Values for an Enzyme and a substrate are: Km=4 uM and kcat=20/min. For an experiment where...
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 an alpha value of 10, which then reduces the Vo down to 240 nM/min. What was the substrate concentration? Give your answer in u .
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Values for an Enzyme and a substrate are: Km=4 uM and
kcat=20/min. For an experiment where...
D-Lactose is the substrate for B-galactosidase. Given Vo = kcat [Et] [S]/km + [S], calculate [S],...
D-Lactose is the substrate for B-galactosidase. Given Vo = kcat [Et] [S]/km + [S], calculate [S], when Km = 4.0 nM, V. = 10.5 M s', kcat = 500 s, and [Et] = 40 uM Calculate the catalytic efficiency. Below is a double-reciprocal plot for an enzyme reaction in the absence and presence of of inhibitor. Give the equation for the line. Calculate Vmax and Km for the enzyme and enzyme plus inhibitor. Which type of inhibition is apparent. 0.10...
Applying the Michaelis-Menten Equation II Another enzyme is found that catalyzes the reaction A <===> B...
Applying the Michaelis-Menten Equation II Another enzyme is found that catalyzes the reaction A <===> B Researchers find that the K for the substrate A is 4 uM, and the kcat is 20 min^-1 (a) In an experiment, [A] = 6 mM, and the initial velocity, Vo was 480 nM min^-1 What was the [Et] used in the experiment? (b) In another experiment, [Et] 0.5uM, and the measured Vo=5uM min^-1 What was the [A] used in the experiment? ( c)...
a. An enzyme has a Vmax of 100 umol/min and a Km of 40 uM. When...
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...
A different experiment yields the following kinetic data Substrate] (mM)Vo (uM/min) no inhibitorVo (uM/min)+7 nM inhibitor...
A different experiment yields the following kinetic data Substrate] (mM)Vo (uM/min) no inhibitorVo (uM/min)+7 nM inhibitor 0.02 0.04 0.10 0.25 1.00 2.50 0.8 2.9 8.6 24 36 50 Plot the data for the kinetics of the enzyme (with and without the inhibitor) in a double reciprocal (Lineweaver-Burk) plot. Keep in mind that the x axis is 1/[S] and the y axis is 1/Vo. If you are using Excel you want to choose the (x,y) scatter plot (without a ne). You...
For her undergraduate project, Jessica studied an enzyme that catalyzes the reaction A B. For substrate...
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...
An enzyme catalyzes the reaction M↽−−⇀N . An enzyme catalyzes the reaction M = N. The...
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...
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
Given the enzyme system just described (Km = 8.20*10-5M and kcat = 40.0 s-1, [Eltot =...
Given the enzyme system just described (Km = 8.20*10-5M and kcat = 40.0 s-1, [Eltot = 2.00*10-7 M), suppose we introduce a classical noncompetitive inhibitor with K1 = 1.20*10-7M. If the concentration of inhibitor introduced is 3.0 um, what will the effective kcat be? Express your answer in s-1.
For her undergraduate project, Jessica studied an enzyme that catalyzes the reaction A↽−−⇀B. For substrate A,...
For her undergraduate project, Jessica studied an enzyme that catalyzes the reaction A↽−−⇀B. For substrate A, she determined that ?m=2.5 μM and ?cat=35 min−1. 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 Michaelis–Menten kinetics. 1)...
A compound is a strong competitive inhibitor with an a = 10. The total enzyme concentration...
A compound is a strong competitive inhibitor with an a = 10. The total enzyme concentration in the reaction is 10 nM. What substrate concentration reduces the velocity to 250 nM/min, knowing that this enzyme is characterized by a specificity constant of 2.1 x 105 M-1s-1 and a kcat of 25/min?
D-Lactose is the substrate for B-galactosidase. Given Vo = kcat [Et] [S]/km + [S], calculate [S], when Km = 4.0 nM, V. = 10.5 M s', kcat = 500 s, and [Et] = 40 uM Calculate the catalytic efficiency. Below is a double-reciprocal plot for an enzyme reaction in the absence and presence of of inhibitor. Give the equation for the line. Calculate Vmax and Km for the enzyme and enzyme plus inhibitor. Which type of inhibition is apparent. 0.10...
A different experiment yields the following kinetic data Substrate] (mM)Vo (uM/min) no inhibitorVo (uM/min)+7 nM inhibitor 0.02 0.04 0.10 0.25 1.00 2.50 0.8 2.9 8.6 24 36 50 Plot the data for the kinetics of the enzyme (with and without the inhibitor) in a double reciprocal (Lineweaver-Burk) plot. Keep in mind that the x axis is 1/[S] and the y axis is 1/Vo. If you are using Excel you want to choose the (x,y) scatter plot (without a ne). You...
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...
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
Given the enzyme system just described (Km = 8.20*10-5M and kcat = 40.0 s-1, [Eltot = 2.00*10-7 M), suppose we introduce a classical noncompetitive inhibitor with K1 = 1.20*10-7M. If the concentration of inhibitor introduced is 3.0 um, what will the effective kcat be? Express your answer in s-1.
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