![QUESTION 8 The plot below indicates: 30 nM 1/vo 20 nM 10 nM 1/[S] 1. The effect of increasing concentrations of competitive i](http://img.homeworklib.com/questions/98afd380-7726-11eb-bb83-55c908376c6f.png?x-oss-process=image/resize,w_560)
Homework Answers
The correct option for this question is:- option(2).
The plot is a Lineweaver-Burk plot and it indicates the effect of increasing concentrations of non-competitive inhibitor on the kinetics of an enzyme.
- Non-competitive inhibiton is a type of inhibiton where the inhibitor reduces the activity of the enzyme and binds equally well to the enzyme whether or not it has already bound to the substrate.
- Eg. Of non-competitive inhibiton :- In the enzyme catalysed reactions of glycolysis, phosphoenol pyruvate is converted into pyruvate catalyzed by pyruvate kinase. Alanine is anaminoacid which is synthesized from pyruvate also inhibits the enzyme pyruvate kinase during glycolysis. Alanine is a non-competitive inhibitor, therefore it binds away from the active site to the substrate in order for it to still be the final product.
When a noncnon-competi inhibitor is added, Vmax is changed while the Km remains unchanged.
According to Lineweaver-Burk plot, Vmax is
reduced during the addition of a non-competitive inhibitor which is
shown in the plot by a change in both the slope and the
y-intercept. 
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QUESTION 8 The plot below indicates: 30 nM 1/vo 20 nM 10 nM 1/[S] 1. The...
Please show me how to solve these and I'll give thumbs up. Thank you! Question 6...
Please show me how to solve these and I'll give thumbs up.
Thank you!
Question 6 O out of 10 points An enzyme catalyzed reaction was performed in the presence and in the absence of an inhibitor. The Lineweaver Burk plot showed competitive kinetics with X- intercepts of -10mM -1 and -3.5mM-1 in the presence and absence of the inhibitor respectively. If the inhibitor concentration used was 2micro molar (UM), calculate KI for the inhibitor enzyme binding? Question 7 O...
Onate #2 6 8 26. The effect of an inhibitor (II) on the rate of a single-su investigated and gave...
onate #2 6 8 26. The effect of an inhibitor (II) on the rate of a single-su investigated and gave the following results (three experiments at three different inhibitor concentrations [I]-0.0, 0.20 and 0.40 mmol/L): bstrate enzyme-catalyzed reaction was Substrate concentration [S] Inhibitor concentration [I] 0.0 mmol/L mmol / L) Rate of reaction (vo) (umol /L/s) 0.05 0.10 0.20 0.40 0.60 0.033 0.055 0.083 0.126 Substrate concentration [S (mmol / L) Inhibitor concentration [I] 0.20 mmol/L Rate of reaction (Vo)...
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....
1- Describe two attributes that are different for hemoglobin and myoglobin; focus on characteristics that are...
1- Describe two attributes that are different for hemoglobin and myoglobin; focus on characteristics that are a direct consequence of differences in tertiary and/or quaternary structure. For each attribute, be sure to specifically explain how 3° and 4° organization impacts the described differences. 2- The kinetics of an enzyme (E)-catalyzed reaction are measured over a range of substrate (S) concentrations in the presence or absence of a reversible inhibitor (I). The total enzyme concentration is 10 uM, and the inhibitor...
16. At right is a graph obtained from a series of enzyme kinetics assays. The 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 uM/s. 5 4.5 4 a) What is the KM? KM: v. (mM/s) 3.5 3 2.5 2 1.5 1 0.5 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 inhibitor (write them below) and draw an appropriate curve...
16. At right is a graph obtained from a series of enzyme kinetics assays. The 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 uM/s. a) What is the KM? 5 4.5 4 3.5 3 2.5 2 KM: 3mm V. (mM/s) 1.5 1 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 inhibitor (write them below) and draw an appropriate curve...
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...
high and detecting one product at a time Question 77 The initial velocity of an enzyme...
high and detecting one product at a time Question 77 The initial velocity of an enzyme reaction (vo) describes A) The concentration of the enzyme at maximal velocity B) The concentration of substrate at maximal velocity C) The concentration of both at the start of the reaction D) The rate of the reaction when the substrate and enzyme are first med Question 78 What is the shape of a typical plot of initial rate vs substrate concentration tres kinetics? A)...
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...
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...
Please show me how to solve these and I'll give thumbs up.
Thank you!
Question 6 O out of 10 points An enzyme catalyzed reaction was performed in the presence and in the absence of an inhibitor. The Lineweaver Burk plot showed competitive kinetics with X- intercepts of -10mM -1 and -3.5mM-1 in the presence and absence of the inhibitor respectively. If the inhibitor concentration used was 2micro molar (UM), calculate KI for the inhibitor enzyme binding? Question 7 O...
onate #2 6 8 26. The effect of an inhibitor (II) on the rate of a single-su investigated and gave the following results (three experiments at three different inhibitor concentrations [I]-0.0, 0.20 and 0.40 mmol/L): bstrate enzyme-catalyzed reaction was Substrate concentration [S] Inhibitor concentration [I] 0.0 mmol/L mmol / L) Rate of reaction (vo) (umol /L/s) 0.05 0.10 0.20 0.40 0.60 0.033 0.055 0.083 0.126 Substrate concentration [S (mmol / L) Inhibitor concentration [I] 0.20 mmol/L Rate of reaction (Vo)...
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....
16. At right is a graph obtained from a series of enzyme kinetics assays. The Vmax for this enzyme and substrate is 4.5 uM/s. 5 4.5 4 a) What is the KM? KM: v. (mM/s) 3.5 3 2.5 2 1.5 1 0.5 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 inhibitor (write them below) and draw an appropriate curve...
16. At right is a graph obtained from a series of enzyme kinetics assays. The Vmax for this enzyme and substrate is 4.5 uM/s. a) What is the KM? 5 4.5 4 3.5 3 2.5 2 KM: 3mm V. (mM/s) 1.5 1 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 inhibitor (write them below) and draw an appropriate curve...
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...
high and detecting one product at a time Question 77 The initial velocity of an enzyme reaction (vo) describes A) The concentration of the enzyme at maximal velocity B) The concentration of substrate at maximal velocity C) The concentration of both at the start of the reaction D) The rate of the reaction when the substrate and enzyme are first med Question 78 What is the shape of a typical plot of initial rate vs substrate concentration tres kinetics? A)...
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...
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