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In a population in Hardy- Weinberg equilibrium 64% of the individuals show recessive trait. What is the percentage of heterozygotes in the individuals?

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40%

32%

20%

answer is C.

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Detailed Solution

$q^{2} = \frac{64}{100} = 0.64$

p + q = 1

q = 0.8
p = 0.2

No. of Heterozygotes $2 pq = 2 \times 0.8 \times 0.2$

$= 0.32 = \frac{32}{100}$

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