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Chapter 6: Problem 8

Applying the Michaelis-Menten Equation I An enzyme has a \(V_{\max }\) of \(1.2 \mu \mathrm{M} \mathrm{s}^{-1}\). The \(K_{\mathrm{m}}\) for its substrate is \(10 \mu \mathrm{M}\). Calculate the initial velocity of the reaction, \(V_{0}\), when the substrate concentration is a. \(2 \mu \mathrm{M}\) b. \(10 \mu_{M}\) c. \(30 \mu_{\mathrm{M}}\).

Short Answer

Expert verified
a. 0.2, b. 0.6, c. 0.9 \(\mu \mathrm{M} \mathrm{s}^{-1}\)

Step by step solution

01

Understand the Michaelis-Menten Equation

The Michaelis-Menten equation is used to describe the rate of enzyme-catalyzed reactions. The equation is given by: \[ V_0 = \frac{V_{\max} [S]}{K_m + [S]} \]where \( V_0 \) is the initial velocity, \( V_{\max} \) is the maximum rate, \( K_m \) is the Michaelis constant, and \([S]\) is the substrate concentration.
02

Plug Values for Case (a)

For part (a), we have \( [S] = 2 \mu \mathrm{M} \), \( V_{\max} = 1.2 \mu \mathrm{M} \mathrm{s}^{-1} \), and \( K_m = 10 \mu \mathrm{M} \). Plug these into the equation:\[ V_0 = \frac{1.2 \times 2}{10 + 2} = \frac{2.4}{12} \]Calculate \( V_0 \):\[ V_0 = 0.2 \mu \mathrm{M} \mathrm{s}^{-1} \]
03

Plug Values for Case (b)

For part (b), use \( [S] = 10 \mu \mathrm{M} \), \( V_{\max} = 1.2 \mu \mathrm{M} \mathrm{s}^{-1} \), and \( K_m = 10 \mu \mathrm{M} \):\[ V_0 = \frac{1.2 \times 10}{10 + 10} = \frac{12}{20} \]Calculate \( V_0 \):\[ V_0 = 0.6 \mu \mathrm{M} \mathrm{s}^{-1} \]
04

Plug Values for Case (c)

For part (c), use \( [S] = 30 \mu \mathrm{M} \), with \( V_{\max} = 1.2 \mu \mathrm{M} \mathrm{s}^{-1} \), and \( K_m = 10 \mu \mathrm{M} \):\[ V_0 = \frac{1.2 \times 30}{10 + 30} = \frac{36}{40} \]Calculate \( V_0 \):\[ V_0 = 0.9 \mu \mathrm{M} \mathrm{s}^{-1} \]

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Key Concepts

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

Enzyme Kinetics
Enzyme kinetics is the study of how enzymes interact with substrates to accelerate chemical reactions. Enzymes are proteins that act as catalysts; they speed up reactions without being consumed. Understanding enzyme kinetics is crucial because it lets us predict how fast a reaction will occur when an enzyme and its substrate are mixed under different conditions. The Michaelis-Menten equation is a key tool used to describe these reactions.

- **Enzyme Activity**: Refers to the rate at which an enzyme catalyzes a reaction. The activity depends on various factors like enzyme concentration, substrate concentration, temperature, and pH. - **Michaelis-Menten Kinetics**: This model simplifies the complex process of enzyme-substrate interaction into a simple mathematical equation, helping to predict the reaction rate. In summary, enzyme kinetics provides a mathematical framework to analyze and predict how enzymes behaves, enabling more efficient use of enzymes in research and industry.
Reaction Rate
The reaction rate is a measure of how quickly the reactants are transformed into products in a chemical reaction. In the context of enzyme kinetics, the reaction rate can also refer to the initial velocity (\( V_0 \)) of an enzyme-catalyzed reaction.

- **Initial Velocity (\( V_0 \))**: This is the rate of reaction at the very start and is particularly important because it helps determine the behavior of the enzyme under different substrate concentrations.- **Factors Affecting Reaction Rate**: - Substrate concentration - Enzyme concentration - Temperature - pHThe reaction rate is crucial in understanding how different conditions affect the speed of the enzymatic reaction, allowing scientists to optimize these conditions for desired outcomes.
Substrate Concentration
The substrate concentration (\([S]\)) is a key player in enzyme kinetics. It refers to the amount of substrate available to bind with the enzyme.- If substrate concentration is low, there may be free enzyme molecules unoccupied.- When substrate concentration is high, most enzyme molecules may be bound to substrate.This relationship is quantitatively described by the Michaelis-Menten equation. As substrate concentration increases, the reaction rate also increases, but at a diminishing rate. Eventually, a saturation point is reached where increasing substrate concentration no longer affects the rate of reaction, because the enzyme is working at its maximum capacity (\( V_{\text{max} }\)).In enzyme kinetics, understanding how changes in substrate concentration affect the reaction rate is crucial for determining the efficient usage of substrates in biological and industrial processes.

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Most popular questions from this chapter

Applying the Michaelis-Menten Equation III A research group discovers a new version of happyase, which they call happyase *, that catalyzes the chemical reaction HAPPY \(\rightleftharpoons\) SAD. The researchers begin to characterize the enzyme. a. In the first experiment, with \(\left[E_{t}\right]\) at \(4 \mathrm{~nm}\), they find that the \(V_{\max }\) is \(1.6 \mu \mathrm{M} \mathrm{s}^{-1}\). Based on this experiment, what is the \(k_{\text {cat }}\) for happyase*? (Include appropriate units.) b. In the second experiment, with \(\left[E_{t}\right]\) at \(1 \mathrm{~nm}\) and [HAPPY] at \(30 \mu \mathrm{M}\), the researchers find that \(V_{0}=300 \mathrm{nM} \mathrm{s}^{-1}\). What is the measured \(K_{\mathrm{m}}\) of happyase* for its substrate HAPPY? (Include appropriate units.) c. Further research shows that the purified happyase * used in the first two experiments was actually contaminated with a reversible inhibitor called ANGER. When ANGER is carefully removed from the happyase * preparation and the two experiments are repeated, the measured \(V_{\max }\) in (a) is increased to \(4.8 \mu \mathrm{M} \mathrm{s}^{-1}\), and the measured \(K_{\mathrm{m}}\) in (b) is now \(15 \mu_{\mathrm{M}}\). Calculate the values of \(a\) and \(\alpha^{\prime}\) for ANGER. d. Based on the information given, what type of inhibitor is ANGER?

Protection of an Enzyme against Denaturation by Heat When enzyme solutions are heated, there is a progressive loss of catalytic activity over time due to denaturation of the enzyme. A solution of the enzyme hexokinase incubated at \(45{ }^{\circ} \mathrm{C}\) lost \(50 \%\) of its activity in \(12 \mathrm{~min}\), but when incubated at \(45^{\circ} \mathrm{C}\) in the presence of a very large concentration of one of its substrates, it lost only \(3 \%\) of its activity in \(12 \mathrm{~min}\). Suggest why thermal denaturation of hexokinase was retarded in the presence of one of its substrates.

Intracellular Concentration of Enzymes To approximate the concentration of enzymes in a bacterial cell, assume that the cell contains equal concentrations of 1,000 different enzymes in solution in the cytosol and that each protein has a molecular weight of 100,000 . Assume also that the bacterial cell is a cylinder (diameter \(1.0 \mu \mathrm{m}\), height \(2.0 \mu \mathrm{m}\) ), that the cytosol (specific gravity \(1.20\) ) is \(20 \%\) soluble protein by weight, and that the soluble protein consists entirely of enzymes. Calculate the average molar concentration of each enzyme in this hypothetical cell.

Rate Enhancement by Urease The enzyme urease enhances the rate of urea hydrolysis at \(\mathrm{pH} 8.0\) and \(20{ }^{\circ} \mathrm{C}\) by a factor of \(10^{14}\). Suppose that a given quantity of urease can completely hydrolyze a given quantity of urea in \(5.0 \mathrm{~min}\) at \(20^{\circ} \mathrm{C}\) and \(\mathrm{pH} 8.0\). How long would it take for this amount of urea to be hydrolyzed under the same conditions in the absence of urease? Assume that both reactions take place in sterile systems so that bacteria cannot attack the urea.

Estimation of \(V_{\max }\) and \(\boldsymbol{K}_{\mathrm{m}}\) by Inspection Graphical methods are available for accurate determination of the \(V_{\max }\) and \(K_{\mathrm{m}}\) of an enzyme-catalyzed reaction. However, these quantities can sometimes be estimated by inspecting values of \(V_{0}\) at increasing [S]. Estimate the \(V_{\max }\) and \(K_{m}\) of the enzyme-catalyzed reaction for which the data in the table were obtained. \begin{tabular}{cc} {\([\mathbf{S}](\mathrm{M})\)} & \(V_{0}(\mu \mathrm{M} / \mathrm{min})\) \\ \hline \(2.5 \times 10^{-6}\) & 28 \\ \(4.0 \times 10^{-6}\) & 40 \end{tabular} \begin{tabular}{ll} \(1 \times 10^{-5}\) & 70 \\ \(2 \times 10^{-5}\) & 95 \\ \(4 \times 10^{-5}\) & 112 \\ \(1 \times 10^{-4}\) & 128 \\ \(2 \times 10^{-3}\) & 140 \\ \(1 \times 10^{-2}\) & 139 \\ \hline \end{tabular}
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