Question

Carbonic anhydrase of erythrocytes (Mr 30,000) has one of the highest turnover numbers known. It catalyzes the reversible hydration of H2O + CO2 ⇄H2CO3 This is an important process in the transport of CO2 from the tissues to the lungs. If 10.0 g of pure carbonic anhydrase catalyzes the hydration of 0.30 g of CO2 in 1 min at 37 C at Vmax, what is the turnover number (kcat ) of carbonic anhydrase (in units of min-1 )?

Answer 1

Explanation:

According to the given data, the turnover number can be calculated as follows.

      Turnover number = $K_{cat} = \frac{V_{max}}{\text{Concentration of enzyme}}$

     $V_{max} = \frac{\text{Moles of CO_{2} hydrolyzed}{second}$

Therefore, moles of $CO_{2}$ hydrolyzed is as follows.

Moles of $CO_{2}$ hydrolyzed = $\frac{Mass of CO_{2}}{Molar mass of CO_{2}}$

                 = $\frac{0.30}{44}$

                 = 0.00682 moles

Now, moles of $CO_{2}$ hydolyzed per second is calculated as follows.

Moles of $CO_{2}$ hydolyzed per second = $\frac{0.00682}{60}$

             = $1.137 \times 10^{-4} moles/second = V_{max}$

And,

Moles of enzyme = $\frac{Mass}{\text{Molar mass}}$

                       = $\frac{10.0 \mu g}{30000}$

                       = $3.33 \times 10^{-10}$ moles

Therefore, the value of $K_{cat}$ is as follows.

    $K_{cat} = \frac{1.137 \times 10^{-4} moles}{3.333 \times 10^{-10} moles}$

               = $0.3411 \times 10^{6}$ per second

               = $0.3411 \times 60 \times 10^{6}$ per minute

               = $20.466 \times 10^{6}$ per minute

Thus, we can conclude that the turnover number ($K_{cat}$) of carbonic anhydrase (in units of $min^{-1}$) is $20.466 \times 10^{6}$ per minute.