To determine: The reason behind not being the concentration of finding Km and Vmax on the constant spaced x-axis of the graph.
Introduction: Enzymes are “biocatalysts” which takes part in a process and alters the
To suggest: An improvement in the graph.
Introduction: Enzymes are “biocatalysts” which takes part in a process and alters the rate of reaction. The enzymes work on the entity known as “substrates” and after having interaction with the substrate is converted into the “product.” The Km of an enzyme can be described as the amount of substrate required for the purpose of saturation. Whereas, when the rate of reaction is at the top, the condition is known as Vmax.
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Becker's World of the Cell (9th Edition)
- The KM for the reaction of chymotrypsin with N-acetylvaline ethyl ester is 8.8 × 10−2 M, and the KM for the reaction of chymotrypsin with N-acetyltyrosine ethyl ester is 6.6 × 10−4 M. (a) Which substrate has the higher apparent affi nity for the enzyme? (b) Which substrate is likely to give a higher value for Vmax?arrow_forwardTable 1 shows the kinetic data that have been obtained for glucoamylase from Aspergillus niger at different temperatures in the production of glucose from maltodextrin. The rates of reaction measured are using an enzymes concentration of 0.003 μmolml-1. Outline and describe the reaction scheme of the production of glucose from maltodextrin by using glucoamylase.arrow_forwardThe KM for the reaction of chymotrypsin with Substrate A is 8.8 x 10-4 M, while the KM for the reaction of chymotrypsin with Substrate B is 8.7 x 10-3 M. Which of the following statements are likely true? Chymotrypsin has a higher apparent affinity for Substrate A. The Vmax would be higher in the presence of Substrate A. The kcat would be higher with Substrate A. The V max would be higher in the presence of Substrate B. Two of the above are true.arrow_forward
- In an uncatalyzed reaction, KF(uncatalyzed) = 2.5 X 10-8/s and Keq is 5 X 10². For the same reaction in the presence of an enzyme, KF(catalyzed) = 2.5 X 10²/s. Remember that Keq = KF/KR = k₁/k-1. (a) What is the rate enhancement of the catalyzed reaction in the forward direction? (b) What is the value for KR(uncatalyzed)?arrow_forwardThe 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.arrow_forwardThe enzyme β-methylaspartase catalyzes the deamination of β-methylaspartate. For this aspartate reaction in the presence of the inhibitor hydroxymethylaspartate (3.8 M), determine KM and whether the inhibition is competitive or noncompetitive (KI = 1.0 M). [S], M V w/o inhibitor, M/s V w/ inhibitor, M/s 1x10-4 0.0259 0.0098 5x10-4 0.0917 0.040 1.5x10-3 0.136 0.086 2.5x10-3 0.150 0.120 5x10-3 0.165 0.142 In the ABSENCE of inhibitor: The Lineweaver-Burke equation is 1V=1V= __________ (1[S])(1[S]) + __________, and the KM is __________ M. In the PRESENCE of inhibitor: The Lineweaver-Burke equation is 1V=1V= ____________ (1[S])(1[S]) + ___________, and the KM is ___________ M. The type of inhibition is ____________. Round-off all answers to two (2) significant figures.arrow_forward
- A biochemist is trying to determine the type of proteases they have isolated from walrus blubber. The three proteases and their relative activities in the presence of the indicated non-specific irreversible inhibitors are shown in the table below: Protease a Protease B Protease y Given this data, please answer the following question: The catalytic site of Protease ß contains an important: C R H + lodoacetate Normal Activity Normal Activity No Activity U + Tetranitromethane Normal Activity No Activity Normal Activityarrow_forwardThe Lineweaver-Burk plot, which illustrates the reciprocal of the reaction rate (1/v) versus the reciprocal of the substrate concentration (1/[S]), is a graphical representation of enzyme kinetics. This plot is typically used to determine the maximum rate, Vmax, and the Michaelis constant, Km, which can be gleaned from the intercepts and slope. Identify each intercept and the slope in terms of the constants Vmax and Km. What term is represented by C? Linewegver-Burk Pt 1/Vmax O A. В. -1/Km Km/Vmax C.arrow_forwardDuring 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.arrow_forward
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