A) 0.4
B) 0.5
C) 0.6
D) 0.7
Correct Answer: C
Solution :
According to Hardy-Weinberg principle \[{{(p+q)}^{2}}={{p}^{2}}+2pq+{{q}^{2}}=1\] 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 (AA) \[{{P}^{2}}\] 360 out of 1000 individual. or \[{{p}^{2}}=36\] 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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