Biochemistry—Catalysis and Energetics Chymotrypsin increases the rate of peptide bond hydrolysis by approximately 10^10. At physiological temperature, this rate enhancement corresponds most closely to what reduction in activation free energy (ΔΔG‡)?

Introduction / Context:Rate enhancements by enzymes can be translated into decreases in activation free energy using the relationship between rate constants and activation barriers. This question connects kinetic observables (fold-increase in rate) with thermodynamic barriers (ΔG‡) at a given temperature.

Given Data / Assumptions:

• Rate enhancement ≈ 10^10 relative to uncatalyzed reaction.
• Temperature near 298 K for estimation.
• Relationship: ΔΔG‡ ≈ −RTln(k_cat/k_uncat).
• R ≈ 8.314 J/mol·K; RT at 298 K ≈ 2.48 kJ/mol.

Concept / Approach:Use the exponential dependence of rate on activation energy. A 10^10 enhancement implies a substantial lowering of the barrier. Compute ΔΔG‡ with RT and the natural logarithm of the rate ratio to estimate the energetic advantage provided by chymotrypsin’s catalytic strategy (transition-state stabilization, covalent catalysis, general acid-base catalysis, and precise positioning).

Step-by-Step Solution:

Verification / Alternative check:Sensitivity analysis: if T varies modestly around 298 K, RT shifts slightly (about 2.5 kJ/mol), giving ΔΔG‡ still near 57–59 kJ/mol, confirming that 58 kJ/mol is the best match among options.

Why Other Options Are Wrong:

• 40 or 49 kJ/mol: underestimate the barrier reduction implied by a 10^10 rate boost.
• 88 kJ/mol: overestimates the barrier reduction.
• 25 kJ/mol: far too small for 10 orders of magnitude in rate.

Common Pitfalls:Using log base 10 directly without converting to natural log; forgetting to multiply by RT in kJ/mol rather than J/mol; mixing enthalpic and entropic interpretations.

Final Answer:58 kJ/mol