Enzyme Kinetics using Absorbance to make a Lineweaver Burk plot

So here’s the deal, I have an exam tomorrow and one of the topics is enzyme kinetics. I need to know how to derive a Lineweaver Burk plot using an absorbance over time graph for both a control and in the presence of an inhibitor.

So, in the practice lab, I have five concentrations for the control: 6.25uM, 12.5uM, 25uM, 50uM, 100uM. I also have four for the inhibitor: 25uM, 50uM, 100uM, 200uM.

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I’ve plotted the absorbance readings over time (at 15 second intervals for 120 seconds) for each concentration.

I need to know in detail:

How do I find the rate, in this case v? (It just says to plot 1/v not 1/V0 or 1/Vi which is usually the case). Do I simply use the change in absorbance between the two points, which have the greatest difference and multiply this figure by 4 (it says the formula is change in absorbance divided by change in time and that they want it to be represented as change in Absorbance/minute)? Or do I just take the initial absorbance from the final absorbance (at 120 seconds) and divide it by 2 to get the the rate/v?

To get 1/v I assume I just divide 1 by the rate for each substrate concentration.

Also, for the X axis of the Lineweaver Burk plot, do I simply plot 1/6.25uM, 1/12.5uM etc.? I know this sounds silly, but the reason I ask is that when I do this I get 0.16, 0.08, 0.04 etc., where as in lecture example, the X axis had negative 100,00 all the way up to 300,000!

I also need to calculate the Vmax and Km values, which I’m guessing are just the reciprocals of the X and Y Axes?

I can give you the actual figures if you like, both the ones I observed experimentally and the ones that were in the example. To start, I’ve just attached the example Lineweaver Burk plot (on page 4 of the pdf). I have a feeling this is really simple, and I’ve done Lineweaver Burk plots years ago that used concentration instead of absorbance, but what the lecturers’ example shows doesn’t make a lot of sense to me. Solution Preview
1. For large values on ‘x’ axis, most likely the unit is 1/M not 1/uM. Since 1uM = 10^(-6)M, 1 / M = 1×10^6 /M. So you can end up with values of 10^6 on the x-axis.

2. The point at which the line intersects the ‘y’ axis is equal = 1 / Vmax. So, the reciprocal of the y-intercept is the …

Solution Summary
The enzyme kinetics using absorbance to make a Lineweaver Burk plot is examined.

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