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Phosphoenolpyruvate (PEP) is converted to pyruvate by the enzyme pyruvate kinase. The standard free energy change (delta Gº') is -31.4 kJ/mol. After eating a candy bar, Billy's cellular concentrations are pyruva = 0.04 mM and PEP = 12 mM. What is the delta G of this reaction at 37 %

Phosphoenolpyruvate (PEP) is converted to pyruvate by the enzyme pyruvate kinase. The standard free energy change (delta Gº') is -31.4 kJ/mol. After eating a candy bar, Billy's cellular concentrations are pyruva = 0.04 mM and PEP = 12 mM. What is the delta G of this reaction at 37 %

Biochemistry
Biochemistry
9th Edition
ISBN:9781319114671
Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Chapter1: Biochemistry: An Evolving Science
Section: Chapter Questions
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**Phosphoenolpyruvate (PEP) Conversion to Pyruvate: Thermodynamics** Phosphoenolpyruvate (PEP) is converted to pyruvate by the enzyme pyruvate kinase. The standard free energy change (∆G°') for this reaction is -31.4 kJ/mol. **Problem Statement:** After consuming a candy bar, Billy's cellular concentrations are as follows: - Pyruvate = 0.04 mM - PEP = 12 mM **Question:** What is the ∆G of this reaction at 37°C? **Answer Choices:** - A) -33.3 kJ/mol - B) +6.3 kJ/mol - C) -46.1 kJ/mol - D) -2.4 kJ/mol - E) -16.7 kJ/mol **Explanation:** To solve for the actual free energy change (∆G) of the reaction under non-standard conditions, you can use the equation: \[ \Delta G = \Delta G^\circ' + RT \ln(Q) \] where: - \( R \) is the gas constant (8.314 J/mol·K), - \( T \) is the temperature in Kelvin (310 K for 37°C), - \( Q \) is the reaction quotient \(\left(\frac{[\text{products}]}{[\text{reactants}]}\right)\). **Calculation Steps:** 1. Determine the reaction quotient \( Q = \frac{[\text{pyruvate}]}{[\text{PEP}]} \). 2. Substitute the values into the equation to find \(\Delta G\). **Note:** Detailed calculations should be performed to determine the exact ∆G value, exploring how shifts in concentration from standard conditions affect the energy of the reaction.
Transcribed Image Text:**Phosphoenolpyruvate (PEP) Conversion to Pyruvate: Thermodynamics** Phosphoenolpyruvate (PEP) is converted to pyruvate by the enzyme pyruvate kinase. The standard free energy change (∆G°') for this reaction is -31.4 kJ/mol. **Problem Statement:** After consuming a candy bar, Billy's cellular concentrations are as follows: - Pyruvate = 0.04 mM - PEP = 12 mM **Question:** What is the ∆G of this reaction at 37°C? **Answer Choices:** - A) -33.3 kJ/mol - B) +6.3 kJ/mol - C) -46.1 kJ/mol - D) -2.4 kJ/mol - E) -16.7 kJ/mol **Explanation:** To solve for the actual free energy change (∆G) of the reaction under non-standard conditions, you can use the equation: \[ \Delta G = \Delta G^\circ' + RT \ln(Q) \] where: - \( R \) is the gas constant (8.314 J/mol·K), - \( T \) is the temperature in Kelvin (310 K for 37°C), - \( Q \) is the reaction quotient \(\left(\frac{[\text{products}]}{[\text{reactants}]}\right)\). **Calculation Steps:** 1. Determine the reaction quotient \( Q = \frac{[\text{pyruvate}]}{[\text{PEP}]} \). 2. Substitute the values into the equation to find \(\Delta G\). **Note:** Detailed calculations should be performed to determine the exact ∆G value, exploring how shifts in concentration from standard conditions affect the energy of the reaction.
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