The rate of transformation of S, have been determined in the presence of twocompounds M and N. Results are given in the graphs (below).2- Determine if compounds M and N are activators or inhibitors. Justify.3- In the case of an inhibitor, specify the type of inhibition. Justify.(2)1/Vo1[M] = 0[M] = 2 mM1/[S]o1/V₂[N] = 2 mM[N] = 01/[S]o

Enzyme Inhibitors are the substances that can bind to enzyme and inhibit the enzyme activity . On…

Answered: The rate of transformation of S, have…

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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Match the different names for inhibition mechanisms (1-5) with a description of their properties 7a-7d: 1. competitive inhibitor. 2. allosteric inhibitor also known as non-competitive inhibitor. 3. un-competitive inhibitor. 4. affinity label also known as active site directed covalent (irreversible) enzyme inhibitor. 5. Kcat inhibitor, also known as a mechanism-based covalent (irreversible) enzyme inhibitor. 4a. An enzyme inhibitor in which a substrate or competitive inhibitor is modified so that it contains a chemically reactive electrophile which can bind to and subsequently react with the enzyme active site: 4b. An enzyme inhibitor that contains latent reactive group that upon binding followed by catalytic turnover at the enzyme active site produces a reactive electrophile that reacts covalently with the enzyme: 4c. A reversible inhibitor that competes with the substrate for binding to the enzyme active site: 4d. A reversible inhibitor that can bind independently of substrate to its…
A. Lineweaver-Burk plot of the enzyme with increasing amounts of substrate in the absence or the presence of the inhibitor is shown below. Graph A : x-intercept Graph B : x-intercept = - 0.012, y-intercept = 0.8 Graph C : x-intercept = - 0.027, y-intercept = 0.8 Graph D : x-intercept = - 0.039, y-intercept = 0.8 - 0.007, y-intercept = 0.8 Graph A 4 Graph B Graph C Graph D 1 -0,04 -0,02 0,00 0,02 0,04 1/[Substrate] (uM) (i) Which graph indicates an enzymatic reaction without inhibitor? (ii) Which type of inhibitor is it? Briefly explain. (iii) Which graph indicates the highest concentration of inhibitor? (iv) Calculate the Vmax and Km of the graph showing an enzymatic reaction with the lowest concentration of inhibitor. Show the steps of calculation and unit in your answers. Keep 2 decimal places in your answers. 1/Rate (umol/min)
The graphs 3 and 4 representing 1/Vo = f(1/[S]o) have been done in the presence of a competitive (CI) and noncompetitive inhibitor (NCI). a- For each figure, determine from the relative position of the straight lines which one is obtained in presence of an inhibitor. b- Indicate which graph corresponds to the competitive inhibition and which one the noncompetitive inhibition. Justify your answer. c- Complete the graphs by indicating which values can be determined from the arrows. 3 1/No 1/[S]⁰ (4) 1/No 1/[S]o
Below is a plot of Vo vs. [S] for a specific allosteric enzyme under different conditions. Which of the following best describes the graph? 4 (S] O Adding a positive modulator to #2 would result in curve 3. Curve 1 represents maximum inhibition. Line 4 is valid exclusively for curves 1 and 2. Adding a positive modulator to #1 would result in curve 2. O Curve 3 represents the effect of a negative modulator added to curve 2.
Noncompetitive inhibition can often be explained by which of the following models? the induced fit model the lock and key model both (a) and (b) neither (a) nor (b)
In mixed inhibition as shown below, please draw a lineweaver-burk plot when Kl is greater than KI'. Please draw plot with and without inhibitor and label axis and intercepts. Please explain what is happening to Km and Vmax. In mixed inhibition as shown below, please draw a lineweaver-burk plot when Kl is greater than KI'. Please draw plot with and without inhibitor and label axis and intercepts. Please explain what is happening to Km and Vmax. 1/v 1/[S1]
Data from enzyme inhibition are used to determine a Kmapp and Vmax PP. Comparison of these values with assays run without inhibitor are used to understand how the inhibition is occurring. This is useful for better understanding the active site as well as the practical aspect of pharmaceutical drugs. Below are idealized Line-Weaver Burke plots of different types of inhibitors. Comnetitive Uncomnetitive Mixed +Inh +Inh 4Inh Anh Inh Anh [S] [S] [S] a. How does the value of Vmax for the enzyme compare to the Vmax PP of the inhibited enzyme for: i. Competitive ii. Uncompetitive iii. Mixed b. How does the value of Km for the enzyme compare to the Km PP of the inhibited enzyme for: i. Competitive ii. Uncompetitive iii. Mixed c. For each situation in Model 1, consider an inhibitor that is better than the one shown on the graph. Answer the following questions for each type of inhibition: i. How would the KmPP change? ii. How would the Vmax PP change?
On the figure below are shown three Lineweaver-Burk plots for enzyme reactions that have been carried out in the presence, or absence, of an inhibi- tor. Indicate what type of inhibition is predicted based on each Lineweaver- Burk plot. For each plot indicate which line corresponds to the reaction without inhibitor and which line corresponds to the reaction with inhibitor present. 1 1 1 [S] [S] [S]
Give a possible set of bacterial factors (or lack thereof) and what properties they have that would explain an increase in the 50% inhibitory concentration for tetracycline by 1,000-fold in an organism.
An enzyme is present at a concentration of 1 nM and has a Vmax of 2 µM s-'. The Km for its primary substrate is 4 µM. Calculate kcat- kcat Calculate the apparent Vmax and apparent Km of this enzyme in the presence of sufficient amounts of an uncompetitive inhibitor to generate an a' of 2. Assume that the enzyme concentration remains at 1 nM. apparent Vmax µM s-1 apparent Km = µM
A schematic representation of the enzyme IspD complexed to inhibitor 3, and a series of inhibitors 3-5 are shown below. Ala202 lle240 mwww NH NH Val263 ОН www HN N- lle177 HN 'N' CI 3 X = N 4 X = C-CN 5 X = C-COO IC50 274 µM IC50 140 nM IC50 35 nM NH2 HN Val266 N -N O-H---- N HN %3D Arg157 HN wwww lle265 Explain why structure 4 is a more potent inhibitor (lower IC50 value) than inhibitor 3 and why structure 5 is a much weaker inhibitor (higher IC50 value) than 3 and 4.
In some organisms, isoleucine can be synthesized in a multi-step procedure (a series of enzymatic reactions), beginning with a molecule of threonine. Keeping that in mind explain the experimental results below. Amount of endproduct Activity of threonine (isoleucine) deaminase None Low Medium High Very high High Moderate Very low

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