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# primer dilutions

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### #1 molbio1234

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Posted 23 April 2012 - 04:12 PM

hello,

I have:
F primer @ 100uM
R primer @ 100uM

2uL taken from each and placed into 8ul water, to make 10uL final volume primer mix. Each primer is (2/10)*100 = 20uM in the mix.

How to calculate the concentration of the mix? Will it be (4/10)*200uM = 80uM or (4/10)*100uM = 40uM

if I have :
F primer @ 100uM
R primer @ 100uM

i add 10uL of each into a new tube and mix. Then I will have F and R primer each at 50uM in the mix. The primer mix F/R is 100uM. Is that right?

If I have:
10uM F/R primer mix, is it possible to determine the concentration of F and R in the mix, is it simply 5uM each?

### #2 pcrman

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Posted 23 April 2012 - 05:06 PM

i add 10uL of each into a new tube and mix. Then I will have F and R primer each at 50uM in the mix. The primer mix F/R is 100uM. Is that right?

Right. For each primer, the concentration is 50 uM, for the total oligos, the concentration is still 100 uM.

If I have: 10uM F/R primer mix, is it possible to determine the concentration of F and R in the mix, is it simply 5uM each?

Yes, the concentration for each is 5 uM.

### #3 molbio1234

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Posted 23 April 2012 - 06:57 PM

Hi pcrman,

thanks. So is the first one, 40uM then?

### #4 pcrman

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Posted 23 April 2012 - 07:44 PM

I am a little confused about your first question, you said to take 2 ul from each and add to 8 ul. then the total will be 12 ul.

I assume you mean take 1 ul from each and add to 8 ul so the total will be 10 ul. If that is the case, the final concentration for each oligo will be 10 uM and the total oligo concentration will be 20 uM.

### #5 molbio1234

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Posted 24 April 2012 - 02:57 AM

yes that is what i meant. thank you pcrman.