教学课件高等有机化学lecture 26

1、Lecture 26 - Perfection in CatalysisE. KwanChem 1061. What is the mechanism of this reaction in glycolysis?Perfection in CatalysisEugene E. KwanNovember 4, 2011.OHOOOOPHOOPHOHOP HHHOOHenediolateGAPDHAPTIMOOPHOOPB HB: HScope of Lecture2. How can we derive its energy profile? Can the enzyme be improve

2、d any further?energybarrier 2-TTIMdetection of an enediolmechanism of proton transferbarrier 4perfection in catalysisbarrier 2-Hbarrier 1QM/MMmodelingsolvent isotope effectsbarrier 3efficiency and perfectionexchange-conversion isotope effectsdeuterium KIEsE-enediolateE-GAPE-DHAPHelpful References1.

3、Perfection in Enzyme Catalysis: The Energetics of Triose- phosphate Isomerase. Knowles, J.R.; Albery, J.W.Acc. Chem. Res. 1977, 10, 105-111.E + GAPE + DHAP3. How does the active site work?2. Efficiency and Evolution of Enzyme Catalysis. Albery, W.J.; Knowles, J.R. Angew. Chem. Int. Ed. Engl. 1977, 1

4、6, 285- 293.3. Enzyme Catalysis - Not Different, Just Better. Knowles, J.R.Nature. 1991, 350, 6314, 121-124.4. Triosephosphate Isomerase: A Highly Evolved Biocatalyst.Cell. and Mol. Life Sciences. 2010, 67, 3961-3982.5. Catalysis and Specificity in Enzymes. Cui, Q.; Karplus, M.Adv. Protein Chem. 200

5、3, 66, 315-372.Lecture 24 - First-Order KineticsE. KwanChem 106The proposed mechanism involves deprotonation to form an enediolate, followed by reprotonation. The pro-R hydrogen on DHAP is stereospecifically removed; labeling that hydrogen with tritium results in a product with 3-6% of the original

6、reactivity.Triosephosphate Isomerase (TIM)Perfection in Enzyme Catalysis: The Energetics of Triose- phosphate Isomerase. Knowles, J.R.; Albery, J.W. Acc. Chem. Res. 1977, 10, 105-111.OHOOTIM is an essential enzyme in glycolysis:glucose+ATPglucose-6-phosphateOOPHOOPHOHOPHHHenediolateGAPDHAPB HB:It is

7、 known that neither GAP nor DHAP exchange protium/tritium with solvent very rapidly. However, the result above shows that the new proton mostly comes from solvent. Therefore, there must be an intermediate that can exchange quickly with solvent. This is the enediolate.Equilibrium and Kinetic Isotope

8、EffectsQ: What are the relative barriers of this process?fructose-6-phosphate+ATPOHOOOHOHPOOOHOHPHOOKnowles performed an ingenious series of experiments that basically determined the entire potential energy surface. The review article above is a must-read for any mechanistic chemist, and the materia

9、l I go through below is some of the most complicated stuff youll see in the entire course.OHfructose 1,6-bisphosphateOHOTIMOOPHOOPThe approach I will take is to show you the energy surface as a diagram, and then explain how various experiments support it. First, in additions to the simple steps show

10、n above, one must also consider binding and release of substrate and product from the enzyme. This gives the following scheme:Hglyceraldehyde3-phosphate (GAP)- 2 ATPdihydroxyacetone phosphate (DHAP)E-enediolateE + DHAPE-DHAPThe key point here is that the cleavage of fructose 1,6-bis- phosphate produ

11、ces DHAP and GAP. If TIM did not exist,then glycolysis would only produce one ATP per glucose instead of two.E + GAPE-GAPAt physiologic conditions, this process is slightly endoergic overall (about 2 kJ/mol). Interestingly, glycolysis seems to be a constant-concentration, variable-flux process.Q: Ho

12、w does TIM work?Lecture 26 - Perfection in CatalysisE. KwanChem 106Triosephosphate Isomerase (TIM)enediolate exchange equilibrium is fastOOHere is the proposed energy diagram.I have exaggerated all ofHOOPHOOPthe energy differences for clarity, but one should be aware thatthese barriers are all very

13、similar in height. This is connected to the fact that this enzyme is incredibly efficient. The rate of enolization of DHAP in pH 7.0 buffer is 6 x 10-7 s-1, but an astounding 109 times faster with a rate constant of 2x103 s-1 in the active site of TIM. The activation free energy for converting GAP t

14、o DHAP with TIM corresponds to a rate of 4x108 s-1, which is basically diffusion-controlled!solid lines = protium, dashed lines = tritiumHHB TB HIt is important to note that proton/tritium exchange is very fast once the enediolate stage is reached.Because the enzyme and substrate are awash in a sea

15、of solvent, the isotopic composition of the starting material does not affect the position of this equilibrium at all. From here, barrier 4 is rate-limiting. Because barrier 4 is the same height whether E-GAP is tritiated or not, there is no discrimination. (This makes sense because release does not

16、 involve cleavage or formation of a C-H/T bond.)Experiment 2: Add unlabeled GAP, run reaction under irreversible conditions in T2O, and ask how much tritium ends up in the product DHAP.energybarrier 2-Tbarrier 4barrier 2-Hbarrier 1barrier 3Observation 2: Product has only about 11% of the radioactivi

17、ty that the solvent has.Implication 2: Barrier 3 is lower than barriers 2-H and 4.Proceeding from the reverse direction (i.e., from right to left) means that we traverse barriers 4 and 3 to reach the ene- diolate, whereupon exchange is fast. Now, the system has a Curtin-Hammett choice between passin

18、g through barrier 2-H as a protium species or barrier 2-T as a tritium species. It prefers the lower 2-H barrier, so the product has much more proton in it than tritium. Overall, the 9x difference corresponds to just over a kcal in primary KIE (6-20x is the expected range for a tritium KIE).Overall,

19、 relevant barrier from E+GAP to H product is barrier 2-H, but barrier 2-T from E+GAP to T product. Whether we see a KIE or not depends on whether the selectivity- determining step is after before the turnover limiting step.E-enediolateE-GAPE-DHAPE + GAPE + DHAPExperiment 1: Add unlabeled DHAP, run r

20、eaction under irreversible conditions in T2O, and ask how much tritium ends up in the product GAP.Observation 1: Product has the same radioactivity as the solvent does.Implication 1: Release of GAP from the enzyme is rate-limiting.The starting material contains all protium. Therefore, it traverses b

21、arriers 1 and 2-H to become ene-diolate. Here, we must consider exchange between ene-diolate and solvent:Lecture 26 - Perfection in CatalysisE. KwanChem 106For a more detailed look at this, consider a plot of exchange vs. conversion, shown below in cartoon form:energybarrier 2-Tbarrier 4barrier 2-Hb

22、arrier 1exchange(isotope in remaining starting material as % of activity in solvent)barrier 3E-enediolateE-GAPE-DHAPE + GAPE + DHAPNote that barrier 3 undoubtedly has a large primary H/T KIE as well. However, since it is lower than barriers 2 and 4, the KIE is not observable in these experiments.Exp

23、eriment 3: Add unlabeled GAP, run reaction under irreversible conditions in T2O, and ask how much tritium ends up in recovered GAP as a function of conversion. This is called an exchange-conversion experiment, and was famously used to show that a tetrahedral intermediate exists in ester hydrolysis.O

24、bservation 3: At all conversions, there is a 3:1 preference for conversion to DHAP over exchange.Implication 3: This measures the gap between barriers 2-H and barrier 4.Starting from E+GAP, we pass through barriers 3 and 4 to Enz(H) + enediolate. This is in rapid equilibrium with Enz(T) + enediolate

25、. To put tritium in the starting material, we must return to reactants via Enz(T) + enediolate. The highest barrier in this reverse direction is barrier 4. If we try to find the easiest path towards product, then it is via Enz(H) + enediolate and barrier 2-H. Once again, we have a Curtin-Hammett sce

26、nario at the enediolate stage.conversion(% extent of reaction)It is useful to consider what this graph would look like if the barriers were much farther apart in energy. For example, imagine that barrier 4 were very high. That means that as soon as we get over barrier 4 to the enediolate, we must go

27、 to product, regardless of whether it is tritiated or not.In contrast, if barrier 4 were very low, then we would go from starting materials directly to enediolate/tritiated enzyme.There would be an effective pre-equilibrium between GAP and the ene-diolate. The ene-diolate and tritium in the ocean of

28、 solvent would exchange immediately, and return back to starting materials.Note that this analysis depends on running the reaction under irreversible conditions.Now, we consider what happens if we run the experiment from left to right: i.e., place unlabeled DHAP and recover DHAP at different convers

29、ions.if barrier 4 were very lowif barrier 4 were very highLecture 26 - Perfection in CatalysisE. KwanChem 106Experiment 5: Add 2-d-GAP and run the reaction from right to left under irreversible conditions.Observation 5: The rate (kcat, Km) are is unchanged. There is no deuterium in the product.Impli

30、cation 5: The barrier heights in the energy diagram are correct.From d-GAP we end up at the enediolate, where the deuterium label rapidly washes into the solvent. At that point, the system goes over barrier 2-H to product, regardless of what label it started out with. Since the rate-determining barr

31、ier is 4,energybarrier 2-Tbarrier 4barrier 2-Hbarrier 1barrier 3E-enediolateE-GAPE-DHAPE + GAPregardless of whether we started with deuterated GA the rate is not affected.Experiment 6: Add 1R-d-DHAP and run the reaction from left to right under irreversible conditions.Observation 6: kcatH/kcatD = 2.

32、9.ot,E + DHAPExperiment 4: Add unlabeled DHAP, run reaction under irreversible conditions in T2O, and ask how much tritium ends up in recovered DHAP as a function of conversion.Observation 4: At low conversions, there is a 3:1 preference for conversion to GAP over exchange. At high conversions, the

33、recovered DHAP is completely tritiated.Implication 4: The initial slope measures the gap between barriers 2-T and barrier 4.Consider what happens at low conversion. All the starting material contains protium, so we go over barriers 1 and 2-H to ene-diolate. We can go to products via barriers 3 and 4

34、. Along this path, the highest barrier is barrier 4. To put tritium back into the starting material, we must go over barrier 2-T.At higher conversions, the pool of starting material becomes enriched in tritium. Since tritiated starting material must leave slower (via barrier 2-T) than protium starti

35、ng material (via barrier 2-H), there is an additional factor for enriching the starting material in tritium. In the previous experiment, the largest barrier between starting material and enediolate was barrier 4, which does not depend on whether H or T =is present; therefore, this additional factor

36、was not in force.Implication 6: The rate difference arises from a primary KIE, even though the step is not rate-limiting!This is by far the weirdest observation. The way to think about this is about the flux of molecules into the Enz/enediolate pool and the flux out of the pool. The flux into the po

37、ol depends on the barrier height difference between 2-H and 2-D. The flux out of the pool does not depend on which label the system started with.You might be tempted to think that this measures the barrier height between barrier 4, which is turnover-limiting for the proton substrate, and barrier 2-D

38、, which is turnover-limiting for the detuerated substrate. But here, the barriers are so close inLecture 26 - Perfection in CatalysisE. KwanChem 106Q: What does the energy profile look for an enzyme that is as efficient as possible?Let us consider a generic scheme for any enzymatic reaction:k1energy

39、barrier 2-Dbarrier 4barrier 2-Hbarrier 1k2kk3E + SE-SE-PE + Pbarrier 3kk-1-2-3This says that the substrate has to bind to the enzyme, get turned into product, and then be released from the enzyme. Let us consider the corresponding energy diagram for an undeveloped, proto-enzyme:E-enediolateE-GAPE-DH

40、APTS2E + GAPE + DHAPenergy that it is incomplete to think of any one step as rate- limiting. If we assume that barriers 1 and 3 are not important, and it is valid to treat E+DHAP and Enz(H/D)+enediolate as in equilibrium, then:TS3TS1E-Prate=k4Enz(H/D)-enediolateK=Enz(H/D)-enediolate/ EDHAPE-SE + PE

41、+ Srate(H) = k4KEDHAP a K = k2H/k-2HThe overall rate here is controlled by TS2. Now, lets imagine improving the uniform binding. We imagine making random mutations to the bulk of the enzyme that will improve its binding to the intermediates and transition states equally well.What energies can this i

42、mpact? TS1 and TS3 are fixed, because they represent the free energy barrier for bringing the enzyme and substrate or enzyme and product together (mostly diffusion controls this). However, the internal thermodynamics can change. The fact that E-S is higher in energy than E+S has to do with a balance

43、 between the loss of translational and rotational entropy on binding, and then gain of enthalpic (e.g., hydrogen bonding) or entropic (expulsion of water molecules) interactions. The structure of the enzyme will determine the heights of ES, TS2, and EP.rate(D) a K = k D/kH2-2This latter equation is

44、surprising! The idea is that deuterated starting material can go over the 2-D barrier into the pool, but if it wants to go back to starting materials, it must first undergo fast, irreversible exchange, and then go back over the 2-H barrier backwards.How Much Better Can It Get?Efficiency and Evolutio

45、n of Enzyme Catalysis. Albery, W.J.; Knowles, J.R. Angew. Chem. Int. Ed. Engl. 1977, 16, 285-293.As Knowles puts it:Chemical kinetics without catalysis is like skiing without skis.Lecture 26 - Perfection in CatalysisE. KwanChem 106So, lets say we make a lot of mutations and improve the binding a lot

46、 so that TS2 comes way down. Are we doing any better?TS2TS3TS3TS1TS1TS2E + PE + PE + SE + SE-PE-SThis is harder than improving the uniform binding from an evolutionary perspective, since we now have to distinguish between relatively similar internal intermediates. We can also try to improve the cata

47、lysis of the elementary step as well. For example, we can switch from a monofunctional catalysis mode to a bifunctional one. This is the hardest change becauseit requires very precise positionings of the functional groups, and might require a number of simultaneous changes.What happens when the enzy

48、me becomes fully optimized? It works as fast as diffusion and the barriers all become similar in height. Improving TS2 (or the height of any other internaltransition states) will no longer bring much improvement in rate:E-PE-SThe blue arrow is the old barrier and the red arrow is the new barrier, wh

49、ich is actually higher! Why did things get worse?Although we have lowered the energy of TS2, we have simultaneously made ES very stable.The best we can do with uniform binding is:TS2TS3TS2TS3TS1TS1E + PE + PE + SE-PE-SE + SE-PThis is what has happened in TIM. One could imagine that we could try to d

50、o better with the diffusion limit by using super- enzyme aggregates. But this limits natures ability to share intermediates between pathways, and so most enzymes seem to be independent, freely diffusing catalysts.Q: How efficient are most enzymes?E-SThe next thing we can do is to improve the interna

51、l thermo- dynamics. We can improve the differential binding. We can try to bind E-P more tightly to and drive the reaction harder (the Hammond postulate). As it turns out, the best scenario is when E-S and E-P become equal in energy (see Knowles Biochem 1976, 15 5631):Lecture 26 - Perfection in Cata

52、lysisE. KwanChem 106The Moderately Efficient Enzyme. Milo et al. Biochemistry2011, 50, 4402-4410.Sk1EPk2k-1ESRecall that under Michaelis-Menten conditions, we have:kcat: the same as k2; how fast the enzyme can turn substrates into productsKM: the Michaelis constant, a measure of the binding affinity

53、 of the ES complex; under the steady state approximation,it is (k-1 + k2)/k1. The smaller the constant, the tighter the substrate is bound.The ratio of kcat/KM can be thought of as a measure of enzyme efficiency: you can either have a fast rate of turnover (large kcat)or have the enzyme bind the sub

54、strate very well (small KM). the diffusion limit kcat/KM becomes 109 1/s/M or so.AtSo is TIM an outlier, or are all enzymes perfect? As it turns out, enzymes involved in primary metabolism are much better optimized than those involved in secondary metabolism:primary metabolism - carbohydrates and en

55、ergy (CE)e.g. glycolysis, citric acid cycle, pyruvate, etc.primary metabolism - amino acids, fatty acids, nucleotides (AFN)intermediate metabolism - biotin, folate, thiamine, etc.secondary metabolism - flavonoids, caffeine, retinol, etc.There seems to be a much higher evolutionary pressure on the en

56、zymes responsible for energy:Lecture 26 - Perfection in CatalysisE. KwanChem 106Q: How does TIM actually work?So far, we have only treated the enzyme as an abstract entity, with associated rate constants and energies. But, what does it look like from an atomistic perspective?TIM is a dimeric enzyme of two identical subunits, each of which is about 250 residues long. Each