On the figure below, the post-synaptic neuron is innervated by 6 pre-synaptic ax
ID: 168788 • Letter: O
Question
On the figure below, the post-synaptic neuron is innervated by 6 pre-synaptic axons (labelled A-F) from other neurons. Suppose that V 70 mv, To 50 mV, average velocity of propagation of a potential is 25 nm/msec, and that the potential degrades (returns to Vm) at an average rate of 1 mV/10 nm. Complete the table below. naptic Distance from Initial Is this Time it takes Fina potential Neuron axon hillock Potential mV) potential an to reach axon (mVI when (nml EPSP or IPSP? hillock (msecl axon hillock is reached 35 22 45 16 88 19 60 11 40 15 13 If using the initial potential as the criteria, will the post-synaptic neuron fire an action potential? lfusing the final potential as the criteria, will the postsynaptic neuron fire an action potential? ICLOCExplanation / Answer
Excitatory postsynaptic potential is positive potential because it gets from calcium ions movement. Inhibitory postsynaptic potential requires when a neuron needs to restore its original state. Neuron repolarization needs negative potential to lower down the excitatory potential.
Therefore, in the given question, all positive Initial potentials are EPSP and negative Initial potentials are IPSP.
To calculate final potential we use following formula: Vm+ Ip [Ip= Initial potential]
For -22mv Initial potential= final potential = -70+ (-22) = -92mV
Total distance covered by -22mv Ip = Ip x 10nm = 22 x 10 = 220nm. [10nm derived from the average rate]
Time has taken to reach the axon hillock= distance/ velocity = 220nm / 25 nm msec-1 = 8.8 msec
If we have at least +71mv total postsynaptic potential then the neuron can generate action potential. For that, we have -22 +16+19-11+7+13 = 33 mV, which is not sufficient for excitation of membrane.
According above, calculate the same through final potential.
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