60. The first step in the biosynthesis of valine begines with enzyme catalyzed condensation of two molecules of pyruvic acid. If an equimolar mixture of 13CH3COCOOH and 12CH3COCOOH are used as substrates for the reaction, which one of the following would represent the correct isotope incorporation pattern of the pro-R and pro-S diastereotopic methyl group in valine? (1) 50% 13CH3 (pro-R), 12CH3 (pro-S) 50% 12CH3 (pro-R), 13CH3 (pro-S) (2) 75% 13CH3 (pro-R), 12CH3 (pro-S) 25% 12CH3 (pro-R), 13CH3 (pro-S) (3) 25% 13CH3 (pro-R), 12CH3 (pro-S) 25% 12CH3 (pro-R), 13CH3 (pro-S) 25% 13CH3 (pro-R), 13CH3 (pro-S) 25% 12CH3 (pro-R), 12CH3 (pro-S) (4) 75% 12CH3 (pro-R), 13CH3 (pro-S) 25% 13CH3 (pro-R), 13CH3 (pro-S)
  1. The first step in the biosynthesis of valine begines with enzyme catalyzed condensation of two molecules of pyruvic acid. If an equimolar mixture of 13CH3COCOOH and 12CH3COCOOH are used as substrates for the reaction, which one of the following would represent the correct isotope incorporation pattern of the pro-R and pro-S diastereotopic methyl group in valine?
    (1) 50% 13CH3 (pro-R), 12CH3 (pro-S)
    50% 12CH3 (pro-R), 13CH3 (pro-S)
    (2) 75% 13CH3 (pro-R), 12CH3 (pro-S)
    25% 12CH3 (pro-R), 13CH3 (pro-S)
    (3) 25% 13CH3 (pro-R), 12CH3 (pro-S)
    25% 12CH3 (pro-R), 13CH3 (pro-S)
    25% 13CH3 (pro-R), 13CH3 (pro-S)
    25% 12CH3 (pro-R), 12CH3 (pro-S)
    (4) 75% 12CH3 (pro-R), 13CH3 (pro-S)
    25% 13CH3 (pro-R), 13CH3 (pro-S)

    Background: Valine Biosynthesis and Pyruvate Condensation

    • Valine is synthesized from two pyruvate molecules via acetohydroxyacid synthase (AHAS) catalyzing their condensation to form acetolactate.

    • The two pyruvate molecules contribute differently to the two methyl groups (pro-R and pro-S) in the final valine structure.

    • Because the two pyruvate molecules are chemically identical but isotopically distinguishable (one labeled, one unlabeled), the distribution of ^13C in the methyl groups depends on which pyruvate molecule contributes to which methyl group.

    Isotope Incorporation Logic

    • Using an equimolar mixture of ^13C-labeled and ^12C unlabeled pyruvate means each pyruvate molecule has a 50% chance of being labeled.

    • The condensation reaction is random and symmetrical with respect to the two pyruvate molecules.

    • Therefore, four possible combinations for the two methyl groups arise:

      1. pro-R methyl from ^13C pyruvate, pro-S methyl from ^12C pyruvate

      2. pro-R methyl from ^12C pyruvate, pro-S methyl from ^13C pyruvate

      3. both pro-R and pro-S methyls from ^13C pyruvate

      4. both pro-R and pro-S methyls from ^12C pyruvate

    • The probabilities for these are based on independent incorporation:

      • Both methyls ^13C: 0.5 × 0.5 = 25%

      • Both methyls ^12C: 0.5 × 0.5 = 25%

      • Mixed methyls (one ^13C, one ^12C): 0.5 × 0.5 × 2 = 50% (because order matters)

    Expected Distribution of Methyl Groups

    • 50% of valine molecules have one methyl group labeled and the other unlabeled, with equal chance for pro-R or pro-S methyl group to be labeled.

    • 25% have both methyl groups labeled (^13C).

    • 25% have both methyl groups unlabeled (^12C).

    This corresponds to option (3) in the query.

    Summary Table of Isotope Incorporation

    Methyl Group Labeling Percentage Description
    pro-R ^13CH3, pro-S ^12CH3 25% Mixed labeling with pro-R methyl labeled
    pro-R ^12CH3, pro-S ^13CH3 25% Mixed labeling with pro-S methyl labeled
    pro-R ^13CH3, pro-S ^13CH3 25% Both methyl groups labeled
    pro-R ^12CH3, pro-S ^12CH3 25% Both methyl groups unlabeled

    Conclusion

    When an equimolar mixture of ^13C-labeled and ^12C unlabeled pyruvic acid is used in valine biosynthesis, the isotope incorporation pattern in the pro-R and pro-S methyl groups of valine is:

    (3) 25% 13CH3 (pro-R), 12CH3 (pro-S); 25% 12CH3 (pro-R), 13CH3 (pro-S); 25% 13CH3 (pro-R), 13CH3 (pro-S); 25% 12CH3 (pro-R), 12CH3 (pro-S)

4 Comments
  • Aakansha sharma Sharma
    September 24, 2025

    (3) 25% 13CH3 (pro-R), 12CH3 (pro-S); 25% 12CH3 (pro-R), 13CH3 (pro-S); 25% 13CH3 (pro-R), 13CH3 (pro-S); 25% 12CH3 (pro-R), 12CH3 (pro-S

    • Varsha Tatla
      September 27, 2025

      Option 3 will be correct

  • Kajal
    October 1, 2025

    Option 3 is correct

  • Khushi Vaishnav
    October 7, 2025

    3 is the correct answer

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