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Statistical physics of neurons

L3 · Chapter 4

Splitting the gating variables into single channels

In the first three chapters the conductances changed smoothly. On a real membrane the current flows through individual protein channels, each either open or shut, and the moments when they switch are random. Hodgkin and Huxley could not see single channels when they wrote their equations, but their gating variables read naturally as probabilities. This chapter writes that reading as a Markov chain and works out how much the number of open channels fluctuates on a small patch.

A gating variable is a probability

Read nn as the probability that one subunit sits in its open position. A subunit opens at rate αn\alpha_n and closes at rate βn\beta_n, and the average of many subunits obeys the first-order equation of chapter 2. A potassium channel has four such subunits and conducts only when all four are open, with probability n4n^4.

Counting only how many subunits are open, a potassium channel has five states, from fully shut C0C_0 to open OO. From C0C_0 any of four subunits can open, so the rate is 4αn4\alpha_n; from OO any of four can close, at rate 4βn4\beta_n. A sodium channel has three mm subunits and one hh subunit, 4×2=84\times2=8 states in all, of which one is open.

Each channel hops among its states independently. As the number of channels goes to infinity, the open fraction returns to the deterministic Hodgkin–Huxley equations. Goldwyn and Shea-Brown use this description as the reference against which cheaper approximations are tested.

C0 ⇌4αnβn C1 ⇌3αn2βn C2 ⇌2αn3βn C3 ⇌αn4βn OC_0\ \underset{\beta_n}{\overset{4\alpha_n}{\rightleftharpoons}}\ C_1\ \underset{2\beta_n}{\overset{3\alpha_n}{\rightleftharpoons}}\ C_2\ \underset{3\beta_n}{\overset{2\alpha_n}{\rightleftharpoons}}\ C_3\ \underset{4\beta_n}{\overset{\alpha_n}{\rightleftharpoons}}\ O
The five-state Markov chain of the potassium channel. The subscript counts open subunits; rates above the arrows go right and below go left, all multiplied by the temperature factor ϕ\phi of chapter 2.

Source: J. H. Goldwyn and E. Shea-Brown · A. L. Hodgkin and A. F. Huxley

At a given voltage the open count is binomial

If the voltage follows a prescribed curve, the channels do not influence one another and each is open with the same probability pp. A sum of NN independent yes-or-no outcomes is binomial (Eqs. (10) to (13) of Goldwyn and Shea-Brown).

The relative fluctuation is (1−p)/(Np)\sqrt{(1-p)/(Np)}, which falls as N−1/2N^{-1/2}. That is why the deterministic equations hold for a large patch and fail for a small one.

How dense are the channels? Goldwyn and Shea-Brown put 6000 sodium and 1800 potassium channels in 100 square micrometres, that is 60 and 18 per square micrometre. Dividing the Hodgkin–Huxley maximum conductances by these densities gives every channel a conductance of γ=120 mS cm−2/60 μm−2=20 pS\gamma=120\ \mathrm{mS\,cm^{-2}}/60\ \mu\mathrm{m^{-2}}=20\ \mathrm{pS}.

Our simulation freezes the voltage every 0.01 ms, flips each subunit with the exact probability of a two-state process, and then advances the voltage with the frozen conductances. We tested it under a 60 mV clamp: on a 10 square micrometre patch over 1000 trials, the sodium open fraction 0.25 ms after the step is 0.1046±0.00040.1046\pm0.0004 against a binomial prediction of 0.1047, and the variance is 1.00±0.041.00\pm0.04 times the binomial variance. Across three patch sizes, two step sizes and six times, sodium and potassium both stay within 2.6 standard errors.

P(Nopen=k)=(Nk)pk(1−p)N−k,⟨Nopen⟩=Np,Var⁡Nopen=Np(1−p),pNa=m3h, pK=n4P(N_{\mathrm{open}}=k)=\binom{N}{k}p^{k}(1-p)^{N-k},\qquad \langle N_{\mathrm{open}}\rangle=Np,\quad \operatorname{Var}N_{\mathrm{open}}=Np(1-p),\quad p_{\mathrm{Na}}=m^{3}h,\ p_{\mathrm{K}}=n^{4}
Valid when the voltage is fixed or follows a given curve. When the voltage itself fluctuates with the channels, the channels influence one another through it and the binomial law is only approximate.

Source: J. H. Goldwyn and E. Shea-Brown · A. L. Hodgkin and A. F. Huxley

Worked example: resting noise on a 100 square micrometre patch

Problem. At 6.3 °C and rest, how many sodium and potassium channels are open on average on a 100 square micrometre patch, and with what standard deviation? How long does a sodium channel stay open on average, and how far does one opening lift the patch voltage?

Model. At rest m=0.0529m=0.0529, h=0.596h=0.596 and n=0.318n=0.318. The open probabilities are pNa=m3hp_{\mathrm{Na}}=m^3h and pK=n4p_{\mathrm{K}}=n^4. There are four ways out of the open state, any of three mm subunits closing or the hh subunit closing, and the mean dwell time is the inverse of the total exit rate. The patch capacitance is CMA=1 μF cm−2×10−6 cm2=1 pFC_MA=1\ \mu\mathrm{F\,cm^{-2}}\times10^{-6}\ \mathrm{cm^2}=1\ \mathrm{pF}.

Numbers. pNa=8.8×10−5p_{\mathrm{Na}}=8.8\times10^{-5}, so 6000 channels have 0.530.53 open on average with a standard deviation of 0.73. pK=0.0102p_{\mathrm{K}}=0.0102, so 1800 channels have 18.3 open with a standard deviation of 4.3, a 23% fluctuation. The exit rate from the open state is 3×4+0.047=12.0 ms−13\times4+0.047=12.0\ \mathrm{ms^{-1}}, so a sodium channel stays open for 0.083 ms0.083\ \mathrm{ms} on average. One open sodium channel carries 20 pS×115 mV=2.3 pA20\ \mathrm{pS}\times115\ \mathrm{mV}=2.3\ \mathrm{pA}, delivers 0.19 fC0.19\ \mathrm{fC} per opening and lifts the voltage by about 0.19 mV0.19\ \mathrm{mV}.

Explanation. At rest the sodium channels are mostly all shut, and now and then one opens and nudges the voltage. Shrink the patch to 10 square micrometres and the same opening is worth 1.9 mV; at 1 square micrometre it is over ten millivolts. With threshold only 6.5 mV away, a few openings that happen to come close together are enough to start a spike.

Check. The mean potassium current is 18.3 channels times 20 pS times 12 mV, or 4.4 pA, which per square centimetre is exactly the 4.40 µA/cm² of chapter 1. The deterministic equations describe the average, and the single-channel picture leaves it unchanged.

tˉopen=13βm+βh,Δu≈γ (ENa−u) tˉopenCMA\bar t_{\mathrm{open}}=\frac{1}{3\beta_m+\beta_h},\qquad \Delta u\approx\frac{\gamma\,(E_{\mathrm{Na}}-u)\,\bar t_{\mathrm{open}}}{C_MA}
Left: the mean dwell time in the open state of a sodium channel. Right: during one opening the sodium current charges the patch capacitance, ignoring other currents over that short time. AA is the patch area.

Source: J. H. Goldwyn and E. Shea-Brown · A. L. Hodgkin and A. F. Huxley

Try it in the Lab

Open channels jump inside the binomial band

  1. Predict first: when a 1000 square micrometre patch fires, how far will the number of open sodium channels stray from its average curve? And on a patch ten times smaller?
  2. Choose the Many channels: back to Hodgkin–Huxley preset and look at the trial and the band of two standard deviations in the open sodium channel chart. Then drag the patch rim to shrink the area to about 10 square micrometres.
  3. Explain: the band width grows as the square root of the open count, so its relative width shrinks as the square root of the channel count. The smaller the patch, the rougher the curve and the more easily it leaves the average path.

Exercises with explained answers

  1. On a 10 square micrometre patch at rest, how many potassium channels are open on average? How large is the relative fluctuation?

    Show the answer

    With 180 channels, ⟨N⟩=180×0.0102=1.83\langle N\rangle=180\times0.0102=1.83 and the standard deviation is 1.83×0.99=1.35\sqrt{1.83\times0.99}=1.35, a 74% fluctuation.

  2. Under a long clamp at 60 mV, how many of 1800 potassium channels are open on average, and how large is the relative fluctuation?

    Show the answer

    pK=0.8954=0.642p_{\mathrm{K}}=0.895^4=0.642, so 1155 are open on average, with a standard deviation of 1800×0.642×0.358=20.3\sqrt{1800\times0.642\times0.358}=20.3, a 1.8% fluctuation. For a fixed number of channels the fluctuation in count is largest near an open probability of one half, but the relative fluctuation is still small.

  3. Why does a sodium channel have 8 states? Starting from the state with all m subunits shut and the h subunit open, what is the rate toward opening?

    Show the answer

    Zero to three of the mm subunits can be open, 4 cases, and the hh subunit open or shut, 2 cases: 4×2=84\times2=8. With all three mm subunits shut, any of them can open, at rate 3αm3\alpha_m.

  4. Are the channels on one unclamped patch really independent?

    Show the answer

    Not entirely. Each channel’s switching rates depend only on the voltage, and in that sense they are independent; but all of them share one voltage, so a channel that opens changes the voltage and with it the rates of all the others. The binomial law is exact only when the voltage is given, which is the condition Goldwyn and Shea-Brown stress.

Related reading on this site

Sources and what we recomputed

  1. A. L. Hodgkin and A. F. Huxley, A quantitative description of membrane current and its application to conduction and excitation in nerve, Journal of Physiology 117, 500–544 (1952), PMC1392413

    We read every page, pp. 500 to 544, from the scanned pages in PubMed Central. Sign convention on pp. 505 and 506: V is the displacement from rest with depolarisation negative, and inward current is positive. Rate constants: Eqs. (12), (13), (20), (21), (23) and (24). Eq. (26) and Table 3 on pp. 518 to 520: the full membrane equation and its constants at 6.3 °C, with the leak potential chosen so that the current is zero at rest. Numerical method on p. 523. Figs. 12, 13 and 21: threshold between 6 and 7 mV. Table 4 on p. 527: spike shape. Table 5 on pp. 532 and 533: sodium and potassium movements. Fig. 20: refractory period. Fig. 22: anode break. P. 541: their warning that the equations are an empirical description. We recomputed Tables 4 and 5 for the membrane action potentials, not the propagated one.

  2. J. H. Goldwyn and E. Shea-Brown, The what and where of adding channel noise to the Hodgkin–Huxley equations, arXiv:1104.4823v1; PLoS Computational Biology 7, e1002247 (2011)

    Version 1, 25 April 2011, read in full. Eq. (1): the Hodgkin–Huxley equation with depolarisation positive. The Markov-chain description of independent channels and its deterministic limit. Eqs. (10) to (13): for a fixed voltage history, the number of open channels is binomial with mean and variance set by the gating variables. Figs. 1 to 3: 100 µm² holds 6000 sodium and 1800 potassium channels, 10 µm² holds 600 and 180, with a spontaneous spike from channel noise alone. Their comparison of current, subunit and conductance noise. The label of Eq. (13) in v1 says sodium where the formula is the potassium variance.

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