Question

Refer to the figure showing the conduction of an action potential traveling from left to right in a myelinated axon. Note the action potential at node 2. Which statement about section 2 of the axon is true?

a. A loss of signal will occur as the action potential travels to each adjacent node.

b. The speed of the action potential in this axon depends mostly on its diameter.

c. Na⁺ channels will inactivate at this node and enter a refractory state.

d. Na⁺ channels remain open here until the action potential has reached the axon terminal.

e. Depolarization is caused here by the opening of voltage-gated K⁺ channels.

Answer 1

D. Na+ channels remain open here until the action potential has reached the axon terminal.

The action travels from one node to another beacause the saltatory conduction mechanism of Nodes of ranvier that allows rapid propagation of the impulse,it prevents loss of signal. The speed of the action potentail depends on the size of the axons but not on the diameter of axon.At the refractory period, Na+ starts recovering from the inactivation. When channel are open , it allows the Na+ to enter which changes the electrochemical gradient and eventually it will produce the membrane potential. Opening of the voltage gated K+ channels  promotes negative potential inside in resting phase.