Map Genetics: A Conceptual Approach 6th Edition Pieros MHE/Freeman presented by

Question

Map Genetics: A Conceptual Approach 6th Edition Pieros MHE/Freeman presented by Sapling Learning 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 ci 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. Incorrect. Thenumbero map units,8m.u equal to the recombination rate, 8% Calculate the total humber of recombinants by multiplying the percentage of recombinants as a fraction of one 0.08, by the total number of plaques Number 460Pa plaques Because ct mt is one of the two recombinant classes it is expected to b half of the total recombinants S. Calculate the expected number of ci m plaques Number 460 plaques Calculate the expected number of ct m plaques Continued below Number 40 plaques e Previous # Try Again O Next Exit r

Explanation / Answer

To understand this question. let us first determine what is map units! Genetic map unit or m.u. is defined as the distance between two genes for which 1 of 100 offspring through meiosis is recombinant, i.e. a recombinant frequency of 0.01 = 1 m.u.

From the definition itself, frequency of recombinants for 8 m.u. shall be 0.08. Now in this case, parental types are - c-m+ and c+m- and recombinant genotypes shall be - c+m+ and c-m-. Now total number of recombinant plaques shall be 0.08 * 1000 = 80. Half of this shall be of c+m+ and half of this shall be of c-m- genotype. Therefore, both of their number is 40 each. Now the remaining are 920 plaques. Simple dividing them between the two parental genotypes, c-m+ and c+m- both are 460 each.

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