Question

a. H. H. Weetall and N. B. Havewala report the following data for the production of dextrose from corn starch using both soluble and immobilized (azo-glass beads) glucoamylase in a fully agitated CSTR system. 1\. Soluble data: \(T=60^{\circ} \mathrm{C},\left[\mathrm{S}_{0}\right]=168 \mathrm{mg}\) starch \(/ \mathrm{ml},\left[\mathrm{E}_{0}\right]=11,600\) units, volume \(=1000 \mathrm{ml}\). 2\. Immobilized data: \(T=60^{\circ} \mathrm{C},\left[\mathrm{S}_{0}\right]=336 \mathrm{mg} \operatorname{starch} / \mathrm{ml},\left[\mathrm{E}_{0}\right]=46,400\) units initially, immobilized, volume \(=1000 \mathrm{ml}\). \begin{tabular}{ccc} \hline & \multicolumn{2}{c}{ Product concentration (mg dextrose/ml) } \\ \cline { 2 - 3 } Time (min) & Soluble & Immobilized \\ \hline 0 & \(12.0\) & \(18.4\) \\ 15 & \(40.0\) & 135 \\ 30 & \(76.5\) & 200 \\ 45 & \(94.3\) & 236 \\ 60 & \(120.0\) & 260 \\ 75 & \(135.5\) & 258 \\ 90 & \(151.2\) & 262 \\ 105 & \(150.4\) & 266 \\ 120 & \(155.7\) & 278 \\ 135 & \(160.1\) & 300 \\ 150 & \(164.9\) & 310 \\ 165 & \(170.0\) & 306 \\ 225 & \(-\) & 316 \\ 415 & \(-\) & 320 \\ \hline \end{tabular} Determine the maximum reaction velocity, \(V_{m}(\mathrm{mg} / \mathrm{ml}-\mathrm{min}\) - unit of enzyme) and the saturation constant, \(\mathrm{K}_{M}(\mathrm{mg} / \mathrm{ml})\). b. The same authors studied the effect of temperature on the maximum rate of the hydrolysis of corn starch by glucoamylase. The results are tabulated next. Determine the activation energy ( \(\Delta E \mathrm{cal} / \mathrm{g}\) mole) for the soluble and immobilized enzyme reaction. \begin{tabular}{ccc} \hline & \multicolumn{2}{c}{\(V_{\max }\left(\mathrm{m} \mathrm{mol} / \mathrm{min} 10^{6}\right)\)} \\ \cline { 2 - 3 } \(\mathrm{T},{ }^{\circ} \mathrm{C}\) & Soluble & Azo- immobilized \\ \hline 25 & \(0.62\) & \(0.80\) \\ 35 & \(1.42\) & \(1.40\) \\ 45 & \(3.60\) & \(3.00\) \\ 55 & \(8.0\) & \(6.2\) \\ 65 & \(16.0\) & \(11.0\) \\ \hline \end{tabular} c. Using these results, determine if immobilized enzyme is diffusion limited. [Courtesy of A. E. Humphrey from "Collected Coursework Problems in Biochemical Engineering" compiled by H. W. Blanch for 1977 Am. Soc. Eng. Educ. Summer School.]

Short answer

Obtain \( V_{m} \) and \( K_{M} \) from Lineweaver-Burk plots for soluble and immobilized enzymes. Use Arrhenius plots to find \( \Delta E \). Assess diffusion limitation by comparing immobilized and soluble \( V_{m} \) values.

Step-by-step solution

- Determine the maximum reaction velocity, \(V_{m}\)

To determine the maximum reaction velocity (\( V_{m} \)), plot the product concentration against time for both the soluble and immobilized enzyme data. Identify the linear portion of the curve at the beginning of the reaction, where the rate is highest and constant. Linear regression on these initial rates can give the maximum initial reaction velocity.

- Determine the substrate concentration, \([S]\)

Given the substrate concentration (\( [S_{0}] \)) for both soluble and immobilized enzymes:- Soluble: \( [S_{0}] = 168 \text{ mg/ml} \)- Immobilized: \( [S_{0}] = 336 \text{ mg/ml} \)

- Utilize Michaelis-Menten equation

The Michaelis-Menten equation is utilized to describe the kinetics of enzyme-catalyzed reactions:\[ V = V_{m} \frac{[S]}{K_{M} + [S]} \]

- Rearrange for Lineweaver-Burk Plot (Double Reciprocal Plot)

Convert the Michaelis-Menten equation into its double reciprocal form (Lineweaver-Burk plot) for better analysis:\[ \frac{1}{V} = \frac{K_{M}}{V_{m}[S]} + \frac{1}{V_{m}} \]Plot \( \frac{1}[V] \) against \( \frac{1}[S] \) for both soluble and immobilized enzymes.

- Determine \(K_{M}\) and \(V_{m}\) from Lineweaver-Burk Plot

The y-intercept of the Lineweaver-Burk plot gives \( \frac{1}{V_{m}} \) and the slope gives \( \frac{K_{M}}{V_{m}} \). Use these values to calculate \( K_{M} \) and \( V_{m} \) for both enzymes.

- Determine the Activation Energy \(\Delta E\)

Using Arrhenius equation, the activation energy \( \Delta E \) can be determined from the temperature dependency of \( V_{m} \):\[ k = Ae^{-\frac{\Delta E}{RT}} \]Take natural logarithm for linearization:\[ \ln V_{m} = \ln A - \frac{\Delta E}{R} \frac{1}{T} \]Plot \( \ln V_{m} \) versus \( \frac{1}{T} \) to determine the slope, which corresponds to \( -\frac{\Delta E}{R} \). Calculate \( \Delta E \) for both enzyme systems.

- Determine if Immobilized Enzyme is Diffusion Limited

Compare the \( V_{m} \) values for both soluble and immobilized enzymes across temperatures. A significant difference at higher temperatures suggests diffusion limitations in the immobilized enzyme due to reduced substrate accessibility.

Key concepts

Michaelis-Menten Kinetics

Michaelis-Menten kinetics is a model that describes how the rate of an enzyme-catalyzed reaction is dependent on the concentration of a substrate. This model is defined by the Michaelis-Menten equation:

\[ V = V_{m} \frac{[S]}{K_{M} + [S]} \]

Here,

• \( V \) represents the reaction rate.
• \( V_{m} \) is the maximum reaction velocity.
• \( [S] \) is the substrate concentration.
• \( K_{M} \) is the Michaelis constant, indicating the substrate concentration at which the reaction rate is half of its maximum value.

This equation helps to understand how an enzyme performs under different substrate concentrations. It’s essential in bioprocess engineering for designing and optimizing enzyme reactions.

Lineweaver-Burk Plot

The Lineweaver-Burk plot is a double reciprocal graph of the Michaelis-Menten equation. It transforms the hyperbolic curve of the Michaelis-Menten equation into a straight line for easier analysis. The equation for this plot is:

\[ \frac{1}{V} = \frac{K_{M}}{V_{m}[S]} + \frac{1}{V_{m}} \]

In this linear form,

• The y-intercept represents \( \frac{1}{V_{m}} \).
• The slope represents \( \frac{K_{M}}{V_{m}} \).

By plotting \( \frac{1}{V} \) against \( \frac{1}{S} \), you can easily determine the values of \( V_{m} \) and \( K_{M} \). This plot is particularly useful for analyzing enzyme kinetics when experimental data are available.

Activation Energy

Activation energy (\( \Delta E \)) is the minimum energy required for a chemical reaction to occur. For enzyme-catalyzed reactions, the Arrhenius equation describes the temperature dependence of the reaction rate:

\[ k = Ae^{-\frac{\Delta E}{RT}} \]

Here,

• \( k \) is the rate constant.
• \( A \) is the pre-exponential factor.
• \( R \) represents the universal gas constant.
• \( T \) is the absolute temperature.

Taking the natural logarithm of both sides linearizes the equation:

\[ \ln V_{m} = \ln A - \frac{\Delta E}{R} \cdot \frac{1}{T} \]

By plotting \( \ln V_{m} \) versus \( \frac{1}{T} \), the slope of the line gives \( -\frac{\Delta E}{R} \), allowing for determination of the activation energy.

Immobilized Enzyme

Immobilized enzymes are enzymes that are attached to an inert, insoluble material. These can be used in various industrial processes where they offer several advantages:

• They are reusable, reducing the cost of enzyme replacements.
• They exhibit enhanced stability in reaction environments.
• They can be easily separated from the reaction mixture, simplifying product purification.

However, immobilization can also introduce issues like diffusion limitations, where the substrate's movement to the enzyme's active site is restricted. This limitation can significantly impact the enzyme's effectiveness, especially in large-scale bioprocesses.

Soluble Enzyme

Soluble enzymes are enzymes that are in solution rather than being immobilized. They offer benefits such as:

• High catalytic efficiency due to unrestricted diffusion of substrates and products.
• Homogeneity in the reaction mixture, ensuring uniform enzyme activity.

Despite these advantages, soluble enzymes can be harder to recover and reuse, making them more expensive for industrial applications. They may also show less stability under extreme conditions compared to immobilized enzymes.

Diffusion Limitation

Diffusion limitation occurs when the rate of an enzymatic reaction is restricted by the slow movement of substrate molecules to the enzyme's active site. This is often a challenge with immobilized enzymes because the substrate must diffuse through the matrix holding the enzyme.

Key indications of diffusion limitations include:

• A decrease in reaction rate at higher substrate concentrations.
• A significant difference in maximum reaction velocities (\( V_{m} \)) between soluble and immobilized enzymes, especially at higher temperatures.

Detecting and mitigating diffusion limitations is crucial for optimizing bioprocesses involving immobilized enzymes.