A population of 200 mice contains 168 brown mice. Brown is dominant to gray. How much of the population would be h o m o z y g o u s dominant? A. 84% B. 36% C. 6% D. 48%

If the population of 200 mice contains 168 brown mice, and brown is dominant to gray, then 36 percent (36%) of the population would be homo-zygous dominant.

In a population at Hardy–Weinberg (HW) equilibrium, the allele frequencies of a given gene in the population will remain constant generation after generation if there is no mutation, no migration, no natural selection, and if there is random mating. The genotype frequencies will also remain the same if the population is at HW equilibrium.According to the HW equilibrium, p² indicates the frequency of the homo-zygous dominant genotype 'BB' (phenotype: brown mice), q² indicates the frequency of the homo-zygous recessive genotype 'bb' (phenotype: gray mice), and 2pq represents the frequency of the heterozygous genotype Bb (phenotype: brown mice).  Moreover, the sum of the allele frequencies for B and b alleles at the target locus must be equal to 1, so p + q = 1.In this case, the frequency of the homo-zygous recessive genotype is equal to the number of individuals that don't exhibit the dominant phenotype, so

q² (bb) = (200 - 168)/200 = 32/200 = 0.16 >> q: √ 0.16 = 0.4

p = 1 - q = 1 - 0.4 =  0.6 >> p² (BB) = 0.6 x 0.6 = 0.36

In conclusion, at HW equilibrium, the frequency of the homo-zygous dominant genotype 'BB' (p²) will be 0.36 (36%).

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