Question

A researcher is studying the rII locus of phage T4. Three rII- strains are obtained: A, B, C, and D. In the first experiment, E. colistrain K(λ) is coinfected with two rII- strains simultaneously and the results are recorded.

Infection with A and B phage = plaques form

Infection with A and C phage = plaques form

Infection with B and C phage = no plaques form

Infection with B and D phage = no plaques form

Infection with C and D phage = no plaques form

In a second experiment, coinfections are performed first in E. coli strain B, then the progeny phage are used to infect E. colistrain K(λ).

Progeny of A and B phage = plaques form

Progeny of B and C phage = plaques form

Progeny of C and D phage = plaques form

Progeny of B and D phage = no plaques from

What conclusions are consistent with these data? (Select all that apply.)

	A.	A, B, C, and D carry mutations in the same gene.
	B.	Strains B and C both carry the same mutation.
	C.	Strains B and D both carry the same mutation.
	D.	Strains B, C, and D carry mutations in the same gene.
	E.	Strain A carries a mutation in a different gene than strains B, C, and D.
	F.	Strains A and B carry mutations in the same gene.

Answer 1

E.coli Kl will allows the growth of Wild type phages while not allowing rII mutants. E.coli strand B allows both wild type and rII mutants to grow. However, wild type E.coli B produces smaller plaques while rII mutants produce larger plaques. The rII mutants are conditional mutants that can grow only in permissive condition (E.coli strain B). Complementation, restoration of wild type function, occurs when the two mutation are present on different genes. Thus, there is always wild type copy present in the mutant strains for the two genes. If the two mutations are on same gene, then no wild type copy of the gene is present when the two strains infect one host cells. Hence, no complementation occurs, leading to no formation of plaques. Complementation will indicate that the two phages will together have some wild type function.

First infection of rII mutants in E.coli K:

1. When A and B infect E.coli K, there is complementation occurring and hence, the plaques are seen. Complementation occurs when the two rII mutation are in two different phage chromosomes and infect one host cell (E.coli K). Thus, A and B are different mutations are on different gene.

2. When A and C infect E.coli K, there is plaques formed. Hence, these mutations complement each other. They are different rII mutations present on different genes.

3. When B and C infect E.coli K cell, there is no plaque formation. Hence, these two mutations are present on the same gene.

4. When B and D infect E.coli K, there is no complementation, giving no plaques. Hence, these two mutations occur on same gene.

5. When C and D infect the same E.coli K, no plaques are seen as these mutations are in same gene, causing no complementation.

Second infection of progeny of coinfection of E. coli B plated on E.coli Kl: Recombination occurs when there is exchange of genetic material between chromosomes at homologous regions. When the two mutants A and B infect E.coli B, there will be recombination seen. There will be formation of wild type recombinants, double mutant recombinants and parental phages. As E.coli B allows growth of both wild type recombinant and double mutant recombinant phages, the progeny obtained will contain both types of recombinant progeny. When these progeny infect E.coli K, only recombinants that restore wild type function will grow. Thus, if the mutations are on different genes, wild type recombinant will form. However, if the mutations are on same gene, recombination cannot occur to form wild type recombinants. Wild type recombinants progeny can form plaques on E.coli K.

If the mutations are on same gene but different regions, wild type recombinants will form, causing plaques on E.coli K. However, if they are on same mutations in same gene, there will be no plaque formation in the E.coli K as recombination will not lead to wild type recombinant progeny.

1. A and B do form recombinant wild type progeny as they are present on different genes of rII. Hence, they give plaques on E.coli K.

2. A and C are present on different genes. Hence, they form wild type recombinant phages and thus, give plaques on E.coli K.

3. C and D have mutations on different regions of the same gene. This leads to wild type recombinants because they are on different regions of rII gene.

4. B and D have the same mutation on same gene rII locus (A or B). Thus, wild type functions recombinants are not formed.

Thus, we can conclude that

A has a different mutation on different gene than B, C and D. Strain B and D have the same mutation on same gene on same strand. Hence, wild type recombinants cannot be formed.  Strains C and D have mutations in same gene but different regions on different strands. Thus, wild type recombinants can be formed. Strains B, C and D have the mutations in same gene. Hence, recombination could restore wild type function. B and C do not have same mutation as C and D are different mutations, while B and D have same mutation. Hence, B cannot have same mutation as C.

Right choice:

C. Strains B and D both carry the same mutation.

D. Strains B, C, and D carry mutations in the same gene.

E. Strain A carries a mutation in a different gene than strains B, C, and D.