Ultimately, if you make a mutant, you need to sequence the DNA to confirm it. You have to be able to grow the mutant, PCR the mutated region, and then sequence the PCR products.
To help me understand better, could you lay out the procedures you went through to construct your mutant including the vectors you used and what antibiotic resistance genes are encoded by them?
By what you describe above, you get growth on media, but in the presence of kan+amp, there is no growth. Couldn't that simply be WT P. aeruginosa growing?
I assume you're transforming the P. aer by conjugation. Following conjugation, what if any antibiotics are you using in your plates? If you simply plate the conjugation products onto no-antibiotic-media, you're likely to just have a ton of WT growing. Interspersed in that is likely some single and double crossovers.
However, if your first plating following conjugation is onto kan-amp, you are selecting in the first round for single crossover events. At least that way, the plasmid and construct are in the genome. After that, without sacB, things get hairy. You then need to select for the second crossover event, which sacB is used for. In your case, there is no selection, so it amounts to a whole bunch of screening on your part. From the colonies you'd get on the kan amp plates, you'd want to patch those onto no-antibiotic-media. That will take away the pressure to maintain the kan-amp resistance provided by the integrated plasmid. That plasmid then could loop out (the 2nd recombination). Your results then are the 50% WT 50% mutant. Of those colonies that grow on the plate, you'd have to patch to kan-amp media to verify the plasmid is gone, and then of the colonies that don't grow on kan-amp, screen them to determine which reverted to WT and which maintained the mutation.
It ends up being a lot of extra work to find those mutants. Throw in that it sounds like you're trying to make a mutation in a required gene, then at every step you also need to provide the substrate to allow the bacteria to grow.
What the sacB does is prevent those kan-amp colonies from growing after you try to induce the 2nd crossover event. That way, what you're left with are the final double crossovers not contaminated with a bunch of single crossovers.
I am checking gene essentiality. For eg. a virulent gene is responsible for pathogenicity of the bacterium which in turn explains that the virulent gene is responsible for survival of the pathogen. I have a gene which is found to be "always"
essential for PAO1, so knocking out the gene should result in killing the bacteria.
I designed my experiment with pFS100 suicide vector kanR,AmpR (a derivative of pGP704). Other strains used for transformation E.coli MC1061 and SM10lambda phir. After conjugation i should see the following outcome which i have summarized in the table below,
Growth on Centrimide ||| Growth on Cent+kan+amp
Wild Type (WT) - G (Growth) ||| NG (No Growth)
WT recombine with
recomb plasmid but
gene NON ESSENTIAL - G ||| G
WT recombine with
recomb plasmid and
gene ESSENTIAL - NG ||| NG
No recombination - G ||| NG
For my gene i found it to be essential and hence i find no growth in both the above combination(I get the 3rd combo as outcome) suggesting that i have knocked out the virulent gene that is responsible for pathogen survival. These are the only controls i used to show that i have knocked out the gene.
There is a paper in NAR link given below which also checks the essentiality of the genes and they say "A total of 347 candidate reading frames were subjected to disruption analysis, with 113 presumed to be essential due to lack of recovery of antibiotic-resistant colonies
" (in abstract) meaning they have knocked out the essential gene whereas non essential genes will keep growing in antibiotics.http://www.pubmedcen...bmedid=12136097
Also read the section "Streptococcus pneumoniae competent cell preparation and gene disruption" where they explain " Plates were then examined and colony numbers compared to controls. The lytA non-essential controls typically gave 150–200 colonies per transformation and the ftsZ positive controls gave 0 colonies
per transformation. Experimentally, most non-essential genes gave 200–300 colonies per transformation (lytA was found to be on the low end of the spectrum) and essential genes gave results
similar to ftsZ. An occasional single chloramphenicol-resistant colony was seen for some experiments, presumably due to contamination or non-homologous integration of the construct. For verification, all essential genes were subjected to a second independent gene disruption experiment
Do the above control expts enough for me to prove that i have knocked out the gene or do i know need to find a negative selection marker??
Edited by cloneboy, 26 August 2009 - 07:24 AM.