Km = {(Vmax([S]))/(V)} – [S]
• Vmax = 100 (units)
• At [S] = 20 M
• Velocity = 25 (units)
calculate Km
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Km = {(Vmax([S]))/(V)} – [S] • Vmax = 100 (units) • At [S] = 20 M...
The velocity of an enzyme is 20% Vmax. What is the ratio of [S] to KM under this condition?
Kcat =30.0 s-1
Km= 0.005 M
Operating at 1/4 Vmax
What is [S] ?
The solutions manuel doesn't explain the problem well. Whete
does the 0.33 km come from?
For part 2 : Plug in [S]= 1/2 Km, 2 Km, and 10 Km
Where does the 1.5 Km come from?
Answer (a) Here we want to find the value of [S] when Vo = 0.25 V max. The Michaelis-Menten equation is V = Vmax[S]/(Km + [SD so V = Vmax...
Name Page Number Date Vmax 1. Estimate Vmax and Km using the velocity vs. substrate concentration plot - KM 2. Calculate the Vmax and Km using the Lineweaver-Burk plot. KM Vmax (S), uM 0.08 0.12 0.54 1.23 1.82 2.72 Reaction Velocity (UM/min) 0.15 0.21 0.70 1.1 1.3 15 4.94 10.00 1.8
Find Km, Vmax, Ki and Kcat values.
Find Vmax, Km and Ki values according to the graph
350 y = 51.843x + 99.634 300 y = 45.555x + 93.809 250 y = 33.086x + 88.634 200 No In 150 100 10 pl • 20 ul ........ Linea -50 Linea -100 Benze Linea 1/[S] mM Benze
An enzyme has a Km of 64.9 mM, determine the fraction (ratio) of Vmax, I.e., v/Vmax, that would be obtained at each substrate concentration below. a. 3.25 mM b. 64.9 mM c. 694 mM d. 6.49 M e. 64.9 M
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. 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...
1. Show, using the Michaelis-Menten equation, that when [S] >>> Km, vo = Vmax. Show, using the M-M equation that when [S] <<<Km, vo =[S][Et]kcat/Km. 2. What is Vmax? Provide both a mathematical and written description of Vmax? How can Vmax be experimentally altered? How can we use Vmax to determine the turnover number (kcat) of an enzyme-catalyzed reaction? What is the major challenge of determining Vmax from an Michaelis-Menten plot?
a. An enzyme has a Vmax of 100 umol/min and a Km of 40 uM. When substrate concentration is 40 uM what is the initial reaction rate? b. An enzyme with a Vmax of 100 umol/minute and a Km of 10 uM was reacted with a irreversible active site specific inhibitor. After reaction with the inhibitor, the enzyme was assayed using a 2 mM concentration of substrate, and it gave a reaction rate of 20 umol/min. What percentage of the...
Determine the kinetic parameters, Km & Vmax and calculate
k2.
Penicillin is hydrolyzed and thereby rendered inactive by penicillinase, an enzyme present in some penicillin-resistant bacteria. The mass of this enzyme is 29.5 kD. The amount of penicillin hydrolyzed in 1 minute in a 10 mL solution containing 100 g of purified penicillinase was measured as a function of the concentration of penicillin. Assume that the concentration of penicillin does not change appreciably during the assay. (a) Plot v versus....