Question

CALCULATIONS [2 points each] You are asked to perform a PCR experiment in the lab. However, in this experiment you must add each ingredient separately (unlike the "master-mix" we used in our own experiment). In this experiment, each PCR reaction tube contained the following ingredients added with a micropipetor... 10 HL of 5x Buffer 6 μ1 of 25 mM MgCl2 solution μL of 10 mM dNTP mix 1.5 μL oligonucleotide primer mix 0.5 HL Taq DNA Polymerase stock solution 28 HL ddH20 (or nuclease free H20) 3 HL DNA Template 50 μL Total Volume of reaction A. what is the final concentration of MgCl2 in your PCR reaction tube? B. Assuming the "stock" solution of Taq DNA Polymerase was provided at a concentration of 5,000 Units per ml, how many Units are present in your PCR reaction tube? C. In order to visualize your DNA in the agarose gel, you added 2.5 HL Ethidium bromide (from a stock solution of 5 mg/ml) to the agarose gel. If the volume of your agarose gel was 50 ml, what is the final concentration of Ethidium bromide in your agarose gel?

Answer 1

A) Final concentration of MgCl₂ in the PCR reaction tube = C2

Initial concentration of magnesium chloride C1 = 25 mM

Initial volume of magnesium chloride V1 = 6 ul

Final volume of solution after mixing V2 = 50 ul

C1V1 = C2V2

25 mM x 6 ul = C2 x 50 ul

C2 = 150 mM/50 = 3 mM

Final concentration of MgCl₂ in the PCR reaction tube = C2 = 3 mM

B) Stock solution (Initial concentration) of Taq DNA polymerase, C1 = 5000 units /ml

0.5 ul of Taq DNA polymerase is added to the above mixture

10^-3 l (1 ml) contain ----> 5000 units

0.5 x 10^-6 l ---> ==> ==   0.5 x 10^-6 x 5000 units/10^-3 = 2.5 units of Taq DNA polymerase

No. of units of Taq DNA polymerase added in the PCR reaction tube = 2.5 units

C) Initial volume of ethidium bromide, V1 = 2.5 ul = ( 1ul = 0.0001 ml ) = 2.5/1000 = 0.0025 ml

Initial (stock solution) concentration of ethidium bromide, C1 = 5 mg/ml

Final volume of solution after addition of ethidium bromide to the agarose gel, V2 = 50 ml

Final concentration of ethidium bromide in agarose gel = C2

C1 V1 = C2 V2

5 mg/ml x 0.0025 ml = C2 x 50 ml

C2 = 0.00025 mg/ml

Final concentration of ethidium bromide after adding it into the agarose gel = 0.00025 mg/ml or 0.25 ug/ml