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Question

Which option is correct for the correctly matched groups for the column I, column II and column III
Column IColumn IIColumn IIIA. Resting membrane potential(i) NA+channel get open(p) Na+ and K+ pumps are responsible for itB. Action potential(ii) Na+channel is closed(q) Last for very short timeC. Depolarisation(iii) Na+ions are more on outer side of membrane(r) K+ions move on outersideD. Repolarisation(iv) Na+ions are more on inner side of membrane(s) Positive charge on innerside of membrane

A
(A - iii - p) (B - iv - q) (C - i - s) (D - ii - r)
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B
(A - ii - s) (B - i - r) (C - iii - p) (D - iv - q)
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C
(A - iv - q) (B - iii - p) (C - i - p) (D - i - s)
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D
(A - iv - p) (B - iii - q) (C - ii - r) (D - i - r)
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Solution

The correct option is A (A - iii - p) (B - iv - q) (C - i - s) (D - ii - r)
The generation of a nerve impulse is the temporary removal of the resting potential in the neuron.
The correct matchings are
(i) Resting membrane potential: It is the state of electrical potential difference across the resting plasma membrane.
The state of resting membrane is called polarised state which occurs due to Na+ and K+ pumps in which (Na^+\) is more on the outer side of the membrane.
(ii) Action potential: It is the state which lasts for a very short period of time. Na+ ions get greatly increased on the inner side of the membrane at the point of stimulation when the stimulus of adequate strength is applied to a polarised membrane.
(iii) Depolarization: It is a positive charge in a cell membrane making it more positive or less negative. In this Na+ ion channel opens due to which more positive charge gets accumulated on the inner side of membrane.
(iv) Repolarisation: It is a state of re - establishment of polarity, where resting potential is restored at the site of excitation. It is followed by a rise of impermeability to K+, i.e. within a fraction of a second, Na+ influx stops and K+ out flow begins (K+ ions moves on the outer side).

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