6. The V, and KM values for an unusual hexokinase found in Try- max 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 Vmax (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?
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- 2. The two diagrams to the right il- lustrate plots of steady-state ki- netic studies to characterize the in- FB 0.8- nH = 3.5 0.6- teraction of heart muscle phos- phofructokinase-1 with a non-phy- siological, synthetic substrate fruc- tose-6-sulfate. Because the kcat is smaller than that for the natural 0.4- 0.2- 10 μΜ 20 μΜ 48 μΜ substrate, higher enzyme concen- trations could be used. The results show the influence of increasing 2- 3.2- 0.4- concentrations of ATP on the initial -0.6- > velocity of the enzyme catalyzed reaction in the presence of no AMP (•), 10 µM AMP (•), 20 µM AMP (-), and 48 µM AMP (-). -0.8- 4 12 20 28 36 44 52 60 68 76 84 92 1.2 1.6 2.0 2.4 [AΤP (μΜ) log[ATP] (µM) (a) how ATP interacts with the enzyme in the case of no AMP (•). Write the reaction in words catalyzed by the enzyme for the alternative substrate, describe (b) with respect to the binding of AMP and ATP to the allosteric effector sites on the enzyme. Explain the physical significance of the…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. Toconazole9. 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 1/Va 00 80 70 60 1/V 50- 40 30- 20- 10 F0008
- 2. The two diagrams to the right il- lustrate plots of steady-state ki- 5FA 0.8 netic studies to characterize the in- nH = 3.5 0.6- teraction of heart muscle phos- phofructokinase-1 with a non-phy- siological, synthetic substrate fruc- tose-6-sulfate. Because the kcat is 0.4- 0.2- smaller than that for the natural 10 μΜ 20 μΜ 48 μΜ substrate, higher enzyme concen- trations could be used. The results show the influence of increasing 0.2 0.4 concentrations of ATP on the initial -0.6- > velocity of the enzyme catalyzed reaction in the presence of no -0.8 4 12 20 28 36 44 52 60 68 76 84 92 1.2 2.0 2.4 AMP (•), 10 µM AMP (•), 20 µM AMP (-), and 48 µM AMP (). [ΑΤPΙ (μM) log[ATP] (µM) (а) how ATP interacts with the enzyme in the case of no AMP (•). Write the reaction in words catalyzed by the enzyme for the alternative substrate, describe (b) with respect to the binding of AMP and ATP to the allosteric effector sites on the enzyme. Explain the physical significance of the displacement of the…-1/aka 1/v. Increasing a=1 (no inhibitor) Slope = a KµV…... Slope ak 7-15 satino inhibitori B D MELIN H W MIN 061 d'Nat 1/V a = 2 a=1.5 Increasing (1) o1 (no inhibitor) Slope = KNALLE9. 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 1/V 70 60- 40- 30- 20- 10- 10. For your answer in #9, explain briefly: a. 1. how / and S bind to the E as shown by the Lineweaver Burk plot. b. the significance of the following obtained values for Km and Vmax. c. effect in slope and x-intercept F000 F0000 Fo008 F0009 F000 Fo00
- 3. Subtilisin (Mol. weight 27,600) is a protease that can catalyze hydrolysis of certain amino acid esters and amides. For the synthetic substrate N-acetyl-L-tyrosine ethyl ester (Ac-Try-OEt), subtilisn exhibits Km and kcat values of 0.15 M and 550 s-¹, respectively. a) What is the Vmax when the subtilisin concentration is 0.4 mg/ml? b) Indole is a competitive inhibitor of subtilisin with a Ki of 0.05 M. What is the Vmax for Ac-Try-OEt hydrolysis by 0.4 mg/ml subtilisin in the presence of 6.25 mM indole? c) What is the Vo when 0.4 mg/ml subtilisin is incubated with 0.25 M Ac-Try-OEt and 1.0 M indole?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.what possible outcome may be produced when the molecular weight of an alcohol antiseptic is increase to a C20, will it still be effective? Explain your answer
- 2. The two diagrams to the right il- lustrate plots of steady-state ki- netic studies to characterize the in- teraction of heart muscle phos- phofructokinase-1 with a non-phy- siological, synthetic substrate fruc- tose-6-sulfate. Because the kcat is smaller than that for the natural 5 0.8- NH = 3.5 0.6- 0.4- 0.2- 10 μΜ 20 μΜ 48 µM substrate, higher enzyme concen- trations could be used. The results show the influence of increasing 2 0.2- 0.4- concentrations of ATP on the initial -0.6- > velocity of the enzyme catalyzed reaction in the presence of no -0.8F 4 12 20 28 36 44 52 60 68 76 84 92 .2 2.0 2.4 ΑMP () , 10 μΜ AMP (+ ) 20 μΜ AMP (-), and 48 µM AMP (-). [AΤP] (μΜ) log[ATP] (µM) (a) . Write the reaction in words catalyzed by the enzyme for the alternative substrate, describe how ATP interacts with the enzyme in the case of no AMP (•). (b) ! with respect to the binding of AMP and ATP to the allosteric effector sites on the enzyme. Explain the physical significance of the displacement…5) In an experiment to investigate the inhibition of the enzyme-glucosidase the following data for the rates of reaction with glucopyranoside for various substrate concentrations was obtained. By constructing a Leaver-Burk plot, determine the value of the Michaelis constant. [S]/ (10-6 mol dm-3) v/ (10-3 mol dm-3 s-1) 1.00 2.00 3.00 4.00 16.7 33.3 41.1 49.8The Mechanism of Action of Cyclophosphamide and its toxicity reference :http://dx.doi.org/10.3390/scipharm88040042 (reseach paper DOI plz answer in detail do not give short answer )