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 isB. the value of KiNo Inhibitor:With Inhibitor:Substrate Concentration (g/L) 0.86Reaction Velocity (g/L-s)Reaction Velocity (g/L-s)0.1790.1612.35 3.80 4.90 6.10 18.40.398 0.567 0.666 0.726 0.9500.322 0.407 0.443 0.495 0.639

Here, we are given the initial substrate concentration ([S0] ) in g/L and the initial rate of the…

Answered: 2. An enzyme-catalyzed reaction is…

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
1. Make a Lineweaver-Burk plot and use the plot to complete the information in the table and the following questions. a. Is it possible for the enzyme to overcome the effect of the inhibitor in question from the chart. Explain. b. What prevents this enzyme from being an even more catalytically efficient enzyme? c. What do single molecule data indicate about the validity of ensemble data?d. What is the reason that humans are insensitive to sulfa drugs?
(a) You are given the following experimental measurements for the effects of an inhibitor on the initial velocity of an enzymatic reaction. Plot the data and determine what type of inhibition is being observed. Label all axes. [S] (MM) Vo (mM • Vo with I present (mM ⚫s-¹) 1 2.8 3 6.0 1.0 2.5 4 7.0 3.1 8 9.9 12 11.5 4.8 5.7 (b) Briefly explain how varying [S] can be used to determine whether an inhibitor is uncompetitive or competitive. Be sure to include in your explanation why this method would work.
5. By using Excel or GoogleSheets. graph the Lineweaver-Burk plots for the behavior of an enzyme for which the following experimental data are available. What are the Km and Kwax values for the inhibited and uninhibited reactions? Is the inhibitor competitive or noncompetitive? [S] (mM) V, No Inhibitor (mmol min-) V, Inhibitor Present (mmol min-') 1 × 10-4 5 × 10-4 1.5 x 10-3 2.5 x 10-3 5 x 10-3 0.026 0.010 0.092 0.136 0.040 0.086 0.150 0.120 0.165 0.142
4. Below is a Michaelis-Menten plot for a wild-type (WT) and mutant (V105A) enzyme isolated from the bacterium Staphylococcus aureus. The enzyme is involved in carbohydrate metabolism and is a potential biocatalyst for the large-scale production of rare sugars. (8) mg V₂ μmol s1 1.50- 1.25- 1.00- 0.75- 0.50- 0.25- 0.00 0 100 200 300 [s], mM 400 500 WT (a) Estimate the KM and Vmax for the wild-type and mutant enzyme from the graph. V105A (b) Calculate the keat and keat/KM for the wild-type and mutant enzyme based on your estimated values in (a) if the total enzyme concentration is 0.5 µmol/mg. (c) Is the mutant enzyme a more or less efficient catalyst than the wild-type enzyme? Briefly explain.
4. a. Calculate the KM (Michaelis constant) and the vmax (the maximum initial rate) for both substrates (sphingosine and ATP). Show your work, and be careful about units. b. threo-dihydrosphingosine, a stereoisomer of sphingosine, is an inhibitor of sphingosine kinase. What kind of inhibitor (competitive, uncompetitive, noncompetitive) is threo-dihydrosphingosine? Citing information from the Lineweaver-Burke plots, explain how you can tell.
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)
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 =
4. a. Use the data in the graph above to estimate a KM value for the enzyme in the presence of these metabolites, and enter them into the table below. b. Classify these metabolites as either activators or inhibitors, and explain your rationale below.

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