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

To evaluate the proposed pathway, I’ll examine both the individual steps and the overall thermodynamics of the sequence.
Option 1: This pathway will not occur. This is incorrect. A pathway can still proceed if the overall free energy change is favorable, even if some individual steps are unfavorable. The sum of ΔG°′ values determines the net feasibility, not a blanket statement that the pathway cannot occur.
Option 2: The first two reactions of the pathway will occur, but once W is formed the pathway will halt. While the first two steps (U→V and V→W) have negative ΔG°′ values (–15 and –2 kJ/mol), suggesting they are favorable, this argument ignores subsequent steps. A later step with a positive ΔG°′ could be unfavorable, but a subsequent highly favorable step can still drive the overall process if the net ΔG°′ is negative. Therefore asserting that the pathway halts after W is formed is not justified by the data alone.
Option 3: This pathway may occur because the overall ΔG°′ is negative. This is the key point: summing all five steps gives ΔG°′ total = –15 + (–2) + 5 + 6 + (–7) = –13 kJ/mol, which is negative. A negative overall ΔG°′ indicates the pathway is energetically favorable as a whole, so it could occur under standard conditions, assuming no kinetic or regulatory barriers prevent it.
Option 4: This pathway may occur because the largest ΔG°′ is a negative number. This statement is misleading. The largest magnitude of ΔG°′ does not by itself determine feasibility; what matters for spontaneity is the sum of all steps. There are positive ΔG°′ steps (+5 and +6 kJ/mol) that raise the energy of those steps, and a single largest negative step (–15 kJ/mol) does not guarantee overall feasibility when balanced by the positives. Hence this rationale is flawed.
In summary, when you consider the complete sequence, the net ΔG°′ is negative, making the path energetically favorable overall, while the other options rely on incorrect interpretations of how individual steps relate to the whole pathway.

类似问题

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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).