In drosophila, the genes crossveinless-c and stubble are linked, about 7 map units apart on chromosome 3. cv-c is a recessive mutant allele of crossveinless-c (cv-c+ is wild type), while sb is a dominant mutant allele of stubble (sb+ is wild type). a dihybrid female drosophila with genotype cv-c sb+/cv-c+sb is testcrossed. the proportion of phenotypically wild-type individuals in the progeny of the testcross will be:
a. 0.035.
b. 0.070.
c. 0.350.
d. 0.465
e. 0.930

A. 0.035.

Explanation:

Female : cv-cSb+/cv-c+Sb

Male : cv-cSb/cv-cSb ( since test cross involves mating with a recessive parent )

Their progeny:

cv-cSb+/cv-cSb : parental

cv-c+Sb/cv-cSb : parental

cv-c+Sb+/cv-cSbc : recombinant

cv-cSb/cv-cSb : recombinant

Wild type phenotype (cv-c+Sb+/cv-cSbc) is a recombinant. Distance between the two genes is 7mu. Since map distance is equal to recombination percentage, there will be 7% recombinants in the progeny. Out of total 7% recombinants, 3.5% will be cv-c+Sb+/cv-cSbc( wild type) and rest 3.5% will be cv-cSb/cv-cSb (mutant type).

Hence, proportion of phenotypically wild-type individuals in the progeny of the test cross will be 3.5% or 0.035.