Cambridge International AS & A Level Biology 9700: Inheritance (by subtopic) – Practice Questions

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 – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

This set works through topic 16, Inheritance, of the Cambridge International AS & A Level Biology 9700 syllabus for examination in 2025, 2026 and 2027, one subtopic at a time: 16.1 Passage of information from parents to offspring (questions 1–2), 16.2 The roles of genes in determining the phenotype (questions 3–9) and 16.3 Gene control (question 10). It is A Level content, examined on Paper 4; chi-squared can also appear on Paper 5.

These differ from the topic-level inheritance practice questions. For explanations, use the inheritance study guide and revision notes.

Critical value of χ² at p = 0.05 with 3 degrees of freedom: 7.82.

Questions

1. A species of insect has a diploid number of 8.

(a) State the number of chromosomes in a gamete, and explain why gametes are made by a reduction division. [2] (b) Explain what is meant by a homologous pair of chromosomes. [2] (c) Calculate the number of chromosome combinations possible in one insect’s gametes by independent assortment alone. [1] (d) Calculate the number of chromosome combinations possible in a zygote, and name the process this depends on. [2] (e) Explain why the real number of genetically different gametes is far larger than your answer to (c). [2]

2. Four animal cells in meiosis, seen in photomicrographs, are described.

• P: paired homologous chromosomes lined up across the equator.
• Q: two cells, each with single chromatids moving to opposite poles.
• R: one cell, with a group of chromosomes of two chromatids at each pole, nuclear envelopes forming and the cell surface membrane pinching in.
• S: two cells, each with single chromosomes of two chromatids on the equator.

(a) Name the stage shown by each cell. [4] (b) Describe what happens to the nuclear envelope and spindle between stages R and S. [2] (c) State how stage R would differ in a plant cell. [1]

3. In peas, the allele Le (tall) is dominant to le (dwarf).

(a) Explain the terms locus and test cross. [2] (b) A tall pea plant of unknown genotype is test crossed. Use genetic diagrams to show the two possible outcomes and what each shows. [3] (c) Explain why a Lele plant is as tall as a LeLe plant. [2]

4. The HBB gene has two codominant alleles, Hbᴬ and Hbˢ.

(a) Explain how the Hbˢ allele leads to sickle cell anaemia. [4] (b) The ABO gene has alleles Iᴬ and Iᴮ (codominant) and Iᴼ (recessive). A group AB man and a group O woman plan children. Draw a genetic diagram and calculate the probability that their first two children are both group A. [3]

5. Haemophilia A is caused by a recessive allele of the F8 gene, on the X chromosome. A woman without haemophilia, whose father had haemophilia, has children with a man who has haemophilia.

(a) Explain why the woman must be a carrier. [1] (b) Draw a genetic diagram and state the probability that a daughter of this couple has haemophilia. [3] (c) Explain how the recessive F8 allele causes haemophilia. [2]

6. Explain how each allele affects the phenotype.

(a) The recessive TYR allele, which causes albinism. [2] (b) The dominant HTT allele, which causes Huntington’s disease; explain why one copy is enough. [4]

7. In fruit flies, normal wings (N) are dominant to vestigial wings (n), on an autosome. Red eyes (Xᴿ) are dominant to white eyes (Xʳ). A true-breeding red-eyed, normal-winged female was crossed with a white-eyed, vestigial-winged male.

(a) Give the genotypes of the F1 females and males. [2] (b) The F1 were interbred. Draw a genetic diagram to show the F2 phenotypes and their ratio in each sex. [6]

8. In a plant, tall stem (T) is dominant to dwarf (t), and purple flowers (P) are dominant to white (p). A plant from a cross of TTPP × ttpp was test crossed with ttpp. The offspring were 143 tall purple, 137 dwarf white, 42 tall white and 38 dwarf purple.

(a) Calculate the expected number in each class if the genes are not linked. [1] (b) Calculate χ². Show your working. [3] (c) State the degrees of freedom and what the χ² value shows. [2] (d) Explain these results, using a genetic diagram of the test cross. [4]

9. In a plant, the dominant allele C codes for an enzyme that makes purple pigment. Gene I, on a different chromosome, has a dominant allele I coding for a protein that inhibits this enzyme. Alleles c and i code for non-functional proteins.

(a) A white IICC plant is crossed with a white iicc plant. State and explain the F1 phenotype. [2] (b) An F1 plant is test crossed. Draw a genetic diagram and give the phenotype ratio. [4] (c) Name this type of gene interaction. [1]

10. In E. coli, the lac operon codes for enzymes that take up and break down lactose.

(a) Distinguish between a structural gene and a regulatory gene, using the lac operon as an example. [2] (b) β-galactosidase is an inducible enzyme. Explain how this differs from a repressible enzyme. [2] (c) Explain how adding lactose leads to the production of β-galactosidase. [3] (d) State what is meant by a transcription factor. [2] (e) Give one difference between how lactose switches on the lac operon and how gibberellin switches on genes for stem growth. [1]

Answers

1. (a) 4 [1]; gametes must be haploid so fertilisation restores the diploid number rather than doubling it [1]. (b) One maternal and one paternal chromosome [1] with the same genes at the same loci, though alleles may differ [1]. (c) 4 pairs: 2⁴ = 16 [1] (d) 16 × 16 = 256 [1]; random fertilisation [1]. (e) Crossing over in prophase I exchanges sections between non-sister chromatids [1], making new allele combinations on chromatids [1]. Examiner insight: For (b), “two identical chromosomes” scores nothing: the mark depends on “same genes, same loci”.

2. (a) P metaphase I [1]; Q anaphase II [1]; R telophase I [1]; S metaphase II [1]. (b) In prophase II the nuclear envelopes break down [1]; new spindles form and fibres attach to the centromeres [1]. (c) A cell plate forms instead of the membrane pinching in [1]. Examiner insight: Paired chromosomes on the equator always means metaphase I; single chromatids moving means anaphase II.

3. (a) Locus: the position of a gene on a chromosome [1]. Test cross: crossing an individual of dominant phenotype with a homozygous recessive one to find its genotype [1]. (b) LeLe × lele: all Lele, all tall [1]. Lele × lele: Lele and lele, 1 tall : 1 dwarf [1]. Any dwarf offspring means Lele; all tall means LeLe [1]. (c) One Le allele codes for a functional enzyme that makes active gibberellin (GA₁) [1], enough for normal stem elongation [1]. Examiner insight: In (b), a diagram without the conclusion linking each ratio to a genotype loses the final mark.

4. (a) HBB codes for the β-globin polypeptide of haemoglobin [1]. In Hbˢ a base substitution changes glutamic acid to valine [1]. At low oxygen concentration haemoglobin S forms fibres, so red cells become sickle-shaped [1], carry less oxygen and block capillaries [1]. (b) IᴬIᴮ × IᴼIᴼ gives 1 IᴬIᴼ (group A) : 1 IᴮIᴼ (group B) [1]. P(one child group A) = ½ [1]; P(both) = ½ × ½ = ¼ [1]. Examiner insight: Write codominant and multiple alleles as a gene letter with superscripts; plain A, B and O can cost the genotype mark.

5. (a) Her father gave her his only X, Xʰ, and she has no haemophilia, so she is XᴴXʰ [1]. (b) XᴴXʰ × XʰY; gametes Xᴴ, Xʰ and Xʰ, Y [1]; offspring XᴴXʰ carrier female, XʰXʰ female with haemophilia, XᴴY unaffected male, XʰY male with haemophilia [1]; probability a daughter has haemophilia = ½ [1]. (c) The allele codes for non-functional factor VIII [1], a blood-clotting protein, so blood clots slowly [1]. Examiner insight: “A daughter has haemophilia” (½) differs from “a child is a daughter with haemophilia” (¼).

6. (a) The allele codes for non-functional tyrosinase [1], so tyrosine is not converted to melanin, giving pale skin, hair and eyes [1]. (b) HTT codes for the protein huntingtin [1]. The disease allele has an expanded CAG repeat, giving the protein an extra-long glutamine chain [1]. This damages neurones in the brain [1]. One copy is enough because the abnormal protein causes damage even when normal huntingtin is also made [1]. Examiner insight: Name the gene, the protein and its effect separately; never call the gene “huntingtin”.

7. (a) Females NnXᴿXʳ [1]; males NnXᴿY [1]. (b) Female gametes NXᴿ, NXʳ, nXᴿ, nXʳ [1]; male gametes NXᴿ, NY, nXᴿ, nY [1].

	NXᴿ	NY	nXᴿ	nY
NXᴿ	NNXᴿXᴿ	NNXᴿY	NnXᴿXᴿ	NnXᴿY
NXʳ	NNXᴿXʳ	NNXʳY	NnXᴿXʳ	NnXʳY
nXᴿ	NnXᴿXᴿ	NnXᴿY	nnXᴿXᴿ	nnXᴿY
nXʳ	NnXᴿXʳ	NnXʳY	nnXᴿXʳ	nnXʳY

Correct genotypes [1]. Females 3 normal red : 1 vestigial red [1]. Males 3 normal red : 3 normal white : 1 vestigial red : 1 vestigial white [1]. White eyes occur only in males, as every female gets Xᴿ from her father [1]. Examiner insight: The Y chromosome carries no eye-colour allele; writing Yʳ loses the gamete mark.

8. (a) 360 ÷ 4 = 90 per class [1]. (b) (O − E)² = 2809, 2209, 2304, 2704 [1]; (O − E)²/E = 31.21 + 24.54 + 25.60 + 30.04 [1]; χ² = 111.4 [1]. (c) 4 − 1 = 3 degrees of freedom [1]; 111.4 > 7.82, so the difference is significant [1]. (d) The genes are linked; the parent is TP/tp [1]. Tall purple and dwarf white are parental classes from gametes TP and tp [1]. Tall white and dwarf purple are recombinants from gametes Tp and tP [1], made by crossing over between the two loci in prophase I, so these classes are small [1]. Examiner insight: “Linked” alone is not enough; explain the small classes by crossing over.

9. (a) IiCc, white [1]: the I allele makes the inhibitor, so no purple pigment forms [1]. (b) IiCc × iicc; gametes IC, Ic, iC, ic and ic [1]; offspring IiCc, Iicc, iiCc, iicc [1]; only iiCc is purple, the rest white [1]; 3 white : 1 purple [1]. (c) Epistasis (gene I is epistatic to gene C) [1]. Examiner insight: A memorised ratio with no diagram scores little; work it out from the pathway.

10. (a) A structural gene codes for a functional protein, e.g. lacZ (β-galactosidase) [1]; a regulatory gene codes for a protein controlling other genes, e.g. lacI (repressor) [1]. (b) An inducible enzyme is made only when its substrate is present [1]; a repressible enzyme is made until its product builds up and stops its synthesis [1]. (c) Without lactose the repressor binds to the operator [1]. Lactose binds to the repressor, changing its shape so it leaves the operator [1]; RNA polymerase binds to the promoter and transcribes lacZ, lacY and lacA [1]. (d) A protein that binds to DNA [1] and increases or decreases the rate of transcription in eukaryotes [1]. (e) Lactose changes the repressor’s shape, whereas gibberellin causes DELLA proteins to be broken down, releasing transcription factors [1]. Examiner insight: The repressor binds to the operator, not the promoter; swapping them loses the mark, and cAMP is not required.

Where marks are usually lost

• Calling homologous chromosomes “identical”.
• Mixing up anaphase I (homologues separate) and anaphase II (chromatids separate).
• Using plain dominant/recessive letters for codominant alleles.
• Giving the Y chromosome an allele, or not splitting a sex-linked ratio by sex.
• Stopping at “significant” without explaining the cause.
• Swapping the operator and the promoter.

Next steps

• A Level diagnostic
• Course hub and printable checklist
• Book a free trial class

Official syllabus

Cambridge International AS & A Level Biology 9700 syllabus for examination in 2025, 2026 and 2027 (Version 1), Cambridge University Press & Assessment – topic 16, Inheritance: 16.1, 16.2 and 16.3.