For FYI they are one questions couldn’t take the picture together so please do a

Question

For FYI they are one questions couldn’t take the picture together so please do all! h Prin of 30 Incorrect | 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. She collects the progeny phages, cultures a sample of them on plated bacteria, and observes 1000 total plaques. Calculate the expected number of c' m plaques. Number plaques Calculate the expected number of c m plaques. Number plaques Calculate the expected number of c m plaques. Continued below Number plaques

Explanation / Answer

Given that the gene for minute plaque and clear plaque is 8 m.u. which means that the maximum percentage of recombinant can be 8%.

Given that the there was the cross between c+ m- and c-m+ strains which gives total 1000 progenies.

The genotype of parental progenies will be c+ m- and c-m+

The genotype of recombinant progenies will be c+ m+ and c-m-

So the total recombinant will be 1000*8% = 80 out of which half will have c+ m+ and half will have c-m- i.e. 40 each.

Total parental prognosis will be 1000-80 = 920.

So the total parental will be 920 out of which half will have c- m+ and half will have c-m+ i.e. 460 each.

c+m+ = 40 plaque

c-m- = 40 plaque

c+m- = 460 plaque

c-m+ = 460 plaque

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