Question

In an excitable cell, which of the following events will make the cell more likely to fire an action potential?

Na+ leak channels close

K+ leak channels close

Voltage-gated Na+ channels open

K+ channels open when the membrane potential (Vm) is -90 mV

Answer 1

The opening of voltage-gated Na+ channels will make the cell more likely to fire an action potential, causing depolarization that can reach the threshold needed to trigger this event.

Step-by-step explanation:

In an excitable cell, the event that will make the cell more likely to fire an action potential is when voltage-gated Na+ channels open. The opening of these channels allows the influx of Na+ ions, leading to depolarization of the cell's membrane. Once the threshold of approximately -55 mV is reached or surpassed, an action potential is triggered, which is an all-or-nothing event where the membrane potential reverses from around -70mV to about +40mV.

This depolarization is then followed by the repolarization phase, wherein voltage-gated K+ channels open allowing K+ ions to leave the cell. Consequently, repolarization restores the negative membrane potential. Other events such as the closing of Na+ or K+ leak channels can affect the membrane potential, but they do not directly trigger an action potential.