If you apply the Boltzmann curve to a protein temperature denaturation, you will only get an estimate of the Tm (mid-denaturation temperature). Of course, the width, dx, estimated from that will be related to the unfolding enthalpy, because the unfolding enthalpy is related to the slope at the inflection point in the unfolding trace, and dx is also related to that slope.

In fact, that slope is equal to:

slope=(A-B)/4dx, in the Boltzmann curve,

slope=0.25(sN+sU+(mN+mU)Tm)dH(Tm)/(RTm2)+0.5(mN+mU), in the unfolding trace,

where sN and mN (intercept and slope) define the native region in the thermal denaturation trace, sU and mU (intercept and slope) define the unfolded region in the denaturation trace, dH(Tm) is the unfolding enthalpy, and R is the gas constant. If, instead of the raw signal, the fraction of folded or unfolded protein is considered, the expression for the slope gets simplified:

slope=0.25dH(Tm)/(RTm2)

Using these expressions you can relate dx and dH(Tm).

Thank Adrian for your efforts, I will use it. Can you provide me some article where this equation or dx relates unfolding enthalpy?

I think you do not need articles for that.

From the Boltzmann curve it is easy to calculate the derivative at x0.

From the expression for the spectroscopic signal along the thermal denaturation curve:

S=PNSN+PUSU

where PN and PU are the fractions of folded and unfolded protein:

PN=1/(1+K)

PU=K/(1+K)

with K=exp(-dG/RT), the equilibrium constant for unfolding,

and SN and SU are the intrinsic signal for the folded and unfolded protein:

SN=sN+mNT

SU=sU+mUT

you can calculate the derivative of S at Tm. In particular, working directly with PN or PU, their derivatives at Tm are:

dPU/dT|Tm = –dPN/dT|Tm =0.25dH(Tm)/(RTm2)

where I have considered:

dK(T)/dT=K(T)dH(T)/(RT2)

Thus, the broader the thermal transition, the smaller dx and the smaller dH(Tm). And the narrower the thermal transition, the larger dx and the larger dH(Tm).

Thanks Adrian, I will look into it.

There was an error in a previous post. The slope of the spectroscopic signal during the thermal unfolding at Tm is given by:

slope=0.25(-sN+sU+(-mN+mU)Tm)dH(Tm)/(RTm2)+0.5(mN+mU)

that is, there were two "-" signs missing in the first term (-sN and -mN).

In addition, a minor but sometimes important detail. In a spectrocopic thermal denaturation the Tm does not coincide with the location of the inflection point or the point for maximal slope. Both points (Tm and temperature for maximal slope) are usually very close, but sometimes (when the unfolding heat capacity is large and the unfolding enthalpy is small) the difference can be significant.

A similar results applies to differential scanning calorimetry: Tm does not coincide with the temperature at the maximum in the heat capacity curve. Both points (Tm and temperature for maximum in the unfolding trace) are usually very close, but sometimes (when the unfolding heat capacity is large and the unfolding enthalpy is small) the difference can be significant.