Control of Cell Volume and Membrane Potential

James Sneyd

Auckland University, New Zealand

Basic reference: Keener and Sneyd, Mathematical Physiology (Springer, 1998)

A nice cell picture

• 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

Typical ionic concentrations (in mM)

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

The cell at steady state

3 Na+

2 K+

Cl-

Ca2+

We need to model

• pumps and exchangers

• ionic currents

• osmotic forces

OsmosisP1 P2

waterwater +Solvent(conc. c)

At equilibrium:

€

P1 + kcT = P2

Note: equilibrium only. No information about the flow.

The cell at steady state

3 Na+

2 K+

Cl-

Ca2+

We need to model

• pumps and exchangers

• ionic currents

• osmotic forces

I’ll talk about this a lot more in my next talk.

Na,K-ATPase

CalciumATPase

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.

• Not all that many pump models that I know of. It doesn't seem to be a popular modelling area. I have no idea why.

A Simple ATPase

E E•ATP E•ATP•L

Ee•ATP•LEe•ATPEe•ADP•P

Inside

Outside

Li

Lo

ATP

ADP + P

k1 k2

k3

k4k5

k6

k-1 k-2

k-3

k-4k-5

k-6

€

J = k1[ATP ][E] − k−1[E • ATP ]

=

[ATP ][L e ]

[ADP][P][L i]− K1K2K3K4K5K6

a nasty function of the rate constants

=

[ATP ][L e ]

[ADP][P][L i]−

[ATP ]eq[L e ]eq

[ADP]eq[P]eq[L i]eq

a nasty function of the rate constants

Note how the flux is driven by how far the concentrations are away from equilibrium

flux

Reducing this simple model

E E•ATP E•ATP•L

Ee•ATP•LEe•ATPEe•ADP•P

Li

Lo

ATP

ADP + P

k1 k2

k3

k4k5

k6

k-1 k-2

k-3

k-4k-5

k-6

X3

Y3Y2Y1

X1 X2

Na+-K+ ATPase (Post-Albers)

E(N3)

Ee•K2

E(K2)

internal

occluded

external

Ei•K2 Ei•N3Ei

Ee•N3Ee

ATP

P

2Ki+

3Nai+

ADP

2Ke+

3Nae+

1 2

3

4

56

7

8

X1

Y2Y1

Z1 Z2

X3X2

Y3

Simplified Na+-K+ ATPase

X

Z2

Y

Z1

internal

occluded

external

3Nai+ 2Ki

+

2Ke+

3Nae+

ATP

P

ADPk-3

� 3� -8

� 7

� -4

k8

k-7k4

The cell at steady state

3 Na+

2 K+

Cl-

Ca2+

We need to model

• pumps

• ionic currents

• osmotic forces

The Nernst equation

€

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 is the crucial question

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

€

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

€

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

⎛

⎝ ⎜

⎞

⎠ ⎟

€

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

Poisson-Nernst-Planck equations.

PNP equations.

The short-channel limit

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

€

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

€

⇒ 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.

€

Then 1

λ2

d2φ

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

c1 = c2 ⇒ 2dc1

dx= −J1 − J2

€

⇒ 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.

A Model of Volume Control

Putting together the three components (pumps, currents and osmosis) gives.....

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+ (horrors!). Cl- just equilibrates passively.€

−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

Note how this is a reallycrappy pump model

Charge and osmotic balance

€

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

€

≤−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). Just solve.

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)

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

€

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 squid

Don't believe people thattell you that this is a smallsquid

Hodgkin Huxley

The reality

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.

€

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

gNa + gK

€

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 just to assume that the internal and external ionic concentrations are constant.

• Justification: Firstly, it takes only small currents to get large voltage deflections, and thus only small numbers of ions cross the membrane. Secondly, 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

€

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.

€

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 just 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.

€

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 still easy 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.

Basic enzyme kinetics

Law of mass actionGiven a basic reaction

A + B Ck1

k-1

we assume that the rate of forward reaction is linearly proportional to the concentrations of A and B, and the back reaction is linearly proportional to the concentration of C.

€

d[A]

dt= k−1[C] − k1[A][B]

Equilibrium

Equilibrium is reached when the net rate of reaction is zero. Thus

€

k−1[C] − k1[A][B] = 0

€

K1[C] = [A][B], K1 =k−1

k1

= eΔG0 / RTor

This equilibrium constant tells us the extent of the reaction, NOT its speed.

change in Gibb’sfree energy

Enzymes

• Enzymes are catalysts, that speed up the rate of a reaction, without changing the extent of the reaction.

• They are (in general) large proteins and are highly specific, i.e., usually each enzyme speeds up only one single biochemical reaction.

• They are highly regulated by a pile of things. Phosphorylation, calcium, ATP, their own products, etc, resulting in extremely complex webs of intracellular biochemical reactions.

Basic problem of enzyme kinetics

Suppose an enzyme were to react with a substrate, giving a product.

S + E P + E

If we simply applied the law of mass action to this reaction, the rate of reaction would be a linearly increasing function of [S]. As [S] gets very big, so wouldthe reaction rate.

This doesn’t happen. In reality, the reaction rate saturates.

Michaelis and Menten

In 1913, Michaelis and Menten proposed the following mechanism for a saturating reaction rate

S + E k1

k-1

C k2 P + E

Complex. product

• Easy to use mass action to derive the equations.• There are conservation constraints.

Equilibrium approximation

€

k−1c = k1se

And thus, since

€

c + e = e0

€

c =e0s

Ks + s

Thus

€

V = k2c =k2e0s

Ks + s=

Vmaxs

Ks + s

reaction velocity

Pseudo-steady state approximation

€

(k−1 + k2)c = k1se

And thus, since

€

c + e = e0

€

c =e0s

Km + s

Thus

€

V = k2c =k2e0s

Km + s=

Vmaxs

Km + s

reaction velocityLooks very similar to previous, but is actually quite different!

Basic saturating velocity

s

V

Vmax

Km

Vmax/2

Lineweaver-Burke plots

€

1

V=

1

Vmax

+Km

Vmax

1

s

Plot, and determine the slope and intercept to get the required constants.

Cooperativity

S + E k1

k-1

C1k2 P + E

S + C1 k3

k-3

C2k4 P + E

Enzyme can bind two substrates molecules at different binding sites.

or

E C1 C2

E E

S S

S S

P P

Pseudo-steady assumption

€

c1 =K2e0s

K1K2 + K2s + s2

c1 =e0s

2

K1K2 + K2s + s2

V = k2c1 + k4c2 =(k2K2 + k4s)e0s

K1K2 + K2s + s2

Note the quadraticbehaviour

Independent binding sites

€

k1 = 2k3 = k+

2k−1 = k−3 = k−

2k2 = k4

E C1 C2

E E

S S

S S

P P

2k+ k+

2k-k-

€

V = 2k2e0s

K + sJust twice the single binding rate, as expected

Positive/negative cooperativity

Usually, the binding of the first S changes the rate at which the second S binds.

• If the binding rate of the second S is increased, it’s called positive cooperativity

• If the binding rate of the second S is decreased, it’s called negative cooperativity.

Hill equation

In the limit as the binding of the second S becomes infinitely fast, we get a nice reduction.

€

Let k3 → ∞, and k1 → 0, while keeping k1k3 constant.

€

V =(k2K2 + k4s)e0s

K1K2 + K2s + s2→

Vmaxs2

Km2 + s2

Hill equation, withHill coefficient of 2.

This equation is used all the time to describe a cooperative reaction. Mostly use of this equation is just a heuristic kludge.

VERY special assumptions, note.

Another fast equilibrium model ofcooperativity

E C1 C2

E E

S S

S S

P P

Let C=C1+C2

€

V = k2c1 + k4c2 =k2K3 + k4s

s + K3

⎛

⎝ ⎜

⎞

⎠ ⎟c = ϕ (s)c

k-1

k1 k3

k-3

k2 k4

S + E k1

k-1

C s)P + E

Monod-Wyman-Changeux model

A more mechanistic realisation of cooperativity.

Equilibrium approximation

Don’t even think about a pseudo-steady approach. Waste of valuable time.

€

Y =r1 + 2r2 + t1 + 2t2

2(r0 + r1 + r2 + t0 + t1 + t2)

K1r1 = 2sr0,K

which gives

€

Y =sK1

−1(1+ sK1−1) + K2

−1[sK3−1(1+ sK3

−1)]

(1+ sK1−1)2 + K2

−1(1+ sK3−1)2

occupancy fraction

and so on for all the other states

Note the sigmoidal character of this curve

Reversible enzymes

Of course, all enzymes HAVE to be reversible, so it’s naughty to put no back reaction from P to C. Should use

S + E k1

k-1

Ck2

P + Ek-2

I leave it as an exercise to calculate that

€

V =e0(k1k2s − k−1k−2 p)

k1s + k−1p + k−1 + k2

Allosteric modulation

substrate binding

inhibitorbinding at adifferent site

this state canform no product

(Inhibition in this case, but it doesn’t have to be)

X

Y Z

Equilibrium approximation

€

(e0 − x − y − z)s − K1x = 0

(e0 − x − y − z)i − K3y = 0

ys − K1z = 0

and thus

x =e0K3

K3 + i

s

K1 + s

V = k2x =Vmax

1+ i /K3

s

K1 + s

X

Y Z

Could change these rate constants, also.

Inhibition decreases theVmax in this model