If 1000 f2 offspring are obtained, how many flies of each phenotype are expected?

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 ++/++

The answer is energy. the arrows are showing how exited and energized the stones are.