This calculator helps to evaluate the quantity of primers and nucleotides for
optimal PCR reaction and the necessary number of cycles. Estimation is quite
basic, so it is best to use it as a guide only. Default parameters correspond to
the amplification of 2Kbp fragment from 0.5µg of human DNA. It is assumed that:
- Reaction conditions are close to normal, so it is not necessary to worry
about the course of the reaction (e.g. too much polymerase can lead to
unspecific amplification; too much primers can lead to primer-dimers).
- A, T and G, C are equivalently presented in PCR product.
- There are no primer-dimers.
- Taq polymerase does not loose activity during the reaction.
- length of the PCR product is "L" [kbp];
- dNTP's concentration is "c" [mM];
- primers quantity is "q" [pmol];
- quantity of Taq polymerase is "a" [u];
- reaction volume is "V" [µl];
- elongation time is "t" [min];
- template quantity is "mo";
- maximal yield is the minimum from two evaluations:
if all nucleotides will be consumed:
mn = 4[nucleotides] x c[mmol/l] 324.5[g/mol] x V[µl] = 1300cV [ng]
if all primers will be consumed:
mp = q[pmol] x 2[strands] 324.5[g/mol] x L[kbp] = 650qL [ng]
- maximum quantity of PCR product per one cycle depends on two factors:
mcycle = 10[nmol] x 324.5[g/mol] x a [u] t[min] / 30[min] = 108at [ng]
- Taq polymerase velocity: 2-4[kbp/min];
- Taq polymerase activity (1 u is the amount of enzyme, that incorporate 10nmol of all four dNTP’s in 30 min at 72oC).
- the number of cycles, which are necessary for synthesis of "mmax" PCR-product is:
mmax = 2n x mo => n = ln(mmax/mo)/ln2
- the relationship of mass and mole quantities is:
m[µg] = 649[g/mol] x q[µmol] x L[kbp] x 1000