The action potential

We refer to the intracellular and extracellular potential as \(V_i\) and \(V_e\) respectively. Note that a potential must be measured realtive to someting and we assume that \(V_i\) and \(V_e\) are measured relative to some common ground lead. The membrane potential is thus the potiential differece between the intracellular and extracellular space

\[ V = V_i - V_e\]

The resting membrane potential is typically around -70 mV (ref??). This means that the potential in the intrecellular space is lower than in the extrecellular space.

As been noted, the membrane potential arises from a separatio of the intrecellular and extracelluar space by a membrane. The intrecellular medium, called the cytoplasm and the extracellular medium contains different concentraion of various ions. The most important ions are sodium (Na\(^{+}\)), potassium (K\(^{+}\)) and Calcium (Ca\(^{2+}\)).

The membrane, which is about 5nM (ref??) is composed of a lipid bilayer, i.e two layers of lipids which have one hydrophilic end pointing outwards, and one hydrophobic end pointing inwards. The membrane is virtually impermable to water and ions, meaning that it can seprate positive ions on one side and negative ions on the other side, which leads to an electric field across the membrane. From a modeling perspective we can think of this as electrical capasitor.

Ion channels are pores in the lipid bilayer, which allows specific ions to flow through the membrane. For example ion channels that are mostly permeable to potassium are called potassium channels. Some of these ion channles can exist in different states; they can be open meaning that ions can flow through or closed, meaning that ions cannot flow through. Such channels are usually referred to as activate channels, wheras channels than does not change the pemeability is reffered to as passive states.

Ion pumps is a different protien stucture that primarly pumps specific ions in and out of the cell. One example is the sodium-potassium pump which pumps potassium (2 K\(^{+}\)) into the cell sodium (3 Na\(^{+}\)) out of the cell.

Electrodiffusion

Ions that flow through the membrane can lead to an electric field, i.e an electric force resulting for different charges in the intrecellular and extracellular space. At the same time, concentration differences leads to diffusion of ions for the space with higher concentration to the space with lower concentration. This combination of electric forces and diffusion is called electrodiffusion.

In the following we will look at a specific ion (or moelcule) \(X\) with an intracellular concentration \(X_i\) and extracellular concentration \(X_e\).

Diffusion

Diffusion is the process that moves particle from regions of high concentrations to region of low concentration. A steady state solution is obtained when all regions have equal concentration. Given an arbitrary molecule \(X\), with different concentrations \([X]\) in the intracellular and extracellular space will have a flux of iouns described by Ficks law

\[ J_{X, \text{diff}} = -D_X \frac{\mathrm{d}[X]}{\mathrm{d}x}. \]

Here \(J_{X, \text{diff}}\) is the flux, \(\frac{\mathrm{d}[X]}{\mathrm{d}x}\) is the concentration gradient and \(D_X\) is the diffusion coefficient determining how easy it is for ions to flow through the membrane. The negative sign indicates that the flux is in the opposite direction to that in which the concentration gradient increases.

Electric forces

The contribution of flow of ions throug the membrance coming from the electric field is given by Planck’s equation

\[ J_{X, \text{drift}} = -u \frac{z_X}{|z_X|}[X] \frac{\mathrm{d}V}{\mathrm{d}x}, \]

where \(u\) is the mobility of the ion and \(z_X\) is the valence so that \(\frac{z_X}{|z_X|}\) is +1 for cations and -1 for anions. The diffusion coeffeicient can be further be written as

\[ D_X = \frac{ u R T }{F |z_X|} \]

with R being the gas constant, T the temperature and F is Faraday’s constant, so that the flux becomes

\[ J_{X, \text{drift}} = -D \frac{z_X F}{RT}[X] \frac{\mathrm{d}V}{\mathrm{d}x}, \]

The Nernst-Planck equation

The total flux of ions of type \(X\) across the membrane is therefore

\[ J_X = J_{X, \text{diff}} + J_{X, \text{drift}} = -D_X \left( \frac{\mathrm{d}[X]}{\mathrm{d}x} + \frac{z_X F}{RT}[X] \frac{\mathrm{d}V}{\mathrm{d}x} \right) \]

The Nernst equation

The Nernst equation can be found by saying the flux coming from the electric field should balance the flux coming from the diffusion, i.e

\[-D_X \left( \frac{\mathrm{d}[X]}{\mathrm{d}x} + \frac{z_X F}{RT}[X] \frac{\mathrm{d}V}{\mathrm{d}x} \right) = 0 \]

or

\[\frac{1}{[X]} \frac{\mathrm{d}[X]}{\mathrm{d}x} + \frac{z_X F}{RT} \frac{\mathrm{d}V}{\mathrm{d}x} = 0\]

Integrating \(x\) for \(0\) to \(L\) gives

\[ln ([X]) \; \big|_{X_i}^{X_e} = \frac{z_X F}{RT} V \; \big|_{V_i}^{V_e} = \frac{z_X F}{RT} (V_i - V_e) = \frac{z_X F}{RT} V \]

where now \(V\) denotes the membrane potential. We want thus write the membrane potential as

\[ V = E_X = \frac{RT}{z_X F} ln \left( \frac{[X]_e}{[X]_i} \right)\]

This final equation is called Nernst equation. (Note that the same equation can also be derrived using Gibbs free energy)

For a given ion X with a know intracellular an extracellular concentration we can eailty compute the value of the membrane potential that will lead to a zero net flux across the membrane. We usually call this potential the Nernst potential.

from math import log
R = 8314.47 # J*mole**-1*K**-1"
T = 310 # K
F = 96485.3415 # C*mM**-1

# These values are taken from the TenTussher 2006 model for Human Ventricular Tissue
Na_i = 11.6 # mM
Na_e = 140.0 # mM
E_Na = (R * T / (1 * F)) * log(Na_e / Na_i)
print(f"Resting potential for Na^+: {E_Na:.2f} mV")

K_i = 138.3 # mM
K_e = 5.4 # mM
E_K = (R * T / (1 * F)) * log(K_e / K_i)
print(f"Resting potential for K^+: {E_K:.2f} mV")
Resting potential for Na^+: 66.53 mV
Resting potential for K^+: -86.63 mV

So assuming that the ion channels for the sodium is open, there will be a flow of ions into the cell when the membrane potential is below 66.5 mV. If the membrane potential is higher than 66.5 mV then sodium will flow out of the cell. For potassium there will be a flow of potassium out of the cell if the membrane potential is larger than -86.6 mV.

From flux to Current

While a flux is the number of moles of an ions through a cross-section og unit area, a current is a the amount of positive charge flowing per unit time. A current density is the amount of charge flowing per unit of time per uning ot cross-sectional area. We typically denote the current for ion \(X\) by a \(I_X\) and the current denisty by \(i_X\).

The current denisty \(i_X\) is propotional to the flux \(J_X\) and can be written as

\[ i_X = F z_X J_X \]

where \(F\) is Faraday’s constant adn z_X is the valence.

More than one ion - The Goldman–Hodgkin–Katz Equations

In general there can be more then one ion flowing through the membrane, meaning that the flux for ion \(X\) must be found by taking into cosideration the other ions as well (because of the electric field only cares about the charge). There for \( J_X = J_{X, \text{diff}} + J_{X, \text{drift}} \) is not neccesary zero. However we can find a steady state solution by taking this into account by solving for a given membrane potential \(V\) and a membrane of thickness \(L\). In this case we can write $\( \frac{\mathrm{d}V}{\mathrm{d}x} = - \frac{V}{L}\)$, to get

\[J + D_X \left( \frac{\mathrm{d}[X]}{\mathrm{d}x} - \frac{z_X F}{RT}[X] \frac{V}{L} \right) = 0\]

or equivalently

\[ \frac{\mathrm{d}[X]}{\mathrm{d}x} - \frac{z_X F}{RT}[X] \frac{V}{L} + \frac{J_X}{D_X} = 0\]

whose solutions is

\[\exp \left( -\frac{z_XVFx}{RTL} \right) [X](x) = \frac{J_XRTL}{D_Xz_XVF} \left[ \exp \left( -\frac{z_XVFx}{RTL} \right) - 1 \right] + [X]_i.\]

We can therefore find that the flux can be written as

\[ J_X = \frac{D_X}{L}\frac{z_XVF}{RT}\frac{[X]_i - [X]_e \exp \left( -\frac{z_XVF}{RT} \right)}{1- \exp \left( -\frac{z_XVF}{RT} \right)}\]

The corresponding current density is thus

\[i_X = F z_X J_X = P_X \frac{z_X^2F^2}{RT}V\frac{[X]_i - [X]_e \exp \left( -\frac{z_XVF}{RT} \right)}{1- \exp \left( -\frac{z_XVF}{RT} \right)}\]

where \(P_X = \frac{D_X}{L}\) is the permeability of the membrane to ion X.

%matplotlib notebook
import matplotlib.pyplot as plt
import numpy as np

R = 8314.47 # J*mole**-1*K**-1"
T = 310 # K
F = 96485.3415 # C*mM**-1

# These values are taken from the TenTussher 2006 model for Human Ventricular Tissue
Na_i = 11.6 # mM
Na_e = 140.0 # mM
z_Na = 1

# Set this to one - it will only scale the current anyway
P_Na = 1e-6 #

# Select membrane potential between -85 and 50 mV
V = np.linspace(-200, 200)

i_Na = P_Na * (((z_Na * F)**2) / (R*T)) * V * ((Na_i - Na_e * np.exp(-(z_Na * V * F)/(R*T)))/(1 - np.exp(-(z_Na * V * F)/(R*T))))

fig, ax = plt.subplots()
ax.plot(V, i_Na)
ax.grid(True)
ax.set_xlabel("Membrane potential (mV)")
ax.set_ylabel("Sodium current density (uA/cm^2)")
plt.show()

From this we can read that there will be a current into the cell as long as the membrane potential is below 66.5 and for larger values there will be a very small current out of the cell. The same experiment for the potassium current is displayed below.

R = 8314.47 # J*mole**-1*K**-1"
T = 310 # K
F = 96485.3415 # C*mM**-1

K_i = 138.3 # mM
K_e = 5.4 # mM

z_K = 1

# Set this to one - it will only scale the current anyway
P_K = 1e-6 #

# Select membrane potential between -85 and 50 mV
V = np.linspace(-200, 200)

i_K = P_K * (((z_K * F)**2) / (R*T)) * V * ((K_i - K_e * np.exp(-(z_K * V * F)/(R*T)))/(1 - np.exp(-(z_K * V * F)/(R*T))))

fig, ax = plt.subplots()
ax.grid(True)
ax.plot(V, i_K)
ax.set_xlabel("Membrane potential (mV)")
ax.set_ylabel("Potassium current density (uA/cm^2)")
plt.show()

As we said earlier, there are usually more ions that are separated by the same membrane. For example say that we have sodium (Na\(^{+}\)), potassium (K\(^{+}\)) and cloride (Cl\(^{-}\)). Then the total current denity would be

\[i = i_{Na} + i_K + i_{Cl}.\]

We can now try to find the membrane potential needed to have a net zero current.

\[0 = i_{Na} + i_K + i_{Cl} = \frac{F^2}{RT}V \left( P_{Na}\frac{[Na]_i - [Na]_e \exp \left( -\frac{VF}{RT} \right)}{1- \exp \left( -\frac{VF}{RT} \right)} + P_{K}\frac{[K]_i - [K]_e \exp \left( -\frac{VF}{RT} \right)}{1- \exp \left( -\frac{VF}{RT} \right)} + P_{Cl}\frac{[Cl]_i - [Cl]_e \exp \left( +\frac{VF}{RT} \right)}{1- \exp \left( +\frac{VF}{RT} \right)} \right)\]

(note the sign change in the exponential for Cl\(^{-}\)). This means that

\[ P_{Na}\frac{[Na]_i - [Na]_e \exp \left( -\frac{VF}{RT} \right)}{1- \exp \left( -\frac{VF}{RT} \right)} + P_{K}\frac{[K]_i - [K]_e \exp \left( -\frac{VF}{RT} \right)}{1- \exp \left( -\frac{VF}{RT} \right)} + P_{Cl}\frac{[Cl]_i - [Cl]_e \exp \left( +\frac{VF}{RT} \right)}{1- \exp \left( +\frac{VF}{RT} \right)} = 0\]

We collect the terms and take the natural logatrithm to obtain

\[ V = -\frac{RT}{F} \ln \left( \frac{P_{Na}[Na]_i + P_{K}[K]_i + P_{Cl}[Cl]_e}{P_{Na}[Na]_e + P_{K}[K]_e + P_{Cl}[Cl]_i} \right).\]
from math import log
R = 8314.47 # J*mole**-1*K**-1"
T = 310 # K
F = 96485.3415 # C*mM**-1

# These values are taken from the TenTussher 2006 model for Human Ventricular Tissue
P_Na = 1.0
Na_i = 11.6 # mM
Na_e = 140.0 # mM
E_Na = (R * T / (1 * F)) * log(Na_e / Na_i)
print(f"Resting potential for Na^+: {E_Na:.2f} mV")

P_K = 1.0
K_i = 138.3 # mM
K_e = 5.4 # mM
E_K = (R * T / (1 * F)) * log(K_e / K_i)
print(f"Resting potential for K^+: {E_K:.2f} mV")

# Total
V_rest = -(R * T / (1 * F)) * log((P_Na * Na_i + P_K * K_i) / (P_Na * Na_e + P_K * K_e))
print(f"Total resting potential is {V_rest} mV")
Resting potential for Na^+: 66.53 mV
Resting potential for K^+: -86.63 mV
Total resting potential is -0.8142309550910197 mV