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 mMd.Draw the estimated Michaelis Menten Curve of Inhibitor ABC and the curve without the inhibitor.

Vmax km Competitive remains same increases Non Competitive Decreases remains same…

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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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
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
The accompanying figure shows three Lineweaver–Burk plots for enzymereactions that have been carried out in the presence, or absence, of aninhibitor. Indicate what type of inhibition is predicted based on eachLineweaver–Burk plot. For each plot indicate which line corresponds to thereaction without inhibitor and which line corresponds to the reaction withinhibitor present.
5.50 1/V, min/umol 5.00 4.50 4.00 y = 0.9474x + 2.6649 y = 0.9997x + 2.032 0.00 1.00 2.00 2.50 3.00 1/[S], uM -1 Looking at the double reciprocal plot for an enzyme in the absence of inhibitor and in the presence of two concentrations of inhibitor, what would be the Vmax for the uninhibited enzyme? (bottom graph) Equation is given. Choose the one best answer. 3.50 3.00 2.50 2.00
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
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
D 1. 0.25 Short answer questions, write answers on blank paper. A plot of enzyme activity with and without an inhibitor present gave the following plot. 0.2 0.15 1/V v in mM/s 0.1 0.05 0 0 0.05 0.1 y=0.67x+0.02 0.15 y=0.2x+0.02 0.2 0.25 0.3 1/[S]. [S] in mM Without the inhibitor, what are the values of Vmax and Km? In the presence of the inhibitor, which of these items has changed and what is (are) the new value(s)? What type of inhibitor is present (explain why you have arrived at this conclusion)? Be sure to include units as necessary--answers that do not contain required units will be marked incorrect
can you explain what is happening in the attached graph? Figure 3: Lineweaver-Burke plot of alkaline phosphatase activity without inhibitor
Please explain your reasoning for every answer
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
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
You are evaluating the kinetics of an enzyme catalyzed reaction containing 5.5 μM total enzyme and 11.2 μM substrate. At this substrate concentration, you determine that the Vo = 88.6 μmol mL-¹. s-¹. If the Vmax 833.3 mM s the KM is: . == " 30.9 μΜ 10.4 μΜ Ο 124.6 μΜ 234 μΜ 94.5 μM

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