Adapted from Molecular Driving Forces - Statistical Thermodynamics in Chemistry and Biology by Ken A. Dill and Sarina Bromberg, Garland Science, Taylor & Francis Group, New York and London (2003). Hereafter, we will abbreviate the names of the authors as D&B.(a) The protein below has four

Question

Adapted from Molecular Driving Forces - Statistical Thermodynamics in Chemistry and Biology by Ken A. Dill and Sarina Bromberg, Garland Science, Taylor & Francis Group, New York and London (2003). Hereafter, we will abbreviate the names of the authors as D&B.

(a) The protein below has four distinguishable binding sites (α, β, γ, and δ) for the ligand L. Find the protein equilibrium binding population for a case where the ligand-protein association and dissociation constants are equal. Specifically, calculate the number of distinct arrangements, W, and the entropy (in units of [latex]k_{B}[/latex]) when:

1. No ligands are bound ([latex]N_{L}[/latex] = 0).
2. One ligand is bound ([latex]N_{L}[/latex] = 1).
3. Two ligands are bound ([latex]N_{L}[/latex] = 2).
4. Three ligands are bound ([latex]N_{L}[/latex] = 3).
5. Four ligands are bound ([latex]N_{L}[/latex] = 4).
6. Which states have the highest entropy?
7. Which states have the lowest entropy?
8. If the binding constants are not equal, namely, if ligand L has a higher probability of being bound, say:

[latex]P_{ ext {bound }}=75 \% ext { and } P_{ ext {unbound }}=25 \%[/latex]

Calculate the probability distribution for all the available states.

(b) Consider a zipper that has N links. Each link can be in two states, where state 1 means that the zipper is closed and has energy of zero (the zipper ground state) and state 2 means that the zipper is open and has energy ε (the zipper excited state). The zipper can only unzip from the left end, and the ith link cannot open unless all the links to its left (1, 2, 3... i-1) are already open.

HINT: [latex]\sum_{n=0}^{k} a r^{n}=a \frac{1-r^{k+1}}{1-r}, ext { when } r<1[/latex].

1. Think carefully about what microstates are allowed, and derive an explicit expression for the Canonical partition function, Q, for the zipper.
2. If the average energy of the zipper is [latex],[/latex] find the average number of open links, [latex],[/latex] in the low-temperature limit, [latex]\varepsilon / k_{B} T \gg 1[/latex]. (Do not leave [latex],[/latex] in your final expression, but instead, calculate it.)

Accepted Answer

(a1) [latex]N_{L}=0 \quad \Rightarrow \quad W=\frac{4 !}{0 ! 4 !}=1[/latex], only one arrangement is possible.

[latex]S=k_{B} \ln W=k_{B} \ln (1)=0[/latex]

(a2) [latex]N_{L}=1 \Rightarrow W=\frac{4 !}{1 ! 3 !}=4[/latex], four ways to arrange on α, β, γ, or δ.

[latex]S=k_{B} \ln W=k_{B} \ln (4)[/latex]

(a3) [latex]N_{L}=2 \Rightarrow W=\frac{4 !}{2 ! 2 !}=6[/latex], six ways to arrange: αβ, αγ, αδ, βγ, βδ, or γδ.

[latex]S=k_{B} \ln W=k_{B} \ln (6)[/latex]

(a4) [latex]N_{L}=3 \Rightarrow W=\frac{4 !}{3 ! 1 !}=4[/latex], four ways to arrange: the vacant spot is on α, β, γ, or δ.

[latex]S=k_{B} \ln W=k_{B} \ln (4)[/latex]

(a5) [latex]N_{L}=4 \quad \Rightarrow \quad W=\frac{4 !}{4 ! 0 !}=1[/latex], only one arrangement is possible.

[latex]S=k_{B} \ln W=k_{B} \ln (1)=0[/latex]

(a6) The states of highest entropy have [latex]N_{L}=2[/latex].

(a7) The states of lowest entropy have [latex]N_{L}=0 ext { or } N_{L}=4[/latex].

(a8) The probability distribution associated with binding [latex]N_{L}[/latex] ligands with probability p and not binding [latex]\left(M-N_{L} ight)[/latex] ligands with probability (1-p) onto M sites on the protein is given by:

[latex]p\left(N_{L}, M ight)=\frac{M !}{N_{L} !\left(M-N_{L} ight) !} p^{N_{L}}(1-p)^{M-N_{L}}[/latex] (20)

Using Eq. (20), it follows that:

(i) For [latex]N_{L}=0, p(0,4)=\frac{4 !}{0 ! 4 !}(0.75)^{0}(0.25)^{4}=3.91 imes 10^{-3}[/latex].

(ii) For [latex] N_{L}=1, p(1,4)=\frac{4 !}{1 ! 3 !}(0.75)^{1}(0.25)^{3}=4.69 imes 10^{-2}[/latex].

(iii) For [latex]N_{L}=2, p(2,4)=\frac{4 !}{2 ! 2 !}(0.75)^{2}(0.25)^{2}=0.211[/latex].

(iv) For [latex]N_{L}=3, p(3,4)=\frac{4 !}{3 ! 1 !}(0.75)^{3}(0.25)^{1}=0.422[/latex].

(v) For [latex]N_{L}=4, p(4,4)=\frac{4 !}{4 ! 0 !}(0.75)^{4}(0.25)^{0}=0.316[/latex].

The resulting probability distribution is plotted in Fig. 1.

(b1) The possible states of the zipper are determined by the open link number i. Each state of the zipper has an energy of iε, and the Canonical partition function can be obtained by summing [latex]e^{-E_{i} / k_{B} T}[/latex] for every possible state. Specifically:

[latex]Q=\sum_{i=0}^{N} e^{-i \varepsilon / k_{B} T}=e^{0}+e^{-\varepsilon / k_{B} T}+e^{-2 \varepsilon / k_{B} T}+\ldots+e^{-N \varepsilon / k_{B} T}[/latex] (21)

Equation (21) is a finite geometric series (recall that [latex]\left.e^{-\varepsilon / k_{B} T}<1 ight)[/latex] and can be summed according to the HINT given in the Problem Statement [latex]\left(\sum_{n=0}^{k} a r^{n}=a \frac{1-r^{k+1}}{1-r} ight.[/latex], for r < 1). Specifically:

[latex]Q=\frac{1-e^{-(N+1) \varepsilon / k_{B} T}}{1-e^{-\varepsilon / k_{B} T}}[/latex] (22)

(b2) The average energy of the zipper can be obtained by differentiating lnQ with respect to [latex]\beta=1 / k_{B} T[/latex]. Specifically, from Eq. (22), it follows that:

[latex]\ln Q=\ln \left(1-e^{-(N+1) \beta \varepsilon} ight)-\ln \left(1-e^{-\beta \varepsilon} ight)[/latex] (23)

[latex]\langle E angle=-\left(\frac{\partial \ln Q}{\partial \beta} ight)[/latex] (24)

[latex]\begin{aligned}\langle E angle &=-\left(\frac{1}{1-e^{-(N+1) \beta \varepsilon}} \frac{\partial}{\partial \beta}\left(1-e^{-(N+1) \beta \varepsilon} ight)-\frac{1}{1-e^{-\beta \varepsilon}} \frac{\partial}{\partial \beta}\left(1-e^{-\beta \varepsilon} ight) ight) \&=-\left[\frac{(N+1) \varepsilon e^{-(N+1) \beta \varepsilon}}{1-e^{-(N+1) \beta \varepsilon}}-\frac{\varepsilon e^{-\beta \varepsilon}}{1-e^{-\beta \varepsilon}} ight]\end{aligned}[/latex]

For [latex]e^{-\beta \varepsilon}<<1[/latex], the two denominators will both approach 1. Hence, we obtain:

[latex]\begin{aligned}\langle E angle &=-\left[(N+1) \varepsilon e^{-(N+1) \beta \varepsilon}-\varepsilon e^{-\beta \varepsilon} ight] \&=-\left[(N+1) e^{-N \beta \varepsilon}-1 ight] \varepsilon e^{-\beta \varepsilon}\end{aligned}[/latex]

Again, using the fact that [latex]e^{-\beta \varepsilon}<<1[/latex] (the term in brackets goes to 1) yields:

[latex]\langle E angle \cong \varepsilon e^{-\beta \varepsilon}[/latex] (25)

Because [latex]\langle E angle=\varepsilon\langle i angle[/latex], it follows that:

[latex]\varepsilon\langle i angle=\varepsilon e^{-\beta \varepsilon}=>\langle i angle=e^{-\beta \varepsilon}[/latex] (26)

Question 24.2: Adapted from Molecular Driving Forces - Statistical Thermody...

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