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 map units apart. The geneticist mixes phages with genotype c+m+ and genotype c-m- and uses the mixture to infect bacterial cells. She collects the progeny phages and cultures a sample of them on plated bacteria. A total of 1000 plaques are observed. What numbers in the different types of plaques (c+m+, c-m-, c+m-, c-m+) should she expect to see?

your knowledge is highly appreciated :) thank you!

Explanation / Answer

It is shown that clear plaques, c, and minute plaques, m, are seen in the mutated bacteriophages. When phages with two genetic traits are mixed up then the reulting progeny would be of following c+m+ x   c-m- The distance between the two genes is 8 map units apart. Therefore it corresponds to a percentage recombination of 8% between the two genes. Two types of recombinant plaque phenotypes would be observed, cm+ and c+m. So, they both constitute 80 progeny of total 1000. cm+ = 40 c+m = 40 The remaining 92% corresponds to wild type and double mutany phage, c m and c+ m+ . S0, 920 of 1000 progeny constitute them. c m = 460 c+m+ = 460 Total progeny are c m    = 460 c+m+ = 460 c m    = 460 c+m+ = 460 cm+ = 40 c+m   = 40 cm+ = 40 c+m   = 40

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