A geneticist isolates two mutations in bacteriophage. one causes clear plaques (c), other causes minute plaques (m). Previous mapping experiments have shown that the genes responsible for these two mutations are 8 map units apart. he mixes phages with genotype c+m+ and c-m- and uses the mixture to infect bacterial cells. He collects the progeny cells and cultures them on a lawn of bacteria; and observes 1000 plaques. What numbers of the different types of plaques (c+m+, c-m-, c+m-, c-m+) should he expect to see?
1) Plaque phenotype produced by c+m+, 460; by c-m-, 460; by c+m-, 40 and by c+m-, 40.
Explanation: The last two types are recombinants. The distance between recombinants is 8mu or the RF is 8%. 8% of 1000 is 80. Therefore, the two recombinants are each 40.
1000 -80 = 920/ 2 = 460