Question

7. You are making a PCR reaction mixture with a total volume of 50uL.( THIS IS IN MICROLITERS NOT MILILITERS AS ANOTHER QUESTION ON CHEGG HAS ALREADY BEEN POSTED BUT IT STATES THE TOTAL VOLUME AS 50 mL.)

The primer set concentration is 1μl primer/10μl reaction volume. The plasmid stock solution concentration = 0.01μg/μl, of which you need 100ng. How much of each PCR reaction component will you add, given the following?

5x Mastermix __________

Primer set __________

DNA template __________

H₂O _________

Answer 1

PCR reaction mixture total volume = 50 l

Primer set concentration = 1 l primer/10 l reaction volume Therefore primer set volume = 5 X 1 l = 5l

Plasmid stock solution concentration = 0.01 g/l We need 100 ng plasmid (DNA template) 1 g = 1000 ng . Therefore 100 ng = 0.1 g So, we need 0.1 g of plasmid DNA stock solution for our reaction mixture. Given that, 0.01 g plasmid is present in 1 l plasmid stock 1 g plasmid is present in {1/0.01} l plasmid stock 0.1 g plasmid is present in {(1 X 0.1) /0.01 } l plasmid stock = 10 l plasmid stock Therefore, we need 10 l of plasmid stock solution for our PCR reaction mixture. DNA template = plasmid stock solution = 10 l

PCR master mix is needed in 1X concentration for the final reaction mixture. Let us assume , we need , V1l of 5X master mix solution to make it 1X in 50 l final reaction mixture solution. We know, V1S1 = V2S2 Here,  V1 = unknown. V2 = 50 l . S1 = 5X . S2 = 1X Therefore V1 X 5X = 50 l X 1X V1 = 10 l 5X PCR master mix = 10 l

H2O = {Total Reaction volume- (5X master mix + Primer set + DNA template)} H2O = {50 - (10+ 5 +10)} l = 25 l H2O = 25 l

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