lineweaver-burk equation is
= +
now lineweaver-burk plot at no inhibitor
bestline equation is
y = 0.10 x + 0.50
so, = 0.50
or, Vmax = (1/.0.50) = 2.00 nmol/min.
= 0.10
or, Km = Vmax * 0.10 = (2 *0.10) = 0.20 M
now the plot in presence of inhibitor
best line equation is
y = 0.20 x + 0.5
or, Vmax = (1/.0.50 ) = 2.00 nmol/min.
= 0.20
or, Km = Vmax * 0.20 = (2 *0.20) = 0.40 M
Therefore, in presence of inhibitor Vmax is unchanged and Km has increased, so type of inhibition is competitive .
Competitive inhibitor : The structure of inhibitor is similar to substrate so it competes with the substrate so such kind of inhibition is overcome by increasing the substrate concentration.
and Vmax 2 points for interpreting the gapl Name (can be on graph paper or Excel. It is easier on Excel). This is w...
Name (Can be on graph paper or Excel. It is easier on Excel This is worth 10 points. 3 points for determining the correct data, 3 points for the graph, 2 points each for the Km and Vmax (be sure to remember units 1. Use this experimental data to plot a lineweaver burke graph 2. Use the graph determine them and Vmax of the reaction 3. Be sure to keep the same number of sig figs in your answer, and...
please graph all 3 lines and explain the vmax&km How to: Lineweaver Burke 1. The following data was determined for an enzyme in the absence of an inhibitor and in the presence of two different inhibitors (V2 and V3). Determine the V. and K for the enzyme (1) Plot the data and determine the type of inhibition for each inhibitor (S) mm 1 V2 4.3 5.5 V1 12 20 29 2 relliate 150b
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....
CHEM3250 Assignment-Enzyme Inhibition Consider the data below for an enzyme catalyzed reaction. The rate of the reaction has been determined with and without an inhibitor. A total concentration of enzyme of 20 uM was used in the experiment. SHOW WORK AND UNITS!!! Without Inhibitor With Inhibitor [substrate] (mM)Rate of formation of te of formation of product product (mM/min) mM/min) 6.67 5.25 0.49 7.04 38.91 1.0 2.2 6.9 41.8 44.0 1.5 3.5 1 a) On the same graph, plot the data...
9. The following data were obtained from an enzyme kinetics experiment. Graph the data using a Lineweaver-Burk plot and determine the values for Km and Vmax. [S] (UM) V(nmol/min) 0.20 0.26 1.43 1.67 2.08 3.33 0.33 1.00
11. In Excel, prepare Lineweaver-Burk plots for the behavior of an enzyme for which the following experimental data are available: V, umol/min umol/min (No Inhibitor) S], mM (Inhibitor Present) 3.66 5.12 6.18 6.98 7.60 4.58 6.40 7.72 8.72 9.50 3.0 5.0 7.0 9.0 11.0 a. What are the KM and Vmax values for the inhibited and uninhibited reaction 5 pts. each reaction) b. Is the inhibitor competitive or noncompetitive? (5 pts.) Micheli-Menten) EQUATIONS: VV
Problem Set #3 Due in class 10/10/18 se in a classic 1. Sucrose (common table sugar) is hydrolyzed to glucose and fructo experiment in kinetics. The reaction is catalyzed by the enzyme invertase. Using the following data, determine by the Lineweaver-Burke method A) the Vmax and Km of the enzyme and B) whether the inhibition of this reaction by 2 M urea is competitive or noncompetitive. V, no inhibitor (arbitrary units) V, inhibitor present (same arbitrary units) Sucrose concentration 0.0292...
Prostaglandins are a class of eicosanoids (fatty acid derivatives) with a variety of extremely potent actions on vertebrate tissues. They are responsible for producing fever and inflammation and its associated pain. Prostaglandins are derived from the 20-carbon fatty acid, arachidonic acid, in a reaction catalyzed by the enzyme, prostaglandin endoperoxide synthase. This uses oxygen to convert arachidonic acid to PGG2 (a prostaglandin). Ibuprofen inhibits this enzyme and is a fever reducer and anti-inflammatory agent. The kinetic data is given below...
What is the Km of this graph if 1/2 the Vmax with inhibitor- yellow dot- is 1.96. And Kmax 1/2 with out inhibitor(grey dot) is 2.34. V/Inhibitor V/Reaction Rate (mol/min) V/No Inhinitor km(no inhibitor) Km/w/inhinitor) 0.00033 0.000033 0.000133 0.000133 0.00133 0.002 0.000667 [S] (mol/liters)
5) (14 marks) The following kinetic data were obtained for an enzyme in the absence of inhibitor (1), and in the presence of an inhibitor at 5 mM concentration (2). Assume[ET] is the same in each experiment. [S] (MM) (1) v(umol/mL sec) 12 (2) v(umol/mL sec) 4.3 1 8 2 4 20 29 14 21 8 35 12 40 26 a. Using a graphing program (excel or sigmaplot) construct a lineweaver burke plot representing the uninhibited reaction and the inhibited...