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Evolution

Question
CBSEENBI12002364
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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

  • 0.4

  • 0.5

  • 0.6

  • 0.7

Solution

C.

0.6

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 = 0.161/2 = 0.4
As p+q = 1
so, p is 0.6

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