Question

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

Answer 1

A. 0.035.

Step-by-step 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.