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Consider a randomly-mating population that is not evolving. If the population consists of 16% homozygous recessive...

Consider a randomly-mating population that is not evolving. If the population consists of 16% homozygous recessive individuals, what percentage of the population is heterozygous? A. 84% B. 50% C. 42% D. 48%
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Answer #1

Homozygous recessive individual parcentage is 16% . So let the whole population be 100 . Then recessive individual will be 16 . Now we can say according to Hardy Weinberg equilibrium q2 = 0.16 so q = 0.4 so, p= 0.6 and 2pq= (0.4 x 0.6)x2 = 0.48 so parcentage of heterozygous population will be 48 or 48 % population is heterozygous

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