Question

A research associate working for Intragene Therapeutics prepares an investigative PCR reaction to screen human tissue samples for specific mutations. He uses 55 ng of genomic DNA from one tissue sample in a single PCR reaction, to amplify a 1,246 bp fragment. The PCR protocol requires a final concentration for two primers of 1.5 uM each, and a final concentration of 180 uM for dNTPs, in a total PCR reaction volume of 65 uL. During the PCR process, the fragment is amplified for 33 cycles. He quantifies the amount of PCR products to be 74 ng/uL.

1. If his purified genomic DNA template stock has a concentration of 1.53 mg/mL, and he dilutes it 1:100, how many microliters of this dilution does he use? Answer to one decimal place.

2. He decides to dilute his DNA template stock so that he can pipet a 15 uL volume that contains 55 ng of genomic DNA into his PCR reaction. If he starts with 1 uL of the DNA template stock, what is the total volume of the dilution he needs to make? Answer to the nearest whole number.

3. How many copies of the fragment are produced in his PCR reaction? Assume that the target sequence is present in only one copy in the genome, as in a haploid human genome. Answer to the nearest whole number.

4. What is the amount of amplification of the PCR product in his reaction ‐ that is, the fold increase in the amount of amplified DNA segment? Answer in decimal notation.

ANSWERS (I just need to know how to solve these):

1. 3.6 microliters

2. 417 mircoliters

3. 16,667 copies of target sequence

4. 210,652,721 fold increase

Answer 1

1. To calculate the volume of DNA, the following formula can be used.

V1 x C1 = V2 x C2

Where,

V1 = volume of stock solution

V2 = final volume of working solution

C1 = concentration of stock solution

C2 = concentration of working solution

Initial concentration of DNA = 1.53mg/ml = 1530µg/1000µl = 1.53µg/µl

Dilution = 1:100

i.e 1µl of DNA with a concentration of 1.53µg/µl is diluted with 99µl of water.

i.e 1.53µg/100µl = 1530ng/100µl = 15.3ng/µl

C1 = 15.3ng/µl

C2 = 55ng in 65µl = 0.8461ng/µl

V1 = has to be calculated

V2 = 65µl

V1 x C1 = V2 x C2

V1 x 15.3ng/µl = 65µl x 0.8461ng/µl

V1 = (65µl x 0.8461ng/µl) / 15.3ng/µl

V1 = 54.9965/15.3 µl

V1 = 3.6µl

2. V1 x C1 = V2 x C2

Where,

V1 = volume of stock solution

V2 = final volume of working solution

C1 = concentration of stock solution

C2 = concentration of working solution

C1 = has to be calculated

C2 = 55ng in 65µl = 0.8461ng/µl

V1 = 15µl

V2 = 65µl

V1 x C1 = V2 x C2

15µl x C1 = 65µl x 0.8461ng/µl

C1 = (65µl x 0.8461ng/µl)/ 15µl

C1 = 54.9965 /15 ng/µl

C1 = 3.6664ng/µl

We need a solution with 3.6664ng/µl of DNA.

We have a stock solution of 1.53mg/ml = 1.53µg/µl = 1530ng/µl

V1 x C1 = V2 x C2

Where,

C1 = 1530ng/µl

C2 = 3.6664ng/µl

V1 = 1µl

V2 = has to be calculated

1µl x 1530ng/µl = V2 x 3.6664ng/µl

V2 = (1µl x 1530ng/µl) / 3.6664ng/µl

V2 = 1530 / 3.6664 µl

V2 = 417 µl

Dilute 1µl of stock DNA with the concentration 1.53mg/ml with 416µl of water to get a total volume of 417µl with a concentration of 3.6664ng/µl.

Thus, when 15µl of solution with a DNA concentration of 3.6664ng/µl is taken, we will have 55ng of DNA (15µl x 3.6664ng/µl).