The Y150F (Tyr 150 to Phe) mutant of B-lactamase catalyzes the reaction with a keat of 3.1 * 10³ s¹ at pH = 7.0 and 37 °C. Assuming this decrease in keat is due to the loss of transition state stabilization by the Y150F mutant, calculat the AAG* between the wild-type and Y150F enzymes. Show your work and state any assumptions you make in solving this problem.
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- 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 answeA 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.The enzyme lysozyme kills certain bacteria by attacking a sugar called N-acetylglucosamine (NAG) in their cell walls. At an enzyme concentration of 2 × 10-6 M, the maximum rate for substrate (NAG) reaction, found at high substrate concentration, is 1 × 10-6 mol L¯'s1. The rate is reduced by a factor of 2 when the substrate concentration is reduced to 6 x 10-6 M. Determine the Michaelis-Menten constants Km and k, for lysozyme.
- Shown below is a proposed mechanism for the cleavage of sialic acid by the viral enzyme neuraminidase. The kcat for the wild-type enzyme at pH =6.15, 37 °C is 26.8 s-1.(a) Describe the roles of the following amino acids in the catalytic mechanism: Glu117, Tyr409, and Asp149. List all of the following that apply:general acid/base catalysis (GABC), covalent catalysis, electrostaticstabilization of transition state.(b) Based on the information shown in the scheme, would you expect mutation of Glu 117 to Ala to have a greater effect on KM or kcat?(c) For the R374N mutant at pH = 6.15, 37 °C, kcat is 0.020 s-1, and KMis relatively unaffected. Based on this result, it seems that R374 is morecritical for catalysis than for substrate binding. Explain how R374 stabilizesthe reaction transition state more than the substrate (i.e., what feature of this reaction would explain tighter binding to the transition state vs. substrate?).Explain why each of the following data sets from a Lineweaver–Burk plot are not individually ideal for determining KM for an enzymecatalyzed reaction that follows Michaelis–Menten kinetics.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
- During the early stages of an enzyme purification protocol, when cells have been lysed but cytosolic components have not been separated, the reaction velocity-versus-substrate concentration is sigmoidal. As you continue to purify the enzyme, the curve shifts to the right. Explain your results. This is an allosteric enzyme and you must use a Lineweaver-Burk plot to determine KM and Vmax correctly. This is an enzyme that displays Michaelis-Menten kinetics and you purify away an inhibitor. This is an allosteric enzyme and during purification you purify away an activator. This is an allosteric enzyme displaying a double-displacement mechanism and during purification you purify away one of the substrates: This is an enzyme that displays Michaelis-Menten kinetics, and you must use a Lineweaver-Burk plot to determine KM and Vmax correctly.#1 Specify the role each of the following amino acids play within the crystal structure and/or active site for Be as specific as possible, with pictures (and mechanistic arrows) as necessary. His11 Arg140 Glu89 Trp68 #2 Provide a step-wise mechanism for the reaction Bisphosphoglycerate mutase catalyzes, using the amino acids responsible for aiding in catalysis. You do not need to add surrounding amino acids that aid in substrate specificity. (drawn out)In 1966, Ferdinand showed that a random-order ternary-complex mechanism for a two-substrate enzyme-catalysed reaction can lead to sigmoidal kinetics being observation in the absence of cooperative binding. Discuss this scenario so that it is clear why a plot of Vo versus [AXo] at constant [Bo] will be sigmoidal.
- Although graphical methods are available for accurate determination of the Vmax and Km of an enzyme-catalyzed reaction, sometimes these quantities can be quickly estimated by inspecting values of V0 at increasing [S]. Estimate the Vmax and Km of the enzyme-catalyzed reaction for which the following data were obtained:Given the following data, calculate Keq for the denaturation reaction of the protein β-lactoglobin at 25oC: ΔH° = –88 kJ/mol ΔS° = 0.3 kJ/mol. The free energy of hydrolysis of ATP in systems free of Mg2+ is −35.7 kJ/mol. When the concentration of this ion is 5 mM, ΔG°observed is approximately −31 kJ/mol at pH 7 and 38°C. Suggest a possible reason for this effect.Residue Asn 204 in the glucose binding site of hexokinase IV was mutated, in two separate experiments, to either Ala or Asp. The Asn → Ala mutant had a KM nearly 50-fold greater than the wild-type enzyme, and the Asn → Asp mutant had a 140-fold greater KM value than the wild-type enzyme. These mutations impact the intermolecular interactions between the enzyme and the glucose substrate.The amide functional group of the Asn side chain can form with the hydroxyl groups of the glucose substrate and can potentially function as either a . The methyl group of Ala cannot participate in hydrogen bond formation, which explains the in glucose affinity as indicated by the higher KM for the mutant enzyme. The side chain of Asp could potentially serve as a , but…