mmm tricky question. keep in mind that your x intercept is (-)1/Km.
For example:
if Km is 2, your x intercept would be -0.5
if Km is 3, your x intercept would be -0.33 which is more than -0.5 (will be to the right of the uninhibited x intercept)
So as Km increases, -1/Km also increases - this is what you see in the graph, the enzyme competitive inhibitor complex has an increased x-intercept
This also makes intuitive sense if you think about it. A competitive inhibitor competes for the binding of your enzyme. This means that your Km (the concentration of substrate at which your enzyme is going (1/2)Vmax) will be larger, because you need more substrate to compete against your inhibitor.
I think your problem was that you were thinking about -1/Km in terms of magnitude, not actual value. Let me know if that makes sense
On a similar topic, regarding mixed inhibitors, I know that they can either bind to enzyme where they increase km or they can bind to the ES complex where they decrease km but either way it would decrease vmax. So would the graph would look like this. Is this right?
On a similar topic, regarding mixed inhibitors, I know that they can either bind to enzyme where they increase km or they can bind to the ES complex where they decrease km but either way it would decrease vmax. So would the graph would look like this. Is this right?
you're correct in that there isn't a fixed y or x intercept. although i'm not entirely certain that your graph is correct (I don't know whether it is or not). Enzyme kinetics is not my strongest topic but here's a Line-Burk plot of what mixed inhibition looks like. alpha is just the factor by which (S) must be increased to overcome the inhibitor. I don't know if you can graph the (inhibitor-enzyme) and (inhibitor ES complex) on the same graph because they are different complexes. For example, could the affinity/Vmax/Km change due to substrate binding? I don't know you've got a good question here.<