PCR Optimization


PCR product length:   kbp     
dNTP's concentration   mM each 
primers quantity:   pmol    
quantity of Taq polymerase:   u       
reaction volume:   µl      
elongation time:   min :  sek    
template quantity:       or   
Proportion of nucleotides and primers:
%  will be consumed

Maximal yield:    
concentration:      or   

Taq polymerase can synthesize in one cycle:       DNA;
concentration:      or   

Maximal length of PCR product (from 2kbp/min)  kbp

For the ideal doubling in each cycle, it would be necessary to perform cycles.
But, amplification will be linear after cycles, due to the restriction by polymerase activity.

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";


  1. 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]
  2. maximum quantity of PCR product per one cycle depends on two factors:
    1. Taq polymerase velocity: 2-4[kbp/min];
    2. Taq polymerase activity (1 u is the amount of enzyme, that incorporate 10nmol of all four dNTP’s in 30 min at 72oC).
         mcycle = 10[nmol] x 324.5[g/mol] x a [u] t[min] / 30[min] = 108at [ng]
  3. the number of cycles, which are necessary for synthesis of "mmax" PCR-product is:
         mmax = 2n x mo     =>     n = ln(mmax/mo)/ln2
  4. the relationship of mass and mole quantities is:
         m[µg] = 649[g/mol] x q[µmol] x L[kbp] x 1000