4 In Drosophila the two genes w and sn are Xlinked and 25 ma

4. In Drosophila, the two genes w and sn are X-linked and 25 map units apart. A female fly of genotype w+sn+/w sn is crossed to a male from a wild-type line. What percent of male progeny will be w+ sn? A) 0% B) 12.5% C) 25% D) 37.5% E) 50%

Solution

w+-sn+/w-sn X w+-sn+/Y

Male-produced will be four types and receive its X-chromosome from mother.

1) w+-sn+/Y (parental types)

2) w-sn/Y (parental types)

3) w+-sn/Y (recombinant type)

4) w-sn+/Y (recombinant type)

The distance between w+-sn+ is 25 cM. It means that for female, with the given genotype, the total gametes will have 75% of the parental types with 37.5% for each of the two genotypes. It also means that 25% will be the recombinant types in the population and each genotype will be represented by 12.5% in proportion. Since a male receives X-chromosome from mother, so this proportion holds true for this case.

The question asks for the amount of recombinant type in w+-sn/Y in the male population. If we consider that total F1 male population is 100% then the proportion discussed above holds true. So, the w+-sn/Y will be represented by 12.5%

The correct option is B).