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SDS PAGE = SDS polyacrylamide gel electrophoresis

s odium d odecyl s ulfate, SDS (or SLS): CH 3 -(CH2) 11 - SO 4 -- CH 3 -CH 2 -CH 2 -CH 2 -CH 2 -CH 2 -CH 2 -CH 2 -CH 2 -CH 2 -CH 2 -CH 2 -SO 4 --. SDS PAGE = SDS polyacrylamide gel electrophoresis. SDS. All the polypeptides are denatured and behave as random coils

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SDS PAGE = SDS polyacrylamide gel electrophoresis

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  1. sodium dodecyl sulfate, SDS (or SLS): CH3-(CH2)11- SO4-- CH3-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-SO4-- SDS PAGE = SDS polyacrylamide gel electrophoresis SDS All the polypeptides are denatured and behave as random coils All the polypeptides have the same charge per unit length All are subject to the same electromotive force in the electric field Separation based on the sieving effect of the polyacrylamide gel Separation is by molecular weight only SDS does not break covalent bonds (i.e., disulfides) (but can treat with mercaptoethanol for that) (and perhaps boil for a bit for good measure)

  2. Disulfides between 2 cysteines can be cleaved in the laboratory by reduction, i.e., adding 2 Hs (with their electrons) back across the disulfide bond. One adds a reducing agent: mercaptoethanol (HO-CH2-CH2-SH). In the presence of this reagent, one gets exchange among the disulfides and the sulfhydryls: Protein-CH2-S-S-CH2-Protein  + 2 HO-CH2CH2-SH  ---> Protein-CH2-SH + HS-CH2-Protein   +  HO-CH2CH2-S-S-CH2CH2-OH The protein's disulfide gets reduced (and the S-S bond cleaved), while the mercaptoethanol gets oxidized, losing electrons and protons and itself forming a disulfide bond. 

  3. e.g., “p53” Molecular weight markers (proteins of known molecular weight) P.A.G.E. 12 18 48 80 110 130 160 140

  4. Molecular sieve chromatography (= gel filtration, Sephadex chromatography) Sephadex bead

  5. Molecular sieve chromatography Sephadex bead

  6. Molecular sieve chromatography Sephadex bead

  7. Molecular sieve chromatography Sephadex bead

  8. Molecular sieve chromatography Sephadex bead

  9. Plain Fancy 4oC (cold room)

  10. Larger molecules get to the bottom faster, and …. Non-spherical molecules get to the bottom faster ~infrequent orientation Non-spherical molecules get to the bottom faster

  11. Handout 4-3: protein separations

  12. Largest and most spherical Lowest MW Winners: Largest and least spherical Similar to handout 4-3, but Winners &native PAGE added Most chargedand smallest Winners:

  13. Enzymes = protein catalysts

  14. Each arrow = an ENZYME Each arrow = an ENZYME

  15. H2 + I2 H2 + I2 2 HI 2 HI + energy Chemical reaction between 2 reactants “Spontaneous” reaction: Energy released Goes to the right H-I is more stable than H-H or I-I here i.e., the H-I bond is stronger, takes more energy to break it That’s why it “goes” to the right, i.e., it will end up with more products than reactants i.e., less tendency to go to the left, since the products are more stable

  16. Atom pulled completely apart (a “thought” experiment) 2H + 2I say, 100 kcal/mole say, 103 kcal/mole Change in Energy (Free Energy) H2 + I2 { -3 kcal/mole 2 HI Reaction goes spontaneously to the right If energy change is negative: spontaneously to the right = exergonic: energy-releasing If energy change is positive: spontaneously to the left = endergonic: energy-requiring

  17. H2 + I2 2 HI H2 + I2 2 HI H2 + I2 2 HI Different ways of writing chemical reactions H2 + I2 2 HI H2 + I2 2 HI

  18. But: it is not necessary to break molecule down to its atoms in order to rearrange them 2H + 2I say, 100 kcal/mole say, 103 kcal/mole Change in Energy (Free Energy) H2 + I2 { -3 kcal/mole 2 HI

  19. + I I I I I I I I I I (H2 + I2) H H H H H H H H H H Transition state (TS) + (2 HI) Reactions proceed through a transition state + Products

  20. 2H + 2I ~100 kcal/mole Change in Energy H-H | | I-I (TS) Say, ~20 kcal/mole H2 + I2 Activation energy { -3 kcal/mole 2 HI

  21. HHII (TS) Allows it to happen Energy needed to bring molecules together to form a TS complex Change in Energy (new scale) Activation energy determines speed = VELOCITY = rate of a reaction H2 + I2 { 3 kcal/mole 2 HI Net energy change: Which way it will end up. the DIRECTION of the reaction, independent of the rate 2 separate concepts

  22. Direction We need it to go in the direction we want Speed We need it to go fast enough to have the cell double in one generation Concerns about the cell’s chemical reactions

  23. Example Biosynthesis of a fatty acid 3 glucose’s 18-carbon fatty acid Free energy change: ~ 300 kcal per mole of glucose used is REQUIRED So: 3 glucose 18-carbon fatty acid So getting a reaction to go in the direction you want is a major problem (to be discussed next time)

  24. Direction We need it to go in the direction we want Speed We need it to go fast enough to have the cell double in one generation Catalysts deal with this second problem, which we will now consider Concerns about the cell’s chemical reactions

  25. The velocity problem is solved by catalysts The catalyzed reaction The catalyst takes part in the reaction, but it itself emerges unchanged

  26. HHII (TS) Activation energy without catalyst TS complex with catalyst Change in Energy Activation energy WITH the catalyst H2 + I2 2 HI

  27. Reactants in an enzyme-catalyzed reaction = “substrates”

  28. Reactants (substrates) Active site or substrate binding site (not exactly synonymous, could be just part of the active site) Not a substrate

  29. Substrate Binding Unlike inorganic catalysts, enzymes are specific

  30. Small molecules bind with great specificity to pockets on ENZYME surfaces Too far

  31. Unlike inorganic catalysts, enzymes are specific                                       succinic dehydrogenase HOOC-HC=CH-COOH <-------------------------------> HOOC-CH2-CH2-COOH +2H fumaric acid                                                     succinic acid NOT a substrate for the enzyme: 1-hydroxy-butenoate:    HO-CH=CH-COOH (simple OH instead of one of the carboxyls) Maleic acid Platinum will work with all of these, indiscriminantly

  32. + • Enzymes work as catalysts for two reasons: • They bind the substrates putting them in close proximity. • They participate in the reaction, weakening the covalent bonds • of a substrate by its interaction with their amino acid residue side groups (e.g., by stretching).

  33. Dihydrofolate reductase, the movie: FH2 + NADPH2  FH4 + NADP or: DHF + NADPH + H+  THF + NADP+ Enzyme-substrate interaction is often dynamic. The enzyme protein changes its 3-D structure upon binding the substrate. http://chem-faculty.ucsd.edu/kraut/dhfr.mpg

  34. Substrate  Product (reactants in enzyme catalyzed reactions are called substrates) S  P Velocity = V = ΔP/ Δ t So V also = -ΔS/ Δt (disappearance) From the laws of mass action: ΔP/ Δt = - ΔS/ Δt = k1[S] – k2[P] For the INITIAL reaction, [P] is small and can be neglected: ΔP/ Δt = - ΔS/ Δt = k1[S] So the INITIAL velocity Vo = k1[S] back reaction Chemical kinetics O signifies INITIAL velocity

  35. Vo = ΔP/ Δ t P vs. t Slope = Vo

  36. Effect of different initial substrate concentrations on P vs. t 0.6 [S4] [S3] 0.4 P [S2] 0.2 [S1] 0.0 t

  37. Effect of different initial substrate concentrations 0.6 [S4] [S3] 0.4 [S2] 0.2 [S1] 0.0 t Dependence of Vo on substrate concentraion Vo = the slope in each case P Vo = k1[S] Slope = k1 Considering Vo as a function of [S] (which will be our usual useful consideration):

  38. Now, with an enzyme: We can ignore the rate of the non-catalyzed reaction (exaggerated here to make it visible)

  39. Enzyme kinetics (as opposed to simple chemical kinetics) Vo independent of [S] Vo proportional to [S] Can we understand this curve?

  40. Michaelis and Menten mechanism for the action of enzymes (1913)

  41. Assumption 1. E + S <--> ES: this is how enzymes work, via a complex Assumption 2. Reaction 4 is negligible, when considering INITIAL velocities (Vo, not V). Assumption 3. The ES complex is in a STEADY-STATE, with its concentration unchanged with time during this period of initial rates.  (Steady state is not an equilibrium condition, it means that a compound is being added at the same rate as it is being lost, so that its concentration remains constant.) Michaelis-Menten mechanism X

  42. Steady state is not the same as equilibrium System is at equilibrium Constant level No net flow System is at “steady state” Constant level Plenty of flow E + S ES E + P

  43. k3 [Eo] [S] Vo = Km + [S] Michaelis-Menten Equation(s) See handout 5-1 at your leisure for the derivation (algebra, not complicated, neat) k3[Eo][S] Vo = [(k2+k3)/k1] +[S] If we let Km = (k2+k3)/k1, just gathering 3 constants into one, then: =

  44. Otherwise, the rate is dependent only on S k3 [Eo] [S] Vo = Km+ [S] All the k‘s are constants for a particular enzyme Rate is proportional to the amount of enzyme At high S (compared to Km), Rate is constant Vo = k3Eo At low S (compared to Km), rate is proportional to S: Vo ~ k3Eo[S]/Km

  45. At high S, Vo here = k3Eo, = Vmax = So the Michaelis-Menten equation can be written: k3 [Eo] [S] Vmax [S] Simplest form Vo = Vo = Km + [S] Km + [S]

  46. Understanding Vmax: ( the maximum intital velocity achievable with a given amount of enzyme ) Now, Vmax = k3Eo So: k3 = Vmax/Eo = the maximum (dP/dt)/Eo, = the maximum (-dS/dt)/Eo k3 = the TURNOVER NUMBER the maximum number of moles of substrate converted to product per mole of enzyme per second; the maximum number of molecules of substrate converted to product per molecule of enzyme per second Turnover number (k3) then is: a measure of  the enzyme's catalytic power.

  47. Succinic dehydrogenase: 19 (below average) Most enzymes: 100 -1000 The winner: Carbonic anhydrase (CO2 +H20 H2CO3) 600,000 That’s 600,000 molecules of substrate, per molecule of enzyme, per second. Picture it! You can’t. Some turnover numbers (per second)

  48. Consider the Vo that is 50% of Vmax Km ? Vmax/2 is achieved at a [S] that turns out to be numerically equal to Km So Km is numerically equal to the concentration of substrate required to drive the reaction at ½ the maximal velocity Try it: Set Vo = ½ Vmax in the M.M. equation and solve for S.

  49. k2 ES E + S k1 Another view of Km: Consider the reverse of this reaction (the DISsociation of the ES complex): The equilibrium constant for this dissociation reaction is: Kd = [E][S] /[ES]= k2/k1 = = (It’s the forward rate constant divided by the backward rate constant. See the Web lecture if you want to see this relationship derived)

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