Question

You have a phenomenal day in lab and collect 35 µg of TAS2R38 PCR product in 50 µl of nuclease free water. How many µl of this stock would you use to make 30 µl of 2.5 ng/µl DNA final concentration for the DNA sequencing core? Round to the nearest tenth. (Show work for partial credit.)

Answer 1

As I have35 µg of TAS2R38 PCR product in 50 µl of nuclease free water which in concentration term we can represent as = (35/50) µg/ul = 0.7 µg/ µl = 700ng/ µl    (as we know 1 µg = 1000ng)

therefore the concentration of stock= 700ng/µl DNA

now we are making 30 µl of 2.5 ng/µl DNA final concentration

use formula V1 x S1=V2 x S2

where V1, V2 is the volume of stock and working solution and S1, S2 is the strength or concentration of stock and working solution

in your case, we need to find out V1

therefore, V1= (V2 x S2)/S1 = (30 x 2.5)/700= 0.10714µl ( nearest tenth =0.1µl)