Amale and a female mouse are each from pure-breeding albino strains. they have a litter of 10 pups, all of which have normal pigmentation. the f1 pups are crossed to one another to produce 56f2 mice, of which 31 are normally pigmented and 25 are albino.

You are forgetting the question but I can help you understand the exercise.

For this exercise it's important to know that there are more than one gene that causes albinism. In this exercise, one of the parents is albino because of one gene( gene A) and the other parent is albino because of another gene( gene B)
Albinism only happens if it's inherited both recessive allelles.
The male and female mouse follow those rules, so the cross between them is:
AAbb x aaBB

All their pups, the generation F1, is:AaBb
Since the disease is recessive (that means: aa), they don't have albinism.

After they cross each other: AaBb x AaBb (see the image attached)

The answer would be c, white eyes are homozygous recessive. in the early 1900s, morgan conducted several experiments with vinegar  flies (drosophila). one  sample revealed an anomaly: a  white-eyed, single male fly. he attempted to breed this fly with the typical red-eyed females. the offspring were all red-eyed. using flies from this new generation (brother-sister pairs)  produced some red-eyed and a few white-eyed flies. the white-eyed were all male. in later experiments, he was able to produce white-eyed females. this proved that white eyes are not lethal to either sex (ruling out a and b). since the percentage of white eyes to red eyes is so low, white eyes are not a dominant trait. it would be a recessive trait.