Impact of the activation rate of the hyperpolarization- activated current $\hbox {I}_{\mathrm{h}}$ on the neuronal membrane time constant and synaptic potential duration

Abstract

The temporal dynamics of membrane voltage changes in neurons is controlled by ionic currents. These currents are characterized by two main properties: conductance and kinetics. The hyperpolarization-activated current ($\hbox {I}_{\mathrm {h}}$) strongly modulates subthreshold potential changes by shortening the excitatory postsynaptic potentials and decreasing their temporal summation. Whereas the shortening of the synaptic potentials caused by the $\hbox {I}_{\mathrm {h}}$ conductance is well understood, the role of the $\hbox {I}_{\mathrm {h}}$ kinetics remains unclear. Here, we use a model of the $\hbox {I}_{\mathrm {h}}$ current model with either fast or slow kinetics to determine its influence on the membrane time constant ($\tau _{m})$ of a CA1 pyramidal cell model. Our simulation results show that the $\hbox {I}_{\mathrm {h}}$ with fast kinetics decreases $\tau _{m}$ and attenuates and shortens the excitatory postsynaptic potentials more than the slow $\hbox {I}_{\mathrm {h}}$. We conclude that the $\hbox {I}_{\mathrm {h}}$ activation kinetics is able to modulate $\tau _{m}$ and the temporal properties of excitatory postsynaptic potentials (EPSPs) in CA1 pyramidal cells. To elucidate the mechanisms by which $\hbox {I}_{\mathrm {h}}$ kinetics controls $\tau _{m}$, we propose a new concept called “time scaling factor”. Our main finding is that the $\hbox {I}_{\mathrm {h}}$ kinetics influences $\tau _{m}$ by modulating the contribution of the $\hbox {I}_{\mathrm {h}}$ derivative conductance to $\tau _{m}$.

Appendix A. Membrane time constant is modulated by rate of activation of \(\hbox {I}_{\mathrm {h}}\)

A single compartment neuron with one leak current ($\hbox {I}_{\mathrm {L}})$ and one $\hbox {I}_{\mathrm {h}}$ has the membrane equation:

$$\begin{aligned} C\frac{dV}{dt}= -I_{L}- I_{h} \end{aligned}$$

(A1)

And assuming that the voltage temporal evolution of the capacitor charging can be described by a single exponential function when a constant current step is injected [32]:

$$\begin{aligned} V\left( t \right) =B\left( 1-\exp \left( -\frac{t}{\tau _{m}} \right) \right) +V_{0} \end{aligned}$$

(A2)

where B and V are constants and $\tau _{m}$ is the membrane time constant. Then differentiating equation (A2) in time and multiplying by C:

$$\begin{aligned} C\frac{dV}{dt}=CB\left( \frac{e^{-\frac{t}{\tau _{m}}}}{\tau _{m}} \right) \end{aligned}$$

(A3)

Isolating the exponential term in Eq A2 we get that $e^{-\frac{t}{\tau _{m}}}= 1- \frac{V\left( t \right) -V_{0}}{B}$ , and substituting this in Eq (A3)

$$\begin{aligned} C\frac{dV}{dt}=CB\left( \frac{1- \frac{V\left( t \right) -V_{0}}{B}}{\tau _{m}} \right) \end{aligned}$$

(A4)

Equaling A4 and A1 and differentiating with respect to V:

$$\begin{aligned} \frac{C}{\tau _{m}}= \frac{\partial I_{L}}{\partial V}+\frac{\partial I_{h}}{\partial V} \end{aligned}$$

(A5)

Isolating $\uptau _{\mathrm {m}}$:

$$\begin{aligned} \tau _{m}= \frac{C}{\frac{\partial I_{L}}{\partial V}+\frac{\partial I_{h}}{\partial V}} \end{aligned}$$

(A6)

Differentiating each current, we obtain:

$$\begin{aligned} \tau _{m}= \frac{C}{{\bar{g}}_{L}+{\overline{g_{h}}}A_{h}+ {\overline{g_{h}}}\left( V- E_{h} \right) \frac{\partial A_{h}}{\partial V}} \end{aligned}$$

(A7)

For the sake of simplicity, we will only investigate the cases where $\Delta \hbox {A}_{\mathrm {h}}$ is small due to small $\Delta \hbox {V}$. Under this condition, ${\overline{g_{h}}}A_{h}\approx {\overline{g_{h}}}A_{h}^{\infty }$, where this term corresponds to the $\hbox {I}_{\mathrm {h}}$ chord conductance ($\hbox {g}_{\mathrm {h}}$). Replacing in the equation:

$$\begin{aligned} \tau _{m}= \frac{C}{{\bar{g}}_{L}+{\overline{g_{h}}}A_{h}^{\infty }+ {\overline{g_{h}}}\left( V- E_{h} \right) \frac{\partial A_{h}}{\partial V}} \end{aligned}$$

(A8)

The partial derivative ($\frac{\partial A_{h}}{\partial V})$ in the third term of the denominator has well known analytical solution only for the extreme cases when $\tau _{h }\rightarrow \infty$ (infinitely slow kinetics) and $\tau _{h} = 0$ (instantaneous kinetics). When $\tau _{h} \rightarrow \infty$, $\hbox {A}_{\mathrm {h}}$ does not change with V, then $\hbox {A}_{\mathrm {h}} = \hbox {A}_{\mathrm {h}}^{{\infty }}$ (V) and $\frac{\partial A_{h}}{\partial V} = 0$, then:

$$\begin{aligned} \tau _{m}(\tau _{h}\rightarrow { \infty })= \frac{C}{{\bar{g}}_{L}+{\overline{g_{h}}}A_{h}^{\infty }} = \frac{C}{{\bar{g}}_{L}+g_{h}} \end{aligned}$$

(A9)

On the other hand, when $\tau _{h} = 0$, $\frac{\partial A_{h}}{\partial V}= \frac{\partial A_{h}^{\infty }}{\partial V}$, then:

$$\begin{aligned} \tau _{m}= \frac{C}{{\bar{g}}_{L}+{\overline{g_{h}}}A_{h}^{\infty }+ {\overline{g_{h}}}\left( V- E_{h} \right) \frac{\partial A_{h}^{\infty }}{\partial V}}= \frac{C}{{\bar{g}}_{L}+G_{h}}\nonumber \\ \end{aligned}$$

(A10)

where $\hbox {G}_{\mathrm {h}}$ is the slope conductance ($\hbox {G}_{\mathrm {h}})$. Concluding, we can state that $\uptau _{\mathrm {m}}$ is determined by the steady state slope conductance of the instantaneous current and chord conductance of the infinitely slow current.