22. Wild type T₄ bacteriophage can grow on Band K strains of E. coli forming small plaques. rll mutants of T₄bacteriophage cannot grow on E. coli strain K (non- permissive host), but form large plaques on E. coli strain B (permissive host). The following two experiments were carried out:

Experiment l:
E. coli K cells were simultaneously infected with two rllmutants (a^– and b^–). Several plaques with wild typemorphology were formed.

Experiment II:
E. coli B cells were simultaneously infected with thesame mutants as above. T4 phages were isolated fromthe resulting plaques and used to infect E. coli K cells.Few plaques with wild type morphology were formed.
Which one is the correct conclusion made regarding therll mutants. a^– and b^– from the above experiments?
(1) The mutations a^– and b^– belong to two differentcistrons (experimentl) and there is norecombination between them (experiment ll).
(2) The mutations a^– and b^– belong to two differentcistrons (experiment l) and they recombined(experiment ll).
(3) The mutations a^– and b^– belong to two differentcistrons (experiment ll) and they recombined(experiment l).
(4) The mutants a^– and b^– belong to the same cistron(experiment l) and they did not recombine(experiment ll).

Step-by-step reasoning

Background:

• Wild-type T4: grows on both E. coli B and K, forming small plaques.

• rII mutants:

  • Cannot grow on K (non-permissive) → no plaques unless wild-type function is present.

  • Grow on B (permissive) and form large plaques.

Two key phenomena in Benzer’s T4 system:

• Complementation (functional rescue in a mixed infection without DNA exchange).

• Recombination (physical exchange of genetic material creating true wild-type genomes).

Experiment I: coinfection of E. coli K with rII a– and b–

• Individually, a– or b– cannot form plaques on K.

• However, “several plaques with wild-type morphology” appear.

• On non-permissive K, plaques can only form if phage genomes in the infected cell collectively provide all required rII functions.

• Since each mutant alone fails, but together they allow growth, a– and b– must each supply a different rII function.
  ⇒ They complement each other → mutations are in different cistrons (different genes).

Experiment II: coinfection of E. coli B with the same mutants, then test progeny on K

• On permissive B, both mutants grow and produce many phages; within those mixed infections, recombination can occur.

• Progeny phage are then plated on K.

• Only phages with a fully wild-type rII region can form plaques on K.

• “Few plaques with wild-type morphology” are observed, meaning rare true wild-type recombinant genomes (not just mixed complementation) are present.
  ⇒ This shows that recombination occurred between a– and b–.

Therefore:

• Experiment I demonstrates that a– and b– are in different cistrons (complementation).

• Experiment II demonstrates that they recombined.

Option-wise explanation

1. Different cistrons (experiment I) and no recombination (experiment II) – incorrect

• Experiment II clearly shows new wild-type phages arising after growth in B and then plating on K; these must be recombinants.

2. Different cistrons (experiment I) and they recombined (experiment II) – correct

• Matches complementation on K and recombination detected via progeny testing.

3. Different cistrons (experiment II) and recombined (experiment I) – incorrect

• Complementation (not recombination) is inferred from experiment I; different cistrons are shown there, not in II.

4. Same cistron (experiment I) and no recombination (experiment II) – incorrect

• If they were in the same cistron, coinfection on K would not restore growth (no wild-type plaques). Experiment I shows the opposite, and experiment II shows recombinants.

Hence, the correct conclusion is option (2).