Which of the following is true about a mixed type inhibition? a. None of these is true b. A Lineweaver-Burk plot will give parallel lines c. The lines of a Lineweaver-Burk graph will cross in the top left quadrant d. The KM will change but not the Vmax
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Which of the following is true about a mixed type inhibition?
a. None of these is true
b. A Lineweaver-Burk plot will give parallel lines
c. The lines of a Lineweaver-Burk graph will cross in the top left quadrant
d. The KM will change but not the Vmax
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- 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]In pure noncompetitive inhibition: a. Where on the enzyme does the inhibitor bind? b. Does the inhibitor bind to E, ES or both? c. What is the effect of I on Vmax? d. What is the effect of I on Km?An experiment was carried out to measure the reaction rate of hydrolysis of acetylcholme (substrate) with serum enzymes (Eadie, 1949). In the experiment, two experiments were conducted, namely experiment 1 without using a prostigmine inhibitor and experiment 2 using a prostigmine inhibitor at 1.5 x 10^-7 mol/l. the data obtained are: a. Is prostigmine competitive or noncompetitive inhibitor? b. determine the value of km and rmax for the two experiments, compare
- Below is a plot of Vo vs. [S] for a specific allosteric enzyme under different conditions. Which of the following best describes the graph? 4 (S] O Adding a positive modulator to #2 would result in curve 3. Curve 1 represents maximum inhibition. Line 4 is valid exclusively for curves 1 and 2. Adding a positive modulator to #1 would result in curve 2. O Curve 3 represents the effect of a negative modulator added to curve 2.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)If the new higher KM value is 0.1 mM resulting in the new plot red curve is due to presence of enzyme inhibitor is the inhibitor reversible or irreversible?
- During an SAR effort to identify tyrosine kinase inhibitors, it was found that compound 2 was significantly more potent and selective than compound 1. To what might you attribute this improvement in activity and selectivity? Explain. IC-5 micromolar IC-0.1 mromolarIn an enzymatic reaction, the equations that correspond with and without inhibition according to linewear-burk are calculated, these are with inh: 3.5x + 6 = y without inh: 3.5 + 10 = y. calculate alpha factorWhich of the following statements are correct about the drug Gleevec (select all that apply)? A. Only inhibits activity of transit amplifying cells B. Gleevec is a natural product that was identified in a rare plant C. Negative side effects of drug are inhibition of PDGF and KIT D. Mutations in ABL gene cause resistance to Gleevec E. Inhibits the phosphorylation activity of tyrosine kinase
- 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??1. Which expression below shows hemoglobin bound to a proton a. HbH+ b. HbO₂H+ c. HbBPG d. None of the above 2. Does hemoglobin, from you answer choice in question number 1, exist in the high affinity state or the low affinity state? What is hemoglobin's heterotropic ligand, based upon your answer choice in question number 1? 3. What type of the four catalysis reactions referenced in class is depicted by Asp52 in the diagram to the right? Mur2Ae H AcN H AcÑ Glu Foo Asp82 (1) HH CH₂OH Glu 00 ⁰0 Asp2 H НО OH Н GlcNAc CH₂OH H OH H H NAC First productWhat will you be looking for in a graph of 1/Vo against 1/[AXo] at different concentrations of [Bo]? A. Whether the inhibition pattern is competitive or mixed B. Whether substrate inhibition is taking place C. Whether B is an inhibitor or not D. Whether the slope of the lines change or not