1. What is the keat of this enzyme? 2. At which concentration of the substrate this enzyme reaches initial reaction velocity (Vo) of 500 nMis (assuming (otal is the same 10 nMy? 3. What is the maximum reaction velocity (Vmax) of the reaction catalyzed by this enzyme at (Etotail4 nM? of the items from the answer list should NOT be used.
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- Velocity (mmol/minute) [S], (mM) No inhibitor Inhibitor 3 10.4 4.1 5 14.5 6.4 10 22.5 11.3 30 33.8 22.6 90 40.5 33.8 The kinetics of an enzyme are measured as a function of substrate in the presence and the in absence of 2mM inhibitor (I). What are the values of Vmax and KM in the absence of inhibitor? In its presence? In its presence? What is the type of inhibition?Question: Determine the Km and Vmax for this enzyme/substrate combination. [Substrate] (mM) V0 (mM/min) 0.25 0.183 0.50 0.356 1.00 0.665 2.50 1.45 5.00 2.35 What is the concentration of substrate necessary to achieve a turnover rate of 1.00 mM/min?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 M
- Initial rate data for an enzyme that obeys Michaelis–Menten kinetics areshown in the following table. When the enzyme concentration is 3 nmolml-1, a Lineweaver–Burk plot of this data gives a line with a y-intercept of0.00426 (μmol-1 ml s). (a) Calculate kcat for the reaction.(b) Calculate KM for the enzyme.(c) When the reactions in part (b) are repeated in the presence of 12 μM ofan uncompetitive inhibitor, the y-intercept of the Lineweaver–Burk plotis 0.352 (μmol-1 ml s). Calculate K′I for this inhibitor.The Michaelis‑Menten equation models the hyperbolic relationship between [S] and the initial reaction rate ?0V0 for an enzyme‑catalyzed, single‑substrate reaction E+S↽−−⇀ES⟶E+PE+S↽−−⇀ES⟶E+P. The model can be more readily understood when comparing three conditions: [S]<<?m[S]<<Km, [S]=?m[S]=Km, and [S]>>?m[S]>>Km. Match each statement with the condition that it describes. Note that "rate" refers to initial velocity ?0V0 where steady state conditions are assumed. [Etotal][Etotal] refers to the total enzyme concentration and [Efree][Efree] refers to the concentration of free enzyme.List 4 major types of inhibition modes and clearly indicate the effect on Vmax and KM for each mode?2. What is the effect of each of the 4 types of inhibitors on the initial rate of an enzyme catalyzed reaction?3. A potent inhibitor effectively inhibits an enzyme catalyzed reaction. What kind of a Ki value you would expect for a potent inhibitor?4. Considering PNPP → PNP reaction, would you expect to see more intense or pale color for the reaction that contain the inhibitor? Explain.
- Question- For every 1 TAG metabolized, how many water molecules would be produced by the electron transport chain? (show calculations)An uncatalyzed reaction has keq=50. in the presence of an appropriate enzyme.the forward rate of the reaction increased by 20-fold.what is the equilibrium constant in the presence of the enzyme?Question:- For a simple enzymatic reaction that involves only one substrate and follows Michaelis–Menten kinetics, the changes in the concentrations of substrate, product, free enzyme, and the enzyme–substrate complex over the course of the reaction are depicted by solid curves in the graph below. Which curve (1 to 4) corresponds to [ES]? Write down the number as your answer.
- If an enzyme catalyzed reaction has a KM of 5mM and a Vmax of 60 nm/sec, the substrate concentration at 30 nM/sec is? Thank you.MATHEMATICAL You do an enzyme kinetic experiment and calculate a Vmax of 100 mol of product per minute. If each assay used 0.1 mL of an enzyme solution that had a concentration of 0.2 mg/mL, what would be the turnover number if the enzyme had a molecular weight of 128,000 g/mol?MATHEMATICAL Using the data in Table 15.1, calculate the value of G for the following reaction. Creatinephosphate+GlycerolCreatine+Glycerol-3-phosphate Hint: This reaction proceeds in stages. ATP is formed in the first step, and the phosphate group is transferred from ATP to glycerol in the second step.