1Recall, what is ion channel conductance?
The ease with which ions flow through an open channel under a given electrical driving force
2What characterizes a linear (ohmic) ion-channel conductance?
Linear (ohmic) ion channels have a conductance that is independent of membrane potential, producing a straight current-voltage (I-V) relation: I = G × (Vm − Erev). For the specific channel type being considered, I is its current, G is its conductance and Erev is its reversal potential (for a channel selective for one ion, Erev is that ion's equilibrium potential). Current changes with Vm because the driving force, Vm − Erev, changes, but the key is that G remains constant.
3What characterizes a nonlinear ion-channel conductance?
In non-linear ion channels (i.e. voltage-gated channels), the conductance depends on membrane potential (G = f(Vm)), producing a nonlinear I-V relation. In voltage-gated channels, membrane potential changes the open probability (the fraction of time a channel spends open), which then changes the population's mean conductance and enables nonlinear computations and action-potential generation.
4How are voltage-gated channels organized and what functions do their structural regions perform?
Four homologous subunits (as in K+ channels where they are different proteins) or domains (as in Na+ channels where they are four domains of the same protein) surround a central aqueous, ion conducting pore. Each subunit/domain contains six transmembrane alpha helices, S1-S6. The S1-S3 domains are form the lipophilic part of the channel, they lie towards the outside of the protein and interact with the phospholipid bilayer so they contain mostly nonpolar amino acids. The S5-S6 regions are the pore-lining regions: they form the pore and contains the selectivity filter (the filter is formed by the pore loop between S5 and S6 --> see the yellow non-transmembrane loop in the image). S4 is the relatively mobile voltage-sensing region and contains many arginine residues, which are positively charged at physiological pH.
5Describe the unique properties of the S4 domain of a voltage-gated ion channel
The S4 region forms a unit that is relatively mobile within the ion channel and contains many arginine amino acids, which are positively charged at physiological pH. These properties allow S4 to move in response to the electric field across the plasma membrane (the direction of the electric field influences the direction of movement). S4 is mechanically coupled to a voltage gate, so its movement then influences gate opening and closing.
6How does a change in membrane potential control the opening of a voltage-gated ion channel?
The electric field across the membrane acts on S4's positive charges. At hyperpolarized potentials, the field points from outside to inside because the cell interior is more negative, which pushes the positive charges inward, favouring inward movement of S4. S4 is mechanically coupled to a voltage gate, so this movement favours gate closure and low open probability. Depolarization weakens the inward electric field, which reverses direction if Vm becomes positive. This favours outward movement of S4, changing the protein's conformation and favouring gate opening and higher open probability
7Is a voltage-gated channel always open when the membrane potential is positive?
No, the voltage gate itself opens and closes stochastically (randomly). Movement of S4 changes which state is favoured: depolarization favours opening, while hyperpolarization favours closure. Voltage therefore controls the open probability (fraction of time the channel spends open) without determining the exact timing of each opening and closing. A high open probability does not mean the channel remains continuously open.
8What happens if membrane potential is rapidly changed to a positive value while ion flux through the channel pore of a voltage-gated channel is prevented?
Even in the absence of ion flux, a very brief, fast capacitative current (the gating current) can be measured. This indicates movement of the positive charges on the S4 voltage-sensing domain as it moves outwards, with fast components on a microsecond timescale. These charges move with the protein, even though no ions pass through the pore. The current is brief because the initial net charge movement ends, even if the positive voltage is maintained.
9How does open probability change with membrane potential in a linear vs voltage-gated ion channel?
In a linear channel, the open probability is the same regardless of membrane potential, because we have a constant conductance (linear I-V relationship). Voltage-gated ion channels have a low open probability at negative potentials (voltage gate is closed due to the inward electric field pushing S4 region inwards) and a high open probability at positive potentials. Thus, the open probability of a voltage-gated channel is a function of the membrane potential: P(Vm)
10What is the equation for the mean conductance of a population of identical ion channels, and how is it derived?
G = g × N × Popen, where G is mean conductance, g is the conductance of one open channel, N is the number of channels and Popen is their open probability. The mean reflects averaging over stochastic openings and closings across time or across many independently gating channels. For a single voltage-gated channel changes in membrane potential changes Popen rather than its individual conductance g.
11How do open probability and single-channel current compare at +60 and −60 mV for an ohmic channel versus a voltage-gated channel, assuming Erev = 0 mV?
An ohmic channel has a constant Popen, so if Popen is 0.5, the channel spends half its time open at both +60 and −60 mV, carrying equal-magnitude outward and inward currents, respectively. A depolarization-activated spends less time open at negative voltages because its Popen is lower, so the instantaneous magnitude of current is equal at + 60 mV and -60 mV but its mean inward current at -60 mV is smaller in magnitude than its mean outward current at +60 mV.
** For both channel types, the current while open is i = g × (Vm − Erev) because ONCE A CHANNEL IS OPEN IT FOLLOWS OHMIC LAWS, the thing that changes between the two channel types is the PROBABILITY of opening.
12How to plot conductance vs voltage, G(V), for ohmic and voltage-gated channels?
Macroscopic conductance is G = NgP, where P = open probability. For ohmic channels, P is constant and therefore G is constant. So just draw a straight horizontal line: G is the same for all values of V. For voltage-gated channels, Popen (and therefore G) is a function of the membrane potential, it increases sigmoidally with membrane potential, so conductance also follows a sigmoid. It begins near zero at very negative potentials, rises steeply starting around the activation range of approximately −50 to −40 mV, and then saturates at its maximal value. This maximum need not correspond to Popen=1 the professor used approximately 0.5 as a rough illustrative value.
13How to plot current vs voltage, I(V), for ohmic and voltage-gated channels?
Current is always I = G (Vm − Erev), so both curves cross zero at Erev. For ohmic channels, G is constant, so I is a straight line with an x-intercept of Erev and with slope G. For voltage-gated channels, we convolve the sigmoid G(V) (or equivalently the open probability) with the single-channel I-V relationship of the ion channel, which depends on the specific ion and its reversal potential. A voltage-gated channel's I(V) curve passes through zero whenever the membrane potential is at the reversal potential for the its specific ion, as well as wherever G (or P)=0 (whenever the channels are closed, e.g. at low potentials up to about -40 or -50 mV). Elsewhere, the curve takes the sign of (Vm-Erev) and the magnitude of G. E.g. when the membrane potential is slightly negative (like -40 mV which is below ENa but above Ek), there is an inward current of Na+ but an outward current of K+ -> so the I(V) curve at those potentials is negative for Na+ but positive for K+. Once P is maximal (at around Vm = 0), the curve becomes near-linear since P no longer changes much with Vm.
14Why is the voltage-gated Na+ I-V relation biphasic but the K+ one not?
Ena≈ +60 mV, Ek ≈ −90 mV, and both channels activate above about −40 or -50 mV (as channels start to open and Popen incerases). For Na+, this activation happens while Vm is still below Ena, so an inward current is activated from −50 mV, reaches a trough, crosses zero at Ena, and becomes outward above it: biphasic. For K+, this activation happens at potentials that are already above Ek, so the current is outward as soon as it activates and simply increases into the linear regime: no biphasic phase.
** Once P is maximal, the I(V) curve is near-linear
15What is the state of most voltage-gated ion channels (specifically K+ and Na+ channels) at resting membrane potential?
Most are closed (in the range of around -70 to -50 mV) -> very low open probability, so they will make very little contribution to the overall conductance of the membrane. At more positive potentials, the open probability increases sharply and the voltage-gated ion channels then begin to conduct and can contribute to the overall conductance of the cell membrane.
16What specifically causes a change in conductance of voltage-gated ion channels?
A change in the overall open probability, NOT a change in the single channel conductance. Voltage-gated ion channels open and close stochastically. Whenever a channel is actually open, its single-channel conductance, g, is constant and it behaves like an ohmic channel: the current through the open channel is i = g × (Vm − Erev), giving a linear I-V relationship. What changes with membrane potential is the probability that the channel is open. Depolarization increases Popen for depolarization-activated channels, so more channels are open on average. This increases the mean conductance of the channel population according to G = N × g × Popen. Therefore, voltage changes the overall conductance by changing Popen, not by changing the conductance of each open channel.
17What is the input resistance of a cell, and what is its typical value?
The total resistance of its cell membrane, measured as the ratio of the voltage change produced to the injected current. Roughly 50 MΩ --> corresponding to an input conductance of around 20 nS
18What is input conductance?
The total conductance of the cell, measured from the electrode as G = I/ΔV. It is the sum of every conductance that is open at that moment, whatever its type. Given by G_input = G_leak + G_gated
19What is leak conductance?
The contribution to cell conductance from non-gated, continuously open channels, dominated by K+ leak conductance that keeps membrane potential around -70 mV. At resting membrane potential, since gated channels are closed, the leak conductance is equivalent to the input conductance.
20What are the typical single-channel conductance of a voltage-gated Na+ channel and input conductance of a typical neuron?
A single voltage-gated Na+ channel is roughly 20 pS. The input conductance of a cell is around 20 nS, i.e. a factor of 1,000 larger.
21What determines the membrane potential when several conductances are present? What needs to happen for voltage-gated sodium channels have a considerable impact upon the overall membrane potential?
Recall that Vm is a weighted average of the reversal potentials, weighted by each conductance's share of the total, so something like Vₘ = (GₖEₖ + GₙₐEₙₐ + G꜀ₗE꜀ₗ) / Gtotal. Whichever conductance dominates Gtotal pulls Vm towards its own reversal potential. For Na+ to make a considerable impact, the overall sodium conductance must get close to the total conductance of the cell of around 20 nS, which is 1000 times greater than the single channel conductance of a voltage-gated Na+ channel of 20 pS. So we would need something like 1000 sodium channels to open to make a substantial impact upon the membrane potential.
22How many voltage-gated Na+ channels must open to significantly depolarize the cell, and why?
For G_Na to make a considerable impact on Vm it must be of the same order of magnitude as the leak conductance, i.e. approach Gtotal. Since 20 nS / 20 pS = 1,000, around 1,000 Na+ channels must open. This is why most neurons have high densities of voltage-gated Na+ and K+ channels: it lets them change Vm rapidly in a voltage-dependent manner.
23What happens if a cell with voltage-gated Na+ channels is depolarized past −50 mV?
At rest (−70 mV) and up to just negative of −50 mV, the channels are largely closed (P is very low) so almost no current flows. Depolarizing a little past −50 mV causes the open probability to increase, and because Vm is below E_Na, this results in an inward current of Na+. That inward current brings more positive charge inside the cell and depolarizes the membrane potential further, which increases the conductance further, giving a larger inward current, which depolarizes even more, and so on. Thus this creates an explosive positive feedback loop (see image) in which a small depolarization sets off a positive feedback loop that takes Vm rapidly towards E_Na (though Vm does not necessarily reach E_Na exactly, it depends on the strength of the leak conductances and what other conductances are open).
24What happens if a cell with voltage-gated K+ channels is depolarized past −50 mV?
Below -50 mV the conductance is basically zero (like in Na+ channels), so no current flows. Past around −50 mV, the open probability increases, leading to an outward K+ current because -50 mV is ABOVE E_K of -90 mV. This causes the membrane potential to become more negative. Thus, for a given Vm, perturbing it with a small depolarization will activate the voltage-gated K+ channels that will then hyperpolarize the membrane potential and bring it back down towards E_K, so this is a stabilizing influence (though again Vm does not necessarily reach E_K exactly, it depends on what other ocnductances are open).
25T/F: voltage-gated potassium channels always cause hyperpolarization under physiological condtions
True: Opening voltage-gated potassium channels will always hyperpolarise a typical neuron under normal physiological conditions. The opening of potassium channels will drive the membrane potential towards the potassium reversal potential (but not necessarily reaching it -> this depends upon what other conductances are open). The potassium reversal potential is the most negative of all the physiological ions (more negative then resting membrane potential) and therefore increasing this conductance always causes hyperpolarisation.
26How quickly does the open probability of a voltage-gated channel change when the membrane potential changes?
Changes in open probability of a voltage-gated channel depend on changes to the 3D conformation of the channel protein itself, and this does not occur instantaneously. Whereas the transitions of individual ion channels between open and closed states are stochastic and happen on a microsecond timescale, the voltage-dependent change in open probability occurs on slightly longer timescales.
27In a patch-clamp recording, what happens to a single voltage-gated Na+ channel when the membrane potential is suddenly stepped from −70 mV to 0 mV? (including timings)
At −70 mV the open probability is very low and the channel is closed. When Vm jumps to 0 mV, initially nothing happens. Then after a short delay of about 10 μs, the channel opens, stays open for a brief period of about ~200 μs, closes again, and remains closed for the rest of the depolarization to 0 mV. The precise opening and closing times are stochastic, and the individual channel can open and close multiple times within the "open" period, so they vary from trial to trial, but the overall pattern is highly reproducible. If you average over many trials, you get the macroscopic response shown in the bottom of the image.
28What does the macroscopic Na+ current look like during a prolonged step to 0 mV, and why?
Averaging hundreds of single-channel trials, or summating across a whole cell membrane (with multiple channels) in one voltage step, shows that after a brief delay (10 μs), an inward current is activated, the Na+ conductance remains high for around 200 μs, then the current falls as the channels enter the inactivated state, even though the depolarization is maintained.
The channel inactivates rapidly, so a prolonged step to 0 mV produces only a fast, brief, transient Na+ conductance rather than a sustained one.
29In a patch-clamp recording, what happens to a single voltage-gated K+ channel when the membrane potential is suddenly stepped from −70 mV to 0 mV? (including timings)
At −70 mV the open probability is very low and the channel is closed. When Vm jumps to 0 mV, initially nothing happens. Then after a longer delay of about 200 μs, the channel opens (with brief stochastic closings and reopening) and its open probability remains high for as long as the depolarization lasts.
30What does the macroscopic K+ current look like during a prolonged step to 0 mV, and why?
Averaging hundreds of single-channel trials, or summating across a whole cell membrane (with multiple channels) in one voltage step shows that after a longer delay (200 μs), then an outward current is activated that is sustained for as long as the depolarization lasts.
31How do the gating kinetics of voltage-gated Na+ and K+ channels differ in response to the same step to 0 mV?
Both channel types are activated at similar membrane potentials that are higher than about -50 to -40 mV, but the time courses of their activation are very different. Na+ conductance activates fast (~10 μs delay) but transient (~200 μs), whereas K+ activates slowly (~200 μs delay) but is sustained. The same voltage step gives a fast, brief, transient inward Na+ conductance (leading to an inward current) and a delayed, long-lasting outward K+ conductance (leading to an outward current)
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see image: an inward current is negative while an outward current is positive; make sure you know how to draw the current traces
32How does the state of the voltage-gated Na+ channel differ before the depolarization to 0 mV and after the transient current during that same depolarization?
Before the step, at −70 mV, the channel is closed: the open probability is very low, but depolarization can open it. After the transient current, during the maintained depolarization, it is inactivated: it remains closed for the duration of the step even though the membrane is still depolarized, because the channel inactivates rapidly on the timescale of hundreds of microseconds. To conduct again it must de-inactivate, which requires hyperpolarizing the membrane.
33What happens to an inactivated voltage-gated Na+ channel, and how do you get it to conduct again?
The Na+ conductance fully inactivates after just a few hundred microseconds of depolarization, meaning the depolarization can no longer open the channel, but it doesn't remain inactivated forever. Hyperpolarizing the membrane again, say to −70 mV, causes the channel to de-inactivate, so it is ready to be opened by a subsequent depolarization and will again give a transient Na+ conductance. De-inactivation is not instantaneous and takes some time, and the more hyperpolarized the membrane, the faster it occurs.
34How do you measure a voltage-gated Na+ channel's recovery from inactivation, and what is the time course?
Use a two-pulse voltage-clamp protocol (each pulse is a voltage step commanded by the patch clamp amplifier while we record the current). Step from approximately −70 mV to 0 mV to activate the transient Na+ current and measure the peak of the current. Then return to −70 mV for a variable interval Δt, and then apply the same second step to 0 mV and measure its peak current. At short intervals, such as 0.5 ms, the second current is small because few channels have recovered. At longer intervals, such as 0.8 ms and beyond, progressively more current returns.Then you can plot I2/I1 (the ratio of the second current peak to the first current peak) against Δt. The ratio rises exponentially towards 1 as the channels recover from inactivation, with a recovery time constant of approximately 1-2 ms, the time when the current returns to 63% of its maximum fully recovered value
35How are voltage-gated Na+ channels encoded, and what is their subunit structure?
Voltage-gated Na+ channels are formed by a pore-forming alpha subunit and a much smaller, tightly associated accessory beta subunit. The entire alpha subunit is encoded by a single gene: each gene encodes one long string of amino acids with four homologous domains that are all part of the same string (see image). These four domains each contain six membrane-spanning regions (alpha helices) and fold together in the plasma membrane to form the conducting channel pore. There are 9 genes encoding the alpha subunit, Nav1.1 to Nav1.9, giving 9 different isoforms, plus 4 types of beta subunit, which together provide some diversity.
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In previous videos the prof refers to the four alpha domains as "subunits" themselves. The key point is that four homologous parts come together around the pore, but standard terminology calls them domains, not subunits, because they are four regions of a single protein rather than separate proteins.
36What functional diversity arises among voltage-gated Na+ channels?
The 9 types of alpha subunits and 4 types of beta subunits can combine in different combinations to give rise to channels with slightly different single-channel conductance, opening kinetics, and voltage-dependence. They also differ in the cell types in which they are expressed and in their subcellular localization, with some small subregions of neurons having very high channel densities. Some subtypes show much less inactivation than discussed so far and can form a more persistent Na+ current.
37How diverse are K+ channels, and which ones are not voltage-gated?
There are approximately 80 genes in the mammalian genome encoding K+ channels, and not all are voltage-gated. Examples of non-voltage-gated ones: BK and SK channels which are gated primarily by intracellular Ca2+ concentration; GIRK channels (GPCRs) which are controlled by G-proteins; and tandem pore K+ channels which are involved in setting the resting membrane potential.
38How does the subunit organization of voltage-gated K+ channels differ from that of Na+ channels?
Unlike the Na+ channel, where one gene encodes the whole channel, each K+ channel gene encodes only a single subunit, which each subunit having six transmembrane regions. Four such subunits assemble to form a channel, and different gene products can mix together, generating even more diversity than Na+ channels.
39What else varies among voltage-gated K+ channels (besides being voltage-gated vs non voltage gated), and why does it matter?
They vary in the genes encoding their subunits, their cellular and subcellular localization, the voltage-dependence of activation, and the kinetics. Some show a degree of inactivation, but typically on a much longer timescale than voltage-gated Na+ channels. This diversity causes a great deal of the diversity in how membrane potential changes in different cell types and in different compartments of the same cell.
40Depolarisation induces a time-dependent activation of voltage-gated Na+ and K+ conductances. Which statement is correct, and why are the others wrong?
1. The single-channel conductance depends strongly on the time-course of Vm.
2. The open probability depends only upon voltage.
3. Voltage-gated channels open and close under precise temporal control of Vm.
4. Voltage-gated channels open and close stochastically on the microsecond timescale, with a slower underlying change in open probability governed by Vm.
4 is correct.
1 is wrong: g is a fixed property of the open pore, so all voltage and time dependence sits in P.
2 is wrong because of "only": P is also time-dependent, which is why at a constant 0 mV the Na+ conductance rises then inactivates and the K+ conductance rises only after a delay.
3 is wrong because of "precise": voltage sets the probability per unit time of opening, not when any given channel opens. Identical steps give different traces each time due to stochastic openings and closings, and only the average over many channels or trials is reproducible.
41What is the duration of an action potential?
Approximately 1 ms
42How were Hodgkin and Huxley able to record the membrane potential during an action potential?
They used the giant axon of the squid, which is more than a thousand times larger in diameter than most axons: a typical mammalian axon is around 100 nm, while the squid giant axon is close to 1 mm. This made it possible to insert an electrode inside the axon (Vinside), with another electrode outside in the sea water bath (Voutside), and record Vm through a differential amplifier.
43What shape did Hodgkin and Huxley's intracellular recording of Vm over time show during an action potential?
A stereotypical waveform lasting around 1 ms, with a very steep rising positive phase and a very steep falling phase. Importantly, the AP overshot 0 mV and reached around +40 mV before repolarising. This was the first time someone had recorded the intracellular membrane potential correlate of the spike event previously seen only from outside cells as small extracellular blips.
44How did Hodgkin and Huxley go from recording Vm to measuring the currents underlying the action potential?
They used voltage clamping. So, they already had their measured Vm from a differential amplifier comparing Vinside and Voutside. Then they fed that Vm into a second differential amplifier that compares it with a command voltage, which H&H specify. This second amplifier injects whatever current is needed into the axon to abolish the difference, allowing Vm to be held at the command voltage. They then measured the injected current which is a readout of the membrane current. This let them impose voltage-clamp steps across time and measure the resulting membrane currents."
45In a voltage-clamp experiment, why is the injected current a readout of the membrane current?
Since Vm is held constant, no charge accumulates on the membrane capacitance (because remember increasing membrane potential charges the capacitor while decreasing it removes charge). so in this case every charge crossing the membrane through ion channels has to be supplied by the amplifier. The injected current is therefore equal and opposite to the membrane current.
46Holding the squid axon at −75 mV and using a voltage-clamp technique with command voltages (steps) of −50, −25, 0, +40 and +75 mV, what membrane currents did H&H record? What did they conclude from this?
At −50 mV, very little current was activated. At −25 mV and 0 mV, a rapid inward current appeared and then rapidly disappeared, giving way to a delayed outward current. At +40 mV the inward current was almost gone and the delayed outward current was even bigger. At +75 mV both the early and delayed components were outward. H&H concluded that the waveform could be accounted for by two different conductances: a fast, transient conductance and a delayed, long-lasting one.
47How did Hodgkin and Huxley turn their voltage-clamp current traces into the I-V relations of the Na+ and K+ conductances?
For each command voltage they measured the current twice: at an early time point (peak of the fast component) and at a late time point (plateau of the delayed component). Plotting each recorded current value against the command voltage gives two curves. The early one activates positive to −50 mV, is inward, and reverses around +40 mV: the Na+ conductance. The late one activates positive to −50 mV and rises roughly linearly, outward throughout: the K+ conductance.
48What kind of model did Hodgkin and Huxley build to describe the currents they had measured?
A simple quantitative phenomenological model. Their voltage-clamp data showed currents that depended on both voltage and time (green currents in image show different traces for different voltages and also that the current flow changes over the timecourse), so they needed equations capturing both: differential equations describing how the Na+ and K+ conductances evolve in time, with voltage-dependent rate constants. The equations were empirical, chosen to fit the data they had measured.
49How did Hodgkin and Huxley express the voltage- and time-dependence of the conductances in their equations?
Through gating particles, which are their postulated channel elements that must be in a permissive position for the channel to conduct. Each particle sits in either a permissive or a non-permissive state, and the variable attached to it (n, m, or h) is the probability of being permissive, with the particle moving between the two states with voltage-dependent rate constants alpha and beta.
50What form did Hodgkin and Huxley's equation for the K+ current take, and what does each term represent?
I_K = n⁴ · g_Kₘₐₓ · (Vm − Ek), because current is always conductance times driving force (which for K+ is Vm−Ek). g_Kₘₐₓ is the maximal K+ conductance (presumably relating to the number of K+ channels present in the membrane). n is the probability that one gating particle is permissive, and since the channel conducts only when all four of its gating particles (subunits) are permissive, the open probability is n⁴. n⁴ is the voltage- and time-dependent term in the equation, since the particles move between states with voltage-dependent rate constants.
51How did Hodgkin and Huxley model the open probability of the K+ conductance?
using their particle-gating scheme (card 49): For K+ the gating particle has probability n of being in the activated (permissive) state and 1−n of being in the impermeable state, with rate constant alpha for the transition into the permissive state and beta for the transition out of it. This gives a first order differential equation for n across time, dn/dt = (1−n)·α_n − n·β_n, which describes how open probability changes over time. Since this is a voltage-dependent channel, alpha and beta are themselves voltage dependent, described by empirical equations that relate to expectations from Boltzmann statistics (see image).
52Why is n raised to the fourth power in the K+ conductance equation? What is the interesting part of this?
Empirically, H&H found that the fourth power was needed for the equations to fit the experimental data. After H&H's equations were derived, it was discovered that each K+ channel is in fact composed of four subunits, each with its own positively charged S4 voltage-sensing domain that moves depending on the membrane potential and causes the gate to open. So n⁴ can be read as the probability that all four subunits are in the activated state, with alpha and beta relating to the movement of these four independent gating particles
53How does the Hodgkin and Huxley equation for the Na+ current differ from the K+ one, and why?
Na+ conductances are more complicated because they have both voltage-gated activation and voltage-gated inactivation. The equation they derived was I_Na = m³ · h · g_Namax · (V − E_Na), where m is the activation particle and h the inactivation particle, each governed by its own voltage-dependent alpha and beta rate constants (so alpha_m, beta_m, alpha_h, beta_h), each following a first order differential equation with Boltzmann-type voltage dependence. Fitting the data required the third power of m multiplied by h, with h not raised to any power.
54How did Hodgkin and Huxley model the open probability of the Na+ conductance?
With two particle-gating schemes instead of one, because the Na+ channel has both voltage-gated activation and voltage-gated inactivation. An activation particle has probability m of being in the permissive state and 1−m of being in the non-permissive state, and an inactivation particle has probability h of being permissive and 1−h of being non-permissive, each with its own rate constant alpha for the transition into the permissive state and beta for the transition out of it. Each gives a first order differential equation across time, dm/dt = (1−m)·α_m − m·β_m and dh/dt = (1−h)·α_h − h·β_h, and all four rate constants are voltage dependent, again with Boltzmann-type empirical equations (see image). To fit the experimental data the conductance depended on m³·h, with h not raised to any power.
55Why is m raised to the third power in the Na+ conductance equation while h is not raised to any power?
The exponent encodes a delay. After a voltage step, activation does not rise immediately: the current stays near zero briefly, then bends upward, giving an S-shaped time course. Cubing m produces this, since early on all three factors are small and their product is smaller still. Inactivation decays as a single exponential with no such delay, which one h already describes.
56How did Hodgkin and Huxley go from these equations to a reconstructed action potential?
They fitted their voltage-clamp data to obtain values for the constants in the differential equations, then placed the resulting conductances into the circuit model of the cell: a capacitor in parallel with Na+, K+ and leak conductances, each in series with its reversal potential, between Vm inside and 0 mV outside. Current conservation in that circuit gives C·dV/dt = −(I_Na + I_K + I_leak), so at each small time step they updated m, h and n from Vm, computed the currents, and advanced Vm accordingly, iteratively tracing out the AP trajectory. The reconstruction agreed well with their measured waveform.
57How did H&H use their model to determine the ionic basis of the action potential?
Their reconstructed AP model computes I_Na and I_K separately at every time step, each from its own conductance and driving force, so at any instant you can see which one accounts for the total membrane current. During the upstroke g_Na is high and I_Na is large and inward while I_K is still near zero, so they concluded that the depolarization is caused by the Na+ current. During the downstroke h has fallen so I_Na is gone, while n has risen so g_K and I_K are large and outward, so they concluded that the repolarization is caused by the K+ current.
58What is the action potential threshold?
The value of the membrane potential, typically between -40 and -50 mV, at the earliest time point at which the action potential is initiated, defined experimentally as the point of maximal curvature in the Vm trajectory (see image: threshold is the peak of the second derivative plot). It is the moment where an autocatalytic step is reached and the voltage-gated Na+ conductance goes into its positive feedback loop, where an increase in Vm increases the Na+ conductance, driving the explosive rise of the AP. Importantly, the Na+ current must be larger than other hyperpolarizing currents such as the K+ conductance to be able to drive the membrane potential towards ENa.
59Injecting current into a cell, what is the difference between a subthreshold depolarization and one that triggers an AP?
Below threshold, injecting current charges the capacitance of the cell and depolarizes Vm, with the size of the depolarization depending on the leak conductance. When the current is turned off, the capacitance discharges and Vm returns to rest. Injecting more current gives a larger depolarization, but if it stays just below threshold nothing further happens. A tiny bit more current reaches threshold, where the Na+ current exceeds the opposing hyperpolarizing currents and the positive feedback loop takes over, so Vm is now driven by the Na+ conductance rather than by the injected current, driving it towards ENa. So it is an all-or-nothing response: anything below threshold doesn't trigger the feedback loop, reaching it does.
60What complete sequence of events underlies an action potential?
From resting membrane potential up to approximately −50 mV, the incoming current mainly charges the membrane capacitance and gradually depolarizes the cell.
Once Vm reaches the threshold of approximately −50 to −40 mV, the open probabilities of voltage-gated Na+ and K+ channels begin to increase. The Na+ channels activate much faster, so inward Na+ current depolarizes the cell further. This activates still more Na+ channels, producing an explosive positive-feedback loop and the rising phase of the AP.
As Na+ conductance becomes dominant, Vm is driven towards ENa, which is approximately +60 mV. However, the AP usually peaks at only approximately +30 to +40 mV because other conductances remain present and the Na+ channels begin to inactivate.
Meanwhile, the more slowly activating K+ channels have opened. As the Na+ channels inactivate and K+ conductance becomes dominant, outward K+ current drives Vm back towards EK, producing the falling or repolarization phase.
The K+ channels close relatively slowly, so K+ conductance remains elevated briefly and produces an afterhyperpolarization. As the K+ channels close, Vm returns to resting membrane potential. The entire AP usually lasts approximately 1 ms, although this varies between cell types.
61What is the absolute refractory period and why does it occur?
Just after an AP, the Na+ channels have inactivated on the rising phase and remain inactivated for a few ms as we saw before. During this period it is impossible to fire another AP, because the voltage-gated Na+ channels are not just closed but inactivated and cannot be reopened by depolarization.
62What happens to the AP threshold potential within the few ms after an action potential is fired?
The voltage-gated Na+ channels recover from inactivation with a time constant of a few ms, so after a few ms most but not quite all of the Na+ current is available again. With a lower total available Na+ conductance, a stronger depolarization is therefore required for the inward Na+ current to overcome the opposing K+ and leak currents. Thus, the threshold is slightly increased: an AP can still be triggered but requires a stronger depolarization. (This phase is standardly called the relative refractory period, though he doesn't use the term.) As more Na⁺ channels recover, threshold returns towards its usual value.
63How does Ap threshold depend on the trajectory of the membrane potential towards threshold?
If we introduce a very steep trajectory, the AP threshold is lower than with a very flat membrane potential trajectory. As we slowly approach the threshold, the voltage-gated Na+ channels have time to inactivate, which means there is less of the Na+ channel available to drive the explosive rise of AP. Overall, the AP threshold is not absolute but varies by a small amount, but is typically between −50 and −40 mV.
64Two cells both depolarize to threshold, one quickly and one slowly. How does this affect their AP threshold potentials?
The fast one (with a steeper Vm trajectory towards threshold) has a lower threshold, the slow one has a threshold a couple of mV higher (requiring stronger depolarization). This is because approaching slowly gives the Na+ channels more time to inactivate, so fewer are available and Vm must climb further before the Na+ current exceeds the opposing hyperpolarizing currents.
65How do the firing patterns of excitatory and inhibitory neurons in mouse neocortex differ?
Injecting current into a typical inhibitory cell produces APs at very high frequency, up to several hundred Hz (red in the image). The excitatory cell fires at a much lower rate (black in the image), and although increasing the current can raise the firing frequency a little, but its maximal rate is 10 times slower than that of inhibitory cells which can fire 10x faster.
66How do the individual AP waveforms of excitatory and inhibitory neurons differ, and why?
The upstroke is relatively similar and very fast in both, driven by activation of the voltage-gated Na+ conductance. The repolarization phase is where they differ: the inhibitory cell expresses a specific type of voltage-gated K+ channel that leads to very rapid repolarization, so the whole AP lasts roughly half a ms, while the excitatory cell has a different K+ channel and its AP is almost twice as long, about 1 ms. In image: excitatory waveform is black and inhibitory waveform is red.
67Why can inhibitory neurons fire at much higher rates than excitatory neurons?
Inhibitory neurons express K⁺ channels that produce faster repolarization and a shorter AP. Because Na⁺ channels de-inactivate faster the more hyperpolarized the membrane is, driving the membrane back to negative potentials sooner means Na⁺ channels become available again sooner. Their absolute and relative refractory periods are therefore shorter, allowing another AP to occur sooner. In excitatory cells, the availability of Na+ channels takes longer so longer delays are needed for the cell to fire again.
68What factors set the AP threshold of a given cell or part of a cell? i.e. what does AP threshold depend on?
The condition is that the Na+ conductance must be larger than the K+ conductance, i.e. the Na+ current must be larger than the opposing hyperpolarizing currents. What matters is therefore how much of each conductance is available at a given voltage, which depends on three properties of both the voltage-gated Na+ and K+ channels: their densities, the Vm at which they activate and the dynamics of that activation (different subtypes have their sigmoid activation curve positioned slightly differently along the voltage axis), and the Vm and dynamics of inactivation (if enough Na+ channels are inactivated, the K+ conductance out-competes them and threshold is never reached). In the mammalian nervous system, the density of voltage-gated Na+ channels is the most relevant factor.
69Why can't the action potential simply spread passively down the axon?
The thin cables of dendrites and axons are associated with considerable spatial and temporal filtering (as seen last week), with a length constant of around 0.5-1 mm and membrane time constants of several ms. Axons can travel many cm or even meters, so over these distances basically no signal would be transferred passively (i.e. the signal would already decay to 37% of its initial value after only 1mm). The action potential is an all-or-none, digital signal, and its regenerative explosive activation of voltage-gated sodium channels allows it to propagate down long axons without losing signal.
70Where is the action potential initiated in most mammalian neurons, and why there?
At the axon initial segment, located approximately 50 μm down the axon from the cell body. This area has a very high density of voltage-gated Na+ channels, many times more than anywhere else in the cell membrane. They are clustered there by molecular scaffolds through binding to a protein called Ankyrin G, which binds the cytoskeleton and anchors the channels at that location, which is what maintains their high density. Since threshold requires the Na+ conductance to outpace the K+ conductance, having more Na+ channels means the Na+ conductance builds faster, so this is the point of the neuron that is most easily excited, with the lowest threshold for AP initiation. (In image: black = dendrites and soma, red = axon.)
71If the AP is initiated at the axon initial segment, how does the rest of the neuron come to be depolarized?
The initial segment is in the middle of the cell, not at one end (about 50 um down the axon from the soma), so the AP spreads outward in both directions by sequential recruitment of voltage-gated Na+ channels, which are distributed across all neuronal membranes. Down the axon the channel density is high enough for the AP to propagate faithfully all the way to the axon terminal. In the other direction the AP spreads back through the soma (which is where a somatic recording picks it up) and into the dendrites, which have a lower Na+ channel density. So somewhere along the dendrites, sodium channel density drops below a critical number and the AP stops propagating.
72Do APs only travel in one direction?
Not technically. They are initiated in the middle of the cell rather than at one end, so the AP can propagate in two directions from the axon initial segment: forwards down the axon and backwards through the soma into the dendrites. The AP stops propagating somewhere in the dendrites once the Na+ channel density drops too low. What it cannot do is travel back through membrane it has already passed through, because those voltage-gated Na+ channels are inactivated (the absolute refractory period).
73Why does the action potential only travel in one direction along the axon?
In the stretch of axon the AP has just passed through, the voltage-gated Na+ channels have become inactivated (since they are only activated for around 200 μs) and the delayed activation of the K+ conductance has taken over, hyperpolarizing the membrane potential. That stretch is therefore refractory for a period of some milliseconds, the K+ conductance is open, and no further APs can be initiated there. In the stretch ahead of the AP the Na+ channels are still available, so the axial spread of current from the excited region can only drive a new regenerative spike forwards, propagating the voltage waveform in that direction.
74In a snapshot of an axon through which an AP is propagating, what is happening at each part?
The excited area (middle segment in the image) is depolarized, its voltage-gated Na+ channels are open, and a large Na+ current flows in. Part of that current flows axially into the neighbouring area (right segment in the image), thereby depolarizing it, activating its voltage-gated Na+ conductance and driving a new regenerative spike there, propagating the waveform forwards. On the tail (left segment in the image), the Na+ channels have inactivated (since they are only activated for around 200 μs) and the delayed K+ conductance has taken over, hyperpolarizing the membrane. That tail is therefore refractory for a period of some ms, so no further AP can be initiated there for about 1-2 ms until the Na+ channels de-inactivate (recall recovery constant of 1-2 ms)
75What cable properties lead to more rapid and reliable AP propagation? (i.e. if we have two neurons with different rates of propagation, what properties would the faster one possess)?
1. High membrane resistance, so there is less leak current (more resistant to leak currents leaving the cell), meaning a longer length constant
2 Low axial resistance, so there is greater axial current flow (less resistance along the current's path) and again a longer length constant. Larger diameter axons have lower axial resistance (like the squid giant axon=APs can travel very fast).
3. Low membrane capacitance, so less of the axial current is used charging the membrane and more of it propagates forward to regeneratively activate the next voltage-gated Na+ channels.
76How fast do action potentials propagate, and why is that a problem?
Around 1 m/s = 1 mm/ms, in a typical mammalian neuron. This allows local processing within a 1 mm³ cube of brain on the millisecond timescale, which is compatible with behaviour. However, if the signal needs to propagate farther, this can be quite slow. e.g., controlling a leg involves distances of around 1 m, which would take around 1 s: far too slow for normal movement. The solution to this problem is myelination to increase membrane resistance and reduce membrane capacitance.
77Why is the squid giant axon so large, and what does that mean for AP propagation?
The squid has an escape reflex it needs to activate very rapidly: the giant axon fires an AP that contracts muscles driving the squid away from a predator. Its diameter is close to 1 mm, more than a thousand times larger than a typical mammalian axon at around 100 nm. That enormous diameter gives a very low axial resistance, so the length constant is long and current spreads far, raising propagation speed to around 25 m/s instead of the usual 1 m/s, allowing the squid to very rapidly escape death.
78What is myelination and what does it do electrically?
Myelination is the process whereby specialized glial cells (oligodendrocytes in the CNS and Schwann cells in the PNS) wrap very thin lipid processes (myelin) many times around selected axons, building up a thick insulating sheath. Lipid is a good electrical insulator (a dielectric), so the sheath increases axonal membrane resistance by a factor of around 5,000. The myelinated stretches also contain very few ion channels, since these are concentrated at the nodes of Ranvier instead, which further reduces leak. It also decreases axonal capacitance by a factor of around 50, because the myelin increases the separation between the intracellular and extracellular conductors. The higher resistance increases the length constant and the lower capacitance reduces the charge needed to change Vm, so signals travel further and are more efficient at depolarizing the membrane.
79What are the nodes of Ranvier?
In a myelinated axon, the AP is regeneratively amplified at points known as nodes of Ranvier, which are small gaps between myelinated segments. A node is molecularly very similar to the axon initial segment, with a high density of voltage-gated Na+ channels clustered and held in place by the scaffolding molecule Ankyrin G and the cytoskeleton, so the threshold for AP initiation there is very low.
80Why is myelination alone not enough, so that nodes of Ranvier are needed?
Myelin improves passive current spread, since the high membrane resistance and low capacitance mean current travels far along the axon with little loss and little filtering. But the sheath covers the membrane, so there are no accessible voltage-gated Na+ channels along that stretch and the AP cannot be regenerated there. Passive spread alone decays with distance, so over a long axon the signal would die out without regeneration at intervals.
81How does saltatory propagation work?
An AP fires at one node of Ranvier, and between it and the next node there is about 1 mm of myelin. Along that stretch the membrane resistance is high and the membrane capacitance is low, so there is a faithful transfer of current axially with little loss and little filtering. That current reaches the next node, which has a very low threshold, and fires a new AP there. So the current spreads continuously along the axon, but the AP itself only occurs at the nodes and therefore appears to jump between them. The jump is fast because the capacitance is low, giving propagation speeds of around 100 m/s.
82What is patch-clamp recording and how does it work?
A glass pipette with a tip size of about 1 μm, filled with solution and connected to an amplifier, is pressed against the membrane of a cell. Gentle suction pulls the membrane onto the glass, forming a gigaseal: a seal of over 1 GΩ between the pipette tip and the ECF that forces essentially all current to pass through the membrane under the tip rather than leaking around it, making it possible to measure tiny currents. Leaving that patch intact records the currents through the few channels in the patch of membrane under the tip. Applying further brief suction pulses breaks the patch, so the pipette becomes continuous with the cell interior and records the whole cell instead. The amplifier can be used in voltage-clamp mode, holding the voltage and measuring current, or in current-clamp mode, injecting current and measuring the voltage.
83What are the four steps for obtaining in vitro whole-cell patch clamp recordings from brain slices?
Prepare the brain slices, image the neurons, prepare the recording electrodes, and perform the whole-cell patch-clamp recording.
84How are the brain slices prepared (step 1), and why is each solution used?
The mouse is anesthetized, the brain extracted and placed in ice-cold sucrose-based slicing solution. The solution is modified ACSF (artificial CSF -> salt solution to match ionic composition of CSF) to keep cells healthy during cutting. The brain is sliced into 300 μm sections with a vibrotome. Slices are then incubated for 30 minutes in 35 °C warm sucrose solution, which lets the cells that were damaged by the blade at the slice surface float away, leaving healthy tissue behind. Finally they are transferred to room-temperature ACSF. All solutions are aerated with 95% O2 and 5% CO2 to hold the pH at 7.3.
85What are the components of ACSF?
(concs in mM): 125 sodium chloride, 25 sodium bicarbonate, 25 D-glucose, 2.5 potassium chloride, 1.25 sodium phosphate, 2 calcium chloride and 1 magnesium chloride (aerated with 95% oxygen and 5% carbon dioxide).
86How are the neurons imaged (step 2)?
The slice is placed in a recording chamber superfused with ACSF at 35 °C (solution flows continuously over it to keep it oxygenated). It is viewed first at 4X to find the brain region and then at 60X to visualize individual cells, using high-contrast infrared video microscopy: infrared light penetrates thick tissue better than visible light, so living cells inside the slice can be seen.
87How are the electrodes prepared (step 3)?
Electrodes are made by heating a borosilicate glass capillary (a thin glass tube) in a pipette puller, which pulls both ends apart while the middle melts, thinning it down to a tip of approximately 1 μm. The pipette is filled with intracellular solution containing a red fluorophore (Alexa-594) so the recorded cell can be imaged afterwards.
88What are the three configurations of a patch-clamp recording (step 4), and how do you move from one to the next?
Note: they are not three alternatives but three successive stages you pass through to reach the whole-cell recording.
1. Search: the pipette is in the bath, not yet touching a cell, with positive pressure applied inside it (around 15 mBar) so solution flows outward and keeps the tip clean. You can see this working when the pipette touches the slice surface and the tissue is visibly pushed away from the tip.
2. Cell-attached: the pipette is pressed against the cell body until a small dimple appears, then the positive pressure is removed and gentle suction applied, so the membrane seals onto the pipette tip forming a gigaseal. The cell interior is still closed off.
3. Whole-cell: further brief suction pulses break the membrane patch under the tip, so the pipette is continuous with the inside of the cell and can record its membrane potential. This is the end goal.
89How do you know which configuration you are in during a patch-clamp recording?
You measure the current response to a test voltage step.
1. Search: a large current response, since the open tip has low resistance (only around 5 MΩ).
2. Cell-attached: very little current, since the gigaseal (over 1 GΩ) leaves current no path except the tiny patch of membrane under the tip.
3. Whole-cell: large capacitance transient currents, because the voltage step now charges the whole cell's membrane capacitance.
90Once in whole-cell configuration, how is the membrane potential measured and how are action potentials evoked?
The amplifier is switched from voltage-clamp to current-clamp mode, which holds the injected current at a set value and measures Vm. The switch is necessary because in voltage clamp Vm is held fixed, so no AP could occur; current clamp lets Vm move freely. Current is then injected: hyperpolarizing current hyperpolarizes the cell, and sufficient depolarizing current evokes action potentials, with more current giving more action potentials. So current clamp is what you use to measure AP waveforms, and voltage clamp is what you use to measure ionic currents (as Hodgkin and Huxley did).
91Under good voltage clamp, what role do capacitive currents play in the measured current?
None, once the step has settled. Capacitive current is I = C·dV/dt, so it only flows while Vm is changing. Under good voltage clamp the amplifier holds Vm fixed at the command value, so dV/dt = 0 and the capacitive current is zero: everything the amplifier injects is going through ion channels, and the measured current is purely ionic (since recall that the total transmembrane current = I_ionic + !_capacitive --> charge crosses the membrane through ion channels (ionic current) and by charging or discharging the membrane capacitance (capacitive current)). This is why voltage clamp is the technique for studying channels.
92Approximate values for:
-Resting Vm
-Voltage-gated Na+ and K+ channel activation range
-AP threshold
-AP peak
-ENa and EK
-Neuronal input resistance and input conductance
-Single voltage-gated Na+ channel conductance
-Number of Na+ channels needed for GNa to approach the cell's input conductance
-Na+ channel activation delay and current duration
-K+ channel activation delay and current duration
-Na+ recovery-from-inactivation time constant
-Typical, inhibitory-neuron and excitatory-neuron AP durations
-Axon initial segment location
-Passive length constant and membrane time constant
-Distance between nodes of Ranvier
-Typical mammalian and squid giant axon diameters
-Typical mammalian, squid giant axon and myelinated axon propagation speeds
-Effects of myelin on membrane resistance and capacitance
-Brain-slice thickness
-Patch-pipette tip diameter and open-tip resistance
-Gigaseal resistance
Voltages
-Resting Vm: −70 mV
-Channel activation/AP threshold: −50 to −40 mV
-AP peak: +30 to +40 mV
-ENa: +60 mV; EK: −90 mV
Electrical scales
-Cell: Rin ≈ 50 MΩ; Gin ≈ 20 nS
-Single Na+ channel: ≈ 20 pS
-Na+ channels needed to approach Gin: ≈ 1,000
Timing
-Na+: ≈ 10 μs activation delay; ≈ 200 μs before inactivation
-K+: ≈ 200 μs activation delay; sustained during depolarization
-Na+ recovery time constant: ≈ 1-2 ms
-AP duration: typical/excitatory ≈ 1 ms; inhibitory ≈ 0.5 ms
Axons
-Axon initial segment: ≈ 50 μm from soma
-Length constant: ≈ A few hundred microns to about 1 mm; membrane time constant: several ms
-Nodes of Ranvier: ≈ 1 mm apart
-Mammalian axon: ≈ 100 nm diameter; ≈ 1 m/s
-Squid giant axon: ≈ 1 mm diameter; ≈ 25 m/s
-Myelinated axon: ≈ 100 m/s
-Myelin: Rm increases ≈ 5,000-fold; Cm decreases ≈ 50-fold
Patch clamp
-Brain slice: ≈ 300 μm
-Pipette tip: ≈ 1 μm; open-tip resistance ≈ 5 MΩ
-Gigaseal: >1 GΩ