![]() The post-synaptic cell will only send the message along if it gets enough excitatory input to depolarize across the threshold, and open the voltage-gated ion channels in its axon. At the peak action potential, K + channels open and the cell becomes (c) hyperpolarized. A nerve impulse causes Na + to enter the cell, resulting in (b) depolarization. There might even be competition among the neurons, with a single post-synaptic neuron receiving glutamate from one pre-synaptic neuron and GABA from its neighbor. Figure 42.2.2: The (a) resting membrane potential is a result of different concentrations of Na + and K + ions inside and outside the cell. This is why the brain uses neuron networks to send many signals to a single cell, or why a neuron may have to fire a couple times before it can pass the message along. An action potential moves from the axon to the soma. To get an electrical signal started, the membrane potential has to change. Without any outside influence, it will not change. Neuroglia are solely responsible for creating action potentials. Resting membrane potential describes the steady state of the cell, which is a dynamic process that is balanced by ion leakage and ion pumping. When action potentials reach the axon terminal, they cause the release of neurotransmitters. Most of the time, all the vesicles released from an action potential aren’t enough to trigger an action potential in the following cell. Which statement is true regarding action potentials (nerve impulses) Is typically only generated in the dendrite portion of a neuron. However, a single vesicle of neurotransmitter isn’t enough to depolarize the cell body. If the neurotransmitter is inhibitory, it will hyperpolarize the cell body. If the neurotransmitter is excitatory, the influx of positive ions will depolarize (bring closer to zero) the cell body. Each vesicle packet released from the pre-synaptic neuron will contain a set amount of neurotransmitters, which will then bind to some of the receptors on the post-synaptic cell. ![]()
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