For example, ethanol is metabolized into acetaldehyde by the enzyme, alcohol dehydrogenase. In the absence of ADH, the rate of the reaction would be less than 0. Please read the Duke Wordpress Policies. Contact the Duke WordPress team.
The Alcohol Pharmacology Education Partnership. Schwartz-Bloom, Ph. Fulton T. Crews, Ph. Linda J. Porrino, Ph. David P. Friedman, Ph. Leslie Morrow, Ph. Kathleen K. Sulik, Ph. Search Search. Enzymes as Catalysts Enzymes are proteins that have a specific function. These enzymes, termed hydrolases, break single bonds by adding the elements of water. Other hydrolases function as digestive enzymes, for example, by breaking the peptide bonds in proteins. Formation or removal of a double bond with group transfer.
The functional groups transferred by these lyase enzymes include amino groups, water, and ammonia. Dehydratases remove water, as in fumarase fumarate hydratase :. Deaminases remove ammonia, for example, in the removal of amino groups from amino acids:. Isomerization of functional groups. In many biochemical reactions, the position of a functional group is changed within a molecule, but the molecule itself contains the same number and kind of atoms that it did in the beginning.
In other words, the substrate and product of the reaction are isomers. The isomerases for example, triose phosphate isomerase, shown following , carry out these rearrangements. Single bond formation by eliminating the elements of water. Hydrolases break bonds by adding the elements of water; ligases carry out the converse reaction, removing the elements of water from two functional groups to form a single bond.
Synthetases are a subclass of ligases that use the hydrolysis of ATP to drive this formation. Another way to look at enzymes is with an initial velocity plot. The rate of reaction is determined early in the progress curve—very little product is present, but the enzyme has gone through a limited number of catalytic cycles.
In other words, the enzyme is going through the sequence of product binding, chemical catalysis, and product release continually. This condition is called the steady state.
For example, the three curves in Figure represent progress curves for an enzyme under three different reaction conditions. In all three curves, the amount of enzyme is the same; however, the concentration of substrate is least in curve a , greater in curve b , and greatest in curve c. The progress curves show that more product forms as more substrate is added.
The slopes of the progress curves at early time, that is, the rate of product formation with time also increase with increasing substrate concentration. These slopes, called the initial rates or initial velocities, of the reaction also increase as more substrate is present so that:. The more substrate is present, the greater the initial velocity, because enzymes act to bind to their substrates. Figure 2. A plot of the initial velocities versus substrate concentration is a hyperbola Figure.
Why does the curve in Figure flatten out? Because if the substrate concentration gets high enough, the enzyme spends all its time carrying out catalysis and no time waiting to bind substrate. In other words, the amount of substrate is high enough so that the enzyme is saturated, and the reaction rate has reached maximal velocity, or V max.
Note that the condition of maximal velocity in Figure is not the same as the state of thermodynamic equilibuium in Figures 1 and 2. Figure 3. Although it is a velocity curve and not a binding curve, Figure is a hyperbola. Just as myoglobin is saturated with oxygen at high enough pO 2 , so an enzyme is saturated with substrate at high enough substrate concentration, designated [S].
The equation describing the plot in Figure is similar in form to the equation used for O 2 binding to myoglobin:. K m is the Michaelis constant for the enzyme binding substrate.
The Michaelis constant is analogous to, but not identical to, the binding constant for the substrate to the enzyme. V max is the maximal velocity available from the amount of enzyme in the reaction mixture.
If you add more enzyme to a given amount of substrate, the velocity of the reaction measured in moles of substrate converted per time increases, because the increased amount of enzyme uses more substrate. This is accounted for by the realization that V max depends on the total amount of enzyme in the reaction mixture:. In other words, the K m is numerically equal to the amount of substrate required so that the velocity of the reaction is half of the maximal velocity. In the last case, the enzyme is said to be under first order conditions, because the velocity depends directly on the concentration of substrate.
Alkaline phosphatase catalyzes a simple hydrolysis reaction:. Phosphate ion, a product of the reaction, also inhibits it by binding to the same phosphate site used for binding substrate.
When phosphate is bound, the enzyme cannot bind substrate, so it is inhibited by the phosphate. How to overcome the inhibitor? Because the substrate and the inhibitor bind to the same site on the enzyme, the more substrate that binds, the less inhibitor binds. When is the most substrate bound to the enzyme?
Under V max conditions. Phosphate ion reduces the velocity of the alkaline phosphate reaction without reducing V max.
If velocity decreases, but V max doesn't, the only other thing that can change is K m. Because more substrate is required to achieve V max , K m must necessarily increase.
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