You are working on an enzyme that obeys standard Michaelis-Menten kinetics. Based on the following reaction expression, what is the Km value for this enzyme? E+SESE + P . . . K₁ = 880.8 M-¹5-1 k.₁ = 42.18 S-1 k₂ = 56.29 S-1
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- Consider an enzyme that follows standard Michaelis-Menten kinetics and has the following kinetic constants: %3| k2 = 1.5 x 10? s1 Еo 3D 1 х 104 М = 1 x 104 M a. What is the value of the maximum rate VM? b. Prepare a hand-drawn quantitative plot on graph paper (not a simple sketch nor an EXCEL-generated graph) of the enzymatic reaction rate versus substrate concentration (like Fig 3-3) using the kinetic parameters given above. Be sure to label the axes and include numeric values on the axes. C. Based upon your hand drawn saturation plot, at what substrate concentration is the enzymatic reaction rate 75% of Vm?You have obtained experimental kinetic data for two versions of the same enzyme, a wild‑type and a mutant differing from the wild‑type at a single amino acid. The data are given in the table. ?maxVmax(μmol min−1) ?MKM(mM) Wild‑type 100 10 Mutant 1 0.1 Compare the kinetic parameters of the two versions using the data in the table. Assuming a two-step reaction scheme in which ?−1k−1 is much larger than ?2,k2, which of the following statements are correct? The wild‑type version requires a greater concentration of substrate to achieve ?maxVmax. The wild‑type version has a higher affinity for the substrate. The mutant version has a higher affinity for the substrate. The mutant version requires a greater concentration of substrate to achieve ?maxVmax. Calculate the initial velocity of the reaction catalyzed by the wild‑type enzyme when the substrate concentration is 10 mM. ?0=V0=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
- 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?An enzyme has a V of 1.2 uM s The Km for its substrate is 10 µM. max Calculate the initial reaction velocity, Vo, for each substrate concentration, [S]. Calculate Vo when [S] is 2 µM. Vo = µM s Calculate Vo when [S] is 10 µM. Vo = µM s-1 Calculate Vo when [S] is 30 µM. Vo = µM s-1The Lineweaver-Burke plot was originally developed in order to "linearize" the data obtained from enzyme kinetics experiments, in order to facilitate the determination of kinetic parameters. Why is it not considered to be an accurate method for this purpose? It is very difficult to draw a straight line on a computer. It is very difficult to calculate the variables required for the "x" and "y" axis. It is more accurate to use the standard "V versus [S]" plot to determine Vmax and KM- The plot weights the least accurate data points the most heavily. It is no longer considered to be acceptable to extrapolate from known data.
- The rate constants of an enzyme-catalyzed reaction, obeying the Michaelis-Menten kinetics, have been determined : E + S K₁ = 2 x 108 M-¹ S-¹ -1 -1 -1 K-₁= 1 x 10³ S K₂ = 5 x 10³ S-1 K₁ 1 K-1 ES K₂ E +P 1- Determine the Michaelis constant Km of the enzyme. 2- Determine the catalytic constant (kcat) of the enzyme. 3- Determine the catalytic efficiency of the enzyme.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.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?
- 1. A Lineweaver-Burk Plot is shown below. 30 25 Curve A y = 3.1207x + 2.4978 20 15 Curve B y = 1.0003x + 2.3602 10 5 -3 1 5 7 11 1/[Catechol] (mM1) With these curves, determine the following enzyme parameters. Show all pertinent solutions. a. Km of Curve A and Curve B b. Vmax of Curve A and Curve B c. Assuming that one of these curves corresponds to the kinetics of one enzyme and one substrate, which curve represents the effect of an inhibitor? Why do you say so? d. What type of inhibition is exhibited by your answer in question c? Why do you say so? 1/V, (units of activity 1)If a michaelis-type enzyme has an apparent Km of 2*10^-8 and apparent Vmax of 1*10^-4 moles/min in the presence of 1*10^-3 M uncompetitive inhibitor (Ki=1*10^-7 M), what is the true Km of the enzymeIf you want to determine the KM for lactate, what protocol do you set up? Discuss the significance of the following kinetic parameters that are used to characterize enzyme activity: KM, Vmax, kcat, and kcat / KM.