Glucose-6-phosphate <-->fructose-6-phosphate The equilibrium constant Keq for th

Question

Glucose-6-phosphate <-->fructose-6-phosphate

The equilibrium constant Keq for this reaction at 25 degrees Celsius is .5M

(a) Assume that you incubate a solution containing .15M glucose-6-phosphate overnight at 25 degrees Celsius with the enzyme phosphoglucoisomerase that catalyzes this reaction. How many milimoles of fructose-6-phosphate will you recover from 10mL of the incubation mixture the next morning assuming you have an appropriate chromatographic procedure to separate fructose-6-phosphate from glucose-6-phosphate?

Explanation / Answer

1. millimoles of f6p? Keq = [f6p]/[g6p] Keq = 0.5M [g6p] = .15 M V = .010 L .5 = [x]/[.15-x] x = amount f6p formed = amount of g6p consumed = .050 M f6p .050M x .010L = n moles of f6p 5x10^-4 moles f6p = .5 millimoles f6p

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