The hydrolysis of ATP proceeds via the following reaction: ATP + H_20 rightarrow
ID: 1051694 • Letter: T
Question
The hydrolysis of ATP proceeds via the following reaction: ATP + H_20 rightarrow HPO_4^2- Delta G degree = -31 kJ mol^-1 a) Calculate Kaq for the reaction at 37 degree C. [R = 8.314 J.K^d, mol^-1] b) The first step in glycolysis has the following overall reaction: Glucose + HPO_4^2- rightarrow-6-PO_4 + H_20 Delta G degree = 14 kJ.mol^1 Calculate K_eq under standard conditions (37 degree C) for this reaction. (Ignore H_2O) In a coupled reaction wit ATP, the reaction becomes: Glucose + ATP rightarrow Glucose-6-PO_4 + ADP i) Calculate Delta G degree for te coupled reaction ii) Calculate K_aq at 37 degree C for this reaction. d) Under stendy state conditions in the cell: [ATP]/[ADP][HPO_4^2-] - 500/1 i. Determine the ratio: [Glucose-6-PO_4/[Glucose][HPO_4^2-] ii. By how much has the equilibrium position for [Glucose-6-PO_4]/[Glucose][HPO_4^2-] reaction been shifted by coupling the phosphorylation of glucose to ATP hydrolysis? GTP functions as a phosphagem like ATP in certain reactions. GTP is made in the Citric Acid Cycle by coupling the following two hydrolysis reactions: suceinylCoA + H_2O rightarrow succinate + CoA Delta G^8' = -93. kJ mol^-1 GDP + HPO_4^2- rightarrow GTP + H_2O Delta G^8+ = + 31 kJ mol^-1 a) Determine Delta G degree for the coupled reaction in which suceinylCoA hydrolysis drives GTP synthesis. b) How many moles of GTP could theoretically be made by coupling GTP synthesis to suceinylCoA hydrolysis? c) Calculate K_eq for this coupled reaction at 37 degree C. Under steady state conditions. Delta G for this reaction is -2.9 kJ mol^-1, calculate the ratio of [products][] under these conditions at 37 degree C. In a muscle cell, in which subcellular organelle does the regeneration of creations phosphate from creations happen? ATP is used as the phosphorylation agent - how car this unfavorable reaction occur?Explanation / Answer
Go = -RT ln Keq where R = 8.314 J/mol/K
Keq = e -(Gorxn/RT)
1)
a) Given that
Go = -31 kJ/mol = -31000 J/mol
T = 37oC = 37 + 273 K = 310 K
Keq = e -(Gorxn/RT)
= e -( -31000/ 8.314 x 310)
= 167360
Therefore,
Keq = 167360
b)
Go = 14 kJ/mol = 14000 J/mol
T = 37oC = 37 + 273 K = 310 K
Keq = e -(Gorxn/RT)
= e -(14000/ 8.314 x 310)
= 4.37 x 10-3
Therefore,
Keq = 4.37 x 10-3
c)
ATP + H2O ----------> ADP + HPO42- Go = -31 kJ/mol
Glucose + HPO42- -------------> Glucose -6-PO4- + H2O Go = 14 kJ/mol
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Add above two reactions and cancel similar terms on both sides,
ATP + Glucose -------------> Glucose -6-PO4- + ADP
Go = -31 kJ/mol + 14 kJ/mol
= -17 kJ/mol
Go = -17 kJ/mol
Hence,
T = 37oC = 37 + 273 K = 310 K
Keq = e -(Gorxn/RT)
= e -( -17000/ 8.314 x 310)
= 732.1
Therefore,
Keq = 732.1
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