1. A sigmoidal plot of substrate concentration ([S]) verses reaction velocity (V) may indicate:(A) Michaelis-Menten kinetics(C) Competitive inhibition2. The kinetic effect of purely competitive inhibitor of an enzyme:(A) Increases K without affecting Vma (B) Decreases K without affecting Vmas(C) Increases Vma without affecting Km (D) Decreases Vmas without affecting Km3. In reversible non-competitive enzyme activity inhibition:(A) Vma is increased(C) Vma is decreased4. In enzyme kinetics Vma reflects:(A) The amount of an active enzyme(C) Half the substrate concentration5. In non-competitive enzyme activity inhibition, inhibitor:(A) Increases Km(C) Does not affect Km(B) Co-operative binding(D) Non-competitive inhibition(B) K is increased(D) Concentration of active enzyme is reduced(B) Substrate concentration(D) Enzyme substrate complex(B) Decreases K(D) Increases Km

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Answered: 1. A sigmoidal plot of substrate…

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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The typical Michaelis-Menten equation mathematically describes the overall rate of the reaction as V (this is because biologists don't like math). What does V actually mean? (write the definition of V in differential equation form). V= d( )/dt Reaction rate Substrate concentration V max ·½V TURK
1. ( . tion can be represented by the so-called Michaelis equation ) The velocity of an enzyme catalyzed reac- Vmax [S] Vo = Km Vo = max kcat [Eo] [So] (Км + [Sol) V max where [So] is the initial substrate concentration, [Eo] is the total enzyme concentration, kcat is the cataly- tic rate constant or turnover number, and KM repre- sents the Michaelis constant. The graph on the right illustrates the dependence of the velocity of the reaction as a function of substrate concentration. Ex- Km [S] (mm) plain why the maximal velocity can be represented by the expression Vmax = Kcat [Eo] at very high substrate concentrations. Vo (µm/min)
5. For a Michaelis-Menten enzyme, k1 = 5.2 ⅹ 108 M-1 s-1, k-1 = 3.1 ⅹ 104 s-1, and k2 = 3.4 ⅹ 105 s-1. a) Write out the reaction, showing k1, k-1, and k2. Calculate Ks and Km. Does substrate binding approach equilibrium or the steady state? Justify your answer. b) What is kcat for this reaction? Justify your answer. c) Calculate Vmax for the enzyme. The total enzyme concentration is 25 pmol L-1, and each enzyme has two active sites. d) What substrate concentration would be required for the reaction in (c) to reach half of Vmax. Justify your answer mathematically. e) A second Michaelis-Menten enzyme has k1 = 4.2 ⅹ 107 M-1 s-1, k-1 = 6.1 ⅹ 104 s-1, and k2 = 5.3 ⅹ 102 s-1. Which enzyme is most efficient? 6. A pharmaceutical company is trying to develop a
The process of protein decomposition by pepsin enzymes in stomach is assumed and modeled as a batch reactor. If the protein concentration over time is measured as follows: Write the relationship between time (t), protein concentration (CA). Assuming that the enzyme reaction follows Michaelis-Menten kinetics, write the required constant (Vmax, Km), and the initial concentration of the protein is CA0.
6. An enzyme catalyzed reaction has a KM of 1 mM and a Vmax of 5 nM/s. What is the reaction velocity when the substrate concentration is (a) 0.25 mM (b) 1.5 mM, and (c) 10 mM?
Answer #2
1) Is the ratio of the forward rate constant and reverse rate constant changed by the presence of an enzyme catalyst? 2) 4) What is the simplest mathematical relationship between substrate concentrations [S], initial velocity (Vo) of an enzyme catalyzed reaction, the maximal velocity of the reaction (Vmax), and the ½ maximal Vo (i.e. Km) ? (Hint: what is the Michaelis-Menten Equation? 3) Graphically illustrate the most useful derivation of the Michaelis-Menton equation (that darn linearized, double-reciprocal)
You measure the kinetics of an enzyme as a function of substrate concentration using the Lineweaver-Burke equation. The enzyme concentration is maintained constant at a level of 1 μM. From these data, determine the equation of the line, Vmax, KM, and kcat for the enzyme. [S], μM Velocity, μM/min 2 2.9 3 3.8 4 4.4 5 5.0 6 5.4 7 5.8 8 6.2 9 6.4 10 6.7 The Lineweaver-Burke equation is 1V=1V= ____________(1[S])(1[S]) + ___________. Therefore, the Vmax is ____________ μM/min, KM is ____________ μM, and kcat is ____________ /min. Round-off all answers to four (4) decimal places.

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