The influence of sodium and potassium dynamics on excitability, seizures, and the stability of persistent states: I. Single neuron dynamics

Abstract

In these companion papers, we study how the interrelated dynamics of sodium and potassium affect the excitability of neurons, the occurrence of seizures, and the stability of persistent states of activity. In this first paper, we construct a mathematical model consisting of a single conductance-based neuron together with intra- and extracellular ion concentration dynamics. We formulate a reduction of this model that permits a detailed bifurcation analysis, and show that the reduced model is a reasonable approximation of the full model. We find that competition between intrinsic neuronal currents, sodium-potassium pumps, glia, and diffusion can produce very slow and large-amplitude oscillations in ion concentrations similar to what is seen physiologically in seizures. Using the reduced model, we identify the dynamical mechanisms that give rise to these phenomena. These models reveal several experimentally testable predictions. Our work emphasizes the critical role of ion concentration homeostasis in the proper functioning of neurons, and points to important fundamental processes that may underlie pathological states such as epilepsy.

Appendix

1.1 Current to concentration conversion:

In order to derive the ion concentration dynamics, we begin with the assumption that the ratio of the intracellular volume to the extracellular volume is β = 7.0 (Somjen 2004). This corresponds to a cell with intracellular and extracellular space of 87.5% and 12.5% of the total volume respectively. For the currents across the membrane, conservation of ions requires

$$\Delta c_{\text{i}} {\text{Vol}}_{\text{i}} = - \Delta c_{\text{o}} {\text{Vol}}_{\text{o}} ,$$

where c and Vol represent ion concentration and volume respectively, Δ indicates change, and the subscripts i, o correspond to the intra- and extracellular values. The above equation leads to

$$\Delta c_{\text{i}} = - \Delta c_{\text{o}} \left( {\frac{{{\text{Vol}}_{\text{o}} }}{{{\text{Vol}}_{\text{i}} }}} \right) = - \frac{{\Delta c_{\text{o}} }}{\beta }.$$

Let I be the current density in units of μA/cm² from the Hodgkin–Huxley model. Then, the total current i _total= IA entering the intracellular volume produces a flow of charge equal to ΔQ = i _totalΔt in a time Δt, where A is the membrane area. The number of ions entering the volume in this time is therefore ΔN = i _totalΔt/q where q is 1.6 × 10⁻¹⁹ coul. The change in concentration Δc _i = ΔN/N _A Vol _i depends on the volume of the region to which the ions flow, where Avogadro’s number N _A converts the concentration to molars. The rate of change of concentration, or concentration current dc _i/dt = i _c,i, is related to the ratio of the surface area of the cell to the volume of the cell as follows

$$i_{{\text{c,i}}} = \frac{{\Delta c_{\text{i}} }}{{\Delta t}} = \frac{{\Delta N}}{{\Delta tVol_{\text{i}} N_{\text{A}} }} = \frac{{i_{{\text{total}}} }}{{qVol_{\text{i}} N_{\text{A}} }} = \frac{{IA}}{{qVol_{\text{i}} N_{\text{A}} }} = \frac{I}{\alpha }.$$

For a sphere of radius 7 μm, α = 21 mcoul/M cm². An increase in cell volume would result in a smaller time constant and therefore slower dynamics.

For the outward current the external ion concentration is therefore given as

$$i_{{{\text{c,o}}}} = \beta i_{{{\text{c,i}}}} = \frac{{\beta I}}{\alpha } = 0.33I.$$

1.2 Equations for reduced model:

The reduced model uses empirical fits of the average membrane currents of the Hodgkin-Huxley model neuron, as described in the main text. The fits are given below.

$$\begin{array}{*{20}l} {I_{{\text{K}}\infty } = \alpha _{\text{K}} \left( {g_1 g_2 g_3 + {\text{g}}_{{\text{lK}}} } \right)} \hfill \\ {I_{{\text{Na}}\infty } = \alpha _{{\text{Na}}} \left( {g_1 g_2 g_3 + {\text{g}}_{{\text{lNa}}} } \right)} \hfill \\ {{\text{K}}_{{\text{o/i}}} {\text{ = }}{{\left[ {\text{K}} \right]_{\text{o}} } \mathord{\left/ {\vphantom {{\left[ {\text{K}} \right]_{\text{o}} } {\left[ {\text{K}} \right]_{\text{i}} }}} \right. \kern-\nulldelimiterspace} {\left[ {\text{K}} \right]_{\text{i}} }}} \hfill \\ {{\text{Na}}_{{\text{i/o}}} {\text{ = }}{{\left[ {{\text{Na}}} \right]_{\text{i}} } \mathord{\left/ {\vphantom {{\left[ {{\text{Na}}} \right]_{\text{i}} } {{\text{[Na]}}_{\text{o}} }}} \right. \kern-\nulldelimiterspace} {{\text{[Na]}}_{\text{o}} }}} \hfill \\ {g_1 = 420.0\left( {{\text{1 - A}}_{\text{1}} \left( {{\text{1 - B}}_{\text{1}} {\text{exp}}\left( {{\text{ - }}\mu _{\text{1}} {\text{Na}}_{{{\text{i}} \mathord{\left/ {\vphantom {{\text{i}} {\text{o}}}} \right. \kern-\nulldelimiterspace} {\text{o}}}} } \right)} \right)^{{{\text{1}} \mathord{\left/ {\vphantom {{\text{1}} {\text{3}}}} \right. \kern-\nulldelimiterspace} {\text{3}}}} } \right)} \hfill \\ {g_2 = {\text{exp}}\left( {\sigma _{\text{2}} {{{\text{(1}}{\text{.0 - }}\lambda _2 {\text{K}}_{{{\text{o}} \mathord{\left/ {\vphantom {{\text{o}} {\text{i}}}} \right. \kern-\nulldelimiterspace} {\text{i}}}} {\text{)}}} \mathord{\left/ {\vphantom {{{\text{(1}}{\text{.0 - }}\lambda _2 {\text{K}}_{{{\text{o}} \mathord{\left/ {\vphantom {{\text{o}} {\text{i}}}} \right. \kern-\nulldelimiterspace} {\text{i}}}} {\text{)}}} {{\text{(1}}{\text{.0 + exp( - }}\mu _{\text{2}} {\text{Na}}_{{{\text{i}} \mathord{\left/ {\vphantom {{\text{i}} {\text{o}}}} \right. \kern-\nulldelimiterspace} {\text{o}}}} {\text{))}}}}} \right. \kern-\nulldelimiterspace} {{\text{(1}}{\text{.0 + exp( - }}\mu _{\text{2}} {\text{Na}}_{{{\text{i}} \mathord{\left/ {\vphantom {{\text{i}} {\text{o}}}} \right. \kern-\nulldelimiterspace} {\text{o}}}} {\text{))}}}}} \right)} \hfill \\ {g_3 = \left( {{{\text{1}} \mathord{\left/ {\vphantom {{\text{1}} {\left( {{\text{1 + exp}}\left( {\sigma _{\text{3}} \left( {{\text{1}}{\text{.0 + }}\mu _3 {\text{Na}}_{{{\text{i}} \mathord{\left/ {\vphantom {{\text{i}} {\text{o}}}} \right. \kern-\nulldelimiterspace} {\text{o}}}} {\text{ - }}\lambda _3 {\text{K}}_{{{\text{o}} \mathord{\left/ {\vphantom {{\text{o}} {\text{i}}}} \right. \kern-\nulldelimiterspace} {\text{i}}}} } \right)} \right)} \right)^{\text{5}} }}} \right. \kern-\nulldelimiterspace} {\left( {{\text{1 + exp}}\left( {\sigma _{\text{3}} \left( {{\text{1}}{\text{.0 + }}\mu _3 {\text{Na}}_{{{\text{i}} \mathord{\left/ {\vphantom {{\text{i}} {\text{o}}}} \right. \kern-\nulldelimiterspace} {\text{o}}}} {\text{ - }}\lambda _3 {\text{K}}_{{{\text{o}} \mathord{\left/ {\vphantom {{\text{o}} {\text{i}}}} \right. \kern-\nulldelimiterspace} {\text{i}}}} } \right)} \right)} \right)^{\text{5}} }}} \right)} \hfill \\ {g_4 = \left( {{{\text{1}} \mathord{\left/ {\vphantom {{\text{1}} {\left( {{\text{1 + exp}}\left( {\sigma _{\text{4}} \left( {{\text{1}}{\text{.0 + }}\mu _4 {\text{Na}}_{{{\text{i}} \mathord{\left/ {\vphantom {{\text{i}} {\text{o}}}} \right. \kern-\nulldelimiterspace} {\text{o}}}} {\text{ - }}\lambda _4 {\text{K}}_{{{\text{o}} \mathord{\left/ {\vphantom {{\text{o}} {\text{i}}}} \right. \kern-\nulldelimiterspace} {\text{i}}}} } \right)} \right)} \right)^{\text{5}} }}} \right. \kern-\nulldelimiterspace} {\left( {{\text{1 + exp}}\left( {\sigma _{\text{4}} \left( {{\text{1}}{\text{.0 + }}\mu _4 {\text{Na}}_{{{\text{i}} \mathord{\left/ {\vphantom {{\text{i}} {\text{o}}}} \right. \kern-\nulldelimiterspace} {\text{o}}}} {\text{ - }}\lambda _4 {\text{K}}_{{{\text{o}} \mathord{\left/ {\vphantom {{\text{o}} {\text{i}}}} \right. \kern-\nulldelimiterspace} {\text{i}}}} } \right)} \right)} \right)^{\text{5}} }}} \right)} \hfill \\ {{\text{g}}_{{\text{lK}}} {\text{ = A}}_{{\text{lK}}} {\text{exp}}\left( {{\text{ - }}\lambda _{{\text{lK}}} {\text{K}}_{{{\text{o}} \mathord{\left/ {\vphantom {{\text{o}} {\text{i}}}} \right. \kern-\nulldelimiterspace} {\text{i}}}} } \right)} \hfill \\ {{\text{g}}_{{\text{lNa}}} {\text{ = A}}_{{\text{lNa}}} } \hfill \\ \end{array} $$

where

$$\begin{aligned}& \alpha _{\text{K}} = 1.0,\alpha _{{\text{Na}}} = 1.0,A_1 = 0.75,B_1 = 0.93,\mu _1 = 2.6,\lambda _2 = 7.41, \\& \sigma _{\text{2}} = 2.0,\mu _2 = 2.6,\sigma _{\text{3}} {\text{ = 35}}{\text{.7, }}\mu _3 = 1.94,\lambda _3 {\text{ = 24}}{\text{.3,}}\sigma _{\text{4}} {\text{ = 0}}{\text{.88,}} \\& {\text{ }}\mu _4 = 1.48,\lambda _4 {\text{ = 24}}{\text{.6,A}}_{{\text{lNa}}} = 1.5,{\text{A}}_{{\text{lK}}} = 2.6,\lambda _{lK} {\text{ = 32}}{\text{.5}} \\ \end{aligned} $$