Lecture 11.
Diffusion in biological systems
Zhanchun Tu ( 涂展春 )
Department of Physics, BNU
Email: [email protected]
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 region
by 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
Random Walk (Redux)
●
1D random walk
L L L L L L L L L L L L L L L
0
L---Length of each step
x
x0=0---start point
xn---position after the n-th step
knL---displacement of the n-th step with P(kn=1)=P(kn=-1)=1/2
x n= xn −1 k n L
2
N
2
t
2
⇒ 〈 x 〉= NL ⇔ 〈 x 〉=2 Dt
2
t=N  t , D=L /2  t
●
Diffusion & Random Walk
What's meaning of the ‹...›
Particle 1
L L L L L L L L L L L L L L L
Particle 2
L L L L L L L L L L L L L L L
0
0
x
...
...
Particle M
x
L L L L L L L L L L L L L L L
0
x
M
1
2
2
〈 x t 〉=lim
x
∑
i t=2 Dt
M ∞ M i =1
Equivalent to release simultaneously M particles at origin and then
let them do independently random walk (or diffuse to other places)
Fick's law
●
Assumption-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
●
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.
Density of number
c  x , t =
N  x ,t 
A×L
Flux
1
[N  x−L , t −N  x , t ]
2
∂c
j=
=−D
 t× A
∂x
2
D=L /2  t
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.
Continuity equation
[    ]
L
L
 N  x , t= j x− − j x
2
2
∂j
=−
L×A× t
∂x
∂c
∂j
=−
∂t
∂x
⇒
A t
●
Diffusion equation and its solution
∂c
j =−D
∂x
∂c
∂j
=−
∂t
∂x
2
∂c
∂ c
=D 2
∂t
∂x
(Diffusion equation)
Pulse solution
2
∂c
∂ c
=D 2 (Diffusion equation)
∂t
∂x
c  x , 0= x  (initial condition)
1
−x /4 Dt
c  x , t=
e
 4  Dt
2
c(x,t)
0.5
0.4
Problem: Does the solution satisfy the
diffusion equation and initial condition?
t=0.25/D
0.3
0.2
0.1
-6
-4
-2
t=1/D
t=10/D
2
4
6
x
●
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)
ffriction
f
fext
friction
=v
a= f ext − f
v
friction
/ m= f ext −v /m
i:v f ext / ⇒ a0 ⇒ v  ⇒ v  f ext / 
t
ii: v f ext /⇒ a0 ⇒ v  ⇒ v  f ext / 
v drift ≡ f ext / ⇔ f ext = f
friction
The process to reach the drift velocity in micrometer scale
dv /dt =a= f ext − v/ m=v drift −v/ m⇒
− t /m
v t =v drift [v 0−v drift ]e
The larger ζ/m, v reaches more quickly to vdrift
Stokes formula:
-3
-1 -1
ηwater≈10 kg m s
=6   R
Rpollen≈ 1μm
η--viscosity
−t / c
e
0.35
0.3
0.25
−7
 c ≡m/≃10 s
0.2
0.15
0.1
0.05
2
3
v→vdrift very quickly such that we can omit the process!
4
5
t/τc
●
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
2
1−3⇒ x=v 0x  t1/2 a  t =v0x  t1/2 f ext / m t
4⇒ 〈 v0x 〉=0
〈 x 〉 f ext
v drift ≡
=

t
=2 m/ t
2
●
Einstein relation
2
D=L /2  t
=2 m/ t
2
 
L
 D=m
t
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〈 v 2x 〉
1
1
2
m〈 v x 〉= k B T
2
2
 D=k B T
Hypothesis
Heat is disordered molecular motion
!
Testable prediction
Einstein relation
●
Mean field thought---more rigorous treatment
The collisions of water molecules are
simplified as a white noise (force)
x
〈  t〉=0, 〈  t t ' 〉=g t −t ' 
m dv x / dt =−v x  t 
Newton's law==>
Solution:
−t / c
v x =v 0 e
t
(Langevin equation)
1 t − t−s / 
 ∫0 e
 s ds ,  c ≡m/ 
m
c
−t / c
x t=∫0 v x dt=v 0  c [1−e
1 t
− t −s /
] ∫0 [1−e
]  s ds

c
2
x
2 −2 t / c
0
v =v e
1
 2
m
[∫ e
t
− t−s / c
0
2 v 0 −t / t −t− s/ 
 s ds 
e ∫0 e
  s ds
m
2
]
c
c
Assume that v0 is independent of Γ(s): 〈 v 0  s〉=〈 v 0 〉 〈   s〉=0
1
2
2
−2 t /
〈 v x 〉=〈 v 0 〉 e
 2
m
c
=
〈
2
− t−s / 
e
 s ds ]
∫
[0
t

c

〉
g
g
−2 t / 
2
 〈 v 0 〉−
e
2 m
2 m
The observed time scale
2
x
〈 v 〉=g / 2 m
1
1
2
m〈 v x 〉= k B T
2
2
c
t≫ c
g /2 =k B T
2
2
0
2
1 t
−t −s /
]  2 {∫0 [1−e
] s ds }

−t / c 2
2
c
x t=v  [1−e
c
t
v0 c
−t /
− t−s / 
2
[1−e
]∫0 [1−e
]  s ds

c
2
2
0
2
c
1
] 2

−t / c 2
〈 x t〉=〈 v 〉  [1−e

= 〈 v 20 〉−
c

− t− s/  c
0
}〉
2
]  s ds
g
g
 2c [1−e−t /  ] 2 2 [ tc 1−e−t /  ]
2 m

c
c
t≫ c
The observed time scale
2
〈{∫ [1−e
t
2
〈 x t〉= g / t ≡2 Dt ⇒
⇒ D=g /2
2
g / 2 =k B T
 D=k B T
Diffusion equation in a potential
Revised Fick's law
●
Assumption-Large number particles
● Uniform in y, z direction
● Nonuniform in x direction
● Jump distance L per time Δt
● Jump probability to left ≠ right
●
P (k x =1)≠ P( k x =−1)
kx=1 step right, kx= -1 step left
h(x)
x
Heuristic views:
(1) h'(x)=0, i.e., h(x)=const.
h(x)
x n= xn −1 k n L
P(kn=1)=P(kn=-1)=1/2
x
(2) h'(x)≠0, i.e., h(x)≠const.
x n= xn −1 k n L
1
P  k n = {1− f [ k n L h'  x n−1 ]}
2
f : odd function , mono−increasing
Assume the step L is small enough, then up to the linear term,
1
P k n = [1− k n L h'  x n−1 ]
2
How many particles will pass through
x-L/2 to the left in Δt?
1
P k x =−1 N  x , t= [1 L h '  x ] N  x , t
2
Simultaneously, How many particles will
pass through x-L/2 to the right?
1
P k x−L =1 N  x−L , t = [1− L h '  x− L] N  x−L , t 
2
Keep the leading term, we have
Pk x− L =1 N  x−L , t−P k x =−1 N  x , t 
∂c
j=
=−D
−2 D  h '  x c
A t
∂x
●
Diffusion equation in a potential
∂c
j =−D
−2 D  h '  x c
∂x
∂j
Continuity equation ∂c
=−
unchanged
∂t
∂x
[
∂c
∂2 c
∂h ' c
=D
2 
2
∂t
∂x
∂x
]
●
Determine γ
Equilibrium state, no flux:
∂c
j =−D
−2 D  h '  x c=0
∂x
−2  h  x
⇒ c x =const.×e
∂c
∂j
=− =0
∂t
∂x
We have known, in equilibrium state, Boltzmann distribution
−
c  x=const.×e
h  x
k BT
Compare both distributions, we obtain  =1/2 k B T
[
k x L h '  x
1
P k x =±1= 1−
2
2 k BT
[
∂c h'  xc
j =−D

∂x
kBT
]
]
Smoluchowski equaiton
[
2
∂c
∂ c
1 ∂ h' c
=D

2
∂t
∂ x kBT ∂ x
]
c(x)
§3.6 Biological applications
of diffusion
Limit on bacterial metabolism
●
Idealized model
O2 is dissolved in the water, with
Lake
bacterium
R
a concentration c0:
c ∞=c0
Bacterium immediately gobbles
up ( 吞噬 ) O2 at its surface:
c  R=0
Problem: Find the concentration profile c(r)
and the maximum consumed number of O2 per time
●
Solution
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:
0
r
R
I r =I
Flux of O2:
Fick's law:
BCs:
j r =−I /4  r
2
independent of r 
c r =c 0 1−R/r 
j r =−D dc / dr
c  R=0
&
c ∞=c0
I =4  Dc0 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
∝R
2
Metabolic rate (ref. Lecture 4)
The maximum consumed number of O2 per time
Q̇
I =4  Dc0 R
I
Q̇≤I
R≤Rmax
Rmax
R
There is an upper limit to the size of a bacterium!
Membrane potentials
●
V
Drift and diffusion: ionic solution under E-field
+
+
+
+
x
+
E =− V /l
f drift =qE=−q  V /l
v drift = f drift /=−q  V /l 
 D=k B T


Dq  V
v drift =−
lk B T
Dq  V
j drift =c v drift =−
c
lk B T
∂c
j diffusion =− D
∂x
j= j drift  j diffusion
[  ]
∂c
qV
j=−D

c
∂x
lk B T
(Nernst-Planck formula)
●
Equilibrium state
[  ]
j=0 at  V = V eq


q  V eq
q  V eq
∂c
dc
j=−D

c =0 ⇒
=−
c
∂x
lk B T
dx
lk B T
q  V eq  ln c 
q  V eq
d ln c
⇒
=−
⇒
=−
dx
lk B T
l
lk B T
q  V eq
 ln c=−
k BT
(Nernst relation)
Membrane potentials maintains a concentration jump in equilibrium!
Discuss: Nernst relation and Boltzmann distribution.
FRAP
●
1D model for FRAP
c
Governing equation
Cell wall
c0
Cell wall
∂c
∂2 c
=D 2
∂t
∂x
Initial conditions
c0 , (-1<x<-1/2)
c(x,0)=
0 , (-1/2<x<1/2)
-1
-0.5
0.5
1
x
t=0, just after photobleaching
c0 , (1/2<x<1)
Boundary conditions
∂c
∂x
∣
x=−1
=
∂c
∂x
∣
x=1
=0
the flux of fluorescent molecules vanishes at
the boundaries of the one-dimensional cell
Solutions
[
∞
1
2 −1n−1 −D n  t
c  x ,t =c 0 −∑
e
cos n  x
2 n=1
n
2
[
2
∞
c0
8 −D n 
N f t =∫−1/ 2 c  x ,t dx=
1−∑ 2 2 e
2
n =1 n 
1/ 2
2Nf/c0
●
2
2
t
]
]
recovery curve
§. Summary & further reading
Summary
●
Diffusion & Random walks
●
Friction and diffusion
–
Einstein relation  D=k B T
–
Fick's law
–
●
j=−D
[
∂c h'  xc

∂x
kBT
Diffusion equation
2
t
〈 x 〉=2 Dt
]
[
2
∂c
∂ c
1 ∂ h' c
=D

∂t
∂ x2 k B T ∂ x
]
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