1. Calculation of AG from an Equilibrium Constant Calculate the standard free-energy change for eachof the following metabolically important enzyme-catalyzed reactions, using the equilibrium constants givenfor the reactions at 25 °C and pH 7.0, R = 8.315 J/molKAspartate aminotranterase(a)Glutamate + oxaloacetate 2 aspartate + a-ketoglutarate Keg-6.8Triosephosphate isomerase(b)Dihydroxvacetonephosphate glyceraldehyde3phosphate Keg-0.0475phosphofructokinasee fructose1,6-bisphosphate+ADP Keg=254(c)Fructose6-phosphate + ATP

Answered: Image /qna-images/answer/4faf7ae6-4a6b-46ae-9851-0dea45521e30.jpg

1. Calculation of AG from an Equilibrium Constant Calculate the standard free-energy change for each of the following metabolically important enzyme-catalyzed reactions, using the equilibrium constants given for the reactions at 25 °C and pH 7.0, R = 8.315 J/mol K Aspartate amingtranferase (a)Glutamate + oxaloacetate 2 aspartate +a-ketoglutarate Keg-6.8 Triosephosphate isomerase (b)Dihydroxvacetonephosphate glyceraldehyde3phosphate Keg-0.0475 phosphofructokinase + ATP = fructose1,6-bisphosphate+ADP Keg-254 (c)Fructose6-phosphate

1. Calculation of AG from an Equilibrium Constant Calculate the standard free-energy change for each of the following metabolically important enzyme-catalyzed reactions, using the equilibrium constants given for the reactions at 25 °C and pH 7.0, R = 8.315 J/mol K Aspartate amingtranferase (a)Glutamate + oxaloacetate 2 aspartate +a-ketoglutarate Keg-6.8 Triosephosphate isomerase (b)Dihydroxvacetonephosphate glyceraldehyde3phosphate Keg-0.0475 phosphofructokinase + ATP = fructose1,6-bisphosphate+ADP Keg-254 (c)Fructose6-phosphate

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter11: Chemical Kinetics: Rates Of Reactions

Section: Chapter Questions

Problem 83QRT

See similar textbooks

Similar questions

One step in glycolysis is the conversion of fructose-6-phosphate to fructose-1,6-bisphosphate. The direct reaction of fructose-6-phosphate with inorganic phosphate fructose-6-phosphate + Pi ⇌ fructose-1,6-bisphosphate + H₂O is very unfavorable, with ∆G'°= 16.3 kJ/mol and K'eq = 1.39×10⁻³. However, in glycolysis the reaction is coupled to ATP hydrolysis: fructose-6-phosphate + ATP ⇌ fructose-1,6-phosphate + ADP and the reaction if favorable. What is the equilibrium [fructose-1,6-bisphosphate]/[fructose-6-phosphate] ratio in the coupled reaction when [ATP] = [ADP]? For ATP hydrolysis ∆G'°= -30.5 kJ/mol at 25°C.
What is the equilibrium constant expression for the Ksp of Fe2S3?
Calculate the standard free energy change (ΔG°') for the reaction catalyzed by the enzyme aspartate amino transferase (K'eq = 6.8) using the given equilibrium constant for the reactions at 25°C and pH 7.0.
The equilibrium constant, Keq, for the isomerization of dihydroxyacetone phosphate (DHAP) to glyceraldehyde 3-phosphate (GAP) is 0.0475 under standard conditions (298 K, 1 atm, pH 7). Calculate ∆G ̊’ for the reaction. What is ∆G’ for the isomerization reaction when the initial concentrations of DHAP and GAP are 2.2 ́ 10-4 M and 2.9 ́ 10-6 M, respectively? Assume a temperature of 298 K. The same enzyme catalyzes the reverse reaction, the isomerization of GAP to form DHAP. Assume the concentration of GAP remains constant at 2.9 ́ 10-6 M. What is the maximum concentration of DHAP that allows the reaction to proceed spontaneously in this direction?
An enzymatic assay was performed using the enzyme phophofructokinase-1, which catalyses the committed step of the glycolysis pathway. What is the final concentration of the fructose 6-phosphate substrate in a test tube containing the following: 250 µL of 150pM fructose 6-phophate, 100 µL 20 mg.ml-1 phosphofructoskinase-1 and 150 µL 100mM buffer pH 7.5
How can the Michaelis constant (Km) be identified and used to represent the affinity between the enzyme and the substrate?
Given that the Ksp for Ag2S is 6×10^-51, Kf=1.1×10^5 for AgCl2-, and Ka1 and Ka2 for H2S are 9.5×10^-8 and 1×10^-19, respectively, calculate the equilibrium constant for the following reaction .
3) The enzyme aldolase catalyzes the conversion of fructose-1,6-diphosphate (FDP)to dihydroxyacetone phosphate (DHAP) and glyceraldehide-3-phosphate (G3P). Thereaction isFDP ⇄ DHAP + G3P , ∆rG0(298.15 K) = 23.8 kJ mol−1In red blood cells the concentration of these species are [FDP] = 35 µM, [DHAP] =130 µM, and [G3P] = 15 µM. (Remember that 1.0 µM = 1.0 × 10−6 mol L−1. Thestandard state for reactions in solution can be taken as c0 = 1.0 mol L−1).Calculate ∆rG in a red blood cell at 25 0C. Will the reaction occur spontaneouslyin the cell at this temperature?
Dihydrogen phosphate, H₂PO₄⁻(aq), undergoes the following acid base equilibria in blood: H₂PO₄⁻(aq) ⇌ H⁺(aq) + HPO₄²⁻(aq)⇌ H⁺(aq) + PO₄³⁻(aq). The pKa values for the first and second steps are 7.2 and 12.4, respectively. The pH of human blood is 7.37 and this is tightly regulated. Which of the following statements most accurately describes the concentrations of the different species for the above reactions in blood. (Hint, think of the titration curves for H₂PO₄⁻(aq) and HPO₄²⁻(aq) and their respective pKa's.) A) [H₂PO₄⁻(aq)] is much greater than [HPO₄²⁻(aq)] and [PO₄³⁻(aq)] B) [HPO₄²⁻(aq)] is much greater than [H₂PO₄⁻(aq)] and [PO₄³⁻(aq)]. C) [H₂PO₄⁻(aq)] and [HPO₄²⁻(aq)] are both present and greater than [PO₄³⁻(aq)]. D) [PO₄³⁻(aq)] is much larger than the other concentrations, [H₂PO₄⁻(aq)] and [HPO₄²⁻(aq)]
Calculate the specificity constant for an enzyme-catalyzed reaction if the total enzyme concentration is 15 nM, Vmax is 9 M/s, and Km is 150 M. A. Has the enzyme achieved catalytic perfection? (Y or N)
What is the function of the enzyme cis-trans isomerase?
Consider the formation of a dimeric protein 2P ->P2 At 25.0 °C , the enthalpy of the reaction is 17 kj/mol and the entropy of the reaction is 65 j/kmol. Is the dimerization favored at this temperature ? Thermodynamically speaking, explain/support your answer. Show work.