Vmax [S] Vo = Km + [S] Eadie-Hofstee plot Lineweaver-Burk (L-B) plot v=Vm-Km si [S] Km 1 Vm Vm [S] The equations above apply for Michaelis-Menten enzyme kinetics but are presented in three different formats. For uncompetitive inhibition The Michaelis-Menten equation becomes Vo-Vmax(S/(Km+a°[S) Put the Minchaelis-Menten equation for uncompetitive inhibition in the Eadie-Hofstee format. The Y-intercept of this plot will be

Vmax [S] Vo = Km + [S] Eadie-Hofstee plot Lineweaver-Burk (L-B) plot v=Vm-Km si [S] Km 1 Vm Vm [S] The equations above apply for Michaelis-Menten enzyme kinetics but are presented in three different formats. For uncompetitive inhibition The Michaelis-Menten equation becomes Vo-Vmax(S/(Km+a°[S) Put the Minchaelis-Menten equation for uncompetitive inhibition in the Eadie-Hofstee format. The Y-intercept of this plot will be

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Below which is false about the Michaelis constant Km in Michaelis-Menten kinetics? k₂ k₁ E+SESE + P k₁ k₂ A higher K indicates a higher affinity between the substrate and the enzyme. Km is effectively equal to the Dissociation Constant (k_₁/k₁) for ES. When [Substrate] = Km, Vo = 1/2 V max When [Substrate] = Km [ES] = [Efree] = [Etotal]
A, +e- > A; + 2e A; +e- K » 2A A +A-K A, + A Write down the rate equation for the species. Do not include a diffusion term.
4. Graphing the results from kinetics experiments with enzyme inhibitors The following kinetic data were obtained for an enzyme in the absence of inhibitor (1), and in the presence of two different inhibitors (2) and (3) at 5 mM concentration. Assume [Er] is the same in each experiment. [S] (mM) 1 2 4 8 12 1 2 4 8 12 [S] (mm) a. Determine Vmax and Km for the enzyme. b. Determine the type of inhibition and the Kıfor each inhibitor. 1/[S] M-1 Plot these data as double-reciprocal Lineweaver-Burk plots and use your graph to answers a. and b. 1000 500 250 125 83.3 Answer: The data may be analyzed using double-reciprocal variables. For each [S] and corresponding u, we will calculate 1/[S] and 1/v. sed to get mo Helmien (1) v (μmol/mL sec 12 20 29 35 40 1/v (1) v (1) μmol/mL se mL-sec/umol C 12 20 29 35 40 Please explain how to get K 1/v 20 × 104 10 × 104 -400 8.33x104 5.00x104 3.45x104 2.86x104 2.50x104 Condition No inhibitor 0 5 mM inhibitor (2) 5 mM inhibitor (3) Plots of 1/v vs. 1/[S]…
An orange juice is thermally processed for determining ascorbic acid degradation kinetics. After different temperature experiments, the following ascorbic acid concentration values were obtained: Ascorbic Acid (mg/L) Processing Time (second) 0 15 30 60 90 120 Temperature (oC) 75 558.3 546.5 535 526.5 519.6 508.6 85 608.6 595.2 587.4 572.7 557.5 542.3 95 612.6 601.5 588.2 546.1 521.8 509.5 According to the results; Determine ascorbic acid degradation order Derive the equations related to the degradation order (general equation and t1/2) Calculate k-values at the constant processing temperatures Calculate activation energy (Ea) of the degradation reaction Calculate half-life time of ascorbic acid for all processing (For better evaluation, add figures for the addressed questions, if necessary)
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If 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, calculate the new Vmax. Calculate the new Km of the above lactase reaction, with inhibitor X. Calculate the slope on a Lineweaver-Burk plot (Km / Vmax) for the lactase reaction with inhibitor X. Inhibitor X exerts which of the following effects on the above enzyme (lactase)?
A protein catalyzes the following reaction and had a Michaelis-Menten constant of Km=25 mM and a turnover number of 4.0 x 107 s-1. The total enzyme concentration is 0.012 µM and the initial substrate concentration is 4.76 µM. The enzyme has a single active site. Calculate the values of Vmax and V0 for this enzyme. 2H2O2 (aq) -----> 2H2O(l) + O2(g) Vmax= Vo=
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