Isomerization of a Phosphorylated Glucose Molecule
Not what you're looking for? Search our solutions OR ask your own Custom question.
The isomerization of a phosphorylated glucose molecule from 1-p to 6-p form with standard free energy change=-1.74kcal/mole. Find the equilibrium concentration ratio of glucose-p in the two isometric states.
© BrainMass Inc. brainmass.com March 6, 2023, 1:28 pm ad1c9bdddfhttps://brainmass.com/biology/cell-biochem/isomerization-phosphorylated-glucose-molecule-27177
Solution Preview
The relationship between the free energy change (DG) and the equilibrium concentration ratio (K) is:
DG = -RT ln K
T = absolute temperature. 298 K is standard but 310 K is ...
Solution Summary
Answer is found using the formula DG = -RT ln K, representing the relationship between the free energy change (DG) and the equilibrium concentration ratio (K).
$2.49