A Lineweaver-Burk plot generates a line with the following formula: y = 0.3x + 0.4. What is the Km of this enzyme?
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A Lineweaver-Burk plot generates a line with the following formula: y = 0.3x + 0.4. What is the Km of this enzyme?
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- 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.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 stepPlease note the following Lineweaver-Burk plot for the enzyme Virbraniumase reacting with a substrate: -0.4 . 1/V 5 4 3 2 -0.2 0 0.2 y = 5.2781x + 1.3338 R² = 0.9967 0.4 0.6 0.8 1/[S] 1 Based on the information provided, what is the Vmax for this reaction?
- If a data from an enzyme experiment is plotted as a Lineweaver - Burk plot, and the Vmax is 0.02 sec/mol, and x-intercept is -2.5 mM-1, then what is the KM value?From 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.The Lineweaver-Burke plot was originally developed in order to "linearize" the data obtained from enzyme kinetics experiments, in order to facilitate the determination of kinetic parameters. Why is it not considered to be an accurate method for this purpose? It is very difficult to draw a straight line on a computer. It is very difficult to calculate the variables required for the "x" and "y" axis. It is more accurate to use the standard "V versus [S]" plot to determine Vmax and KM- The plot weights the least accurate data points the most heavily. It is no longer considered to be acceptable to extrapolate from known data.
- Calculate 1/[S] and 1/V to complete the table. Use this data to draw a Lineweaver-Burke plot, with lines for ‘no inhibitor’ and ‘with inhibitor’. Be sure to label your axes and lines. What kind of inhibitor is it? Estimate Vmax and Km for the uninhibited reaction.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.1. A Lineweaver-Burk Plot is shown below. 30 25 Curve A y = 3.1207x + 2.4978 20 15 Curve B y = 1.0003x + 2.3602 10 5 -3 1 5 7 11 1/[Catechol] (mM1) With these curves, determine the following enzyme parameters. Show all pertinent solutions. a. Km of Curve A and Curve B b. Vmax of Curve A and Curve B c. Assuming that one of these curves corresponds to the kinetics of one enzyme and one substrate, which curve represents the effect of an inhibitor? Why do you say so? d. What type of inhibition is exhibited by your answer in question c? Why do you say so? 1/V, (units of activity 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.In your acid phosphatase enzyme kinetics lab, you constructed a Lineweaver-Burk plot. Lefs assume that you graphicaly obtain 1/Vmax - 9.33 (umoliey and -1Km--0.012 M. Which of the following is correct? A) Vmax = 9.33 (no units) B) Km = 0.012 (no units) C) Vmax = (1/Vru" = 0.107 umolis D) Km =-14-1K = 83.3 pM E) Both C) and D) are correctThe commercial sample of carboxyl esterase contained 15.0 mg of enxyme in 0.30 mL of suspension. In your experiment, you add 0.090 mL of this suspension to the sample tube that contains a total volume of 3.0 mL. How many mg of carboxyl esterase enzyme do you have in the sample tube?