Enzymes are biological catalysts that speed up many reactions essential for the human body. Their chemistry is the same as catalysts. They speed up reactions by lowering the activation energy of reactions. This allows reactions to easily overcome the energy barrier and create the necessary products from reactants.

Competitive inhibition decreases enzyme activity by binding to the active site of the enzyme. Recall that active sites are sites on enzymes where the substrates bind. Upon binding to the enzyme, the substrates undergo changes that facilitate and speed up the chemical reaction. A competitive inhibitor binds to this active site and prevents the substrate from binding. With no binding, the substrate will not undergo the necessary changes and, subsequently, the chemical reaction. A key characterisitic of competitive inhibitors is that the bond between the inhibitor and the active site is reversible. This means that the chemical bonds involved here are weak, reversible noncovalent bonds such as hydrogen bonds and van der Waals forces. Covalent bonds are very strong and are usually found in irreversible interactions.

Allosteric inhibitors are molecules that bind to enzymes at their allosteric site(s). In this way, the allosteric inhibiton is very similar to, and is a subset of, another type of enzyme inhibition, noncompetitive inhibition.

Recall that competitive inhibition can be overcome by increasing substrate concentration. Competitive inhibitors alter the Michaelis constant, , but maintain the  (maximum reaction rate). Inhibitors act to decrease the reaction rate. To figure out the effect of competitive inhibitors on the Michaelis constant, we need to look at the Michaelis-Menten equation.

where  is reaction rate,  is maximum reaction rate,  is substrate concentration, and  is the Michaelis constant. Since reaction rate is inversely proportional to the , competitive inhibitors will increase  and, thereby, decrease reaction rate.

To answer this question we need to first figure out the equation for slope and x-intercept of Lineweaver-Burk plot. The Linweaver-Burk plot is a graphical way to plot the Michaelis-Menten equation. It is defined as the reciprocal of Michaelis-Menten equation. Michaelis-Menten equation is as follows.

where  is reaction rate,  is maximum reaction rate,  is substrate concentration, and  is the Michaelis constant. Taking the reciprocal of this gives us

The slope, therefore, is . The x-intercept can be found by plugging in zero for the Y value (the reaction rate, ). The x-intercept is .

 The question states that the slope is  and the x-intercept is . Using the equation for x-intercept we can solve for .

Using the equation for slope we can solve for 

Recall that the addition of a noncompetitive inhibitor alters the  but not the ; therefore,  is still  after the addition of noncompetitive inhibitior.

Note that if we were asked to solve for the , we would have had to use the new slope () and the same  value ().

Noncompetitive inhibitors bind to enzymes and prevent the formation of the enzyme-substrate complex. These inhibitors bind to a location other than the active site. Upon binding, the inhibitors alter the conformation of the active site and prevent the binding of substrate. They form covalent bonds with the enzyme; therefore, these are irreversible inhibitors and are hard to remove.  and  are altered by both types of inhibitors (competitive and noncompetitive). Competitive inhibitors alter the  whereas the noncompetitive inhibitors alter the .

This implies that competitive inhibition can be overcome by increasing substrate concentration whereas noncompetitive inhibition cannot. Molecularly this makes sense. Competitive inhibitors bind reversibly to the active site and prevent binding of substrate. If we were to drastically increase its concentration, substrate will compete with and remove the competitive inhibitor from the active site. Noncompetitive inhibitors, on the other hand, alter the conformation of the active site, making it hard for substrates to bind to the active site; therefore, the substrate will not be able bind, regardless of the substrate concentration.  

Inhibitors are molecules that prevent the action of catalysts. They bind to catalysts and prevent substrate binding, thereby halting the catalytic action. Since catalysts increase the speed of a reaction, addition of an inhibitor will lower the speed of the reaction. This does not mean that the reaction will stop proceeding; it simple means that it will take longer for the reaction to complete (reach equilibrium).

Remember that a catalyst speeds up both the forward and the reverse reaction; therefore, inhibitors will slow down both reactions. As mentioned, inhibitors will only slow down the reaction. The amount of products produced (equilibrium) will not change, although it will take longer for products to form.

There are two main types of inhibitors. Competitive inhibitors bind to the active site of the catalyst and prevent substrate from binding. This phenomenon causes a decreasing in the binding affinity of substrate and catalyst. However, competitive inhibitors can be overcome by adding excess substrates. The substrates will dissociate the competitive inhibitor and carry out the reaction; therefore, the reaction can still be carried out at a faster rate and the maximum rate of reaction is not altered.

Noncompetitive inhibitors bind to the catalyst at an allosteric site. They alter the conformation of the active site and prevent substrate binding. They cannot be overcome by addition of excess substrate; therefore, they lower the maximum rate of reaction.

To solve this question we need to use the definition of equilibrium constant. The equilibrium constant, , for this reaction is

For a given  value, the concentration of C and D would be the same. The stoichiometric coefficients of C and D are both 1; therefore, the amount of C and D produced would be the same and would depend on the limiting reagent. We do not know what the limiting reagent is (it could be A or B) and cannot determine the absolute concentrations of C and D; however, we can determine the relative concentrations. The concentration of C and D are the same and the ratio is 1:1.

The reaction quotient, , is calculated the same way as equilibrium constant; however, the concentrations used are derived from a nonequilbrium state. Consider the reaction below.

If this reaction is NOT in equilibrium, then the reaction quotient is defined as

If it’s at equilibrium then equilibrium constant is defined as

Note that reaction quotient cannot be calculated when reaction is in equilibrium.

The question states that the reaction quotient is twice as much as the equilibrium constant. Since we can calculate reaction quotient, this reaction is not in equilbrium. The numerator for  is higher than the numerator of  (because reaction quotient is higher). There are more products created than reactants in the current nonequilbrium state and, therefore, the forward reaction is happening faster than the reverse reaction. 

Chemical equilibrium is the state during which the rate of forward reaction equals the rate of reverse reaction. Equilibrium properties of a reaction determine how much products is produced. It’s important to remember that enzymes alter the rate of a reaction; however, they DO NOT alter the equilibrium. This means that enzymes speed up a reaction; however, they do not increase the amount of products produced.

ONLY increasing or decreasing temperature will alter equilibrium constant. Other factors such as concentration, volume, and pressure do not change it. Changing these other factors might shift the reaction left or right to bring the reaction back to equilibrium; however, it does not change the equilibrium constant. This means that these other factors can change the individual concentration of reactants and products at equilibrium, but they will not change the ratio of products to reactants.

As mentioned, at equilibrium the rates of forward and reverse reactions are equal. This means that new products and reactants are constantly being produced; however, there is no net production of these molecules.