9. For the following aspartate reaction in the presence of inhibitor, Km = 0.00065 M. Determine Vmax in both reactions and in the reaction without inhibitor, the Km. Identify whether the inhibition is competitive, non-competitive or uncompetitive.
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- For the following aspartase reaction in the presence of the inhibitor hydroxymethylaspartate, determine Km and whether the inhibition is competitive or noncompetitive. You have to plot thegraph on the graph paper and also by using excel.[S] V, No Inhibitor V, Inhibitor Present(molarity) (arbitrary units) (same arbitrary units) 1 x 10-4 0.026 0.0105 x 10-4 0.092 0.0401.5 x 10-3 0.136 0.0862.5 x 10-3 0.150 0.1205 x 10-3 0.165 0.1429. For the following aspartate reaction in the presence of inhibitor, Km = 0.00065 M. Determine Vmax in both reactions and in the reaction without inhibitor, the Km. Identify whether the inhibition is competitive, non-competitive or uncompetitive. 110 100- 90- 70 60- 40- 30- 20- 10- F0000I 0008 Foo0 Foo0A schematic representation of the enzyme IspD complexed to inhibitor 3, and a series of inhibitors 3-5 are shown below. Ala202 lle240 mwww NH NH Val263 ОН www HN N- lle177 HN 'N' CI 3 X = N 4 X = C-CN 5 X = C-COO IC50 274 µM IC50 140 nM IC50 35 nM NH2 HN Val266 N -N O-H---- N HN %3D Arg157 HN wwww lle265 Explain why structure 4 is a more potent inhibitor (lower IC50 value) than inhibitor 3 and why structure 5 is a much weaker inhibitor (higher IC50 value) than 3 and 4.
- The KM values for the reaction of chymotrypsin with two different substrates are given in the table below. Considering this information, which substrate has the lower apparent affinity for the enzyme? Which substrate is likely to give a lower value for Vmax? Substrate N-acetylvaline ethyl ester N-acetyltyrosine ethyl ester KM (M) 8.8 X 10-² 6.6 X 10-4 N-acetylvaline ethyl ester has the lower apparent affinity for the enzyme; it will also likely to give a lower Vmax: N-acetyltyrosine ethyl ester has the lower apparent affinity for the enzyme; it will also likely to give the lower V₁ max. N-acetylvaline ethyl ester has the lower apparent affinity for the enzyme; N- acetyltyrosine ethyl ester is likely to give the lower Vmax: N-acetyltyrosine ethyl ester has the lower apparent affinity for the enzyme; N- acetylvaline will likely to give the lower Vmax. None of the above statements are correct.. The following data, presented by G. Bowes and W. L. Ogre in J. Biol. Chem. (1972) 247:2171–2176, describe the relative rates of incorpo- ration of CO, by rubisco under N, and under pure O,. Decide whether O, is a competitive or uncompetitive inhibitor. [CO,] (mM) Under N2 Under O, 0.20 16.7 10 0.10 12.5 5.6 0.067 8.3 4.2 0.050 7.1 3.26. The Vmax and KM values for an unusual hexokinase found in Try- pansoma cruzi (the causative agent of Chagas disease) are shown in the presence and absence of a bisphonate inhibitor (structure shown). Without inhibitor With inhibitor Км (mM) 90 125 Vn max (umol min-1. mL-1) 0.30 0.12 ОН O=P-O- A bisphonate compound a. What type of inhibitor is bisphonate? b. The parasite hexokinase, unlike the mammalian enzyme, is not inhibited by glucose-6-phos- phate but is inhibited by pyrophosphate (PP;). Is this observation con- sistent with your answer to part a? c. Might bisphonate be a good candidate for a drug to treat the disease?
- 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.For the following aspartate reaction in the presence of inhibitor, Km = 0.00065 M. Determine Vmax in both reactions and in the reaction without inhibitor, the Km. Identify whether the inhibition is competitive, non-competitive or uncompetitive. ( see attached picture ) how I and S bind to the E as shown by the Lineweaver Burk plot. the significance of the following obtained values for Km and Vmax. effect in slope and x-interceptThe 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.
- Match the different names for inhibition mechanisms (1-5) with a description of their properties 7a-7d: 1. competitive inhibitor. 2. allosteric inhibitor also known as non-competitive inhibitor. 3. un-competitive inhibitor. 4. affinity label also known as active site directed covalent (irreversible) enzyme inhibitor. 5. Kcat inhibitor, also known as a mechanism-based covalent (irreversible) enzyme inhibitor. 4a. An enzyme inhibitor in which a substrate or competitive inhibitor is modified so that it contains a chemically reactive electrophile which can bind to and subsequently react with the enzyme active site: 4b. An enzyme inhibitor that contains latent reactive group that upon binding followed by catalytic turnover at the enzyme active site produces a reactive electrophile that reacts covalently with the enzyme: 4c. A reversible inhibitor that competes with the substrate for binding to the enzyme active site: 4d. A reversible inhibitor that can bind independently of substrate to its…22) answer the following question.. Refer to the kinetic scheme for competitive inhibition and the structures shown below to E+S ES E +P -co- CO- 1 2 EI Compound 1 was determined to act as a competitive inhibitor through standard inhibition studies. Structural studies did not show any resemblance to the transition state. Compound 2 was also determined to act as a competitive inhibitor. Structural studies showed that it does resemble the transition-state. The K, constant is used to assess relative affinity of inhibitors for enzymes. That is, each compound has its own K, value. We can interpret K, the same way we do with Ka values. True or False: K, > Kµ2. Briefly explain your answer.Which of these heterocyclic drugs is likely to be the least soluble in water? Use the Fsp³ parameter to decide. OH Tramadol Chemical Formula: C16H25NO2 YOUR OW Pantoprazole Torasemide Chemical Formula: C16H15F2N3O4S Chemical Formula: C16H20N4O3S Temazepam -OH Chemical Formula: C16H13CIN₂O2 Tioconazole Chemical Formula: C16H13C3N₂OS A. Tramadol B. Pantoprazole C. Torasemide D. Temazepam E. Toconazole