Question

If the yeast genomic insert is 1000bp, will the PCR generate the 100ng needed for a sequencing reaction? (Use: the average molecular weight of a base pair = 650g/mole and Avogadro's number).

Answer 1

As given for 650g insert we need = 6.0221409E+23 molecules,

For 100 ng we will need = (6.0221409E+23/(650*1.00E+07) molecules

                                   = 9.26E+13 molecules

At the end of PCR for 30 cycles we get 1.07E+9 molecules from 1 single DNA molecule. Therefore, if we start PCR with 86285.47 or 86286 (9.31E-08 ng) DNA molecules, we will be able to generate 100 ng PCR product.