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- Catalane in un enzyme that speeds up the decomposition ot hydrogen peroxide H0) to waler and axvgen. Students conducted two investigationa to determine the ideal conditions for the function of catalane. One wetigation compared catalase activity at diferent values of pH. The other imvestigation compared catalane activity al diferent temperatures. Enayme Activity ve pH Enzyme Aetivity vs Temperature Teate C) According to the data in the graphs, which pH and temperature combination provides the BEST conditions for catalase to tunction? Anury lag acalculate the reaction velocity at saturating substrate concentrations. Your numerical answer is assumed to be in units of M sec-1. [S] = 100 mM k1 = 10 sec-1 k2 = 3000 sec-1 k-1 = 20 sec-1 [E]T = 1 \muμMAn enzyme following Michaelis-Menten kinetics was found to have highest activity at 37° C and pH 7. If maximum velocity for the enzyme was 2.4 x 10 mol. Lits with initial enzyme concentration of 2.4 nmol, then its turnover frequency will be O 10°mol/s 10 mol/s 10-2mol/s O None of these
- I need help with parts d-g a.) For Ado: catalytic efficiency = 4.18 x 106 M-1s-1 For ATP: catalytic efficiency = 12.1 x 107 M-1s-1 b.) Both specificity constants are below the diffusion limit so neither substrate is limiting the reaction rate due to diffusion c.) Line weaver Burk equation for AdoK enzyme with Ado substrate: y = 0.68x + (2.04 x 104), where y = 1/vo and x = 1/[Ado]Carbonic anhydrase catalyzes the hydration of CO. CO2 + H2O ¬ H½CO3 The Km of this enzyme for CO, is 1.20×104 µ.M. When [CO,] = 3.60×104 µM, the rate of reaction was 4.50 umol·mL! sec-1 a What is Vmax for this enzyme? umol·mL-!sec-!when saturated with substrate, an enzyme has a maximum initial rate of 110mumoles of substrate converted to product per second. At a substrate concentration of 100mu M, the same enzyme converts substrate to product at a rate of 0.010mmoles/ sec. Assuming that Michaelis - Menten kinetics are followed, calculate the reaction rate when substrate concentration is 2x10^-3M.
- Calculate the Km of the enzyme with these parameters. kcat = 130s^-1 Vo = 3.0 μMs-1 S = 10 μM Et = 0.09 µMIndicate the hinge region, GK kinaseselectivity pocket and explain "Solvent AccessibilitySurface" and indicate the salt forming group at ATPbinding site13. Calculate the equilibrium constant K'eg, for each of the following reactions at pH 7.0 and 25°C. glucose + Pi a. Glucose 6-phosphate + H20 enz. Glucose 6-phosphatase; AG'O=-13.8kJ/mol
- Qis an analog of substrate A that binds to enzyme X and produces the following kinetics: [A] Vo (umole/ml/min) [Q] = 0 [Q] = 0.5 pM [Q] = 2 µM 1 µM 10 7 4 3 μΜ 20 16 10 10 µM 35 31 23 30 μΜ 43 41 36 80 µM 47 46 43 a) Plot the data in Lineweaver Burk plot form (Hint: there should be three plots of data on your graph) and determine the following: the KM and Vmax of the enzyme X in tbo absence of Q, and the KMapp the Vmaxapp at each concentration of Q. Vmax b) What type of inhibition is this? c) Calculate the inhibition constant (K) for Compound QWhich of the following statements regarding Michaelis-Menten kinetics is TRUE? The ES complex can move toward product or toward substrate. As Vmax is approached, KM approaches 1. Vmax is reached when [S] is low. When Vmax is reached, the concentration of the E is high.1. Explain why the maximum initial reaction rate cannot be reached at low substrate concentrationsPredict or describe the absorbance or enzyme activity at: pH = 2 pH = 14 Explain your predictions 2. Explain why the maximum initial reaction rate cannot be reached at low substrate concentrations.