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DNA detection limit of the PCR - Calculating detection limits (Mar/08/2011 )

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

-Lankan-

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

-ivanbio-

ivanbio on Wed Mar 9 01:56:57 2011 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


Thanks a lot. It does help indeed. May I know a referance for this equation please?

-Lankan-

ivanbio on Wed Mar 9 01:56:57 2011 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:

I made a web aplication for calculating copy number, just the average weight I used is 660 and not 649 grams per mole, that number seems to vary a bit depending on the way of calculation (do you know is there any recent change in consensus about what is the right number?).
However it gives the same copy number (mine is rounded to 2 x 105 copies per Ál).


Lankan on Wed Mar 9 15:23:10 2011 said:


Thanks a lot. It does help indeed. May I know a referance for this equation please?

I'm not sure that there is any reference, it is just basic chemical calculation, transforming concentration in grams to moles, and then moles to copies. You can google other copy number calculators apart from mine, that uses same equation.

-Trof-

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

-alpo-