Foundations I Fall, 2016 Ionic Bases of Cellular Potentials Ionic - - PowerPoint PPT Presentation

Foundations I Fall, 2016 Ionic Bases of Cellular Potentials

Ionic Bases of Cellular Potentials

All living cells have membrane potentials

The resting membrane potential comes about from an unequal distribution of permeant ions inside and outside the cell The unequal distribution is made possible by the existence of a cell membrane

• I. Physical Bases of The Resting Membrane

Potential

Cell loses contents when damaged Intracellular injections of dyes spreads throughout cell but not outside Some types of molecules enter cell from outside but others won't

Early observations suggest that cells possess a surface barrier

Ernest Overton (1910)

Highly lipophillic substances entered relatively easily But some hydrophillic substances also entered, but slowly pores ~ 6Å

The Porous Lipid Skin Model

ergo, hypothesis

the penetration of uncharged molecules (through the membrane) is governed by their size and lipid solubility the penetration of charged substances (through the pores) is governed by their size and charge.

Gorter and Grendel (1925)

surface area of monomolecular layer of extracted lipids ~ 2X that

• f cells they came from

"chromocytes [red blood cells] are covered by a layer of fatty substances that is two molecules thick" (Gorter and Grendel 1925).

So the membrane must be a bilayer

Electron microscopic evidence of a bilayer

Singer-Nicholson Fluid Mosaic Lipid Bilayer (1972)

The bilayer is not symmetric Most of the mass is not lipid about 1/2 protein, 1/3 lipid and the rest carbohydrate At body temperature the lipid bilayer has the consistency of

• live oil

polar heads hydrophobic tails

Surface Proteins

glycoproteins - short sugars - Cell surface antigens

Intrinsic Proteins

receptors, ion channels, etc.

Proteoglycans

long polysaccharides - cell-cell recognition

Cytoplasmic Proteins

cytoskeleton

actin ankyrin spectrin

Biophysical and Thermodynamic Origins

• f the Resting Membrane Potential

The Principle of Electrical Neutrality (PEN)

In a chemical compound or in a solution, the sum of cation charges must equal the sum of anion charges

In 1911 Donnan studied the conditions under which equilibrium is established between two electrolytic solutions separated by a semipermeable membrane-- that is, by a membrane through which the solvent and some, but not all, of the dissolved ions can pass. In the absence of such a membrane, the solvent and every species of dissolved ion will diffuse freely from each solution into the

• ther, until the composition of the two solutions becomes the same.

The semipermeable membrane, however, prevents the transfer of at least one ionic species, and the preservation of electrical neutrality limits the diffusion

• f that species' oppositely charged partner. Nevertheless, some movement of

mobile ions does occur, and the compositions of the solutions change; as a result, the final distribution of the ionic species is unequal, and there is a measurable difference in the electric potential of the solutions on each side of the membrane. This was predicted some years earlier by J.W. Gibbs on thermodynamic grounds and thus the equilibrium is known as the Gibbs- Donnan equilibrium.

Frederick G. Donnan

[

• ut

K +

] [

in

K + ] = [

in

Cl− ]

[

• ut

Cl−

]

[

• ut

K +

]*[

• ut

Cl−

] = [

in

K + ]*[

in

Cl− ]

• r

Calculation of the Gibbs-Donnan Equilibrium

But... What about the water?

At Gibbs-Donnan Equilibrium, the system is in electrical and chemical equilibrium, but not in osmotic equilibrium...

Control of cellular volume

• 1. Make cell impermeable to water
• 2. Limit volume with a rigid cell wall
• 3. Balance the osmotic pressure

add an impermeant cation (Na+) outside

X X

• II. Ionic Basis of the Resting Membrane Potential

The plasma membrane is semi-permeable (high permeability to K+ and Cl-, low permeability to most everything else) so ions distribute themselves across membrane in order to satisfy the PEN and the Gibbs- Donnan Equilibrium, at which point the diffusional forces are exactly balanced by the electrical gradient across the membrane. How is the equilibrium point described?

Take a squid giant axon

in

• ut

mM mM K+ 400 20 Na+ 50 440 Cl- 52 560 proteins- 385

Let’s deal with one ion, K+

in

flux =

in

r [K+]out

• ut

flux

=

• ut

r

[K+]in

where r is the rate constant

netflux=

in

r [K+]out −

• ut

r

[K+]in = ρK

where ρ is the net flux, or permeability

if the membrane were not charged, then

• ut

r

in

r

= and since in a squid,

[K+]out [K+]in = 20 x

then the flux should be 20 x the flux

in

• ut

flux is net rate of movement and is proportional to concentration

But since the membrane is negatively charged, it is harder for K+ to move out than it is to move in. Thus the two rate constants differ depending on valence, membrane potential and temperature.

in

r = ρzV'

zV'

(e

−1)

• ut

r

= ρzV'

−zV'

(1−e

)

where z = valence and V’ = thermodynamic potential, the membrane potential divided by (RT/F), e.g., V’ =Vm/(RT/F)

where F= the faraday constant and R is the gas constant

At room temperature RT/F =25 so for a cell at-75 mV, V’=-75/25 = -3.02

in

• 3.02

e

using equation above for r ,

= 0.0488

r

in = 3.02ρ/(0.0488-1) = 3.175ρ

i.e., the inward flow of a positively charged ion is more than 3 times greater than it would be if there were no membrane potential

netflux= zV' (

zV'

−1) ρ[K+]out − zV'

−zV '

(1 e

) ρ[K+]in

e −

multiply by to get denominators the same

zv

e

'

netflux= zV' (

zV'

−1) ρ[K+]out − zV'

zV'

e

zV '

(e

−1) ρ[K+]in

e

[K+]out = [K+]in

zv'

e

at equilibrium, net flux=0 so

zV'= ln[K+]out [K+]in

take ln

zVmF RT = ln[K+]out [K+]in

substitute back for V’

Vm= RT zF ln[K+]out [K+]in

solve for Vm

What conditions obtain at equilibrium?

[K+]out [K+] =

zV'

e

rearrange

Nernst Equation

in

• ut

mM mM mV K+ 400 20

• 75

Na+ 50 440 +55 Cl- 52 560

• 60

Ca++ 0.0001 10 +125

Squid

in

• ut

mM mM mV K+ 140 5

• 90

Na+ 15 145 +61 Cl- 4 110

• 89

Ca++ 0.0001 10 +136 Typical Mammalian Cell E Eq E Eq

The PEN is false!

By how much? A net difference of 600 charges/µm gives a membrane potential of 10 mV

2

The electrostatic force is so much stronger than the diffusional force (10 times) that for a cell with a resting membrane potential of

• 80 mV, for every 100,000 intracellular cations there are ~100,001

anions

18

The violation of the PEN only occurs within +/- 1 µm of the cell membrane, held in place by the membrane capacitance. The rest of the outside and inside of the cell has no charge.

The PEN is true!

In a small axon the ratio is even less - perhaps 0.000000007

The existence of a membrane potential indicates that

Julius Bernstein

1902 “~The resting membrane potential is due to the selective permeability of the membrane to potassium. It should obey the Nernst equation.”

Text

But there is a difference between the calculated membrane potential based

• n the Nernst equation and

what is observed in real neurons at physiological concentrations of extracellular K+ What does that difference mean?

Text

And what can one conclude from the nature

• f the deviation?

slope = 58 mV/log[K ]

+

The neuron at rest is not in equilibrium. It is in steady- state and requires energy input to stay there.

Na-K ATPase

Na-K ATPase

In some cells the the transport ratio is 1:1 But in most cells, the ratio is Na+>K+. The ratio is often given as 3:2 Na+:K+, but in nerve and muscle estimates range from 4:3 to 5:1.

Na+ -K+ ATPase

In other words, the pump generates a current - it is said to be electrogenic

Vm= ( RT zF )(r[K+]out + b[Na+]out r[K+]in + b[Na+]in)

The contribution of the pump to the resting membrane potential is given by: where r is the pump ratio and b is the ratio of ρNa+:ρK+

Na+ -H+ transporter Ca++ - Mg++ - ATPase HCO 3

• :Cl
• (NDAE)

Other pumps and transporters

K+ - Cl- cotransporter (KCC2)

Slower than movement through channels but still very fast

So how does one predict the resting membrane potential? The Nernst equation only handles one ion at a time

Nernst Equation?

Vm= RT zF ln[K+]out [K+]in

Constant Field Equation

netflux= ρKzV' [K+]out −[K+]in

zV '

e

zV '−1

e

start with net flux for an ion, same as for Nernst

K

i = FρKzV'[K+]out −[K+]in

zV'

e

zV'

e

−1

Na

i

= FρNazV' [Na+]out −[Na+]in

zV'

e

zV'

e

−1

Cl

i

= FρClzV'[Cl−]out −[Cl−]in

zV'

e

zV'

e

−1

turn flux (moles/cm2/sec) into current by multiplying by Faraday constant - do this for each ion

ρK([K+]out −[K+]in

zV '

e

) + ρNa([Na+]out −[Na+]in

zV'

e

) + ρCl([Cl−]in −[Cl−]out

zV'

e

) = 0

at equilibrium the membrane potential is not changing (constant field) so the sum of all the currents must equal 0 rearranging gives

ρK([K+]out + ρNa([Na+]out + ρCl([Cl−]in =

zV'(ρK([K+]in + ρNa([Na+]in + ρCl ([Cl−]out

e

zV'

e

= ρK ([K+]out + ρNa([Na+]out +ρ ([Cl−]

Cl in

ρK ([K+]in + ρNa([Na+]in + ρCl([Cl−]out

rearrange

substitute back V’=Vm/(RT/F) and

zVm/ RT

e

= ρK ([K+]out + ρNa([Na+]out + +ρCl([Cl−]in ρK ([K+]in + ρNa([Na+]in + ρCl([Cl−]out

Take ln and solve for Vm

Vm= RT zF ρK ([K+]out + ρNa([Na+]out +ρCl([Cl−]in ρK ([K+]in + ρNa([Na+]in + ρCl([Cl−]out

The Constant Field, or Goldman-Hodgkin-Katz Equation

If one assumes ρK:ρNa:ρCl at rest = 1:0.04:0.45 (for squid axon) then Vm=58*log (20+17.6+23.4)/(400+2+252) Vm=-59.7 mV

How important is permeability to Cl- to resting membrane potential?

Vm=58*log (20+17)/(400+2) Vm=-60.09 mV

How can you figure this out? Simply remove the Cl- terms from the equation and solve again Although [Cl-]o is critical to synaptic potentials, it plays essentially very little role in the squid resting membrane potential, but a more significant role in mammalian neurons.

Dendritic Electrotonus

Steady-state Electrotonus Transient Electrotonus

• n the role of crusty old guys in the history of science...

"Heavier-than-air flying machines are impossible." (1895) "There is nothing new to be discovered in physics now. All that remains is more and more precise measurement." (1900)

Lord Kelvin (1824-1907)

Professor Willliam Thompson (~1860)

but earlier...

Originated mathematical analogy between flow of heat in solid bodies and the flow of electricity in core conductors (1842) In 1855, Prof. William Thompson (Lord Kelvin) presented to the Royal Society a theoretical analysis of factors responsible for attenuation of signals in the transatlantic telegraph cable that was then being planned. An undersea cable is similar to a nerve fiber. It has a conducting core covered with an insulating sheath, and is surrounded by sea water. As the insulation is not perfect, there will be a finite leakage resistance through the

• insulation. The main quantitative difference is that the core of the cable is made of copper, which is a much

better conductor than the salt solution inside a neuron, and the cable covering is a much better insulator than the cell membrane.

The “Cable Equations”

Wilfred Rall

RALL W. Membrane time constant of motoneurons.

• Science. 1957 Sep 6;126(3271):454.

RALL W. Branching dendritic trees and motoneuron membrane

• resistivity. Exp Neurol. 1959 Nov;1:491-527.

RALL W. Membrane potential transients and membrane time constant of motoneurons. Exp Neurol. 1960 Oct;2:503-32. RALL W. Electrophysiology of a dendritic neuron model. Biophys J. 1962 Mar;2(2 Pt 2):145-67. RALL W. Theory of physiological properties of dendrites. Ann N Y Acad Sci. 1962 Mar 2;96:1071-92.

2006

The father of modern cable analysis of dendritic electronus and compartmental modeling

Define: electrotonic potential V=Vm=Er where Vm is the instantaneous value of the membrane potential and Er is the resting potential Since the conductor (dendrite) is assumed to be isopotential, Er does not vary with distance. is assumed to be the same as Vi V

m

(the potential just inside the membrane)

dV dx = d

i

V dx

i.e., the gradient of the electrical potential along the dendrite is equal to the gradient of the internal potential

v = ir

Ohm’s law

(1)

dV dx = −I i

r

substitute into (1) then rearrange; note that I is axial current and r is axial resistance

i

I = −1

i

r • dV

dx

rearrange

(2)

m

i = dI

dx

From Kirchoff’s law (current flowing into a point = current flowing out of it) the current flowing out through the membrane equals the change in axial current per unit length

(3) Thus the change in V is given by:

m

i = V

m

r

m

i = −d

dx[−1

i

r • dV

dx ]= 1 i r [ d2V dx2 ]

substituting (2) into (3)

V = rm ri

• d 2V

dx2

A fundamental equation of linear cable theory

Given a steady state input at x=0, the spread of V is proportional to the second spatial derivative of the potential times the ratio of r and r

m i

λ2( d2V dx

2 )−V = 0

The equation is usually rearranged as follows:

where

λ = rm ri

λ is called the length constant or characteristic length

(5)

Integrating (5) gives the exponential function

V = V0e

−x λ rearranging gives

V V0 = e

− x λ

V V0 = e−1 = 1 e = 0.37

What happens when x=λ?

At x= λ the potential has decayed to 1/e or ~37% of its peak steady state value 0.37

1/e=

N.B.

In the discussion of the decay of potential along the cable we used r and r not R and R . There is a potential problem with units

i m m i i 2 i 2 m

The difference is that while R and R are expressed in units of Ω, r is expressed as specific internal resistance/cross-sectional area, or R (Ω cm)/πa = Ω/cm and r is specific membrane resistance/ circumference, or R (Ω cm )/2πa (cm). Dividing r by r gives

i m

m

r

i

r

=

m

R

i

R

⋅ a 2

This means that is proportional to the radius of the dendrite, and the larger the radius the larger the ratio and the farther the electrotonic spread.

m

r

i

r

λ = rm ri =

m

R

i

R

⋅ d 4

Length constant depends on diameter

2,000 Ωcm

2

8,000 Ωcm

2

32,000 Ωcm

2

Effect of specific membrane resistance (Rm) on λ

D

G = I V

= 1

m

r

i

r

D

G = π

2 1

m

R

i

R

d

3 2

d

32 = i

d

32

∑

Rall, 1959

input conductance of a dendritic segment

Transient Electrotonus where τ = r c

m m

λ2( d2V dx

2 )−V = 0

The cable equation described previously, is actually a simplification of the more general equation that obtains when the voltage has been applied for a long time, and so time is neglected. The more general form of the equation is

V = λ2∂2V ∂x2 −τ∂V ∂t

The ordinary differential equation (ODE) has been replaced by a partial differential equation (PDE) because now both time and space can vary independently

V = V e

−t τ

The solution of the cable equation in the time domain gives the following exponential function Does this look familiar?

Peeling Exponents to estimate tau

The voltage response to current input into any cable structure can be expressed as an infinite sum of exponentials.

V(x,t) =V∞(x)+ B0e−t /τ 0 + B0e−t /τ1 + B0e−t /τ 2 +........

slope = -1/tm

So how do you find tau experimentally?

plot logV against t

So inject a pulse and measure the time course of the tail (or onset)

y mV x ms