(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?

Short Answer

Expert verified
a. The standard free energy change for the pyruvate kinase reaction is -31.4 kJ/mol. b. The equilibrium constant for this reaction depends on the temperature, but assuming physiological temperature it can be calculated using the provided equation. c. The ratio of [pyruvate]/[phosphoenolpyruvate] at equilibrium can also be computed using the equilibrium constant and given concentrations of ATP and ADP.

Step by step solution

01

Calculate the standard free energy change for the pyruvate kinase reaction

The standard free energy change for the pyruvate kinase reaction can be obtained by subtracting the free energy change for ATP hydrolysis from that of PEP hydrolysis. This can be represented as: \(\Delta G'_{\text{{reaction}}} = \Delta G'_{\text{{PEP Hydrolysis}}} - \Delta G'_{\text{{ATP Hydrolysis}}}\). Substituting provided values from the problem, we get: \(\Delta G'_{\text{{reaction}}} = -61.9 \text{{kJ/mol}} - (-30.5 \text{{kJ/mol}}) = -31.4 \text{{kJ/mol}}\)
02

Calculate the equilibrium constant for the pyruvate kinase reaction

The equilibrium constant for the reaction can be obtained from the standard free energy change using the relationship: \( K_{eq} = e^{\Delta G'_{\text{{reaction}}}/-RT }\), where \(\Delta G'_{\text{{reaction}}}\) is the standard free energy change for the reaction (in J/mol), R is the universal gas constant (8.3145 J/mol/K), and T is the temperature. Assuming that the temperature is at physiological temperature (37 degrees Celsius or 310.15 K), we can substitute these values into the equation to calculate the equilibrium constant. Therefore, \( K_{eq} = e^{-31.4 \times 10^3/ (8.3145 \times 310.15)} \).
03

Calculate the equilibrium ratio of [pyruvate]/[phosphoenolpyruvate]

The equilibrium constant is defined by the ratio of product concentrations to reactant concentrations at equilibrium. For the pyruvate kinase reaction, this means: \( K_{eq} = ([\text{{Pyruvate}}][\text{{ATP}}])/([\text{{ADP}}][\text{{phosphoenolpyruvate}}]) \). We can rearrange this equation to solve for the desired ratio, \( [\text{{Pyruvate}}]/[\text{{phosphoenolpyruvate}}] \), based on the given concentrations of ATP and ADP. Solving for the specific case given in the problem, we find that \( [\text{{Pyruvate}}]/[\text{{phosphoenolpyruvate}}] = (K_{eq}\times[\text{{ADP}}])/[\text{{ATP}}] = (e^{-31.4 \times 10^3/ (8.3145 \times 310.15)} \times 1 \text{{mmol}})/8 \text{{mmol}} \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Free Energy Change
The concept of standard free energy change, represented as \( \Delta G^{\circ \prime} \) in biochemistry, is crucial for understanding the energetics of reactions in living organisms. It measures the amount of energy change that occurs during a reaction under standard conditions, which include a temperature of 25°C, 1 atmosphere of pressure, and all reactants and products at 1 M concentration, except for protons whose concentration is set at pH 7. A negative value of \( \Delta G^{\circ \prime} \) indicates that the reaction releases energy and can occur spontaneously, while a positive value would require energy input.

In biochemical pathways, the hydrolysis of compounds like phosphoenolpyruvate (PEP) and ATP is a common source of this energy. The magnitude of \( \Delta G^{\circ \prime} \) informs us about the potential of these compounds to drive biochemical reactions, with larger negative values indicating a higher energy release upon hydrolysis. Therefore, by analyzing \( \Delta G^{\circ \prime} \) values, students can predict the direction and spontaneity of reactions in metabolism.
Equilibrium Constant
The equilibrium constant, \( K_{eq} \) is a ratio that reflects the relative concentrations of products and reactants in a chemical reaction at equilibrium. It's a dimensionless number that gives us a sense of how far a reaction will go to completion before reaching a state where the forward and reverse reactions occur at the same rate. A large \( K_{eq} \) (much greater than 1) indicates that, at equilibrium, the reaction favors the formation of products; whereas a small \( K_{eq} \) (much less than 1) implies that the reactants are favored.

In the context of the pyruvate kinase reaction, knowing \( K_{eq} \) allows us to predict the extent to which phosphoenolpyruvate will be converted to pyruvate and ATP during cellular respiration. It informs researchers and students about how efficient this enzymatic step is under standard biological conditions and how different conditions might alter the reaction's equilibrium.
ATP Hydrolysis
ATP hydrolysis is a fundamental reaction in biochemistry where the molecule adenosine triphosphate (ATP) is broken down into adenosine diphosphate (ADP) and an inorganic phosphate (Pi). This reaction is known for releasing a significant amount of energy, which is then used by the cell to perform various functions, including muscle contraction, nerve impulse propagation, and chemical synthesis.

The standard free energy change for ATP hydrolysis is about \( -30.5 \text{ kJ/mol} \). Given its central role in cellular energy transfer, ATP is commonly referred to as the 'energy currency' of the cell. Understanding the energetics of ATP hydrolysis not only clarifies how cells harness energy for physiological processes but also ties into broader concepts like energy coupling and enzyme kinetics.
Phosphoenolpyruvate Hydrolysis
Phosphoenolpyruvate (PEP) hydrolysis is another biochemical reaction that releases energy and is used by cells to drive endergonic processes. During glycolysis, PEP is converted to pyruvate by the enzyme pyruvate kinase, and this step involves the transfer of a phosphate group from PEP to ADP, forming ATP.

The standard free energy change for the hydrolysis of PEP is significantly negative (\( -61.9 \text{ kJ/mol} \) ), making it an even more potent source of energy than ATP hydrolysis. The comparison of these energy changes is beneficial in understanding the energy dynamics within the cell. It provides insight into why PEP is such an effective phosphate donor and plays a pivotal role in the final steps of glycolytic ATP production.

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Most popular questions from this chapter

Fructose bisphosphate aldolase in animal muscle is a class I aldolase, which forms a Schiff base intermediate between substrate (for example, fructose- 1,6 -bisphosphate or dihydroxyacetone phosphate and a lysine at the active site (see Figure 18.12 ). The chemical evidence for this intermediate comes from studies with aldolase and the reducing agent sodium borohydride, \(\mathrm{NaBH}_{4}\). Incubation of the enzyme with dihydroxyacetone phosphate and \(\mathrm{NaBH}_{4}\) inactivates the enzyme. Interestingly, no inactivation is observed if \(\mathrm{NaBH}_{4}\) is added to the enzyme in the absence of substrate. Write a mechanism that explains these observations and provides evidence for the formation of a Schiff base intermediate in the aldolase reaction.

Genetic defects in glycolytic enzymes can have serious consequences for humans. For example, defects in the gene for pyruvate kinase can result in a condition known as hemolytic anemia. Consult a reference to learn about hemolytic anemia, and discuss why such genetic defects lead to this condition.

(Integrates with Chapter \(3 .)\) Triose phosphate isomerase catalyzes the conversion of dihydroxyacetone-P to glyceraldehyde-3-P. The standard free energy change, \(\Delta G^{\circ}\) ', for this reaction is \(+7.6 \mathrm{kJ} / \mathrm{mol}\). However, the observed free energy change \((\Delta G)\) for this reaction in erythrocytes is \(+2.4 \mathrm{kJ} / \mathrm{mol}\) a. Calculate the ratio of [dihydroxyacetone-P]/ [glyceraldehyde-3-P] in erythrocytes from \(\Delta G\) b. If [dihydroxyacetone-P] \(=0.2 \mathrm{m} M\), what is [glyceraldehyde-3-P]?

As noted on page 556 , the galactose- 1 -phosphate uridylyltransferase reaction proceeds via a ping-pong mechanism. Consult Chapter 13 page \(406,\) to refresh your knowledge of ping-pong mechanisms, and draw a diagram to show how a ping-pong mechanism would proceed for the uridylyltransferase.

Discuss the cycling of NADH and NAD \(^{+}\) in glycolysis and the related fermentation reactions.

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