Question at position 18 During graph section labeled "B", what is occurring?Na+ is entering the cell.Cl- is entering the cell.Na+ is leaving the cell.K+ is entering the cell.Clear my selection

Question

BlackTom AI · Accepted Answer

Voltage-gated Na+ channels open and Na+ rushes into the cell, producing the rapid depolarization seen at B.

1) Relevant knowledge points and concepts
- Resting membrane potential is typically around −70 to −80 mV (label A on the graph).
- An action potential has phases: depolarization (rapid rise to positive values), repolarization (fall back toward negative), and after-hyperpolarization (below resting).
- Depolarization is produced by inward flow of positive charge, most commonly Na+ through voltage-gated Na+ channels; repolarization is produced by outward K+ flow through voltage-gated K+ channels. Inward Cl− would hyperpolarize or oppose depolarization.

2) Step-by-step reasoning using the graph
- At A the membrane is at a negative resting potential (~−70 to −80 mV).
- The membrane potential rises rapidly through 0 mV and peaks near +30 to +40 mV at the top of the spike (section B–C). This rapid positive-going change is the hallmark of depolarization produced by an inward positive current.
- The most physiologically relevant inward positive current during the rising phase is Na+ moving into the cell through voltage-gated Na+ channels that open after threshold is reached (dashed line near −55 mV).

3) Intermediate steps / key values
- Resting ≈ −70 to −80 mV (A).
- Threshold for opening voltage-gated Na+ channels ≈ −55 mV (dashed line E region).
- Peak at B ≈ +30 to +40 mV — requires an inward positive ion flux (Na+ influx).

4) Why the correct answer is correct
- Na+ entry is the mechanism that rapidly drives the membrane from negative to positive values. The graph’s rapid upward deflection to a positive membrane potential is exactly the effect of Na+ influx through voltage-gated channels.

5) Why the other options are incorrect
- Cl− entering the cell would bring negative charge in and hyperpolarize (make the membrane more negative), not produce the positive spike.
- Na+ leaving the cell would carry positive charge out and hyperpolarize or prevent depolarization; it cannot produce the upward spike.
- K+ entering the cell would also bring positive charge inward but physiologically K+ movement during an action potential is outward (efflux) producing repolarization; inward K+ is not the driver of the rapid positive peak.

6) Conclusion
All evidence from the membrane potential trace and known ion channel behavior confirms that during section B Na+ is entering the cell, which matches the provided correct answer.

Question at position 18 During graph section labeled "B", what is occurring?Na+ is entering the cell.Cl- is entering the cell.Na+ is leaving the cell.K+ is entering the cell.Clear my selection单项选择题

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