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# dNTP Quantity

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### #1 pinokyo

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Posted 10 April 2009 - 09:55 AM

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

### #2 Dr Teeth

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Posted 10 April 2009 - 10:29 AM

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.

Science is simply common sense at its best that is rigidly accurate in observation and merciless to fallacy in logic.
Thomas Henry Huxley

### #3 phage434

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Posted 10 April 2009 - 06:01 PM

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.

### #4 pinokyo

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Posted 11 April 2009 - 08:30 AM

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