1.

Question : In population of 1000 individuals 360 belongs to aa. Based on this date, frequency of allele A in the population is

Answer»

0.4
0.5
0.6
0.7

Solution :According to hardy-Weinberg principle
`(p+Q)^(@)=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 individual 'Aa'
`(A A)[^(2)=36` out of 100.
`q^(2)=160` out 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.


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