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

Question

The genes for mahogany eyes and ebony body are approximately 25 map units apart on chromosome 3 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 mated to mahogany, ebony males. Of 1000 offspring, what would be the expected phenotypes, and in what numbers would they be expected? Answer: mahogany = 375; ebony = 375; wild type = 125; mahogany-ebony = 125 I just need an explanation showing me how to do this problem.

Explanation / Answer

In Drosophila + indicates wild-type allele for any gene, m is mahogany and e is ebony. Female parents are m+/m+ and males are +e/+e. F1 are m+/+e, all wild type. F1 females are crossed with me/me males - the test cross. Offspring will be : non recombinant m+/me, mahogany wild type or +e/me wild type ebony. OR recombinant me/me mahogany ebony or ++/++ 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 m+/me and 375 +e/me. 25% of 1000 is 250 split between two recombinants =125 of each. Therefore you would expect 135 me/me and 125 ++/++

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