Question

MATHEMATICAL You do an enzyme kinetic experiment and calculate a \(V_{\max }\) of \(100 \mu\) mol 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 mmber 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 known values

Given: - Maximum velocity, \(V_{\text{max}}\) = 100 µmol/min - Volume of enzyme solution per assay = 0.1 mL - Concentration of enzyme solution = 0.2 mg/mL - Molecular weight of the enzyme = 128,000 g/mol These values will be used in the calculations.

Calculate total enzyme amount used per assay

The total amount of enzyme used in each assay is given by the volume of enzyme solution multiplied by its concentration:\(\text{Total enzyme amount (mg)} = \text{Volume} \times \text{Concentration} = 0.1 \text{ mL} \times 0.2 \text{ mg/mL}= 0.02 \text{ mg} \)

Convert enzyme amount to moles

To calculate the moles of enzyme used, convert milligrams to grams, then use the molecular weight:\(\text{Total enzyme amount (g)} = \frac{0.02 \text{ mg}}{1000} = 0.00002 \text{ g} \)\(\text{Moles of enzyme} = \frac{\text{Total enzyme amount (g)}}{\text{Molecular weight (g/mol)}} = \frac{0.00002 \text{ g}}{128,000 \text{ g/mol}} = 1.5625 \times 10^{-10} \text{ mol} \)

Calculate turnover number (kcat)

The turnover number, \(k_{\text{cat}}\), is the number of substrate molecules converted to product per enzyme molecule per unit time:\(k_{\text{cat}} = \frac{V_{\text{max}}}{[\text{Enzyme}]} = \frac{100 \text{ µmol/min}}{1.5625 \times 10^{-10} \text{ mol}} = 6.4 \times 10^8 \text{ min}^{-1} \)

Key concepts

Turnover Number

The turnover number, often represented as \(k_{cat}\), is a crucial concept in enzyme kinetics. It indicates the maximum number of substrate molecules an enzyme molecule can convert into product per unit time when the enzyme is fully saturated with substrate. The turnover number can be computed with the formula: \(k_{cat} = \frac{V_{max}}{[E]}\), where \(V_{max}\) is the maximum reaction velocity and \[E\] is the enzyme concentration.

This calculation provides insight into the efficiency and effectiveness of an enzyme. In simple terms, the higher the turnover number, the more efficient the enzyme. For example, if \(k_{cat}\) is 6.4 × 10⁸ min⁻¹, it means each enzyme molecule converts 640 million substrate molecules into the product each minute, given all enzyme molecules are working at full capacity.

Vmax (Maximum Velocity)

The maximum velocity, or \(V_{max}\), represents the highest rate of reaction achieved by the enzyme when the substrate concentration is extremely high and all enzyme active sites are occupied. In the provided exercise, \(V_{max}\) is 100 µmol of product per minute.

Logically, this value helps us understand how quickly an enzyme can work when it is completely saturated with the substrate. For calculating the turnover number, \(V_{max}\) is a fundamental piece of data, as it reflects the enzyme's peak activity under optimal conditions. Determining \(V_{max}\) normally involves plotting the reaction rate against various substrate concentrations, creating a hyperbolic curve approaching this value.

Molecular Weight

The molecular weight of an enzyme is a measurement of the mass of one mole of its molecules. In this exercise, the enzyme has a molecular weight of 128,000 g/mol. Knowing the molecular weight helps to convert the mass of the enzyme into moles, which is essential for many kinetic calculations.

The molecular weight is vital for understanding the size and scale of the enzyme. By calculating enzyme concentration in molarity (moles per liter), we can accurately assess the number of enzyme molecules present, which is crucial when determining the turnover number.
For example, converting 0.02 mg of the enzyme into grams and then to moles was necessary to find the exact number of enzyme molecules working in the reaction, culminating in the calculation of the turnover number.

Enzyme Concentration

The concentration of the enzyme in a solution is another critical factor in enzyme kinetics. It is typically expressed in terms of mass per volume, but for detailed studies, converting this to molarity is helpful. In our exercise, the enzyme concentration is given as 0.2 mg/mL.

This parameter allows us to calculate the total enzyme amount used in the assay. From here, by knowing the volume of the enzyme solution used (0.1 mL), we can determine the total mass of the enzyme, which we then convert to moles considering the molecular weight.
Understanding this allows us to determine how many enzyme molecules are present, which directly ties into calculating the turnover number, depicting the efficiency of the enzyme. Ultimately, enzyme concentration aids in appreciating how enzyme molecules interact with the substrate and overload the reaction velocity.