Discussion
10 micrograms of the enzyme carbonic anhydrase (molecular weight = 30,000 g/mole) removes 300 milligrams of carbon dioxide per minute from the cells. The turnover number of the enzyme is
Answer: B
The turnover number is \(2 \times 10^{7}\ \text{min}^{-1}\).
Turnover number is the number of substrate molecules converted per enzyme molecule per unit time, so work in moles of each:
\[ \begin{aligned} n_{\text{CO}_2} &= \frac{0.300\ \text{g}}{44\ \text{g/mol}} = 6.82 \times 10^{-3}\ \text{mol/min} \\ n_{E} &= \frac{10 \times 10^{-6}\ \text{g}}{30000\ \text{g/mol}} = 3.33 \times 10^{-10}\ \text{mol} \\ k_{cat} &= \frac{6.82 \times 10^{-3}}{3.33 \times 10^{-10}} \approx 2.0 \times 10^{7} \end{aligned} \]
Carbonic anhydrase is famously one of the fastest enzymes, so a value of order \(10^{7}\) per minute (about \(10^{5}\) to \(10^{6}\) per second) is expected. The two \(10^{10}\) values are a thousand or more times too large, the size of error a microgram-milligram slip produces, and 20 per minute is far too slow for this enzyme.
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