Enzyme X can be inhibited by two distinct inhibitors, only one of which is acting as a competitive inhibitor, while the other one is non-competitive. The three plots below (designated A, B, and Cy showMichaelis-Menten kinetics for the reactions carried out with the same concentr ation of the enzyme and either in the absence of any inhibitors or in the presence of one of the two inhibitors. Match the typeof reaction (uninhibited, competitive inhibition, non-competitive inhibition) with the corresponding plot?Plot APlot BPlot CSubstrate concentration1.Plot Av LUninhibited roaction2. Plot Cv Inhibition by a competitive inhibitor3Plot Bv Inhibition by a non-competitive inhibitorA Moving to another question will save this response.lup

Enzymes are the proteins or RNA which catalyses the rate of reaction without undergoing any change…

Answered: Enzyme X can be inhibited by two…

Biochemistry

6th Edition

ISBN:9781305577206

Author:Reginald H. Garrett, Charles M. Grisham

Publisher:Reginald H. Garrett, Charles M. Grisham

Chapter25: Nitrogen Acquisition And Amino Acid Metabolism

Section: Chapter Questions

Problem 22P

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The accompanying figure shows three Lineweaver–Burk plots for enzymereactions that have been carried out in the presence, or absence, of aninhibitor. Indicate what type of inhibition is predicted based on eachLineweaver–Burk plot. For each plot indicate which line corresponds to thereaction without inhibitor and which line corresponds to the reaction withinhibitor present.
Please fully explain (use illustrate where appropriate) the Modes of Enzyme Catalysis exemplified by the serine protease: Chymotrypsin. In your answer discuss employing the illustration whenever possible: the overall reaction mechanism, stability of the reaction transition state, proximity and orientation effects, acid-base catalysis, and covalent catalysis.
The protein catalase catalyzes the reaction 2H,O,(aq) — 2H,O(l) + O,(g) and has a Michaelis-Menten constant of KM = 25 mM and a turnover number of 4.0 × 107 s¯¹. The total enzyme concentration is 0.010 µM and the initial substrate concentration is 4.83 µM. Catalase has a single active site. Calculate the value of Rmax (often written as Vmax) for this enzyme. Rmax Calculate the initial rate, R (often written as V), of this reaction. R = ×10 mM.s-1 mM-s-1
The KMof the enzyme for the substrate adenosine is 3 × 10ꟷ5M. The product inosine acts as an inhibitor of the reaction, with an inhibition constant (KI, the dissociation constant for enzyme-inhibitor binding) of 3 × 10ꟷ4M. However, a transition state analog,Inhibits the reaction with KIof 1.5 × 10ꟷ13M. Explain why 1,6-dihydroinosine serves as a better inhibitor of adenosine deaminase than inosine. Elaborate on your answe
A particular enzyme-catalyzed reaction has an apparent Vmax = 9.00 nmol s-1 and α' = 3.00 when 2.00 µmol L-1 inhibitor X is present and uncompetitively inhibiting the reaction. Calculate Vmax for the uninhibited reaction in nmol s-1.
You will perform the protocol below for the calf intestinal alkaline phosphatase (CIP) provided. For each reaction, your final enzyme concentration should be 10 nM CIP. Note: Enzymes purchased are typically labelled with their “units of activity” (U), as this relates to how much enzyme is needed to catalyze a reaction. The 100 nM CIP provided has approximately 3 U/mL and was diluted 1 in 1,000 from a 500 U/mL purchased enzyme. 1) Create a table (similar to the one below) to help you determine and keep track of what to add to each of the cuvettes in which your reactions will be measured. The five different concentrations of PNPP should be: 25, 50, 100, 200, 300 μM. Each reaction will be in a final volume of 1 mL and contain 10 nM alkaline phosphatase. Concentrations of stock solutions: 1.0 mM PNPP, 100 nM calf intestinal phosphatase
The KM values for the reaction of chymotrypsin with two different substrates are given in the table below. Considering this information, which substrate has the lower apparent affinity for the enzyme? Which substrate is likely to give a lower value for Vmax? Substrate N-acetylvaline ethyl ester N-acetyltyrosine ethyl ester KM (M) 8.8 X 10-² 6.6 X 10-4 N-acetylvaline ethyl ester has the lower apparent affinity for the enzyme; it will also likely to give a lower Vmax: N-acetyltyrosine ethyl ester has the lower apparent affinity for the enzyme; it will also likely to give the lower V₁ max. N-acetylvaline ethyl ester has the lower apparent affinity for the enzyme; N- acetyltyrosine ethyl ester is likely to give the lower Vmax: N-acetyltyrosine ethyl ester has the lower apparent affinity for the enzyme; N- acetylvaline will likely to give the lower Vmax. None of the above statements are correct.
(b) You are investigating the effects of several agents on the activity of alcohol dehydrogenase. The enzyme activity data are shown in the table below. Construct a [substrate] vs. activity plot and a double-reciprocal plot for this enzyme. Be sure to label all axes. Determine the Vmax and KM for AD from the graphs in each type of plot. AD activity (nM/min) AD activity + agent A (nM/min) AD activity + agent B (nM/min) [Alcohol] (nM) 0.1 14 2 0.5 50 7 8. 1.0 65 10 30 2.0 72 12 45 4.0 80 14 62 8.0 85 15 75 32.0 90 16 90
Because of the position of arsenic in the periodic table, arsenate (AsO43- ) is chemically similar to inorganic phosphate and is used by phosphaterequiring enzymes as an alternative substrate. Organic arsenates are quite unstable, however, and spontaneously hydrolyze. Arsenate is known to inhibit ATP production in glycolysis. Identify the target enzyme, and explain the mechanism of inhibition.
Most of the enzyme reactions followed the mathematical kinetic plots suggested by Michaelis-Menten plots: (a) Lineweaver-Burk equation is a manipulation of the Michaelis-Menten equation. Give the equation and state why it is necessary. (b) Is high or low Vmax preferable for an enzyme reaction? Give reason.
Xanthine oxidase (XO) is the enzyme that catalyzes the synthesis of uric acid, which in excess uses gouty arthritis. The inhibition of this enzyme is therefore critical in its treatment. A student searcher is investigating the inhibitory effects of kaempferol (Kmp) and chlorogenic acid (Cha) on XO hich uses xanthine (Xan) as substrate. Table 1 below shows the enzyme kinetic data. Construct the neweaver-Burk plot complete with the linear regression analvsis. Fill in the needed information on Table Table 1. Enzyme Kinetic Data Velocity, mM/s Kmp 0.0351 [S], mM Cha Хan 0.492 0.0678 0.0615 0.211 0.0531 0.0261 0.0451 0.087 0.0298 0.0157 0.0211 0.048 0.0195 0.0091 0.0142 0.029 0.0127 0.0067 0.0081 Enzyme Kinetic Parameter: Parameters Vmax |Ku | Type of Inhibition Mode of Binding Xanthine Kaempferol Chlorogenic acid NA NA Line Weaver Burk Plot:
The Lineweaver-Burk plot, which illustrates the reciprocal of the reaction rate (1/v) versus the reciprocal of the substrate concentration (1/[S]), is a graphical representation of enzyme kinetics. This plot is typically used to determine the maximum rate, Vmax, and the Michaelis constant, Km, which can be gleaned from the intercepts and slope. Identify each intercept and the slope in terms of the constants Vmax and Km. What term is represented by C? Linewegver-Burk Pt 1/Vmax O A. В. -1/Km Km/Vmax C.

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