The effect of an inhibitor on an enzyme was tested and the experiment gave the results below. If necessary, plot the data (Lineweaver-Burk) and determine, by inspection of the graph, what type of inhibition is involved. [S] (µM) 0.40 0.66 1.00 2.00 A. Competitive B. Non-competitive C. D. E. V (μmol/min) without Inhibitor 0.45 0.68 0.88 1.25 V (μmol/min) with 25 nM inhibitor 0.32 0.47 0.61 0.85 Uncompetitive There is no inhibitory effect at either 25 nM or 50 nM. Cannot conclude anything from the data or my graph. V (μmol/min) with 50 nM inhibitor 0.18 0.28 0.35 0.50
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- Determine the Ki for the inhibitor at 30 °Cand decide what type of inhibitor is being used. Eo T I S V (g/I) (°C) (mmol/ml) (mmol/ml) (mmol/ml-min) 1.6 30 0.1 2.63 1.6 30 0.033 1.92 1.6 30 0.02 1.47 1.6 30 0.01 0.96 1.6 30 0.005 0.56 1.6 49.6 0.1 5.13 1.6 49.6 0.033 3.70 1.6 49.6 0.01 1.89 1.6 49.6 0.0067 1.43 1.6 49.6 0.005 1.11 0.92 30 0.1 1.64 0.92 30 0.02 0.90 0.92 30 0.01 0.58 0.92 30 0.6 0.1 1.33 0.92 30 0.6 0.033 0.80 0.92 30 0.6 0.02 0.57You are evaluating the kinetics of an enzyme catalyzed reaction containing 5.5 μM total enzyme and 11.2 μM substrate. At this substrate concentration, you determine that the Vo = 88.6 μmol mL-¹. s-¹. If the Vmax 833.3 mM s the KM is: . == " 30.9 μΜ 10.4 μΜ Ο 124.6 μΜ 234 μΜ 94.5 μMA 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).
- In an enzyme experiment, when the enzyme is added only to the substrate, the Vmax value is 0.828 and the Km value is 17.76. When the substrate and activator are added to this enzyme, the Vmax is 0.964 and the Km value is 10.6066. When substrate and inhibitor are added to the enzyme, the Vmax value is 0.550 and the Km value is 11.34. Estimate the effect of the substance added. Are these substances suitable activators/inhibitors for the laccase enzyme?2) Below are kinetic data at three different inhibitor concentrations. Graph the data in an appropriate form to determine whether the inhibitor is competitive or uncompetitive. a (μm) 0.4 0.67 1.00 2.00 v (μM/min) (no inhibitor) 0.22 0.29 0.32 0.40 v (μM/min) (10 nM inhibitor) 0.21 0.26 0.30 0.36 v (μM/min) (20 nM inhibitor) 0.20 0.24 0.28 0.32An 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.
- 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 organismsConsider the following experimental data from another experiment: [S] 1.5 2.00 2.50 5.00 10.00 V (No inhibitor) mmol ml¹ min¹ 0.167 0.204 0.232 0.313 0.385 V (inhibitor) mmol ml¹¹ min¹¹ 0.115 0.143 0.167 0.250 0.333 Calculate Km and V max and determine whether this inhibitor is competitive, non-competitive or uncompetitive.An enzyme catalyzes a reaction with a Km of 9.50 mM and a Vmax of 2.10 mM · s-1. Calculate the reaction velocity, vo, for each substrate concentration. [S] = 3.25 mM vo : mM · s-1 [S] = 9.50 mM mM · s-1 Vo :
- 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.Calculate the Km of the enzyme with these parameters. kcat = 130s^-1 Vo = 3.0 μMs-1 S = 10 μM Et = 0.09 µM3.An enzyme catalyzed reaction is studied and the following kinetic analysis is obtained: |[S), mM 0.050 0.075 v, (µM min") 0.93 1.264 1.77 2.14 3.7 0.125 0.175 0.935 a. Using Excel, make a fully labeled Lineweaver-Burk plot and determine the Km and Vmax for the enzyme b. The reactions were set up dissolving 1 mg of the enzyme (MW= 100000 Da) in 100 ml of final reaction buffer. Determine the turnover number for the enzyme assuming 1 active site exists per enzyme molecule? c. Determine the catalytic efficiency for the enzyme d. The same reactions are performed in presence of an inhibitor A and the resulting velocities determined: v plus inhibitor, (µM min) 0.272 0.37 0.518 0.626 |1.08 |[S), mM 0.050 0.075 0.125 0.175 0.935 Plot these data on the same graph as above and determine the new Km and Vmax and the type of inhibitor (competitive, non-competitive). e. Can the effects of the inhibitor be over-ridden by adding more substrate? Why?