Question

Today we amplified 50 ng of Bos taurus (calf) DNA by PCR. This amount of DNA contains about 15,000 molecules of the insulin gene [50 ng DNA= 2.5 x 10^-20mol; (2.5x10^-20)x(6.023×10²³) = 1.5 x 10⁴molecules]. We performed PCR for 35 cycles to amplify the amount of this gene.

      a. What is the theoretical fold amount of DNA amplified by 35 cycles of PCR (remember the 2^Nformula)?

      b. How many molecules of the insulin gene would, therefore, be present after PCR?

Answer 1

A) the initial amount of DNA template present is 1.5 x 10⁴ molecules . After 35 cycles the number of DNA molecules amplified will be 1.5 x 10⁴ x 2³⁵ = 5.15 x 10¹⁴ DNA molecules.

B) 50ng of DNA which is equal to 1.5 x 10⁴ ( 15000)  DNA molecules contain 15000 insulin gene molecules. Therefore the 5.15 x 10¹⁴ amplified  DNA molecules would contain the same amount i.e 5.15 x 10¹⁴ insulin gene molecules.