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In a population of 800 individuals, which is in Hardy-Weinberg equilibrium 672 are dominant individuals. What is the number of heterozygous dominant individuals?

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384

128

316

288

answer is A.

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Detailed Solution

Total population = 800

Dominant individuals = 672

Recessive individuals = 800-672 = 128

q² = $\frac{128}{800}$ = 0.16

q = 0.4

p + q = 1

p = 1 - q = 1 - 0.4 = 0.6

2pq = 2 $\times$ 0.6 $\times$ 0.4 = 0.48

Heterozygous individuals = 800 $\times$ 0.48 = 384

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