The genes for the recessive traits of mahogany eyes and ebony body are approximately 25 map units apart on chromosome III in Drosophila. Assume that a mahogany-eyed female was mated to an ebony-bodied male and that the resulting F1 phenotypically wild-type females were then mated to mahogany, ebony males. Of 1000 offspring, what would be the expected phenotypes, and in what numbers would they be expected?

mahagony eyes, normal body = 375

normal eyes, ebony body = 375

mahagony eyes, ebony body = 125

normal eyes, normal body = 125

Explanation:

Distance between genes in map units = recombination frequency (RF) between them

So, RF between mahagony and ebony genes = 25%

So, out of 1000 offspring,

Parental types = 750

There are two parental types: mahagony eyes, normal body and normal eyes, ebony body.

So, 375 will have mahagony eyes, normal body and 375 will have normal eyes, ebony body.

Recombinants = 250

There are two types of recombinants : mahagony eyes, ebony body and normal eyes, normal body.

So, 125 will have mahagony eyes, ebony body and 125 will have normal eyes, normal body.

a. an idea which cannot be disproven!

explanation: