Lineweaver-Burk plots of enzyme kinetics for the reaction, S <-> P, has the following features: 1/v is zero when 1/[S] equals -40 liter mole^-1; 1/[S] is zero when 1/v equals 2.0 x 10^5 min mole^-1. What are the Vmax and Km?
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Lineweaver-Burk plots of enzyme kinetics for the reaction, S <-> P, has the following features: 1/v is zero when 1/[S] equals -40 liter mole^-1; 1/[S] is zero when 1/v equals 2.0 x 10^5 min mole^-1. What are the Vmax and Km?
Vmax = 5 umol min^-1, Km = 2.5 mM?
Vmax = 5 mmol min^-1, Km = 25 M?
Vmax = 5 umol min^-1, Km = 25 mM?
Vmax = 5 mol min^-1, Km = 2.5 mM?
Vmax = 5 mol min^-1, Km = 25 mM?
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- Lineweaver-Burk plots of enzyme kinetics for the reaction, S <-> P, has the following features: 1/v is zero when 1/[S] equals -40 liter mole^-1; 1/[S] is zero when 1/v equals 2.0 x 10^5 min mole^-1. What are the Vmax and Km?For an enzyme that displays Michaelis-Menton kinetics, what is thereaction velocity, V (as a percentage of V max , observed at the followingvalues?[S] = K M[S] = 0.5K M[S] = 0.1K M[S] = 2K M[S] = 10K MFrom a kinetics experiment, the Vmax was determined to be 450µM∙min-1. For the kinetic assay, 0.1mL of a 0.05mg/mL solution of enzyme was used, and the enzyme has a molecular weight of 125,000 g/mole. Assume a reaction volume of 700µL. Calculate the kcat (in sec-1) for the enzyme.
- From a kinetics experiment, kcat was determined to be 295sec-1. For the kinetic assay, 0.3mL of a 0.25mg/mL solution of enzyme was used, and the enzyme has a molecular weight of 125,000 g/mole. Assume a reaction volume of 3mL. Calculate Vmax (µM∙min-1) for the enzyme and catalytic efficiency ( in M-1sec-1) for the enzyme. The Km for the enzyme was determined to be 2.55 x 10-2M.For a Michaelis-Menten enzyme, k1 = 5.2 ⅹ 108 M-1 s -1 , k-1 = 3.1 ⅹ 104 s -1 , and k2 = 3.4 ⅹ 105 s -1 . a) Write out the reaction, showing k1, k-1, and k2. Calculate Ks and Km. Does substrate binding approach rapid equilibrium or the steady state? Show work justify b) What is kcat for this reaction? Show work justify c) Calculate Vmax for the enzyme. The total enzyme concentration is 25 pmol L-1 , and each enzyme has two active sites.From a kinetics experiment, Kcat was determined to be 55sec^-1. For the kinetic assay, 0.05mL of a 0.05mg/mL solution of enzyme was used, and the enzyme has a molecular weight of 30,000g/mole. Assume a reaction volume of 3mL. Calculate Vmax (um*min^-1) for the enzyme and catalytic efficiency (in M^-1sec^-1) for the enzyme. The Km for the enzyme was determined to be 8.3*10^-2M.
- Assume that the experiments performed in the absence of inhibitors were conducted by adding 5 μL of a 2 mg/mL enzyme stock solution to an assay mixture with a total volume of 1 mL. Take into account that XYZase is a monomeric enzyme with a molecular mass of 45,000 Daltons. Hint: To calculate the ???? in units of per second (s−1), you must first determine the ???? in micromoles per second (μmol/sec). Please explain step by stepExplain 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.An enzyme that follows Michaelis-Menten kinetics has a KM value of 10.0 μM and a kcat value of 206 s−1 . At an initial enzyme concentration of 0.0100 μM , the initial reaction velocity was found to be 1.07×10−6 μM/s . What was the initial concentration of the substrate, [S] , used in the reaction ? Express your answer in micromolar to three significant figures..
- Give the Vmax and Km for this enzyme that obeys Michaelis-Menten kinetics. Use 2 sig figs.SUBSTRATE CONCENTRATION [S] µM INITIAL VELOCITY V0 s-1 10 0.13 25 0.27 50 0.45 100 0.65 150 0.77 200 0.85 300 0.94 500 1.03 (i) a) Construct an empty table with the following column headings: Substrate concentration [S] and initial velocity (Vi) where [S] has the unit µM, and Vi has the unit mM/s. (ii) The table provided is the enzyme kinetic data for your mutated enzyme, whereby Vi was expressed using the unit ∆A(405 nm)/s. Using the standard curve, express Vi with the unit mM/s rather than ∆A(405 nm)/s. Place your answer in the table above alongside the appropriate [S]. Hint: To answer this question you need to use this standard curve equation=0.0419x (The slope of the line is= 0.0419) (iii) The unmutated form of your protein has a Km of 25 µM and a Vmax of 43 mM/s. The enzyme kinetic data for your enzyme with the amino acid substitution should now be displayed in the table above. Based on these data, what is Vmax? Km? and…If you want to determine the KM for lactate, what protocol do you set up? Discuss the significance of the following kinetic parameters that are used to characterize enzyme activity: KM, Vmax, kcat, and kcat / KM.