Notes

Lecturer: Valerio Mante

Ohmic Conductances

Current through the membrane passes through a particular channel type as a function of the voltage across the membrane. The slope in the following graph is the conductance through this channel. We say conductance is ohmic when we have passive properties, thus, $I\ = \ g\ \bullet \ V$. Different channel types have different resting potentials (as shown below). Example active conductances (mammals, approx. 37C degrees)

• Triggered by neurotransmitter, voltage, ...
• Time-dependent

Synaptic Currents Synapses are injecting (external) current. Assume dendrite are passive cables that just conduct current. The dendrites tend to have much less active conductances respect to the axons. This way, we have three models:

• Single Compartment Model: voltage has only temporal dependency $V\ = \ V(t)$.
• Cable Equation: depends on time and location (analytical solutions) $V\ = \ V(x,t)$.
• Multicompartment Model: (numerical solutions) $V\ = \ V(x,\ t)$. These approaches offer a trade-off between realism and complexity.

Development of the Cable Equation The key formalism that was used successfully to justify what is happening is the cable equation. The cable equation was derived by Thomson and had practical relevance for transatlantic telegraph cable. In our discussion, we will consider cables as good approximations of dendrites.

We want to find an expression for $V\ = \ V(x,\ t)$, i.e., we want to derive the equation for the voltage of the cell as a function of time and spatial dimension. We will use two steps:

• Single-Compartment Model (iso-potential): $V\ = \ V(t)$.
• Cable equation: $V\ = \ V(x,\ t)$.

Single-Compartment Model It is a very simple model of a neuron.

The capacitive current might end on the surface of the membrane and increase/decrease the charges on the membrane. Otherwise, it could flow out as leak current. What would happen? In the beginning (if it is positive) will charge the inside of the membrane, so the potential will increase, as the potential increases and gets more different from the resting potential will eventually lead to a new equilibrium in which the cells is more depolarized, to the point that the current flowing out of the membrane is balanced with the current flowing in.

Deriving V(t)

Steady-State Solution We inject a current and the membrane potential will increase to an asymptote, which can be derived as shown below. Assuming that we will continue to inject the current constantly. If we double the current, we double the voltage difference that we get and the height of the asymptote, will depend on the properties of the resistance. Hence $V_{inf}$ will increase if $R_m$ increases (less leak) or $I_e$ increases (more input).

Input Resistance Two neurons with the same concentration of channels, differing only regarding the size: the smaller neuron will have more resistance, thus, for the same amount of injected current it will have smaller voltage change. Furthermore, two neurons, one myelinated and another unmyelinated. The myelinated has less resistance thus it needs less current input to achieve the same voltage change. Also, you will need a larger current to create a certain amount of depolarization in a neuron with more channels compared with one of the same dimensions but with fewer channels.

General Solution $\tau_{m}$ is the memory of cell approx. 10 to 100 ms, Neuron forgets after $\tau$. Longer memory requires few mechanisms, for example, charge in synapse or recurrency. In general, we will have a time dependency, so the potential as a function of time will be given by:

$$V(t) = V_{\infty} + (V(0) - V_{\infty}) \bullet e^{- \frac{\ \tau}{\tau_{m}\ }}$$

Where $V(0) = V(t = 0) = initial\ condition$, $V_{\infty} = R_{m}I_{e} + E_{m}$ and $\tau_{m} \approx \ 10\ to\ 100\ ms$.

Implications Time constant $\mathbf{\tau}_{\mathbf{m}}$:

• Typically $\tau_{m} \approx \ 10\ to\ 100\ ms$, i.e., if current is injected in a cell, the neuron will forget about it quickly.
• The time-scale of change in the cell (slow compared to a computer).
• The short-term "memory" of the cell (short compared to an organism).
• Activity "forgotten" after $\tau_{m}$.
• Longer memory: other mechanisms (e.g., plasticity, ...)
• Slower response: recurrent connectivity ("reverberating activity").

Spatial and Temporal Summation

• Spatial Summation: what happens if two synapses open at the same time? We derive the spatial summation. Simultaneous inputs $(\delta t = 0)$ sum linearly. If $I_{e}\ \rightarrow \ k \bullet I_{e}$ then $V_{\infty} \rightarrow \ k \bullet V_{\infty}$. It is a linear equation, which means that when currents come from different parts of the dendrites, they will sum in the soma. So, the overall effect will be the sum of single inputs. These are only effects before the threshold for an AP is reached, then something non-linear happens.
• Temporal Summation: biological neurons, since they are laggish/slow and it takes time, currents injected subsequently in time, can still sum up. Sequential inputs sum if $\delta t < \tau_{m}$.

Integrate and Fire Neurons When an Integrate and Fire neuron achieve the threshold, it generates an action potential and right after, reset it.

Idealized synapse: If $I_{e} > 0$ then $V > E_{m}$ leading to EPSP (depolarization). If $I_{e} < 0$ then $V < E_{m}$ leading to IPSP (hyperpolarization).

Equivalent Circuits We can draw the electric circuit that captures the basic properties of neurons.

Then we add a synapse to the circuit:

Deriving the Cable Equation So far, ions flow (in out) to achieve Erest. But, what if $V = V(t) \neq E_{rest}$ most of the time?

Longitudinal current In the cable equation we use the same variables as before but we need to express $I_L$: longitudinal current. Now, we have a current that flows inside the membrane (for instance, from left to right).

So cable equation derives from conservation of energy and conservation of charge.

Case 1: Infinite Cable & Constant Current

Case 2: Infinite Cable & Current Pulse

Passive Currents in a Branching Neuron

The Big Picture