Past exam of the biological physics course of the University of Oxford / 2021 / 1

At the initial negative membrane potential, most voltage-gated potassium channels have a closed activation gate. Depolarization moves charged residues in each voltage-sensing domain, particularly the S4 helices, through the electric field. Coupling between the voltage-sensing domains and the pore opens the intracellular gate. A classical voltage-gated potassium channel has four subunits, each with six transmembrane helices; S5, the pore loop and S6 contribute to the central pathway and selectivity filter. The rising open probability produces increasing outward potassium electric current, since the imposed positive membrane potential is above the potassium Nernst equilibrium potential. Here outward electric current is positive: $I_K=N\gamma p_O(V-E_K)$, where $N$ is channel number and $\gamma$ is single-channel electrical conductance.

The electric current subsequently falls because the rapidly inactivating Shaker form exhibits ball-and-chain inactivation. An intracellular N-terminal domain, tethered to the channel, blocks the open pore. The ion channel can therefore stop conducting while the membrane potential remains depolarized and the activation machinery remains activated. This is ion-channel inactivation, rather than reversal of the voltage step, loss of the ionic driving force, or ordinary gate closure. Repolarization favors removal of the block and recovery of channel availability. Slower C-type inactivation involving the outer pore can coexist, but is not the principal explanation of the rapid decay described here.

Several independent observations connect this functional explanation to structure. Small transient gating electric currents report movement of sensor electric charge before ionic conduction; charge-neutralization studies of S4/S2 change the measured gating charge and activation. Structural studies resolve separate sensor and pore regions.^[ref][ref] Historical N-terminal removal abolished rapid ion-channel inactivation while preserving activation, and the corresponding free peptide restored block from the cytoplasmic side. These results separate the activation gate from the tethered blocker and identify which side of the cell membrane contains the blocking domain.^[ref][ref]

Source-gap TODO: the supplied PDF's figure invokes an undefined image object, so its actual pulse duration, pulse sequence, current scale, peak amplitude, decay time and any tail-current waveform cannot be read. The account above uses only the stated change from $-60$ to $+20\mathrm{mV}$ and the stated rise and rapid decay; no numerical kinetic fit or reconstruction of the missing trace is claimed.

For two-state ion-channel gating, write $C{\rightleftarrows }_{k_c}^{k_o}O$. Gain minus loss of open probability gives the ordinary differential equationThus the ratio of the transition rates determines the final open probability, while their sum determines the response speed. Both given channels have $p_{\infty }=1/2$. For an initially closed population, with $t$ measured in milliseconds,The mean individual open and closed lifetimes are $1/k_c$ and $1/k_o$: both are $2\mathrm{ms}$ for A and $0.2\mathrm{ms}$ for B. With the same channel number, single-channel electrical conductance and driving force, the steady macroscopic electric currents are equal, but B reaches that value ten times faster. If both populations were already at $p_O=1/2$, neither would show a mean relaxation under these fixed rates: a response also requires an initial condition or a change in rates.

The stationary variance of the number open is $N/4$ for independent channels. Its autocovariance is $(N/4)e^{-|t|/\tau }$, so B's fluctuations decorrelate ten times faster even though their instantaneous variance is unchanged. Faster gating allows an ion channel population to follow a brief stimulus and contribute to rapid action potential timing. Slower gating integrates over brief changes and can provide a delayed opposing electric current, useful in setting action potential duration or frequency selectivity. Those examples describe consequences of kinetics; the stipulated voltage-independent rates alone do not define a complete excitable neuron.

Add a nonconducting inactivated state to the continuous-time Markov chain:For example, ${\dot{p}}_O=k_o{p}_C-(k_c+k_i)p_O+k_r{p}_I$ and ${\dot{p}}_I=k_i{p}_O-k_r{p}_I$. Depolarization first increases $p_O$, then transfers population into $I$ by ball-and-chain inactivation. A small recovery rate during the pulse permits a sustained decline of $I_K=N\gamma p_O(V-E_K)$. Repolarization favors recovery, eventually restoring the closed but available population. Several sequential closed states can represent the cooperative movements of four voltage-sensing domains; three states are the minimal explanation of rise followed by decay, rather than a quantitative fit to the missing trace.

Loss of outward potassium electric current weakens repolarization and can broaden an action potential. Reduced residual potassium electrical conductance can reduce the subsequent hyperpolarization and permit faster repeated firing. Conversely, prolonged depolarization can maintain sodium ion-channel inactivation, delay recovery and lower the maximum firing frequency or cause depolarization block. The sign of the frequency change depends on the other ion channels and recovery kinetics; rapid potassium ion-channel inactivation alone does not fix it.

The passive upper limiting membrane potential for a purely sodium-selective pathway is its Nernst equilibrium potential. Equal inside and outside electrochemical potentials, using ideal concentrations, requireThis is the maximum absolute inside-minus-outside voltage under the stated passive assumptions. If the reference resting membrane potential is $-60\mathrm{mV}$, the corresponding increase is about $121\mathrm{mV}$.

Actual action potentials usually peak below $E_{\mathrm{Na}}$ because voltage-gated sodium channels undergo ion-channel inactivation, while outward potassium and leak electric currents oppose inward sodium electric current. Finite membrane capacitance and finite available time also prevent instantaneous equilibration. At $V=E_{\mathrm{Na}}$, the sodium driving force itself is zero.