Question

A population of 200 mice contains 168 brown mice. Brown is dominant to gray. How much of the population would be homozygous?A. 52%B. 48%C. 100%D. 84%

Answer 1

To answer this question we need two equations and allele frequency.

Hardy-Waynberg: p²+2pq+p²=1

Allele frequency, being p= dominant and q= recessive, so p+q=1

Now we have set our alleles and equation we proceed to obtain proportions

Total population= 200 individuals =100%

Brown mice (dominant phenotype)= 168 individuals = 84%

gray mice (recessive phenotype)= 32 individuals= 16%

So we have that

p² + 2pq= 0.84

q² = 0.16

So we obtain q

q=√0.16 = 0.4

Now by the allele frequency formula, we obtain p

p=1-0.4 = 0.6

Now we use hardy weinberg to obtain dominant homozygous and heterozygous

p²+2pq+p²=1

0.6²+2(0.6)(0.4)+0.4²=1

0.36+2(0.48)+0.16=1

Now we have the proportion of homozygous represented by p² and p² we add one to each other and we have:

0.36+0.16= 0.52

Therefore the homozygous proportion is 52% option A.