# PRACTICE SET 1. Free energy changes and equilibrium constants

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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 + α-ketoglutarate K eq = 6.8 (b) Dihydroxyacetone phosphate glyceraldehyde-3-phosphate K eq = (c) Fructose-6-phosphate + ATP F-1,6-bisphosphate + ADP K eq = 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 (b) Lactose + H 2 O glucose + galactose (c) Malate fumarate + H 2 O G o = kj/mol G o = kj/mol 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. 1

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 i is described by the equation Glucose + P I glucose-6-phosphate + H 2 O G o = 13.8 kj/mol (a) Calculate the equilibrium constant for the above reaction. In the rat hepatocyte, the physiological concentrations of glucose and P i are maintained at approximately 4.8 mm. What is the equilibrium concentration of glucose-6-phosphate obtained by the direct phosphorylation of glucose by P i? Does this route represent a reasonable metabolic route for the catabolism of glucose? Explain. (b) In principle, at least, one way to increase the concentration of glucose-6-phosphate is to drive the equilibrium reaction to the right by increasing the intracellular concentrations of glucose and P i. Assuming a fixed concentration of P i at 4.8 mm, how high would the intracellular concentration of glucose have to be to have an equilibrium concentration of glucose-6-phosphate of 250 µm (normal physiological concentration)? Would this route be a physiologically reasonable approach, given that the maximum solubility of glucose is less than 1M? (c) The phosphorylation of glucose in the cell is coupled to the hydrolysis of ATP; that is, part of the free energy of ATP hydrolysis is utilized to effect the endergonic phosphorylation of glucose: 2

4 11. The synthesis of the activated form of acetate (acetyl-coa) is carried out in an ATPdependent process: Acetate + CoA + ATP acetyl-coa + AMP + PP i (a) The G o for the hydrolysis of acetyl-coa to acetate and CoA is 32.2 kj/mol and that for hydrolysis of ATP to AMP and PP i is 30.5 kj/mol. Calculate G o for the ATPdependent synthesis of acetyl-coa. (b) Almost all cells contain the enzyme inorganic pyrophosphatase, which catalyzes the hydrolysis of PP i to P i. What effect does the presence of this enzyme have on the synthesis of acetyl-coa? Explain. 12. Are All Metabolic Reactions at Equilibrium? (a) Phosphoenolpyruvate is one of the two phosphate donors in the synthesis of ATP during glycolysis. In human erythrocytes, the steady-state concentration of ATP is 2.24 mm, that of ADP is 0.25 mm, and that of pyruvate is mm. Calculate the concentration of phosphoenolpyruvate at 25 o C, assuming that the pyruvate kinase reaction is at equilibrium in the cell. (b) The physiological concentration of phosphoenolpyruvate in human erythrocytes is mm. Compare this with the value obtained in (a). What is the significance of this difference? Explain. 13. Calculate the free energy change at standard conditions for the following reaction: Acetaldehyde + NADH + H + Ethanol + NAD + 4

5 Answer Key 1. G o = -RT ln K eq R=8.315 J/mol o K T=298 o K a (298) (ln 6.8) J/mol = kj/mol b (298) (ln ) 7. 5 kj/mol c (298) (ln 254) 13.7 kj/mol 2. a. 13,800 = (298) ln K eq K eq = 262 b. 15,900 = (298) ln K eq K eq = 612 c. +3,100 = (298) ln K eq K eq = Keq = [Glc-6-P] = = 21.3 [Glc-1-P] G o = -RT ln K eq R=8.315 J/mol o K T=298 o K G o = (298) ln 21 = -7.5 kj/mol 4. rxn 1 Glc-6-P + H 2 O Glc + Pi rxn 2 ATP + Glc ADP + Glc-6-P net rxn ATP + H 2 O ΑDP + Pi G o for rxn 1= (298) ln 270 = kj/mol G o for rxn 2= (298) ln 890 = kj/mol G o for net rxn = (-16.8) = kj/mol 5. a. G = (298) ln 1.97 = kj/mol b. Since concentrations are not 1M, use 5

6 G = G o + RT ln [Glc-6-P] [F-6-P] G = -1, (298) ln = -4.4 kj/mol c. G o is used for standard conditions, G is used for other concentrations. 6. Glc-1-P Glc-6-P G = -7.3 kj/mol Glc-6-P F-6-P G = +1.7 kj/mol = -5.6 kj/mol G o = -RT ln K eq -5,600 = (298) ln K eq K eq = a. 13,800 = (298) ln Keq Keq= 3.8 x x 10-3 = [Glc-6-P] [0.0048] [0.0048] [Glc-6-P] = M Concentration too low, not physiological b. 3.8 x 10-3 = [250 x10-6 ] [0.0048] [Glc] Glc = 13.7 M Not soluble at this concentration! c. G o = (-30.5) = kj/mol -16,700 = (298) ln Keq K= = [ ] [ ] [ ] [Glc] [Glc] = = 116 µm Yes, concentration is physiologically reasonable. 6

7 8. a. Phosphocreatine + H 2 O creatine + Pi G o = -43 kj/mol ADP + Pi ATP + H 2 O G o = kj/mol Phosphocreatine + ADP creatine + ATP G o = kj/mol b. Pi + fructose fructose-6-phosphate + H 2 O G o = kj/mol ATP + H 2 O ADP + Pi G o = kj/mol ATP + fructose ADP + fructose-6-phosphate G o = kj/mol 9. a. 20,000 = (298) ln Keq Keq = M b. ATP + H 2 0 ADP + Pi G o = kj/mol when coupled to the reaction gives G o =-10.5 kj/mol -10,500 = (298) ln K K = 69.2 c. Keq = [ADP] [Pi ] [Y] 69.2= [ ] [ ] [Y] [ATP] [X] [ ] [X] [Y] = [X] 10. get G o from the handbook Phosphocreatine + H 2 O creatine + Pi ADP + Pi ATP + H 2 O Phosphocreatine + ADP creatine + ATP G o = -43 kj/mol G o = kj/mol G o = kj/mol G = G o + RT ln [products ] [reactants] = -12, (298) ln [1 x 10-3 ] [2.6 x 10 3 ] [4.7x 10 3 ] [0.2 x 10 3 ] = -10 kj/mol 11. a. G = = +1.7 J/mol b. It catalyzes the hydrolysis of pyrophosphate and drives the reaction towards the formation of acetyl CoA. 7

8 12. a. PEP + H 2 O PYR + Pi G o = kj/mol ADP + Pi ATP + H 2 O G o = kj/mol G o = = kj/mol -31,400 = (298) ln Keq Keq = 3.19 x 10 5 Keq = [pyruvate] [ATP] [PEP] [ADP] 3.19 x 10 5 = [0.051 x 10-3 ] [2.24 x 10-3 ] PEP = 1.43 x 10-9 [PEP] [0.25 x 10-3 ] b mm is 16,000 x higher - Reaction is not at equilibrium. 13. The redox reaction can be written as two half-reactions: 1. Acetaldehyde + 2H + + 2e - Ethanol 2. NAD + + 2H + + 2e - NADH + H + See Table 3 for reduction potentials E o. E o = E o Oxidant E o Reductant E o = 0.20 V ( 0.32 V) E o = V G o = nf E o G o = kj mol -1 V V G o = kj mol -1 8

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