B. Alberts, A. Johnson, J. Lewis и др. - Molecular Biology of The Cell (6th edition) (1120996), страница 55
Текст из файла (страница 55)
Thus, it is convenient to rewrite theMichaelis–Menten equation asV =pre-steadystate:ES formingkcat [Eo][S]Vmax [S]Km + [S]143THE DOUBLE-RECIPROCAL PLOTTHE SIGNIFICANCE OF Km, kcat, and kcat /KmA typical plot of V versus [S] for an enzyme that followsMichaelis–Menten kinetics is shown below. From this plot,neither the value of Vmax nor of Km is immediately clear.V = steady state velocity ofproduct formation (µmole/second)[S] =12345678806040A comparison of kcat/Km for the same enzyme withdifferent substrates, or for two enzymes with theirdifferent substrates, is widely used as a measure of enzymeeffectiveness.200042[S]68mmole/literTo obtain Vmax and Km from such data, a double-reciprocalplot is often used, in which the Michaelis–Menten equationhas merely been rearranged, so that 1/V can be plottedversus 1/ [S].1/V=KmVmaxAs described in the text, Km is an approximate measure ofsubstrate affinity for the enzyme: it is numerically equal tothe concentration of [S] at V = 0.5 Vmax.
In general, a lowervalue of Km means tighter substrate binding. In fact, forthose cases where kcat is much smaller than k–1, the Km willbe equal to Kd, the dissociation constant for substratebinding to the enzyme (Kd = 1/Ka; see Figure 3–44).We have seen that k cat is the turnover number for theenzyme.
At very low substrate concentrations, where[S] << Km, most of the enzyme is free. Thus we can think of[E] = [Eo], so that the Michaelis–Menten equation becomesV = kcat/Km [E][S]. Thus, the ratio kcat/Km is equivalent to therate constant for the reaction between free enzyme andfree substrate.1[S]For simplicity, in this Panel we have discussed enzymesthat have only one substrate, such as the lysozyme enzymedescribed in the text (see p. 144). Most enzymes have twosubstrates, one of which is often an active carriermolecule—such as NADH or ATP.A similar, but more complex, analysis is used to determinethe kinetics of such enzymes—allowing the order of substratebinding and the presence of covalent intermediates alongthe pathway to be revealed.+ 1/ VmaxSOME ENZYMES ARE DIFFUSION LIMITED[S] =864321The values of kcat, Km, and kcat /Km for some selectedenzymes are given below:enzymeacetylcholinesterase1/V (second/µmole)0.04– 0.5–1[S]–fumarase0.030.02slope = KM / Vmax0.011Vmax– 0.25=catalase01Km0.251[S]0.50.75liter/mmolesubstratekcat(sec–1)Km(M)kcat/Km(sec–1M–1)acetylcholine1.4x1049x10–51.6x108H2O24x10714x107fumarate8x102–65x101.6x108Because an enzyme and its substrate must collide beforethey can react, kcat /Km has a maximum possible value that is1.0limited by collision rates.
If every collision forms anenzyme–substrate complex, one can calculate from diffusiontheory that kcat /Km will be between 108 and 109 sec–1M–1, inthe case where all subsequent steps proceed immediately.Thus, it is claimed that enzymes like acetylcholinesterase andfumarase are “perfect enzymes,” each enzyme havingevolved to the point where nearly every collision with itssubstrate converts the substrate to a product.144Chapter 3: ProteinsFigure 3–47 Enzymatic acceleration of chemical reactions by decreasingthe activation energy.
There is a single transition state in this example.However, often both the uncatalyzed reaction (A) and the enzyme-catalyzedreaction (B) go through a series of transition states. In that case, it is thetransition state with the highest energy (ST and EST) that determines theactivation energy and limits the rate of the reaction. (S = substrate;P = product of the reaction; ES = enzyme–substrate complex; EP = enzyme–product complex.)Enzymes Can Use Simultaneous Acid and Base CatalysisFigure 3–48 compares the spontaneous reaction rates and the correspondingenzyme-catalyzed rates for five enzymes. Rate accelerations range from 109 to1023.
Enzymes not only bind tightly to a transition state, they also contain preciselypositioned atoms that alter the electron distributions in the atoms that participatedirectly in the making and breaking of covalent bonds. Peptide bonds, for example, can be hydrolyzed in the absence of an enzyme by exposing a polypeptide toeither a strong acid or a strong base. Enzymes are unique, however, in being ableto use acid and base catalysis simultaneously, because the rigid framework of theprotein constrains the acidic and basic residues and prevents them from combining with each other, as they would do in solution (Figure 3–49).The fit between an enzyme and its substrate needs to be precise.
A smallchange introduced by genetic engineering in the active site of an enzyme cantherefore have a profound effect. Replacing a glutamic acid with an aspartic acidin one enzyme, for example, shifts the position of the catalytic carboxylate ion byonly 1 Å (about the radius of a hydrogen atom); yet this is enough to decrease theactivity of the enzyme a thousandfold.STAenergyBecause this tight binding greatly lowers the energy of the transition state, theenzyme greatly accelerates a particular reaction by lowering the activation energythat is required (Figure 3–47).activation energyfor uncatalyzed reactionESTSESBPEPprogressof reactionactivation energyfor catalyzed reactionMBoC6 m3.46/3.43Lysozyme Illustrates How an Enzyme WorksTo demonstrate how enzymes catalyze chemical reactions, we examine an enzymethat acts as a natural antibiotic in egg white, saliva, tears, and other secretions.Lysozyme catalyzes the cutting of polysaccharide chains in the cell walls of bacteria.
The bacterial cell is under pressure from osmotic forces, and cutting even asmall number of these chains causes the cell wall to rupture and the cell to burst.A relatively small and stable protein that can be easily isolated in large quantities,lysozyme was the first enzyme to have its structure worked out in atomic detail byx-ray crystallography (in the mid-1960s).The reaction that lysozyme catalyzes is a hydrolysis: it adds a molecule of waterto a single bond between two adjacent sugar groups in the polysaccharide chain,thereby causing the bond to break (see Figure 2–9).
The reaction is energeticallyfavorable because the free energy of the severed polysaccharide chain is lowerhalf-time for reaction106 years1 year1 min1msec1µsecOMP decarboxylasestaphylococcal nucleaseadenosine deaminasetriosephosphateisomerasecarbonicanhydraseUNCATALYZEDCATALYZEDFigure 3–48 The rate accelerationscaused by five different enzymes.(Adapted from A. Radzicka andR. Wolfenden, Science 267:90–93, 1995.)PROTEIN FUNCTION145+NHOOSLOWNHHCOFASTHCHCNHHHOOFASTHCHHCHOHHVERYFASTCNHO(B)acid catalysis(C)HCHOHHOC(A) no catalysisCNHHO+NHOOCbase catalysis(D)both acid andbase catalysesthan the free energy of the intact chain.
However, there is an energy barrier to thereaction, and a colliding water molecule can break a bond linking two sugars onlyif the polysaccharide molecule is distorted into a particular shape—the transitionstate—in which the atoms around the bond have an altered geometry and electron distribution. Because of this requirement, random collisions must supply avery large activation energy for the reaction to take place.
In an aqueous solutionat room temperature, the energy of collisions almost never exceeds the activationenergy. The pure polysaccharide can therefore remain for years in water withoutbeing hydrolyzed to any detectable degree.This situation changes drastically when the polysaccharide binds to lysozyme.The active site of lysozyme, because its substrate is a polymer, is a long groove thatMBoC6 m3.49/3.45holds six linked sugars at the sametime. As soon as the polysaccharide binds toform an enzyme–substrate complex, the enzyme cuts the polysaccharide by adding a water molecule across one of its sugar–sugar bonds.
The product chains arethen quickly released, freeing the enzyme for further cycles of reaction (Figure3–50).An impressive increase in hydrolysis rate is possible because conditions arecreated in the microenvironment of the lysozyme active site that greatly reducethe activation energy necessary for the hydrolysis to take place. In particular, lysozyme distorts one of the two sugars connected by the bond to be broken from itsnormal, most stable conformation. The bond to be broken is also held close to twoamino acids with acidic side chains (a glutamic acid and an aspartic acid) thatparticipate directly in the reaction. Figure 3–51 shows the three central steps inthis enzymatically catalyzed reaction, which occurs millions of times faster thanuncatalyzed hydrolysis.Other enzymes use similar mechanisms to lower activation energies andspeed up the reactions they catalyze.
In reactions involving two or more reactants,the active site also acts like a template, or mold, that brings the substrates togetherin the proper orientation for a reaction to occur between them (Figure 3–52A). Aswe saw for lysozyme, the active site of an enzyme contains precisely positioned++(A)SFigure 3–49 Acid catalysis and basecatalysis.
(A) The start of the uncatalyzedreaction that hydrolyzes a peptide bond,with blue shading used to indicate electrondistribution in the water and carbonylbonds. (B) An acid likes to donate a proton(H+) to other atoms. By pairing with thecarbonyl oxygen, an acid causes electronsto move away from the carbonyl carbon,making this atom much more attractive tothe electronegative oxygen of an attackingwater molecule. (C) A base likes to takeup H+.
By pairing with a hydrogen of theattacking water molecule, a base causeselectrons to move toward the water oxygen,making it a better attacking group for thecarbonyl carbon. (D) By having appropriatelypositioned atoms on its surface, an enzymecan perform both acid catalysis and basecatalysis at the same time.+EESEPE+P(B)Figure 3–50 The reaction catalyzed by lysozyme. (A) The enzyme lysozyme (E) catalyzes the cutting of a polysaccharidechain, which is its substrate (S).
![](https://s.studizba.com/z.php?f=/templates/si/i/noavatar.png&w=100&h=100&t=1"){kind=avatar}
