Dihybrid crosses, autosomal linkage and epistasis - AQA A-Level Biology (Paper 2) Exam Questions

Practise 19 exam-style questions on Dihybrid crosses, autosomal linkage and epistasis for AQA A-Level Biology (Paper 2). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with the Study Helper on hand if you get stuck.

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19 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

In chickens, a dominant allele, I, of one gene prevents feather pigment from forming, so chickens with at least one I allele have white feathers. A second gene, on a different chromosome, also affects pigment: chickens with the genotype cc cannot make pigment and also have white feathers. Only chickens with the genotype ii and at least one C allele have coloured feathers.

Two chickens that were heterozygous for both genes (IiCc) were crossed.

Calculate the ratio of white chickens to coloured chickens expected among the offspring. Show your working.

Model answer

Coloured chickens must be ii (probability 1/4) and have at least one C (probability 3/4).
Probability coloured = 1/4 × 3/4 = 3/16
Probability white = 1 - 3/16 = 13/16
Ratio white : coloured = 13 : 3

Explanation

Work out the rarer phenotype first: only iiC_ chickens are coloured, which is 3 of the 16 combinations. Every other combination (I_C_, I_cc and iicc) is white, so 9 + 3 + 1 = 13 are white. Two genes affecting the same characteristic in this way is another type of epistasis, giving 13 : 3 instead of 9 : 3 : 3 : 1.

Mark scheme

1 mark
Coloured chickens (iiC_) = 1/4 × 3/4 = 3/16 (so white = 13/16); or a correct final answer of 13 : 3 with no working
A correct answer of 13 : 3 (white : coloured) with no working scores 2 marks. Accept for 1 mark: 3 : 13 (inverted) OR 3/16 coloured or 13/16 white worked out correctly but no final ratio OR a 16-box grid or list of genotypes with one error that leads to an incorrect ratio.
1 mark
13 : 3 (white : coloured)
Accept 13/16 white and 3/16 coloured stated as a ratio. Reject 12 : 3 : 1 or 9 : 3 : 4.

Question 2

3 marks

Explain what is meant by each of the following terms:
- dihybrid cross
- recombinant offspring
- epistasis.

Model answer

A dihybrid cross is a cross that follows the inheritance of two different genes at the same time.
Recombinant offspring have a combination of alleles that was not present together in either parent, produced by crossing over (or by independent segregation).
Epistasis is when the alleles of one gene mask or affect the expression of the alleles of a different gene.

Explanation

A monohybrid cross follows one gene; a dihybrid cross follows two genes, each with its own alleles. Recombinant offspring are those with new combinations of alleles: for linked genes they come from gametes produced by crossing over, so they are usually the smallest classes in a test cross. Epistasis is an interaction between two different genes; do not confuse it with dominance, which is between alleles of the same gene, or with epigenetics.

Mark scheme

1 mark
Dihybrid cross: a cross involving / following the inheritance of two genes (two characteristics controlled by different genes)
Accept 'a cross between individuals that are heterozygous for two genes'. Reject 'a cross involving two alleles'.
1 mark
Recombinant offspring: offspring with a combination of alleles/characteristics not found together in either parent (or on either chromosome of a parent), produced by crossing over / independent segregation
Accept 'offspring with new combinations of alleles produced by crossing over'. Insufficient: 'offspring that are different from their parents'.
1 mark
Epistasis: (the alleles of) one gene affect / mask / prevent the expression of (the alleles of) another gene
Accept 'interaction between two genes in which one gene controls whether the other is expressed'. Reject 'one allele masks another allele of the same gene' (dominance) and 'epigenetics'.

Question 3

5 marks

In guinea pigs, the allele for black coat (B) is dominant to the allele for brown coat (b), and the allele for short hair (L) is dominant to the allele for long hair (l). The two genes are on different chromosomes.

A guinea pig that is heterozygous for both genes was crossed with a brown, long-haired guinea pig.

Give the genotypes of the parents, the gametes, and the genotypes, phenotypes and ratio of the offspring expected from this cross.

Model answer

Parents: BbLl (black, short) × bbll (brown, long)
Gametes: BL, Bl, bL, bl and bl
Offspring genotypes: BbLl, Bbll, bbLl, bbll
Phenotypes: black short-haired, black long-haired, brown short-haired, brown long-haired
Ratio: 1 : 1 : 1 : 1

Explanation

The genes are on different chromosomes, so the double heterozygote produces four types of gamete in equal numbers by independent segregation. The homozygous recessive parent produces only bl gametes, so the offspring phenotypes show directly which alleles came from the heterozygous parent: 1 : 1 : 1 : 1. Each gamete must contain one allele of each gene. A common error is to give the 9 : 3 : 3 : 1 ratio, which comes from crossing two double heterozygotes.

Mark scheme

1 mark
Parents: BbLl × bbll
Max 2 if the genotypes given for the parents are wrong. In that case award only point 2 (gametes that follow correctly from the student's parental genotypes) and point 3 (offspring genotypes that follow correctly from the student's gametes). If the parents' genotypes are correct but the gametes are wrong, point 3 may still be awarded for offspring genotypes that follow correctly from the student's gametes. Points 4 and 5 are not consequential: they need the correct phenotypes and ratio for this cross. Accept the alleles within a genotype in any order (for example bBlL), and a Punnett square written out in words or rows.
1 mark
Gametes: BL, Bl, bL and bl (from BbLl) AND bl (from bbll)
Reject gametes containing two alleles of the same gene (e.g. Bb) or only one allele.
1 mark
Offspring genotypes: BbLl, Bbll, bbLl and bbll
1 mark
Phenotypes: black short-haired, black long-haired, brown short-haired and brown long-haired
Both characteristics must be given for each phenotype, matched to the genotypes or in the same order.
1 mark
Ratio: 1 : 1 : 1 : 1
Accept 25% of each. Reject 9 : 3 : 3 : 1.

Question 4

6 marks

A plant breeder wants to produce a variety of a crop plant that is resistant to a fungal disease and flowers early. The allele for resistance (R) is dominant to the allele for being susceptible (r), and the allele for early flowering (E) is dominant to the allele for late flowering (e). The genes are on different chromosomes.

The breeder crossed two plants that were both heterozygous for both genes (RrEe).

Calculate the probability that an offspring plant has the genotype RREE.

The breeder chose one offspring plant that was resistant and flowered early. Calculate the probability that this plant has the genotype RREE.

Give the genotype of the plant that the breeder should cross with this plant to find out whether it has the genotype RREE.

The breeder carried out this cross. All 8 offspring were resistant and flowered early. The breeder concluded that the chosen plant has the genotype RREE. Evaluate this conclusion.

Model answer

Probability of RR = 1/4 and of EE = 1/4, so probability of RREE = 1/4 × 1/4 = 1/16

Resistant, early-flowering plants are R_E_: 9 of the 16 combinations, and only 1 of these is RREE, so probability = 1/9

The breeder should cross it with an rree plant.

The result supports the conclusion: an RREE plant gives only RrEe offspring, which are all resistant and early flowering, and no offspring showed a recessive phenotype. However, there were only 8 offspring. A plant with one recessive allele (RREe or RrEE) produces a resistant, early-flowering offspring with a probability of 1/2 each time, so the probability of 8 such offspring is (1/2)⁸ = 1/256 (about 0.004). The conclusion is very likely to be correct, but it is not certain; more offspring would make it more certain.

Explanation

Once you know a plant has both dominant phenotypes, the probability is out of those 9 combinations, not out of 16 - this is why 1/9 differs from 1/16. A test cross with a double homozygous recessive plant reveals the genotype, because any recessive allele in the tested plant can show up in the offspring. With a small number of offspring, a heterozygote could produce only dominant offspring by chance: multiply the probabilities to see how likely this is.

Mark scheme

1 mark
1/16 / 0.0625 (a correct answer with no working scores this mark)
Points 1 and 2 are calculations: a correct answer of 1/16 with no working scores point 1, and a correct answer of 1/9 with no working scores point 2. Accept 6.25% or 1 in 16.
1 mark
1/9 / 0.11 (a correct answer with no working scores this mark)
Accept 0.111 or 11%. Do not accept 1/16 here: the plant is already known to be resistant and early flowering (9 equally likely combinations, one of which is RREE).
1 mark
rree / (a plant) homozygous recessive for both genes
Accept 'a susceptible, late-flowering plant'. Points 4 to 6 evaluate the conclusion.
1 mark
(Supports) an RREE plant gives only RrEe (resistant, early-flowering) offspring, and a plant with an r or e allele could have given some susceptible or late-flowering offspring, but none appeared
Accept 'no offspring had a recessive phenotype, as expected for RREE'.
1 mark
(Limitation) there were only 8 offspring, so a plant with one recessive allele (RREe or RrEE) could have produced 8 resistant, early-flowering offspring by chance
Insufficient: 'more offspring are needed' with no reason.
1 mark
The probability of this for an RREe or RrEE plant is (1/2)⁸ = 1/256 / 0.0039 (and (1/4)⁸ for RrEe), so the conclusion is very likely, but not certainly, correct
Accept 0.004 or 0.39%. Both a correct probability and a judgement are needed.

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