Suppose that in a population the frequency of a particular autosomal recessive d

Question

Suppose that in a population the frequency of a particular autosomal recessive disease is 1/400. Assume the presence of only a dominant allele (A) and a recessive allele (a) in the population and that the population is at Hardy–Weinberg equilibrium. What is the frequency of heterozygotes in the population?

Suppose that in a population the frequency of a particular autosomal recessive disease is 1/400. Assume the presence of only a dominant allele (A) and a recessive allele (a) in the population and that the population is at Hardy–Weinberg equilibrium. What is the frequency of homozygotes for the dominant allele in the population?

Explanation / Answer

According to Hardy-Weinberg equilibrium principle - allele and genotype frequencies in a population remain constant from generation to generation in the absence of other evolutionary influences.

That is, p2 + 2pq + q2 = 1

where p2 is frequency of homozygous dominant people in a population

q2 is frequency of homozygous recessive people in a population

pq is frequency of heterozygous people in a population.

Here p = A (dominant allele), q = a (recessive allele)

Population frequency is 1/400

So, A = 0.0025, A2 =

Therefore, frequency of homozygotes dominant allele in the population = p2 = A2 = 0.00000625

Frequency of homozygous recessive is A + a = 1

a = 1 - A

= 1- 0.0025

= 0.9975.

That is frequency of heterozygous people in a population = 2pq = 2Aa

= 2 x 0.0025 x 0.9975

= 0.0049 = 0.005.

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