505_Quiz_1 - K eq = 19 Fructose 6-phosphate Glucose...

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Quiz 1 9/4/06 Biochemistry 411: Section 505 1. Calculate the equilibrium constant K eq for the reaction: glucose-1-phosphate + H 2 O glucose + P i at pH 7, T = 25 o C, R = .008315 kJ/mol, Δ G’° = -13.2 kJ/mol Δ G’°= -RTln K eq -13.2= -(.008315)(298)ln K eq K eq = 205.9 2. Given Half Reactions: Glucose-1-phosphate  Glucose 6-phosphate
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Unformatted text preview: K eq = 19 Fructose 6-phosphate Glucose 6-phosphate G = -1.7 kJ/mol Calculate the G of the total reaction: Glucose-1-phosphate Fructose 6-phosphate G1P G6P F6P G= -(.008315)(298)ln 25= -7.3kJ/mol Total G = -7.3+1.7= -5.6 kJ/mol...
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This note was uploaded on 03/27/2010 for the course BICH 441 taught by Professor Staff during the Spring '08 term at Texas A&M.

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