
During the conduction of nerve impulse the repolarization occurs with the
(a)Influx of ${ K }^{ + }$ ions
(b)Efflux of ${ K }^{ + }$ ions
(c)Efflux of ${ Na }^{ + }$ ions
(d)Influx of ${ Na }^{ + }$ ions
Answer
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Hint: Repolarization is the process by which negative resting potential is restored in the neuron. In this stage, voltage-gated sodium channels close and voltage-gated potassium channels open.
Complete answer:
At the time of conduction of nerve impulse, the repolarization occurs with the efflux of the ${ K }^{ + }$ ions. Just after the depolarization phase of an action potential which changes the membrane potential to a positive value, it changes and returns to a negative value. This is called repolarization. In this process, ${ K }^{ + }$ ions move outside and ${ Na }^{ + }$ stays inside the membrane. After the impulse is traveled, the cell membrane returns to resting potential.
Additional Information: The generation and conduction of nerve impulses are carried out in several steps or changes in the axon membrane. These are
-Polarization: It occurs at the resting stage and in this step ${ Na }^{ + }$ gets transported outside the intracellular compartment to the outside fluid.
-Depolarization: The influx of the ${ Na }^{ + }$ causes depolarization in the cell membrane and makes the interior of the membrane positive.
-Repolarization: It occurs due to the efflux of ${ K }^{ + }$ to the extracellular fluid due to K-channels.
-Hyper-polarization: The excessive efflux of the ${ K }^{ + }$ ions causes the hyperpolarization stage. It further causes a more negative membrane potential which gets stabilized by the efflux of ${ Na }^{ + }$ by the activity of the ${ Na }^{ + }$ -${ K }^{ + }$ pump.
So, the correct answer is ‘Efflux of ${ K }^{ + }$ ions’.
Note: Repolarization occurs between +30 and +40mV. At this range, the sodium gates close and potassium gates open. Before the potassium channels close completely, the membrane potential drops past the resting potential around -80 mV and hyperpolarization starts.
Complete answer:
At the time of conduction of nerve impulse, the repolarization occurs with the efflux of the ${ K }^{ + }$ ions. Just after the depolarization phase of an action potential which changes the membrane potential to a positive value, it changes and returns to a negative value. This is called repolarization. In this process, ${ K }^{ + }$ ions move outside and ${ Na }^{ + }$ stays inside the membrane. After the impulse is traveled, the cell membrane returns to resting potential.
Additional Information: The generation and conduction of nerve impulses are carried out in several steps or changes in the axon membrane. These are
-Polarization: It occurs at the resting stage and in this step ${ Na }^{ + }$ gets transported outside the intracellular compartment to the outside fluid.
-Depolarization: The influx of the ${ Na }^{ + }$ causes depolarization in the cell membrane and makes the interior of the membrane positive.
-Repolarization: It occurs due to the efflux of ${ K }^{ + }$ to the extracellular fluid due to K-channels.
-Hyper-polarization: The excessive efflux of the ${ K }^{ + }$ ions causes the hyperpolarization stage. It further causes a more negative membrane potential which gets stabilized by the efflux of ${ Na }^{ + }$ by the activity of the ${ Na }^{ + }$ -${ K }^{ + }$ pump.
So, the correct answer is ‘Efflux of ${ K }^{ + }$ ions’.
Note: Repolarization occurs between +30 and +40mV. At this range, the sodium gates close and potassium gates open. Before the potassium channels close completely, the membrane potential drops past the resting potential around -80 mV and hyperpolarization starts.
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