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Asked by meenatoofansingh3 | 30 Apr, 2021, 16:14: PM

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According to Hardy-Weinberg equilibrium,

p $^{2}$ +2pq + q $^{2}$ = 1

where p $^{2}$ is the frequency of homozygous genotype AA

q $^{2}$ is the frequency of genotype aa

pq is the frequency of genotype Aa

There are 1000 individuals, out of which 360 belong to genotype AA.

p $^{2}$ =360/1000 = 0.36

Then p = 0.6.

Answered by Sheetal Kolte | 03 May, 2021, 13:17: PM

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NEET neet - Biology

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Asked by meenatoofansingh3 | 30 Apr, 2021, 16:14: PM

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NEET neet - Biology

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