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Notebook 1 — One channel

SWC ENC 2026 · ephys-one module

Before we can count channels we need to understand what one of them does. The answer is simple and strange: an ion channel is a random switch. It sits closed, then at some unpredictable moment pops open and passes a fixed current, then at another unpredictable moment snaps shut. All the statistics we use later come from this picture, so we build it carefully here.

In this notebook you will:

1. Simulate a single channel as a coin flip at every time step and see what its current looks like.
2. Measure its open probability and see how it depends on voltage.
3. Measure how long it stays open or closed (dwell times) and see that they are exponentially distributed.
4. Watch what happens when many repeats of a voltage step are averaged — a preview of Notebook 2.

import numpy as np
import matplotlib.pyplot as plt
import picopatch as pp

1. A channel is a switch with two rates

The simplest kinetic scheme has two states, Closed and Open, and two numbers:

• $\alpha$, the opening rate: the probability per second that a closed channel opens;
• $\beta$, the closing rate: the probability per second that an open channel closes.

Rates have units of 1/s. $\alpha = 150/\mathrm{s}$ means that, in a tiny interval $\Delta t$, a closed channel opens with probability $\alpha\,\Delta t$ — e.g. with $\Delta t = 20\ \mu s$ that's $150 \times 0.00002 = 0.003$, a 0.3% chance per tick. Nothing about the channel's history matters: it has no memory, just two coins.

fig, ax = plt.subplots(figsize=(6, 2.6))
pp.plotting.plot_scheme(ax, pp.two_state_fixed(alpha=150, beta=150), V=0,
                        rate_names=[(r"\alpha", r"\beta")],
                        title="the two-state scheme: closed ⇌ open")
plt.show()

Exercise 1 (~8 min). Complete simulate_channel. Walk through time in steps of dt; at each step, if the channel is closed, open it with probability alpha*dt; if open, close it with probability beta*dt. Record the state (0 = closed, 1 = open) at every step. rng.random() gives a uniform number in [0, 1), so "with probability q" is rng.random() < q.

Check / unstuck. With α = β = 150/s the channel should spend about half its time open and flip every few milliseconds. Stuck? pp.simulate_two_state(alpha, beta, dt, n_steps, rng) does exactly this.

def simulate_channel(alpha, beta, dt, n_steps, rng):
    state = np.zeros(n_steps, dtype=int)      # 0 = closed, 1 = open
    s = 0
    for k in range(n_steps):
        if s == 0 and rng.random() < alpha * dt:
            s = 1
        elif s == 1 and rng.random() < beta * dt:
            s = 0
        state[k] = s
    return state

dt = 1 / pp.FS                                # 20 microseconds
n_steps = int(0.2 / dt)                       # 200 ms
rng = np.random.default_rng(0)
state = simulate_channel(alpha=150, beta=150, dt=dt, n_steps=n_steps, rng=rng)
i_unit = 1.0                                  # pA through the open channel
current = i_unit * state
t_ms = np.arange(n_steps) * dt * 1e3

pp.plotting.plot_single_channel(t_ms, current, i=i_unit,
                                title="one simulated channel, alpha = beta = 150/s")
plt.show()
print(f"fraction of time open: {state.mean():.2f}")

fraction of time open: 0.41

That square, hopping trace is what a real single-channel recording looks like (apart from the noise). The channel is either passing i or passing nothing; there is no in-between. Change the rates and the look changes: a large $\alpha$ means short closed times, a large $\beta$ means short open times.

fig, axes = plt.subplots(3, 1, figsize=(11, 6), sharex=True)
for ax, (a, b) in zip(axes, [(150, 600), (150, 150), (600, 150)]):
    st = pp.simulate_two_state(a, b, dt, n_steps, rng=1)
    pp.plotting.plot_single_channel(t_ms, i_unit * st, ax=ax, i=i_unit,
                                    title=f"alpha = {a}/s, beta = {b}/s   →   open {st.mean():.0%} of the time")
for ax in axes[:-1]:
    ax.set_xlabel("")
plt.tight_layout(); plt.show()

2. Open probability

The open probability $p$ is the fraction of time the channel spends open. From the traces above you can guess the rule: the channel is open more when it opens faster ($\alpha$ big) or closes slower ($\beta$ small). The exact result is

$$p = \frac{\alpha}{\alpha + \beta}.$$

So with $\alpha = \beta$, $p = 0.5$; with $\alpha = 600, \beta = 150$, $p = 0.8$.

▸ Go deeper: where α/(α+β) comes from (optional)

Let $p(t)$ be the probability the channel is open at time $t$. In a small interval $\Delta t$ it gains probability from closed channels opening, $\alpha \Delta t\,(1-p)$, and loses probability from open channels closing, $\beta \Delta t\, p$. So

$$\frac{dp}{dt} = \alpha (1 - p) - \beta p .$$

At steady state $dp/dt = 0$, giving $\alpha(1-p) = \beta p$, i.e. $p_\infty = \alpha / (\alpha + \beta)$. Away from steady state the solution is

$$p(t) = p_\infty + (p_0 - p_\infty)\, e^{-t/\tau}, \qquad \tau = \frac{1}{\alpha + \beta},$$

an exponential relaxation with time constant $\tau$. We'll see this curve in section 4, when we average many sweeps.

Voltage dependence

Voltage-gated channels have rates that depend on the membrane potential $V$. We use the dependence from the paper:

$$\alpha(V) = \alpha_0\, e^{+2(V - 45)/25}, \qquad \beta(V) = \alpha_0\, e^{-2(V - 45)/25},$$

with $V$ in mV and $\alpha_0 = 150/\mathrm{s}$. Depolarising (making $V$ more positive) speeds up opening and slows down closing, so $p$ rises with $V$. At $V = 45$ mV the two rates are equal and $p = 0.5$. Here are the two rates, and the open probability they imply, as functions of voltage:

V = np.linspace(0, 100, 200)
scheme = pp.two_state()                       # the voltage-dependent two-state channel
p_inf = np.array([scheme.open_probability(v) for v in V])

fig, axes = plt.subplots(1, 2, figsize=(11, 3.6))
axes[0].semilogy(V, pp.alpha_rate(V), label="alpha (opening)", color="tab:orange")
axes[0].semilogy(V, pp.beta_rate(V), label="beta (closing)", color="0.3")
axes[0].set_xlabel("V (mV)"); axes[0].set_ylabel("rate (1/s)"); axes[0].legend()
axes[0].set_title("rates vs voltage")
axes[1].plot(V, p_inf, color="k"); axes[1].axvline(45, color="0.7", ls=":")
axes[1].set_xlabel("V (mV)"); axes[1].set_ylabel("open probability p = α/(α+β)")
axes[1].set_title("open probability vs voltage")
plt.tight_layout(); plt.show()

Exercise 2 (~6 min). Simulate one channel held at each of 25, 45 and 65 mV for 2 s, measure the fraction of time it is open, and compare to $\alpha/(\alpha+\beta)$. Use pp.alpha_rate(V) and pp.beta_rate(V) for the rates at a voltage, and your simulate_channel (or pp.simulate_two_state) for the trace.

Check / unstuck. You should get roughly 0.04, 0.5 and 0.96. With only 2 s of data expect a little scatter. Stuck? scheme.open_probability(V) gives the theory value; pp.simulate_single(scheme, pp.constant_protocol(V, 2000)) the trace.

n_2s = int(2.0 / dt)
measured, theory = [], []
fig, axes = plt.subplots(3, 1, figsize=(11, 6), sharex=True)
for ax, v in zip(axes, [25, 45, 65]):
    a, b = float(pp.alpha_rate(v)), float(pp.beta_rate(v))
    st = simulate_channel(a, b, dt, n_2s, np.random.default_rng(2))
    measured.append(st.mean()); theory.append(a / (a + b))
    tt = np.arange(n_2s) * dt * 1e3
    pp.plotting.plot_single_channel(tt[:10000], i_unit * st[:10000], ax=ax, i=i_unit,
                                    title=f"V = {v} mV:  measured p = {st.mean():.2f},  alpha/(alpha+beta) = {a/(a+b):.2f}")
for ax in axes[:-1]:
    ax.set_xlabel("")
plt.tight_layout(); plt.show()

plt.figure(figsize=(5.5, 3.6))
plt.plot(V, p_inf, color="k", label="theory  α/(α+β)")
plt.plot([25, 45, 65], measured, "o", color="tab:red", ms=8, label="measured from 2 s traces")
plt.xlabel("V (mV)"); plt.ylabel("open probability"); plt.legend(); plt.show()

3. How long does it stay open? Dwell times

Look back at any trace: the open stretches have all sorts of lengths, some very short, a few long. The duration of one stretch is a dwell time. Because the channel has no memory — the chance of closing in the next tick is $\beta\,\Delta t$ no matter how long it has been open — short dwells are always more common than long ones, and the histogram of dwell times is an exponential with mean $1/\beta$ (for open dwells) or $1/\alpha$ (for closed dwells).

This is the single-channel way of measuring the rates: the mean open time is $1/\beta$.

▸ Go deeper: why dwell times are exponential (optional)

For a channel that is open now, the probability of still being open a time $t$ later is the probability of surviving $t/\Delta t$ ticks each with escape chance $\beta\Delta t$:

$$S(t) = (1 - \beta\Delta t)^{t/\Delta t} \;\xrightarrow{\Delta t \to 0}\; e^{-\beta t}.$$

The density of dwell times is $-dS/dt = \beta e^{-\beta t}$: an exponential with mean $1/\beta$. Same argument with $\alpha$ for closed dwells. If you histogram dwell times on a log y-axis you get a straight line of slope $-\beta$.

Exercise 3 (~8 min). Complete dwell_times: given a 0/1 state array, return the durations (in seconds) of every open stretch and every closed stretch. One approach: find where the state changes (np.diff(state) != 0), and the gaps between successive change points are the dwell lengths in samples; whether each stretch is open or closed is given by the state at its start. Drop the first and last stretch (we didn't see them begin or end).

Check / unstuck. At 45 mV (α = β = 150/s) both mean dwell times should be about 1/150 s = 6.7 ms. Stuck? pp.dwell_times(state, dt).

def dwell_times(state, dt):
    change = np.flatnonzero(np.diff(state) != 0) + 1    # indices where a new stretch starts
    starts, stops = change[:-1], change[1:]             # interior stretches only
    lengths = (stops - starts) * dt
    is_open = state[starts] == 1
    return lengths[is_open], lengths[~is_open]

a45, b45 = float(pp.alpha_rate(45)), float(pp.beta_rate(45))
n_10s = int(10.0 / dt)
st45 = pp.simulate_two_state(a45, b45, dt, n_10s, rng=3)   # 10 s of data for a smooth histogram
open_dw, closed_dw = dwell_times(st45, dt)
print(f"{len(open_dw)} open dwells, mean {open_dw.mean()*1e3:.2f} ms   (1/beta = {1e3/b45:.2f} ms)")
print(f"{len(closed_dw)} closed dwells, mean {closed_dw.mean()*1e3:.2f} ms (1/alpha = {1e3/a45:.2f} ms)")

755 open dwells, mean 6.84 ms   (1/beta = 6.67 ms)
754 closed dwells, mean 6.40 ms (1/alpha = 6.67 ms)

fig, axes = plt.subplots(1, 2, figsize=(11, 3.6))
for ax, dw, rate, name, color in [(axes[0], open_dw, b45, "open", "tab:orange"),
                                  (axes[1], closed_dw, a45, "closed", "0.3")]:
    bins = np.linspace(0, 40, 41)
    ax.hist(dw * 1e3, bins=bins, density=True, color=color, alpha=0.6, label=f"{name} dwell times")
    tt = np.linspace(0, 40, 200)
    ax.plot(tt, rate * np.exp(-rate * tt / 1e3) / 1e3, color="k", lw=2,
            label=f"exponential, mean {1e3/rate:.1f} ms")
    ax.set_xlabel("dwell time (ms)"); ax.set_ylabel("probability density"); ax.legend()
    ax.set_title(f"{name} dwells: short ones are the most common")
plt.tight_layout(); plt.show()

4. A voltage step, repeated: the ensemble average

Now the experiment we'll use for the rest of the module. Hold the membrane at −80 mV, where the channel is almost always closed, then step to +65 mV, where it is open 96% of the time. The channel doesn't open instantly: it takes a random latency to its first opening, then flickers as before. Each repeat of the step — each sweep — looks different.

But average many sweeps and the randomness washes out, leaving a smooth curve: the open probability as a function of time, $p(t)$, rising from ~0 to 0.96 with the time constant $\tau = 1/(\alpha+\beta)$ at the new voltage. That smooth average is exactly what a patch with many channels records all at once — the subject of Notebook 2.

protocol = pp.step_protocol(v_hold=-80, v_step=65, t_pre_ms=2, t_step_ms=12)
states = pp.simulate_single(scheme, protocol, n_sweeps=200, rng=4)    # (sweeps, samples)
open_1 = states == 1

fig, axes = plt.subplots(3, 1, figsize=(10, 7), sharex=True,
                         gridspec_kw=dict(height_ratios=[1, 3, 2]))
pp.plotting.plot_voltage(axes[0], protocol.t_ms, protocol.V)
# raster: a bar wherever the channel is open, one row per sweep
pp.plotting.plot_open_raster(axes[1], open_1, protocol.t_ms, n_show=40,
                             title="40 sweeps of ONE channel: orange = open")
axes[2].plot(protocol.t_ms, open_1.mean(axis=0), color="k", lw=1.5, label="fraction of sweeps open (200 sweeps)")
axes[2].plot(protocol.t_ms, pp.exact_open_probability(scheme, protocol), color="tab:red", ls="--",
             label="theory p(t)")
axes[2].set_xlabel("time (ms)"); axes[2].set_ylabel("open probability"); axes[2].legend(fontsize=9)
plt.tight_layout(); plt.show()
tau_ms = 1e3 / (float(pp.alpha_rate(65)) + float(pp.beta_rate(65)))
print(f"relaxation time constant at +65 mV: tau = 1/(alpha+beta) = {tau_ms:.2f} ms")

relaxation time constant at +65 mV: tau = 1/(alpha+beta) = 1.29 ms

Where we are. One channel: two rates, a memoryless random switch, open probability $\alpha/(\alpha+\beta)$ that rises with voltage, exponential dwell times, and an ensemble average that relaxes exponentially after a step. In Notebook 2 we put $N$ of these switches in one patch and ask what their sum looks like — and, crucially, how much it fluctuates.

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