During a laboratory experiment, you discover that an enzyme-catalyzed reaction has a deltaG of -20kcal/mol. If you double the amount of enzyme in the reaction, what will be the deltaG for the new reaction?

Question

BlackTom AI · Accepted Answer

Understanding the question requires recognizing the role of enzymes in a reaction. Enzymes accelerate the rate at which equilibrium is reached but do not alter the thermodynamic driving force, i.e., the Gibbs free energy change (deltaG) of the reaction itself.
Option 1: 40kcal/mol. This would suggest that doubling enzyme changes the sign or magnitude of deltaG to a positive value or reverses the energy landscape. In reality, enzyme concentration does not change deltaG, so this option misinterprets enzyme effect as a change in thermodynamics rather than kinetics.
Option 2: -20kcal/mol. This matches the original deltaG of the reaction. Since deltaG is a thermodynamic property determined by the difference in free energy between products and reactants under standard conditions (and not by the amount of enzyme present), doubling enzyme does not alter this value.
Option 3: -10kcal/mol. This option implies a partial reduction in the magnitude of deltaG, which would require a change in the intrinsic thermodynamics of the reaction, not simply a change in enzyme concentration. Enzymes cannot halve deltaG just by being present in greater amounts.
Option 4: 20kcal/mol. A positive deltaG would indicate a non-spontaneous or endergonic reaction under the given conditions, which contradicts the original thermodynamics of the system. Enzyme quantity does not flip the sign of deltaG.
Option 5: -40kcal/mol. This would double the magnitude in the negative direction, suggesting a stronger thermodynamic drive due to more enzyme, which is incorrect because deltaG is not dependent on enzyme amount.
In summary, the only statement consistent with fundamental biochemistry is that deltaG remains unchanged by simply doubling the amount of enzyme, since enzymes affect rate (kinetics) but not the intrinsic thermodynamic potential of the reaction.

During a laboratory experiment, you discover that an enzyme-catalyzed reaction has a deltaG of -20kcal/mol. If you double the amount of enzyme in the reaction, what will be the deltaG for the new reaction?单项选择题

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During a laboratory experiment, you discover that an enzyme-catalyzed reaction has a deltaG of -20kcal/mol. If you double the amount of enzyme in the reaction, what will be the deltaG for the new reaction?单项选择题

类似问题

The Gibbs free energy for Step 3 of glycolysis, is ∆G = –5.3 kcal/mole. In terms of energy changes, how would you classify this reaction?

The ∆Gº is a _______________ value often written in textbooks. The _______________ is used in cellular conditions because varying temperature and a high amount of _______________ may make even a ______________ ∆Gº reaction spontaneous. (overthrowing the natural favorability/non-favorability).

The transaminase transfer of an amine group from alanine to α-ketoglutarate to form pyruvate and glutamate has a change in Gibb’s free energy (ΔG) of 1.9 kJ/mol. What does this ΔG value tell you about the reaction?

The de-phosphorylation of phosphoenolpyruvate (PEP) to form pyruvate has a change in Gibb’s free energy (ΔG) of -61.9 kJ/mol. How would you describe the reaction?

The free energy change of hydrolysis of sucrose into glucose and fructose is ΔG = - 7 kcal/mol. When the enzyme sucrase is added to the reaction mixture, what happens to the ΔG of the reaction?

The transaminase transfer of an amine group from alanine to α-ketoglutarate to form pyruvate and glutamate has a change in Gibb’s free energy (ΔG) of 1.9 kJ/mol. What does this ΔG value tell you about the reaction?

更多留学生实用工具

希望你的学习变得更简单

The ∆Gº is a _______ value often written in textbooks. The _ is used in cellular conditions because varying temperature and a high amount of _ may make even a ______ ∆Gº reaction spontaneous. (overthrowing the natural favorability/non-favorability).