Figure I shows the Michaelis Menten plot of initial reaction velocity (as percentage of Vmax)versus [S] (concentration) for the carbonic anhydrase reaction in the absence and presence ofthe inhibitor acetazolamide. Carbonic anhydrase participates in regulation of the pH andbicarbonate content of a number of body fluids.100No inhibitorAcetazolamide0.20.40.60.81[S] (mM)Figure 1(i)Compare Vmax and Km of the enzyme without inhibitor and in the presence ofacetazolamide. Determine the type of inhibition shown by acetazolamide. Explainyour answer.(ii)Name TWO (2) other types of inhibitions besides the inhibition shown byacetazolamide in Qla)(i). List down the kinetic properties of these inhibitions.Sketch a graph of I/V versus 1/[S] showing plots in the absence of an inhibitor and inthe presence of the types of inhibitors mentioned in Qla)(ii).(iii)V (% of Vmax

Michaelis menten constant, Km is the substrate concentration required to produce half maximum…

Answered: Compare Vmax and Km of the enzyme…

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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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.
Drug D reversibly binds to enzyme E and inhibits its activity toward substrate S. D binds equally well whether or not S is bound. Sketch a graph of the expected 1/v vs. 1/[S] relationship for: A) The enzyme reacting with S in the absence of drug D, B) The enzyme reacting with S in the presence of a small amount of drug D, and C) The enzyme reacting with S in the presence of a large amount of drug D.
Describe experimental enzyme inhibition and how it leads to a deeper understanding of enzyme mechanisms. Explain mechanism-based enzyme inhibition. Describe enzyme inhibition by a transition-state analog. For fluoroacetate and hypoglcyin A, explain how each of these mechanism-based inhibitors is converted to the actual enzyme inhibitor and how this inhibitor then inhibits the relevant enzyme..
The KM values for the reaction of chymotrypsin with two different substrates are given in the table below. Considering this information, which substrate has the lower apparent affinity for the enzyme? Which substrate is likely to give a lower value for Vmax? Substrate N-acetylvaline ethyl ester N-acetyltyrosine ethyl ester KM (M) 8.8 X 10-² 6.6 X 10-4 N-acetylvaline ethyl ester has the lower apparent affinity for the enzyme; it will also likely to give a lower Vmax: N-acetyltyrosine ethyl ester has the lower apparent affinity for the enzyme; it will also likely to give the lower V₁ max. N-acetylvaline ethyl ester has the lower apparent affinity for the enzyme; N- acetyltyrosine ethyl ester is likely to give the lower Vmax: N-acetyltyrosine ethyl ester has the lower apparent affinity for the enzyme; N- acetylvaline will likely to give the lower Vmax. None of the above statements are correct.
Phosphoglycerate-mutase catalyses the isomerization of 3-phospho-glycerate to 2-phospho-glycerate. The enzyme has a molecular weight of 28,556 g/mol. In assays using 1 milligram of enzyme per assay the Km for 3-phosphoglycerate was 1.2 x 10^-3M and the Vmax was 840 μmole min^-1. What would be the Vmax in units of mole s^-1?
Please provide an explanation for how to work through this practice problem: Using table 12.1, calculate the free energy change for the synthesis of ATP from cAMP and inorganic phosphate. (Note: cAMP is hydolyzed to AMP, and the free engery of hydrolysis for ATP and ADP is approximately equal.) Table 12.1: (Compound: Change in Free Energy in kJ/mol) cAMP: -50.4 Creatine phosphate: -43.3 ATP: -30.5 Glucose 6-phosphate: -13.9 AMP: -9.2
From a kinetics experiment, kcat was determined to be 295sec-1. For the kinetic assay, 0.3mL of a 0.25mg/mL solution of enzyme was used, and the enzyme has a molecular weight of 125,000 g/mole. Assume a reaction volume of 3mL. Calculate Vmax (µM∙min-1) for the enzyme and catalytic efficiency ( in M-1sec-1) for the enzyme. The Km for the enzyme was determined to be 2.55 x 10-2M.
Neuraminidase (NA) catalyzes a hydrolysis of the glycosidic linkages of sialic acids using a mechanism shown in Fig. (a). and Fig. (b) shows competitive inhibitors of NA Considering the mechanism of NA, why do zanamivir and oseltamivir inhibit NA so potently?
The Michaelis-Menten rate equation for reversible mixed inhibition is written as Vo = Vmax [S] aKm + a' [S] where Vo is initial velocity, Vmax is maximum velocity, [S] is substrate concentration, a represents the effect of the inhibitor bound to free enzyme (E), a' represents the effect of the inhibitor bound to the enzyme-substrate complex (ES), and Km is the Michaelis constant that represents the [S] at which the reaction reaches/Vm Vmax 2α' Derive an expression for the effect of a reversible inhibitor on apparent Km from the previous equation. Use the alphabet tab to enter a and the basic tab to enter the prime sign in your answer. = Apparent, or observed, Km is equivalent to the [S] at which Vo max. apparent Km =
Amino acid alanine
You are working on an enzyme that obeys standard Michaelis-Menten kinetics. Based on the following reaction expression, what is the Km value for this enzyme? E+SESE + P . . . k₁ = 880.8 M-¹5-1 k.₁ = 42.18 S-1 k₂ = 56.29 S-1
Chymotrypsin has the highest affinity for which of the following substrates: Table. The values of KM and kcat for some Enzymes and Substrates Enzyme Chymotrypsin Ки (М) 4.4 x 10-1 8.8 x 10-2 6.6 x 104 Kcat (S-1) 5.1 x 10-2 1.7 x 10-1 1.9 x 102 Substrate N-acetylglycine ethyl ester N-acetylvaline ethyl ester N-acetyltyrosine ethyl ester Catalase H2O2 2.5 x 10-2 1.0 x 107 Urease Urea 2.5 x 10-2 4.0 x 105 OA. N-acetylglycine ethyl ester OB. N-acetylvaline ethyl ester OC. N-acetyltyrosine ethyl ester D. Urea

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