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

The enzyme invertase catalyzes the conversion of sucrose, a disaccharide, to invert sugar, a mixture of glucose and fructose. When the concentration of invertase is \(4.2 \times 10^{-7} \mathrm{M}\) and the concentration of sucrose is \(0.0077 \mathrm{M}\) invert sugar is formed at the rate of \(1.5 \times 10^{-4} \mathrm{M} / \mathrm{s}\). When the sucrose concentration is doubled, the rate of formation of invert sugar is doubled also. (a) Assuming that the enzyme-substrate model is operative, is the fraction of enzyme tied up as a complex large or small? Explain. (b) Addition of inositol, another sugar, decreases the rate of formation of invert sugar. Suggest a mechanism by which this occurs.

Short Answer

Expert verified
(a) The fraction of enzyme tied up as a complex is small. This is because the reaction rate doubles when the concentration of sucrose doubles, indicating that the enzyme is not saturated with the substrate, and therefore, only a small fraction of the enzyme is present in the enzyme-substrate complex. (b) The addition of inositol likely causes competitive inhibition. Inositol may compete with sucrose to bind with the enzyme invertase due to structural similarities. This competition decreases the effective concentration of sucrose binding to the enzyme, resulting in a slower rate of invert sugar formation.

Step by step solution

01

(a) Determine the fraction of enzyme tied up as a complex

Since the rate of formation of invert sugar doubles when the concentration of sucrose doubles, this implies that the reaction is still under the influence of substrate concentration, which means the enzyme is not saturated with the substrate. This further means that there is a relatively small fraction of enzyme tied up as the enzyme-substrate complex. Because if it was large, the doubling of substrate concentration wouldn't have doubled the rate as the enzyme would have been saturated.
02

(b) Suggest a mechanism for the effect of inositol on the reaction rate

As the addition of inositol, another sugar, decreases the rate of formation of invert sugar, one possible mechanism by which this occurs may be competitive inhibition. In this case, inositol may compete with sucrose to bind with the enzyme invertase due to structural similarities, thus decreasing the effective concentration of sucrose that can bind to the enzyme, resulting in a slower rate of invert sugar formation. This competition for the enzyme's active site would result in a decrease in the rate of the reaction involving sucrose and invertase.

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.

Invertase
Invertase is a specific enzyme that plays a vital role in the chemical digestion of sucrose. It catalyzes the breakdown of sucrose into its constituent monosaccharides, glucose and fructose. This enzymatic reaction is essential because it helps organisms, such as plants and microorganisms, convert complex sugars into simpler ones that are more readily absorbed or utilized. The process is often compared to unlocking a door, where invertase acts as the key that opens up sucrose and releases its simpler components.
  • Invertase is commonly used in the food industry to produce invert sugar from sucrose.
  • Invert sugar has a different sweetness profile, making it valuable in culinary applications.
  • It improves stability and adds a sheen to products like confectionery.
The activity of invertase, like many enzymes, is influenced by various factors such as temperature, pH levels, and substrate concentration. Understanding these factors is essential for optimizing its use in scientific and industrial processes.
Enzyme-Substrate Complex
The enzyme-substrate complex is a temporary molecular structure formed when an enzyme binds to its substrate. This binding occurs at a specific region on the enzyme known as the active site. The active site is uniquely shaped to fit the substrate, facilitating a reaction that converts it into the product. In the context of invertase and sucrose, the enzyme-substrate complex is necessary for the conversion of sucrose into glucose and fructose.
  • The rate of a reaction is influenced by the concentration of enzyme-substrate complexes.
  • If too few complexes form, the reaction rate is slow; if many form, the rate is faster.
  • The relationship between substrate concentration and reaction rate helps determine enzyme efficiency.
In the original exercise, it's highlighted that the fraction of enzyme tied up in the complex is relatively small. This conclusion is drawn because doubling the substrate results in a proportional increase in the rate of reaction, indicating that the enzymes are not saturated and can form additional complexes efficiently.
Competitive Inhibition
Competitive inhibition occurs when molecules similar in structure to the substrate compete for the enzyme's active site. This process can decrease the overall rate of the reaction by blocking the actual substrate from binding. In the original problem, inositol acts as a competitive inhibitor by mimicking the structure of sucrose, thereby occupying the binding sites on invertase that would otherwise facilitate the production of invert sugar. Competitive inhibition is an important concept because it shows how different molecules can influence enzyme activity:
  • It highlights the specificity of enzymes and the importance of structural similarity.
  • It demonstrates how enzymes can be regulated by other molecules in biological systems.
  • Understanding this process is crucial for drug design and the control of metabolic pathways.
By introducing a competitive inhibitor, the reaction speed is reduced without permanently disabling the enzyme. This form of inhibition can often be overcome by increasing the concentration of the substrate, eventually outcompeting the inhibitor for binding sites.

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

Hydrogen sulfide \(\left(\mathrm{H}_{2} \mathrm{~S}\right)\) is a common and troublesome pollutant in industrial wastewaters. One way to remove \(\mathrm{H}_{2} \mathrm{~S}\) is to treat the water with chlorine, in which case the following reaction occurs: $$ \mathrm{H}_{2} \mathrm{~S}(a q)+\mathrm{Cl}_{2}(a q) \longrightarrow \mathrm{S}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) $$ The rate of this reaction is first order in each reactant. The rate constant for the disappearance of \(\mathrm{H}_{2} \mathrm{~S}\) at \(28^{\circ} \mathrm{C}\) is \(3.5 \times 10^{-2} \mathrm{M}^{-1} \mathrm{~s}^{-1}\). If at a given time the concentration of \(\mathrm{H}_{2} \mathrm{~S}\) is \(2.0 \times 10^{-4} \mathrm{M}\) and that of \(\mathrm{Cl}_{2}\) is \(0.025 \mathrm{M}\), what is the rate of formation of \(\mathrm{Cl}^{-} ?\)

The gas-phase reaction \(\mathrm{Cl}(g)+\mathrm{HBr}(g) \longrightarrow \mathrm{HCl}(g)+\) \(\mathrm{Br}(g)\) has an overall enthalpy change of \(-66 \mathrm{~kJ}\). The activation energy for the reaction is \(7 \mathrm{~kJ}\). (a) Sketch the energy profile for the reaction, and label \(E_{a}\) and \(\Delta E\). (b) What is the activation energy for the reverse reaction?

The rates of many atmospheric reactions are accelerated by the absorption of light by one of the reactants. For example, consider the reaction between methane and chlorine to produce methyl chloride and hydrogen chloride: Reaction 1: \(\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{Cl}(g)+\mathrm{HCl}(g)\) This reaction is very slow in the absence of light. However, \(\mathrm{Cl}_{2}(g)\) can absorb light to form \(\mathrm{Cl}\) atoms: $$ \text { Reaction 2: } \mathrm{Cl}_{2}(g)+h v \longrightarrow 2 \mathrm{Cl}(g) $$ Once the \(\mathrm{Cl}\) atoms are generated, they can catalyze the reaction of \(\mathrm{CH}_{4}\) and \(\mathrm{Cl}_{2}\), according to the following proposed mechanism: Reaction 3: \(\mathrm{CH}_{4}(g)+\mathrm{Cl}(g) \longrightarrow \mathrm{CH}_{3}(g)+\mathrm{HCl}(g)\) $$ \text { Reaction 4: } \mathrm{CH}_{3}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{Cl}(g)+\mathrm{Cl}(g) $$ The enthalpy changes and activation energies for these two reactions are tabulated as follows: $$ \begin{array}{lll} \hline \text { Reaction } & \Delta H_{\mathrm{ran}}^{\circ}(\mathrm{k} \mathrm{J} / \mathrm{mol}) & \mathrm{E}_{a}(\mathrm{~kJ} / \mathrm{mol}) \\ \hline 3 & +4 & 17 \\ 4 & -109 & 4 \\ \hline \end{array} $$ (a) By using the bond enthalpy for \(\mathrm{Cl}_{2}\) (Table \(8.4\) ), determine the longest wavelength of light that is energetic enough to cause reaction 2 to occur. \(\ln\) which portion of the electromagnetic spectrum is this light found? (b) By using the data tabulated here, sketch a quantitative energy profile for the catalyzed reaction represented by reactions 3 and 4. (c) By using bond enthalpies, estimate where the reactants, \(\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g)\), should be placed on your diagram in part (b). Use this result to estimate the value of \(E_{a}\) for the reaction \(\mathrm{CH}_{4}(g)+\mathrm{Cl}_{2}(g) \longrightarrow\) \(\mathrm{CH}_{3}(g)+\mathrm{HCl}(g)+\mathrm{Cl}(g) .\) (d) The species \(\mathrm{Cl}(g)\) and \(\mathrm{CH}_{3}(g)\) in reactions 3 and 4 are radicals, that is, atoms or molecules with unpaired electrons. Draw a Lewis structure of \(\mathrm{CH}_{3}\), and verify that it is a radical. (e) The sequence of reactions 3 and 4 comprise a radical chain mechanism. Why do you think this is called a "chain reaction"? Propose a reaction that will terminate the chain reaction.

Metals often form several cations with different charges. Cerium, for example, forms \(\mathrm{Ce}^{3+}\) and \(\mathrm{Ce}^{4+}\) ions, and thallium forms \(\mathrm{Tl}^{+}\) and \(\mathrm{Tl}^{3+}\) ions. Cerium and thallium ions react as follows: $$ 2 \mathrm{Ce}^{4+}(a q)+\mathrm{Tl}^{+}(a q) \longrightarrow 2 \mathrm{Ce}^{3+}(a q)+\mathrm{Tl}^{3+}(a q) $$ This reaction is very slow and is thought to occur in a single elementary step. The reaction is catalyzed by the addition of \(\mathrm{Mn}^{2+}(a q)\), according to the following mechanism: $$ \begin{aligned} \mathrm{Ce}^{4+}(a q)+\mathrm{Mn}^{2+}(a q) & \longrightarrow \mathrm{Ce}^{3+}(a q)+\mathrm{Mn}^{3+}(a q) \\ \mathrm{Ce}^{4+}(a q)+\mathrm{Mn}^{3+}(a q) & \longrightarrow \mathrm{Ce}^{3+}(a q)+\mathrm{Mn}^{4+}(a q) \\ \mathrm{Mn}^{4+}(a q)+\mathrm{Tl}^{+}(a q) &\left.\longrightarrow \mathrm{Mn}^{2+}(a q)+\mathrm{T}\right]^{3+}(a q) \end{aligned} $$ (a) Write the rate law for the uncatalyzed reaction. (b) What is unusual about the uncatalyzed reaction? Why might it be a slow reaction? (c) The rate for the catalyzed reaction is first order in \(\left[\mathrm{Ce}^{4+}\right]\) and first order in \(\left[\mathrm{Mn}^{2+}\right]\). Based on this rate law, which of the steps in the catalyzed mechanism is rate determining? (d) Use the available oxidation states of \(\mathrm{Mn}\) to comment on its special suitability to catalyze this reaction.

Consider the following reaction: $$ \mathrm{CH}_{3} \mathrm{Br}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{Br}^{-}(a q) $$ The rate law for this reaction is first order in \(\mathrm{CH}_{3} \mathrm{Br}\) and first order in \(\mathrm{OH}^{-}\). When \(\left[\mathrm{CH}_{3} \mathrm{Br}\right]\) is \(5.0 \times 10^{-3} \mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]\) is \(0.050 \mathrm{M}\), the reaction rate at \(298 \mathrm{~K}\) is \(0.0432 \mathrm{M} / \mathrm{s}\) (a) What is the value of the rate constant? (b) What are the units of the rate constant? (c) What would happen to the rate if the concentration of \(\mathrm{OH}^{-}\) were tripled?

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