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Question:

At a particular locus, frequency of 'A' allele is 0.6 and that of 'a' is 0.4. What would be the frequency of heterozygotes in a randomly mating population at equilibrium?

0.24

0.48

0.16

0.36

Solution:

Hardy-Weinberg equation is p² + 2pq + q² = 1. Here p is the frequency of the "A" allele and q is the frequency of the "a" allele in the population. Hence, frequency of heterozygous genotype = 2pq = 2 x 0.6 x 0.4 = 0.48. So, the correct answer is 0.48