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dNTP calculation - (Apr/29/2005 )

For example I have a bottle with 5mM and 1.25 ml dNTP. According to the protocol i need 200 uM each. Do you know how much volume should i use?

-seasons-

that depends on the final volume of your PCR.

to get from 5mM to 200uM concentration you will have to dilute the dNTPs 1:25. So the volume you should use will be 1:25 of the final volume of your PCR reaction..


good luck!!

Nick tongue.gif

-methylnick-

QUOTE (methylnick @ Apr 29 2005, 04:56 PM)
that depends on the final volume of your PCR.

to get from 5mM to 200uM concentration you will have to dilute the dNTPs 1:25. So the volume you should use will be 1:25 of the final volume of your PCR reaction..


good luck!!

Nick tongue.gif



It is a 50ul pcr reaction. The question is when it says 200 uM each does it mean 200 uM for each nucleotide or is that the final concentration of all nucleotides which is needed in the reaction? unsure.gif

-seasons-

that means the total concentration of dNTP is 800┬ÁM

-fred_33-

200uM of each dNTP

-methylnick-

an easy way to calculate the volumes required for any dilution you want to do is as follows

C1*V1=C2*V2

where concentration 1 (C1) is the concentration you want, Volume 1 (V1) is the volume you want and C2 is the concentration you have already, therefore you can work out V2.

for example, you have a 50 ul reaction (V1) and you want 200 uM dNTPs (C1) and your starting dNTP concentration is 5 mM = 5000 uM (C2). Make sure your units are all the same otherwise you will be out by factors of the unit change.

so... 50 uL*200 uM =5000 uM *V2, therefore V2= (50*200)/5000

=10000/5000
=2uL.

As you can see you will use 2 uL of dNTP mix per reaction.

Bob

-bob1-