(e) What kind of inhibitor is it likely to be? Plot the data in Lineweaver-Burk plot form to support your answer.
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- Please find the Vmax and Km values of the enzyme without/minus an inhibitor, and the enzyme with/plus an inhibitor. (The Vmax may be the same value).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]oNeed typed solution only and full explanation too ASAP...
- A. Lineweaver-Burk plot of the enzyme with increasing amounts of substrate in the absence or the presence of the inhibitor is shown below. Graph A : x-intercept Graph B : x-intercept = - 0.012, y-intercept = 0.8 Graph C : x-intercept = - 0.027, y-intercept = 0.8 Graph D : x-intercept = - 0.039, y-intercept = 0.8 - 0.007, y-intercept = 0.8 Graph A 4 Graph B Graph C Graph D 1 -0,04 -0,02 0,00 0,02 0,04 1/[Substrate] (uM) (i) Which graph indicates an enzymatic reaction without inhibitor? (ii) Which type of inhibitor is it? Briefly explain. (iii) Which graph indicates the highest concentration of inhibitor? (iv) Calculate the Vmax and Km of the graph showing an enzymatic reaction with the lowest concentration of inhibitor. Show the steps of calculation and unit in your answers. Keep 2 decimal places in your answers. 1/Rate (umol/min)One way of identifying a drug target in a complex cellular extract is to use an affinity approach, i.e. fix the drug to a resin (agarose etc) and use it to "pull down "" the target from the extract. What potential problems do you think may be encountered with attempting this approach?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??
- 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?In mixed inhibition as shown below, please draw a lineweaver-burk plot when Kl is greater than KI'. Please draw plot with and without inhibitor and label axis and intercepts. Please explain what is happening to Km and Vmax. In mixed inhibition as shown below, please draw a lineweaver-burk plot when Kl is greater than KI'. Please draw plot with and without inhibitor and label axis and intercepts. Please explain what is happening to Km and Vmax. 1/v 1/[S1]a. Estimate KM and Vmax for the uninhibited reaction from the first graph. Whatdifficulties do you find in getting accurate values?b. Make a Lineweaver-Burk (double reciprocal) plot to determine KM and Vmax again.What advantages do you see with the second method? c. Use the Lineweaver-Burk method and the table of data for the inhibitors to determine the kind of inhibition for each inhibitor.
- If the data from an enzyme experiment is plotted as a Lineweaver-Burk plot, and the Vmax is 0.02 mol/sec, and x-intercept is –2.5 mM then what is the KM value? Show yourwork/reasoning.Use 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=If the higher value of KM resulting in the new plot ( red curb ) is due to the presence of an enzyme inhibitor is inhibitor reversible or irreversible? And why?