Question

The size of the human genome is 3.2x109bp. The 3.2kb long fragment is to be amplified by Polymerase Chain Reaction (PCR) using the human genome as a template. How many cycles of PCR is necessary for the amplified target sequence at least 50% (by mass) of the total DNA in the reaction mixture. Explain how you arrived to this number.

Answer 1

• 20 cycles of PCR is necessary for the amplification target sequence

Mass is proportional to sequence length.

Let one base pair have mass M.

Thus,

$3.2*10^9 bp$ will have mass $(3.2*10^9) * M$

And the 3.2kb target will have a mass of

For the total amount of amplified target sequence to constitute 50% (by mass) of the total DNA in the reaction mixture, it has to be amplified to such an extent that the total mass of the amplicon matter is also $(3.2 * 109) * M$

PCR amplification happens geometrically, doubling the DNA target with each cycle.

Thus, if the number of cycles in N,

Then,

$(3.2 * 10^9) x M = 3.2 x 1000 x M x (2)N\\\\or 10^6 = (2)^N\\\\or 6 = N log2\\\\or N = \frac{6}{log2} = 20$

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