Blog of Andrés Aravena
Bioinfo:

# 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 temperature1.

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 equilibrium2 $\frac{[A][B]}{[AB]} =\frac{k_F}{k_R}=\frac{K \exp\left(\frac{E_F}{RT}\right)}{K\exp\left(\frac{E_R}{RT}\right)}$

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 $$\Delta S$$.

1. Note that energy has a positive sign.↩︎

2. check order of ratios↩︎

Originally published at https://anaraven.bitbucket.io/blog/2022/bioinfo/dna-thermodynamics.html