How do I measure 1.0pmole and 0.05pmole of primer for PCR?

Question

I have a stock concentration of 100uM, these volumes are so small they are confusing me - I know it is possible because my PCR protocol calls for 1.0 and 0.05 pmol of primers. Any help? THANKS!

Answer 1

If you have a stock concentration of 100uM, you have 100*10^6 pmol/L in your solution because 1uM = 10^6 pmol. If you only want 1 pmol in your PCR reaction, you need to figure out how many microliters that is. You may want to double check the math, but I have that for 1 pmol, you'll want 0.01uL of your stock solution. For 0.05 pmol, you'll need to use 0.0005uL. Depending on how many reactions you're setting up, you may want to make a master mix, so you don't have to measure out those tiny volumes.

Answer 2

As dixichc615 said 100 uM is 100*10^6 pmole/L= 100 pmole/ uL

I always make a dilution so that my primers have
10 uM= 10 pmole/ uL and use that for the PCR. In your case you need even more dilute solution so that you can pipet accurately.

You should make a 1 uM solution (1 pmol/ul). Then you would pipet 1 ul for the 1pmole case

For the 0.05 pmole you need to dilute even more. E.g. make a 0.05 uM (0.05 pmole/ul) and pipet 1 ul from that.Use the 1 uM as a staring solution to prepare the 0.05 uM (1/20 dilution)

DON'T dilute all of your stock. If you will set up n tubes for PCR calculate how much volume of the dilute solutions you need for n+1 tubes and use the suitable volume of concentrated solution (M1*V1=M2*V2). However if you have just a few tubes, you will have to prepare 100 ul of 1uM by mixing 1ul of 100uM and 99ul water in order to have volumes which can be accurately and reliably pipetted. As for the 0.05 uM the least volume I'd prepare is 20 ul by mixing 1ul of 1 uM and 19 ul of water.
Don't use less than 1 ul from any stock solution (when doing dilutions) because the error will be huge.

Your protocol is using very small quantities of primers (I use 20-30 pmole for standard PCR). What exactly are you doing?