Q.Explain the process of conduction of nerve impulses upto development of action potential.

Conduction of a nerve impulse runs through a fixed sequence of membrane events.

Step 1. Resting potential. At rest, the membrane is polarised: Na+ predominates outside, K+ and negatively charged proteins predominate inside, actively maintained by the Na+-K+ pump (3 Na+ out for every 2 K+ in) against their gradients, producing a steady resting potential of about -70 mV.

Step 2. Stimulus and depolarisation. A stimulus above threshold opens the voltage-gated Na+ channels (K+ channels stay shut), so Na+ rushes in; the membrane potential rises rapidly from -70 mV, reversing polarity.

Step 3. Action potential. The membrane potential peaks at about +30 to +60 mV, with the axoplasm now electropositive relative to the extracellular fluid -- this peak is the action potential, and this rising local depolarisation immediately triggers the same event in the adjoining patch of membrane, propagating the impulse forward.

Step 4. Repolarisation. At the peak, Na+ gates close and K+ gates open; K+ rushes out (faster than Na+ had entered), and the Na+-K+ pump becomes active again, restoring the resting ion distribution and the -70 mV resting potential, after a brief refractory period.

Step 5. Propagation. This cycle of depolarisation-then-repolarisation repeats continuously along the axon; in medullated (myelinated) fibres, the impulse jumps from node of Ranvier to node of Ranvier (saltatory conduction, ~120 m/s), far faster than the continuous conduction (no node-jumping) seen in non-medullated fibres.

Present the sequence as a chain: resting potential -> stimulus -> depolarisation -> action potential -> repolarisation -> propagation, naming the ion movement responsible for each step.

• Reversing which ion moves in depolarisation (Na+ in) versus repolarisation (K+ out).
• Forgetting to explain why medullated fibres conduct faster (saltatory conduction, jumping node to node) rather than just stating that they are faster.