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In fruit flies, a gene for body colour has a dominant allele for grey body, G, and a recessive allele for black body, g - AQA - A-Level Biology - Question 6 - 2019 - Paper 1
Question 6
In fruit flies, a gene for body colour has a dominant allele for grey body, G, and a recessive allele for black body, g. A gene for eye colour has a dominant allele... show full transcript
Worked Solution & Example Answer:In fruit flies, a gene for body colour has a dominant allele for grey body, G, and a recessive allele for black body, g - AQA - A-Level Biology - Question 6 - 2019 - Paper 1
Step 1
Give the full genotype of the fly numbered 6 in Figure 4.
Answer
The full genotype of the fly numbered 6 is GgXrY. This indicates that the fly has a grey body (due to the dominant allele G) and red eyes (due to the dominant allele R), with the presence of the Y chromosome indicating that it is male.
Step 2
Give one piece of evidence from Figure 4 to show that the allele for grey body colour is dominant.
Answer
In Figure 4, the offspring numbered 3 and 4 both display a grey body phenotype despite being produced from a black-bodied parent (fly 2). This demonstrates that the grey body allele (G) is dominant, as it overrides the black body allele (g).
Step 3
Explain one piece of evidence from Figure 4 to show that the gene for body colour is not on the X chromosome.
Answer
Evidence from Figure 4 includes that both male and female offspring exhibit a black body phenotype, which is observed in flies 2, 6, and 9. If the gene for body colour were on the X chromosome, we would expect to see different inheritance patterns between males and females, specifically that all male offspring would show the same phenotype influenced by the mother's X-linked genes.
Step 4
Step 5
Use the Hardy–Weinberg equation to calculate the percentage of flies heterozygous for gene G.
Answer
Given that 64% of the population are grey-bodied flies (dominantly expressing gene G), we denote the frequency of the dominant allele as p. Hence, p² + 2pq + q² = 1. Here, p² represents the homozygous dominant (64%), so p = √0.64 = 0.8. Therefore, q = 1 - p = 0.2. The frequency of heterozygous individuals is given by 2pq = 2(0.8)(0.2) = 0.32, which corresponds to 32% of the population being heterozygous for gene G.