Generation and conduction of Nerve Impulse Questions in English

(B) Chronaxie is a physiological term used in neurophysiology to describe the excitability of a tissue (like nerve or muscle).
It is defined as the minimum time required to excite a tissue using a current that is twice the strength of the rheobase.
Rheobase is the minimum current intensity required to produce an action potential when the duration of the stimulus is infinite.
Therefore,chronaxie provides a measure of the excitability of the tissue; a shorter chronaxie indicates higher excitability.

(D) The reversal potential,also known as the equilibrium potential,is the membrane potential at which there is no net flow of a particular ion across the membrane.
For a single ion species,the reversal potential is determined by the Nernst equation.
Depending on the concentration gradient and the charge of the ion,the reversal potential can be positive (e.g.,for $Na^+$) or negative (e.g.,for $K^+$).
Therefore,the reversal potential is not fixed as always positive,negative,or neutral; it depends on the specific ion and its electrochemical gradient.

(A) For a stimulus to be effective in initiating a nerve impulse,it must reach a minimum intensity known as the $threshold$ $stimulus$.
If the stimulus intensity is less than the $threshold$ $value$,it fails to trigger the opening of voltage-gated $Na^+$ channels,and thus,no action potential is generated.
Therefore,the value of the potential must not be less than the $threshold$ $value$ to start the conduction of an impulse.

(B) The velocity of nerve impulse conduction is directly proportional to the diameter of the nerve fibers.
As the diameter of the nerve fiber increases,the internal resistance to the flow of ions decreases,which allows the action potential to propagate more rapidly along the axon.
Therefore,thick nerve fibers conduct impulses faster than thin nerve fibers.

(C) The conduction of a nerve impulse along a nerve fibre is initiated by the process of depolarisation.
When a stimulus is applied to the nerve fibre,the membrane permeability for $Na^+$ ions increases significantly.
This leads to a rapid influx of $Na^+$ ions into the intracellular space,causing the membrane potential to shift from a resting state (negative) to a depolarised state (positive).
Therefore,the sudden increase in permeability to $Na^+$ ions is essential for the generation and propagation of an action potential.

(D) The propagation of action potential is significantly faster in myelinated nerve fibres compared to non-myelinated ones.
This is due to a process called saltatory conduction,where the impulse 'jumps' from one node of Ranvier to the next.
Additionally,a larger fibre diameter reduces internal resistance,allowing for faster signal transmission.
Therefore,both a large fibre diameter and the presence of a myelin sheath contribute to rapid conduction.

(B) In a resting state,the axonal membrane is polarized,meaning the inside is negatively charged relative to the outside.
When a stimulus is applied,the permeability of the membrane to $Na^+$ ions increases,causing an influx of $Na^+$.
This leads to depolarization,where the inside of the membrane becomes positively charged.
Following this,the membrane permeability to $K^+$ increases,and $K^+$ ions move out,leading to repolarization,where the membrane potential returns to its original negative state.

(A) During the transmission of a nerve impulse (depolarization),the permeability of the axonal membrane to $Na^+$ ions increases significantly.
This leads to a rapid influx of $Na^+$ ions into the intracellular space.
Simultaneously,the permeability to $K^+$ ions decreases,but as the action potential progresses toward repolarization,$K^+$ ions flow out of the cell.
Therefore,the primary event during the initiation of the impulse is the inward flux of $Na^+$ and the outward flux of $K^+$.

(D) The conduction of a nerve impulse along an axon is associated with the generation and propagation of an action potential.
An action potential is a rapid sequence of changes in the voltage across the membrane,characterized by a wave of depolarization that travels down the nerve fibre.
This process is often compared to a spark moving along a fuse,where the depolarization at one point triggers the depolarization of the adjacent segment of the axon membrane.

(A) The conduction of a nerve impulse is primarily driven by changes in the permeability of the axonal membrane to specific ions.
When a stimulus is applied,the membrane permeability to $Na^+$ ions increases significantly,leading to an influx of $Na^+$ into the cell (depolarization).
Subsequently,the permeability to $K^+$ ions increases,allowing $K^+$ to flow out of the cell (repolarization).
These rapid changes in membrane permeability create the action potential necessary for impulse conduction.

(B) During the resting state,the axonal membrane is polarized,meaning it is positively charged on the outside and negatively charged on the inside.
When a stimulus is applied,the permeability of the membrane to $Na^+$ ions increases,leading to a rapid influx of $Na^+$ ions.
This process is called depolarization.
During depolarization,the polarity of the membrane reverses,making the inside of the cell positively charged and the outside negatively charged.

(C) The generation of an action potential in a neuron is primarily driven by the rapid influx of $Na^+$ ions into the cell.
When a stimulus reaches the threshold level,voltage-gated $Na^+$ channels open,allowing $Na^+$ ions to rush into the axon down their electrochemical gradient.
This rapid influx causes depolarization,which determines the rate of rise (slope) and the peak amplitude of the action potential.
Therefore,the concentration of $Na^+$ ions in the extracellular fluid is the primary determinant of these parameters.

(A) The resting membrane potential is the electrical potential difference across the plasma membrane of a resting neuron.
It is maintained by the unequal distribution of ions,primarily $Na^+$ and $K^+$,and the activity of the $Na^+-K^+$ pump.
The resting membrane potential typically ranges from $-40\, mV$ to $-90\, mV$.
The average resting membrane potential value is approximately $-70\, mV$,which falls within the range provided in option $A$.

(D) The speed of nerve impulse conduction depends on the diameter of the nerve fiber and the presence of a myelin sheath.
In mammals,myelinated motor nerve fibers conduct impulses at a high velocity,typically ranging from $60 \, m/s$ to $120 \, m/s$.
Among the given options,$100 \, m/s$ falls within this physiological range.
Therefore,the correct option is $D$.

(A) The resting membrane potential of a nerve cell is typically $-70 \ mV$. When a stimulus is applied,the membrane depolarizes,and the potential rises to reach a peak value of approximately $+30 \ mV$ to $+45 \ mV$ during the action potential. Among the given options,$+45 \ mV$ is the standard value representing the peak of the action potential.

(B) The electrical potential difference across the resting plasma membrane of a neuron is called the resting potential.
In a resting state,the axonal membrane is more permeable to $K^+$ ions and nearly impermeable to $Na^+$ ions.
Consequently,the inside of the axon remains negatively charged relative to the outside,typically measured at approximately $-70 \ mV$.

(D) During the depolarization phase of a nerve impulse,the membrane potential becomes positive inside due to the rapid influx of $Na^+$ ions.
However,the question asks what maintains the depolarized state or the ionic balance during the action potential cycle.
Actually,the influx of $Na^+$ causes depolarization,while the efflux of $K^+$ ions is responsible for repolarization.
In the context of standard multiple-choice questions regarding nerve impulse conduction,the movement of $K^+$ ions is the primary factor regulating the membrane potential after the initial spike.
Thus,the correct option is $D$.

(B) In a resting state,the axonal membrane is polarized,meaning the inside is negative relative to the outside. When a stimulus is applied,the membrane becomes permeable to $Na^+$ ions,causing depolarization where the inside becomes positive. This is followed by repolarization,where the membrane becomes permeable to $K^+$ ions,causing the inside to return to its original negative state. Therefore,the charge changes from negative to positive and back to negative.

(B) In a resting state,the axonal membrane is polarized,meaning the inside is negatively charged relative to the outside. When a stimulus is applied to a nerve fibre,the permeability of the membrane to $Na^+$ ions increases significantly. This leads to a rapid influx of $Na^+$ ions into the cell. As a result,the inside of the membrane shifts from a negative potential to a positive potential. This change in electrical potential is known as depolarization. Therefore,upon stimulation,the inside of the membrane becomes positively charged.

(B) The transmission of a nerve impulse along a neuron is primarily dependent on the movement of ions across the neuronal membrane.
Specifically,the resting membrane potential and the generation of an action potential rely heavily on the influx of $Na^+$ (sodium) ions into the cell.
Therefore,sodium is the essential mineral required for nervous conduction.

(A) The resting membrane potential of a neuron is primarily maintained by the differential permeability of the axonal membrane to ions.
Specifically,the membrane is much more permeable to potassium ions $(K^+)$ than to sodium ions $(Na^+)$.
Due to the action of the $Na^+/K^+$ pump,$K^+$ ions diffuse out of the cell down their concentration gradient,leaving behind negatively charged proteins and organic anions inside the cell.
This efflux of $K^+$ ions creates an electrical charge difference across the membrane,making the inside of the neuron electrically negative relative to the outside.

(A) The speed of a nerve impulse is significantly higher in medullated (myelinated) nerve fibers compared to non-medullated (unmyelinated) nerve fibers.
In medullated nerve fibers,the impulse travels via saltatory conduction,where the action potential jumps from one node of Ranvier to the next.
This process makes the transmission of the nerve impulse approximately $20$ times faster than in non-medullated nerve fibers,where the impulse must travel along the entire length of the axon membrane.

(D) The initiation of a nerve impulse occurs when a stimulus is applied to a neuron. This stimulus causes the opening of voltage-gated $Na^+$ channels. Consequently,$Na^+$ ions rush into the axoplasm from the extracellular fluid. This rapid influx of $Na^+$ ions causes the membrane potential to change from a resting state (negative) to a depolarized state (positive),which is the fundamental step in the generation of an action potential.

(D) The generation and conduction of a nerve impulse involve a sequence of electrochemical changes across the neuronal membrane.
$1$. When a stimulus is received,the membrane undergoes $Depolarisation$ due to the influx of $Na^+$ ions.
$2$. This is immediately followed by $Repolarisation$ due to the efflux of $K^+$ ions to restore the resting membrane potential.
$3$. Therefore,a nerve impulse is a self-propagating wave consisting of both $Depolarisation$ and $Repolarisation$ processes.

(D) In myelinated nerve fibres,the action potential jumps from one node of $Ranvier$ to the next. This type of nerve impulse transmission is known as saltatory conduction. This process significantly increases the speed of nerve impulse transmission compared to unmyelinated fibres.

(B) The conduction of a nerve impulse is an electrochemical process.
It involves the movement of ions (like $Na^+$ and $K^+$) across the neuronal membrane,which creates an electrical potential difference (electrical component).
Additionally,this process is regulated by chemical changes and the release of neurotransmitters at the synapse (chemical component).
Therefore,it is classified as an electrochemical phenomenon.

(C) The conduction of a nerve impulse involves the movement of ions across the neuronal membrane, which requires energy to maintain the resting membrane potential and to restore ion gradients after an action potential.
This energy is provided by the hydrolysis of $ATP$ $(Adenosine Triphosphate)$, which is the primary energy currency of the cell.
$ATP$ powers the $Na^+/K^+$-ATPase pump, which actively transports $Na^+$ ions out of the cell and $K^+$ ions into the cell, thereby maintaining the electrochemical gradient necessary for nerve impulse conduction.

(B) When a neuron is not conducting any impulse,the axonal membrane is comparatively more permeable to $K^+$ ions and nearly impermeable to $Na^+$ ions.
Consequently,the outer surface of the axonal membrane possesses a positive charge,while its inner surface becomes negatively charged.
This electrical potential difference across the resting plasma membrane is called the resting potential.

(B) The resting membrane potential of a neuron is maintained by the selective permeability of the membrane to $K^+$ ions and the action of the $Na^+-K^+$ pump.
When a stimulus is applied,the permeability of the axonal membrane to $Na^+$ ions increases significantly.
This leads to a rapid influx of $Na^+$ ions into the axon.
As a result,the inner side of the membrane becomes positively charged relative to the outer side,which is known as depolarization or the generation of an action potential.
Therefore,the reversal of polarity occurs because more $Na^+$ ions enter the axon compared to $K^+$ ions.

(B) The refractory period is a short interval of time following an action potential during which the nerve fiber is unresponsive to a new stimulus.
During the absolute refractory period,the voltage-gated $Na^+$ channels are inactivated and cannot be reopened,meaning the neuron cannot generate another action potential regardless of the stimulus strength.
Therefore,the nerve fiber does not conduct any impulse during this phase.

(A) The action potential in a neuron is primarily generated by the rapid influx of sodium ions $(Na^+)$ into the cell.
When a stimulus reaches the threshold level,voltage-gated $Na^+$ channels open,allowing $Na^+$ to rush into the neuron down its electrochemical gradient.
This influx causes the membrane potential to change from its resting state (negative) to a positive value,a process known as depolarization.
Therefore,the correct answer is $Na^+$.

(B) The speed of nerve impulse conduction in mammalian nerve fibers varies depending on the diameter of the axon and the presence of a myelin sheath.
In myelinated nerve fibers,the conduction velocity can reach up to $100-120 \ m/s$.
Therefore,among the given options,$100 \ m/s$ is the most appropriate representation for the speed of nerve impulse conduction in mammalian nerves.

(B) Statement $(a)$ is incorrect because the electrical potential difference across the resting plasma membrane is called the 'resting potential',not 'action potential'.
Statement $(b)$ is correct because the rapid influx of $Na^+$ ions into the neuron leads to depolarization,which generates the action potential.
Statement $(c)$ is correct because a nerve impulse is defined as the action potential that travels along the axon.
Therefore,statements $(b)$ and $(c)$ are correct.

(A) During the repolarization phase of an action potential in a nerve fiber,the following events occur:
$1$. The voltage-gated $Na^+$ channels close,preventing further influx of sodium ions into the cell.
$2$. The voltage-gated $K^+$ channels open,allowing potassium ions to flow out of the cell.
This efflux of $K^+$ ions restores the negative resting membrane potential inside the cell.
Therefore,the correct events are $b$ ($Na^+$ channels close) and $d$ ($K^+$ channels open).

(B) When a stimulus is applied to a nerve fiber,the membrane potential changes.
Depolarization occurs due to the rapid influx of $Na^+$ ions into the axoplasm through voltage-gated $Na^+$ channels.
This sudden increase in the permeability of the membrane to $Na^+$ ions causes the inside of the membrane to become positively charged relative to the outside,leading to the generation of an action potential.

(B) In the resting state,the axonal membrane is comparatively more permeable to $K^+$ ions and nearly impermeable to $Na^+$ ions.
Due to the concentration gradient,$K^+$ ions move out of the cell through the leakage channels,while the membrane remains relatively impermeable to $Na^+$ ions.
This selective permeability,along with the action of the $Na^+-K^+$ pump,maintains the resting membrane potential.

(C) When a nerve fiber is at rest,the axonal membrane is more permeable to $K^+$ ions and nearly impermeable to $Na^+$ ions. This maintains a resting potential where the inner surface is negatively charged.
When a stimulus is applied,the membrane becomes freely permeable to $Na^+$ ions,causing a rapid influx of $Na^+$ into the cell.
This sudden influx of $Na^+$ ions reverses the polarity of the membrane,making the inner surface positively charged relative to the outer surface. This state is known as depolarization.

(B) The generation of a nerve impulse (action potential) is initiated by the rapid influx of sodium ions $(Na^+)$ into the neuron.
When a stimulus is applied to a site on a polarized membrane,the permeability of the membrane to $Na^+$ ions increases significantly.
This leads to a rapid influx of $Na^+$ ions,which causes the reversal of the polarity at that site,known as depolarization.
This change in membrane potential is the fundamental step in generating a nerve impulse.

(B) Saltatory conduction is the propagation of action potentials along myelinated axons from one node of Ranvier to the next node,increasing the conduction velocity of action potentials.
In myelinated nerve fibers,the myelin sheath acts as an electrical insulator.
Because the myelin sheath prevents ion flow across the membrane,the action potential can only occur at the nodes of Ranvier,where the membrane is exposed.
This 'jumping' of the impulse from node to node is known as saltatory conduction.

(B) 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 to $Na^+$ ions increases rapidly.
This leads to a rapid influx of $Na^+$ ions from the extracellular fluid into the intracellular fluid,causing a reversal of polarity at that site,which is known as depolarization or the action potential.
Therefore,the action potential results from the movement of $Na^+$ from the extracellular fluid into the intracellular fluid.

(D) The correct answer is $D$.
In a resting state,the axonal membrane is selectively permeable.
It is highly permeable to $K^+$ and nearly impermeable to $Na^+$.
Additionally,the membrane is impermeable to negatively charged proteins present in the axoplasm.
Therefore,the statement that the membrane is highly permeable to negatively charged proteins is incorrect.

(A) During the depolarization phase of an action potential,the voltage-gated $Na^+$ channels open rapidly,leading to a massive influx of $Na^+$ ions into the neuron. This causes the membrane permeability for $Na^+$ to increase significantly,far exceeding that of $K^+$. Therefore,$Na^+$ has the highest permeability during the generation of the nerve impulse.

(A) In the resting state of a neuron,the axonal membrane is significantly more permeable to $K^+$ ions and nearly impermeable to $Na^+$ ions.
Due to the concentration gradient,$K^+$ ions diffuse out of the cell through the leakage channels.
This movement of $K^+$ ions out of the cell contributes to the maintenance of the resting membrane potential.

(C) When a stimulus is applied to a nerve fiber,the permeability of the axonal membrane to $Na^+$ ions increases significantly.
This leads to a rapid influx of $Na^+$ ions into the intracellular space,causing depolarization of the membrane.
This change in membrane potential results in the generation of an action potential or nerve impulse.
Therefore,the membrane becomes more permeable to $Na^+$ ions during the initiation of a nerve impulse.

(D) $1$. At the resting stage,the inner surface of the axonal membrane is negatively charged due to the presence of $K^+$ ions and negatively charged proteins.
$2$. When a stimulus is applied,the membrane becomes permeable to $Na^+$ ions,causing an influx of $Na^+$. This leads to depolarization,where the inner surface becomes positively charged.
$3$. Subsequently,the membrane permeability to $Na^+$ decreases and $K^+$ permeability increases,leading to repolarization where the inner surface returns to a negatively charged state.
$4$. Thus,the sequence of charges on the inner surface is negative (resting),positive (depolarized),and negative (repolarized).

(C) The resting membrane potential is the electrical potential difference across the plasma membrane of a resting neuron.
In a typical mammalian neuron,the resting membrane potential is approximately $-70 \, mV$.
This state is maintained by the unequal distribution of ions ($Na^+$ and $K^+$) across the membrane,primarily due to the activity of the $Na^+-K^+$ pump and the differential permeability of the membrane to these ions.

(A) In myelinated nerve fibers,the myelin sheath acts as an insulator,preventing ion flow across the membrane.
As a result,the action potential cannot travel continuously along the entire length of the axon.
Instead,the action potential 'jumps' from one Node of Ranvier to the next.
This type of rapid,saltatory (jumping) conduction is known as Saltatory conduction,which significantly increases the speed of nerve impulse transmission.

(D) $1$. At the resting state,the axonal membrane is polarized,meaning the inner side is negatively charged relative to the outer side.
$2$. 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.
$3$. Following this,the membrane becomes permeable to $K^+$ ions,which move out,leading to repolarization,where the inner side becomes negatively charged again.
$4$. Thus,the sequence of charge on the inner side is negative (resting),positive (depolarized),and negative (repolarized).

(B) During the repolarization phase of an action potential,the membrane potential returns to its resting state.
This process is triggered by the inactivation of voltage-gated $Na^+$ channels,which stops the influx of $Na^+$ ions.
Simultaneously,voltage-gated $K^+$ channels open,allowing $K^+$ ions to diffuse out of the neuron.
This efflux of positive $K^+$ ions restores the negative charge inside the membrane relative to the outside,thus completing repolarization.

(B) The correct answer is $B$. Neurons follow the 'All-or-None' law. This means that if a stimulus is strong enough to reach the threshold,an action potential is generated. The action potential does not vary in magnitude regardless of the strength of the stimulus. Therefore,neurons do not exhibit 'Graded potentials' in the same way that receptor potentials do; instead,they fire action potentials of a constant amplitude once the threshold is crossed.