The genome of virus SV40 is a circular double‐stranded DNA molecule 5 kb long
(i.e., it contains 10,000 bases)
A 1 kb region is amplified by PCR for four cycles. (Assume that specific
amplification of the target region occurs.) What fraction of the total DNA
does the target sequence constitute?
During PCR, the region of interestis doubled with each cycle. The end result of four cycles is 24, or 16 copies of the region of interest. 1 kb will produce 16 kb after 4 cycles. The total DNA after 4 cycles will be 16 kb + 4 kb (5kb – 1 kb of target DNA) = 20 kb. The fraction of target sequence will be 16 kb/20 kb=4/5 or 80%.