Generation and conduction of Nerve Impulse Questions in English

(A) The $Na^+ - K^+$ pump is an active transport mechanism that maintains the resting membrane potential in neurons.
$1$. Statement $(A)$ is incorrect because the pump transports $3\, Na^+$ ions $OUTSIDE$ the cell and $2\, K^+$ ions $INSIDE$ the cell.
$2$. Statement $(B)$ is correct; it uses $ATP$ to maintain the ionic gradient via active transport.
$3$. Statement $(C)$ is correct; it moves ions against their respective concentration gradients (pumping $Na^+$ out where it is already high and $K^+$ in where it is already high).
Since only statement $(A)$ is incorrect,the answer is $A$ only.

(D) The resting membrane potential of a neuron is the electrical potential difference across the plasma membrane when the neuron is at rest.
In a typical resting neuron,the intracellular fluid is negatively charged relative to the extracellular fluid.
This potential difference is maintained primarily by the $Na^+/K^+$ pump and the differential permeability of the membrane to ions.
The value of this resting membrane potential is approximately $-70 \ mV$.

(A) During the resting state,the axonal membrane is more permeable to $K^+$ ions and nearly impermeable to $Na^+$ ions.
When a stimulus is applied at a site on the polarized membrane,the permeability of the membrane to $Na^+$ ions increases significantly.
This leads to a rapid influx of $Na^+$ ions into the cell,which causes the reversal of the polarity at that site,known as depolarization.
Therefore,the correct answer is that $Na^+$ ions move rapidly into the cell.

(C) During the action potential,the depolarization phase is caused by the rapid influx of $Na^+$ ions into the neuron.
Following this,the repolarization phase occurs to restore the resting membrane potential.
Repolarization is primarily caused by the closure of voltage-gated $Na^+$ channels and the opening of voltage-gated $K^+$ channels.
As a result,$K^+$ ions move out of the neuron (efflux),which makes the inside of the membrane negative again relative to the outside.

(B) The conduction of a nerve impulse is faster in myelinated nerve fibers because of saltatory conduction.
In myelinated fibers,the myelin sheath acts as an electrical insulator.
This forces the action potential to 'jump' from one Node of Ranvier to the next,which significantly increases the speed of transmission compared to continuous conduction in non-myelinated fibers.

(D) In the resting state,the axonal membrane of a neuron is significantly more permeable to potassium ions $(K^+)$ due to the presence of leak channels,while it remains nearly impermeable to sodium ions $(Na^+)$.
This differential permeability,combined with the action of the $Na^+-K^+$ pump,helps maintain the resting membrane potential.

(B) : Action potential is the change in electrical potential that occurs across a plasma membrane during the passage of a nerve impulse.
As an impulse travels in a wavelike manner along the axon of a nerve,it causes a localized and transient switch in electric potential across the membrane from $-60 \ mV$ (resting potential) to $+45 \ mV$.
This depolarization is primarily due to the rapid opening of voltage-gated sodium channels.
As a result,these channels permit the massive influx of $Na^+$ ions from the extracellular fluid into the intracellular fluid via diffusion,leading to the generation of the action potential.

(D) The correct answer is $D$.
$1$. At the resting state,the inner side of the axonal membrane is negatively charged relative to the outer side due to the differential permeability of the membrane to $K^+$ and $Na^+$ ions.
$2$. When a stimulus is applied,the membrane becomes permeable to $Na^+$ ions,causing an influx of $Na^+$. This leads to depolarization,where the inner side becomes positively charged.
$3$. Subsequently,the membrane becomes permeable to $K^+$ ions,which move out,leading to repolarization,where the inner side returns to its original negative state.
$4$. Thus,the potential changes from negative to positive and back to negative during the transmission of an action potential.

(A) During the resting state of a neuron,the axonal membrane is significantly more permeable to potassium ions $(K^+)$ due to the presence of many open potassium leak channels.
Conversely,it is nearly impermeable to sodium ions $(Na^+)$ because most sodium channels are closed.
This selective permeability,combined with the action of the $Na^+/K^+$ pump,maintains the resting membrane potential,which is typically around $-70 \text{ mV}$.

(D) During the resting state of a neuron,the axonal membrane is more permeable to $K^+$ ions and nearly impermeable to $Na^+$ ions.
When a stimulus is applied to the site on the polarized membrane,the permeability of the membrane to $Na^+$ ions increases rapidly.
This influx of $Na^+$ ions from the outside to the inside of the axon occurs due to the concentration gradient (high concentration outside,low inside) and the electrochemical gradient created by the negatively charged proteins inside the axoplasm.
Therefore,all the mentioned factors contribute to the transport of $Na^+$ ions during depolarization.

(A) During the resting state of a neuron,the axonal membrane is significantly more permeable to $K^+$ ions than to $Na^+$ ions.
This is due to the presence of a large number of 'leak' channels for $K^+$ in the membrane,which allow $K^+$ to diffuse out of the cell down its concentration gradient.
Conversely,the membrane is nearly impermeable to $Na^+$ ions and negatively charged proteins present in the axoplasm.
This differential permeability,along with the action of the $Na^+-K^+$ pump,maintains the resting membrane potential.

(D) In the resting state,the axonal membrane of a neuron is significantly more permeable to potassium ions $(K^+)$ due to the presence of leak channels.
Conversely,the membrane is nearly impermeable to sodium ions $(Na^+)$.
This differential permeability,along with the action of the $Na^+-K^+$ pump,helps maintain the resting membrane potential,where the inside of the axon is negatively charged relative to the outside.

(A) The sodium-potassium pump ($Na^+/K^+$-ATPase) is an active transport mechanism that maintains the resting membrane potential of a neuron.
It functions by pumping $3Na^+$ ions out of the cell and $2K^+$ ions into the cell against their respective concentration gradients.
This process requires energy in the form of $ATP$ hydrolysis.
Therefore,the correct transport ratio is $3Na^+$ outwards for $2K^+$ into the cell.

(C) The resting membrane potential is the electrical potential difference across the plasma membrane of a resting neuron.
It is primarily maintained by the active transport of ions via the $Na^+-K^+$ pump (an ion pump).
This pump actively transports $3$ $Na^+$ ions out of the cell and $2$ $K^+$ ions into the cell,consuming $ATP$ in the process.
Additionally,the differential permeability of the membrane to $K^+$ and $Na^+$ ions contributes to this potential.

(B) When a stimulus is applied to a site on the polarized membrane,the permeability of the membrane to $Na^+$ ions increases rapidly.
This leads to a massive influx of $Na^+$ ions into the axon from the extracellular fluid.
As a result,the inner side of the membrane becomes positively charged relative to the outer side,which is known as depolarization.
This rapid influx of $Na^+$ ions is significantly greater than the efflux of $K^+$ ions,causing the reversal of polarity (action potential).

(C) In the resting state of a neural membrane,the axonal membrane is significantly more permeable to potassium ions $(K^+)$ and nearly impermeable to sodium ions $(Na^+)$.
Due to the concentration gradient,$K^+$ ions tend to diffuse out of the cell,while $Na^+$ ions are prevented from entering.
However,if we consider the permeability and the electrochemical gradient,the resting membrane potential is primarily maintained by the leakage of $K^+$ out of the cell.
If the question asks which ion is allowed to enter the cell during specific conditions or if we look at the permeability of the membrane,$Na^+$ is generally restricted,but the question implies the movement across the membrane.
Given the options provided and the standard physiological context,$Na^+$ is the ion that would enter the cell if the membrane permeability were to change (as in an action potential),but in the resting state,the membrane is selectively permeable to $K^+$.
However,based on standard textbook questions of this type,the correct answer is $Na^+$ entering the cell when the membrane becomes permeable during depolarization.

(B) $1$. In the resting state,the axolemma is polarized,meaning the inside is negative relative to the outside (resting membrane potential is approximately $-70 \ mV$).
$2$. When a stimulus is applied,the membrane becomes permeable to $Na^+$ ions,causing depolarization,where the inside becomes positive relative to the outside (action potential).
$3$. Following this,the membrane undergoes repolarization,where $K^+$ ions move out,restoring the original negative charge inside the axolemma.
$4$. Therefore,the sequence of potential change is negative $\rightarrow$ positive $\rightarrow$ negative.

(D) During the conduction of a nerve impulse,the resting membrane potential is disturbed by a stimulus.
This stimulus causes the voltage-gated $Na^+$ channels to open,leading to a rapid influx of $Na^+$ ions from the extracellular fluid $(ECF)$ into the intracellular fluid $(ICF)$.
This process is known as depolarization,which generates the action potential.
Therefore,the propagation of the action potential is primarily driven by the movement of $Na^+$ ions from the $ECF$ to the $ICF$.

(D) In the resting state of a neuron,the axonal membrane is significantly more permeable to potassium ions $(K^+)$ due to the presence of leakage channels.
Conversely,the membrane is nearly impermeable to sodium ions $(Na^+)$.
This differential permeability,along with the action of the $Na^+-K^+$ pump,maintains the resting membrane potential where the inside of the axon is negatively charged relative to the outside.

(D) Nerve conduction,or the transmission of a nerve impulse along an axon,primarily relies on the movement of $Na^+$ and $K^+$ ions across the neuronal membrane.
During the resting state,the axonal membrane is more permeable to $K^+$ ions and nearly impermeable to $Na^+$ ions.
When a stimulus is applied,the membrane becomes permeable to $Na^+$ ions,leading to a rapid influx of $Na^+$ into the cell,which causes depolarization.
Following this,the membrane becomes permeable to $K^+$ ions,leading to an efflux of $K^+$ out of the cell,which causes repolarization.
Thus,$Na^+$ and $K^+$ are the essential ions for the generation and conduction of nerve impulses.

(C) The resting membrane potential is generated due to the differential permeability of the axonal membrane to $Na^+$ and $K^+$ ions and the active transport of these ions by the $Na^+-K^+$ pump.
This pump requires metabolic energy in the form of $ATP$ (chemical energy),not electrical energy,to maintain the unequal distribution of ions against their concentration gradients.
Therefore,the Assertion is correct,but the Reason is incorrect because it mentions electrical energy instead of chemical energy $(ATP)$.

(D) The axonal membrane of a neuron in a resting state is significantly more permeable to potassium ions $(K^+)$ and nearly impermeable to sodium ions $(Na^+)$. Therefore,the Assertion $(A)$ is incorrect.
In a resting state,the neuron is polarized and does not conduct any nerve impulse. Therefore,the Reason $(R)$ is correct.

(N/A) $ \Rightarrow $ Neurons are excitable cells because their membranes are in a polarised state.
Different types of ion channels are present on the neural membrane. These ion channels are selectively permeable to different ions.
When a neuron is not conducting any impulse, i.e., at rest, the axonal membrane is comparatively more permeable to $ K^{+} $ and nearly impermeable to $ Na^{+} $.
- Similarly, the membrane is impermeable to negatively charged proteins present in the axoplasm.
The axoplasm inside the axon contains a high concentration of $ K^{+} $ and negatively charged proteins and a low concentration of $ Na^{+} $.
In contrast, outside the axon, a low concentration of $ K^{+} $ and a high concentration of $ Na^{+} $ form a concentration gradient.
These ionic gradients across the resting membrane are maintained by the active transport of $ 3 Na^{+} $ outwards and $ 2 K^{+} $ inside the cell.
As a result, the outer surface of the axonal membrane is positively charged and its inner surface becomes negatively charged; therefore, it is polarised.
The electrical potential difference across the resting membrane is called the resting potential.
When a stimulus is applied at point $ A $ on the polarised membrane, it becomes freely permeable to $ Na^{+} $. $ Na^{+} $ influx is followed by the reversal of the polarity, i.e., the outer membrane is negatively charged and the inner side is positively charged.
The polarity at the site $ A $ is reversed and hence depolarised. The electrical potential difference across the plasma membrane at the site $ A $ is called the action potential, which is in fact termed as a nerve impulse.
At sites immediately ahead, the axon membrane has a positive charge on the outer surface and a negative charge on its inner surface. As a result, a current flows on the inner surface from site $ A $ to site $ B $. On the outer surface, current flows from site $ B $ to site $ A $ to complete the circuit of current flow. Hence, the polarity at the site is reversed and an action potential is generated at site $ B $.
Thus, the impulse (action potential) generated at site $ A $ arrives at site $ B $.
The sequence is repeated along the length of the axon, and consequently, the impulse is conducted.
The rise in the stimulus-induced permeability to $ Na^{+} $ is short-lived. It is quickly followed by the rise in permeability to $ K^{+} $. Within a fraction of a second, $ K^{+} $ diffuses outside and restores the resting potential of the membrane.

(N/A) Resting potential is the electrical potential difference across the resting axonal membrane.
Neurons are excitable cells because their membranes are in a polarised state.
Different types of ion channels are present on the neural membrane,which are selectively permeable to different ions.
When a neuron is not conducting any impulse (resting state),the axonal membrane is comparatively more permeable to $K^{+}$ and nearly impermeable to $Na^{+}$.
Similarly,the membrane is impermeable to negatively charged proteins present in the axoplasm.
The axoplasm inside the axon contains a high concentration of $K^{+}$ and negatively charged proteins,and a low concentration of $Na^{+}$.
In contrast,outside the axon,there is a low concentration of $K^{+}$ and a high concentration of $Na^{+}$,forming a concentration gradient.
These ionic gradients across the resting membrane are maintained by the active transport of $3 Na^{+}$ outwards and $2 K^{+}$ inwards by the $Na^{+}-K^{+}$ pump.
As a result,the outer surface of the axonal membrane is positively charged and its inner surface becomes negatively charged; thus,it is polarised.
When a stimulus is applied at point $A$ on the polarised membrane,it becomes freely permeable to $Na^{+}$. The influx of $Na^{+}$ causes a reversal of polarity (depolarisation),where the outer membrane becomes negatively charged and the inner side becomes positively charged.
The electrical potential difference across the plasma membrane at the site $A$ is called the action potential,which is termed a nerve impulse.
At sites immediately ahead (e.g.,site $B$),the axon membrane still has a positive charge on the outer surface and a negative charge on its inner surface.
As a result,a current flows on the inner surface from site $A$ to site $B$,and on the outer surface,current flows from site $B$ to site $A$ to complete the circuit.
Consequently,the polarity at site $B$ is reversed,and an action potential is generated there.
This sequence is repeated along the length of the axon,conducting the impulse.
The stimulus-induced permeability to $Na^{+}$ is short-lived and is quickly followed by a rise in permeability to $K^{+}$. Within a fraction of a second,$K^{+}$ diffuses outside,restoring the resting potential of the membrane.

(N/A)

Resting potential	Action potential
$1$. Nerve fiber is polarized.	$1$. Nerve fiber is depolarized (not polarized).
$2$. Inner side of the plasmalemma is negatively charged,and the outer side is positively charged.	$2$. Inner side of the plasmalemma is positively charged,and the outer side is negatively charged.

(N/A) $(1)$ At the end of a branched axon,a knob-like swollen structure called the synaptic knob is present. It contains secretory vesicles that produce neurotransmitter substances,such as acetylcholine.
Acetylcholine is released into the synaptic cleft and binds to receptors on the postsynaptic membrane,allowing the conduction of nerve impulses across the synapse.
$(2)$ For the maintenance of the resting membrane potential,ion pumps and ion channels play a significant role. Through the $Na^+-K^+$ pump,$3 Na^+$ ions are actively transported out of the neuroplasm,while $2 K^+$ ions are transported inside. This active transport,combined with the selective permeability of the membrane,results in a higher concentration of $K^+$ ions inside the neuron compared to the extracellular fluid.

(N/A) $(1)$ In the excited region,$Na^+$ ion channels open. At this time,the interior of the neurilemma contains negatively charged proteins and a lower concentration of $Na^+$. Consequently,a large amount of $Na^+$ enters the cell through ion channels,resulting in the formation of a positive charge inside the neurilemma.
$(2)$ In myelinated nerve fibres,the medullary sheath is not continuous. At regular intervals,it forms nodes of Ranvier. The action potential (conduction of nerve impulse) jumps from one node to the next,which is why it is called saltatory conduction.

(N/A) $(1)$ When a nerve fiber is stimulated by touch,smell,or chemical changes,its polarized state changes,which is known as a nerve impulse or action potential.
$(2)$ The electrical potential difference across the resting plasma membrane of a neuron is known as the resting potential.

(N/A) During the resting state,the axonal membrane is selectively permeable to $K^+$ ions and nearly impermeable to $Na^+$ ions.
This is maintained by the $Na^+-K^+$ pump,which actively transports $3$ $Na^+$ ions outwards for every $2$ $K^+$ ions into the neuron.
As a result,the outer surface of the axonal membrane possesses a positive charge,while the inner surface becomes negatively charged due to the presence of negatively charged proteins and high $K^+$ concentration inside.
This electrical potential difference across the resting plasma membrane is called the resting potential.

(A) nerve fiber is said to be polarised when it is in a resting state.
In this state,the axonal membrane is more permeable to $K^+$ ions and nearly impermeable to $Na^+$ ions.
Consequently,the outer surface of the axonal membrane becomes positively charged due to the higher concentration of $Na^+$ ions outside.
Simultaneously,the inner surface becomes negatively charged due to the presence of negatively charged proteins and $K^+$ ions inside.
This difference in electrical charge across the membrane is known as the resting membrane potential,and the fiber is described as polarised.

(B) When a stimulus is applied at a site on the polarized membrane,the permeability of the membrane to $Na^+$ ions increases. This leads to a rapid influx of $Na^+$ ions,causing a reversal of polarity at that site,which is known as depolarization. The electrical potential difference across the plasma membrane at this depolarized site is called the action potential.

(A) The resting membrane potential of a relaxed muscle fibre or a neuron is approximately $-70 \; mV$.
This potential difference is maintained due to the unequal distribution of ions across the plasma membrane, resulting in a negative charge inside the cell relative to the outside.

(A) In a resting nerve fibre (a nerve fibre that is not conducting an impulse),sodium ions $(Na^{+})$ predominate in the extracellular fluid,whereas potassium ions $(K^{+})$ predominate in the intracellular fluid (within the fibre).
This results in the fact that the resting membrane has a very low permeability for $Na^{+}$ ions,whereas it has a significantly higher permeability for $K^{+}$ ions.

(A) The myelin sheath of myelinated nerve fibres prevents the flow of ions between the extracellular fluid and the axoplasm.
Exchange of ions can occur only at the nodes of $Ranvier$.
Therefore,the action potential jumps from node to node and passes along the myelinated axon faster than the series of smaller local currents in a non-myelinated axon.
This process is called saltatory conduction.

(D) During the conduction of a nerve impulse,the resting membrane potential is disturbed by a stimulus.
This stimulus causes the opening of voltage-gated $Na^{+}$ channels.
As a result,there is a rapid influx of $Na^{+}$ ions from the extracellular fluid into the intracellular fluid.
This sudden change in membrane potential is known as the action potential.

(A) The maintenance of ionic gradients across the resting membrane is achieved by the $Na^{+}-K^{+}$ pump.
This pump utilizes energy in the form of $ATP$ to actively transport $3 Na^{+}$ ions out of the cell and $2 K^{+}$ ions into the cell.
Since this process moves ions against their concentration gradient,it is classified as the active transport of ions.

(A) Repolarization is the process of restoring the resting membrane potential of a neuron after depolarization.
During depolarization,the membrane is highly permeable to $Na^+$. As the action potential reaches its peak,the voltage-gated $Na^+$ channels close (Process $II$).
Simultaneously,the voltage-gated $K^+$ channels open (Process $IV$),allowing $K^+$ ions to flow out of the nerve fibre.
This efflux of $K^+$ ions makes the interior of the membrane negative again relative to the exterior,thereby restoring the resting potential.
Therefore,the correct processes are the closing of $Na^+$ channels and the opening of $K^+$ channels.

(A) The correct answer is $A$.
When a neuron is not conducting any impulse (resting stage),the axonal membrane is comparatively more permeable to potassium ions $(K^+)$ and nearly impermeable to sodium ions $(Na^+)$.
Additionally,the membrane is impermeable to negatively charged proteins present in the axoplasm.
Consequently,the axoplasm inside the axon contains a high concentration of $K^+$ and negatively charged proteins,and a low concentration of $Na^+$.
In contrast,the fluid outside the axon contains a low concentration of $K^+$ and a high concentration of $Na^+$,which establishes the concentration gradient.

(B) When a neuron is in a resting state,the axonal membrane is more permeable to $K^{+}$ ions and nearly impermeable to $Na^{+}$ ions.
Additionally,the membrane is impermeable to the negatively charged proteins present in the axoplasm.
Therefore,the axoplasm inside the axon contains a high concentration of $K^{+}$ and negatively charged proteins,and a low concentration of $Na^{+}$.
In contrast,the extracellular fluid outside the axon contains a low concentration of $K^{+}$ and a high concentration of $Na^{+}$.
Statement $I$ is True ($K^{+}$ and proteins are high inside).
Statement $II$ is True ($Na^{+}$ is low inside).
Statement $III$ is True ($K^{+}$ is low outside).
Statement $IV$ is False (the fluid outside contains high $Na^{+}$,not $Mg^{++}$ or negatively charged proteins).
Thus,the correct sequence is $I-$ True,$II-$ True,$III-$ True,$IV-$ False.

(C) The process of nerve impulse conduction occurs in the following sequence:
$1$. Initially,a stimulus causes a disturbance to the membrane at site $A$,which leads to the opening of voltage-gated $Na^+$ channels,resulting in the rapid influx of $Na^+$ ions into the nerve fibre $(II)$.
$2$. Due to the influx of $Na^+$ ions,the polarity of the membrane at site $A$ is reversed (depolarization),where the outer surface becomes negatively charged and the inner side becomes positively charged,generating a nerve impulse $(I)$.
$3$. Immediately ahead of the depolarized site $A$,the membrane at site $B$ is still in a resting state,having a positive charge on the outer surface and a negative charge on the inner surface. Consequently,a current flows on the inner surface from site $A$ to site $B$ $(IV)$.
$4$. To complete the circuit,current flows on the outer surface from site $B$ to site $A$. This causes the polarity at site $B$ to reverse,generating an action potential at site $B$. This sequence repeats along the axon,conducting the impulse $(III)$.
Therefore,the correct sequence is $II \rightarrow I \rightarrow IV \rightarrow III$.

(C) In a resting state,the axonal membrane is significantly more permeable to $K^+$ ions compared to $Na^+$ ions.
This is due to the presence of a large number of $K^+$ 'leaky' channels that remain open,allowing $K^+$ to diffuse out of the neuron.
Conversely,the membrane is nearly impermeable to $Na^+$ ions and negatively charged proteins present in the axoplasm.

(B) The ionic gradients across the resting membrane are partially due to the opening of $K^+$ leaky channels,but they are primarily maintained by the $Na^+-K^+$ pump.
This pump continuously transports $3Na^+$ ions outwards in exchange for $2K^+$ ions inwards,utilizing energy in the form of $ATP$ to maintain the resting membrane potential.

(A) An action potential is a rapid sequence of changes in the voltage across a membrane.
It is the fundamental mechanism by which information is transmitted along the axon of a neuron.
Therefore,the transmission of a nerve impulse is essentially the propagation of an action potential along the nerve fiber.

(C) When a stimulus is applied to the axonal membrane,the $Na^+$ voltage-gated channels open at that specific site. This leads to a rapid influx of $Na^+$ ions from the extracellular fluid into the intracellular space,causing depolarization of the membrane.

(B) The depolarisation of the neuron is primarily caused by the rapid influx of $Na^+$ ions into the cell.
When a stimulus is applied,the voltage-gated $Na^+$ channels open,allowing $Na^+$ to move down its electrochemical gradient into the neuron.
This influx of positive charge makes the inner side of the neuronal membrane less negative (or more positive) compared to the resting state.
This change in membrane potential is known as depolarisation.

(C) When the inside of the neuronal membrane becomes more negative than the resting membrane potential,it is known as hyperpolarisation.
This occurs due to the prolonged opening of $K^+$ channels,which leads to an increased efflux of $K^+$ ions out of the cell,making the interior more negative.

(C) The generation of an action potential in the nervous system occurs due to the rapid influx of $Na^+$ ions into the axoplasm through voltage-gated sodium channels.
This influx causes depolarization of the axonal membrane.
The Reason statement is incorrect because the influx of $Na^+$ ions is driven by the electrochemical gradient,not by the efflux of $K^+$ ions.
In fact,during the depolarization phase,$K^+$ channels are typically closed.

(A) The myelin sheath acts as an electrical insulator around the axon.
Because of this insulation,ionic channels (voltage-gated $Na^+$ channels) are largely absent in the internodal regions covered by the myelin sheath.
Consequently,depolarization can only occur at the nodes of Ranvier,where these channels are highly concentrated.
This forces the action potential to 'jump' from one node to the next,a process known as saltatory conduction,which significantly increases the speed of nerve impulse transmission.

(C) The resting membrane potential is generated due to the differential permeability of the axonal membrane to $K^+$ and $Na^+$ ions and the presence of negatively charged proteins inside the axoplasm.
During the resting state,the axonal membrane is significantly more permeable to $K^+$ ions and nearly impermeable to $Na^+$ ions.
This leads to the efflux of $K^+$ ions,creating a negative charge inside the membrane.
To maintain this unequal distribution of ions,neurons utilize metabolic energy in the form of $ATP$ (chemical energy),not electrical energy,via the $Na^+-K^+$ pump.
Therefore,the Assertion is correct,but the Reason is incorrect.