Hi. We are toying with the idea of doing some whole genome work with E. coli colonies. Given that the Ion Torrent only requires an input library of 26 picomolar, I suspect that there will be sufficient DNA in a single E. coli colony. Does this sound reasonable? The Ion Torrent fragmentation kit requires an input of 50 - 100 ng only. Cheers.
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26 pM as the input for emPCR, but you'll probably need a higher concentration to get a reliable estimate of concentration from a Bioanalyzer or Qubit. My rule of thumb was to have a dilution factor as a high-2 or low-3 digit number (say, 50 to 500). Too low and the concentration measurement might be unreliable. Too high and it's probably over amplified and therefore less representative of the starting material. The more concentrated, the better for long term storage/archival.
But to your original question, it might be just enough. Life Tech posits ~0.017 pg DNA/E coli cell. Therefore ~3M cells for 50 ng @ 100% awesomeness, that's ~ 21 cell divisions [log_2(3M)]. On antibiotics, that's probably ~14-21 hours growth.
So yeah, that might just work. Someone please check my math.
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I wonder how Life Tech arrived at that weight of DNA? I always think of 1 pg = 1 gigabase (and http://en.wikipedia.org/wiki/Genome_...pairs_.28bp.29 backs that up.) So 0.017 pg would be a 17 Mb genome, or nearly 4X more than it actually is (4.6 Mb). Plasmids? Replication? Anyway, it is possible (see http://jac.oxfordjournals.org/conten...kt494.full.pdf for a nextera protocol).Providing nextRAD genotyping and PacBio sequencing services. http://snpsaurus.com
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At first I thought, "must be all the mitochondrial DNA" then somewhat soon after realized that wasn't likely!Providing nextRAD genotyping and PacBio sequencing services. http://snpsaurus.com
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