![/A mer 2017 Assignment 1 10 points Due at the beginning of Class: July. 19 Summer 410 or a competitive inhibitor. If the [l) -2K, and [S)-2Km. What is the reaction velocity, v )For a 2 pts) ressed as a 35) (2 pts) For a noncompetitive inhibitor. If the [] -2K,and (S)-2K., What is the reaction velocity, v 35.) (2 pts) (expressed as a percentage of Vanun (expressed as a percentage of Vmax 36.) (2 pts) For an uncompetitive inhibitor. If the I) 2K 36) (2 pts) For an uncompetitive inhibitor. If the [l] = 2K, and [S]-2Km. what is the reaction velocity, v (expressed as a percentage of Vmax 37.) 3 pts) The K for a noncompetitive inhibitor of a particular enzyme is 32.5 μM How much inhibitor would be required to reach 90% inhibition of the Vmax for the enzyme catalyzed reaction?](http://img.homeworklib.com/questions/7659e6e0-3502-11eb-9de1-dbeb4cbea547.png?x-oss-process=image/resize,w_560)
Homework Answers
34) Competitive Inhibition: the inhibitor competes with the
substrate for binding to the enzyme.
Ki represents the dissociation constant for the inhibitor; Km
represents the dissociation constant for the substrate.
Ki = [E][I]/[EI]
Km = [E][S]/[ES]
These equations can be used to arrive at modified Michaelis-Mentin
equation:
v = (Vmax*[S])/(Km*(1 + [I]/Ki) + [S])
Substituting the given expressions for [I] and [S] in the above
equation, we get:
v = Vmax*(2/5)
v = 0.4Vmax
Expressing as a percentage, v = 40% of Vmax
35) Non-competitive Inhibition: This occurs when I binds to both E and ES. For this case it is assumed that dissociation constants for both the reactions are same. The modified Michaelis-Menten for this is given below:
v = (Vmax*[S])/((Km + [S])*(1 + [I]/Ki))
Substituting the given expressions for [I] and [S] in the above
equation, we get:
v = Vmax*(2/9)
v = 0.222Vmax
Expressing as a percentage, v = 22.2% of Vmax
36) Uncompetitive Inhibition: this type of inhibitor
kinetics is occasionally observed, primarily with multisubstrate
enzymes. It occurs when the inhibitor only binds to the ES complex.
The modified Michaelis-Menten for this is given below:
v = (Vmax*[S])/(Km + (1 + [I]/Ki)*[S])
Substituting the given expressions for [I] and [S] in the above
equation, we get:
v = Vmax*(2/7)
v = 0.286Vmax
Expressing as a percentage, v = 28.6% of Vmax
Add Answer to:
For a competitive inhibitor. If the [I] = 2K_ and [S] = 2K_m. What is the...
18. The following kinetic scheme that shows the Inhibitor () only binding to the enzyme-substrate (ES)...
18. The following kinetic scheme that shows the Inhibitor () only binding to the enzyme-substrate (ES) complex is typic of what type of enzyme inhibition? E + SEES -E + PSN ESI no reaction A) Irreversible B) Competitive C) Noncompetitive D) Uncompetitive or acetylcholinesterase 19. DIFP acts as an inhibitor of the enzyme chymotrypsin. A) Irreversible B) Competitive C) Uncompetitive D) Mixed
Determine the type of inhibition that has occurred in a first order Michaelis-Menten enzyme catalyzed reaction that has...
Determine the type of inhibition that has occurred in a first order Michaelis-Menten enzyme catalyzed reaction that has yielded the following data. Vi is the velocity in the presence of inhibitor, V is the velocity when run without inhibitor and [S] refers to the substrate concentration. [S] V Vi 5.00 0.29 0.16 1.25 0.27 0.15 0.45 0.23 0.13 0.22 0.19 0.10 0.13 0.14 O.08 Uncompetitive No inhibition at all O None the above ONon-competitive O Competitive
Determine the type of...
21. When you measure an enzyme reaction both without and with several concentrations of an inhibitor,...
21. When you measure an enzyme reaction both without and with several concentrations of an inhibitor, the calculated Vmay for all of the reactions remains constant. However, KM is changing with addition of the inhibitor to apparent higher values (Km apparent). What kind of inhibition is this likely to be? a. Competitive inhibition. b. Uncompetitive inhibition. c. Mixed inhibition. d. Pure noncompetitive inhibition.
A plot of 1/V versus 1/[S], called a Lineweaver-Burk or double-reciprocal plot, is a useful tool...
A plot of 1/V versus 1/[S], called a Lineweaver-Burk or double-reciprocal plot, is a useful tool for identifying the type of enzyme inhibition. Modify each graph by dragging the endpoints to show the various types of enzyme inhibition. Competitive inhibition What is the inhibition mechanism for the competitive inhibitor? with inhibitor UNI The inhibitor binds to both free enzyme and enzyme-substrate complexes with identical binding constants. The inhibitor binds only to free enzyme. The inhibitor binds to both free enzyme...
Question 28 1 pts Determine the type of inhibition that has occurred in a first order...
Question 28 1 pts Determine the type of inhibition that has occurred in a first order Michaelis-Menten enzyme catalyzed reaction that has yielded the following data. Vi is the velocity in the presence of inhibitor, V is the velocity when run without inhibitor and [S] refers to the substrate concentration s V 4.00 0.28 0.28 .33 0.27 0.25 0.56 0.24 0.20 0.20 0.18 0.13 0.10 0.13 0.08 O Non-competitive O Competitive O None of the above No inhibition at all...
An enzyme binds to a competitive inhibitor with Kd = 1.2 × 10-6 M at pH 7.0 and 25°C.
An enzyme binds to a competitive inhibitor with Kd = 1.2 × 10-6 M at pH 7.0 and 25°C.(a)At what inhibitor concentration will 75% of the enzyme be bound to the inhibitor if there is no substrate present? (b) This enzyme has a Km of 4.0 × 10-5 M and a Vmax of 50 μM/s. At a substrate concentration of 3.0 × 10-4 M, calculate (i) the velocity of reaction in the presence of the inhibitor at 4.8 x 10-5 M (ii) the degree of...
An enzyme has a Km of 4.7X10^-5 M. If the Vmax of the preparation is (22...
An enzyme has a Km of 4.7X10^-5 M. If the Vmax of the preparation is (22 micromoles X liters^-1 Xmin^-1). What velocity would be observed in the presence of 2X10^-4 M substrate and 5X10^-4M of a. a competitive inhibitor b. a noncompetitive inhibitor c. an uncompetitive inhibiter Ki in all three cases is 3X10^-4M. What is the degree of inhibition in all three cases?
4) (5 points) What fraction of Vmax is observed at [S] = 5 KM? 5) (20...
4) (5 points) What fraction of Vmax is
observed at [S] = 5 KM?
5) (20 points) For the following data:
[S] (μM)
V0 (no inhibitor)
V0 (2.45 μM inhibitor present)
2.1
0.031
0.020
4.2
0.06
0.045
13
0.138
0.09
20
0.153
0.13
52
0.170
0.135
a) Construct a 1/v (y-axis) versus 1/[S]
(x-axis) plot in the space below.
b) Is the inhibition competitive, noncompetitive, or
uncompetitive?
c) Calculate KM, KMapp,
Vmax, and Vmaxapp....
A. Choose anly one correcet answer for each of the following questions (4 pts cach). AL Some and ...
A. Choose anly one correcet answer for each of the following questions (4 pts cach). AL Some and KM values are shown below for enzyme-substrate pairs. Which of the following enzymes is most efficient in converting the substrate into the product? b) kes.-4x10s s", KM-0.026 M d)k,,-5.7 x1o's", K-2x10s M c)人at-900 s", KM-2.5 × 10.5 M A2. Which of the following enzyme reaction mechanisms has multiple substrates? a) induced-fit e) Michaclis-Menton b) random sequential d) reversible covalent modification e) None...
Problem 2: (Enzyme Kinetics) A competitive inhibitor I interferes with an enzyme-catalyzed reaction according to the...
Problem 2: (Enzyme Kinetics) A competitive inhibitor I interferes with an enzyme-catalyzed reaction according to the mechanism: E+S →ES, rate constant = ki, ES → E+S, rate constant = k-1, ES → E+P, rate constant = k2, E + EI, rate constant = k3. EI → E + I, rate constant = k-3. Assuming that the concentrations of S and I are much larger than the total enzyme concentration, derive an expression for the initial rate of appearance of product,...
18. The following kinetic scheme that shows the Inhibitor () only binding to the enzyme-substrate (ES) complex is typic of what type of enzyme inhibition? E + SEES -E + PSN ESI no reaction A) Irreversible B) Competitive C) Noncompetitive D) Uncompetitive or acetylcholinesterase 19. DIFP acts as an inhibitor of the enzyme chymotrypsin. A) Irreversible B) Competitive C) Uncompetitive D) Mixed
Determine the type of inhibition that has occurred in a first order Michaelis-Menten enzyme catalyzed reaction that has yielded the following data. Vi is the velocity in the presence of inhibitor, V is the velocity when run without inhibitor and [S] refers to the substrate concentration. [S] V Vi 5.00 0.29 0.16 1.25 0.27 0.15 0.45 0.23 0.13 0.22 0.19 0.10 0.13 0.14 O.08 Uncompetitive No inhibition at all O None the above ONon-competitive O Competitive
Determine the type of...
21. When you measure an enzyme reaction both without and with several concentrations of an inhibitor, the calculated Vmay for all of the reactions remains constant. However, KM is changing with addition of the inhibitor to apparent higher values (Km apparent). What kind of inhibition is this likely to be? a. Competitive inhibition. b. Uncompetitive inhibition. c. Mixed inhibition. d. Pure noncompetitive inhibition.
A plot of 1/V versus 1/[S], called a Lineweaver-Burk or double-reciprocal plot, is a useful tool for identifying the type of enzyme inhibition. Modify each graph by dragging the endpoints to show the various types of enzyme inhibition. Competitive inhibition What is the inhibition mechanism for the competitive inhibitor? with inhibitor UNI The inhibitor binds to both free enzyme and enzyme-substrate complexes with identical binding constants. The inhibitor binds only to free enzyme. The inhibitor binds to both free enzyme...
Question 28 1 pts Determine the type of inhibition that has occurred in a first order Michaelis-Menten enzyme catalyzed reaction that has yielded the following data. Vi is the velocity in the presence of inhibitor, V is the velocity when run without inhibitor and [S] refers to the substrate concentration s V 4.00 0.28 0.28 .33 0.27 0.25 0.56 0.24 0.20 0.20 0.18 0.13 0.10 0.13 0.08 O Non-competitive O Competitive O None of the above No inhibition at all...
4) (5 points) What fraction of Vmax is
observed at [S] = 5 KM?
5) (20 points) For the following data:
[S] (μM)
V0 (no inhibitor)
V0 (2.45 μM inhibitor present)
2.1
0.031
0.020
4.2
0.06
0.045
13
0.138
0.09
20
0.153
0.13
52
0.170
0.135
a) Construct a 1/v (y-axis) versus 1/[S]
(x-axis) plot in the space below.
b) Is the inhibition competitive, noncompetitive, or
uncompetitive?
c) Calculate KM, KMapp,
Vmax, and Vmaxapp....
A. Choose anly one correcet answer for each of the following questions (4 pts cach). AL Some and KM values are shown below for enzyme-substrate pairs. Which of the following enzymes is most efficient in converting the substrate into the product? b) kes.-4x10s s", KM-0.026 M d)k,,-5.7 x1o's", K-2x10s M c)人at-900 s", KM-2.5 × 10.5 M A2. Which of the following enzyme reaction mechanisms has multiple substrates? a) induced-fit e) Michaclis-Menton b) random sequential d) reversible covalent modification e) None...
Problem 2: (Enzyme Kinetics) A competitive inhibitor I interferes with an enzyme-catalyzed reaction according to the mechanism: E+S →ES, rate constant = ki, ES → E+S, rate constant = k-1, ES → E+P, rate constant = k2, E + EI, rate constant = k3. EI → E + I, rate constant = k-3. Assuming that the concentrations of S and I are much larger than the total enzyme concentration, derive an expression for the initial rate of appearance of product,...
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