AQA A-level Biology: Inheritance and Chi-squared (7402) – 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 follow AQA A-level Biology (7402) section 3.7.1, Inheritance, from the AQA AS and A-level Biology specification for first teaching 2015. 3.7.1 is A-level only content (not AS), assessed in A-level Paper 2 and Paper 3. Explanations are in the study guide and the revision notes. Course pages: AQA A-level Biology hub, the printable checklist and the free diagnostics. Where a question gives alleles, critical values or data, use only those. Marks are a practice guide, not an official tariff.

Questions

1. (a) State what is meant by the terms genotype and phenotype. [2] (b) Two clones of a plant have the same alleles. One is grown in full light and one in shade, and the shaded plant is paler. Explain the difference. [1]

2. In a freshwater fish, scale colour is controlled by one gene. Allele Fᴳ gives gold scales and Fᴮ gives blue scales. They are codominant, and heterozygotes have gold-and-blue patched scales. (a) Draw a genetic diagram for a cross between two patched fish. [4] (b) Give the numbers of each phenotype expected among 200 offspring. [1]

3. Wing colour in a butterfly has three alleles. Cᵇ (brown) is dominant to Cʸ (yellow), which is dominant to Cʷ (white). A brown butterfly heterozygous for brown and white is crossed with a yellow butterfly heterozygous for yellow and white. (a) Give the genotypes and phenotype ratio of the offspring. [3] (b) What is the probability that a brown offspring carries the yellow allele? [1]

4. In a gourd, disc fruit (D) is dominant to sphere (d), and striped skin (S) is dominant to plain (s). The genes are on different chromosomes. A plant DdSs is crossed with a plant ddss. (a) State the expected phenotype ratio and the numbers among 480 offspring. [2] (b) Two DdSs plants are crossed. Calculate the probability that an offspring has disc fruit and plain skin. [2] (c) Explain why the ddss plant is used in the first cross. [1]

5. In fruit flies, red eyes (Xᴿ) are dominant to white eyes (Xʳ). The gene is on the X chromosome. (a) Draw a genetic diagram for a cross between a heterozygous red-eyed female and a white-eyed male. [4] (b) A red-eyed male is crossed with a white-eyed female. Predict the eye colour of the sons and daughters and explain. [2]

6. In chickens, females are ZW and males ZZ. Barred plumage (Zᴮ) is dominant to non-barred (Zᵇ). (a) Draw a genetic diagram for a cross between a heterozygous barred male and a non-barred female. [4] (b) Explain why no female chicken can be heterozygous for this gene. [1]

7. In a beetle, dark elytra (D) are dominant to pale (d), and long antennae (A) are dominant to short (a). A heterozygous beetle was test crossed with a pale, short-antennae beetle. Of 700 offspring: 301 dark long, 288 pale short, 56 dark short and 55 pale long. (a) Identify the parental classes and give the arrangement of alleles in the heterozygous parent. [2] (b) Calculate the percentage of recombinant offspring. [1] (c) Explain how the recombinants arose. [2] (d) A chi-squared test against 1 : 1 : 1 : 1 gives 327 (degrees of freedom 3, critical value 7.81 at p = 0.05). State the conclusion. [2]

8. Flower colour in a plant needs two enzymes in sequence: a colourless precursor is converted by enzyme 1 (made from allele C) to a colourless intermediate, then by enzyme 2 (made from allele P) to purple pigment. Alleles c and p give no functional enzyme. Two plants CcPp are crossed. (a) Explain why 7 of every 16 offspring are expected to be white. [3] (b) Give the expected numbers among 320 offspring. [1] (c) Name the type of gene interaction shown. [1] (d) Explain why the ratio is not 9 : 3 : 3 : 1. [1]

9. A cross between two heterozygous gourd plants (two genes, independent assortment) gave 640 offspring: 352 disc striped, 135 disc plain, 108 sphere striped and 45 sphere plain. Use a chi-squared test (critical value 7.81 at p = 0.05, 3 degrees of freedom) to test whether the results fit 9 : 3 : 3 : 1. [6]

10. A student test crossed a heterozygous plant with a homozygous recessive plant and expected 1 : 1. The 160 offspring were 87 with the dominant phenotype and 73 with the recessive phenotype. (a) Calculate χ². [2] (b) Use 3.84 as the critical value (p = 0.05) and state the conclusion. [2] (c) The student converted the counts to percentages before testing. Explain why this is not valid. [1]

11. Plan an investigation, using fruit flies, to find out whether two genes are linked. Include how you would analyse the results. [8]

Answers

1. (a) Genotype: the genetic constitution of an organism (its alleles) [1]; phenotype: the expression of the genetic constitution and its interaction with the environment [1]. (b) The genotype is the same, so the difference is caused by the environment (light affects pigment or growth) [1]. Examiner insight: Define phenotype as the expression of the genes plus the environment, because an answer that gives only appearance leaves out why identical clones can differ.

2. (a) Parental genotypes FᴳFᴮ × FᴳFᴮ, with a key [1]. Gametes Fᴳ and Fᴮ from each [1]. Offspring FᴳFᴳ, FᴳFᴮ (twice), FᴮFᴮ [1]. Phenotypes gold : patched : blue = 1 : 2 : 1 [1]. (b) 50 gold, 100 patched, 50 blue [1]. Examiner insight: A key and labelled rows (parents, gametes, offspring) are the easiest way to show the examiner each step of a diagram.

3. (a) Gametes Cᵇ, Cʷ and Cʸ, Cʷ [1]. Offspring CᵇCʸ, CᵇCʷ, CʸCʷ, CʷCʷ [1]. Phenotypes 2 brown : 1 yellow : 1 white [1]. (b) 1/2 (CᵇCʸ is one of the two brown genotypes) [1]. Examiner insight: With three alleles, write the dominance order in your key; the ratio is 2 : 1 : 1 because two genotypes share one phenotype.

4. (a) The ddss parent makes only ds gametes, so the ratio is 1 : 1 : 1 : 1 [1] with 120 of each phenotype (disc striped, disc plain, sphere striped, sphere plain) [1]. (b) P(disc) = 3/4 and P(plain) = 1/4 [1]. 3/4 × 1/4 = 3/16 [1]. (c) It produces only recessive gametes, so the phenotype of each offspring shows which alleles came from the other parent [1]. Examiner insight: For probability, show each single-gene probability and the multiplication so that a slip in the final fraction is still easy to follow.

5. (a) Parents XᴿXʳ × XʳY [1]. Gametes Xᴿ, Xʳ and Xʳ, Y [1]. Offspring XᴿXʳ, XʳXʳ, XᴿY, XʳY [1]. Red female, white female, red male, white male = 1 : 1 : 1 : 1 [1]. (b) Daughters all XᴿXʳ, red eyed [1]. Sons all XʳY, white eyed, as a son receives the Y from his father and his only X from his mother [1]. Examiner insight: Write X and Y in the genotypes rather than just letters, so it is clear the trait is sex-linked.

6. (a) Parents ZᴮZᵇ × ZᵇW [1]. Gametes Zᴮ, Zᵇ and Zᵇ, W [1]. Offspring ZᴮZᵇ, ZᵇZᵇ, ZᴮW, ZᵇW [1]. Barred male, non-barred male, barred female, non-barred female = 1 : 1 : 1 : 1 [1]. (b) A female has only one Z chromosome, so she has only one allele of this gene [1]. Examiner insight: Read which sex has the two different sex chromosomes before you start; in birds it is the female, the reverse of humans.

7. (a) Parental classes: dark long and pale short [1]. The heterozygote has D and A on one chromosome and d and a on the homologue [1]. (b) (56 + 55) ÷ 700 × 100 = 15.9% [1]. (c) Crossing over between chromatids of homologous chromosomes in prophase I of meiosis [1], exchanging alleles to give gametes with new combinations (Da and dA) [1]. (d) 327 is greater than 7.81, so the difference from 1 : 1 : 1 : 1 is significant, so the null hypothesis is rejected [1]. The genes do not assort independently, which is consistent with linkage [1]. Examiner insight: Name both the pattern in the data (large parental classes) and the cause (crossing over) when asked to explain linkage.

8. (a) Purple needs at least one C and one P, which happens in 9 of 16 offspring [1]. The other 7 (3 C_pp, 3 ccP_, 1 ccpp) lack a functional enzyme [1], so the pathway is blocked and no pigment forms [1]. (b) 180 purple and 140 white [1]. (c) Epistasis [1]. (d) Alleles at one locus affect expression of the other locus, so the phenotype classes merge [1]. Examiner insight: Link the genotypes to the pathway: examiners look for the missing enzyme and the blocked step, not only the counting of boxes.

9. Null hypothesis: any difference between the counts and the 9 : 3 : 3 : 1 expectation is due to chance [1]. Expected: 360, 120, 120, 40 [1]. χ² = (352−360)²/360 + (135−120)²/120 + (108−120)²/120 + (45−40)²/40 [1] = 0.178 + 1.875 + 1.200 + 0.625 = 3.88 [1]. Degrees of freedom = 3 and 3.88 < 7.81 [1], so the difference is not significant: the results fit 9 : 3 : 3 : 1 and any difference is probably due to chance [1]. Examiner insight: Finish with a conclusion about the genetics; a bare comparison of two numbers is less convincing.

10. (a) Expected 80 and 80 [1]. χ² = 49/80 + 49/80 = 1.23 [1]. (b) Degrees of freedom = 1 and 1.23 < 3.84 [1], so the difference is not significant and the results fit a 1 : 1 ratio [1]. (c) The value of χ² depends on the number of individuals counted, and percentages hide the sample size [1]. Examiner insight: Give the expected values from the total and the ratio, not the observed counts.

11. Any eight of: cross a true-breeding (homozygous) dominant fly with a homozygous recessive fly to obtain heterozygous offspring [1]; test cross the heterozygous females with double-recessive males [1]; rear all flies under the same conditions, such as temperature and food, so phenotype reflects genotype [1]; count offspring in each of the four phenotype classes [1]; use a large sample or replicate crosses [1]; state the null hypothesis of independent assortment, expected 1 : 1 : 1 : 1 [1]; calculate χ² with 3 degrees of freedom and compare with 7.81 at p = 0.05 [1]; if significant and the parental classes are larger, the genes are linked and recombinants arose by crossing over [1]. Examiner insight: A plan scores better when it names the cross, what is counted, the expected ratio and how the result is interpreted, in that order.

Where marks are usually lost

• Omitting the key, or using different letters for alleles of the same gene.
• Writing gametes with two alleles of the same gene.
• Stating a ratio without the phenotypes.
• Treating X-linked and autosomal crosses alike; sons and daughters can differ.
• Using observed values as expected values in chi-squared.
• Taking degrees of freedom as sample size minus one.
• Concluding that a non-significant result proves a ratio.
• Describing epistasis as dominance between genes instead of interaction between loci.

Next steps

• Inheritance and chi-squared study guide
• Inheritance and chi-squared revision notes
• AQA A-level Biology hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

AQA AS and A-level Biology specification (7401 and 7402), for first teaching 2015, section 3.7.1 (version 1.5), published by AQA.