Transmission of nerve impulse

Transmission of nerve impulse (Transmission of nerve impulse.com)

               

 Resting potential

1. When at rest the inside of the membrane has a –ve electrical potential compared to the outside.
2.  The difference in potential is called the resting potential and is typically about between -70 mv to -80 mv (-90mv)
3. In this resting state the axon is said to be polarized.
4.  This is maintained because the neuron has an internal composition which is different to the outside.
5. Na+ and k+ are transported across the membrane against their concentration gradients by active transport.
6. Carrier proteins pick up Na+ ions and transport then to outside.
7. At the same time k+ ions are transported into the axon.
8. This is known as Na-k pump and relies on ATP from respiration.
9. Inside the axon there are large numbers of –ve charged organic ions which cannot move out of axon.
10. The Na+ ions are passed out faster than the k+ ions are brought in.
11. Approx. 3 Na+ ions leave for every two K+ ions that enter.
12. K+ ions can also diffuse back out quicker than Na+ ions can diffuse back in.
13. Net result is that the outside of membrane is +ve compared to the inside.

The action potential (depolarization)

1. A nerve impulse can be initiated by mechanical, chemical, thermal or electrical stimulation.
2.  When axon is stimulated the resting potential charges.
3. It chare from -70mv inside the membrane to +40mv.
4. For a very brief period the inside of the axon becomes +ve and the outside negative.
5. The change in potential is called action potential and about 3 millisecond.
6. When an action potential occurs the axon is said to be depolarized.
7. When the resting potential is re-established the axon membrane is said to be     repolarised.
8. When the membrane depolarizes charges occur in the membrane to the permeability of both Na+ and K+ ions.
9.  When the axon is stimulated channels opens on its cell surface which allows Na+ ions to pass through.
10. Na+ ions flood in by diffusion.
11. The Na +ions create a +ve charge of +40mv inside the membrane reversing the resting potential and casting the action potential.

          Repolarization

1. K-channels open in the membrane and K+ ions diffuse out along their concentration gradient.This start of repolarization.
2.  At the same time sodium channels in the membrane close preventing any further influx of Na + ions.The resting potential is re-established as the outside of the membrane becomes positive again compared to the inside
3. .So many k+ ions leave that the charge becomes more negative that it was originally.
4. The potassium channels close and the Na-K pump start again.
5. Normal concentrations of Na and K ions are re-established.
6. The membrane is once again at its resting potential.

                Direction of impulse

1. In the resting axon, there is a high conc. Of Na+ ions outside and a high conc. Of K+ ions inside. But the net effect is that the outside is positive compared to the inside give the resting potential.
2.  The axon is stimulated producing an action potential setting up local circuits on the axon membrane.
3. Na+ ions rush into the axon a diffusion gradient depolarising the membrane causing an action potential.
4. As the action potential passes along the axon K+ ions diffuse along  a concentration gradient starting off the process of repolarization.
5. Na-K pump is re-established fully repolarization the membrane.