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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 \(\mathrm{Dp}=4 \mathrm{~mm}\) ?

Short Answer

Expert verified
Maximum rate of hydrolysis is 14 mmoles/h. For \( D_p = 2 mm \), \( \phi = 0.659 \) and \( \eta = 0.88 \). For \( D_p = 4 mm \), \( \phi = 1.318 \) and \( \eta = 0.657 \).

Step by step solution

01

Determine the maximum rate of hydrolysis (\textrm{V}_{m})

Using the Michaelis-Menten equation, the rate of reaction is given by: \[ v = V_{m} \frac{[S]}{K_{m}' + [S]} \] Given, \( v = 10 \text{ mmoles} \cdot\text{h}^{-1} \), \( [S] = 0.5 \text{ mM} \), \( K_{m}' = 0.2 \text{ mM} \). Rearrange the equation to solve for \( V_{m} \): \[ 10 = V_{m} \frac{0.5}{0.2 + 0.5} \], \[ V_{m} = 10 \cdot \frac{0.7}{0.5} = 10 \cdot 1.4 = 14 \text{ mmoles} \cdot\text{h}^{-1} \].
02

Determine the Thiele modulus (\textrm{\phi})

The Thiele modulus is defined as: \[ \phi = \frac{r}{3} \sqrt{\frac{V_{m}}{D_{e}K_{m}'}} \] where \( r \) is the bead radius. Given, \( r = \frac{2 \text{ mm}}{2} = 1 \text{ mm} = 0.1 \text{ cm} \), \( D_{e} = 1.5 \times 10^{-5} \text{ cm}^{2}/\text{s} \), and \( K_{m}' = 0.2 \text{ mM} \). Substituting these values we get, \[ \phi = 0.1 \text{ cm} \cdot \sqrt{\frac{14 \text{ mmoles}/\text{h}}{3 \times 1.5 \times 10^{-5} \text{ cm}^{2}/\text{s} \cdot 0.2 \text{ mM}}} \]. Convert \text{h} to \text{s} (1 hour = 3600 s): \[ \phi = 0.1 \text{ cm} \cdot \sqrt{\frac{14}{3 \times 1.5 \times 10^{-5} \cdot 0.2 \cdot 3600}} \], \[ \phi = 0.1 \cdot \sqrt{\frac{14}{0.324}} \approx 0.1 \cdot 6.59 \approx 0.659 \].
03

Determine the effectiveness factor (\eta)

For a spherical bead, the effectiveness factor \( \eta \) can be approximated by: \[ \eta = \frac{1}{\phi} \left[ \text{tanh}(\phi) \right] \] For \( \phi = 0.659 \), calculate \text{tanh}(0.659): \[ \eta = \frac{1}{0.659} \left[ \text{tanh}(0.659) \right] \approx 1.52 \left[ 0.578 \right] \approx 0.88 \].
04

Calculate values for a particle size of \text{Dp} = 4 mm

First, recalculate the radius: \( r = \frac{4 \text{ mm}}{2} = 2 \text{ mm} = 0.2 \text{ cm} \). Find the new Thiele modulus \( \phi \): \[ \phi = 0.2 \cdot \sqrt{\frac{14}{0.324}} \approx 0.2 \cdot 6.59 \approx 1.318 \]. Find the effectiveness factor \( \eta \): \[ \eta = \frac{1}{1.318} \left[ \text{tanh}(1.318) \right] \approx 0.76 \left[ 0.866 \right] \approx 0.657 \].

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Key Concepts

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

Michaelis-Menten Kinetics
Michaelis-Menten kinetics is a model that describes how enzyme-catalyzed reactions occur. It explains the relationship between the rate of the reaction and the concentration of the substrate. Essentially, this model shows that as substrate concentration increases, the reaction rate also increases until it reaches a maximum rate, denoted as \(V_{max}\). The formula for Michaelis-Menten kinetics is: \[ v = \frac{V_{max}[S]}{K_{m} + [S]} \] where \(v\) is the reaction rate, \(V_{max}\) is the maximum reaction rate, \(K_{m}\) is the Michaelis constant, and \( [S] \) is the substrate concentration. By understanding this equation, we can determine the efficiency of an enzyme and how quickly it processes a given substrate. For example, in our exercise, we can calculate the maximum rate of hydrolysis \(V_{max}\) using the given values of reaction rate and substrate concentration.
Thiele Modulus
The Thiele modulus is a dimensionless number used in catalysis to describe the ratio of the reaction rate to the rate of diffusion within porous catalysts. It is given by the formula: \[ \phi = \frac{r}{3} \sqrt{\frac{V_{m}}{D_{e} \cdot K_{m}}} \] where \(r\) is the radius of the bead, \(V_{m}\) is the maximum rate of hydrolysis, \(D_{e}\) is the diffusivity, and \(K_{m}\) is the Michaelis constant. The Thiele modulus helps in understanding how effectively substrates can reach the active sites of the enzyme when it is immobilized in a matrix. For a small Thiele modulus, diffusion is rapid compared to the reaction rate. For larger values, the reaction rate is high compared to diffusion, indicating that diffusion limitations need to be considered.
Effectiveness Factor
The effectiveness factor, \( \text{η} \), quantifies the efficiency of an immobilized enzyme. It is defined as the actual reaction rate over the rate if there were no diffusion limitations. The effectiveness factor can be estimated using the Thiele modulus with the formula: \[ \eta = \frac{1}{\phi} [ \tanh(\phi) ] \] This factor ranges from 0 to 1, where an \( η \) close to 1 indicates that diffusion is not a limiting factor, and one close to 0 shows significant diffusion limitations. By calculating \( \text{η} \) for different bead sizes, one can assess the impact of particle size on the enzyme's effectiveness.
Diffusivity
Diffusivity \(D_{e}\) measures how easily a substrate can move through a medium. In the context of enzyme immobilization, it is critical as higher diffusivity allows better access of substrates to the enzyme. The unit of diffusivity is \( \text{cm}^{2}/\text{s} \) and it plays a crucial part in calculating the Thiele modulus. For example, in our exercise, we use the diffusivity of urea through \(Ca-alginate\) beads to understand how it affects the overall reaction kinetics.
Ca-alginate Beads
Ca-alginate beads are commonly used for enzyme immobilization. They are created by mixing alginate with calcium ions, forming a gel-like structure. The beads provide a medium where enzymes can be entrapped and are beneficial due to their biocompatibility, porosity, and simplicity in preparation. By immobilizing urease in these beads, it is easier to handle the enzyme, and the reaction can be controlled effectively. The bead diameter also influences the reaction rate, diffusivity, and the effectiveness factor, as seen in examples comparing different bead sizes.

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Most popular questions from this chapter

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.

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 $$ \frac{d(\mathrm{P})}{d t}=k_{2}[\mathrm{E}] $$

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 $$ v=\frac{d[\mathrm{P}]}{d t}=\frac{\left(v_{s} / K_{m}\right)[\mathrm{S}]-\left(v_{p} / K_{p}\right)[\mathrm{P}]}{1+\frac{[\mathrm{S}]}{K_{m}}+\frac{[\mathrm{P}]}{K_{p}}} $$ $$ \text { where } K_{m}=\frac{k_{-1}+k_{2}}{k_{1}} \text { and } K_{p}=\frac{k_{-1}+k_{2}}{k_{-2}} \text { and } V_{x}=k_{2}\left[\mathrm{E}_{0}\right], V_{p}=k_{-1}\left[\mathrm{E}_{0}\right] \text {. } $$

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?

The following data were obtained from enzymatic oxidation of phenol by phenol oxidase at different phenol concentrations. \(\begin{array}{lrccllllcrrr}\mathrm{S}(\mathrm{mg} / \mathrm{l}) & 10 & 20 & 30 & 50 & 60 & 80 & 90 & 110 & 130 & 140 & 150 \\ v(\mathrm{mg} / \mathrm{l}-\mathrm{h}) & 5 & 7.5 & 10 & 12.5 & 13.7 & 15 & 15 & 12.5 & 9.5 & 7.5 & 5.7\end{array}\) a. What type of inhibition is this ? b. Determine the constants \(V_{\mathrm{m}}, K_{m}\) and \(K_{s i-}\) c. Determine the oxidation rate at \([\mathrm{S}]=70 \mathrm{mg} / 1\).

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