The following data were obtained in a study of an enzyme that is known to follow the Michaelis-Menten kinetics. Approximately, what is the value of K for this enzyme?
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- 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?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 MLineweaver-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?
- 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=41 The following data describe an enzyme-catalyzed reaction (hydrolysis of carbobenzoxyglycyl-L-tryptophan) Plot these results using the Lineweaver-Burk method, and determine values for KM and Vmax Velocity (mM.sec-) 0 024 0 036 0 053 0 060 0 061 0 062 Substrate Concentration (mM) 25 50 10 0 15 0 200 25 0 42 If the KM of an enzyme for its substrate remains constant as the concentration of the inhibitor increases, what can be said about the mode of inhibition and why? 43 Calculate the turnover number for an enzyme, assumıng Vmax IS 05 M sec1 and the concentration of the enzyme used is 0 002 M Why is it useful to know this? 44 Dıscuss the mechanism of the Bohr effect that occurs during the interactions of Hb with oxygen under physiological conditions in the lungs and tissues Make use of relevant graphs and diagrams to explain your answerShown below are Km, and Vmax values obtained for an enzyme A which catalyze the transformation of the following substrates. Enzyme concentration used was 0.01 M. Km, mM 0.02 Vmax, mM/min 5.3 Substrate 1 2 1.5 13.7 3 2.6 100 4 0.1 25 0.05 62 1. Which substrate have the highest affinity for the enzyme? Explain. 2. Which will show higher efficiency of converting the substrate to the product? Show solutions and еxplain.
- 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 SubmitConsider 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?The initial rates of an enzyme-catalyzed reaction have been determined for five different [S] (see table below). [S]o (mol.L-¹) 1.0 x 10 1.5 x 10-4 2.0 x 10-4 -4 5.0 x 10 7.5 x 10-4 -1 Vo (μmol.L-¹ min-¹) 28 35 42 63 75 1- Determine graphically the value of Km (first with the Michaelis-Menten representation and then with the Lineweaver-Burk representation, then compare by reporting on the Michaelis-Menten representation the values found by the Lineweaver-Burk representation) 2- Calculate the concentration of a competitive inhibitor 1, with a K₁ value of 2.4.104 M, whose action would quadruple the apparent value of Km.
- 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.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?126 nmol of enzyme catalyzes the conversion of 2.8 ugram of substrate (molar mass- 350 g/mol) in t second at 37C at Vman what is the turnover number kcat for the enzyme in units of s R Show all your work to recenve credit. For the toolbar, press ALTF10 (P) or ALTFNF10 (Mac). %3B .... BIUS Paragraph Arial 14px