B. cn + = wild type eyes (dull red), cn = cinnabar (bright red) eyes. w + = wild

Question

B. cn+ = wild type eyes (dull red), cn = cinnabar (bright red) eyes.

w+ = wild type eyes (pigmented)   w = white eyes (unpigmented).

Cinnabar and white are each recessive to their respective wild type allele. The cn allele is autosomal, while the w allele is X-linked. The white eye trait is epistatic to cinnabar, that is, all w w homozygous flies have white eyes, regardless of the cn genotype. A wild-type eyed female of genotype cn+cn; w+w is crossed to a cn+cn; w Y white eyed male.

What progeny ratios for eye color will be produced among the male and female offspring?

Note: The wild-type allele of each gene is dominant only to its own mutant allele. Masking epistasis is NOT a form of dominance. Be careful to clarify your thinking about the difference between epistasis and dominance. Epistasis will not necessarily be defined on the exam.

Explanation / Answer

Cn+ Cn Cn+ Cn = 1, Cn+ Cn w+ w = 2, w Y w+ w =1

therefore Dull red : white eye = 3:1

Since homozygous ww will have white eyes and it is X linked, both male and female

have 50% chance of getting white eye

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