508_Quiz_1 - eq = 19 Fructose 6-phosphate Glucose...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Quiz 1 9/7/06 Biochemistry 411: Section 508 1. Calculate the equilibrium constant K eq for the reaction: Malate fumarate + H 2 0 at pH 7, T = 25 o C, R = .008315 kJ/mol, Δ G’° = 3.1 kJ/mol Δ G’°=-RTln K eq 3.1= -(.008315)(298)ln K eq K eq = .286 2. Given Half Reactions: Glucose-1-phosphate  Glucose 6-phosphate K
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: eq = 19 Fructose 6-phosphate Glucose 6-phosphate Δ G’°= -1.7 kJ/mol Calculate the K eq of the total reaction: Glucose-1-phosphate Fructose 6-phosphate G1P G6P F6P K eq =19 Δ G’°=1.7kJ/mol 1.7=-RTln K eq K eq =.5035 K eq total = 19*.5035= 9.6...
View Full Document

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.

Ask a homework question - tutors are online