Vz2- is a hypothetical ion with an equilibrium potential of -80mV. There are no leak channels for this ion in the cell membrane.
If a channel for this ion existed on a cardiac myocyte (contractile cell), and opened in response to an event, in which direction would Vz2- move while the cell is at rest?

Question

BlackTom AI · Accepted Answer

The membrane potential will be driven toward the ion’s equilibrium potential (-80 mV). Because the resting potential of a ventricular/contractile myocyte is more negative (≈ -85 to -90 mV), opening a Vz2- channel will cause Vz2- to move out of the cell, producing an outward current.

1) Relevant knowledge points and concepts
- Equilibrium (Nernst) potential: the membrane potential at which there is no net flux of that ion.
- Electrochemical driving force: ions move to bring Vm toward their equilibrium potential.
- For an anion (negative charge), efflux (moving out) makes the inside less negative (depolarizes); influx makes it more negative (hyperpolarizes).

2) Step-by-step reasoning
- Given: E(Vz2-) = -80 mV. Typical resting Vm for a cardiac contractile cell ≈ -85 to -90 mV.
- Compare Vm and E: Vm is more negative than E (e.g., -90 mV vs -80 mV). The membrane will move toward -80 mV if the channel opens.

3) Calculation (intermediate steps)
- Compute the difference E − Vm: E − Vm = (−80 mV) − (−90 mV) = +10 mV.
- This +10 mV means the equilibrium potential is 10 mV more positive than the current Vm; opening the channel will change Vm upward by driving charge movement that makes the inside less negative.

4) Why the correct answer is correct
- To depolarize from −90 toward −80 mV the cell must lose negative charge (or gain positive). For a negatively charged ion (Vz2-), that means net efflux (anion moving out). Net movement of negative charge out is described in this context as an outward current for that ion’s flux. Thus Vz2- moves outward.

5) Why other choices are incorrect
- Inward movement would push Vm further negative (hyperpolarize) and move Vm away from E(= −80 mV), so it contradicts the electrochemical gradient.
- No movement is incorrect because opening a channel for an ion whose E differs from Vm produces flux until Vm approaches E.

6) Conclusion
Because resting Vm is more negative than E(Vz2-), Vz2- will flow out of the cell, producing an outward current. This confirms the correct answer: b. Outward current.

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