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    plot a Lineweaver-Burk problem

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    A) Using the data given in table below plot a Lineweaver-Burk plot

    b) Determine the Km and Vmax for the three sets of data

    c) Which inhibitor would the most effective at high substrate concentrations?

    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

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    https://brainmass.com/biology/biological-chemistry/plot-a-lineweaver-burk-problem-90304

    Solution Preview

    Please see attached files

    Solution
    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 ...

    Solution Summary

    The solution provides 3-page explanations, instructions, and drawings in the attached files.

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