Thoughts on protein conformation change.
Systematic descriptions of the energy landscapes help us to intrinsically quantify conformations and their changes. Here I will give a brief summary of pioneered works of Prof. Jin Wang et al.
For analytical derivation
For a protein sequence, its total number of conformations is assumed to be $\Omega_0$. Then the conformational entropy with energy $E$ is:
\[\begin{aligned} S(E) &= K_B\ln(\underbrace{\Omega_{0} n(E)}_{\text{the number of conformations with energy} E}) \\ &= K_B \ln \Omega_0 + K_B \ln n(E)\\ &= \underbrace{K_B \ln \Omega_0}_{S_0} - K_B \frac{(E-\bar{E})^2}{2\Delta E^2} \color{gray}{\underbrace{- K_B \frac{\ln(2\pi\Delta E^2)}{2}}_{\text{dropped}}} \end{aligned}\]Since we have the thermodynamic relation:
\[\frac{\partial S}{\partial E} = \frac{1}{T}\]we can derive the most probable energy as a function of $T$ as:
\[\begin{aligned} \frac{-2K_{B}(E-\bar{E})}{2 \Delta E^{2}}&= \frac{1}{T}\\ &\Downarrow \\ E(T) &= \bar{E} - \frac{\Delta E^{2}}{K_{B}T} \\ \end{aligned}\]So the entropy at the most probable energy as a function of $T$ is:
\[S(T) = S(E(T)) = S_{0} - \frac{\Delta E^{2}}{2K_{B}T^{2}}\]From the thermodynamic expressions of the energy $E(T)$ and entropy $S(T)$, the Helmholtz free energy of the system as a function of $T$ can be expressed as:
\[\begin{aligned} F(T) &= E(T) - TS(T)\\ &=\bar{E} - \frac{\Delta E^{2}}{K_{B}T} - TS_{0} + \frac{\Delta E^{2}}{2K_{B}T}\\ &= \bar{E} - TS_{0} - \frac{\Delta E^{2}}{2K_{B}T}\\ \end{aligned}\]With the observations of naturally occurring proteins, we can assume that a natural protein normally has a unique ground state (with energy $E_N$) at which both the energy and entropy variance (roughness) are zero. Then the free energy of the native state equals $E_N$.
A first-order transition between native state and non-native state is expected at the temperature $T_{f}$ (folding transition temperature) where they have equal free energy. Thus we have:
\[\begin{aligned} E_{N} &= \bar{E} - T_{f}S_{0} - \frac{\Delta E^{2}}{2K_{B}T_{f}} \\ \underbrace{\bar{E}-E_{N}}_{\delta E} &= T_{f}S_{0} + \frac{\Delta E^{2}}{2K_{B}T_{f}} \\ &\Downarrow \\ T_{f} &= \frac{\delta E}{2S_{0}}(1+\sqrt{1 - \frac{2S_{0}\Delta E^{2}}{K_{B} \delta E^{2}}}) \end{aligned}\]$T_{f}$ should be larger than $T_{g}$ so as not to be trapped in the frozen state and make the system able to reach the native state. So we are focusing on $\frac{T_{f}}{T_{g}}$ and find out that:
\[\begin{aligned} \frac{T_{f}}{T_{g}} &= \underbrace{\sqrt{\frac{K_{B}}{2S_{0}}}\frac{\delta E}{\Delta E}}_{\Lambda} + \sqrt{\frac{K_{B}\delta E^{2}}{2S_{0}\Delta E^{2}}-1} \\ &= \Lambda + \sqrt{\Lambda^2 - 1} \\ \end{aligned}\]So we have the Optimal Foldability Criterion
The probability $P(E)$ of sampling any conformation at a finite temperature $T$ with energy $E$ is:
\[P(E) = \frac{n(E)\exp[-\frac{E-\bar{E}}{K_{B}T}]}{Z}\]Thus, at a particular $T$ larger than $T_{g}$, the probability of the system in its unique native state and the non-native state are $P_N$ and $P_{D}$ respectively:
\[\begin{aligned} P_{N}&=\frac{\exp[-\frac{E_{N}-\bar{E}}{K_{B}T}]}{Z}\\ P_{D}&=\sum_{E>E_{N}}\frac{n(E)\exp[-\frac{E-\bar{E}}{K_{B}T}]}{Z}\\ \end{aligned}\]So the thermodynamic stability of the native ground state is quantified as:
\[\begin{aligned} \Delta G &= -K_{B}T\ln(\frac{P_N}{P_D})\\ &= E_N + K_{B}T\ln\left[\sum_{E>E_N} n(E) \exp(\frac{-E}{K_B T}) \right] \end{aligned}\]NOTE: Still on writing.
Cited as:
@online{zhu2022funneled,
title={Conformation 101 - Funneled energy landscape},
author={Zefeng Zhu},
year={2022},
month={May},
url={https://naturegeorge.github.io/blog/2022/05/funneled-energy-landscape/},
}