Suppose that in a parent (P1) population of this rodent, you find?

Question

In a particular species of rodent, there is one gene, B/b with incomplete dominance, that determines fur color.

Suppose that in a parent (P1) population of this rodent, you find

384 black individuals (BB)
123 gray individuals (Bb)
24 white individuals (bb)

a) What are the genotype frequencies in the P1 population?
b) What are the allele frequencies in the P1 population?
c) Assuming Hardy-Weinberg Equilibrium, what percent of the offspring population (F1) will have gray fur?

Explain, in words, how you calculated the answer for (c).

Answer 1

384 BB means 384*2 = 768 genes for black
123 gray Bb means 123 black and 123 white alleles
24 white means 24*2=48 white

Totals
Black, B, 768+123=891 black alleles
White, b, 123+48+ 171 white alleles
Total alleles = 891+171=1062 alleles

a. Frequencies in the parental population
384+123+24 = 531 individuals
Frequency of genotypes in P1:
BB: 384/531= 0.72 BB
Bb: 123/531 = 0.23 Bb
bb: 24/531 = 0.05 bb

b. Allelic frequencies in the parental population:
Black, B: 891/1062 = 0.84 black
White, b: 171/1062 = 0.16 white

c. Assuming Hardy-Weinberg, the grays should be 2pq. The previous calculations show that p is 0.84. q is 0.16.

2 pq = 2*0.84*0.16 = 0.2688 or 27% of the offspring should have gray fur.