A geneticist isolates two mutations in a bacteriophage One m

A geneticist isolates two mutations in a bacteriophage. One mutation causes clear plaques and the other produces minute plaques (m). Previous mapping experiments have established that the genes responsible for these two mutations are 8 m.u. apart. The geneticist mixes phages with genotypes c^+ m^- and uses the mixture to infect bacterial cells. She collects the progeny plaques and culture a sample of them on plated bacteria. A total of 1000 plaques were observed. Complete the table below. What are the expected numbers of the different types of plaques (c^+ m^+, c^- m^-, c^+ m^-)? Three different Hfr strains arc crossed with separate samples of an F^- strain, and the following mapping data are obtained using interrupted conjugation:

Solution

You have crossed c+m- X c-m+

You will get four classes here c+m-, c-m+, c+m+ and c-m-.

In the above four classes, the c+m- and c-m+ are parental combinations.

In the above four classes, the c+m+ and c-m- are non parental combinations or recombinants.

You only said that these two genes are separated by 8 m.u, so the possibility of recombination is 8%.

Number of recombinants=1000x8/100=80

So you will get 80 recombinants, 40 will be c+m+ and 40 will be c-m-

Left over is 1000-80=920. 920 parental combinations.

Plague Phenotype

Plague Genotype

Expected count

Clear Plague

c+m-

460

Minute Plague

c-m+

460

Clear Plague

c+m+

40

Minute Plague

c-m-

40

Plague Phenotype	Plague Genotype	Expected count
Clear Plague	c+m-	460
Minute Plague	c-m+	460
Clear Plague	c+m+	40
Minute Plague	c-m-	40