V, (μmol-min-¹) 0.35 1/V₂ 0.3 0.25 0.2 0.15 0.1 0.05 0 0 50 100 150 200 250 300 350 400 450 [S] (nm) The enzyme kinetics data were also used to create a double-reciprocal plot of - versus Vo [S] 12 10 8 6
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- An enzyme that follows Michaelis-Menten kinetics has a KM value of 3.00 µM and a keat value of 181 s1. At an initial enzyme concentration of 0.0100 µM, the initial reaction velocity was found to be 1.07 x 10-0 µM/s. What was the initial concentration of the substrate, S, used in the reaction ? Express your answer in micromolar to three significant figures. > View Available Hint(s) ? [S] !! µM SubmitAn 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 :You 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 μM
- You are working on an enzyme that obeys standard Michaelis-Menten kinetics. You have determined the Vmax to be 0.1 mol/sec and the Km to be 2.5 mM. What would the rate of the reaction be when the substrate concentration is 20 mM? 0.09 MS-1 O 0.133 Ms-1 O 0.18 Ms ¹ 9 Ms-1 O 0.018 Ms-1 0.2 MS-1In your acid phosphatase enzyme kinetics lab, you constructed a Lineweaver-Burk plot. Lefs assume that you graphicaly obtain 1/Vmax - 9.33 (umoliey and -1Km--0.012 M. Which of the following is correct? A) Vmax = 9.33 (no units) B) Km = 0.012 (no units) C) Vmax = (1/Vru" = 0.107 umolis D) Km =-14-1K = 83.3 pM E) Both C) and D) are correctYou 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=
- An enzyme catalyzes a reaction with a K of 7.50 mM and a Vmax of 4.15 mMs. Calculate the reaction velocity, o, for each substrate concentration. [S] = 1.75 mM MM-s-1 [S] = 7.50 mM [S] = 11.0 mM DO mM-s mM-sQ is an analog of substrate A that binds to enzyme X and produces the following kinetics:[A] V0 (µmole/ml/min) [Q] = 0 [Q] = 0.5 µM [Q] = 2 µM1 µM 10 7 43 µM 20 16 1010 µM 35 31 2330 µM 43 41 3680 µM 47 46 43a) Plot the data in Lineweaver Burk plot form (Hint: there should be three plots of dataon your graph) and determine the following: the KM and Vmax of the enzyme X in theabsence of Q, and the KMapp the Vmaxapp at each concentration of Q. b) What type of inhibition is this?c) Calculate the inhibition constant (Ki) for Compound QThe following data were obtained for a reaction that catalyzed by an enzyme: Initial concentration of 0.141 0.109 0.077 0.040 0.028 0.020 0.016 the enzyme [S], mol L- Relative velocity, V 22.0 20.5 19.0 12.5 9.0 7.0 6.0 For this reaction, the rate is found experimentally to follow the Michaelis-Menten equation: Vmax[S] Vo = Км+[S] (1) Transform the above equation into a straight-line equation. By plotting a suitable graph, determine the Michaelis constant (KM) and the maximal velocity (Vmax) for the above reaction. Determine the Vmax value when [S] = KM.
- L(24 points) Explain How is the Michaelis constant defined. and what does a low or high value for Km tell you? What is the difference between the velocity and initial velocity of an enzyme reaction? What determines the efficiency of an enzyme reaction, and what terms are used to describe it? 2. (50 points) About how to obtain kinetic data experimentally Lisa decides to obtain values for the Km and Vmax of an enzyme she has just isolated from liver cells (it is now pure), using a Michaelis Menten plot. Describe in detail what kinds of measurements she would have to make, and what she would need to plot on graphs in order to estimnte the values for Km and Vmax. (Show the kinds of graphs she would have to plot, and how these will allow her to estimate Km and Vmax.) Describe how she would be able to obtain Vmax experimentally and from the Michaelis Menten plot - what conditions are needed and what would be measured). Also, describe what she would have to do to obtain the turnover number of…An enzyme catalyzes a reaction at a velocity of 20 micromole/min when the concentration of substrate (S) is 0.01 M. The Km for this 1 X 10^-5 M. Assuming that Michaelis-Menten Kinetics are followed, what will the reaction velocity be when the concentration of S is 1.0 X 10^-6 M?During a test of kinetics of an enzyme-catalyzed reaction, the following data were recorded: E, () T(C) I (mmol/ml) S (mmol/ml) V (mmol/ml-min) 1.6 1.6 30 0.1 2.63 1.92 147 30 0.033 1.6 30 0.02 0.01 0.005 16 30 0.96 0.56 5.13 1.6 30 1.6 49.6 0.1 1.6 49.6 0.033 3.70 1.89 1.6 1.6 1,6 49.6 0.01 0.0067 49.6 1.43 49.6 0.005 LII 0.92 30 0.1 1.64 0.02 0.01 0.1 0.90 0.92 0.92 30 30 0. 0.58 0.92 30 0.6 0.92 0.92 30 30 0.6 0.6 0.033 0.02 1.33 0.80 0.57 a. Determine the Michaelis-Mentcn constant for the reaction with no inhibitor present at 30°C and at 49.6°C. b. Determine the maximum velocity of the uninhibited reaction at 30°C and an enzyme concentration of 1.6 g/l. c. Determine the KI for the inhibitor at 30*C and decide what type of inhibitor is being used.