Question

If this trisomic male (ey+ ey-, gw+ gw-) is crossed with a (ey− ey−, gw− gw−) female, what proportion of the progeny will be phenotypically wild-type?

Answer 1

In an X-linked cross between a white-eyed male and a heterozygous red-eyed female fruit fly, 50% of the offspring would be phenotypically wild-type, exhibiting red eyes, while the other 50% would be phenotypically non-wild-type, exhibiting white eyes.

Step-by-step explanation:

In an X-linked cross between a white-eyed male fruit fly (ey-) and a female fruit fly that is heterozygous for red eye color (ey-), the offspring would exhibit a 1:1 ratio of phenotypically wild-type to non-wild-type individuals.

Specifically, 50% of the offspring would be phenotypically wild-type, meaning they would have red eyes. These individuals would inherit the wild-type allele (ey+) from the female parent and the wild-type or non-wild-type allele (ey+ or ey-) from the male parent.

The other 50% of the offspring would be phenotypically non-wild-type, meaning they would have white eyes. These individuals would inherit the non-wild-type allele (ey-) from both parents.

Answer 2

The Ey and the gw abbreviations stand for eyeless and gawky genes in the fruit fly Drosophila. A wild-type individual has at least one dominant allele for both of these genes.
In the cross presented below, you can see that 25% (shown in red) of the progeny would be wild-type and the rest of the progeny would show at least one recessive trait (eyeless or gawky)