Question

In a population of 1000 individuals, 360 have genotype $AA$, 480 have $Aa$ and the remaining 160 have $aa$. 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

According to Hardy-Weinberg principle, $(p+q{)}^2=p^2+2pq+q^2=1$ where, $p=$ The frequency of allele ${}^{\prime }{A}^{\prime },$ $q=$ The frequency of allele $‘a^{\prime },$
$p^2=$ The frequency of individual ‘AA,’ $q^2=$ The frequency of individual ${}^{\prime }a{a}^{\prime },$ $2pq=$ The frequency of individual $Aa$ $\left(AA\right)p^2=360$ out of 1000 individual or $p^2=0.36$
So, $p=\sqrt{0.36}=0.6$. So, the frequency of allele $A$ in the population is 0.6.