The initial rates of an enzyme-catalyzed reaction have been determined for fivedifferent [S] (see table below).[S]o (mol.L-¹)1.0 x 101.5 x 10-42.0 x 10-4-45.0 x 107.5 x 10-4-1Vo (μmol.L-¹ min-¹)28354263751- Determine graphically the value of Km (firstwith the Michaelis-Mentenrepresentation and then with the Lineweaver-Burk representation, then compareby reporting on the Michaelis-Menten representation the values found by theLineweaver-Burk representation)2- Calculate the concentration of a competitive inhibitor 1, with a K₁ value of2.4.104 M, whose action would quadruple the apparent value of Km.

Enzymes kinetics - is the study of the reaction rate of enzyme catalyzed reactions. The reaction…

Answered: The initial rates of an…

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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at what substrate concentration is v= 5.0 mM s^-1 for the enzyme catalyzed hydrolysis of trehalose?
Given that the AG" values for the hydrolysis of glucose 1-phosphate and glucose 6-phosphate are approximately -21 kJ/mol and –14 kJ/mol, respectively, which statement is true regarding the isomerization shown? glucose 1-phosphate → glucose 6-phosphate O The reaction can likely proceed in either direction and will therefore depend on the concentrations of glucose 1-phosphate and glucose 6-phosphate. O Only the anabolic direction will occur because the free energy change is positive. O The reaction will proceed in both directions simultaneously. This reaction can only proceed in one direction and must be the rate limiting reaction for this point in metabolism.
You are working on an enzyme that obeys standard Michaelis-Menten kinetics. You have determined the Vmax to be 0.1 mol/sec and the Km to be 2.5 mM. What would the rate of the reaction be when the substrate concentration is 20 mM? 0.09 MS-1 O 0.133 Ms-1 O 0.18 Ms ¹ 9 Ms-1 O 0.018 Ms-1 0.2 MS-1
Vo 60 50 40 30 20 mm/sec 0100 0 O 0 0 50 100 150 200 250 300 Substrate (mM) Use the Michaelis-Menten plot to answer this question. What is the estimated value of Vmax of the enzyme catalyzed reaction (square data points)? (choose the one best answer) 10
The equilibrium constant for the hydrolysis of the peptide alanylglycine (Gly-Ala in the reaction from Part B) by a peptidase is K = 9.0 × 10² at 310 K. Calculate AG for this reaction. Express the Gibbs free energy to three significant figures. AG = Submit ΠΑΠΙ ΑΣΦ Request Answer ? kJ/mol Keq [Gly] [Ala] [Gly-Ala]
Following is a Dixon plot for PNPP inhibition by inorganic phosphates. What answer choice shows the correct combination of the mode of inhibition and Ki of the EI complex? - Mixed inhibition, Ki= -21.2 uM - Mixed inhibition, Ki= 1/16.5 uM - Competitive inhibition, Ki= 16.5 uM - Competitive inhibition , Ki= -1/16.5 uM - None of the above
. The optimal conditions for salivary lysozyme (hydrolyzing glycoproteins of bacterial wall) are 37 C - temperature and pH is 5.2. Explain the decrease in this enzyme activity if the temperature will rise up to 60 °C and pH will be changed to 8.0. To answer the question: a) draw the graph of the velocity dependency on temperature and pH; b) calculate the relative enzyme activity if 10 mg of lysozyme catalyzes the formation of 5 uM of the product per 2 minutes. Concidor NH3: 5.
What is the Keq of Reaction #3? Show your work & clearly label your answer. Is Reaction #3 favorable under mammalian cellular conditions when the concentration of dGTP is 0.2 M, dCMP is 20 mM, dGMP-dCMP is 7 mM, and PPi is 10 mM? Use △G to support your answer.
ver is given. A +B C+ D 1) Calculate AG for the above react un and indicáte whether the reaction is favorable or unfavorable [A] = 0.9 M 20°C AG° = 4 KJ/mol %3D [B] = 15mM [C] = 4mM [D] = 3 M R= 8.314 J/moleK %3D %3D 4.7 and explain how you know.
Calculate the Km of the enzyme with these parameters. kcat = 130s^-1 Vo = 3.0 μMs-1 S = 10 μM Et = 0.09 µM
When 10 micrograms of an enzyme with a molecular mass of 80,000 Daultons (grams/mol) is added to a solution containing its substrate at a concentration 100 times the Km, it catalyzes the conversion of 65 micromoles of substrate into product in 3 minutes. What is the enzyme's turnover number (in units of min-1)?

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