Here's my plan for making the mutant strain knocked out for both HI0660 and HI0659:
I'll start with the two single-knockout plasmids that the RA made by recombineering. Both were made from the same parent plasmid containing a chromosomal segment (green) containing both genes and about 500 bp of flanking DNA on each side. In the left plasmid the HI0660 gene has been replaced by a SpcR cassette (orange). In the right plasmid the HI0659 gene has been replaced by the same cassette.
I'll cut both plasmids with the same two restriction enzymes. SpeI cuts in the vector, close to the left end of the insert, and SacII cuts in the SpcR cassette, close to its right end. Then I'll inactivate the enzymes (with heat or phenol extraction), mix the two digested DNAs, ligate the mixture and transform it into E. coli, selecting for AmpR and maybe SpcR.
The single fragments won't be able to self-ligate because the enzymes give different sticky ends, but 10 different bi-molecular ligation products are possible. The plasmid I want (A+D) is shown below. Three of the others won't be able to replicate (A+A, A+C, C+C), and three others will contain inverted duplications of the vector (B+B, B+D, D+D) and thus probably be unstable; they'll also be much larger than A+D and not SpcR. The other unwanted combinations will also be larger than A+D (A+B, C+D and B+C) so I should be able to easily distinguish them.
Once I identify the plasmid I want I'll just transform our wildtype H. influenzae strain with its insert DNA and select for SpcR. Then I can find out whether deleting HI0660 eliminates the need for HI0659.
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what about gel-purification?!
ReplyDeleteGel purifying fragments A and D (before ligating them) would increase my chances of getting the plasmid I want, but I'm lazy so I might first try ligating the whole mixture. I'd be betting that the gel purification would be more work than the plasmid screening, but I could easily be wrong.
ReplyDelete"plasmid screening" involves at least running a gel. so yes, you are wrong.
ReplyDelete