Q.37 The 𝚫𝐆′𝐨for the malate dehydrogenase catalyzed step of Krebs cycle is +𝟕. 𝟏𝐤𝐜𝐚𝐥/𝐦𝐨𝐥𝐞.Nevertheless, the conversion of malate to oxaloacetate in vivo proceeds spontaneously because the subsequent reaction that consumes oxaloacetate has a 𝚫𝐆′𝟎 of
(A) −3.0kcal/mole
(B) +3.0kcal/mole
(C) −7.7kcal/mole
(D) +7.7kcal/mole

The correct answer is (C) −7.7 kcal/mole.
A strongly negative 𝚫G′° for the subsequent step (citrate synthase reaction) pulls the malate dehydrogenase reaction forward in vivo, despite its positive 𝚫G′°.

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Concept Overview

In the Krebs (TCA) cycle, the oxidation of malate to oxaloacetate by malate dehydrogenase has a standard free energy change 𝚫G′° of +7.1 kcal/mole, which means that under standard conditions the reaction favors malate, not oxaloacetate. In living cells, however, this step proceeds spontaneously forward because it is tightly coupled to the next reaction: condensation of oxaloacetate with acetyl‑CoA by citrate synthase, which is highly exergonic (strongly negative 𝚫G′°). This coupling makes the overall free energy change negative, driving the sequence forward.

For the overall process (malate → oxaloacetate → citrate) to be spontaneous, the 𝚫G′° of the citrate synthase step must be more negative in magnitude than +7.1 kcal/mole, so that the sum is negative. That corresponds to about −7.7 kcal/mole, which matches option (C) as the best choice.

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Why Option (C) −7.7 kcal/mole Is Correct

• The malate dehydrogenase reaction:
  malate + NAD⁺ ⇌ oxaloacetate + NADH + H⁺
  has 𝚫G′° = +7.1 kcal/mole, so by itself it is endergonic and does not proceed spontaneously under standard conditions.

• The next step, catalyzed by citrate synthase, consumes oxaloacetate and forms citrate from oxaloacetate and acetyl‑CoA. This step is highly exergonic, with a large negative 𝚫G′°.

• For the combined reaction (malate → oxaloacetate → citrate) to have overall negative free energy, the 𝚫G′° of the second step must be more negative than −7.1 kcal/mole. A value of about −7.7 kcal/mole makes the net 𝚫G′° negative and thus spontaneous in vivo.

• Therefore, among the given values, −7.7 kcal/mole is the only one that is both sufficiently negative and consistent with the known strongly exergonic character of the citrate synthase step.

Hence, the correct answer is (C) −7.7 kcal/mole.

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Explanation of All Options

Option (A) −3.0 kcal/mole

• A 𝚫G′° of −3.0 kcal/mole for the citrate synthase step would be exergonic but weakly so.

• If we add the two steps:

  • Malate dehydrogenase: +7.1 kcal/mole

  • “Citrate synthase” (hypothetical here): −3.0 kcal/mole

  • Net: +4.1 kcal/mole

• The combined 𝚫G′° would still be positive, meaning the overall process would not be spontaneous under standard conditions.

• This option cannot explain why malate is effectively pulled forward to oxaloacetate in vivo, so it is incorrect.

Option (B) +3.0 kcal/mole

• A 𝚫G′° of +3.0 kcal/mole means the subsequent reaction is also endergonic.

• Adding:

  • Malate dehydrogenase: +7.1 kcal/mole

  • “Citrate synthase” (hypothetical): +3.0 kcal/mole

  • Net: +10.1 kcal/mole

• Now the overall process is even more non‑spontaneous, which completely contradicts the observed smooth operation of the TCA cycle in cells.

• A positive 𝚫G′° for the step that consumes oxaloacetate cannot drive the malate → oxaloacetate reaction forward, so this option is clearly wrong.

Option (C) −7.7 kcal/mole

• This value is strongly negative, indicating a highly exergonic reaction.

• Adding:

  • Malate dehydrogenase: +7.1 kcal/mole

  • Citrate synthase (given): −7.7 kcal/mole

  • Net: −0.6 kcal/mole

• The combined 𝚫G′° becomes slightly negative, which is enough to make the overall coupled process spontaneous under physiological conditions.

• This captures the key biochemical principle: an unfavorable step (positive 𝚫G′°) can proceed if tightly coupled to a more favorable step (sufficiently negative 𝚫G′°).

• Therefore, this is the correct option.

Option (D) +7.7 kcal/mole

• A 𝚫G′° of +7.7 kcal/mole for the subsequent reaction means it is even more endergonic than the malate dehydrogenase step.

• Adding:

  • Malate dehydrogenase: +7.1 kcal/mole

  • “Citrate synthase” (hypothetical): +7.7 kcal/mole

  • Net: +14.8 kcal/mole

• This would make the combined sequence strongly non‑spontaneous, which is incompatible with the continuous operation of the Krebs cycle.

• A positive and large 𝚫G′° for the step that consumes oxaloacetate would actually oppose malate oxidation instead of pulling it forward, so this option is incorrect.

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Key Takeaways for Exam Preparation

• A reaction with positive 𝚫G′° can still proceed in vivo if it is coupled to a subsequent reaction with a more negative 𝚫G′°, making the overall process exergonic.

• In the Krebs cycle:

  • Malate dehydrogenase step: +7.1 kcal/mole (unfavorable alone)

  • Citrate synthase step: large negative 𝚫G′°, approximated here by −7.7 kcal/mole

• This coupling explains why malate → oxaloacetate proceeds spontaneously in living cells, making option (C) the correct answer.