# dNTP Quantity

### #1

Posted 10 April 2009 - 09:55 AM

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

Posted 10 April 2009 - 10:29 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

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

Posted 10 April 2009 - 06:01 PM

### #4

Posted 11 April 2009 - 08:30 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

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