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Cell # 1 forms an excitatory synapse onto cell # 2 , when cell #1 fires a single action potential...


Cell # 1 forms an excitatory synapse onto cell # 2 , when cell #1 fires a single action potential it releases glutamate onto
Cell # 1 forms an excitatory synapse onto cell # 2 , when cell #1 fires a single action potential it releases glutamate onto the membrane of cell #2 resulting in a 5 millivolt depolarization, a 5 mV excitatory postsynaptic potential, when cell #1 fires three action potentials in rapid succession this causes a 15 mV depolarization in cell #2 resulting in cell #2 reaching threshold and firing its own action potential. Cell #1 Cell #2 +40 Vm IN Cell #2 0 (mV) The picture at right depicts the connection between Cell #1 and #2, and the plot below it is an intracellular recording in Cell #2 as it responds to activity from Cell #1. threshold -65 t ttt Cell #1 activity tine ime A) If the extracellular concentration of potassium were increased what effect would this have on the resting membrane potential of cell #27(circle one; 3pts) hyperpolarize depolarize remain the same Given this new state, what effect would an increase in extracellular K+ have on cell #2's response if cell #1 started firing action potentials? (circle one; 3pts) Cell # 2 would be more likely to fire an AP less likely to fire an AP unaffected B) The glutamate receptors on cell #2 are ligand-gated ion channels. what ion are these channels permeable to when activated? (3pts) C) If extracellular calcium (Ca++) was removed how would this affect cell #1's ability to excite cell #27(4pts) D) Draw an arrow in the diagram above to indicate the precise location in cell #2 where the action potential is first generated. (3pts)
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A) hyperpolarize

When cell 2 is in its resting membrane potential (-70 mV), it means positively charged ions are more outside the cell than inside of the cell. Now, if you increase the concentration of extracellular potassium, it adds more positive charges outside. So, the difference (-70 mV) rises more (suppose -90 mV). This is hyper polarized state.

So, cell 2 would be less likely to fire an AP, as it is in a hyperpolarized state. Because hyper polarized state inhibits action potentials.

B) Sodium ions.

These channels permit influx of sodium ions that depolarizes postsynaptic membrane (that is membrane of cell 2).

C) If extracellular calcium was removed, during action potential less calcium would entered the presynaptic terminal. As calcium influx strengthen the synaptic transmission, less calcium in this case would result into less synaptic transmission. It will reduce the ability of cell 1 to excite cell 2.

D)

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