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Chapter 36: Problem 16

The enzyme fumarase catalyzes the hydrolysis of fumarate: Fumarate \((a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{L}\) -malate \((a q)\). The turnover number for this enzyme is \(2.5 \times 10^{3} \mathrm{s}^{-1},\) and the Michaelis constant is \(4.2 \times 10^{-6} \mathrm{M}\). What is the rate of fumarate conversion if the initial enzyme concentration is \(1 \times 10^{-6} \mathrm{M}\) and the fumarate concentration is \(2 \times 10^{-4} \mathrm{M} ?\)

Short Answer

Expert verified
The rate of fumarate conversion is approximately \(2.447 \times 10^{-3} M s^{-1}\).

Step by step solution

01

Calculate Vmax

To find the maximum rate at which the enzyme can work, use the formula: \(V_{max} = k_{cat}[E]\). We are given the turnover number (\(k_{cat}\)) and initial enzyme concentration ([E]): \(2.5 \times 10^{3} s^{-1}\) (Turnover number) \(1 \times 10^{-6} M\) (Initial enzyme concentration) Now, calculate Vmax: \(V_{max} = (2.5 \times 10^{3} s^{-1})(1 \times 10^{-6} M) = 2.5 \times 10^{-3} M s^{-1}\)
02

Apply the Michaelis-Menten Equation

We have found the value of \(V_{max}\) in the previous step. Now, we will use the Michaelis-Menten equation to find the rate of the reaction (v): \(v=\dfrac{V_{max}[S]}{K_{M}+[S]}\) We are given the Michaelis constant (\(K_{M}\)) and fumarate concentration ([S]): \(4.2 \times 10^{-6} M\) (Michaelis constant) \(2 \times 10^{-4} M\) (Fumarate concentration) Now substitute the known values into the equation and solve for v: \(v = \dfrac{(2.5 \times 10^{-3} M s^{-1})(2 \times 10^{-4} M)}{(4.2 \times 10^{-6} M) + (2 \times 10^{-4} M)}\)
03

Calculate the Rate of Fumarate Conversion

Now that we have plugged in our values to the Michaelis-Menten equation, we can solve for the rate of the reaction (v): \(v = \dfrac{(2.5 \times 10^{-3} M s^{-1})(2 \times 10^{-4} M)}{(4.2 \times 10^{-6} M) + (2 \times 10^{-4} M)} = \dfrac{5 \times 10^{-7} M^2 s^{-1}}{2.042 \times 10^{-4} M}\) Now, divide the numerator by the denominator to find v: \(v = \dfrac{5 \times 10^{-7} M^2 s^{-1}}{2.042 \times 10^{-4} M} = 2.447 \times 10^{-3}\, Ms^{-1}\) The rate of fumarate conversion is approximately \(2.447 \times 10^{-3} M s^{-1}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Enzymatic Kinetics
Enzymatic kinetics is a branch of biochemistry that studies the rates at which enzymatic reactions occur and the factors affecting these rates. One key aspect of enzymatic kinetics is understanding how enzymes, which are biological catalysts, speed up reactions without being consumed in the process.

For any given enzyme-catalyzed reaction, the reaction rate increases with the concentration of substrate up to a certain point. This behavior can be described by the Michaelis-Menten equation, an essential mathematical model in this field. The equation relates the rate of the reaction to the concentration of substrate and defines the maximum rate and affinity of the enzyme for the substrate through the parameters Vmax and Km, respectively.

By providing clarity on the characteristics of enzymatic kinetics, students can better interpret an enzyme's efficiency and its biological significance within various physiological contexts, leading to a profound comprehension of metabolic pathways.
Fumarase Catalysis
Fumarase is an important enzyme in the citric acid cycle, catalyzing the hydration of fumarate to L-malate. This enzyme's activity is an example of how a catalytic process is orchestrated in the cell to facilitate metabolic reactions.

As part of its function in cellular respiration, fumarase operates under enzymatic kinetics principles, where the conversion rate of fumarate can be determined using the Michaelis-Menten equation. Understanding fumarase's catalytic role provides insights into how metabolic pathways are interconnected and regulated, which can be especially relevant when studying diseases where energy metabolism is disrupted.

A detailed look into how fumarase operates at the molecular level helps students grasp the complex interplay between enzyme structure and function, ultimately painting a clearer picture of its role in bioenergetics.
Turnover Number
The turnover number, often represented by the symbol kcat, refers to the maximum number of substrate molecules that one enzyme molecule can convert to product per unit of time, when the enzyme is fully saturated with substrate. This value is crucial for gauging an enzyme's efficiency.

The turnover number serves as a yardstick for comparing different enzymes or the same enzyme under various conditions. In the provided exercise, a turnover number of 2.5 x 10^3 s^-1 implies that each fumarase enzyme can convert 2,500 fumarate molecules into L-malate every second at full saturation, highlighting the enzyme's potency in catalysis.

Grasping this concept helps students understand why even small amounts of highly efficient enzymes are sufficient for sustaining life's biochemical reactions, showing the exquisite fine-tuning of biological systems.
Michaelis Constant
The Michaelis constant, Km, signifies the substrate concentration at which the reaction rate is half of the maximum velocity (Vmax). It provides an estimate of the substrate's affinity for the enzyme; a lower Km indicates a higher affinity, meaning that the enzyme reaches its half-maximum velocity at a lower concentration of substrate.

In the context of the exercise, a Km of 4.2 x 10^-6 M implies fumarase doesn't need a high concentration of fumarate to operate near its peak efficiency. A thorough understanding of Km allows students to predict how changes in substrate concentration can affect the velocity of enzymatic reactions, which is a fundamental skill for biotechnological applications and the design of therapeutic interventions.

Exploring the significance of the Michaelis constant can therefore deepen a student's appreciation for the delicate balance and regulation inherent in biological systems, as well as the precision required for effective enzyme function.

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