An enzyme catalyzes a reaction with a Km of 9.00 mM and a Vmax of 3.95 mM·s–1. Calculate the reaction velocity, v0, for the following substrate concentrations:
A. 1 mM
B. 9 mM
C. 11 mM
An enzyme catalyzes a reaction with a Km of 9.00 mM and a Vmax of 3.95...
An enzyme catalyzes a reaction with a Km of 5.00 mM and a Vmax of 1.85 mM s-1. Calculate the reaction velocity, vo. for the following substrate concentrations. a) 3.00 mM Number mM.s b) 5.00 mM Number mM.s c) 10.5 mM Number
An enzyme catalyzes a reaction with a Km of 8.00 mM and a Vmax of 4.45 mM s-1. Calculate the reaction velocity, Vo, for the following substrate concentrations a) 1.00 mM Number mM.s-1 b) 8.00 mM Number mM s-1 c) 11.0 mM Number mM.s-1
An enzyme catalyzes a reaction with a Km of 9.50 and a Vmax of 1.75 An enzyme catalyzes a reaction with a K_m of 9.50 mM and a of 1.75 mM- s^-1. Calculate the reaction velocity, V_0, for the following substrate concentrations. 2.50 mM 9.50 mM 12.0 mM
An enzyme catalyzes a reaction with a Km of 8.50 mM and a Vmax of 2.70 mM 5-1. Calculate the reaction velocity, vo, for each substrate concentration. [S] = 2.25 mM Vo: 0.5 mM.s-1 [S] = 8.50 mM 1.70 mMs-1 [S] = 13.0 mM [S] = 13.0 mM Vo: 2.05 mm. s-1
An enzyme catalyzes a reaction with a K of 6.50 mM and a Vmax of 2.75 mM s velocity, vo. for the following substrate concentrations a) 2.50 mM Calculate the reaction Number mM. sl b) 6.50 mM Number c) 14.0 mM Number mM.sl
Score: 3318/3/00 15 of 37 > An enzyme catalyzes a reaction with a km of 8.50 mM and a Vmax of 2.45 mM.s. Calculate the reaction velocity, to, for each substrate concentration. S = 3.75 mm mMs-1 S = 8.50 mM mM.s-1 [S] = 10.5 mM about us Careers Privacy policy terms of use contact us help
An enzyme catalyzes a reaction with a K m of 9.00 mM and a V max of 2.40 mM ⋅ s − 1 . Calculate the reaction velocity, v 0 , for each substrate concentration. [ S ] = 3.50 mM v 0 : mM ⋅ s − 1 [ S ] = 9.00 mM v 0 : mM ⋅ s − 1 [ S ] = 12.0 mM v 0 : mM ⋅ s − 1
An enzyme catalyzes the reaction M↽−−⇀N . An enzyme catalyzes the reaction M = N. The enzyme is present at a concentration of 3.5 nM, and the Vmax is 1.9 uM -. The Km for substrate M is 2.9 uM. Calculate kcat kcat = 542.86 What values of Vmax and Km would be observed in the presence of sufficient amounts of an uncompetitive inhibitor to generate an a' of 1.5? apparent Vmax = 0.526 UM s-1 apparent Km = 1.2
An enzyme catalyzes the reaction M N. The enzyme is present at a concentration of 3.5 nM, and the Vmax is 2.9 PM s-. The Km for substrate M is 6.5 MM. Calculate kcat kcat = 1 What values of Vmax and Km would be observed in the presence of sufficient amounts of an uncompetitive inhibitor to generate an a' of 1.3? apparent Vmax = UM S-1 apparent Km = UM
9. Applying the Michaelis-Menten Equation I An enzyme catalyzes the reaction A = B. The enzyme is present at a con- centration of 2 nm, and the Vmax is 1.2 ums". The Km for substrate A is 10 um. Calculate the initial velocity of the reaction, Vo, when the substrate concentration is (a) 2 um, (b) 10 um, (C) 30 um.