Neuron X has standard resting permeabilities, and standard intracellular and extracellular concentrations of sodium, potassium, chloride and calcium. When depolarised to +58mV by a strong, excitatory synaptic input, what direction will Na+ move across the membrane if its selective channel were opened?

(there are only 3 options in this question)

Question

Neuron X has standard resting permeabilities, and standard intracellular and extracellular concentrations of sodium, potassium, chloride and calcium. When depolarised to +58mV by a strong, excitatory synaptic input, what direction will Na+ move across the membrane if its selective channel were opened?

(there are only 3 options in this question)

BlackTom AI · Accepted Answer

At +58 mV the membrane potential equals the electrochemical equilibrium for Na+, so there is no net Na+ movement.

1) Relevant knowledge points and concepts
- Nernst (equilibrium) potential: Eion = (RT/zF) ln([ion]out/[ion]in) (or in mV: ≈ (61.5 mV/z)·log10([out]/[in]) at 37°C). Eion is the membrane voltage at which the electrical and chemical driving forces for that ion balance.
- Driving force for an ion = Vm − Eion. If driving force = 0, net flux = 0 (no net movement), even though individual ions still undergo random thermal motion.
- For Na+, z = +1 and with “standard” physiological concentrations E_Na is approximately +58 to +62 mV depending on exact values used.

2) Step-by-step reasoning
- Determine E_Na using standard concentrations.
- Compare the given membrane potential Vm (+58 mV) to E_Na.
- If Vm = E_Na, the driving force is zero ⇒ no net flux through an open Na+ selective channel.

3) Calculations (intermediate steps)
- Using the common form at 37°C: E_Na ≈ 61.5 mV · log10([Na+]out/[Na+]in).
- Example with typical values [Na+]out ≈ 145 mM, [Na+]in ≈ 15 mM:
  log10(145/15) = log10(9.67) ≈ 0.986
  E_Na ≈ 61.5 × 0.986 ≈ +60.6 mV (close to +58 mV; slight differences reflect different “standard” intracellular Na+ reported).
- Because the question states Vm = +58 mV (the standard E_Na in this context), Vm ≈ E_Na → driving force = 0.

4) Why the correct answer is correct
- With Vm equal to Na+’s equilibrium potential, electrical and concentration gradients cancel; opening a Na+ channel produces no net directional Na+ current. That is exactly the “No movement” option.

5) Why the other options are incorrect
- “Into cell” would be correct only if Vm were more negative than E_Na (Vm < E_Na) so the electrochemical gradient favored inward Na+ flow.
- “Out of cell” would require Vm to be more positive than E_Na (Vm > E_Na), making the electrical force push Na+ out. Neither condition holds here.

6) Conclusion
Since Vm equals Na+’s equilibrium potential, there is no net movement of Na+ when its selective channel opens; this confirms the provided answer: No movement.

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