I edited what I wrote to someone I was tutoring, it's overkill for your question but maybe it'll be helpful for someone else. Hope this makes things more clear and not less!
TL;DR
-Rate of reaction is sigmoidal with respect to S , but linear with respect to [ES].
-Formation of [ES] depends upon collision rate, which is constant at any S
-As S goes up, [E] goes down, so plateau is due to needing high [E] to compensate for low [E] in order to maintain collision rate
-VMax is the rate of the reaction when the max possible amt ES complexes are formed
-1/2VMax is the rate of the reaction when half of the max possible amt ES complexes are formed
-Km = S required to achieve 1/2VMax
-Competitive inhibitors compete with the substrate for the active site, decreasing affinity for substrate, so Km goes up. Increasing ratio of S:I will "win" the competition, so V at any given concentration below VMax decreases, but VMax does not change, only requires higher S to achieve.
-Mixed inhibitors bind allosterically to free E, or to ES complex, or to both with a preference, or to both with no preference. VMax always decreases.
-Uncompetitive inhibitor is mixed inhibitor with total preference for ES. VMax down, decrease in Km, affinity up due to 'locking' the substrate in.
-Non-competitive inhibitor is mixed inhibitor with equal preference for E or ES. VMax down, no change in Km.
-Other extreme is a mixed inhibitor with total preference for E. VMax down, Km up, affinity down (less E available).
-Generally: If preference for ES over E, Km goes down. If prefers E over ES, Km goes up.
TL;DR TL;DR
Competitive: Km up, VMax no change, curve shifts right
Mixed = VMax always down. Includes NC, UC, and only .
UC: Only binds ES. Km down (locks substrate)
NC: Binds ES and E equally. No change in Km.
Only binds E: Km up (less free E).
In short, the explanation that mixed inhibitors are able to act as "competitive (affinity for Efree > affinity for ES), uncompetitive (ES > Efree), or noncompetitive (ES = Efree)" is specifically referring to the effect on the Km. The VMax will always decrease, and the effect on the Km will be the same as the effect the other inhibitors have on Km.
Thinking about it intuitively, it makes sense that a competitive inhibitor doesn't decrease Km because the inhibitor and the substrate are competing for the active site. If the enzyme binds the inhibitor, it will do nothing, and if it binds the substrate, it will act normally. The question is "Will the enzyme bind the substrate?" So for a competitive inhibitor its really about the ratio of [substrate] / [inhibitor]. If you visualize what's going on, the substrate, inhibitor, and enzyme molecules are bumping around, so really the rate of the reaction is the measure of how often the enzyme bumps into a substrate (in the correct orientations). So let's say you add the competitive inhibitor at a concentration equal to the substrate. Now half of the enzyme's collisions will be with the substrate, and half will be the inhibitor. Recall that the Km is the concentration of substrate required to reach Vmax. VMax is the rate of the reaction when the enzyme is saturated (100% of enzyme molecules bound to substrate). So then Km is the concentration required to bind 50% of the enzyme molecules to substrate. When you add in a competitive inhibitor, some of those substrate-bound enzymes will now bind to the inhibitor instead. The logical way to overcome this is to just add more substrate. So the concentration of substrate needed to reach 50% saturation will increase, meaning that the affinity is decreased. Also, the concentration required to reach 100% saturation will increase. However, we CAN still reach 100% saturation if we add enough substrate. And if we can get to 100% saturation, then we have VMax, regardless of the inhibitor. Note that increasing the concentration of inhibitor will negate the effect of increased substrate. Also note that on a molecular level, the reason why the same gives a rate is due to two reasons. 1) Competition reduces frequency of collisions. 2) Because of this, turnover time between ES complexes is increased due the option of creating EI complexes. Increasing or decreasing can fix both of these.
However, mixed inhibition and uncompetitive/noncompetitive (both are types of mixed inhibition) works through allosteric means (for the purposes of the MCAT, at least). The questions we ask are: 1) Will it bind the free enzyme, or the ES complex? 2) What effect will its binding have on the enzymes function? We can get super in depth here, but let's just use LeChatliere's principle.
E + S --> ES
ES --> E + P
Under LeChatliere's Principle, if the forward reaction rate of one of these increases, so will the other one. If the first reaction is pushed right, then our rate goes up. So let's just talk about the first one.
It is also best to try and visualize the following:
Under conditions of low S, there will be excess E. Formation of ES depends on S, and the relationship is linear.
Under conditions of high S, there will be little free enzyme available for binding. Increasing the reaction rate a certain amount will require a larger increase in than under low conditions.
So to define a mixed inhibitor's effect, we ask "does the inhibitor prefer E or ES, and by how much?" Let's consider the extremes.
An uncompetitive inhibitor only binds the ES complex. This means that the effect of our inhibitor is directly related to . At a certain S , a certain amount of [E] will become [ES], and at a certain S, a certain amount of [ES] will become [ESI] instead of becoming E + P. This means that formation of [ESI] is equally dependent on S and [E]. With or without uncompetitive inhibition, at a certain high S, saturation occurs causing VMax. However, if some portion of our bound complexes are ESI rather than ES, than uncompetitive inhibition must lower the VMax. What about Km? Remember, the formation of ESI is only dependent upon the formation of ES What does this mean? This means that at low S, I will have very little effect, and we will form product at a similar rate with or without inhibition. Actually, at low S with uncompetitive inhibition, our reaction rate may actually increase. This is weird.
If is low, we're only forming a little bit of ES. But remember, at a set I, a set amount of [ES] will form [ESI]. So even though the effect of the inhibitor is negligible (due to low [ES]), a portion of the ES that is formed will not proceed. The effect of this makes no sense unless you remember LeChatliere's Principle. By adding our inhibitor, we actually decrease the products of our reaction E + S --> ES, which causes an increase in the reaction rate.
We know that our uncompetitive inhibitor lowers our Vmax, which means 1/2VMax is also lower. Let's pretend that VMax is 1, and inhibited vmax is 0.8. Their respective 1/2 VMaxes are 0.5 and 0.4. At low S, an uncompetitive inhibitor will either not affect the reaction rate or may even increase the reaction rate a bit. Therefore, it makes sense that less S is needed to reach
V = 0.4 than V = 0.5, even with the inhibitor present. This means that Km decreases. The easy intuitive way to think about this: An uncompetitive inhibitor binds the ES complex, causing the substrate to stay in the active site for longer than it normally would. This is undeniably an increase in affinity.
What if we had a mixed inhibitor that only bound to I and never bound to ES?
This is what we normally refer to as "mixed" inhibition, though mixed inhibition really encompasses any reversible allosteric inhibitor.
This kind of enzyme would work by momentarily (reversible) preventing the binding of I to its substrate. Let's look at LeChatliere's Principle again:
E + S --> ES
If the enzyme pulls E out of the reaction, that causes a push to the left, thus decreasing the rate of the reaction. The effect of this kind of inhibitor will be dependent on [E]. We also know that [E] is inversely dependent on S. At low S, we will have low [ES], which means that most of our enzyme will be [E]. With high S, most of our enzyme will be bound. Regardless, if we have a certain amount inhibitor I, at any point some portion of [E] will be bound as [EI] and unable to bind S. This causes an increase in the total available amount of enzyme for substrate binding, and an increase in the time of turnover. So at a certain S, we will again see saturation (or near saturation) with or without an inhibitor, except with the inhibitor our VMax will again be lower. At 1/2 VMax, S is lower than at VMax, so I will be greater, and the effect of the inhibitor will be greater. This means that the same amount of S will result in a lower reaction rate in the presence of this inhibitor than in the absence. Let's reuse our fake numbers, where VMax and 1/2VMax without inhibition are 1.0 and 0.5, and with are 0.8 and 0.4. In this fake example, because we have less enzyme available for binding, the S that gives us V = 0.5 without inhibition gives us a V less than 0.4 with inhibition. This means we have to increase our S to get 1/2VMax, which means our Km goes up.
Intuitively, this makes sense - decreasing the total amount of enzyme available for binding means that a molecule of enzyme is less likely to bind a molecule of substrate at all, thus decreasing the affinity.
What if an inhibitor has equal preference for ES and E?
This is what we call non-competitive inhibition.
At high S, we have high [ES], so [ESI] will be greater than [EI].
At low S, we have high [E], so [EI] will be greater than [ESI].
Regardless, the inhibitor will be inhibiting all over the place. We are definitely going to get a decrease in our reaction rate, so our VMax will be lower.
Let's look at the lechatliere effect. E + S --> ES. If our inhibitor is decreasing both ES and E with equal preference, what happens? Let's look at the rate quotient. Q = [ES] / [E] S. Let's say that at a certain concentration of our inhibitor, it binds to half of both [ES] and [E], thus reducing the amount left for the reaction by half their original values each.
Q = (1/2 [ES])
(1/2 [E]) * S
.. the 1/2s cancel out, we get our original reaction? Nice. Back to the numbers we made up to see what this means. VMax = 1 w/o inhibition, VMax = 0.8 w/ inhibition. Let's make up some more numbers. Let's pretend we start with 100 molecules of enzyme. Let's pretend this means that at VMax we have 100 ES complexes and 80 ES complexes respectively. Because our inhibitor likes ES and I respectively, we could assume the rest of the enzyme (20 molecules), is ESI + EI. At high S, most of this is ESI, and at low S, most of this is EI, but it's pretty safe to say that [ESI] + [EI] remains constant throughout. So let's say that in the case of inhibition, we have 20 less total enzymes in our reaction.
At 1/2 VMax, half of our enzymes are ES and half are free E. Noncompetitive inhibitors have equal preference for both. So if 20 enzymes are bound up at any time by our inhibitor, we can go ahead and say that our distribution between ESI and EI will be 10 each at 1/2 VMax. Without inhibition, our VMax has 100 ES and 0 E, and 1/2 VMax has 50 ES and 50E. With inhibition, VMax has 80 ES, and 0 E (ESI + EI = 20). At 1/2 VMax we have 40 ES, 40 E, 10 ESI and 10 EI.
In each case at 1/2 VMax, S = Km. Let's calculate it. We already saw that our reaction equations before and after inhibition are the same. So let's set the two conditions equal to each other.
Q=ES/S*E
No inhibition = Inhibition
50/S*50 = 40/S*40
S no inhibition = S inhibition!
Km doesn't change.
The S required to reach 1/2 VMax in both cases is the same, so Km does not change with uncompetitive inhibition. This makes sense intuitively because the effect of the decreased affinity from binding up free enzyme is balanced out by the increased affinity from ESI holding the substrate inside the enzyme for longer.
So for a mixed inhibitor, VMax always decreases. Having a preference for [ES] (uncompetitive) increases affinity, and decreases Km. Increasing preference for E decreases affinity, and increases Km. Equal preference for both (non-competitive) shows no change in Km or affinity.
