A metabolic pathway consisting of five reactions U→V→W→X→Y→Z has been proposed. The ΔG°′ for each reaction is shown below. Which of the following statements is TRUE regarding this proposed pathway?
U→V ΔG°′ = –15kJ/mol
V→W ΔG°′ = –2kJ/mol
W→X ΔG°′ = +5kJ/mol
X→Y ΔG°′ = +6kJ/mol
Y→Z ΔG°′ = –7kJ/mol

Question

BlackTom AI · Accepted Answer

Question restatement: A metabolic pathway with five sequential reactions U→V→W→X→Y→Z has given standard free energy changes, ΔG°′, for each step: U→V is –15 kJ/mol, V→W is –2 kJ/mol, W→X is +5 kJ/mol, X→Y is +6 kJ/mol, and Y→Z is –7 kJ/mol. Which statement about this proposed pathway is true? The provided answer claims: “This pathway may occur because the overall ΔG°′ is negative.”

First, compute the overall ΔG°′ for the entire pathway by summing the ΔG°′ values of all steps: ΔG°′ total = (–15) + (–2) + (+5) + (+6) + (–7) = –13 kJ/mol. This indicates that, under standard conditions, the overall reaction sequence would be exergonic.

Understanding the implications: In metabolic pathways, a negative overall ΔG°′ means that, as a whole, the pathway tends to proceed spontaneously in the forward direction from U to Z when reactant and product concentrations are at standard-state conditions. A key nuance is that individual steps can have positive or negative ΔG°′ values. Positive ΔG°′ steps (endergonic steps) can still occur within a pathway if the preceding steps release enough free energy to drive them, typically through coupled reactions and favorable metabolite concentrations.

Option-by-option analysis (conceptual since no explicit options were provided):
- If an option stated that the pathway can occur because the overall ΔG°′ is negative, that aligns with thermodynamic principles: a negative cumulative ΔG°′ supports feasibility of the forward progression of the pathway.
- An option suggesting that the pathway cannot occur because some individual steps have positive ΔG°′ would be misleading. Positive ΔG°′ steps do not preclude the overall pathway from proceeding if the net ΔG°′ is negative and if the cellular context provides coupling and appropriate metabolite levels.
- An option asserting that the pathway’s feasibility depends solely on the sign of each single step would be incorrect, because the emergent property is the sum of all steps, not the sign of each step in isolation.
- An option claiming that a positive total ΔG°′ would allow the pathway would be inconsistent with thermodynamics, as a positive overall ΔG°′ indicates nonspontaneity under standard conditions without coupling.

Conceptual takeaway: The crucial factor is the sum of the ΔG°′ values for all steps. A negative overall ΔG°′ indicates that, thermodynamically, the forward progression of the pathway is favorable, provided cellular conditions permit coupling and the relevant concentrations. Conversely, a positive overall ΔG°′ would imply that the pathway is not favored energetically unless significant coupling or energetic input changes the effective conditions.

Note on data gaps: The question as provided lists the correct-sounding statement but does not present alternative answer choices to evaluate. The analysis above focuses on the thermodynamic rationale for why a negative overall ΔG°′ supports pathway feasibility and why reliance on individual step signs alone can be misleading.

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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).

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?

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).