Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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

Expert verified
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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The rate of a first-order reaction is followed by spectroscopy, monitoring the absorbance of a colored reactant at \(520 \mathrm{nm}\). The reaction occurs in a \(1.00-\mathrm{cm}\) sample cell, and the only colored species in the reaction has an extinction coefficient of \(5.60 \times 10^{3} \mathrm{M}^{-1} \mathrm{~cm}^{-1}\) at \(520 \mathrm{nm}\). (a) Calculate the initial concentration of the colored reactant if the absorbance is 0.605 at the beginning of the reaction. (b) The absorbance falls to 0.250 at \(30.0 \mathrm{~min}\). Calculate the rate constant in units of \(\mathrm{s}^{-1}\). (c) Calculate the half-life of the reaction. (d) How long does it take for the absorbance to fall to \(0.100 ?\)

The oxidation of \(\mathrm{SO}_{2}\) to \(\mathrm{SO}_{3}\) is accelerated by \(\mathrm{NO}_{2} .\) The reaction proceeds according to: $$ \begin{array}{l}{\mathrm{NO}_{2}(g)+\mathrm{SO}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{SO}_{3}(g)} \\ {2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)}\end{array}$$ (a) Show that, with appropriate coefficients, the two reactions can be summed to give the overall oxidation of \(S O_{2}\) by \(\mathrm{O}_{2}\) to give \(S O_{3} .(\mathbf{b})\) Do we consider \(N O_{2}\) a catalyst or an intermediate in this reaction? (c) Would you classify NO as a catalyst or as an intermediate? { ( d ) } Is this an example of homogeneous catalysis or heterogeneous catalysis?

(a) For a generic second-order reaction \(\mathrm{A} \longrightarrow \mathrm{B}\) , what quantity, when graphed versus time, will yield a straight line? (b) What is the slope of the straight line from part line? (b) What is the slope of the straight line from part (a)? (c) Does the half-life of a second-order reaction increase, decrease, or remain the same as the reaction proceeds?

(a) Can an intermediate appear as a reactant in the first step of a reaction mechanism? (b) On a reaction energy profile diagram, is an intermediate represented as a peak or a valley? (c) If a molecule like \(C l_{2}\) falls apart in an elementary reaction, what is the molecularity of the reaction?

Indicate whether each statement is true or false. (a) If you compare two reactions with similar collision factors, the one with the larger activation energy will be faster. (b) A reaction that has a small rate constant must have a small frequency factor. (c) Increasing the reaction temperature increases the fraction of successful collisions between reactants.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free