In a population of 200 individuals at Hardy-Weinberg equilibrium, 72 are homozygous recessive for the character of eye color (cc).One hundred individuals from this population die from a fatal disease.Thirty-six of the survivors are homozygous recessive.In the original population, the frequency of the dominant allele was
A) 0.16.
B) 0.36.
C) 0.40.
D) 0.48.
E) 0.60.

Question

Quizplus · Accepted Answer

According to the Hardy-Weinberg equilibrium, the frequency of the homozygous recessive genotype is q2, where q is the frequency of the recessive allele.

In the original population, there are 72 individuals with the homozygous recessive genotype, so q2 = 72/200 = 0.36.

Taking the square root of 0.36, we get q = 0.6, the frequency of the recessive allele in the original population.

Since there are only two alleles in this gene pool, the frequency of the dominant allele (p) can be calculated as 1 - q = 1 - 0.6 = 0.4.

Therefore, the answer is C.

In a Population of 200 Individuals at Hardy-Weinberg Equilibrium, 72