An enzyme catalyzes the reaction A ® B. The initial rate of the reaction was measured as a function of the concentration of A (substrate). The following data were obtained: [10 pts.]
[A], micromolar V0, nmoles/min
0.05 0.08
0.1 0.16
0.5 0.79
1 1.6
5 7.3
10 13
50 40
100 53
500 73
1,000 76
5,000 79
10,000 80
20,000 80
Answer the following questions:
What is the Km of the enzyme for the substrate A?
What is the value of V0 when [A] = 43?
What is the value of the y-intercept of the line?
What is the value of the x-intercept of the line?
Hint: Plot 1/ V0 vs. 1/[A], and a straight line will be obtained (for c and d).
Please attached a copy of your graph for a full point for this question (in excel format)
An enzyme catalyzes the reaction A ® B. The initial rate of the reaction was measured...
2. (20 points) An enzyme catalyzes the reaction A | B. The initial rate of the reaction was measured as a function of the concentration of A. The following data were obtained: 0.08 |(A), micromolar V.,. nmoles/min 0.05 0.1 0.5 5 10 SO 100 500 1.000 5,000 10,000 20,000 2012 a) b) What is the of the enzyme for the substrate A? What is the K of the enzyme for the substrate A?
An enzyme catalyzes the reaction A ⇌ B. The enzyme is present at a concentration of 2 nM, and the Vmax is 1.2 μM s−1. The Km for substrate A is 10 μM. Calculate the initial velocity of the reaction, V0, when the substrate concentration is (a) 2 μM, (b) 10 μM, (c) 30 μM.
An enzyme catalyzes a reaction with a Km of 9.00 mM and a Vmax of 3.95 mM·s–1. Calculate the reaction velocity, v0, for the following substrate concentrations: A. 1 mM B. 9 mM C. 11 mM
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
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
9. Applying the Michaelis-Menten Equation I An enzyme catalyzes the reaction A = B. The enzyme is present at a con- centration of 2 nm, and the Vmax is 1.2 ums". The Km for substrate A is 10 um. Calculate the initial velocity of the reaction, Vo, when the substrate concentration is (a) 2 um, (b) 10 um, (C) 30 um.
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...
For an enzyme catalyzed reaction, the initial rate R. was determined at each initial concentration of substrate [S]. The following data were obtained [S], u M/1 Ro, u M/1 2.5 9.8 20.2 31.7 41.2 50.2 60.1 74.3 (a) Use these data to determine Km and Rmax by means of a Line- weaver-Burke-analysis. (b) Analyze the data using the Eadie-Hofstee procedure to determine Km and Rmax. (c) Analyze the data using the Hanes-Wolff procedure to determine Km and Rmax
The key factor that controls the initial rate of an enzyme catalysed reaction (Vo) is the concentration of the substrate of the reaction ([S]). In Damon's Michaelis-Menten experiment, the highest concentration of substrate used was 500 UM. What do you think will happen to the reaction velocity if higher concentrations of substrate were used? Select one: a. Vo will reach a plateau at higher (S) values O b. Vo will increase exponentially as (S) is increased O c. Vo will...
The enzyme catalase catalyzes the decomposition of hydrogen peroxide. The following data are obtained regarding the rate of reaction as a function of substrate concentration using concentration of catalase equals to 3.5 x 10^–9 M : Analyzing the data using the Lineweaver-Burk plot: we calculated the S = 0.698 s , and I = 26.6 M^-1 s , Km for this enzyme is: [H202lo (M) Initial Rate (Ms1) 1.38 x 10-3 0.001 0.002 2.67 x 10-3 0.005 6.00 x 103...