Transcript Introduction and review of Matlab

Discussion topic for week 6 : Nerve impulses &
ion channels
•
Potassium channels conduct the K+ ions but reject the smaller
size Na+ ions with a selectivity ratio exceeding 1/1000.
How do they achieve this feat?
Problem of nerve impulses (Nelson, chap. 11, 12)
How do we send signals from brain to muscles in milliseconds?
The axons, which are the pathways for signals, form leaky cables in a
conducting environment (salt solution). Thus compared to a copper wire,
sending electric signals across axons is a difficult problem, requiring a
novel solution.
Besides the leaky cable problem, ions in a salt solution move very slowly
due to small D, which greatly reduces the signal transmission speed
<x2> = 2Dt, with D~10-9 m2/s, and L = 1 m, gives
t ~ 5 x 108 s ~ 16 years!
Also ion concentration and energy are quickly dissipated (cf. pulse solution).
Even if there were an applied potential difference between the neurons and
muscles (solving the latter problem), it wouldn’t help much with the speed
vd 
f

 f
D
, ( D  kT , Einstein relation)
kT
eV D eV D


 40  10 9 m/s for 1 eV  40 kT
L kT kT L
L kT L2
t

 2.5  10 7 s  1 year
vd eV D
Action potential basics:
Diffusion limits the distance scale of signal propagation to μm or less.
The only obvious place where diffusion of ions could make an observable
difference is across membranes - ion flow could change the potential
difference across the axon membrane.
Experimental facts:
1. Na+ concentration is high outside cells and low inside, and vice versa
for K+ ions.
2. There are channels on the membrane that, when open, selectively
conduct either Na+ or K+ ions.
3. There are ion pumps (called sodium-potassium pump) on the
membrane that help to maintain this concentration difference.
In each cycle, Na-K pump uses 1 ATP molecule to pump 3 Na+ ions
out of the cell and 2 K+ ions in to the cell.
Maintenance and propagation of the action potential:
1. Change of membrane voltage opens the sodium channels.
2. Na+ ions flow into the cell, collapsing the membrane potential (−60 mV).
3. This triggers the opening of the potassium channels, while the sodium
channels shut down stochastically.
4. K+ ions flow outside the cell, restoring the membrane potential.
The potassium channels shut down, returning the system to step 1.
5. This process is repeated along the axon, which propagates the action
potential.
out
in
Lessons from squid giant axon:
Squid giant axon played a critical role in understanding nerve signals.
It has a diameter 1 mm which is 100 times larger than a typical axon’s.
Cells have been known to maintain a potential difference with outside for
a long time (Galvani vs Volta, ~1800). What is the source of this ΔV?
Two observations:
1. Cells are electro-neutral
but have more KCl inside
2. K+ is more permeable
than ClThus K+ will leak out of the
cell until Nernst equilibrium
is reached
Concentration profiles
of K+ and Cl ions
across a cell membrane
(assuming only K+ is
permeable)
Corresponding
electrostatic potential
(from PB eqn.)
Nernst potential:
V  V2  V1
 VNernst  
kT c2
ln
ze c1
Realistic case:
Inside the cells there are negatively charged impermeant macromolecules
whose charge density is equivalent to c_ = 125 mM of excess electrons.
The three major ions, Na+, K+ and Cl have the concentrations outside (1):
c1Na = 140 mM, c1K = 10 mM, c1Cl = 150 mM
From electro-neutrality, the concentrations inside (2) must satisfy
c2Na + c2K - c2Cl – c_ = 0
In equilibrium, each permeant species must separately satisfy the Nernst
relation with the same potential difference V
kT c2 Na
kT c2 K kT c2Cl
V   ln
  ln

ln
e
c1Na
e
c1K
e
c1Cl
c1Na c1K c2Cl


c2 Na c2 K c1Cl
(Gibbs-Donnan relations)
Let
x  c2 Na 1 M
c2 K  c1K
c2 Na
c1Na

c2Cl  c1Cl
c1Na
c2 Na

c2 K 0.01

x
1 M 0.14
c2Cl
1
 0.15  0.14
1M
x
Substitute in the electro-neutrality equation
1
0.021
15 2
x x
 0.125  0 
x  0.125 x  0.021  0  x  0.21
14
x
14
Inside: c2Na = 210 mM, c2K = 15 mM, c1Cl = 100 mM
Donnan potential: V  
kT 210
ln
 25  0.4  10 mV
e 140
(Vobs  60 mV )
Observed (expected) concentrations in squid giant axon (in mM):
c2 (in)
c1 (out)
Vnernst (mV)
K
400 (15)
20 (10)
75 (-10)
Na
50 (210)
440 (140)
54
“
Cl
52 (100)
560 (150)
59
“
All the potentials are different! Hence the cell is not in equilibrium.
Na concentration and voltage differs most from the Donnan equilibrium.
Ion pumps in membranes actively transport Na+ out and K+ in, and thus
maintain this imbalance in concentrations.
In one cycle, the Na-K pump hydrolizes one ATP molecule moving 3 Na+
ions out and 2 K+ ions in. Work done for each ion:
W(Na+) = e (60 + 54) = 114 meV, W(K+) = e (-60 + 75) = 15 meV
Thus the total work done is W = 3 x 114 + 2 x 15 = 372 meV = 15 kT
Cf. ATP hydrolysis liberates 19 kT, so only 4 kT is lost as heat.
Experimental demonstration of the active ion pumps in the membrane
Flux of Na ions out of an axon
stops when toxins are introduced
Toxins block the pump stopping
the transport of Na outside.
(Hodgkin-Keynes, 1955)
The rate of ATP hydrolysis catalyzed
by the Na-K pump as a function of the
interior Na and exterior K concentration.
Lack of either stops ATP consumption.
(Skou, 1957; Nobel 1997)
Crystal structure of potassium channel (MacKinnon, 1998; Nobel 2003).
Reveals the mechanisms of selectivity and voltage gating
Selectivity filter has the
right size to bind the
K+ ions but too large
for the smaller Na+ ions
(More details in the video)
Voltage gated
ion channels:
life’s transistors
Crystal structure of sodium-potassium pump (Poul Nissen et al. Dec. 2007)
F0-F1 ATPase: a molecular rotor in mitochondria
12 nm
Exploits the proton gradient to
synthesize ATP.
The work done by transporting
3 protons across the

11.4 nm
d



membrane is converted to
chemical energy by
3.3 nm
synthesizing ATP.
8.2 nm
b

a c12
9.2 nm
Nerve impulses:
Response of axons to a weak stimulus:
Injecting positive charges in an axon changes the membrane voltage to a
more positive value (depolarization). This stimulus spreads along the axon
like a pulse solution—its amplitude is dissipated within a few cm.
Response of axons to a strong stimulus: Action potential
Unless the stimulus is strong enough to change the membrane potential by
about 10 mV, it dies down. Above that threshold, it triggers an action
potential which propagates along the axon without any loss in amplitude.
Time course of an
action potential
Showing how the
membrane potential
and the corresponding
membrane current
change in time.
• An initial stimulus
opens Na channels
• 1-3 inward Na current
• K channels open;
an outward K current
gradually drives the
potential back to the
resting potential
Trigger of action potential in
giant squid axon:
a) A stimulus of 56 mV
depolarizing potential is applied
b) Total membrane current
c) Inward Na current
d) Outward K current
Note that the K current starts
flowing when the Na current
(and the membrane potential)
is at a maximum.
Both the Na and K channels
open in response to changes in
the membrane potential
(analogous to transistors).
Modeling action potential:
Analogy with electrical circuits
Representing each type of current with a different circuit element,
we can write for each one:
V  V2  V1  I i Ri  Vi Nernst
Recall that VNernst > ΔV for Na+ and VNernst < ΔV for K+
Capacitive current:
Ic 
dq
dV
C
dt
dt

jc 
I c C dV
dV

 CA
A A dt
dt
Cable model
Cable equation
dV

I x ( x)  I x ( x  dx)   jr  C A
dt

dI x
dV 

 2a jr  C A

dx
dt 


2a dx

(*)
V ( x  dx / 2)  V ( x  dx / 2)   I x dRx
x
x
Rx   
A a 2
I x  a 2
a 
2
d 2V
dx 2
dV
dx

dRx 
dx
a 2
(3 W1m1 conductiv.
of axoplasm)
Diff. wrt to x and substitute in (*)
dV 

 2a jr  C A

dt


If we assume Ohm’s law for radial conductance:
jr  (V  V 0 ) g tot 
  jr  vg tot
v  V ( x, t )  V 0 
a 
2
( Rr  1 / G  1 / Agtot )
d 2v
dv 


2

a
g
v

C
 tot

A
2
dt


dx
C A dv 
a d 2 v 



v


2
2 g tot dx
g tot dt 

2axon
d 2v
dv

v
2
dt
dx
axon 
a
2 g tot
(~1 cm)
Linear cable equation

CA
g tot
(~2 ms)
Axon’s space and time const.
Solution of the cable equation:
v( x, t )  e t  w( x, t )
2axone t 
d 2w
dx 2
 e t 
dw 2axon d 2 w

dt
 dx 2
v( x, t )  ce
t 
e
dw  t 
 e w  e t  w
dt 
2axon
Diffusion equation with D 

 x 2 4 Dt
4Dt
Decaying pulse solution
The decaying pulse solution is degraded quickly.
Need to give up linearity of g(Na) to sustain the pulse
0
g Na  g Na
 Bv 2
Solution of the linear cable equation compared to the pulse solutions
c=1
1 cm
D=0.05 m2/s
4cm
Solid lines: linear cable equation plotted at x = 1, 2, 3, 4 cm
Dashed lines: pulse solutions for the same parameters
The radial current density is modified to




Nernst
jr   V  Vi Nernst g i0  V  VNa
Bv 2
i


0
Nernst
 (V  V 0 ) g tot
 V  V0  V0  VNa
Bv 2
0
 vg tot
 (v  H ) Bv 2
1
0
jr  0  v  0  v1, 2  H  H 2  4 g tot
B
2


Non-linear cable equation:
2axon
d 2v
dx 2

dv
1

v(v  v1 )(v  v2 )
dt v1v2
Leads to a traveling wave solution with velocity: vaxon
axon 2v1  v2 

  1

v2  v1 
How do neurons communicate?
Structure of GltPh (Glutamate transporter)
from Pyrococcus horikoshii