2 (10 marks)The KM and kcat of a carboxypeptidase isoform were found to be 2.00mM and 150s-1 respectively for substrate A.
2 (10 marks)The KM and kcat of a carboxypeptidase isoform were found to be 2.00mM and...
54. What is the catalyze reaction rate? Km=2 mM Kcat= 3 s-1 At the enzyme concentration=10 nM and substrate concentration= 3 mM
Glucokinase is a liver enzyme with a kcat of 33/second and a Km for Glucose of 7 mM. What is the initial reaction rate (in mM product / second) of 0.1 mM glucokinase in the presence of 50 mM glucose? Please show your work clearly.
Values for an Enzyme and a substrate are: Km=4 uM and kcat=20/min. For an experiment where [S]=6mM, it was found that Vo=480nM/min. What was the enzyme concentration? Give your answer in nM. Using the same kcat and Km as the previous question, if [Et]=0.5uM gives a Vo=5 uM/min, what wat the [S]? Give your answer in uM. reaction is run with the kcat=20/min and the Km=4uM. Use the enzyme concentration from question 1 above. A very strong inhibitor is added creating...
If KM increases to 4 times its original value in the presence of 2 mM of a competitive inhibitor what is the Ki for that inhibitor?
An enzyme that follows Michaelis-Menten kinetics has a KM value of 20.0 μM and a kcat value of 211 s−1. At an initial enzyme concentration of 0.0100 μM, the initial reaction velocity was found to be 1.07×10−6 μM/s. What was the initial concentration of the substrate, [S], used in the reaction ? Express your answer in micromolar to three significant figures.
CHEM3250 Assignment-Enzyme Inhibition Consider the data below for an enzyme catalyzed reaction. The rate of the reaction has been determined with and without an inhibitor. A total concentration of enzyme of 20 uM was used in the experiment. SHOW WORK AND UNITS!!! Without Inhibitor With Inhibitor [substrate] (mM)Rate of formation of te of formation of product product (mM/min) mM/min) 6.67 5.25 0.49 7.04 38.91 1.0 2.2 6.9 41.8 44.0 1.5 3.5 1 a) On the same graph, plot the data...
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)
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...
For her undergraduate project, Jessica studied an enzyme that catalyzes the reaction A B. For substrate A, she determined 30 min that Km 3.0 HM and kcat Jessica graduated and her project has been passed on to you. Unfortunately, Jessica was so busy that she sometimes forgot to record all of the details of an assay in her lab notebook. Your mentor suggests that you try to back calculate some of the missing concentration values. Assume that the enzyme follows...
HUUR UUU UUILUUITUL LUIUILINU 2 PV, 10 pul, posta 10. Applying the Michaelis-Menten Equation II An en- zyme catalyzes the reaction M = N. The enzyme is present at a concentration of 1 nm, and the Vmax is 2 um s-1. The Kn for substrate M is 4 um. (a) Calculate kcat. (b) What values of Vmax and Km would be observed in the presence of sufficient amounts of an uncompetitive inhibitor to generate an a' of 2.0?