SÍNTESIS DE INVESTIGACIONES ARQUEOLÓGICAS

v (K V m m ax[S [S ] ])

Velocity depends on substrate concentration when [S] is low but does not depend on substrate concentration when [S] is high.

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Equations are useless by themselves—what’s important is what the equation tells you about how the enzyme behaves. You should be able to escape biochemistry having learned only three equations. This is one of them. The Michaelis-Menten equation describes the way in which the velocity of an enzyme reaction depends on the substrate concentration. Memorize it and understand it because the same equation (with only the names of the symbols changed) describes receptor binding and oxygen binding to myoglobin. In the cell, the enzyme is exposed to changes in the level of the substrate. The way the enzyme behaves is to increase the velocity when the substrate concentration increases. The dependence of enzyme velocity on the concentration of substrate is the first line of meta- bolic regulation. In cells, enzymes are often exposed to concentrations of the substrate that are near to or lower than the Km. Some enzymes have evolved to have Km’s near the physiological substrate concentration so that changes in the substrate levels in the cell cause changes in the veloc- ity of the reaction—the more substrate, the faster the reaction.

Mathematically, the Michaelis-Menten equation is the equation of a rectangular hyperbola. Sometimes you’ll here reference to hyperbolic kinetics; this means it follows the Michaelis-Menten equation. A number of other names also imply that a particular enzyme obeys the Michaelis- Menten equation: Michaelis-Menten behavior, saturation kinetics, and hyperbolic kinetics.

The initial velocity is measured at a series of different substrate con- centrations. In all cases the concentration of substrate used is much higher (by thousands of times, usually) than that of the enzyme. Each substrate concentration requires a separate measurement of the initial velocity. At low concentrations of substrate, increasing the substrate con- centration increases the velocity of the reaction, but at high substrate con- centrations, increasing the initial substrate concentration does not have much of an effect on the velocity (Fig. 8-4).

The derivation of the equation is found in most texts and for the most part can be ignored. What you want to understand is how and why it works as it does. v d d [P t ] d d [S t ] K V m m ax[S [S ] ]

When there is no substrate present ([S] 0), there is no velocity— so far, so good. As the substrate concentration [S] is increased, the reac- tion goes faster as the enzyme finds it easier and easier to locate the substrate in solution. At low substrate concentrations ([S] Km), dou- bling the concentration of substrate causes the velocity to double.

8 Enzyme Kinetics _• 105 _•

As the initial substrate concentration becomes higher and higher, at some concentration, the substrate is so easy to find that all the enzyme active sites are occupied with bound substrate (or product). The enzyme is termed saturated at this point, and further increases in substrate con- centration will not make the reaction go any faster. With [S] Km, the velocity approaches Vmax.

The actual velocity of the reaction depends on how much of the total amount of enzyme is present in the enzyme–substrate (ES) complex. At low substrate concentrations, very little of the enzyme is present as the ES complex—most of it is free enzyme that does not have substrate bound. At very high substrate concentrations, virtually all the enzyme is

At high [S], v doesn't change much when [S] changes

Vmax

[S]

velocity

At low [S], v increases linearly as [S] increases

K_m = [S] where v = 1/2 V_max 1/2 Vmax

Figure 8-4

SUBSTRATE CONCENTRATION affects the velocity of an enzyme-

catalyzed reaction. Almost all enzyme-catalyzed reactions show saturation behavior. At a high enough substrate concentration, the reaction just won’t go any faster than Vmax. The substrate concentration required to produce a velocity

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in the ES complex. In fact, the amount of enzyme in the ES complex is just given by the ratio of v/Vmax, that is, ES/(ES E) v/Vmax [S]/ (Km[S]).

The Vmax is a special point. At Vmax, the velocity does not depend on the concentration of substrate. Most assays are performed at substrate concentrations that are near saturating (the word near is usually used because Vmax, like Nirvana, is approached, not reached). For practical people, though, 99 percent of Vmax is as good as Vmax. The Vmax and v have exactly the same units. The Vmax conceals the dependence of the velocity on the concentration of enzyme. It’s buried in there. If Vmax is expressed in units of micromolar per minute (M/min), then doubling the enzyme concentration doubles Vmax; in contrast, if Vmax(and v) are given in units of micromoles per minute per milligram [mol/(min mg), i.e., specific activity], the normalized velocity and Vmax won’t depend on enzyme concentration.

The turnover number, or kcat(pronounced “kay kat”), is another way of expressing Vmax. It’s Vmax divided by the total concentration of enzyme (Vmax/Et). The kcat is a specific activity in which the amount of enzyme is expressed in micromoles rather than milligrams. The actual units of kcat are micromoles of product per minute per micromole of enzyme. Frequently, the micromoles cancel (even though they’re not exactly the same), to give you units of reciprocal minutes (min1). Notice that this has the same units as a first-order rate constant (see later, or see Chap. 24). The kcat is the first-order rate constant for conversion of the enzyme–substrate complex to product. For a very simple mechanism, such as the one shown earlier, kcatwould be equal to k3. For more complex

k

Turnover number—another way of expressing Vmax.

Micromoles of product made per minute per micromole of enzyme (Vmax/Et). The kcatis the first-order rate constant for the conversion of the enzyme–substrate complex to product.

V

This is the velocity approached at a saturating concentration of sub- strate. Vmaxhas the same units as v.

8 Enzyme Kinetics _• 107 _•

mechanisms kcatis actually a collection of sums and products of rate con- stants for individual steps of the mechanism.

If [S] Km, the Michaelis-Menten equation says that the velocity will be one-half of Vmax. (Try substituting [S] for Km in the Michaelis- Menten equation, and you too can see this directly.) It’s really the rela- tionship between Km and [S] that determines where you are along the hyperbola. Like most of the rest of biochemistry, Km is backward. The larger the Km, the weaker the interaction between the enzyme and the substrate. Kmis also a collection of rate constants. It may not be equal to the true dissociation constant of the ES complex (i.e., the equilibrium constant for ES sE S).

The Km is a landmark to help you find your way around a rectan- gular hyperbola and your way around enzyme behavior. When [S] Km (this means [S] Km Km), the Michaelis-Menten equation says that the velocity will be given by v(Vmax/Km)[S]. The velocity depends lin- early on [S]. Doubling [S] doubles the rate.

At high substrate concentrations relative to Km ([S] Km), The Michaelis-Menten equation reduces to v Vmax, substrate concentration disappears, and the dependence of velocity on substrate concentration approaches a horizontal line. When the reaction velocity is independent of the concentration of the substrate, as it is at Vmax, it’s given the name

zero-order kinetics.

k

/K

The specificity constant, kcat/Km, is the second-order rate constant for the reaction of E and S to produce product. It has units of M1min1.

SPECIAL POINTS

[S] Km v Vmax[S]/Km [S] Km v Vmax [S] Km v Vmax/2

K

This is the concentration of substrate required to produce a veloc- ity that is one-half of Vmax.

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4_{If you’re reading this section because you want to understand how rate accelerations are actu-} ally determined, proceed; however, this information will be pretty low on the trivia sorter list. The term kcat/Kmdescribes the reaction of any enzyme and substrate at low substrate concentration. At low substrate concentration, the veloc- ity of an enzyme-catalyzed reaction is proportional to the substrate con- centration and the enzyme concentration. The proportionality constant is kcat/Km and v (kcat/Km)[S] [E]T. If you’re real astute, you’ll have noticed that this is just a second-order rate equation and that the second- order rate constant is kcat/Km.

The term kcat/Kmlets you rank enzymes according to how good they are with different substrates. It contains information about how fast the reaction of a given substrate would be when it’s bound to the enzyme (kcat) and how much of the substrate is required to reach half of Vmax. Given two substrates, which will the enzyme choose? The quantity kcat/Kmtells you which one the enzyme likes most—which one will react faster.

The term kcat/Kmis also the second-order rate constant for the reac- tion of the free enzyme (E) with the substrate (S) to give product. The kcat/Km is a collection of rate constants, even for the simple reaction mechanism shown earlier. Formally, kcat/Km is given by the pile of rate constants k1k3/(k2 k3). If k3 k2, this reduces to k1, the rate of encounter between E and S. Otherwise, kcat/Km is a complex collection of rate constants, but it is still the second-order rate constant that is observed for the reaction at low substrate concentration.

To impress you, enzymologists often tell you how much faster their enzyme is than the uncatalyzed reaction. These comparisons are tricky. Here’s the problem: Suppose we know that the reaction S P has a first- order rate constant of 1 103 min1 (a half-life of 693 min). When an enzyme catalyzes transformation of S to P, we have more than one reaction:

k1 _k

3

E S ∆ES ¡E P

k2

Which of these reactions do we pick to compare with the noncatalyzed reaction? We can’t pick k1, because that’s a second-order reaction. You