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

Step 1: Recall what turnover number means.
The turnover number, \(k_{cat}\), is the number of substrate molecules one enzyme molecule converts to product per unit time when the enzyme is fully saturated with substrate. It comes from the Michaelis-Menten relation
\[ V_{max} = k_{cat}[E]_{T} \]
where \([E]_T\) is the total enzyme concentration.

Step 2: Rearrange for \(k_{cat}\).
\[ k_{cat} = \frac{V_{max}}{[E]_T} \]

Step 3: Convert both quantities to consistent units.
\(V_{max} = 1800\ \mu\text{mol L}^{-1}\text{min}^{-1} = 1800 \times 10^{-6}\ \text{mol L}^{-1}\text{min}^{-1} = 1.8 \times 10^{-3}\ \text{mol L}^{-1}\text{min}^{-1}\).
\([E]_T = 1.5\ \mu\text{M} = 1.5 \times 10^{-6}\ \text{mol L}^{-1}\).

Step 4: Compute \(k_{cat}\) per minute, then convert to per second.
\[ k_{cat} = \frac{1.8 \times 10^{-3}}{1.5 \times 10^{-6}}\ \text{min}^{-1} = 1200\ \text{min}^{-1} \]
Since there are 60 seconds in a minute,
\[ k_{cat} = \frac{1200}{60}\ \text{s}^{-1} = 20\ \text{s}^{-1} \]

Final Answer:
The turnover number of the enzyme is \[ \boxed{20\ \text{s}^{-1}} \]