Primer concentration - (Jul/21/2011 )

Hi,


I´m a little confused caused by Primer concentration dilution.
Our stock concentration is 20µM, working con. Is 2µM.
How do I dilute this:
Primer from vendor: 100µM (dilutet by me)
Now I take 20µL PrimerF and 20µL PrimerR and fill up to 200µL with water. That´s our 20µM stock.
To get the working concentration I take 1µL stock + 9µL water-> 2µM.

But I can dilute 2µL PrimerF and 2µL PrimerR and fill up to 200µL with water. That´s our 2µM stock.
So I can get directly my 2µM concentration.

Now I need 500nM working concentration and I use 2µL PrimerF and 2µL PrimerR and fill up to 640µL.

Make that sense?


Greetings,
Mathias

To be sure we understand each other, there are some definitions:
1 µM equals 1 pmol/µl
Stock concentration is what you keep safe for diluting to working concentration, better rather concentrated because that prevents degradation. 100 µM or 200 µM is usually used as a stock concentration.
We use 100 µM.

Working concentration is what you pipet from to PCR, in volume corresponding to the frequency of usage (for 10 reactions every week make less than for 10 reaction every day) and concentration corresponding to the desired final concentration in PCR reaction (ideally pipet around 1 - 2 µl of working concentration to a single reaction, if you do really many samples at once, the volume can be smaller and concentration higher).
We use 10 µM each to use 1 µl of it for single reaction.

Final concentration in PCR reaction is what you get when you pipet your working concentration to the desired PCR volume. That is usualy recomended to be 0.3 - 1 µM, with 0.5 µM (500 nM) of each as a starting concentration.
As we pipet 1 µl of our 10 µM each (10 pmol/µl) primers to 20 µl reaction, final concentration is 10/20 = 0.5 uM of each.

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Now your stock concentration (20 µM) is pretty low, I would suggest to use higher. Working concentration seems also low for classic PCR, but you may have reason for it. But as you say you need 0.5 µM working concentration I suspect you maybe confuse working concentration and final concentration in PCR reaction. So what concentration (final) do you need in your PCR?

Your calculations of stock concentrations are wrong. If you have 20 µl of 100µM primer in 200 µl, it's 10µM, not 20µM. You need to calculate 20 µM for each, not for both together (or use the primers separately, not mixed together). Correct is to take 40 µl of each primer and fill up to 200. The calculation of 2µM and 500nM concentration has the same mistake, so it's also wrong. And I wouldn't recommend to have 640 µl of working concentration unless you use up this amount say in one month. It's better to have smaller volumes of working concentration so you don't store it too long and for changing to freshly made working concentration in case you suspect contamination.

I'm confused as to what your final concentration needs to be also.

Use C1V1=C2V2. If you're still confused try using the "Dilute a stock solution" calculator on http://www.graphpad.com/quickcalcs/molarityform.cfm
It will give you the answer, but then also try and work it out for yourself so it makes sense.

Hi,


ok, you are right. My mistake was that I calculate 20µM for both primer together not each.

Yes indeed.

I try again:

Stock: 100µM
Working: 10µM


I think I get it.

For further dilutions I take 20µL 100µM in 200µL to get 10µM, right? And 40µL 100µM in 200µL for 20µM.
And for both primer in a dilution to get a 10µM working concentration: 20µl 100µM F-Primer and 20µl 100µM R-Primer and fill up with 160µl water to 200µl total vol. Right?
And for both primer in a dilution to get a 20µM working concentration: 40µl 100µM F-Primer and 40µl 100µM R-Primer and fill up with 120µl water to 200µl total vol.

The more I think about the more confusing is it.

Thanks a lot!!!

muc77 on Fri Jul 22 05:03:13 2011 said:

Yes, that's right.

If it helps you, for this type of calculations I usually imagine the concentration as number of moles floating in corresponding volume (I do actually see them floating in my mind, but that's not required ). Since you mostly work in µl volumes, I "convert" 10 µM to 10 pmol/µl and imagine 10 pmoles floating in one µl. So if you take 20 µl of this stock, you just calculate 10*20=200 that's the number of pmoles floating in 20 µl I took out. Then I look at the final volume of dilution. I have the final volume say 200 µl, so there will be 200 pmoles I took earlier floating in 200 µl and to get back the concentration I divide it by the number of µls, 200/200=1 that is 1 pmole in µl which equals 1 µM.