How do nerve impulses travel within a nerve cell?

The membrane of all nerve cells maintains a potential difference of -70mV by pumping positive Na+ ions out of the cell. This is called the resting potential. A nerve impulse is a change in this potential - an increase up to +30mV - caused by an influx of these Na+ ions, this is called depolarisation. When a new area of the nerve cell membrane is reached by this depolarisation, if the potential change is large enough (above the threshold potential ~45mV) , it allows many voltage-gated Na+ protein channels to open. This allows a new influx of Na+ ions, which increases the potential in this region. In myelinated motor neurons, these regions are the Nodes of Ranvier which lie between Schwann cells, leading to large gaps between nodes, while in unmyelinated relay neurons, these regions may simply lie next to each other continuously.
Once a region of a nerve cell has been depolarised, it then uses Na+/K+ pumps (carrier proteins) to return to the resting potential, such that it can transmit another impulse. The time between depolarisation, and the readiness to transmit another signal is called the refractory period.

Answered by Felix W. Biology tutor

1636 Views

See similar Biology A Level tutors

Related Biology A Level answers

All answers ▸

Explain why an artery may be described as an organ.


What is meant by the term "tertiary structure" of a protein? Describe the forms of bonding that could be involved.


What is meant by the "fluid-mosaic" model of cell membrane structure?


Describe how the Meselsohn- Stahl experiment was set up. Describe how the results of the Meselsohn- Stahl experiment provide evidence for semi- conservative replication


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2024

Terms & Conditions|Privacy Policy