A) Using the data given in table below plot a Lineweaver-Burk plot
d) Explain you choice
Table 1 :
Kinetics data for an enzyme catalysed reaction with and without the presence of inhibitors :
[S] mmol/L : 1.67 - 2.50 - 5.00 - 10.00
Initial velocity of reaction (micromol/min) :
without inhibitior : 0.100 - 0.125 - 0.167 - 0.200
with inhibitor A : 0.083 - 0.106 - 0.149 - 0.185
with inhibitor B : 0.068 - 0.083 - 0.105 - 0.122© BrainMass Inc. brainmass.com October 24, 2018, 8:12 pm ad1c9bdddf
Please see attached files
A. The Lineweaver-Burk plot is a plot of 1/V versus 1/S as demonstrated in the below equation. Please see the attached section on theory.
I have plotted in the Excel file. Please see attached.
B. I fitted 3 trendline equations for each case as below:
No inhibitor: y = 10x + 4
According to the above relationship, Vmax = ¼ = 0.25 (μmol/min)
KM = 10*Vmax = 2.5
Inhibitor A: y = 13.4x + 4.1
Vmax = 1/4.1 = 0.24 (μmol/min)
KM = 13.4*0.24 = 3.2
Inhibitor A: y = 13.0x + 6.9
Vmax = 1/6.9 = 0.14 (μmol/min)
KM = 13*0.14 = 1.8
C and D.
Note that V approaches Vmax very slowly so that it is extremely difficult to get experimental data that clearly shows what the value Vmax is, and consequently what KM is. Hence at high substrate concentrations, we would ...
The solution provides 3-page explanations, instructions, and drawings in the attached files.
Lineweaver-Burk, Eadie-Hofstee, and Woolf Plots: Example Problem
Using the data in the attached file, prepare Lineweaver-Burk, Eadie-Hofstee, and Woolf plots. To determine the effect of errors in concentration, prepare Lineweaver-Burk and Eadie-Hofstee plots of the data where the concentration of the lowest 3 concentration of substrates are off by 10, 6, and 4% (the most error is associated with the lowest concentration). You choose the direction of the errors.View Full Posting Details