Question

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?

Answer 1

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%.