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 wild-type and mahogany-ebony phenotypes, and in what numbers would they be expected?

1) wild-type = 375; mahogany-ebony = 375
2) wild type = 125; mahogany-ebony = 125
3) wild-type = 375; mahogany-ebony = 125
4) wild-type = 125; mahogany-ebony = 375

Answer 1

In the F2 generation, out of 1000 offspring, we can expect a ratio of 375 wild-type phenotype and 375 mahogany-ebony phenotype.

Step-by-step explanation:

In this case, the genes for mahogany eyes and ebony body are located approximately 25 map units apart on chromosome 3 in Drosophila. When a mahogany-eyed female is mated to an ebony-bodied male, the resulting F1 generation phenotypically appears as wild-type females. These F1 females are then mated to mahogany, ebony males to produce the F2 generation.

Based on the given information, out of 1000 offspring in the F2 generation, we can expect 375 to have the wild-type phenotype and 375 to have the mahogany-ebony phenotype. Therefore, the correct answer is option 1) wild-type = 375; mahogany-ebony = 375.