Control of Cell Volume and Membrane Potential

Basic reference: Keener and Sneyd, Mathematical Physiology

• The cell is full of stuff. Proteins, ions, fats, etc.

• Ordinarily, these would cause huge osmotic pressures, sucking water into the cell.

• The cell membrane has no structural strength, and the cell would burst.

Basic problem

• Cells carefully regulate their intracellular ionic concentrations, to ensure that no osmotic pressures arise

• As a consequence, the major ions Na+, K+, Cl- and Ca2+ have different concentrations in the extracellular and intracellular environments.

• And thus a voltage difference arises across the cell membrane.

• Essentially two different kinds of cells: excitable and nonexcitable.

• All cells have a resting membrane potential, but only excitable cells modulate it actively.

Basic solution

The cell at steady state

3 Na+

2 K+

Cl-

Ca2+

We need to model

• pumps

• ionic currents

• osmotic forces

Typical ionic concentrations

Squid Giant Axon Frog Sartorius Muscle

Human Red Blood Cell

Intracellular

Na+ 50 13 19

K+ 397 138 136

Cl- 40 3 78

Extracellular

Na+ 437 110 155

K+ 20 2.5 5

Cl- 556 90 112

Active pumping

• Clearly, the action of the pumps is crucial for the maintenance of ionic concentration differences

• Many different kinds of pumps. Some use ATP as an energy source to pump against a gradient, others use a gradient of one ion to pump another ion against its gradient.

• A huge proportion of all the energy intake of a human is devoted to the operation of the ionic pumps.

Na+-K+ ATPase

Na+-K+ ATPase

Na+•C•P K+•C•P

Na+•C K+•C

P

Na+K+

ADP

Inside the cell

Outside the cell

K+ Na+

Step 1

Step 2

Step 3

Step 4ATP

OsmosisP1 P2

waterwater +Solvent(conc. c)

At equilibrium:

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P1 + kcT = P2

Note: equilibrium only. No information about the flow.

The Nernst equation

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Vi −Ve =RT

Fln

[S]e

[S]i

⎛

⎝ ⎜

⎞

⎠ ⎟

Note: equilibrium only. Tells us nothing about the current. In addition, there is very little actual ion transfer from side to side.

We'll discuss the multi-ion case later.

[S]e=[S’]e[S]i=[S’]i

Vi Ve

Permeable to S,not S’

(The Nernst potential)

Only very little ion transferspherical cell - radius 25 m

surface area - 8 x 10-5 cm2

total capacitance - 8 x 10-5 F (membrance capacitance is about 1 F/cm2)

If the potential difference is -70 mV, this gives a total excess charge on the cell

membrane of about 5 x 10-12 C.

Since Faraday's constant, F, is 9.649 x 104 C/mole, this charge is equivalent to

about 5 x 10-15 moles.

But, the cell volume is about 65 x 10-9 litres, which, with an internal K+

concentration of 100 mM, gives about 6.5 x 10-9 moles of K+.

So, the excess charge corresponds to about 1 millionth of the background K+

concentration.

Electrical circuit model of cell membrane

C

outside

inside

Iionic C dV/dt

€

CdV

dt+ Iionic = 0

Vi −Ve = V

How to model this

How to model Iionic

• Many different possible models of Iionic

• Constant field assumption gives the Goldman-Hodgkin-Katz model

• The PNP equations can derive expressions from first principles (Eisenberg and others)

• Barrier models, binding models, saturating models, etc etc.

• Hodgkin and Huxley in their famous paper used a simple linear model

• Ultimately, the best choice of model is determined by experimental measurements of the I-V curve.

Two common current models

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INa = gNa (V −VNa )

INa = PNa

F 2

RTV

[Na+]i −[Na+]e exp −VF

RT( )

1− exp −VF

RT( )

⎡

⎣

⎢ ⎢

⎤

⎦

⎥ ⎥

GHK model

Linear model

These are the two most common current models. Note how they both have the same reversal potential, as they must.

(Crucial fact: In electrically excitable cells gNa (or PNa) are not constant, but are functions of voltage and time. More on this later.)

Electrodiffusion: deriving current models

x=0 x=L

[S1+] = [S2

-] = ci[S1

+] = [S2-] = ce

S1

S2

Inside Outside

f (0) = V f ( ) = 0L

cell membrane

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d2φ

dx 2= −λ2(c1 − c2), λ2 = stuff × L2

J1 = −D1

dc1

dx+

F

RTc1

dφ

dx

⎛

⎝ ⎜

⎞

⎠ ⎟

J2 = −D2

dc2

dx−

F

RTc2

dφ

dx

⎛

⎝ ⎜

⎞

⎠ ⎟

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c1(0) = c i, c1(L) = ce

c2(0) = c i, c2(L) = ce

φ(0) = V , φ(L) = 0

Boundary conditions

Poisson equation andelectrodiffusion

The short-channel limit

If the channel is short, then L ~ 0 and so ~ 0.

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Then d2φ

dx 2= 0, which implies that the electric field,

dφ

dx, is constant through the membrane.

dφ

dx= v ⇒

dc1

dx− vc1 = −J1

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⇒ J1 = vc i − cee

−v

1− e−v

⇒ I1 =D1F

2

LRTV

c i − ce exp −VF

RT( )

1− exp −VF

RT( )

⎛

⎝

⎜ ⎜

⎞

⎠

⎟ ⎟

This is the Goldman-Hodgkin-Katz equation.

Note: a short channel implies independence of ion movement through thechannel.

The long-channel limit

If the channel is long, then 1/L ~ 0 and so 1/ ~ 0.

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Then 1

λ2

d2φ

dx 2= c1 − c2, which implies that c1 ≈ c2 through the membrane.

c1 = c2 ⇒ 2dc1

dx= −J1 − J2

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⇒ c1 = c i + (ce − c i)x

⇒ φ = −v

v1

lnc i

ce

+ 1−c i

ce

⎛

⎝ ⎜

⎞

⎠ ⎟x

⎡

⎣ ⎢

⎤

⎦ ⎥ v1 = nondimensional Nernst potential of ion 1

⇒ J1 =ce − c i

v1

(v − v1)

This is the linear I-V curve.The independence principle is not satisfied, so no independent movement ofions through the channel. Not surprising in a long channel.

Volume control: The Pump-Leak Model

3 Na+

2 K+

X-

Cl-

Na+ is pumped out. K+ is pumped in. So cells have low [Na+] and high [K+] inside. For now we ignore Ca2+(for neurons only!). Cl- equilibrates passively.

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−d

dt(qwN i) = gNa V −

RT

Fln

Ne

N i

⎛

⎝ ⎜

⎞

⎠ ⎟

⎡

⎣ ⎢

⎤

⎦ ⎥+ 3pq

−d

dt(qwK i) = gK V −

RT

Fln

Ke

K i

⎛

⎝ ⎜

⎞

⎠ ⎟

⎡

⎣ ⎢

⎤

⎦ ⎥− 2pq

−d

dt(qwCi) = gCl V +

RT

Fln

Ce

Ci

⎛

⎝ ⎜

⎞

⎠ ⎟

⎡

⎣ ⎢

⎤

⎦ ⎥

cell volume

[Na]ipump rate

Charge and osmotic balance

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qw(N i + K i − Ci) + zxqX = qwe (Ne + Ke − Ce )

N i + K i + Ci +X

w= Ne + Ke + Ce

charge balance

osmotic balance

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≤−1

• The proteins (X) are negatively charged, with valence zx.• Both inside and outside are electrically neutral.• The same number of ions on each side.

• 5 equations, 5 unknowns (internal ionic concentrations, voltage, and volume). Solve and analyze.

Steady-state solution

If the pump stops, the cell bursts, as expected.The minimal volume gives approximately the correct membrane potential.In a more complicated model, one would have to consider time dependence also. And the real story is far more complicated.

RVD and RVI Okada et al., J. Physiol. 532, 3, (2001)

RVD and RVI Okada et al., J. Physiol. 532, 3, (2001)

Lots of interesting unsolved problems

• How do organsims adjust to dramatic environmental changes (T. Californicus)?

• How do plants (especially in arid regions) prevent dehydration in high salt environments? (They make proline.)

• How do fish (salmon) deal with both fresh and salt water?

• What happens to a cell and its environment when there is ischemia?

Ion transport

• How can epithelial cells transport ions (and water) while maintaining a constant cell volume?

• Spatial separation of the leaks and the pumps is one option.

• But intricate control mechanisms are needed also.

• A fertile field for modelling. (Eg. A.Weinstein, Bull. Math. Biol. 54, 537, 1992.) The KJU model.

Koefoed-Johnsen and Ussing (1958).

Steady state equations

€

PNavN i − Nme−v

1− e−v

⎛

⎝ ⎜

⎞

⎠ ⎟+ 3qpN i = 0

PKvK i − Kse

−v

1− e−v

⎛

⎝ ⎜

⎞

⎠ ⎟− 2qpN i = 0

PClvCi − Cse

v

1− ev

⎛

⎝ ⎜

⎞

⎠ ⎟= 0

w(N i + K i − Ci) + zX = 0

N i + K i + Ci +X

w= Ns + Ks + Cs

Note the different current and pump models

electroneutrality

osmotic balance

Transport control

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Nm

Ns

<1+

3pq

PNa

1+2pq

PK

Simple manipulations show that a solution exists if

Clearly, in order to handle the greatest range of mucosal to serosal concentrations, one would want to have the Na+ permeability a decreasing function of the mucosal concentration, and the K+ permeability an increasing function of the mucosal Na+ concentration.

As it happens, cells do both these things. For instance, as the cell swells (due to higher internal Na+ concentration), stretch-activated K+ channels open, thus increasing the K+ conductance.

Inner medullary collecting duct cells

A. Weinstein, Am. J. Physiol. 274 (Renal Physiol. 43): F841–F855, 1998.

IMCD cellsReal men deal with real cells, of course.

Note the large Na+ flux from left to right.

Active modulation of the membrane potential: electrically excitable cells

Hodgkin, Huxley, and the Giant Squid Axon

Don't believe people thattell you that HH worked on aGiant squid axon

Hodgkin Huxley

The reality

It was a squidgiant axon!

Resting potential• No ions are at equilibrium, so there are continual background currents. At steady-state, the net current is zero, not the individual currents.• The pumps must work continually to maintain these concentration differences and the cell integrity.• The resting membrane potential depends on the model used for the ionic currents.

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gNa (V −VNa ) + gK (V −VK ) = 0 ⇒ Vsteady =gNaVNa + gKVK

gNa + gK

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PNa

F 2

RT

⎛

⎝ ⎜

⎞

⎠ ⎟V

[Na+]i −[Na+]e exp(−VF

RT)

1− exp(−VF

RT)

⎛

⎝ ⎜

⎞

⎠ ⎟+ PK

F 2

RT

⎛

⎝ ⎜

⎞

⎠ ⎟V

[K+]i −[K+]e exp(−VF

RT)

1− exp(−VF

RT)

⎛

⎝ ⎜

⎞

⎠ ⎟= 0

⇒ Vsteady =RT

Fln

PNa [Na+]e + PK [K+]e

PNa [Na+]i + PK [K+]i

⎛

⎝ ⎜

⎞

⎠ ⎟

linear current model (long channel limit)

GHK current model (short channel limit)

Simplifications

• In some cells (electrically excitable cells), the membrane potential is a far more complicated beast.

• To simplify modelling of these types of cells, it is simplest to assume that the internal and external ionic concentrations are constant.

• Justification: First, small currents give large voltage deflections, and thus only small numbers of ions cross the membrane. Second, the pumps work continuously to maintain steady concentrations inside the cell.

• So, in these simpler models the pump rate never appears explicitly, and all ionic concentrations are treated as known and fixed.

Steady-state vs instantaneous I-V curves

• The I-V curves of the previous slide applied to a single open channel

• But in a population of channels, the total current is a function of the single-channel current, and the number of open channels.

• When V changes, both the single-channel current changes, as well as the proportion of open channels. But the first change happens almost instantaneously, while the second change is a lot slower.

€

I = g(V , t)φ(V )

I-V curve of singleopen channel

Number of open channels

Example: Na+ and K+ channels

K+ channel gating

S0 S1 S2

2

2

S00 S01

S10 S11

€

dx0

dt= βx1 − 2αx0

dx2

dt= αx1 − 2βx2

x0 + x1 + x2 =1

€

x0 = (1− n)2

x1 = 2n(1− n)

x2 = n2

dn

dt= α (1− n) − βn

Na+ channel gating

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x21 = m2h

dm

dt= α (1− m) − βm

dh

dt= γ(1− h) −δh

2

2

S00 S01

S10 S11

S02

S12

2

2

S i j

inactivation activation

activation

inactivation

Experimental data: K+ conductanceIf voltage is stepped up and held fixed, gK

increases to a new steady level.

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gK = g K n4

dn

dt= α (V )(1− n) − β (V )n

τ n (V )dn

dt= n∞(V ) − n

time constant

steady-state

four subunits

Now fit to the data

rate of rise gives n

steady state gives n∞

Experimental data: Na+ conductanceIf voltage is stepped up and held fixed, gNa

increases and then decreases.

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gNa = g Nam3h

τ h (V )dh

dt= h∞(V ) − h

τ m (V )dm

dt= m∞(V ) − m

time constant

steady-state

Four subunits.Three switch on.One switches off.

Fit to the data is a little more complicated now, but the same in principle.

Hodgkin-Huxley equations

∞

∞

∞€

CdV

dt+ g K n4 (V −VK ) + g Nam3h(V −VNa ) + gL (V −VL ) + Iapp = 0

τ n (V )dn

dt= n∞(V ) − n

τ m (V )dm

dt= m∞(V ) − m, τ h (V )

dh

dt= h∞(V ) − h

generic leak

applied current

much smaller thanthe others

inactivation(decreases with V)

activation(increases with V)

An action potential

• gNa increases quickly, but then inactivation kicks in and it decreases again.

• gK increases more slowly, and only decreases once the voltage has decreased.

• The Na+ current is autocatalytic. An increase in V increases m, which increases the Na+ current, which increases V, etc.

• Hence, the threshold for action potential initiation is where the inward Na+ current exactly balances the outward K+ current.

The fast phase plane: I

€

CdV

dt+ g K n0

4 (V −VK ) + g Nam3h0(V −VNa ) + gL (V −VL ) + Iapp = 0

τ m (V )dm

dt= m∞(V ) − m

n and h are slow, and so stay approximately at their steady states while V and m change quickly

The fast phase plane: II

As n and h change slowly, the dV/dt nullcline moves up, ve and vs merge

in a saddle-node bifurcation, and disappear.

Vs is the only remaining steady-state,

and so V returns to rest.

h0 decreasing

n0 increasing

The fast-slow phase planeTake a different cross-section of the 4-d system, by setting m=m∞(v), and using the useful fact that n + h = 0.8 (approximately). Why? Who knows. It just is. Thus

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CdV

dt+ g K n4 (V −VK ) + g Nam∞

3 (0.8 − n)(V −VNa ) + gL (V −VL ) + Iapp = 0

τ n (V )dn

dt= n∞(V ) − n

Oscillations

When a current is applied across the cell membrane, the HH equations can exhibit oscillatory action potentials.

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CdV

dt+ Iionic + Iapplied = 0

V

Iapplied

HB

HB

• Simplified models - FHN, Morris Lecar, Mitchell-Schaffer-Karma…

• More detailed models - Noble, Beeler-Reuter, Luo-Rudy, … .

• Forced oscillations of single cells - APD alternans, Wenckebach patterns.

• Other simplified models - Integrate and Fire, Poincare oscillator

• Networks and spatial coupling (neuroscience, cardiology, …)

Where does it go from here?