number of molecules in PCR - (Mar/18/2010 )
Hello there,
I have 25 ul of a PCR product of 177 bps at a concentration of 200ng/ul. This means that I have 5 ug of DNA in a tube. Now how do i calculate the number of fragments in this? I understand that it weights about 660g/mole per base pair. So accordingly, 177bps times 660g/mole per bp=1.16X10 5 g/ mole.
Now the point is I have to do a nested PCR using the above PCR product such that my template has ~800 million fragments. So how many micro litres do I need to take from the above PCR product to use as template in my nested PCR? I would appreciate any help in this.
Thanks
Franco
I think you need to use Avogadro's number to convert from moles to molecules.
6.022x10^23 molecules/mole
I once created an online aplication to calculate copy number.
The result for 200 ng/ul of 177bp fragment is 1 x 1012 copies per µl.
That means you should dilute your sample 1000x and take 0.8 ul if I'm correct.
Trof on Apr 14 2010, 11:20 AM said:
The result for 200 ng/ul of 177bp fragment is 1 x 1012 copies per µl.
That means you should dilute your sample 1000x and take 0.8 ul if I'm correct.
Neat tool. Bookmarked it. Thanks Trof.