B/B 07 · Membrane potentials and action potentials
Ions, Toxins and the Action Potential
Use the Nernst equation and channel blockers to explain each phase of a neuron’s action potential.
Passage
A neuron’s membrane potential depends on the concentrations of ions on each side of the membrane and on the permeability of the membrane to each ion. The equilibrium potential for an ion X with charge +1 at body temperature is given approximately by the Nernst equation, E = 61 log₁₀([X]outside/[X]inside) mV. At rest, the membrane is much more permeable to K⁺ than to Na⁺, through leak channels. When the membrane is depolarized to a threshold, voltage-gated Na⁺ channels open, Na⁺ enters, and the potential rises rapidly toward the equilibrium potential for Na⁺. The Na⁺ channels then inactivate and voltage-gated K⁺ channels open, which repolarizes the membrane.
Two features of voltage-gated Na⁺ channels shape the action potential. They open quickly when the membrane depolarizes, so that depolarization causes more depolarization, and after a short time they enter an inactivated state in which they cannot open again until the membrane has repolarized. The resulting refractory period makes action potentials travel in one direction and limits how fast a neuron can fire. The Na⁺/K⁺ pump, which uses ATP to move three Na⁺ out and two K⁺ in, restores the ion gradients over longer periods, but it contributes little to the potential during a single action potential.
A researcher studies a single neuron from a mammal. The ion concentrations inside and outside the cell are shown in Table 1, together with the equilibrium potentials calculated from them.
Table 1. Ion concentrations and equilibrium potentials.
| Ion | Outside (mM) | Inside (mM) | Equilibrium potential (mV) |
|---|---|---|---|
| Na⁺ | 150 | 15 | +61 |
| K⁺ | 5 | 150 | −90 |
The neuron is bathed in a solution that resembles the fluid outside neurons in the body. Solution changes are made by perfusing the chamber, and each recording is made after the new solution has been flowing for five minutes. The resting potential in each condition is measured before the stimulus, and the stimulus is the same size in every condition.
Next, the researcher records the membrane potential with a microelectrode and stimulates the cell with a brief current. She measures the resting potential, the peak potential reached during an action potential, and how long the action potential lasts. She repeats the experiment after changing the solution outside the cell or adding a drug. Tetrodotoxin (TTX) blocks voltage-gated Na⁺ channels, and tetraethylammonium (TEA) blocks voltage-gated K⁺ channels. Table 2 shows the results.
Table 2. Effects of changing the solution or adding drugs.
| Condition | Resting potential (mV) | Peak potential (mV) | Duration of action potential (ms) |
|---|---|---|---|
| Control | −70 | +40 | 1 |
| TTX added | −70 | No action potential (stimulus depolarizes to −45 mV) | — |
| TEA added | −70 | +40 | 5 |
| Extracellular Na⁺ reduced to 15 mM | −70 | −15 | 1 |
| Extracellular K⁺ raised to 20 mM | −45 | No action potential | — |
For the last condition, the researcher notes that the Nernst equation predicts a new equilibrium potential for K⁺ of about −53 mV when [K⁺] outside is 20 mM. In every condition, the pump that exchanges Na⁺ and K⁺ was working normally during the short period of the recording.
Questions
Question 1 of 6
Why is the resting potential of −70 mV closer to the equilibrium potential for K⁺ than to the one for Na⁺?
The pump moves 3 Na⁺ out for 2 K⁺ in, and its contribution is small. The resting potential mainly reflects ion permeability.
The reverse is true. K⁺ is at 150 mM inside and 5 mM outside.
K⁺ leak channels are open at rest, and most voltage-gated Na⁺ channels are closed.
Why the answer is what it is
A membrane that is mainly permeable to one ion sits near that ion’s equilibrium potential. The resting membrane is much more permeable to K⁺, so the resting potential (−70 mV) is near E(K) (−90 mV) and far from E(Na) (+61 mV).
Review: The Action Potential · Membrane Potentials & Equilibrium Gradients · Resting Membrane Potential · Neurophysiology
Question 2 of 6
Why does TTX prevent an action potential?
If K⁺ channels were open, the resting potential would move toward −90 mV. It stayed at −70 mV.
K⁺ is higher inside the cell, and when K⁺ channels open, K⁺ leaves the cell.
The resting potential was unchanged (−70 mV). The stimulus depolarized the cell to −45 mV but it did not trigger an action potential.
Why the answer is what it is
With Na⁺ channels blocked, depolarization from the stimulus cannot open enough Na⁺ channels to begin the regenerative upstroke. The membrane reaches only −45 mV and does not fire.
Review: The Action Potential · Membrane Potentials & Equilibrium Gradients · Resting Membrane Potential · Neurophysiology
Question 3 of 6
TEA lengthens the action potential from 1 ms to 5 ms without changing the peak. What is the best explanation?
The peak and the upstroke were unchanged, so the upstroke does not depend on these channels.
The peak is the same, so there is no evidence of increased Na⁺ entry.
The pump works too slowly to affect a single action potential, and the recording is only a few milliseconds long.
Why the answer is what it is
The falling phase of the action potential relies on Na⁺ channel inactivation and the opening of voltage-gated K⁺ channels. With the K⁺ channels blocked, repolarization is slower and the action potential is longer.
Review: The Action Potential · Membrane Potentials & Equilibrium Gradients · Resting Membrane Potential · Neurophysiology
Question 4 of 6
When extracellular Na⁺ is reduced from 150 mM to 15 mM, the peak falls from +40 mV to −15 mV. What explains this change?
The resting potential remained at −70 mV.
The duration of the action potential is unchanged (1 ms), so repolarization is not affected.
The cell still fires, but the peak is lower. The Na⁺ gradient determines how far the potential can rise.
Why the answer is what it is
The peak approaches E(Na). With [Na⁺]out reduced to 15 mM, equal to the inside concentration, the ratio is 1 and E(Na) = 61 × log 1 = 0 mV. The peak (−15 mV) falls accordingly.
Review: The Action Potential · Membrane Potentials & Equilibrium Gradients · Resting Membrane Potential · Neurophysiology
Question 5 of 6
With 20 mM extracellular K⁺, the resting potential is −45 mV and the neuron does not fire. What is the best explanation?
E(K) becomes less negative (about −53 mV), not more negative.
The resting potential changed because of the concentration gradient, not because channels were blocked.
The intracellular K⁺ concentration does not change appreciably in this time, and the shift is caused by the change in the gradient.
Why the answer is what it is
Raising [K⁺]outside from 5 to 20 mM reduces the gradient for K⁺ and shifts E(K) from −90 mV to about −53 mV. The resting potential follows, and after a prolonged depolarization the voltage-gated Na⁺ channels inactivate, so no action potential occurs.
Review: The Action Potential · Membrane Potentials & Equilibrium Gradients · Resting Membrane Potential · Neurophysiology
Question 6 of 6
Based on Table 1, what is the equilibrium potential of K⁺ if the extracellular concentration is 15 mM and the intracellular concentration is 150 mM? (Use log₁₀(0.1) = −1.)
This would be the result of using a ratio of 0.01. The ratio is 15/150 = 0.1.
This would require equal concentrations on the two sides of the membrane.
The sign is reversed. K⁺ is higher inside the cell, so its equilibrium potential is negative.
Why the answer is what it is
E(K) = 61 × log₁₀(15/150) = 61 × log₁₀(0.1) = 61 × (−1) = −61 mV.
Review: The Action Potential · Membrane Potentials & Equilibrium Gradients · Resting Membrane Potential · Neurophysiology