Blog of Andrés Aravena

DNA melting temperature

13 December 2022 by Andrés Aravena, Ph.D.

Arrhenius says that the reaction rate is \[k= K\exp\left(\frac{E_a}{RT}\right)\] where \(E_a\) is the activation energy, \(K\) is the pre-exponential constant (kind of the spontaneous rate), \(R\) is the gas constant (to convert units) and \(T\) is the absolute temperatureNote that energy has a positive sign.


The reaction is \[A+B \leftrightarrow AB\] We have indeed two reactions: a forward reaction with rate \(k_F\), and a reverse one with rate \(k_R\).

So, in equilibriumcheck order of ratios

\[\frac{[A][B]}{[AB]} =\frac{k_F}{k_R}=\frac{K \exp\left(\frac{E_F}{RT}\right)}{K\exp\left(\frac{E_R}{RT}\right)}\]

which simplifies to \[\frac{[A][B]}{[AB]} =\exp\left(\frac{E_F-E_R}{RT}\right) =\exp\left(\frac{\Delta E}{RT}\right)\]

We want to find the conditions where 50% of all diners are formed. We will study that case later. For now, and to simplify the notation, let’s call this concentration ratio \(C.\) \[C=\exp\left(\frac{\Delta E}{RT}\right)\] Therefore \[\log(C)= \frac{\Delta E}{RT}\]

Now, the energy is formed by two parts \[\Delta E = \Delta H - T\Delta S\] Replacing this in the previous formula, we get \[RT\log(C)= \Delta H - T\Delta S\] and therefore \[T=\frac{\Delta H}{\Delta S + R\log(C)}.\]

Finding \(C\)

The concentrations of \(A\) and \(B\) at every time are related to the initial concentrations \([A_{ini}]\) and \([B_{ini}]\) by the expression \[[A]=[A_{ini}]-[AB], [B]=[B_{ini}]-[AB]\]

Therefore, \[\begin{aligned} [A][B]&=([A_{ini}]-[AB])([B_{ini}]-[AB])\\ &=[A_{ini}] [B_{ini}]-[AB]([A_{ini}]+[B_{ini}])+[AB]^2 \end{aligned}\] and then the ratio between the two sides of the reaction is \[\frac{[A][B]}{[AB]}=\frac{[A_{ini}] [B_{ini}]}{[AB]}-[A_{ini}]+[B_{ini}]+[AB].\]

To simplify, let’s assume that initially \(B\) is in lower concentration that \(A\), that is \([A_{ini}]≥[B_{ini}].\)

We want to find the conditions where 50% of all diners are formed. This is achieved when \([AB]=[B_{ini}]/2,\) thus in that moment \[\begin{aligned} [A]=&[A_{ini}]-[B_{ini}]/2\\ [B]=&[B_{ini}]-[B_{ini}]/2=[B_{ini}]/2 \end{aligned}\]

Let us now think about \(C.\) We have at equilibrium \[C=\frac{[A][B]}{[AB]}=\frac{([A_{ini}]-[B_{ini}]/2)(B_{ini})/2}{[B_{ini}]/2}= [A_{ini}]-[B_{ini}]/2\] It is useful to remember that the total concentration is \[C_{ini}=[A_{ini}]+[B_{ini}].\] Thus, we have \[C=C_{ini}-3[B_{ini}]/2.\]

There are two important cases. If the initial concentration of \(A\) is much larger than the concentration of \(B\) (like primers in a PCR reaction), then \[C\approx[A_{ini}]\approx C_{ini}.\]

If A and B have similar initial concentrations (\([A_{ini}]\approx[B_{ini}]\)), then \[C=[A_{ini}]/2\approx C_{ini}/4.\]

The formula used for the melting temperature of dimers is then \[T_m=\frac{\Delta H}{\Delta S + R\log(C_{ini}/F)}\] Where \(F=1\) for asymmetric initial concentrations, and \(F=4\) when the initial concentrations are similar.

Now the challenge is to find \(\Delta H\) and \(\Delta S\).

Originally published at