from the question, or listing phenotypes instead of alleles, did not explain anything.

Source for the examiner insights on this page: Cambridge International AS & A Level Biology 9700 June 2024 Principal Examiner Report for Teachers, Paper 9700/42 section, paraphrased.

Try the real question next: Cambridge International AS & A Level Biology 9700, June 2024, Paper 42, Question 3(a).

2. (a) Fᴾ and Fᵂ are codominant [1]. Order of dominance: Fᴾ = Fᵂ > Fᴳ [1].

(b) Parental phenotypes: purple × white. Parental genotypes: FᴾFᴳ × FᵂFᴳ (both parents must carry Fᴳ, because some offspring are FᴳFᴳ) [1].

Gametes: Fᴾ and Fᴳ from the purple parent; Fᵂ and Fᴳ from the white parent [1].

Offspring genotypes and phenotypes: FᴾFᵂ striped; FᴾFᴳ purple; FᵂFᴳ white; FᴳFᴳ gold, each genotype clearly linked to its phenotype [1].

Ratio: 1 striped : 1 purple : 1 white : 1 gold [1].

Examiner insight (Cambridge 9700 June 2024 examiner report, Paper 42, Question 3(b)): The main errors were not clearly linking each offspring genotype to its phenotype, listing phenotypes in a different order from the ratio, and writing the allele superscripts the same size as the gene letter.

Try the real question next: Cambridge International AS & A Level Biology 9700, June 2024, Paper 42, Question 3(b).

3. (a) Females ee and males EE [1].

(b) If the gene were on the X chromosome, cross A would be XᵉXᵉ × XᴱY, so every son would receive Xᵉ from his ebony mother and would have an ebony body [1]. Instead, all the males in cross A are grey, and both crosses give the same result in both sexes, so the phenotype does not depend on which parent was grey [1].

(c) 3 grey : 1 ebony in the males and 3 grey : 1 ebony in the females [1].

Examiner insight (Cambridge 9700 June 2024 examiner report, Paper 42, Question 3(d)(ii)): Only the strongest candidates gained credit, for example by describing similar numbers of males and females for each phenotype. Many wrote about the overall ratio of phenotypes, which on its own does not show whether a gene is on the X chromosome.

Try the real question next: Cambridge International AS & A Level Biology 9700, June 2024, Paper 42, Question 3(d)(ii).

4. Expected numbers: 300 black and 100 white [1].

χ² = Σ (O − E)² ÷ E = (290 − 300)² ÷ 300 + (110 − 100)² ÷ 100 = 0.333 + 1.000 = 1.33 [1].

Degrees of freedom = 2 − 1 = 1; 1.33 is less than the critical value of 3.84 [1].

So the difference between observed and expected results is not significant; it is due to chance, and the results fit the 3 : 1 ratio [1].

Tip: Always state the degrees of freedom and compare your χ² value with the critical value before writing a conclusion in words.

Try the real question next: Cambridge International AS & A Level Biology 9700, June 2024, Paper 42, Question 3(d)(iii).

5. Any four: a gene codes for a protein, and the protein determines the phenotype [1]; the TYR gene codes for the enzyme tyrosinase, which converts tyrosine into melanin (via DOPA) [1]; albinism is caused by a recessive allele of TYR, so a person with albinism is homozygous recessive [1]; this allele codes for a non-functional tyrosinase (or no tyrosinase), because a change in the base sequence changes the primary structure and shape of the enzyme [1]; so little or no melanin is made, giving very pale skin and hair and pale eyes that are sensitive to light [1].

Examiner insight (Cambridge 9700 June 2024 examiner report, Paper 42, Question 4(a)): It was common to mix up the name of the gene with the name of its protein. Others wrote as if the gene was missing. The gene is always present: which allele is present decides whether the protein works.

Try the real question next: Cambridge International AS & A Level Biology 9700, June 2024, Paper 42, Question 4(a).

6. Any four: lele is homozygous recessive [1]; the le allele codes for a non-functional (or much less active) enzyme, the one that makes active gibberellin (GA₁) from an inactive precursor [1]; so the plant has little or no active gibberellin [1]; so there is less cell elongation in the stem (shorter internodes), giving a short stem [1]; spraying on GA₁ bypasses the faulty enzyme, so the stem cells elongate normally [1].

Examiner insight (Cambridge 9700 June 2024 examiner report, Paper 42, Question 4(b)): Relatively few answers stated that the genotype is homozygous recessive, and some missed the step that the allele codes for a non-functional enzyme. Some wrote about germination instead of stem elongation.

Try the real question next: Cambridge International AS & A Level Biology 9700, June 2024, Paper 42, Question 4(b).

7. Any four: DELLA proteins bind to a transcription factor (PIF) [1]; this stops the transcription factor binding to the promoter of genes needed for stem growth [1], so RNA polymerase cannot bind and the genes are not transcribed (no mRNA) [1]; normally, gibberellin binds to a receptor, and this leads to the DELLA proteins being broken down, releasing the transcription factor [1]; in the mutant, DELLA is not broken down even when gibberellin binds to its receptor, so the growth genes stay switched off, there is little cell elongation and the plant stays short [1].

Examiner insight (Cambridge 9700 June 2024 examiner report, Paper 42, Question 4(c)): Some answers confused this with repressor proteins in prokaryotes binding to an operator. Others wrongly said DELLA binds directly to the promoter.

Try the real question next: Cambridge International AS & A Level Biology 9700, June 2024, Paper 42, Question 4(c).

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Where marks are usually lost

• Saying “multiple alleles” without stating how many alleles one gene has.
• Not linking each offspring genotype to its phenotype when setting out a cross, or writing alleles that cannot be told apart.
• Discussing the phenotype ratio when the question is about whether a gene is sex-linked: compare males with females.
• Writing that a gene is “missing” or “not coded for”, instead of saying a recessive allele codes for a non-functional protein.
• Describing DELLA as a prokaryotic repressor that binds to an operator.