The effect of dendritic voltage-gated conductances on the neuronal impedance: a quantitative model

Abstract

Neuronal impedance characterizes the magnitude and timing of the subthreshold response of a neuron to oscillatory input at a given frequency. It is known to be influenced by both the morphology of the neuron and the presence of voltage-gated conductances in the cell membrane. Most existing theoretical accounts of neuronal impedance considered the effects of voltage-gated conductances but neglected the spatial extent of the cell, while others examined spatially extended dendrites with a passive or spatially uniform quasi-active membrane. We derived an explicit mathematical expression for the somatic input impedance of a model neuron consisting of a somatic compartment coupled to an infinite dendritic cable which contained voltage-gated conductances, in the more general case of non-uniform dendritic membrane potential. The validity and generality of this model was verified through computer simulations of various model neurons. The analytical model was then applied to the analysis of experimental data from real CA1 pyramidal neurons. The model confirmed that the biophysical properties and predominantly dendritic localization of the hyperpolarization-activated cation current I _h were important determinants of the impedance profile, but also predicted a significant contribution from a depolarization-activated fast inward current. Our calculations also implicated the interaction of I _h with amplifying currents as the main factor governing the shape of the impedance-frequency profile in two types of hippocampal interneuron. Our results provide not only a theoretical advance in our understanding of the frequency-dependent behavior of nerve cells, but also a practical tool for the identification of candidate mechanisms that determine neuronal response properties.

Appendices

Appendix A: Derivation of the impedance of a single-compartment model with multiple voltage-gated conductances

We start from Eqs. (6) and (7), and make the approximations $m_{\infty }^{(j)}(V_{m})\approx m_{\infty }^{(j)}(V_{0})+b_{j}(V_{m}-V_{0})$ and $\tau _{m}^{(j)}(V_{m})\approx \tau _{j}$. If we use these approximations in Eq. (7), we get

$$ \frac{\mathit{dm}_{j}}{\mathit{dt}}=\frac{m_{\infty }^{(j)}\big(V_{0}\big)+b_{j}\big(V_{m}-V_{0}\big)-m_{j}}{\tau _{j}} $$

(28)

In order to compute the impedance represented by the voltage-gated current I _j , we introduce a formalism based on complex numbers: oscillatory variable X will be characterized by the complex number $\bar{{X}}=A_{X}e^{i\phi _{X}}$ where A _X is the amplitude and ϕ _X is the phase (at time 0) of the oscillation of X; the actual value of X at time t is then given by $X(t)=\Re (\bar{{X}}e^{i\omega t})=\Re (A_{X}e^{i(\omega t+\phi _{X})})=A_{X}\cos (\omega t+\phi _{X})$, where ω = 2πf, f is the oscillation frequency, and $\Re (z)$ is the real part of complex number z. We will assume that our single-compartment model undergoes a small-amplitude sinusoidal voltage fluctuation given by $V_{m}=V_{0}+A_{V}e^{i\omega t}$. If the dynamics of the current is approximately linear for small-amplitude voltage fluctuations (which holds in almost all cases), then the resulting current will also be sinusoidal at the same frequency, $I_{j}=I_{j}^{0}+A_{j}e^{i(\omega t+\phi _{j})}$, and the complex impedance represented by this current is $Z_{j}=\frac{V_{m}-V_{0}}{I_{j}-I_{j}^{0}}=\frac{A_{V}}{A_{j}}e^{-i\phi _{j}}$. As a first step in the calculation of Z _j , we compute the response of gating variable m _j to the voltage fluctuation, $m_{j}=m_{j}^{0}+A_{m}e^{i(\omega t+\phi _{m})}$(we have dropped the index ’j’ from A _m and ϕ _m to simplify notation). Substituting the appropriate complex expressions into Eq. (28), and noting that $\frac{\mathit{dm}_{j}}{\mathit{dt}}=A_{m}i\omega e^{i(\omega t+\phi _{m})}$, we get $\tau_{j}A_{m}i\omega e^{i(\omega t+\phi _{m})}=m_{\infty }^{(j)}(V_{0})+b_{j}A_{V}e^{i\omega t}-m_{j}^{0}-A_{m}e^{i(\omega t+\phi _{m})}$, from which, since the equation must hold for all t, we can infer both $m_{j}^{0}=m_{\infty }^{(j)}(V_{0})$ and, more importantly, $i\omega \tau _{j}A_{m}e^{i\phi _{m}}=b_{j}A_{V}-A_{m}e^{i\phi_{m}}$ or, rearranging terms, $A_{m}e^{i\phi _{m}}=\frac{b_{j}A_{V}}{1+i\omega \tau _{j}}$. We are now in a position to calculate I _j :

$$ \begin{array}{rll} I_{j}&=&\bar{{g}}_{j}m_{j}\big(V_{m}-E_{j}\big)\\ &=&\bar{{g}}_{j}\big(m_{j}^{0}+A_{m}e^{i\big(\omega t+\phi _{m}\big)}\big)\big(V_{0}+A_{V}e^{i\omega t}-E_{j}\big)\\ &\approx & \bar{{g}}_{j}m_{j}^{0}\big(V_{0}-E_{j}\big)+\bar{{g}}_{j}\big(V_{0}-E_{j}\big)A_{m}e^{i\big(\omega t+\phi _{m}\big)}\\ &&+\,\bar{{g}}_{j}m_{j}^{0}A_{V}e^{i\omega t}\\ &=&\bar{{g}}_{j}m_{j}^{0}\big(V_{0}-E_{j}\big)+\bar{{g}}_{j}\big(V_{0}-E_{j}\big)\frac{b_{j}A_{V}}{1+i\omega \tau _{j}}e^{i\omega t}\\ &&+\,\bar{{g}}_{j}m_{j}^{0}A_{V}e^{i\omega t} \end{array} $$

(29)

where, in the second step, we neglected the second-order term proportional to A _V A _m , and, in the last step, we made use of the expression we derived for $A_{m}e^{i\phi _{m}}$ above. Since we also have $I_{j}=I_{j}^{0}+A_{j}e^{i(\omega t+\phi _{j})}$, we see comparing terms (and defining $g_{j}^{0}=\bar{{g}}_{j}m_{j}^{0}$) that $I_{j}^{0}=g_{j}^{0}(V_{0}-E_{j})$ and $A_{j}e^{i\phi _{j}}=A_{V}\left[g_{j}^{0}+\frac{\bar{{g}}_{j}b_{j}(V_{0}-E_{j})}{1+i\omega \tau _{j}}\right]$. It follows that

$$ Z_{j}=\frac{A_{V}}{A_{j}}e^{-i\phi _{j}}=\frac{1}{g_{j}^{0}+\frac{\bar{{g}}_{j}b_{j}(V_{0}-E_{j})}{1+i\omega \tau _{j}}}. $$

(30)

Appendix B: Derivation of the impedance of a neuronal model with a semi-infinite active dendrite

1.1 B.1 Steady-state membrane potential distribution

In order to determine the steady-state voltage distribution in the dendrite, we start from Eqs. (11) and (12). Approximating the steady-state value of the gating variable as a linear expansion around V _∞, as embodied by Eq. (14), we replace Eq. (12) by

$${\frac{\partial m}{\partial t}=\frac{m_{{\infty }}\left(V_{{\infty }}\right)+b\big(V_{m}-V_{\infty }\big)-m}{\tau _{{m}}\left(V_{{m}}\right)}} $$

(31)

The steady-state solution of this equation is m = m _∞(V _∞) + b(V _m  − V _∞). We substitute this expression into Eq. 11, and set its left-hand-side to 0, obtaining the steady-state equation

$$ \begin{array}{rll} \frac{d}{4R_{A}}\frac{d^{2}V_{0}}{\mathit{dx}^{2}}&=&g_{l}\big(V_{0}-E_{l}\big)\\&&+\,\bar{{g}}_{d} \big(V_{0}-E_{d}\big)\big[m_{\infty }\big(V_{\infty }\big)+b\big(V_{0}-V_{\infty }\big)\big] \end{array} $$

(32)

where V ₀(x) is the steady-state voltage distribution along the dendrite. Making use of Eq. (13), which defined V _∞, Eq. (32) can be transformed into

$$ \begin{array}{rll} \frac{d}{4R_{A}}\frac{d^{2}V_{0}}{\mathit{dx}^{2}}&=&g_{l}\big(V_{0}-V_{\infty }\big)+\bar{{g}}_{d}\big(V_{0}-V_{\infty }\big)m_{\infty }\big(V_{\infty}\big)\\ &&+\bar{{g}}_{d}\big(V_{\infty }-E_{d}\big)b\big(V_{0}-V_{\infty}\big)\\ &&+\bar{{g}}_{d}\big(V_{0}-V_{\infty }\big)b\big(V_{0}-V_{\infty }\big) \end{array} $$

(33)

Of the four terms on the right-hand-side of Eq. (33), the first three depend linearly on V ₀ − V _∞, while the fourth term depends on it quadratically. We will neglect this quadratic term in the following calculations based on our earlier assumption that V ₀ − V _∞ is relatively small; in the current context, it may be justified by noting that the fourth term in Eq. (33) relates to the second term as b(V ₀ − V _∞) relates to m _∞(V _∞), a ratio which is small if V ₀ − V _∞ is small. In fact, we should neglect the fourth term also for the sake of the consistency of our approach since we have already neglected factors that would otherwise contribute second-order terms to Eq. (33) when we linearized the expression for m _∞(V _∞).

Now let us define V′(x) = V ₀(x) − V _∞, and substitute it into Eq. (33). We get

$$ \frac{d}{4R_{A}}\frac{d^{2}V'}{\mathit{dx}^{2}}=\big[g_{l}+\bar{{g}}_{d}m_{\infty }(V_{\infty })+\bar{{g}}_{d}b\big(V_{\infty }-E_{d}\big)\big]V' $$

(34)

a second-order linear differential equation, and the corresponding boundary conditions become $\left(\frac{\mathit{dV}'}{\mathit{dx}}\right)_{x=\infty }=0$ and V′(0) = V ₀(0) − V _∞, where V ₀(0) is the holding potential at the soma. The solution of this equation is $V'(x)=(V_{0}(0)-V_{\infty })e^{-x/\lambda }$, with λ given by Eq. (16).

Finally, returning to the original variables, we obtain the solution represented by Eq. (15).

1.2 B.2 Input impedance

In order to obtain the time-dependent solution, we start from Eqs. (11) and (12). Let us define V(x, t) = V _m (x, t) − V ₀(x) and m′(x, t) = m(x, t) − m _∞(V ₀(x)). Plugging these expressions into Eq. (11) and noting that, following from the definition of V ₀(x), $\frac{d}{4R_{A}}\frac{d^{2}V_{0}}{\mathit{dx}^{2}}+g_{l}(E_{l}-V_{0})+m_{\infty }(V_{0})(E_{d}-V_{0})=0$, we get

$$ \begin{aligned}[b] C\frac{\partial V}{\partial t}={}&\frac{d}{4R_{A}}\frac{\partial ^{2}V}{\partial x^{2}}-g_{l}V-\bar{{g}}_{d}\big(V_{0}-E_{d}\big)m' \\ &-\bar{{g}}_{d}Vm_{\infty}\big(V_{0}\big)-\bar{{g}}_{d}Vm' \end{aligned} $$

(35)

Once again, we neglect the second-order last term containing Vm′ (as it should be much smaller in magnitude than either of the preceding two terms). The last significant approximation we make is that we assume that the time constant of the voltage-gated conductance can be considered uniform across the relevant membrane potential range, i.e., ${\tau _{{m}}\left(V_{{m}}\right)=\tau _{d}}$. With these simplifications and the approximation of Eq. (14), we arrive at (using the new variables)

$$ \begin{aligned}[b] C\frac{\partial V}{\partial t}={}&\frac{d}{4R_{A}}\frac{\partial ^{2}V}{\partial x^{2}}-g_{l}V\\ &-\bar{{g}}_{d}\big(V_{0}-E_{d}\big)m'-\bar{{g}}_{d}Vm_{\infty }\big(V_{0}\big) \end{aligned} $$

(36)

and

$$ \frac{\partial m'}{\partial t}=\frac{\mathit{bV}-m'}{\tau _{d}} $$

(37)

We will use Eqs. (36) and (37) to derive the impedance of this quasi-active semi-infinite dendrite.

Note that Eqs. (36) and (37) define a linear system of differential equations; therefore, in response to periodic sinusoidal perturbation (such as a sinusoidal current injection or sinusoidal variation in the membrane potential at the soma), all state variables (both V and m′) at all locations vary in a sinusoidal manner at the frequency of the input. Thus, for a given input frequency, the solution is completely characterized by providing the amplitude and phase of the oscillation of the state variables (relative to the amplitude and phase of the input) at each point along the dendrite. The calculations are again more straightforward if we employ the formalism based on complex numbers that we introduced in Appendix A. If we now substitute $V=\bar{{V}}e^{i\omega t}$ and $m'=\bar{{m}}e^{i\omega t}$ (where $\bar{{V}}(x)$ and $\bar{{m}}(x)$ are complex numbers which depend on x) into Eqs. (36) and (37), and note that the temporal differentiation on the left-hand-side is now straightforward, we get the following system of ordinary differential and algebraic equations:

$$ \begin{aligned}[b] i\omega C\bar{{V}}={}&\frac{d}{4R_{A}}\frac{d^{2}\bar{{V}}}{\mathit{dx}^{2}}-\big[g_{l}+\bar{{g}}_{d}m_{\infty }(V_{0})\big]\bar{{V}}\\ &-\bar{{g}}_{d}\big(V_{0}-E_{d}\big)\bar{{m}} \end{aligned} $$

(38)

and

$$ i\omega \tau _{d}\bar{{m}}=b\bar{{V}}-\bar{{m}} $$

(39)

From Eq. (39), we obtain $\bar{{m}}=\frac{b}{1+i\omega \tau _{d}}\bar{{V}}$, which can be substituted into Eq. (38). We arrive at the second-order ordinary differential equation

$$ \begin{array}{rll} \frac{d}{4R_{A}}\frac{d^{2}\bar{{V}}}{\mathit{dx}^{2}}&=&i\omega C\bar{{V}}+\big[g_{l}+\bar{{g}}_{d}m_{\infty }(V_{0})\big]\bar{{V}}\\&& +\,\frac{\bar{{g}}_{d}b\big(V_{0}-E_{d}\big)}{1+i\omega \tau _{d}}\bar{{V}} \end{array} $$

(40)

The only remaining complexity hidden in the notation of Eq. (40) is that V ₀ is actually a function of x. However, we have already calculated V ₀(x), which is given by Eq. (15). In order to obtain ${m_{\infty }\left(V_{{0}}\right)}$, which is also a function of x, we combine Eq. (14) with Eq. (15), getting $m_{\infty }(V_{0}(x))\approx m_{\infty }(V_{\infty })+b(V_{0}(x)-V_{\infty })=m_{\infty }(V_{\infty })+b(V_{0}(0)-V_{\infty })e^{-x/\lambda }$. If we now insert these expressions into Eq. (40) and rearrange terms, we get

$$ \begin{array}{rll} \frac{d}{4R_{A}}\frac{d^{2}\bar{{V}}}{\mathit{dx}^{2}}\!&=&\!\left\{g_{l}+i\omega C+\bar{{g}}_{d}m_{\infty }(V_{\infty }) \!+\!\frac{\bar{{g}}_{d}b(V_{\infty }-E_{d})}{1+i\omega \tau _{d}}\right.\\ &&+\left.\bar{{g}}_{d}b(V_{0}(0)\!-\!V_{\infty })\left[1\!+\!\frac{1}{1+i\omega \tau _{d}}\right]e^{-x/\lambda }\right\}\!\bar{{V}} \end{array} $$

(41)

We solved this second-order ordinary differential equation for $\bar{{V}}(x)$, taking into account the boundary conditions $\left(\frac{d\bar{{V}}}{\mathit{dx}}\right)_{x=\infty }=0$ and $\bar{{V}}(0)=A_{0}$ (where A ₀ is the amplitude of the somatic voltage fluctuation, which is assumed to be real for simplicity), using the MATLAB Symbolic Math Toolbox (version 3.2.2). Following algebraic manipulations involving Bessel function identities, we arrived at a numerically stable form of the solution:

$$ \bar{V}(x)=A_{0}\cdot {\frac{I_{2\lambda \sqrt{A}}\left(2\lambda \sqrt{B}e^{-x/2\lambda }\right)}{I_{2\lambda \sqrt{A}}\left(2\lambda \sqrt{B}\right)}} $$

(42)

where I _ν (z) is the modified Bessel function (of order ν) of the first kind, and λ, A, and B are given by Eqs. (16), (18), and (19), respectively. The axial current at x = 0 (which is equal to the current flowing between the dendrite and the soma) can be computed as

$$ \begin{array}{rll} I&=&-{\frac{d^{2}\pi}{4R_{A}}}\left(\frac{d\bar{{V}}}{\mathit{dx}}\right)_{x=0} \\ &=&-{\frac{d^{2}\pi}{4R_{A}}}A_{0}\left[\sqrt{A}-\sqrt{B}\frac{I_{2\lambda \sqrt{A}-1}\left(2\lambda \sqrt{B}\right)}{I_{2\lambda \sqrt{A}}\left(2\lambda\sqrt{B}\right)}\right] \end{array} $$

(43)

The (complex) impedance of the dendrite as seen from the soma is then given by

$$ Z_{d}=\frac{\bar{V}(0)}{I}=-{\frac{4R_{A}}{d^{2}\pi }}A_{0}\left[\left(\frac{d\bar{{V}}}{\mathit{dx}}\right)_{x=0}\right]^{-1} $$

(44)

which, in this case, results in the expression shown in Eq. (17).

The soma-dendritic transfer impedance can be calculated as $K^{(0)}_{\mathit{sx}}=\bar{V}(x)/I$ in the case of infinite somatic impedance (when all of the current injected at location 0 flows into the dendrite), resulting in the expression given by Eq. (24) in the main text. However, this result is straightforward to generalize to an arbitrary somatic impedance. We first note that the current I flowing into the dendrite from the soma is related to the external somatic current injection I _e as $I = \frac{Z_s}{Z_s+Z_d} I_e$, where Z _s is the total somatic impedance, given by Eq. (9) divided by the somatic membrane area, and Z _d is given by Eq. (17). In this general case, the transfer impedance can be calculated as

$$ K_{sx} = \bar{V}(x)/I_e = \frac{Z_s}{Z_s+Z_d} \bar{V}(x)/I = \frac{Z_s}{Z_s+Z_d} K^{(0)}_{\mathit{sx}} $$

(45)