Please correct answer and don't use hand raiting If the middle line (darker one) is the rate of an enzyme reaction in the presence of anuncompetitive inhibitor, which line number represents the reaction without the inhibitor?Explain your answer.[S]1234Draw a Lineweaver-Burk plot of the graph in the previous question. Values are notnecessary just comparative shapes of lines. (Label axis and y intercept)

An uncompetitive inhibitor binds only to the enzyme-substrate complex, which leads to a decrease in…

Answered: If the middle line (darker one) is the…

Chemistry for Today: General, Organic, and Biochemistry

9th Edition

ISBN:9781305960060

Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. Hansen

Publisher:Spencer L. Seager, Michael R. Slabaugh, Maren S. Hansen

Chapter20: Enzymes

Section: Chapter Questions

Problem 20.45E

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The Michaelis-Menten equation models the hyperbolic relationship between [S] and the initial reaction rate Vo for an enzyme-catalyzed, single-substrate reaction E + SES →→ E + P. The model can be more readily understood when comparing three conditions: [S] > Km. Match each statement with the condition that it describes. Note that "rate" refers to initial velocity Vo where steady state conditions are assumed. [Etotal] refers to the total enzyme concentration and [Efree] refers to the concentration of free enzyme. [S] > Km [ES] is much higher than [Efree]. This condition rarely occurs for most in vivo enzymes. Not true for any of these conditions The rate is directly proportional to [S].
Q1: the following table gives the results of enzyme-catalyzed reaction. The enzyme concentration [E] = 0.00012 mmol/L Use the information of substrate concentration [S] and rate Vo for both experiments, with inhibitor and without inhibitor to answer the following questions (NOTE: [E] = 0.00012 mmol/L) [S], mmol/L 0.710 0.400 0.310 0.098 0.066 0.040 Experiment 11 without inhibitor VO, mmol/L-min 0.200 0.180 0.160 0.120 0.100 0.070 Experiment 2 with inhibitor [1] = 0.033 mmol/L VO, mmol/L-min 0.180 0.150 0.110 0.070 0.050 0.040 1- Calculate Km and Vmax for both experiments using Michaelis-Menten equation. [S] vs Vo 2- Calculate Km and Vmax for both experiments using Lineweaver-Burk plot, 1/[S] vs 1/vo 3- Calculate Kcat for the enzyme (without inhibition) 4- Based on the values of Km and Vmax, with inhibitor and without inhibitor, determine the type of the Inhibition (competitive, non-competitive or uncompetitive)
Q1: the following table gives the results of enzyme-catalyzed reaction. The enzyme concentration [E] = 0.00012 mmol/L Use the information of substrate concentration [S] and rate Vo for both experiments, with inhibitor and without inhibitor to answer the following questions (NOTE: [E] = 0.00012 mmol/L) [S], mmol/L 0.710 0.400 0.310 0.098 0.066 0.040 Experiment 1 without inhibitor VO, mmol/L-min 0.200 0.180 0.160 0.120 0.100 0.070 Experiment 2 with inhibitor [1] = 0.033 mmol/L VO, mmol/L-min 0.180 0.150 0.110 0.070 0.050 0.040 1- Calculate Km and Vmax for both experiments using Michaelis-Menten equation. [S] vs Vo 2- Calculate Km and Vmax for both experiments using Lineweaver-Burk plot, 1/[S] vs 1/vo 3- Calculate Kcat for the enzyme (without inhibition) 4- Based on the values of Km and Vmax, with inhibitor and without inhibitor, determine the type of the Inhibition (competitive, non-competitive or uncompetitive) Note:- show all your work to answer the above questions. Upload your answers…
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ENZYME CATALYSIS-lab ENZYME CONCENTRATION What chemical reaction is being catalyzed in the experiment? Label the substrate(s), enzyme, and product(s).
1. Reaction rate data of an enzyme, E, and a substrate, S, with increasing inhibitor concentrations are listed in the table below (Cs and C₁ represents the concentrations of the substrate and the inhibitor, respectively). Reaction Rates (μmol.Li¹.s¹) [CS] (mmol/L) [CI]=0 [CI]=0.25 [CI]=0.50 mmol/L mmol/L mmol/L 0.10 0.79 0.57 0.45 0.15 1.11 0.84 0.66 0.20 1.45 1.06 0.86 0.30 2.00 1.52 1.22 (a) Calculate Vmax (mmol.m³.s-¹). (b) Calculate Kм (g.mol.m³). (c) Describe the type of this inhibition.
Past paper
ucose-cocp-2000+100g ge Find Cyoy the value of the [HgCl2] (M) [C2042-1 (M) | Rate (Mis) 0.10 0.10 1.3.10-7 0.10 0.20 5.2.10-7 0.20 0.20 1.0-10-6 A) 1.4 x 10-8 1/M2-s B) 1.3 x 10-7 1/M2-s C) 1.4 x 10-5 1/M2-s D) 1.3 × 10-4 1/M2-s
1. For an enzyme to catalyze a reaction, it must preferentially stabilize the transition state (TS) of the reaction. If an enzyme instead preferentially stabilizes the enzyme-substrate complex (ES), it fails to catalyze the reaction. Justify this observation using a reaction coordinate diagram (free energy (G) vs. reaction coordinate).
Reaction rate data of an enzyme, E, and a substrate, S, with increasing inhibitor concentrations are listed in the table below (Cs and C₁ represents the concentrations of the substrate and the inhibitor respectively). Reaction Rates (umol.L¹.s"¹) [CS] (mmol/L) [CI]=0 [CI]=0.25 [CI]=0.50 mmol/L mmol/L mmol/L 0.10 0.79 0.57 0.45 0.15 1.11 0.84 0.66 0.20 1.45 1.06 0.86 0.30 2.00 1.52 1.22 (a) Calculate Vmax (mmol.m³.s¹). (b) Calculate KM (g.mol.m³). (c) Describe the type of this inhibition. (33 points)
You performed a kinetic study of the enzyme lipase and you obtained the following double reciprocal (Lineweaver-Burk) plot: 4 y = 4.3511x + 1.8889 %3D 1 -0.5 -0.3 -0.1 0.1 0.3 0.5 0.7 0.9 1/[S] When you performed the same kinetic study in the presence of an inhibitor, you obtained a new line that had y-intercept 3 and x-intercept -0.3. Based on this information select all correct statements about this inhibitor. This is mixed inhibitor. When this inhibitor is present apparent Km of enzyme decreases. This inhibitor prevents substrate from binding to the enzyme. This is uncompetitive inhibitor. When this inhibitor is present apparent Vmax of enzyme decreases. When this inhibitor is present apparent Km of enzyme increases. This inhibitor can bind to either enzyme or enzyme-substrate complex.
How does Kcat influence the velocity of MM enzymes ? A. Linear relationship B. Inverse relationship C. No effect D. Only effects velocity if total enzyme concentration is low v = Vmax [S] KM + [S] Turnover rate kcat = Vmax [ET] Speed (substrates product/sec)

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