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o2blea o2blea
wrote...
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12 years ago
A 2 kbp (1 kilo base pairs = 1,000 bp) DNA fragment is to be amplified by PCR from the genome of nematode C. elegans. The total genome size is 105 kbp (108 base
pairs).
a. Prior to amplification, what fraction of the total DNA does the target sequence constitute?
b. What fraction does it constitute after 10 cycles of PCR? c. After 25 cycles of PCR?
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wrote...
12 years ago
This is a nice fun question! But I will assume that the 105 should be 10exp5 (100,000) and the 108 should be 10exp8. Otherwise it does not make a lot of sense.


a. That would be 2/100000 = 2 * 10exp-5
b. In every cycle, the number of genes is doubled. So after 1 cycle you have 2x the number of amplicons, after 2 cycles you have 4x, after 3 cycles you have 8, etc. It grows by the power of 2 (2n). After 10 cycles you have 2 to the power of 10 = 1024 copies of that gene. Now the ratio is 2048/100000 =  0.02048 (remember that the size is 2 kb). After 25 cycles you have 2^25 = 33 554 432 copies, so now you have 67108864/100000=671, so almost 700 x more than the total DNA!
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