Membrane potential after exposure to glutamate

Question

Neurons were kept in a physiological solution. During the resting phase, the membrane potential in the axoplasm of neurons was negative compared to the extracellular space and a potential difference of -70 mV was observed in this phase. Neurons were then treated in two different experiments with either gamma-amino butyric acid (GABA; an inhibitory neurotransmitter) or glutamate (an excitatory neurotransmitter) and the membrane potentials were recorded. Choose the correct statement/s:

(A) The resting membrane potential of -70 mV would not change with either GABA or glutamate treatments.

(B) The membrane potential would be even more negative than resting phase with GABA treatment.

(C) The membrane potential would be positive when the neuron was exposed to glutamate.

(D) The membrane potential would be more negative than resting potential after glutamate treatment.

I feel, since, glutamate is excitatory, so resting potential should decrease when exposed to glutamate and increase when exposed to GABA. So (A)&(D) automatically gets eliminated and (B) is correct answer. I am confused at (C)

Answer 1

I believe C is also correct. The excitatory neurotransmitters open $\ce{Na}$ channels which makes the cell membrane more permeable to sodium as compared to potasium and therefore the equilibrium membrane potential would be much closer to the nernst potential for sodium. Now, since the concentration of sodium outside is much greater than inside, the nernst potential for sodium is positive with respect to the inside of the membrane. This makes the overall potential also positive.

If you want further clarifications regarding the calculation of the equilibrium and the nernst potential across the membrane, feel free to ask.

ADDENDUM

The approximate concentrations of sodium and potassium inside and outside a cell are :-
$$\ce {Na+_{in}}=10 \text{mmoll}^{-1}$$ $$\ce {K+_{in}}=105 \text{mmoll}^{-1}$$ $$\ce {Na+_{out}}=140 \text{mmoll}^{-1}$$ $$\ce {K+_{out}}=5 \text{mmoll}^{-1}$$

The Nernst potential is given by $E=-\frac{RT}{nF}\ln\frac{\text{in}}{\text{out}}$

Which for the given concentrations comes out to be $+67.7mV$ for sodium and $-78.1mV$ for potassium. During a sufficiently strong action potential, the relative permeability of sodium with respect to potassium becomes $10$. Using the GHK equation(ignoring the contribution of chlorine) gives us a net potential of $+47.5mV$.

The graph of the action potential clearly indicates that at the peak of the AP, the membrane potential is positive. Source : Graph of action potential

Answer 2

It is more correct to call it "resting potential difference" (like your question), because electrical potential is relative, not absolute.

That phrasing exposes a crucial point: The difference of what? Cell cytoplasms are negatively charged (to remember this, it helps to remember that protons are usually pumped out of the cytosol either into the periplasm, vesicles, mitochondrial intermemberane space, or outside of the cell). If you subtract the potential of the outside from the inside, you'll get +70. If you do it the other way around, you'll get -70. By convention, the (+) probe of the voltmeter is stuck inside the cell, and the (-) is stuck outside, so we end up with the "official" figure of -70 mV.

Nature obviously does not like this potential difference, and wants to neutralize it by pushing current across the membrane. Luckily the membrane is not very conductive, and the cell can expend energy to undo the effects of any leaking and prevent the potential [difference] from drifting toward 0 (it would actually drift toward a number above 0 because potential difference isn't the only factor, there is also the concentration difference).

So the cell is like a battery that keeps itself charged. It also has a threshold, and it will only empty itself if it has discharged at least to a certain point. That point depends on the cell, but a typical value is -55 mV (so closer to equilibrium point than the resting -70 mV).

What will an inhibitory chemical like GABA do? It will pull the cell further away from the threshold, so it's harder to overcome it. Bringing -70 mV to -90 mV would be lowering it (because the number goes down).

What will an excitatory chemical like glutamate do? It will bring the cell closer to the threshold, so it's easier to overcome it. Bringing -70 mV to -55 mV would be raising it (because the number goes up).

The question, unfortunately, does not specify whether the action potential fires. Typically, the peak of the AP is +40 mV. In theory, you could have a threshold say, at +20 mV, and then perhaps the potential could go positive (eg. +10 mv) and stay there. But I really, really doubt you could find a cell with a threshold above 0. If the threshold is below zero, then the cell will reach a positive potential difference (if you use enough glutamate to elicit the AP), but it will only stay there momentarily before collapsing back to -80 mV (the refractory state).

Answer 3

I don't think any of the answers here are quite right, so I'll give a try:

(A) The resting membrane potential of -70 mV would not change with either GABA or glutamate treatments.

The answer here is: "It depends".

For glutamate, if the neuron only had NMDA receptors that were not "open" at -70 mV, there would be no change in resting membrane potential. If there were at least some AMPA receptors, though, the membrane potential would change to more positive.

For GABA, if the cell has only GABAa receptors, then the resting membrane potential wouldn't change, because the reversal potential for GABAa is right there at -70 mV. But if the cell had at least some GABAb receptors, their reversal potential is around -100 mV, and so would cause a change in the resting membrane potential to something more negative than -70 mV.

(B) The membrane potential would be even more negative than resting phase with GABA treatment.

Again, if GABAa receptors only, then no, it would be no more negative and would in fact stay the same at -70 mV. If there are GABAb receptors, then yes, the resting membrane potential would be even more negative.

(C) The membrane potential would be positive when the neuron was exposed to glutamate.

Maybe--it depends on whether the neuron is allowed to have an action potential. Glutamate receptors generally reverse around 0 mV. 0 mV is neither positive nor negative. If there were no action potential in this neuron, we should therefore not say the membrane would be "positive" (though this is splitting hairs, I suppose). However it could be said to be "more positive than -70 mV". However, if the cell were allowed to have an action potential, then yes, the membrane potential would shoot up to about +20 mV, so could be said to be positive.

(D) The membrane potential would be more negative than resting potential after glutamate treatment.

No. Again, glutamate receptors' reversal potential is about 0 mV, so it would only move the membrane potential to more positive values, close to or right at 0 mV.