Transcript Computational Biology
V1: V2: V3: V4: V5: V6: V7: V8:
Lecture of Graduiertenkolleg - Helms
Reaction rate theory Kramer‘s theory Reaction rates for electronic transitions Potential and free energy landscapes Lattice optimization Optimization methods in protein folding Protein folding with molecular dynamics Manipulating potential energy landscapes with forces 1. Lecture SS 2006 GK 1276 1
V1: Chemical Kinetics & Transition States
see chapter 19 in book of K. Dill
Aim
: describe kinetics of processes on energy landscapes (e.g. chemical reactions). - detailed balance - mass action law - temperature effect, Arrhenius law - concept of
transition state/activation barrier
-value analysis transition state theory - effect of catalysts 1. Lecture SS 2006 GK 1276 2
Rate theory
Rate theory provides the relevant information on the long-time behavior of systems with different metastable states important for understanding of many different physical, chemical, biological, and technical processes.
Arrhenius (1889) Wigner (1932), Eyring (1935) Transition State Theory (TST) Pechukas (1976): proper definition of the transition state Chandler (1978): the activation energy is a free energy Kramers (1940): effect of friction on reaction rates Pollak (1986): link of Kramers‘ expression to TST 1. Lecture SS 2006 GK 1276 Pollak, Talkner, Chaos 15, 026116 (2005) 3
Reaction rates are proportional to concentrations
Lets consider a simple kinetic process, the interconversion between 2 states,
A
k r B 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 ?
d
k f d dt
k f dt
k r
k r
The two equations are coupled.
One can solve them by matrix algebra … 1. Lecture SS 2006 GK 1276 4
Excursion: coupled differential equations
Matrix diagonalisation can be used to solve coupled ordinary differential equations. For example, let
x(t)
and
y(t)
be differentiable functions and
x'
and
y'
their time derivatives. The differential equations are relatively difficult to solve:
x
y
4
x
2
x
y y
but
u'
=
ku
for
k =
const is easy to solve. The solution is
u = Ae kx
where
A
= const.
translate the ODEs into matrix form
x y
4 2 1 1
x y
By diagonalizing the square matrix, we get
x y
1 1 1 2 3 0 0 2 1 1 1 2 1
x y
www.algebra.com
1. Lecture SS 2006 GK 1276 5
Excursion: coupled differential equations
By diagonalizing the square matrix, we get
x y
1 1 1 2 3 0 0 2 1 1 1 2 1
x y
We then put
u v
1 1 1 2 1
x y
It follows that
u
v
1 1 1 2 1
x y
Thus
u
v
3 0 0 2
u v
The solutions of this system are found easily: With
u v
1 1 1 2 1
x y
u
Ce
3
t v
De
2
t
1 1 1 2
u v
x y
with some constants C and D.
1. Lecture SS 2006
x
Ce
3
t
De
2
t y
Ce
3
t
2
De
2
t
GK 1276 6
Reaction rates are proportional to concentrations
With this technique, we could solve our system of coupled diff. equations.
In the special case that
k r
<<
k f
, the first equation simplies to
d dt
k f
dA A
0
t
k f dt
ln
A
k f t
A
e
k f t
If [
A(t)
]
+
[
B(t)
] = constant, then
constant
A
e
k f t
1. Lecture SS 2006 GK 1276 7
At equilibrium, rates obey detailed balance
The principle of
detailed balance
says that the forward and reverse transitions must be identical for an elementary reaction at equilibrium:
k f
eq
k r
eq
[A] eq and [B] eq : equilibrium concentrations.
To see that this is a condition of equilibrium follows from inserting into
d d dt
dt
k f k f
k r
k r
resulting in
d
0 ,
d
0
dt dt k f
eq
k r
eq
Taken from Dill book 8 1. Lecture SS 2006 GK 1276
At equilibrium, rates obey detailed balance
The detailed balance condition relates the rate coefficients
k f
equilibrium constant
K
:
K
eq eq
k f k r
and
k r
to the 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:
eq eq
k AI k IA
,
eq eq
k IB k BI
,
eq eq
k BA k AB
Let‘s consider a 3-state mechanism with
k IA
0,
k BI
0,
k AB
0.
Taken from Dill book 9 1. Lecture SS 2006 GK 1276
At equilibrium, rates obey detailed balance
This results in the mechanism shown right.
The only conditions for equilibrium are:
k k IB AI
k k IB BA
Taken from Dill book 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 .
eq eq
k AI k IA
,
eq eq
k IB k BI
,
eq eq
k BA k AB
( 1 )
eq
k AI k IA
eq
, ( 3 ) ( 2 )
eq k AB k BA
k AI k IA
eq
k IB k BI B eq
A eq k AB k BA
or 1
k IA k AB k BI k AI k BA k IB
, or 1
k AI k BA k IB k IA k AB k BI
1. Lecture SS 2006 GK 1276 10
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
the backward reaction cannot be
B
I
A.
B
, 1. Lecture SS 2006 GK 1276 11
The mass action laws describe mechanisms in chemical kinetics
Suppose the following reaction leading from reactants A, B, and C to product P:
aA
bB
cC
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 the stoichiometry in the same way that equilibrium constants do“.
d
dt
k f
c
Although mass action is in agreement with many experiments, there are exceptions. These require a quantum mechanical description.
1. Lecture SS 2006 GK 1276 12
Energy Barriers
Where do energy barriers come from?
Why do reactions have activation barriers?
Different processes are characterized by similar energy barriers that are due to very different mechanisms.
Effects on energy barriers - Chemical reactions - protein:ligand association - temperature - pH - protein:protein association - protein:membrane association - protein:DNA association - during protein folding - D 2 O vs. H 2 O - viscosity - time scales of protein dynamics - vesicle budding - virus assembly 1. Lecture SS 2006 GK 1276 13
History of Energy Barriers of Chemical Reactions
1834 Faraday: chemical reactions are not instantaneous because there is an electrical barrier to reaction 1889 Arrhenius: reactions follow Arrhenius law with an activation barrier
E a
.
Rate
rate
0
e
E kT a
Bodenstein: reactions occur via a series of elementary steps where bonds break and form. Bodenstein showed that Arrhenius‘ law is applicable only to elementary reactions. Overall reactions often show deviations.
1935 Polanyi & Evans: bonds need to stretch during elementary reactions.
The stretching causes a barrier. Bonds also break.
Physical causes of barriers to chemical reactions
bond stretching and distortion orbital distortion due to Pauli repulsions (not more than 2 electrons may occupy one orbital) quantum effects special reactivity of excited states 1. Lecture SS 2006 GK 1276 14
Energy Barriers of Chemical Reactions
When chemical bonds need to be „broken“, - nuclei need to move only over small distances, electrons need to redistribute „settle down“ in different orbitals intermediate state has high energy One can compute the energy barriers by electronic structure methods.
However, these calculations do not explain
why
the barriers arises.
Chemists like to think in concepts and rules and like to separate these.
How fast can a reaction proceed? Even within one bond vibration.
Such processes need to be activated.
E.g. bond length should stretch far beyond equilibrium distance.
This is possible by statistical fluctuations and by coupling with other modes (large energy becomes concentrated in this mode).
1. Lecture SS 2006 GK 1276 15
Energy Barriers of Protein:Ligand Interaction
Step 1: Protein and Ligand are independently solvated (left picture below) Step 2: The Ligand may preorganize into its binding conformation (costs usually 3 kcal/mol) Step 3: The Ligand approaches the binding pocket of the protein.
System partly looses 6 degrees of freedom (CMS of ligand: 3 translation, 3 rotation) No collective modes!
Step 4: The Ligand enters the binding pocket of the protein Waters are displaced from binding pocket.
Sometimes: simultaneous conformational changes of protein/receptor
Receptor
Bound and associated H 2 O
Ligand
Displaced H 2 O 1. Lecture SS 2006 GK 1276 16
Energy Barriers of Protein:Protein Interaction
Steps involved in protein-protein association: - random diffusion (1) + hydrodynamic interaction - electrostatic steering (2) - formation of encounter complex (3) (Possible: large-scale conformational changes of one or two proteins) - dissociation or formation of final complex via TS (4)
Origin of Barrier (4)
: System partly looses 6 degrees of freedom (CMS: 3 translation, 3 rotation) Desolvation: large surface patches need to be partially cleared from water Induced fit of side chains at interface potential entropy loss 1. Lecture SS 2006 Effects of hydrodynamic Interactions: (left) effect of translation (right) effect of rotation GK 1276 17
Energy Barriers of Protein:Membrane Interaction
Membrane surface either carries a net negative charge (mixture of neutral and anionic lipids) or has a partially negative character.
wikipedia.org
Cloud of positive counter ions accumulates near membrane to compensate membrane charge.
Membrane surface is not well defined and quite dynamic, ondulations.
1. Lecture SS 2006 GK 1276 18
Energy Barriers of Protein:Membrane Interaction
Neutron scattering: four layers of ordered water molecules above membrane – also found by MD simulation. Water layers significantly weaken membrane potential.
Lin, Baker, McCammon, Biophys J, 83, 1374 (2002) Interaction potential of protein:membrane systems is largely unknown.
1. Lecture SS 2006 GK 1276 19
Energy Barriers of Protein:DNA Interaction
DNA backbone carries strong permanent negative charge.
is surrounded by cloud of positive ions and coordinating water molecules Protein must displace this cloud and must form very polar interactions with DNA backbone.
Spatial distribution functions of water, polyamine atoms and Na + ions around a CA/GT fragment. View of the minor groove. Data are for systems with 30 diaminopropane 2+ (
A
), 30 putrescine 2+ (
B
), 20 spermidine 3+ (
C
) and 60 Na + (
D
), averaging the MD trajectories over 6 ns, with three decamers with three repeated CA/GT fragments in each decamer. Water (oxygen, red; hydrogen, gray) is shown for a particle density >40 p/nm 3 (except for the Na/15 system, where this value is 50 p/nm3); Spherical distribution function of the polyamine N + atoms (blue) and Na + (yellow) ions are drawn for a density >10 p/nm 3 ; polyamine carbon and hydrogen atoms not shown. Korolev et al. Nucl Acid Res 31, 5971 (2003) 1. Lecture SS 2006 GK 1276 20
Energy Barriers of Protein:DNA Interaction
Recognition shows
Faster-than-diffusion paradox
(similar to Anfinsen paradox for protein folding).
Maximal rate achieveable by 3D diffusion:
10 8 M -1 s -1
This would correspond to target location
in vivo
on a timescale of only a few seconds, when each cell contains several tens of TFs.
However: Experimentally found (LacI repressor and its operator on DNA):
10 10 M -1 s -1
.
Suggests that dimensionality of the problem changes during the search process. While searching for its target site, the protein periodically scans the DNA by sliding along it. This is best done if the TF is only partially folded and only adopts its folded state when it recognizes its binding site.
Slutsky, Mirny, Biophys J 87, 4021 (2004) 1. Lecture SS 2006 GK 1276 21
Time scales of protein dynamics
Motion Relative vibration of bonded atoms Elastic vibration of globular region Rotation of side chains at surface Torsional libration of buried groups Relative motion of different globular regions (hinge bending) Rotation of medium-sized side chains in interior Allosteric transitions Local denaturation Protein folding Spatial extent (nm) 0.02 to 0.05
1 to 2 0.5 to 1 0.5 to 1 1 to 2 0.5
0.5 to 4 0.5 to 1 3 to 5 Log 10 of characteristic time (s) -14 to – 13 -12 to –11 -11 to –10 -14 to –13 -11 to –7 -4 to 0 -5 to 0 -5 to 1 -5 to 2 Adapted from http://www.dbbm.fiocruz.br/class/Lecture/d22/kolaskar/ask-11june-4.ppt
1. Lecture SS 2006 GK 1276 22
Energy Barriers during protein folding
Peptide chain must organize into particular 3D fold: large entropy loss.
Formation of secondary structure elements formation of hydrogen bonds.
This is almost cancelled by loss of hydrogen bonds with solvent molecules.
Burial of hydrophobic surface free energy gain due to hydrophobic effect.
Charged active site residues must be buried in hydrophobic protein interior often electrostatically unfavorable 1. Lecture SS 2006 GK 1276 23
Energy Barriers during vesicle budding
Vesicle budding (right, above) does not occur spontaneously: would be too dangerous for cell. Distorting plane membrane costs deformation energy binding of coat proteins reduces energy cost and gives natural membrane curvature (right, below).
SNARE proteins help to overcome energy cost for fusion of membranes.
1. Lecture SS 2006 GK 1276 24
Energy Barriers during formation of virus capsid
Many individual particles combine into one larger particle big loss of translational and rotational degrees of freedom Much hydrophobic surface gets buried between assembling proteins free energy gain according to hydrophobic effect (primarily solvent entropy) Electrostatic attraction: probably not very significant for binding affinity but important for specificity.
1. Lecture SS 2006 GK 1276 25
What is the effect of pH on energy barriers?
At different pH, titratable groups will adopt different protonation states.
e.g. at low pH, Asp and Glu residues will become protonated salt-bridges (Asp – Lys pairs) in which residues were involved will break up.
proteins unfold at low and high pH.
1. Lecture SS 2006 GK 1276 26
What is the effect of D
2
O on energy barriers?
It is a common strategy to compare the speed of chemical reactions in H 2 O and in D 2 O.
The electronic energy profile of the barrier is the same.
But the deuteriums of D 2 O have a higher mass than the hydrogens of H 2 O.
Their zero-point energies are lower they need to overcome a higher effective energy barrier all chemical reactions involving proton transfer will be slowed down, typically by a factor of 1.4
This is called the „kinetic isotope effect“ (KIE).
1. Lecture SS 2006 GK 1276 27
What is the effect of viscosity on energy barriers?
Adding co-solvents in the solvent to increase the viscosity should, in principle, slow down conformational transitions.
It is often problematic that the co-solvent will also change the equilibrium, e.g. between folded and unfolded states of a protein.
1. Lecture SS 2006 GK 1276 28
What is the effect of temperature on energy barriers?
In general, higher temperature will enormously speed up activated processes.
However, we often need to consider free energy barriers instead of energy barriers. The free energy barriers often change considerably with temperature.
E.g. in a MD simulation of a protein at 500K, the protein residues will more easily overcome individual torsional energy barriers.
But, after a certain time, the whole protein will unfold.
1. Lecture SS 2006 GK 1276 29
Reaction rates depend on temperature
Consider a binary reaction in the gas phase:
A
B k
2
P
Suppose that
d dt
k
2 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.
1. Lecture SS 2006 GK 1276 30
Arrhenius and activated molecules
1889 Arrhenius: found temperature dependence of the rates of inversion of sugar in the presence of acids.
Arrhenius cites van‘t Hoff (1884) for suggesting e –A/T dependence.
Rate
rate
0
e
E a kT E a
: activation energy Svante Arrhenius 1859 – 1927 Noble price 1903 Arrhenius postulated that this relationship indicates the existence of an „activated sugar“ whose concentration is proportional to the total concentration of sugar, but is exponentially temperature dependent.
Arrhenius
is the
father of rate theory
1. Lecture SS 2006 GK 1276 31
Arrhenius equation
1889, S. Arrhenius started from the van‘t Hoff equation for the strong dependence of the equilibrium constant
K
on temperature:
d
ln
K dT
h kT
2 and proposed that
k f
and
k r
also have van‘t Hoff form
d
ln
dT k f
E kT a
2 and
d
ln
dT k r
E a
kT
2 where
E a E a
and
E‘ a
and
E‘ a
have units of energy that are chosen to fit exp. data.
are called
activation energies
.
1. Lecture SS 2006 GK 1276 32
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.
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 overcome on their way 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 °.
Taken from Dill book 1. Lecture SS 2006 GK 1276 33
Population at different temperatures
From
K
k f k r
it follows
h
E a
E
'
a
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.
1. Lecture SS 2006 GK 1276 Taken from Dill book 34
Arrhenius plots
The figures show examples of H 2 + I 2 2HI H 2 chemical systems showing 2HI (open circles) + I 2 (full circles) Arrhenius behavior.
d
ln
k f
Integrating
dT
over temperature
T
E a kT
2 gives:
ln
k f k f
E kT a
B
e
E a kT
B k f
e
E a kT
e B k f
Ae
E a kT
with
A
e B
Taken from Dill book Diffusion of carbon in iron 1. Lecture SS 2006 GK 1276 35
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.
H
3
HCN
H
2
H
2
CN
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.
1. Lecture SS 2006 GK 1276 36
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.
Energy surface for
D + H 2
HD + H
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.
1. Lecture SS 2006 GK 1276 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.
Taken from Dill book 37
Eyring theory
The
Eyring-theory
or
transition state theory (Theorie des Übergangszustandes)
is a molecular reaction theory. It uses molecular descriptors like the partition function and describes the absolute rate of chemical reactions.
The
reactants
are separated from the
products
by an activation barrier. The reaction from the reactants via the transition state to the products proceeds along a trajectory = the reaction coordinate.
[taken from Dill book]
Transition state
: point of highest potential energy along this reaction coordinate.
Activated complex
: atomic arrangement in the transition state.
The main assumptions of TST are: - the
activated complex
exists in an equilibrium with the reactants - All molecules that reach the
transition state
from the reactant states leave it in direction of the products. Recrossings are not allowed.
1. Lecture SS 2006 GK 1276 38
Wigner and Eyring: The transition state method
In the 1930s it was well established that reaction rates
k
should be written in the form:
k
e
E a kT
where is a prefactor with the dimensions of s -1 unimolecular reactions and s -1 cm -3 for for bimolecular reactions.
In „The activated complex in chemical reactions“ (1935), Eyring gave a heuristic derivation of an expression for the prefactor based on the assumption of an equilibrium between the activated complex and reactants.
[taken from Dill book] To obtain the time constant, he postulated, that at the saddle point, any quantum state perpendicular to the reaction coordinate reacts with the same universal time constant kT/2 ħ.
The rate is then given by the product of this universal time constant with the ratio of the partition function of the activated complex to the partition function of the reactants.
1. Lecture SS 2006 GK 1276 Henry Eyring (1901-1981) 39
Wigner and Eyring: The transition state method
1932 Pelzer & Wigner: estimated rate of conversion of parahydrogen into normal hydrogen. To compute the reaction rate, they use a thermal equilibrium distribution in the vicinity of the saddle point of the PES and estimate the unidirectional classical flux in the direction from reactants to products They ignore the possibility of recrossings of the saddle point noting that their probability at room temperature would be rather small.
How can one define the „activated complex“?
Eyring‘s definition is questionable.
Wigner‘s definition leaves no ambiguity: the best dividing surface is that which minimizes the unidirectional flux from reactants to products.
1932, Wigner also derived an estimate for the tunneling contribution to the thermal flux of particles crossing a barrier.
1. Lecture SS 2006 GK 1276 Eugene Wigner (1902-1995) Noble price 1963 40
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.
1. Lecture SS 2006 GK 1276 Taken from Dill book 41
Calculating rate coefficients from TST
Let us consider the reaction
A
B k
2
P
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 ‡ : A
B
‡ ‡
k
‡
P
Key assumption of TST
: step (1) can be expressed as an equilibrium between the reactants A and B and the transition state (AB) ‡
K
‡ ‡
, with even though (AB) ‡ is not a true equilibrium state.
1. Lecture SS 2006 GK 1276 42
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 ‡
d dt
for the second product-forming step
k
‡
‡
k
‡
K
‡ Because the quantitiy
K ‡
is regarded as an equilibrium constant, it can be expressed in terms of the molar partition functions:
K
‡
q
‡
q A q B
e
D
‡
kT
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.
1. Lecture SS 2006 GK 1276 43
relation between
value analysis and TST
Later we will characterize the effect of a protein mutant by its -value
G 0
k B T
ln
k
G
0
mut k wt
G
0
G mut U
F
G wt U
F
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
.
k wt
e
G wt
‡
kT k mut
e
G mu t
‡
kT k mut k wt
e
G mu t
‡
kT e
G wt
‡
e
G
‡
kT kT
;
e
G mu t
‡
G wt
‡
kT
G
‡
G mut
‡
G wt
‡ 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.
1. Lecture SS 2006 GK 1276 44
Catalysts speed up chemical reactions
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
.
A
B
P k c
ABC Free energy barrier
G ‡
reduced by a catalyst C. is GK 1276 Linus Pauling 1935 Taken from Dill book 45 1. Lecture SS 2006
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
)
k c k
0
ABC
‡ ‡ This ratio 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.
1. Lecture SS 2006 GK 1276 46
Speeding up reactions by intramolecular localization or solvent preorganization
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.
1. Lecture SS 2006 GK 1276 Taken from Dill book 47
Speeding up reactions by intramolecular localization or solvent preorganization
Enzymes can reduce the activation barrier by having a site with pre-organized dipoles.
1. Lecture SS 2006 GK 1276 Taken from Dill book 48
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 the ‚product‘, but there can be many different ‚reactants‘, such as the many open configurations of a denatured protein.
http://www.dillgroup.ucsf.edu/ 1. Lecture SS 2006 GK 1276 49
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.
1. Lecture SS 2006 GK 1276 50