![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](http://img.homeworklib.com/questions/d2c54d20-e508-11eb-b266-49f2c28fda1a.png?x-oss-process=image/resize,w_560)
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D-Lactose is the substrate for B-galactosidase. Given Vo = kcat [Et] [S]/km + [S], calculate [S],...
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 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
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
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...
Can help me with these questions please. 30. Sphingosine-1-phosphate (S1P) is important for cell survival. The...
Can help me with these questions please. 30. Sphingosine-1-phosphate (S1P) is important for cell survival. The synthesis of S1P from sphingosine and ATP is catalyzed by the enzyme sphingosine kinase. The velocity of the sphingosine kinase reaction was measured in the presence and absence of threo-sphingosine, a stereoisomer of sphingosine that inhibits the enzyme. The results are shown below. Sphingosine (uM) vo (mg /min) vo (mg /min) with inhibitor 2.5 32.3 8.5 3.5 40 11.5 5.0 50.8 14.6 10 72...
)* (Vmax). 4. When [S] = KM, Vo=( A) 0.5 B) KM C) 0.75 D) kcat...
)* (Vmax). 4. When [S] = KM, Vo=( A) 0.5 B) KM C) 0.75 D) kcat E) [S] 5. The overall transformation shown in the following reaction: E s. Es p + E For the reaction, the steady state assumption A) ES breakdown occurs at the same rate as ES formation B) [P]>>[E] C) implies that ki-k-1 D) implies that k., and k2 are such that the [ES] = k1[ES] E) [S] = [P]
a. what are the values of Vmax and Km in the abscence if the inhibitor what...
a. what are the values of Vmax and Km in the abscence if the
inhibitor what are the values of Vmax and Km in the presence of the
inhibitor?
b. what type of inhibition is it?
c. what is the dissociation constant (Ki) of the
inhibition?
***d. graph a linear scatter plot including equation.
Homework (CHE 407) The initial velocity for an enzyme-catalyzed reaction is measured at various initial substrate concentration [S]o, in the absence and in the presence of...
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...
por favor mostrar el proceso A protease (Mr = 21,500) hydrolyzes a heptapeptide substrate with a...
por favor mostrar el proceso
A
protease (Mr = 21,500) hydrolyzes a heptapeptide substrate with a
1000 s kcat. -3 -1 When the initial substrate concentration is 0.1
M, the v0 is 5.31 x 10 Ms-1
a) Calculate vmax and this hydrolysis when the protease
concentration is 0.2 mg / mL of this protease.
b) Calculate Km with the previous data.
c) In the presence of an inhibitor, it was found that vmax =
9.3 x 10-3 M / s....
need B C and D done please please please help!!! 1. You measure the initial rate...
need B C and D done please please please help!!!
1. You measure the initial rate of an enzyme reaction as a function of substrate concentration in the presence and absence of an inhibitor. The following data was obtained: V. (-) Inhibitor (mm/min) (+) Inhibitor (mM/min) 17 [S] (MM) 0.0001 0.0002 0.0005 0.001 0.002 Please submit calculations and graph for full credit! Note: You are required to use Excel to generate the Lineweaver-Burk plot (a) (10 points) Create a Lineweaver-Burk...
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
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
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...
)* (Vmax). 4. When [S] = KM, Vo=( A) 0.5 B) KM C) 0.75 D) kcat E) [S] 5. The overall transformation shown in the following reaction: E s. Es p + E For the reaction, the steady state assumption A) ES breakdown occurs at the same rate as ES formation B) [P]>>[E] C) implies that ki-k-1 D) implies that k., and k2 are such that the [ES] = k1[ES] E) [S] = [P]
a. what are the values of Vmax and Km in the abscence if the
inhibitor what are the values of Vmax and Km in the presence of the
inhibitor?
b. what type of inhibition is it?
c. what is the dissociation constant (Ki) of the
inhibition?
***d. graph a linear scatter plot including equation.
Homework (CHE 407) The initial velocity for an enzyme-catalyzed reaction is measured at various initial substrate concentration [S]o, in the absence and in the presence of...
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...
por favor mostrar el proceso
A
protease (Mr = 21,500) hydrolyzes a heptapeptide substrate with a
1000 s kcat. -3 -1 When the initial substrate concentration is 0.1
M, the v0 is 5.31 x 10 Ms-1
a) Calculate vmax and this hydrolysis when the protease
concentration is 0.2 mg / mL of this protease.
b) Calculate Km with the previous data.
c) In the presence of an inhibitor, it was found that vmax =
9.3 x 10-3 M / s....
need B C and D done please please please help!!!
1. You measure the initial rate of an enzyme reaction as a function of substrate concentration in the presence and absence of an inhibitor. The following data was obtained: V. (-) Inhibitor (mm/min) (+) Inhibitor (mM/min) 17 [S] (MM) 0.0001 0.0002 0.0005 0.001 0.002 Please submit calculations and graph for full credit! Note: You are required to use Excel to generate the Lineweaver-Burk plot (a) (10 points) Create a Lineweaver-Burk...
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