If the middle line (darker one) is the rate of an enzyme reaction in the presence of an uncompetitive inhibitor, which line number represents the reaction without the inhibitor? Explain your answer. [S] 1 2 3 4 Draw a Lineweaver-Burk plot of the graph in the previous question. Values are not necessary just comparative shapes of lines. (Label axis and y intercept)

If the middle line (darker one) is the rate of an enzyme reaction in the presence of an uncompetitive inhibitor, which line number represents the reaction without the inhibitor? Explain your answer. [S] 1 2 3 4 Draw a Lineweaver-Burk plot of the graph in the previous question. Values are not necessary just comparative shapes of lines. (Label axis and y intercept)

Chemistry & Chemical Reactivity

10th Edition

ISBN:9781337399074

Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Chapter14: Chemical Kinetics: The Rates Of Chemical Reactions

Section14.7: Reaction Mechanisms

Problem 1.2ACP

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Pre-Lecture Problems: Predict products for each enzyme-catalyzed reaction given the enzyme classification. hydrolase H2N -CH- CH OH C- CH3 H. OH lyase + 02 oxidoreductase H. oxidoreductase + NADH, H+ H.
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…
8. Data on the speeds of a Michaelis-Menten enzyme reaction v (in micromolar/minute) as a function of the substrate concentration S (in micromolar) has been collected. Your goal is to find a model that fits this data. a) What type of transformation do you need to do for the data so that the transformed data looks like a linear function? b) Assume that the appropriate transformation on the data has been performed and that a linear regression line has been fitted to the transformed data and that its equation is given by Y = 0.501147.X +0.454986 Compute the parameters Vmax and KM and write the model. and KM = v(S) = micromolar minute c) What is the context domain of this function? d) Predict the speed of the reaction (in micromolar/minute) when the concentration is 12.45 micromolar. Round your answer to two decimal places. v(S) = micromolar minute e) What is the maximal speed of the reaction? micromolar minute
Chemistry Calculate the substrate concentration of an enzyme catalyzed reaction whose rate is equal to half of its maximum velocity. Also evaluate Michaelis Menten constant and substrate concentration when the rate is equal to its maximum velocity.
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In the "Catalysis by Preferential Transition State Binding" mechanism, the substrate perfectly fits with the enzyme binding site after its geometry gets changed. Who does it? And how?
5. In workshop and on the problem set we looked at a graph of rate of reaction vs. substrate concentration for an enzyme-catalyzed reaction in the presence of reversible inhibitors. On the graph below, draw a curve that shows what happens in the enzyme-catalyzed reaction in the presence of an irreversible inhibitor. Explain how you came up with your curve. You may assume the concentration of the inhibitor is less than the concentration of the enzyme. rate substrate concentration
Enzyme kinctics a. How to determine Km and Vmax of a-glucosidase enzyme isolated from weathered rice b. Curcumin is expected to inhibit the activity of the a-glucosidase enzyme. How to determine the inhibitory activity of curcumin against a -glucosidase? c. If the results of experiment number 3b are known that curcumin can inhibit a -glucosidase, how to determine the type of inhibitor?
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