P9-11B The following data on baker's yeast in a particular medium at 23.4°C were obtained in the presenceand in the absence of an inhibitor, sulfanilamide. The reaction rate (-3) was measured in terms of theoxygen uptake rate Qo, obtained as a function of oxygen partial pressure.(a) Assume the rate Qo, follows Michaelis-Menten kinetics with respect to oxygen. Calcu-late the lo, maximum (i.e., Vmax), and the Michaelis-Menten constant KM-(Ans.: V = 52.63 µL O,/h-mg cells.)(b) Using the Lineweaver-Burk plot, determine the type of inhibition sulfanilamide that causes the O₂uptake to change.Po,0505355.0Qo, (no sulfanilamide)0023.533.037542.043.0Qo, (20 mg sulfanilamide/mLadded to medium)Page 428 / 9930.0.25.630.836.439.640.0Po₂oxygen partial pressure, mmHg. Qo, = oxygen uptake rate,µL of O₂ per hour per mg of cells.(c) List ways you can work this problem incorrectly.(d) Apply one or more of the six ideas in Preface Table P-4, page xxviii, to this problem.Q+

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Answered: P9-11 The following data on baker's…

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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TABLE 3-LACTATE PRODUCTION IN FORTIFIED HEMOLYSATES OF HUMAN ERYTHROCYTES* Substrate Glucose Glucose Lactate production† No. of experiments pH 6 7.1 2.03 ± 0.91 6 7.8 4.76 ± 1.09 7-1 10-73 +1-88 5 7.8 12.34 ±2.92 5 7.0 7-15±0.73 5 7-7 (b)( ) In mature erythrocytes (red blood cells) the end product of glycolysis is lactate because of the absence of mitochondria. On the right is a table comparing the rate of lac- tate production in hemolysates (lysed cells) of human RBCs as a function of pH with dif- ferent substrates introduced into the glyco- lytic pathway. The hemolysate was fortified with 30 μmoles substrate, 7.5 μmoles MgCl2, 10 μmoles disodium phosphate, 1.5 μmoles NAD and 5 μmoles ATP in a volume of 5 mL. The rate of lactate production is given as μmoles of lactate/g Hb/hr at 37° C, buffered to either pH 7.1 or 7.8, as indicated. According to the results in the table which glycolytic enzyme is rate-limiting? Explain. Glucose-6-phosphate Glucose-6-phosphate Fructose-1,6-diphosphate…
How does the reaction catalyzed by GAPDH from T. tenax presented here differ from the reaction carried out in E. coli?
Based on the Lineweaver-Burke plot attached. Kinetic data were generated in the (1) absence of any inhibitor, (2) presence of 15 µM of a reversible inhibitor, or (3) presence of 20 µM of a second (distinct) reversible inhibitor. Purified enzyme concentration was 5 µM. The y-intercept of Lines (A) and (B) is 0.9 sec/uM; the y-intercept of Line (C) is 0.3 sec/uM. The slope of Line (A) is 1.8 sec; the slope of Lines (B) and (C) is 0.6 sec. QUESTION: Line (A) represents ____________. A. an enzyme-catalyzed reaction in the absence of any inhibitor. B. an enzyme-catalyzed reaction in the presence of a competitive inhibitor. C. an enzyme-catalyzed reaction in the presence of an uncompetitive inhibitor. D. an enzyme-catalyzed reaction in the presence of a noncompetitive, or mixed, inhibitor.
Kinetic Parameters of Enzyme-Catalyzed Reactions TABLE 12-1 The Values of KM, Keat, and Keat/KM for Some Enzymes and Substrates Enzyme Substrate KM (M) 9.5 x 10-5 1.2 x 10-² 2.6 x 10-2 2.5 x 10-2 4.4 x 10-1 8.8 x 10-2 6.6 x 10-4 Acetylcholinesterase Carbonic anhydrase Catalase Chymotrypsin Fumarase Urease Acetylcholine CO₂ HCO₁ H₂O₂ N-Acetylglycine ethyl ester N-Acetylvaline ethyl ester N-Acetyltyrosine ethyl ester Fumarate Malate Urea 5.0 x 10-6 2.5 x 10-5 2.5 x 10-2 Keat (S-¹) 1.4 x 104 1.0 × 106 4.0 × 105 1.0 X 107 5.1 x 10-2 1.7 × 10-1 1.9 X 10² 8.0 x 10² 9.0 × 10² 1.0 X 104 Keat/KM (M¹s¹) 1.5 × 108 8.3 x 107 1.5 x 107 4.0 X 108 1.2 x 10-1 1.9 2.9 × 105 1.6 × 108 3.6 X 107 4.0 X 105 Which enzyme is the most catalytically efficient? Which substrate does chymotrypsin bind to most tightly (assume k_₁ >> K₂)? Is fumarate or malate a better substrate of fumarase? Is it possible to have a kcat/KM of greater than 1 x 10⁹ M-¹ s-¹? Why or why not?
82.16
What is the principle of LAP? Question 3 options: A) phosphatase activity liberates naphthol and immediately coupled with a diazonium salt which forms an insoluble, visible pigment at sites of phosphatase activity B) naphthol couples with fast garnet and forms insoluble maroon deposits at sites of enzyme activity C) peroxidase present in the primary granules of the granulocytic line transfers hydrogen from various dye substrates to hydrogen peroxidase which yields a colored derivative of the dye which is localized at the site of the enzyme activity D) ferric iron deposits in tissue (present mostly as ferric iron within the storage proteinferritin) then react with the soluble ferrocyanide in the stain
CHEM151/251 Biochemistry 1 a. Determine Km 1. The following data were obtained for a competitive inhibition study in which the [I] = 3 µM for each determination of vo in the presence of inhibitor. The Vmax = 200 μM P/min for both data sets. 200 Vo (UM P/min) 8 8 8 8 8 8 8 8 8 180 160 140 120 100 40 Name (Print)/ID #: the absence of inhibitor. Participation Question # 10 No Inhibitor 50 +Inhibitor 100 [Substrate] (UM) 150 b. Determine Km, app for the data obtained in presence of inhibitor. 200 c. Calculate the value for Ki. Note: a = 1 + [I]/Ki and a¹ = 1 + [I]/Ki*. for the data obtained in
An uncatalyzed reaction progresses at a rate of 20micromoles per minute while the same reaction in the presence of an enzyme progresses at a rate of 100micromoles per second. What is the rate enhancement achieved by the presence of the enzyme? At 25oC, how much energy is required to produce a 10-fold rate enhancement? At the same temperature, how much energy is required to produce a million-fold rate acceleration?

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