Transcript No Slide Title

Transfer of charged molecules
Electric field does work on ion as it passes
between solutions.
[Na+](1)
[Cl-](1)
f1
DµA = µA(solution 2) - µA(solution 1) + ZFV
[Na+](2)
[Cl-](2)
f2
V = (f2 - f1) potential difference in volts between the two
solutions.
F = Faraday = 96,485 (eV-1) = charge of mole of electrons.
Z = charge of the ion (e.g. ±1)
2
w   f dx
f  ZFE  ZF
2
w   ZF
1
f
x
f
dx  ZFf 2  f1 
x
Note: this is true for any f(x). Convenient to
write:
µA,tot = µA + ZFf
Electrical Potential (f)
1
f1
f
E
x
f2
Position (x)
Electrostatics in Water
Counter charges
(in solution)
Charged surface
s = charge density
+
+
+
+
+
+
- - - - - -
y
y = f(s)
Potential
Chemical potentials of molecules on the surface are
influenced by the surface potential, y.
Gouy-Chapman Theory
c  x 
zey 


 kT 
Ie
c  x 
zey 


 kT 
Ie

 2y  ze 


c

x


c
 x
2

x
w


e z  e z
sinh z  
2
 x 


zey 0  
4kT 
1 
LD 
y  x 
tanh
tanh
e
 ze 
 4kT 




 2y 2zeI 
zey 



sinh
 kT 
x 2   w 
+
+ L
+ + + +
- - - - - -
D
LD 
y0
kT w
2Iz 2 e 2
Debye length
Potential
LD 
10
nm
I
where I is givin in mM
Surface Potential
y
s
 E 
x
w
at x  0
1
+
+ L
+ + + +
- - - - - -
8IkT  2
y
zey 
 sinh 
 
2kT 
x
  w 
D
s
y0
Potential
szeL D 
2kT 
1 


y 0 s  
sinh
 ze 
2kT w 
Gouy Equation
 x 


zey 0  
4kT 
1 
LD 
y  x 
tanh
tanh
e
 ze 
 4kT 




Linearized Poisson-Boltzmann Equation
 2y 2zeI 
zey 



sinh
 kT 
x 2   w 
y
x
 E 
x 0
s
w
2
 2y 
1 

 y

2
LD 
x
y  x  y 0
y
x
 
y 0
LD
y 0 s  
sLD
w
x 0
 xL
D
e
+
+ L
+ + + +
- - - - - -
Good for y0 ≤ 25 mV
D
s

y0
Potential
LD 
LD 
kT w
2Iz 2 e 2
10
nm
I
Debye length
where I is givin in mM
Mobility and Chemical Potential
• Molecular motion and transport disucssed in TSWP Ch. 6. We can
address this using chemical potentials.
• Consider electrophoresis:
– We apply an electric field to a charged molecule in water.
– The molecule experiences a force (F = qE)
– It moves with a constant velocity (drift velocity = u)
• It obtains a speed such that the drag exactly opposes the electrophoretic force. No
acceleration = steady motion.
– u = qE·1/f where f is a frictional coefficient with units of kg s-1.
• qE is a force with units of kg m s-2 giving u with the expected ms-1 velocity units.
– Think of (1/f) as a mobility coefficient, sometimes written as µ.
– u = mobility · force
– We can determine the mobility by applying known force (qE) and
measuring the drift velocity, u.
Mobility and Chemical Potential
• Gradient of the chemical potential is a force.
• Think about gradient of electrical potential energy:
qy 
electrical potential energy ( q = charge, y = electrical potential)
qy  q
y
 qE where E is the electric field and qE is a FORCE
x
• Extending this to the total chemical potential:

 Force
x
1 
u   
Drift velocity obtained from gradient of the chemical potential
 f x
 
– where f is a frictional coefficient
– (1/f) as a mobility coefficient
Mobility and Chemical Potential: Example
+
    kTln c  zef
0
If concentration (c) is constant
througout:

f
 ze  qE
x
x
And the drift velocity is
u
1 
qE
  
f
 f x
+
Potential (f)
Write down chemical potential
as a function of position in this
electrophoresis:
-
E
E
f
x
Position (x)
What if c is not constant?
Can the entropy term give rise to an
effective force that drives motion?
This is diffusion, and we can derive
Fick’s Law (TSWP p. 269) from
chemical potentials in this way.
Brownian Motion
Brownian trajectory
1
• Each vertex represents measurement of position
“Random walk”
• Time intervals between measurements constant
µm
• After time (t) molecule moves distance (d)
t=0
• 2-dimensional diffusion:
<d2> = <x2> + <y2> = 4Dt
d
• 3-dimensional diffusion:
t = 0.5 s
0
0
µm
A lipid will diffuse around a 10 µm
diameter cell:
 10 
 4Dt  t  61s
 2 
2
1
<d2> = <x2> + <y2> + <z2> = 6Dt
• In cell membrane, free lipid diffusion:
D ~ 1 µm2/s
Diffusion: Fick’s First Law
1
• Jx = Flux in the x direction
µm
• Flux has units of #molecules / area
(e.g. mol/cm2)
• Brownian motion can lead to a net flux of
molecules in a given direction of the concentration
is not constant.
• Ficks First Law:
0
0
µm
1
c 
J x  D
x 
Derivation of Fick’s First Law from
Entropy of Mixing
1 x   10  kTln c1 x 
• Chemical potential of component 1 in mixture.
1   
u1  x     
 f  x 
• Net drift velocity (u) related to gradient of
chemical potential by mobility (1/f) where f is
frictional coefficient.
1   1 c1

   kT
 f  c1  x   x
1  c1
J1x  x   c1  x u1  x     kT
 f  x
1 
 kT  D
 f 
• Flux (J1x) is simply concentration times net drift.
• Einstein relation for the diffusion coefficient.
• Entropy is the driving force behind diffusion.
D  kT
where   mobility (not chemical potential)
Fick’s Second Law: The Diffusion Equation
• Consider a small region of space (volume for 3D,
area for 2D)
• Jx(x) molecules flow in and Jx(x+dx) molecules
flow out (per unit area or distance per unit time).
N 
 J x  x   J x  x  dx dy
 t 
N
J x  x
c  J x  x   J x  x  dx 

t 
dx
J x  x  dx 
J 
   x 
 x 
 2 c 
c 
 D 2 
t 
x 
since c  N / dxdy
definition of the derivative
c 
since J x   D
x 
Equilibrium Dialysis Example
At equilibrium:
O2(out) = O2(in, aq)
[Mb·O2]/[Mb][O2(aq)] = Keq
If we are able to asses the total ligand
concentration in the dialysis bag:
[O2(aq)] + [Mb·O2] = [O2 (in, total)]
Then [Mb·O2] = [O2 (in, total)] - [O2(aq)]
H2O (l)
O2(aq), N2(aq)
etc.
H2O (l)
O2(aq), N2(aq)
Mb(aq), Mb·O2(aq)
(these are measurable)
Can compute Keq.
If we have direct a probe for Mb·O2, then we
don’t need the dialysis, can read of
concentrations and compute Keq.
Dialysis can also be used to exchange solution
(eg. change [salt])
Semipermeable
membrane
(cellulose): allows water and
dissolved small solutes to pass,
blocks passage of large proteins
such as myoglobin (Mb)
Scatchard Equation
General version: M + A
Keq 
M·A
c M
Keq = [M·A]/([M][A])
c A  bound 
 c A (bound)c A  outside
Simplify by introducing n, the average number of ligand molecules (A)
bound to the macromolecule (M) at equilibrium:
n
n
 A
n
1 n A
 K eq 1 n 
For one ligand binding
site per macromolecule
Scatchard plot
NKeq
n
 A
 Keq N  n 
Scatchard equation
N independent binding
sites per macromolecule.
n/[A]
Keq 
c A bound 
cM
Slope = -Keq
n
N
Cooperative Binding
For a macromolecule with multiple binding sites, binding to one site can influence
binding properties of other sites.
Failure of data plotted in a Scatchard plot to give a straight line indicates
cooperative or anticooperative binding among binding sites.
Cooperative = binding of second ligand is made easier
Anticooperative = binding of second ligand is made more difficult
Hemoglobin is a favorite example of a protein with cooperative binding behavior.
Binds up to 4 O2
1
Cooperative: most O2 released in tissue
while binding O2 maximally in the lungs
Binding curve shows characteristic
sigmoidal shape
Myoglobin
f
p50 = 1.5 Torr
Hemoglobin
p50 = 16.6 Torr
f = fraction of sites bound
0
0
PO2(Torr)
40
Hill Plot
n N
f

 K eq A
1 n N  1 f 
Scatchard equation (non-cooperative
binding)
f
 K An
1 f 
For cooperative binding.
n = Hill coefficient
K = a constant, not the Keq for a single ligand
 f 
  n log  A  log K
log 
1 f 
log[f /(1-f)]
1.5
-1.5
-0.5
Slope of each line give the Hill cooperativity
coefficient.
Myoglobin
n = 1.0
Slope = 1
Slope = N
Hemoglobin
n = 2.8
log[P02] (Torr)
+2
no cooperativity
maximum (all-or-nothing)
cooperativity
See Example 5.4 (TSWP p. 204 - 207) for a detailed
study of Hemoglobin