# Question 16084

Apr 10, 2015

The turnover number for your enzyme is ${\text{90.9 min}}^{- 1}$.

The turnover number, or ${K}_{\text{cat}}$, expresses how many substrate molecules are converted to product by the enzyme per unit time. Moreover, the turnover number is calculated when the enzyme has maximum efficiency $\to$ ${V}_{\text{max}}$.

${K}_{\text{cat" = V_"max}} / \left(\left[E\right]\right)$, where

$\left[E\right]$ - the total concentration of enzyme present.

Since ${V}_{\text{max}}$ uses $\mu \text{mol}$ and the concentration of the enzyme is given in $\text{nmol}$, I'll convert the latter to $\mu \text{mol}$

121cancel("nmol") * (10^(-3)mu"mol")/(1cancel("nmol")) = 121 * 10^(-3)mu"mol"#

Once again, plug and play

${K}_{\text{cat" = (11.0cancel(mu"mol") * "min"^(-1))/(121 * 10^(-3)cancel(mu"mol")) = "90.9 min}}^{- 1}$