2specific activity is the amount of product formed by an enzyme in a given amount of time under given conditions per milligram of enzyme.The rate of a reaction is the concentration of substrate disappearing (or product produced) per unit time (mol L − 1s − 1)The enzyme activity is the moles converted per unit time (rate × reaction volume). Enzyme activity is a measure of quantity of enzyme present. The SI unit is the katal, 1 katal = 1 mol s-1, but this is an excessively large unit. A more practical value is 1 enzyme unit (EU) = 1 μmol min-1 (μ = micro, x 10-6).The specific activity is the moles converted per unit time per unit mass of enzyme (enzyme activity / actual mass of enzyme present). The SI units are katal kg-1, but more practical units are μmol mg-1 min-1. Specific activity is a measure of enzyme efficiency, usually constant for a pure enzyme.If the specific activity of 100% pure enzyme is known, then an impure sample will have a lower specific activity, allowing purity to be calculated.The % purity is 100% × (specific activity of enzyme sample / specific activity of pure enzyme). The impure sample has lower specific activity because some of the mass is not actually enzyme.
3Enzyme ActionEach enzyme has a unique three-dimensional shape that binds and recognizes a group of reacting molecules called substrates.The active site of the enzyme is a small pocket to which the substrate directly binds.Some enzymes are specific only to one substrate; others can bind more than one substrate.
5Models of Enzyme Action Early theory: lock-and-key model. Active site (lock) had the same shape as the substrate (key). Only the right shape key could bind.Current theory: induced fit model. Active site closely resembles but does not exactly bind the substrate.Allows for more flexibility in type of substrateAlso explains how the reaction itself occurs. As the substrate flexes to fit the active site, bonds in the substrate are flexed and stressed -- this causes changes/conversion to product.
6Molecular Recognition How does an enzyme bind a substrate, reduce the activation barrier, and produce a product?Lock & Key HypothesisInduced Fit Hypothesisvs.
7C. Factors Affecting Enzyme Activity Enzyme activity is defined as how fast an enzyme catalyzes its reaction.Many factors affect enzyme activity:Temperature: most have an optimum temp around 37oCpH: most cellular enzymes are optimal around physiological pH, but enzymes in the stomach have a lower optimum pHConcentration of enzyme and substrate: have all of the enzyme molecules been used up, even though substrate is still available?
8Energy of activation: ΔG‡ Effect of catalysisΔG‡ΔGcat‡A → BEffect of tempΔGT1‡ΔGT2‡(T1 > T2)A → B
10Rate acceleration: mechanisms E + SESEX‡E + PStabilization of the transition state: covalent bonds, metals, acid-base, and proximity.Destabilization of ES: strain, charge, electrostaticsReduced entropy in ES formation.
11Rate acceleration: mechanisms hydrolysis of a β-glycosidic bond yielding a unit of α-glucose
12Major factors: pH, ions, & temp At pH ~ 7 amino acids exist as zwitterions.The R group determines pH.aspartic acid [pKa = 4.0]arginine [pKa = 12.5]
13Major factors: pH, ions, & temp ionic strengthtemperaturebarley α-amylase activity plotted as a function of pH
14Major factors: pH, ions, & temp ionic strengthtemperatureHaving the correct ions is important. Why?barley α-amylase isozyme 1[crystallized with Ca2+ (green)]
16Michaelis-Menten Kinetics E + S ↔ ES → E + Pk1k-1k2Assumptions: Steady-state of the intermediate complex ES Neglect back rxn from product (k-2; not shown) Conservation of mass ([ET] = [E] + [ES])Vmax = k2[ET]ν =Vmax [S]Km + [S]where:Km =(k-1 + k2)k1
18Michaelis-Menten Kinetics Many types of inhibition can be included in the MM model as well as multiple substrates and steps:Inhibition:competitive (rev)noncompetitive (rev)mixed (rev)irreversibleReaction Schemes:single substratemultiple substratesingle displacementdouble disp (ping-pong)
20Control of Enzyme Activity We don’t always need high levels of products of enzyme-catalyzed reactions around. What kind of control system is used to regulate amounts of enzyme and products?Two main methods: zymogens, and feedback control.
21Zymogens Many enzymes are active as soon as they’re made. However, some are made in an inactive form and stored. This inactive form is called a zymogen or proenzyme.To become active, the body needs only to cleave off a small peptide fragment.
22Feedback ControlSome enzymes (allosteric enzymes) bind molecules called regulators (different from the substrate) that can affect the enzyme either positively or negativelyPositive regulator: speeds up the reaction by changing the shape of the active site -- substrate binds more effectivelyNegative regulator: slows down reaction by preventing proper substrate binding, again, by changing enzyme shapeFeedback control: the end product acts as a negative regulator. If there is enough of the end product, it will slow down the first enzyme in a pathway.
23The kinetics of enzyme catalysis: Steady state kinetics
24A hyperbolic curve between V0 and [S] was revealed by in vitro studies using purified enzymes It was the initial velocity (rate), V0, that was measured, so the change of [S] could be ignored.The catalysis was assumed to occur as:The enzyme will become saturated at high [S]: the V0 will not be affected by [S] at high [S].
25The effect on V0 of varying [S] is measured when the enzyme Vmax is extrapolated from the plot:V0 approaches but never quite reaches Vmax.The effect on V0 of varying [S] ismeasured when the enzymeconcentration is held constant.Hyperbolic relationship betweenV0 and [S]
26A mathematical relationship between V0 and [S] was established (Michaelis and Menten, 1913; Briggs and Haldane, 1925)k 1k 2()E + S ES E + PFormation of ES is fast and reversible.The reverse reaction from PS (k-2 step) was assumed to be negligible.The breakdown of ES to product and free enzyme is the rate limiting step for the overall reaction.ES was assumed to be at a steady state: its concentration remains constant over time.Thus V0 = k2[ES]
27k1([Et]-[ES])[S]=k-1[ES] + k2[ES] Steady-state assumption:Rate of ES formation=rate of ES breakdownk1([Et]-[ES])[S]=k-1[ES] + k2[ES]([Et] is the total enzyme concentration.)Solve the equation for [ES]:Km is called theMichaelis constant.V0 = k2[ES]
28The maximum velocity is achieved when all the enzyme is saturated by substrate, i.e., when [ES] =[Et].Thus Vmax =k2[Et]The Michaelis-Menten Equation
29The Michaelis-Menten Equation nicely When [S] >> KmWhen [S] << KmThe Michaelis-Menten Equation nicelydescribes the experimental observations.The substrate concentration at which V0 is half maximal is Km
30The Vmax and Km values of a certain enzyme can be measured by the double reciprocal plot (i.e., the Lineweaver-Burk plot).
32The Michaelis-Menten equation, but not their approximated mechanism applies to a great many enzymes Most enzymes (except the regulatory enzymes) have been found to follow the Michaelis-Menten kinetics, but their actual mechanisms are usually more complicated (by having more intermediate steps) than the one assumed by Michaelis and Menten.The values of Vmax and Km alone provide little information about the number, rates, or chemical nature of discrete steps in the reaction.
33The actual meaning of Km depends on the reaction mechanism ForIf k2 is rate-limiting, k2<<k-1, Km = k-1/k 1Km equals to the dissociation constant (Kd) of the ES complex;Km represent a measure of affinity of the enzyme for its substrate in the ES complex.
35PLOT EADIE-HOFSTEE DAN HANES - WOOLF Plot Lineweaver – Burk mempunyai sedikit kelemahan, yaituSering kali pada saat mengekstrapolasi grafik untuk menentukan harga -1/Km ternyata akan memotong sumbu 1/[S] di luar grafik yang dibuatPada konsentrasi substrat yang terlalu rendah, maka akan diperoleh hasil yang kurang akuratAwal dari kelinearannya sering kurang jelas dibanding dengan plot lain, terutama plot Eadie – Hofstee, padahal hal ini sangat penting pada penentuan mekanisme reaksi
36Plot Eadie-Hofstee dan Hanes diturunkan dari persamaan Lineweaver-Burk dengan mengalikan kedua sisi persamaan dengan faktor vo Vmax sehingga akan diperoleh persamaan garis lurus selanjutnya dipergunakan untuk menghitung Vmax dan Km
37Dengan cara penurunan yang mirip, Hanes-Woolf mengalikan perasamaan Lineweaver-Burk dengan [So] maka diperoleh:Plot Eadie – Hofstee dan Hanes banyak digunakan pada studi kinetik enzim, namun demikian studi enzim secara umum masih menggunakan plot Lineweaver – Burk.
46Comparison of MethodsLineweaver-Burk: supposedly gives good estimate for Vmax, error is not symmetric about data points, low [S] values get more weightEadie-Hofstee: less bias at low [S]Hanes-Woolf: more accurate for Vmax.When trying to fit whole cell data – I don’t have much luck with any of them!
47PERSAMAAN HALDAN UNTUK REAKSI REVERSIBEL Reaksi enzimatis dalam sel sering berlangsung secara reversibel.Reaksi substrat tunggal, S P, berlangsung melalui pembentukan satu kompleks intermediate, arah ke kanan dianggap sebagai kompleks ES dan sebaliknya kompleks EPE + S ES/EP P + E
48Persamaan MM arah kekanan pada [Eo] tetap dengan laju awal vf dan Vsmax Persamaan MM ke arah sebaliknya pada [Eo] tetap dengan laju awal vb dan VPmax
49Perumusan Haldan hubungan antara konstanta laju dan kesetimbangan reaksi pada reaksi kesetimbangan adalahKarena
50Maka:Bila konstanta kesetimbangan diketahui, maka persamaan tersebut dapat digunakan untuk memvalidasi konstanta laju yang diperolehSecara umum Km dari arah reaksi metabolisme penting akan sedikit lebih kecil dari arah sebaliknya. Namun arah metabolisme dipengaruhi juga oleh [S] dan [P] dalam sel
51KINETIKA REAKSI CEPATKinetika keadaan steady penting bagi enzimologis yang memungkinkan untuk menentukan Km dan KcatBilangan peredaran (turn over number), Kcat pada beberapa enzim pada tingkat 100 s-1, yaitu 100 molekul produk dihasilkan setiap detik tiap molekul enzim. Hal ini berarti bahwa tahapan yang paling lambat dari suatu mekanisme reaksi mempunyai waktu hidup hanya beberapa detik
52Untuk reaksi sederhana order pertama A BMaka laju reaksi pada t adalah-d[A]/dt = d[B]/dt = k[A]Integrasi dari persamaan tersebut memberikan persamaanln[Ao] – ln [A] = kt[A] = [Ao]e –kt
53konsentrasiABwaktuKurva teoritis perubahan reaksi A menjadi B
54Persamaan integrasi dari sebagian besar mekanisme reaksi akan lebih rumit . Untuk reaksi substrat tunggal yang membentuk suatu kompleks intermediate dengan konsentrasi awal dari [S] >>> [E] dinyatakan sebagai:E + S ES E + PLaju penambahan [ES] pada waktu t (pada periode awal dimana pembentukan [P] diabaikan dinyatakan sebagai:d[S]/dt = [E][S] [ES] [ES]d[S]/dt = ([Eo] - [ES])([So] – [ES] – [P]) [ES] [ES]karena : [So]>>[Eo], maka ([So]-[ES]-[P] ~ [So]Jadi : d[S]/dt = ([Eo] - [ES])([So] ) [ES] [ES]k2k1k-1k-2k1k-1k2k1k-1k2k1k-1k2
55Integrasi persamaan tersebut akan menunjukkan perubahan ES terhadap waktu E + S ES EP E + P dimana [So] >> [Eo]Konstr.SPEESInduction periodSteady state phaseWaktuTransient or pre steady state phase
56Bagian linier dari grafik [P] vs t menunjukkan fase steady state dari reaksi dengan slope = k2[E][So]/([So]/Km) yang didapat dari mensubstitusi (k-1 + k2)/k1 dengan Km dari persamaan integrasinyaJika bagian grafik steady state linier, ekstrapolasinya akan memotong sumbu t pada t = 1/(k1[So]+Km) dan disebut sebagai periode induksi.Kurva perubahan konsentrasi akan lebih rumit bila reaksinya seperti;Dengan tahap laju yang menentukan (rate-limiting-step) adalah EP menjadi E dan P.E + S ES EP E + P
57E + S ES EP E + P dimana EP menjadi E dan P sebagai penentu Konstr.EPEPESWaktu
58Vmax is determined by kcat, the rate constant of the rate-limiting step Vmax = kcat[Et]kcat equals to k2 or k3 or a complex function of both, depending on which is the rate-limiting step.kcat is also called the turnover number: the number of substrate molecules converted to product in a given unit of time per enzyme molecule when the enzyme is saturated with substrate.
5940,000,000 molecules of H2O2 are converted to H2O and O2 by one catalase molecule within one second!
60The kinetic parameters kcat and Km are often studied and compared for different enzymes Km often reflects the normal substrate concentration present in vivo for a certain enzyme.The catalytic efficiency of different enzymes is often compared by comparing their kcat/Km ratios (the specificity constant).kcat/Km is an apparent second-order rate constant (with units of M-1S-1), relating the reaction rate to the concentrations of free enzyme and substrate.
61The value of kcat/Km has an upper limit (for the perfected enzymes) It can be no greater than k1.The decomposition of ES to E + P can occur no more frequently that E and S come together to form ES.The most efficient enzymes have kcat/Km values near the diffusion-controlled limit of 108 to 109 M-1S-1.
62Catalytic perfection (rate of reaction being diffusion-controlled) can be achieved by acombination of different values of kcat and Km.
63Rate enhancement is often used to describe the efficiency of an enzyme kcatcatalyzeduncatalyzedkcatRate enhancement: ratio of the rates ofthe catalyzed and the uncatalyzed reactions.
64Rate enhancement by selected enzymes Uncatalyzed rate(kun, s-1)Catalyzedrate(kcat, s-1)Rateenhancement(kcat/kun)Nonenzymatichalf-lifeEnzyme
65Enzyme-catalyzed reactions of two or more substrates can also be analyzed by the Michaelis-Menten approachEach substrate will have one characteristic Km value.Noncovalent ternary complex (with two substrates bound to the enzyme concurrently) may or may not be formed for the bisubstrate reactions depending on the mechanism.Steady-state kinetics can often help distinguish these two mechanisms.
66In those enzyme-catalyzed bisubstrate reactions where a ternary complex is formed, the two substrates may either bind in a random sequence or in a specific order.
67For those reactions where ternary complex is formed: Maintaining the concentrationof one substrate (S2) constant,the double reciprocal plotsmade by varying the concentrationof the other substrate (S1) willintersect.
68No ternary complex is formed in the Ping-Pong (or double displacement) mechanism: The first substrate is converted to a product that leaves the enzyme active site before the second substrate enters.
69As [S2] increases, Vmax increases, For enzymes having Ping-Pong mechanisms (ternarycomplex not formed).Maintaining the concentrationof one substrate (S2) constant,the double reciprocal plotsmade by varying the concentrationof the other substrate (S1) will notintersect.As [S2] increases, Vmax increases,as does the Km for S1.S1
70Rates of individual steps for an enzyme-catalyzed reaction may be obtained by pre-steady state kineticsThe enzyme (of large amount) is used in substrate quantities and the events on the enzyme are directly observed.Rates of many reaction steps may be measured independently.Very rapid mixing and sampling techniques are required (the enzyme and substrate have to be brought together in milliseconds and measurements also be made within short period of time).
71“Rapid kinetics” or “pre-steady- state kinetics”is applied to the observation of rates of systemsthat occur in very short timeintervals (usually ms or sub-msscale ) and very low productconcentrations. This periodcovers the time from the enzymeencountering its target (eithera substrate, inhibitor or someother ligands) to the point ofsystem settling to equilibrium.The concentration of ES will risefrom zero to its steady-state value.(ms or sub-ms)
72The turnover number of carbonic anhydrase: Carbonic anhydrase of erythrocytes (Mr 30,000) has one of the highest turnover numbers among known enzymes, it catalyses the reversible reaction of CO2:H2O + CO2 -> H2CO3This is an important process in the transport of CO2 from the tissues to the lungs. If 10μg of pure carbonic anhydrase catalyses the hydration of 0.30g of CO2 in 1min at 37°C at Vmax, and the reaction volume is 1ml. What is the turnover number (Kcat) of carbonic anhydrase expressed in units of per min and per sec)? Mr of CO2 is 44.