Question

explain the mechanizam transmission of nerve impulses​

Answer 1

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• The transmission of a nerve impulse along a neuron from one end to the other occurs as a result of electrical changes across the membrane of the neuron.
• It is these large, negatively charged ions that contribute to the overall negative charge on the inside of the cell membrane as compared to the outside.

Answer 2

Answer:

This the mechanism transmission of nerve impulses​

• In the resting state, the membrane of the nerve cell is more permeable to the  K+ ions and impermeable to Na+ ions. Also only 2 K+ ions move inside for 3 Na+ ions to come out of cytoplasm of axon called axoplasm.  

• The membrane of nerve cell is also impermeable to negatively charged proteins present inside the axoplasm. Thus, it does not let them go out. As a result, the axoplasm has more concentration of negatively charged proteins than positively charged  Na+and K+. This makes the over all charge inside the axoplasm negative and outside the membrane becomes positive due to more number of Na+.  Hence, the outer surface of the membrane possesses positive charge and the inner surface possesses negative charge. This  potential difference is called Resting Potential.

• When stimulus is applied at a particular site (say site A), of the nerve cell that site of the membrane becomes freely permeable to the influx of Na+. This causes increase in the overall positive charge inside the axoplasm and thus polarity is reversed.  

• The membrane is now depolarized, and the potential difference across the site is called Action Potential (nerve impulse).

• At a site (B) that is adjacent to the depolarized site (A), the outer surface of the membrane has positive charge and the inner surface has negative charge.

• Depolarized site (A): Positive (in) and Negative (out)

• Adjacent site (B): Positive (out) and Negative (in)

• Due to the potential difference between two points (point A and point B) inside as well as outside, current starts flowing on both sides.

• Current flow on the inner surface is from site A to B

• Current flow on the outer surface is from site B to A

• This completes the circuit of current flow. Due to this circuit, the impulse travels to point B and polarity of site B is reversed and action potential is generated at site B in the same way as at point A. Hence, the impulse travels to site B, and we say that this impulse is conducted. This sequence is repeated along the length of the axon.

• Stimulus-induced permeability to Na+ is short-lived, and is quickly followed by permeability to K+. When this happens, the resting potential is restored, and once more, the site can be stimulated.

Explanation: