Recall that the change in  is only effected by the binding affinity to the enzyme substrate complex. We can represent the effect on the  by this inhibitor by the following equation. 

Since the apparent  decreased by a factor of 3, we know that  is 3.

Solve for  in the following equation:

Above,  is of the enzyme substrate complex, which is . 

Solve.

This is an example of feedback inhibition, as feedback inhibition is a mechanism in which a molecule binds to an enzyme to decrease its activity. The mechanism is now balanced. Blocking an enzyme typically helps correct a metabolic imbalanace or assists in destruction of a pathogen. In this case the ATP binds to a site other than the protein's active site, and since it blocks PFK-1, a feedback inhibition has occurred.

Competitive inhibitors bind to the active site of the target enzyme. K_m is the substrate concentration at which the reaction rate is at half V_max. A competitive inhibitor can be outcompeted by adding additional substrate; thus V_max is unaffected, since it can be accomplished with enough additional substrate. However, since we need to add additional substrate to compete with the inhibitor to get the reaction to the same rate, our K_m increases.

The graph shown in the question stem is a Lineweaver-Burk plot, otherwise known as a double-reciprocal plot. In this plot,  is plotted along the -axis and  is plotted along the -axis. Furthermore, the -intercept in such a graph is equal to , and the -intercept is equal to .

From the graph shown in the question stem, we can see that there are two lines, each with different slopes. Each line corresponds to a certain concentration of inhibitor. (Note that one of the lines corresponds to no inhibitor, or a concentration of 0.)

It is evident that the two lines intersect each other along the -axis, right on the  value. Consequently, we can conclude that the inhibitor in this case is not having any effect on the value of  for the enzyme. Despite the different slopes for the two reactions, both of them have a common -intercept but differing -intercepts. This means that we can conclude the inhibitor in question must be competitive, since the result will be a rise in the  value for the enzyme-catalyzed reaction, but will have no effect on the  value.

This scenario is a classic example of allosteric inhibition. When cyanide binds to a site on cytochrome c oxidase other than the active site, cytochrome C oxidase becomes deactivated, stopping oxidative phosphorylation and causing cells to die since they cannot produce ATP anymore.

In competitive inhibition, the inhibitor binds the active site of the enzyme, competing with the substrate for this binding site. The  of a competitively inhibited enzyme remains unchanged, but the  increases. This means that a higher concentration of substrate is required to bring the reaction rate to . However, since this is competitive inhibition, and the maximum velocity is unchanged, we can overcome this increase in  and achieve maximum velocity if we saturate the enzyme with substrate.

Competitive inhibition involves the substrate's access to the active site. In the case of competitive inhibition, the inhibitor blocks the substrate from the active site. As a result, the  is unchanged, but the  is increased. Recall that  is the substrate concentration at which the reaction rate is . Additionally, the reaction rate will increase with increased concentration of competitive inhibitor and substrate, because they are competing for the active site, causing an increase in reaction rate.

Uncompetitive inhibitors can only bind the ES complex, whereas competitive and non-competitive inhibitors do not require the enzyme to be complexed with the substrate. , which describes the maximum reaction velocity of the enzyme, is decreased because the inhibitor slows the dissociation of the substrate from the enzyme, thereby slowing the rate at which the enzyme can interact with other substrate molecules. 

All types of inhibitors will induce a change in the  of an enzyme except for competitive inhibitors. This is because competitive inhibitors have no effect on the enzyme-subtrate complex. The  may still be reached, but by adding more substrate, since the  is increased by a competitive inhibitor.

This question is presenting us with a situation in which a chemical is being added to a mixture of enzyme and substrate, and its effects on the kinetic parameters of the reaction are observed. We're told that the  and  for this reaction both become reduced. We then are asked to identify which term best describes the added chemical.

To begin with, let's take note that all of the answer choices are some kind of inhibitor. Thus, we know that the chemical we're adding to the mixture is an inhibitor of some type. The challenge is in identifying which type of inhibition is happening. For this question, we'll need to have familiarity with each type of inhibition in order to identify the correct answer.

Let's start with what we know. Both the  and the  are being decreased. Right away, we can rule out competitive inhibition because the  should remain the same.

We should also be able to rule out feedback inhibition right off the bat, as this kind of inhibition involves the products of a reaction putting a halt on the reaction that led to its production.

Next, we can also realize that two of the answer choices are so similar that they are actually saying nearly the same thing. Mixed inhibition is a case in which the inhibitor binds to the enzyme regardless of whether substrate is also bound to the enzyme. However, with mixed inhibition, the inhibitor shows greater affinity for either the free enzyme or the enzyme-substrate complex. In such a case, the  for the reaction is expected to fall, but the  can either increase or decrease.

Noncompetitive inhibition is a special type of mixed inhibition, in which the inhibitor binds both the free enzyme and the enzyme-substrate complex with equal affinity. In such a situation, the  of the reaction will fall, but the  will remain unchanged.

And finally, we look at uncompetitive inhibition, which is the correct answer. In this type of inhibition, the inhibitor binds only to the enzyme-substrate complex rather than the free enzyme. It does so by binding to an allosteric site, which is distinct from the active site to which substrate binds. Thus, there is no way to out-compete the inhibitor by adding more and more substrate, as can be done in competitive inhibition. The end result of this is that the  becomes irrecoverably lowered. And since this value becomes less, the substrate concentration needed to obtain half of that reduced value (the ) also becomes decreased.