1 / 37

Overview of Lecture Topics by RC

Overview of Lecture Topics by RC. 1) Substrate diffusion and catalysis 2) Biological reaction kinetics 3) Biological reactions that change the environment. Lec 1 Substrate diffusion and catalyis. What are bacteria? The advantage of being small.

kynton
Télécharger la présentation

Overview of Lecture Topics by RC

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Overview of Lecture Topics by RC 1) Substrate diffusion and catalysis 2) Biological reaction kinetics 3) Biological reactions that change the environment

  2. Lec 1 Substrate diffusion and catalyis • What are bacteria? The advantage of being small. • What do bacteria do? Catalyse exergonic redox reactions. • Substrate diffusion to the cell: Can we predict the random diffusion of a molecule? Not of one, but of many molecules. • What is the kinetics of substrate diffusion (1st order kinetics, Fick’s law). • What is diffusion driven by? Order, Concentration gradient, • How come that many molecules seem to have a behaviour but an individual molecule does not? • What is the principle of catalysing redox reaction? The enzyme does not bind to the substrate.

  3. Catalysis15 min Lecture block

  4. What do bacteria do? • Catalyse exergonic redox reactions • Exergonic, (downhill) reactions loose Gibbs Free Energy. (DG=negative) • Bacteria utilise a portion of the free energy released for growth processes and multiplication S G ΔG P

  5. What does ΔG (Gibbs Free Energy Change) mean? • Spontaneous reactions are downhill reactions • Energy of substrates is higher than of products • Products are more stable than Substrate (stable = low energy) • The driving force = hill height = difference in G = Delta (Δ) G ΔG = G (prod.) – G (substr.) • The reaction is driven by the loss in G  Change in G is a negative value (e.g. Δ G = -256 kJ/mol) S G ΔG P

  6. Does the Δ G allow to predict the reaction rate ? • No • But for ΔG = zero, reaction rate will be zero • For ΔG = positive, reaction would go backwards (PS) • The rate is determined by the activation energy (AE) • In biological reactions the AE is largely determined by the presence of enzymes (lower AE) • Now how do they do that? AE S G ΔG P

  7. The ΔGo of redox reaction is related to the ΔEoof e- donor and acceptor • ΔEo is the difference in redox potential of the half reactions • Couple with lower Eo will become e- donor • If not reaction would reverse • ΔGo = n F Δ Eo • Microbes that use electron donors of a very low Eo (e.g. H2 and an electron acceptor of a very high level e.g. O2) have a lot of free energy available for growth e- don. -Eo ΔE e- acc.

  8. What is enzyme catalysis? The rate of a spontaneous reaction is increased. If the reaction is already spontaneous, why do we need an enzyme? The activation energy needs to be overcome. Spontaneous = downhill = exergonic Enzymes can not catalyse uphill = endergonic reactions If catalysed uphill reactions proceed in the reverse direction (Products  Substrates) G G Reaction path

  9. How do enzymes catalyze ? A. By lowering the activation energy barrier! B. By binding to the substrate like a lock to a key??? This is the simplified textbook explanation, but how can we visualise it? How can an enzyme convert a substrate to product by binding to it? Binding means stabilising (lower Energy level) and hence slowing down reaction rates. Example: Antibody binding to antigen. Antibody is a protein designed by the immune system to bind and “neutralise” a foreign substances (the antigen).

  10. Why don’t antigens catalyze? Will an antibody against the substrate be able to catalyse the reaction? No What is the difference between an antigen and an enzyme. Both bind, one catalyses the other does not. Does the enzyme have perhaps a special mechanism (lever) that can break the substrate into product(s) ? No P S

  11. S T P How do enzymes catalyse ? Example: Substrate = stick, Product = broken stick For the stick to be broken it must go through a transition state (T) How does the enzyme (stickase, of course) catalyse by binding to the stick? Binding to the stick stabilises rather than activates the substrate  not making reaction easier The simplified concept of the enzyme binding to the substrate does not make sense Let’s have a closer look at the energy diagram for clues.

  12. How do enzymes catalyse ? S T P High Energy= Unlikely, reactive, activated Low Energy= Likely to exist, stable needed to lower the activation energy level: a way to make the transition (T) state more likely S G P Reaction path Note T not an intermediate just a “deformed molecule” that makes forward and backward reaction equally likely

  13. How do enzymes catalyse ? S T P Example: Substrate = stick, Product = broken stick For the stick to be broken it must go through a transition state (T) The enzyme (stickase, of course) binds to T stabilises T makes T more likely higher quantities of T available higher likelyhood for P to form (P cannot form when T is in ultra low concentration) Enzymes catalyse by binding to the transition state and making it more likely.

  14. How do enzymes catalyse ? S T P Significant product formation depends on availability of T. Non catalysed reactions are slow because [T] is low By binding to T enzymes increase the chances of T to exist, hence speed up the reaction. Binding does not mean, holding on to T but releasing T rapidly, either as S or as P. S G P Reaction path

  15. How do enzymes catalyse ? S T P The enzyme substrate complex (ES) is not a transitional state but a true, existing, defined intermediate Intermediates are in a “valley” Transition states are on a “peak” S ES G P Reaction path

  16. How far will the catalysed reacton go? With decreasing substrate concentration the energy content of the substrates sinks. Increasing product concentrations lift the energy content of the products. The reaction continues until the difference in G (Delta G) is zero. Then the energy level of substrate equals that of the product Reaction is at equilibrium Rate of backwards reaction equals that of forward reaction. The ratio [P]/[S] now represents the equilibrium constant keq. ` S G P Reaction path

  17. The dynamic equilibrium • The ratio [P]/[S] now represents the equilibrium constant keq. • An original very exergonic reaction needs lots of P to accumulate until equilibrium is reached • at equilibrium [P]/[S] is very high (e.g. 10,000) • endergonic reactions have low keq (e.g. 0.00001) The enzyme does not affect the spontaneity or reversibility of reaction but the energetics does. Not surprisingly the reaction driving force is related to keqΔG =RT ln keq ` P S G Reaction path

  18. Binding Energy T Where does the energy come from to overcome activation energy? An enzyme/substrate to be more precise enzyme/T complex forms hydrogen bonds and hydrophobic interaction bonds. The binding energy released is the energy source of lowering the activation energy (analogy of magnets in stickase) H bonds of T with H2O are replaced by bonds with E (dry bonding)

  19. T What is the effect of activation energy on reaction rate? The overal rate constant (k) of the reaction depends directly on the activation energy: k= (B/P*T)*e -DG(activ.)/RT B/P=Bolzman/Planck constant E.g. lowering the activation energy by 5.7kJ will increase rate 10 fold

  20. Microbial ChemistryLec 2 2006

  21. T What is the effect of activation energy on reaction rate? After we have concluded: that the enzyme (E) catalyses the substrate (S) conversion by forming an Enzyme-substrate complex Let us see how we can derive the kinetic behaviour of the enzyme reaction from first principles. The widely accepted enzyme kinetics model is the Michaelis Menten model.

  22. k1 k2 E + S ES E + P k-1 Foundation of Michaelis-Menten Kinetics For the overall enzyme reaction a number of rate constants need to be considered: The rate of conversion of E and S to ES is k1 k-1 is the rate constant for ES going back to E and S k2 is the rate for conversion of ES to E and P In enzyme assays P is negligible (startup velocity of reaction) k-2 is not included. (Rate constants mean first order kinetics rate constants as explained below.)

  23. k1 k2 E + S ES E + P k-1 Foundation of Michaelis Menten kinetics Rate constant of first order reaction predicts that the rate is proportional to the substrate concentration The rate for ES to go to E +P is given by: ES (mM) * k2 (h-1)  (mM/h)

  24. k1 k2 E + S ES E + P k-1 Foundation of Michaelis Menten kinetics At substrate saturation no free enzyme is available ([E]=0) overall rate is determined by k2 ( kcat = k2) (btw: kcat= vmax/(total enzyme concentration (Et, E+ES)) The ratio of ES formation over ES disintegration is km (formula?) km = (k2 + k-1)/k1  enzyme with low km: ES formation faster than disintegration

  25. k1EtS k-1+ K2 + k1S k-1+ K2 = km k1 k1 k2 E + S ES E + P k-1 Derivation of MM kinetics from first principles At steady state: rate of ES disintegration =rate of ES formation k-1*ES + K2*ES= k1*ES(Et=E+ES) k-1ES + K2ES = k1(Et-ES)S multiply out right k-1ES + K2ES = k1EtS - k1ESS k1ESS k-1ES + K2ES + k1ESS = k1EtS bracket out ES ES (k-1+ K2 + k1S ) = k1EtS solve for ES ES = k1EtS / (k-1+ K2 + k1S ) ES =

  26. k1EtS k-1+ K2 + k1S Et S Et S vmax S k2Et S EtS km+S km+S km+S km+S k-1+ K2 = km k1 k-1+ K2 vo +S k1 = k2 k1 k2 E + S ES E + P k-1 cancel k1 = ES = ES = (as vo = k2 *ES  ES = vo/k2 )  (as vmax =k2*Et)  vo = vo = (k2 is also called kcat)

  27. S vo = vmax km + S The Michaelis Menten Model has been derived: (students don’t need to be able to derive it but to know the final equation) Vo = the initial velocity (no products present) S = [substrate] km = half saturation constant, also called kS vmax = maximum velocity under substrate saturation when overall reaction only depends on k2

  28. S vo = vmax km + S Microbial Reaction Kinetics: The Michaelis Menten (MM) model is not only useful for enzymatic reactions but overall microbial reactions such as algal blooms or microbial growth in bioreactors. After we have seen where the most widely used biological kinetic mode comes from… let us compare 2 types of microbial kinetics (MM and exponential kinetics) with traditional chemical kinetics (zero and first order).

  29. Kinetics15 min Lecture block

  30. S vo = vmax km + S Microbial Reaction Kinetics: The substrate (reactant) of bioprocesses can disappear according to different time courses, such as linear disappearance, first order, exponential, mixed kinetics (such as Michaelis Menten) or incompletely such as for product inhibition or approaching of the thermodynamic equilibrium. Students should be able to sketch the different types and explain which kinetic types occur in which processes.

  31. Zero Order kinetics S P • constant velocity • velocity independent of S • v = K (M/s) • Ex: limited access to S (O2 diffusion to food) • water loss • enzyme reactions at high S V Time V S

  32. First Order Kinetics P • velocity decreases over time (parallel to [S]) • velocity is determined by and hence proportional to S • v = S * K (s-1) • K is the rate constant • Ex: Most chemical reactions • Radioactive decay (half time) V S Time V S

  33. Exponential Kinetics S P • velocity exponentially increasing over time • independent of S but related to P • Can give appearance of sudden increase in P • v ~ P * K (s-1) • a product must enhance reaction velocity (e.g. heat, chain reaction) Biological examples: • bacterial spoilage (e.g. milk) multiplication of catalyst • Auto-oxidation of fats (radicals propagation) V Time V P

  34. Phase 1 Phase 2 Michaelis Menten Kinetics S P • Two reaction phases: • 1: zero order, [Enzyme] limiting • 2: first order, [S] limiting • Why do we get first and zero order if S or E is limiting? • [S] changes, [E] does not change! • v = vmax * S / (s + kS) Examples: • Most biochemical reactions • Microbial reaction in the environment when S is low (pollution, groundwater, soil, ocean) V Time Phase 2 Phase 1 V S

  35. Kinetics Summary S • At least 4 different types of reaction can occur in biological environments • Development of reaction rate can increase, decrease or stay the same. • There are clear mechanistic reasons for reaction behaviour • Significance: When knowing the reaction kinetics the behaviour biological material (e.g. food, bioreactors, body) can be predicted • 2nd order reaction (dependence on two substrates) is similar to 1st order and neglected here Time 1 Exp Enz V 0 Time

  36. Decide Order or Kinetics S 1. Decide which order the reaction kinetics is: • v constant, S decrease linear  Zero order • v increasing  exponential • v continuously slowing (1st, 2nd, 3rd order) • v constant , then slowing  MM kinetics Time 1 Exp Enz V 0 Time

  37. Possible Lec 3 Biological Reactions that Change the Environment The two big reactions: Photsynthesis and Respiration match each other What do bacteria do without oxygen ? Examples of anaerobic respiration. Stoichiometry of anaerobic respirations Impact of anaerobic respirations

More Related