00:01
Suppose a fragment of human dna with a 3 .5 kb length is to be amplified with pcr.
00:08
The total genome size is 4 times 10 to the 6th power kb.
00:11
Prior to application, what fraction of the total dna does the target sequence constitute? so the fraction or percentage of the target sequence is going to be 3 .5 kb divided by the total genome size 4 times 10 to the 6th power kb times 100 percent.
00:46
So this is 0 .875 times 10 to negative 6 times 100 percent.
01:01
So 87 .5 times 10 to the negative 6 percent.
01:07
Or we can, yeah, so that's the percentage of target dna that is before the amplification.
01:19
Now what fraction does it constitute after the 15th cycle of pcr? so now the target dna after the 15th cycle of pcr equals the original piece 3 .5 kb times 2 to the nth power.
01:42
This n actually is number of cycles.
01:44
Since this time is 15 cycles, i'll replace this n with 15.
01:53
So this is 3 .5 times 32 ,768.
02:01
So you get 114 ,688 kb...
