The Potassium Exchange values
We use the measurement by Cole and Curthis 40mS/cm squared was their measure of Potassium ions leaving the membrane
$$ \Delta Q = Idt = GA \Delta E dt $$The potassium concentration is 0.155 moles per litre. Where $G$ is the conductance per unit area, $A$ the membrane surface, $E$ voltage deflection Remember that the conductance is the reciprocal of the resistance, and $V = IR \implies I = \frac{V}{R} = GV$
Assuming for the radius $r$ of our cylindrical axon $r = 1\mu$m $(A = 2\pi r l, V = \pi r^2 l)$, the potassium concentration $[\text{K}^+]_i = 0.155 \text{ M/l}$, the action potential duration $dt = 1 \text{ ms}$, the action potential amplitude $\Delta E = 0.1 \text{ V}$, we get using Faraday’s constant $F \simeq 10^5 \text{ C/M}$ that:
$$ \frac{\Delta Q}{Q} = \frac{2G \Delta E dt}{[\text{K}^+]_i F r} \simeq 0.5\%. $$So we discover that the main part of potassium stays inside.
Typical membrane potentials
$$ V_{T} = \frac{RT}{F} $$Where $R$ is the gas constant, $T$ the temperature, $F$ Faraday’s constant. Is the thermal voltage formula. at body temperature we have: $$ \begin{align*} V_T &= \frac{(8.314 \ \text{J/(mol·K)}) \cdot (310 \ \text{K})}{96,485 \ \text{C/mol}}\ V_T &= \frac{2,577.34 \text{ V·C}}{96,485 \text{ C}} = 0.0267 \text{ V} = 26.7 \text{ mV} \end{align*}
$$ $$V_T = \frac{(8.314)(310)}{96,485} \approx 26.7 \text{ mV} $$
Typical living creatures range -3 to 2 values.
Leaky Integrate and fire neuron
Equivalent circuit model of a membrane patch 🟩–
$$ I(t) = \frac{u(t) - u_{\text{rest}}}{R} + C \frac{du(t)}{dt} $$$$ u(t) = u_{\text{rest}} + \Delta u \exp\left( - \frac{t - t_{0}}{\tau_{m}} \right) $$If the entire surface of an axon were insulated, there would be no place for current to flow out of the axon and action potentials could not be generated.
Which is called the free solution, useful to model passive membrane models.
$$ v = \frac{V - V_{L}}{V_{\theta} - V_{0}}, i_{app} = \frac{I_{app}}{g_{L}(V_{\theta} - V_{0})} $$$$ \tau \frac{dv}{dt} = -v + i_{app} $$$$ f = \frac{1}{\tau} \left[ \log \left( 1 + \frac{1}{i_{app} - v_{\theta}} \right) \right] ^{-1} $$When $i_{app} > v_{\theta}$.
Saltatory Condution Hypothesis
This model is useful to model myelinated membranes.
The Cable Equation 🟥++
I’ll solve the cable equation step by step to derive the solution V(x) shown in the slide.
Let’s start with the fundamental equations from the circuit model:
- From voltage drop: $$\Delta V = -j_ir_i\Delta x = \frac{\partial V}{\partial x} = -j_ir_i$$
- From current: $$\Delta j_i = -j_m\Delta x$$
- And membrane current: $$j_m = V/r_m$$
- Combining these equations: First, take derivative of the voltage equation with respect to x: $$\frac{\partial}{\partial x}(\frac{\partial V}{\partial x}) = -\frac{\partial j_i}{\partial x}r_i$$
- $$\frac{\partial j_i}{\partial x} = -j_m = -\frac{V}{r_m}$$
- $$\frac{\partial^2 V}{\partial x^2} = \frac{r_i}{r_m}V$$
- $$\frac{\partial^2 V}{\partial x^2} - \frac{r_i}{r_m}V = 0$$
-
This is a second-order linear differential equation. Let’s solve it:
- The characteristic equation is: $$m^2 - \frac{r_i}{r_m} = 0$$
- Define $$\lambda = \sqrt{\frac{r_m}{r_i}}$$ (this is the electrotonic length)
- Then general solution is: $$V(x) = Ae^{x/\lambda} + Be^{-x/\lambda}$$
- $$V(x) = Be^{-x/\lambda}$$
- $$V(x) = e^{-x/\lambda}$$
is the electrotonic length.
This solution shows how voltage decays exponentially along the passive membrane, with λ determining the characteristic length scale of the decay.
Both resistivities are dependend on the area. (inverse of that).
Integrating Synapses
$$ C \frac{dV_i}{dt} = -g_{L_i} (V_i - V_L) - \sum_j g_{ij} (V_i - V_{ij}) + I_{\text{app}} $$where:
- $V_{ij}$: reversal potential of synapses
- $V_{ij} > V_{\theta}$: excitatory,
$V_{ij} < V_{\theta}$: inhibitory - $g_{ij}$: synaptic conductance
- ’tug of war between batteries'
Which is a simple extension of the integrate and fire model with more currents defined by the $g$ value.
Leaky integrator model
The Leaky is the threshold reset equation with the equivalent circuit model for the neuron. Threshold resetting means putting the $V$ value back at a certain $V$ value when it is above a certain threshold.
$$ f \approx \frac{[I_{app} - gL(V_{1/2} - V_L)]^+}{C(V_\theta - V_0)} $$Where we have a numerator in the Relu.
Rate based approximator
The firing rate is not the slowly responding variable, it is instantaneous. The synaptic activation slowly reacts to the firing rate. This is the rate-based approximation of the membrane conductance.
$$g_{syn} \leftarrow g_{syn} + \frac{\alpha}{\tau}$$$$\tau \frac{dg_{syn}}{dt} + g_{syn} = 0$$$$\tau \frac{dg_{syn}}{dt} + g_{syn} = \alpha\sum_j \delta(t-t_j)$$$$\simeq \alpha F$$Where F represents the firing rate, as indicated in the image. The method uses averaging over the fast variable.
Firing rate of a Neuron
After it fired the first time, making it fire again requires a charge difference of $C(V_{\theta} - V_{0})$. We need to solve the capacitor recharge equation to get that value:
$$ \begin{align*} T &\;=\; \int_{V_0}^{V_\theta} \frac{C\,dV}{I_{\text{app}} \;-\; g_L\,[\,V - V_L\,]}\\ &=\; \int_{V_0}^{V_\theta} \frac{C\,dV}{I_{\text{net}} \;-\; g_L\,V} \;=\; \frac{C}{g_L}\;\ln \!\biggl[\, \frac{I_{\text{net}} \;-\; g_L\,V_0}{\,I_{\text{net}} \;-\; g_L\,V_\theta} \biggr].\\ &\implies f \;=\; \frac{1}{T} \;=\; \biggl[\;\frac{C}{g_L}\;\ln \!\Bigl(\! \frac{I_{\text{net}} \;-\; g_L\,V_0}{\,I_{\text{net}} \;-\; g_L\,V_\theta} \Bigr)\biggr]^{-1}. \\ & = \biggl[\;\frac{C}{g_L}\;\ln \!\Bigl(\! 1 +\frac{g_L(V_{\theta} - V_0)}{\,I_{\text{net}} \;-\; g_L\,V_\theta} \Bigr)\biggr]^{-1} \end{align*} $$Which is our firing rate.
$$ f\;\approx\; \frac{1}{T} \;=\; \frac{\bigl[I_{\text{app}} - g_L\,\bigl(V_{1/2} - V_L\bigr)\bigr]^+} {C\,\bigl(V_\theta - V_0\bigr)}, $$Probabilistical Models for Synapses
End Plate Potentials
Synapses change slightly the potentials in the post synaptic neuron. Sometimes there are mini potentials that are not enough to trigger an action potential.
These potential amplitudes could be modeled by a binomial distribution. All the MEPP are unimodal about 0.4mV. While the EPP are multimodal.
$$ P_{k} = \binom{n}{k} p^{k} (1-p)^{n-k} $$And this gives a good fit of it.