Could someone explain to how solve these types of questions please. A. Mutations

Question

Could someone explain to how solve these types of questions please.

A. Mutations rII1 and rII2 are known to fall into the rIIA and rIIB cistrons (genes) respectively. Cells of E. coli B are infected simultaneously with phage strains rII1 and rII2 at a high moi. The progeny phage are collected and used to infect cells of E. coli K. This infection is also carried out at a high m.o.i., and the resulting phage are collected. These progeny are diluted and plated on both E. coli B and E. coli K. Do you expect more plaques on strain B, on strain K, an equal amount on both or is it impossible to tell? (circle one) E. coli B E. coli K Equal on both Can't tell B. rII3 and rII4 are two non-overlapping deletion mutations of rIIB cistron (gene). Mutations rII3 and rII4 are used to infect E. coli B at high m.o.i and the progeny phage collected. These phage are then used to infect strain K at very low multiplicity of infection (m.o.i.). The resulting phage are diluted and plated on both E. coli B and E. coli K. Do you expect more plaques on strain B, on Strain K, an equal amount on both or is it impossible to tell? (circle one) E. coli B E. coli K Equal on both Can't tell

Explanation / Answer

4. a) R mutants could be spotted easily because they would produce larger plaques rather than the smaller plaques characteristic of the wild type virus. Since cells of E.coli B is infected first with both the strains it will show larger plaques. Answer: E.coli B

b) Complementation test: a schematic view of rII complementation. Two different mutants of rII are used to simultaneously infect E. coli K (mixed infection). Normally, an rII mutant cannot lyse E. coli K or generate progeny phages. However, if the two different mutants can complement each other, then lysis and phage growth will result. If the two rII mutants cannot complement each other, then no lysis or phage growth will result.

Here complementation is happening so the plaques will be seen in E.coli B and K - both in equal amount.

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