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I have a few issues with this question.

An enzyme assay is performed and the kinetic data graphed. The y-intercept of a Lineweaver-Burk plot is 0.6 min/ΔA. The substrate being assayed has a molar absorbance coefficient (ε) of 4500 L/mol.cm. What is the V max of the enzyme in mmol/L.min?

My working:

Vmax = 1 / 0.6min/ΔA = 1.66ΔA/min (the units switch back I believe?)

And then 1.66ΔA/min / 4.5L/mmol.cm to get 0.37 mmol/L.min?

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  • $\begingroup$ This looks correct to me, although it odd not to relate the activity to the amount of enzyme that was added to the reaction - the Vmax that you have calculated is a property of the exact assay conditions used. $\endgroup$
    – Alan Boyd
    Commented Aug 26, 2014 at 16:03

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see here.

y-intercept = $\frac{1}{V_{max}}$

To convert absorbance to concentration, plot the standard curve and get the linear relationship between the two quantities. Make sure that the line that you fit (for standard curve) passes through origin.

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    $\begingroup$ No need for a standard curve if you have an extinction coefficient, as in this case. $\endgroup$
    – Alan Boyd
    Commented Aug 26, 2014 at 15:56

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