Question

Cold and Hot Denaturation of Proteins.

Answer 1

This is yet another post, in which I am going to present some theory I have been working with and asking a question, to which the biologist usually like to answer. I will however warn the biologist reading and say there are chemistry and mathematics ahead.

When I was reading about denaturation of proteins, I got the conclusion that the denaturation most relate to the Gibbs free energy; that is the spontaneous change from a native folded state to a unfolded state.
The standard equation for the Gibbs free energy is
\[\Large \Delta G = \Delta H-T \Delta S\]
This equation relates the thermodynamic state functions to the Gibbs free energy at constant pressure. The state functions however relates to the specific value of \(T\). Only by assuming that the heat capacity is equal to 0, this equation can be used to calculate \(\Delta G\) as a function of \(T\). I will however assume that the heat capacity is not 0 and therefor need to extend the equation above. We define the change in heat capacity at constant pressure as
\[\Large \frac{ \partial \Delta H }{ \partial T }=\Delta C _{p}\]
To evaluate the enthalpy as a function of temperature we therefor integrate the expression above
\[\large \partial \Delta H=\int\limits_{T_{R}}^{T} \Delta C_{P} ~ \partial T \to \Delta H(T)=\Delta H(T_{R})+\Delta C_{p}(T-T_{R})\]
In order to look at the entropy as a function of temperature, we will use the thermodynamic definition of entropy from the Boltzmann formula
\[\Large \partial \Delta S=\frac{ \partial \Delta q _{rev} }{ T }=\frac{ \Delta C_{p} }{ T } \partial T\]
To evaluate the entropy as a function of temperature we therefor integrate the expression above
\[\large \partial \Delta S=\int\limits_{T_{R}}^{T}\frac{ \Delta C_{p} }{ T } \partial T \to \Delta S(T)=\Delta S(T_{R})+ \Delta C_{p} \ln \left( \frac{ T }{ T _{R} } \right)\]
Insert the entropy and entropy as a function of temperature into the standard equation for the Gibbs free energy
\[\large \Delta G(T)=\Delta H(T_{R})+\Delta C_{p}(T-T_{R})-T \left\{ \Delta S(T_{R})+ \Delta C_{p} \ln \left( \frac{ T }{ T _{R} } \right) \right\}\]
This rearranges into
\[\large \Delta G(T)=\Delta H(T_{R})-T \Delta S(T_{R})+ \Delta C_{p}\left\{ T-T_{R}-R \ln \left( \frac{ T }{ T_{R} } \right) \right\}\]
To plot this function, all we need are the enthalpy, entropy and heat capacity at a reference temperature. I’ve found a literature value for pepsinogen [found within the graph]. Plotting these data with the equation derived, we get the following graph (See attachment 1)

If solving for the temperature when the Gibbs free energy is equal to zero we get the two solutions
\[\Large T_{1}=253 ~ \textrm {K}=-20.2 ^{\circ} \textrm{C} ~~~ \vee ~~~ T _{2}=340 ~ \textrm {K}=66.9 ^{\circ} \textrm{C} \]
According to the definition of the Gibbs free energy we should therefore be able to conclude that pepsinogen should denature at -20 °C and 67 °C; a cold and hot denaturation?

Answer 2

Oh..... way beyond me lol

Answer 3

A part of my question is: Have we ever experimentally witnessed this?

Answer 4

The thermodynamics part looks good, the upper temperature 66\(^{\circ}\)C is the unfolding temperature according to lit. Look at the table on page 21
http://www.biochem.oulu.fi/Biocomputing/juffer/Teaching/PhysicalBiochemistry/PhysBiochem-protein-folding.pdf

I'm not sure about the lower point; a typical solution would have frozen before that temperature was reached, and if you used 50% ethylene glycol would that not make the artificial system unrealistic?

Answer 5

I don't know really. I would still believe there is a rotational, translational, vibrational, electrical ect. contribution to the free energy at 253 K?

Answer 6

i'm sure there is energy, i'm not sure though that you could freeze a solution at like -15\(^{\circ}\)C then thaw it and maintain the biological activity. Which would mean that denaturation occurred before the expected temperature.

Answer 7

but again, i think thats because of the physical properties of the solvent.

Answer 8

@Frostbite : After looking at the equations you have presented, I have this to say:

Your statement that the Gibbs free energy (as defined for a constant pressure based on the equation above) is correct. However, your subsequent statement that "only by assuming that the heat capacity is equal to 0, this equation can be used to calculate ΔG as a function of T" is incorrect, since heat capacity (by definition) relates to the amount of heat energy required to raise the temperature of a substance by 1 Kelvin. To claim that the heat capacity = 0 is equivalent to saying that you need 0J of heat energy to raise the temperature of the substance by 1 Kelvin, which is ridiculous and in contradiction with the fundamentals of thermodynamics. This is not the basis by which the Gibb's free energy equation (such as you have drawn out above) was formulated. In addition, it is perfectly reasonable to say that the heat capacity of the system≠0, in which case this equation would be a simple linear function of ΔG with respect to T, with -ΔS as the gradient and ΔH as the y-intercept.
Your stated equation [that the change in heat capacity at constant pressure, ΔCp = (δΔH)/(δT) ] is incorrect as well. By definition, ΔH=CΔT ⇒ Cp=(ΔH/ΔT). For an adiabatic process, assuming a variable heat capacity in the system at constant pressure (as you have stated above), we have:
Isobaric change in heat capacity per unit change in temperature, ΔCp/ΔT = (δ/δT)(ΔH/ΔT)
Your subsequent evaluations are incorrect as well, since the solution of a first order differential equation is not as how you portrayed it. I'd suggest that you refer to some college- or university-level mathematics texts to see how a first order differential equation is solved and incorporate that into your solution and proof as above.

Take care.

Answer 9

He didnt assume that \(C_P\) was zero, though.

"I will however assume that the heat capacity is not 0 and therefor need to extend the equation above"

Answer 10

@aaronq : Yes, but the very fact that he made the statement "Only by assuming that the heat capacity is equal to 0, this equation can be used to calculate ΔG as a function of T" is an incorrect perception in itself.

Answer 11

Thats true, good point.

Answer 12

@aaronq : You're welcome.

Answer 13

Responds and corrections to original post.