3.3.2 Melting temperatures

The melting temperature of a DNA hairpin is defined as the temperature at which half of the molecules are in the native state (i.e., the double helix state) and half are denaturated (i.e., the two strands are split). It is implicitly assumed that the hybridization of the molecule is well described by a two-state model, in which the base-pairs are either fully bonded or disrupted (no partial disruption of bases is considered). This model is convenient for oligos shorter than 100 bp and the predicted melting temperatures deviate from the experimental measurements if the DNA is longer (the so-called polymeric DNA).

The NN model is widely used to predict melting temperatures of DNA duplexes. In order to predict melting temperatures, the enthalpy (Eq. 3.2) and the entropy (Eq. 3.3) of formation of the DNA hairpin must be known. The melting temperature of a non-self-complementary duplex3.1 is deduced from the the equation that relates the equilibrium constant with the free energy of formation of the hairpin,

$\displaystyle \Delta G = \Delta G^0+R T \ln K$ (3.6)

where $ \Delta G$ is the free energy of formation of the hairpin at the desired experimental conditions, $ \Delta G^0$ is the standard free energy calculated with Eq. 3.4, $ R$ is the gas constant, $ T$ is the temperature and $ K$ is the equilibrium constant. The hybridization reaction can be written as

$\displaystyle \textrm{ssDNA}_\mathrm{1}\textrm{+ssDNA}_\mathrm{2} \rightleftharpoons \textrm{dsDNA}$ (3.7)

where ssDNA $ _\mathrm{1}$ and ssDNA $ _\mathrm{2}$ are the two complementary strands and dsDNA is the hybridized duplex. The equilibrium constant of the hybridization reaction is given by,

$\displaystyle K=\frac{\textrm{[dsDNA]}}{\textrm{[ssDNA}_\mathrm{1}\textrm{]}\cdot\textrm{[ssDNA}_\mathrm{2}]}$ (3.8)

where [ssDNA $ _\mathrm{1}$], [ssDNA $ _\mathrm{2}$] and [dsDNA] are the concentrations (in molar units M=mols$ \cdot $l$ ^{-1}$) of the single strands and the duplex, respectively. By definition, the melting temperature is the one in which [dsDNA] = [ssDNA $ _\mathrm{1}$] = [ssDNA $ _\mathrm{2}$]. If the sample has a total concentration of strands equal to $ c_T$, it is expected that at the melting temperature the strands will be distributed according to,

 & \textrm{ssDNA}_1 & + & \textrm{ssDNA...
...\textrm{dsDNA} \\ 
 c_T= & x & + & x & + & 2 x \\ 
 \end{array}\end{displaymath} (3.9)

where $ x$ is the concentration of strands at equilibrium (the concentration of strands is double for [dsDNA]). From this, we deduce that $ c_T=4x$ at the melting temperature and the concentrations are equal to [ssDNA$ _1$] = [ssDNA$ _2$] = [dsDNA] = $ c_T/4$. Note that one molecule of dsDNA contributes twice to the total concentration of ssDNA, so we have

 c_T&=&[\textrm{ssDNA}_1] & + & [\tex...
...&c_T/4 & + & c_T/4 & + & 2 \; \times & c_T/4~. \\ 
 \end{array}\end{displaymath} (3.10)

By introducing these concentrations into Eq. 3.8 we find $ K=(c_T/4)^{-1}$ at equilibrium. At the melting temperature ($ T_m$) the hybridization reaction is in equilibrium, so $ \Delta G=0$. Considering this and introducing Eq. 3.4 into Eq. 3.6 it is possible to write,

$\displaystyle 0 \equiv \Delta G = \Delta H^0-T_m\Delta S^0 +R T_m \ln \left( c_T/4 \right)^{-1}$ (3.11)

which leads to the following result after solving for $ T_m$:

$\displaystyle T_m=\frac{\Delta H^0}{\Delta S^0+R\ln[c_T/4]}$ (3.12)

JM Huguet 2014-02-12