MCAT Biochemistry Review

Chapter 2: Enzymes

2.3 Enzyme Kinetics


Enzyme kinetics is a high-yield topic that can score us several points on Test Day. Just as the relief our student derives from squeezing a stress ball depends on a number of factors, such as size and shape of the ball and his or her baseline level of stress, enzyme kinetics are dependent on factors like environmental conditions and concentrations of substrate and enzyme.

The concentrations of the substrate, [S], and enzyme, [E], greatly affect how quickly a reaction will occur. Let's say that we have 100 stress balls (enzymes) and only 10 frustrated students (substrates) to derive stress relief from them (high enzyme concentration relative to substrate). Because there are many active sites available, we will quickly form products (students letting go and feeling relaxed); in a chemical sense, we would reach equilibrium quickly. As we slowly add more substrate (students), the rate of the reaction will increase; that is, more people will relax in the same amount of time because we have plenty of available stress balls for them to squeeze. However, as we add more and more people (and start approaching 100 students), we begin to level off and reach a maximal rate of relaxation. There are fewer and fewer available stress balls until finally all active sites are occupied. Unlike before, inviting more students into the room will not change the rate of the reaction. It cannot go any faster once it has reached saturation. At this rate, the enzyme is working at maximum velocity, denoted by vmax. The only way to increase vmax is by increasing the enzyme concentration. In the cell, this can be accomplished by inducing the expression of the gene encoding the enzyme. These concepts are represented graphically in Figure 2.4.

Figure 2.4. Michaelis–Menten Plot of Enzyme Kinetics As the amount of substrate increases, the enzyme is able to increase its rate of reaction until it reaches a maximum enzymatic reaction rate (vmax); once vmax is reached, adding more substrate will not increase the rate of reaction.


For most enzymes, the Michaelis–Menten equation describes how the rate of the reaction, v, depends on the concentration of both the enzyme, [E], and the substrate, [S], which forms product, [P]. Enzyme–substrate complexes form at a rate k1. The ES complex can either dissociate at a ratek2 or turn into E + P at a rate k3:

Equation 2.1

Note that in either case, the enzyme is again available. On Test Day, the concentration of enzyme will be kept constant. Under these conditions, we can relate the velocity of the enzyme to substrate concentration using the Michaelis–Menten equation:

Equation 2.2

Some important and Test Day-relevant math can be derived from this equation. When the reaction rate is equal to half of vmaxKm = [S]:

Km can therefore be understood to be the substrate concentration at which half of the enzyme's active sites are full (half the stress balls are in use). Km is the Michaelis constant, and is often used to compare enzymes. Under certain conditions, Km is a measure of the affinity of the enzyme for its substrate. When comparing two enzymes, the one with the higher Km has the lower affinity for its substrate because it requires a higher substrate concentration to be half-saturated. The Km value is an intrinsic property of the enzyme–substrate system and cannot be altered by changing the concentration of substrate or enzyme.


We can assess an enzyme's affinity for a substrate by noting the Km. A low Km reflects a high affinity for the substrate (low [S] required for 50% enzyme saturation). Conversely, a high Km reflects a low affinity of the enzyme for the substrate.

For a given concentration of enzyme, the Michaelis–Menten relationship generally graphs as a hyperbola, as seen in the Michaelis–Menten plot in Figure 2.4. When substrate concentration is less than Km, changes in substrate concentration will greatly affect the reaction rate. At high substrate concentrations exceeding Km, the reaction rate increases much more slowly as it approaches vmax, where it becomes independent of substrate concentration.


The Lineweaver–Burk plot is a double reciprocal graph of the Michaelis–Menten equation. The same data graphed in this way yield a straight line as shown in Figure 2.5. The actual data are represented by the portion of the graph to the right of the y-axis, but the line is extrapolated into the upper left quadrant to determine its intercept with the x-axis. The intercept of the line with the x-axis gives the value of  . The intercept of the line with the y-axis gives the value of  . The Lineweaver–Burk plot is especially useful when determining the type of inhibition that an enzyme is experiencing because vmax and Km can be compared without estimation.

Figure 2.5. Experimentally Determined Lineweaver–Burk (Double Reciprocal) Plot Used to Calculate the Values of Km and vmax


Certain enzymes do not show the normal hyperbola when graphed on a Michaelis–Menten plot (v vs. [S]), but rather show sigmoidal (S-shaped) kinetics owing to cooperativity among substrate binding sites, as shown in Figure 2.6. Cooperative enzymes have multiple subunits and multiple active sites. Subunits and enzymes may exist in one of two states: a low-affinity tense state (T) or a high-affinity relaxed state (R). Binding of the substrate encourages the transition of other subunits from the T state to the R state, which increases the likelihood of substrate binding by these other subunits. Conversely, loss of substrate can encourage the transition from the R state to the T state, and promote dissociation of substrate from the remaining subunits. Think of cooperative enzyme kinetics like a party. As more people start arriving, the atmosphere becomes more relaxed and the party seems more appealing, but as people start going home the party dies down and more people are encouraged to leave so the tense hosts can clean up. Enzymes showing cooperative kinetics are often regulatory enzymes in pathways, like phosphofructokinase-1 in glycolysis. Cooperative enzymes are also subject to activation and inhibition, both competitively and through allosteric sites.

Figure 2.6. Cooperative Enzyme Kinetics


This cooperative binding of hemoglobin, which acts as a transport protein rather than an enzyme, results in a characteristic sigmoidal binding curve that is an MCAT favorite.

MCAT Concept Check 2.3:

Before you move on, assess your understanding of the material with these questions.

1.    What are the effects of increasing [S] on enzyme kinetics? What about increasing [E]?

·        Increasing [S]:

·        Increasing [E]:

2.    How are the Michaelis–Menten and Lineweaver–Burk plots similar? How are they different?

·        Similarities:

·        Differences:

3.    What does Km represent? What would an increase in Km signify?

4.    What do the x- and y-intercepts in a Lineweaver–Burk plot represent?

·        x-intercept:

·        y-intercept:

5.    What is enzyme cooperativity?