The genet for mahogany eyes and ebony body are approximately 25 map units apart

Question

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

Explanation / Answer

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).

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