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.

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nucleic acid molecules are incredibly complex, containing the code that guarantees the four different nucleotide units comprise dna, the nucleic acid of interest to the genealogist. and one of four nitrogen-containing molecules called nucleotides