Assume that the genes for tan body and bare wings are 15 map

Assume that the genes for tan body and bare wings are 15 map units apart on chromosome II in Drosophila. Assume also that a tan-bodied, bare-winged female was mated to a wild-type male and that the resulting F1 phenotypically wild-type females were mated to tan-bodied, bare-winged males to obtain an F_2 generation. (Remember the two genes are 15 mu apart on the same chromosome. In order to determine how many offsprings of each type are produced, you must determine if the F1 females carries both mutations on one chromosome, and wildtype alleles on the other or if she carries one mutation and one wildtype allele on each chromosome.) If 1000 offspring, what would be the expected phenotypes, and in what numbers would they be expected? Enter your answer as whole numbers, without decimal places. For example if you determine there will be 35 of a particular offspring, enter the number 35 in the appropriate blank. 425 wild type ___________ tan body ____________ bare wings _____________ tan body, bare wings

Solution

Wild type 425

Tan 75

Bare 75

Tan bared 425

Number of recombinant is directly proportional to the distance between the two genes in the map unit.

Here \"tan\" and \"bare\" are the crossover phenotypes. They make up 15% of the offspring, which is consistent with the claim that the 2 genes are 15 map units apart.