Because most of the cells in our competent-cell props aren't actually competent, I'm going to transform the cells with a mixture of the other strain's DNA and a short fragment carrying an antibiotic resistance allele. By selecting for this allele I'll make sure that all the cells I send actually did take up DNA.
I have two fragments I can use; both are about 2.5 kb long, produced by PCR from genomic DNA (the postdoc is making them for me right now). One carries novobiocin resistance and the other carries nalidixic acid resistance. I think I should use both, in separate transformations, as this will control for the slight possibility that the unmapped gene they're looking for is in the selected fragment.
How much of each DNA should I use? I want a saturating amount of the chromosomal DNA, which I have already prepared; 1 µg in a 1 ml transformation should be fine. I should use a lot less of the fragment, because I want most of the DNA the cells take up to be chromosomal. But I need to use a substantial amount, as I want to have at least thousands and preferably hundreds of thousands of transformants. Transformation frequencies with pure fragments are typically high, with saturation reached at about 20 ng (I think). If the transformantion frequency with this fragment was 1%, I would expect to get about 10^7 transformants from 1 ml (10^9) cells, so using 1-10 ng of fragment DNA should be lots.
I can do several controls to check how much of the chromosomal DNA the antibiotic-resistant transformants had taken up. The chromosomal DNA carries a kanamycin cassette, so the best control will be to select for KanR NovR and KanR NalR double transformants. The frequency with which the NovR or NalR cells carry KanR is an estimate of the fraction of cells that carry any particular donor segment. It's an underestimate for SNP alleles (with homologs in both strains), and a good estimate for the efficiency of transformation of heterologous genes flanked by homologous segments.
Other controls that might not be worth the trouble:
- Transforming cells with both PCR fragments would give another estimate of SNP transformation frequencies.
- Transformations that contain only the PCR fragment would tell me how efficiently the chromosomal DNA is competing with the fragment for uptake. I think that more competition means that the transformants will contain more segments of the other strain's DNA.
Once I've resususpended the colonies I'll check the OD600 to estimate the cell density, dilute them down to OD600 = ~0.3 (~10^9 cfu/ml), and add glycerol and freeze multiple aliquots at -80 °C. I'll also plate dilutions to check the cfu/ml. I might as well also plate on kanamycin plates, to confirm that these cells do carry the expected segments of chromosomal DNA. If Everything checks out then on Monday I'll just pack the frozen cells in dry ice and ship them out.