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 random mating population at equilibrium?
(1) 0.48
As per NCERT, Hardy–Weinberg equilibrium gives p2 + 2pq + q2 = 1, where 2pq is the frequency of heterozygotes.
Here p = 0.6, q = 0.4, so
2pq = 2 0.6 0.4
= 0.48
∴ Frequency of heterozygotes (Aa) = 0.48
Answer: (1)
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