On sampling a population, the frequency of homozygous recessive genotype (aa) is 36%. Calculate the following:
i) The frequency of dominant allele
ii) The frequency of dominant homozygous genotype
iii) The frequency of heterozygous genotype
A
i - 0.6, ii - 0.12, iii - 0.36
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B
i - 0.4, ii - 0. 16, iii - 0.48
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C
i - 0.4, ii - 0.9, iii - 0.54
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D
Data insufficient to calculate
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Solution
The correct option is B i - 0.4, ii - 0. 16, iii - 0.48 According to Hardy-Weinberg equation, p2+2pq+q2=1andp+q=1
Assuming that ‘p’ is the frequency of occurrence of the dominant allele ‘A’ and ‘q’ is the frequency of occurrence of the recessive allele ‘a’, we could say that the frequency of occurrence of the genotype aa=36% ⇒q×q=36 ⇒q2=0.36 ⇒q=√0.36 ⇒q=0.6
Therefore the frequency of recessive allele a = 0.6
Applying the equation p+q = 1 ⇒p=1−q ⇒p=1−0.6=0.4
Therefore, the frequency of dominant allele A = 0.4
The frequency of the homozygous genotype (AA)=p2 p2=(0.4×0.4)=0.16
The frequency of the heterozygous genotype (Aa)=2pq 2pq=2×0.4×0.6=0.48