Question

The enzyme invertase catalyzes the conversion of sucrose, a disaccharide, to invert sugar, a mixture of glucose and fructose When the concentration of invertase is \(4.2 \times 10^{-7} M\) and the concentration of sucrose is \(0.0077 M\), invert sugar is formed at the rate of \(1.5 \times 10^{-4} \mathrm{M} / \mathrm{s} .\) When the sucrose concentration is doubled, the rate of formation of invert sugar is doubled also. (a) Assuming that the enzyme- substrate model is operative, is the fraction of enzyme tied up as a complex large or small? Explain. (b) Addition of inositol, another sugar, decreases the rate of formation of invert sugar. Suggest a mechanism by which this occurs.

Short answer

(a) The fraction of the enzyme tied up as a complex is small. This is because the reaction rate is directly proportional to the concentration of sucrose, indicating that the reaction rate is not limited by the enzyme-substrate complex formation. (b) The decrease in reaction rate due to the addition of inositol can be attributed to one or more of these potential mechanisms: competitive inhibition, non-competitive inhibition, or formation of a complex with sucrose that reduces its availability for the reaction.

Step-by-step solution

Part A: Determine the Fraction of Enzyme Tied Up as a Complex

Given that when the concentration of sucrose is doubled, the rate of formation of invert sugar is also doubled, this indicates a first-order reaction with respect to sucrose concentration. This means that the rate of reaction is directly proportional to the concentration of sucrose. Mathematically, this can be represented as: Rate \(= k[E][S]\) where Rate is the rate of reaction, \(k\) is the rate constant, \([E]\) is the enzyme concentration, and \([S]\) is the substrate (sucrose) concentration. Since we know the rate of reaction and the concentrations of invertase and sucrose, we can rearrange the equation to find the rate constant, \(k\): Rate \(= k[E][S] \Rightarrow k = \frac {Rate}{[E][S]}\) Plugging in the given values for the reaction: \(k = \frac {1.5 \times 10^{-4} M/s}{(4.2 \times 10^{-7} M)(0.0077 M)}\) Calculating the rate constant \(k\): \(k \approx 49.36 M^{-1}s^{-1}\) Now that we have the rate constant, let's analyze the effect of the enzyme-substrate complex to determine if the fraction of the enzyme tied up as a complex is large or small. Considering that the rate of the reaction is directly proportional to the concentration of sucrose, this implies that the reaction rate is not limited by the enzyme-substrate complex formation. In other words, only a small fraction of the enzyme is tied up as a complex at any one moment in time, allowing the reaction to proceed at a higher rate as sucrose concentration increases.

Part B: Suggest a Mechanism for the Decrease in Reaction Rate due to Inositol

There are several mechanisms by which the addition of inositol could decrease the rate of formation of invert sugar. Some possible mechanisms are: 1. Inositol may act as a competitive inhibitor, binding to the active site of the enzyme invertase and blocking the binding of sucrose. This reduces the rate of the reaction since the enzyme is not free to catalyze the conversion of sucrose to invert sugar. 2. Inositol may act as a non-competitive inhibitor, binding to an allosteric site on the enzyme invertase. This causes a conformational change in the enzyme's active site, decreasing its affinity for sucrose and consequently reducing the rate of the reaction. 3. Inositol may interact with sucrose, forming a complex that has a lower affinity for the enzyme invertase. This would decrease the effective concentration of sucrose available to bind to the enzyme, thus decreasing the rate of reaction. In summary, the decrease in reaction rate due to the addition of inositol can be attributed to one or more of these potential mechanisms: competitive inhibition, non-competitive inhibition, or formation of a complex with sucrose that reduces its availability for the reaction.

Key concepts

Enzyme-Substrate Complex

Understanding the enzyme-substrate complex is crucial for grasping the basics of enzyme kinetics. An enzyme-substrate complex is a temporary molecule formed when an enzyme comes into perfect contact with its substrate. This happens when the substrate binds to the active site of the enzyme, a specific region where the chemical reaction takes place.

The formation of this complex is essential for the catalysis of reactions within the body, as it lowers the activation energy required for the reaction, allowing it to proceed faster. In our example with the enzyme invertase, when it binds with sucrose (the substrate), an enzyme-substrate complex is formed. This complex is a pivotal point for the conversion of sucrose into invert sugar. The formation of the complex is usually a fast step followed by a slower step where the substrate is converted into the products.

Reaction Rate

The reaction rate in enzyme kinetics refers to the speed at which the reaction occurs, typically measured as a change in concentration of substrate or product over time. The rate can be influenced by factors such as enzyme and substrate concentration, temperature, pH, and the presence of inhibitors.

In our example, the rate of conversion of sucrose into invert sugar was measured as an increase in the concentration of invert sugar over time. Doubling the sucrose concentration also doubled the reaction rate, signifying that the reaction is first-order with respect to sucrose concentration. A linear relationship between the reaction rate and substrate concentration usually indicates that the enzyme is operating under conditions where it is not saturated by the substrate, meaning that not all enzyme molecules have formed complexes with substrate molecules.

Competitive Inhibitor

A competitive inhibitor is a substance that competes with the substrate for binding to the active site of an enzyme. When a competitive inhibitor is present, it can temporarily block the active site, preventing the substrate from binding and forming an enzyme-substrate complex—thus slowing down the rate of the reaction. Competitive inhibitors are often structurally similar to the substrate, which allows them to fit into the active site.

For instance, inositol, when added to a solution of invertase and sucrose, likely acts as a competitive inhibitor. It resembles sucrose enough to bind to invertase, thereby reducing the overall activity of the enzyme and slowing down the reaction rate. The effect of the inhibitor can be overcome by increasing the concentration of the substrate, so increasing the amount of sucrose could restore the reaction rate.

Rate Constant

The rate constant, denoted as 'k' in chemical kinetics, is a proportionality factor that connects the rate of the reaction to the concentrations of the reactants. It is a measure of the intrinsic speed at which a chemical reaction occurs, and it is specific to a particular reaction at a given temperature. The units of the rate constant vary depending on the order of the reaction.

In the provided exercise, the rate constant for the reaction catalyzed by invertase is derived from the given reaction rate and concentrations of invertase and sucrose. Mathematically, the rate constant is found using the equation \( k = \frac{\text{Rate}}{[E][S]} \) where \( [E] \) and \( [S] \) are the respective concentrations of enzyme and substrate. This constant is key to understanding how different parameters will affect the rate at which the enzyme can catalyze the reaction.