# Lecture 3 - For sodium: E Na = (58/+1) log (450/50) = 58...

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Nernst equation: E I = (RT/zF) ln ([I o ]/[I i ]) Converted for constants and room temperature to log base 10, answer will be in mV: E I = (58/z) log ([I o ]/[I i ])

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For potassium: E K = (58/+1) log (20/400) = 58 log (1/20) = -58 log 20 = -58 (1.3) = -75 mV
For potassium: E K = (58/+1) log (20/400) = 58 log (1/20) = -58 log 20 = -58 (1.3) = -75 mV

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For sodium: E Na = (58/+1) log (450/50) = 58 log (9) = +55 mV For chloride: E Cl = (58/-1) log (560/40) = -58 log (14) = -67 mV
2.7 Evidence that the resting potential is determined by K + concentration gradient.

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For potassium: E K = (58/+1) log (20/400) = 58 log (1/20) = -58 log 20 = -58 (1.3) = -75 mV
2.7 Evidence that the resting potential is determined by K + concentration gradient.

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Unformatted text preview: For sodium: E Na = (58/+1) log (450/50) = 58 log (9) = +55 mV For chloride: E Cl = (58/-1) log (560/40) =-58 log (14) = -67 mV A current flows in response to closing the switch. Current is constant. Current measured in units called amperes (1coulomb/sec). The relationship of current and voltage is dictated by Ohm’s law, V=IR Resistance is measured in units called ohms In Neurobiology we often use the inverse of resistance, called conductance, i.e. V=I/g; units here are siemens. V R = IR V C = V B (1-e-tau/RC ) 2.8 The role of sodium in the generation of an action potential. (Part 1) 2.8 The role of sodium in the generation of an action potential. (Part 2)...
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## This note was uploaded on 02/01/2010 for the course BIO 365R taught by Professor Draper during the Spring '08 term at University of Texas at Austin.

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Lecture 3 - For sodium: E Na = (58/+1) log (450/50) = 58...

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