Which of the following equations accurately shows enzyme-catalyzed conversion of substrate to product? OE+S- ESP E+P O E+S EP E+P OE+S+ ES →E + P OE+S ES E+P
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- Which of the following are true when enzymes and substrate interact? kı E + S ● ES- -P k-1 the rate of the formation of ES = k1 [ES] the rate of the formation of the product = k2[ES] k1 [ES] = k-1 [ES] the rate of breakdown of ES to reform the enzyme and substrate = K-1[E][S] the rate of formation of ES = k1[E][S]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."In the Lineweaver Burke plot shown below, which of the following is true? With inbito Wthout nhitor O Km is increasing. Vmax is decreasing Km is decreasing, Vmax is decreasing Km is increasing, Vmax is increasing Km is decreasing, Vmax is in QUESTION 35 As per the Enzyme commission classification, the tranferases belong to which of the following Class? O 2 O 4 O 3 Click Save and Submit to save and submit. Click Save All Answers to save all answers. 78 F Mostly cloudy
- 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.Please note the reaction expression below. Which of the following rate constants describes the breakdown of the enzyme-substrate complex? There may be more than one answer. k₂ E+S OK-2 U k₂ 0 k₁ OK.1 k3 k.3 SES EPE+P K.3 K.₂ k.In an enzymatic reaction, the equations that correspond with and without inhibition according to linewear-burk are calculated, these are with inh: 3.5x + 6 = y without inh: 3.5 + 10 = y. calculate alpha factor
- In step four of proteolytic cleavage via chymotrypsin, water is introduced into the active site. Label on the image below what catalytic mechanisms are depicted to enhance the overall rate of catalysis . CH2 (N) Ser CH 195-CH,-0 H. His N- 57 Asp 102 ACYL-ENZYME- H,0 COMPLEXJ. C. Servaites, in Plant Physiol. (1985) 78:839–843, observed that Rubisco from tobacco leaves collected before dawn had a much lower specific activity than the enzyme collected at noon. This difference persisted despite extensive dialysis, gel filtration, or heat treatment. However, precipitation of the predawn enzyme by 50% (NH4)2SO4 restored the specific activity to the level of the noon-collected enzyme. Suggest an explanation.Lysozyme catalyzes a "bi-bi" reaction, which means there are (how many) reactants and (how many) products. List, in order, the reactants that bind and the products that are released during a lysozyme-catalyzed reaction cycle -- be succinct but be specific. 1. First reactant = 2. First product = 3. Second reactant = 4. Second product %3D
- An allosteric enzyme that follows the concerted mechanism has a T/R ratio of 500 in the absence of substrate. Suppose that a mutation reversed the ratio. How would this mutation affect the relation between the rate of the reaction and substrate concentration? The mutant enzyme would behave like an enzyme that obeys Michaelis Menton kinetics. The mutant enzyme would have a smaller vmax There would be no difference in the mutant enzyme in terms of substrate binding and catalysis. More than one answer is correct. The mutant enzyme would display cooperativity more than the wild type. MacBook Air 888 F5 F4 F3 F2 %23 2$ %In the scheme below which represents the mechanism of action for a large number of enzymes: A+B⟺AB⟶C The steady state approximation is reached when: d[AB]/dt≈0 k2≫k1 k−1≫k1 k−1=k1The equil ibrium constant for the attachment of a substrate to the active site of an enzyme was measured as 200.In a separate experiment, the rate constant for the secondorder attachment was found to be 1.5 x 108 dm3 mol-1 s- 1.What is the rate constant for the loss of the unreacted substrate from the active site?