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answrs_2002_mdtm

# answrs_2002_mdtm - 9.09J/7.29J Cellular Neurobiology Spring...

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7.29 J 9.09 Cellular Neurobiology Answers to Midterm Test 20 March, 2002 Question 1. This question asks you to go around one of Hodgkin and Huxley’s predictive cycles. Half the problem is tightly analogous to a homework problem. (a) First step: Given conductances and voltage forces, calculate the resultant current. Use Ohm’s law for membranes. I = g Na (V m -E Na ) + g K (V m -E K ) I = (2 mS/cm 2 ) [-50mV - (+50mV)] + 5mS/cm 2 [-50mV – (-80mV)] I = (2 x 10 -3 S/cm 2 ) (-100 x 10 -3 V) + (5 10 -3 S/cm 2 ) (30 x 10 -3 V) I = (-2 x 10 --4 Amps/cm 2 ) + (1.5 x 10 --4 Amps/cm 2 ) I = -5 x 10 --5 Amps/cm 2 Note that the sodium (negative) current is the larger term; therefore the net current is negative = inward = depolarizing . Second step: Use the differentiated definition of capacitance to get the voltage slope. I = C dV/dt dV/dt = I/C dV/dt = (-5 x 10 --5 Amps/cm 2 )/(10 -6 Farad/cm 2 ) dV/dt = (-5 x 10 --5 Coulombs/sec/cm 2 )/(10 -6 Farad/cm 2 ) dV/dt = -50 Volts/sec = 50 millivolts/millisec in depolarizing direction. V’= V + V V = (50 millivolts/millisec) x (1 millisec) V’ = -50mV + 50 mV V’ = 0 mV (b) Three principal approximations: (i) Top of overshoot approximares E Na (ii) Bottom of undershoot approximates E K (iii) the conductances g Na , g K and the voltage driving forces don’t change much over one millisecond. This is the most questionable of the three approximations. 1 9.09J/7.29J - Cellular Neurobiology, Spring 2005 Massachusetts Institute of Technology Department of Brain and Cognitive Sciences Department of Biology Instructors: Professors William Quinn and Troy Littleton

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Question 2. The triangle represents a differential amplifier . In Hodgkin and Huxley’s day it was a system of tubes and resistors. Nowadays it’s transistors and integrated circuits in a box that you buy.
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