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Chapter 3: Problem 12

You are working for company A and you join a research group working on immobilized enzymes. Harry, the head of the lab, claims that immobilization improves the stability of the enzyme. His proof is that the enzyme has a half- life of 10 days in free solution, but under identical conditions of temperature, \(\mathrm{pH}\), and medium composition, the measured half-life of a packed column is 30 days. The enzyme is immobilized in a porous sphere \(5 \mathrm{~mm}\) in diameter. Is Harry's reasoning right? Do you agree with him? Why or why not?

Short Answer

Expert verified
Yes, Harry's reasoning is correct because the enzyme's half-life increases from 10 days in free solution to 30 days when immobilized.

Step by step solution

01

Understand the Problem

We need to evaluate if the head of the lab's reasoning that immobilization improves enzyme stability is correct based on the given half-life values.
02

Compare Half-lives

Compare the half-life of the enzyme in free solution (10 days) with the half-life of the enzyme immobilized in a packed column (30 days).
03

Analyze the Comparison

Since the half-life of the enzyme is higher when immobilized in the packed column (30 days) compared to the enzyme in free solution (10 days), this indicates greater stability in the immobilized form.
04

Conclusion

Harry's reasoning is correct. The extended half-life in the immobilized form shows increased stability of the enzyme compared to its free solution form under identical conditions.

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Key Concepts

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

enzyme stability
Enzyme stability refers to the ability of an enzyme to remain active and functional over time. When enzymes lose their activity, they are said to have degraded. This process can be influenced by factors such as temperature, pH, and the presence of inhibitors.

In this exercise, we're looking at how immobilization affects enzyme stability.

Immobilization involves attaching enzymes to a solid support or encapsulating them in a gel. By doing this, the enzyme's environment is controlled and isolated from external changes.

Harry claims that immobilization improves enzyme stability. His proof is based on the half-life data of the enzyme in free solution versus its immobilized form.

The enzyme's half-life in free solution is 10 days, while its half-life when immobilized in a packed column is 30 days. This significant increase in half-life illustrates better stability in the immobilized state.

Stability can be crucial in industrial applications where enzymes need to function for extended periods. By increasing stability, enzymes can be reused, reducing costs and improving efficiency.
half-life comparison
Half-life is the time required for half of the enzyme's activity to diminish. If an enzyme has a half-life of 10 days, it means that after 10 days, only half of its original activity remains.

To evaluate Harry's claim, we need to compare the half-life of the enzyme in two forms: free solution and immobilized.

In free solution, the enzyme has a half-life of 10 days. When immobilized in a packed column, the half-life extends to 30 days. This indicates that the enzyme retains its activity three times longer when immobilized.

This comparison directly supports Harry's assertion that immobilization enhances enzyme stability.

Enzyme half-life is a crucial metric in bioprocess engineering, as it affects the overall efficiency and cost-effectiveness of enzyme-based processes.
bioprocess engineering
Bioprocess engineering combines biology with engineering principles to develop processes for manufacturing products such as pharmaceuticals, food, and biofuels.

Enzymes play a pivotal role in bioprocess engineering due to their ability to catalyze reactions efficiently and specifically.

In this context, enzyme stability and half-life are key factors. More stable enzymes with longer half-lives reduce the frequency of enzyme replacement, cutting costs and improving process sustainability.

Immobilized enzymes are particularly valuable in bioprocess engineering. Immobilization not only improves stability but also facilitates enzyme recovery and reuse.

By introducing stability through immobilization, bioprocesses become more efficient and economically viable.

Understanding the relationships between enzyme stability, half-life, and immobilization can lead to better design and optimization of bioprocesses, driving advancements in industries that rely on these biological catalysts.

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Most popular questions from this chapter

An enzyme ATPase has a molecular weight of \(5 \times 10^{4}\) daltons, a \(K_{M}\) value of \(10^{-4} M\), and a \(k_{2}\) value of \(k_{2}=10^{4}\) molecules ATP/min molecule enzyme at \(37^{\circ} \mathrm{C}\). The reaction catalyzed is the following: $$ \mathrm{ATP} \stackrel{\text { ATPase }}{\longrightarrow} \mathrm{ADP}+\mathrm{P}_{i} $$ which can also be represented as $$ \mathrm{E}+\mathrm{S} \underset{k_{1}}{\stackrel{A_{1}}{\longrightarrow} \mathrm{ES} \stackrel{k_{2}}{\longrightarrow} \mathrm{E}+\mathrm{P} $$ where \(S\) is ATP. The enzyme at this temperature is unstable. The enzyme inactivation kinetics are first order: $$ \mathrm{E}=\mathrm{E}_{0} e^{-k_{\alpha} r} $$ where \(\mathrm{E}_{0}\) is the initial enzyme concentration and \(k_{d}=0.1 \mathrm{~min}^{-1}\). In an experiment with a partially pure enzyme preparation, \(10 \mu \mathrm{g}\) of total crude protein (containing enzyme) is added to a \(1 \mathrm{ml}\) reaction mixture containing \(0.02 \mathrm{M}\) ATP and incubated at \(37^{\circ} \mathrm{C}\). After 12 hours the reaction ends (i.e., \(t \rightarrow \infty\) ) and the inorganic phosphate \(\left(\mathrm{P}_{i}\right)\) concentration is found to be \(0.002 M\), which was initially zero. What fraction of the crude protein preparation was the enzyme? Hint: Since \([\mathrm{S}]>>K_{m}\), the reaction rate can be represented by $$ \frac{d(\mathrm{P})}{d t}=k_{2}[\mathrm{E}] $$

The following data were obtained for an enzyme-catalyzed reaction. Determine \(\mathrm{V}_{\max }\) and \(K_{m}\) by inspection. Plot the data using the Eadie- Hofstee method and determine these constants graphically. Explain the discrepancy in your two determinations. The initial rate data for the enzyme- catalyzed reaction are as follows: \begin{tabular}{cc} \hline\({[\mathrm{S}] }\) \(\mathrm{mol} / 1\) & \(v\) \(\mu \mathrm{mol} / \mathrm{min}\) \\ \hline \(5.0 \times 10^{-4}\) & 125 \\ \(2.0 \times 10^{-4}\) & 125 \\ \(6.0 \times 10^{-5}\) & 121 \\ \(4.0 \times 10^{-5}\) & 111 \\ \(3.0 \times 10^{-5}\) & \(96.5\) \\ \(2.0 \times 10^{-5}\) & \(62.5\) \\ \(1.6 \times 10^{-5}\) & \(42.7\) \\ \(1.0 \times 10^{-5}\) & \(13.9\) \\ \(8.0 \times 10^{-6}\) & \(7.50\) \\ \hline \end{tabular} Do these data fit into Michaelis-Menten kinetics? If not, what kind of rate expression would you suggest? Use graphical methods.

Consider the following reaction sequence: $$ \mathrm{S}+\mathrm{E} \underset{k_{2}}{\rightleftharpoons}(\mathrm{ES})_{1} \rightleftharpoons_{k_{4}}^{k_{3}}(\mathrm{ES})_{2} \stackrel{k_{g}}{\longrightarrow} \mathrm{P}+\mathrm{E} $$ Develop a suitable rate expression for production formation \(\left[v=k_{3}(\mathrm{ES})_{2}\right]\) by using (a) the equilibrium approach, and (b) the quasi-steady-state approach.

The enzyme, urease, is immobilized in Ca-alginate beads \(2 \mathrm{~mm}\) in diameter. When the urea concentration in the bulk liquid is \(0.5 \mathrm{~m} M\) the rate of urea hydrolysis is \(v=10 \mathrm{mmoles-1- \textrm {h } .}\) Diffusivity of urea in \(\mathrm{Ca}\)-alginate beads is \(D_{e}=1.5 \times 10^{-5} \mathrm{~cm}^{2} / \mathrm{sec}\), and the Michaelis constant for the enzyme is \(K_{m}^{\prime}=0.2 \mathrm{~m} M\). By neglecting the liquid film resistance on the beads (i.e., \(\left.\left[\mathrm{S}_{0}\right]=\left[\mathrm{S}_{\mathrm{s}}\right]\right)\) determine the following: a. Maximum rate of hydrolysis \(V_{m}\), Thiele modulus \((\phi)\), and effectiveness factor \((\eta)\). b. What would be the \(V_{w}, \phi\), and \(\eta\) values for a particle size of \(\mathrm{Dp}=4 \mathrm{~mm}\) ?

Consider the reversible product-formation reaction in an enzyme-catalyzed bioreaction: $$ \mathrm{E}+\mathrm{S} \stackrel{k_{1}}{\rightleftharpoons}(\mathrm{ES}) \stackrel{k_{2}}{\stackrel{k_{3}}{\sum}} \mathrm{E}+\mathrm{P} $$ Develop a rate expression for product-formation using the quasi-steady-state approximation and show that $$ v=\frac{d[\mathrm{P}]}{d t}=\frac{\left(v_{s} / K_{m}\right)[\mathrm{S}]-\left(v_{p} / K_{p}\right)[\mathrm{P}]}{1+\frac{[\mathrm{S}]}{K_{m}}+\frac{[\mathrm{P}]}{K_{p}}} $$ $$ \text { where } K_{m}=\frac{k_{-1}+k_{2}}{k_{1}} \text { and } K_{p}=\frac{k_{-1}+k_{2}}{k_{-2}} \text { and } V_{x}=k_{2}\left[\mathrm{E}_{0}\right], V_{p}=k_{-1}\left[\mathrm{E}_{0}\right] \text {. } $$
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