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In guinea pigs, black fur is an autosomal dominant trait, and white is the alternative recessive trait. A Hardy-Weinberg population of 400 guinea pigs was found to contain 64 white individuals. In this population, what percentage of BLACK animals is expected to be HETEROZYGOUS?

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Answer:

49.14%

Step-by-step explanation:

The frequency of the heterozygous individuals in the population is therefore: 64/400 = 0.32

Using the Hardy Weingburg equation of a population is equilibrium = p2 + 2pq + q2 = 1

q2 = 0.32; Therefore q = 0.5657 –> alellic frequencies

If p + q = 1; Then p 1 – q = 1 – 0.5657 = 0.4343

Heterozygous individuals are represented by 2pq = 2 * 0.4343 * 0.5657 = 0.4914

0.4914 * 100 = 49.14% of the population which translates to 0.4914 * 400 = 196.56 (197) individuals

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