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Chapter 14: Problem 101

One of the many remarkable enzymes in the human body is carbonic anhydrase, which catalyzes the interconversion of carbonic acid with carbon dioxide and water. 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, which catalyzes the interconversion of the substrate, carbonic acid, with carbon dioxide and water. The turnover number is \(10^7\) molecules of \(\mathrm{CO}_{2}\) per second.

Step by step solution

01

Define Key Terms

Before we can identify the components in the description, let's define the key terms. 1. Enzyme: A protein that acts as a catalyst, speeding up a specific biochemical reaction. 2. Substrate: The molecule upon which an enzyme acts. 3. Turnover number: The number of substrate molecules an enzyme can convert into product per unit time.
02

Identify the Enzyme

In the given description, the enzyme is mentioned directly: "One of the many remarkable enzymes in the human body is carbonic anhydrase." Thus, the enzyme in this case is carbonic anhydrase.
03

Identify the Substrate

The description provides information about the biochemical reaction that carbonic anhydrase catalyzes: "catalyzes the interconversion of carbonic acid with carbon dioxide and water." Therefore, the substrate for the enzyme carbonic anhydrase is carbonic acid.
04

Identify the Turnover Number

The turnover number is mentioned in the description: "The enzyme catalyzes the dehydration (release to air) of up to \(10^7 \mathrm{CO}_{2}\) molecules per second." In this case, the turnover number is \(10^7\) molecules of \(\mathrm{CO}_{2}\) per second. In conclusion: Enzyme: Carbonic anhydrase Substrate: Carbonic acid Turnover number: \(10^7\) molecules of \(\mathrm{CO}_{2}\) per second.

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Most popular questions from this chapter

The gas-phase reaction of \(\mathrm{NO}\) with \(\mathrm{F}_{2}\) to form NOF and \(\mathrm{F}\) has an activation energy of \(E_{a}=6.3 \mathrm{~kJ} / \mathrm{mol}\) and a frequency factor of \(A=6.0 \times 10^{8} \mathrm{M}^{-1} \mathrm{~s}^{-1}\). The reaction is believed to be bimolecular: $$ \mathrm{NO}(g)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{NOF}(g)+\mathrm{F}(g) $$ (a) Calculate the rate constant at \(100^{\circ} \mathrm{C}\). (b) Draw the Lewis structures for the \(\mathrm{NO}\) and the NOF molecules, given that the chemical formula for NOF is misleading because the nitrogen atom is actually the central atom in the molecule. (c) Predict the structure for the NOF molecule. (d) Draw a possible transition state for the formation of NOF, using dashed lines to indicate the weak bonds that are beginning to form. (e) Suggest a reason for the low activation energy for the reaction.

The reaction \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) is second order in \(\mathrm{NO}\) and first order in \(\mathrm{O}_{2}\). When \([\mathrm{NO}]=0.040 \mathrm{M}\) and \(\left[\mathrm{O}_{2}\right]=0.035 \mathrm{M}\), the observed rate of disappearance of \(\mathrm{NO}\) is \(9.3 \times 10^{-5} \mathrm{M} / \mathrm{s}\). (a) What is the rate of disappearance of \(\mathrm{O}_{2}\) at this moment? (b) What is the value of the rate constant? (c) What are the units of the rate constant? (d) What would happen to the rate if the concentration of NO were increased by a factor of \(1.8\) ?

(a) Most heterogeneous catalysts of importance are extremely finely divided solid materials. Why is particle size important? (b) What role does adsorption play in the action of a heterogeneous catalyst?

A flask is charged with \(0.100 \mathrm{~mol}\) of \(\mathrm{A}\) and allowed to react to form \(B\) according to the hypothetical gas-phase reaction \(\mathrm{A}(\mathrm{g}) \longrightarrow \mathrm{B}(\mathrm{g})\). The following data are collected: \begin{tabular}{lccccc} \hline Time (s) & 0 & 40 & 80 & 120 & 160 \\ \hline Moles of A & \(0.100\) & \(0.067\) & \(0.045\) & \(0.030\) & \(0.020\) \\ \hline \end{tabular} (a) Calculate the number of moles of \(\mathrm{B}\) at each time in the table. (b) Calculate the average rate of disappearance of \(\mathrm{A}\) for each \(40-\mathrm{s}\) interval, in units of \(\mathrm{mol} / \mathrm{s}\). (c) What additional information would be needed to calculate the rate in units of concentration per time?

(a) For a second-order reaction, what quantity, when graphed versus time, will yield a straight line? (b) How do the half-lives of first-order and second- order reactions differ?
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