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

0.035

Explanation:

cv+ is the wild-type dominant allele over cv, therefore:

Sb is a dominant mutant allele over wild-type Sb+, therefore:

It's the cross between the heterozygous female with a homozygous recessive male. Remember that cv and Sb+ are the recessive alleles.

 X

-The male produces only 1 type of gamete: cv Sb+

-The female produces 4 types of gametes:

The genes are linked and separated by 7 map units. A distance of 7 mu means that 7% of the resulting gametes will be recombinant. Because there are 2 possible recombinant gametes, each of them will appear in 3.5% of the cases.

The genotypes and proportions of the offspring resulting from the test cross can be seen in the Punnett Square. The phenotypically wild-type individuals will have the genotype cv+ Sb+ / cv Sb+ (heterozygous for crossveinless and homozygous recessive for Stubble) and a 0.035 proportion.

the endomembrane system i think