The Lineweaver-Burke plots of a reaction without inhibitor and one with non-competitive inhibitor will have the same 1. Vmax 2. Km 3. Km/Vmax 4. 1/Vmax
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The Lineweaver-Burke plots of a reaction without inhibitor and one with non-competitive inhibitor will have the same
1. Vmax
2. Km
3. Km/Vmax
4. 1/Vmax
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- Using the attachment, Answer the following questions: Prepare a double reciprocal plot with all three experiments (lines) on the same graph. Use your graph, and then answer items 2, 3, and 4 below. 1. Calculate the Vmax or apparent Vmax for all three sets of data. Likewise, calculate the Km or apparent Km for each set. 2. For Inhibitor X, what is the mode/type of inhibition? 3. For Inhibitor Z, what is the mode/type of inhibition? By comparison of the apparent Vmax to the control Vmax, what is the value of α’, as defined in class? If Ki’ = 10 mM for this inhibitor, then what must the inhibitor concentration [Z] be?You are saying that the inhibitor is competitive inhibitor. But according to data Vmax for reaction (with no inhibitor) is 4,17mM/min and Vmax for reaction (with inhibitor)=2,31mM/min. Then you show that Km for reaction with no inhibitor is 1.66. Then I calculate further for Km for reaction with inhibitor by using MM equation and I get 0,9. So both the Vmax and Km is reduced in reaction with inhibitor. That must mean the inhibitor is not competitive but non-competitive inhibitor. Or is it me that got it wrong??A junior student performed an experiment to found Ki values for 3 different competitive inhibitors. The results are 5, 1 and 0,2 . Please answer the questions:a. Which inhibitor has higher affinity for the enzyme?b. If the student uses equal concentrations of inhibitor in each trial, which inhibitor gives the smallest Km value?
- An enzyme-catalyzed reaction has a KM of 20.0 mmol L-1 and Vmax of 17.0 pmol s-1. When a mixed inhibitor is added, the apparent KM is 50.0 mmol L-1 and the apparent Vmax is 5.20 pmol s-1. Calculate α.You are trying to come up with a drug to inhibit the activity of an enzyme thought to have a role in liver disease. In the laboratory the enzyme was shown to have a Km of 1.0 x 10-6 M and Vmax of 0.1 micromoles/min.mg measured at room temperature. You developed a competitive inhibitor. In the presence of 5.0 x 10-5 M inhibitor, the apparent Km of the enzyme was found to be 1.5 x 10-5 M. What is the Ki of the inhibitor?The graphs 3 and 4 representing 1/Vo = f(1/[S]o) have been done in the presence of a competitive (CI) and noncompetitive inhibitor (NCI). a- For each figure, determine from the relative position of the straight lines which one is obtained in presence of an inhibitor. b- Indicate which graph corresponds to the competitive inhibition and which one the noncompetitive inhibition. Justify your answer. c- Complete the graphs by indicating which values can be determined from the arrows. 3 1/No 1/[S]⁰ (4) 1/No 1/[S]o
- Refer to the following graph when answering questions. These words may be used to fill: competitive, uncompetitive, noncompetitive, irreversible. -1 mg , μmol min-1 v-1 0.2 0.1 THA -4-2 CBA + 2 4 [S]-1, (mM-1) Word 1, The value of Km for the enzyme depicted by curve A is 2, The value of Vmax for the enzyme depicted by curve A is 3, Curve B depicts the effect of an inhibitor on the system described by curve A. This inhibitor is a inhibitor. 4, Curve C depicts the effect of a different inhibitor of the system described by curve A. This second inhibitor is a inhibitor. DUse the relationships revealed by a Lineweaver-Burk plot and the table of enzyme performance to calculate the Vmax and Km of the enzyme with no inhibitor. with inhibitor A, and with inhibitor B. [S] (uM) Vo (umol/min); w/ no inhib. Vo (umol/min); w/ inhib. A Vo (umol/min); w/ inhib. B 10 6.3 5.1 4.0 40 18.4 15.8 11.8 100 29.9 27.0 19.1 150 34.7 32.0 22.2 No inhib. Vmax= Km= Inhib. A Vmax= Km= Inhib. B Vmax= Km=A с Wan WWW GHEDE MAK am2 Increasing CE a-1 (no inhibitor) Slope #KY... ...." 7-15 7 a>c²=1 [no inhibitori Slope R 101 9-19 aver-Burk plots in this figure represent the activities of enzymes in the 4-5 n. 155 B D MIN -a'/KM 0.8- 1/V >~- -1/K 0 01 m(app) d'Nmax a=2 a=1.5 1[S] per mM inhibitor a=10 Slope K
- Match each inhibitor with its effect on Michaelis-Menten reactions. Group of answer choices Vmax and apparent Vmax are equal and the apparent Km is greater than Km. The apparent Vmax is less than Vmax and the apparent Km and Km are equal The apparent Vmax and the apparent Km are both lowered to the same degree. The apparent Vmax is less than Vmax and the apparent Km can be either greater than or less than Km.A medicinal chemist is trying to determine the mechanism of action of inhibitors she has synthesized. The relative change in KM and Vmax upon incubation of the targeted enzyme with each inhibitor is shown in the table below. Inhibitor A Inhibitor B Inhibitor C Using this data, the mechanism of action of Inhibitor C is: Uncompetitive TS‡ analog Mixed Inhibition Competitive Кмарр- Км 0 Non-competitive app - Vmax <0 <0 0 VmaxCalculate 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.
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