A geneticist isolates two mutations in a bacteriophage. One mutation causes clea

Question

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

Plaque Phenotype

Plaque Genotype

Expected Count

Explanation / Answer

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%

Plaque Phenotype

Plaque Genotype

Expected Count

c+ m–

46%

c- m+

46%

c+ m+

4%

c- m–

4%

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