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?
- 0.48
- 0.24
- 0.36
- 0.16
Answer
(1) 0.48
Solution
As per NCERT, Hardy–Weinberg equilibrium: p2 + 2pq + q2 = 1, where 2pq is the frequency of heterozygotes.
⇒ p = 0.6, q = 0.4
⇒ 2pq = 2 0.6 0.4
= 0.48
∴ Frequency of heterozygotes (Aa) = 0.48
Answer: (1)