The equilibrium constant Kc for the reaction fructose-1,6-diphosphate ⇋ glyceraldehyde-3-phosphate + dihydroxyacetone phosphate is 8.9 x 10-5 M at 250C and the behavior is assumed to be ideal. Calculate .for the process (standard state: 1M) Suppose that we have a mixture that is initially 0.010 M in fructose-1,6-diphosphate and 1.0 x 10-5 M in both glyceraldehyde-3-phosphate and dihydroxyacetone phosphate. What is ?
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The equilibrium constant Kc for the reaction
fructose-1,6-diphosphate ⇋ glyceraldehyde-3-phosphate + dihydroxyacetone phosphate
is 8.9 x 10-5 M at 250C and the behavior is assumed to be ideal.
Calculate .for the process (standard state: 1M)
Suppose that we have a mixture that is initially 0.010 M in fructose-1,6-diphosphate and 1.0 x 10-5 M in both glyceraldehyde-3-phosphate and dihydroxyacetone phosphate. What is ?
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- The equilibrium constant Kc for the reaction fructose-1,6-diphosphate ⇋ glyceraldehyde-3-phosphate + dihydroxyacetone phosphate is 8.9 x 10-5 M at 250C and the behavior is assumed to be ideal. Calculate the delta G for the process Suppose that we have a mixture that is initially 0.010 M in fructose-1,6-diphosphate and 1.0 x 10-5 M in both glyceraldehyde-3-phosphate and dihydroxyacetone phosphate. What is the delta GAssume that the reaction Arabinose +Pi <> Arabinose-6-P has a delta G of +13.0 kJ/mol and the reaction XTP <> XDP + Pi has a delta G of -25.7 kJ/mol. Calculate the delta G for the combined reaction Arabinose + XTP <> Arabinose-6-P + ADP, in kJ/mol to four significant figures.The phosphoryl group transfer potentials for glucose-1-phosphate and glucose-6-phosphate are 20.9 kJ/mol and 13.8 kJ/mol, respectively. (a) What is the equilibrium constant for the reaction shown below at 25 °C? (b) If a mixture was prepared containing 1 m glucose-6-phosphate and 1 x 10-3 M glucose-1-phosphate, what would be the thermodynamically favored direction for the reaction?
- The Keq (25C) of the reaction below is 635.67. Fructose 1,6-biphosphate <-->fructose -6-phosphate + Pi. a) What is the standard Gibbs free energy change for this reaction? b) if the concentrationof fructose 1,6 biphosphate is adjusted to 0.85 M and that of fructose 6 phosphate and phosphate adjusted to 0.055 M, what is the actual free energy changeThe breakdown of a complex carbohydrate into its monomeric unit can be described as an enzyme-catalyzed reaction that has a deltaG(∆G) of -90kcal/mol. Suppose you were to add triple the amount of enzyme into this reaction - what would be the ∆G of this reaction?The standard state free energy of hydrolysis of acetyl phosphate isΔG° = -42.3 kJ/mol. Acetyl-P + H2O acetate + Pi Calculate the free energy change for the acetyl phosphate hydrolysis in a solution of 2 mM acetate, 2 mM phosphate and 3 nM acetyl phosphate.
- The biochemical standard free energy change for the reaction: A → B is −15.0 kJ/mol. What is the equilibrium constant at 25°C and pH 7? What is the free energy change for the reaction A→B at 37°C when [A]=10.0 mM and [B]=0.100 mM? (Give your answer in kJ/mol)Consider the following chemical equation whose delta(G) = 9kcal/mol: AC + BD ---> AB + CD what are the reactants and what are the products is this reaction spontaneous? How do you know? Is energy released or consumed by this reaction? If an enzyme, which catalyzes this reaction is added, what will happen to delta (G) If this reaction is coupled to another reaction, whose delta(G) is -12 kcal/mol, what will be the net delta(G) value? will the overall reaction be spontaneousThe turnover number for an enzyme that approximates Michaelis-Menten kinetics is known to be 500 min^-1. From the results shown in the table, enumerate Km and total amount of enzyme present. What is the Km for this enzyme? What is the Vmax for this enzyme? And what is the [E]T for this enzyme?
- The decomposition of crystalline N2O5 N2O5(s) → 2NO2(g) + 1/2O2(g) is an example of a reaction that is thermodynamically favored, even though it absorbs heat. At 25 °C we have the following values for the standard state enthalpy and free energy changes of the reaction: ∆H° = +109.6 kJ/mol ∆G° = -30.5 kJ/mol (a) Calculate ∆S ° at 25 °C. (b) Why is the entropy change so favorable for this reaction?The turnover number of the enzyme fumarase that catalyzes the reaction, Fumarate + H20 ===→ L-malate, is 2.5 x 103 S - l and Km = 4.0 X 10- 6 mol/L. Calculate the rate of conversion of fumarate to L-malate if the fumarase concentration is 1.0 x 1 0 - 6 mol/L and the fumarate concentration is 2.04 x 10- 4 mol/L.The formation of maltose, a disaccharide, from two glucose molecules, is not energetically favorable. However, if this reaction is coupled with the hydrolysis of ATP, the reaction occurs more favorably. Maltose + H2O = 2 Glucose , ΔG'o = -15.5 KJ/mol or -3.7 kcal/mol a. Determine if the coupled reaction will occur spontaneously at standard state through calculating the Gibbs Free Energy of Reaction. b. Calculate the equilibrium constant for each individual reaction, and for the coupled reaction (using free energy of reaction). Show that the equilibrium constant for the coupled reaction equals the equilibrium constants for the individual reactions multiplied together. c. If the reaction medium contains the following chemical species at their given concentrations (298 K and 1.0 atm, pH = 7.0), will the reaction proceed in the forward or the reverse direction? [Maltose] = [Glucose] = 10.0 mM; [ATP] = 5.0 mM; [ADP] = [Pi] = 20 mM