Electrochemical Potentials 1 Biophysics Prof. Radmacher

Membrane Potentials 1. Gibb's free energy and chemical potential We consider one ionic species in a system with two compartments (A and B) separated by a permeable membrane.

A B

nA nB

The concentration of ions in each compartment is nA and nB, respectively. The flux into compartment A is ΔnA, the flux into compartment B is ΔnB. Since the number of particles is conserved we define: 1. Δn := ΔnA = - ΔnB By transporting an ion from compartment A to B we change the free energy of the entire system. The total change in free energy is the difference in change of free energy for each compartment 2. ΔG := ΔGA - ΔGB = - ( µA - µB ) * Δn = - µAB * Δn Here µA and µB are the chemical potential for each compartment, which is basically the change in free energy when changing the particle number. µAB is the total chemical potential of the system.

3.

€

µi =∂Gi

∂ni with i = A,B If the total chemical potential µAB is positive, the change in free energy ΔG will be negative, and hence there will be a spontaneous flux of particles from B to A, Δn < 0. The particles will flow to the compartment with lowest chemical potential, this is equivalent to the situation in gravitational or electrical fields, where the particles will also move to the location of lowest potential, i.e. lowest potential energy.

Electrochemical Potentials 2 Biophysics Prof. Radmacher

Chemical Potential as driving force for diffusion We can therefore consider the chemical potential difference as the driving field for a force, which is chemical in nature. (In analogy to an electrical potential, as source of an electrical force) We now consider a chemical potential, which is dependent on position. The above system with two compartments was a simple implementation of this general idea. This chemical force can be defined as the derivative of the chemical potential. We need to divide by Avogadro's number, since the chemical potential is defined as energy per mole, the force is the force per particle:

4.

€

Fchemical = −1/NA *∂µ∂x

NA is Avogadro's number (~ 6*1023 per Mole) The particles (ions) being pulled by the chemical force will be slowed down by friction in the medium. There will be a steady state where friction force and chemical force will be identical.

5.

€

Ffriction = −Fv = −kTDv

here F is the friction coefficient and v is the velocity of the particle, k is Boltzmann's constant, T is the absolute temperature and D is the diffusion constant. Often the quantity D/kT is called mobility u. 6. u := D/kT

7.

€

Ffriction = Fchemical

−Fv = −1/NA *∂µ∂x

v = −1F* 1NA

* ∂µ∂x

= −u* 1NA

* ∂µ∂x

We define the current j as concentration c times velocity v: j = v * c We can rearrange the above equation and obtain: 8. j = - c * 1/NA * D/kT * ∂µ/∂x = - c * D/RT * ∂µ/∂x The chemical potential is a function of the particle number and in the case of an ideal gas or an ideal solution we can write: 9. µ = µ0 + R T * ln(c) 10. ∂µ/∂x = R T * 1/c * ∂c/∂x

Electrochemical Potentials 3 Biophysics Prof. Radmacher

We can combine eq. 10 and eq. 8 to obtain: 11. j = - c * D/RT * ∂µ/∂x = - c * D/RT * R T * 1/c * ∂c/∂x = - D * ∂c/∂x This is Fick's first law describing diffusion. So, the driving force behind diffusion is local differences in chemical potential, resulting from the dependence of chemical potential on concentration. Particles diffusing freely are trying to minimize their chemical potential, which happens to be possible at locations of lowest chemical energy, which is at locations of lowest concentration. The electrochemical potential In the case of neuronal membranes the particles carry an electrical charge and there may also be difference in electrical potential, e.g. across the membrane. So we need to extend the chemical potential by adding an electrical contribution, thus defining the electrochemical potential: 12. µelchem = µ0 + µchem + µelectrical = µ0 + R T ln(c) + z F Φ where z is the valency, F is Faraday's number ( F = e * NA, where e is the unit charge, F = 96500 C/Mol) and Φ is the electrical potential. So, we rewrite eq. 8 by using eq. 12:

13.

€

j = −c * DRT

* RT ∂(ln(c))∂x

+ zF ∂Φ∂x

= −1NA

* u*c * RT ∂(ln(c))∂x

+ zF ∂Φ∂x

In equilibrium, no net current is flowing: j = 0, thus equation 13 results in:

14.

€

RT ∂(ln(c))∂x

+ zF ∂Φ∂x

= 0

which can be interpreted as a force balance between chemical and electrical forces.

Electrochemical Potentials 4 Biophysics Prof. Radmacher

Nernst Potential Now, we can come back to our simple system with two compartments, now being separated by a membrane of thickness d. Equation 14 can be simplified:

15.

€

RT Δ(ln(c))∂x

= −zF ∂Φ∂x

RT Δ(ln(c))∂d

= −zF ΔΦd.

RTΔ(ln(c)) = −zFΔFΦRT(lncA − lncB ) = −zFΔΦ

RT ln cAcB

= −zFΔΦ

16.

€

ΔΦ = −RTzFln(cAcB)

where the prefactor RT/F is approximately 59 mV at room temperature. In the case of a cell membrane, compartment A can be associated with the interior of the cell, compartment B with the extra-cellular space.

17.

€

ΔΦ = −RTzFln(c intracextra

)

If z is positive (e.g. K+ ions) the potential will be negative, if there is a larger amount of ions inside the cell than outside. In this case positive ions from the interior will follow to the exterior charging the exterior more positive, or the interior more negative. So, the potential is the potential of the interior with reference to the exterior. For negative ions the sign will be positive in this case.

Electrochemical Potentials 5 Biophysics Prof. Radmacher

Donnan Potential Let's discuss the following situation: we have kations and anions in the intracllular and extracellular space, and in addition charged macro-molecules in the intracellular space.

We use the following abbreviations:

(1)

€

cKo concentration of kations in the extracellular spacecKi concentration of kations in the intracellular spacecAo concentration of anions in the extracellular spacecAi concentration of anions in the intracellular space

cM concentration of macromolecules (only in intracellular space)

zK valency of kationszA valency of anionszM valency of macromolecules

In equilbrium the Nernst equation has to be fulfilled for bot, anions and kations:

Electrochemical Potentials 6 Biophysics Prof. Radmacher

(2)

€

ΔΦ = −RTzKF

ln cKo

cKi

ΔΦ = −RTzAF

ln cAo

cAi

In addition charge neutrality has to be maintainded in the intracellular and extracellular space:

(3)

€

zKcKo + zAcA

o = 0zKcK

i + zAcAi + zMcM = 0

Let's assume for simplying the equations: zA= 1; zK= -1

(4)

€

ΔΦ =RTFln cK

o

cKi

ΔΦ = −RTFln cA

o

cAi

(5)

€

cKo − cA

o = 0cKi − cA

i + zMcM = 0

Equation (4) can be simplified:

(6)

€

RTFln cK

o

cKi = −

RTFln cA

o

cAi

ln cKo

cKi = ln cA

i

cAo

cKo

cKi =

cAi

cAo

(7)

€

cKo *cA

o = cKi *cA

i We define the contration outside c0, since it identical for anions and kations: (8)

€

co = cKo = cA

o and rearrange the condition for charge neutrality in the cell interior: (9)

€

cAi = cK

i + zMcM So, equation (7) reads:

(10)

€

co *co = cKi * cK

i + zMcM( )0 = cK

i 2 + cKi zMcM − c

o2

Electrochemical Potentials 7 Biophysics Prof. Radmacher

This quadratic equation can be solved for ciK:

(11)

€

cKi = −

zMcM2

+zMcM2

2

+ co2

with eq. 9 we get the concentration of anions in the cell interior:

(12)

€

cAi =

zMcM2

+zMcM2

2

+ co2

Since the intracellular concentrations ci

A and ciK are different from the extracellular

concnetrations co there will be a membrane potential.

Electrochemical Potentials 8 Biophysics Prof. Radmacher

Planck Potential Our above example considering only one ionic species is not realistic, since we need at least one species of anions and one species of kations to maintain electro-neutrality. Thus we will extend our model to two ionic species; i.e. a salt in solution. In the general case our salt will consist of anions A and kations K with valencies zA and zK and stochiometry νA and νK, resulting in the following chemical reaction during dissolving: 1.

€

Kν KAν A

⇒νKKzK + νAA

zA Here zA and zK are positive numbers. So, we have to put a "minus" sign in front of equations were the sign of the electrical charge is relevant. This somewhat cluttered nomenclature becomes more clear, when considering an example: 2. MgCl2 -> Mg2+ + 2 Cl- For each ionic species we need to consider its current driven by its chemical potential and by the identical electrical potential.

3a.

€

jA = −1NA

* uA *cA * RT ∂(ln(cA )∂x

− zAF∂Φ∂x

3b.

€

jK = −1NK

* uK *cK * RT ∂(ln(cK )∂x

+ zKF∂Φ∂x

(Nota bene: there is a minus sign in front of zA since it is a positive number, however the electrical force is negative). Charge neutrality requires: 4. cK * zK = cA * zA

Maintenance of charge neutrality also requires that the currents are in balance: 5. jK * zK = jA * zA

The currents are particle currents, and since zK and zA are positive numbers, this equation says, that the there is the same number of anions flowing from inside to outside, than kations are. The sum of the corresponding electrical currents is zero, although, or actually because particles are flowing. By using eq. 3a and 3b we rewrite eq. 5

Electrochemical Potentials 9 Biophysics Prof. Radmacher

6.

€

−zK1NA

* uk *cK * RT ∂(ln(cK ))∂x

+ zKF∂Φ∂x

= −zA1NA

* uA *cA * RT ∂(ln(cA ))∂x

− zAF∂Φ∂x

Using eq. 4 this reduces to:

7.

€

uk * RT ∂(ln(cK ))∂x

+ zKF∂Φ∂x

= uA * RT ∂(ln(cA ))

∂x− zAF

∂Φ∂x

We simplify this equation be defining the salt concentration cs as the amount of salt added. The following relation holds: 8a. cK = νK * cS

8b. cA = νA * cS 9a. ln(cK) = ln(νK) + ln(cS)

9b. ln(cA) = ln(νA) + ln(cS)

10a.

€

∂(ln(cK ))∂x

=∂(ln(nK ))

∂x+∂(ln(cS ))

∂x= 0 +

∂(ln(cS ))∂x

=∂(ln(cS ))

∂x

10b.

€

∂(ln(cA ))∂x

=∂(ln(nA ))

∂x+∂(ln(cS ))

∂x= 0 +

∂(ln(cS ))∂x

=∂(ln(cS ))

∂x

We further simplify eq. 7 by using eq. 10:

11.

€

uk * RT ∂(ln(cS ))∂x

− zKF∂Φ∂x

= uA * RT ∂(ln(cS ))

∂x+ zAF

∂Φ∂x

Rearranging of eq. 11 yields:

12.

€

∂Φ∂x

=uK − uA

zA * uA + zA * uK

* RTF* ∂(ln(cS )

∂x

As above with the Nernst potential we make the transition to a membrane of thickness d, obtaining the final equation for the Planck potential:

13.

€

ΔΦ =uK − uA

zA * uA + zA * uK

* RTF* ln cextracellular

c intracellular

Thus, the membrane potential depends on the salt concentration and on the mobilities of the two species.

Electrochemical Potentials 10 Biophysics Prof. Radmacher

Goldmann equation In the above chapter we have considered the effect of different ionic species and the influence of (passive) mobilities of these ions through the membrane. In addition, we find in biological system active pumps transporting ionic currents through membrane. Thus we need to extent equation by 3 above including active currents:

1.

€

ji = jiactive −

1NA

* ui *ci * RT ∂(ln(ci))∂x

+ ziF∂Φ∂x

Here, the index i denotes the ionic species, which can be anionic or kationic and we don't want to restrict us to the case of two ionic species. For each ionic species we have to write an equation of the form of eq. 1, which all are coupled through the electrical potential. A lengthy derivation with many tricks and side tours, which are not really interesting from the biophysical point of view leads to the following solution:

2.

€

ji − jiactive = −

Dd* zi *F *VM

RT* cinner *e

zi *F*VMRT − couter

ezi *F*VM

RT −1

where all variables are defined as above. VM is the potential across the membrane. The two compartments are now labeled inner and outer to make the application to membrane potentials in neuronal cells more obvious. Since this equation is somewhat hard to read, we introduce renormalized variables: The permeability is defined as: 2. P = D/d The reduced membrane potential is defined as:

3.

€

v =zi *F *VM

RT

The reduced Nernst potential is given by:

4.

€

w = ΔΦzFRT

= −RTzFln(c intracextra

) * zFRT

= −ln(c intracextra

) = ln(cextrac intra

)

Equation 2 can now be rewritten in the following form:

5.

€

ji − jiactive = −P *v * cinner *e

v − couterev −1

6.

€

ji − jiactive = −P *v *cinner

ev − ew

ev −1

Electrochemical Potentials 11 Biophysics Prof. Radmacher

7.

€

ji = jiactive − P *cinner

ev − ew

ev −1*v

ji = jiactive − λ(v ) *v

This has some reminiscence of the Ohmic law, where the current is proportional to the voltage v applied and the conductance λ. In our case λ depends on the voltage, thus we have an non-linear conductance, which needs to be discussed in more detail. Let's discuss a few special cases: 1. w = 0 => Nernst potential zero, no difference in ionic concentration of both sides

8.

€

ji = jiactive − P *cinner

ev − e0

ev −1*v

ji = jiactive − P *cinner *v

This is ohmic behaviour since the conductance λ is constant, it does not depend on the membrane potential. 2. v >> w, membrane potential much larger than Nernst potential. This corresponds to a large external bias. This is analogous to small w, we can disregard w, and obtain again an Ohmic behaviour. 9. ji = jiactive - P * cinner *v. 3. v = w, Membrane potential is equal to Nernst Potential. ev – ew = 0 => l = 0 10. j = jactive How can we extend the model to several ionic species. We obtain equation 7 for each ionic species, which are again coupled via the membrane potential shared by all species. The Nernst potential is different for each species.

11.

€

ji = jiactive

i∑ − P *cinner

i

i∑ ev − ewi

ev − l*v

The total current is the sum over all active currents of each species plus all electrochemical currents of each species, where the concentration inside the membrane and the Nernst potential of each species (which is a function of the concentration inside and outside) play a role. In neuronal cells three ionic species are important: Na, K, Cl. The active current of Cl is negligible. The K and Na are coupled, since they are generated by the Na-K-antiporter, thus they are identical, but opposite in sign. Thus the sum of all active currents is zero.

Electrochemical Potentials 12 Biophysics Prof. Radmacher

What is the steady state potential? 12. j = 0 A lengthy derivation equivalent to what we have learned in the case of the Planck potential:

13.

€

Vss =RTF* ln

Picouteri + Pjcinner

j

j∑

i∑

Picinneri + Pjcouter

j

j∑

i∑

where i runs over all cationic and j over all anionic species. In the case of neurons, where only Na, K, and Cl are present, eq. 13 reduces to:

14.

€

VSS =RTF* ln PNacouter

Na + PKcouterK + PClcinner

Cl

PNacinnerNa + PKcinner

K + PClcouterCl