Question

In a population of 1000 individuals, 360 belong to genotype AA, 480 to Aa, and the remaining 160 to aa. Based on this data, the frequency of allele A in the population is:

• A. 0.4
• B. 0.5
• C. 0.6
• D. 0.7

Answer 1

The correct option is C 0.6

Applying the Hardy-Weinberg binomial expression to the above given
$p^2+2\mathrm{pq}+q^2=1$
$\mathrm{AA}+\mathrm{Aa}+\mathrm{aa}=1$
$360+480+160=1000$
$0.36+0.48+0.16=1$
i.e. ${(0.6)}^2+2\times 0.6\times 0.4+{(0.4)}^2=1$
Thus, $p^2=0.36={(0.6)}^2$
Hence p = 0.6. The frequency of the allele A in the population is 0.6 or 60%.