10 answer this question, please reference ine Problem Solving video: Michaens-Menten Equations and Lineweaver-BurkPlots.Double-reciprocal, or Lineweaver-Burk, plots can reveal the type of enzyme inhibition exhibited by an inhibitor.Competitive inhibitors increase the KM without affecting the Vmax. Noncompetitive inhibitors decrease the Vmax withoutaffecting the KM. Uncompetitive inhibitors decrease both the KM and Vmax1->competitivenoncompetitiveuncompetitivecannot be determined+ InhibitorNo inhibitor.1[S]Based on the Lineweaver-Burk plot, what type of enzyme inhibition is exhibited by the inhibitor?

Answered: Double-reciprocal, or 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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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 Michaelis-Menten rate equation for reversible mixed inhibition is written as Vo Vmax [S] aKm + a' [S] where Vo is initial velocity, Vmax is maximum velocity, [S] is substrate concentration, a represents the effect of the inhibitor bound to free enzyme (E), a' represents the effect of the inhibitor bound to the enzyme-substrate complex (ES), and Km is the Michaelis constant that represents the [S] at which the reaction reaches Vmax Apparent, or observed, K is equivalent to the [S] at which Vo max Derive an expression for the effect of a reversible inhibitor on apparent K from the previous equation. Use the alphabet tab to enter a and the basic tab to enter the prime sign in your answer. apparent Km =
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…
Consider the given data for an enzyme-catalyzed reaction. Determine the Vm, Km and the type of inhibition based on the given data below Substrate concentration, uM 30 50 100 300 900 slope y-intercept Complete the table below (include correct units). Experiment A Vm Km Experiment A (Initial velocity without inhibitor, uM-min) Type of Inhibition: 10.4 14.5 22.5 33.8 40.8 Experiment B (Initial velocity with inhibitor, uM-min) 5.1 7.3 13.3 25.7 37.2 Experiment B
The Michaelis-Menten rate equation for reversible mixed inhibition is written as Vo = Vmax [S] aKm + a' [S] where Vo is initial velocity, Vmax is maximum velocity, [S] is substrate concentration, a represents the effect of the inhibitor bound to free enzyme (E), a' represents the effect of the inhibitor bound to the enzyme-substrate complex (ES), and Km is the Michaelis constant that represents the [S] at which the reaction reaches/Vm Vmax 2α' Derive an expression for the effect of a reversible inhibitor on apparent Km from the previous equation. Use the alphabet tab to enter a and the basic tab to enter the prime sign in your answer. = Apparent, or observed, Km is equivalent to the [S] at which Vo max. apparent Km =
The above graphs (Lineweaver-Burk) were plotted using data from an experiment that investigated changes in initial velocity, V0, (in mmol/min) as a function of substrate concentration, [S], (in mmol/L) for an enzyme-catalysed reaction. The initial velocity was measured at a fixed concentration of enzyme at different concentrations of substrate in the absence of any inhibitor and was then repeated in the presence of a competitive inhibitor at two different concentrations. Use the graph to determine the maximal velocity of the reaction the Km value of the uninhibited reaction the Km value of the reaction with (i) the lower level of inhibitor and (ii) the higher level of the inhibitor. TEMPLATE FOR ANSWERS Vmax of uninhibited reaction Intercept on y-axis = = __________________ Vmax = __________________ Km value of the uninhibited reaction Intercept on x-axis = =…
please help me understand this practice problem with as much details as possible. An enzyme reversibly binds a substrate. The rate constant (kcat) for catalytic conversion of enzyme- substrate complex into product and free enzyme is 1.0 s-1. The rate constant for dissociation of enzyme- substrate complex (k-1) is 9.0s^-1. 1) For each substrate binding event, what is the probability that substrate is converted into product rather than dissociated without conversion? 2) What is the probability that 20 consecutive binding events result in no product formation? 3) If you evaluated 1,000 different sets of 20 consecutive binding events, what would be the average number of product molecules formed (per 20 binding events)? 4) What is the probability that 20 consecutive binding events result in a number of product molecules at least equal to this average?
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?
Note the Michaelis Menton kinetics results of inhibition by inhibitor A and by B, separately. Normal enzyme Inhibitor A Convert these to lineweaver burke in graphs below. -5+ -4 Inhibitor B -3+ -2 Effect of Inhibitor A. Draw uninhibited first and then draw the result- ing inhibition for comparison. What kind of an inhibitor is A? How can you tell? Effect of Inhibitor B. Draw uninhibited first and then draw the result- ing inhibition for comparison. What kind of an inhibitor is B? How can you tell?
What is the impact of the lower value Vmax on the affinity for enzyme for substrate? And what is impact of the lower V max on the amount of product formed ? If the lower value of black resulting in the new plot (red curve) is due to presence of enzyme inhibitor is the inhibitor reversible or irreversible ? And why?
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
List 4 major types of inhibition modes and clearly indicate the effect on Vmax and KM for each mode?2. What is the effect of each of the 4 types of inhibitors on the initial rate of an enzyme catalyzed reaction?3. A potent inhibitor effectively inhibits an enzyme catalyzed reaction. What kind of a Ki value you would expect for a potent inhibitor?4. Considering PNPP → PNP reaction, would you expect to see more intense or pale color for the reaction that contain the inhibitor? Explain.

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