The mitochondrial enzyme fumarase is involved in the Krebs Cycle, and catalyzes the reversible hydration of fumarate to (S)-malate. What is the ΔGo′ (in kJ/mol) for the forward reaction catalyzed by mitochondrial fumarase at an average adult body temperature of 37 oC? Equilibrium concentrations of the reactants and products are given below for each of the components. You must show your calculations.

The mitochondrial enzyme fumarase is involved in the Krebs Cycle, and catalyzes the reversible hydration of fumarate to (S)-malate. What is the ΔGo′ (in kJ/mol) for the forward reaction catalyzed by mitochondrial fumarase at an average adult body temperature of 37 oC? Equilibrium concentrations of the reactants and products are given below for each of the components. You must show your calculations.

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter16: Thermodynamics: Directionality Of Chemical Reactions

Section: Chapter Questions

Problem 83QRT: Another step in the metabolism of glucose, which occurs after the formation of glucose6-phosphate,...

See similar textbooks

Similar questions

Another step in the metabolism of glucose, which occurs after the formation of glucose6-phosphate, is the conversion of fructose6-phosphate to fructose1,6-bisphosphate(bis meanstwo): Fructose6-phosphate(aq) + H2PO4(aq) fructose l,6-bisphosphate(aq) + H2O() + H+(aq) (a) This reaction has a Gibbs free energy change of +16.7 kJ/mol of fructose6-phosphate. Is it endergonic or exergonic? (b) Write the equation for the formation of 1 mol ADP fromATR for which rG = 30.5 kJ/mol. (c) Couple these two reactions to get an exergonic process;write its overall chemical equation, and calculate theGibbs free energy change.
Aerobic respiration links the oxidation of glucose with the production of ATP form ADP.Given the following thermodynamic data, C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) ΔrH° = −2803 kJ/mol-rxn ATP + H2O(l) → ADP + HPO42-(aq) ΔrH° ≈ −24 kJ/mol-rxn calculate the enthalpy of reaction for the aerobic production of 25 ATP shown below.C6H12O6(s) + 6O2(g) + 25ADP + 25HPO42-(aq)→ 6CO2(g) + 25ATP + 31H2O(l) ΔrH° = ?
The hydrolysis of ATP to form ADP requires energy to break the bonds initially, but releases moreenergy as bond are remade making the reaction exothermic overall. The Gibbs free energy for thisreaction is -30.5 kJ/mole, meaning that it is also exergonic (ΔG=-). This energy can be used by otherbiological systems if this reaction is coupled with another less favorable one. In this case, the creation ofglucose-6-phosphate, C6H10O9P, this is a pivotal intermediate in many biochemical pathways. PO43- + C6H12O6 ⇌ H2O + C6H10O9P ΔG° = +13.8 kJ/moleATP + H2O ⇌ ADP + PO43- ΔG° = -30.5 kJ/mole A) What is the overall reaction that takes place? B) Calculate the ΔG° for this reaction C) Will this be a successful coupled reaction?
Calculate the equilibrium constant at 37oC for the reactionGlucose + 6O2 + 36ADP + 36Phospate → 6CO2+ 42H2O +36ATP ... ... ...Entropy change of Go= − 423 kcal/mole of Glucose
Combustion of glucose ( C6H12O6 ) is the main source of energy for animal cells: C6H12O6 ( s ) + 6O2 ( g ) → 6CO2 ( g ) + 6H2O ( l ) =ΔGrxn37°C−2872.kJ This energy is generally stored as ATP (adenosine triphosphate) molecules, which can release it when convenient by hydrolysis ("water-assisted decomposition") into ADP (adenosine diphosphate) molecules and a phosphate anion, often given the symbol Pi in biochemistry: ATP ( aq ) → ADP ( aq ) + Pi ( aq ) =ΔGrxn−35 to 70kJ The actual amount of free energy released by the hydrolysis of ATP varies, depending on the exact conditions inside the cell. Suppose under certain conditions hydrolysis of ATP actually releases −51.3/kJmol . Calculate the maximum number of ATP s that could be created from ADPs and Pi by the combustion of a molecule of glucose. Your answer must be a whole number.
The protein lysozyme, an enzyme that breaks down bacterial cell walls, unfolds at a transition temperature of 75.5◦C and the standard enthalpy of transition ∆H⊖ - TRS was measured at +509 kJ mol−1.Calculate the standard entropy of transition. What is the significance of a positive ∆S TRS
Calculate the standard free-energy changes for the following reactions at 25°C and interpret the values obtained from the calculations. Write your answers in a separate sheet of paper.a. CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)b. 2MgO(s) → 2Mg(s) + O2(g) Substance ∆G°f (kJ/mol) CH4 -50.8 O2 0 CO2 -394.4 H2O -237.2 MgO -569.6 Mg 0
1. Describe how you would calculate ∆G0′ for the reaction: glucose+ 6 O2 = 6 CO2 + 6 H2O. If you were told that this reaction is highlyexergonic, what would be the arithmetic sign (negative or positive)of the ∆G0′ you would expect for this reaction?
In each of the following pairs, which compound would you expect to have the higher standard molar entropy:CO21g2 or CO1g2?
For the series of reactions below, what is the overall reaction and the value of ∆Go’(kJ/mol) for this set of reactions? Is the reaction spontaneous? (1) 2 Pi+ H2O ➔Pyrophosphate+ H2O (with 5 mM Mg2+) (2) (Mg)Pyrophosphate + AMP ➔(Mg)ATP (3) (Mg)ATP + creatine ➔creatine-phosphate + (Mg)ADP
In glycolysis, glucose is broken down to yield energy. If the equilibrium constant of the isomerization of glucose-6-phosphate to fructose-6-phosphate is 0.5, would this reaction proceed in the forward or reverse direction? What is the free energy of the reaction? Using standard conditions (1 M) as an initial state, what are the equilibrium concentrations?
What terms would best describe the above coupled reaction? (the DGo for ATP hydrolysis into ADP + inorganic phosphate is -7.3 kcal/mole, and the DGo for sucrose synthesis from glucose + fructose is +5.5 kcal/mole) it is non-spontaneous and endothermic (because the overall DGo is positive) it is spontaneous and exothermic (because the overall DGo is positive) it is non-spontaneous and exothermic (because the overall DGo is positive) it is spontaneous and exothermic (because the overall DGo is negative) it is non-spontaneous and endothermic (because the overall DGo is negative)