If 10 µg of pure carbonic anhydrase catalyzes the hydration of 0.30 g of CO₂ in 1 min at 37°C at $$V_{max},$$ what is the turnover number $$(k_{cat})$$ of carbonic anhydrase (in units of $$min^{-1}$$)? ($$M_{r}$$ of carbonic anhydrase is 30,000)​

Correct option is B

Step 1: Calculate moles of enzyme

Mass of enzyme = 10 µg = 10 × 10⁻⁶ g
Molar mass of enzyme = 30,000 g/mol

Moles of enzyme = (10 × 10⁻⁶) / 30,000
Moles of enzyme = 3.33 × 10⁻¹⁰ mol

Step 2: Calculate moles of CO₂ converted

Mass of CO₂ = 0.30 g
Molar mass of CO₂ = 44 g/mol

Moles of CO₂ = 0.30 / 44
Moles of CO₂ = 6.82 × 10⁻³ mol

Step 3: Turnover number (kcat)

kcat = (moles of substrate converted per minute) / (moles of enzyme)

kcat = (6.82 × 10⁻³) / (3.33 × 10⁻¹⁰)
kcat ≈ 2.05 × 10⁷ min⁻¹
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Final Answer:
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Turnover number, kcat = 2.05 × 10⁷ min⁻¹