lecture 14 membranes continued diffusion membrane transport
Post on 22-Dec-2015
223 views
TRANSCRIPT
From S. Feller
Lipid Bilayers are dynamic
distributions of phosphate and carbonyl groups and lateral pressure profiles
surface pressure
= 70 dyne/cm
compressed monolayer
surface pressure of the crowding surfactant balances part of the surface tension, thus the apparent surface tension to the left of the barrier is smaller
Lipids at air-water interface
Irving Langmuir
w- surf = surf
surf
w - surface tension of pure water
surf - surface tension in the
presence of surfactant
surf
surf – surface pressure of the surfactant
dipalmitoyl phosphatidylcholine (DPPC)
monolayer-bilyer equivalence pressure 35-40 dyn/cm
Differential Scanning Calorimeter (DSC): Phase transition for DPPC (Dipalmitoyl phosphatidylcholine)
http://employees.csbsju.edu/hjakubowski/classes/ch331/lipidstruct/oldynamicves.html
For DOPC (oleyl)…-18°C
For DPPC (palmytoyl)…+41°C
S = H/Tm
Biochim Biophys Acta. 2005 Dec 30;1746(3):172-85.
DOPC/DPPC
POPC…palmitoyl, oleyl
http://www.nature.com/emboj/journal/v24/n8/full/7600631a.html
Phospholipid/ganglioside
Lateral Phase Separation
Diffusion is a result of random motion which simply maximizes entropy
Einstein treatment:
c1 c2
l l
butl
CC
dx
dc 12
C
distance
negative slope
therefore: butdx
dcDJ net (Fick’s law)
dx
dc
t
lJnet
2
2
1
tlCJ /2
11 tlCJ /
2
12
tlCCJnet /)(2
121
Dtl 22 Dtl 2 (one dimension)
Diffusion = random walkti
me
X, distance
2
2
x
cD
t
c
Diffusionequation
x
cDJ
Fick’s law
flux gradient
rate
Dt
x
Dttxp
4exp
4
1),(
2
Dt22 Variance
2
21exp
2
1)(
x
xp
Normal distribution Random walk in one dimension
D = diffusion coefficientt = time 0.06 0.04 0.02 0 0.02 0.04 0.06
0
20
40
60
80
100
p1 x( )
p2 x( )
p3 x( )
x, cm
t = 1 s
t = 10 s
t = 100 s
D = 10-5 cm2/s
Dt2
root-mean-square (standard)deviation
x = deviation from the origin
Dtx 2
Replace:
where
0 1 2 3 40
0.5
x,
1.0
0.607
area inside 1 = 0.68
If we step 1 sigma () away from the origin, what do we see?
conce
ntr
ati
on
observer
Dt
x
Dttp
4exp
4
1)(
2
Dt
x
Dtxp
4exp
4
1)(
2
x = x1, x2, x3t = t1, t2, t3
t, s
0 0.005 0.01 0.015 0.02 0.0250
20
40
60
80
100
x, cm
= 0.0045 cm
= 0.014 cm
= 0.045 cm
Dtx 2t1 = 1 s
t2 = 10 s
t3 = 100 s
An observer sees that the
concentration first increases and then
decreases
1 is a special point where the concentration of the diffusible substance reaches its maximum
0 20 40 60 80 1000
10
20
30
40
50
60
t = 1 s
t = 10 s
t = 100 s
x = 0.0045 cm
x = 0.014 cm
x = 0.045
D = 10-5 cm2/s