The genet for mahogany eyes and ebony body are approximately

The genet for mahogany eyes and ebony body are approximately 25 map units apart on chromosome 3 in Drosophila. Assume that a mahogany eyed gray bodied female was mated to an red eyed ebony-bodied male and that the resulting phenotypically wild type (red eyed, gray bodied) females were mated to mahogany, ebony males to obtain an F_2 generation. Which of the following phenotypes would be expected in the F_2 offspring? Select all that apply. red eyed, gray bodied mahogeny eyed, gray bodied red eyed, ebony bodied mahogeny eyed, ebony bodied

Solution

Answer:

In Drosophila, red eye (M) and gray body (E) indicates the wild-type alleles, m is mahogany and e is ebony.

Female parents are mE/mE and males are Me/Me.

F1 are mE/Me, all wild type. F1 females are crossed with me/me males - the test cross.

Offspring will be : non recombinant mE/me, mahogany wild type or Me/me wild type ebony. OR

recombinant me/me mahogany ebony or ME/ME wild type.

As the two genes are 25 map units apart, the percentage of recombinants will be 25% and therefore percentage parental types will be 75%.

75% 1000 is 750. There are two parental types, so you would expect 375 of each. Therefore, you would expect 375 mE/me (mahogeny eyed, gray bodied) and 375 Me/me (red eyed, ebony bodied).

25% of 1000 is 250 split between two recombinants =125 of each. Therefore you would expect 125 me/me (mahogeny eyed, ebony bodied) and 125 ME/ME (red eyed, gray bodied).