theoretical biophysics

8/12/2019 Theoretical Biophysics

1/73

Theoretical Biophysics

Stephan W. Grill 1,2 and Frank Julicher 2

transcribed by Falk Hoffmann and Fabian Rost

1Max-Planck Institute for Molecular Cell Biology and Genetics,

Pfotenhauerstr. 108, 01307 Dresden, Germany 2Max-Planck-Institute for the Physics of Complex Systems,

N othnitzerstrasse 38, 01187 Dresden, Germany

1


2/73

Contents

A. Lecture: Theoretical Biophysics 4

I. Biopolymers 5

A. Freely jointed chain 5

1. General facts 5

2. Comparison with the diffusion of a Brown particle 6

3. Statistical mechanics 8

4. Freely jointed chain with external force 12

B. Bead-and-Spring Model - Gaussian chain 14

1. Small forces 14

2. Large forces 15

3. Effect of self-avoidance 17

C. Experiments 19

1. Magnetic Tweezers 19

2. Optical Tweezers 20

D. Semiexible Polymers 21

1. Stiff/Semiexible polymers 22

2. Effective stiffness 27

II. Biomembranes 29

A. Differential geometry (geometry of surfaces) 30

1. Curves in R2 30

2. Surfaces in R3 31

3. Curvature of a surface 32

4. Formalism of vectors and tensors 33

5. Tensor of curvature 34

6. Additional surface elements that are independent of parametrization: 35

B. Bending energy of a uid membrane 36

1. Expand bending energy 36

2. Theorem of Gauss-Bonnet 37

3. Biological implications 40

2


3/73

C. Monge-representation 40

1. Tensor of curvature in Monge-representation 42

D. Persistence length of a membrane 44

1. Partition function 45

2. Continuum limit 47

3. Correlation functions 48

4. Periodic boundary conditions 50

5. Persistence length of a membrane 51

E. Stretching elasticity of a membrane 52

III. Dynamics of biological molecules 56

A. One-dimensional harmonic oscillator with uctuations 591. Linear response function 59

2. Correlation function 65

3. Fluctuation-dissipation-theorem 66

B. Stochastic motion in periodic potentials 66

1. Periodic boundary conditions 67

2. Kramers rules 68

3. Transition state picture 69

4. Detailed balance 70

5. Effective diffusion 70

6. Effective mobility 71

C. Experiments 72

3


4/73

A. Lecture: Theoretical Biophysics

Cytoskeleton: actin, microtubules

Organells: nucleous, golgi, mitochondria, en-

dosoms, endoplasmatic reticulum(ER)

Molecular building blocks of the cell

small molecules

ions Lipids DNA/RNA Proteins

H2O cell membrane chromosomes enzymes

Ca++ Golgi, ER ribosomes actin, tubulin

ATP mRNA motor proteins

microRNAs ion channels

Statistical physics of biomolecules exible

semiexible

Statistical physics of membranes

Dynamics, non-equilibrium processes Langevin equation, Fokker-Planck equation

detailed balance, uctuation dissipation theorem active processes, dynamic uctuations

motor proteins

Provide the basic tools required to work in theoretical biophysics

4


5/73

I. BIOPOLYMERS

A. Freely jointed chain

1. General facts

Biopolymers are long chain molecules. An easy model to describe them is

the freely jointed chain. In two dimensions it is similar to a bike chain.

N : angle to the horizontal

a: segment length

RN = R0 + a

N

i=1

cosi

sin i

R = RN R0 = aN

i=1

cosi

sin i

The probability for the conguration i has to be normalized

20 d1 . . . 20 dN P (1, . . . , N ) = 1No angle is preferred, so the probability is

P (1, . . . , N ) = 1(2)N

The rst expectation value for the end-to-end-distances:

< R > = 20 d1 . . . 20 dN RP = a 20 d1 . . . 20 dN N i=1 cosisin i 1(2)N = 0because

2

0d cos =

2

0d sin = 0

The next expectation value is

< ( R)2 > = a2 20 d1 . . . 20 dN N i,j =1 (cos i cos j + sin i sin j ) 1(2)N < ( R)2 > = Na2 20 d1 . . . 20 dN 1(2)N = Na2

5


6/73


7/73

Diffusion equationP t

= D P

We can solve this equation with a Greens function. In two dimensions we get the following

solution:P (R, t ) =

14Dt

exp (R R0)2

4DtA repetition how to solve a gaussian integral:

dx e A2 x2 = 2A dx x2e A2 x2 = 2A 1A

The total probability must be normalized

dx dy P (R, t ) = 14Dt dx exp (x x0)24Dt dy exp (y y0)24Dt = 1With A = 12Dt we get

< (R R0)2 > = 1

4Dt

dx (x x0)2 exp x x04Dt

2A

dy (y y0)2 exp y y04Dt

2A

+ dx exp (x x0)24Dt dy exp (y y0)24Dt< (R R0)2 > =

2A

= 4 Dt (3)

Now we look to the distribution of the end-to-end-distance

R = RN R04Dt a

2N

P N ( R) = 1a 2N

e( R ) 2

a 2 N

2.2. Three dimensions:

7


8/73

< ( R)2 > = Na2

P N ( R) = 3

2a 2N

32

e3( R ) 2

2a 2 N

P (R, t ) = 1

4Dt

32

exp (R

R0)2

4Dt

< R2 > = 6Dt (4)

6Dt = Na2

4Dt = 23

Na 2

Now we want to discuss the situation with an external force.

Therefore we need statistical mechanics.

3. Statistical mechanics

3.1. Basics

Microscopical conguration: n

Microcopic state energy: E n

The Micro state energy underlays a probability distribution.

Boltzmann distribution: P n = 1Z e E n

Partition function: Z = n e E n = 1kB T Free Energy: F =

kB T ln Z

Inner Energy: U = < E > = 1Z n E n e E n

F = U T S dF = dU T dS SdT

8


9/73

The First Law of thermodynamics:

dU = dW + TdS

dQ

dF = dW SdT Entropy: dW = 0: S = F T Therefore:

S = F T

= kB ln Z kB T

Z T

n

E n e E n

Z

ln Z = 1Z

S = kB ln Z

F T + 1

TZ n

E n e E n

U T TS = F + U

This comes from information theory. The expectation value of ln P says something about

the entropy. The idea is:

S = kB n P n ln P nP n =

1Z

e E n

S = kBn

e E n

Z (E n ln Z )

S = kB ln Z + 1ZT

n

E n e E n = F T

(5)

3.2. Statistical mechanics with external force

Steric expansion in micro condition nLn Force-Ensemble

Partiton function:

Z =n

exp( (E n fL n ))

9


10/73


11/73

Now we look to the effective stiffness. Lets assume a linear stiffness(spring):

F (T, < L > ) 12

K (< L > L0)2 (10)with L0 = < L > |f =0Lets calculate spring constant from the partition function!

F < L >

= f = K (< L > L0)

2F

< L > 2 = f

< L > = L0= K

f < L > = L 0

= < L >

f f =0

1

= 2H f 2 f =0

1

2H f 2 f =0

= 1K

(11)

We can use the thermodynamic potential H to calculate K.

H =

kB T ln Z (f )

H f

= 1Z n

Ln exp( (E n fL n ))

2H f 2

= Z 2 n

Ln exp( (E n fL n ))2

Z n

L2n exp( (E n fL n )) 2H f 2

= (< L 2 > < L > 2) f =0=

1K

< L 2 > < L > 2= < (L< L > )2 > = kB T 2H f 2

Temperature and stiffness set the scale of uctuation (Kubo-relation) For f = 0:

< L 2 > < L > 2= kB T

K (12)

The Equipartition theorem:

12

K (< L 2 > < L > 2) = 12

kB T (13)

11


12/73

4. Freely jointed chain with external force

n = 1, . . . , N

E n = 0

L = ai

cos i

We write the partition function

Z = 20 d1 . . . 20 dN exp af i cos i= 20 d eaf cos N

I (A) = 2

0d eA cos

A = af

Z = I (af )N

H (T, f ) = kB T N ln I (af )dI dA

= I = 20 d cos eA cos d2I dA2 = I =

2

0 d cos2

eA cos

We look at f=0 A=0: Relaxed state!I (0) = 2

I (0) = 0

I (0) =

< L > = H f

= kB TNaI I

= 0(= L0)

2H f 2 f =0

= kB TN 2a2I I

= 12

Na 2

kB T

< R 2x > = < L2 > = kB T

2H f 2

= 12

Na 2

< R2 > = < R 2x > + < R2y > = N a

2

12


13/73

So we have the following results for equipartition:

K = 2kB T

Na 2 (14)

< L 2 > = kB T

K =

1

2Na 2 (15)

That is the entropic elasticity.

4.1. Effective stiffness

Entropic feather:

K = 1 2 H f 2

= 2kB T

Na 2

F (T, < L > ) 12K < L >

2

F = U

U =0 T S

S = F T

= K T

< L > 2

dQ = T dS

S

kB

Na2 < L >

2

Rubber elasticity: Network of entropic feathers

T K contractionT K expansionthermical expansion coefficient is negative

f K (< L > L0) (16)

stretching leads to cooling!

13


14/73

B. Bead-and-Spring Model - Gaussian chain

1. Small forces

contour length s = aN = anM

Rn = x

yN = nM

Entropic elasticity k = 2k

BT

na 2

Z = d2R1 . . . d2RM exp k2kB T i (R i+1 R i)2 + i (R i+1 R i)f Z = d2 R1 . . . d2 RM exp k2 i ( R i)2 + i R if Z = dx dy exp k2 (x2 + y2) + xf M

Again have to solve a gaussian integral:

dx exp A2 x2 + Bx = 2A eB 22AZ = 2k eM f 22k = 2kB T k e M 2kk B T f 2H =

M 2k

f 2 + H 0

1 2 H f 2

= K = kM

= 2kB T

Na 2 (17)

14


15/73

1.1. Effective stiffness

Bead-and-Spring model is only correct for small forces.

L2 a2N 2

L f K = Na2f

2kB T aN af

2kB T 1a =

2kB T f

f 2kB T

a

Blob picture

L = M

2 = na 2

N = nM

n = 2

a2

L = N n

= Na2

=

fNa 2

2kB T =

f K

2. Large forces

f kB T

a (18)

af 1

Z = 20 d eaf cos N This integral can be explicitely solved with the Bessel function:

I nAi

= 12i n 20 d exp[(A cos() + in )]

15


16/73

with n = 0 and A = af :

20 d eA cos = 2 I 0(A)Z = 2I

0

af

i

N

Sidenote about the Bessel function

The Bessel function looks like

I n (x) = 12i n 20 d exp[i(x cos() + n)]

For non-integer n it is combined by the following expressions:

J n (x) =x2

n

m =0

(

1)m

m!(m + n + 1)x2

2m

J n (x) =x2

n

m =0

(1)mm!(m n + 1)

x2

2m

I n (x) = c1J n (x) + c2J n (x)

with complex c1 and c2.

For integer n J n (x) and J n (x) are not independent. So we need the Neumann function

N n (x) = J n (x)cos n J n (x)sin nxand the Hankel functions

H (1)n (x) = J n (x) + iN n (x)

H (2)n (x) = J n (x) iN n (x)I n (x) = c1H (1)n (x) + c2H

(2)n (x)

with complex c1 anc c2.

But we want to look for an easier way to solve the integral by using a saddle-point

approximation.

With the gaussian integral:

e x2 = 16


17/73

Saddle-Point-Approximation(large A, small ):

20 d eA cos d exp A 1 22

d exp A 1 2

2 eA

2A

Z 2af

N 2

eaNf

H kB T N af 12

ln af

2 = Naf +

12

kB T N ln af

2

< L > = H f Na

kB TN 2f

(19)

3. Effect of self-avoidance

For the gaussian chain: = 12 . That are normal uctuations in the absence of force.

Effective Free Energy F :

dF = d W S dT isothermal :dT =0= d W

17


18/73

System can perform work until F at minimum (d F =0)

F = kB T ( R Na )2

Na 2F

R =

2kB T

Na2 ( R

Na )

2F R2

= K = 2kB T

Na 2

System will progress until F R = 0

R2 = N a2

Add an interaction potential V int ( R)

F ( R) = kB T ( R Na )2

Na 2 + V int ( R)

V int ( R) ddx (x)2We have introduced the density (x) = N ( R )d .

F ( R) = kB T ( R Na )2

Na 2 +

N 2

( R)d

We minimize the free energy:

F R

= 2kB T

Na 2 ( R Na ) d

N 2

( R)d+1 = 0

Scaling Arguments:

RN N

12 N 2

Rd+1 0 R N

12

N 3

Rd+1 0 Rd+2 N

12 Rd+1

0 for d small

N 3

0 for d large

0

d small:

Rd+2 N 3 R N

3d +2

Flory : F = 3d + 2

18


19/73

d large:

Rd+2 N 12 Rd+1

R N 1

2

Gaussian chain : = 12

d F

1 1 1

2 343 35 0,588

4 1

212

5 3712

. . . . . . . . .

C. Experiments

Now we want to have a look of some experiments with our models Gaussian chain, Freely

jointed chain and Bead and spring.

Measure elasticity of DNA.

Magnetic Tweezers, Optical Tweezers(Lasertrap)

1. Magnetic Tweezers

Paramagnetic beads, several mTurn on force by turning on magnetic eld and measure the length the molecule of DNA

attains.

19


20/73


21/73

Measured elastcity of single-stranded DNA:

Gaussian chain, s=100nt 30nm, a=1nt 0.3nm (N=sa =100)

k = 2kB T

Na 2 =

2kB T sa

8 pN nm30nm 0.3 nm 0.8

pNnm

Measured: 0.3 pNnma =

2kB T sk

8 pN nm2

30nm 0.3 pN 0.8 nm 2.5ntA hinge every 2-3 bases!

DNA is a semiexible polymer!

D. Semiexible Polymers

E [i] =N 1

i=1

k2

i+1 i2

L = aN

i=1

cos i

Z = 20 d1 . . . 20 dn exp N 1i=1 k2(i+1 i)2 af N

i=1

cos i

Continuum limit:

E [i] =N 1

i=1

a ka2

K (i+1 i)2a2

i (s); s = iaE [(s)] = Lmax0 ds K 2 2

= dds

K. . .Bending rigidity

s. . .Arc length

R i R(s)dRds

s = R; R2 = s2

(dRds

)2 = 1

21


22/73

1. Stiff/Semiexible polymers

s. . .Arc length

c. . .curvature

R = t; t = cn; n = ct; t = cos

sin ; n = sin

cos ; c =

Bending Energy:

E = Lmax0 ds K 2 c2R(s) =

s

0ds t(s) =

s

0ds cos(s )

sin (s )

Z = D(s) exp Lmax0 ds K 2 2Z = D(s) exp Lmax0 ds K 2 2 + f cos

Correlation function:

D(s s ) = < t (s)t(s ) > = < cos ((s) (s )) >D(s s ) =

1Z D(s) cos((s) (s ))exp K 2kB T Lmax0 ds 2

We have to solve a gaussian path integral

Gaussian path integral:

I (B) = Ds(x) exp 12 dD x dD y s(x)K (x, y)s(y) + dD x B(x)s(x)Discreetion: x xi ; s i = s(xi); B i = a

D

B(xi); K kl = a2D

K (xk , x l)

I [B i] = i ds i exp 12 kl skK kl s l k Bksk

22


23/73

K kl . . . symmetric matrix

eigen vectors u(i)k and eigen values (i)

l

K kl u(i)l =

(i)u(i)k

l

u(i)l u( j )l = ij

i

u(l)k u(i)l = kl

Variable transformation sk = l ukl s l; ukl = u(k)lUnitar matrix l U kl (U T )lm = S km ; det U kl = 1I [B i] =

i

ds i det sk

s l 1exp

1

2 k (k)s 2k +

k

skBk

Bk =l

ukl B l

I [B i] =k dsk exp 12 (k)s 2k Bksk

ds e 2 s2 + Bs = 2 eB 22I [B i] =

k 2

(k)e

B 2k2 ( k )

12

k

Bk1

(k)Bk =

12

lm

B lK 1lm Bm

I [B i] = (2)

N 2

det K ij exp12

ij

B iGij B j

Gij

= ( K 1)ij

23


24/73

Continuum limit: Bi = aD B(xi); K ij = a2D K (xi , x j ); Gij = G(xi , x j ); l K kl Glm = km

dD x K (y, x)G(x, z ) = (D )(y z )aD

l

K (xk , x l)G(xl, xm ) = kl

aD

I [B] = N det K exp

12

P ij B i G ij B j

dD x dD y B(x)G(x, y)B(y)

I [B] = N det K exp

12 dD x dD yB (x)G(x, y)B(y)

dD x K (y, x)G(x, z ) = (D )(y z )

det K ij =k

(k) = expk

ln (k) = eTr(ln K ij )

Continuum limit: det K ij exp dk (k)I [B]I [0]

= exp12 dD x dD y B(x)G(x, y)B(y)

D(s1 s2) = < cos((s1) (s2)) > = < exp(i((s1) (s2))) >D(s1 s2) =

1Z D(s) exp 12 ds K kB T 2 cos((s1) (s2))

D(s1 s2) = 1Z D(s) exp Lmax0 ds K 2 2 + Lmax0 ds B (s)(s)

I [B] = D(s) exp Lmax0 ds K 2 2 Lmax0 ds B (s)(s)B(s) = i (s s1) i (s s2)

D(s1 s2) = I [B]I [0]

24


25/73

I [B] = D(s) exp 12 ds ds (s)K (s, s )(s ) + ds B (s)(s)K (s, s ) =

K kB T

2s (s s )

ds K (s, s )G(s , s ) = (s s )I [B] = N det K exp 12 ds ds B(s)G(s, s )B(s )I [B]I [0]

= exp12 ds ds B(s)G(s, s )B(s )

G(s, s ) = G(s s ) = dq 2 G(q )exp( iq (s s ))K (s, s ) =

dq

2K (q ) exp(iq (s

s ))

G(q ) = ds G(s, 0)e iqs (s) =

12 dq eiqs

(q q ) = 12 ds exp(i(q q )s )

ds

dq 2

K (q )exp( iq (s s ))

K (s,s ) dq

2G(q )exp( iq (s s ))

G(s,s )=

dq

2 exp(iq (s s ))

12 dq dq (q q )K (q )G(q )exp( i(qs q s )) = dq 2 exp(iq (s s ))

K (q )G(q ) = 1

K (q ) = ds K (s, 0)e iqs = ds K kB T (s)e iqs = Kq 2kB T G(q ) =

kB T

Kq 2

G(s s ) = dq 2 kB T K exp(iq (s s ))q 2

25


26/73

J = 12 ds ds B(s)G(s, s )B(s )

J = ( i)212 ds ds ( (s s1) (s s2))G(s, s )( (s s1) (s s2))

J =

1

2(G(s1, s1)

G(s1, s2)

G(s2, s1) + G(s2, s2))

J = 12

kB T K dq 2 2 exp(iq (s1 s2)) exp(iq (s1 s2))q 2

J = kB T K dq 2 1 cos

u

qsq 2J =

kB T K |s| du2 1 cos(u)u2

12J =

kB T 2K |s|

< t (s1)t(s2) > = I [B]I [0]

= e J = exp kB T 2K |s1 s2|

With persistence length = 2K kB T and K = 12 kB T ([K] = energy * length):

< t (s1)t(s2) > = e |s1 s2 |

Explicit values are:

DNA: 50nmActin: 16 mMicrotubuley: 2 mm

Now we calculate the end-to-end-distance: R = R(s) R(0) = s

0 ds t(s )

26


27/73

< R2 > = s0 ds s0 ds < t (s )t(s ) >=

s

0ds

s

0ds exp |

s s |

= 2 s

0ds

s

0ds exp

s s

= 2 s0 ds exp s s s0= 2 s0 ds (1 e s )

< R2 > = 2|s|+ 2 2(e |s | 1) (20)

|s

|:

< R2 > 2|s| a2N N =

sa 2a = a

22 = a

|s| :< R2 > |s|2 + O

s3

< R2 >

2

|s

|+ 2 2 1

|s|

+ |s3|

33 + . . .

1

= 12 |s| 21|s|+ |s|2 13|s|3

+ . . . |s|2 O |s|3

2. Effective stiffness

F (T, < L > ) 12

K (L L0)2L0 = < L >

|f =0

K = 2H f 2 f =0

1

2H f 2

= (< L 2 > < L > 2)

27


28/73

small forces, freely jointed boundaries:

1K

= 2

< R2 >

K = kB T

< R2

> =

kB T

2|s|+ 2 2(e |s |

1)(21)

compare with K f jc = 2kB T Na 2

28


29/73

II. BIOMEMBRANES

Amphiphile moleculespolar head (hydrophilic)

chain (hydrophobic)

spontaneous formation of micelles beyond a critical concentration

spherical

cylindrical

Lipidspolar head

two chains

formation of micelles is sterically hindered

spontaneous formation of a bilayer instead membrane

Hydrophobic chains are in contact to water at

boundary: Energetic cost!

E = l

with the line tension and the length of

boundary lFormation of closed structures: Vesicles

29


30/73

Holes close spontanously

E = 2r

Therefore: Lipid membranes are extremely stable!

Biological membranes

Lipid bilayer + proteins More than 200 types of lipidsBilayer is attached to the cytoskeletonAway from thermal equilibrium

Physical properties: morphology, mechanical properties, thermal uctuations

Fluid membranes

Relative position of lipids is not xed

Rapid Diffusion (D 1 10 m2

s )

Number of molecules is constant

surface is constant

Incompressible surface, no shear stress within bilayer ( water)Bending of the plane surface costs energy

Only important degree of freedom is the geometry (or shape) of surface Theory of uc-tuating geometries

Physics is invariant with respect to the choice of parametrisation of the surface

A. Differential geometry (geometry of surfaces)

1. Curves in R 2

t: tangent vector

n: normal vector

R(s); s: arc length

30


31/73

Measure a distance:

( R)2 = ( R(s2) R(s1))2 dRds

s2

= t2( s)2

s is an arc length if t2 = 1

R2 = s2; t = R; t2 = R2

= 1

t = cnn = ct

c: curvature

t n = cn20 c; c = t n = R

2nd derivative

n

Circle:

x = R cos sRy = R sin sR

t = R = sin sR

cos sR; t2 = 1: parametrization is correct

t = R = 1R

c

cos sRsin sR

n

2. Surfaces in R 3

ui : parametrization (compare to arc length s)

Problem: There is no arc length parametrization for a surface!

31


32/73

innitesimal displacement d ui

dR = Ru i

dui = eidui

dui as contravariant vectors, i covariant

tangent vector (not normalized)

ei = Ru i

= iR

distance on a surface

ds2 = (d R)2 = ei e j duidu jgij = ei e j

Metric tensor

We can now measure arc length using the metric tensor!

Local reference system ei = iR; e1 and e2; that is non-orthogonal

a = a ieinotation will become clear

later...Some things depend on parametrization chosen, some things do not.

Physics (e.g. Energy, curvature) cannot depend on the parametrization.

What are the parametrization-independent objects in this formalism?

3. Curvature of a surface

Tangent plane e1e2Normal vector n

How does the normal vector/tangent plane change with a displacement d ui?

32


33/73

in = C ji e j (22)

ie j = D ij n + kij ek (23)lives in tangent plane

lives everywhere

Note C ji and D ij describe a curvature

C ki gkj = D ij related to the tensor of curvature

Christoffel-symbol (23): kij = ek ( ie j ) igki Divergence of a vectorRelationship between C ji and D ij

n ei = 0 i(n e j ) = ( in) e j + ( ie j ) n = 0(22) : ( in) e j = C ki ek e j = C ki gkj

(23) : ( ie j ) n = D ij n2 + kij ek n = D ij

C ki gkj = D ij

C ki is the tensor of curvature

D ij = ( ie j ) ne j = j R

D ij =

n

i j R (24)

curvature: 2 nd derivative!

How can we extract the curvature from C ji and D ij ?

Information is there: Geometry of the surface is locally characterized by C ji (and gij to

dene a metric).

4. Formalism of vectors and tensors

eis are not normalized and not orthogonal

Introduce a conjugate base e j

33


34/73

ei e j = i jTwo coordinate systems: a = ai

contravariantei =

a i

covariant

ei

Coordinate transformation:

a i = a ei = a j e j ei = a j ji = aia i = a ei = a j e j ei = a j i j = a i

but also

a i = a ei = a j e j ei = a j g jiThe metric tensor transforms coordinates from contra- to covariant bases!

a i = a j g ji ; a j = aigij ; gij = ( gij ) 1

Scalar product:

a b = aib j ei e j = aib j gij = aib j ei e j = aib j ji = aibiParametrization-independent objects

Scalar: a b = aibiVector: a = aiei = a iei

Tensor: a = aij

(eie j ) = a

i j (eie

j

)Example: T a = T i j (eie j ) a = T i j (eie j akek) = T i j a j ei is a vector

5. Tensor of curvature

C ji = n gkj

symmetric

i k

symmetric

R

C ji is symmetric and can be diagonalized

2 Eigenvalues c1 and c2

2 orthogonal Eigenvectors

independent of parametrizationTwo scalars of curvature

(i) Mean curvature H = 12 T r(C ji ) =

12 (c1 + c2)

34


35/73

(ii) Gaussian curvature K = det( C ji ) = c1c2

c1 and c2 the principal curvatures. The corresponding two axes of principal curvature are

orthogonal.

C ii = 2 H ; C ji C i j = (2 H )2

2K

c1 > 0

c2 > 0

H > 0

K > 0

c1 < 0

c2 < 0

H < 0

K > 0

c1 > 0

c2 < 0

K < 0

saddle point

6. Additional surface elements that are independent of parametrization:

Normal vector n

n = e e2| e1 e2 | ; n2 = 1

Surface element d A

ndA = ( 1Rdu1) ( 2Rdu2) = ( 1R 2R)du1du2 = ( e1 e2)du1du2

(ndA)2 = d A2 = ( e1

e2)2(du1du2)2 = ( g11g22

g12g21)

[(e1 e2 )2 = e21 e22 (e1 e2 )2 ](du1du2)2

(ndA)2 = d A2 = det gij (du1du2)2

g = det gij

dA = gdu1du2 independent of parametrization!A = gdu1du2

35


36/73

B. Bending energy of a uid membrane

1. Expand bending energy

E = dA f (C ji (R))where C ji is the tensor of curvature and f (C

ji ) is a scalar, which is independent of the

parametrization. We now develop f to second order terms:

f a0 + a1C kk + a2(C kk )2 + a3C ki C ik + O (C ji )3

So we now have 4 parameters and 2 scalars of curvature H and K :

C kk = 2H

C ki C ik = (2 H )

2 2K With a0 = + 2 (c0)

2, a1 = c0, a2 = + 2 and a3 = 2 , f is typically written as:f = +

2

(2H c0)2 + K 2 Scalars of curvature:

H = 12 C kk mean curvature

K = det C ji gaussian curvature

4 Parameters:

bending rigidity

gaussian rigidity

surface tension

c0 spontaneous curvature

With that the bending energy looks like

E = A

constant+ dA 2(2H c0)2 + K discuss nextThe rst term A is constant, when A is constant. We are now going to discuss the last

term, which leads us to the Theorem of Gauss-Bonnet.

36


37/73

2. Theorem of Gauss-Bonnet

A

dA K = 4 n (25)

The integrated surface must be closed. n = 1g, whereg is the topological genus (or the number of holes) of

the surface.

Gauss interpreted it in the following way: As above d A is:

dA n = ( 1R 2R)du1du2dA = g du1du2

g = det ( iR j R) = det gijNow lets look at the surface created by n(u1, u2). If the surface you integrate over is closed,

the surface created by all vectors n(u1, u2) is a sphere with radius 1. In anology to d A we

determine the surface element of this unit sphere d As2 :

dAs2 n s2 = ( 1n 2n)du1

du2

dAs2 = k du1du2k = det ( in j n)

One can see that

A

dAs2 n = 4n (26)

37


38/73

Be aware that n is the normal vector and n is the one from equation (25). We now determine

k:

in j n = C ki ek C l j el = C ki gkl C l jk = det ( C ki gkl C l j ) = K 2 g

We used equations (22) and (23). So now:

4n = A

dAs2 n = A

k du1du2 = A

g du1du2 dAK =

A

dA K

A closed integral of gaussian curvature is topological invariant.

38


39/73

Making equation (26) plausible

Lets look at a torus as an example. There exist two parallel normal vectors n and n for

each direction of n. So the surfaces created by n are two spheres with radius 1.

But it is not clear if d As2 is parallel

or antiparallel to n. We will check

this for the two normal vectors n and

n and the parametrization shown in

the gure on the left.

dAs2 = ( 1n 2n)du1du

2

nu 1

du1 = n(u1 + d u1, u2) n(u1, u2)

We have two different situations for n and n :

n and n s2 parallel n and ns2 antiparallel

So one sphere gives a surface area of 4 whereas the other gives 4:

dAs2 n = 4 4 = 0

39


40/73

3. Biological implications

E = A

constant

+

A

dA2

(2H c0)2 +

A

dA K

constantIn a typical biological reaction, the number of holes in a given closed membrane does notchange and A dA K is constant. Exceptions are the budding or fusion of vesicles, wherethis term needs to be considered. Furthermore we assume A to be constant, so the rst termis constant, too.

E = A

dA2

(2H c0)2

For asymmetric bilayers c0 = 0:

solvent asymmetric composition asymmetric

In all other cases c0 = 0

E = A dA 2 (2H )2we conclude that the bending energy E only depends on a single parameter .

C. Monge-representation

Now lets choose a paramatrization!

40


41/73

z = h(x, y); u1 = x; u2 = y; R =

x

y

h(x, y)

The base vectors therefore are:

ei = iR

ex =

1

0

xh

; ey =

0

1

yh

The normal vector is:

n = e1 e2|e1 e2| = 1 1 + (h)2

x h

yh1

Where:

h x h

yh; (h)

2 = ( x h)2 + ( yh)2

The metric tensor is:

gij = ei e jgij =

1 + ( x h)2 ( xh)( yh)

( xh)( yh) 1 + ( yh)2 = ij + ( ih)( j h)

Also

g = det gij = 1 + ( x h)2 1 + ( yh)2 ( x h)2( yh)2= 1 + ( x h)2 + ( yh)2 = 1 + (h)

2

The conjugate metric tensor gij can be written as follows:

gij = ( gij ) 1 = 1g

1 + ( yh)2 ( x h)( yh)( x h)( yh) 1 + ( x h)2

= ij ( ih)( j h)1 + (h)

2

Where we used:a b

b c

1

= 1

deta b

b c

c bb a

41


42/73

We now calculate the conjugate base vectors:

e j = gij ei ex = g11ex + g

12ey = 1g

1 + ( yh)21

0

x h

1g

( xh)( yh)

0

1

yh

ex =

1g

1 + ( yh)2

( x h)( yh)

x h

ey = g21ex + g22ey = 1g

( xh)( yh)1 + ( x h)2

yh

Now we proove that ei e j = i jex ey =

1g

(( x h)( yh) + ( xh)( yh)) = 0ex ex =

1g

(1 + ( yh)2 + ( x h)2)

1+( h)2 = g= 1

1. Tensor of curvature in Monge-representation

With equation (24):

C ij = n i j R = 1

g x h yh

1

0

0

i j h

= i j h g

C ji = C ik gkj =

i kh g kj

( kh)( j h)g

= i j h g +

1g

32

k

( i kh)( kh)( j h)

Bending Energy

E = d2x 2(2H )2 2H = Tr(C ji)We now assume weak uctuations: ( h)

2 1.

g = 1 + O (h)2 2H = h 1 + O (h)2= 2x +

2y

42


43/73

E = dx dy 2 ( h)2Partition function of a symmetric bilayer:

Z = DRVol( g) exp 2 dA (2H )2DR(u1, u2) Sum over all surfaces and parametrizations of each surfaceVol(g) Measure of all parametrizations of a surface conrmation, volume of

the group of reparametrizations

Z is invariant with respect to the

parametrization!

In Monge-representation it is:

Z = Dh exp 2 d2x ( h)2

43


44/73

Sidenote about partial integration

b

a

u v dx = b

a

v u dx + [uv]ba

1st note:

dx 2

( 2x h)

u( 2x h)

v=

dx 2

( 3x h)

v( x h)

u+ ( xh)

u( 2x h)

v

=

dx 2

h( 4x h) + ( x h)( 2x h) h 3x h

0, if h( )=0=

dx

2h( 4x h)

2nd note:

d2y 1kB T ( y y + m) (x y) h(y) 4xP .I.= 1kB T ( x x + m)h(x) dx (x)f (x) = dx f (x) (x) = f (0)

D. Persistence length of a membrane

Compare to the persistence length of a polymer = 2K kB T calculated from the correlation

function < t (s1)t(s2) > .

To be able to calculate this quantity, we will place the membrane between two surfaces

and later move the surfaces apart from each other.

eliminates translation in h-

direction

Introduce a repulsive potential V(h):

44


45/73

V tot = V (h + d) + V (d h)

12

mh 2 + const. m = 2

2

h 2V (h)

h= d

E = d2x

2( h)2 + m

2 h2

2xP .I.= d2x h 2 + m2 hThis is valid for periodic boundary conditions.

1. Partition function

Z (B) = Dh exp d2x 12h Kh + BhB(x) is an external force density. Periodic boundary conditions:

K = kB T

+ mkB T

With:

K (x, y) = kB T

y y + mkB T

(x y)One can write K as:

Kh = d2y K (x, y)h(y)Therefore:

Z (B) = Dh exp d2x d2y h(x)K (x, y)h(y) d2x B(x)h(x)45


46/73

Reminder: Gaussian Path Integral (discrete case):

Z (B1, , Bn ) =n

i=1 dh i exp 12h iK ij h j B j h j=

2

N

det K ij exp 12BiGij B j

Gij = ( K 1)ij

Free Energy:

H (B1, , Bn ) = kB T ln Z =

kB T 2

ln (det K ij ) kB T

2 BiGij B j + const

Mean height (compare with Freely jointed chain: < L > = H F ):< h i > =

H B i

= kB T

2 Gij B j

= 0, if B = 0

Correlation function:

< h ih j > = 1kB T

2H B iB j

= Gij

Gij determines the correlations.H B i is a functional derivative in the continuum case.

46


47/73

Functional Derivative Example: Minimize the bending energy of a surface

E [h(x)] = d2x 2 ( x h(x))2E h(x) + h(x)

E [h(x)] + E

E = d2x E h(x) h(x)Discrete:

E =i

E h i

h i

E [h(x) + h(x)] = d2x 2 [ x {h(x) + h(x)}]2= d

2x

2( x h(x))

2

E [h(x)]+ d

2x [ x h(x)] [ x h(x)] E E = d2x [ x h(x)] [ x h(x)]

2xP .I.= d2x ( x x h(x)) Eh ( x ) h(x)

E h(x)

= x x h(x)

For the surface with the minimal bending energy:

E h(x)

= 0

2. Continuum limit

< h (x) > = H B(x) B =0

= kB T 2 d2y G(x, y)B(y) B =0 = 0

< h (x)h(y) > = (2) H

B(x)B(y) B =0= G(x y)

Determine the propagator G = K 1

kB T

x x + mkB T

G(x y) = (x y)

47


48/73

Fourier space:

G(x) = 1(2)2 d2q G(q )e iqx

(x) = 1

(2)2 d2q e iqx

kB T

q 4 + mkB T

G(q ) = 1

G(x) = kB T (2)2 d2q e iqxq 4 + m

If m = 0, G diverges for q 0. This is why we placed the membrane between two surfaces!

3. Correlation functions

< h (x)h(0) > = G(x) = kB T (2)2 d2q e iqxq 4 + m

< h(x) h(0)2 > = C (x) = < h (x)2 + h(0)2 > 2 < h (x)h(0) >

= < h (x)h(x) > + < h (0)h(0) > 2G(x)C (x) = 2 G(0) 2G(x)

What does C (x) look like?

So now lets calculate C (x). Therefore we introduce the lenth scale :

= m

14

48


49/73

G(x) = kB T

1

(2)2 d2q e iqxq 4 + 4G(x) =

kB T

1(2)2

0

dq q 2

0

d e iqx cos

q 4 + 4

= kB T

1

2

0

dq qJ 0(qx)q 4 + 4

= kB T 2

2kei

x

J o(y) is a Bessel-function, kei(y) is the Thompson-function. There are approximations for

the Thompson-function:

kei (x) 4

+ 4

(a + bln x)x2 x 0kei(y)

1y

12

e y 2 sin

y 2 y

< h (x)2 > = kB T

8 2

C (x) = 2 G(0) 2G(x)= < h(x) h(0)

2 > = kB T 22 d2q 1e iqxq 4 + 4

C (x) = kB T

4a + bln

x

x2 x

C (x) = kB T

4 2 x

49


50/73

C (x)L2

x

L

L kB T

12

L3

4. Periodic boundary conditions

Finite system size:

q min = 2L

q max = 2

a

a: microscopic length, thickness of the membrane, size of molecules. a4nm.

Because undulations on lengths smaller than the size of the microscopic constituent are

unphysical, we integrate from q min to q max .

G(x) = k

BT

(2)2 Q d2q

e iqx

q 4

< h 2 > = G(0) = kB T (2)2 qmaxqmin dq q 1q 4 = kB T 2 12q 2

2a

2L

= 1(2)3

kB T

(L2 a2)

50


51/73

5. Persistence length of a membrane

< t (s1)t(s2) > = e |s 1 s 2 | p

p: units of length.

At which length do normal vectors de-correlate?

< (h)2 > 1 ?

Under what conditions is < (h)2 > 1, if we change the system size? This will give us

some kind of persistence length!

< h(x)h(y) > = xy < h (x)h(y) >

= 2x G(x y)=

kB T (2)2 Q d2q q 2e iq(x y)q 4

< h(x)2 > =

kB T (2)2 Q d2q 1q 2

= kB T 2 qmaxqmin dq 1q

= kB T 2

ln La

We need the mircroscopic cut-off! Gradient grows with larger system size L and will become

larger than one at some size. Persistence length is estimated as L = p, where < (h)2 > = 1.

p ae2

k B T

kB T 10 for typical lipids, a4nm p ae60 = 4 1017m astronomical number!

Persistence length of membranes is nite but very, very large: Membranes are at.

51


52/73

But: Do not take thermodynamic limit (do not make system size innite as people

typically do in the statistical thermodynamic limit), because the membrane will fold!

E. Stretching elasticity of a membrane

: isotropic tension

dA = g dx dyg = 1 + (h)

2

A = d2x 1 + (h)2 A0 + 12 d2x (h)2A0 = L2: projected area.

Remember polymers:

Z (f ) = DR exp ds 2c2 + Lf H (f ) = kB T ln Z (f )

< L > = H f

Legendre transformation:

F (< L > ) = H + < L > f

f = F

< L >

= f

< L > =

2F < L > 2

= 2H f 2

1

52


53/73

Do the same for membranes!

E = d2x 2( h)2 A0 A0 = A const 12 d2x (h)2

=

d2x

2( h)2 +

2(

h)2

A

Z ( ) = Dh exp d2x 2 ( h)2 + 2(h)2 + AH ( ) = kB T ln Z ( )

H

= kB T

Z Z

= A + < d2x 12(h)2 >

d2x

A 0< 12 ( h)

2 >

= < A 0 > average of the projected area

F (< A 0 > ) = H + Z < A 0 >F

< A 0 > =

Stretch molecules:

= < A 0 > Z

< A 0 > (units of energy / surface area)

= < A 0 > 2F

< A 0 > 2 =

< A 0 > 2 H 2

< A 0 > = A

1

2

A = < A 0 > 1 + 12

< (h)2 >

< A 0 > A 1 12

< (h)2 >

Total area A is larger then the projected are < A 0 > .

< (h)2 > =

kB T (2)2 d2q q 2q 4 + q 2 = kB T (2)2 qmaxqmin dq q 1q 2 +

53


54/73

q min = 2L

q max = 2

a

< A 0 > = A 1 kB T 8

ln 4

2

a 2 +

4 2L 2 +

: Energy

area : Energy

L2

: Energy

area

< A 0 >Z =0

= A 1 kB T 4

ln La

For L 2 a 2 :

< A 0 > A 1 kB T 8

ln 4

2

a 2 +

4 2L 2 +

= A 1 + kB T 8

ln a2

4 2

Grows logarithmically with !

54


55/73

A0 dlP V = A0

V 4R 2 R A 8R R

= RP

2

Measure A0 and and plot ln vs. < A 0 > . This should be a line with slope kB T 8 determined!

1 = 1

< A 0 >

< A 0 >

=0=

kB T

2 qmax

qmin

d2q 1

(2)2

q 4

(q 4

)2

= 323 2

kB T 1

L2 a2Stiffness at = 0: diverges when L = a, because there are no uctuations then.

55


56/73

III. DYNAMICS OF BIOLOGICAL MOLECULES

Time-dependent stochastic processes.

Diffusion: Ficks second law:

t P = D P With:

=

2

R 2

P = P (R, t )

Solution for 2 dimensions:

P (R, t ) = 14Dt exp

(R

R

0)2

4Dt

Compare to gaussian polymer (freely jointed chain):

P (R, N ) = 1a 2N

exp (R R0)2

a2N

Random path of a diffusing particle:

a = v0 t; N = t t

4Dt a2

N

RN = R0 + aN

i=1

cosi

sini

R(t i + t) = R(t i) + v0cosi

sini

i t

i = random angle, i = random velocity.

< i > = v0 20 di2 cosisini = 0< i k > =

v2022 20 di 20 dk (cosi cos k + sin i sin k)

= v20 ik

56


57/73

Continuum-limit: stochastic differential equation. t 0, D = const .dRdt

= (t) for Diffusion

What are the stochastic properties of (t)?

< (R R0)2 > = N a2 = t t

v20 t 2 = 4 Dt

D = 14

v20 t v20 =

4D t

a2 = v20 t2 = 4 D t

< (t i) (tk) > = 4 D ik t

Continuum-limit:

< (t) (t ) > = 4 D (t t )dRdt

= (t)

< (t) > = 0

< (t) (t ) > = 4D (t

t )

Vector components in any dimension:

< (t) (t ) > = 2 D (t t )Diffusion in a potential energy landscape:

W (R); =

: Mobility, : Random force

dRdt

= W R

+ < (t) (t ) > = 2D (t t )dRdt

= W R

Drift+

Diffusion< (t) (t ) > = 2

D2

(t t )

57


58/73

Back to probability P (R, t ):

t P + R

J = 0

J =

D

P

Diffusion

RW P

DriftWith that you get the Smoluchowsky-Equation: t P = D P

Diffusion+

R

R

W P

DriftStationary State: t P = 0 R J = 0Steady state: J = 0

P (R) = N e

W ( R )D

Compare to the Boltzmann-Distribution:

P (R) = N e W (R ) =

1kB T

= D

This gives the Einstein-Relation :

D = kB T

Equation of Diffusion:

t P = D P

Stochastic differential equation:

t R = (t) < (t) > = 0

< (t) (t ) > = 2 D (t t )

Stochastic motion in a potential energy landscape, with random force (Langevin-Equation):

t R = W + (t) < (t) (t ) > = 2kB T

(t t )

Smoluchowsky-equation describes diffusion in a potential energy landscape:

t P = D P + (P W )

58


59/73

Einstein-relation:

D = kB T

External force:

W (x) = V (x) f ext xHarmonic Oscillator with force:

W (x) = 12

kx2 f ext x

A. One-dimensional harmonic oscillator with uctuations

m 2t x + 1 t x + kx = + f ext < (t)(t ) > = 2kB T 1 (t t )

Correlation function

< x (t)x(t ) > = C (t t )

1. Linear response function

The response-function (t) is convoluted with the external force to give < x (t) > .

< x (t) > = t dt (t t )f ext (t ) + O(f 2ext ) t dt (t t )f ext (t )= dt (t t )f ext (t )

Where (t) = (t)(t).

Fourier-representation of (t) and f ext (t):

(t) = 12 d ()e it

f ext (t) = 12 d f ()e it

59


60/73

With that < x (t) > is:

< x (t) > = dt d2 d2 ()e i(t t ) f ( )e i t=

d

2 d () f ( )e it

dt

2 e( )t

( )= d2 () f () =x()e it

< x (t) > = d x()e it x() = () f ()We look at the mean response < x (t) > , which should be independent of uctuations, = 0.

With that and the fourier-representation of < x (t) > the equation of motion for < x (t) >

looks like:

d2 m2 i 1 + k () f ()e it = d2 f ()e it () =

1

m2 i 1 + k () =

1m( 1)( 2)

Solve:2 +

im

km

= 0

Solution:

1/ 2 = i2m

12 4 km 1(m)2

With:

= 12 4 km 1(m)2

= 12m

One can write the solutions as:

1 = i2 = i

60


61/73

(t) = d

2 ()e it

t < 0 : (t) = 0

t > 0 : (t) = (t) = d2 ()e itRemember that (t) = (t)( t).

61


62/73

Note about Residuals:

dx f (x) = 2 i Res[f (x); a]a single pole:

Res[f (x); a] = limxa(x a)f (x)a multiple pole, order m:

Res[f (x); a] = 1

(m 1)! limxa

dm 1

dxm 1(x a)m f (x)

Only the rst term of a Laurent expansion survives, i.e. expand f (x) around x = a to order1x .

Explanation: Lets look at the pole 1zn :

dz 1zn

z = ei

dz = ieid

dz 1z n = 20 d ieie in = 20 d ie i(n 1)

This integral is always zero, except for n = 1:

if n = 1 : dz 1z n = i 12

0= 2 i

t > 0 : (t) = + i 1

m(1 2)e 1 t +

1m(2 1)

e 2 t

Note the sign change because of clockwise integration instead of counter clockwise (mathe-

matically positive).

(t) = i 1

2m e

1t

+ 12m e

i2

t

= i2m

e t i t e t+ i t

Note: ex+ iy = ex(cos y + i sin y)

(t) = i2m

e t [cos( t) + i sin( t) cos( t) i sin( t)]

62


63/73

(t) = 1m

sin( t)e t for t > 0

(t) = 1m

sin( t)e t (t) t

(t) = (t) + i (t)

(t) real (t) imaginaryIn principal there are two ways to write (t):

1)

(t) = (t)antisymmetric, imaginary

(t) = (t)symmetric, real

2)

(t) = (t)symmetric, imaginary

63


64/73

(t) = (t)antisymmetric, real

We now check if both ways are possible:

() = dt (t)eit ; (t) real() = dt (t)e it = () () complex

If (t) = (t) () = () : () realIf (t) = (t) () = () : () imaginary

We know: () = 1 m 2 i + k has real and imaginary part.

We still dont know which one is symmetric and which one is antisymmetric.

could it be that (t) is symmetric and (t) is antisymmetric?

Lets do the inverse transform:

() = ()

real: (t )= ( t )+ i ()

imaginary: (t)= (t )For (t) = (t) + i (t) to be real, (t) needs to be imaginary!

(t) symmetric, (t) antisymmetric.

Given a particular response function, causality

invokes a specic dissipation function ().

64


65/73

2. Correlation function

f ext = 0

C ( ) = < x (t)x(t + ) >

C ( ) = d2 C ()e i C ( ) = C ( ) real

C () = C () = C () real and symmetric

More precisely:

C ( ) 12T

T

T dt x(t)x(t + )

= 12T T T dt d2 d2 x()x( )e it i (t+ )

= 12T T T dt d2 d2 x()x( )e it (+ ) i lim

T

12T T T dt e it (+ ) = 2 ( + )

C ( ) = d2 C ()e i With C () = < x()x() > . Now:

x() = ()( + f ext ) f ext = 0

x() = ()()

Therefore:

C () = < x()x() >= < ()() ()() > () = ()= < ()() > ()()= < ()() > | ()|2 < ()() > = 2

kB T

= 2kB T

1

|k m2 i |2

65


66/73

We now calculate ():

() = 12i

[ () ()]=

1

2i

1

k m2 i

1

k m2 + i

=

1

|k m2 i |2This leads to the:

3. Fluctuation-dissipation-theorem

C () = 2kB T ()

Relationship between the complex part of the linear response function (the dissipation func-

tion) and the correlation function!

(t) = (t)kB T

dC (t)dt

B. Stochastic motion in periodic potentials

One-dimensional:

W (x + a) = W (x)

Smoluchowsky-Equation:

t P + x J = 0

J = 1

x P + ( x W f ext )P P denotes the probability density for the particle to be at position x at time t; P (x, t )

= (W f ext x)J 1 = x P P x

66


67/73

Periodic system:

P (0) = P (a)

a

0dx P = 1

Nonequilibrium steady state (stationary state):

J = const. t P = 0

P (x) = N e ( x) J

e ( x) x0 dy e( y)

a0 dx P = 1 1 = N ao dx e ( x) J a0 dx e ( x) x0 dy e( y)N =

1 + J a0 dx e ( x) x0 dy e( y)

ao dx e ( x)

1. Periodic boundary conditions

P (0) = P (a)

Ne (0) = Ne ( a) J

e ( a) a0 dy e( y)

= ( a)

(0) =

f ext

a

N (e 1) = J

a

0dy e( y)

Solve for J:

(e 1) 1 + J

a0 dx e ( x) x0 dy e( y) = J a0 dy e( y) a0 dx e ( x)(1 e ) =

J a0 dx e( x) x0 dy e( y)e x0 dy e( y) + a0 dy e( y)

I (x)

I (x) = x+ a

ady e

x

0dy e +

a

0dy e =

x+ a

0dy e

x

0dy e =

x+ a

xdy e

J =

1 e

a0 dx e ( x) x+ ax dy e( y)v(x) = J P (x)

< v > = a J

67


68/73

2. Kramers rules

< v > = v = akB T

a

0 dx e ( x)

x+ a

x dy e( y)

(1 e )

Denition:

r = kB T

a0 dx e ( x) x+ ax dy e( y)r + = r

r = re f ext ak B T

v = ar (1 ef ext ak B T )

v = a(r + r )

E (x) = W (x) f ext x; = E Saddle point approximation:

x+ ax dy e+ E (y) e+ E B dx e 2 |E B | (x xB )2 = 2 |W B

|

eE B

W B = E B . This integral is over one period and does not depend on x in this approximation!

a0 dx e E (x) e E A dx e 2 E A (x xA )2 = 2W A e E AW A = E A . Stationary rate:

68


69/73

r 2

W A |W B |12

e E (f ext )

E = E B E A

r depends on second derivatives at minimum/maximum and on E (f ext )!

3. Transition state picture

Expand E (f ext ) to rst order in f ext :

E (f ext ) E 0 af ext + O(f 2

ext ) E 0 = W f ext =0

0 1

v0 = ar 0e W 0k B T

Where we dened r 0:

r0 = 2 (W A |W B |)

12

= 0 : v = v0 1 e f ext ak B T

= 12

: v = v0 ef ext a2k B T e

f ext a2k B T

= 1 : v = v0 ef ext ak B T 1

Motion in a continuous free energy landscape discrete hopping modelSystem spends most of the time at minima and moves to adjacent minima according to

Kramers Rules. This is not only true for periodic energy landscapes.

69


70/73

4. Detailed balance

At equilibrium:

P N = N e W N k B T

or:

P ArAB = P B r BA

Therefore:r ABr BA

= eW A W B

k B T

Periodic potential:

v = a(r + r ) r+

r = e

f ext ak B T

5. Effective diffusion

70


71/73

t P n = (r + + r )P n + r + P n 1 + r P n +1=

1a

(J n +1 J n ) xJ J n = a(r P n r + P n 1)

Detailed balance: J n = 0 at equilibrum!

J n = vP n + P n 1

2 DP n P n 1

a vP D x P

ar = v2

Da

ar + = v2

+ Da

v = a(r+ r )

D = 12

a2(r + r )

v a2

(W A |W B |)12 e W 0 (1 e f ext a )

D

a2

4 (W

A |W

B |)

12 e W 0 (1 + e f ext a )

6. Effective mobility

0 = vf ext f ext =0

D0 = Df ext =0

0 = a2

2kB T

(W A

|W B

|)

12 e W 0 =

D0

kB T Einstein Relation:

0kB T = D0

D0 Fluctuations Correlation function0 friction 1 Dissipation Response function

71


72/73

C. Experiments

Single molecule, one protein at a time.

Example: RNA polymerase.

Copies DNA into messanger RNA. mRNA is then translated into protein by the ribosome.

Optical trap:

A bead in a laser trap can be described by an overdamped harmonic oscilator with uctua-

tions:

Overdamped: m = 0

1 t x + kx = f extNeed to determine the stiffness of the trap for calibrated measurements!

Measure at f ext = 0: thermal uctuations of bead.

Determine: C ( ) = < x (t)x(t + ) > and C () from experimental data.

Spectral Density C ()

C ()m =0

= 2kB T

1

k i 2

= 2kB T

1

k2 + 22D = kB T

= 2D

2k2 + 2

Lorentz-function:

72


73/73

We know and D, so we can very accurately determine k by tting the data to this function!

Stiffness of DNA:

Semiexible polymer: K DN A = kB T 2 |s |+2 2 (e|s | 1) for small forces.

Motion of the polymerase along the DNA:Interaction potential is periodic with a period of 1 basepair!

Periodic Potential

Discrete hopping model with Kramers Rules

and external force!

THE END

theoretical biophysics

Documents