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 2.A step change in voltage will ‘instantly’ charge the membrane capacitance so we don’t have to see capacity current Squid axon: Rest: R m = 5000  cm 2 Active: R m = 25  cm 2 1.A voltage change causes a conductance (1/R) change want to study conductance

out in + ++ out in Unclamped membrane I I ionic = I C Inject current to charge C m Clamped membrane + + I injected = -I ionic

External plates (isopotential) Axial wire: current injection

Voltage clamp of axon: capacity transient RmRm CmCm i R = V m /R m i C = dV m /dt dV m /dt = 0, i C = 0 i C = ∞ dV m /dt = ∞ i R = V m /R m All capacity current All resistive current Small depolarization

Voltage clamp lets you measure membrane “current-voltage” relations 1/g 1/g ion V ion I V slope = g I V V ion Nernst I = gV I = g ion (V m -V ion )

1/g(V) Nerves have voltage-dependent conductances where g(V) V ~0 mV threshold I V rectification

In a real nerve: 1/g K (V) VKVK + - 1/g Na (V) V Na + - I K = g K (V m -V K ) I Na = g Na (V m -V Na ) IKIK V peak I Na V -30 mV V K = ~ -85 mV ~ -35 mV V Na

g K = I K (V m ) / (V m -V K )g Na = I Na (V m )/ (V m -V Na ) From the I-V relationships we can determine g(V) gKgK VmVm ~ -30 mV g Na VmVm ~ -30 mV peak Hodgkin and Huxley used the voltage clamp to measure g Na (V) and g K V) The action potential results from voltage-dependent changes in g Na and g K

Voltage clamp of axon

Problem: how to go from membrane current to channel molecular properties? Step 1: separate total membrane current into I Na and I K Step 2: calculate g Na and g K Step 3: determine value of g Na and g K at each voltage

Separation of Na + and K + current I total I Na-free I Na = I total - I Na-free (Hodgkin & Huxley, 1952a) Assumptions: I K amplitude or time course doesn’t change after Na + removal I Na time course doesn’t change when Na + reduced I Na can be measured at a time when there is no I K

The Na + and K + conductance g Na = I Na /(V-E Na )g K = I Na /(V-E Na ) (Hodgkin & Huxley, 1952a) weak depolarization strong depolarization

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?

Conductance gets bigger with depolarization reaches saturating value at positive V: goes from 0  maximum Conductance is a continuous function of voltage (current discontinuous) Conductance turns on faster with depolarization Put into simple membrane model Properties of membrane conductance

I C + I Na + I K = 0

I C = CdV m /dt I Na = g Na (V m -V Na ) I K = g K (V m -V K ) So, Solve for V m g Na (V m -V Na ) = - g K (V m -V K ) V m = g Na V Na + g K V K g Na + g K V Na + (g K /g Na ) V K 1 + (g K /g Na ) When g K >> g Na, V m = V K When g Na >> g K, V m = V Na = I C = 0 in steady-state

A very simple action potential: S1S1 S2S2 CmCm + + S 1 (Na channel switch) S 2 (K channel switch) off on VmVm 0 V Na VKVK in  = 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 V m 3. Determine the time course of gate opening at each V m

Na and K conductance: voltage dependence I ion = g(V) (V m -V ion ) the conductance can be rewritten: g(V) = N o x  N o is the number of open channels  is the single-channel conductance N o depends on voltage and time N o (V m,t) = N total x P o (V m,t) P o is the probability a channel is open I ion = N total x  x P o (V m,t) x (V m -V ion )

   Na + out in Na +

g = N T  p o (V) normalized conductance g(V)/g max = N T  N T  p o (max) If p o (max) = 1 and  constant with V g(V)/g max = p o (V) p o (V) 1. Isolate the “gate” (N T and  drop out) p o (V) is the fraction of open gates

The Na + and K + conductance g Na = I Na /(V-E Na )g K = I Na /(V-E K ) g K-maximum g Na-maximum g K (V)g Na (V)

2. Determine the fraction of open gates at each V m limiting slope indicates steep voltage dependence Defined by: position on x-axis (V 1/2 ) steepness (slope) g(V)/g max

What information can we get from the P o versus V m relation? We need to make a simple model for ion channel gating

out in GG big  G = slow rate small  G = fast rate

out in out in  V = + depolarization

outin  V reduces barrier height going from in  out) Net rate (in  out) ~ exp-[∆G -  zFV]/RT ∆G chemical part zFV electrical part 0 1 

 = exp-[∆G -  zFV]/RT  = exp-[∆G + (1-  )zFV]/RT C O  (V)  (V) Fraction of open channels =   exp(-zFV/RT) = Voltage-dependent rate constants g/g max =

Slope of Boltzmann equation gives valence of gating charge z = valence of gating charge

All that effort just to get “z”? C O  (V)  (V) Use the 2-state model to design experiments:

out in  V = positive out in  V = negative When V m negative, positive voltage step measures   All channels closed All channels open

out in  V = positive When V m, positive, negative voltage step measures   out in  V = negative All channels open All channels closed

Deactivation of the Na and K conductance (Hodgkin & Huxley, 1952b) What happens when the membrane potential is suddenly returned to rest when g>0? deactivation rate ~  activation

 out in  Depolarization

Know steady-state value of g Na and g K at each V m Want to determine the time course of g Na and g Na at each V m

Total number of channels in membrane: N T = N C + N O dN o /dt = N C  (V) - N o  (V) or dN o /dt = (N T -N o )  (V) - N o  (V) Define N o (V,t) = N T x P(V,t) N T dP/dt = N T (1-P)  (V) - N T P  (V) dP/dt = (1-P)  (V) - P  (V) At steady state dP/dt = 0 P = Isolate probability of opening What happens when we change V to a new value?  (V)+  (V)  (V) Know P o f(V), want to find P o f(t) C O  (V)  (V) What are limiting cases?

VoVo V P o (V o, t=0)  (V o )  (V o ) P ∞ (V, t ∞ )  (V)  (V) P o =  (V o )+  (V o )  (V o ) P ∞ =  (V)+  (V)  (V) define  =  (V)+  (V) 1 we can write dP/dt as: dP/dt = P ∞ - P  Solution of differential eqn. for the voltage step is: P (V, t) = P ∞ - ( P ∞ - P o ) exp-t/  (verify for homework) what does this look like? what is P at start of step? 3. Determine the time course of gate opening at each V m

P (V, t) = P ∞ - (P ∞ - P o ) exp-t/  When P o = 0 and P ∞ = 1 P(V, t) = (1- exp-t/  ) When P 0 = 1 and P ∞ = 0 P(V, t) = exp-t/  time P ∞ = 1 P ∞ = 0 P

Insert P(V,t) in expressions for I Na and I K (and we are almost done, except for one minor problem) Use: m(V,t) for Na current n(V,t) for K current

exp-t/  1-exp-t/  (1-exp-t/  ) 4 (exp-t/  ) 4 = n= n 4 it don’t fit:

closedopenclosed n n n n A picture of H-H n 4 gating Each n particle opens along (1-exp-t/  ) So 4 n particles opens (1-exp-t/  ) 4

P (V, t) = P ∞ - ( P ∞ - P o ) exp-t/  Simple equation used by Hodgkin & Huxley to explain ionic current of squid axon under voltage clamp g K /g max = n ∞ VmVm VKVK -80 I K = g K-max n 4 (V-V K ) time course of I K : n is the probability of opening ~ -35 n (V, t) = n ∞ - (n ∞ - n o ) exp-t/  Time course of the K + conductance why did they use n 4 ?

Time course of Na + current inactivation activation I Na = g Na-max m 3 h (V-V Na ) m goes from 0 to 1 h goes from 1 to 0

Inactivation  out in  depolarization

Measuring steady-state inactivation of the Na + current 1. Apply test pulse (constant V m ) to open Na + channels 2. Precede each test pulse by a prepulse to a different V m 3. Normalize I Na with prepulse to I Na without prepulse (Hodgkin & Huxley, 1952c)

I Na-max with prepulse to -90 mV Two-pulse experiment to measure steady-state inactivation I Na with prepulse to -30 mV I Na = N T  p o (V)(V-V Na ) I Na-peak (0 mV)-with prepulse I Na-peak (0 mV)-without prepulse = N T-with prepulse /N T-without prepulse Normalization removes everything but N T

Prepulse voltage I with prepulse I no prepulse

Now describe I Na in terms of m and h

g K /g max = m ∞ VmVm -80 ~ -30 h∞h∞ ~ -65 VmVm m = m ∞ - ( m ∞ - m o ) exp-t/  m h = h ∞ - ( h ∞ - h o ) exp-t/  h 1 1 V = h m3m3 m3hm3h I Na = g Na-max m 3 h (V-V Na )

In practice, we measure: m ∞ from normalized g Na /g Na-max h ∞ from steady-state inactivation measured with two-pulse voltage clamp method  m from fit to rising phase of I Na  h from fit to declining phase of I Na

Where do the parameters in the HH equations come from?  measured from time course of current activation (or inactivation) g(V)/g max = N T  n ∞ (V) N T  n ∞ (max) n ∞ measured from normalized conductance-voltage relation  n = n ∞ /  n  n = (1- n ∞ ) /  n rate constants calculated from  n =  n +  n 1 where n ∞ =  n +  n nn = n ∞ (V)

Hodgkin-Huxley used m 3, h, n 4 to reconstruct action potential I Na + I K + I L + I C = 0 g = maximum conductance Solve for dV m /dt

-Calculate dV/dt -approximate V 1 =V(t+∂t) - ∂t ~ 0.01 ms (integration step) dV m /dt = - g Na-max m 3 h(V-V Na ) + g K-max n 4 (V-V K ) CmCm dn/dt = n ∞ - n nn dm/dt = m ∞ - m mm dh/dt = h ∞ - h hh Get V m by numerical integration

Reconstruction of the action potential calculated measured depolarization Theory predicts: all or none behavior, peak, & threshold Late repolarization steeper

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: Does it apply to other excitable cells? Do ions move through pores or by transporters? Is there a current associated with voltage-dependent gating? What is the mechanism of activation and inactivation? How to measure N T and 

Inactivation of K + current by internal perfusion of C x

g K /g max = n ∞ VmVm VKVK -80 ~ -35

Conclusions: Inactivation depends on voltage - gets faster with depolarization Inactivation starts after a delay The delay is correlated with channel opening COI 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