Question

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?

A. E. coli B
B. E. coli K
C. Equal on both
D. Can't tell

Answer 1

the answer is A. E. coli B

Step-by-step explanation:

The multiplicity of infection (MOI) refers to the ratio between the numbers of viruses used to infect E. coli cells and the numbers of these E. coli cells. Benzer carried out several experiments in order to define the gene in regard to function. Benzer observed that E. coli strains with point mutations could be classified into two (2) complementary classes regarding coinfection using the restrictive strain as the host. With regard to his experiments, Benzer observed that rII1 and rII2 mutants (rapid lysis mutants) are complementary when they produce progeny after coinfect E. coli K (where neither mutant can lyse the host by itself). The rII group of mutants studied by Benzer does not produce plaques on E. coli K strains that carry phage λ (lysogenic for λ), but they produce plaques on E. coli B strains. This study showed that rIIA and rIIB are different genes and/or cistrons in the rII region.