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?

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obliquitybathurst obliquitybathurst · Answer 1 · 2020-04-16T09:26:13+02:00

Answer:

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.