Question

Applying the Michaelis-Menten Equation IV Researchers discover an enzyme that catalyzes the reaction \(\mathrm{X} \rightleftharpoons \mathrm{Y}\). They find that the \(K_{\mathrm{m}}\) for the substrate \(\mathrm{X}\) is \(4 \mu \mathrm{M}\), and the \(k_{\text {cat }}\) is \(20 \mathrm{~min}^{-1}\). a. In an experiment, \([\mathrm{X}]=6 \mathrm{mM}\), and \(V_{0}=480 \mathrm{nM} \mathrm{min}^{-1}\). What was the \(\left[\mathrm{E}_{\mathrm{t}}\right]\) used in the experiment? b. In another experiment, \(\left[\mathrm{E}_{\mathrm{t}}\right]=0.5 \mu \mathrm{M}\), and the measured \(V_{0}=5 \mu \mathrm{M} \mathrm{min}^{-1}\). What was the \([\mathrm{X}]\) used in the experiment? c. The researchers discover that compound \(Z\) is a very strong competitive inhibitor of the enzyme. In an experiment with the same \(\left[E_{t}\right]\) as in (a), but a different \([\mathrm{X}]\), they add an amount of \(\mathrm{Z}\) that produces an \(a\) of 10 and reduces \(V_{0}\) to \(240 \mathrm{nM} \mathrm{min}^{-1}\). What is the \([\mathrm{X}]\) in this experiment? d. Based on the kinetic parameters given, has this enzyme evolved to achieve catalytic perfection? Explain your answer briefly, using the kinetic parameter(s) that define catalytic perfection.

Short answer

a. \([E_t] = 24 \text{ nM}\) b. \([X] = 4 \mu M\) c. \([X] = 40 \mu M\) d. No, the enzyme is not catalytically perfect.

Step-by-step solution

Understanding the Michaelis-Menten Equation

The Michaelis-Menten equation is given by \[ V_0 = \frac{V_{max} [S]}{K_m + [S]} \] where \( V_0 \) is the initial reaction velocity, \( V_{max} \) is the maximum velocity, \([S]\) is the substrate concentration, and \(K_m\) is the Michaelis-Menten constant.

Solving Part A

We know that in terms of enzyme kinetics, \( V_{max} = k_{cat} [E_t] \). We are given \( V_0 = 480 \text{ nM min}^{-1} \), \([X]\) is significantly larger than \(K_m\), so \( V_0 \approx V_{max} \). Therefore, \([E_t] = \frac{V_0}{k_{cat}} = \frac{480}{20} \). \([E_t] = 24 \text{ nM}.\)

Solving Part B

Here, \([E_t] = 0.5 \mu M\) and \( V_0 = 5 \mu M \text{ min}^{-1} \). First, calculate \( V_{max} = k_{cat} [E_t] = 20 \times 0.5 = 10 \mu M \text{ min}^{-1} \). Use the Michaelis-Menten equation to solve for \([X]\): \( 5 = \frac{10 [X]}{4 + [X]} \). Solving this equation gives \([X] = 4 \mu M\).

Solving Part C

With a competitive inhibitor, the modified Michaelis constant \( K_m' = aK_m \) and \( a = 10 \). Thus, \( K_m' = 40 \mu M \). The reaction velocity \(V_0 = 240 \text{ nM min}^{-1}\) needs to be calculated with the equation \( V_0 = \frac{k_{cat} [E_t] [X]}{K_m' + [X]} \). Given \([E_t] = 24 \text{ nM}\). Solving \( 240 = \frac{480 [X]}{40 + [X]} \) gives \([X] = 40 \mu M\).

Understanding Catalytic Perfection

An enzyme achieves catalytic perfection when the catalytic efficiency \( \frac{k_{cat}}{K_m} \) is close to the diffusion limit (about \(10^8\) to \(10^9\) \(M^{-1} s^{-1}\)). For this enzyme, \( \frac{k_{cat}}{K_m} = \frac{20 \text{ min}^{-1}}{4 \mu M}\), which is \(5 \times 10^3 \text{ min}^{-1} \mu M^{-1}\) or \(8.33 \times 10^4 \text{ M}^{-1} \text{ s}^{-1}\). This value is not close to catalytic perfection.

Key concepts

Enzyme Kinetics

Enzyme kinetics is a branch of biochemistry focusing on the rates of enzyme-catalyzed reactions. One of the fundamental equations in enzyme kinetics is the Michaelis-Menten equation. This equation helps us understand how enzyme activity is affected by substrate concentration.
The equation itself is: \[ V_0 = \frac{V_{max} [S]}{K_m + [S]} \]Where:

• \( V_0 \) is the initial velocity of the reaction, representing the speed of product formation when the reaction begins.
• \( V_{max} \) stands for the maximal velocity, which the reaction approaches as substrate concentration increases.
• \([S]\) is the concentration of the substrate.
• \(K_m\) is the Michaelis constant, reflecting the substrate concentration at which the reaction rate is half of \(V_{max}\).

Understanding enzyme kinetics is crucial for determining how different conditions affect enzyme activity, such as changes in substrate concentration or enzyme inhibitors.

Catalytic Efficiency

Catalytic efficiency is a measure of how efficiently an enzyme converts a substrate into a product. It's typically expressed as the ratio of the enzyme's turnover number \((k_{cat})\) to the Michaelis constant \((K_m)\). This ratio, \(\frac{k_{cat}}{K_m}\), indicates how quickly the enzyme reacts with the substrate compared to how well it binds to the substrate.
When enzymes have high catalytic efficiency, they are generally considered more effective catalyzers. Enzymes approaching 'catalytic perfection' operate at rates limited only by how fast substrates diffuse into the enzyme's active site.

• An enzyme with a \(k_{cat}/K_m\) near \(10^8\) to \(10^9\) \(M^{-1} s^{-1}\) is near the theoretical diffusion limit and thus regarded as achieving catalytic perfection.

For example, the enzyme referenced here has a catalytic efficiency of \(8.33 \times 10^4\) \( M^{-1} s^{-1} \), which is not near the limit of catalytic perfection, indicating room for evolutionary improvement to increase its efficiency further.

Competitive Inhibition

Competitive inhibition is a type of enzyme inhibition where an inhibitor competes with the substrate for binding to the active site on the enzyme. This type of inhibition can be reversible and typically increases the apparent value of the Michaelis constant, \(K_m\), without affecting \(V_{max}\) because the inhibitor only impedes substrate binding, not the catalytic event itself.

In the presence of a competitive inhibitor:

• The apparent Michaelis constant increases to \(K_m' = aK_m\), where \(a\) is a factor indicating the degree of inhibition.
• The inhibitor effectively raises the substrate concentration needed to achieve the same rate as in the uninhibited reaction.

In the given problem, compound \(Z\) acts as a competitive inhibitor with \(a = 10\), showing its strong inhibition by significantly increasing \(K_m\) and reducing the reaction velocity \(V_0\). This demonstrates how competitive inhibitors can be used experimentally to study enzyme kinetics and how they affect enzymatic reactions.

Enzyme-Substrate Complex

The enzyme-substrate complex is a transient molecular assembly where the enzyme interacts with the substrate prior to catalysis. This complex is crucial for understanding the enzymatic mechanism as it represents the first step in the reaction mechanism leading to product formation.
The formation of this complex can be represented by the equation:\[ E + S \rightleftharpoons ES \rightarrow E + P \]Where:

• \(E\) is the enzyme, \(S\) is the substrate, \(ES\) is the enzyme-substrate complex, and \(P\) is the product.
• The substrate binds to the active site of the enzyme to form the complex, which subsequently can convert to product, releasing the enzyme.

Understanding this complex is essential in enzyme kinetics as it provides insights into the factors affecting reaction rates, including substrate concentration and enzyme modifications. The formation of the enzyme-substrate complex is the determinant step for reaction specificity and catalytic efficiency.