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Question : Two neurons, A and B, synapse onto a third neuron, C.If neurotransmitter from A opens ligand-gated channels permeable to Na^+ and K^+ and neurotransmitter from B opens ligand-gated Cl^- channels, which of the following statements is true ?

Answer»

An action potential in NEURON A causes a depolarisation in neuron B.
An action potential in neuron B causes a depolarisation in neuron C.
Simultaneous action potentials in A and B will cause hyperpolarisation of neuron C.
Simultaneous action potentials in A and B will cause less depolarisation of neuron C than if only neuron A fired an action potential.

Solution :Once neurotransmitters are released from the pre-synaptic axon terminal, they diffuse across the cleft.This results in binding of neurotransmitter to receptor and OPENING or closing of specific ion channels in the post-synaptic plasma membrane.Usually at an EXCITATORY snapse, channels in the post-synaptic cell that are permeable to `Na^+, K^+` and other small positive ions open, but `Na^+` flux dominates, because it has the largest electrochemical gradient.At inhibitory synapses, channels to `Cl^-` or `K^+` open.In the given SITUATION, the neuron B sends inhibitory pre-synaptic signals onto post-synaptic neuron C which would summate (subtract from ) with the AMPLITUDE of excitatory pre-synaptic signal caused by neuron A.Therefore depolarisation will be of lower amplitude.


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