Question

Applying the Michaelis-Menten Equation II An enzyme is present at a concentration of \(1 \mathrm{~nm}\) and has a \(V_{\max }\) of \(2 \mu \mathrm{M} \mathrm{s}^{-1}\). The \(K_{\mathrm{m}}\) for its primary substrate is \(4 \mu \mathrm{M}\). a. Calculate \(k_{\text {cat }}\). b. Calculate the apparent (measured) \(V_{\max }\) and apparent (measured) \(K_{\mathrm{m}}\) of this enzyme in the presence of sufficient amounts of an uncompetitive inhibitor to generate an \(\alpha^{\prime}\) of 2 . Assume that the enzyme concentration remains at \(1 \mathrm{~nm}\).

Short answer

\( k_{\text{cat}} = 2000 \text{s}^{-1}, V_{\max}^{\text{app}} = 1 \mu \text{M s}^{-1}, K_m^{\text{app}} = 2 \mu \text{M} \).

Step-by-step solution

Understand the given data

We have an enzyme concentration of \( 1 \text{ nM} \), \( V_{\max} = 2 \mu \text{M s}^{-1} \), and \( K_m = 4 \mu \text{M} \). The presence of an uncompetitive inhibitor creates an \( \alpha'\) value of 2. We need to calculate the turnover number \( k_{\text{cat}} \), and the apparent \( V_{\max} \) and \( K_m \) with the inhibitor.

Calculate \( k_{\text{cat}} \)

The turnover number, \( k_{\text{cat}} \), is calculated using the formula:\[ k_{\text{cat}} = \frac{V_{\max}}{[E]} \]Substitute the given values:\[ k_{\text{cat}} = \frac{2 \mu \text{M s}^{-1}}{1 \times 10^{-3} \mu \text{M}} = 2000 \text{s}^{-1} \]

Recall the effect of uncompetitive inhibition

In uncompetitive inhibition, the apparent \( V_{\max} \) and \( K_m \) are both reduced by the factor \( \alpha' \). This is because the inhibitor binds to the enzyme-substrate complex.

Calculate the apparent \( V_{\max} \, (V_{max}^{app}) \)

The apparent \( V_{\max} \) in the presence of an uncompetitive inhibitor is given by:\[ V_{\max}^{\text{app}} = \frac{V_{\max}}{\alpha'} \]Substitute the given values:\[ V_{\max}^{\text{app}} = \frac{2 \mu \text{M s}^{-1}}{2} = 1 \mu \text{M s}^{-1} \]

Calculate the apparent \( K_m \, (K_m^{app}) \)

The apparent \( K_m \) in the presence of an uncompetitive inhibitor is:\[ K_m^{\text{app}} = \frac{K_m}{\alpha'} \]Substitute the given values:\[ K_m^{\text{app}} = \frac{4 \mu \text{M}}{2} = 2 \mu \text{M} \]

Summarize the results

We calculated the turnover number \( k_{\text{cat}} = 2000 \text{s}^{-1} \). In the presence of the uncompetitive inhibitor, the apparent \( V_{\max} = 1 \mu \text{M s}^{-1} \) and the apparent \( K_m = 2 \mu \text{M} \).

Key concepts

Enzyme kinetics

Enzyme kinetics is the study of how fast chemical reactions are catalyzed by enzymes. It provides insights into the enzyme's function and efficiency in biological reactions. At the heart of enzyme kinetics is the Michaelis-Menten equation:\[V = \frac{V_{\max}[S]}{K_m + [S]}\]where:

• \(V\) is the reaction rate.
• \(V_{\max}\) is the maximum rate of the reaction when the enzyme is saturated.
• \([S]\) is the substrate concentration.
• \(K_m\) is the Michaelis constant, a measure of how easily the substrate binds to the enzyme.

Enzyme kinetics is crucial for understanding how different factors, such as enzyme concentration, temperature, pH, and inhibitors, affect the speed of a reaction.
This field allows scientists to predict how biochemical pathways operate in different conditions, thereby designing drugs and treatments.

Uncompetitive inhibition

Uncompetitive inhibition is a form of enzyme inhibition where the inhibitor only binds to the enzyme-substrate complex. This binding prevents the complex from releasing the product, lowering both the apparent maximum reaction rate (\(V_{\max}\)) and the Michaelis constant (\(K_m\)).In uncompetitive inhibition, the inhibitor binds after the substrate has already bound, making this type of inhibition dependent on the substrate concentration. This results in reduced values of both \(V_{\max}\) and \(K_m\) by a factor \(\alpha'\), defined as:

• \(V_{\max}^{\text{app}} = \frac{V_{\max}}{\alpha'}\)
• \(K_m^{\text{app}} = \frac{K_m}{\alpha'}\)

The factor \(\alpha'\) is greater than 1, indicating the presence of the inhibitor. Understanding uncompetitive inhibition is vital for drug design, as many inhibitors act by targeting the enzyme-substrate complex to control metabolic pathways.

Turnover number

The turnover number, also known as \(k_{cat}\), is a measure of the catalytic activity of an enzyme. It represents the number of substrate molecules converted into product per enzyme molecule per unit of time when the enzyme is fully saturated with substrate.The formula for turnover number is:\[k_{cat} = \frac{V_{max}}{[E]}\]where \([E]\) is the enzyme concentration.
This rate constant is crucial for understanding how effectively an enzyme catalyzes a reaction.
It allows for the comparison of catalytic efficiencies between different enzymes or different conditions. A high \(k_{cat}\) value indicates a highly efficient enzyme, which is particularly useful in various applications such as biotechnology, drug discovery, and synthetic biology.

Enzyme-substrate complex

The enzyme-substrate complex is a temporary molecular structure formed when an enzyme binds to its substrate. This binding is a key step in catalyzing a biochemical reaction. The lock-and-key model originally described this interaction, suggesting that the substrate fits into the enzyme like a key fits into a lock. However, the induced fit model is more accurate. It describes how the enzyme changes shape slightly to accommodate the substrate, which increases the reaction's efficiency. The formation of the enzyme-substrate complex can be represented as follows:\[E + S \rightleftharpoons ES \rightarrow E + P\]where

• \(E\) is the enzyme.
• \(S\) is the substrate.
• \(ES\) is the enzyme-substrate complex.
• \(P\) is the product.

Understanding the enzyme-substrate complex is vital in studying enzyme kinetics because it dictates how efficiently a reaction will proceed.
It provides insights into how different inhibitors can affect enzyme activity by altering this complex formation.