It is desirable to amplify a DNA sequence by PCR the fragment to amplify measuri

Question

It is desirable to amplify a DNA sequence by PCR the fragment to amplify measuring 150 base pairs. For the reaction one starts from 100 nanograms of a genomic DNA of 3 x 10 ^ 9 bp. The PCR yield is 100%

What mass (in grams) of 150 base pair fragments in the tube at the beginning of the experiment?

What is the amplification factor for PCR to obtain 10 picograms of the 150 base pair fragment? (justify the result)

How many cycles of PCR should be performed? give an integer or integer, the margin of error is a few percent

Explanation / Answer

Since PCR performs exponential amplification of the target DNA i.e. by a factor of 2n, where n is the number of cycles for which PCR is performed.

Here, the initial amount of DNA is 100 ng or 100-9g or 10-8g.

The final amount of DNA is 10 pg or 10-12g.

Thus, the amplification factor would be give by:

Amplification factor : Initial amount of DNA / Final amount of DNA

Amplification factor : 10-8 / 10-12 or 104 times or 10,000 folds amplification.

The number of PCR cycles required for this amplification would be given by:

2n = 10000 or 104

Thus, solving for n, we get:  

nlog2 = 4log10

nlog2 = 4*1

0.301n = 4

or n = 4/0.301 or 13.289 (nearly 14)

Thus, nearly 14 cycles of PCR are required for this amplification.

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