The figure below shows the dependence of the enzyme's rate, v (in uM/min), as a function of substrate concentration, S (in mM). Also shown is t dependence of the rate in the presence of an inhibitor, present at a concentration of 0.2 mM. Based on this information, the Vmax of the enzyme is approximately: 1/v 1- 0.5 1/[S] O A. 0.25 pM/min O B. 0.5 pM/min O C. 1.0 pM/min O D. 2 uM/min
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- An enzyme with a single active site has a Michaelis-Mentin constant (KM ) of 25 mM and aturnover number of 4.0 x 107 s –1 .(a) Calculate the initial rate of the reaction in μM/s if the total enzyme concentration is0.016 μM and the initial substrate concentration is 4.32 μM.(b) Calculate the value of RmaxThe equilibrium constant for the attachment of a substrate to the active site of an enzyme was measured as 200. In a separate experiment, the rate constant for the secondorder attachment was found to be 1.5 x 108 dm3 mol-1 s- 1. What is the rate constant for the loss of the unreacted substratefrom the active site?A An enzyme that follows Michaelis-Menten kinetics has a KM value of 16.0 uM and a kcat value of 181 s-1. At an initial enzyme concentration of 0.0100 uM, the initial reaction velocity was found to be 1.07 x 10- uM/s. What was the initial concentration of the substrate, [S], used in the reaction ? Express your answer in micromolar to three significant figures
- Relation between Reaction Velocity and Substrate Concentration: Michaelis-Menten Equation. a)At what substrate concentration would an enzyme with a kcatof 30.0 s-1and a Kmof 0.0050 M operate at one-quarter of its maximum rate? b)Determine the fraction of Vmax that would be obtained at the following substrate concentrations [S]:1/2Km,2Km,and10Km. c)An enzyme that catalyzes the reaction X ßàY is isolated from two bacterial species. The enzymes have the same Vmax, but different Km values for the substrate X. Enzyme A has a Km of 2.0 μM, while enzyme B has a Km of 0.5 μM. The plot below shows the kinetics of reactions carried out with the same concentration of each enzyme and with [X]=1 μM. Which curve corresponds to which enzyme?An enzyme-catalyzed reaction has a Km of 1.4 mM and a Vmax of 7 nM/s. What is the initial velocity when the substrate concentration is 0.6 mM?An enzyme catalyzes a reaction with a KmKm of 7.507.50 mM and a VmaxVmax of 3.00 mM⋅s−1.3.00 mM⋅s−1. Calculate the reaction velocity, v0,v0, for each substrate concentration. a. [S]=4.00mM*s^-1 b. [S]=7.5mM*s^-1
- The kinetics of an enzyme are measured as a function of [S] in the presence and absence of 2 mM I. Compute Km and Vmax in the absence and presence of I S] (uM) V (uM/min) without I with I 3 10.4 4.1 5 14.5 6.4 10 22.5 11.3 30 33.8 22.6 90 40.5 33.8An enzyme-catalyzed reaction has a Km of 1 mM and a Vmax of 4 nmole/L sec-1. The reaction velocity (nmole/L sec-1) when the substrate concentration is 0.25 mM is: A. 1.25 B. 10.0 C. 5.0 D. 0.50 E. 1.0 F. 100 G. 150 H. 55 I. 75A penicillinase inactivates a penicillin. An enzyme with Mw = 30,000 has only a single active site, kcat = 2,000 s-1 and KM = 5.10-5 M. In response to a treatment with 5 μmol of penicillin, a 1 mL suspension of bacteria releases 0, 5 µg of enzyme. A) What concentration of penicillin would cause the enzyme to react at half the maximum rate? B) A modified penicillin acts as a competitive inhibitor. If the affinity of E for penicillin and for modified penicillin are the same, what concentration of inhibitor reduces the rate of penicillin loss 5-fold with low Cs in the medium?
- A protein catalyzes the following reaction and had a Michaelis-Menten constant of Km=25 mM and a turnover number of 4.0 x 107 s-1. The total enzyme concentration is 0.012 µM and the initial substrate concentration is 4.76 µM. The enzyme has a single active site. Calculate the values of Vmax and V0 for this enzyme. 2H2O2 (aq) -----> 2H2O(l) + O2(g) Vmax= Vo=The following table indicates the rates at which a substrate reacts as catalyzed by an enzyme that follows the Michaelis-Menten mechanism: (1) in the absence of inhibitor; (2) and (3) in the presence of 10 mM concentration, respectively, of either of two inhibitors. Assume [E]T is the same for all reactions.(a) Determine KM and Vmax for the enzyme. For each inhibitor determine the type of inhibition and KI and/or K’I. What additional information would be required to calculate the turnover number of the enzyme? (b) For [S] = 5 mM, what fraction of the enzyme molecules have a bound substrate in the absence of inhibitor, in the presence of 10 mM inhibitor of type (2), and in the presence of 10 mM inhibitor of type (3)?An enzyme has a Km equal to 1.0 x 10-4 M (molar) for a certain substrate S. When [S] = 1.0 x 10-4 M, 2% of the substrate is transformed into product in 1 minute. Which of the following statements are correct? Hint: use the Michaelis-Menten equation; Remember that a reaction rate can be expressed in molar units / min (M / min). Select one or more than one: a) When [S] = 1.0 x 10-4 M, the initial reaction rate, V0, is 2.0 x 10-7 M / min b) For this enzyme Vmax = 4.0 x 10-7 M / min c) For this enzyme Vmax is equal to 2.002 x 10 -7 M / min. d) When [S] 0 = 1.0 x 10-3 M, the initial speed is 3.64 x 10-7 M / min. e) When [S] = 0.01 M, the initial speed is 4.0 x 10-3 M / min