Homework Answers
Ans:- To solve this you must be aware of Michaelis -Menten equation and the assumption made while deriving it.
The right answers are 2) and 4) ,explanation is provided in the picture below.
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Question 5 (1 point) Saved Suppose a cell that had a total enzyme concentration of 1,...
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
3. The Michaelis-Menten Graph also shows the theoretical maximum rate of the enzyme (Vmax), the point...
3. The Michaelis-Menten Graph also shows the theoretical maximum rate of the enzyme (Vmax), the point where the enzyme is working at its maximum rate (Vmax/2), and amount of substrate needed to bind half of the active sites (Km). Label these points on the graph. Vmax represents: Vm Vmax/2 represents: Reaction velocity v Vmax 2 Km represents: Kim Substrate concentration (5)
112 marks] 3. The relationship between initial velocity (V.) and substrate concentration of most of the...
112 marks] 3. The relationship between initial velocity (V.) and substrate concentration of most of the enzyme- catalized reactions are explained by Michaelis-Menten equation. IMPORTANT: Show the calculations and indicate the units for all your answers. a. For an enzyme which follows the Michaelis-Menten enzyme kinetics, Km is 50 mmol L. Calculate the substrate concentration required to obtain the initial velocity (V.) equivalent to 90% of the maximum velocity (Vmax). b. The Vmax of the above reaction is 250 mmol...
Consider the enzyme-catalyzed reaction with Vmax=164 (μmol/L)min−1 and KM=32μmol/L. Part A If the total enzyme concentration...
Consider the enzyme-catalyzed reaction with Vmax=164 (μmol/L)min−1 and KM=32μmol/L. Part A If the total enzyme concentration was 6 nmol/L, how many molecules of substrate can a molecule of enzyme process in each minute? Express your answer to three significant figures. kcat kcat = 2.73×104 min−1 Part B Calculate kcat/KM for the enzyme reaction.
The Michaelis-Menten equation is often used to describe the kinetic characteristics of an enzyme-catalyzed react...
The Michaelis-Menten equation is often used to describe the kinetic characteristics of an enzyme-catalyzed reaction. S Where v is the velocity or rate, Vmax is the maximum velocity, Km is the +IST Michaelis- Menten constant, and I5 s the substrate concentration. K + S v (uM/min) a) A graph of the Michaelis-Menten equation is a plot of a reaction's initial velocity (Vo) at different substrate concentrations ([S]) 300 Vmax 250 1/2 Vmax First, move the line labeled "Vmax to a...
Need help with number 13! I already asked about number 12. The inverse velocity and inverse substrate concentration relationship for an enzyme-catalyzed reaction is given below V Vmax Vmax S For the...
Need help with number 13! I already asked about number
12. The inverse velocity and inverse substrate concentration relationship for an enzyme-catalyzed reaction is given below V Vmax Vmax S For the hydration of CO2 catalyzed by carbonic anhydrase, it was determined experimentally that (dm s mol 4023.9+ 39.934 at a total enzyme IS] concentration of 2.32 × 10-y mol-dm- What is the value of the Michaelis constant KM for this enzymatic reaction? (B). 9.92x103 mol dm3 (D). 100.8 mol...
An enzyme catalyzes a reaction with a Km of 8.50 mM and a Vmax of 2.70...
An enzyme catalyzes a reaction with a Km of 8.50 mM and a Vmax of 2.70 mM 5-1. Calculate the reaction velocity, vo, for each substrate concentration. [S] = 2.25 mM Vo: 0.5 mM.s-1 [S] = 8.50 mM 1.70 mMs-1 [S] = 13.0 mM [S] = 13.0 mM Vo: 2.05 mm. s-1
The amino acid asparagine can promote cancer cell proliferation. Treating patients with the enzyme asparaginase is...
The amino acid asparagine can promote cancer cell proliferation.
Treating patients with the enzyme asparaginase is sometimes used as
a chemotherapy treatment. Asparaginase hydrolyzes asparagine to
aspartate and ammonia.
Considering the provided Michaelis–Menten curves for two
different asparaginase enzymes, complete the passage. The arrow
indicates the concentration of asparagine in the human body.
< Question 2 of 4 > Asparaginase 1 Asparaginase 2 [S] The Vmax of asparaginase 1 is equal to the Vmax of asparaginase 2. At the substrate...
To determine the kinetic characteristics of an enzyme you used 1 nmol/L of enzyme in a...
To determine the kinetic characteristics of an enzyme you used 1 nmol/L of enzyme in a series of assays where you measured the rate of reactions as you varied the concentration of substrate in each assay (Table A). Estimate from a Michaelis-Menten plot approximate values for Vmax, KM, Kcat, and the specificity constant for this enzyme and substrate. (The only information that is given is this paragraph and the table below). Table 1: [S] (μM) v (μmol/L/min) 0 0 5...
Km is: A. the concentration of an inhibitor needed to give 1/2 Vmax B. a measure...
Km is: A. the concentration of an inhibitor needed to give 1/2 Vmax B. a measure of the rate of product formation C. related to the affinity of the enzyme for its substrate D. a measure of the activation energy of the enzyme-catalyzed reaction E. the velocity equal to one-half of the maximum velocity
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
3. The Michaelis-Menten Graph also shows the theoretical maximum rate of the enzyme (Vmax), the point where the enzyme is working at its maximum rate (Vmax/2), and amount of substrate needed to bind half of the active sites (Km). Label these points on the graph. Vmax represents: Vm Vmax/2 represents: Reaction velocity v Vmax 2 Km represents: Kim Substrate concentration (5)
112 marks] 3. The relationship between initial velocity (V.) and substrate concentration of most of the enzyme- catalized reactions are explained by Michaelis-Menten equation. IMPORTANT: Show the calculations and indicate the units for all your answers. a. For an enzyme which follows the Michaelis-Menten enzyme kinetics, Km is 50 mmol L. Calculate the substrate concentration required to obtain the initial velocity (V.) equivalent to 90% of the maximum velocity (Vmax). b. The Vmax of the above reaction is 250 mmol...
The Michaelis-Menten equation is often used to describe the kinetic characteristics of an enzyme-catalyzed reaction. S Where v is the velocity or rate, Vmax is the maximum velocity, Km is the +IST Michaelis- Menten constant, and I5 s the substrate concentration. K + S v (uM/min) a) A graph of the Michaelis-Menten equation is a plot of a reaction's initial velocity (Vo) at different substrate concentrations ([S]) 300 Vmax 250 1/2 Vmax First, move the line labeled "Vmax to a...
Need help with number 13! I already asked about number
12. The inverse velocity and inverse substrate concentration relationship for an enzyme-catalyzed reaction is given below V Vmax Vmax S For the hydration of CO2 catalyzed by carbonic anhydrase, it was determined experimentally that (dm s mol 4023.9+ 39.934 at a total enzyme IS] concentration of 2.32 × 10-y mol-dm- What is the value of the Michaelis constant KM for this enzymatic reaction? (B). 9.92x103 mol dm3 (D). 100.8 mol...
An enzyme catalyzes a reaction with a Km of 8.50 mM and a Vmax of 2.70 mM 5-1. Calculate the reaction velocity, vo, for each substrate concentration. [S] = 2.25 mM Vo: 0.5 mM.s-1 [S] = 8.50 mM 1.70 mMs-1 [S] = 13.0 mM [S] = 13.0 mM Vo: 2.05 mm. s-1
The amino acid asparagine can promote cancer cell proliferation.
Treating patients with the enzyme asparaginase is sometimes used as
a chemotherapy treatment. Asparaginase hydrolyzes asparagine to
aspartate and ammonia.
Considering the provided Michaelis–Menten curves for two
different asparaginase enzymes, complete the passage. The arrow
indicates the concentration of asparagine in the human body.
< Question 2 of 4 > Asparaginase 1 Asparaginase 2 [S] The Vmax of asparaginase 1 is equal to the Vmax of asparaginase 2. At the substrate...
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