10) You are working in a research lab and the postdoc you work with asks you to

Question

10) You are working in a research lab and the postdoc you work with asks you to set up a PCR reaction. You are given the following components as stock solutions. Indicate how much of each component you need in a 50 uL reaction. (12 pts). Volumc Buffer Forward primer Reverse primer MgCl2 Taq polymerase 5X SuM 5uM 25mM 5 units/ul 1X. 250nM 250nM lmM 1.25 units. You need to add 50 ng of DNA to the reaction and you are given a stock of 10 ng/u. How much of the DNA stock do you add? How much water do you need to bring the final volume to 50uL

Explanation / Answer

We have to use the following formula to solve the given question

V1C1 = V2C2

Where

V1 = Stock volume

C1 = Stock concentration

V2 = Final volume

C2 = Final concentration

Now we shall calculate the required volume for each reagent for a 50 microlitre PCR reaction

a. Buffer

V1 = ? (X), C1 = 5X, V2 = 50 microlitres, C2 = 1X

V1C1 = V2C2

(X) x 5 = (50 x 10-6) x 1

X = [(50 x 10-6) x 1] / 5

X = 10 microlitres

b. Forward primer

V1 = ? (X), C1 = 5 micromolar, V2 = 50 microlitres, C2 = 250 nM

V1C1 = V2C2

(X) x 5 x 10-6 = (50 x 10-6)x (250 x 10-9)

X = [(50 x 10-6)x (250 x 10-9] / 5 x 10-6

X = (12500 x 10-15) / 5 x 10-6

X = 2500 x 10-9

X = 2.5 x 10-6 = 2.5 microlitres

c. Reverse primer

The initial concentration of both forward and reverse primer are same. So, we must use 2.5 microlitres of reverse primer stock in the PCR reaction.

d. MgCl2

V1 = ? (X), C1 = 25 mM, V2 = 50 microlitres, C2 = 1 mM

V1C1 = V2C2

(X) x 25 x 10-3 = (50 x 10-6)x (1 x 10-3)

X = [(50 x 10-6)x (1 x 10-3) ] / 25 x 10-3

X = (50 x 10-9) / 25 x 10-3

X = 2 x 10-6 = 2 microlitres

e. Taq polymerase

5 units is present in 1 microlitre. We want 1.25 units in 50 microlitre.

So, we must find out how many units of DNA polymerase are there in 1 microlitre in a 50 microlitre PCR reaction

For that, we have to divide 1.25 by 50. That is 1.25/50 = 0.025. So, this is the final concentration of DNA polymerase which we need for the PCR.

V1 = ? (X), C1 = 5 units, V2 = 50 microlitres, C2 = 0.025 units

V1C1 = V2C2

(X) x 5= (50 x 10-6)x (0.025)

X = (1.25 x 10-6) / 5

X = 0.25 microlitres

f. DNA

We have 10 ng/ microlitre DNA and we need 50 ng/microlitre in the 50 microlitre PCR mix.

So, we have to find out how much DNA is there in a 1 microlitre of PCR mix.

For that, we have to divide 50 ng of DNA by 50 microlitres. 50/50 = 1 ng/ microlitre. So, the final required concentration is 1 ng/microlitre.

V1 = ? (X), C1 = 10 x 10-9 , V2 = 50 microlitres, C2 = 1 x 10-9

V1C1 = V2C2

(X) x 10 x 10-9= (50 x 10-6)x (1 x 10-9)

X = [(50 x 10-6)x (1 x 10-9)] / 10 x 10-9

X = 5 microlitres

Now we shall draw the table with the stock concentration, final concentration and volume

If you like my answer, then please rate it.

Component Stock concentration Final concentration Volume Buffer 5X 1X 10 microlitres Forward primer 5 micromolar 250 nM 2.5 microlitres Reverse primer 5 micromolar 250 nM 2.5 microlitres MgCl2 25 mM 1 mM 2 microlitres Taq polymerase 5 units/ microlitre 1.25 units 0.25 microlitres Water ---- ----- 32.75 microlitres Total ---- ----- 50 microlitres

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