Question

If the homozygous recessive genotype makes of 9% of the population and the heterozygous genotype makes up 42% of the population, what is the frequency of the dominant allele in that population?

Answer 1

Answer on Question #46586, Biology, Other

Question. If the homozygous recessive genotype makes up 9% of the population and the heterozygous genotype makes up 42% of the population, what is the frequency of the dominant allele in that population?

Solution. The structure of the population is: f(AA)+f(Aa)+f(aa)=1.

If the frequency of aa genotype is 0,09 and frequency of Aa genotype is 0,42 thus the frequency of AA genotype is: 1-0,09-0,42=0,49.

f(AA) =0,49.

The frequency of the dominant allele in that population we may calculate by the next way:

pA = f(AA) + 0,5 * f(Aa).

(0,5 - because the heterozygous have only one dominant allele).

Thus pA=0,49+0,5*0,42=0,7.

Answer. The frequency of the dominant allele in the population is 0,7 or 70%.

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