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Q. 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 :-

AIPMTAIPMT 1980Evolution

Solution:

According to Hardy-Weinberg principle (p+q)2 = p2 + 2pq + q2 = 1
where, p = The frequency of allele 'A'
q = The frequency of allele 'a'
p2 = The frequency of individual 'AA'
q2 = The frequency of individual 'aa'
2pq = The frequency of individual Aa (AA) P2 360 out of 1000 individual. or p2=36 out of 100.
q2 = 160 out of 1000 or q2 = 16 out of 100.
So, $q = \sqrt{0.16} = 0.4 $
As p + q = 1 So, p is 0.6