Question

In myelinated neurons, the ion exchange of an action potential does not actually occur in myelinated regions of the axon. The nodes of Ranvier are sites where the action potential regenerates.

a. Explain why an action potential will not regenerate if the nodes are too greater than 1.5 mm apart. (the answer not because they are too far apart).
b. Explain why a demyelinating disease such as multiple sclerosis also causes the action potential to die out and not reach the axon terminal of the neuron

Answer 1

The distance between the nodes of Ranvier plays a crucial role in the regeneration of the action potential in myelinated neurons. Multiple sclerosis, a demyelinating disease, disrupts the propagation of the action potential by causing a loss of myelin, leading to the action potential dying out and not reaching the axon terminal.

Step-by-step explanation:

In myelinated neurons, the action potential regenerates at the nodes of Ranvier. The distance between these nodes is crucial for proper propagation of the action potential. If the nodes are too far apart, the depolarization would have fallen off too much for voltage-gated Na+ channels to be activated at the next node, thus preventing the action potential from regenerating.

In demyelinating diseases like multiple sclerosis, the loss of myelin disrupts the Saltatory conduction process. Without myelin, the action potential is unable to regenerate at the nodes of Ranvier, leading to a loss of propagation and the action potential not reaching the axon terminal of the neuron.