Question

In a population of 1000 individuals 360 belongs 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

According to Hardy - Weinberg principle
$(p+q{)}^2=p^2+3pq+q^2=1$
where, p = The frequency of allele 'A'
q = The frequency of allele 'a'
$p^2$ = The Frequency of individual 'AA'
$q^2$ = The frequency of individual 'aa'
2pq=The frequency of indi vidual Aa
(AA)$P^2$ = 360 out of 1000 individual
or $P^2$ =36 out of 100.
$q^2$ = 160 our of 1000 or $q^2$ = 16 out of 100.
So , q =$\sqrt{0}.16=0.4$
As p + q = 1
So, p is 0.6