Molecular Biology problem!?

T

Troubled Man

New member
A reaction has been designed to be carried out in a 2ml total volume. It requires a final NAD concentration of 3mM. If the protocol allows 60 µl for the NAD addition, what concentration would you make up the stock solution?
 
That's a rule of three problem. Nothing to do with molecular biology. It's easy.

3mM NAD / 1000ml
(that's your final concentration, 3mM per one litre)

Now you want 2 ml of that, so you do a rule of three

( 3mM NAD / 1000ml ) = ( x / 2ml )
(you multiply your 3 by 2 and divide that by 1000)

That will give you the number of total moles you need in your 2ml. You take that amount and then calculate by another rule of three how many moles you need in your 60µl,

( x / 2ml ) = ( y / 60µl )

( x * 60µl ) / 2 ml

then from that amount do another rule of three and calculate your molarity (concentration) over the volume you want. Obviously you'll have to have all your units on the same level (e.g. convert them either all to ml and mM or to µl and M.. whatever floats your boat)
 
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