dilution question help - (Feb/25/2009 )

Say I have two DNA samples. One measures 200 ng/ul in concentration and the other measures 350 ng/ul.

My target concentration for both is 10ug, final volume is 200ul. How do I do this?

Stock A: 200ng/ul

need 10ug in 200ul, therefore final concentration is 10ug/200ul

CV=CV

stock concentration x required stock volume = final concentration x final volume
200ng/ul x V = 10ug/200ul x 200ul
V = 10ug / 200ng/ul
V = 10ug / 0.2ug/ul
V= 50ul

Therefore the required stock volume is 50ul + 150ul buffer/water



Stock B: 350ng/ul

stock concentration x required stock volume = final concentration x final volume
350ng/ul x V = 10ug/200ul x 200ul
V = 10ug / 350ng/ul
V = 10ug / 0.35ug/ul
V= 28.6ul

Therefore the required stock volume is 28.6ul + 171.4ul buffer/water


Hope this may help.

Thanks for the formula.

Say I have DNA that measured 1 mg/ml concentration and my target concentration is 100ug, 50ul final, how would I do this? I tried using it with the formula below, but it doesn't work very well. How can I get 100ug of 1 mg/ml somehow?

1mg/ml = 1ug/ul

(1ug/ul)*V = (100ug/50ul)* 50ul
(1ug/ul)*V=(2ug/ul)*50ul
V = 100ug/(1ug/ul)
V = 100ul

claritylight on Feb 25 2009, 07:23 PM said:

The final concentration is 2ug/ul which is 2mg/ml... twice as much as the stock concentration!

You may need a higher concentrated stock solution.

Minnie Mouse on Feb 25 2009, 10:32 PM said:

Ok, I try with 100ng. If I do the calculation again:

1ug/ul = 1000ng/ul

(1000 ng/ul)*V = (100ng/50ul)*50ul
(1000 ng/ul)*V = (2ng/ul)*50ul
V = (100ng)/(1000 ng/ul)
V = 0.1 ul

Is that correct?

claritylight on Feb 26 2009, 09:09 AM said:

yes, but you cannot measure 0.1ul accurately.

You may need to take 1ul of stock solution and diluted it into a 9ul of buffer/water (1:10 dilution)
Then take a 1ul of this solution and add 49ul of buffer/water.

Hope this may help.

yes thank you!!!