OxfordAQA International AS and A-level Biology: Inheritance and genetic crosses (9610) – 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 sections 3.3.6.1 and 3.3.6.2 of the OxfordAQA International AS and A-level Biology (9610) specification, Version 5.1 (Unit 3, listed as International A-level only). Read the study guide and revision notes first. Course pages: hub and checklist. Not sure where to start? Try the free diagnostics. Where a question says “assume”, the allele pattern is given for the exercise. Critical values of chi-squared at p = 0.05 are 3.84 for df 1, 5.99 for df 2, 7.81 for df 3 and 9.49 for df 4.

Questions

1. Genetically identical cuttings of one plant are grown, one in bright light and one in shade. The shaded plant is taller and paler. State what is meant by genotype and by phenotype [2], and explain the difference between the cuttings [2].

2. In Andalusian fowl, assume black feathers (Fᴮ) and splashed-white feathers (Fᵂ) are codominant, and the heterozygote is blue-grey.

(a) Draw a fully labelled genetic diagram for a cross between two blue-grey birds. [4] (b) A blue-grey bird is crossed with a splashed-white bird and produces 60 chicks. State the expected number of each phenotype. [2]

3. A man of blood group B, whose mother was group O, has children with a woman of group AB.

(a) State the genotypes of the parents and their gametes. [2] (b) State the probability that a child is group A. [1] (c) Find the probability that both of their first two children are group B. [2]

4. Assume beetle wing-case colour has three alleles in a dominance series, Cᵈ (dark) > Cᵐ (mid) > Cˡ (light). A dark beetle CᵈCᵐ is crossed with a mid beetle CᵐCˡ.

(a) Give the genotypes and phenotypes of the offspring and their ratio. [3] (b) Explain why no light offspring are produced. [1]

5. Assume in zebrafish that stripes (S) are dominant to spots (s) and long fins (L) to short fins (l), on different chromosomes. A striped long-finned fish SsLl is test-crossed.

(a) Give the expected phenotype ratio. [2] (b) Two SsLl fish are crossed. Calculate the probability that an offspring is striped with short fins. [2] State how many of 320 offspring you would expect to be spotted with long fins. [1]

6. Two SsLl zebrafish are crossed and give 480 offspring: 255 striped long, 98 striped short, 84 spotted long and 43 spotted short. Use a chi-squared test to decide whether the results fit a 9 : 3 : 3 : 1 ratio. [6]

7. Assume a rare condition is caused by a dominant allele (Xᴴ) on the X chromosome. An affected man (XᴴY) and an unaffected woman have children.

(a) Draw a genetic diagram showing the genotypes and phenotypes of their children. [3] (b) Account for the fact that no son is affected. [1] (c) Now consider an affected woman (XᴴXʰ) and an unaffected man. What is the chance that a daughter of this couple is affected? Give your reasoning. [2]

8. Assume in chickens silver plumage (Zᴾ) is dominant to gold (Zᵖ). Birds have ZZ males and ZW females. Hatchery workers cross gold males with silver females.

(a) Draw a labelled genetic diagram for the cross. [3] (b) Explain how the offspring can be sexed at hatching. [1] (c) Explain why this would not work if the gene were autosomal. [1]

9. Assume in a moth that striped wings (S) and feathery antennae (F) are dominant to plain (s) and thread-like (f), and the loci are on one chromosome. A moth SF / sf is crossed with sf / sf. Of 1000 offspring, 462 are striped feathery, 444 plain thread-like, 44 striped thread-like and 50 plain feathery.

(a) Identify the parental and recombinant classes. [2] (b) Explain how the recombinant offspring arise. [2] (c) Calculate the recombination frequency. [1] (d) Use a chi-squared test to decide whether the genes assort independently. [4]

10. Assume pigment in a fish needs allele P; pp fish are white. With P, allele R gives dark and rr gives pale. Two PpRr fish are crossed and give 240 offspring: 112 dark, 58 pale and 70 white.

(a) Explain how this cross gives a 9 : 3 : 4 ratio. [4] (b) Test whether the results fit 9 : 3 : 4. [4]

11. Assume gourd skin is white when allele W is present. In plants without W, a Y allele gives yellow skin and yy gives green skin. Two WwYy plants are crossed and give 288 offspring: 227 white, 49 yellow and 12 green.

(a) State the expected ratio and the expected numbers. [2] (b) Test whether the results fit the expected ratio. [3] (c) Explain why W is epistatic to Y. [2]

Answers

1. Genotype is the genetic constitution (the alleles) [1]; phenotype is the expression of the genotype and its interaction with the environment [1]. The cuttings share a genotype [1], so the difference is due to the environment, here light intensity [1]. [4] Examiner insight: Naming the environmental factor and linking it to the identical genotype tends to gain more credit than saying “the environment affects it”.

2. (a) Key: Fᴮ black, Fᵂ splashed white [1]. Parents FᴮFᵂ × FᴮFᵂ; gametes Fᴮ or Fᵂ from each [1]. Offspring FᴮFᴮ : 2 FᴮFᵂ : FᵂFᵂ [1]. Phenotypes 1 black : 2 blue-grey : 1 splashed white [1]. (b) FᴮFᵂ × FᵂFᵂ gives 1 : 1 [1], so 30 blue-grey and 30 splashed white [1]. [6] Examiner insight: Show the key, parental genotypes, gametes, offspring genotypes and phenotypes as separate steps; a bare ratio can leave nothing to credit.

3. (a) Man Iᴮi (mother ii gave him i) and woman IᴬIᴮ [1]; gametes Iᴮ or i, and Iᴬ or Iᴮ [1]. (b) Offspring IᴬIᴮ, Iᴬi, IᴮIᴮ and Iᴮi, so group A (Iᴬi) is 1/4 [1]. (c) Group B (IᴮIᴮ or Iᴮi) is 2/4 = 1/2 [1]; 1/2 × 1/2 = 1/4 [1]. [5] Examiner insight: Explaining how you got the man’s genotype from his mother’s group shows reasoning that a bare genotype does not.

4. (a) Gametes Cᵈ or Cᵐ, and Cᵐ or Cˡ [1]. Offspring CᵈCᵐ dark, CᵈCˡ dark, CᵐCᵐ mid, CᵐCˡ mid [1]; 2 dark : 2 mid, which is 1 : 1 [1]. (b) A light beetle needs CˡCˡ, and the dark parent has no Cˡ allele [1]. [4] Examiner insight: With a dominance series, state the order first and apply it to each genotype before giving phenotypes.

5. (a) SsLl × ssll: gametes SL, Sl, sL, sl and sl only [1]; 1 striped long : 1 striped short : 1 spotted long : 1 spotted short [1]. (b) Striped (3/4) and short (1/4) [1]: 3/16 [1]. Spotted long-finned is 1/4 × 3/4 = 3/16, and 3/16 × 320 = 60 [1]. [5] Examiner insight: Multiplying single-gene probabilities is quicker and less error-prone than counting 16 boxes; show each fraction.

6. Null hypothesis: no significant difference from 9 : 3 : 3 : 1; differences are due to chance [1]. Expected 270, 90, 90, 30 [1]. χ² = 225/270 + 64/90 + 36/90 + 169/30 = 0.833 + 0.711 + 0.400 + 5.633 [1] = 7.58 [1]. df = 3, critical value 7.81 [1]. 7.58 is less than 7.81, so the difference is not significant and the null hypothesis is not rejected: results fit 9 : 3 : 3 : 1 [1]. [6] Examiner insight: A result close to the critical value still needs a clear comparison and a statement about the null hypothesis; avoid “proves” or “accepts”.

7. (a) Man XᴴY, woman XʰXʰ [1]; gametes Xᴴ or Y, and Xʰ [1]. Daughters XᴴXʰ affected; sons XʰY unaffected [1]. (b) Sons inherit the Y from their father and an X from their mother, who has no Xᴴ [1]. (c) Probability 1/2 [1], because each daughter gets the mother’s Xᴴ or Xʰ with equal chance, and the father gives Xʰ [1]. [6] Examiner insight: For sex-linked answers, say which parent each chromosome comes from; “males get it from their mothers” without the Y-chromosome explanation is incomplete.

8. (a) Gold male ZᵖZᵖ × silver female ZᴾW [1]; gametes Zᵖ, and Zᴾ or W [1]; sons ZᴾZᵖ silver, daughters ZᵖW gold [1]. (b) All silver chicks are male and all gold are female [1]. (c) An autosomal allele would be inherited equally by both sexes, so colour would not show sex [1]. [5] Examiner insight: Give the sex of each colour class and say why, because in birds the female is the sex with one Z.

9. (a) Parental: striped feathery and plain thread-like [1]. Recombinant: striped thread-like and plain feathery [1]. (b) Crossing over between non-sister chromatids of homologous chromosomes in prophase I [1] exchanges alleles between the loci, giving SF and sf together with Sf and sF gametes [1]. (c) (44 + 50)/1000 = 9.4% [1]. (d) Expected 250 each [1]. χ² = 179.8 + 169.7 + 160.0 + 150.5 = 660 [1]. df = 3, critical value 7.81, so significant [1]; reject the null hypothesis; the genes are linked [1]. [9] Examiner insight: Show the expected value for a test cross as 1 : 1 : 1 : 1 and finish with a biological conclusion about linkage.

10. (a) pp is epistatic: it prevents pigment whatever the R locus holds [1]. PpRr × PpRr gives 9 P_R_ dark [1] and 3 P_rr pale [1]; the 3 ppR_ and 1 pprr merge into 4 white [1]. (b) Expected 135, 45, 60 [1]. χ² = 529/135 + 169/45 + 100/60 = 3.919 + 3.756 + 1.667 = 9.34 [1]. df = 2, critical value 5.99, so significant [1]; reject the null hypothesis: the results do not fit 9 : 3 : 4 [1]. [8] Examiner insight: The merged class must be explained by genotype; saying “recessive epistasis” without showing which genotypes combine tends to leave marks unclaimed.

11. (a) 12 white : 3 yellow : 1 green [1]; 216, 54 and 18 [1]. (b) χ² = 121/216 + 25/54 + 36/18 = 0.560 + 0.463 + 2.000 = 3.02 [1]. df = 2, critical value 5.99 [1]. Not significant, so the results fit 12 : 3 : 1 [1]. (c) W masks the effect of the Y locus [1]: white plants (9 W_Y_ and 3 W_yy) look the same whatever the Y alleles [1]. [7] Examiner insight: Epistasis answers are strongest when they say one gene masks the expression of another gene, rather than one allele dominating another.

Where marks are usually lost

• Missing the key or writing a diagram without parent genotypes.
• Writing two alleles of the same gene in one gamete.
• Giving a male two X-linked alleles.
• Reading a 1 : 2 : 1 ratio as dominance.
• Using percentages, or leaving out a class, in chi-squared.
• Giving degrees of freedom as the number of categories.
• Writing “accepted” or “proved” instead of “not rejected”.
• Not stating what a significant result suggests, such as linkage or epistasis.

Next steps

• Revise: revision notes and study guide
• Course: OxfordAQA A-level Biology hub and checklist
• Check your level with the free diagnostics
• Book a free trial class: trial

Official syllabus

Source: OxfordAQA, International AS and A-level Biology (9610), Version 5.1, International AS exams May/June 2017 onwards and International A-level exams May/June 2018 onwards, sections 3.3.6.1 and 3.3.6.2.