IChO 2009 · Theory problem 3

Protein Folding

The unfolding reaction for many small proteins can be represented by the equilibrium:

Figure 1

You may assume that the protein folding reaction takes place in a single step. The position of this equilibrium changes with temperature; the melting temperature \(T_m\) is defined as the temperature at which half of the molecules are unfolded and half are folded.

The intensity of the fluorescence signal at a wavelength of 356 nm of a \(1.0\ \mu\mathrm{M}\) (\(\mathrm{M}=\mathrm{mol\,dm^{-3}}\)) sample of the protein Chymotrypsin Inhibitor 2 (CI2) was measured as a function of temperature over the range 58 to 66 °C:

Temperature / °C 58 60 62 64 66
Fluorescence intensity (arbitrary units) 27 30 34 37 40

A \(1.0\ \mu\mathrm{M}\) sample in which all of the protein molecules are folded gives a fluorescence signal of 21 units at 356 nm. A \(1.0\ \mu\mathrm{M}\) sample in which all of the protein molecules are unfolded gives a fluorescence signal of 43 units.

3.1 Assuming that the fluorescence intensity from each species is directly proportional to its concentration, calculate the fraction, \(x\), of unfolded molecules present at each temperature.

3.2 Give an expression for the equilibrium constant, \(K\), in terms of \(x\), and hence calculate the value of \(K\) at each temperature.

3.3 Estimate the value of \(T_m\) for this protein (to the nearest 1 °C).

Assuming that the values of \(\Delta H^\circ\) and \(\Delta S^\circ\) for the protein unfolding reaction are constant with temperature then:

\[ \ln K=-\frac{\Delta H^\circ}{RT}+C \]

where \(C\) is a constant.

3.4 Plot a suitable graph and hence determine the values of \(\Delta H^\circ\) and \(\Delta S^\circ\) for the protein unfolding reaction.

[If you have been unable to calculate values for \(\Delta H^\circ\) and \(\Delta S^\circ\), you should use the following incorrect values for the subsequent parts of the problem:

\(\Delta H^\circ=130\ \mathrm{kJ\,mol^{-1}}\); \(\Delta S^\circ=250\ \mathrm{J\,K^{-1}\,mol^{-1}}\).]

3.5 Calculate the equilibrium constant for the unfolding reaction at 25 °C.

[If you have been unable to calculate a value for \(K\), you should use the following incorrect value for the subsequent parts of the problem: \(K=3.6\cdot10^{-6}\)]

The first order rate constant for the CI2 protein folding reaction can be determined by following the fluorescence intensity when a sample of unfolded protein is allowed to refold (typically the pH of the solution is changed). The concentration of protein when a \(1.0\ \mu\mathrm{M}\) sample of unfolded CI2 was allowed to refold was measured at a temperature of 25 °C:

time / ms 0 10 20 30 40
concentration / \(\mu\mathrm{M}\) 1 0.64 0.36 0.23 0.14

3.6 Plot a suitable graph and hence determine the value of the rate constant for the protein folding reaction, \(k_f\), at 25 °C.

[If you have been unable to calculate the value for \(k_f\), you should use the following incorrect value for the subsequent parts of the question: \(k_f=60\ \mathrm{s^{-1}}\).]

3.7 Determine the value of the rate constant for the protein unfolding reaction, \(k_u\), at 25 °C.

3.8 At 20 °C the rate constant for the protein folding reaction is \(33\ \mathrm{s^{-1}}\). Calculate the activation energy for the protein folding reaction.

Official English text of the problem.

Answers

  • 3.1 At \(58,60,62,64,66\,^\circ\mathrm C\), respectively: \[ \boxed{x=0.273,\ 0.409,\ 0.591,\ 0.727,\ 0.864.} \]

  • 3.2 \[ \boxed{K=\frac{x}{1-x}} \] giving, in the same temperature order, \[ \boxed{K=0.375,\ 0.692,\ 1.44,\ 2.67,\ 6.33.} \]

  • 3.3 \(\boxed{T_m\approx61\,^\circ\mathrm C}\).

  • 3.4 Plot \(\ln K\) against \(1/T\): \[ \boxed{\Delta H^\circ\approx+327\ \mathrm{kJ\,mol^{-1}},\qquad \Delta S^\circ\approx+978\ \mathrm{J\,K^{-1}\,mol^{-1}}.} \]

  • 3.5 \(\boxed{K(25\,^\circ\mathrm C)\approx7.0\times10^{-7}}\).

  • 3.6 Plot \(\ln([U]_t/[U]_0)\) against time: \[ \boxed{k_f\approx49.6\ \mathrm{s^{-1}}\approx50\ \mathrm{s^{-1}}.} \]

  • 3.7 \(\boxed{k_u\approx3.5\times10^{-5}\ \mathrm{s^{-1}}}\).

  • 3.8 \(\boxed{E_a\approx59\ \mathrm{kJ\,mol^{-1}}}\), or approximately \(60\ \mathrm{kJ\,mol^{-1}}\) using the graph-rounded \(k_f=50\ \mathrm{s^{-1}}\).

Solution

Let \(F\) and \(U\) denote folded and unfolded chymotrypsin inhibitor 2 (CI2), respectively. These are conformational states of the same protein. Its amino-acid sequence is not supplied, so a molecular SMILES cannot be specified from the data; ordinary SMILES would not describe the distinction between its folded and unfolded conformations.

The numerical results below use least-squares straight-line fits to all five points, retaining unrounded fit parameters for subsequent calculations. We use \(R=8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}\).

3.1 — Fraction of unfolded molecules

If the unfolded fraction is \(x\), the folded fraction is \(1-x\). Since fluorescence contributions are proportional to concentration and additive, the measured intensity \(I\) is

\[ I=21(1-x)+43x=21+22x. \]

Thus,

\[ \boxed{x=\frac{I-21}{22}}. \]

Temperature / °C \(I\) \(x\)
58 27 \(6/22=0.2727\)
60 30 \(9/22=0.4091\)
62 34 \(13/22=0.5909\)
64 37 \(16/22=0.7273\)
66 40 \(19/22=0.8636\)

3.2 — Equilibrium constants

For the unfolding equilibrium, using the state labels defined above,

\[ \ce{F <=> U}, \]

the equilibrium constant is

\[ K=\frac{[U]}{[F]}. \]

Writing the total protein concentration as \(c_{\mathrm{tot}}\),

\[ [U]=xc_{\mathrm{tot}},\qquad [F]=(1-x)c_{\mathrm{tot}}, \]

so

\[ \boxed{K=\frac{x}{1-x}=\frac{I-21}{43-I}}. \]

Temperature / °C \(K\)
58 \(6/16=0.375\)
60 \(9/13=0.6923\)
62 \(13/9=1.4444\)
64 \(16/6=2.6667\)
66 \(19/3=6.3333\)

These equilibrium constants are dimensionless.

3.3 — Melting temperature

At the melting temperature,

\[ x=\frac12,\qquad K=1. \]

The corresponding fluorescence intensity is

\[ I_m=\frac{21+43}{2}=32. \]

This lies between the intensities of 30 at 60 °C and 34 at 62 °C. Linear interpolation gives

\[ T_m=60+\frac{32-30}{34-30}(62-60) =61\,^\circ\mathrm C. \]

Therefore, to the nearest degree,

\[ \boxed{T_m=61\,^\circ\mathrm C}. \]

3.4 — Enthalpy and entropy of unfolding

The thermodynamic relations

\[ \Delta G^\circ=\Delta H^\circ-T\Delta S^\circ \]

and

\[ \Delta G^\circ=-RT\ln K \]

give

\[ \ln K=-\frac{\Delta H^\circ}{R}\frac1T +\frac{\Delta S^\circ}{R}. \]

Thus, plot \(\ln K\) on the vertical axis against \(1/T\), in \(\mathrm{K^{-1}}\), on the horizontal axis, using absolute temperatures.

\(T\) / K \(1/T\) / \(\mathrm{K^{-1}}\) \(\ln K\)
331.15 0.00301978 −0.98083
333.15 0.00300165 −0.36772
335.15 0.00298374 0.36772
337.15 0.00296604 0.98083
339.15 0.00294855 1.84583

The points lie approximately on a descending straight line. A least-squares fit gives

\[ \ln K= -\left(3.93049\times10^4\ \mathrm K\right)\frac1T +117.65318. \]

The slope is \(-\Delta H^\circ/R\), so

\[ \Delta H^\circ =-R(\text{slope}) \approx (8.314)(3.93049\times10^4) \ \mathrm{J\,mol^{-1}}. \]

Hence,

\[ \boxed{\Delta H^\circ\approx+327\ \mathrm{kJ\,mol^{-1}}}. \]

The intercept is \(C=\Delta S^\circ/R\), giving

\[ \Delta S^\circ=RC \approx(8.314)(117.65318) \ \mathrm{J\,K^{-1}\,mol^{-1}}, \]

and therefore

\[ \boxed{\Delta S^\circ\approx+978\ \mathrm{J\,K^{-1}\,mol^{-1}}}. \]

Unfolding is endothermic and has a positive standard entropy change. As a consistency check, the fitted line gives

\[ T_m=\frac{\Delta H^\circ}{\Delta S^\circ} \approx334.07\ \mathrm K =60.92\,^\circ\mathrm C, \]

consistent with part 3.3.

3.5 — Equilibrium constant at 25 °C

At \(25\,^\circ\mathrm C\),

\[ T=298.15\ \mathrm K. \]

Using the fitted equation directly avoids rounding errors in extrapolation:

\[ \ln K =-\frac{39304.939}{298.15}+117.65318 =-14.1762. \]

Therefore,

\[ K=e^{-14.1762}=6.97\times10^{-7}, \]

so

\[ \boxed{K(25\,^\circ\mathrm C)\approx7.0\times10^{-7}}. \]

The equilibrium strongly favours the folded protein.

3.6 — Folding rate constant

The decreasing concentrations in the kinetic table refer to unfolded protein. Since the equilibrium unfolded fraction at 25 °C is only

\[ x_{\mathrm{eq}}=\frac{K}{1+K}\approx7.0\times10^{-7}, \]

reverse unfolding is negligible over these measurements. The first-order folding equation is therefore

\[ -\frac{d[U]}{dt}=k_f[U]. \]

Integration gives

\[ [U]_t=[U]_0e^{-k_ft}, \]

or

\[ \ln\left(\frac{[U]_t}{[U]_0}\right)=-k_ft. \]

Plot \(\ln([U]_t/[U]_0)\) vertically against \(t\), in seconds, horizontally.

\(t\) / s \([U]_t/[U]_0\) \(\ln([U]_t/[U]_0)\)
0.000 1.00 0.00000
0.010 0.64 −0.44629
0.020 0.36 −1.02165
0.030 0.23 −1.46968
0.040 0.14 −1.96611

The graph is approximately a straight line through the origin with a negative slope. A least-squares fit gives

\[ \ln\left(\frac{[U]_t}{[U]_0}\right) =0.01038-\left(49.556\ \mathrm{s^{-1}}\right)t. \]

Consequently,

\[ \boxed{k_f=49.6\ \mathrm{s^{-1}}\approx50\ \mathrm{s^{-1}}}. \]

For an exact reversible treatment, the equilibrium concentration would be subtracted before taking logarithms, and the decay constant would be \(k_f+k_u\). The correction is negligible at the precision of these data.

3.7 — Unfolding rate constant

At equilibrium, the unfolding and folding rates are equal:

\[ k_u[F]_{\mathrm{eq}}=k_f[U]_{\mathrm{eq}}. \]

Therefore,

\[ \frac{k_u}{k_f} =\frac{[U]_{\mathrm{eq}}}{[F]_{\mathrm{eq}}} =K, \]

so

\[ k_u=Kk_f. \]

Using the unrounded results,

\[ k_u=(6.9718\times10^{-7})(49.556\ \mathrm{s^{-1}}) =3.4549\times10^{-5}\ \mathrm{s^{-1}}. \]

Thus,

\[ \boxed{k_u\approx3.5\times10^{-5}\ \mathrm{s^{-1}}}. \]

3.8 — Activation energy for folding

The Arrhenius equation is

\[ k_f=Ae^{-E_a/(RT)}. \]

Assuming a constant pre-exponential factor over this temperature interval,

\[ \ln\left(\frac{k_2}{k_1}\right) =\frac{E_a}{R}\left(\frac1{T_1}-\frac1{T_2}\right). \]

Here,

\[ T_1=293.15\ \mathrm K,\quad k_1=33\ \mathrm{s^{-1}}, \]

\[ T_2=298.15\ \mathrm K,\quad k_2=49.556\ \mathrm{s^{-1}}. \]

Hence,

\[ E_a= \frac{R\ln(k_2/k_1)} {1/T_1-1/T_2} = \frac{ (8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}) \ln(49.556/33) }{ (1/293.15-1/298.15)\ \mathrm{K^{-1}} }. \]

This gives

\[ E_a\approx5.91\times10^4\ \mathrm{J\,mol^{-1}}, \]

or

\[ \boxed{E_a\approx59\ \mathrm{kJ\,mol^{-1}}}. \]

Using \(k_f=50\ \mathrm{s^{-1}}\) rounded from the graph gives approximately \(60\ \mathrm{kJ\,mol^{-1}}\), consistent with graphical precision.

The solution was checked against the official solution.