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

From Figure 4, it can be observed that flies 3 and 4, which are grey-bodied, produced offspring (fly 6) with black-bodied phenotype only when paired with the black-bodied fly. This demonstrates that the grey body allele (G) is dominant over the black body allele (g).

In Figure 4, both male and female flies can have grey bodies, as evidenced by flies 1, 2, and 3. If the body colour gene were on the X chromosome, males would not exhibit the recessive trait (black body) as homozygous. Since males can be grey-bodied, it indicates that the gene is autosomal, not linked to the X chromosome.

Genotypes of parents: GgXr × ggXY

Genotypes of offspring: 1 GgXRY (Grey-bodied, red-eyed) 1 GgXrY (Grey-bodied, white-eyed) 1 GgXRY (Grey-bodied, red-eyed) 1 ggXRY (Black-bodied, red-eyed) 1 ggXrY (Black-bodied, white-eyed)

Phenotypes of offspring:

• 3 Grey-bodied (1 GgXRY + 1 GgXrY)
• 2 Black-bodied (1 ggXRY + 1 ggXrY)

Ratio of phenotypes: Grey-bodied : Black-bodied = 3:2.

Given that 64% (or 0.64) of the flies are grey-bodied, we can denote the frequency of the dominant allele (p) as follows:

Using the equation: $p^2 + 2pq + q^2 = 1$

Where:

• $p^2$ = frequency of homozygous grey-bodied (G)
• $2pq$ = frequency of heterozygous (Gg)
• $q^2$ = frequency of homozygous black-bodied (gg)

Let’s set $q^2 = 0.36$ (since 1 - 0.64 = 0.36):

Now, taking the square root gives us: q = rac{1}{q} = rac{0.6}{1}

Then, $p = 1 - q = 0.4$.

Finally, to find the heterozygous genotype ( hence, $2pq = 2(0.4)(0.6) = 0.48$) which shows that 48% of the flies are expected to be heterozygous for gene G.