Question

2. You are performing PCR under the following conditions: 1. Each PCR reaction contains 50 pl final volume II. Each PCR reaction contains: 1x PCR buffer, 1.5 mM MgCl2, 0.2 mM dNTP mix, 0.2 MM sense primer, 0.2 uM anti-sense primer, 1.5 units of Taq polymerase and 2 ul of DNA template. [all concentrations are final] III. Stock solutions of the following reagents are available: 10x PCR buffer, 25 mm MgCl2, 10 mM dNTP mix, 10 M of sense primer, 10 PM of anti-sense primer, 5 unitul Taq polymerase and PCR quality water You want to prepare a "PCR Master Mix" (containing all of the PCR reagents, except for the DNA template) such that each PCR reaction tube can be prepared by pipetting 48 pl of the Master Mix and 2 pl of the DNA. Assuming that you want to make sufficient Master Mix for exactly 20 PCR reactions (you will actually only need Master Mix for 19 PCR reactions), how would you make the master mix? Show your calculations.

Answer 1

Master mix for 20 reaction (final PCR volume= 20*50= 1000 ul (40 template DNA+960 master mix)

10x Buffer- 100 ul

dNTPs- 20 ul

Primer sense- 20 ul

Primer anti-sence- 20 ul

MgCl2- 60 ul

Taq Polymerase- 6 ul

DNase free water- 734 ul

Total volume= 960 ul

Therefore, 960/20= 48 ul of master mix+ 2 ul template DNA.