Question

1. The diagram below shows a happy family of mice, with two parents and four offspring.
However, one of the offspring mice is an imposter. The black-fur parent has the genotype FF Ee,
and the white-fur parent has the genotype ff Ee. Given these parent genotypes, which offspring
mouse could not have been born to these parents?
• A. Offspring A (Ff Ee)
• B. Offspring B (FfEE)
• C. Offspring C (Ff ee)
• D. Offspring D (ff EE)

Answer 1

Answer:

The diagram is not attached but the question can be solved with the question alone, since the necessary information was provided.

• D. Offspring D (ff EE)

Explanation:

This is a cross involving two genes in mice. The two parents mice are crossed to produce four offsprings. The genotypes of the parents mice are FFEe (black fur) and ffEe (white fur).

These two parents will produce the following allelic combinations of gametes;

FFEe- FE, Fe, FE, Fe

ffEe- fE, fe, fE, Fe

Using these gametes in a pumnet square (see attached image), 16 possible offsprings will be produced with only three (3) distinct genotypes viz:

FfEE (4/16)

FfEe (8/16)

Ffee (4/16)

Hence, based on this question, all of the offsprings except offspring D (ffEE) is likely to be produced. This means that the offspring with genotype ffEE is the imposter as the parents mice with genotypes FFEe and ffEe could not have given birth to such offspring.