Question

1. You are interested in cloning a gene from the B. sanfranciscus genome, so you design PCR primers that should amplify a 1 kilobase pair (kbp) PCR product that contains the gene of interest.  After amplification, you will see if the PCR  was successful by loading the entire reaction onto an agarose gel and performing electrophoresis to see if a product of the expected size was generated.  To visualize the DNA, you will stain the gel with a fluorescent dye called ethidium bromide, which fluoresces when it binds to DNA. The sensitivity of ethidium-bromide-stained DNA is 10 nanograms (i.e. – there must be at least 10 ng of DNA in the band in the gel to emit a detectable amount of light).

If your PCR reaction initially contained 30 B. sanfranciscus genomes, how many cycles of PCR will required before there is a detectable amount of amplified product?  You can assume:  a) there is 1 copy of the gene per genome, b) the PCR occurs with perfect efficiency and therefore the amount of product doubles after each cycle, and, c) that the molecular weight of a 1 kbp molecule of DNA is 6.5 x 10⁵ Daltons. Express your answer in the number of complete (not fractional) cycles and show your work.

Answer 1

We need to amplify a 1 kb region, which exists in the B. sanfranciscus genome, and there are 30 copies of the genome in the reaction, which also means there are 30 copies of the 1 kb region.

Now, Mass of 1 copy of the 1 kb region = 6.5 x 10⁵ Dalton

Since the detection limit of EtBr during Agarose Gel electrophoresis is 10 ng, we need at least 10 ng of the 1 kb DNA region to be able to detect it at the appropriate size after PCR.

Now,

1 Da = 1/12th the Mass of a Carbon-12 atom = 1.6605 * 10^-24 g

=> Mass of one 1 kb copy = 6.5 x 10⁵ Dalton = 6.5 x 10⁵ x 1.6605 * 10^-24 g

Total mass required = 10 ng = 10 * 10^-9 g

Number of DNA copies required = Total Mass required/Mass of one copy
   = 10 * 10^-9 g / ( 6.5 x 10⁵ x 1.6605 x 10^-24 g) = 9.26505 * 10⁹

Now, each cycle produces two copies.

Number of cycles required to make 9.26505 * 10⁹ copies = 30 * 2ⁿ

=> 2ⁿ = 9.26505 * 10⁹/30
=> 2ⁿ = 3.08835×10⁸
=> n $\approx$ 28.2023

Therefore, at least 29 cycles are required to produce detectable number of copies.