Imagine that a newly discovered species of fish normally has three eyes (E), but the rare mutation (ee) causes "diopthalmic" fish with two eyes to be born. In addition, eye color in these fish is inherited similarly to humans, with black eyes (B) being dominant to blue (bb). Assuming these two genes are carried on separate chromosomes, what must the genotypes of two black-eyed, three-eyed parents be if they have a blue-eyed, diopthalmic son

Heterozygous with BbEe and BbEe

Explanation:

The genotypes of the two parents must be heterozygous.

From the illustration, E codes for three eyes while ee causes diopthalmic. B causes black eye while bb codes for blue eye.

Black eye genotypes could be BB or Bb.

Three eye genotype could be EE or ee.

A black eyed, three-eyed couple produced blue-eyed (bb), diopthalmic son (ee). From the law of inheritance, the son inheritance the alleles for each trait from the two parents, one from each. Hence, it means that each of the parent posses recessive alleles in their genotypes, that is they are heterozygous for the two traits.

BbEe   x   BbEe

9 BBEE black-eyed, three-eyed

3 B_ee Black-eyed, diopthalmic

3 bbEe Blue-eyed, three-eyed

1 bbee blue-eyed, diopthalmic

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