Question

1. The exponential nature of PCR allows spectacular increases in the abundance of a DNA sequence being amplified. Consider a 10-kbp DNA sequence in a genome of 10^10 base pairs.

2. What fraction of the genome is represented by this sequence? That is, what is the fractional abundance of this sequence in this genome? Calculate the fractional abundance of this target sequence after 10, 15, and 20 cycles of PCR , starting with DNA representing the whole genome and assuming that no other sequences in the genome undergo amplification in the process.

Express your answers using three significant figures separated by commas.

Answer 1

The fraction of the genome that a 10 kbp DNA sequence occupies in a 10¹⁰ base pair genome is :

[(10*10³)/10¹⁰] * 100

= 0.0001%

After 10 PCR cycles there would be 2⁹ copies of the DNA sequence

Therefore after 10 cycles the percentage of the target sequence would be:

[(2⁹*10*10³)/10¹⁰]*100

=0.0512%

After 15 cycle there would be 2¹⁴ copies of the target sequence

Therefore after 15 cycles the percentge of the target sequence will be:

[(2¹⁴*10*10³)/10¹⁰] * 100

= 1.6384%

After 20 cycles of PCR, there would be 2¹⁹ copies of the target sequence

Therefore the percentage of the target sequence would be:

[(2¹⁹*10*10³)/10¹⁰]*100

= 52.4288%