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The initial reaction velocity for an enzyme reaction reaches a maximum at high substrate concentration because the free enzyme can no longer regenerate at the end of each reaction cycle. True/False

Short Answer

Expert verified
False

Step by step solution

01

Understand the Concept of Enzyme Reaction Velocity

The initial reaction velocity ( v_0 ) of an enzyme-catalyzed reaction refers to the rate of the reaction when the enzyme first encounters the substrate. It's crucial to understand that at low substrate concentrations, the enzyme sites are not saturated, leading to an increase in v_0 as substrate concentration increases.
02

Define Saturation Point and Maximum Velocity

As substrate concentration increases, enzyme sites become saturated, leading to a maximum reaction velocity, known as V_{max} , where adding more substrate does not increase the velocity because all enzyme active sites are occupied temporarily.
03

Analyze the Given Statement in Context

The statement claims that the maximum velocity occurs because the enzyme cannot regenerate at the end of each cycle, which does not accurately describe why V_{max} is reached. At high substrate concentrations, the enzyme's active sites are all occupied, leading to saturation, not due to the inability of enzyme regeneration itself.
04

Conclusion and Judgment

Based on the definition of V_{max} and enzyme kinetics, the initial reaction velocity reaching a maximum at high substrate concentration is due to saturation of enzyme active sites, not because the enzymes cannot regenerate. Therefore, the factual content of the statement is incorrect.

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Key Concepts

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

Reaction Velocity
The term "reaction velocity" or "initial velocity" in enzyme kinetics represents how fast a reaction occurs under specific conditions. When we talk about enzyme-catalyzed reactions, its velocity denotes the rate at which substrate ( S ) is converted into product ( P ) with the aid of an enzyme. When a reaction first begins, the reaction velocity is often called the initial velocity ( v_0 ). Initially, as more substrate molecules are available, the reaction velocity tends to increase. This is because there are more substrate molecules available to interact with the enzyme's active sites. But this relationship doesn't stay linear forever; eventually, the reaction velocity stabilizes as the system reaches other limiting factors. Understanding reaction velocity is key to grasping how enzymes function in varying conditions and how they can ultimately impact biological processes.
Substrate Concentration
Substrate concentration is a measure of how much substrate is present in a solution for the enzyme to act upon. It's important for determining how quickly a reaction proceeds, especially in the early stages of enzyme action. At low substrate concentrations, the reaction velocity is directly proportional to the substrate level because there are plenty of active sites available on the enzyme for the substrate to occupy. As a result, any increase in substrate leads to a corresponding rise in reaction velocity. However, as the substrate concentration becomes higher, you'll notice a diminishing increase in reaction velocity. This is because the enzymes begin to approach a saturation point where most of their active sites are occupied. When nearly all enzymes are busy, adding more substrate doesn't make a significant difference in reaction speed.
Enzyme Saturation
Enzyme saturation occurs when all active sites of the available enzymes are fully occupied by substrate molecules. Think of it like a busy restaurant where all tables (enzyme active sites) are occupied by diners (substrate molecules). Adding more diners doesn’t speed up the service because there are no free tables left. This saturation leads to a plateau in reaction velocity, meaning it can't increase any further, no matter how much more substrate is available. The reaction is functioning at its full capacity. Enzyme saturation is a critical factor in understanding how enzymes work and why reactions can't exceed certain speeds past a particular substrate concentration. It highlights the enzyme's maximum working capacity in converting substrates to products.
Maximum Velocity
Maximum velocity, denoted as V_{max} , is the peak rate of an enzyme-catalyzed reaction given that all active sites of the enzyme molecules are occupied by substrate. At this point, the addition of more substrate will no longer increase the velocity of the reaction. Achieving V_{max} is analogous to hitting the speed limit of a car on a given road; no matter how much more you press the pedal, the speed won't increase once you've reached the limit. Max velocity demonstrates the enzyme's efficiency and productivity when fully saturated with substrate. It is an important parameter for comparing different enzymes or even the same enzyme under different conditions. Understanding V_{max} helps in biochemical applications, for instance, in drug development and industrial enzyme processes, by highlighting the limits and capabilities of enzymatic reactions.

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Most popular questions from this chapter

Why is triose phosphate isomerase considered to be an example of a "perfect enzyme"?

The table below lists initial velocities measured for an enzymatic reaction at different substrate concentrations in the presence and absence of an inhibitor. The enzyme concentration is identical in both reactions. a. Graph a Lineweaver-Burk plot for each set of data. b. What are the values of \(V_{\max }\) and \(K_{M}\) for each experiment? c. What is the inhibition mechanism ? d. If the concentration of inhibitor is \(100 \mathrm{nM}\), what is the value of \(K_{1}\) ?

Which of the following is not a commonly observed feature of proteases? a. The catalytic triad in the active site. b. Exclusively hydrophobic residues in the active site. c. A cysteine residue in the active site. d. Metal ions coordinated in the active site. e. A pair of acidic residues in the active site.

Given the following three data tables of substrate concentrations and initial velocities for enzymes that obey Michaelis-Menten kinetics, estimate \(K_{\mathrm{M}}\) for each enzyme in molar units. a. \begin{tabular}{|c|c|} \hline\([\mathbf{S}](\mathrm{mM})\) & \(v_{0}\left(\mathrm{mM} \cdot \mathrm{sec}^{-1}\right)\) \\ \hline 1 & \(266.7\) \\ \hline 3 & \(553.8\) \\ \hline 5 & \(705.9\) \\ \hline 50 & \(1121.5\) \\ \hline 500 & \(1191.7\) \\ \hline 5000 & \(1199.2\) \\ \hline \end{tabular} b. \begin{tabular}{|c|l|} \hline IS] (nM) & \(v_{0}\left(\mathrm{mM}^{2} \cdot \mathrm{min}^{-1}\right)\) \\\ \hline 4 & \(123.5\) \\ \hline 5 & \(137.4\) \\ \hline 6 & \(148.5\) \\ \hline 10 & \(177.3\) \\ \hline 100 & \(240.2\) \\ \hline 1000 & \(249.0\) \\ \hline \end{tabular} C. \begin{tabular}{|c|l|} \hline\([\mathrm{S}](\mathrm{mM})\) & \(\nu_{0}(\) Mohour \(-1)\) \\ \hline 1 & \(0.00\) \\ \hline 10 & \(0.01\) \\ \hline 100 & \(0.07\) \\ \hline 200 & \(0.10\) \\ \hline 1000 & \(0.17\) \\ \hline 5000 & \(0.19\) \\ \hline \end{tabular}

An experiment with \(10 \mathrm{nM}\) of an enzyme obeying Michaelis-Menten kinetics yields a \(V_{\max }\) of \(7 \times 10^{-3}\) Mesec 1 . a. What is the turnover number \(\left(k_{2}\right)\) ? b. The experiment is repeated in the presence of a noncompetitive inhibitor and the \(V_{\max }\) is reduced to \(5 \times 10^{-4} \mathrm{M}_{2} \mathrm{Sec}^{-1}\). What fraction of the enzyme is bound to the inhibitor?

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