Question

You do an enzyme kinetic experiment and calculate a \(V_{\max }\) of 100 \mumol of product per minute. If each assay used \(0.1 \mathrm{mL}\) of an enzyme solution that had a concentration of \(0.2 \mathrm{mg} /\) \(\mathrm{mL},\) what would be the turnover number if the enzyme had a molecular weight of \(128,000 \mathrm{g} / \mathrm{mol}\) ?

Short answer

The turnover number is \(6.4 \times 10^{8} \text{min}^{-1}\).

Step-by-step solution

Identify the Given Data

The maximum reaction rate, or \(V_{\text{max}}\), is 100 \(\text{\textmu mol/min}\). The enzyme solution concentration is \(0.2 \text{mg/mL}\), and the volume of the enzyme used in each assay is \(0.1 \text{mL}\). The enzyme's molecular weight is \(128,000 \text{g/mol}\).

Calculate the Total Mass of Enzyme Used

Multiply the concentration of the enzyme solution by the volume used: \[ \text{Total mass of enzyme} = 0.2 \text{mg/mL} \times 0.1 \text{mL} = 0.02 \text{mg} \]

Convert Mass of Enzyme to Moles

Use the molecular weight to convert the mass of enzyme used to moles: \[ \left( \text{Total moles of enzyme} = \frac{0.02 \text{mg}}{128,000 \text{g/mol}} \right) \times \frac{1 \text{g}}{1000 \text{mg}} = 1.5625 \times 10^{-10} \text{mol} \]

Calculate the Turnover Number (\text{k}_{\text{cat}})

The turnover number \(k_{cat}\) is calculated by dividing \V_{\text{max}}\ by the total moles of enzyme: \[ k_{cat} = \frac{V_{\text{max}}}{[\text{Total moles of enzyme}]} = \frac{100 \text{\textmu mol/min}}{1.5625 \times 10^{-10} \text{mol}} \right = 6.4 \times 10^{8} \text{min}^{-1} \]

Key concepts

turnover number

The turnover number, often denoted as \( k_{cat} \), represents the number of substrate molecules each enzyme molecule converts to product per unit time when the enzyme is fully saturated with substrate. This is an important measure because it provides insight into the efficiency of the enzyme. To calculate the turnover number, you need both the maximum reaction rate \( V_{max} \) and the total moles of enzyme present. The formula for turnover number is: \[ k_{cat} = \frac{V_{max}}{[\text{Total moles of enzyme}]} \]. In the given exercise, it was calculated that \( k_{cat} = 6.4 \times 10^{8} \text{ min}^{-1} \), meaning each enzyme molecule catalyzes the conversion of 6.4 billion substrate molecules per minute.

molecular weight

The molecular weight of an enzyme is the mass of one mole of enzyme molecules. It is typically expressed in grams per mole (g/mol) and can be found using various laboratory techniques such as gel electrophoresis or mass spectrometry. Knowing the molecular weight is crucial for converting the mass of an enzyme to moles, which is necessary for enzyme kinetics calculations. In the given exercise, the molecular weight of the enzyme was provided as 128,000 g/mol. This information was used to convert the enzyme mass of 0.02 mg to moles, resulting in \( 1.5625 \times 10^{-10} \) mol.

enzyme concentration

Enzyme concentration refers to the amount of enzyme present in a given volume of solution, usually expressed in terms of mass per volume, such as mg/mL. It is a key variable in enzyme kinetics experiments because changes in enzyme concentration can affect the reaction rate. To utilize enzyme concentration effectively, you may need to perform serial dilutions or prepare stock solutions to reach the desired concentration for your assays. In our exercise, the enzyme concentration given was 0.2 mg/mL and the volume used was 0.1 mL. Using these values, we calculated the total mass of enzyme as 0.02 mg. This value was essential in further calculations for determining the turnover number.

Vmax calculation

The maximum reaction rate \( V_{max} \) is the rate of an enzymatic reaction when the enzyme is fully saturated with substrate. It indicates the total capacity of the enzyme to convert substrate to product. For calculating \( V_{max} \), you need to monitor the reaction rate at increasing substrate concentrations until the rate no longer increases, indicating that all enzyme active sites are occupied. In the exercise, \( V_{max} \) was provided as 100 \( \text{\textmu mol/min} \). This value was used alongside the total moles of enzyme to determine the turnover number, giving us a complete understanding of the enzyme's efficiency under the provided conditions.