PRACTICE SET 1-1 - PRACTICE SET 1 Free energy changes and...

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1 PRACTICE SET 1 Free 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 + # -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’ eq for each of the following reactions at pH 7.0 and 25 o C, using the $ G o ’ values given: (a) Glucose-6-phosphate + H 2 O % glucose + P I $ G o ’= -13.8 kJ/mol (b) Lactose + H 2 O % glucose + galactose $ G o ’= -15.9 kJ/mol (c) Malate % fumarate + H 2 O $ G o ’= +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 -3 M 9.6 X 10 -2 M Calculate K’ eq and $ G o ’ for this reaction at 25 o C. 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 $ G o ’ can be calculated indirectly, however, from the equilibrium constants of two other enzymatic reactions having less favorable equilibrium constants: Glucose-6-phosphate + H 2 O % glucose + Pi K’ eq = 270 ATP + glucose % ADP + glucose-6-phosphate K’ eq = 890 Using this information, calculate the standard free energy of hydrolysis of ATP. Assume a temperature of 25 o C.
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2 5. Consider the following interconversion, which occurs in glycolysis: Fructose-6-phosphate ! glucose-6-phosphate K’ eq = 1.97 (a) What is $ G o ’ for the reaction (assuming that the temperature is 25 o C)? (b) If the concentration of fructose-6-phosphate is adjusted to 1.5 M and that of glucose-6-phosphate is adjusted to 0.5M, what is $ G? (c) Why are $ G o ’ and $ G different? 6. Glucose-1-phosphate is converted into fructose-6-phosphate in two successive reactions: Glucose-1-phosphate % glucose-6-phosphate $ G o ’= -7.3 Glucose-6-phosphate % fructose-6-phosphate $ G o ’= +1.7 Calculate the equilibrium constant, K’ eq , for the sum of the two reactions at 25 o C: Glucose-1-phosphate % fructose-6-phosphate 7. ATP-Dependent Chemical Coupling The phosphorylation of glucose to glucose-6- phosphate is the initial step in the catabolism of glucose. The direct phosphorylation of glucose by P
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PRACTICE SET 1-1 - PRACTICE SET 1 Free energy changes and...

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