Stochastic amplification of calcium-activated potassium currents in Ca²⁺ microdomains

Abstract

Small conductance (SK) calcium-activated potassium channels are found in many tissues throughout the body and open in response to elevations in intracellular calcium. In hippocampal neurons, SK channels are spatially co-localized with L-Type calcium channels. Due to the restriction of calcium transients into microdomains, only a limited number of L-Type Ca²⁺ channels can activate SK and, thus, stochastic gating becomes relevant. Using a stochastic model with calcium microdomains, we predict that intracellular Ca²⁺ fluctuations resulting from Ca²⁺ channel gating can increase SK2 subthreshold activity by 1–2 orders of magnitude. This effectively reduces the value of the Hill coefficient. To explain the underlying mechanism, we show how short, high-amplitude calcium pulses associated with stochastic gating of calcium channels are much more effective at activating SK2 channels than the steady calcium signal produced by a deterministic simulation. This stochastic amplification results from two factors: first, a supralinear rise in the SK2 channel’s steady-state activation curve at low calcium levels and, second, a momentary reduction in the channel’s time constant during the calcium pulse, causing the channel to approach its steady-state activation value much faster than it decays. Stochastic amplification can potentially explain subthreshold SK2 activation in unified models of both sub- and suprathreshold regimes. Furthermore, we expect it to be a general phenomenon relevant to many proteins that are activated nonlinearly by stochastic ligand release.

Appendix

In the discussion section of this paper, we stated that, as the pulse duration is made infinitesimally small relative to the time constants of the channel, the delay in channel activation cancels out the two factors that generally lead to stochastic amplification. In this section, we prove this analytically by solving the differential equations for an arbitrary length Markov process. Extending Eqs. (16) and (17) to arbitrary length chains yields:

(22)

$$ p_o(t) = s^N(t) \qquad \frac {d s(t)}{dt} = \alpha [{\rm Ca}]_i (1-s) - \beta s $$

(23)

This N + 1 length Markov process corresponds to a K_Ca channel with N independent Ca²⁺ binding sites that all must be open for the channel to activate. A sample input to the system, [Ca²⁺] _i , as well as the system’s response, is depicted in Fig. 11. This represents a single iteration of the calcium pulse protocol that is repeated indefinitely. The total length is nT ₀. The input consists of a pulse of height nA ₀ and length T ₁, followed by a delay of length T ₂ = nT ₀ − T ₁, during which there is no calcium present. The calcium pulse height is allowed to vary, scaled by the integer n ≥ 1 and the corresponding period nT ₀ is scaled accordingly to produce a fixed mean calcium value of A ₀ T ₁/T ₀. Integer n must be finite, and T ₀ > T ₁ > 0. The time course of the system’s open probability, p _o (t), is obtained by solving Eq. (23). This solution is given by x(t) during the rising phase (T ₁) and y(t) during the falling phase (T ₂), taking the form:

$$ x(t) = (x_\infty + (x_0 - x_\infty) e^{-t / \tau_x})^N $$

(24)

$$ y(t) = (y_\infty + (y_0 - y_\infty) e^{-t / \tau_y})^N $$

(25)

$$ = (y_0 e^{-t / \tau_y})^N $$

(26)

Fig. 11

Calcium pulse protocol used for minimal model. Parameter n is an integer ≥ 1 that controls the height of the calcium pulse. Period, nT ₀ is scaled accordingly so that the mean calcium value remains constant at A ₀ T ₁/T ₀. Traces x(t) and y(t) depict examples of the rise and falling p _o (t) time courses, respectively

In these equations, τ _x  = 1/(αn A ₀ + β) and τ _y  = 1/β; also, x _∞ = αn A ₀ τ _x and y _∞ = 0. Enforcing continuity at the boundaries gives $x(T_1) = y_0^N$ and $y(T_2) = x_0^N$. We next solve this system of equations to determine x ₀ and y ₀:

$$ y_0^N = x(T_1) = (x_\infty + (x_0 - x_\infty) e^{-T_1 / \tau_x})^N $$

(27)

$$ x_0^N = y(T_2) = (y_0 e^{-T_2 / \tau_y})^N $$

(28)

Note that x ₀ and y ₀ are both real and positive. Substituting Eq. (28) into Eq. (27), after raising both to the power 1/N, gives:

$$ y_0 = x_\infty + (y_0 e^{-T_2/\tau_y} - x_\infty) e^{-T_1 / \tau_x} $$

(29)

$$ = x_\infty \frac{1-e^{-T_1/\tau_x}}{1-e^{-T_1/\tau_x} e^{-T_2/\tau_y}} $$

(30)

Substituting Eq. (30) back into Eq. (28) yields:

$$ x_0 = x_\infty \frac{1-e^{-T_1/\tau_x}}{1-e^{-T_1/\tau_x} e^{-T_2/\tau_y}}e^{-T_2/\tau_y} $$

(31)

Now, we proceed to estimate the mean value of p _o (t), $\overline{p_o}$, for the case when T ₁/τ _x and T ₂/τ _y are infinitesimally small. Our goal is to show that $\overline{p_o}$ is independent of pulse height, and therefore unaffected by stochastic amplification. We have $\overline{p_o} = (\overline{x} T_1 + \overline{y} T_2) / (T1 + T2)$. Setting u = T ₁/τ _x and w = T ₂/τ _y , and letting w = γu allows us to solve for $\overline{x}$ as u and w approach zero:

$$ \overline{x} = \lim\limits_{u\to 0} \frac{1}{T_1}\int_0^{T_1} x(t) \! \, \mathrm{d}t $$

(32)

$$ = \lim\limits_{u\to 0} \frac{1}{T_1}\int_0^{T_1} (x_\infty + (x_0 - x_\infty)e^{-t/\tau_x} )^N \! \, \mathrm{d}t $$

(33)

Substituting t′ = t/τ _x gives:

$$= \lim\limits_{u\to 0} \frac{\tau_x}{T_1}\int_0^{T_1/\tau_x} (x_\infty + (x_0 - x_\infty)e^{-t'} )^N \! \, \mathrm{d}t' $$

(34)

$$= \lim\limits_{u\to 0} \frac{1}{u}\int_0^{u} (x_\infty + (x_0 - x_\infty)e^{-t'} )^N \! \, \mathrm{d}t' $$

(35)

$$= \lim\limits_{u\to 0} (x_\infty + (x_0 - x_\infty)e^{-u} )^N \cdot \lim\limits_{u\to 0} \frac{1}{u} \int_0^{u} \! \, \mathrm{d}t $$

(36)

$$ = (x_\infty + (\lim\limits_{u\to 0} x_0 - x_\infty)(1) )^N \cdot (1) $$

(37)

$$= (\lim\limits_{u\to 0} x_0)^N $$

(38)

Now, it just remains to solve for lim _u→0 x ₀. Substituting in from Eq. (31), invoking w = γu, and applying L’Hopital’s rule provides:

$$ \lim\limits_{u\to 0} x_0 = \lim\limits_{u\to 0} x_\infty \frac{1-e^{-T_1/\tau_x}}{1-e^{-T_1/\tau_x - T_2/ \tau_y}}e^{-T_2/\tau_y} $$

(39)

$$ = \lim\limits_{u\to 0} x_\infty \frac{1-e^{-u}}{1-e^{-u(1+\gamma)}}e^{-\gamma u} $$

(40)

$$ = \lim\limits_{u\to 0} x_\infty \frac{e^{-u}}{(1+\gamma)e^{-u(1+\gamma)}}e^{-\gamma u} $$

(41)

$$ = x_\infty \frac{1}{1+\gamma}(1) $$

(42)

$$ = x_\infty \frac{u}{u+w} $$

(43)

$$ = x_\infty \frac{T_1/\tau_x}{T_1/\tau_x+T_2/\tau_y} $$

(44)

We now use our earlier definitions τ _x  = 1/(αn A ₀ + β), τ _y  = 1/β, x _∞ = αn A ₀ τ _x , and T ₂ = nT ₀ − T ₁. Substituting these back into the equation for $\overline{x}$ (38) provides a form for $\overline{x}$ independent of n:

$$ \lim\limits_{u\to 0} x_0 = \big(\alpha n A_0 \tau_x\big) \frac{T_1/\tau_x }{\big(T_1\big(\alpha n A_0 + \beta\big)+\big(n T_0 - T_1\big)\beta\big)} $$

(45)

$$ = \big(\alpha n A_0 \tau_x\big) \frac{T_1/\tau_x }{\alpha n A_0 T_1+n T_0 \beta} $$

(46)

$$ = \frac{\alpha n A_0 T_1 }{\alpha n A_0 T_1+n T_0 \beta} $$

(47)

$$ = \frac{\alpha A_0 T_1 }{\alpha A_0 T_1+ T_0 \beta} $$

(48)

Thus, substituting back into the equation for $\overline{x}$ (38) gives:

$$ \overline{x} = \left(\frac{\alpha A_0 T_1 }{\alpha A_0 T_1+ T_0 \beta}\right)^N $$

(49)

Note that the integer n has cancelled out, making $\overline{x}$ independent of pulse height. Now it just remains to show that $\overline{y}$ is also independent of n. One can solve for $\overline{y}$ by analogy with the solution for x(t):

$$ \overline{y} = \lim\limits_{u\to 0} \frac{1}{T_2}\int_0^{T_2} y(t) \! \, \mathrm{d}t $$

(50)

$$ = \lim\limits_{u\to 0} \frac{1}{T_2}\int_0^{T_2} \big(y_0 e^{-t/\tau_y} \big)^N \! \, \mathrm{d}t $$

(51)

$$ = \lim\limits_{u\to 0} \frac{\tau_y}{T_2}\int_0^{T_2/\tau_y} \big(y_0 e^{-t'} \big)^N \! \, \mathrm{d}t' $$

(52)

$$ = \lim\limits_{u\to 0} \frac{1}{\gamma u}\int_0^{\gamma u} \big(y_0 e^{-t'} \big)^N \! \, \mathrm{d}t' $$

(53)

$$ = \big(\lim\limits_{u\to 0} y_0\big)^N $$

(54)

Since $y_0 = x_0 / (e^{-T_2/\tau_y})$ from Eq. (28), we can substitute this in and adopt the expression for x ₀ from Eq. (48):

$$ \overline{y} = \big(\lim\limits_{u\to 0} x_0 / \big(e^{-T_2/\tau_y}\big)\big)^N $$

(55)

$$ = \big(\lim\limits_{u\to 0} x_0 \big)^N $$

(56)

$$ = \left(\frac{\alpha A_0 T_1 }{\alpha A_0 T_1+ T_0 \beta}\right)^N $$

(57)

$$ = \overline{x} $$

(58)

Thus, in the limit where T ₁/τ _x and T ₂/τ _y become infinitesimally small, $\overline{x}=\overline{y}=\overline{p_o}$, where $\overline{p_o}$ is the mean value of the channel’s open probability for the experiment duration:

$$ \overline{p_o} = \big(\overline{x} T_1 + \overline{y} T_2\big) / (T1 + T2) $$

(59)

$$ = \left(\frac{\alpha A_0 T_1 }{\alpha A_0 T_1+ T_0 \beta}\right)^N $$

(60)

Thus p _o is independent of the variable that scales pulse height, n, and stochastic amplification does not occur. As mentioned above, the mean value of our calcium pulse input protocol is A ₀ T ₁ / T ₀. If one applies this calcium value as a constant input to the system, the system will equilibrate to its steady-state p _∞:

$$ p_\infty = s_\infty^N $$

(61)

$$ = \left(\frac{\alpha [{\rm Ca}]_i}{\alpha [{\rm Ca}]_i + \beta}\right)^N $$

(62)

$$ = \left(\frac{\alpha A_0 T_1 / T_0}{\alpha A_0 T_1 / T_0 + \beta}\right)^N $$

(63)

$$ = \left(\frac{\alpha A_0 T_1}{\alpha A_0 T_1 + T_0\beta}\right)^N $$

(64)

$$ = \overline{p_o} $$

(65)

Therefore, as T ₁ and T ₂ become infinitesimally small relative to τ _x and τ _y , stochastic amplification no longer occurs. The mean value of the system’s open probability, $\overline{p_o}$, converges to the system’s steady-state response (p _∞) to the mean calcium input ([Ca] _i  = A ₀ T ₁ / T ₀). This was seen for the case of the sAHP channel in Fig. 9(b). Note that τ _x  = 1/(αn A ₀ + β) and, thus, T ₁/τ _x  = T ₁(αn A ₀ + β). In order for the condition T ₁/τ _x → 0 to hold, n must be finite. This explains why, in Fig. 9(b), a slight amount of stochastic amplification can be seen for pulse height 9.6 μM.