Keg = 270 ATP + glucose → ADP + glucose 6-phosphate Keq = 890 Glucose 6-phosphate + H20 → glucose + Pi

Anatomy & Physiology
1st Edition
ISBN:9781938168130
Author:Kelly A. Young, James A. Wise, Peter DeSaix, Dean H. Kruse, Brandon Poe, Eddie Johnson, Jody E. Johnson, Oksana Korol, J. Gordon Betts, Mark Womble
Publisher:Kelly A. Young, James A. Wise, Peter DeSaix, Dean H. Kruse, Brandon Poe, Eddie Johnson, Jody E. Johnson, Oksana Korol, J. Gordon Betts, Mark Womble
Chapter24: Metabolism And Nutrition
Section: Chapter Questions
Problem 32CTQ: Explain how glucose is metabolized to yield ATP.
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A direct measurement of the standard free-energy change associated with the hydrolysis of ATP is technically demanding because the minute amount of ATP remaining at equilibrium is difficult to measure accurately. The value of ΔG′° can be calculated indirectly, however, from the
equilibrium constants of two other enzymatic reactions having less favorable equilibrium constants:
Using this information for equilibrium constants determined at 25 °C, calculate the standard free energy of hydrolysis of ATP.

Keg = 270
ATP + glucose → ADP + glucose 6-phosphate Keq = 890
Glucose 6-phosphate + H20 → glucose + Pi
Transcribed Image Text:Keg = 270 ATP + glucose → ADP + glucose 6-phosphate Keq = 890 Glucose 6-phosphate + H20 → glucose + Pi
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ISBN:
9781938168130
Author:
Kelly A. Young, James A. Wise, Peter DeSaix, Dean H. Kruse, Brandon Poe, Eddie Johnson, Jody E. Johnson, Oksana Korol, J. Gordon Betts, Mark Womble
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OpenStax College