Generation and conduction of Nerve Impulse Questions in English

(A) The electrical potential difference across the plasma membrane of a resting neuron is known as the resting membrane potential.
In a resting state,the axonal membrane is more permeable to potassium ions $(K^+)$ and nearly impermeable to sodium ions $(Na^+)$.
Consequently,the interior of the axon contains a high concentration of $K^+$ and negatively charged proteins,while the exterior contains a high concentration of $Na^+$.
This ionic gradient,maintained by the $Na^+-K^+$ pump,results in the inner surface of the membrane being negatively charged relative to the outer surface,typically measured at approximately $-70 \ mV$.

(B) polarized nerve fiber refers to the state where the resting membrane potential is maintained across the axonal membrane.
In this state,the outer surface of the axonal membrane is positively charged,while the inner surface is negatively charged.
This condition is known as the resting state,where the neuron is not conducting any impulse.
Therefore,the correct answer is the resting membrane potential.

(D) The action potential in a neuron is generated by the rapid influx of $Na^{+}$ ions into the cell.
When a stimulus is applied to a neuron,the voltage-gated $Na^{+}$ channels open,allowing $Na^{+}$ ions to rush into the intracellular space.
This influx causes depolarization of the neuronal membrane,shifting the membrane potential from its resting state (approximately $-70 \ mV$) to a positive value (approximately $+30 \ mV$ to $+45 \ mV$).
Therefore,the correct answer is $Na^{+}$.

(B) In a resting state,the axonal membrane is significantly more permeable to potassium ions $(K^{+})$ and nearly impermeable to sodium ions $(Na^{+})$.
Additionally,the membrane is impermeable to the negatively charged proteins present in the axoplasm.
This differential permeability,maintained by the sodium-potassium pump,is responsible for the resting membrane potential.

(A) In a resting state,the axonal membrane is more permeable to $K^+$ ions and nearly impermeable to $Na^+$ ions.
Due to the action of the $Na^+-K^+$ pump,which transports $3$ $Na^+$ ions outwards for every $2$ $K^+$ ions inwards,the axoplasm contains a high concentration of $K^+$ and negatively charged proteins,and a low concentration of $Na^+$.
This ionic gradient is responsible for maintaining the resting membrane potential.

(C) In a resting state,the axonal membrane is more permeable to $K^+$ ions and nearly impermeable to $Na^+$ ions.
Due to the action of the $Na^+-K^+$ pump,which actively transports $3$ $Na^+$ ions outwards for every $2$ $K^+$ ions brought into the cell,the extracellular fluid (outside the axon) maintains a high concentration of $Na^+$ ions and a low concentration of $K^+$ ions.
Conversely,the axoplasm (inside the axon) has a high concentration of $K^+$ ions and a low concentration of $Na^+$ ions.
Therefore,the correct option is $C$.

(C) The diagram represents the depolarization of a nerve fiber membrane.
At the site marked $P$,the membrane potential changes from negative to positive inside due to the rapid influx of sodium ions $(Na^{+})$ through voltage-gated sodium channels.
This influx of $Na^{+}$ ions causes the depolarization of the axonal membrane,which is the initial step in the generation of an action potential.

(A) The sodium-potassium pump is a membrane protein that maintains the resting membrane potential of a neuron.
It functions via active transport,which requires energy in the form of $ATP$.
For every cycle of the pump,$3 Na^+$ ions are transported out of the cell,and $2 K^+$ ions are transported into the cell against their respective concentration gradients.
Therefore,option $A$ is the correct description.

(B) nerve impulse is essentially an electrical signal that travels along the axon of a neuron.
This electrical signal is technically known as an $Action \ potential$.
When a neuron is at rest,it maintains a resting membrane potential,but the actual impulse that propagates information is the rapid change in this potential,which is the action potential.
Therefore,the correct answer is $Action \ potential$.

(A) resting nerve membrane is said to be in a polarized state. In this state,the outer surface of the axonal membrane possesses a positive charge,while the inner surface possesses a negative charge. This is due to the differential distribution of ions,specifically $Na^+$ and $K^+$,across the membrane,maintained by the $Na^+-K^+$ pump. Option $A$ correctly depicts this arrangement.

(B) When a neuron is at rest,the axonal membrane is more permeable to $K^{+}$ ions and nearly impermeable to $Na^{+}$ ions. This maintains a negative charge inside the membrane.
When a stimulus is applied at a site on the membrane,the permeability to $Na^{+}$ ions increases rapidly.
This leads to a rapid influx of $Na^{+}$ ions into the cell,causing the inner surface of the membrane to become positively charged relative to the outer surface.
This reversal of polarity is known as depolarization,which generates an action potential.

(A) In a resting neuron,the axonal membrane is significantly more permeable to potassium ions $(K^+)$ than to sodium ions $(Na^+)$.
This high permeability to $K^+$ is due to the presence of numerous 'leak' channels for potassium in the axonal membrane.
Conversely,the membrane is nearly impermeable to $Na^+$ ions during the resting state.
This selective permeability,combined with the action of the $Na^+-K^+$ pump,helps maintain the resting membrane potential (approximately $-70 \ mV$).

(B) The correct answer is $(B)$.
During the resting stage,the axonal membrane is comparatively more permeable to $K^+$ ions and nearly impermeable to $Na^+$ ions.
Furthermore,the membrane is nearly impermeable to the negatively charged proteins present in the axoplasm.
This differential permeability,along with the action of the $Na^+-K^+$ pump,maintains the resting membrane potential.

(A) During the resting state of a neuron,the axonal membrane is in a polarized state.
In this state,the membrane is significantly more permeable to potassium ions $(K^+)$ due to the presence of leak channels.
Conversely,the membrane is nearly impermeable to sodium ions $(Na^+)$ because the voltage-gated sodium channels are closed.
This differential permeability,along with the action of the $Na^+-K^+$ pump,maintains the resting membrane potential.

(A) In a resting nerve fibre,the axonal membrane is more permeable to $K^{+}$ ions and nearly impermeable to $Na^{+}$ ions.
Consequently,the concentration of $K^{+}$ ions is high inside the axoplasm,while the concentration of $Na^{+}$ ions is high in the extracellular fluid (outside the membrane).
This differential distribution of ions,along with the presence of negatively charged proteins and nucleic acids inside the cell,maintains the resting membrane potential,where the outer surface is positively charged and the inner surface is negatively charged.
Therefore,the correct distribution is high $Na^{+}$ outside and high $K^{+}$ inside the axon.

(A) The correct sequence of events during the conduction of a nerve impulse is as follows:
$1$. Initially,the neuron is in a resting state where the intracellular fluid is electronegative with a potential difference of $-70 \ mV$ $(iii)$.
$2$. $A$ stimulus causes a disturbance to this resting potential $(v)$.
$3$. This leads to the rapid influx of $Na^{+}$ ions inside the axon,causing depolarization $(i)$.
$4$. As a result of the influx of positive charges,the extracellular fluid becomes relatively electronegative compared to the interior $(iv)$.
$5$. Finally,the $Na^{+}$ gates close and $K^{+}$ gates open to repolarize the membrane after the refractory period $(ii)$.
Thus,the correct sequence is $iii \rightarrow v \rightarrow i \rightarrow iv \rightarrow ii$. However,looking at the provided options,the most logical progression starting from the resting state is $iii \rightarrow v \rightarrow i \rightarrow iv \rightarrow ii$. Since this specific sequence is not listed,we re-evaluate the options. Option $A$ represents the cycle of repolarization and subsequent stimulation. Given the standard progression of an action potential,the correct sequence is $iii \rightarrow v \rightarrow i \rightarrow iv \rightarrow ii$. If we must choose from the provided options,there appears to be a mismatch in the starting point,but $A$ represents the cyclic nature of the process.

(C) 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 followed by the reversal of the polarity at that site,which is called depolarization.
During this phase,the $Na^{+}$ voltage-gated channels open,while the $K^{+}$ voltage-gated channels remain closed.

(D) Statement $i$ is correct: The resting membrane potential is typically $-70 \text{ mV}$.
Statement $ii$ is incorrect: Voltage-gated $Na^{+}$ and $K^{+}$ channels operate independently (sequentially) and are self-closing.
Statement $iii$ is correct: During the peak of an action potential,the membrane potential reaches $+30$ to $+60 \text{ mV}$.
Statement $iv$ is incorrect: In medullated nerve fibres,the action potential undergoes saltatory conduction,jumping from one node of Ranvier to the next,rather than traveling as a continuous wave of depolarization.
Statement $v$ is correct: The resting potential is maintained by the selective permeability of the membrane and the closure of voltage-gated $Na^{+}$ and $K^{+}$ channels.
Therefore,statements $i, iii,$ and $v$ are correct.

(D) When a stimulus is applied to the membrane of a neuron,the permeability of the membrane to $Na^{+}$ ions increases significantly.
This leads to a rapid influx of $Na^{+}$ ions into the intracellular space from the extracellular fluid.
This process is known as depolarization,which is the initial phase of an action potential.
Therefore,the correct answer is $Na^{+}$.

(C) The resting membrane potential of a neuron is maintained by the differential permeability of the membrane to ions.
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 rapid influx of $Na^+$ ions into the intracellular space,which causes the reversal of the polarity at that site,a process known as depolarization.
Therefore,the correct answer is $Sodium$ $(Na^+)$ ions.