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 2

In Figure 4, flies numbered 3 and 4 produce fly numbered 9, which is black-bodied. This indicates that the parental flies must have at least one grey-bodied parent, demonstrating that the grey body allele is dominant over the black body allele.

In Figure 4, fly numbered 5 is a grey-bodied male and produces both grey-bodied and black-bodied offspring (numbered 9 and 10) with a female that must carry the recessive allele for black body. If body colour were on the X chromosome, we would not observe male offspring with different body colours.

Phenotypes of parents: Grey-bodied, white-eyed female × Black-bodied, red-eyed male. Genotypes of parents: $GgX^r × ggY$.

Genotypes of offspring: $GgX^r, GgY, ggX^r, ggY$.

Phenotypes of offspring: Grey-bodied, white-eyed; Grey-bodied, red-eyed; Black-bodied, white-eyed; Black-bodied, red-eyed.

Ratio of phenotypes: 1:1:1:1.

Given that 64% of the flies are grey-bodied, it indicates that they include both homozygous dominant ($Gg$) and heterozygous ($GG$) flies. Using the Hardy-Weinberg equation, we have: Let $p$ be the frequency of the dominant allele (G) and $q$ the frequency of the recessive allele (g); thus, $p^2 + 2pq + q^2 = 1$, where $p^2$ is the homozygous dominant proportion, $2pq$ is the heterozygous proportion, and $q^2$ is the homozygous recessive proportion.

Here, $q^2 = 0.36 ightarrow q = 0.6$, therefore, $p = 1 - q = 0.4$. The percentage of heterozygous flies (2pq) is calculated as: $2pq = 2(0.4)(0.6) = 0.48$ So, the percentage of heterozygous flies is $48\%$.