Question

Suppose a geneticist isolates two bacteriophage mutants. One mutation causes clear plaques (c), 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 genotype c+ m− and genotype c− m+ and uses the mixture to infect bacterial cells. The geneticist collects the progeny phages, cultures a sample of them on plated bacteria, and observes 1000 total plaques. What numbers in the different types of plaques (c+m+, c-m-, c+m-, c-m+) should she expect to see?

Answer 1

Answer:

The mentioned parental types are c+m- and c-m+. Thus, the recombinants will be c+m+ and c-m-.  

Now, the given distance between c and m is 8 map units. Thus, the recombinant frequency is 8% or 0.08.  

The total recombinants from 1000 plaques will come out to be 80,  

Thus, the recombinants of each type will be 40.  

Total parental type will be 920, and therefore, each parental type count will be 460.  

Thus, expected c+m- = 460, expected c-m+ = 460, expected c+m+ = 40 and expected c-m- = 40.