A population consists of 9 percent white sheep and 91 percent black sheep what is the frequency of the black wool allele if the black wool allele is dominant and the white wool allele is recessive?

Answer 1

If you assume hardy-weinburg equilibrium, then:

let B = frequency of black allele (dominant)

b = frequency of white allele (recessive)

BB (or B^2) = frequency of homozygous black sheep

2Bb = frequency of heterozygous black sheep

bb (or b^2) = frequency of white sheep

Since 9% of the sheep are white, the frequency of white sheep is 0.09, or bb = 0.09, so b=.3, which means B = 1-b = 1-.3 = 0.7

You should check to make sure that the hardy-weinburg assumption holds:

BB = 0.49

2Bb = 0.42

And BB + 2Bb = 0.91, which is the frequency of black sheep. ?The hardy-weinburg assumption is valid!

Answer 2

If 91% of the population consists of black sheep, and the black wool allele is dominant, we can assume that all black sheep carry at least one copy of the black wool allele. Therefore, the frequency of the black wool allele in the population is 100%.