Question

You are interested in mapping the genetic distance between two phage genes. You know from your genetics course that relative gene distance in both bacteria and eukaryotes can be determined by finding the recombination frequency of the two genes.

You infect a bacterial cell with two phages (h+r+ and hr) and you plate the progeny phages onto a bacterial lawn containing both strains of bacteria susceptible to the h mutant phenotype.  

a) Describe all the possible plaque phenotypes you would see and the genotypes associated with each.

b) Which two phenotypes are the most abundant?

c) If the two most abundant plaque phenotypes together total 140 out of 200 plaques, what is the genetic distance between these two genes in map units?

d) What would happen if the double mutant phage (hr) had failed to coinfect the bacterial cell? Would your plate look the same? Why or why not?

Answer 1

A- possible plaque phenotypes will be a double wild-type, the double mutant, one wildtype one mutant, one mutant one wildtype h+r+, hr, h+r and hr+.

the possible plaque genotype will be h+r+, hr, h+r and hr+. Out of these four phenotypes h+r+ and hr will be the parental types and h+r and hr+ will be the recombinants types.

B- The most abundant form will be  h+r+, hr.

C- the genetic distance between the gene is given by the total number of recombinants*100/ total number of progenies.

Given that the out of 200 plaque 140 was the plaque which was parental types.

So total recombinant plaque will be 200-140 = 60

So genetic distance will be 60*100/200 = 30 map unit.

D- If the double mutant phage (hr) had failed to coinfect the bacterial cells, then, in that case, we will get only one type of plaque on the plate which will be the double wildtype. The plate will not be the same as we found in the first case. We will get only one type of plaque. In the first case, we got four different phenotypes on the plate