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

2185 Views

See similar Biology A Level tutors

Related Biology A Level answers

All answers ▸

Explain how a nerve impulse causes the chambers of the heart to contract (6).


Why does anaerobic respiration have a lower ATP yield than aerobic respiration? (4 marks)


What is the process of DNA replication?


What is the difference between DNA and RNA?


We're here to help

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