The Resting Membrane and Passive Signals

In a made-up neuron resting at −71 mV, about 200 picoamperes of potassium current leave the cell all the time, and about 200 picoamperes of sodium current come in. Neither ion is at equilibrium, yet the voltage does not move.

That standoff is the resting potential. Building on neurons and glia at work, this lesson computes it from the leaks, then asks how fast and how far a small injected current changes the voltage, which sets up the action potential and synaptic integration. The worked problem’s numbers are invented and describe no real neuron.

The idea: driving force is the distance from equilibrium

Each ion has an equilibrium potential, E_ion, where the electrical force on it balances its concentration gradient; the Nernst equation, derived in transport across membranes, gives it, and here E values are supplied. Vm is inside minus outside.

The driving force on an ion is how far the membrane sits from that ion’s equilibrium potential: Vm − E_ion. If channels for that ion are open and behave like a resistor (ohmic), the current they carry is

I_ion = g_ion × (Vm − E_ion)

where g is the conductance, in nanosiemens (nS) for a small cell. By convention, outward current (positive charge leaving) is positive and inward current negative, so a positive driving force on a cation gives outward current. Think of a ball on a slope whose bottom lies at E_ion: driving force is the height, conductance the slipperiness.

Worked problem, part 1. A cell at 37 °C has [K+] of 140 mM inside and 4 mM outside, and [Na+] of 15 mM inside and 145 mM outside. You are given EK = −95 mV and ENa = +60.6 mV. It rests at −71.2 mV (part 2 shows where that comes from).

K+ driving force: Vm − EK = −71.2 − (−95) = +23.8 mV. Positive, so K+ current is outward.

Na+ driving force: Vm − ENa = −71.2 − 60.6 = −131.8 mV. Negative and large, so Na+ current is inward.

The currents: suppose that near rest the cell has gK = 8.47 nS and gNa = 1.53 nS of open leak channels, each treated as ohmic. Then IK = 8.47 nS × 23.8 mV = +202 pA, and INa = 1.53 nS × (−131.8 mV) = −202 pA (nS times mV gives pA). They cancel. That is what rest means: zero net current, not zero current for each ion.

Left alone, those leaks would run the gradients down. The Na+/K+ pump (three Na+ out, two K+ in per ATP) puts the ions back. Being electrogenic, it adds a direct hyperpolarization that is small in many cells but varies with cell type and pump activity, and its main job is keeping the gradients. How a K+ channel excludes Na+ belongs to ion channels as receptors.

Many leaks, one voltage: the GHK idea

A real membrane passes several ions, so it sits between their equilibrium potentials, nearest the ion it passes most easily. The Goldman-Hodgkin-Katz (GHK) voltage equation makes that precise, with P the relative permeability of each ion:

Vm = (RT/F) × ln[(PK[K]o + PNa[Na]o + PCl[Cl]i) / (PK[K]i + PNa[Na]i + PCl[Cl]o)]

Chloride’s inside concentration goes on top because its charge is negative. The equation comes from Goldman’s 1943 constant-field theory, which assumes a uniform electric field through the membrane, and it gives the voltage at which total current is zero: a steady state kept up by the pump, not an equilibrium.

Worked problem, part 2. Take RT/F = 26.72 mV (37 °C), PK = 1, PNa = 0.04, PCl = 0, and the concentrations above.

Top: 4 + 0.04 × 145 = 4 + 5.8 = 9.8 mM.

Bottom: 140 + 0.04 × 15 = 140 + 0.6 = 140.6 mM.

Vm = 26.72 × ln(9.8 / 140.6) = 26.72 × ln(0.0697) = 26.72 × (−2.664) = −71.2 mV.

The small sodium leak holds the cell 24 mV positive of EK. Now double PNa to 0.08. The top becomes 4 + 11.6 = 15.6 and the bottom 140 + 1.2 = 141.2, so Vm = 26.72 × ln(0.1105) = 26.72 × (−2.203) = −58.9 mV. Raising one ion’s permeability moves Vm toward that ion’s equilibrium potential, here by 12 mV toward ENa.

Two cautions. PK : PNa = 1 : 0.04 is the ratio Hodgkin and Katz used for the resting squid giant axon in 1949 (with PCl about 0.45); other cells differ, so there is no universal resting potential. And chloride is often close to equilibrium at rest, though in many neurons transporters hold it away from it.

The membrane as an RC circuit

The bilayer is a thin insulator between two conducting solutions, so it stores charge like a capacitor (see membrane structure and function), and the leak channels are resistors in parallel with it. For a compact cell whose whole surface sits at one voltage (isopotential), that is the entire model.

Input resistance (Rin) is the steady voltage change divided by the small steady current that caused it: Rin = ΔV / I. Capacitance (C) is the charge stored per volt. In current clamp you set the injected current and record the voltage. A current step first flows mostly into the capacitor, charging it, and only then increasingly through the resistor, so the voltage does not jump. It follows

ΔV(t) = I × Rin × (1 − e^(−t/τ)), with τ = Rin × C

The membrane time constant τ is the time to reach 63% of the final change, since 1 − e^−1 = 0.632. It is not the time to finish: 2τ gives 86% and 3τ 95%. The initial slope is I / C. When the current stops, the deflection decays toward rest with the same τ.

Worked problem, part 3. Rin = 100 MΩ and C = 200 pF. Inject −100 pA from 0 to 60 ms.

τ = (100 × 10^6 Ω) × (200 × 10^−12 F) = 0.020 s = 20 ms.

Final deflection: −100 pA × 100 MΩ = −10 mV.

At 20 ms: −10 × (1 − e^−1) = −10 × 0.632 = −6.32 mV.

At 60 ms (3τ): −10 × (1 − e^−3) = −10 × 0.950 = −9.50 mV, still short of −10.

At 80 ms, 20 ms after the step ends: the decay starts from −9.50 mV, so −9.50 × e^−1 = −3.50 mV.

Initial slope: −100 pA / 200 pF = −0.5 mV per ms.

In absolute terms: −71.2 mV, then −80.7 mV at 60 ms, then −74.7 mV at 80 ms. Inputs arriving within about one τ of each other overlap and add, which is temporal summation. In a branched cell the transient is a sum of exponentials (Rall’s 1969 theoretical analysis), so one exponential is an approximation.

The cable: how far a passive signal spreads

A dendrite or axon is a thin tube of salty cytoplasm in leaky membrane. Injected current can flow along the core (axial resistance) or out through the membrane, like water in a garden hose full of small holes. For a uniform, effectively infinite passive cable with a steady current at one point, the voltage falls exponentially with distance:

V(x) = V0 × e^(−x/λ), with λ = √(rm / ri)

Here rm is the membrane resistance of a unit length of cable (Ω·cm) and ri its axial resistance per unit length (Ω/cm). The length constant λ is the distance over which a steady deflection falls to 37% (e^−1) of its value at the source: smaller, not gone. A leakier membrane or a narrower core shortens λ; with membrane and cytoplasm unchanged, λ grows as the square root of diameter. Capacitance does not appear: it sets how fast the steady state arrives, not its shape in space.

Uniform membrane leak per unit area, relative to the lesson’s cable 1× Diameter, relative to the lesson’s cable 1× Measuring cursor: distance from the electrode 0.50 mm (drag it on the plot, or use this slider)

Part 4 cable Part 5: leak doubled Design problem: leak 0.625× Diameter 4×

Cursor reading

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Lesson’s arithmetic

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λ from the rule (relative to 0.5 mm)

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In this sketch: infinite uniform passive cable, steady state only, voltage held at the source, no active channels; leak (per unit area; cytoplasm unchanged) and diameter in relative units, the lesson’s cable (λ = 0.5 mm) is 1× for both.

V(x) = V0 × e^(−x/λ): at one λ the deflection is 37% of its value at the source, smaller, not gone. Leakier or narrower shortens λ; with membrane and cytoplasm unchanged, λ grows as the square root of diameter.

Worked problem, part 4. An infinite uniform passive cable with λ = 0.5 mm is held at a steady −10 mV deflection at one point.

At 0.5 mm: −10 × e^−1 = −3.68 mV.

At 1 mm (2λ): −10 × e^−2 = −10 × 0.135 = −1.35 mV.

In human temporal-lobe cortex resected from epilepsy patients, Beaulieu-Laroche and colleagues (2018) found that distal inputs to layer 5 pyramidal neurons were attenuated more on the way to the soma than in rat layer 5 neurons: a result for those cells, not all human neurons.

Worked problem, part 5. Instead of changing PNa, double the uniform leak conductance, so membrane resistance halves, with C and geometry unchanged. Rin falls to 50 MΩ, so τ = 50 MΩ × 200 pF = 10 ms. Because λ scales with √rm, λ = 0.5 / √2 = 0.354 mm. The leakier cell is faster, but its signals reach less far, and the step now gives only −5 mV. Whether rest moves depends on the added leak’s reversal potential.

Reading a current-clamp trace

One current step gives three numbers: Rin = ΔV / I, τ (the time to 63%), and C = τ / Rin. Small hyperpolarizing steps, which stay away from threshold, plotted as steady ΔV against I give Rin as the slope; a straight line says the membrane is passive over that range.

Check the trace against the model. A pure RC response rises smoothly to a plateau. If the voltage overshoots and sags back during a hyperpolarizing step, a conductance is turning on with voltage, often the HCN current. If the voltage-current plot bends, Rin depends on voltage. Either way a single RC fit will mislead you.

To predict a permeability change, ask which ion and which way. More K+ conductance moves rest toward EK and lowers Rin and τ; more Na+ conductance depolarizes toward ENa and also lowers Rin. Closing a leak raises Rin and τ and moves rest away from that leak’s reversal potential. A conductance that reverses at rest lowers Rin without moving Vm (shunting).

Design problem

In a made-up current-clamp recording from a neuron in a brain slice, a −50 pA step gives a steady −5 mV deflection with τ = 20 ms. You apply a drug. Rest depolarizes by 8 mV, and the same step now gives −8 mV with τ = 32 ms. What did the drug do, and how would you test your reading?

One way to do it

Before the drug, Rin = 5 mV / 50 pA = 100 MΩ and C = 20 ms / 100 MΩ = 200 pF. After it, Rin = 8 mV / 50 pA = 160 MΩ and C = 32 ms / 160 MΩ = 200 pF.

Capacitance, and so membrane area, did not change; τ rose only because Rin did. Resting conductance fell from 10 nS to 6.25 nS, so the drug closed about 3.75 nS. Closing a conductance moves rest away from its reversal potential, and rest depolarized, so the closed conductance reversed below rest: a K+ leak is the likeliest candidate. If that leak was uniform over the dendrites, λ would rise by √1.6, about 1.26 times.

To test it, voltage-clamp the cell and subtract currents recorded at several voltages with and without the drug. A K+ leak should give a drug-sensitive current that is linear, reverses near EK, and shifts its reversal when external K+ is raised. If that current bends with voltage, or the drug removed a sag seen in the control trace, the target is more likely a voltage-gated conductance such as HCN than a passive leak, though a leak can also curve slightly when ion concentrations differ on the two sides of the membrane.