Lecture 11.

Diffusion in biological systems

Zhanchun Tu (涂展春 )

Department of Physics, BNU

Email: tuzc@bnu.edu.cn

Homepage: www.tuzc.org

Main contents

● Diffusion in the cell

● Diffusive dynamics

● Biological applications of diffusion

§11.1 Diffusion in the cell

Brownian motion● Brown (1828)

– Pollen micro-grains (1μm) suspended in water do a peculiar dance

– The motion of the pollen has never stopped

– Totally lifeless particles do exactly the same thing

● Einstein (1905)– Quantitative explanation from

thermal motion of water molecules

Active vs passive transport● Opening of ion channel & diffusion of ions

When channel is open, ions diffusion inward-- Passive transport (no energy input)

● Diffusive & directed motions of RNA polymerase

Passive: Free RNA polymerase molecule diffusing in a bacterial cell

Active: One-dimensional motion of RNA polymerase along DNA

Energy source: NTP

● Patterns of E. coli swimming at different scales

At low magnification, the swimming

movement of a single bacterium

appears to be a random walk

At higher magnification, it is clear

that each step of this random walk

is made up of very straight, regular

movements

Biological distances measured in diffusion times

● Diffusion time as a function of the length

globular protein

t=1010s=300y

● Diffusion is no effective over large cellular distances

nerve cell

Time scale of the active transport by virtue of molecular motors is shortened by many orders of magnitude than passive diffusion

Experimental techniques: measuring diffusive dynamics

● Fluorescence recovery after photobleaching(光漂白 )

Fluorescently labeled molecule

Photobleached regionby laser pulse

Fluorescently labeled molecules diffuse into the region

FRAP experiment showing recovery of a GFP-labeled protein confined

to the membrane of the endoplasmic reticulum

The boxed region is photobleached at time instant, t = 0.

In subsequent frames, fluorescent molecules from

elsewhere in the cell diffuse into the bleached region.

● Fluorescence correlation spectroscopy

(1) Measure the fluorescence intensity in a small region of the cell as a

function of time

(2) By analyzing the temporal fluctuations of the intensity through the use

of time-dependent correlation functions, the diffusion constant and

other characteristics of the molecular motion can be uncovered

● Single-particle tracking

Record trajectory {x(t),y(t),z(t)} of a particle Calculate Diffusion constant

§11.2 Diffusive dynamics

● 1D random walk

0

L L L L L L L L L L L L L L L x

L---Length of each step

xn---position after the n-th step

x0=0---start point

knL---displacement of the n-th step with P(k

n=1)=P(k

n=-1)=1/2

xn= xn−1k n L

Random Walk (Redux)

⇒⟨ xN2⟩=NL2 ⇔⟨ xt

2⟩=2 Dt

t=N t , D=L2/2 t

Diffusion & Random Walk

⟨ x t2⟩=lim

M∞

1M∑i=1

M

x i2t =2 Dt

● What's meaning of the ‹...›

0

L L L L L L L L L L L L L L L x

0

L L L L L L L L L L L L L L L x

0

L L L L L L L L L L L L L L L x

...

Particle 1

Particle 2

...

Particle M

Equivalent to release simultaneously M particles at origin and then let them do independently random walk (or diffuse to other places)

● Fick's law

● Large number particles● Uniform in y, z direction● Nonuniform in x direction● Jump distance L per time Δt● Same jump probability to left and right

Assumption--

N(x,t): the number of particles in the box between x-L/2 and x+L/2 at time t. (1/2)N(x,t) particles will pass through x-L/2 to the left in Δt.

Simultaneously, (1/2)N(x-L,t) particles pass through x-L/2 to the right.

Flux

j=

12[N x−L , t −N x , t ]

t×A=−D

∂c∂ x

Density of number

c x , t =N x ,t

A×LD=L2

/2 t

● Continuity equation

Question: why is there a flux since

each particle has the same jump

probability to left and right?

Reason for net flow: there are

more particles in one slot than in

the neighboring one due to

nonuniform.

N x , t =[ jx− L2 − jxL

2 ]A t

=−∂ j∂ x

L×A× t ⇒

∂c∂ t=−

∂ j∂ x

● Diffusion equation and its solution

j=−D∂c∂ x

∂c∂ t=−

∂ j∂ x

∂c∂ t=D

∂2c∂ x2

(Diffusion equation)

Pulse solution

∂c∂ t=D

∂2c∂ x2

c x , 0=x

(Diffusion equation)

(initial condition)

c x , t =1

4Dte−x2

/4 Dt

Problem: Does the solution satisfy the diffusion equation and initial condition?

-6 -4 -2 2 4 6

0.1

0.2

0.3

0.4

0.5

x

c(x,t)

t=0.25/D

t=1/D

t=10/D

● Fluctuation

Molecular thermal motion (random) ==> Diffusion equation

Question: Why is diffusion equation deterministic?

Answer: Average on ensemble (collection of large number repeated systems).

Available to describe a system of large number particles.

Finite particle system: its behavior deviated from the prediction of diffusion equation. This deviation is called statistical fluctuation.

Few particle system: Statistical fluctuations will be significant, and the system's evolution really will appear random, not deterministic.

Nanoworld of single molecules: need to take fluctuations seriously

Einstein relation● A ball in viscous fluid (macroscopic analysis)

fext

ffriction

f friction=v

a= f ext− f friction/m= f ext−v /m

i :v f ext /⇒a0⇒ v⇒ v f ext /

ii: v f ext /⇒a0⇒ v ⇒ v f ext /t

v

vdrift≡ f ext / ⇔ f ext= f friction

The process to reach the drift velocity in micrometer scale

dv /dt=a= f ext− v/m=vdrift−v/m⇒

v t =vdrift[v 0−vdrift ]e− t /m

The larger ζ/m, v reaches more quickly to vdrift

Stokes formula: =6 R η--viscosity

ηwater

≈10-3kg m-1s-1R

pollen≈ 1μm

c≡m/≃10−7 s

2 3 4 5

0.05

0.1

0.15

0.2

0.25

0.3

0.35

t/τc

e−t / c

v→vdrift

very quickly such that we can omit the process!

● A ball in viscous fluid (microscopic analysis)

fext

Assumption:(1) The collisions occur exactly once per Δt(2) In between kicks, the ball is free of random

influences, so a=fext

/m

(3) v0 is the starting value just after a kick

(4) Each collision obliterates (抹去 ) all memory of the previous step

1−3⇒ x=v0x t1/2a t 2=v0x t1/2 f ext /m t 2

4⇒ ⟨v0x ⟩=0

vdrift≡⟨ x ⟩ t

=f ext

=2 m / t

● Einstein relation

=2 m / t

D=L2/2 t

D=m L t

2

In our discussion, we have confined L and ∆t to be constant. In fact, they are also stochastic quantities. Strict derivation gives

D=m⟨ L t

2

⟩=m⟨vx2⟩

12

m⟨vx2⟩=

12

k B TD=k B T

HypothesisHeat is disordered molecular motion

Testable predictionEinstein relation!

● Mean field thought---more rigorous treatment

The collisions of water molecules are simplified as a white noise (force)

m dvx /dt=−v xt

x ⟨ t ⟩=0, ⟨ t t ' ⟩=g t−t '

Newton's law==>

Solution:

v x=v0 e−t / c

1m∫0

te− t−s / c s ds , c≡m/

x t =∫0

tvx dt=v0 c[1−e

−t / c ]1∫0

t[1−e

− t−s / c] sds

(Langevin equation)

v x2=v0

2 e−2 t / c1

m2 [∫0

te− t−s / c s ds ]

2

2 v0

me−t / c∫0

te− t− s/ c sds

⟨vx2⟩=⟨v0

2⟩e−2 t / c

1

m2 ⟨[∫0

te− t−s / c s ds ]

2

⟩

=g

2 m⟨v0

2⟩−

g2m e−2 t / c

Assume that v0 is independent of Γ(s): ⟨v0 s⟩=⟨v0⟩ ⟨ s⟩=0

The observed time scale t≫ c

⟨vx2⟩=g / 2 m

12

m⟨vx2⟩=

12

k B Tg /2=k B T

x2t =v02 c

2[1−e−t / c]2

1

2 {∫0

t[1−e

−t−s / c ] s ds}2

2v0 c

[1−e

−t / c ]∫0

t[1−e

− t−s / c ] sds

⟨ x2t ⟩=⟨v0

2⟩c

2[1−e

−t / c ]2

1

2 ⟨{∫0

t[1−e

− t− s/ c] sds}2

⟩=⟨v0

2⟩−

g2 m c

2[1−e−t / c]

2

g

2 [tc 1−e−t / c]

The observed time scale t≫ c

⟨ x2t ⟩=g /2

t

⇒D=g /22

g /2=k B T

≡2 Dt⇒

D=k B T

Diffusion equation in a potential

● Revised Fick's law

● Large number particles● Uniform in y, z direction● Nonuniform in x direction● Jump distance L per time Δt● Jump probability to left ≠ right

Assumption--

x

h(x)P (k x=1)≠P(k x=−1)

kx=1 step right, k

x= -1 step left

x

h(x)

Heuristic views:

(1) h'(x)=0, i.e., h(x)=const.

xn= xn−1k n L

P(kn=1)=P(k

n=-1)=1/2

(2) h'(x)≠0, i.e., h(x)≠const.

xn= xn−1k n L P k n=12{1− f [ k n L h' xn−1]}

f : odd function , mono−increasing

Assume the step L is small enough, then up to the linear term,

P k n=12[1− k n L h' xn−1]

How many particles will pass through

x-L/2 to the left in Δt?

Simultaneously, How many particles will

pass through x-L/2 to the right?

Pk x=−1N x , t =12[1 L h ' x ]N x , t

Pk x−L=1N x−L , t =12[1− L h ' x−L]N x−L , t

Keep the leading term, we have

j=Pk x−L=1N x−L , t −P k x=−1N x , t

A t=−D

∂c∂ x−2 D h ' x c

● Diffusion equation in a potentialj=−D

∂c∂ x−2D h ' xc

∂c∂ t=−

∂ j∂ x

Continuity equation unchanged

∂c∂ t=D[ ∂

2 c

∂ x22

∂h ' c∂ x ]

● Determine γ

Equilibrium state, no flux:

j=−D∂c∂ x−2D h ' xc=0

⇒ cx =const.×e−2 h x

∂c∂ t=−

∂ j∂ x=0

We have known, in equilibrium state, Boltzmann distribution

c x=const.×e−

h xk BT

Compare both distributions, we obtain =1/2k B T

j=−D[ ∂c∂ x

h' xck B T ]

∂c∂ t=D[ ∂

2 c

∂ x2

1k B T

∂h' c∂ x ]

Pk x=±1=12 [1− k x L h ' x

2k BT ] Smoluchowski equaiton

c(x)

§3.6 Biological applications of diffusion

Limit on bacterial metabolism● Idealized model

Lake

R

bacterium

c R=0

O2 is dissolved in the water, with

a concentration c0:

Bacterium immediately gobbles

up (吞噬 ) O2 at its surface:

c ∞=c0

Problem: Find the concentration profile c(r)

and the maximum consumed number of O2 per time

● Solution

R0

r

I(r): the number of O2 per time passing

through the fictitious (假想的 ) spherical shell

Assume: quasi-steady state, oxygen does not pile up (累积 ) anywhere:

I r =I independent of r

Flux of O2: j r =−I /4 r 2

Fick's law: j r =−D dc /drBCs: c ∞=c0c R=0 &

c r =c01−R/r

I=4Dc0 R

maximum consumed number of O2 per time

● Limit on the size of the bacterium

The number of O2 per time that a living body needs

Q̇∝M 2 /3∝R2

Metabolic rate (ref. Lecture 4)

The maximum consumed number of O2 per time

I=4Dc0 R

Q̇≤I

RmaxR

Q̇

I

R≤Rmax

There is an upper limit to the size of a bacterium!

Membrane potentials● Drift and diffusion: ionic solution under E-field

E=−V /l f drift=qE=−qV /l

vdrift= f drift /=−qV /l

jdrift=c vdrift=−Dq Vlk B T c

V + +

++ +

x

D=k B Tvdrift=−

Dq Vlk B T

jdiffusion=−D∂c∂ x

j= jdrift jdiffusion

j=−D[ ∂c∂ x qV

lk B T c](Nernst-Planck formula)

● Equilibrium state j=0 at V=V eq

j=−D[ ∂c∂ x qV eq

lk B T c]=0⇒d cd x=− q V eq

lk BT c

⇒d ln cd x

=−q V eq

lk B T⇒ ln c

l=−

qV eq

lk BT

ln c=−q V eq

k BT(Nernst relation)

Membrane potentials maintains a concentration jump in equilibrium!

Discuss: Nernst relation and Boltzmann distribution.

FRAP● 1D model for FRAP

∂c∂ t=D

∂2 c

∂ x2

Initial conditions

Governing equation

Boundary conditions

c(x,0)=

c0 , (-1<x<-1/2)

c0 , (1/2<x<1)

0 , (-1/2<x<1/2)

c

-1 1-0.5 0.5x

t=0, just after photobleaching

c0

Cell w

all

Cell w

all

∂ c∂ x∣x=−1

=∂c∂ x∣x=1

=0 the flux of fluorescent molecules vanishes at the boundaries of the one-dimensional cell

● Solutionsc x ,t =c0[ 12−∑n=1

∞2 −1n−1

ne−D n22 t cos n x ]

N f t =∫−1/2

1/2c x ,t dx=

c0

2 [1−∑n=1

∞8

n2

2e−Dn2 2 t] recovery curve

2Nf/c

0

§. Summary & further reading

● Diffusion & Random walks

● Friction and diffusion

– Einstein relation

– Fick's law

– Diffusion equation

D=k B T

∂c∂ t=D[ ∂

2 c

∂ x2

1k B T

∂h' c∂ x ]

j=−D[ ∂c∂ x

h' xck B T ]

⟨ x t2⟩=2 Dt

Summary

● Biological applications of diffusion– Bacterial metabolism

– Permeability of membranes, membrane potentials

– Electrical conductivity of solutions

– FRAP

Further reading

● Reichl, A modern course in statistical physics● Phillips, Physical biology of the cell, Ch13● Nelson, Biological physics, Ch4