Posted 19 March 2009 - 10:34 AM
I take 5 ul of dNTPs (1mM stock solution) to put into a PCR reaction --> I suppose this is 0.25 mM of each dATP, dCTP, dGTP and dTTP (¿?)
PCR reaction = 50 ul
According to my calculations (C1V1=C2V2) --> C2 = [(1000 pmol/ul)(5 ul)] / 50 ul = 100 pmol/ul
That is, the concentration of dNTPs in the 50 ul PCR reaction is 100 uM. This in turn means that I have 25 uM of each individual dNTP (i.e., dATP, dCTP........dTTP) in the 50 ul PCR reaction.
Is all this correct or am I completely ¿?
I am comparing two protocols, and 25 uM of each dNTP looks a bit low.
THANKS EVERYBODY FOR YOUR HELP !!!!!!!!!!!!!! GOD BLESS YOU ALL !!!!!!
Posted 19 March 2009 - 11:02 AM
Posted 19 March 2009 - 12:21 PM
Is your mix comercial or home made?
I usually use between 100-200uM of dNTPs (mix) and always make the stock mix at 10mM which is 10mM of dATP, 10mM dCTP, 10mM dGTP and 10mM dTTP.
In your case, if the stock is 1mM I'm pretty sure is 1mM dATP, 1mM dCTP, 1mM dGTP and 1mM dTTP, if this is the case you are adding 100uM of each dNTP which should be good enough.
check the supplier or whoever made the stock.
hope this helps,