4
$\begingroup$

In the textbook that I'm using, it explains that bacteria does not digest its own chromosomal DNA because the sites that would be cut by its own endonuclease are methylated. Is there a similar mechanism involved for protecting its own plasmids? If so, how are plasmids cut while serving as vectors?

$\endgroup$
  • $\begingroup$ You're asking about engineered plasmids, as opposed to natural ones? $\endgroup$ – canadianer May 15 '17 at 20:55
  • $\begingroup$ Sure, but I guess it can apply to both; for example, if an engineered plasmid is inserted into E. Coli, what prevents E. Coli's own endonuclease from digesting the new plasmid? $\endgroup$ – John Smith May 15 '17 at 21:14
  • $\begingroup$ Welcome to SE Biology. Your question seems to imply that you think that restriction enzymes are involved in recombination, which is not so. Perhaps I misunderstand, but I do not see what relevance recombination could have. You will need to clarify if your question is not to be closed as unclear. $\endgroup$ – David May 15 '17 at 21:49
  • $\begingroup$ I'm sorry; I have no idea why I said recombination there at the end. I think I was thinking of ligation during DNA cloning and got confused because of "recombinant DNA." $\endgroup$ – John Smith May 16 '17 at 0:32
4
$\begingroup$

Laboratory strains used for the purposes of cloning have been genetically engineered to address this issue, typically by deleting genes of the various restriction-modification systems. There are four broad classes of restriction modification systems, which I will discuss individually. Unless individually referenced, most information below is based on the following supplier technical notes:

Promega: What are the effects of the bacterial DNA restriction-modification systems on cloning?

New England Biolabs: Restriction of Foreign DNA by E. coli K-12


Type I

In Escherichia coli, this system consists of the hsdR, hsdM and hsdS genes and is capable of both restricting (digesting) unmethylated DNA and methylating hemimethylated DNA. Knockout of hsdR and/or hsdM prevents restriction and methylation, respectively. This allows engineering of strains for specific purposes. For example, a common E. coli strain used for cloning, DH5α, has the genotype hsdR17 (rK–, mK+) which abrogates restriction activity but maintains methylation activity. This means that exogenous DNA will not be restricted, but will be methylated and therefore able to be subsequently transformed to r+ strains without worry of degradation, if the need may arise.


Type II

This is the class of restriction enzyme commonly used in the lab to cut DNA because they generally recognize short, palindromic sequences and cut within or near the recognition site. The E. coli background strains K and B, from which most lab strains are derived, do not contain Type II enzymes. The Type II enzyme EcoRI, for example, was isolated from the E. coli RY13 strain (Yoshimori, 1971; PhD thesis).


Type III

Similarly to Type II, the K and B strains do not possess this restriction-modification system, which was first characterized in E. coli 15T-.


Type IV

This system differs from the others in that it cleaves methylated and hemimethylated DNA, though it does not recognize methylation by Type I or Type II modification systems, nor Dam or Dcm methylation. In E. coli, this system consists of the mcrA, mcrBC and mrr genes and is most problematic when attempting transformation of DNA from organisms with ubiquitous cytosine or adenine methylation (eg plants and mammals). For example, the lab strain T7 Express, which is derived from the common strain BL21, has the genotype Δ(mcrC-mrr)114::IS10 which deletes mcrBC, hsdRMS and mrr.


Note on Dam and Dcm Methylases

These bacterial enzymes methylate adenine and cytosine residues, respectively, but are not directly involved in the restriction-modification systems discussed above. However, some Type II enzymes used for cloning are sensitive to this methylation and so lab strains are available with these enzymes deleted.

$\endgroup$
  • $\begingroup$ Thank you for coming back and making the detailed edit; I appreciate the explanation! $\endgroup$ – John Smith May 17 '17 at 19:14
  • 1
    $\begingroup$ @JohnSmith You're welcome. If this answered your question, you can click the checkmark next to the answer to mark it as accepted (as well as upvote). $\endgroup$ – canadianer May 17 '17 at 19:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.