IChO 2005 · Theory problem 8

Protein Folding

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Most proteins exist usually only in two forms, the native form (N) and the unfolded form (U) when they are thermally or chemically denatured, without appreciable concentrations of other stable intermediates in equilibrium with the native and unfolded forms. For these proteins, the folding-unfolding equilibrium can be described by the following simple chemical equation:

Figure 1

where N and U denote the folded state (native state) and the unfolded state (denatured state) of the protein, respectively. \(K(T)\) is the equilibrium constant for the process at absolute temperature \(T\).

8.1 What is the equilibrium constant for the process when the native and denatured states are present in equal proportions at equilibrium?

8.2 What is the standard free energy change of the process \((\Delta G^\circ(T))\) when the native and denatured states are present in equal proportions at equilibrium? Express your answer in SI units.

8.3 If \((C_{\mathrm N})_{\mathrm{eq}}\) and \((C_{\mathrm U})_{\mathrm{eq}}\) denote the equilibrium concentrations of \(N\) and \(U\) in solution, respectively, and \(C\) is the total concentration of the protein, the fraction of the total protein that is unfolded under the equilibrium condition is given by \(f_{\mathrm U}=(C_{\mathrm U})_{\mathrm{eq}}/C\). Deduce an expression for \(f_{\mathrm U}\) in terms of the equilibrium constant \(K\). Show all work on the answer sheet.

When a protein is denatured by increasing the temperature of the solution, the fraction of the unfolded protein increases with temperature, as shown in the following Figure.

Figure 2

The mid-point of the denaturation curve is given by \(f_{\mathrm U}=\tfrac12\) and \(T=T_{1/2}\). The latter is often referred to as the denaturation temperature. At temperatures higher than \(T_{1/2}\), \(f_{\mathrm U}\) increases above \(\tfrac12\), but at temperatures lower than \(T_{1/2}\), \(f_{\mathrm U}\) decreases below \(\tfrac12\).

8.4 What is the sign of \(\Delta G^\circ(T)\) at temperatures below and above \(T_{1/2}\)? Select your answer from the following choices.

a) Negative both below and above \(T_{1/2}\).

b) Positive both below and above \(T_{1/2}\).

c) Positive below \(T_{1/2}\), but negative above \(T_{1/2}\).

d) Negative below \(T_{1/2}\), but positive above \(T_{1/2}\)..

8.5 How does the standard Gibbs free energy change for the process vary when the temperature (i) increases above \(T_{1/2}\) and (ii) decreases below \(T_{1/2}\)? Select your answer from the following choices.

a) Decrease in both cases.

b) Increase in both cases.

c) Increases above \(T_{1/2}\), but decreases below \(T_{1/2}\).

d) Decreases above \(T_{1/2}\), but increases below \(T_{1/2}\).

The kinetics of unfolding and refolding of a protein has recently become an intense area of study. We could rewrite the chemical equation for the process as follows:

Figure 3

where \(k_f\) and \(k_b\) denote the forward and backward reaction rate constants, respectively., assuming that both the forward and reverse processes are elementary steps that follow first-order kinetics.

8.6 For the simple chemical equation and elementary kinetic steps used to describe the protein folding-unfolding process outlined above, what is the relationship between equilibrium constant \(K\) and the rate constants \(k_f\) and \(k_b\)?

8.7 Derive a rate law for the overall process, that is \(\mathrm dC_{\mathrm U}/\mathrm dt\) in terms of only rate constants, \(C_{\mathrm U}\) and \((C_{\mathrm U})_{\mathrm{eq}}\).

Official English text of the problem.

Answers

  • 8.1: \(\boxed{K=1}\), dimensionless.
  • 8.2: \(\boxed{\Delta G^\circ(T)=0\ \mathrm{J\,mol^{-1}}}\).
  • 8.3: \(\boxed{f_{\mathrm U}=\dfrac{K}{1+K}}\).
  • 8.4: (c) — positive below \(T_{1/2}\), negative above \(T_{1/2}\).
  • 8.5: (d) — relative to its value at \(T_{1/2}\), \(\Delta G^\circ\) decreases on heating above the midpoint and increases on cooling below it.
  • 8.6: \(\boxed{K=\dfrac{k_f}{k_b}}\).
  • 8.7: \(\boxed{\dfrac{\mathrm dC_{\mathrm U}}{\mathrm dt}=(k_f+k_b)\big[(C_{\mathrm U})_{\mathrm{eq}}-C_{\mathrm U}\big]}\).

Solution

Throughout, the forward process is unfolding, \(\ce{N -> U}\). The labels denote the native and unfolded conformations of the same protein. Its molecular structure is unspecified, so no unique SMILES can be assigned; ordinary SMILES also does not describe the folding conformation.

8.1

For the two-state equilibrium \(\ce{N <=> U}\), the equilibrium constant in the concentration model used here is

\[ K(T)=\frac{(C_{\mathrm U})_{\mathrm{eq}}} {(C_{\mathrm N})_{\mathrm{eq}}}. \]

Equal proportions mean equal concentrations, giving

\[ \boxed{K(T)=1}. \]

The ratio, and hence \(K\), is dimensionless.

8.2

The standard reaction Gibbs energy and equilibrium constant satisfy

\[ \Delta G^\circ(T)=-RT\ln K(T). \]

Using \(K=1\),

\[ \boxed{\Delta G^\circ(T)=-RT\ln 1 =0\ \mathrm{J\,mol^{-1}}}. \]

8.3

Conservation of the total protein concentration gives

\[ C=(C_{\mathrm N})_{\mathrm{eq}}+(C_{\mathrm U})_{\mathrm{eq}}. \]

Since

\[ (C_{\mathrm U})_{\mathrm{eq}} =K(C_{\mathrm N})_{\mathrm{eq}}, \]

the unfolded fraction is

\[ \begin{aligned} f_{\mathrm U} &=\frac{(C_{\mathrm U})_{\mathrm{eq}}} {(C_{\mathrm N})_{\mathrm{eq}}+(C_{\mathrm U})_{\mathrm{eq}}}\\ &=\frac{K(C_{\mathrm N})_{\mathrm{eq}}} {(1+K)(C_{\mathrm N})_{\mathrm{eq}}}\\ &=\boxed{\frac{K}{1+K}}. \end{aligned} \]

Equivalently,

\[ K=\frac{f_{\mathrm U}}{1-f_{\mathrm U}}. \]

8.4

Below the denaturation midpoint,

\[ f_{\mathrm U}<\frac12 \quad\Longrightarrow\quad K=\frac{f_{\mathrm U}}{1-f_{\mathrm U}}<1. \]

Thus \(\ln K<0\), and because \(R,T>0\),

\[ \Delta G^\circ(T)=-RT\ln K>0. \]

Above the midpoint,

\[ f_{\mathrm U}>\frac12 \quad\Longrightarrow\quad K>1 \quad\Longrightarrow\quad \Delta G^\circ(T)<0. \]

Therefore the answer is (c):

\[ \boxed{ \begin{cases} \Delta G^\circ(T)>0,&T<T_{1/2},\\ \Delta G^\circ(T)=0,&T=T_{1/2},\\ \Delta G^\circ(T)<0,&T>T_{1/2}. \end{cases}} \]

8.5

At \(T_{1/2}\), equal populations give \(K=1\) and \(\Delta G^\circ=0\).

(i) Heating above \(T_{1/2}\) increases the unfolded fraction, giving \(K>1\). Consequently, \(\Delta G^\circ=-RT\ln K\) becomes negative: it decreases from its midpoint value.

(ii) Cooling below \(T_{1/2}\) decreases the unfolded fraction, giving \(K<1\). Consequently, \(\Delta G^\circ\) becomes positive: it increases from its midpoint value.

Thus the intended choice is

\[ \boxed{\text{(d)}}. \]

8.6

The first-order forward and backward reaction rates are

\[ r_f=k_fC_{\mathrm N}, \qquad r_b=k_bC_{\mathrm U}. \]

At equilibrium these rates are equal:

\[ k_f(C_{\mathrm N})_{\mathrm{eq}} =k_b(C_{\mathrm U})_{\mathrm{eq}}. \]

Rearranging,

\[ \boxed{ K=\frac{(C_{\mathrm U})_{\mathrm{eq}}} {(C_{\mathrm N})_{\mathrm{eq}}} =\frac{k_f}{k_b}}. \]

Both first-order rate constants have SI units \(\mathrm{s^{-1}}\), so their ratio is dimensionless.

8.7

At fixed temperature, the net rate of formation of unfolded protein is

\[ \frac{\mathrm dC_{\mathrm U}}{\mathrm dt} =k_fC_{\mathrm N}-k_bC_{\mathrm U}. \]

The total protein concentration is conserved:

\[ C_{\mathrm N}+C_{\mathrm U}=C. \]

Substituting \(C_{\mathrm N}=C-C_{\mathrm U}\),

\[ \frac{\mathrm dC_{\mathrm U}}{\mathrm dt} =k_fC-(k_f+k_b)C_{\mathrm U}. \]

At equilibrium, the derivative vanishes, so

\[ k_fC=(k_f+k_b)(C_{\mathrm U})_{\mathrm{eq}}. \]

Eliminating \(C\) yields the requested rate law:

\[ \boxed{ \frac{\mathrm dC_{\mathrm U}}{\mathrm dt} =(k_f+k_b)\left[(C_{\mathrm U})_{\mathrm{eq}}-C_{\mathrm U}\right]}. \]

This correctly predicts an increase in \(C_{\mathrm U}\) when it is below equilibrium and a decrease when it is above equilibrium. With concentrations in \(\mathrm{mol\,m^{-3}}\), the rate has units \(\mathrm{mol\,m^{-3}\,s^{-1}}\).

The solution was checked against the official solution.