C7 Q12: The Eadie-Hofstee plot illustrated shows enzyme activity in the absence of an inhibitor.Adjust one or both points so that the line shows enzyme activity in the presence of a competitive inhibitor.20191817161514131211100987 65432100 1 2 3 4 5 6 7 8V9 10 11 12 13 14 15 16 17 18 19 20[S] A plot of V₁ on the y-axis versus V₁/ [S] on the x-axis is known as an Eadie-Hofstee plot. Like the Lineweaver-Burk plot,the Eadie-Hofstee plot is derived from a transformation of the Michaelis-Menten equation. The Michaelis-Mentenequation expresses the relationship between the following terms:Vo, the rate of reaction;• Vmax, the maximal rate;[S], the substrate concentration; andKM, the Michaelis constant.Rearrange the Michaelis-Menten equation to obtain the Eadie-Hofstee equation, giving Vo as a function of Vo/[S].●In an Eadie-Hofstee plot, what is the significance of the x-intercept?Vmax/KMVmax1/[S]O-KMThe Eadie-Hofstee plot illustrated shows enzyme activity in the absence of an inhibitor.Adjust one or both points so that the line shows enzyme activity in the presence of a competitive inhibitor.2019181716151413121110ܘܢ9

A graphical depiction employed in enzyme kinetics to examine enzyme-catalyzed reactions and…

Answered: A plot of V₁ on the y-axis versus V₁/…

Principles Of Radiographic Imaging: An Art And A Science

6th Edition

ISBN:9781337711067

Author:Richard R. Carlton, Arlene M. Adler, Vesna Balac

Publisher:Richard R. Carlton, Arlene M. Adler, Vesna Balac

Chapter1: Basic Mathematics

Section: Chapter Questions

Problem 7RQ

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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?
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.
What are the benefits of measuring the initial rate of a reaction Vå for use in kinetic studies? (This is a multi-select question). [ES] can be measured accurately. changes in [S] are negligible, so the value of [S] is known. changes in Km are negligible, so Km can be treated as a constant. V₁ = Vmax. --> A negligible amount of product has formed, so that the back reaction P -- need not be considered. ES
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 Activity
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)?
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:
Calculate the K of A given K₁ = HA(aq) + H2O(l) = H3O+ (aq) + A¯¯(aq) K = 5.90×108
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.
An enzyme has the following values: vmax's; 0.028 M s-' and 0.021 M s-' (I), and km's; 0.00198 M and 0.00197 M (I), when the total concentration of enzyme is 10-º M. The (I) values were determined with 1 microM inhibitor. Select the value closest to the k, for this enzyme. (hint 1 microM = 10-6 M).
Please 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?
Would you expect a change in column temperature to have an effect on a reversed-phase chromatographic separation of a steroid mixture? Explain the reason for your answer in (1) above

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