Draw Lineweaver-Burk plots for the behavior of an enzyme for which the following experimental dataare available. What are the Km and Vmax values for the inhibited and uninhibited rxns? Is the inhibitorcompetitive or noncompetitive?[S] (M)10,000 1 x 1042.000-5 x 1046666731.5 x 10-3400- 2.5 x 10-³2005x103V, No Inhibitor (Arbitrary units)0.026 38,4610.092 10.87371000.136 7.35Wak0.150 -670.165 6.06Lslope: 0033yint=5057полестеum.x=0.198km 6.5x 10-40.0100.040 2510.086 11.630.120 8.330.142 7.042500V, Inhibitor Present (Arbitrary Units)100↓$op.= .009yint: 5.21900x12km: 1.73 y/oconf.V10-3321000

For a one-substrate enzyme-catalyzed reaction, the Michaelis-Menton equation shows the quantitative…

Answered: 10,000 2.000 66667 Draw Lineweaver-Burk…

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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Similar questions

Lisa decides to obtain values for the Km and Vmax of an enzyme that was isolated from liver cells.. Using the Michaelis Menten plot. In what kind of measurements are needed and what would be needed to plot on a graph to estimate Km and Vmax?
If the higher value of KM resulting in the new plot ( red curb ) is due to the presence of an enzyme inhibitor is inhibitor reversible or irreversible? And why?
Use the relationships revealed by a Lineweaver-Burk plot and the table of enzyme performance to calculate the Vmax and Km of the enzyme with no inhibitor. with inhibitor A, and with inhibitor B. [S] (uM) Vo (umol/min); w/ no inhib. Vo (umol/min); w/ inhib. A Vo (umol/min); w/ inhib. B 10 6.3 5.1 4.0 40 18.4 15.8 11.8 100 29.9 27.0 19.1 150 34.7 32.0 22.2 No inhib. Vmax= Km= Inhib. A Vmax= Km= Inhib. B Vmax= Km=
Draw the Michaelis-Menten Plot and Lineweaver-Burke Plot of an enzyme in thefollowing situation. An enzyme, Enzyme A, has an optimal temperature at 34.5oC. The enzyme kinetic parameters were recorded to be Km = 15.0 μM, Vmax = 12.0 μM/s. The condition for an enzyme kinetic experiment has the temperature raised to 36.5oC. What will happen to the results? Provide at least five data sets prior to raising the temperature and after raising the temperature and plot the graphs as overlaying figures for comparison. Explain youranswer using a minimum of 200 words.
Explain why each of the following data sets from a Lineweaver–Burk plot are not individually ideal for determining KM for an enzymecatalyzed reaction that follows Michaelis–Menten kinetics.
Why can’t an enzyme have a kcat/KM value greater than 109 M−1 ∙ s−1?
Using the appropriate graph and table above, explain what the R48C mutation appears to be doing to the enzyme’s function. Discuss the kinetic parameter changes and their meaning in this context, not the structure of the enzyme, which was not given to you.
You obtain a calculated Vmax of 4.26uM/s and a Km of 122.5uM from a kinetics experiment performed using 0.5uM enzyme. What is the catalytic efficiency of this enzyme?
Although graphical methods are available for accurate determination of the Vmax and Km of an enzyme-catalyzed reaction, sometimes these quantities can be quickly estimated by inspecting values of V0 at increasing [S]. Estimate the Vmax and Km of the enzyme-catalyzed reaction for which the following data were obtained:
For an enzyme that displays Michaelis-Menton kinetics, what is thereaction velocity, V (as a percentage of V max , observed at the followingvalues?[S] = K M[S] = 0.5K M[S] = 0.1K M[S] = 2K M[S] = 10K M
Assume that the experiments performed in the absence of inhibitors were conducted by adding 5 μL of a 2 mg/mL enzyme stock solution to an assay mixture with a total volume of 1 mL. Take into account that XYZase is a monomeric enzyme with a molecular mass of 45,000 Daltons. Hint: To calculate the ???? in units of per second (s−1), you must first determine the ???? in micromoles per second (μmol/sec). Please explain step by step
You have obtained experimental kinetic data for two versions of the same enzyme, a wild‑type and a mutant differing from the wild‑type at a single amino acid. The data are given in the table. ?maxVmax(μmol min−1) ?MKM(mM) Wild‑type 100 10 Mutant 1 0.1 Compare the kinetic parameters of the two versions using the data in the table. Assuming a two-step reaction scheme in which ?−1k−1 is much larger than ?2,k2, which of the following statements are correct? The wild‑type version requires a greater concentration of substrate to achieve ?maxVmax. The wild‑type version has a higher affinity for the substrate. The mutant version has a higher affinity for the substrate. The mutant version requires a greater concentration of substrate to achieve ?maxVmax. Calculate the initial velocity of the reaction catalyzed by the wild‑type enzyme when the substrate concentration is 10 mM. ?0=V0=

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