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 of equilibrium?

• A. $0.16$
• B. $0.36$
• C. $0.48$
• D. $0.24$

Answer 1

The correct option is C $0.48$

Hardy-Weinberg equation is $p^2+2pq+q^2=1$. 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 option C.