Question

7.You want to set up a PCR reaction. The protocol suggests a 50 µL reaction volume containing 1X buffer, 200 µM dNTPs, 0.5 µM each primer (forward and reverse), 250 ng template DNA, and 1 U enzyme.

You have the following reagents: 5 X buffer, 10 mM dNTPs, 10 µM each primer (forward and reverse), 125 ng/µL template DNA, and 2 U/µL enzyme.

How many µL of water, reaction buffer, dNTPs, each primer, template DNA and enzyme would you combine to set up that reaction? (note that the order I’ve listed these components is the order you’d add them to the reaction)

Answer 1

Ans. #I. Buffer:

Using              C1V1 (5X stock buffer) = C2V2 (final PCR mix, 1X, 50 uL)

            Or, 5X x V1 = 1X x 50.0 uL

            Or, V1 = (1X x 50.0 uL) / 5X

            Hence, V1 = 10.0 uL

Hence, required volume of 5X stock buffer = 10.0 uL

#II. dNTPs

Using              C1V1 (10 mM stock dNTPs) = C2V2 (final PCR mix, 200 uM, 50 uL)

            Or, 10 mM x V1 = 200 uM x 50.0 uL                                  ; [1 mM = 1000 uM]

            Or, 10000 uM x V1 = 200 uM x 50.0 uL

            Or, V1 = (200 uM x 50.0 uL) / 10000 uM

            Hence, V1 = 1.0 uL

Therefore, required volume of 10 mM stock dNTPs solution = 1.0 uL

#III. Primers

Using              C1V1 (10 uM stock dNTPs) = C2V2 (final PCR mix, 0.5 uM, 50 uL)

            Or, 10 uM x V1 = 0.5 uM x 50.0 uL                       

            Or, V1 = (0.5 uM x 50.0 uL) / 10.0 uM

            Hence, V1 = 2.5 uL

Therefore, required volume of 10 uM stock primer solution = 2.5 uL

#IV. Required volume of stock template DNA =

Required amount of template DNA/ Concertation of stock template soln.

= 250.0 ng / (125.0 ng/ uL)

= 2.0 uL

#V. Required volume of stock enzyme solution=

Required amount of enzyme/ Concertation of stock enzyme soln.

= 1.0 U / (2.0 ng/ uL)

= 0.5 uL

#VI. Required volume of water = 50.0 uL – (summed vol. of all other ingredients I-V)

                        = 50.0 uL – (10.0 uL + 1.0 uL + 2.5 uL + 2.0 uL + 0.5 uL)

                        = 50.0 uL – 16.0 uL

                        = 34.0 uL