Enzyme speed via turnover number ratio Steps:

• Recall standard turnover number for catalase: 2,884,000 s⁻¹ (high for H₂O₂ breakdown).
• Identify turnover number for phosphatase: 1,000 s⁻¹.
• Compute ratio of catalase to phosphatase: 2,884,000 ÷ 1,000 = 2,884.
• The ratio shows catalase converts substrate 2,884 times faster under optimum conditions. Why C is correct:
• Turnover number (k_cat) defines maximum reaction rate per enzyme molecule; ratio equals speed factor per Michaelis-Menten kinetics. Why the others are wrong:
• A: Underestimates ratio, possibly confusing with carbonic anhydrase (60,000 ÷ 1,000 ≈ 60, halved).
• B: Still underestimates, perhaps dividing by 10 incorrectly.
• D: Overestimates, likely extraneous multiplication by 10. Final answer: C