membrane transport - slide 6

membrane transport - slide 6 - = (8.314 x 310 x (-2.324)...

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Example: Glucose transport C 1 = 1 mM, C 2 = 66 mM, T = 310K , Δ V = -50 mV Example: Na+ transport C 1 = 143 mM, C 2 = 14 mM, T = 310K, Δ V = -50 mV Δ G = RT ln ( C 2 / C 1 ) + Z F Δ V\ = ( 8.314 J/(molK) )( 310K ) ln (66mM/1mM) + 0 = 8.314 x 310 x 4.190 J/mol = 10.8 kJ/mol Δ G = RT ln ( C2 / C1 ) + Z F Δ V = (8.314 J/(molK))(310K) ln (14 mM / 143 mM) + (1)(96.5 kJ/(mol V))(-50mV)
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Unformatted text preview: = (8.314 x 310 x (-2.324) J/mol) + (96,500 x ( -.050 ) J/mol) = -5990 J/mol + -4825 J/mol = -10.8 kJ/mol (transport from high to low concentration is favorable, especially if aided by the electrical potential) (transport from low concentration to high is unfavorable)...
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This note was uploaded on 05/06/2010 for the course BIOC 460 taught by Professor Ziegler during the Spring '07 term at University of Arizona- Tucson.

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