If Km1 > Km2, then the affinity of the enzyme to the substrate with Km1 is: Select one: a. Lower than for Km2 O b. Higher than for Km2 c. None of these d. Equal to Km2
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- -Inhibitor +Inhibitor [S] (mM) V0&νβσπ;(μmol/sec). νο&νβσπ: &ν βσπ (μmol/sec) 0.0001 33 17 0.0005 71 50 0.001 83 67 0.005 96 91 0.01 98 95 What is the Vmax of this enzyme WITH iinhibitor?-Inhibitor +Inhibitor [S] (mM) V0&νβσπ; (μmol/sec). V0&νβσπ:&νβ σπ: (μmollsec) 0.0001 33 17 0.0005 71 50 0.001 83 67 0.005 96 91 0.01 98 95 What is the Km of this enzyme WITH iinhibitor?Under what conditions does Km represent the true binding affinity (i.e. Kd) of the substrate to the enzyme? 1. when kcat << k-1 2. when k-1 >> k1 3. when k1 >> kcat 4. when k1 = k-1
- 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)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?When k2 >> k-1, KM approximates the affinity of the enzyme•substrate complex.The (circle one) higher / lower the KM, the more tightly the enzyme binds to its substrate
- n a particular enzyme, an alanine residue is located in a cleft where the substrate binds. A mutation that changes this residue to glycine has little effect on activity; however, another mutation, which changes the alanine to a glutamate residue, leads to a complete loss of activity. Provide a brief explanation for these observationThe steps of the chymotrypsin mechanisms are listed below (1-7). Put the steps of chymotrypsin mechanism in the correct order. Figure representing chymotrypsin mechanism is given for reference. a.The portion (N-terminal end) of original substrate with the new C terminus diffuses away b. Substrate binding c. His 57 catalyzes removal of H from Ser 195 hydroxyl; Ser 195’s nucleophilic O attacks carbonyl C of substrate; tetrahedral intermediate is formed d. Water binding; water is deprotonated by His 57; resulting OH nucleophilically attacks carbonyl of remaining substrate; tetrahedral intermediate is formed e. His 57 donates H to N of…An enzyme that follows the MWC (concerted) model for allostery has a T/R ratio of 300 in the absence of substrate. Suppose that a mutation were to reverse that ratio (T/R = 1/300 = 3.3 x 10–3 in the absence of any substrate). How would this mutation affect the relation between the rate of the reaction and substrate concentration, i.e., what would a Vo vs. [S] plot look like, and why?
- On the graph provided, sketch a plot of Vo/Vmax vs. [S] for an allosteric enzyme, 1.0 0.8 0.6 0.4 02 [Substrate] A. in the absence of any other ligand besides the substrate B. in the presence of an allosteric inhibitor C. in the presence of an allosteric activator. Clearly LABEL the 3 curves A, B and CIdentify the type of regulation of enzyme activity seen in the following situations - for example, competitive inhibition, allosterism, phosphorylation, zymogen conversion, association-dissociation, feedback inhibition, etc. a. Trypinsogen, which is not catalytically active, is converted to the active enzyme trypsin by removal of a hexapeptide from the N-terminal end. b. The dimer protein kinases is catalytically inactive. Binding of cAMP causes protein kinase dimer to split into its monomer which are active catalysts.The figure below shows the dependence of the enzyme's rate, v (in µM/min), as a function of substrate concentration, S (in mM). Also shown is the dependence of the rate in the presence of an inhibitor, present at a concentration of 0.2 mM. Based on this information, which of the following does this inhibitor most likely interact with? 1/v 1- 05 1/[S) O A. Michaelis complex O B. [E O C. free enzyme O D. free substrate O E. Both "A" and "B."