How to do a primer dilution - (Apr/06/2014 )

Hi guys,

 

I'm trying to reconstitute primers for use in a PCR reaction and I'm getting confused by the calculations

 

My primers are forward 66.7nmol and reverse 71.4 nmol

 

So if I add 667ul of water  to the forward and 714ul to the reverse primer I get a 100uM stock?

 

I need a final concentration of 100 pmol of each primer in a 25ul reaction so how much do I need to dilute my 100uM stock?

 

Any help would be appreciated!

 

 

 

yes its 100µM

 

so you have 100 x 10^-6 mol/liter and you need to have 100 x 10^-12 mol/liter (as a concentration) in 25 µl....

 

you need to dilute 10^6 times thus , but factor in the 25µl...

 

 

100 x 10^-12 mol for 1 liter, means for 25µl => ?

 

 

you should be able to calculate this yourself...

 

 

But are you sure you need 100pM ? Seems a bit low to me.

Yes I need 100 pmol final concentration of primer

 

So I need to dilute the 100uM 1 in a 1,000,000?

Final concentration of primers 100pM for a PCR? Is that some kind of weird PCR that uses 1000x lower final concentration in reaction, than the usual range? (0.1 - 0.5 µM)

 

Or is that pmol/µl or just something different molar concentration than per liter (so 100µM primers are equal to 100pmol/µl primers).

 

In any case, you don't want to dilute your primers to the final concentration, but to a working concentration 10-20-50x or so higher than the target one, you need to add other things in reaction too. The volume of the primers used must be suitably low, but not too low, to be pipetted accurately.

 

For example:

Stock solution of each primer is usually 100µM (100pmol/µl). 

 

You make a working solution of 10µM out of that (dilute 10x).

 

For a 20µl PCR reaction, you use 1 µl of working solution of each primer (and so have a space to add other reagents) to achieve final concentration in reaction 0.5µM of each. 

Thanks for your reply and calculation example.

 

I was taking the volume from a paper I read and it said to add 100pmol of each primer to a 25ul reaction, so I assumed that meant it was the final concentration of primer, or am I not interpreting that correctly?

if it said to add 100pmol to 25µl than you have 100pmol/25µl or 4pmol/µl or 4 x 10^-12mol/10^-6liter or 4 x 10^-6mol/liter or 4µM

So must I dilute some of my 100uM stock to get 4uM and add how much of this do I add to the reaction?

No.

You need 100pmol in your reaction, fix on that (thought that is quite a lot, on the contrary, but anyway..).

 

You have 100uM stock. You can also express this concentration as 100pmol/ul (that's equal to 100uM).

 

That means in one 1ul of stock you will have 100pmol of primer.

 

So, any idea how much you will use in your reaction?

Il need 2500 pmol so if I add 1ul of it to a 25ul reaction it will be diluted to 100pmol?

caroloconnor20 on Mon Apr 7 07:08:27 2014 said:

Il need 2500 pmol so if I add 1ul of it to a 25ul reaction it will be diluted to 100pmol?

no. you need 100 pmoles in the total reaction. that means you have to add 1 ul of the 100 um stock.

 

don't confuse mass with concentration (pmol with pM (pmol/liter)).