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

year1 s

kT 40 eV 1 for m/s

relation Einstein

72

9

105.2

1040

),(,

DL

eVkT

vL

t

LD

kTeV

kTD

LeV

kTDkTD

ff

v

d

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 Cl-

Thus 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:

1

2

12

lnc

c

zekT

V

VVV

Nernst

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

Cl

Cl

K

K

Na

Na

Cl

Cl

K

K

Na

Na

c

c

cc

c

c

c

c

ekT

cc

ekT

c

c

ekT

V

1

2

2

1

2

1

1

2

1

2

1

2 lnlnln

(Gibbs-Donnan relations)

Let

Substitute in the electro-neutrality equation

Inside: c2Na = 210 mM, c2K = 15 mM, c1Cl = 100 mM

Donnan potential:

21.00021.0125.01415

0125.0021.0

141 2 xxx

xxx

xM

c

c

ccc

xM

c

c

ccc

Mcx

Cl

Na

NaClCl

K

Na

NaKK

Na

114.015.0

1

14.001.0

1

1

2

2

112

2

1

212

2

)60(

104.025140210

ln

mV

mV

obsV

ekT

V

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)

c12c12

bb

dd

aa

11.4 nm

8.2 nm

9.2 nm

12 nm

3.3 nm

Exploits the proton gradient to

synthesize ATP.

The work done by transporting

3 protons across the

membrane is converted to

chemical energy by

synthesizing ATP.

F0-F1 ATPase: a molecular rotor in mitochondria

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

dt

dVC

dt

dV

A

C

A

Ij

dt

dVC

dt

dqI A

ccc

Representing each type of current with a different circuit element,

we can write for each one:

Recall that VNernst > ΔV for Na+ and VNernst < ΔV for K+

Capacitive current:

Nernstiii VRIVVV 12

Cable model

Cable equation

dtdV

Cjadx

Vda

dxdV

aI

a

dxdR

a

xAx

R

dRIdxxVdxxV

dtdV

Cjadx

dI

dxadtdV

CjdxxIxI

Ar

x

xx

xx

Arx

Arxx

2

)2/()2/(

2

2)()(

2

22

2

22(m conductiv.

of axoplasm)

Diff. wrt to x and substitute in (*)

(*)

If we assume Ohm’s law for radial conductance:

tot

A

totaxon

axon

tot

A

tot

Atot

totrtotr

g

C

ga

vdtdv

dx

vd

dtdv

g

Cv

dx

vdga

dtdv

Cvgadx

vda

vgjVtxVv

gVVj

2

2

2

),(

)(

2

22

2

2

2

22

0

0

Linear cable equation

Axon’s space and time const.

(~1 cm) (~2 ms)

)/1/1( totr AgGR

Solution of the cable equation:

Dt

ecetxv

dx

wd

dt

dw

wewedt

dwe

dx

wde

txwetxv

Dtxt

axon

ttttaxon

t

4),(

),(),(

4

2

22

2

22

2

Diffusion equation with 2axonD

The decaying pulse solution is degraded quickly.

Need to give up linearity of g(Na) to sustain the pulse

Decaying pulse solution

20 Bvgg NaNa

Solid lines: linear cable equation plotted at x = 1, 2, 3, 4 cm

Dashed lines: pulse solutions for the same parameters

1 cm

4cm

Solution of the linear cable equation compared to the pulse solutions

c = 1

D=0.05 m2/s

The radial current density is modified to

20

200

00

20

)(

)(

BvHvvg

BvVVVVgVV

BvVVgVVj

tot

NernstNatot

NernstNai

i

Nernstir

BgHHvvj totr02

2,1 421

00

Non-linear cable equation:

))((1

2121

2

22 vvvvv

vvdt

dv

dx

vdaxon

Leads to a traveling wave solution with velocity:

1

2

1

2

2

1

v

v

v

vv axonaxon

How do neurons communicate?

Structure of GltPh (Glutamate transporter)

from Pyrococcus horikoshii