Figure 3–46 enzyme kinetics. The rate of an enzyme reaction (V) increases as the substrate concentration increases until a maximum value (Vmax) is reached. At this point all substrate-binding sites on the enzyme molecules are fully occupied, and the rate of reaction is limited by the rate of the catalytic process on the enzyme surface. For most enzymes, the concentration of substrate at which the reaction rate is half-maximal (Km) is a measure of how tightly the substrate is bound, with a large value of Km corresponding to weak binding.
WHY ANALYZE THE KINETICS OF ENZYMES? Enzymes are the most selective and powerful catalysts known. An understanding of their detailed mechanisms provides a critical tool for the discovery of new drugs, for the large-scale industrial synthesis of useful chemicals, and for appreciating the chemistry of cells and organisms. A detailed study of the rates of the chemical reactions that are catalyzed by a purifed enzyme—more specifcally how these rates change with changes in conditions such as the concentrations of substrates, products, inhibitors, and regulatory ligands—allows biochemists to fgure out exactly how each enzyme works. For example, this is the way that the ATP-producing reactions of glycolysis, shown previously in Figure 2–48, were deciphered—allowing us to appreciate the rationale for this critical enzymatic pathway. In this Panel, we introduce the important feld of enzyme kinetics, which has been indispensable for deriving much of the detailed knowledge that we now have about cell chemistry. 142 Panel 3–2: Some of the Methods Used to Study enzymes
STEADY-STATE ENZYME KINETICS Many enzymes have only one substrate, which they bind and then process to produce products according to the scheme outlined in Figure 3–50A. In this case, the reaction is written as Here we have assumed that the reverse reaction, in which E + P recombine to form EP and then ES, occurs so rarely that we can ignore it. In this case, EP need not be represented, and we can express the rate of the reaction—known as its velocity, V, as where [ES] is the concentration of the enzyme–substrate complex, and kcat is the turnover number, a rate constant that has a value equal to the number of substrate molecules processed per enzyme molecule each second. But how does the value of [ES] relate to the concentrations that we know directly, which are the total concentration of the enzyme, [Eo], and the concentration of the substrate, [S]? When enzyme and substrate are frst mixed, the concentration [ES] will rise rapidly from zero to a so-called steady-state level, as illustrated below. E + S E + P ES k1 k–1 kcat V = kcat [ES]
At this steady state, [ES] is nearly constant, so that or, since the concentration of the free enzyme, [E], is equal to [Eo] – [ES], Rearranging, and defning the constant Km as we get or, remembering that V = kcat [ES], we obtain the famous Michaelis–Menten equation As [S] is increased to higher and higher levels, essentially all of the enzyme will be bound to substrate at steady state; at this point, a maximum rate of reaction, Vmax , will be reached where V = Vmax = kcat [Eo]. Thus, it is convenient to rewrite the Michaelis–Menten equation as rate of ES formation k1 [E][S] rate of ES breakdown k–1 [ES] + kcat [ES] = k1 k–1 + kcat [ES] = [E][S] = [Eo] – [ES] [S] k1 k–1 + kcat k1 k–1 + kcat Km + [S] kcat [Eo][S] [ES] = [Eo][S] Km + [S] V = Km + [S] Vmax [S]V =
THE DOUBLE-RECIPROCAL PLOT A typical plot of V versus [S] for an enzyme that follows Michaelis–Menten kinetics is shown below. From this plot, neither the value of Vmax nor of Km is immediately clear.