1. The following data were obtained for a competitive inhibition study in which the [I] = 3 µM for each determination of vo in the presence of inhibitor. The Vmax= 200 μM P/min for both data sets. a. Determine Km Vo (MP) 200 180 160 140 120 100 the absence of inhibitor. No inibitor 50 [Substrate] (M) 150 b. Determine Km, app for the data obtained in presence of inhibitor. c. Calculate the value for Ki. Note: a = 1+ [1]/Ki and a = 1 + [1]/Ki'. for the data obtained in
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- Part 1: Assess the following partial results section below by editing it for brevity by omitting any unnecessary parts (1 point), explain why you decided to remove certain sections (1 point): To evaluate inhibitory effects of the selected molecules, 10mM stock solutions of each molecule were prepared in DMSO. A reaction mixture (200μl) was prepared with the same formula optimized for the enzyme activity assay (0.1 M Tris-HCl ph 8, 0.1 M KCI, 25 mM NaCl, 0.25 mM ATP, and two units of inorganic yeast pyrophosphatase) with 10 µM of the sample molecule. The reaction mixture was incubated for 20 minutes at ambient temperature. Enzymatic reaction was triggered by addition of the substrate B (0.2 mM) and the absorbance of the product was monitored at 290 nm for 10 minutes. Six out of 15 sample molecules showed appreciable inhibition at 10 μM (Figure 5). Three of the molecules, A3, A6, and A7 exhibited more than 50% inhibition of the enzyme activity and were further diluted to find the minimal…A 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.Aerobic degradation of an organic compound by mixed cultureof organism in wastewater can be represented by following reaction. C3H6O3 + a O2 + b NH3 → c C5H7NO2 + d H2o + e CO2 A. Determine a, b, c, d and e, if YX/S = 0.4 d X/g S. B. Determine the yield coefficients YX/O2 and YX/NH3. C. Determine the degree of reductions for the substrate, bacteria and RQ for the organisms
- An enzymatic reaction follows M-M kinetics with Vmax= 2.5 mol m-3s-1and Km = 5 mM.Calculate the time required for 50% conversion of the substrate in a batch reactor if theinitial substrate concentration is0.2 M.Show your calculation steps.Biochemical tests for S.simulans : (+ or -) Catalase ___+___ DNase ___+___ Coagulase ___-___ 7.5% salt ___+__ Mannitol fermentation___-___ 2. Cell morphology and arrangement of S.simulans ____Select true if the statement is CORRECT and false if OTHERWISE 1. Enzymes are catalysts and increase the speed of a chemical reaction without themselves undergoing any permanent chemical change. 2. Catalysis is defined as the acceleration of a chemical reaction 3. if the amount of the enzyme is kept constant and the substrate concentration is then gradually increased, the reaction velocity will decrease. 4. In the Induced-fit Model, if a dissimilar substance which does not fit the site is present, the enzyme rejects it 5. The Michaelis constant Vo is defined as the substrate concentration at 1/2 the maximum velocity. 6. A prosthetic group - an organic substance which is dialyzable and thermostable which is firmly attached to the protein or apoenzyme portion. 7. The rate of an enzyme-catalyzed reaction increases as the temperature is raised beyond optimum temperature. 8. Enzymes can be classified by the kind of chemical reaction catalyzed. 9. The living cell is the site of tremendous…
- A bacterial culture is grown using either octadecane (C18H38) or pentachlorophenol (C6HOCI5)) as the sole source of carbon and energy. The cell yield value is determined by dry weight analysis to be 1.49 for octadecane and 0.05 for pentachlorophenol. Using either octadecane or pentachlorophenol, please describe the steps taken (and the final result) showing what percentage of the substrate carbon will be found as cell mass and as CO2?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.11) Using the data above, construct Lineweaver-Burk plots. First, graph the inverse of the reaction velocity (rate of reaction) data in column II versus the inverse of the methanol concentration in column I. Then, on the same graph, plot the inverse of the reaction velocity (rate of reaction) data in column III versus the inverse of the methanol concentration in column I
- These are SIM media. What does SIM stand for? Describe the reactions in Tubes A, B, C, D, E.Determine the minimum inhibitory concentrations (MIC) for compounds 1, 2, and 3 based on the microdilution results.A purified protein sample was used in a reaction, resulting in an activity of 696.7 nmol min-1. The reaction volume was 145.0 µL and the final volume before loading the plate was 1,050 µL. The total reaction time was 4.25 min. The amount of protein used in the reaction was 4.270 µg. Calculate the specific activity of the sample (in nmol min-1 µg-1).