Second Messenger Cascades  Biophysical Substrate  Kinetics of Two-State Channels

Transitions and Metastable States

Even at steady state, individual channels will make frequent transitions between open and closed states. The frequency of these transitions typically follows a bell-shaped curve of voltage, with the highest frequency of transitions typically occurring when the channel is balanced between open and closed states at $m_\infty(V) = 1/2$.

 The number of channel closings $f_C$ per unit time at equilibrium is simply the product of the closing rate $\beta(V)$ with the number of open channels given by $m_\infty(V)$:
$$\begin{align*}f_C= & \beta(V) \: m_\infty(V), \ \intertext{which a short compu... ...}^{1 \over 2} = r \left[ m_{\infty}(V) - {m_{\infty}(V)}^2 \right],\end{align*}$$
which peaks at $V = V_{1 \over 2}$ (centered at the inflection point of the S-shaped function $m_\infty(V)$), falling off to either side in the fashion of a bell-shaped curve as in fig. 12. At equilibrium, the average number $f_O$ of channel openings per unit time must equal the average number $f_C$ of channel closings.
 
 

Figure 12: The bell-shaped curve represents the number of channel openings (closings) at equilibrium, while the sigmoid represents the steady-state activation curve of the conductance, corresponding to the average fraction of channels in the open state. The bell-shaped curve is proportional to the square root of the sigmoid's derivative.

 

This general shape to the transition frequency is generic, regardless of the details of the kinetics, i.e., the form of $\alpha(V)$ and $\beta(V)$. If we make a different kinetic assumption, namely that the time constant associated with the ion channels is voltage-independent, then the frequency of channel closings becomes

$$f_C = r/s \; {\left[ \displaystyle{d \over dV} m_{\infty}(V... ...ht]}^{1} = r \left[ m_{\infty}(V) - {m_{\infty}(V)}^2 \right].$$ If the neuron is spiking periodically, the voltage $V(t)$ will be periodic. The equilibrium mean rate of channel closings will be $$f_C = r \biggl\langle m(V) - {m(V)}^2 \biggr\rangle .$$

For the sake of argument, suppose the metastable state ${\text{C}}^\ast$ is subject to modification. If the dwell time in the state ${\text{C}}^\ast$is $\tau_{{\text{C}}^*}^{\phantom{2}}$, the ``concentration" of channels in this state at equilibrium is:

$$\begin{align*}[ {\text{C}}^*]& = f_C \; \tau_{{\text{C}}^*}^{\phantom{2}}\ &=... ...}(V) \right]}^{1 \over 2} \; \tau_{{\text{C}}^\ast}^{\phantom{2}}.\end{align*}$$ The rate of modification can thus depend on the frequency of channel closings or openings.

3/7/1998