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Question

# In a human population at Hardy-Weinberg equilibrium, the recessive gene is homozygous in 64% of the population. What will be the frequency of heterozygotes in the population?

A
23%
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B
32%
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C
96%
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D
64%
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Solution

## The correct option is B 32%The frequency of recessive gene is 64% that is 0.64, and it is represented as q2 q= √q2 = √0.64= 0.8 Since p + q = 1 p + 0.8 = 1 p = 1 - 0.8 = 0.2 Using the Hardy-Weinberg equation p2+2pq+q2=1, heterozygotes are represented 2pq Therefore, 2pq= 2×0.8×0.2 2pq = 0.32 or 32% So, 32% of the individuals in the population are heterozygous for the given gene.

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