Question

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 ________.

Answer 1

Answer:

0.035

Explanation:

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

• cv+cv+ and cv+cv cause wild-type phenotype for crossveinless
• cv cv causes the crossveinless phenotype

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

• Sb Sb and Sb Sb+ cause Stubble phenotype
• Sb+ Sb+ causes wild type phenotype for Stubble

Test cross

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

$\frac{cv\ Sb^+}{cv^+\ Sb}$  X $\frac{cv \ Sb+}{cv \ Sb+}$

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

-The female produces 4 types of gametes:

• cv Sb+   ] Parental
• cv+ Sb   ] Parental
• cv Sb     ] Recombinant
• cv+ Sb+ ] Recombinant

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.