Today we will combine the concepts from previous lectures. We will use the RC circuit to model the electrical behavior of the cell membrane. Here we will include a potassium channel along with its Nernst potential, which are incorporated into the chord conductance equation. We will solve the model and examine the role of applied current, cell capacitance, the potassium condutance, and the potassium reversal potential in determining cell behavior. We will examine the current-voltage (IV) curve and then translate this information into a state-space diagram, and will analyze the stability of fixed points in the model using graphical analysis and linearization about the fixed point.

Recall the model for an RC circuit contained a simple resistor. Now we will change the resistor to a potassium conductance in series with its Nernst potential (E_K). Again we use Kirchoff's current law (KCL) and the chord conductance equation for potassium current (I_K), and we apply I_app (a positive stimulus current) from t₀ to t:

The current-balance equation

Σ I_M - I_app = 0

I_K + I_C - I_app = 0

g_K(V_M - E_K) + CV′ - I_app = 0

CV′ = I_app - g_K(V_M - E_K)

From Lecture 4, recall the general expression for charging and discharging RC circuits:

V(t) = V_SS - (V_SS - V₀) EXP(-(t - t₀)/τ)

It is straightforward to show that V_SS = I_app/g_K + E_K, when I_app is non-zero (i.e., from t₀ to t), otherwise V_SS = E_K because I_app = 0. V₀ is determined by initial conditions.

The current-voltage (IV) curve

The IV curve shows V_SS with and without I_app = 50 pA.

I_app causes a positive shift in the IV curve toward the more depolarized V_SS. The slope of the IV curve in this model is simply g_K, since no other ion channels are present.

Phase diagram

The phase diagram is closely related to the IV curve, which we can plot according to:

CV′ = I_app - g_K(V_M - E_K)

V′ = I_app/C - (V_M - E_K)/τ, where we express g_K as 1/R_K and recall the definition τ = R_KC.

We plot the phase diagram: V′ versus V. Note that the slope is 1/τ, which shows why τ is the time constant.

Stability?

The fixed point is V_SS. We note from the graph of V′ vs. V_M that positive deflections from V_SS are associated with V′ < 0, and negative deflections from v_SS are associated with V′ > 0. This graphical analysis indicates that the fixed point is stable.

For our system V′ = f(V), we can also use linearization (Lecture 2) to evaluate the stability in the neighborhood of the fixed point. The derivative f′(V) = -1/τ. Therefore, regardless of V_SS, the fixed point is stable.

You can simulate this model using L6_gKpassive.ode.

The fractional conductance equation and resting membrane potential
In general, cells at rest are dominated by a potassium "leak" conductance that sets the baseline membrane potential. However, other ions can also have small "leak" conductances too. To model their contribution we will add conductances to the RC circuit model in parallel with the existing potassium channel. From basic physics you will recall that conductances in parallel can be added together (i.e., g_total = g₁ + g₂ + ... + g_n). Thus we will add other ionic currents to the current-balance equation. One easy way to determine resting potential under these conditions is to assume that all the channels are ohmic, i.e., in their open state the channel current is a constant function of voltage according to the chord conductance equation. Then you can solve the current-balance equation because at rest V′ = 0 and obtain the expression:

V_M = Σ E_Xg_X/Σ g_X,

where E_X and g_X denote the reversal potential and ohmic conductance for any particular ion channel type.

For the example in the figure consider two "leak" conductances, for potassium and sodium.

V_M = (E_Kg_K + E_Nag_Na)/(g_K+g_Na)

We will plot the sum of the currents as I_leak. There are several interesting principles shown in the IV curve. Notice that the curve for I_K is much steeper than I_Na, which is because at rest the cell has a much larger potassium conductance compared to sodium. The combined IV curve I_leak is steeper than either curve in isolation, because it reflects the sum of the conductances. Lastly the combined reversal potential is weighted more toward E_K because g_K>g_Na.