Question 25.2: Statistical mechanics can be used to predict the dependence ...
Statistical mechanics can be used to predict the dependence of protein folding on temperature. Consider a polypeptide chain consisting of six beads having the energy ladder shown below, where the energy increment of the unfolded states, \varepsilon_{0}, is positive. You are asked to answer the following questions:
(a) What is the degeneracy of each energy state?
(b) What is the Canonical partition function of the polypeptide chain?
(c) What are the probabilities of observing the polypeptide chain in each of the three states as a function of temperature? Discuss the low-temperature and the hightemperature limits.
(a) The degeneracy corresponds to the number of microstates in each energy level.
Examination of Fig. 1 in the Problem Statement shows that:
E_{0}=0, \quad g_{0}=4 (24)
E_{1}=\epsilon_{0}, \quad g_{1}=11 (25)
E_{2}=2 \epsilon_{0}, \quad g_{2}=21 (26)
(b) The partition function of the polypeptide chain, corresponding to the energy ladder in Fig. 1 in the Problem Statement, is given by:
q=\sum_{i} g_{i} \exp \left(-\beta E_{i}\right) (27)
q=g_{0} \exp \left(-\beta E_{0}\right)+g_{1} \exp \left(-\beta E_{1}\right)+g_{2} \exp \left(-\beta E_{2}\right) (28)
Using the information from Fig. 1 and the Problem Statement, as well as using the degeneracies found in Part (a), yields:
q=4 \exp (-\beta(0))+11 \exp \left(-\beta\left(\epsilon_{0}\right)\right)+21 \exp \left(-\beta\left(2 \epsilon_{0}\right)\right) (29)
q=4+11 \exp \left(-\beta \epsilon_{0}\right)+21 \exp \left(-2 \beta \epsilon_{0}\right) (30)
or
q=4+11 \exp \left(-\epsilon_{0} / k_{B} T\right)+21 \exp \left(-2 \epsilon_{0} / k_{B} T\right) (31)
(c) The probabilities of finding the polypeptide chain in the folded state (0) and in the two unfolded states (1 and 2) are given by:
p_{0}(\text { folded state })=\frac{g_{0} \exp \left(-0 / k_{B} T\right)}{q}=4 / q (32)
p_{1}(\text { partially unfolded state })=\frac{g_{1} \exp \left(-\epsilon_{0} / k_{B} T\right)}{q}=11 \exp \left(-\epsilon_{0} / k_{B} T\right) / q (33)
p_{2}(\text { fully unfolded state })=\frac{g_{2} \exp \left(-2 \epsilon_{0} / k_{B} T\right)}{q}=21 \exp \left(-2 \epsilon_{0} / k_{B} T\right) / q (34)
Because each energy level has a different degeneracy \left(g_{0} \neq g_{1} \neq g_{2}\right), the energy level with the largest degeneracy is more likely to be observed at higher temperatures. At high temperature (T \rightarrow \infty),
\lim _{T \rightarrow \infty} q=4+11+21=36 (35)
\lim _{T \rightarrow \infty} p_{0}=4 / 36 \approx 0.1111 (36)
\lim _{T \rightarrow \infty} p_{1}=11 / 36 \approx 0.3056 (37)
\lim _{T \rightarrow \infty} p_{2}=21 / 36 \approx 0.5833 (38)
As expected, p_{2}>p_{1}>p_{0}. We can see that as T increases, the unfolded states (denatured polypeptide) are more likely to be observed, as seen in nature.
At low temperatures, the lowest-energy levels are more likely, because at low temperatures (T \rightarrow 0), it follows that:
\lim _{T \rightarrow 0} q=4 (39)
\lim _{T \rightarrow 0} p_{0}=4 / 4=1 (40)
\lim _{T \rightarrow 0} p_{1}=0 / 4=0 (41)
\lim _{T \rightarrow 0} p_{2}=4 / 4=0 (42)
As expected, p_{0}=1 \text { and } p_{1}=p_{2}=0 \text { as } T \rightarrow 0.