Question

You are characterizing a new enzyme isolated and purified in the laboratory. If the maximum velocity of the enzyme is 1800 \(\mu\)moles L^-1 min^-1 and the total concentration of the enzyme in the reaction mixture is 1.5 \(\mu\)M, then the turnover number of the enzyme is s^-1. (answer in integer)

Hint:

Turnover number kcat = Vmax divided by total enzyme concentration; keep both in the same concentration unit before dividing, then convert time units if needed.

Answer 1

Correct Answer: 20

Let's take the opposite route: convert the rate to a per second basis first, then divide by the enzyme concentration.

The maximum velocity is $V_{max} = 1800\ \mu\text{mol L}^{-1}\text{min}^{-1}$. Since 1 minute has 60 seconds, the rate per second is

\[ V_{max} = \frac{1800}{60}\ \mu\text{mol L}^{-1}\text{s}^{-1} = 30\ \mu\text{mol L}^{-1}\text{s}^{-1} \]

The enzyme concentration is $[E]_T = 1.5\ \mu\text{M} = 1.5\ \mu\text{mol L}^{-1}$, already in the same $\mu$mol/L unit as $V_{max}$, so the units cancel cleanly.

Turnover number is defined as how many substrate molecules a single enzyme molecule converts to product each second when saturated with substrate:

\[ k_{cat} = \frac{V_{max}}{[E]_T} = \frac{30\ \mu\text{mol L}^{-1}\text{s}^{-1}}{1.5\ \mu\text{mol L}^{-1}} = 20\ \text{s}^{-1} \]

Let's summarize:

• Converting the rate to per second units before dividing skips a separate unit conversion step at the end.
• Turnover number carries units of reciprocal time (here, per second) because it counts molecules converted per enzyme molecule per unit time.

So the turnover number of this enzyme is $20\ \text{s}^{-1}$.