SAmple_exam2B

SAmple_exam2B - Part 1: (21 points, correct elimination 4...

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Part 1: (21 points, correct elimination 4 points, incorrect elimination –21 points). The following properties apply to the circuit shown below. ¾ Cell F has no spontaneous activity ¾ Cell F does NOT habituate ¾ Cell F has post-inhibitory rebound (through the cell E to Cell F synapse) ¾ Cell E has spontaneous activity ¾ Cell E habituates through the Cell D to Cell E synapse ¾ This habituation is through the D Æ E synapse so the longer D is on the less effective Cell E’s excitatory neurotransmitter receptor becomes ¾ Cell E does NOT have has post-inhibitory rebound ¾ Cell F’s inhibition onto cell E can completely shut down cell E including inhibiting and turning off spontaneous spikes ¾ Cell E’s spontaneous firing is at a low enough rate that it cannot counteract Cell D Æ F excitation ¾ D Æ F is a shorter length than D Æ E Which of the following can be the activity pattern of the circuit, when D fires as shown in the panels. Excitation Inhibition LEGEND (1) (2) (3) (4) (5) (6)
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Part 2: (22 points, Very little partial credit will be given, do your best to get the correct answer). You find a neuron which at rest has the permeability and concentrations to the following ions. [Na +1 ] out = 145 mM [K +1 ] out = 5 mM [Cl -1 ] out = 120 mM [Br -1 ] out = 1 mM [Na +1 ] in = 15 mM [K +1 ] in = 175 mM [Cl -1 ] in = 12 mM [Br -1 ] in = 10 mM PNA = 100 PK = 4000 PCl = 1000 PBr = 1 On the dendrite of this neuron is a neurotransmitter receptor that when opened is equally permeable to Cl -1 and Br –1 and is not permeable to any other ion type. What is the reversal potential of this receptor?
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Part 3: (20 points correct elimination 4 points, incorrect elimination –20 points). An action potential has been drawn in dotted line to the right. You now modify the axon or the axon’s environment and obtain an action potential shown with the solid line on the graph to the right and below (note
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SAmple_exam2B - Part 1: (21 points, correct elimination 4...

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