Dear James,

Interesting questions and indeed not straightforward to understand!

After carefully checking the files you provided, I realized that there are a couple of issues at the figure legend and also from your annotations in red, let me explain myself better...

It is true that the double reciprocal plot that you're showing us on your post has a similar behavior to the one you obtain while plotting the initial velocities. However, the picture shows what happens to the enzyme when you add an inhibitor to the assay (at specific fixed concentrations) and vary the substrate concentration. As it is clearly shown, the more inhibitor you add, the higher the slope of that curve, which in this case means your enzyme exhibits a lower Vmax... it does make sense, right?

Now, if you want to determine the individual Km values for a two-substrate enzyme, of course you must do a similar approach by fixing one substrate at a time and varying the other substrate; thus, no problem there. However, the higher the fixed concentration of the first substrate is, the lower the slope you'll get; i.e. in these double reciprocal plots, the world is inverted, high y-axis values mean low activities! In other words the intercept of the y-axis representing 1/Vmax will be very low if the enzyme has a high activity. A similar thing happens for the intercepts in the x-axis to determine the -1/Km values. If the latter is true, then in the figures from the paper the fixed concentrations of L-Met are arranged in the opposite way (I'm attaching a theoretical plot from the book of Irwin H. Segel, Biochemical Calculations, 2nd edition).

Now to your question, if we get clear that the x-axis does always represent the substrate we are titrating, the Km and other kinetic values that we can estimate MUST be for that particular substrate. For this reason, most likely authors had an error while writing the figure legend or inverted the panels. If not wrong, Panel A does represent the assay that varies ATP while fixing L-Met at specific concentrations; Panel B the opposite. If also correct, the graph that has the x-axis as 1/ATP is the one that you would need to determine the "real" Km for ATP.

The intercept of all curves at the left side does mean you have an enzyme that has 2 substrates and forms a ternary complex. Finally, why is this intercept useful to get the "real" Km for each substrate? Well, mathematically, that's the x-value that you'd get if you would have a zero concentration of the second substrate (of course not possible); this is just an assumption, very useful though! More info here: https://www.sciencedirect.com/topics/engineering/michaelis-menten-equation

Hope these thoughts are useful!

Best wishes,

Alfredo