PCR Optimization

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The program helps to evaluate the quantity of primers and nucleotides for optimal PCR reaction and the necessary number of cycles. Estimation is quite simple, so it is dangerous to rely on it completely. Default parameters correspond to the amplification of 2kbp fragment from 0.5µg of human DNA. It is supposed, that:

• reaction conditions are close to normal, so it is not necessary to worry about the course of the reaction (too much polymerase can lead to the unspecific amplification; too much primers - to primer dimers);
• A, T and G, C are equivalently presented in PCR product;
• there is no primer-dimers;
• Taq polymerase does not loose activity during reaction.

If:

• length of the PCR product is "L" [kbp];
• dNTP's concentration is "с" [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 "m_o";
  

Then:

1. maximal yield is the minimum from two evaluations:
   if all nucleotides will be consumed:
   m_n = 4[nucleotides] x с[mmol/l] 324.5[g/mol] x V[µl] = 1300сV [ng]
   if all primers will be consumed:
   m_p = 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 72^oC).
      
   m_cycle = 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 "m_max" PCR-product is:
   m_max = 2ⁿ x m_o => n = ln(m_max/m_o)/ln2
4. the relationship of mass and mole quantities is:
   m[µg] = 649[g/mol] x q[µmol] x L[kbp] x 1000

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