One of the many remarkable enzymes in the human body is carbonic anhydrase, which catalyzes the interconversion of carbon dioxide and water with bicarbonate ion and protons. If it were not for this enzyme, the body could not rid itself rapidly enough of the \(\mathrm{CO}_{2}\) accumulated by cell metabolism. The enzyme catalyzes the dehydration (release to air) of up to \(10^{7} \mathrm{CO}_{2}\) molecules per second. Which components of this description correspond to the terms enzyme, substrate, and turnover number?

Short Answer

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In the given description, the enzyme is carbonic anhydrase, the substrates are carbon dioxide (CO2) and water (H2O), and the turnover number is \(10^7\) CO2 molecules per second.

Step by step solution

01

Identify the enzyme

In the description, "One of the many remarkable enzymes in the human body is carbonic anhydrase", the enzyme is clearly stated as carbonic anhydrase.
02

Identify the substrate

The substrate is the substance on which the enzyme acts. In this case, the enzyme "catalyzes the interconversion of carbon dioxide and water with bicarbonate ion and protons". Thus, we can say that the substrates in this case are carbon dioxide (CO2) and water (H2O).
03

Identify the turnover number

The turnover number is the number of molecules of substrate that an enzyme can convert to product per unit time. In the description, it is mentioned that "the enzyme catalyzes the dehydration (release to air) of up to \(10^{7} \mathrm{CO}_{2}\) molecules per second". This means that the enzyme is able to convert up to \(10^7\) CO2 molecules per second, which represents the turnover number.

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Key Concepts

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

Carbonic Anhydrase
Understanding the powerful enzyme called carbonic anhydrase is fundamental in grasping its biological significance. Found in red blood cells, this enzyme is a catalyst for a vital chemical reaction in the body: converting carbon dioxide (CO2) and water (H2O) to bicarbonate ions (HCO3-) and protons (H+). This reaction is essential for maintaining the acid-base balance in the blood and for transporting CO2 from the tissues to the lungs where it can be exhaled.

Carbonic anhydrase operates at an astonishing speed, allowing it to deal with the high amounts of CO2 produced by cellular metabolism. Its efficiency exemplifies the power of enzymes to accelerate reactions that are critical for life but would otherwise occur at imperceptibly slow rates. This enzyme's role in facilitating gas exchange and pH regulation is crucial for overall homeostasis, underpinning the reason why understanding its function is key in the study of physiology and biochemistry.
Enzyme-Substrate Interaction
The concept of enzyme-substrate interaction is a cornerstone of enzyme kinetics, providing insight into how enzymes work at a molecular level. An enzyme's efficacy is defined by its ability to bind specific substrates – the reactants on which enzymes act – to form an enzyme-substrate complex. The specific match between an enzyme's active site and the substrate is often described using the 'lock and key' model, emphasizing the idea that enzymes are highly selective.

For carbonic anhydrase, the substrate molecules are CO2 and H2O. The enzyme's active site perfectly accommodates these molecules to facilitate their conversion into HCO3- and H+. This specificity ensures that the enzyme only catalyzes this particular reaction, highlighting the precision with which biological systems are tuned. Additionally, factors such as pH and temperature can influence this interaction, as they can cause changes in enzyme structure, affecting the efficiency of substrate binding and the rate of the reaction.
Turnover Number
The turnover number is a valuable measure in enzyme kinetics, reflecting an enzyme's catalytic activity – essentially, how many substrate molecules it can convert into product in a given time frame. It's a direct quantification of an enzyme's efficiency. For carbonic anhydrase, the impressive turnover number is up to 10 million (107) CO2 molecules per second, which is one of the highest recorded for any enzyme.

It's important to understand that the turnover number can vary from one enzyme to another, influenced by various factors including enzyme concentration, substrate availability, and the presence of inhibitors or activators. In medical and biotechnological contexts, the turnover number is essential for designing drugs and optimizing industrial processes that involve enzymes. The tireless work of carbonic anhydrase in our bodies, reflected in its high turnover number, is a testament to the power and necessity of enzymes in life-sustaining reactions.

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Most popular questions from this chapter

The decomposition reaction of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in carbon tetrachloride is \(2 \mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\) . The rate law is first order in \(\mathrm{N}_{2} \mathrm{O}_{5}\) . At \(64^{\circ} \mathrm{C}\) the rate constant is \(4.82 \times 10^{-3} \mathrm{s}^{-1}\) (a) Write the rate law for the reaction. (b) What is the rate of reaction when \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=0.0240 M ?(\mathbf{c})\) What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is doubled to 0.0480\(M ?(\mathbf{d})\) What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is halved to 0.0120 \(\mathrm{M} ?\)

(a) Consider the combustion of hydrogen, \(2 \mathrm{H}_{2}(g)+\) \(\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) .\) If hydrogen is burning at the rate of 0.48 \(\mathrm{mol} / \mathrm{s}\) , what is the rate of consumption of oxygen? What is the rate of formation of water vapor? (b) The reaction \(2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NOCl}(g)\) is carried out in a closed vessel. If the partial pressure of \(\mathrm{NO}\) is decreasing at the rate of 56 torr/min, what is the rate of change of the total pressure of the vessel?

The reaction between ethyl bromide \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}\right)\) and hydroxide ion in ethyl alcohol at 330 \(\mathrm{K}\) , \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}(a l c)+\mathrm{OH}^{-}(a l c) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{Br}^{-}(a l c)\) is first order each in ethyl bromide and hydroxide ion. When \(\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}\right]\) is 0.0477 \(\mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]\) is \(0.100 \mathrm{M},\) the rate of disappearance of ethyl bromide is \(1.7 \times 10^{-7} \mathrm{M} / \mathrm{s}\) (a) What is the value of the rate constant? (b) What are the units of the rate constant? (c) How would the rate of disappearance of ethyl bromide change if the solution were diluted by adding an equal volume of pure ethyl alcohol to the solution?

(a) The reaction \(\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\frac{1}{2} \mathrm{O}_{2}(g)\) is first order. At 300 \(\mathrm{K}\) the rate constant equals \(7.0 \times 10^{-4} \mathrm{s}^{-1}\) . Calculate the half-life at this temperature. (b) If the activation energy for this reaction is \(75 \mathrm{kJ} / \mathrm{mol},\) at what temperature would the reaction rate be doubled?

(a) Consider the combustion of ethylene, \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\) \(3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) .\) If the concentration of \(\mathrm{C}_{2} \mathrm{H}_{4}\) is decreasing at the rate of \(0.036 \mathrm{M} / \mathrm{s},\) what are the rates of change in the concentrations of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O} ?(\mathbf{b})\) The rate of decrease in \(\mathrm{N}_{2} \mathrm{H}_{4}\) partial pressure in a closed reaction vessel from the reaction \(\mathrm{N}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) is 74 torr per hour. What are the rates of change of \(\mathrm{NH}_{3}\) partial pressure and total pressure in the vessel?

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