1PRACTICE SET 1Free energy changes and equilibrium constants 1. Calculate the standard free-energy changes of the following metabolically important enzyme-catalyzed reactions at 25°C and pH 7.0 from the equilibrium constants given. (a) Glutamate + oxaloacetate ↔ aspartate + α-ketoglutarate K’eq= 6.8 (b) Dihydroxyacetone phosphate ↔glyceraldehyde-3-phosphate K’eq= 0.0475 (c) Fructose-6-phosphate + ATP ↔F-1,6-bisphosphate + ADP K’eq= 254 2. Calculate the equilibrium constants K’eqfor each of the following reactions at pH 7.0 and 25oC, using the ∆Go’ values given: (a) Glucose-6-phosphate + H2O → glucose + PI ∆Go’= -13.8 kJ/mol (b) Lactose + H2O → glucose + galactose ∆Go’= -15.9 kJ/mol (c) Malate →fumarate + H2O ∆Go’= +3.1 kJ/mol 3. If a 0.1 M solution of glucose-1-phosphate is incubated with a catalytic amount of phosphoglucomutase, the glucose-1-phosphate is transformed to glucose-6-phosphate until equilibrium is established. The equilibrium concentrations are: Glucose-1-phosphate glucose-6-phosphate 4.5 X 10-3M 9.6 X 10-2M Calculate K’eqand ∆Go’ for this reaction at 25oC. 4. 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 ∆Go’ can be calculated indirectly, however, from the equilibrium constants of two other enzymatic reactions having less favorable equilibrium constants: Glucose-6-phosphate + H2O →glucose + Pi K’eq= 270ATP + glucose →ADP + glucose-6-phosphate K’eq= 890 Using this information, calculate the standard free energy of hydrolysis of ATP. Assume a temperature of 25oC.
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