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GUY asked in Science & MathematicsBiology · 2 months ago


The opening of _______ results in ________.

A.   Voltage-gated K+ channels; an action potential

  B. Voltage-gated Na+ channels; membrane repolarization

 C.  Voltage-gated Ca2+ channels; membrane hyperpolarization

 D.  Ligand-gated Cl- channels; an inhibitory postsynaptic potential (IPSP)

 E.  Ligand-gated K+ channels; release of neurotransmitters from the axon terminal

1 Answer

  • 2 months ago
    Favorite Answer

    Hey guy! this all comes back to your simple Nernst and Goldmans equations. Knowing what the net electrostatic and diffusive forces are for each ion determines its flow through a permeable membrane. That being said, NA, Ca2, and CL- tend to be high outside of a cell, and the inside of a cell tends to be a negative -70-80 mV. This resting membrane potential is essentially set by two major contributors. First is the existence of amino acids inside of cells which carry strong negative charges. Next is the relatively high permeability of positively charged K at rest, which will enable attraction of K into the cell by electrostatic forces until the concentration of K is so high that the diffusive repulsive forces (in this case) counteract the attractive electrostatic forces. Due to the high permeability of K, and not much else, the resting membrane potential is essentially regulated by K.

    With that being said, opening Vgated K channels will likely not exert strong effects on membrane potentials because K is already permeable

    Opening Vgated Na channels will result in an influx of Na into a cell due to both electrostatic and diffusive forces, which will lower the net - charge (a term called depolarization). Repolarization refers to re-establishing the resting membrane potential.

    Opening Vgated Ca channels will result in a similar influx into a cell, depolarizing the cell. Hyperpolarization refers to over-shooting the resting membrane potential (say -100 instead of -70) when a membrane potential attempts to reset itself.

    Opening a Ligand-gated Cl channel will result in Cl- entering a cell and adding more negative charge to the cell, which can hyperpolarize a cell from rest. If this occurs due to neurotransmitter release at a synapse, the hyperpolarization is referred to as an inhibitory post-synaptic potential because it moves the membrane potential further from an action potential threshold.

    A mentioned earlier, opening a K channel, even if its a ligand gated channel, will not result in a substantial change in membrane potential from rest however K channels play important roles in re-establishing membrane potentials after action potential. Because release of neurotransmitters from an axon terminal requires Ca (which can occur due to depolarization and action potentials) and further increasing K permeability will not depolarize a cell, opening K channels at rest will not effect Ca influx and therefore not stimulate neurotransmitter release.

    With that being said, the correct answer makes some assumptions that I take issue with, mainly the assumption that the ligand-gated ion channel is always regulated by neurotransmitters and considered an IPSP. Other hyperpolarizing ligand/receptor interactions exist that dont stem from neural conduction. That being said its not really that important for the question and there is only one best answer here.

    I hope I helped! Good luck in your neuroscience studies!

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