Link Between Glycolysis and the TCA Cycle The end product of glycolysis is pyruvate. Before entering the citric acid cycle, pyruvate is converted to which activated molecule?

Biochemistry TCA Cycle Difficulty: Easy
Choose an option
  • A
    Acetic acid
  • B
    Acetyl-CoA
  • C
    Acetaldehyde
  • D
    None of these
  • E
    Oxaloacetate

Answer

Correct Answer: Acetyl-CoA

Explanation

Introduction / Context:Glycolysis occurs in the cytosol and generates pyruvate. To feed the TCA cycle, pyruvate must be transported into mitochondria and converted to an activated two-carbon donor. Identifying the correct intermediate clarifies central metabolism and bioenergetics.

Given Data / Assumptions:

  • Pyruvate dehydrogenase complex (PDC) resides in the mitochondrial matrix.
  • Coenzymes include TPP, lipoamide, FAD, CoA, and NAD+.

Concept / Approach:PDC catalyzes oxidative decarboxylation of pyruvate to acetyl-CoA, releasing CO2 and producing NADH. Acetyl-CoA then condenses with oxaloacetate to form citrate, beginning the TCA cycle. Acetaldehyde formation is a yeast fermentation step, not the aerobic link to TCA; acetic acid is not the activated thioester needed for entry.

Step-by-Step Solution:

Transport pyruvate into mitochondria via the pyruvate carrier.Convert pyruvate to acetyl-CoA by PDC: pyruvate + CoA + NAD+ → acetyl-CoA + CO2 + NADH.Feed acetyl-CoA into TCA by condensation with oxaloacetate to form citrate.

Verification / Alternative check:PDC deficiency causes lactic acidosis and neuro symptoms due to impaired acetyl-CoA supply, underscoring its central role.

Why Other Options Are Wrong:

  • Acetaldehyde: fermentation intermediate in yeast.
  • Acetic acid: lacks CoA activation; not used directly by citrate synthase.
  • Oxaloacetate: a TCA intermediate formed from other routes, not directly from pyruvate for TCA entry under standard conditions (except via pyruvate carboxylase for gluconeogenesis/anaplerosis).

Common Pitfalls:Confusing anaerobic fermentative pathways with aerobic respiration steps.

Final Answer:Acetyl-CoA

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