If the reaction rate has plateaued, this indicates that the enzyme has reached saturation. At this point, every active site on every molecule of enzyme is actively catalyzing the reaction as quickly as it can. The only way to change the reaction rate, at this point, would be to increase the concentration of the enzyme in the solution. Further increasing substrate concentration will have no effect.

We know that the enzyme has not become denatured because the reaction is still occurring. The rate of the reaction is constant during the plateau, and does not drop to zero.

A Lineweaver-Burk plot is a way to graphically represent enzyme kinetics. It is convenient because several portions of the graph readily display important information, such as rate constants. The y-intercept in particular is useful because it represents the reciprocal of the maximum velocity. The x-intercept describes the negative reciprocal of the Michaelis constant. The slope is the quotient of the Michaelis constant over the maximum velocity.

The two plots contain the same information. A Michaelis-Menten plot shows the relationship between initial reaction rate concentration of substrate ( versus ). A Lineweaver-Burk plot shows the relationship between the inverses of these same two variables, however, it is much easier to visualize important data on a Lineweaver-Burk plot. The x-intercept, the y-intercept, and the slope all contain points of interest. A downside of the Lineweaver-Burk plot, however, is that it is more susceptible to inaccuracy if there is some flaw in the accumulated data. 

The only option that will alter the is to add a non-competitive inhibitor. The addition of this inhibitor will affect the amount of free enzyme available to catalyze the reaction, and thus lower the by reducing the effective enzyme concentration.

Addition of a competitive inhibitor will alter the , but not the . Increasing the substrate concentration will have no effect once saturation has been reached. 

During a reaction, the reactants must pass through high-energy transition states before they evolve into the products. The catalyst reduces the free energy of this transition state, thus making it 'easier' for the reactant to undergo the chemical reaction since the activation energy has been lowered. 

The correct answer is uncompetitive inhibition. The formation of a enzyme-substrate complex creates an alternative site on the enzyme for an inhibitor to bind. This mechanism is considered uncompetitive because the inhibitor and substrate are not competing for the same binding site on the enzyme. 

Enzymes exert their effect on the reaction rate by decreasing the energy needed for the reaction to proceed. As a result, the enzyme will decrease the activation energy. It should be noted that the forward reaction rate and reverse reaction rate are both increased by an enzyme. If this were not the case, more product would be made compared to the uncatalyzed reaction, and enzymes do not affect equilibrium constants for reactions.

An enzme is a biological catalyst that increases the rate of a reaction. This is accomplished by lowering the activation energy necessary to start the reaction. The equilibrium of the reaction, however, is not affected. This means that enthalpy and entropy are not affected by an enzyme's presence.

The correct answer is lyases. This class of enzymes only requires one substrate for the forward reaction.

Ligases catalyze the formation of a bond between two molecules, isomerases rearrange the atoms of a molecule, kinases phosphorylate molecules, and transferases transfer or join functional groups from one molecule to another. 

Kinases are enzyme that catalyze the transfer of a phosphate group from ATP to a substrate molecule. The phosphorylation of fructose-6-phosphate to fructose-1,6-phosphate is mediated by a kinase phosphofructokinase.