1

Lecture Set 3:

EE M255, BME M260, NS M206:

NeuroEngineeringLecture Set 3:

Membrane Biophsyics, Action Potentials, and the Hodgkin H le Eq ationsHodgkin-Huxley Equations

Prof. Dejan Markovic

Lecture Topics

• Membrane BiophsyicsI i Fl d C t• Ionic Flow and Current

• Resting Membrane Potential• Goldman Equation• Action-Potential Definitions• Subthreshold Activityy• Origin of the Action Potential• Hodgkin-Huxley Model

3.2

2

Membrane Biophysics

~ 3 nm

3.3

Membrane Structure (Macroscopic)

3.4

3

Membrane Structure (Microscopic)

3.5

Ionic Concentration

• All Excitable Cells– intracellular [K+] exceeds extracellular [K+]

extracellular [Na+] exceeds intracellular [Na+]– extracellular [Na+] exceeds intracellular [Na+]– extracellular [Cl-] exceeds intracellular [Cl-]– Typical values:

Ion Intra- Extra- Intra- Extra-Frog Muscle Squid Axon

Ion Intra- Extra- Intra- Extra-K+ 124 mM 2.2 mM 397 mM 20 mMNa+ 4 mM 109 mM 50 mM 437 mMCl- 1.5 mM 77 mM 40 mM 556 mM

3.6

4

Movement of Charge in Solution• Diffusion (Concentration Gradient Driven)

jdiffusion = −D ⋅∇CFlux:

Diffusivity:

jdiffusion

DC

• Drift (Electric Field Driven)

• Einstein’s Relation:

Concentration: C

jdrift,p =ZpZp

⋅Cp ⋅ vp =ZpZp

⋅Cp ⋅ up ⋅E( )Flux:

Valence:

Concentration:

Velocity:

Mobility:pvpu

pC

jdrift

pZ

• Einstein s Relation:– relates diffusivity and mobility

p

Dp =up ⋅ R ⋅TZp ⋅F C) 27(at mV 8.25energy thermal/

olecoulombs/m 487,96Farady C 27at moleJ/K 314.8constant gas

==⋅

====

FTRFR

3.7

Total Ion Flow

• Nernst-Planck equation– Flux:– Flux:

– Current:

jp = jdiffusion,p + jdrift,p = −Dp ⋅ ∇Cp +Zp ⋅Cp ⋅FR ⋅T

⋅∇V

J = −D ⋅ F ⋅Z( )⋅ ∇C +Zp ⋅Cp ⋅F ⋅∇V

orJp = Dp F Zp( ) ∇Cp + R ⋅T

∇V

or

Jp = − up ⋅ R ⋅T ⋅ZpZp

⋅∇Cp + up ⋅ Zp ⋅Cp ⋅F ⋅∇V

3.8

5

Ionic Conductances• Current due to electric field

– dependence on conductivityJp,drift = −up ⋅ Zp ⋅Cp ⋅F ⋅∇V

Jp = σ p ⋅Edependence on conductivity• Ion Conductivity– However, only a faction disassociates,

decreases from 1 at high – for KCl:

σ p = up ⋅Cp ⋅Fp p

ασ p = α ⋅up ⋅Cp ⋅F

[ ] C F

pC

• Equivalent Conductance

σ = α ⋅ uK + uCl[ ]⋅CKCl ⋅F

Λ = α ⋅ uK + uCl[ ]⋅F Λ = α ⋅ uK +uCl[ ]⋅F ⋅1000

Λ = σ /CKCl( )

3.9

Equivalent Conductance• Main Idea:– At high concentrations the conductivity is higher, but

not proportionally so due to collisions /disassociation– For multiple ions, each contributes to the current.

• for each pth ion:• consider KCl:

• Equivalent conductance:

JKCl = F ⋅CKCl ⋅ uK + uCl( )⋅EσKCl = F ⋅CKCl ⋅ uK + uCl( )

J p = −up ⋅ Zp ⋅Cp ⋅F ⋅∇V

100% disassociation

ΛKCl = F ⋅ uK + uCl( )⋅1000 Note: 1 m3 = 1000 L

JKCl = σKCl ⋅E

• Disassociation:KCl K Cl( )

ΛKCl = α ⋅F ⋅ uK + uCl( )⋅1000σKCl = α ⋅F ⋅CKCl ⋅ uK + uCl( )Concentration of KCl (mM)

0.00010.0010.010.1

Measured Equivalent Conductance148.9146.9141.3128.9

3.10

6

Transference Numbers• The contributions to conductivity

are not equal for all ions in solution• The proportionality factors of theThe proportionality factors of the

contributions of each ion are the transference numbers:

Given:

• Consider KCL again:σKCl = α ⋅F ⋅CKCl ⋅ uK + uCl( )

tK =uK tCl =

uCl t + t = 1tK uK + uCl

tCl uK + uCltK + tCl = 1

Ion Range (equivalents/liters)[KCl] 0.02 - 3.0 tk = 0.49[HCl] 0.01 - 0.2 tH = 0.49[LiCl] 0.01 - 0.2 tLi = 0.49[NaCl] 0.01 - 0.2 tNa = 0.39

3.11

Equilibrium: Nernst Potential• At equilibrium no net current flows

Jp = −Dp ⋅F ⋅Zp ⋅ ∇Cp +Zp ⋅Cp ⋅FR ⋅T

⋅∇V

= 0

Z C F dC Z C F dVthus or

rearranging yields

integration from inside a cell to outside a cell yields

∇Cp = −Zp ⋅Cp ⋅FR ⋅T

⋅∇V

dCpdx

= −Zp ⋅Cp ⋅FR ⋅T

⋅dVdx

dCpCp

= −Zp ⋅FR ⋅T

⋅dV

dCp extra

∫Zp ⋅F

de

∫ lCp

Zp ⋅F ( )

membrane potential:

p

Cp intra∫ = − p

R ⋅T

⋅ dV

i∫ ln p e

Cp i

= − p

R ⋅T

⋅ Ve −Vi( )

Vm = Vi −Ve( )= R ⋅TZp ⋅F

⋅ ln

Cp e

Cp i

Vm =

25 mVZp

⋅ lnCp e

Cp i

=

58 mVZp

⋅ log10

Cp e

Cp i

3.12

7

Resting Potential

Ion Intra- Extra- Intra- Extra-K+ 124 mM 2.2 mM 397 mM 20 mMNa+ 4 mM 109 mM 50 mM 437 mM

Frog Muscle Squid Axon

Na 4 mM 109 mM 50 mM 437 mMCl- 1.5 mM 77 mM 40 mM 556 mM

Vm K( ) =25 mV+1

⋅ ln20397

= −74.7 mV

Vm Na( ) =25 mV+1

⋅ ln43750

= +54.2 mV

V25 mV

l556

65 8 V

Vm K( ) =25 mV+1

⋅ ln2.2124

= −108.8 mV

Vm Na( ) =25 mVZp

⋅ ln109

4

= +82.6 mV

V25 mV

l77

98 5 V

• Actual resting potentials (Squid ~70 mV)– K is nearly equilibriated– Na plays little role in establishing

Vm Cl( ) = −1⋅ ln

40

= −65.8 mVVm Cl( ) = −1

⋅ ln1.5

= −98.5 mV

Vm

3.13

Donnan Equilibrium• Assume K+ and Cl- are uniform &

membrane is impermeable to A-

• Intracellular space has too much K+

K+ Cl-

K+ A-

K+ Cl-

ExtraIntral

Intracellular space has too much K– K+ diffuses out, establishes potential, pulls Cl- too– Nernst potential of K+ and Cl- must be equal

– Although initially , then to get

membrane

Vm K( ) =R ⋅T+1 ⋅F

⋅ ln

K+ e

K+ i

= Vm Cl( ) =

R ⋅T−1 ⋅F

⋅ ln

Cl− e

Cl− i

K+ e

K+ i

=Cl− i

Cl− e

thus

K+ e

K+ i

< 1Cl− i

Cl− e

= 1 K+ i↓ Cl− e

↑

• The equilibrium state = “Donnan Equilibrium”• Multiple ion species redistribute to achieve

equilibrium K+ e

K+ i

=Cl− i

Cl− e

=Na+ e

Na+ i

= r Vm K( ) =R ⋅T+1 ⋅F

⋅ ln r( )

3.14

8

Goldman Equations• To apply Nernst-Planck equations you need to

know variation of and in the membrane• Goldman assumption: assume varies linearlyV

VCp

dV V good for thin membrane• Consider one ion: flux:– Potassium:

dVdx

= V x = d( )−V x = 0( )= −Vm

dd

jp = −Dp ⋅dCpdx

+Zp ⋅Cp ⋅FR ⋅T

⋅dVdx

dCK

dx=jp

−Dp−

+1( )⋅CK ⋅FR ⋅T

⋅−Vm

d

dCKjp

−Dp−CK x( )⋅FR ⋅T

⋅−Vmd

0

d

∫ = dx0

d

∫

d d

– Ion concentration inside both sides of the membrane are related by a partition coefficient • assume same on both sides

dCK

jp−Dp

−CK x( )⋅FR ⋅T

⋅−Vm

d

0

d

∫ = dx0

d

∫ jK =DK ⋅Vm ⋅FR ⋅T ⋅d

CK[ ]x=d − CK[ ]x=0⋅ eVm ⋅F /R⋅T

1− eVm ⋅F /R⋅T

βKCK[ ]x=0

= βK ⋅ K[ ]intraCK[ ]x=d = βK ⋅ K[ ]extra

3.15

Goldman Membrane Potential• Flux: recall:• Current due to potassium ion flow:• Define permeability parameter:

jK =DK ⋅Vm ⋅FR ⋅T ⋅d

CK[ ]x=d − CK[ ]x=0⋅ eVm ⋅F /R⋅T

1− eVm ⋅F /R⋅T

JK = F ⋅ jKPK =

DK ⋅ βK

d

CK[ ]x=0= βK ⋅ K[ ]intra

CK[ ]x=d = βK ⋅ K[ ]extra

• Thus K current is:• Multiple ions:

K d

JK =PK ⋅Vm ⋅F

2

R ⋅TK[ ]e − K[ ]i ⋅ eVm ⋅F /R⋅T

1− eVm ⋅F /R⋅T

J = JK + JNa + JCl = JK =PK ⋅Vm ⋅F

2

R ⋅Tw − y ⋅ eVm ⋅F /R⋅T

1− eVm ⋅F /R⋅T

w = K[ ]e +PNa

PK

Na[ ]e +PCl

PK

Cl[ ]i y = K[ ]i +PNa

PK

Na[ ]i +PCl

PK

Cl[ ]e

• Not all ions can be in equilibrium, but we can solve for steady state

K K K K

w − y ⋅ eVm ⋅F /R⋅T = 0

dVm / dt = 0 → J = 0

Vm =R ⋅TF

⋅ lnwy

=R ⋅TF

⋅ lnPK ⋅ K[ ]e + PNa Na[ ]e + PCl Cl[ ]iPK ⋅ K[ ]i + PNa Na[ ]i + PCl Cl[ ]e

Goldman Equation:(when )J = 0

3.16

9

Action Potentials (“Spikes”)

ReferenceElectrode

3.17

Nonlinear Membrane Behavior

• To activate anall-or-nothinggpolarizationresponse, thestimulus must:– be large enough– be the right polarity– last long enough H l i i D l i ilast long enough

• Hyperpolarization only causes passive response (irregardless of the amplitude)

Hyperpolarizing Depolarizing

3.18

10

Subthreshold Responses

• Passive responses are the same as RC networks

• Can be simulated with fixed networks of lumped RC elements

• Lack of symmetry rane

Pot

entia

l

Threshold

indicates a non-linear componentM

embr Time (ms)

Stimulation Period

3.19

Origin of the Action Potential• Resting potential in steady state given by the

Goldman Equation: dVm / dt = 0 → J = 0

D βR T P ⋅ K[ ] + P Na[ ] + P Cl[ ]

– valid before spike– valid at top of spike

where slope = 0

PK =DK ⋅ βK

dVm =

R ⋅TF

⋅ lnPK ⋅ K[ ]e + PNa Na[ ]e + PCl Cl[ ]iPK ⋅ K[ ]i + PNa Na[ ]i + PCl Cl[ ]e

Recall:

3.20

11

Origin of the Action Potential• Permeability before

spike and at the top of the spike differ

Na+

– Before Peak:– Top of Peak:

PK :PNa :PCl = 1.0 : 0.04 : 0.45P :P :P = 1 0 : 20 0 : 0 45

Vm K( ) =25 mV+1

⋅ ln20397

= −74.7 mV

Vm Na( ) =25 mV+1

⋅ ln43750

= +54.2 mV

Vm Cl( ) =25 mV−1

⋅ ln55640

= −65.8 mV K+

Cl-

– Top of Peak:– Na+ permeability is ~500 times larger at peak– Role of Cl- is also ignored at peak due to small

Vm ≈Vm(K) =R ⋅TF

⋅ lnNa[ ]eNa[ ]i

PK :PNa :PCl = 1.0 : 20.0 : 0.45

PCl

Vm ≈Vm(K) =R ⋅TF

⋅ lnK[ ]eK[ ]i

Rest: Peak:

3.21

Movement of Ions• At rest:– influx of Na+ and efflux of K+

• During rising phase of spike:

EK <Vm < ENa Vm =R ⋅TF

⋅ lnPK ⋅ K[ ]e + PNa Na[ ]e + PCl Cl[ ]iPK ⋅ K[ ]i + PNa Na[ ]i + PCl Cl[ ]e

Vm K( ) =25 mV+1

⋅ ln 20397

= −74.7 mV

V N( ) =25 mV

⋅ ln 437

= +54.2 mV

Na+

– massive influx of Na+

• During falling phase of spike– efflux of K+ to restore

resting potential• A voltage clamp is

d t t d

Vm Na( ) +1ln

50 +54.2 mV

Vm Cl( ) =25 mV−1

⋅ ln 55640

= −65.8 mV

K+

Cl-

used to study conditions at other potentials

3.22

12

Parallel-Conductance Model• Electrical model of a

patch of membrane– membrane potential Vm Cm

gK gNa gCl

Inside Cell

– K, Cl, and Na– Nernst potentials– ion currentsIK = Vm −VK( )⋅ gK

ICl = Vm −VCl( )⋅ gCl

INa = Vm −VNa( )⋅ gNa

if then outward diffusion not balanced by E field, net effluxVm >VK

if then inward diffusion not balanced by E field, net influxVm >VCl

if then inward diffusion not balanced by E field, net effluxVm >VNa

Vm(K) Vm(Na) Vm(Cl)

Outside Cell

– membrane capacitance and associated currentbut at steady state thus

• Membrane potential:

Na Vm VNa( ) gNa if then inward diffusion not balanced by E field, net effluxm Na

ICm= C ⋅

dVm

dtICm

= 0

IK + ICl + INa = Vm −VNa( )⋅ gNa + Vm −VCl( )⋅ gCl + Vm −VK( )⋅ gK = 0

Vm =gK ⋅VK + gCl ⋅VCl + gNa ⋅VNa

gK + gCl + gNa

conductance-weightedaverage of Nernst potentials

3.23

Typical Values

Ion Intra- Extra-K+ 397 mM 20 mM

Squid AxonVm K( ) =

25 mV+1

⋅ ln20397

= −74.7 mV

Vm Na( ) =25 mV+1

⋅ ln43750

= +54.2 mV

Na+ 50 mM 437 mMCl- 40 mM 556 mM

m Na( ) +1 50

Vm Cl( ) =25 mV−1

⋅ ln55640

= −65.8 mV

gK = 0.367 mmhos/cm

gCl = 0.582 mmhos/cm

gNa = 0.010 mmhos/cm

Vm =gK ⋅VK + gCl ⋅VCl + gNa ⋅VNa

gK + gCl + gNa

= −68.0 mV

Parallel Conductance Equation:a weighted average of Nernst potentials

• Results in outflux of K+ due to• Results in influx of Na+ due to

– despite large potential, it acts on a low concentration and low• Assume Cl- is essentially in equilibrium

Vm −VK = 6.7 mV

Vm −VNa = −122.2 mVPNa

3.24

13

Hodgkin and Huxley Equations• Voltage-Clamp Experiments– fix potential, measure dynamic current flow– experimental model: squid axon (enormous)

Vm −Vp

p q ( )– use same concepts discussed above

• Determine conductances

• During voltage-clamp experiments the d i t i t t

gK t( )=IK t( )Vm −VK( ) gNa t( )=

INa t( )Vm −VNa( ) fixedfixed

denominator is a constant– measurements of current provide

information about conductances

3.25

HH: Model for Potassium• Conductance:– where is a fitting parameter

and the power of 4 seems to fit data well

gK t,vm( )= gK ⋅n4 t,vm( )

n

• Form of solution chosen to fit data:

– rate constants only depend onl t d t ti ki ti f i h l

dn t,vm( )dt

= α vm( )⋅ 1− n( )− β vm( )⋅nα vm( ), β vm( ) vm

vm = Vm −Vrest

– related to gating kinetics of ion channels– voltage-clamp tests allow eq. to be solved

• Data indicates that the solution should be of first-order:

n t( )= n∞ − n∞ − n0( )⋅ e− t /τn3.26

14

HH: Fit to Potassium Data• Given• To fit data:

n t( )= n∞ − n∞ − n0( )⋅ e−t /τn

τ n =1

( ) ( )n∞ =α vm( )

( ) ( )

Potassium Conductance(voltage clamped)

10 mS/cm2109 mV

• Perform voltage clamp tests

• Results in best fit of:

n α vm( )+ β vm( )∞ α vm( )+ β vm( )

β vm,i( )=1− n∞ vm,i( ) τ n vm,i( )

α vm,i( )= n∞ vm,i( )τ n vm,i( )

vm,i 88 mV

63 mV

β vm,i( )= 0.125 ⋅ e−vm

80

α vm,i( )= 0.01 ⋅ 10 − vm( )n∞ vm,i( )

e10−vm

10

−1

For in mVvm = Vm −Vrest

38 mV

26 mV

Time (ms)vm

3.27

HH: Model for Sodium • Conductance:– is an activation fitting parameter– is an inactivation fitting parameter

gNa t,vm( )= gNa ⋅m3 t,vm( )⋅h t,vm( )

mh g p

• Forms of solutions chosen to fit data:

– rate constants depend on– voltage-clamp tests allow eq. to be solved

dm t,vm( )dt

= αm vm( )⋅ 1− m( )− βm vm( )⋅m

αm vm( ), αh vm( ), βm vm( ), βh vm( ) vm

dh t,vm( )dt

= αh vm( )⋅ 1− h( )− βh vm( )⋅h

g p q• Data indicates solution should be second order– HH decided to use two first-order solutions insteadm t( )= m∞ − m∞ − m0( )⋅ e− t /τm h t( )= h∞ − h∞ − h0( )⋅ e−t /τh

3.28

15

HH: Fit to Sodium Data• Given:

• Before spike is small, thus

Sodium Conductance(voltage clamped)

10 mS/cm2109 mV

m t( )= m∞ − m∞ − m0( )⋅ e− t /τm h t( )= h∞ − h∞ − h0( )⋅ e−t /τh

τ h =1

αh vm( )+ βh vm( )n∞ =αh vm( )

αh vm( )+ βh vm( ) τm =1

αm vm( )+ βm vm( )m∞ =

αm vm( )αm vm( )+ βm vm( )

gNa m0 ≈ 0Before spike is small, thus • After spike , thus

– result:• Perform voltage clamp tests

R lt i b t fit f

vm,i

88 mV

63 mV

gNa m0 0gNa ≈ 0 h∞ ≈ 0

gNa t,vm( )= gNa ⋅m∞3 ⋅h0 ⋅ 1+ e−t /τm( )3 ⋅ e−t /τh

βm vm,i( )= 1− m∞ vm,i( )τm vm,i( )αm vm,i( )= m∞ vm,i( )

τm vm,i( ) αh vm,i( )= h∞ vm,i( )τ h vm,i( )

βh vm,i( )= 1− h∞ vm,i( )τ h vm,i( )

• Results in best fit of:

Time (ms)

38 mV

26 mV

βm vm,i( )= 4 ⋅ e−vm

18

αm vm,i( )= 0.1 ⋅ 25 − vm( )⋅n∞ vm,i( )

e25−vm

10

−1

For in mV vm = Vm −Vrest

βh vm,i( )= e30−vm

10

+1

−1

αh vm,i( )= 0.07 ⋅ e−vm

20

vm

3.29