Neuron X is at its resting membrane potential of -65 mV.

It expresses a transporter whose activity determines the intracellular concentration of anion "A-".

If the transporter becomes more active, such that the intracellular concentration of anion "A-" increases and the equilibrium potential for "A-" becomes -45 mV, in which direction will A move across the membrane when its selective channel opens?

(there are only 3 options in this question)

Question

Neuron X is at its resting membrane potential of -65 mV.

It expresses a transporter whose activity determines the intracellular concentration of anion "A-".

If the transporter becomes more active, such that the intracellular concentration of anion "A-" increases and the equilibrium potential for "A-" becomes -45 mV, in which direction will A move across the membrane when its selective channel opens?

(there are only 3 options in this question)

BlackTom AI · Accepted Answer

Equilibrium potentials set the voltage toward which ions will drive the membrane when their channels open. The Nernst relation (E = (RT/zF) ln([out]/[in])) defines an ion’s equilibrium potential; for an anion (z = −1) this means a larger intracellular concentration tends to produce a negative E. The net electrochemical driving force on an ion is the difference between the membrane potential (Vm) and that ion’s equilibrium potential (Eion); ion flux will occur so as to move Vm toward Eion.

Step-by-step reasoning and calculation:
1. Given Vm = −65 mV and E_A = −45 mV. Compute the driving-force difference: Vm − E_A = (−65) − (−45) = −20 mV. The magnitude is 20 mV.
2. Interpret the sign: Because Vm is 20 mV more negative than E_A, opening A− channels will cause charge movement that brings Vm up (less negative) toward −45 mV.
3. To depolarize the inside from −65 toward −45 requires loss of negative charge (or gain of positive charge). Therefore negative ions (A−) must move out of the cell when their channels open.
4. Chemical gradient also favors outward movement here: the transporter increased intracellular [A−], so diffusion drives A− from high (inside) to low (outside). Both electrical and chemical forces therefore push A− outward.

Why the correct answer is correct:
- Movement of A− out of the cell will depolarize Vm toward −45 mV, matching the equilibrium potential for A−. This is exactly what the electrochemical gradients predict given Vm < E_A.

Why the other options are incorrect:
- “In”: Inward movement of A− would further hyperpolarize (make Vm more negative), driving Vm away from E_A (contradicts the tendency to move Vm toward E_A).
- “No movement”: Would be true only if Vm = E_A; here Vm (−65 mV) ≠ E_A (−45 mV), so there is a net driving force.

Conclusion: A− will move Out, which matches the provided correct answer.

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