GABA(A) Receptors: Hyperpolarization & Depolarization

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Hyperpolarization of cells leads to a depolarization towards the threshold, which is crucial for action potential generation. When a cell is hyperpolarized, voltage-gated sodium (Na+) channels, which are typically inactivated at resting membrane potential, become available for activation. During depolarization, these channels open, allowing Na+ to flow into the cell. However, they quickly inactivate through a mechanism known as the ball and chain, preventing further ion flow. For these channels to return to a state where they can open again, the cell must undergo hyperpolarization, which effectively resets the inactivation state. Thus, hyperpolarization plays a key role in preparing the cell for subsequent depolarization and action potential firing. The discussion also touches on the role of GABA receptors, which are involved in the hyperpolarization process, but further context on their specific function was not provided.
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In the lecture handout it says "When cells are hyperpolarised the result is a depolarisation towards threshold." What does this mean?:confused:
 
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Not quite sure, can you give us some more context from your notes - where do the GABA receptors fit in?
 
garytse86 said:
In the lecture handout it says "When cells are hyperpolarised the result is a depolarisation towards threshold." What does this mean?:confused:


A voltage gated Na+ channel will transiently open upon depolarization and a rush of Na+ will come into the cell. The channel will then inactivate via the ball and and chain mechanism, (basically the channel is plugged is its in the open conformation but plugged up so no ions can flow through.) In order for the inactivation to go away there must be a decrease in voltage or hyperpolarization. So it turns that channel opening and closing is basically a probability game. At the resting membrane potential some of the Na channels have the inactivation turned on, however upon hyperpolarization these channels will have the inactivation turned off. Now when you put the cell back to the resting potential some of the channels will open and Na will rush in.
 

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