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:

Answer 1

3.5%

Step-by-step explanation:

The genes crossveinless and Stubble are linked and 7mu apart. That means that the frequency of recombination between them during meiosis will be 7%.

The alleles for crossveinless are:

• cv-c+ wildtype, dominant
• cv-c mutation, recessive

The alleles for Stubble are:

• Sb mutant, dominant
• Sb+ wildtype, recessive

A dihybrid female Drosophila with genotype cv-c Sb+/cv-c+Sb is testcrossed (crossed with a homozygous recessive male):

cv-c Sb+/cv-c+Sb X cv-c Sb+/ cv-c Sb+

The male can only produce one type of gametes: cv-c Sb+

The female can produce 4 types of gametes:

• cv-c Sb+ Parental, 46.5%
• cv-c+Sb Parental, 46.5%
• cv-c Sb Recombinant, 3.5%
• cv-c+Sb+ Recombinant, 3.5%

The frequency of recombination between cv-c and Sb is 7%, and 2 recombinant gametes are formed, so each of them will appear 3.5% of the times. The parental gametes will have a frequency of 100%-7%=93%, and there are 2 of them so each will have a frequency of 46.5%.

Only when the recombinant gamete cv-c+Sb+ joins the gamete generated by the male parent will the offspring be wild-type for both genes, so the proportion of phenotypically wild-type individuals in the progeny will be 3.5%.