I have an E. coli cell line I use to express a protein using a two plasmid system. One confers AmpR and one KanR.

For mutagenesis I would like to separate the plasmids to increase efficiency of the mutagenic PCR reaction. I was able to cure the cells of the KanR plasmid because it is low copy whereas the AmpR is high copy so I could easily grow the cells in ampicillin only and I ended up with only the AmpR plasmid. However I was unable to do the same to isolate the KanR plasmid.

My next idea was to run the plasmids through a gel and do an extraction. Since I didn't do a digest it was very hard to differentiate the bands and I ended up getting a negligible yield. I believe the plasmids were simply smeared through the gel due to tertiary structures.

My final hope is to perform a digest on the plasmids using a restriction enzyme that cuts each plasmid once. My concern is that I will not be able to get the stoichiometry correct due to the plasmids both being simulatenously present in vastly different concentrations.

Does anyone have any tips on how else I could isolate the plasmids? Or maybe does anyone have experience running a digest on plasmids with unknown concentrations? I really need help doing this. The mutagenesis project has been going on for months and this isn't even a molecular biology lab... so it's detracting time from main research goals!

  • $\begingroup$ Couldn't you digest with a restriction enzyme that cuts the Amp plasmid one (or preferably many) time(s) but does not cut the Kan plasmid? Then you could gel purify and transform the result, and presumably it would be nearly all Kan $\endgroup$
    – Luigi
    May 28, 2016 at 19:46
  • $\begingroup$ @Luigi I guess my question would still be how would stoichiometry work? If I have anywhere from 10-50x more of one plasmid than another, would that be a major issue? I suppose I could do a serial dilution... $\endgroup$ May 28, 2016 at 19:48
  • $\begingroup$ I don't really why stoichiometry matters here..? You have an excess of the Amp plasmid, but if you cut it all, you'll be left with only Kan. Unless I misunderstand your question $\endgroup$
    – Luigi
    May 28, 2016 at 21:24
  • $\begingroup$ @Luigi I just mean when setting up a digestion or any reaction you prepare the reactants in certain concentrations. For example for digestions they say one 'unit' of enzyme per microgram of DNA. $\endgroup$ May 28, 2016 at 21:37
  • $\begingroup$ In my experience as long as you're within an order of magnitude with restriction enzymes you're probably okay. Since Amp is in excess, you could probably operate using the overall DNA concentration and be just fine. It'd be harder to figure out how much to use to digest the Kan, but I'm not really sure why you want to in the first place $\endgroup$
    – Luigi
    May 28, 2016 at 22:25

1 Answer 1


A retransformation should get you a pure plasmid. Transform the plasmid mix into empty E.coli (don't use too much DNA), plate on Kan / Amp depending on what you need. The probability that one cell took up both plasmids is extremely low, and you could check with colony PCR just to be sure.

Otherwise: ask around, there must be (old) stocks of the single plasmids around, or glycerol stocks of E.coli containing only one of the plasmids.

Third option: take a look at your mutagenesis method. For your needs there might be techniques around that'll work on a mixture of plasmids. If you just need one variant or a single site saturation library you could use Quikchange for example. I don't see a reason why this wouldn't work on a mix of plasmids.


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