in a population of 1,000 mice, there are 400 mice that show the dominant trait of being dark-haired. using hardy-weinberg, determine the number of mice that are in each of the three genotypes for this trait.

If D is dominant and d is recessive, 160 with DD, 360 with Dd, and 480 with dd

Explanation:

if p= the frequency of the dominant allele (D for dark hair, perhaps?)

and q= the frequency of the recessive allele (d for not dark hair?)

remember that p + q = 1

400/1000= frequency of D= 0.4

0.4 + q = 1

q= 0.6 = frequency of d

remember also that p^2 +2pq + q^2 = 1

p^2 = (0.4)(0.4)= 0.16 = DD

q^2 = (0.6)(0.6) = 0.36 = Dd

2pq= (2)(0.4)(0.6)= 0.48 = dd

We now have the frequencies of each genotype and can apply them to the population size.

(0.16)(1000 mice) = 160 mice with DD

(0.36)(1000 mice) = 360 mice with Dd

(0.48)(1000 mice) = 480 mice with dd

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