I am doing a PCR for an intra cellular bacteria (conventioanl PCR)
As I do not have facilities to grow it I got DNA extracted from a culture from a referance lab as my positive control
I know the concentration of DNA in that
I did a dilution titre to establish the detection limit
I know the amount to DNa in a reaction tube needed for the detection
But... I need to know how to convert DNA amounts in nanogrammes to copies/ cfu as in most places the dections limit is given in that manner
Also I need a referance article to cite accpetable detection limits eg. is it 10 ng or 100 ng etc
Thank you very much
DNA detection limit of the PCR
Started by Lankan, Mar 08 2011 04:53 PM
4 replies to this topic
#1
Posted 08 March 2011 - 04:53 PM
#2
Posted 08 March 2011 - 05:56 PM
I am not aware of an article that describes acceptable limits, but to convert DNA amounts to copies you can follow these calculations:
Assuming your bacteria is like E. coli, then the genome size is: 4.6 million base pairs. If the concentration of your solution is 1 nanogram per microliter, then you have 201,715 genomes (or copies) per microliter. You can calculate this as follows:
Micrograms in 1 pmole (ug/pmole) = (length of DNA in bp) x 649 g/mole x 10^-12 moles/pmole x 10^6 ug/g
Copies = (ng/ul of DNA / ug/pmole) x 10^-12 moles/pmole x 10^-3 mg/g x 6.022x10^23 copies per mole
The factors you need to know for this calculations are as follows:
The average base pair weighs 649 grams per mole
Avogadro's number is 6.022x10^23 moles per gram
One gram is equal to 1x10^9 nanograms
Hope this helps
Assuming your bacteria is like E. coli, then the genome size is: 4.6 million base pairs. If the concentration of your solution is 1 nanogram per microliter, then you have 201,715 genomes (or copies) per microliter. You can calculate this as follows:
Micrograms in 1 pmole (ug/pmole) = (length of DNA in bp) x 649 g/mole x 10^-12 moles/pmole x 10^6 ug/g
Copies = (ng/ul of DNA / ug/pmole) x 10^-12 moles/pmole x 10^-3 mg/g x 6.022x10^23 copies per mole
The factors you need to know for this calculations are as follows:
The average base pair weighs 649 grams per mole
Avogadro's number is 6.022x10^23 moles per gram
One gram is equal to 1x10^9 nanograms
Hope this helps
Ivan
Carlsbad, CA
#3
Posted 09 March 2011 - 07:23 AM
ivanbio, on 08 March 2011 - 05:56 PM, said:
I am not aware of an article that describes acceptable limits, but to convert DNA amounts to copies you can follow these calculations:
Assuming your bacteria is like E. coli, then the genome size is: 4.6 million base pairs. If the concentration of your solution is 1 nanogram per microliter, then you have 201,715 genomes (or copies) per microliter. You can calculate this as follows:
Micrograms in 1 pmole (ug/pmole) = (length of DNA in bp) x 649 g/mole x 10^-12 moles/pmole x 10^6 ug/g
Copies = (ng/ul of DNA / ug/pmole) x 10^-12 moles/pmole x 10^-3 mg/g x 6.022x10^23 copies per mole
The factors you need to know for this calculations are as follows:
The average base pair weighs 649 grams per mole
Avogadro's number is 6.022x10^23 moles per gram
One gram is equal to 1x10^9 nanograms
Hope this helps
Assuming your bacteria is like E. coli, then the genome size is: 4.6 million base pairs. If the concentration of your solution is 1 nanogram per microliter, then you have 201,715 genomes (or copies) per microliter. You can calculate this as follows:
Micrograms in 1 pmole (ug/pmole) = (length of DNA in bp) x 649 g/mole x 10^-12 moles/pmole x 10^6 ug/g
Copies = (ng/ul of DNA / ug/pmole) x 10^-12 moles/pmole x 10^-3 mg/g x 6.022x10^23 copies per mole
The factors you need to know for this calculations are as follows:
The average base pair weighs 649 grams per mole
Avogadro's number is 6.022x10^23 moles per gram
One gram is equal to 1x10^9 nanograms
Hope this helps
Thanks a lot. It does help indeed. May I know a referance for this equation please?
#4
Posted 15 March 2011 - 04:23 AM
ivanbio, on 08 March 2011 - 05:56 PM, said:
Assuming your bacteria is like E. coli, then the genome size is: 4.6 million base pairs. If the concentration of your solution is 1 nanogram per microliter, then you have 201,715 genomes (or copies) per microliter. You can calculate this as follows:
However it gives the same copy number (mine is rounded to 2 x 105 copies per µl).
Lankan, on 09 March 2011 - 07:23 AM, said:
Thanks a lot. It does help indeed. May I know a referance for this equation please?
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#5
Posted 07 May 2013 - 12:13 PM
I'm doing RT PCR and look for a protocol to convert ng from starting quantity into CFU (Log CFU/g)
Thank you very much
Thank you very much













