Question

You do an enzyme kinetic experiment and calculate a \(V_{\max }\) of 100 pmol 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 641,025 min^{-1}.

Step-by-step solution

Determine the Amount of Enzyme in the Assay

Each assay uses 0.1 mL of an enzyme solution with a concentration of 0.2 mg/mL. Calculate the amount of enzyme in the assay by multiplying the volume by the concentration: \[ (\text{Volume}) \times (\text{Concentration}) = 0.1 \text{ mL} \times 0.2 \text{ mg/mL} = 0.02 \text{ mg} \text{ enzyme} \]

Convert Enzyme Amount to Moles

To find the amount in moles, use the molecular weight. Given a molecular weight of 128,000 g/mol, convert 0.02 mg to grams: \[ 0.02 \text{ mg} = 0.02 \times 10^{-3} \text{ g} = 2 \times 10^{-5} \text{ g} \] Then find the moles: \[ \text{Moles of enzyme} = \frac{\text{mass}}{\text{molecular weight}} = \frac{2 \times 10^{-5} \text{ g}}{128,000 \text{ g/mol}} = \frac{2 \times 10^{-5}}{128,000} = \frac{2}{128 \times 10^6} = 1.56 \times 10^{-10} \text{ mol} = 1.56 \times 10^{-7} \text{ μmol} \]

Calculate the Turnover Number

Turnover number is calculated by dividing the maximum velocity ( \( V_{\text{max}} \) ) by the moles of enzyme used in the assay: \[ k_{cat} = \frac{V_{\text{max}}}{[\text{Enzyme}]} \] Given a \( V_{\text{max}} \) of 100 pmol/min and 1.56 x 10^-7 μmol of enzyme: \[ k_{cat} = \frac{100 \text{ pmol/min}}{1.56 \times 10^{-7} \text{ μmol}} = \frac{100 \times 10^{-12}}{1.56 \times 10^{-7}} \text{ min}^{-1} = \frac{100}{1.56 \times 10^5} \text{ min}^{-1} = 641025 \text{ min}^{-1} \]

Key concepts

turnover number

The turnover number, often represented as \(k_{cat}\), is a crucial concept in enzyme kinetics. It specifies how many substrate molecules one enzyme molecule can convert to product per minute. This number helps us understand an enzyme's efficiency.
Think of it as how productive a single worker in a factory is over a period of time. The turnover number formula is:
\(k_{cat} = \frac{V_{\text{max}}}{[\text{Enzyme}]}\)
Here, \(V_{\text{max}}\) is the maximum rate of the reaction when the enzyme is saturated with substrate. The enzyme concentration (\([\text{Enzyme}]\)) is the amount of enzyme available to catalyze the reaction.
By understanding the turnover number, scientists can determine how fast an enzyme works. It’s particularly useful in comparing different enzymes to find the most efficient one for a specific process.

molecular weight

Molecular weight (also known as molar mass) is the mass of one mole of a substance. For enzymes, it is typically measured in grams per mole (g/mol).
It's a critical factor in various calculations, such as converting mass to moles. The molecular weight is the sum of the atomic weights of all atoms in the enzyme's molecular formula.
In our problem, the enzyme's molecular weight is 128,000 g/mol. This means that one mole of this enzyme weighs 128,000 grams. To work with smaller amounts, we convert milligrams to grams, then use the molecular weight to determine the enzyme quantity in moles.
Understanding molecular weight helps in scaling reactions and determining the precise amount of enzyme needed for experiments.

enzyme concentration

Enzyme concentration refers to the amount of enzyme present in a solution, typically given in mg/mL. It plays a crucial role in determining the reaction rate.
In our example, the enzyme concentration is 0.2 mg/mL. By multiplying this with the volume of the enzyme solution used (0.1 mL), we find the total amount of enzyme in the assay:
\((\text{Volume}) \times (\text{Concentration}) = 0.1 \text{ mL} \times 0.2 \text{ mg/mL} = 0.02 \text{ mg}\)
After finding the mg of enzyme, we convert this to moles by using the enzyme's molecular weight.
The concentration tells us how packed an enzyme solution is. Higher concentrations mean more enzymes available to catalyze reactions, leading to faster reaction rates, provided there’s enough substrate available.