The RecJExoI-- DNA is made so that a TetR gene interrupts the RecJ gene, and a KanR gene interrupts the ExoI gene. If genes were words, it'd look something like this:
...Re-[TetR]-cJ...These two genes are about 190-thousand base pairs apart on the H.influenzae chromosome.
For the experiment, I plated cells onto plain, Kan, Tet, and Kan+Tet plates. Why? Because
-Wildtype cells that grew on Kan must have taken up ExoI DNA (or both ExoI and RecJ DNAs),
-WT cells growing on Tet must have taken up RecJ DNA (or both ExoI and RecJ DNAs),
-And WTs growing on Kan+Tet means they must have taken up both DNAs.
If the KanR and TetR transformations were independent (i.e. they were never on the same strand of DNA and all cells in the culture were competent), then:
T.Freq. (Kan) x T.Freq. (Tet) = T.Freq. (Kan and Tet)(Simple probability here, folks. The probability of two events occuring is the product of the probability of one and the probability of the other.) This was not the case, since (5.5E-4) x (5.0E-4) doesn't equal (1.0E-6). I saw higher numbers than expected.
Are all the cells in the culture equally competent?
YES: recJ and exoI genes are linked.
NO: congression is at work.
The first one is basic. If the two genes can be found on one intact strand of DNA, then the chances of gaining KanR and TetR are higher than if one cell had to take up two seperate strands of DNA. But this explanation depends on all of the cells in the culture being fully competent.
The second slightly more twisted.
Let's say you observe that 20% of all cells in the culture transform to become KanR and 20% become TetR. 8% become double mutants.
Through calculations, you'd expect (20%)(20%) = 4% of the culture to become double mutants.
Let's say we know that only half the cells in my culture are actually competent. Since transformation frequency is the number of cells out of all competent cells that transform,
instead of 20 KanRs out of 100 total cells (for example), there are actually 20 KanRs out of 50 competent cells, making the actual KanR transformation frequency 40%.
So we now have KanR and TetR transformation frequencies of 40%. We expect that out of 50 competent cells, we'll have (40%)(40%) = 16% KanR+TetR double mutants, or 8 double mutants out of 50 competent cells. Including the other 50 cells that aren't competent, we have a total of 100 cells, but we still have 8 double mutants. So it looks like our "total" transformation frequency for double mutants is 8%.
Using a 48-year old formula by Goodgal and Herriot, the fraction of competent cells for this example is:
T.Freq. (Kan) x T.Freq. (Tet) / T.Freq (Kan and Tet)Since the RecJ and ExoI genes are "just on the brink" of being linked, I'm probably seeing congression. For my experiment:
=(20%) x (20%) / (8%)
(Yes, I engineered the numbers to work out.)
T.Freq. (Kan) x T.Freq. (Tet) / T.Freq (Kan and Tet)I don't know what I'm supposed to think, but that seems pretty low.
=(5.5E-4) x (5.0E-4) / (1.0E-6)
=27.5% of my cells were competent!
Thanks to Josh for helping me understand this!