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.

Short Answer

Expert verified
Equilibrium: \( v = k_3 K_1 K_2 [S][E] \) Quasi-steady-state: \( v = \frac{k_3^2 k_1 [S][E]}{k_4 (k_2 + k_3)} \)

Step by step solution

01

- Define equilibrium constants for sequestration reactions (S and E form complexes)

For the reaction sequence, establish equilibrium constants: For the first equilibrium: \[ K_1 = \frac{[ES]_1}{[S][E]} \] For the second equilibrium: \[ K_2 = \frac{[ES]_2}{[ES]_1} \]
02

- Express complexes in terms of reactants and intermediates under equilibrium approach

Using equilibrium constants, rewrite concentrations of intermediate complexes as: \[ [ES]_1 = K_1 [S][E] \] \[ [ES]_2 = K_2 [ES]_1 = K_2 K_1 [S][E] \]
03

- Develop rate expression using equilibrium concentrations

Under equilibrium, the rate of product formation is: \[ v = k_3 [ES]_2 = k_3 (K_2 K_1 [S][E]) \] Thus, the rate expression is: \[ v = k_3 K_1 K_2 [S][E] \]
04

- Define rate equations under quasi-steady-state approach

Under the quasi-steady-state approach, assume intermediate complexes change slowly: \[ \frac{d[ES]_1}{dt} = k_1[S][E] - k_2[ES]_1 - k_3[ES]_1 = 0 \] \[ \frac{d[ES]_2}{dt} = k_3[ES]_1 - k_4[ES]_2 = 0 \]
05

- Solve quasi-steady-state equations

From \(\frac{d[ES]_1}{dt} = 0\), we get: \[ [ES]_1 = \frac{k_1[S][E]}{k_2 + k_3} \] From \(\frac{d[ES]_2}{dt} = 0\), we get: \[ [ES]_2 = \frac{k_3[ES]_1}{k_4} = \frac{k_3 k_1 [S][E]}{k_4(k_2 + k_3)} \]
06

- Develop rate expression using quasi-steady-state concentrations

Under quasi-steady-state, the rate of product formation is: \[ v = k_3 [ES]_2 = \frac{k_3^2 k_1 [S][E]}{k_4 (k_2 + k_3)} \] Thus, the rate expression is: \[ v = \frac{k_3^2 k_1 [S][E]}{k_4 (k_2 + k_3)} \]

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.

Equilibrium Approach
The equilibrium approach is a method to simplify the rate expression for enzyme-catalyzed reactions. In this method, all intermediate steps are considered to be in equilibrium.
This implies that the forward and reverse reactions occur at the same rate.
This provides a steady concentration of intermediate complexes.
For the given reaction sequence: $$\text{S}+\text{E} \rightleftharpoons (\text{ES})_1 \rightleftharpoons (\text{ES})_2 \rightarrow \text{P}+\text{E}$$ we define equilibrium constants for each step:
  • For the first equilibrium: \( K_1 = \frac{[\text{ES}]_1}{[\text{S}][\text{E}]} \)

  • For the second equilibrium: \( K_2 = \frac{[\text{ES}]_2}{[\text{ES}]_1} \)
Using these, we can express the concentrations of intermediate complexes in terms of reactants and other intermediates.
Thus, we get: \( [\text{ES}]_1 = K_1 [\text{S}][\text{E}] \)
and \( [\text{ES}]_2 = K_2 K_1 [\text{S}][\text{E}] \).
By substituting these into the rate expression for product formation: \( v = k_3 [\text{ES}]_2 = k_3 K_1 K_2 [\text{S}][\text{E}] \).
This provides the rate expression under the equilibrium approach.
Quasi-Steady-State Approach
The quasi-steady-state approach is another method often used in enzyme kinetics.
This approach assumes that the concentrations of intermediate complexes change very slowly over time. As a result, their rate of formation is roughly equal to their rate of breakdown. For the same reaction sequence:
  • \text{d[\text{ES}]_1}/\text{dt} = k_1[\text{S}][\text{E}] - k_2[\text{ES}]_1 - k_3[\text{ES}]_1 = 0

  • \text{d[\text{ES}]_2}/\text{dt} = k_3[\text{ES}]_1 - k_4[\text{ES}]_2 = 0
From these, we first solve for [\text{ES}]_1:
\( [\text{ES}]_1 = \frac{k_1[\text{S}][\text{E}]}{k_2 + k_3} \).
Then, solving for [\text{ES}]_2, we get:
\( [\text{ES}]_2 = \frac{k_3 [\text{ES}]_1}{k_4} = \frac{k_3 k_1 [\text{S}][\text{E}]}{k_4 (k_2 + k_3)} \).
Inserting these into the rate of product formation, we find:
\( v = k_3 [\text{ES}]_2 = \frac{k_3^2 k_1 [\text{S}][\text{E}]}{k_4 (k_2 + k_3)} \).
This provides the rate expression under the quasi-steady-state approach.
Rate Expression
A rate expression provides a mathematical way to describe the speed of a reaction.
For enzyme-catalyzed reactions, this typically involves product formation.
It connects the concentration of substrate and enzyme to the rate at which the product forms.
In our case, for reaction sequence: $$\text{S}+\text{E} \rightleftharpoons (\text{ES})_1 \rightleftharpoons (\text{ES})_2 \rightarrow \text{P}+\text{E}$$ we found two different rate expressions under different assumptions.
Using the equilibrium approach, the rate expression is:
\( v = k_3 K_1 K_2 [\text{S}][\text{E}] \).
Under the quasi-steady-state approach, it's slightly different:
\( v = \frac{k_3^2 k_1 [\text{S}][\text{E}]}{k_4 (k_2 + k_3)} \).
Both expressions highlight how enzyme and substrate concentrations influence the reaction rate uniquely.
Intermediate Complexes
Intermediate complexes are temporary structures formed during a chemical reaction.
In enzyme kinetics, these complexes form between the enzyme and substrate.
They play a critical role in determining the reaction rate.
For our reaction sequence: $$\text{S}+\text{E} \rightleftharpoons (\text{ES})_1 \rightleftharpoons (\text{ES})_2 \rightarrow \text{P}+\text{E}$$, we have two main intermediates: (\text{ES})_1 and (\text{ES})_2.
  • (\text{ES})_1 is the initial complex formed by the enzyme and substrate.
  • (\text{ES})_2 is a secondary complex that eventually breaks down to produce the product.
Understanding these complexes is key to comprehending the overall reaction.
Each step involving these intermediates can be analyzed to derive rate expressions using different kinetic approaches, like the equilibrium or quasi-steady-state methods.
The transitions between these intermediate states determine the reaction's efficiency and rate.

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

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.]

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

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

Michaelis-Menten kinetics are used to describe intracellular reactions. Yet \(\left[\mathrm{E}_{0}\right] \approx\left[\mathrm{S}_{0}\right]\). In in vitro batch reactors, the quasi-steady-state hypothesis does not hold for \(\left[\mathrm{E}_{0}\right] \approx\left[\mathrm{S}_{0}\right]\). The rapid equilibrium assumption also will not hold. Explain why Michaelis-Menten kinetics and the quasi-steady-state approximation are still reasonable descriptions of intracellular enzyme reactions.

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