E-Z Strain Construction

As preliminary data for our genome-wide recombination analysis (outlined in this post from Rosie), we want to sequence the whole genome of a single transformed clone in the next couple of months. The idea is to transform our standard KW20 Rd strain with DNA from one of the other completely sequenced strains (probably 86-028NP, possibly PittGG), select a single transformed colony, and sequence its genome.

This will provide us with all sorts of useful preliminary results:

1. Show that we can indeed handle the type (and amount) of data we’ll be obtaining.
2. Estimate the total amount of donor DNA a single recipient recombines (and fixes) into its genome.
3. Estimate the length of recombination tracts (gene conversions) / the strength of “linkage”.
4. Estimate mosaicism of donor and recipient sequences (mismatch repair).
5. Estimate the transformation rates for different classes of single-nucleotide differences (for example, the number of A->T transformation events observed versus the total A->T differences between the strains)

In particular, item (2) will be crucial for estimating the total amount of sequencing we would need to measure transformation rates per polymorphism across the genome. Simple transformation assays with DNA from the multi-antibiotic resistant MAP7 strain suggest that possibly 20-50kb of DNA may be replaced in a single transformant, but this type of analysis is restricted to only a few different sites in the genome and is very roughly calculated.

The analysis of a single transformed genome will still be preliminary with regards to (3)-(5), for which we will want genome sequences for several independent clones. In the future we are likely to barcode and pool independent transformants, since we expect that a single lane of Illumina sequencing will be overkill for a single Haemophilus genome of less than 2 Mb (250X sequence coverage).

Anyways, one issue with producing the material for this first sequencing experiment is that we need to make sure that the clone we select comes from a cell that was indeed competent and did indeed get transformed. Since only a fraction of cells in a competent culture are competent, we would be wasting a lot of time and money, if we accidentally just re-sequenced our recipient genome.

In order for this to work, we need our donor strain to carry an antibiotic resistance marker. By selecting for recipients that become resistant, we can be sure the clone we select took up DNA that got recombined into the genome. (This may also create a bias for donor alleles near the selected site, due to “linkage”.)

To this end, I am doing the following:

1. Made a couple strains (KW20, 86-028NP, and PittGG) resistant to novobiocin. I just did this. It worked like a charm thanks to the former postdoc having a well-organized lab notebook and a well-organized freezer box containing a tube with a NovR allele of gyrB already prepared for me. This was also my first time doing overnight transformations. I couldn’t believe how easy it was: Add a frozen aliquot of cells and some DNA to some sBHI media, let the cells grow overnight, and plate them the next day. There were plenty of resistant colonies this morning.
2. Prepare DNA from the newly produced 86-028NP NovR and PittGG NovR strains. I’ll do this tomorrow from the overnight cultures I just inoculated.
3. Transform KW20 with this DNA. I’ll use competent cells I already have tomorrow, after my DNA prep.
4. Saturday, assuming I have NovR transformants, I’ll pick and grow up some transformed colonies overnight.
5. Sunday, I can prepare this DNA, and that’ll be what we can send for sequencing!

So if all goes well, we should have our material in a few days! Then we wait. Then the real work begins…

(As a side note, 86-028NP indeed appears to already be resistant to another antibiotic, nalidixic acid. I will check to see if this resistance is transformable when I have the 86-028NP NovR DNA in hand.) (continued...)

Congression versus linkage

Yet again, I did a transformation of Haemophilus influenzae cells with slightly variant H. influenzae genomic DNA. I fed MAP7 DNA-- containing several antibiotic resistance-conferring mutations-- to KW20 competent cells (RR722). I used my third competent cell prep for the third time (experiment 3-3), using frozen stocks. Things looked pretty good, and now I’ve got some real co-transformation numbers to play with to distinguish “congression” from “linkage”...

This time, I selected for four different markers: Resistance to Kan, Nov, Spc, and Nal. As expected, last week’s failure with Spc was due to a mistaken antibiotic concentration. Here’s what the transformation frequencies for each independent marker looked like:

Apparently, all four mutations in MAP7 are point mutations. The good news is that there is indeed variation in transformation rates (~4-fold); the bad news is that the rates I’m getting will require much greater than 10,000X sequence coverage in our planned genome-wide experiment. I may very well need to turn to older methods involving blood in the media to see if I can get rates substantially higher... Or possibly a hyper-rec mutation in the recipient strain.

While Kan and Nov are tightly linked, the other two are unlinked. I should be able to distinguish “congression” from “linkage” by looking at co-transformation rates. I decided to look at co-transformation of the Kan marker relative to the other three, rather than make every possible kind of antibiotic plate for the four markers (which would be 24 kinds of plates).

Here’s what the observed and expected rates of co-transformation looked like (where “expected” was calculated as the product of the independent transformation rates and the ratio of obs/exp is indicated by the number above each pair of bars):
Indeed the co-transformation rate of Kan and Nov dramatically exceeded the expected rate, relative to that of Kan versus Spc or Nal. This is what we expect for linked markers. So the obs/exp ratio I’m seeing for KanSpc and KanNal are presumably due to “congression” rather than “linkage”. That is, not all cells are equally competent in a culture, so I see excess co-transformation as an artifact.

However, if we use the measure Cf to calculate the fraction of competent cells in a culture, we arrive at different answers depending on whether we use the KanSpc or Kan Nal rates.

So, for some definitions (as per Goodgal and Herriot 1961 and others):

If markers behave totally independently:
f(ab) = f(a) * f(b).

Since they don’t, we can calculate the fraction of competent cells as:
Cf = f(a) * f(b) / f(ab).

But then, for two different pairs of unlinked markers:
Cf (KanSpc) = 41%
Cf (KanNal) = 18%

So something more is going on here, since we’re getting different fractions of competent cells, despite it being only a single culture. Thus the assumptions required for the Cf value to work are not entirely valid.

One explanation for this was put forward by Erickson and Copeland 1973, who observed differences in congression for different sets of unlinked markers in B. subtilis. They showed a relationship between co-transformation rates and the position of markers relative to the origin of replication. Thus, co-transformation rates may be influenced by whether or not the transforming markers are recombining into the recipient chromosome before or after the replication fork.

I’ll need more information about the exact identity of the markers I’m using to examine this model with these data. It also would’ve been nice in this case to have SpcNal co-transformation rates.

As an aside, if competence is maximal during DNA replication, this lends support to a conflated DNA repair/food hypothesis for natural competence. Nucleotides are needed during DNA replication, and DNA that’s taken up during this period could serve to faciliate DNA replication, or maybe DNA repair...

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A couple other notes:

Now that I’ve done three experiments with the same competent cells, I can look at how reproducible I am. For exp3-3 and exp3-2, I used frozen stocks, while exp3-1 used fresh cells:
Not stellar, but it looks to me that fresh is better than frozen.

Another note: I might be able to increase relative transformation rate for the genome-wide experiment by selecting for a marker right away. This creates some problems, but it allows me to remove “incompetent” cells from the population and focus only on those cells that were actually able to take up DNA and be transformed, possibly significantly improving observed transformation rates:
Of course, tightly linked markers will look artificially high, but I’d get 2X the Spc transformants and 6X the Nal transformants under this regime. I’ll definitely need to understand congression better, as well as the problem of dead cells before seriously considering this for the bulk transformation experiments.

Another (probably better) possibility would be to somehow fractionate transformable cells from others in a competent culture as has been done with renografin gradients in B. subtilis. (continued...)