A geneticist isolates two mutations in a bacteriophage One m

A geneticist isolates two mutations in a bacteriophage. 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 genotypes c⁺ m^– and c^– m⁺ and uses the mixture to infect bacterial cells. She collects the progeny plaques and cultures 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⁺, c⁺m^–)?

Plaque Phenotype

Plaque Genotype

Expected Count

Solution

Answer:

Distance between the genes = % recombinaiton frequency

8 mu = 8% recombinations

Parental combinations (c+ m– and c– m+) = 92% (each 46%)

Non-parental combinations (recombinants) (c+m+, c–m–)= 8% (each 4%)

Plaque Phenotype

Plaque Genotype

Expected Count

c+ m–

46%

c- m+

46%

c+ m+

4%

c- m–

4%