I was watching Eric Lander's lecture online describing Luria's experiment (see below for link), and I couldn't understand how plaques form. From my understanding,

  1. Bacteriophage infect strain B
  2. Viral DNA is incorporated
  3. Restriction enzymes cut out viral DNA (mostly)
  4. Some viral DNA is methylated before step 3 occurs.

At this point, to replicate, we would need the infected bacteria to produce more bacteriophage to infect nearby virus. However, due to step 3, the newly incorporated viral DNA will be cut out through restriction enzymes. So in my mind, we only have one bacteria with viral DNA (along with its descendants).

So how does the plaque form?



2 Answers 2


This indeed a very interesting experiment. In the lecture, you can see that the obvious explanation for the initial inhability to grow in strain B indicates that the bacteria is resistant to the virus. However, ocassionally, a virus is able to form a plaque in strain B, which initially seems to indicate that it mutated. However, in a second passage from Strain A to Strain B of this initial "mutant", we observe that there are no plaques in Strain B. This indicated that something beyond the DNA (unlike a mutation) was at play, and eventually led to the discovery of these epigenetic markers (methylation) that are made by the machinery of the bacteria (thus, Strain A lacked such machinery, so it maintained the viral DNA but did not mark it as 'viral').

Because these experiments occurr in Petri dishes, where there are millions of cells (billions, in fact), there is always some probability that the restriction machinery does not work. This probability (small, but detectable) of no detection is what is observed in the "original" plaque formed in Strain B. In addtion, when the virus is just added, it is still not marked, so it needs some time to be marked (more cell replication cycles), and during this time the virus can be infective. Either of these possibilities will produce a small fraction of plaques in Strain B.


In the experiment, the genetic material of a single bacteriophage was, by chance, methylated and thus protected from restriction. In order for the phage to propagate, this DNA must be replicated within the cell. Before replication, the phage DNA is methylated on both strands of the recognition palindrome. During replication, the strands are separated and used as templates for DNA synthesis. The methylation is retained in the parent strands, leading to a condition of hemimethylation in the replicated duplex. This hemimethylated DNA is protected from restriction, allowing time for the daughter strand to be methylated (this is the same mechanism by which the bacterial DNA itself is protected after replication). In other words, once the phage DNA is methylated, the bacterial cell no longer recognizes it as foreign and continues to propagate the methylation through subsequent rounds of replication.

Jen-Jacobson L, Engler LE, Lesser DR, Kurpiewski MR, Yee C, McVerry B. 1996. Structural adaptations in the interaction of EcoRI endonuclease with methylated GAATTC sites. EMBO J 15(11):2870-2882.

...our data for the hemimethylated GAmATTC site show that [the rate of cleavage of the unmethylated strand] is reduced only -330-fold relative to an unmodified GAATTC site. ...this would seem to be an inadequate degree of protection, taken alone, during the transient existence of such hemimethylated sites in vivo after DNA replication. Even if a hemimethylated site were to suffer a single-strand nick in the unmethylated DNA strand, however, it is very unlikely that this would progress to double-strand cleavage because the [rate] for the second cut (in the methylated strand) is reduced by 4200-fold. This rate is so slow that the endonuclease would dissociate from the nicked DNA before making such a second-strand cut, and the single-strand nick would then be repaired by DNA ligase.

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Note that the lecture mentions the sequence GAATTC, which is the recognition site of the type II restriction enzyme EcoRI (discussed above). This enzyme is naturally found in E. coli RY13. In their 1952 paper, Luria and Human used E. coli B, which does not have type II restriction modification systems. Rather, the effect observed was probably due to a type I restriction modification system (EcoB in this case). In such systems, hemimethylation not only reduces nuclease activity but also increases methylase activity, thus actively protecting newly replicated DNA.

Vovis GF, Horiuchi K, Zinder ND. 1974. Kinetics of methylation of DNA by a restriction endonuclease from Escherichia coli B. Proc Natl Acad Sci USA 71(10):3810-3813.

In E. coli B, the DNA immediately after replication should possess only [hemimethylated] sites. Such sites must be modified rapidly. If not, another round of DNA replication would yield some completely unmodified [recognition] sites which would make the DNA susceptible to restriction. On the other hand, if bacterial or viral DNA synthesized in a foreign bacterium should enter E. coli B, it would do so with unmodified [recognition] sites. These sites would be expected to be poor substrates for methylation but excellent ones for eliciting restriction, the appropriate protective reaction. The in vitro action of... EcoB, i.e., rapid restriction but very slow methylation of molecules with unmodified [recognition] sites and rapid methylation but no restriction of molecules with [hemimethylated] sites, is entirely consistent with the above in vivo scenario and suggests that this enzyme is responsible for both restriction and modification in vivo.

I don't know if this increased methylase activity on hemimethylated DNA also occurs type II systems, which have separate nuclease and methylase enzymes.


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