User:Zanella84/sandbox

Ionic Flux
To derive the Nernst-Planck equation, we must start from the molar equation for electrochemical potential:

$$ \begin{align} \bar{\mu} &= \mu^\ominus + RT\ln{a}+zFE, \qquad \text{ where } R= 8.314 \text{ } \text{J}\text{ K}^{-1}\text{ mol}^{-1},\ F=96\ 485 \text{ } \text{ C}\text{ mol}^{-1} \\ F_\text{x} &= -{\partial\bar{\mu} \over \partial x} \\ &= -\left({RT \over a} \right) {\partial a \over \partial x} - zF{\partial E \over \partial x} \\ J &= vc = uF_\text{x}c \\ &= -\left({uRTc \over a} \right) {\partial a \over \partial x} - uzFc{\partial E \over \partial x} \\ D (\text{diffusivity}) &= uRT \\ \therefore J &= -\left({Dc \over a} \right) {\partial a \over \partial x} - uzFc{\partial E \over \partial x} \\ \end{align} $$

Activity (a) is more accurate than using concentration (c), as it takes into account electrostatic forces of attraction within the solution. Activities and concentrations are related by the following equation:

a=\gamma c,     where γ is the activity coefficient.

The more dilute a solution is, the less significant these forces become, and the activity approaches the actual concentration.

$$ \lim_{\gamma \to 1}{a} =c $$

If we assume the solution is dilute and the chemical species has no charge (i.e. z = 0), then the equation simplifies into Fick's first law of diffusion:

$$ J = -D {\partial c \over \partial x} $$

Resting Membrane Potential of a Neuron
The resting membrane potential (Vm) can be calculating using the Goldman-Hodgkin-Katz equation, where M represents cations and A are anions:

$$ V_\text{m} = {RT \over F}\ln{\left(\frac{\displaystyle \sum_i P_\text{M} \text{M}^{+}_{\text{out},i} + \displaystyle \sum_j P_\text{A} \text{A}^{-}_{\text{in},j}}{\displaystyle \sum_i P_\text{M} \text{M}^{+}_{\text{in},i} + \displaystyle \sum_j P_\text{A} \text{A}^{-}_{\text{out},j}}\right)} $$

The two main ions which contribute to the resting membrane potential in neurons are sodium (Na+) and potassium (K+).

For sodium:

$$ \begin{align} J &= -uRT{dc \over dx}-uFc{E \over \delta} \\ {dc \over J + uFc{E \over \delta}} &= -{dx \over uRT} \\ \int_{c_\text{out}}^{c_\text{in}} {dc \over J + uFc{E \over \delta}} &= -\int_{0}^{\delta} {dx \over u RT} \\ {\delta \over u FE}\ln{\left({J + uFc_\text{in}{E \over \delta} \over J + uFc_\text{out}{E \over \delta}}\right)} &= -{\delta \over uRT} \\ {J + uFc_\text{in}{E \over \delta} \over J + uFc_\text{out}{E \over \delta}} &= \exp{\left(-{FE \over RT} \right)} \\ J &= uF{E \over \delta}\left[\frac{c_\text{out}\exp{\left(-{FE \over RT} \right)} - c_\text{in}}{1 - \exp{\left(-{FE \over RT} \right)}} \right] \\ P &= u {RT \over \delta} \\ J &= P{FE \over RT}\left[\frac{c_\text{out}\exp{\left(-{FE \over RT} \right)} - c_\text{in}}{1 - \exp{\left(-{FE \over RT} \right)}} \right] \\ \end{align} $$

When a given neuron is at rest, there is no net ionic flux across the membrane:

$$ \begin{align} I_\text{K}+I_\text{Na}+I_\text{Cl} &= 0 \\ &= F(J_\text{K}+J_\text{Na}-J_\text{Cl}) \\ \end{align} $$

This flux terms in brackets yields the following:

$$ {P_\text{K}{FV_\text{m} \over RT}\left[\frac{\left[\text{K}^{+} \right]_\text{out}\exp{\left(-{FV_\text{m} \over RT} \right)} - \left[\text{K}^{+} \right]_\text{in}}{1 - \exp{\left(-{FV_\text{m} \over RT} \right)}} \right]} + {P_\text{Na}{FV_\text{m} \over RT}\left[\frac{\left[\text{Na}^{+} \right]_\text{out}\exp{\left(-{FV_\text{m} \over RT} \right)} - \left[\text{Na}^{+} \right]_\text{in}}{1 - \exp{\left(-{FV_\text{m} \over RT} \right)}} \right]} + {P_\text{Cl}{FV_\text{m} \over RT}\left[\frac{\left[\text{Cl}^{-} \right]_\text{in}\exp{\left(-{FV_\text{m} \over RT} \right)} - \left[\text{Cl}^{-} \right]_\text{out}}{1 - \exp{\left(-{FV_\text{m} \over RT} \right)}} \right]} $$

This means that:

$$ \begin{align} \exp{\left(-{FV_\text{m} \over RT} \right)} \left(P_\text{K}\left[\text{K}^{+} \right]_\text{out} + P_\text{Na}\left[\text{Na}^{+} \right]_\text{out} + P_\text{Cl} \left[\text{Cl}^{-} \right]_\text{in} \right) &= P_\text{K}\left[\text{K}^{+} \right]_\text{in} + P_\text{Na}\left[\text{Na}^{+} \right]_\text{in} + P_\text{Cl}\left[\text{Cl}^{+} \right]_\text{out} \\ -{FV_\text{m} \over RT} &= \ln{\left(\frac{P_\text{K}\left[\text{K}^{+} \right]_\text{in} + P_\text{Na}\left[\text{Na}^{+} \right]_\text{in} + P_\text{Cl}\left[\text{Cl}^{+} \right]_\text{out}}{P_\text{K}\left[\text{K}^{+} \right]_\text{out} + P_\text{Na}\left[\text{Na}^{+} \right]_\text{out} + P_\text{Cl} \left[\text{Cl}^{-} \right]_\text{in}} \right)} \\ \therefore V_\text{m} &= {RT \over F}\ln{\left(\frac{P_\text{K}\left[\text{K}^{+} \right]_\text{out} + P_\text{Na}\left[\text{Na}^{+} \right]_\text{out} + P_\text{Cl} \left[\text{Cl}^{-} \right]_\text{in}}{P_\text{K}\left[\text{K}^{+} \right]_\text{in} + P_\text{Na}\left[\text{Na}^{+} \right]_\text{in} + P_\text{Cl}\left[\text{Cl}^{+} \right]_\text{out}} \right)} \\ \end{align} $$

A typical neuron has a resting Vm of -70 mV. As such, chloride, whose equilibrium potential is roughly the same, does not contribute significantly to the resting Vm, and can be omitted:

$$ \begin{align} V_\text{m} &\approx E_\text{Cl} \\ \therefore V_\text{m} &= {RT \over F}\ln{\left(\frac{\left[\text{K}^{+} \right]_\text{out} + \alpha\left[\text{Na}^{+} \right]_\text{out}}{\left[\text{K}^{+} \right]_\text{in} + \alpha\left[\text{Na}^{+} \right]_\text{in}} \right)}, \qquad \text{where } \alpha = \frac{P_\text{Na}}{P_\text{K}} \\ \end{align} $$

This equation can be refined by taking into account the effect of the sodium/potassium pump, which counteracts sodium and potassium leak and is vital in maintaining a constant resting Vm.

$$ V_\text{m} = {RT \over F}\ln{\left(\frac{\left[\text{K}^{+} \right]_\text{out} + \dfrac{\alpha}{r}\left[\text{Na}^{+} \right]_\text{out}}{\left[\text{K}^{+} \right]_\text{in} + \dfrac{\alpha}{r}\left[\text{Na}^{+} \right]_\text{in}} \right)}, \qquad \text{where } r = \frac{J_\text{Na}}{J_\text{K}} = \frac{3}{2} $$