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Home , Membrane potential, Resting potential

1 – Fundamental units of computation

2 Resting

g Difference in between the interior and exterior of a

Extracellular

Intracellular g Three main players

n Salty fluids on either side of the membrane

n The membrane itself

n that span the membrane

3 Typical Ionic Concentrations

4 Movements of -

Na+ cation Cl- anion

Membrane impermeable to the ions 5 Movements of ions –

Na+ cation Cl- anion

Cathode Anode

6 Establishing equilibrium

Diffusion causes ions to move from region of higher concentration (intracellular side) to a region of lower concentration (extracellular side)

A net accumulation of positive charge on the outside and negative charge on the inside retards movement of K+

An equilibrium is established such that there is no movement of ions and there is a charge difference

rd , Bear et al. 3 edition 7 g

RT [ionout ] E = ln zF [ionin ]

[ionin ], [ionout] are ionic concentration inside and outside the T is temperature () R is z is the valence of the ion F is Faraday’s! constant: it is the magnitude of electric charge per of electrons

Reversal potential at 25C

ENA = 55mV EK = -77 mV 8 ECL = -61 mV Resting membrane potential

g Goldman Hodgkin and Katz equation

RT PK [K out ] + PNa[ Naout ] + PCl [Clin ] ENa, K, Cl = ln F PK [K in ] + PNa[ Nain ] + PCl[Clout ]

Typical resting membrane potential V = -65 mV ! rest

9 Some terminology g

n Change in cellʼs membrane potential making the inside of the cell more positive (or less negative) with respect to the cellʼs outside g Hyper-polarization

n Change in cellʼs membrane potential making the inside of the cell more negative (or less positive) with respect to the cellʼs outside + g Inward current (example due to Na ions)

n Flowing into the cell (considered negative) + g Outward current (example due to K ions)

n Flowing out of the cell (considered positive) g , Spike

n An “all or nothing” event during which the membrane potential

rapidly rises and for a brief moment Vinside-Voutside is positive 10 Action Potential

ENa

Potassium Channels are opened

0 mV

EK

11 Action Potential

ENa

0 mV

Sodium Channels are opened

EK

12 Action Potential Close Open Na Channels K Channels ENa

0 mV

-65 mV

EK

Open Channels

13 Action Potential

rd Neuroscience, Bear et al. 3 edition 14 Action Potential

rd Neuroscience, Bear et al. 3 edition 15 Action Potential

rd Neuroscience, Bear et al. 3 edition 16 Membrane Current and Conductance g Net movements of ion will cause ionic

current Iion g The number of ion channels open will be

proportional to an electrical conductance (gk, gNa; inverse of resistance)

g Ionic current will flow only as long as

membrane potential Vm is not equal to reversal potential Eion

Iion = gion (Vm-Eion)

17 Refractory Period

[From web] 18 Electrical Circuit of a Patch of neuron

[Notes borrowed from Araneda] 19 Passive properties of a Neuron

[Notes borrowed from Araneda] 20 Passive properties of a Neuron

[Notes borrowed from Araneda] 21 Passive properties of a Neuron

[Notes borrowed from Araneda] 22 Electrical Circuits with R and C elements

[Notes borrowed from Araneda] 23 Electrical Structure of Small Passive Neuron

Resistor mimics

A battery to account for resting membrane potential Many Small RC circuits

Since the cell is so small we can assume same membrane potential everywhere (isopotential) [Koch, 1999] 24 Membrane Patch Equation

I inj R C

dV (t) V (t) Injected current = Capacitive current + I = C m + m inj dt R Resistive current

# t $ Vm (t) = V" (1# e ) ! g Show that the solution is: at t = 0 V = 0 25 $ = RC V" = Iinj R

! RC Circuit

26 Equivalent circuit

Current is carried through the membrane either by charging the membrane or by movement of ions through the resistances in parallel with the capacitor.

27 Currents

Total current = Capacitive current + Ionic current

Ionic current = Sodium-Current + Pottasium-Current + Leakage Current (Cl- and other ions)

28 Ionic currents

ENa, Ek and El are equilibrium or reversal potential for respective ion species gNa, gk and gl are ionic conductances

⎧Eion−type < E Iion−type is positive⎫ Notice ⎨E E I is negative⎬ ⎩ ion−type > ion−type ⎭

Larger E-Eion the larger the current due to that ionic species

29 Squid Axon Experiments

30 Technique

31 Hodgkin Huxley Experiments

32 Voltage Clamp

33 Measuring individual ionic currents

34 Voltage Clamp

Outward current is monotonic as you move away from K reversal Potential (-77mV)

RT [ionout ] E ion = ln zF [ionin ] Inward current become smaller as Clamp voltage gets close 35 to Na Reversal Potential (+55mV) and flips sign for larger

! Voltage Clamp (continued …)

36 Measuring individual ionic currents g Selective measurement of the K ion flow alone is possible by utilizing a voltage clamp step corresponding to the Na Nernst potential.

37 Pharmacological agents

Puffer fish

Lips and tongue tingles, hands feel numb and trouble making movements Tetrodotoxin (TTX) – blocks voltage-gated Na channels

38 Pharmacological agents

Scorpion toxin

Scorpion toxin blocks voltage-gated K channels

Another popular blocker is tetraethylammonium (TEA)

39 Hodgkin Huxley g One of the most important contributions of Hodgkin and Huxley was their detailed discussion of membrane kinetics

g The model was not formulated from fundamental principles but rather from theoretical insights and curve fitting

40

Dynamics of K Conductance (3)

How do K conductance change with V? Becomes more rapid at higher depolarization

Does K conductance ever return back to baseline? No!

41 K conductance

Depolarization to 25 mV back to

Inflexion of the curve

42 Whatʼs the best fit?

1 0.9 1-exp(-t/τ) 0.8 0.7 2 0.6 [1-exp(-t/τ)] [1-exp(-t/τ)]4 0.5 [1-exp(-t/τ)]3 0.4 Normalized (mS) K

G 0.3 0.2 0.1 0 0 5 10 15 20 25 30 35 40

Time (ms)

43 Why n4?

44 Dynamics of K conductance (1) g Hodgkin-Huxley assumed that there is only one variable that determines the dynamics of potassium conductance

g K is a constant

4 g What does n represent?

n That K ions cross membrane only when four particles occupy a certain region of the membrane

45 Dynamics of K conductance (1)

www.bem.fi/book/04/04.htm 46

Dynamics of K Conductance (2)

n αn represents of transfer from outside to inside

n βn represents of transfer from inside to outside

g The solution of this first order ODE that

satisfies the boundary condition n=n0 at t=0 is: # n = n " # + $ HW % ( n n t #0 Pblm # n = n n n n exp 0 = " #( " # 0) ' * #0 + $0 & $n ) 1 47 % n = #n + $n

! !

Dynamics of K Conductance (3)

% #t( n = n" #(n" # n0) exp' * & $n )

!

48

Rate constants αn, βn(6)

depolarized hyperpolarized

49

Dynamics of Na Conductance (5)

How does Na conductance change with V? Becomes more rapid at higher depolarization

Does Na conductance ever return back to baseline? Yes!

Are the rising and falling portions of the curves similar? No!

50

Dynamics of Na Conductance (1) g Hodgkin-Huxley assumed that there are two variables that determine the dynamics of sodium conductance

g Na is a constant m represents the proportion of activating molecules on the inside h represents the proportion of inactivating molecules on the outside

3 g What does m h represent?

n That sodium conductance is proportional to the number of sites on the inside of the membrane which are occupied simultaneously by three activating molecules (m3) and not blocked by an inactivating molecule (m3h).

51

Dynamics of Na Conductance (1)

www.bem.fi/book/04/04.htm 52

Dynamics of Na Conductance (2) g The two variables obey the following first order differential equation

g αm or βh represents of transfer from outside to inside g βm or αh represents of transfer from inside to outside

53

Dynamics of Na Conductance (3) g Solving the first order differential equation

g At t=0; m =m0, h =h0

54

Dynamics of Na Conductance (4) g HH noted that in resting state, sodium conductance is negligble, so we can neglect

m0, and inactivation in nearly complete if V< -30mV and hα can also be neglected

where is the value of sodium conductance if h

remained in its resting level h0

55

Dynamics of Na Conductance (5)

56

Rate constants αm, βm(6)

57

How does activation variable m change with V?(7)

58

Rate constants αh, βh(8)

59

How does activation variable h change with V?(9)

60 Putting pieces together – HH model

61 Time courses of HH variables

62 Refractory Period

63 HH model - Summary g Models a patch of membrane g Assumes isopotential i.e. the membrane potential is constant across the patch g The number of channels is large enough that individual gating events are averaged out and the Na and K currents are smooth population currents g The channels are not arranged in anyway which allows local interaction among small number of channels

64 Reducing HH models g State variables of HH eqns : [V, m, n, h] g Most reduction rely on two simplifications

n Time constants are very different

n m changes so fast compared to n and h that you donʼt need a ODE to describe it (removes one ODE)

n n+h is almost a constant; so we can eliminate one more variable

n Now the reduced HH will have state variable [V, w]; V – fast variable (membrane potential) and w – slow variable (potassium activation)

65 Reducing HH models(2)

Original

Reduced

66 Reducing HH models(3)

67 Linear Cable Equation

68 Action Potential Propagation

Transmembrane current across unit length of the axon Vm V !V dV rm m rest m cm im = + cm rm dt

Vrest

Ve=0

im – Transmembrane current per unit length (A/cm) rm – membrane resistance per unit length (Ωcm) cm –membrane capacitance per unit length (F/cm)

69 Action Potential Propagation

Transmembrane current across Δx units of the axon Δx

I (x,t) Ii(x-Δx,t) i ra Δx

x x+Δx x rm/Δx im im im im i Δ m m c

Vrest

= unit transmembrane current (im ) times Δx

The total membrane resistance in a fiber of length Δx is: rm/Δx The total membrane capacitance in a fiber of length Δx is: c Δx m 70 Action Potential Propagation

Plug membrane together: get linear cable

ra Δx Vm(x-Δx,t) Vm(x,t) Vm(x+Δx,t)

r /Δx c Δx … m m …

Vrest

Ve=0

Δx 4R r = a ! / cm a !d 2 R - intracellular resistivity (Ωcm) Rm a rm = !" cm 2 !d Rm - specific membrane resistance (Ωcm ) 2 C - specific membrane capacitance (F/cm ) cm = Cm !!d F / cm m 71 Action Potential Propagation

Plug membrane together: get linear cable

raΔxVm(x-Δx,t) Vm(x,t) Vm(x+Δx,t)

rm/Δx cmΔx … …

Vrest

Note: Constant Ve is an assumption

72 Action Potential Propagation

Plug membrane together: get linear cable

ii(x,t)

ra Δx Vm(x-Δx,t) Vm(x,t) Vm(x+Δx,t)

Δx rm/ c Δx … m … imΔx

Vrest

Vm (x,t)!Vm (x + "x,t) = ra # "x #ii (x,t)

Vm (x,t) "Vm (x + !x,t) !Vm limit !x = 0 = = (x,t) = "r # I (x,t) Eqn 1 !x !x a i

73 Action Potential Propagation

Plug membrane together: get linear cable

Ii(x-Δx,t) Ii(x,t)

ra Δx Vm(x-Δx,t) Vm(x,t) Vm(x+Δx,t)

… … imΔx

By applying Krichhoffʼs current law at node x

im (x,t)!x + Ii (x,t)" Ii (x " !x,t) = 0

$I limit "x = 0 # i (x,t) = i (x,t) Eqn 2 $x m 74

! Linear Cable Equation

Plug membrane together: get linear cable

Ii(x,t)

raΔxVm(x-Δx,t) Vm(x,t) Vm(x+Δx,t)

… … imΔx

#Vm limit "x = 0 (x,t) = $ra % Ii (x,t) #x 2 1 " Vm (x,t) Substitute 2 (x,t) = im (x,t) ra "x !Ii limit !x = 0 " (x,t) = im (x,t) ! !x 75 ! Linear Cable Equation

Plug membrane together: get linear cable

Ii(x,t)

raΔxVm(x-Δx,t) Vm(x,t) Vm(x+Δx,t)

r … m … imΔx

Vrest

If electrical nature of the membrane is constant, then

2 1 " Vm (x,t) Vm (x,t) #Vrest "Vm (x,t) 2 = + cm ra "x rm "t

76

! Linear Cable Equation

Plug membrane together: get linear cable

Ii(x,t)

raΔxVm(x-Δx,t) Vm(x,t) Vm(x+Δx,t)

r … m … imΔx

Vrest

Multiplying both sides by rm we get:

2 2 # Vm (x,t) #Vm (x,t) " = (rm /ra ) " 2 = Vm (x,t) $Vrest +% m #x #t % m = rmcm

λ – steady state space constant 77 τ -Membrane ! Space and Time constants

Rm 2#a Rm a " = rm ri = = Ri 2Ri #a2 g Space constant increases with square root of! the axon radius

R " = r c = m $ C 2#a = R C m m m 2#a m m m g Time constant does not depend on axon size !

78 Steady-State Solutions

Plug membrane together: get linear cable

Ii(x,t)

raΔxVm(x-Δx,t) Vm(x,t) Vm(x+Δx,t)

r … m … imΔx

Vrest

At steady state this term = 0

2 2 # Vm (x,t) #Vm (x,t) " = (rm /ra ) " 2 = Vm (x,t) $Vrest +% m #x #t % m = rmcm

d 2V(x) "2 = V (x) (V = V #V ) dx 2 m rest 79 !

! Steady-State Solutions

Plug membrane together: get linear cable

Iinj(x,t) Ii(x,t)

raΔxVm(x-Δx,t) Vm(x,t) Vm(x+Δx,t)

r … m … imΔx

Vrest

Vm (x,t)!Vrest !Vm (x,t) im (x,t) = + cm ! Iinj (x,t) rm !t

2 1 " Vm (x,t) 2 (x,t) = im (x,t) 80 ra "x

! Steady-State Solutions

Plug membrane together: get linear cable

Iinj(x,t) Ii(x,t)

raΔxVm(x-Δx,t) Vm(x,t) Vm(x+Δx,t)

r … m … imΔx

Vrest

2 1 ! Vm (x,t) 2 (x,t) = im (x,t) ra !x

2 1 ! Vm (x,t) Vm (x,t)!Vrest !Vm (x,t) 2 (x,t) = + cm ! Iinj (x,t) ra !x rm !t 81 Steady-State Solutions

Plug membrane together: get linear cable

Iinj(x,t) Ii(x,t)

raΔxVm(x-Δx,t) Vm(x,t) Vm(x+Δx,t)

r … m … imΔx

Vrest

2 1 ! Vm (x,t) Vm (x,t)!Vrest !Vm (x,t) 2 (x,t) = + cm ! Iinj (x,t) ra !x rm !t

d 2V(x) At steady state 2 ! 2 = V(x)! rmIinj (x,t) (V = Vm !Vrest ) dx 82 Steady-State Solutions

Plug membrane together: get linear cable

Iinj(x,t) Ii(x,t)

raΔxVm(x-Δx,t) Vm(x,t) Vm(x+Δx,t)

r … m … imΔx

Vrest

d 2V(x) "2 = V (x) # r I (x) dx 2 m inj a general solution to this ODE for constant current applied at x V(x) = Aexp(x /") + Bexp(#x /") ! A and B are constants will depend on the boundary conditions 83

! Steady-State Solutions

Plug membrane together: get linear cable

Iinj(x) Ii(x,t)

raΔxVm(x-Δx,t) Vm(x,t) Vm(x+Δx,t)

r … m … imΔx

Vrest

d 2V(x) "2 = V (x) $ r I (x) #x 2 m inj For boundary-value solutions it is often convenient to use another pair of solutions to the second- order ordinary differential equation exp(x) + exp("x) cosh(x) = !V (x) = Aexp(x /") + Bexp(#x /") 2 exp(x) " exp("x) V(x) = Acosh(x / !)+ Bsinh(x / !) sinh(x) = 84 2 ! ! Case 1: Infinite Cable

d 2V(x) "2 = V (x) # r I (x) dx 2 m inj

V(x) = Aexp(x /") + Bexp(#x /") ! For the case in which the cable extends from -∞ to +∞ we require V (-∞) = V (∞) = 0. What should be values of A and B to ! obey this boundary conditions?

For x > 0; A = 0 and B = V0 (where V0 is the membrane voltage at the site of current injection)

Similarly for x<0; A = V0 and B = 0

V(x) = V0 exp(" x /#)

85

! Case 2: Sealed End Cable

d 2V(x) "2 = V (x) # r I (x) dx 2 m inj

V(x) = Acosh(x / !)+ Bsinh(x / !) ! The idea for sealed end is that current should not flow across a boundary at length x = l dV = "ra Iinj B dx At x = 0 "ra Iinj = x=0 # dV B r I = 0 = " a inj # dx x=l

V(x) = Acosh(x / !)+ Bsinh(x / !) At x = l !0 = Asinh(l / !)+ Bcosh(l / !) ! dV(x) A B A = !Bcoth(l / !) = sinh(x /!" ) + cosh(x /") A = r I ! coth(l / !) dx " " a inj 86

! Case 2: Sealed End Cable

d 2V(x) "2 = V (x) # r I (x) dx 2 m inj

V(x) = Acosh(x / !)+ Bsinh(x / !) ! The idea for sealed end is that current should not flow across a boundary at length x = l dV = "ra Iinj dx x=0

dV B = "ra Iinj # = 0 dx x=l A = ra Iinj "coth(l /") V(x) = Acosh(x / !)+ Bsinh(x / !) ! V(x) = raIinj![coth(l / !)cosh(x / !)!sinh(x / !)] ! ! r I ! V(x) = a inj [cosh(l / !)cosh(x / !)!sinh(l / !)sinh(x / !)] sinh(l / !) 87 Case 2: Sealed End Cable

d 2V(x) "2 = V (x) # r I (x) dx 2 m inj V(x) = Acosh(x /") + Bsinh(#x /")

The idea for! sealed end is that current should not flow across a boundary at length x = l ! dV = "ra Iinj dx x=0 dV = 0 B = "ra Iinj # dx x=l A = ra Iinj "coth(l /") V(x) = Acosh(x /") + Bsinh(#x /")

V(x) = raIinj![coth(l / !)cosh(x!/ ! )!sinh(x / !)] ! V (0)cosh[(l ! x) / !] ! ! V(x) = m 0 " x " l sealed x = l cosh(l / !) 88 Vm (0) = raIinj! coth(l / !) Case 3: Killed End Cable

d 2V(x) "2 = V (x) # r I (x) dx 2 m inj

V(x) = Acosh(x /") + Bsinh(#x /") ! The idea here is that one of a boundaries is clamped to resting ! membrane potential dV = "ra Iinj dx x=0 V(l) = 0

V (0)sinh[(l ! x) / !] V(x) = m 0 " x " l killed end x = l s!in h(l / !)

Vm (0) = raIinj! tanh(l / !) 89 Cable Equation Solutions

90 Propagation of AP along an axon

91

92 Saltatory conduction in myelinated axons

93 The Physiology of Excitable Cells. Cambridge: Cambridge Univ. Press, 1971