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Chapter 18: Problem 12

(Integrates with Chapter 3 .) Enolase catalyzes the conversion of 2-phosphoglycerate to phosphoenolpyruvate \(+\mathrm{H}_{2} \mathrm{O}\). The standard free energy change, \(\Delta G^{\circ},\) for this reaction is \(+1.8 \mathrm{kJ} / \mathrm{mol}\). If the concentration of 2 -phosphoglycerate is \(0.045 \mathrm{m} M\) and the concentration of phosphoenolpyruvate is \(0.034 \mathrm{m} M\), what is \(\Delta G\), the free energy change for the enolase reaction, under these conditions?

Short Answer

Expert verified
The free energy change for the enolase reaction under these conditions is approximately 1770 J/mol.

Step by step solution

01

Identify Known Values

From the problem statement, the known values are: \(\Delta G^{\circ} = 1.8 \mathrm{kJ} / \mathrm{mol}\) or 1800 J/mol, the concentration of 2 -phosphoglycerate (reactant) is \(0.045 \mathrm{m} M\), the concentration of phosphoenolpyruvate (product) is \(0.034 \mathrm{m} M\).
02

Convert \( \Delta G^o \) into J/mol

As \(R\) is usually given in J/(mol·K) and the temperature \(T\) is in K, to keep the units consistent, convert the \(\Delta G^{\circ}\) from kJ/mol to J/mol: \(1.8 kJ / mol * 1000 J / kJ = 1800 J / mol\). The provided \(\Delta G^{\circ}\) value is now in the same unit as \(RT\).
03

Use the Gibbs Free Energy Change Formula

The next step is to use the equation \(\Delta G = \Delta G^{\circ} + RT \ln \frac{[products]}{[reactants]}\). Since we are at room temperature which equals approximately 298 K, and \(R = 8.314 J/(mol·K)\), you would substitute these values along with the concentrations of products and reactants into the equation, which gives: \(\Delta G = 1800 J/mol + (8.314 J/(mol·K) * 298 K) * ln(0.034/0.045)\).
04

Solve for \(\Delta G\)

Calculating the above gives a value of approximately 1770 J/mol.

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Most popular questions from this chapter

(Integrates with Chapter \(3 .)\) The standard free energy change \(\left(\Delta G^{\circ \prime}\right)\) for hydrolysis of phosphoenolpyruvate (PEP) is \(-61.9 \mathrm{kJ} / \mathrm{mol}\) The standard free energy change \(\left(\Delta G^{\circ \prime}\right)\) for ATP hydrolysis is \(-30.5 \mathrm{kJ} / \mathrm{mol}\) a. What is the standard free energy change for the pyruvate kinase reaction: ADP \(+\) phosphoenolpyruvate \(\longrightarrow\) ATP \(+\) pyruvate b. What is the equilibrium constant for this reaction? c. Assuming the intracellular concentrations of [ATP] and [ADP] remain fixed at \(8 \mathrm{m} M\) and \(1 \mathrm{m} M\), respectively, what will be the ratio of [pyruvate]/[phosphoenolpyruvate] when the pyruvate kinase reaction reaches equilibrium?

Write the reactions that permit galactose to be utilized in glycolysis. Write a suitable mechanism for one of these reactions.

If \(^{32}\) P-labeled inorganic phosphate were introduced to erythrocytes undergoing glycolysis, would you expect to detect \(^{32} \mathrm{P}\) in glycolytic intermediates? If so, describe the relevant reactions and the \(^{32} \mathrm{P}\) incorporation you would observe.

Regarding phosphofructokinase, which of the following statements is true: a. Low ATP stimulates the enzyme, but fructose- 2,6 -bisphosphate inhibits. b. High ATP stimulates the enzyme, but fructose- 2,6 -bisphosphate inhibits. c. High ATP stimulates the enzyme, but fructose- 2,6 -bisphosphate inhibits. d. The enzyme is more active at low ATP than at high, and fructose- 2,6 -bisphosphate activates the enzyme. e. ATP and fructose- 2,6 -bisphosphate both inhibit the enzyme.

(Integrates with Chapter \(3 .)\) The standard free energy change \(\left(\Delta G^{\circ \prime}\right)\) for hydrolysis of fructose- 1,6 -bisphosphate (FBP) to fructose6-phosphate (F-6-P) and \(\mathrm{P}_{\mathrm{i}}\) is \(-16.7 \mathrm{kJ} / \mathrm{mol}\) \\[ \mathrm{FBP}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \text { fructose- } 6-\mathrm{P}+\mathrm{P}_{\mathrm{i}} \\] The standard free energy change \(\left(\Delta G^{\circ \prime}\right)\) for ATP hydrolysis is \(-30.5 \mathrm{kJ} / \mathrm{mol}\) \\[ \mathrm{ATP}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{ADP}+\mathrm{P}_{\mathrm{i}} \\] a. What is the standard free energy change for the phosphofructokinase reaction: \\[ \text { ATP + fructose- } 6 \text { -P } \longrightarrow \mathrm{ADP}+\mathrm{FBP} \\] b. What is the equilibrium constant for this reaction? c. Assuming the intracellular concentrations of [ATP] and [ADP] are maintained constant at \(4 \mathrm{m}\) Mand \(1.6 \mathrm{m} M\), respectively, in a rat liver cell, what will be the ratio of [FBP]/[fructose-6-P] when the phosphofructokinase reaction reaches equilibrium?
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