![10 Ilv(min MM) 1/[S] (mm) 24. What is the approximate Km for the Control (No [U)? OA. -3.3 MM OB.-2.5 mm, OC.5 mM OD.3.3 mm,](http://img.homeworklib.com/questions/3fc16a10-df55-11ea-bd84-9b50706af6b3.png?x-oss-process=image/resize,w_560)
Homework Answers
1. Answer: Option C is correct
Explanation:
For the control (No inhibitor)
X-intercept = -1/Km
= -1/5
Km = 5 mM
For the 4 mM inhibitor)
Y-intercept = 1/Vmax
= 1/4
= 0.25 mM/min
In the presence of the inhibitor,
Vmax is decreased
Km is increased
= Mixed-inhibition
Add Answer to:
what is the approximage Vmax for [I]=4mM
what type of inhibitor is I?
10 Ilv(min MM)...
QUESTION 21 Consider the following graph and answer questions 21 through 24 below: 10 Calen 02...
QUESTION 21 Consider the following graph and answer questions 21 through 24 below: 10 Calen 02 m Wய பாய) 5 1/[S] (ma!) 24. What is the approximate Km for the Control (No [0])? QUESTION 22 What is the approximate Vmax for [U] = 4 mM? OA 0.25 PM/min, OB. 0.50 LM/min, 02.00 LM/min, OD.3.00 uM/min, E. 4.00 PM/min, OF. None of the above. QUESTION 24 What is the ki or K'? O A. 0.25 mm OB. 0.5 mm OC. 1...
QUESTION 21 Consider the following graph and answer questions 21 through 24 below: 10 Cat N-2...
QUESTION 21 Consider the following graph and answer questions 21 through 24 below: 10 Cat N-2 1/v(min/mm) -5 - 1 0 1 2 3 4 5 6 1/[S] (mM1) TI QUESTION 22 What is the approximate Vmax for [1] = 4 mM? OA. 0.25 uM/min, OB.0.50 MM/min, OC.2.00 uM/min, OD.3.00 uM/min, OE. 4.00 uM/min, F. None of the above. What type inhibitor is I? A.competitive, B.uncompetitive, C.pure-noncompetitive, OD. mixed ( K'j > Ki), O E. all of the above, OF....
show work please! with inhibitor (umoles/min/mg) [S] (mm) v- no inhibitor (umoles/min/mg) V - with inhibitor...
show work please!
with inhibitor (umoles/min/mg) [S] (mm) v- no inhibitor (umoles/min/mg) V - with inhibitor (umoles 3.0 2.29 x 103 1.83 x 103 5.0 or ble 3.20 x 103 Boristeder s obre o 2.56 x 103 7.0 3.86 x 103 USD 3.09 x 103 0 9. 4 .36 x 100W 3.49 x 103 0 11. 4 .75 x 108 3.80 x 103 Draw Lineweaver-Burk plots for the enzyme data shown above. When present, I = 0.200 m a) What...
Scatchard analysis may give information about? O A. the ligand binding constant (Kb) of a protein,...
Scatchard analysis may give information about? O A. the ligand binding constant (Kb) of a protein, OB. the number of ligand binding sites on a protein, the type of homotropic cooperativity with ligand binding, OD. Bmax: O E. all of the above, OF. none of the above. In glycolysis, fructose 1,6-bisphosphate is converted to two products with a AG'º of 23.8 kJ/mol. Under what conditions encountered in a normal cell will the free-energy change (AG) be negative, enabling the reaction...
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...
10.What type of inhibitor is this? How do you know? (2) 11.For your assigned inhibitor 1,...
10.What type of inhibitor is this? How do you know? (2)
11.For your assigned inhibitor 1, what are the apparent Km &
Vmax? (NOTE: apparent Km& Vmax are just the Km & Vmax in
presence of inhibitor, at a given concentration.) (2)
Kinetics experiments were performed on PGI. Enzyme activity
(initial velocity, Vo) was measured at varying concentrations of
Glucose-6-phosphate (G6P). The enzyme concentration used in all
experiments was 1.5 μM.
12.What will be the reaction rate with 0.500 mM...
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...
How do I calculate the apparent vmax? 16. At right is a graph obtained from a...
How do I calculate the apparent vmax?
16. At right is a graph obtained from a series of enzyme kinetics assays. The Vmax for this enzyme and substrate is 4.5 M/s. a) What is the KM? 5 4.5 4 3.5 3 2.5 2 1.5 KM: 3mm ve (MM/s) 1 0.5 0 b) If a pure non-competitive inhibitor was added to the assays, what would the resulting kinetics curve be like? Give a Km and Vmax in the presence of the...
a. What is the Km and Vmax for PFK1 when treated with OmM (represents control for...
a. What is the Km and Vmax for PFK1 when treated with OmM (represents control for enzymatic activity) or with 5mM AMP Show work on the graph draw lines for Vmax and Km. Control: Vmax (AMP), 0.32 0.28 0.2 Km 0.24 0.20 (s)-FTY20 (20LM): 0.18 Vmax 0.12 0.08 0.04 Km 0.00 Fructose-6-phosphate (mM) b. Fructose-6-phosphate is the substrate of the reaction. Based on answer in "a", what type of regulation occurs with and what site on PFK1 is AMP likely...
10. The table below is the results of the study of enzymatic reaction with and without inhibitor. What type of inhibito...
10. The table below is the results of the study of enzymatic reaction with and without inhibitor. What type of inhibitor is it? Explain! Enzyme concentration in all reactions is 1 uM [Inh], μΜ S], mM 0 0.2 0.4 0.7 rate, umol/min 0 0 0 0 0 0.5 23.5 18.8 14.1 7.0 1 32.2 25.8 19.4 9.7 1.5 36.9 29.5 22.1 11.1 2.5 41.8 33.3 25.0 12.5 3.5 44 35.2 26.4 13.2
QUESTION 21 Consider the following graph and answer questions 21 through 24 below: 10 Calen 02 m Wய பாய) 5 1/[S] (ma!) 24. What is the approximate Km for the Control (No [0])? QUESTION 22 What is the approximate Vmax for [U] = 4 mM? OA 0.25 PM/min, OB. 0.50 LM/min, 02.00 LM/min, OD.3.00 uM/min, E. 4.00 PM/min, OF. None of the above. QUESTION 24 What is the ki or K'? O A. 0.25 mm OB. 0.5 mm OC. 1...
QUESTION 21 Consider the following graph and answer questions 21 through 24 below: 10 Cat N-2 1/v(min/mm) -5 - 1 0 1 2 3 4 5 6 1/[S] (mM1) TI QUESTION 22 What is the approximate Vmax for [1] = 4 mM? OA. 0.25 uM/min, OB.0.50 MM/min, OC.2.00 uM/min, OD.3.00 uM/min, OE. 4.00 uM/min, F. None of the above. What type inhibitor is I? A.competitive, B.uncompetitive, C.pure-noncompetitive, OD. mixed ( K'j > Ki), O E. all of the above, OF....
show work please!
with inhibitor (umoles/min/mg) [S] (mm) v- no inhibitor (umoles/min/mg) V - with inhibitor (umoles 3.0 2.29 x 103 1.83 x 103 5.0 or ble 3.20 x 103 Boristeder s obre o 2.56 x 103 7.0 3.86 x 103 USD 3.09 x 103 0 9. 4 .36 x 100W 3.49 x 103 0 11. 4 .75 x 108 3.80 x 103 Draw Lineweaver-Burk plots for the enzyme data shown above. When present, I = 0.200 m a) What...
Scatchard analysis may give information about? O A. the ligand binding constant (Kb) of a protein, OB. the number of ligand binding sites on a protein, the type of homotropic cooperativity with ligand binding, OD. Bmax: O E. all of the above, OF. none of the above. In glycolysis, fructose 1,6-bisphosphate is converted to two products with a AG'º of 23.8 kJ/mol. Under what conditions encountered in a normal cell will the free-energy change (AG) be negative, enabling the reaction...
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...
10.What type of inhibitor is this? How do you know? (2)
11.For your assigned inhibitor 1, what are the apparent Km &
Vmax? (NOTE: apparent Km& Vmax are just the Km & Vmax in
presence of inhibitor, at a given concentration.) (2)
Kinetics experiments were performed on PGI. Enzyme activity
(initial velocity, Vo) was measured at varying concentrations of
Glucose-6-phosphate (G6P). The enzyme concentration used in all
experiments was 1.5 μM.
12.What will be the reaction rate with 0.500 mM...
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...
How do I calculate the apparent vmax?
16. At right is a graph obtained from a series of enzyme kinetics assays. The Vmax for this enzyme and substrate is 4.5 M/s. a) What is the KM? 5 4.5 4 3.5 3 2.5 2 1.5 KM: 3mm ve (MM/s) 1 0.5 0 b) If a pure non-competitive inhibitor was added to the assays, what would the resulting kinetics curve be like? Give a Km and Vmax in the presence of the...
a. What is the Km and Vmax for PFK1 when treated with OmM (represents control for enzymatic activity) or with 5mM AMP Show work on the graph draw lines for Vmax and Km. Control: Vmax (AMP), 0.32 0.28 0.2 Km 0.24 0.20 (s)-FTY20 (20LM): 0.18 Vmax 0.12 0.08 0.04 Km 0.00 Fructose-6-phosphate (mM) b. Fructose-6-phosphate is the substrate of the reaction. Based on answer in "a", what type of regulation occurs with and what site on PFK1 is AMP likely...
10. The table below is the results of the study of enzymatic reaction with and without inhibitor. What type of inhibitor is it? Explain! Enzyme concentration in all reactions is 1 uM [Inh], μΜ S], mM 0 0.2 0.4 0.7 rate, umol/min 0 0 0 0 0 0.5 23.5 18.8 14.1 7.0 1 32.2 25.8 19.4 9.7 1.5 36.9 29.5 22.1 11.1 2.5 41.8 33.3 25.0 12.5 3.5 44 35.2 26.4 13.2
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