Looking for classes? Ksquare Career Institute, Bengaluru →
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 -
A
0.24
B
0.16
C
0.48
D
0.36
Detailed Solution
According to the Hardy-Weinberg principle, in a large, randomly mating population at equilibrium, $p^2 + 2pq + q^2 = 1$.
Here $p$ = frequency of allele A = 0.6 and $q$ = frequency of allele a = 0.4.
$p^2$ = frequency of AA, $q^2$ = frequency of aa and $2pq$ = frequency of the heterozygotes Aa.
$2pq = 2 \times 0.6 \times 0.4$
$2pq = 0.48$
(Check: $p^2 = 0.36$, $q^2 = 0.16$, and $0.36 + 0.48 + 0.16 = 1$.)
Here $p$ = frequency of allele A = 0.6 and $q$ = frequency of allele a = 0.4.
$p^2$ = frequency of AA, $q^2$ = frequency of aa and $2pq$ = frequency of the heterozygotes Aa.
$2pq = 2 \times 0.6 \times 0.4$
$2pq = 0.48$
(Check: $p^2 = 0.36$, $q^2 = 0.16$, and $0.36 + 0.48 + 0.16 = 1$.)