Discussion topic for week 4 : Protein folding

• Levinthal's paradox presents an estimate for the time it would

take for a protein to fold assuming a minimum of two possible

conformations for each pair of amino acids. 

For a 101 residue protein, there are 2^100 ~10^30 possible

conformations.  If it takes 1 ps to sample each conformation, it

would take 10^18 s to sample the whole phase space to find the

absolute minimum of the free energy.  This is longer than the age

of the universe, 5x10^17 s!

How do proteins manage to fold in seconds?

Chemical Forces (Nelson, chap. 8)

Molecular machines in cells use chemical energy to function, and

most of the time they do chemical work, e.g. synthesize proteins.

To deal with this situation we need to consider more than one species

and allow exchange of particles as well as energy.

Let {N, ,…} denote the numbers of each species in a system

The entropy of the system is a function of this set: S(N1, N2, …)

We define the chemical potential for each species as

Recall the definition of the temperature (modified from the fixed N case)

,,NEN

ST

NES

T 1

(availability of particles)

(availability of energy)

When two systems, A and B, exchange only energy, thermal equilibrium

is realized when TA = TB

If they also exchange particles, their chemical potentials for each species

must be equal as well

When two systems are not in chemical equilibrium, entropic forces

arising from the difference in chemical potentials drive the system to

equilibrium.

As a simple example consider an ideal gas. The entropy is given by

(Sakure-Tetrode

formula)

!)!123(

)2(ln

3

2323

NhN

VmEkS

N

NNK

N

,, BA (chemical equilibrium)

Rewrite S using Stirling’s formula

To obtain the chemical potential we need to keep the total energy

fixed, where is the internal energy. This is achieved by

NEE K

322

32

2

2ln

2

3

3

22ln

2

3

11lnln32

3

2

3

2

3ln2

3ln2ln

2

3

c

kT

h

mk

N

V

N

E

h

mk

NhN

VmEkN

S

K

KEK

NNNhN

NNNVNmE

NkS K lnln3

2

3

2

3ln

2

3ln)2ln(

2

3

NKEE ES

NS

NS

K

The last term is just /T. Substituting, we obtain for the chemical pot.

Because we are interested in the number (or concentration) dependence

of the chemical potential, we separate that term

Where c0 is the reference concentration and

is the standard chemical potential

322

322

2ln

2

3

2ln

2

3

c

kT

h

mkT

Tc

kT

h

mkT

Tcc

kT

c

kT

h

mkT

cc

kT

0

0

320

20

ln

2ln

23

ln

The reference concentration is introduced for convenience, it’s choice

has no effect on the chemical potential. Convention:

For gases at STP: c0 = 1 mole/22 L = 0.045 M

For aqueous solutions: c0 = 1 mole/L = 1 M

Notation: [X] = cx/c0 (e.g., [X] = 1 refers to a 1 molar solution)

Rewrite the chemical potential as

For ideal gases, the activity is simply given by the relative concentration.

For solutions, the definition of is more complicated. But if we treat it

as a phenomenological parameter, we can use the same formulas for

dilute solutions.

kTecc 0

0

(activity)

We can generalize the chemical potential by including the potential

energy of the particles in the internal energy:

In the case of charged particles,

is called the electrochemical potential.

Electrochemical equilibrium between two systems is achieved when

Chemical reactions are controlled by the chemical potential of reactants:

high concentration or high internal energy means higher availability.

)(ln 0

0rqV

c

ckT

U 00

)(rqVU

B

ABA

BB

AA

BA

c

ckTVVq

qVc

ckTqV

c

ckT

ln)(

lnln 0

0

0

0

(Nernst relation)

Generalization of the Boltzmann distribution for particle exchange:

Consider a small system “a” which can exchange both energy and

particles with a much larger system B. Fluctuations in EB and NB are

negligible but those in Ea and Na could be large. As we have shown

before, the probability of “a” being in a state with Ea and Na is

proportional to exp[SB(EB)/k]

At equilibrium,

B

Ba

B

atottotB

B

Ba

B

BatottotB

atotatotBBBB

TN

T

ENES

N

SN

E

SENES

NNEESNES

),(

),(

),(),(

BaBa TTT ,

Thus the probability of “a” being in a particular state j with Ej and Nj is

proportional to

Using the grand partition function,

The normalized probability becomes

which is called the grand canonical distribution.

The number of particles “a” contains depends on ; the larger is, the

more particles “a” will have.

kTNEjj

jjeNEP)(

),(

kTNEjj

jjeNEP)(1

),(

Z

j

kTNE jje)( Z

Chemical reactions:

As a simple case, consider a molecule which has two states with internal

energies and (e.g. an isomer). Assume

The chemical potentials are

Chemical equilibrium:

Non-equilibrium cases:

1. Reaction 1 2 proceeds (entropic forces do chem. work)

2. Reaction 2 1 proceeds (chemical en. converted to

heat)

21

21

2,1,,ln 00

0 i

c

ckT iii

ii

kTec

c

c

ckT

c

ckT

1

22

0

21

0

1 lnln

012

21

To summarize, when two systems are at equilibrium:

• Temperatures and chemical pot’s are equal,

• Total entropy S is maximum

• Total free energy (F or G) is minimum

Example: Burning of hydrogen

Free energy change:

At equilibrium:

Assuming an ideal gas behaviour for all three participants, we can write

for the free energy change

2121 , TT

0TG

S

OHOH 222 22

22222 OHOHG

Here Keq is called the equilibrium constant of the reaction, and the ratio

is called the reaction quotient

Often a log scale is used for Keq :

eq

kT

OH

OHKe

cc

ccG OHOH )22(

20

202

02

02

22

20

0001

0

2

0

2

0

0

0

0

0

0

0

222

222

2

2

2

2

2

2

22ln

ln2ln22ln2

OHOHOHOH

OO

HH

OHOH

c

c

c

c

c

ckT

c

ckT

c

ckT

c

ckTG

0

2

2

22

2

c

K

cc

c eq

OH

OH

eqpK

eq KpKK 10log,10

For ideal gases (or dilute solutions), we can use the explicit expression

derived for the standard chemical potential

At low T, reaction favours H2O. As T increases H2 and O2 conc. also inc.

2/3

2

22

0)22(

2/3

320

2

3

320

2

3

320

2)22(

)22(

22

2222

222222

02

02

02

2

222

OH

OHkT

OHOHkT

kTeq

mm

m

kT

hce

c

kT

h

m

c

kT

h

m

c

kT

h

me

eK

OHOH

OHOH

OHOH

320

20 2

ln23

c

kT

h

mkT

From chemical data handbooks, Keq at room temperature is given by:

Clearly almost all the hydrogen will burn. Using the reaction quotient

estimate the number of O2 molecules left from 1 mole of O2 gas

None at all!

183)22( 02

02

02 eeK

kTeq

OHOH

eq

OH

OHK

OH

OH

cc

cc

22

2

22

20

2

22

2

002.0106103][)(

103][][]2[

]2[

23272

273/1831832

2

ANxOn

exexx

Generalization to arbitrary reactions:

Assume n species involved in a reaction; k reactants and m-k products

where k are called the stoichiometric coefficients of the reaction

The free energy difference is

The reaction runs forward when G < 0 and backward if G > 0.

G = 0 corresponds to equilibrium. Again we separate the concentration

dependent part from the rest

mmkkkkG 1111

00111

0

0

01

0

11

]ln[]ln[

lnln

1mmm

mm

m

mXXkT

c

ckT

c

ckTG

mmkkkk XXXX 1111

Setting G = 0, we obtain

where G0 is the standard free energy change

The values of G0 for formation of molecular species can be found in

chemistry handbooks (usually at STP; 298 K and 1 atm)

When more than one reaction occurs at similar rates, there is a

separate mass action rule for each reaction, which implies relations

between the various

0011

0mmG

eq

kTG

k

mk KeXX

XXk

mk

0

1

1

1

1

(Mass action rule)

Reaction Kinetics:

Consider a typical reaction with rate constants k+ and k

Intuitively we expect the forward and backward rates to be proportional to

the concentrations of molecules (first order reaction)

At equilibrium,

The above is true for single step reactions. For more complex reaction

mechanisms, concentration dependence of rates may be different.

2)(,22 XYYX ckrcckr

eqkTG

YX

XY Kek

k

cc

crr

0

22

2)(

XYYXk

k222

(mass action rule)

In general, a reaction is of n’th order in species X if the rate depends on

its concentration as (cX)n.

An alternative 3-step mechanism for the previous reaction, which is

second order in X2 and zeroth order in Y2 :

Each step must be in equilibrium

eqeqeqeqYX

XY

eqXYX

XYeq

YX

XYeq

X

XX

KKKKcc

c

Kcc

cK

cc

ccK

cc

cc

3,2,1,

2

3,

2

2,0

1,0

2

2

22

22

2

2

2

)(

)(,,

)(

)(

XYXYX

XYYX

XXXX

2

2

2

22

222

(slow, rate limiting step)

(fast)

(fast)

(mass action rule is independ. of mechanism)

Product:

Dissociation:

Salts, acids, bases and polar molecules readily dissolve in water because

the loss in potential energy is more than compensated by the interaction

of the charged parts with water molecules (charge-dipole and H-bond)

and gain in entropy.

Example: Dissociation of water

From conductance measurements in pure water:

Mass action rule:

Adding an acid (e.g. HCl) increases [H+] and hence lowers [OH]

Adding a base (e.g. NaOH) increases [OH] and hence lowers [H+]

)32(1010][

]][[ 01427

2kTG

OH

OHHK

w

Mcc OHH710

OHHOH2 (proton + hydroxyl)

In chemistry, the amount of protons in a solution is described by its pH

Pure water has pH = 7, which is called normal pH

Adding acids in water lowers pH. A solution with pH < 7 is called acidic

Adding bases in water raises pH. A solution with pH > 7 is called basic

Common acidic and basic groups in organic molecules:

Carboxyl group

Amine group

protonated deprotonated

Of the 20 amino acids, aspartate and glutamate have acidic side chains

while arginine and lysine (~histidine) have basic side chains.

][log10 HpH

HNHNH

HCOOCOOH

23

Probability of protonation of a side chain

equilibrium:

When

pKpHxP

HK

HKP

H

K

COOHCOO

KCOOH

HCOO

COOHCOOCOOCOOH

COOHP

xpHpK

pHpKeq

eq

eqeq

,

101

1

101

1

10][,10

][1

1

][][][

][]][[

][][1

1

][][

,

,

,,

2/1, PpKpH

Examples:

Aspartic acid: Keq = 103.7 P = 1/(1+103.3) 0 (has charge –e)

Arginine: Keq = 1012.5 P = 1/(1+10) 1 (has charge +e)

The average charge on a side chain is determined by P

Acidic side chain: q = –e (1 – P)

Basic side chain: q = e P

Note that the pH of the solution controls the protonation state of a protein.

In titration experiments, the pH is varied over a wide range, e.g. 1-12.

When pH < pK of all the side chains, all are protonated (max + charge)

As pH increases, and goes through pK of an acidic side chain, q = 0 –e

Beyond pH = 7, basic side chains start deprotonating, q = +e 0

For pH > pK of all the side chains, all are deprotonated (max – charge)

Titration curve of ribonuclease. As pH is raised protein loses protons.

Electrophoresis:

As the titration curve indicates, apart from a critical pH value, proteins

carry a net charge and hence will move under an applied electric field.

This process is called electrophoresis.

A common application is separation of proteins, which is achieved by

setting the pH of the solution at the critical value of the protein we want to

separate and applying an electric field.

Varying pH and measuring the electrophoretic mobility, one can

determine the critical pH value precisely.

A famous example is Pauling’s finding of the cause of sickle-cell anemia.

Patients carry a defective hemoglobin that differs from the normal one by

a single point mutation, Glu Val. Glu has –e (pK = 4.25), Val is neutral.

At pH = 6.9, the two proteins migrate in opposite directions!

Self-assembly of amphiphiles:

How do the cell membranes form?

Amphiphiles: molecules that have both hydrophilic (polar) and

hydrophobic (CH2 chain) parts (detergents, lipids)

Sodium dodecyl sulfate(SDS)

Phosphatidylcholine

When detergents are added to oil-water mixtures, they form a boundary

between the two such that the polar head groups face water and

hydrophobic tails face oil

Oil-water interface stabilized by detergent oil-water emulsion

Micelle formation:

When detergent is added in pure water, they form small spherical objects

just like in the emulsion case. The only difference is that the tails avoid

water by facing each other.

N=5

N=30

Osmotic pressure: P = ckT (McBain, 1944)

Let the number of detergent molecules in a micelle be N, and denote

the concentration of micelles by cN and monomers by c1

The reaction is: (N monomers) (micelle)

Mass action rule (MAR) at equilibrium:

Experimentally measured quantity is the critical micelle concentration, c *

which is defined as,

substitute in MAR

substitute in ctot

For 2c1 << c*, ctot = c1, while for 2c1 >> c*, ctot = NcN

1111 )(1 NeqNtot cNKcNccc

1*11

1***

***1*

)2(1

)2(22

2

Ntot

Neqeq

N

Ntot

cccc

cNKKcNc

cNcccc when

eqN

N Kcc )( 1

Coarse-grained models of lipid aggregation:

United atom models of lipids

Micelle formation (Klein et al. 2004)

Bilayer formation (Marrink et al. 2001)

Cooperative transitions in macromolecules (Nelson, chap. 9)

Biological molecules usually have two distinct conformations:

random coil form of the polypeptide chain and a folded compact form.

Examples:

• Helix-coil transition in a simple amino acid chain

• Full folding of a protein from random coil to a compact 3D structure

• An extreme example is the condensation of DNA, where the full

length of about 1 m is squeezed into a micron size nucleus.

An important parameter in characterizing the elasticity of polymers is the

persistence length, which determines the length scale for bending of the

chain of molecules. Persistence length is typically about 1 nm for

polymers, which are very flexible. In contrast, it is about 100 nm in DNA,

which is relatively very rigid.

Elasticity model of polymers:

If we model polymers as a continuous elastic object, there are three

possible deformations: a) bending, b) stretching, c) twisting (torsion)

Because the covalent bonds in polymers are quite rigid and the torsional

motion is restricted, only the bending deformation is important

kTR

LR

RkTLEdskTLE

Rds

d

ds

ddskTAdE

pp

L

p

tot

44

21

2

1

2

1

1,ˆ

2

1

20

2

2

β

tββ where

(for ¼ circle)

Stretching of DNA (experimental data from DNA of lambda phage)

A force of few pN is sufficient to fully stretch DNA from a random coil.

At 65 pN DNA takes another form, where the backbone is straightened.

Two-state model of DNA stretching (freely jointed chain model in 1D)

Assume DNA consists of N segments of length Ls, which can be oriented

in +z or –z direction. Apply a force f in the z direction to stretch it.

The corresponding potential is U= –fz where z is the DNA length given by

Probability of a particular configuration [σ1,….,σN] is given by the

Boltzmann factor

Where Z is the partition function

11

1

1

),,(

1),,(

1

1

N

Ni is

N

kTfLN

PZ

eZ

P

1,1

i

N

iisLz with

Average DNA extension under a load f

NkTfLkTfL

kTfLkTfL

kTfL

N

iis

kTfL

N

ss

N

Nss

N

Ni is

N

Ni is

N

eedfd

kT

eedfd

kT

edfd

kT

LeZ

zPz

ln

ln

ln

1

),,(

11

11

111

11

1

1

1

1

1

1

1

1

Taking the derivative wrt f gives

Introducing Ltot=NLs

The limiting cases:

1) High force (f>>kT/Ls),

2) Low force (f<<kT/Ls),

At low force, a polymer behaves like a spring, obeying Hooke’s law

kT

fL

L

z

ee

eeNLz

s

tot

kTfLkTfL

kTfLkTfL

sss

ss

tanh

stot

stot

tot

LLkT

kkf

kT

fLLz

Lz

with,

Comparison of theory with experimental data from lambda phage

Long-dash curve: 1D cooperative chain model (includes elastic energy)

Short-dash curve: 3D freely jointed chain model

Ls=104 nm

Ls=35 nm

Helix-coil transition (experimental data from an artificial polypeptide)

• At a critical temperature polypeptide makes a transition from coil to helix

• Transition is sharpened with the number of residues (cooperative effects)

Energetics of helix-coil transition

The free energy change in the transition is given by

From experiments:

Introduce a parameter

which measures favourability of extending the helix formation

confbondtot

coilhelixbond

totbond

SSS

EEE

STEG

0,0

0)33ln(

,0

totbond

conf

SS

kS

E

kT

STE totbond

2

When vanishes, extending the helix by one unit makes no change in

the free energy

Using a cooperative 1D freely jointed model and

The curves in the previous figure are obtained by fitting this expression

to the data points.

m

mbond

m

bond

totbondm

TT

TT

k

E

TTk

E

SET

2

11

2

0

422

1sinh

sinh

e

CC

Protein folding:

Primary sequence determines the folded structure

The free energy gain from folding is about 20 kT

Loss of entropy is compensated by H-bond formation and especially

hydrophobic interactions (Kauzmann, 1950s).

Changes in the environment can lead to denaturation (unfolding) of

proteins. For example, proteins unfold

at both high (T > 50 C) and low (T < 20 C) temperatures

in nonpolar solvents

in the presence of small amounts of surfactants

MD simulations of protein unfolding at high temperatures (Daggett et al.)

Unfolding of

engrailed

homeodomain

Folding time at

298 K, ~1 ms

Unfolding of chymotrypsin inhibitor (Daggett et al.)

Potential

energy

landscapes

for protein

folding:

a) Flat

(i.e. Levinthal)

b) Ant trail

c) Smooth

funnel

d) Rugged

funnel

Rugged protein folding pathways

from lattice calculations (Dill et al)