Question

10. Consider Cell B, which is a normal neuron (i.e. one with standard extra- and intracellular concentrations of most ion species). a) Plot membrane potential vs time for the postsynaptic response of Cell B in response to an inhibitory neurotransmitter being released onto it by another cell, Cell A. Label and number axes. b) Explain how this could result by the neurotransmitter affecting EITHER an anion channel or a cation channel on Cell B. c) What would be the change of the postsynaptic potential of Cell B if you were to instantly depolarize the presynaptic cell (cell A) from resting potential to a membrane potential value exactly equal to E_Ca++? (the Nernst potential for calcium)

Answer 1

a) Plot membrane potential vs time for the postsynaptic response:

Neurotransmitters are the molecules that cause either excitation or inhibition of the neurons. They act on the various ligand-gated channels resulting in their opening. Depending upon the kind of ligand which causes the final action, the neurotransmitter can be divided as

1) Excitatory neurotransmitter

2) Inhibitory neurotransmitter

Inhibitory neurotransmitter: These are the neurotransmitter that results in the opening of Cl- ions gated channels and inhibition of an action, which is a negative response. certain molecules that produce inhibitory response are GABA, glycine, nitric oxide, etc. They result in hyperpolarisation of the cell.

b) Explain how this could result by the neurotransmitter affecting EITHER an anion channel or a cation channel on Cell B:

As the postsynaptic neuron is providing an inhibitory response so it will affect the anion channels like chloride ion channels. They cause hyperpolarisation of the membrane, inhibitory postsynaptic potential develops. Hence further no information pass or we can say no action potential can be generated further.

c) What would be the change of the postsynaptic potential:

Here we need to determine the change that may arise in the postsynaptic membrane which was actually hyperpolarised.. which means its potential was near to -80mV but if we apply a voltage of +135mV that is equal to the potential of E for Ca++ then we can observe the voltage change as +135mV-80mV = +55mv which generate an action potential. so here action potential can be generated by opening the voltage-gated sodium channel. The membrane will pass the signal further.