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Question
A population of 200 fruit flies is in Hardy Weinberg equilibrium. The frequency of the allele (a) 0.4. Calculate the following:
Frequency of the allele (A).
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Solution
In Hardy Weinberg, p2 + 2pq + q2 = 1, where 'p2' is the frequency of the homozygous dominant genotype (AA), '2pq' is the frequency of the heterozygous genotype (Aa), and 'q2' is the frequency of the homozygous recessive genotype (aa).
Given:
q = 0.4
We know p + q = 1
p = 1 − q
= 1 − 0.4
= 0.6
Frequency of allele, p(A) = 1 − 0.4
= 0.6
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