Question2 An enzyme solution has a Vmax of 10 μΜ/sec and a Km of 10 μΜ....
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
a. An enzyme has a Vmax of 100 umol/min and a Km of 40 uM. When substrate concentration is 40 uM what is the initial reaction rate? b. An enzyme with a Vmax of 100 umol/minute and a Km of 10 uM was reacted with a irreversible active site specific inhibitor. After reaction with the inhibitor, the enzyme was assayed using a 2 mM concentration of substrate, and it gave a reaction rate of 20 umol/min. What percentage of the...
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
41. You want to determine the initial velocity (Vo) of an enzyme, which has a Vmax of 0.3??/s and you are using an enzyme concentration of 50mM. The concentration of substrate in the reaction is 8 times greater than the Km. Which value below is the correct initial velocity A. 0.09 HM/s B. 0.27 ??/s C. 0.04 HM/s D. 1??/s E. 0.7 ??/s
The Michaelis-Menten equation is often used to describe the kinetic characteristics of an enzyme-catalyzed reaction. S Where v is the velocity or rate, Vmax is the maximum velocity, Km is the +IST Michaelis- Menten constant, and I5 s the substrate concentration. K + S v (uM/min) a) A graph of the Michaelis-Menten equation is a plot of a reaction's initial velocity (Vo) at different substrate concentrations ([S]) 300 Vmax 250 1/2 Vmax First, move the line labeled "Vmax to a...
An enzyme has a Km for substrate of 10 mM and Vmax of 5 mol L-1 sec-1 at a total enzyme concentration of 1 nM. At [S] = 10 mM, kcat is: A) 2500 per M per sec. B) 5000 per M per sec. C) 1250 per M per sec. D) 2500 per sec. E) 5000 per sec.
An enzyme has a maximum velocity of 48 nmoles substrate converted per sec. In the presence of 4 M inhibitor, Vmax is 36 nmoles substrate converted per sec with no change in Km. Determine the Ki of the inhibitor. Identify the type of inhibitor with explanation.
1. An enzyme has a km of 4.7 X 105M. If the Vmax of the preparation is 224M/min, what velocity would be observed in the presence of 2 X 10^M substrate and 5 X10+M a competitive inhibitor. Ki is 3 X 10^M. What is the degree of inhibition? (10pts)
Determine the kinetic parameters, Km & Vmax and calculate
k2.
Penicillin is hydrolyzed and thereby rendered inactive by penicillinase, an enzyme present in some penicillin-resistant bacteria. The mass of this enzyme is 29.5 kD. The amount of penicillin hydrolyzed in 1 minute in a 10 mL solution containing 100 g of purified penicillinase was measured as a function of the concentration of penicillin. Assume that the concentration of penicillin does not change appreciably during the assay. (a) Plot v versus....
The value of Km for the shown data for a hexokinase-catalyzed reaction is with the unit of . The value of Vmax for. the same reaction is with the unit of . Be sure to give the values with the correct number of significant figures. You might have to construct a kinetic plot. For units, choose one answer from (uM, 1/ UM, HM/second, uM x second, mM, 1/mM, second, 1/second, mM/second, mM x second) vo (mM/sec) Glucose concentration (mm) 0.10...