Question

You have a stock of a DNA solution at 1.2mg/ml, the DNA fragment being 980bp in length. How do you prepare a dilution of this DNA that has a concentration of 0.5uM ?

Answer 1

Given that the DNA fragment is 980 base pair (bp) in length.

The average molecular mass of a single base pair = 650 g/ 1 mol bp

Hence the average molecular mass of the DNA fragment containing 980 bp = 980 mol bp x 650 g /1 mol bp

= 637000 g / mol DNA fragment

Given the initial concentration of DNA, M1 = 1.2 mg/mL = [1.2 mg x (1 g / 1000 mg)] / [mLx(1 L / 1000 mL)]

= 1.2 g / L

= 1.2 g x (1 mol DNA fragment / 637000 g) / L

= 1.88 x 10^-6 mol / L = 1.88 x 10^-6 M = 1.88 uM

Hence the concentration of the stock solution, Ms = 1.88 uM

Now we need to dilute 1.88 uM stock solution to 0.5 uM required solution.

Since in the question the volume of 0.5 uM required is not given, let's consider it be 'V' L

Now suppose the volume of the stock solution(1.88 uM) solution taken be 'Vs' L

Now applying law of dilution

MsxVs = MxV

=> 1.88 uM x Vs = 0.5 uM x V

=> Vs =  (0.5 uM x V) / 1.88 uM = 0.266 V

Hence to prepare V mL of 0.5 uM solution we need to dilute 0.266 V mL of the stock solution to V mL by adding elution buffer or water (answer)

e.g; to get 10 mL of 0.5 uM, we need to take 0.266 x 10 mL = 2.66 mL of the stock solution and add water or elution buffer upto 10 mL mark.