1. 6. Lecture SS 2005Optimization, Energy Landscapes, Protein Folding1 V6: Chemical Kinetics & Transition States see chapter 19 in book of K. Dill Aim: describe kinetics of processes on energy landscapes (e.g. chemical reactions). - temperature effect - detailed balance - mass action law - Arrhenius plots - concept of transition state/activation barrier  transition state theory -  -value analysis revisited - effect of catalysts

2. 6. Lecture SS 2005Optimization, Energy Landscapes, Protein Folding2 Reaction rates are proportional to concentrations Lets consider a simple kinetic process, the interconversion between 2 states, k f and k r : forward and reverse rate coefficients. How do the amounts of A and B change with time t, given the initial amounts at time t = 0 ? The two equations are coupled. One can solve them by matrix algebra …

3. 6. Lecture SS 2005Optimization, Energy Landscapes, Protein Folding3 Excursion: coupled differential equations We can use the method of diagonalisation to solve coupled ordinary differential equations. For example, let x(t) and y(t) be differentiable functions and x' and y' their derivatives. The differential equations are relatively difficult to solve: By diagonalizing the square matrix, we get but u' = ku for a constant k is easy to solve. It has the solution u = Ae kx where A is a constant Remembering this fact, we translate the ODEs into matrix form www.algebra.com

4. 6. Lecture SS 2005Optimization, Energy Landscapes, Protein Folding4 Excursion: coupled differential equations By diagonalizing the square matrix, we get We then put It follows that Thus The solutions of this system are found easily:with some constants C and D. With

5. 6. Lecture SS 2005Optimization, Energy Landscapes, Protein Folding5 Reaction rates are proportional to concentrations With this technique, we could solve our system of coupled diff. equations. If k r << k f, the first equation simplies to If [A(t)] + [B(t)] = constant, then

6. 6. Lecture SS 2005Optimization, Energy Landscapes, Protein Folding6 [A] eq and [B] eq : equilibrium concentrations. To see that this is a condition of equilibrium follows from inserting into resulting in At equilibrium, rates obey detailed balance The principle of detailed balance says that the forward and reverse rates must be identical for an elementary reaction at equilibrium: Taken from Dill book

7. 6. Lecture SS 2005Optimization, Energy Landscapes, Protein Folding7 At equilibrium, rates obey detailed balance The detailed balance condition relates the rate coefficients k f and k r to the equilibrium constant K: Taken from Dill book For more complex systems, the principle of detailed balance gives more information beyond the statement of equilibrium. For a system having more than one elementary reaction, the forward and reverse rates must be equal for every elementary reaction. For this system: Let‘s consider a 3-state mechanism with k IA  0, k BI  0, k AB  0.

8. 6. Lecture SS 2005Optimization, Energy Landscapes, Protein Folding8 These are two independent equations for 3 unknown concentrations  the system has an infinite number of solutions. In mechanism (b), all rates of the Denominator in are zero  mechanism (b) is impossible. At equilibrium, rates obey detailed balance This results in the mechanism shown right. The only conditions for equilibrium are: Taken from Dill book

9. 6. Lecture SS 2005Optimization, Energy Landscapes, Protein Folding9 At equilibrium, rates obey detailed balance The principle of detailed balance says that forward and backward reactions at equilibrium cannot have different intermediate states. That is, if the forward reaction is A  I  B, the backward reaction cannot be B  A. The principle of detailed balance can be derived from microscopic statistical mechanics.

10. 6. Lecture SS 2005Optimization, Energy Landscapes, Protein Folding10 The mass action laws describe mechanisms in chemical kinetics Suppose the following reaction leading from reactants A, B, and C to product P: In general, the initial reaction rate depends on - the concentrations of the reactants - the temperature and pressure - and on the coefficients a, b, and c. Kinetic law of mass action (CM Guldberg & P Waage, 1864): „the reactants should depend on on stoichiometry in the same way that equilibrium constants do“. Although mass action is in agreement with many experiments, there are exceptions. These require a quantum mechanical description.

11. 6. Lecture SS 2005Optimization, Energy Landscapes, Protein Folding11 Reaction rates depend on temperature Consider a binary reaction in the gas phase: Suppose that By definition, the rate coefficient k 2 is independent of [A] and [B]. But k 2 can depend strongly on temperature. The observed dependence of the reaction rate on the temperature is much greater than one would expect from just the enhanced thermal motions of the molecules.

12. 6. Lecture SS 2005Optimization, Energy Landscapes, Protein Folding12 Arrhenius equation 1889, S. Arrhenius started from the van‘t Hoff equation for the strong dependence of the equilibrium constant K on temperature: and proposed that k f and k r also have van‘t Hoff form where E a and E‘ a have units of energy that are chosen to fit exp. data. E a and E‘ a are called activation energies.

13. 6. Lecture SS 2005Optimization, Energy Landscapes, Protein Folding13 Activation energy diagram According to Arrhenius, it is not the average energy of the reactants that determines the reaction rates but only the high energies of the ‚activated‘ molecules. Taken from Dill book There are two plateaus, one for the reactants and one for the products. In between lies an energy maximum (also: transition state or activation barrier) which is the energy that activated molecules must have to proceed from reactants to products. Measuring k f as a function of temperature, and using eq. (1) gives E a. Measuring the reverse rate gives E‘ a. Measuring the equilibrium constant versus temperature gives  h°.

14. 6. Lecture SS 2005Optimization, Energy Landscapes, Protein Folding14 Population at different temperatures From Taken from Dill book it follows The figure shows how activation is interpreted according to the Boltzmann distribution law: a small increase in temperature can lead to a relatively large increase in the population of high-energy molecules.

15. 6. Lecture SS 2005Optimization, Energy Landscapes, Protein Folding15 Arrhenius plots Integrating over temperature T Taken from Dill book gives: H 2 + I 2  2HI (open circles) 2HI  H 2 + I 2 (full circles) Diffusion of carbon in iron The figures show examples of chemical systems showing Arrhenius behavior.

16. 6. Lecture SS 2005Optimization, Energy Landscapes, Protein Folding16 Activated processes Arrhenius kinetics applies to many physical and chemical processes. When should one treat a process as activated? If a small increase in temperature gives a large increase in rate, a good first step is to try the Arrhenius model. E.g. breaking of bonds. Counter example: highly reactive radicals. These can be much faster than typical activated processes and they slow down with increasing temperature. We now describe a more microscopic approach to reaction rates, called transition state theory.

17. 6. Lecture SS 2005Optimization, Energy Landscapes, Protein Folding17 The energy landscape of a reaction An energy landscape defines how the energy of a reacting system depends on its degrees of freedom. E.g. A + BC  AB + C Each reaction trajectory would involve some excursions up the walls of the valleys. When averaged over multiple trajectories, the reaction process can be described as following the lowest energy route, along the entrance valley over the saddle point and out of the exit valley, because the Boltzmann populations are highest along that average route. Taken from Dill book Energy surface for D + H 2  HD + H The transition (saddle) point is denoted by the symbol ‡. It is unstable: a ball placed on the saddle point will roll downhill along the reaction coordinate.

18. 6. Lecture SS 2005Optimization, Energy Landscapes, Protein Folding18 Calculating rate coefficients from TST Let us consider the reaction by transition state theory: Divide the reaction process into two stages: (1) the equilibrium between the reactants ant the transition state (AB) ‡ with ‚equilibrium constant‘ K ‡ (2) a direct step downhill from the TS to the product with rate coefficient k ‡ : Key assumption of TST: step (1) can be expressed as an equilibrium between the reactants A and B and the transition state (AB) ‡, with even though (AB) ‡ is not a true equilibrium state.

19. 6. Lecture SS 2005Optimization, Energy Landscapes, Protein Folding19 Calculating rate coefficients from TST The overall rate is expressed as the number of molecules in the TS, [(AB) ‡ ], multiplied by the rate coefficient k ‡ for the second product-forming step Because the quantitiy K ‡ is regarded as an equilibrium constant, it can be expressed in terms of the molar partition functions: where  D ‡ is the dissociation energy of the TS minus the dissociation energy of the reactants. q (AB)‡ is the partition function of the transition state.

20. 6. Lecture SS 2005Optimization, Energy Landscapes, Protein Folding20 The transition state (left) Contour plot of a reaction pathway (- - -) on an energy landscape for the reaction A + BC  AB + C. The broken line shows the lowest-energy path between reactants and products (right) The transition state is an unstable point along the reaction pathway (indicated by the arrow) and a stable point in all other directions that are normal to the reaction coordinate. Taken from Dill book

21. 6. Lecture SS 2005Optimization, Energy Landscapes, Protein Folding21 Calculating rate coefficients from TST Provided that the unstable degrees of freedom are independent of the stable degrees of freedom, q (AB)‡ is factorable into two components represents the partition function for all of the ordinary thermodynamic degrees of freedom of the TS structure, and q  represents the one nonequilibrium vibrational degree of freedom of the bond along the reaction coordinate. The partition function of a vibration is for a weak vibration with low frequency . Once the system has reached the transition state, it is assumed to proceed as quickly to the product state as the system permits, namely at the frequency of the reaction coordinate vibration,

22. 6. Lecture SS 2005Optimization, Energy Landscapes, Protein Folding22 deviations from TST A factor , the transmission coefficient, is often introduced (k ‡ =   ) to account for observed deviations from the simple rate theory. In condensed-phase medium, or in complex systems,  < 1. This gives the relation between the rate coefficient k 2 and the partition functions:

23. 6. Lecture SS 2005Optimization, Energy Landscapes, Protein Folding23 relation between  value analysis and TST In V5, we characterized the effect of a protein mutant by its  -value  G 0 reflects whether the mutant stabilizes the folded state F over the unfolded state U stronger or weaker than wild-type protein. According to TST, both wild-type and mutant folding proceed via transition states with activation free energies  G ‡ wt and  G ‡ mut. A  -value of 1 means that  G 0 =  G ‡ for this mutant  the mutant has the same effect on the TS structure as on the folded state  this part of the TS structure is folded as in the folded state F.

24. 6. Lecture SS 2005Optimization, Energy Landscapes, Protein Folding24 Catalysts speed up chemical reactions Taken from Dill book Free energy barrier  G ‡ is reduced by a catalyst C. Catalysts affect the rates of chemical reactions; e.g. enzymes accelerate biochemical reactions. Enzymes can achieve remarkable accelerations, e.g. by a factor of 2 x 10 23 for orotine 5‘-phosphate decarboxylase. Linus Pauling proposed in 1946 that catalysts work by stabilizing the transition state. ABC kckc Linus Pauling 1935

25. 6. Lecture SS 2005Optimization, Energy Landscapes, Protein Folding25 Catalysts speed up chemical reactions From transition theory we obtain for the catalyzed reaction rate k c (normalized to the uncatalyzed reaction rate k 0 ) This ration represents the ‚binding constant‘ of the catalyst to the transition state  the rate enhancement by the catalyst is proportional to the binding affinity of the catalyst for the transition state. This has two important implications: (1) to accelerate a reaction, Pauling‘s principle says to design a catalyst that binds tightly to the transition state (and not the reactants or product, e.g.). (2) a catalyst that reduces the transition state free energy for the forward reaction is also a catalyst for the backward reaction.

26. 6. Lecture SS 2005Optimization, Energy Landscapes, Protein Folding26 Speeding up reactions by intramolecular localization or solvent preorganization Taken from Dill book Reactants polarize, so water reorganizes. Two neutral reactants become charged in the transition state. Creating this charge separation costs free energy because it orients the solvent dipoles.

27. 6. Lecture SS 2005Optimization, Energy Landscapes, Protein Folding27 Speeding up reactions by intramolecular localization or solvent preorganization Enzymes can reduce the activation barrier by having a site with pre-organized dipoles. Taken from Dill book

28. 6. Lecture SS 2005Optimization, Energy Landscapes, Protein Folding28 Funnel landscape describe diffusion and polymer folding All the processes described sofar involve well-defined reactants and products, and a well-defined reaction coordinate. But diffusional processes and polymer conformational changes often cannot be described in this way. The starting point of protein folding is not a single point on an energy landscape but a broad distribution. A bumpy energy landscape, such as occurs in diffusion processes, polymer conformational changes, and biomolecule folding. A single minimum in the center may represent ‚product‘, but there can be many different ‚reactants‘, such as the many open configurations of a denatured protein. http://www.dillgroup.ucsf.edu/

29. 6. Lecture SS 2005Optimization, Energy Landscapes, Protein Folding29 Summary Chemical reactions and diffusion processes usually speed up with temperature. This can be explained in terms of a transition state or activation barrier and an equilibrium between reactants and a transient, unstable transition state. For chemical reactions, the transition state involves an unstable weak vibration along the reaction coordinate, and an equilibrium between all other degrees of freedom. Catalysts act by binding to the transition state structure. They can speed up reactions by forcing the reactants into transition-state-like configurations.