OCR GCSE Biology A: Single gene inheritance, genetic crosses, sex determination and the history of genetics (J247) – 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 statements B5.1i to B5.1m of OCR GCSE (9-1) Gateway Science Biology A (J247), specification Version 4.0 (August 2026), for first assessment in 2018: single gene inheritance, Punnett squares, sex determination, multi-gene features and the history of genetics. No statement here is shown in bold, so all questions suit both tiers.

Use the study guide if a topic is new and the revision notes to check facts. The OCR GCSE Biology hub, the printable checklist and the diagnostics help you plan.

Questions

1. Explain the meaning of each term.

(a) Homozygous [1] (b) Heterozygous [1]

2. In a species of mouse, black fur is controlled by a dominant allele (B) and brown fur by a recessive allele (b). Two black mice, both with genotype Bb, are crossed.

(a) Complete a Punnett square for this cross. [2] (b) State the ratio of black to brown offspring. [1] (c) State the probability that an offspring is brown. [1]

3. A student says: “In a heterozygous mouse the recessive allele disappears because the dominant allele dominates it.” Explain why this statement is wrong. [3]

4. A black mouse of unknown genotype is crossed with a brown mouse (bb).

(a) Explain how a brown pup in the litter would show that the black parent has genotype Bb. [2] (b) The black parent has genotype Bb. Calculate the probability that a litter of six pups are all black. [2] (c) A litter of six pups is all black. Suggest what you can conclude about the black parent. [1]

5. Phenylketonuria is a disorder caused by a recessive allele (p). The normal allele is P. Two parents who do not have the disorder have a child who does.

(a) State the genotypes of the parents and explain how you know. [2] (b) Draw a Punnett square for the parents’ next child. [2] (c) Give the probability that the next child is unaffected and carries the recessive allele. [1] (d) Give the probability that the next child has the disorder. [1]

6. A disorder is caused by a dominant allele (D). Both parents have the disorder and both are heterozygous.

(a) Show, using a Punnett square or working, the probability that a child does not have the disorder. [2] (b) Calculate the probability that two children both do not have the disorder. [2]

7. Describe how the sex of a human baby is determined. Use a genetic cross. [3]

8. A couple have four daughters.

(a) A relative says the fifth child is more likely to be a boy. Evaluate this statement. [2] (b) Calculate the probability that all five of their children are girls. [2]

9. A plant breeder crosses two heterozygous tall plants (Tt). Tall (T) is dominant to dwarf (t). The table shows the offspring from two separate crosses.

Cross	Tall offspring	Dwarf offspring	Total
A	10	6	16
B	296	104	400

(a) Calculate the numbers of tall and dwarf offspring expected in each cross. [2] (b) Explain why the results of cross A are further from the expected ratio than the results of cross B. [2] (c) Calculate the ratio of tall to dwarf plants in cross B, in the form x : 1, to 1 decimal place. [1]

10. Human height varies across a wide range, from short to tall, with many values in between.

(a) Explain why a single-gene Punnett square cannot predict the height of a child. [2] (b) Give one other human feature that is controlled by many genes. [1] (c) Name one non-genetic factor that affects height. [1]

11. Gregor Mendel studied inheritance by breeding pea plants in the 1860s.

(a) Give two reasons why pea plants were a good choice for his investigations. [2] (b) His work was not widely accepted until around 1900. Suggest two reasons. [2] (c) Describe two later developments that helped to explain what Mendel’s “units” are. [2]

12. A pure-breeding tall plant (TT) is crossed with a pure-breeding dwarf plant (tt). Tall is dominant.

(a) State the genotype and phenotype of the first generation. [2] (b) Two of the first generation are crossed. Draw a Punnett square for this cross. [2] (c) The cross produces 128 second-generation plants. Calculate how many are expected to be tall and how many dwarf. [2] (d) Calculate how many of the tall plants are expected to be heterozygous. [1]

Answers

1. (a) Having two identical alleles of a gene (for example BB or bb) [1] [1] (b) Having two different alleles of a gene (for example Bb) [1] [1] Examiner insight: Give an example genotype as well as the definition; it shows the idea is understood and protects the answer if the wording is loose.

2. (a) Gametes from each parent B and b shown on the edges [1]; boxes BB, Bb, Bb, bb [1] [2] (b) 3 black : 1 brown [1] [1] (c) 1/4 (or 25%, or 0.25) [1] [1] Examiner insight: A ratio and a probability are different forms of the answer. Check which one the question asks for, and do not give 3 : 1 when 1/4 is wanted.

3. The recessive allele is still present in the heterozygous mouse, so it has not disappeared [1]. Half of the gametes carry it, so it can be passed on to offspring [1]. It is shown in the phenotype only when two copies are present (bb), because one dominant allele is enough to give the dominant feature [1] [3] Examiner insight: Strong answers say the alleles do not interact to remove one another: dominance describes which feature shows in the phenotype, not one allele destroying the other.

4. (a) A brown pup is bb, so it received one b from each parent [1]. The brown parent can only give b, so the black parent must have given b and must be Bb [1] [2] (b) Each pup has a 1/2 chance of being black [1]. (1/2)⁶ = 1/64 (or 0.016, or about 1.6%) [1] [2] (c) The parent is probably BB, but it could still be Bb, because there is a 1/64 chance of this result from a Bb parent [1] [1] Examiner insight: Where an answer asks you to “suggest” a conclusion, a cautious statement that mentions the remaining probability is stronger than a flat claim that the genotype is proved.

5. (a) Both parents are Pp [1]. The child is pp, so each parent must have passed on p, but the parents are unaffected, so each also has P [1] [2]

	P	p
P	PP	Pp
p	Pp	pp

(b) Gametes P and p shown for both parents [1]; correct four boxes as in the square above [1] [2] (c) Two boxes out of four are Pp: 1/2 (50%) [1] [1] (d) One box out of four is pp: 1/4 (25%) [1] [1] Examiner insight: Reasoning from the child back to the parents is a common test. State which alleles each parent must have passed on, not just the genotypes.

6. (a) Hh × Hh gives HH, Hh, Hh, hh [1]. Only hh is unaffected, so the probability is 1/4 [1] [2] (b) The events are independent, so multiply: 1/4 × 1/4 [1] = 1/16 (0.0625, or 6.25%) [1] [2] Examiner insight: Show the multiplication step. A final fraction with no working leaves nothing to give credit for if it is wrong.

7. Females have two X chromosomes (XX) and males have X and Y (XY) [1]. All eggs carry X, while half of the sperm carry X and half carry Y [1]. Punnett square: eggs X and X against sperm X and Y gives XX, XX, XY, XY, so the chance of a boy is 1/2 (50%) and a girl 1/2 [1] [3] Examiner insight: Say that the sperm decides the sex, and label which gametes come from which parent. A square with unlabelled parents is harder to give credit to.

8. (a) The statement is wrong [1]. Each birth is an independent event with a 1/2 chance of a boy, because the sperm is equally likely to carry X or Y, so earlier children do not change the odds [1] [2] (b) (1/2)⁵ [1] = 1/32 (0.03125, or about 3.1%) [1] [2] Examiner insight: Use the word independent and explain it in terms of sperm carrying X or Y; “it is just luck” is too vague to earn the mark.

9. (a) Cross A: 3/4 × 16 = 12 tall and 4 dwarf [1]. Cross B: 3/4 × 400 = 300 tall and 100 dwarf [1] [2] (b) Fertilisation is random, so any set of offspring can differ from the prediction by chance [1]. A small sample (16) is affected more by chance than a large one (400), so the larger sample is closer to 3 : 1 [1] [2] (c) 296 ÷ 104 = 2.846…, so 2.8 : 1 [1] [1] Examiner insight: Refer to the sample sizes in your explanation, using the numbers 16 and 400, so the comparison is clear.

10. (a) Height is controlled by many genes, not one [1]. A single-gene square only predicts a few clear categories, but height shows a continuous range [1] [2] (b) Skin colour [1] [1] (c) Diet (or nutrition, or illness in childhood) [1] [1] Examiner insight: “Genes” alone is not an explanation here; say that many genes act together and say what a single-gene square would predict instead.

11. (a) Any two from: they grow quickly; they produce many offspring so large numbers could be counted; they have clear either-or features; he could control which plants were crossed [1] [1] [2] (b) Any two from: chromosomes and genes had not been discovered, so there was no physical explanation for his units; his work was not widely read; other scientists had not repeated or checked his results [1] [1] [2] (c) Any two from: microscopes showed chromosomes and later work suggested the units are on them; the structure of DNA was worked out in 1953, showing how information can be stored and copied; the human genome was sequenced [1] [1] [2] Examiner insight: History answers need the idea and its effect: do not just name an event, say what it showed or what new evidence it provided.

12. (a) Tt [1]; tall [1] [2] (b) Gametes T and t for both parents [1]; boxes TT, Tt, Tt, tt [1] [2] (c) Tall: 128 × 3/4 = 96 [1]; dwarf: 128 × 1/4 = 32 [1] [2] (d) Two of the three tall boxes are Tt, so 2/3 × 96 = 64 [1] [1] Examiner insight: In part (d), use the tall plants only as your total (96), not all 128, and show the fraction you applied.

Where marks are usually lost

• Writing letters in a Punnett square without saying what they represent.
• Putting both alleles of a parent into one gamete.
• Giving a genotype ratio when asked for phenotypes.
• Answering a probability question with a ratio such as 3 : 1.
• Saying the recessive allele “disappears” or the dominant allele “wins”.
• Treating a past result as changing the next birth’s odds.
• Using total offspring when the question asks only about one phenotype.

Next steps

• Condense the topic with the revision notes.
• Re-read the full explanations in the study guide.
• Go to the OCR GCSE Biology hub or the printable checklist.
• Try all the free 10-minute diagnostics.
• Book a free trial class for help with crosses.

Official syllabus

OCR GCSE (9-1) Gateway Science Biology A (J247), specification Version 4.0 (August 2026), for first assessment in 2018, published by OCR (Oxford Cambridge and RSA). Statements B5.1i to B5.1m.