Introduction to biophysical basis of nerve excitation
axon
Action potential in squid axon
(from Hodgkin, 1957)
Na+ theory
Rising phase of action potential is due to a selective
increase in membrane permeability to Na+
Prediction 1: peak of action potential more or less
follows Nernst potential for Na+
Prediction 2: Rate of rise of the action potential
depends on concentration of external Na+
Get molecular properties of ion channels
by measuring ion channel current with voltage clamp
How many channels are open at each voltage?
How sensitive are channels to changes in voltage?
How fast do channels open (or close) at each voltage?
Construct a model of the action potential based on
ion channel molecular properties
Why a voltage clamp is necessary
Squid axon:
Rest: Rm = 5000 Wcm2
Active: Rm = 25 Wcm2
1. A voltage change causes a conductance (1/R) change
want to study conductance
2. A step change in voltage will ‘instantly’ charge the
membrane capacitance so we don’t have to see
capacity current
Unclamped membrane
Clamped membrane
I
+
+
out
out
++
in
+
Iionic = IC
+
in
-
+
Inject current to charge Cm
Iinjected = -Iionic
QuickTime™ and a
TIFF (Uncompressed) decompressor
are needed to see this picture.
External plates
(isopotential)
Axial wire: current injection
Voltage clamp of axon: capacity transient
dVm/dt = ∞
Small depolarization
dVm/dt = 0, iC = 0
iR = Vm/Rm
iC = dVm/dt
Cm
Rm
iC = ∞
All capacity
current
All resistive current
iR = Vm/Rm
Voltage clamp lets you measure membrane “current-voltage” relations
I
I = gV
slope = g
1/g
V
I
Nernst
1/gion
Vion
I = gion(Vm-Vion)
V
Vion
Nerves have voltage-dependent conductances
g(V)
1/g(V)
where
~0 mV
threshold
I
rectification
V
V
In a real nerve:
1/gNa(V)
1/gK(V)
-
+
VK
+
-
INa = gNa (Vm-VNa)
IK = gK (Vm-VK)
IK
peak INa
VK= ~ -85 mV
-30 mV
VNa
~ -35 mV
V
VNa
V
From the I-V relationships we can determine g(V)
gK = IK (Vm) / (Vm-VK)
gNa
gK
~ -30 mV
gNa = INa(Vm)/ (Vm-VNa)
Vm
~ -30 mV
peak
Vm
Hodgkin and Huxley used the voltage clamp to measure gNa(V) and gK V)
The action potential results from voltage-dependent changes in gNa and gK
Voltage clamp of axon
Problem: how to go from membrane current
to channel molecular properties?
Step 1:
separate total membrane current
into INa and IK
Step 2:
calculate gNa and gK
Step 3:
determine value of gNa and gK
at each voltage
Separation of Na+ and K+ current
(Hodgkin & Huxley, 1952a)
INa-free
Itotal
INa = Itotal - INa-free
Assumptions: IK amplitude or time course doesn’t change after Na+ removal
INa time course doesn’t change when Na+ reduced
INa can be measured at a time when there is no IK
The Na+ and K+ conductance
(Hodgkin & Huxley, 1952a)
strong
depolarization
weak
depolarization
gNa = INa/(V-ENa)
gK = INa/(V-ENa)
So far:
Separate Na+ and K+ currents at each membrane potential
Calculate Na+ and K+ conductance as g = I (V-E)
at each membrane potential
What do we find?
Properties of membrane conductance
Conductance is a continuous function of voltage
(current discontinuous)
Conductance gets bigger with depolarization
reaches saturating value at positive V:
goes from 0  maximum
Conductance turns on faster with depolarization
Put into simple membrane model
+
+
-
IC + INa + IK = 0
IC + INa + IK = 0
IC = CdVm/dt
IC = 0 in steady-state
INa = gNa(Vm-VNa)
IK = gK(Vm-VK)
So,
gNa(Vm-VNa) = - gK(Vm-VK)
Solve for Vm
Vm = gNaVNa + gKVK
gNa + gK
=
When gK >> gNa ,
Vm = VK
When gNa >> gK,
Vm = VNa
VNa + (gK/gNa) VK
1 + (gK/gNa)
A very simple action potential:
S1
S2
Cm
+
+
in
on
S1 (Na channel switch)
off
S2 (K channel switch)
VNa
Vm
0
VK
t = RC
So, have a model for the action potential
but it’s not very good
Voltage-dependent gating
1. Isolate the “gating” process
2. Determine fraction of open gates at each Vm
3. Determine the time course of gate opening at each Vm
Na and K conductance: voltage dependence
Iion = g(V) (Vm-Vion)
the conductance can be rewritten:
g(V) = No x 
No is the number of open channels
 is the single-channel conductance
No depends on voltage and time
No (Vm,t) = Ntotal x Po (Vm,t)
Po is the probability
a channel is open
Iion = Ntotal x  x Po (Vm,t) x (Vm-Vion)
Iion = Ntotal
x

x
Po (Vm,t)
x
Na+ Na+ Na+ Na+ Na+ Na+ Na+
+
Na+ Na+ Na+ Na+ Na
out




in
Na+ Na+ Na+ Na+
(Vm-Vion)
1. Isolate the “gate”
g = NT  po(V)
normalized conductance
g(V)/gmax = NT  po(V)
NT  po(max)
If po(max) = 1
and  constant with V
(NT and  drop out)
g(V)/gmax = po(V)
po(V) is the fraction of open gates
The Na+ and K+ conductance
gK-maximum
gNa-maximum
gNa(V)
gK(V)
gNa = INa/(V-ENa)
gK = INa/(V-EK)
2. Determine the fraction of open gates at each Vm
g(V)/gmax
Defined by:
position on x-axis (V1/2)
steepness (slope)
limiting slope indicates
steep voltage dependence
What information can we get from the Po versus Vm relation?
We need to make a simple model for ion channel gating
out
in
G
big G = slow rate
small G = fast rate
depolarization
out
in
out
in
V=
+
out
in
∆G chemical part
zFV electrical part
0

1
V reduces barrier height
going from in  out)
Net rate (in  out) ~ exp-[∆G - zFV]/RT
Voltage-dependent rate constants
a(V)
C
O
b(V)
a = exp-[∆G - zFV]/RT
b = exp-[∆G + (1-)zFV]/RT
Fraction of open channels =
g/gmax
=
1
a
=
1 + exp(-zFV/RT)
a+b
Slope of Boltzmann equation gives valence of gating charge
z = valence of gating charge
All that effort just to get “z”?
a(V)
C
O
b(V)
Use the 2-state model to design experiments:
When Vm negative, positive voltage step measures a
a
out
in
out
in
V =
positive
V =
negative
All channels closed
All channels open
When Vm, positive, negative voltage step measures b
b
out
out
in
in
V =
positive
V =
negative
All channels open
All channels closed
Deactivation of the Na and K conductance
What happens when the membrane potential
is suddenly returned to rest when g>0?
deactivation rate ~ b
activation
(Hodgkin & Huxley, 1952b)
Depolarization
out

in

Know steady-state value of gNa and gK at each Vm
Want to determine the time course of
gNa and gNa at each Vm
Know Po f(V), want to find Po f(t)
a(V)
Total number of channels in membrane:
C
NT = NC + NO
O
b(V)
dNo/dt = NCa(V) - Nob(V)
dNo/dt = (NT-No)a(V) - Nob(V)
or
Define
No(V,t) = NT x P(V,t)
Isolate probability of opening
NTdP/dt = NT(1-P)a(V) - NTPb(V)
dP/dt = (1-P)a(V) - Pb(V)
What are limiting cases?
At steady state dP/dt = 0
P=
a(V)
a(V)+ b(V)
What happens when we change V to a new value?
3. Determine the time course of gate opening at each Vm
V
Vo
what is P at
start of step?
Po (Vo, t=0)
a (Vo)
b (Vo)
Po =
P∞ (V, t ∞)
a (V)
b (V)
a(Vo)
a(V)
P∞ =
a(V)+ b(V)
a(Vo)+ b(Vo)
define
we can write dP/dt as:
1
t = a(V)+ b(V)
dP/dt =
P∞ - P
t
(verify for homework)
Solution of differential eqn. for the voltage step is:
P (V, t) = P∞ - (P∞ - Po) exp-t/ t
what does this
look like?
P (V, t) = P∞ - (P∞ - Po) exp-t/ t
P∞ = 1
P
When Po = 0 and P∞ = 1
P(V, t) = (1- exp-t/t)
time
When P0 = 1 and P∞ = 0
P(V, t) = exp-t/t
P∞ = 0
Insert P(V,t) in expressions for INa and IK
Use:
m(V,t) for Na current
n(V,t) for K current
(and we are almost done, except for one minor problem)
it don’t fit:
exp-t/t
1-exp-t/t
(exp-t/t)4
(1-exp-t/t)4
=n
= n4
A picture of H-H n4 gating
n
n
n
n
closed
closed
closed
closed
open
Each n particle opens along (1-exp-t/t)
So 4 n particles opens (1-exp-t/t)4
Time course of the K+ conductance
P (V, t) = P∞ - (P∞ - Po) exp-t/ t
Simple equation used by Hodgkin & Huxley to explain ionic
current of squid axon under voltage clamp
gK/gmax = n∞
VK
-80
Vm
~ -35
time course of IK:
n is the probability of opening
IK = gK-max n4 (V-VK)
why did they use n4?
n (V, t) = n∞ - (n∞ - no) exp-t/ t
Time course of Na+ current
INa = gNa-maxm3h (V-VNa)
inactivation
m goes from 0 to 1
h goes from 1 to 0
activation
Inactivation
depolarization
out

in

Measuring steady-state inactivation of the Na+ current
(Hodgkin & Huxley, 1952c)
1. Apply test pulse (constant Vm) to open Na+ channels
2. Precede each test pulse by a prepulse to a different Vm
3. Normalize INa with prepulse to INa without prepulse
Two-pulse experiment to measure steady-state inactivation
INa with prepulse to -30 mV
INa-max with prepulse to -90 mV
INa = NTpo(V)(V-VNa)
INa-peak(0 mV)-with prepulse
INa-peak(0 mV)-without prepulse
= NT-with prepulse/NT-without prepulse
Normalization removes everything but NT
Iwith prepulse
Ino prepulse
Prepulse voltage
Now describe INa in terms of m and h
gK/gmax = m∞
-80
h∞
1
Vm
~ -30
m=
m∞ - (m∞ - mo) exp-t/ tm
h=
h∞ - (h∞ - ho) exp-t/ th
1
Vm
~ -65
INa = gNa-max m3h (V-VNa)
-20
V = -80
1
0
m3
h
m3h
In practice, we measure:
m∞ from normalized gNa/gNa-max
tm from fit to rising phase of INa
h∞ from steady-state inactivation measured with
two-pulse voltage clamp method
th from fit to declining phase of INa
Where do the parameters in the HH equations come from?
1.
t measured from time course of current activation (or inactivation)
2.
n∞ measured from normalized conductance-voltage relation
g(V)/gmax = NT  n ∞(V)
NT  n ∞(max)
3.
= n ∞(V)
rate constants calculated from
an = n∞/ tn
bn = (1- n∞) / tn
where
1
tn = a + b
n
n
n∞ =
an
an + bn
Hodgkin-Huxley used m3, h, n4 to reconstruct action potential
I  Cm dV + g K n 4 (V  VK ) + g Na m3h(V  VNa ) + gl (V  Vl )
dt
g = maximum conductance
INa + IK + IL + IC = 0
Solve for dVm/dt
Get Vm by numerical integration
3
4
dVm/dt = - gNa-maxm h(V-VNa) + gK-max n (V-VK)
Cm
dn/dt = n∞ - n
tn
dm/dt = m∞ - m
tm
dh/dt = h∞ - h
th
-Calculate dV/dt
-approximate V1 =V(t+∂t)
- ∂t ~ 0.01 ms (integration step)
Reconstruction of the action potential
calculated
Late repolarization steeper
depolarization
measured
Theory predicts: all or none behavior, peak, & threshold
Changes in Hodgkin-Huxley parameters during an action potential
Falling phase of action potential due to
Na+ inactivation
K+ activation
Questions raised by Hodgkin-Huxley analysis:
Do ions move through pores or by transporters?
What is the mechanism of activation and inactivation?
Is there a current associated with voltage-dependent gating?
Does it apply to other excitable cells?
How to measure NT and 
Inactivation of K+ current by internal perfusion of Cx
gK/gmax = n∞
VK
-80
~ -35
Vm
Conclusions:
Inactivation depends on voltage - gets faster
with depolarization
Inactivation starts after a delay
The delay is correlated with channel opening
C
O
I
Drug acts only from internal side of membrane
Interpretation of K+ channel inactivation:
External K+ speeds recovery from inactivation
Evidence for “foot in door” mechanism:
drug must first exit before activation gate can close
Evidence for a closed-blocked state