In an enzyme kinetics study, three inhibitors resulted to the following results:Inhibitor ABCInhibitor XYZInhibitor PQRWithout InhibitorVmax40.2 mM/ sec40.3 mM/ sec12.32 mM/ sec65.43 mM/ secKm24.3 mM28.5 mM24.3 mM15.7 mMb. What type of inhibitor is Inhibitor PQR? Why do you say so?

Answered: In an enzyme kinetics study, three…

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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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.
Briefly describe the effects of a linear noncompetitive inhibitors on the kinetic parameters Km and Vmax. breifly discuss the reasons behind the effect of this type of inhibition on the kinetic parameters
Calculate the slope on a Lineweaver-Burk plot (Km / Vmax) for the lactase reaction with inhibitor X. (inhibitor X changes lactase activity to a Vo of 0.10 mM per minute when [S] = 1.0 mM, and a Vo of 0.133333333333 mM per minute when [S] = 2.0 mM) 0.20 per minute 0.50 per minute 1.0 per minute 2.0 per minute 5.0 per minute
In an enzyme kinetics study, three inhibitors resulted to the following results: Inhibitor ABC Inhibitor XYZ Inhibitor PQR Without Inhibitor Vmax 40.2 mM/ sec 40.3 mM/ sec 12.32 mM/ sec 65.43 mM/ sec Km 24.3 mM 28.5 mM 24.3 mM 15.7 mM d. Draw the estimated Michaelis Menten Curve of Inhibitor ABC and the curve without the inhibitor.
Calculate KI' of the inhibitor from the information given. All information may not be needed to calculate. K'm = (29Ki+1.45x10^-10)/Ki Vmax = 11.7 µMs-1 Kcat = 130 s^-1 Vo = 3.0 μMs-1 S = 10 μM Et = 0.09 µM Inhibitor Concentration = 5x10^-12
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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
can you explain what is happening in the attached graph? Figure 3: Lineweaver-Burke plot of alkaline phosphatase activity without inhibitor
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 =
a. What is the Vmax of this enzyme WITHOUT inhibitor? Please show your work. b. What is the Km of this enzyme WITHOUT inhibitor? Please show your work. c. The specificity constant of enzyme X is 8 x 10^7 /(M * seconds) What is the kcat of enzyme X WITHOUT inhibitor? Please show your work d. What was the concentration of enzyme used for measuring the kinetics of enzyme X WITHOUT inhibitor? Please show your work
Compound A is the substrate for two enzymes, El and E2, their reaction rates, r1 and r2,Determine the Km and rmax for both enzymes, with respect to the concentration of A. Which set of data is more likely to be for El and which for E2, and why? Concentration of 0.2 0.6 1.2 3 4 5 6 8 12 15 A (mM) Reaction rate (r.) (mmol/L'min) 3.33 4.29 4.62 4.76 4.84 4.88 4.9 4.92 4.94 4.95 4.96 4.97 Reaction rate (r,) (mmol/L*min) 0.09 0.23 0.38 0.5 0.6 0.67 0.71 0.75 0.8 0.82 0.86 0.80
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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In an enzyme kinetics study, three inhibitors resulted to the following results: Inhibitor ABC Inhibitor XYZ Inhibitor PQR Without Inhibitor Vmax 40.2 mM/ sec 40.3 mM/ sec 12.32 mM/ sec 65.43 mM/ sec Km 24.3 mM 28.5 mM 24.3 mM 15.7 mM b. What type of inhibitor is Inhibitor PQR? Why do you say so?