Cambridge O Level Biology 5090: Inheritance (5090) – 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.

These questions cover Topic 17, Inheritance, of the Cambridge O Level Biology 5090 syllabus for examination in 2026, 2027 and 2028 (version 4): every section from 17.1 Variation to 17.4 Selection. The syllabus is not tiered, so every question applies to every candidate.

Revise first with the study guide and the revision notes. The course hub and the printable checklist list the other topics, and the free 10-minute diagnostics help you decide what to revise next. A calculator is useful for questions 5 and 6. All data are made-up values for the question, not reference figures to learn.

Questions

1. (a) Define variation. [1] (b) State whether each of these shows continuous or discontinuous variation: (i) seed colour in peas, (ii) body length in fish. [2]

2. A student measured the hand span of 30 people in her class and recorded the blood group of each person.

(a) Name the type of graph she should draw for the hand-span data, and give a reason. [2] (b) Suggest why hand span shows a range of values but blood group falls into only four categories. [2] (c) Suggest two ways she could make her hand-span results more reliable. [2]

3. (a) Describe the structure of a DNA molecule. [3] (b) In a sample of DNA, 26% of the bases are T. Calculate the percentage of bases that are G. Show your working. [2] (c) One strand of DNA has the base sequence GACTTA. Write the sequence of the complementary strand. [1]

4. Explain how a gene in the nucleus controls a chemical reaction in the cytoplasm of a cell. [4]

5. In mice, the allele for black fur (B) is dominant to the allele for brown fur (b).

(a) State the genotype of a brown mouse, and state whether it is homozygous or heterozygous. [2] (b) Two heterozygous black mice are crossed. Draw a genetic diagram to show the expected offspring and state the expected phenotypic ratio. [4] (c) Over several litters, the two mice had 104 offspring: 76 black and 28 brown. Calculate the expected numbers of black and brown mice. [2] (d) Explain why the observed numbers differ from the expected numbers. [2]

6. In a plant, the allele for green pods (G) is dominant to the allele for yellow pods (g). A green-podded plant was crossed with a yellow-podded plant. The offspring were 23 green-podded and 17 yellow-podded plants.

(a) Deduce the genotype of the green-podded parent. Explain your answer. [2] (b) Calculate the observed ratio of green to yellow, and state the expected ratio. [2] (c) State what is meant by pure-breeding, and give the genotype of a pure-breeding yellow-podded plant. [2]

7. A woman has blood group A and her husband has blood group B. Their first child has blood group O.

(a) Explain what is meant by codominance, using blood group AB as your example. [2] (b) Deduce the genotypes of the woman and her husband. Explain your answer. [3] (c) State all the blood groups that further children of this couple could have. [1]

8. A couple have three daughters. Use a genetic diagram to explain the chance that their next child will be a boy. [4]

9. (a) Describe what is meant by a gene mutation. [2] (b) Explain how a gene mutation can lead to sickle cell anaemia. [3] (c) Describe the chromosome mutation found in Down’s syndrome. [2] (d) State two factors that increase the rate of mutation. [2]

10. A species of snail lives on dark rocks. Its shell colour varies from light to dark, and shell colour is inherited. Birds eat the snails they can see easily. Explain how natural selection could increase the proportion of dark-shelled snails over many generations. [6]

11. A farmer wants a herd of goats that produce more milk.

(a) Describe how the farmer could use artificial selection to do this. [4] (b) Describe the role of artificial selection in producing one economically important crop plant. [2] (c) The development of MRSA is an example of natural selection. State the feature that varies in the bacterial population and what acts as the selecting agent. [2]

Answers

1. (a) Differences between individuals of the same species [1]. (b) (i) Discontinuous [1]; (ii) continuous [1]. Examiner insight: “Differences between organisms” without “of the same species” is too vague to earn the definition.

2. (a) Histogram [1], because hand span is continuous data / can take any value in a range [1]. (b) Hand span is affected by genes and the environment, giving a range [1]; blood group is controlled by genes only, giving a limited number of phenotypes [1]. (c) Any two from: measure more people (a larger sample) [1]; use the same ruler and the same hand position each time [1]; repeat each measurement and take a mean. Examiner insight: In (a) the reason must refer to the type of data; “because it is easier to read” gains nothing.

3. (a) Any three from: two strands coiled into a double helix [1]; each strand a chain of nucleotides [1]; bases A, T, C, G; bonds between pairs of bases hold the strands together; A pairs with T and C with G [1]. (b) A = 26%, so A + T = 52% and C + G = 48% [1]; G = 48 ÷ 2 = 24% [1]. (c) CTGAAT [1]. Examiner insight: Show the step “C + G = 48%” in (b) so your method is visible even if the final figure slips; a common slip is giving 26% (the T value) or 48% (C + G together).

4. The gene’s sequence of bases determines the sequence of amino acids [1] in a specific protein [1]. The amino acid sequence gives the protein its shape [1]. The protein is an enzyme, whose shape lets it catalyse the reaction in the cytoplasm [1]. Examiner insight: Link each step in order (bases → amino acids → shape → enzyme); naming “protein” without saying it can be an enzyme leaves the reaction unexplained.

5. (a) bb [1]; homozygous [1]. (b)

Parents:   Bb   x   Bb
Gametes:  B, b      B, b

          |  B  |  b
      B   | BB  | Bb
      b   | Bb  | bb

Parental genotypes both Bb [1]; gametes B and b from each parent [1]; offspring BB, Bb, Bb, bb [1]; 3 black : 1 brown [1]. (c) Black = 104 × 3/4 = 78 [1]; brown = 104 × 1/4 = 26 [1]. (d) Fertilisation is random / which gametes fuse is down to chance [1], so the numbers only approximate 3 : 1, especially in a fairly small sample [1]. Examiner insight: In (b) label parents, gametes and offspring; a Punnett square with no gametes shown leaves out a step you are asked to show.

6. (a) Gg (heterozygous) [1], because some offspring are yellow (gg), so each parent must have given a g allele [1]. (b) 23 ÷ 17 = 1.35 : 1 [1]; expected 1 : 1 [1]. (c) Two identical homozygous individuals breed together and all offspring have the same phenotype as the parents [1]; gg [1]. Examiner insight: In (a) the reason must use the yellow offspring as evidence; stating “Gg” with no reason earns only the first mark.

7. (a) Both alleles in a heterozygote are expressed in the phenotype [1]; in group AB, Iᴬ and Iᴮ are both expressed (genotype IᴬIᴮ) [1]. (b) The child is IᵒIᵒ, so each parent must carry Iᵒ [1]. Woman: IᴬIᵒ [1]. Husband: IᴮIᵒ [1]. (c) A, B, AB and O [1]. Examiner insight: Use the Iᴬ, Iᴮ, Iᵒ symbols; writing genotypes as “AO” and “BO” may not be credited.

8.

Parents:   XX (mother)   x   XY (father)
Gametes:   X                 X, Y

          |  X  |  Y
      X   | XX  | XY

Correct parental genotypes XX and XY [1]; gametes X from the mother, X or Y from the father [1]; offspring XX (girl) and XY (boy) in a 1 : 1 ratio [1]; each fertilisation is independent, so the chance of a boy is still 1 in 2 (50%) whatever the earlier children were [1]. Examiner insight: Say that earlier births do not affect the next one; “50%” alone does not explain the chance.

9. (a) A random change [1] in the base sequence of DNA [1]. (b) The base sequence of the gene for haemoglobin is changed [1], so the amino acid sequence of haemoglobin is changed [1], giving haemoglobin a different shape so that red blood cells can become sickle-shaped [1]. (c) A change in chromosome number [1]: 47 chromosomes instead of 46 [1]. (d) Ionising radiation [1]; some chemicals [1]. Examiner insight: In (b) name haemoglobin as the protein affected; “the blood changes shape” is too vague.

10. There is variation in shell colour within the population [1]. Snails produce many offspring, so there is a struggle for survival [1]. Light-shelled snails are seen more easily and eaten by birds [1]. Dark-shelled snails are better adapted, so more survive to reproduce [1]. They pass on their alleles for dark shells to their offspring [1]. Over many generations the proportion of dark-shelled snails increases [1]. Examiner insight: Write “alleles are passed on”; saying the snails “adapt” or “change colour to hide” describes a change in one lifetime, not selection.

11. (a) Select goats that produce the most milk [1]. Cross / breed these goats together [1]. Select offspring that produce the most milk [1]. Repeat over many generations [1]. (b) Humans choose plants with a desirable feature such as high yield or disease resistance and breed from them [1], producing varieties that give more food or income [1]. (c) Resistance to the antibiotic varies [1]; the antibiotic is the selecting agent [1]. Examiner insight: In (a) “repeat over many generations” is a separate step; leaving it out leaves the description incomplete.

Where marks are usually lost

• Drawing a Punnett square without showing the parental genotypes and gametes first.
• Giving a genotypic ratio (1 : 2 : 1) when the question asks for phenotypes.
• Deducing a genotype without quoting the evidence (an offspring with the recessive phenotype).
• Using A, B and O instead of Iᴬ, Iᴮ and Iᵒ for blood group alleles.
• Saying a family with three girls is “more likely” to have a boy next.
• Writing that DNA bases pair A–G or C–T.
• Missing “of the same species” from the definition of variation.
• Describing natural selection as organisms deciding to change.
• Leaving “repeat over many generations” out of artificial selection.

Next steps

• Go back over weak points with the revision notes and the study guide.
• Link mutation back to protein shape in Enzymes, and antibiotic resistance to Disease and immunity.
• See every topic on the course hub and tick off outcomes on the printable checklist.
• Try the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Cambridge O Level Biology 5090 syllabus for examination in 2026, 2027 and 2028 (version 4), Topic 17: Inheritance (17.1 Variation; 17.2 DNA; 17.3 Inheritance; 17.4 Selection), published by Cambridge Assessment International Education.