Question

In a mice population, there are two different genes. One gene codes for fur pigmentation, and the other codes for fur color. In the mice, black fur (B) is dominant to brown fur (b) and pigment (P) is dominant to non-pigment (p). A mouse with a pp genotype has white fur. Two black mice are crossed together and have offspring with black, brown, and white fur. In a litter of 32 mice, how many of these offspring would have white fur?

A. 0
B. 8
C. 16
D. 32

Answer 1

Out of a litter of 32 mice from two black mice crossed together, expectedly 8 offspring would have white fur due to the presence of the homozygous recessive non-pigment allele (pp) in a simple Mendelian 1:3 ratio for the pp genotype.

Step-by-step explanation:

B: 8 mice would have white fur. The epistasis concept in genetics explains why, as in mice, the C gene responsible for pigment production can override the expression of fur color dictated by the A gene. When black mice are crossed, they need to be heterozygous for both pigment (Pp x Pp) and color (Bb x Bb) in order to produce offspring with black, brown, and white fur. If we assume these mice are heterozygous for both traits, then the phenotypic ratio would be 9:3:3:1 for black (BBPp, BbPp, BBPp, BbPp), brown (bbPp, bbPp), and white (Ppbb, ppBB, ppBb, ppbb) fur, respectively.

Since white fur occurs when the non-pigment allele (pp) is present, regardless of the fur color gene B/b, all pp offspring will have white fur. In a cross between two mice heterozygous for pigment production (Pp x Pp), the Mendelian ratio for offspring to have pp genotype is 1 out of 4. Therefore, out of a litter of 32, on average, we would expect 1/4 of the offspring to be white, which is 8 mice.