The steady-state kinetics of an enzyme has been studied in the absence and presence of an inhibitor (Inhibitor A). The initial rate given as a function of substrate concentration in the following table: Substrate Concentration, [S] Initial rate, v, (mmol/L-min) Initial rate, v, (mmol/L-min) (mmol/L) No Inhibitor With Inhibitor A 1.25 1.72 0.98 1.67 2.04 1.17 2.50 2.63 1.47 5.00 3.33 1.96 10.00 4.17 2.38

The steady-state kinetics of an enzyme has been studied in the absence and presence of an inhibitor (Inhibitor A). The initial rate given as a function of substrate concentration in the following table: Substrate Concentration, [S] Initial rate, v, (mmol/L-min) Initial rate, v, (mmol/L-min) (mmol/L) No Inhibitor With Inhibitor A 1.25 1.72 0.98 1.67 2.04 1.17 2.50 2.63 1.47 5.00 3.33 1.96 10.00 4.17 2.38

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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4. An enzyme-substrate complex has a Kd = 100 nM. A competitive inhibitor with which of the following Ki values will be most effective at inhibiting this enzyme? A) K₁ = 10 nM B) K₁ = 100 nM C) K₁ = 1000 nM D) None of the above - only the KM is important Briefly explain:
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Which one of the following best explains the distinctive effect of an uncompetitive inhibitor on Km, according to our coverage? If S binds first, I may never bind and P can be formed. When ESI complexes form, S is trapped and it is as though there is a lower concentration of E.
1. Start with the Michaelis-Menten equation and convert it to a double-reciprocal equation. Show how you would make a linear Lineweaver-Burk plot using this equation. 2. Based on th chart make a Lineweaver-Burk plot and later use the plot to complete the information in the table.
2. An enzyme-catalyzed reaction is carried out at several different substrate concentrations (listed below), either (1) with no inhibitor, or (2) in the presence of an inhibitor, I, at 0.63 g/L. Given the following data, determine: A. what type of inhibitor I is B. the value of Ki No Inhibitor: With Inhibitor: Substrate Concentration (g/L) 0.86 Reaction Velocity (g/L-s) Reaction Velocity (g/L-s) 0.179 0.161 2.35 3.80 4.90 6.10 18.4 0.398 0.567 0.666 0.726 0.950 0.322 0.407 0.443 0.495 0.639
V 23. The graph below is a graph of Vmax (a) Label the graph clearly with both the Vmax and the Km. Estimate the Km from this graph giving the correct units. v/Vmax 1.0 0.5 0.0 0.0 vs [S] for an enzyme. Enzyme Activity vs [Substrate] :0.2 0.4 Substrate (HM): 0.6 (b) If the Vmax = 25 mmoles per minute per µmole of enzyme calculate the Keat and the specificity constant."
3. Enzyme specificity. To determine the specificity of substrate binding for a particular enzyme/protein, structurally related compounds may be used as potential substrates and Km values may be calculated. However, many compounds structurally related to the substrate may bind to the active site but cannot be converted to product. In these instances, the substrate analogs are used as potential competitive inhibitors of substrate binding. Low K, values indicate high affinity of the enzyme for the inhibitor, whereas high K,values indicate low binding affinity. Consider the enzyme xanthine oxidase, which catalyzes the formation of uric acid from the purine bases hypoxanthine or xanthine in humans. The Km for hypoxanthine is 15.0 μM and for xanthine it is 45.0 μM. A few compounds used as competitive inhibitors of the normal substrate hypoxanthine are listed in the table below with their K; values. Comparing the structures of hypoxanthine with the listed substrate analogs, what can you…
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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)
4. Using a fluorescent model substrate, you study the kinetics of an enzyme-catalyzed reaction. You observe the below data. Write an equation describing reaction velocity (v) versus [S]. Define and provide a numerical value for any constants that you include in your equation. 16 14 12 4 2 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 [S] / mM `N' at 50 mM and observe the following In pilot experiments, you add the putative inhibitor imidazole Doesn't affect anything 1 reaction velocities: [S] = 1 mM5 µM min-1 [S] = 2 mM9 9 µM min-1 [S] = 4 mM 911 µM min-1 Is imidazole a competitive or non-competitive inhibitor? Provide a brief explanation. > 9 00 v/ µM min-