biology2u
PhysicsMathsChemistryBiology
Tier
⌕ Search ⌘K
Concept

The Goldman-Hodgkin-Katz equation

T-178Home BU-103Threads information · evolution
Statement

Under three assumptions — that the electric field inside a homogeneous membrane slab is uniform (the constant-field assumption), that each permeant ion crosses by Nernst–Planck electrodiffusion independently of every other ion, and that the membrane is at a steady state carrying zero net transmembrane current — the membrane potential of a cell permeable to the monovalent ions K⁺, Na⁺ and Cl⁻ is fixed by the permeability-weighted concentrations on the two sides, \[ V_m \;=\; \frac{RT}{F}\,\ln\frac{P_{\mathrm K}[\mathrm K^{+}]_{\text{o}}+P_{\mathrm{Na}}[\mathrm{Na}^{+}]_{\text{o}}+P_{\mathrm{Cl}}[\mathrm{Cl}^{-}]_{\text{i}}}{P_{\mathrm K}[\mathrm K^{+}]_{\text{i}}+P_{\mathrm{Na}}[\mathrm{Na}^{+}]_{\text{i}}+P_{\mathrm{Cl}}[\mathrm{Cl}^{-}]_{\text{o}}}, \] where \(V_m=\psi_{\text{i}}-\psi_{\text{o}}\), subscripts \(\text{i}\) and \(\text{o}\) denote the intracellular and extracellular bulk solutions, and \(P_S=\beta_S D_S/L\) is the permeability of species \(S\) (partition coefficient \(\beta_S\), diffusion coefficient \(D_S\), membrane thickness \(L\)). The anion enters with its two concentrations interchanged, only permeability ratios affect \(V_m\), and the same derivation delivers, one step earlier, the GHK current equation for each individual ionic current.

Why it matters

The Nernst equation answers a question about one ion in isolation: at what voltage does this ion stop moving? Real membranes are permeable to several ions at once, no one of which is at equilibrium, so no Nernst potential is the answer to what a voltmeter reads. The resting membrane potential page states the multi-ion result; the Goldman–Hodgkin–Katz equation is where that result comes from, and deriving it is what turns “the resting potential is near \(E_{\mathrm K}\) because K⁺ permeability dominates” from a slogan into a calculation that predicts a number to a few millivolts.

It is also the quantitative bridge between molecular events and voltage. Every channel that opens or closes changes one \(P_S\) and nothing else in the equation, so the action potential, synaptic transmission and sensory transduction are all, at the level of the membrane, stories about permeability ratios moving. Run backwards, the same equation is the standard electrophysiological assay: measure the voltage at which a channel's current reverses, and the permeability ratio of the ions carrying it drops out of the algebra. The GHK current equation, meanwhile, is the standard non-ohmic current–voltage law used for Ca²⁺ channels in biophysical models, where the enormous inward concentration gradient makes a linear driving-force law badly wrong.

Hypotheses
Constant field: the electric potential falls linearly across the membrane, \(d\psi/dx=V_m/L\).This is the assumption that makes the transport equation integrable in closed form. It amounts to neglecting the space charge inside the membrane in Poisson's equation. Where fixed charges line a pore, or where the Debye length is not small compared with \(L\), the field is not uniform and the predicted current–voltage curve has the wrong shape, even though the zero-current potential may survive.
Independence: each ion's flux depends only on its own concentrations, its own charge and the field, not on the presence of other ions.Drop it and channels that hold several ions in single file — most K⁺ channels — are misdescribed: their unidirectional flux ratio is not the one-ion Ussing value, and mixtures of two permeant ions can conduct worse than either alone (the anomalous mole-fraction effect), which no independent-flux theory can produce.
Steady state with zero net current: \(\partial C_S/\partial t=0\) everywhere inside the membrane, and \(\sum_S I_S=0\).The steady-state part makes each \(J_S\) constant across the slab; the zero-current part is what selects \(V_m\). Both fail during an action potential, when the capacitive current \(C_m\,dV_m/dt\) is comparable to the ionic currents, and both fail in the presence of an electrogenic pump, whose current is not in the sum.
Only monovalent permeant ions appear in the voltage equation.A divalent species does not enter linearly: with \(z=+2\) the zero-current condition becomes a quadratic in \(e^{-FV_m/RT}\) (Problem 5), and there is no logarithmic closed form of the GHK shape.
Instantaneous ionic equilibrium at both membrane faces, with a concentration-independent partition coefficient \(\beta_S\), and constant \(D_S\) inside the slab.These make \(C_S(0)=\beta_S[S]_{\text{o}}\) and \(C_S(L)=\beta_S[S]_{\text{i}}\) legitimate boundary conditions and let \(\beta_S\), \(D_S\) and \(L\) collapse into the single measurable \(P_S\). If the interface is rate-limiting, or if binding inside the pore saturates, \(P_S\) becomes a function of concentration and voltage and stops being a material constant.
The bulk solutions are well stirred, dilute enough that concentrations may stand in for activities, and unstirred layers do not deplete.Otherwise the concentrations at the membrane faces are not the bath concentrations that the experimenter measures, and the fitted permeability ratios absorb the error.
Proof

Set up coordinates once and keep them: \(x=0\) at the outer face of the membrane, \(x=L\) at the inner face, so a positive flux \(J_S\) is directed inward. Take \(\psi(0)=0\) and \(\psi(L)=V_m\), which is the physiological sign convention (inside relative to outside). Throughout, \(R=8.314\ \mathrm{J\,mol^{-1}K^{-1}}\) is the gas constant and \(F=96485\ \mathrm{C\,mol^{-1}}\) the Faraday constant.

1
\[ J_S \;=\; -D_S\left(\frac{dC_S}{dx} \;+\; \frac{z_S F}{RT}\,C_S\,\frac{d\psi}{dx}\right) \]
The Nernst–Planck equation: the flux of a dilute charged solute is Fickian diffusion down its concentration gradient plus electrical drift, the mobility of the drift term being fixed by the diffusion coefficient through the Einstein relation, which is why a single constant \(D_S\) governs both terms. B
2
\[ \frac{\partial C_S}{\partial t}=-\frac{\partial J_S}{\partial x}=0 \quad\Longrightarrow\quad J_S \ \text{is independent of}\ x \]
Conservation of matter inside the slab, plus the steady-state hypothesis: nothing accumulates anywhere in the membrane, and no species is created or destroyed there, so whatever enters at \(x=0\) leaves at \(x=L\). This is what makes \(J_S\) a constant of integration rather than an unknown function. A
3
\[ \frac{d\psi}{dx}=\frac{V_m}{L},\qquad u_S\equiv\frac{z_SFV_m}{RT}\qquad\Longrightarrow\qquad \frac{dC_S}{dx}+\frac{u_S}{L}\,C_S \;=\; -\frac{J_S}{D_S} \]
The constant-field hypothesis. Substituting the uniform field into Step 1 and dividing by \(-D_S\) turns a nonlinear electrodiffusion problem into a first-order linear ODE with constant coefficients, whose forcing term is the constant of Step 2. The dimensionless \(u_S\) is the membrane voltage measured in units of \(RT/z_SF\). B
4
\[ \frac{d}{dx}\!\left(C_S\,e^{u_Sx/L}\right)=-\frac{J_S}{D_S}\,e^{u_Sx/L} \quad\Longrightarrow\quad C_S(L)e^{u_S}-C_S(0)=-\frac{J_S L}{D_S\,u_S}\left(e^{u_S}-1\right) \]
Multiply by the integrating factor \(e^{u_Sx/L}\) and integrate from \(0\) to \(L\); the right-hand integral is elementary. No approximation is made here — every simplification was spent in Step 3. A
5
\[ C_S(0)=\beta_S[S]_{\text{o}},\quad C_S(L)=\beta_S[S]_{\text{i}},\quad P_S\equiv\frac{\beta_S D_S}{L} \quad\Longrightarrow\quad \boxed{\,J_S \;=\; P_S\,u_S\,\frac{[S]_{\text{o}}-[S]_{\text{i}}\,e^{u_S}}{e^{u_S}-1}\,} \]
Impose the interfacial boundary conditions and solve Step 4 for \(J_S\). The three unmeasurable membrane quantities \(\beta_S\), \(D_S\) and \(L\) appear only in the combination \(\beta_S D_S/L\), so the theory is testable: permeability is the one membrane parameter the experiment can see. This is the GHK flux equation. B
6
\[ I_S=-z_SFJ_S \;=\; P_S z_S^2\,\frac{F^2V_m}{RT}\cdot\frac{[S]_{\text{i}}-[S]_{\text{o}}\,e^{-z_SFV_m/RT}}{1-e^{-z_SFV_m/RT}} \]
Convert moles to charge and adopt the electrophysiological sign convention that outward current is positive; since \(+x\) points inward, an inward flux of cations is a negative current, hence the minus sign. Multiplying numerator and denominator by \(e^{-u_S}\) and using \(z_SFu_S=z_S^2F^2V_m/RT\) gives the GHK current equation. A
7
\[ I_S=0 \iff [S]_{\text{i}}=[S]_{\text{o}}e^{-z_SFV_m/RT} \iff V_m=\frac{RT}{z_SF}\ln\frac{[S]_{\text{o}}}{[S]_{\text{i}}}=E_S \]
Consistency check: a membrane permeable to one ion only sits at that ion's Nernst potential, so GHK contains Nernst as a special case rather than competing with it. A second check is the limit \(V_m\to0\), where \(u_S/(1-e^{-u_S})\to1\) and Step 5 collapses to Fick's law, \(J_S=P_S([S]_{\text{o}}-[S]_{\text{i}})\). A
8
\[ \text{for } z=-1:\quad I_{\mathrm{Cl}}=P_{\mathrm{Cl}}\frac{F^2V_m}{RT}\cdot\frac{[\mathrm{Cl}^{-}]_{\text{i}}-[\mathrm{Cl}^{-}]_{\text{o}}e^{+FV_m/RT}}{1-e^{+FV_m/RT}} =P_{\mathrm{Cl}}\frac{F^2V_m}{RT}\cdot\frac{[\mathrm{Cl}^{-}]_{\text{o}}-[\mathrm{Cl}^{-}]_{\text{i}}e^{-FV_m/RT}}{1-e^{-FV_m/RT}} \]
Specialise Step 6 to a monovalent anion, where \(z^2=1\) but \(u_{\mathrm{Cl}}=-FV_m/RT\); multiplying numerator and denominator by \(e^{-FV_m/RT}\) puts every ion over the same denominator \(1-e^{-FV_m/RT}\), at the cost of exchanging the anion's inside and outside concentrations. That exchange is the whole reason chloride looks “upside down” in the final formula. B
9
\[ X\equiv e^{-FV_m/RT}:\qquad \frac{F^2V_m}{RT\,(1-X)}\Big[P_{\mathrm K}\big([\mathrm K^{+}]_{\text{i}}-[\mathrm K^{+}]_{\text{o}}X\big)+P_{\mathrm{Na}}\big([\mathrm{Na}^{+}]_{\text{i}}-[\mathrm{Na}^{+}]_{\text{o}}X\big)+P_{\mathrm{Cl}}\big([\mathrm{Cl}^{-}]_{\text{o}}-[\mathrm{Cl}^{-}]_{\text{i}}X\big)\Big]=0 \]
Impose \(I_{\mathrm K}+I_{\mathrm{Na}}+I_{\mathrm{Cl}}=0\). Because Step 8 gave every term the same prefactor, the sum factorises. For \(V_m\neq0\) the prefactor is non-zero, so the bracket must vanish — this is the step where zero net current, not zero current per ion, does the work. B
10
\[ X=\frac{P_{\mathrm K}[\mathrm K^{+}]_{\text{i}}+P_{\mathrm{Na}}[\mathrm{Na}^{+}]_{\text{i}}+P_{\mathrm{Cl}}[\mathrm{Cl}^{-}]_{\text{o}}}{P_{\mathrm K}[\mathrm K^{+}]_{\text{o}}+P_{\mathrm{Na}}[\mathrm{Na}^{+}]_{\text{o}}+P_{\mathrm{Cl}}[\mathrm{Cl}^{-}]_{\text{i}}} ,\qquad V_m=-\frac{RT}{F}\ln X \]
Solve the bracket of Step 9 for \(X\) — it is linear in \(X\) precisely because every ion is monovalent — and invert the definition \(X=e^{-FV_m/RT}\). Reciprocating inside the logarithm removes the minus sign and gives the standard form. Both numerator and denominator are positive sums, so the logarithm is always defined. A
11
\[ I_S=z_SFP_S\big([S]_{\text{i}}\,f(u_S)-[S]_{\text{o}}\,f(-u_S)\big),\qquad f(u)=\frac{u}{1-e^{-u}},\qquad f'(u)=\frac{1-e^{-u}-ue^{-u}}{(1-e^{-u})^2}\ge 0 \]
Existence and uniqueness of the root. Writing \(h(u)=1-e^{-u}-ue^{-u}\) gives \(h(0)=0\) and \(h'(u)=ue^{-u}\), so \(h\) has a global minimum of \(0\) at \(u=0\): \(f\) is increasing, hence \(f(-u_S)\) is decreasing, hence every \(I_S\) is a strictly increasing function of \(V_m\) (for \(z_S<0\) the negative prefactor is cancelled by \(u_S\) decreasing in \(V_m\)). The sum is therefore strictly increasing, runs from \(-\infty\) to \(+\infty\), and crosses zero exactly once: the GHK potential is the unique zero-current voltage, and the removable singularity at \(V_m=0\) is filled by \(f(0)=1\). C
Result
\[ V_m=\frac{RT}{F}\ln\frac{P_{\mathrm K}[\mathrm K^{+}]_{\text{o}}+P_{\mathrm{Na}}[\mathrm{Na}^{+}]_{\text{o}}+P_{\mathrm{Cl}}[\mathrm{Cl}^{-}]_{\text{i}}}{P_{\mathrm K}[\mathrm K^{+}]_{\text{i}}+P_{\mathrm{Na}}[\mathrm{Na}^{+}]_{\text{i}}+P_{\mathrm{Cl}}[\mathrm{Cl}^{-}]_{\text{o}}} \qquad I_S=P_Sz_S^2\frac{F^2V_m}{RT}\cdot\frac{[S]_{\text{i}}-[S]_{\text{o}}e^{-z_SFV_m/RT}}{1-e^{-z_SFV_m/RT}} \]

Reading. The membrane sits at the one voltage where the ionic currents cancel. Each ion pulls \(V_m\) toward its own Nernst potential with a strength proportional to its permeability, and because the pull enters inside a logarithm of summed permeability×concentration products, the winner is whichever ion has the largest \(P_S[S]\) product, not the largest gradient. Only ratios of permeabilities matter for \(V_m\); absolute permeabilities are needed only for the currents.

Scope. Steady state, zero net current, uniform field, independent monovalent ions. \(RT/F=25.3\ \mathrm{mV}\) at 20 °C and \(26.7\ \mathrm{mV}\) at 37 °C; in base-10 form the prefactor \(2.303\,RT/F\) is \(58.2\) and \(61.5\ \mathrm{mV}\) per decade respectively. \(P_S\) has units of velocity (cm s⁻¹), \(I_S\) of current density.

Corollaries & converses
  • Nernst as a limit. If one permeability dominates, \(P_{\mathrm K}\gg P_{\mathrm{Na}},P_{\mathrm{Cl}}\), the other terms cancel between numerator and denominator and \(V_m\to E_{\mathrm K}\). The Nernst equation is the single-ion corner of GHK, not a rival to it.
  • Chloride is a follower. If Cl⁻ is passively distributed, \([\mathrm{Cl}^{-}]_{\text{i}}/[\mathrm{Cl}^{-}]_{\text{o}}\) equals the cation ratio already set by K⁺ and Na⁺, and \(P_{\mathrm{Cl}}\) cancels out of the Result entirely (Problem 3). A cell can have an enormous chloride permeability and still have its resting potential set by K⁺ alone.
  • Ussing flux ratio. Splitting Step 5 into its two additive pieces gives unidirectional influx and efflux whose ratio is \(M_{\text{in}}/M_{\text{out}}=([S]_{\text{o}}/[S]_{\text{i}})\,e^{-z_SFV_m/RT}\), independent of \(P_S\) and of the constant-field assumption. Measured exponents larger than \(1\) are the classic evidence for single-file, multi-ion pores.
  • Goldman rectification. The GHK current equation is not ohmic: when \([S]_{\text{i}}\gg[S]_{\text{o}}\) the outward current at a given driving force exceeds the inward current (Example 2), so a channel with a fixed permeability still shows a curved current–voltage relation.
  • Converse (the standard assay). A measured reversal potential inverts the Result: for two permeant monovalent cations, \(P_{\mathrm{Na}}/P_{\mathrm K}\) follows algebraically from \(E_{\text{rev}}\) and the four concentrations (Problem 4). This bi-ionic measurement is how channel selectivity sequences are determined.
  • Not a conductance average. The chord-conductance formula \(V_m=(g_{\mathrm K}E_{\mathrm K}+g_{\mathrm{Na}}E_{\mathrm{Na}})/(g_{\mathrm K}+g_{\mathrm{Na}})\) is a different model — ohmic channels rather than constant-field ones. The two agree closely for physiological gradients but are not the same equation, and they diverge where rectification is strong.
Fails without
  • Constant field dropped (strong pore charge, or a thin membrane relative to the Debye length): the potential profile inside a real ion channel is set by fixed charges, dipoles and the ions in transit, and Poisson's equation must be solved together with Nernst–Planck rather than replaced by a straight line. The zero-current potential is comparatively robust, but the predicted current–voltage shape is not: constant-field theory gives one universal rectification curve for a given concentration ratio, whereas real channels of identical selectivity show rectification ranging from strongly inward to strongly outward.
  • Independence dropped (multi-ion, single-file pores): in K⁺ channels the selectivity filter holds two or more ions simultaneously and they must move in lockstep. The measured unidirectional flux ratio then follows the Ussing form raised to a power appreciably greater than \(1\) — values near \(2\ldots3\) are reported for K⁺ channels — which is flatly impossible for independent ions. The same failure produces the anomalous mole-fraction effect, where a mixture of two permeant ions carries less current than either pure solution: GHK, being linear in each \([S]\), can never predict a conductance minimum.
  • Zero net current dropped (electrogenic pumps): the Na⁺/K⁺-ATPase exports 3 Na⁺ for every 2 K⁺ it imports, so it carries a standing outward current that is not one of the \(I_S\) terms. The true steady state satisfies \(\sum_S I_S+I_{\text{pump}}=0\), and the measured potential is a few millivolts more negative than the Result predicts; block the pump acutely and the potential jumps by that amount before the gradients have had time to run down.
  • Steady state dropped (fast gating): during the upstroke of an action potential \(P_{\mathrm{Na}}\) changes over a fraction of a millisecond and the capacitive current \(C_m\,dV_m/dt\) is comparable to the ionic currents, so \(\sum_S I_S\neq0\). The GHK potential then describes only the moving target the voltage is chasing, not the voltage itself; the trajectory requires the full Hodgkin–Huxley system.
  • Monovalence dropped (Ca²⁺): with \(z=+2\) the zero-current condition is quadratic in \(X=e^{-FV_m/RT}\) (Problem 5) and has no GHK-shaped logarithmic solution. Inserting \(P_{\mathrm{Ca}}[\mathrm{Ca}^{2+}]\) as though it were another monovalent term is simply a different equation, and it gets the answer wrong wherever calcium's contribution is not negligible.
Common errors
  • “Chloride goes in the numerator with the other outside concentrations.” Its charge is \(z=-1\), so Step 8 exchanges its two concentrations: \([\mathrm{Cl}^{-}]_{\text{i}}\) sits on top with the outside cations. Getting this backwards can shift a computed potential by tens of millivolts.
  • “\(V_m\) is the permeability-weighted average of the Nernst potentials.” That is the chord-conductance formula, a different model. GHK takes the logarithm of a weighted sum of concentrations; the two coincide only approximately.
  • “\(RT/F\) is \(61\ \mathrm{mV}\).” \(61\ \mathrm{mV}\) is \(2.303\,RT/F\) at 37 °C, the prefactor for \(\log_{10}\). With a natural logarithm the number is \(26.7\ \mathrm{mV}\); mixing them inflates every answer by a factor of \(2.303\).
  • “Permeability is conductance.” \(P_S\) is a velocity (cm s⁻¹) and appears in the Result as a pure ratio; conductance \(g_S\) is a current per volt that itself depends on \(V_m\) and on the concentrations. A channel can have a high permeability to an ion that is nearly absent, and thus almost no conductance for it.
  • “At \(V_m\) nothing is moving.” Only the net current is zero. Each \(I_S\) is individually non-zero at the GHK potential (Step 9), which is exactly why gradients run down and the ATPase must keep working — the resting potential is a steady state, not an equilibrium.
  • “Absolute permeabilities are needed.” Scaling every \(P_S\) by the same factor leaves \(V_m\) untouched, so the voltage equation determines ratios only; conversely, the current equation genuinely needs absolute values, and a ratio is not enough.
Discussion

The constant-field integration was published by David Goldman in 1943 as a general treatment of membrane impedance and rectification; Alan Hodgkin and Bernard Katz applied it to the squid giant axon in 1949, and it is their use of it — showing that the axon's potential depends on external Na⁺ in exactly the way a shifting \(P_{\mathrm{Na}}/P_{\mathrm K}\) ratio predicts — that attached their names to the equation and made the ionic hypothesis quantitative. The historical order matters for understanding the assumptions: constant field was a mathematical convenience adopted to obtain any closed form for a membrane of unknown structure, decades before anyone knew that conduction happens through discrete protein pores.

That is also why GHK survives its own falsification. Each hypothesis is known to be false in detail for real channels, yet the voltage equation remains the working tool of the field, because the zero-current potential depends mostly on the ratios of the permeabilities and on the concentration gradients, both of which the model treats correctly, and only weakly on the field profile, which it treats crudely. The current equation is a stronger claim and fails more visibly: it predicts a single universal rectification shape that many channels do not obey.

The modern continuum successor is Poisson–Nernst–Planck theory, which keeps Step 1 but replaces Step 3 by solving Poisson's equation self-consistently for the potential, so the field is computed from the charge distribution rather than imposed. PNP reproduces GHK in the limit of a wide, weakly charged pore and departs from it exactly where GHK is known to fail. Below that lie Brownian and molecular dynamics, which abandon the continuum picture altogether — necessary in a selectivity filter only a few Ångström wide, where the notion of a local concentration of ions inside the pore has no clear meaning. The persistence of GHK alongside these is a good illustration of a model whose value lies in the structure of its answer, not in the literal truth of its premises.

Common misconceptions. That GHK explains why the resting potential exists: it does not, it presupposes the gradients, which are built by the Na⁺/K⁺-ATPase at the cost of ATP; GHK only converts given gradients and permeabilities into a voltage. That the equation applies to a single channel: it is a macroscopic law for a whole membrane, and applying it to one channel requires the further, non-trivial assumption that the same constant-field description holds inside a pore of atomic dimensions. And that it needs the cell to be at equilibrium — it needs the opposite, a non-equilibrium steady state in which every permeant ion is dissipating its own gradient.

Worked examples

Example 1. The squid giant axon in seawater, at 20 °C. Use the standard quoted concentrations \([\mathrm K^{+}]_{\text{o}}=20\), \([\mathrm K^{+}]_{\text{i}}=400\), \([\mathrm{Na}^{+}]_{\text{o}}=440\), \([\mathrm{Na}^{+}]_{\text{i}}=50\), \([\mathrm{Cl}^{-}]_{\text{o}}=560\), \([\mathrm{Cl}^{-}]_{\text{i}}=40\) (all mM), with resting permeability ratios \(P_{\mathrm K}:P_{\mathrm{Na}}:P_{\mathrm{Cl}}=1:0.04:0.45\). Compute the resting potential; then recompute at the peak of the action potential, where the ratios become \(1:20:0.45\), and find the swing.

1
\[ \frac{RT}{F}=\frac{(8.314)(293.15)}{96485}\ \mathrm{V}=0.02526\ \mathrm{V}=25.26\ \mathrm{mV} \]
The thermal voltage at \(T=293.15\ \mathrm{K}\). Units: \(\mathrm{J\,mol^{-1}K^{-1}}\times\mathrm{K}\div\mathrm{C\,mol^{-1}}=\mathrm{J\,C^{-1}}=\mathrm{V}\). A
2
\[ \text{numerator}=(1)(20)+(0.04)(440)+(0.45)(40)=20+17.6+18.0=55.6 \]
Outside cations and the inside anion, weighted by relative permeability (Step 8 of the Proof). Units of mM throughout; because only the ratio of the two sums is taken, the concentration unit and the permeability unit both cancel. A
3
\[ \text{denominator}=(1)(400)+(0.04)(50)+(0.45)(560)=400+2.0+252=654 \]
Inside cations and the outside anion. Note how large the chloride term is: \(252\) of \(654\), yet it moves the answer by only \(2.4\ \mathrm{mV}\) — deleting both chloride terms gives \(-59.9\) in place of the \(-62.3\ \mathrm{mV}\) below. The reason is not that it appears in both sums (so does sodium, which shifts the answer far more) but that \(E_{\mathrm{Cl}}=(25.26)\ln(40/560)=-66.7\ \mathrm{mV}\) already lies close to the potential the two cations set on their own, so chloride has almost nothing to pull against. A
4
\[ V_m=(25.26\ \mathrm{mV})\ln\frac{55.6}{654}=(25.26)\ln(0.08502)=(25.26)(-2.4649)=-62.3\ \mathrm{mV} \]
The measured resting potential of the squid axon is about \(-60\ \mathrm{mV}\); \(E_{\mathrm K}=(25.26)\ln(20/400)=-75.7\ \mathrm{mV}\) and \(E_{\mathrm{Na}}=(25.26)\ln(440/50)=+54.9\ \mathrm{mV}\), so the answer lies between them and much nearer \(E_{\mathrm K}\), as \(P_{\mathrm K}\) dominance requires. A
5
\[ V_m^{\text{peak}}=(25.26)\ln\frac{20+(20)(440)+18.0}{400+(20)(50)+252}=(25.26)\ln\frac{8838}{1652}=(25.26)(1.6771)=+42.4\ \mathrm{mV} \]
Only \(P_{\mathrm{Na}}\) changed, by a factor of \(500\); the potential now sits close to \(E_{\mathrm{Na}}\) instead of \(E_{\mathrm K}\), because sodium's \(P[S]\) product has become the largest term in both sums. This single parameter change is the whole electrical content of the action potential's upstroke. B
\[ V_m^{\text{rest}}=-62.3\ \mathrm{mV},\qquad V_m^{\text{peak}}=+42.4\ \mathrm{mV},\qquad \Delta V=105\ \mathrm{mV} \]

Reading. One permeability ratio, moved from \(P_{\mathrm{Na}}/P_{\mathrm K}=0.04\) to \(20\), carries the membrane potential across \(105\ \mathrm{mV}\) with no change whatever in any ion concentration — which is why an action potential costs so little ionic traffic and can be repeated thousands of times before the pumps must catch up.

Scope. Both values assume a steady state, so they are the asymptotes the real trajectory chases; the true peak falls a little short of \(+42\ \mathrm{mV}\) because K⁺ permeability is already rising before Na⁺ permeability has finished.

Example 2. Goldman rectification, quantitatively. A mammalian neuron at 37 °C has \([\mathrm K^{+}]_{\text{i}}=140\ \mathrm{mM}\), \([\mathrm K^{+}]_{\text{o}}=5\ \mathrm{mM}\) and a resting potassium permeability \(P_{\mathrm K}=1.0\times10^{-6}\ \mathrm{cm\,s^{-1}}\). Compute the K⁺ current density at \(V_m=-70\ \mathrm{mV}\), and compare it with the current at \(V_m=-108\ \mathrm{mV}\), which lies the same \(19\ \mathrm{mV}\) from \(E_{\mathrm K}\) on the other side.

1
\[ \frac{RT}{F}=\frac{(8.314)(310.15)}{96485}=26.73\ \mathrm{mV},\qquad E_{\mathrm K}=(26.73)\ln\frac{5}{140}=(26.73)(-3.332)=-89.1\ \mathrm{mV} \]
The reversal potential, from Step 7 of the Proof. Both test voltages are \(19\ \mathrm{mV}\) from it, one above and one below, so any difference in current magnitude is rectification and nothing else. A
2
\[ P_{\mathrm K}=1.0\times10^{-6}\ \mathrm{cm\,s^{-1}}=1.0\times10^{-8}\ \mathrm{m\,s^{-1}},\qquad [\mathrm K^{+}]_{\text{i}}=140\ \mathrm{mol\,m^{-3}},\quad [\mathrm K^{+}]_{\text{o}}=5\ \mathrm{mol\,m^{-3}} \]
Convert to SI before touching the current equation: \(1\ \mathrm{mM}=1\ \mathrm{mol\,m^{-3}}\). The product \(P\,F\,[S]\) then has units \(\mathrm{m\,s^{-1}}\times\mathrm{C\,mol^{-1}}\times\mathrm{mol\,m^{-3}}=\mathrm{A\,m^{-2}}\), a current density, as required. A
3
\[ u=\frac{FV_m}{RT}=\frac{-70}{26.73}=-2.619,\qquad e^{-u}=e^{2.619}=13.73 \]
Working in the dimensionless voltage keeps the arithmetic clean; \(u\) is just \(V_m\) measured in units of \(26.73\ \mathrm{mV}\). A
4
\[ I_{\mathrm K}=P_{\mathrm K}Fu\,\frac{[\mathrm K^{+}]_{\text{i}}-[\mathrm K^{+}]_{\text{o}}e^{-u}}{1-e^{-u}} =(10^{-8})(96485)(-2.619)\cdot\frac{140-(5)(13.73)}{1-13.73} \]
The GHK current equation with \(z=+1\), using \(z^2F^2V_m/RT=Fu\) to avoid squaring large numbers. A
5
\[ =(-2.527\times10^{-3})\cdot\frac{71.35}{-12.73} =(-2.527\times10^{-3})(-5.605)=1.416\times10^{-2}\ \mathrm{A\,m^{-2}}=1.42\ \mathrm{\mu A\,cm^{-2}} \]
Positive, i.e. outward, as it must be at a voltage \(19\ \mathrm{mV}\) positive of \(E_{\mathrm K}\). The unit conversion is \(1\ \mathrm{A\,m^{-2}}=100\ \mathrm{\mu A\,cm^{-2}}\). A
6
\[ V_m=-108\ \mathrm{mV}:\quad u=-4.041,\ e^{-u}=56.89,\qquad I_{\mathrm K}=(10^{-8})(96485)(-4.041)\cdot\frac{140-284.5}{1-56.89}=-1.008\times10^{-2}\ \mathrm{A\,m^{-2}} \]
Now inward (negative), and smaller in magnitude: \(1.01\) against \(1.42\ \mathrm{\mu A\,cm^{-2}}\) for the same \(19\ \mathrm{mV}\) driving force. An ohmic channel would have given the same magnitude both ways. B
\[ I_{\mathrm K}(-70\ \mathrm{mV})=+1.42\ \mathrm{\mu A\,cm^{-2}},\qquad I_{\mathrm K}(-108\ \mathrm{mV})=-1.01\ \mathrm{\mu A\,cm^{-2}},\qquad \frac{|I_{\text{out}}|}{|I_{\text{in}}|}=1.41 \]

Reading. Constant-field theory makes a channel with a fixed permeability rectify outwards whenever the ion is more concentrated inside, because the current is carried by whichever population the field is sweeping out of the membrane, and that population is set by the concentration at the upstream face. Rectification here is a property of the concentration gradient, not of the protein.

Scope. The effect grows with the concentration ratio and with distance from \(E_S\); for Ca²⁺, where \([\mathrm{Ca}^{2+}]_{\text{o}}/[\mathrm{Ca}^{2+}]_{\text{i}}\) exceeds \(10^{4}\), it is dramatic, and using a linear \(g(V_m-E_{\mathrm{Ca}})\) law instead is qualitatively wrong.

Problems
  1. A mammalian neuron at 37 °C has \([\mathrm K^{+}]_{\text{o}}=5\), \([\mathrm K^{+}]_{\text{i}}=140\), \([\mathrm{Na}^{+}]_{\text{o}}=145\), \([\mathrm{Na}^{+}]_{\text{i}}=15\) (mM) and \(P_{\mathrm{Na}}/P_{\mathrm K}=0.03\). Ignoring chloride, compute \(V_m\) and compare it with \(E_{\mathrm K}\).
    Solution

    Take \(P_{\mathrm K}=1\), \(P_{\mathrm{Na}}=0.03\) and \(RT/F=26.73\ \mathrm{mV}\). Numerator \(=5+(0.03)(145)=5+4.35=9.35\); denominator \(=140+(0.03)(15)=140+0.45=140.45\). The ratio is \(9.35/140.45=0.06657\), and \(\ln(0.06657)=-2.709\), so \[ V_m=(26.73)(-2.709)=-72.4\ \mathrm{mV}. \] For comparison \(E_{\mathrm K}=(26.73)\ln(5/140)=-89.1\ \mathrm{mV}\). A sodium permeability of only 3% of the potassium permeability therefore depolarises the cell by nearly \(17\ \mathrm{mV}\) from \(E_{\mathrm K}\) — the small Na⁺ term is amplified because \([\mathrm{Na}^{+}]_{\text{o}}\) is large where \([\mathrm K^{+}]_{\text{o}}\) is small, so \(P_{\mathrm{Na}}[\mathrm{Na}^{+}]_{\text{o}}=4.35\) is nearly as big as \(P_{\mathrm K}[\mathrm K^{+}]_{\text{o}}=5\) in the numerator, while contributing almost nothing to the denominator.

  2. Continuing Problem 1, a patient's serum potassium doubles from \(5\) to \(10\ \mathrm{mM}\) (hyperkalaemia). Recompute \(V_m\) and state by how much the cell depolarises. Why does the same absolute change in \([\mathrm{Na}^{+}]_{\text{o}}\) matter far less?
    Solution

    Numerator \(=10+4.35=14.35\); denominator is unchanged at \(140.45\). The ratio is \(14.35/140.45=0.10217\), \(\ln(0.10217)=-2.2811\), so \[ V_m=(26.73)(-2.2811)=-61.0\ \mathrm{mV}, \] a depolarisation of \(11.4\ \mathrm{mV}\) from \(-72.4\ \mathrm{mV}\). Adding \(5\ \mathrm{mM}\) to \([\mathrm{Na}^{+}]_{\text{o}}=145\) instead would change the numerator only from \(9.35\) to \(9.35+(0.03)(5)=9.50\), giving \(V_m=-72.0\ \mathrm{mV}\): a shift of \(0.4\ \mathrm{mV}\). The asymmetry is entirely the permeability weighting — a millimole of potassium outside counts \(1/0.03\approx33\) times as much as a millimole of sodium. This is why extracellular potassium is under tight physiological control and why hyperkalaemia is acutely dangerous: the sustained depolarisation inactivates voltage-gated Na⁺ channels and cardiac excitability fails.

  3. Show that if chloride is passively distributed — that is, if \(E_{\mathrm{Cl}}=V_m\) already — then \(P_{\mathrm{Cl}}\) drops out of the Result entirely, whatever its value.
    Solution

    Write \(N=P_{\mathrm K}[\mathrm K^{+}]_{\text{o}}+P_{\mathrm{Na}}[\mathrm{Na}^{+}]_{\text{o}}\) and \(D=P_{\mathrm K}[\mathrm K^{+}]_{\text{i}}+P_{\mathrm{Na}}[\mathrm{Na}^{+}]_{\text{i}}\) for the cation sums, so that without chloride \(e^{FV_m/RT}=N/D\). Chloride at equilibrium means \[ V_m=E_{\mathrm{Cl}}=\frac{RT}{(-1)F}\ln\frac{[\mathrm{Cl}^{-}]_{\text{o}}}{[\mathrm{Cl}^{-}]_{\text{i}}}=\frac{RT}{F}\ln\frac{[\mathrm{Cl}^{-}]_{\text{i}}}{[\mathrm{Cl}^{-}]_{\text{o}}}, \] i.e. \([\mathrm{Cl}^{-}]_{\text{i}}/[\mathrm{Cl}^{-}]_{\text{o}}=e^{FV_m/RT}=N/D\), so \([\mathrm{Cl}^{-}]_{\text{i}}=(N/D)[\mathrm{Cl}^{-}]_{\text{o}}\). Substituting into the full Result, \[ \frac{N+P_{\mathrm{Cl}}[\mathrm{Cl}^{-}]_{\text{i}}}{D+P_{\mathrm{Cl}}[\mathrm{Cl}^{-}]_{\text{o}}} =\frac{N+P_{\mathrm{Cl}}(N/D)[\mathrm{Cl}^{-}]_{\text{o}}}{D+P_{\mathrm{Cl}}[\mathrm{Cl}^{-}]_{\text{o}}} =\frac{(N/D)\big(D+P_{\mathrm{Cl}}[\mathrm{Cl}^{-}]_{\text{o}}\big)}{D+P_{\mathrm{Cl}}[\mathrm{Cl}^{-}]_{\text{o}}}=\frac{N}{D}, \] which is the chloride-free value for every \(P_{\mathrm{Cl}}\ge0\). Physically: a passively distributed anion carries no net current at \(V_m\), so it cannot participate in setting the voltage at which the currents balance. It does still matter dynamically — a large \(P_{\mathrm{Cl}}\) stabilises \(V_m\) against perturbation and shortens the membrane time constant — but it contributes nothing to the steady-state value. Where chloride is actively accumulated or extruded (as by the KCC2 and NKCC1 transporters in neurons), \(E_{\mathrm{Cl}}\neq V_m\), the cancellation fails, and chloride becomes a genuine determinant of the resting potential.

  4. A non-selective cation channel passes both Na⁺ and K⁺. In a cell with \([\mathrm K^{+}]_{\text{i}}=140\), \([\mathrm K^{+}]_{\text{o}}=5\), \([\mathrm{Na}^{+}]_{\text{i}}=15\), \([\mathrm{Na}^{+}]_{\text{o}}=145\) (mM) at 20 °C, the channel's current reverses at \(E_{\text{rev}}=-5\ \mathrm{mV}\). Find \(P_{\mathrm{Na}}/P_{\mathrm K}\).
    Solution

    At the reversal potential the summed GHK currents vanish, so the Result applies to this channel's own permeabilities. Put \(\alpha=P_{\mathrm{Na}}/P_{\mathrm K}\) and divide through by \(P_{\mathrm K}\): \[ e^{FE_{\text{rev}}/RT}=\frac{5+145\alpha}{140+15\alpha}. \] With \(RT/F=25.26\ \mathrm{mV}\) at \(293.15\ \mathrm{K}\), the left side is \(e^{-5/25.26}=e^{-0.19794}=0.8204\). Hence \[ 5+145\alpha=0.8204(140+15\alpha)=114.86+12.31\alpha, \] so \((145-12.31)\alpha=109.86\) and \[ \alpha=\frac{109.86}{132.69}=0.83. \] The channel is very nearly equally permeable to the two cations, which is exactly why its reversal potential sits near \(0\ \mathrm{mV}\) rather than near either \(E_{\mathrm K}=-84\ \mathrm{mV}\) or \(E_{\mathrm{Na}}=+57\ \mathrm{mV}\): a channel with no selectivity between the dominant cations clamps the membrane toward the middle, which is what makes such channels effective excitatory transducers. Note that the measurement returns only the ratio — no absolute permeability can be extracted from a reversal potential alone.

  5. Add calcium. For a membrane permeable only to K⁺ (\(z=+1\)) and Ca²⁺ (\(z=+2\)), show that the zero-current condition is a quadratic in \(X=e^{-FV_m/RT}\), and solve it at 37 °C for \([\mathrm K^{+}]_{\text{o}}=5\), \([\mathrm K^{+}]_{\text{i}}=140\ \mathrm{mM}\), \([\mathrm{Ca}^{2+}]_{\text{o}}=2\ \mathrm{mM}\), \([\mathrm{Ca}^{2+}]_{\text{i}}=100\ \mathrm{nM}\) and \(P_{\mathrm{Ca}}/P_{\mathrm K}=0.01\).
    Solution

    From Step 6 of the Proof, with \(u=FV_m/RT\) so that the K⁺ term carries \(e^{-u}=X\) and the Ca²⁺ term (\(z=2\), \(z^2=4\)) carries \(e^{-2u}=X^2\): \[ \frac{F^2V_m}{RT}\left[P_{\mathrm K}\frac{[\mathrm K^{+}]_{\text{i}}-[\mathrm K^{+}]_{\text{o}}X}{1-X} +4P_{\mathrm{Ca}}\frac{[\mathrm{Ca}^{2+}]_{\text{i}}-[\mathrm{Ca}^{2+}]_{\text{o}}X^{2}}{1-X^{2}}\right]=0. \] Multiply by \(1-X^{2}=(1-X)(1+X)\), which is legitimate for \(V_m\neq0\): \[ P_{\mathrm K}\big([\mathrm K^{+}]_{\text{i}}-[\mathrm K^{+}]_{\text{o}}X\big)(1+X)+4P_{\mathrm{Ca}}\big([\mathrm{Ca}^{2+}]_{\text{i}}-[\mathrm{Ca}^{2+}]_{\text{o}}X^{2}\big)=0, \] and collecting powers of \(X\), \[ \big(P_{\mathrm K}[\mathrm K^{+}]_{\text{o}}+4P_{\mathrm{Ca}}[\mathrm{Ca}^{2+}]_{\text{o}}\big)X^{2} -P_{\mathrm K}\big([\mathrm K^{+}]_{\text{i}}-[\mathrm K^{+}]_{\text{o}}\big)X -\big(P_{\mathrm K}[\mathrm K^{+}]_{\text{i}}+4P_{\mathrm{Ca}}[\mathrm{Ca}^{2+}]_{\text{i}}\big)=0. \] The divalent ion has produced an \(X^{2}\) term, so no logarithm of a ratio of linear sums can solve it: the GHK voltage equation genuinely does not extend to divalents. Numerically, with \(P_{\mathrm K}=1\), \(P_{\mathrm{Ca}}=0.01\), and \([\mathrm{Ca}^{2+}]_{\text{i}}=100\ \mathrm{nM}=1.0\times10^{-4}\ \mathrm{mM}\): \[ a=5+4(0.01)(2)=5.08,\quad b=-(140-5)=-135,\quad c=-\big(140+4(0.01)(10^{-4})\big)=-140.0. \] Then \(X=\dfrac{135\pm\sqrt{135^{2}+4(5.08)(140.0)}}{2(5.08)}=\dfrac{135\pm\sqrt{18225+2844.8}}{10.16}=\dfrac{135\pm145.15}{10.16}\). The negative root is rejected because \(X=e^{-u}>0\), leaving \(X=280.15/10.16=27.57\). Hence \(-u=\ln 27.57=3.3169\) and \[ V_m=-(26.73\ \mathrm{mV})(3.3169)=-88.6\ \mathrm{mV}. \] Compare \(E_{\mathrm K}=(26.73)\ln(5/140)=-89.05\ \mathrm{mV}\): a calcium permeability of 1% of the potassium permeability shifts the resting potential by only \(0.4\ \mathrm{mV}\), because the extracellular calcium concentration \(2\ \mathrm{mM}\) is small next to \([\mathrm K^{+}]_{\text{i}}=140\ \mathrm{mM}\) even after the factor of \(4\) from \(z^{2}\). Calcium is electrically negligible at rest and physiologically decisive as a signal — the enormous \(10^{4}\)-fold gradient matters for the intracellular concentration change it produces, not for the voltage.