OCR A Level Biology A: Variation, genetic diagrams and chi-squared (H420) – 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 section 6.1.2(a) to (d) of the OCR A Level Biology A (H420) specification, Version 4.1 (April 2026), for first assessment in 2025. Revise first with the study guide and revision notes. The chi-squared formula is χ² = Σ (fo − fe)² / fe. Critical values at p = 0.05 are 3.84 (1 degree of freedom), 5.99 (2) and 7.81 (3).

Questions

1. Cuttings from one plant were grown for 14 days, half in light and half in dark. Mean stem length was 21 mm in light and 48 mm in dark; the dark-grown leaves were small and pale. (a) Explain why cuttings from one plant were used. [1] (b) Calculate the percentage increase in mean stem length of the dark-grown cuttings compared with those grown in the light. [2] (c) Explain the pale leaves, and why the long stem is an advantage to a seedling in the dark. [2]

2. A species has 2n = 10. (a) Calculate the number of genetically different gametes one individual can make by independent assortment alone. [2] (b) Calculate the number of different combinations of these gametes that could fuse between two such parents. [1] (c) Describe a process during meiosis, not included in your calculation, that increases variation. [2]

3. Cystic fibrosis is caused by a recessive allele. Two unaffected parents have a child with cystic fibrosis. (a) Draw a genetic diagram to show this, using F and f. [3] (b) Calculate the probability that their next two children both have cystic fibrosis. [2]

4. In the snail Cepaea nemoralis, shell colour is controlled by one gene with three alleles: brown Cᴮ, pink Cᴾ and yellow Cʸ, in the dominance order Cᴮ > Cᴾ > Cʸ. (a) State how many genotypes are possible, and how many of these are homozygous. [2] (b) A pink snail with a yellow parent is crossed with a brown snail with a yellow parent. Draw a genetic diagram and state the phenotype ratio of the offspring. [4]

5. In guinea pigs, rough coat (R) is dominant to smooth (r), and black (B) is dominant to white (b). The genes are on different chromosomes. Two guinea pigs heterozygous for both genes are crossed. (a) State the genotypes of the gametes one parent produces. [1] (b) State the phenotype ratio of the offspring. [2] (c) Of 800 offspring, how many would you expect to be white and smooth? [1] (d) A black rough guinea pig of unknown genotype is test-crossed and gives equal numbers of four phenotypes. State its genotype and the test-cross partner. [2]

6. A rare condition is caused by a dominant allele on the X chromosome (Xᴬ), with Xᵃ the normal allele. (a) An affected man and an unaffected woman have children. Draw a genetic diagram. [3] (b) Explain why none of their sons is affected. [1] (c) A heterozygous affected woman and an unaffected man have a child. Give the probability that the child is affected. [2]

7. In mice, grey coat (G) is dominant to white (g), and long ears (L) to short (l). A mouse GgLl was test-crossed with ggll. Of 400 offspring: 168 grey long, 172 white short, 31 grey short, 29 white long. (a) State the numbers expected in each class if the genes assort independently. [1] (b) Calculate χ². [3] (c) Give the degrees of freedom and interpret the result. [2] (d) Calculate the percentage of recombinant offspring. [1] (e) Explain how the recombinants arose. [1]

8. In summer squash, two genes on different chromosomes control fruit colour. Allele W gives white fruit whatever the second gene. Without W, allele Y gives yellow and yy gives green. Two WwYy plants were crossed and gave 240 offspring: 172 white, 52 yellow and 16 green. (a) Explain what is meant by epistasis. [2] (b) State the genotypes that give white, yellow and green fruit. [2] (c) State the expected number of green fruit. [1] (d) Calculate χ² for a ratio of 12 : 3 : 1. [2] (e) State the conclusion at p = 0.05. [1]

9. Two heterozygous tall pea plants were crossed. The offspring were 182 tall and 78 dwarf. The student expected 3 tall : 1 dwarf. (a) State the null hypothesis. [1] (b) Calculate the expected numbers. [1] (c) Calculate χ². [2] (d) State a conclusion. [2]

10. In 200 plants of one species, flower colour was red in 148 and white in 52. Height was recorded in classes: 20 to 29 cm, 4 plants; 30 to 39, 18; 40 to 49, 52; 50 to 59, 64; 60 to 69, 41; 70 to 79, 15; 80 to 89, 6. (a) State which characteristic shows continuous and which discontinuous variation, with one feature of the data for each. [2] (b) Compare the number of genes controlling each. [2] (c) Explain the role of the environment in each. [2]

11. Two true-breeding white-flowered lines were crossed. All F1 plants were purple. F1 plants crossed with each other gave 512 F2 plants: 301 purple and 211 white. Purple pigment needs two enzymes, made from dominant alleles A and B at different loci. (a) Give the genotypes of the parent lines and the F1. [2] (b) Draw a genetic diagram for the F1 cross, showing the expected ratio. [3] (c) Calculate χ² for the F2 results against the ratio from (b). [3] (d) State and explain the conclusion. [2]

Answers

1. (a) Cuttings are genetically identical, so any difference is due to the environment (light) [1]. (b) (48 − 21) / 21 × 100 [1] = 129% (128.6) [1]. (c) Chlorophyll forms only in light [1]. A long stem raises the shoot tip towards light [1]. [5] Examiner insight: Show the subtraction and the division so your method is visible; “light is needed to make chlorophyll” explains pale leaves better than “no photosynthesis”.

2. (a) Gametes = 2ⁿ with n = 5 [1] = 32 [1]. (b) 32 × 32 = 1024 [1]. (c) Crossing over [1]: in prophase I, sections are exchanged between non-sister chromatids of paired homologues, so chromatids carry new combinations of alleles [1]. [5] Examiner insight: Use n, the haploid number, in 2ⁿ; using 10 is the usual slip.

3. (a) Key: F normal, f cystic fibrosis [1]. Parents Ff × Ff; gametes F or f from each [1]. Offspring FF, Ff, Ff, ff, so one in four affected (ff) [1]. (b) Probability one child affected = 1/4 [1]; both = 1/4 × 1/4 = 1/16 [1]. [5] Examiner insight: Start with a key and the parents’ genotypes; Ff is deduced from the affected child, not from the parents’ phenotype.

4. (a) With three alleles there are 6 genotypes [1], of which 3 are homozygous [1]. (b) Pink parent CᴾCʸ, brown parent CᴮCʸ [1]. Gametes: Cᴾ or Cʸ; Cᴮ or Cʸ [1]. Offspring: CᴮCᴾ, CᴮCʸ (brown), CᴾCʸ (pink), CʸCʸ (yellow) [1]. Ratio 2 brown : 1 pink : 1 yellow [1]. [6] Examiner insight: Say why each parent is heterozygous: each has a yellow parent, so carries Cʸ.

5. (a) RB, Rb, rB, rb [1]. (b) 9 black rough : 3 black smooth : 3 white rough : 1 white smooth [2]. (c) 1/16 × 800 = 50 [1]. (d) Genotype RrBb [1], crossed with rrbb (smooth white) [1]. [6] Examiner insight: Write the ratio against named phenotypes, not as bare numbers.

6. (a) Parents XᴬY × XᵃXᵃ [1]. Gametes Xᴬ or Y; Xᵃ [1]. Offspring XᴬXᵃ (affected daughters) and XᵃY (unaffected sons), all daughters affected [1]. (b) A son receives his father’s Y chromosome, so does not receive Xᴬ [1]. (c) Parents XᴬXᵃ × XᵃY; the mother passes Xᴬ to half of her children [1], so the probability is 1/2 [1]. [6] Examiner insight: Include Y in every male genotype; it is what explains part (b).

7. (a) 400 / 4 = 100 each [1]. (b) Differences 68, 72, 69, 71 [1]; (68² + 72² + 69² + 71²) / 100 = (4624 + 5184 + 4761 + 5041) / 100 [1]; χ² = 196.1 [1]. (c) Degrees of freedom = 3 [1]; 196.1 is far above 7.81, so the difference is significant: the genes are linked [1]. (d) (31 + 29) / 400 × 100 = 15% [1]. (e) Crossing over between the two loci in prophase I of meiosis in the heterozygous parent [1]. [8] Examiner insight: After rejecting the null hypothesis, say what it means for the genes.

8. (a) An allele of one gene masks or suppresses [1] the expression of alleles of a different gene [1]. (b) White W_ (WW or Ww, any Y) [1]; yellow wwY_ and green wwyy [1]. (c) 1/16 × 240 = 15 [1]. (d) Expected 180, 45, 15 [1]; χ² = 64/180 + 49/45 + 1/15 = 0.356 + 1.089 + 0.067 = 1.51 [1]. (e) 2 degrees of freedom: 1.51 is less than 5.99, so not significant; the results fit dominant epistasis (12 : 3 : 1) [1]. [8] Examiner insight: Name which gene masks which, and which allele does the masking.

9. (a) There is no significant difference between the observed and expected numbers [1]. (b) 195 tall and 65 dwarf [1]. (c) 13² / 195 + 13² / 65 = 0.867 + 2.600 [1] = 3.47 [1]. (d) 1 degree of freedom, critical value 3.84; 3.47 is less than 3.84 [1], so not significant: chance may explain the difference and the results are consistent with 3 : 1 [1]. [6] Examiner insight: State the degrees of freedom and the critical value you compare with, and avoid “proves 3 : 1”.

10. (a) Flower colour is discontinuous: two distinct categories [1]. Height is continuous: a range of values with most plants in the middle, peaking at 50 to 59 cm [1]. (b) Flower colour: one gene (or a very few), with different alleles [1]. Height: many genes (polygenic), each with a small additive effect [1]. (c) Flower colour is little affected by environment [1]. Height is strongly influenced by factors such as light and mineral supply, giving a smooth distribution [1]. [6] Examiner insight: Refer to the data given, such as the peak class or the colour counts, not definitions alone.

11. (a) Parents AAbb and aaBB [1]; F1 AaBb [1]. (b) Gametes AB, Ab, aB, ab from each F1 plant [1]. Of 16 combinations, 9 have A and B [1]. Ratio 9 purple : 7 white [1]. (c) Expected 288 purple, 224 white [1]; (13² / 288) + (13² / 224) = 0.587 + 0.754 [1]; χ² = 1.34 [1]. (d) 1 degree of freedom, critical value 3.84; 1.34 is less than 3.84, so not significant and the data fit 9 : 7 [1]. This is epistasis (complementary): each dominant allele makes an enzyme in the pathway, and a plant lacking either enzyme is white [1]. [10] Examiner insight: Link the ratio to the pathway: both enzymes are needed.

Where marks are usually lost

• Using 2n instead of n when calculating gamete types from independent assortment.
• Genetic diagrams without a key, with two alleles of one gene in a gamete, or without the Y in male sex-linked genotypes.
• Giving a ratio without matching each number to a phenotype.
• Taking expected values from the observed numbers instead of the ratio and the total.
• Degrees of freedom taken as the number of classes.
• Stopping at “significant” with no statement about the genes.
• Saying a result proves a ratio.
• Missing crossing over as the explanation for a few recombinants.

Next steps

• Recap with the revision notes and the study guide.
• See the OCR A Level Biology hub and the checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class: /trial/.

Official syllabus

OCR, A Level Biology A (H420) specification, Version 4.1 (April 2026), for first assessment in 2025: section 6.1.2(a) to (d), Patterns of inheritance.