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?

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

Answer 1

Answer: (D)

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.