Calculate kcat for an enzyme where Vmax is 12 U s. mL and the total enzyme concentration is 0.016 U mL1 -1 750 s-1 150 s-1 0.19 s1 0.001 s-1
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Q: [S, mmol L-') Vo (umol L-' min-') 0.5 50 1.0 2.0 3.0 10.0 8||||
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A: Introduction Catalytic constant, also known as the turnover number can be given as kcat=v[Et]where v…
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- For an enzyme that displays Michaelis-Menton kinetics, what is thereaction velocity, V (as a percentage of V max , observed at the followingvalues?[S] = K M[S] = 0.5K M[S] = 0.1K M[S] = 2K M[S] = 10K MGiven the following information, calculate the catalytic efficiency of the enzyme. Step by step please [S] = 100 mM k1 = 10 sec-1 k2 = 3000 sec-1 k-1 = 20 sec-1 [E]T = 1 \muμMcalculate 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 \muμM
- From a kinetics experiment, the Vmax was determined to be 450µM∙min-1. For the kinetic assay, 0.1mL of a 0.05mg/mL solution of enzyme was used, and the enzyme has a molecular weight of 125,000 g/mole. Assume a reaction volume of 700µL. Calculate the kcat (in sec-1) for the enzyme.Substrate Concentration (mol L1) Velocity (mM min-1) 2.500 0.588 1.000 0.500 0.714 0.417 0.526 0.370 0.250 0.256 Determine the values of Km and Vmax for the decarboxylation of a 훃-keto acid given the followingdata. You have to plot the graph by using excel and please include the scope of graphFrom your Lineweaver-Burk plot,the vlaues are: Km Vmax Uninhibited 0.09 mmol/L 3.02 min/mmol Inhibited 6.22 mmol/L 9.98 min/mmol By describing the potential changes in the kinetic parameters, identify and justify the type of inhibitor that was inhibiting the acid phosphatase activity.
- The maximum saturation, 28.3ug of enzyme in 25ml water catalyzes the oxidation of ethanol at a rate of 2.5mm/min. Calculate the kcat in units of seconds if the enzyme has a molar mass of 65kg/mol.The turnover number for an enzyme that approximates Michaelis-Menten kinetics is known to be 500 min^-1. From the results shown in the table, enumerate Km and total amount of enzyme present. What is the Km for this enzyme? What is the Vmax for this enzyme? And what is the [E]T for this enzyme?Lineweaver-Burk plots of enzyme kinetics for the reaction, S <-> P, has the following features: 1/v is zero when 1/[S] equals -40 liter mole^-1; 1/[S] is zero when 1/v equals 2.0 x 10^5 min mole^-1. What are the Vmax and Km? Vmax = 5 umol min^-1, Km = 2.5 mM? Vmax = 5 mmol min^-1, Km = 25 M? Vmax = 5 umol min^-1, Km = 25 mM? Vmax = 5 mol min^-1, Km = 2.5 mM? Vmax = 5 mol min^-1, Km = 25 mM?
- Use the Michaelis-Menten equation to complete the enzyme kinetic data set, whenKm is known to have a value of 1 mmol L−1 [S] (mmol L−1) Vo (μmol L−1 min−1) 0.5 50 1.0 2.0 3.0 10.0Lineweaver-Burk plots of enzyme kinetics for the reaction, S <-> P, has the following features: 1/v is zero when 1/[S] equals -40 liter mole^-1; 1/[S] is zero when 1/v equals 2.0 x 10^5 min mole^-1. What are the Vmax and Km?An enzyme catalysed reaction has a Km of 8 mM and a Vmax of 13 nM.s-1. Use the Michaelis-Menten equation to calculate the reaction velocity when the substrate concentration is 18 mM.