Question

In the peppered moth (Biston betularia), black individuals may be either homozygous (A₁A₁) or heterozygous (A₁A₂), whereas pale gray moths are only homozygous (A₂A₂). Suppose that in a sample of 250 moths from one locality, 108 are black and 142 are gray.

b) Assuming that the locus is in Hardy-Weinberg equilibrium, what are the allele frequencies? Show your work below:

c) Under this assumption, what proportion of the sample is heterozygous? What is the number of heterozygotes?

Answer 1

Here, the black moth colour is the dominant trait.
So, 'A1' is the dominant allele responsible for the black colour and 'A2' is the corresponding recessive allele responsible for the grey colour.
Then, the following are the possible genotypes and corresponding phenotypes -
A1A1 = black color
A1A2 = black color
A2A2 = gray color

Now, let 'p' be the frequency of the dominant allele and 'q' be the frequency of the recessive allele,
Also, 'p2' is the frequency of the homozygous dominant individuals (A1A1)
'q2' is the frequency of the homozygous recessive individuals (A2A2)
and '2pq' is the frequency of heterozygous individuals (A1A2).

Here, p2 + 2pq = 108/250 = 0.432
and q2 = 142/250 = 0.568
So, q = square root of 0.568 = 0.754
Now, according to the hardy weinberg equilibrium, we know that - p + q = 1.
So, p = 1 -q = 1 - 0.754 = 0.246
Hence, p = 0.246 and q = 0.754 .

b)
Allele frequency of A1 allele = 0.246
Allele frequency of A2 allele = 0.754

c)
Frequency of heterzygotes (2*A1A2) = 2*0.754*0.246 =0.37
So, proportion of the heterozygotes = 0.37
Number of heterozygotes = 0.37*250 = 96 (approx)

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