1. How would you set up this experiment so it would provide a selection for restriction mutants?
2. What is the phenotype of hsd-802 for restriction and modification of the phage? Explain your logic.
3. What is the phenotype of hsd-811 for restriction and modification of the phage? Explain your logic.

Wild-type E. coli strain K-12 has a single type I restriction system.

1. How could you isolate a r^- mutant of E. coli K-12?

   ANSWER: A selection for r^- bacteria would be ideal. DNA from a non-modified source (such as E. coli C) will be resticted (digested) when it enters a r⁺ host but will survive and propagate in a r^- host. But, how can you SELECT for the "survival" of the DNA? One approach would be to use a plasmid that encodes an antibiotic resistance as the selectable DNA. For example, you could set up the experiment as follows:

   • Mutagenize K-12 (r⁺ m⁺) cells.
   • Transform the mutagenised cells with plasmid pBR322 (ampicillin resistant) which had isolated fromE. coli C so that it is unmodified. Select for ampicillin resistant survivors. These colonies will include mutants which are defective in restriction of unmodified DNA (r^-).
   • Test any potential r^- mutants by infecting with unmodified phage. The desired mutants should yield plaques with a high EOP.

   An alternative scheme that may work would be to using a phage that carries an antibiotic resistance gene, with selection for antibiotic resistant lysogens in the mutagenized K-12.

2. How could you determine if this mutant was r^-m^- or r^-m⁺?

   ANSWER: Prepare the plasmid from the r^- strain and backcross into a r^- strain versus a r⁺ strain with selection for ampicillin resistance. If the plasmid DNA is modified (i.e. from a r^-m⁺ strain), the number of Ap^r colonies on the r^- and the r⁺ strains will be about the same because the recognition sites for the restriction enzyme are protected on the DNA. In contrast, if the DNA is unmodifed (i.e. from a r^- m^- strain) then there will be far more Ap^r colonies on the r^- recipient than on the r⁺ one because the unprotected plasmid DNA will be digested.

   A similar experiment could be performed by growing phage on the r^- mutant from part a) and then testing for efficiency of plaquing (EOP) on a r^- versus a r⁺ recipient, similar to the approach described for the plasmid experiment above.

ANSWER: There is a selective pressure against phage with sites recognised by restriction enzymes encoded by the host bacterium. The more sites for such an enzyme, the less likely it is that all of the sites are correctly methylated on a given phage genome. Sites which are not methylated are restricted, "killing" that particular phage genome. Any phage genome which accumulates mutations resulting in fewer sites for that particular enzyme will reproduce more efficiently and thus these mutant phage will be more competitive than the wild-type phage. Such mutant phage will have fewer sites than would be expected if the sites occured randomly. Clearly, if the phage genome does not infect a particular bacterial species or strain, there will be no selection pressure exerted against the presence of sites recognised by restriction enzymes from that strain, thus there will be little deviation from the expected number of sites.

Restriction sites may be lost by mutation of one of the base pairs of the endonuclease recognition sequence. Any mutations which do not disrupt the function of the gene in which the recognition site is buried will be tolerated.

ANSWER: There are several possible explanations. The phage would need some mechanism to protect itself against the restriction systems of each host. This mechanism must be phage encoded to ensure that it is independant of the specific host which is infected. The protection may occur in a non-restricting host before entering a restricting host. Furthermore, the protection must be non-specific enough that it works against restriction systems from a variety of strains/ species. Examples of such a non-specific mechanism are: phage encoded proteins that inactivate restriction-modification enzymes, phage encoded methylation systems, and the incorporation of unusual bases in the phage genome. These are only a few general examples -- there are a plethora of tricks used by different phages.

For some more comprehensive reviews on this topic see:

• Bickle and Kruger. 1993. Biology of DNA Restriction. Microbiol. Rev. 57: 434-450.
• Kruger and Bickle. 1983. Bacteriophage survival: Multiple mechanisms for avoiding the deoxyribonucleic acid restriction systems of their hosts. Microbiol. Rev. 47: 345-360.

1. Explain the results for the growth of each phage on each of the five strains.

   ANSWER: This question involves a straightforward application of the principles of restriction/modification of foreign DNA.

   • DNA from phage propagated on rB⁺mB⁺ or rB^-mB⁺ strains will be modified so that it is protected against restriction when it enters rB⁺mB⁺ recipient cells.
   • However the DNA will be restricted by recipient cells expressing the K restriction enzyme, i.e. rK⁺mK⁺.
   • Similarly, the DNA from phage propagated on rK⁺mK⁺ or rK-mK+ strains is protected against restriction when it enters rK⁺mK⁺ recipient cells but will be restricted by recipient cells expressing the B restriction enzyme, i.e. rB⁺mB⁺.
   • DNA from all phage, without regard to their state of modification, will survive in the restriction minus strains rK^-mK⁺ and rB^-mK⁺.
   • DNA from phage propagated in the merodiploid rB⁺mB⁺ / rK^-mK⁺ is modified by both K and B systems and is thus not restricted in any of the E. coli recipient bacteria because of protection against both K and B systems.
   • Growth of the phages on the merodiploid recipient rB⁺mB⁺ / rK-mK+ provides some unexpected results. It is not surprising that K modified phage are restricted in this recipient, which produces functional B restriction endonuclease. What is surprising is that the phage modified for B are also being restricted, despite the fact that there is no K restriction endonuclease being made (rK^- genotype). How does this happen? The K and the B restriction systems are Type I, and thus are encoded by three genes, hsdM, S, and R. The gene S makes a specifictity subunit which recognises specific sequences in the target DNA, while M and R encode for methylase and restriction endonuclease, which have to interact with the specificity subunit to work. The K and B systems are similar enough that in the merodiploid strain, RB endonuclease can associate with the specificity subunit S of the K system and allow cleavage of unprotected K sites. This model is consistent with the observation that DNA from phage propagated in the merodiploid and thus modified for both K and B recognition sites are not degraded upon infection into the merodiploid recipient.

2. Why were no r⁺m^- mutants isolated.

   ANSWER: A bacterium which fails to modify its own DNA but still makes the corresponding restriction endonuclease will rapidly die because it will restrict its own genome.

A 30 minute heat shock temporarily inactivates restriction endonucleases in a variety of bacteria, allowing efficient uptake of foreign DNA by transduction or electroporation [Edwards, R., A. Helm, and S. Maloy. 1999. Increasing DNA transfer efficiency by temporary inactivation of host restriction. BioTechniques, In press]. For example, the efficiency of electroporation of DNA from a Salmonella typhimurium donor into a Salmonella enteritidis recipient is very inefficient unless the S. enteritidis recipient has been heated prior to electroporation (see the table below).

1. Suggest an explanation for the observed EOP under each condition in the S. typhimurium recipient.
2. Suggest an explanation for the observed EOP under each condition in the S. enteritidis recipient.
3. Once DNA transferred from S. typhimurium to the S. enteritidis recipient, it is subsequently transferred to other S. enteritidis strains with an EOP of 1. What does this suggest about the effect of heat shock on the restriction-modification system.

The endonuclease Ceu-I recognizes a 19 bp sequence. How often would you expect this enzyme to cut the 4.8 x 10⁶ bp chromosome of Salmonella typhimurium?

The chromosomes of many enteric bacteria, including Salmonella typhimurium, Escherichia coli, and Klebsiella aerogenes have been digested with Ceu-I and the resulting fragments separated by pulsed field gel electrophoresis. Although the size of the fragments observed differs for the different bacteria, in each case 7 fragments are obtained. Suggest an explanation for these results.