Carbonic anhydrase ( $M = 30000 g mo l^{- 1}$ ) has a turnover number $k_{c a t} = 6.0 \times 1 0^{5} s^{- 1}$ (substrate molecules per enzyme per second at saturation). Assuming saturation throughout, what mass of $CO X_{2}$ ( $M = 44 g mo l^{- 1}$ ) is hydrated by $1.7 mg$ of the enzyme in $1.0 min$ ? Answer in grams to one decimal.

Numerical answer type, answer in g, tolerance 0.1.

Answer: 89.8 g

Moles of enzyme

= 1.7 \times 1 0^{- 3} /30000 = 5.667 e - 08 mol

. Rate

= 6.0 \times 1 0^{5} \times 5.667 e - 08 = 0.0340 mol s^{- 1}

. In 60 s this is 2.040 mol, a mass of

\times 44 = 89.76

g, i.e. 89.8 g.