 # dNTP Quantity - (Apr/10/2009 )

Hi Everybody,
how do we count numbers(how many) of "A" in 1mM dATP solution?
Actually i wanna know that, can we theoritically able to calculate the actual amount of dNTPs(in numbers) required for 30 cycles in PCR? if yes then please tell me.

Thanks

-pinokyo-

pinokyo on Apr 10 2009, 01:55 PM said:

Hi Everybody,
how do we count numbers(how many) of "A" in 1mM dATP solution?
Actually i wanna know that, can we theoritically able to calculate the actual amount of dNTPs(in numbers) required for 30 cycles in PCR? if yes then please tell me.

Thanks

A 1 M solution of anything dATP or otherwise is defined as having 6.023 x10^23 molecules (Avagadro's Number). Therefore, a 1 mM solution has 1000x fewer molecules = 6.023 x10^20 molecules.

-Dr Teeth-

That's the number of molecules in one milli- Mole, not the number in a 1 mM solution, which is 1 milli-mole per liter. You need to tell us the volume to determine the number of molecules. So, a 10 ul reaction with 1 mM dATP would have 10**-5 times fewer molecules, or 6 x 10**15 molecules.

-phage434-

Dr Teeth on Apr 10 2009, 11:59 PM said:

pinokyo on Apr 10 2009, 01:55 PM said:

Hi Everybody,
how do we count numbers(how many) of "A" in 1mM dATP solution?
Actually i wanna know that, can we theoritically able to calculate the actual amount of dNTPs(in numbers) required for 30 cycles in PCR? if yes then please tell me.

Thanks

A 1 M solution of anything dATP or otherwise is defined as having 6.023 x10^23 molecules (Avagadro's Number). Therefore, a 1 mM solution has 1000x fewer molecules = 6.023 x10^20 molecules.

Thanks
Thank u very much

-pinokyo-