Question

Please, I need help with this question. Show work to solve the whole 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 or - stated in a way that is more suited to working this problem – 1 kbp of ds DNA has a mass of 6.5 x 10⁵grams/mole. Express your answer in the number of complete (not fractional) cycles and show your work.

note: If a 1 kbp of DNA amplicon has a molecular weight of 6.5 x 10⁵ Daltons you can assume that this amplicon has 6.5 x 10^5 grams / mole

Answer 1

Given that:

Number of copies of B. sanfranciscus genome in PCR reaction initially = 30
Size of desired product = 1 kbp = 1000 bp
Mass of 1 copy of 1000 bp PCR amplicon of Gene of Interest (GoI) = 6.5 * 105 Daltons
Molecular Weight of GoI fragment = 6.5 * 105 g/mol

Detection limit of Ethidium Bromide(EtBr) in Agarose gel electrophoresis = 10 ng

Since 10 ng of the GoI PCR product are required,

Number of moles of GoI PCR product required
= (Mass of GoI required for EtBr detection)/(Mol. Weight of GoI Fragment)
= (10 ng)/(6.5 * 105 g/mol)
= (10 * 10-9 g)/(6.5 * 105 g/mol)
= (10/6.5) * 10-14 mol
= 1.5385 * 10-14 mol

Therefore,

Number of copies of the 1 kbp DNA molecule required
= No. of Moles * No. of copies per mole
= 1.5385 * 10-14 mol * 6.023 * 1023 mol-1
= 9.2664 * 109

Therefore 9.2664 * 109 copies of the GoI PCR amplicon of size 1 kbp are required,

We can assume that PCR is perfect, and beginning from the 30 initial copies, the first cycle would result in 60 copies (30 * 2); the second cycle would result in 120 copies (30 * 22) and so on.

Therefore,

Number of copies = 30 * 2n; where n is the number of cycles since each cycle doubles the number of copies.

=> 9.2664 * 109 = 30 * 2n
=> 2n = (9.2664 * 109)/30
=> 2n = 3.0888×108
=> n $\approx$ 28.2025

Since we have to report the number of cycles as an integer rather than a decimal number, and since ~ 28.2025 cycles are required for acquiring sufficient number of copies of the GoI, at least 29 PCR cycles would be required for the synthesis of 10 ng of the 1 kb GoI from the original 30 copies of the B. sanfranciscus genome.