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Question

In the nervous system, the action potential is generated at the axon hillock in physiological conditions and it is conducted to the terminal end of axon. The location specific origin of action potential and its direction-specific conduction are explained by a researcher in the following proposed statements:

A. The membrane of axon hillock has highest threshold for the generation of action potential

B. The membrane of axon hillock contains large numbers of voltage-gated Na+ channels and that makes it more excitable

C. The propagating action potential in the middle of the axon cannot generate another action potential in the direction of cell body since a large fraction of voltage-gated Na+ channels in the preceding portion is voltage inactivated

D. As the number of voltage-gated Na+ channels is less in the preceding portion of axonal membrane, the propagating action potential in the middle of the axon cannot generate another action potential in the direction of cell body.

Which one of the following combinations represents both correct explanations?

The correct answer is

B and C

Understanding the action potential, its generation, and its propagation direction is fundamental in studying the nervous system. The action potential is a rapid, transient change in membrane potential that travels along the axon of a neuron.

Action Potential Origin at Axon Hillock

The action potential typically originates at the axon hillock, also known as the trigger zone. This specific location is crucial for initiating the nerve impulse. Let's look at why this is the case:

  • Statement A suggests the axon hillock has the highest threshold for action potential generation. This is incorrect. The axon hillock actually has the lowest threshold for firing an action potential compared to other parts of the neuron, like the dendrites or cell body. A lower threshold means it takes a smaller depolarization to trigger a spike.
  • Statement B states that the membrane of the axon hillock contains large numbers of voltage-gated Na+ channels, which makes it more excitable. This is correct. The high density of voltage-gated Na+ channels at the axon hillock significantly lowers the threshold potential at this site, making it the most excitable region and the usual site of action potential initiation.

Action Potential Conduction Direction

Once generated at the axon hillock, the action potential propagates down the axon towards the terminal end. This propagation is typically unidirectional, meaning it moves away from the cell body and towards the axon terminal. Why doesn't it propagate backward towards the cell body?

  • Statement C explains that the propagating action potential in the middle of the axon cannot generate another action potential in the direction of the cell body because a large fraction of voltage-gated Na+ channels in the preceding portion is voltage inactivated. This is correct. After voltage-gated Na+ channels open and cause depolarization, they quickly enter an inactivated state (the absolute refractory period). During this time, they cannot be opened again by depolarization, preventing the action potential from propagating backward. This ensures the unidirectional flow of the nerve impulse.
  • Statement D proposes that the backward propagation is prevented because the number of voltage-gated Na+ channels is less in the preceding portion of the axonal membrane. While channel density might play some role in overall excitability, the primary reason for unidirectional propagation is the inactivation state of the voltage-gated Na+ channels during the refractory period, not necessarily a significantly lower number of channels in the already excited region compared to the region ahead. Therefore, this statement does not provide the correct primary explanation.

Summary of Explanations

Based on the analysis:

  • Statement A is incorrect (axon hillock has lowest threshold).
  • Statement B is correct (high density of voltage-gated Na+ channels at axon hillock).
  • Statement C is correct (unidirectional propagation due to voltage inactivation of Na+ channels in the refractory period).
  • Statement D is incorrect (primary reason for unidirectional propagation is channel inactivation, not just fewer channels).

Therefore, the correct explanations are statements B and C.

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Important Questions from Nervous system

  1. Which one of the following synaptic vesicle proteins is involved in tethering of the vesicle to the cytoskeletal system in the nerve terminus?

  2. Which one of the following is NOT released by sympathetic preganglionic neurons?

  3. The time taken by synaptic vesicles to travel from the soma of a motor neuron in the spinal cord to its neuromuscular junction in a person’s foot by fast axon transport is about‐

  4. The release of which neurotransmitter from the rods of retina is reduced when light strikes its outer segment?

  5. The explanations for increased conduction velocity of action potentials in a myelinated nerve fibre (MNF) as compared to that of a non‐myelinated fibre (NNF) of same diameter are suggested below:

    A. Much higher number of ions traverse a unit length of an MNF membrane compared to that of an NNF during conduction of action potential as the ionic currents are restricted to the membrane at the Nodes of Ranvier

    B. The energy expenditure required to maintain ionic gradient after conduction of action potential in an MNF is higher than that of an NNF

    C. The action potentials of MNF do not have a hyperpolarizing effect like that of an NNF

    D. The relative refractory period is not extended in an MNF like that of an NNF

    E. The axolemma at the nodes of Ranvier lacks K+ ion channels

    F. MNFs are metabolically less efficient than NNFs

    Choose all correct statements from the following options:

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