We're collaborating with some physicists on a cool laser-tweezers experiment. The tests are done on single cells, so it's important that the cells used for this experiment are as competent as possible. Our collaborator just asked " Is there some quantitative way to go between the transformation frequency of a culture and the fraction of cells that are competent, so we'll know, for a particular batch of cells, how many we need to try?
Yes, and it's a very simple old-fashioned experiment.
Cells that are maximally competent take up at least several DNA fragments. If we give them chromosomal DNA carrying two different antibiotic resistance genes (say novR and nalR), some cells will become resistant to both antibiotics because they took up both genes (double transformants). For this analysis it's important that the two genes be far enough apart on the chromosome that they will be in different DNA fragments. If all the cells are competent, we can predict the frequency of these double transformants; it's the product of the frequencies of single transformants. So if 1% (0.01 or 10^-2) of the cells in this experiment became NovR, and 0.5% (0.005 or 5x10^-3) of the cells became NalR, we predict that 0.005% (0.00005 or 5x10^-5) of the cells will be NovR NalR.
But what if not all the cells were competent? Say only 10% of them were. Then the 1% transformation frequency to NovR was because 10% of the cells had a transformation frequency of 10% and 90% had a transformation frequency of 0. Similarly the 0.5% transformation frequency to NalR would be because 10% of the cells had a transformation frequency of 5% and 90% of the cells had a transformation frequency of 0. Now the predicted frequency of double transformants is different. The 10% that are competent will have a double transformation frequency of 0.5%, and the 90% not competent a frequency of 0, giving a total frequency of 0.05%. This is ten-fold higher than if all the cells were competent.
So, a double transformant frequency higher than expected if all the cells were competent means that only a fraction of the cells were competent. On the other hand, if all cells were only a bit competent, so most cells were only taking up a single fragment, the same analysis would give a double transformation frequency lower than expected. This was all worked out 30 or more years ago.
When we do this analysis on competent H. influenzae cells, we always find that the double transformant frequency is higher than expected if all the cells were competent, which means that only some of the cells are competent, and that the competent cells are taking up multiple DNA fragments. (For reasons having to do with how the DNA is integrated into the chromosome, there's a fudge factor of 2.) When the transformation frequency for the single markers is high, the calculations indicate that about half of the cells are taking up DNA. When the single-transformant frequency is low (in some mutants, and in cells from conditions that only weakly induce competence), fewer cells are competent, but the ones that are are still taking up multiple DNA fragments.
So, we've usually used the single-marker transformation frequency as an indicator of the fraction of the cells that are competent. Maybe for the laser-tweezers experiments we should check again.
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Could you explain the fudge factor of two a little bit? I'm interested to know how/why/if it is exactly two.
ReplyDeleteI think it's somewhere between 0 and 4. It arises partly because only a single strand is replaced. This ccreaters a two-fold dependence of TF on whether the cells divide before being plated. The strand that's recombined in may also be 'corrected' by mismatch repair.
ReplyDeleteI think it's somewhere between 0 and 4. It arises partly because only a single strand is replaced. This ccreaters a two-fold dependence of TF on whether the cells divide before being plated. The strand that's recombined in may also be 'corrected' by mismatch repair.
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