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1 At high S, Vo here = k3Eo, == Vmax So the Michaelis-Menten equation can be written: Vo = k3 [Eo] [S] Km + [S] Vo = Vmax [S] Km + [S] Simplest form Enzyme inhibition: competitive, non-competitive, and allosteric Competitive: A competitive inhibitor resembles the substrate 2 3 A competitive inhibitor can be swamped out at high substrate concentrations Handout 5-3b 4 Vmax will not be affected. + Vo Apparent (measured) Km increases Substrate concentration Inhibitor looks like the substrate And, like the substrate, binds to the substrate binding site Biosynthetic pathway to cholesterol 5 6 Zocor (simvastatin) =“CoA” 7 ½ Vmax w/o inhibitor ½ Vmax with yet more inhibitor Km remains unchanged. Vmax decreases. 8 Non-competitive inhibitor example Substrate still binds OK. But an essential participant in the reaction is blocked (here, by mercury binding a cysteine sulfhydryl) Hg++ 9 snapshot Substrate Non-competitive inhibitor Example: Hg ions (mercury) binding to –SH groups in the active site 10 11 Allosteric inhibition Inhibitor binding site + Active Active = allosteric inhibitor Inactive = substrate Allosteric inhibitor binds to a different site than the substrate, so it need bear no resemblance to the substrate The apparent Km OR the apparent Vmax or both may be affected. The effects on the Vo vs. S curve are more complex and will be ignored here 12 Allosteric inhibitors are used by the cell for feedback inhibition of metabolic pathways Feedback inhibition of enzyme activity, or “End product inhibition” Linear pathway: PQRSTUV Branched pathway: End product End product The first step committed to anend-product is usually inhibited. End product protein 13 Thr deaminase glucose ...... --> --> threonine -----------------> alpha-ketobutyric acid A Substrate B C protein isoleucine (and no other aa) Allosteric inhibitor Also here: Feedback inhibitor (is dissimilar from substrate) Rich medium = provide glucose + all 20 amino acids and all vitamins, etc. 20 minutes!, in a rich medium 60 minutes, in a minimal medium 14 15 Direction Flow of glucose in E. coli of reactions in metabolism Macromolecules Polysaccharides Lipids Nucleic Acids Proteins yn th e tic pa t hw ay monomers bi os intermediates glucose Each arrow = a specific chemical reaction 16 Energy }Freedifference determines the direction of a chemical reagion 17 For the model reaction A + B C + D, written in the left-to-right direction indicated: Consider the quantity called the change in free energy associated with a chemical reaction, or: Δ G Such that: • IF Δ G IS <0: THEN A AND B WILL TEND TO PRODUCE C AND D (i.e., tends to go to the right). • IF Δ G IS >0: THEN C AND D WILL TEND TO PRODUCE A AND B. (i.e., tends to go to the left) • IF Δ G IS = 0: THEN THE REACTION WILL BE AT EQUILIBRIUM: NOT TENDING TO GO IN EITHER DIRECTION IN A NET WAY. ΔG = Δ Go+ 18 RTln([C][D]/[A][B]) • where A, B, C and D are the concentrations of the reactants and the products AT THE MOMENT BEING CONSIDERED. (i.e., these A, B, C, D’s here are not the equilibrium concentrations) • R = UNIVERSAL GAS CONSTANT = 1.98 CAL / DEG K MOLE (R =~2) • T = ABSOLUTE TEMP ( oK ) 0oC = 273oK; Room temp = 25o C = 298o K (T =~ 300) • ln = NATURAL LOG • Δ Go = a CONSTANT: a quantity related to the INTRINSIC properties of A, B, C, and D 19 Also abbreviated form: Δ G = Δ Go+ RTlnQ (Q for “quotient”) Where Q = ([C][D]/[A][B]) Qualitative term: What molecules are in play Quantitative term: How much of each is present Josiah Willard Gibbs (1839 1903) 20 Δ Go STANDARD FREE ENERGY CHANGE of a reaction. If all the reactants and all the products are present at 1 unit concentration, then: Δ G = Δ Go + RTln(Q) = Δ Go + RTln([1][1] / [1][1]) Δ G = Δ Go + RTln(1) or Δ G = Δ Go +RT x 0, or Δ G = Δ Go, when all components are at 1 ….. a special case (when all components are at 1) “1” usually means 1 M So Δ G and Δ Go are quite different, and not to be confused with each other. Δ Go allows us to compare all reactions under the same standard reaction conditions that we all agree to, independent of concentrations. So it allows a comparison of the stabilities of the bonds in the reactants vs. the products. It is useful. AND, It is easily measured. 21 22 Because, • at equilibrium, Δ G = Δ Go + RTln(Q) = 0 and at equilibrium Q = Keq = (a second special case). [C]eq [D]eq [A]eq [B]eq • So: at equilibrium, Δ G = Δ Go + RTln(Keq) = 0 • And so: Δ Go = - RTln(Keq) • So just measure the Keq, • Plug in R and T • Get: ΔGo, the standard free energy change E.g., let’s say for the reaction A + B Keq happens to be: [C]eq[D]eq [A]eq[B]eq 23 C + D, = 2.5 x 10-3 Then Δ Go = -RTlnKeq = -2 x 300 x ln(2.5 x 10-3) = -600 x -6 = +3600 3600 cal/mole (If we use R=2 we are dealing with calories) Or: 3.6 kcal/mole 3.6 kcal/mole ABSORBED (positive number) So energy is required for the reaction in the left-to-right direction And indeed, very little product accumulates at equilibrium (Keq = 0.0025) 24 Note: If ΔGo = +3.6 for the reaction A + B < --- >C + D Then ΔGo = -3.6 for the reaction C + D <--- > A + B (Reverse the reaction: switch the sign) And: For reactions of more than simple 1 to 1 stoichiometries: aA + bB <--> cC + dD, ΔG = ΔGo + RT ln [C]c[D]d [A]a[B]b Some exceptions to the 1M standard condition: Exception #1: • 1) Water: 55 M (pure water) is considered the “unit” concentration in this case instead of 1M The concentration of water rarely changes during the course of an aqueous reaction, since water is at such a high concentration. • So when calulating Go, instead of writing in “55” when water participates in a reaction (e.g., a hydrolysis) we write “1.” • This is not cheating; we are in charge of what is a “standard” condition, and we all agree to this: 55 M H20 is unit (“1”) concentration for the purpose of defining Go. 25 Exception #2: In the same way, Hydrogen ion concentration, [H+]: 10-7 M is taken as unit concentration, by biochemists. since pH7 is maintained (buffered) in most parts of the cell despite a reaction that may produce acid or base. This definition of the standard free energy change requires the designation ΔGo’ However, I will not bother. But it should be understood we are always talking about ΔGo’ in this course. 26 27 Summary ΔG = Go + RTln(Q) This combination of one qualitative and one quantitative (driving) term tells the direction of a chemical reaction in any particular circumstance ΔGo = - RTln(Keq) The ΔGo for any reaction is a constant that can be looked up in a book. 28 Hydrolysis of ATP: ATP + HOH ADP + HPO4-- ATP, a small molecule in the cell that helps in the transfer of energy from a place where it is generated to a place where it is needed. Acid anhydride (new functional group) adenine ribose e A-R e H H O O | | HO - P - OH HO - P - OH || || O O Dehydration between 2 acids 29 The hydrolysis of ATP: :: AMP ADP ATP + HOH ADP + Pi The Go of this reaction is about -7 kcal/mole. Energy is released in this reaction. This is an exergonic reaction Strongly to the right, towards hydrolysis, towards ADP 30 O O O || || || A-R-O-P-O-P-O-P-O- + HOH | | | O- O- O- ATP Adenosine triphosphate O O || || A-R-O-P-O-P-O- + | | O- O- ADP Adenosine diphosphate The Go of this reaction is about -7 kcal/mole. Energy is released in this reaction. This is an exergonic reaction Strongly to the right, towards hydrolysis, towards ADP O || -O-P-O| O- Pi Inorganic phosphate “High energy” bonds: • Go of a least ~ -7 kcal/mole is released upon hydrolysis • Designated with a squiggle (~) often • ATP = A-P-P~P • Rationalized by the relief of electrical repulsion upon hydrolysis: ΔGo = -7 kcal/mole 31 32 Hydrolysis of ADP AMP ΔGo = -7 kcal/mole ADP A-R-P~P~P Not a high energy bond ATP + HOH + ATPase Keq ~ 100,000 ADP + Pi + heat Prob set 4 33 • The cell often uses the hydrolysis of ATP to release energy. • The released energy is used to drive reactions that require energy. • How does this work ?? 34 E.g., A reaction that requires energy, an endergonic reaction: glucose + Pi glucose-6-phosphate OP03-- +Pi Glucose + Pi --> glucose-6-P + H2O; ATP + Δ Go = +3.6 kcal/mole. H2O --> ADP + Pi Glucose + Pi --> G6P + H2O Δ Go = -7 Keq= 2.5 x 10 -3 kcal/mole Δ Go = +3.6 kcal/mole ATP + H2O+ Glucose + Pi ADP + Pi + G6P + H2O Δ Go = -3.4 kcal/mole overall Glucose + ATP --> G6P + ADP Δ Go = -3.4 kcal/mole overall = net sum of the two considered reactions Δ Go’s of multiple reactions are additive ATP + H2O --> ADP + Pi Δ Go = -7 kcal/mole Glucose + Pi --> G6P + H2O Δ Go = +3.6 kcal/mole Glucose + ATP --> G6P + ADP Δ Go = -3.4 kcal/mole overall = net sum of the two considered reactions Enzymes needed … ATPase? Glucose phosphorylase? No. Just get 7 kcal/mole as heat. But: Hexokinase Glucose + AR-P-P-P glucose-6-P + AR-P-P Glucose + ATP glucose-6-P04 + ADP, Δ Go = -3.4 kcal/mole A coupled reaction. A new reaction, ATP not simply hydrolyzed. Coupling of reactions is one of two ways the cell solves the problem of getting a reaction to go in the desired direction. The second: later. 35 36 Flow of glucose in E. coli Macromolecules Polysaccharides Lipids Nucleic Acids Proteins ATP ATP monomers ATP ATP ay ATP ATP yn th e ATP tic pa t ATP hw ATP bi os intermediates glucose Each arrow = a specific chemical reaction 37 • So does this solve the direction problem? Only for a second … • Where does this ATP come from, if we are E. coli growing in minimal medium… • Glucose is the only carbon source. • Need to make ATP from glucose, and this path TAKES energy. • But need only to regenerate ATP from ADP: ATP ATP Via GLYCOLYSIS, e.g. Handout 7-1a 38 Regeneration of ATP from ADP Two solutions: • 1) Photosynthesis Metabolic breakdown of an organic compound • 2) Catabolism of organic comounds (e.g., glucose) 39 Glucose catabolism overview/preview • 1- GLYCOLYSIS (6C 3C) • 2- KREBS CYCLE (3C – 1C, CO2 release) • 3- ELECTRON TRANSPORT CHAIN (oxygen uptake, water release) Glycolysis, in detail, as: A) Basic mechanism of energy metabolism (getting energy by glucose breakdown.) B) An example of a metabolic pathway. 40 Handout 7-2 41 The first 5 steps of glycolysis dihydroxyacetone pnosphate -OPO3 5 PO 3 e l as o Ald OH, +ATP +ATP Fructose-6-P G-6-P Glucose Fructose- +H OH -H Isomerization Phosphorylation 1,6-di-P Phosphorylation 1 2 3 Handout 7-3 4 gyceraldehyde -3-phosphate 2 moles of G3P for every mole of glucose 42 43 Acid anhydride functional group (if add HOH get 2 acids) Glyceraldehyde-3- phosphate 1,3,-diphospho-glyceric acid Abbreviations for the phosphate group Need an acceptor for these electrons: An oxidizing agent. Handout 7-4a 44 NAD: The acceptor of the electrons (and one proton) NAD = nicotinamide adenine dinucleotide Niacin (Vit. B3) NAD Handout 7-4c 45 Loose end #2 = NAD Loose end #1 was ATP Top carbon has been oxidized 46 ATP debt paid in full -2 ATP + 4 ATP = + 2 ATP “glycolysis” ends here Handout 7-2 47 1 glucose + 2 ADP + 2 Pi + 2 NAD Δ Go = 2 pyruvate + 2 ATP + 2 NADH2 -18 kcal/mole So overall reaction goes essentially completely to the right. 48 (Handout 7-3) Handout 7-4b 49 •Δ G = Δ Go + RTln([products] [reactants]) pull pull Handout 7-4b 50 The second way the cell gets a reaction to go in the desired direction: 1) First way was: a coupled reaction (i.e., a different reaction) . One of two ways the cell solves the problem of getting a reaction to go in the desired direction Glucose + ATP glucose-6-P04 + ADP, Δ Go = -3.4 kcal/mole 2) The second way: • Removal of the product of an energetically unfavorable reaction • Uses a favorable downstream reaction • “Pulls” the unfavorable reaction • Operates on the second term of the Δ G equation. • Δ G = Δ Go + RTln([products]/[reactants]) • So glucose pyruvic acid • ADP ATP, as long as we have plenty of glucose • Are we all set? • No…. What about the NAD?.. We left it burdened with those electrons. • Soon all of the NAD will be in the form of NADH2 • Very soon • Glycolysis will screech to a halt !! • Need an oxidizing agent in plentiful supply to keep taking those electron off the NADH2, to regenerate NAD so we can continue to run glucose through the glycolytic pathway. 51 52 Oxidizing agents around for NAD: 1) Oxygen Defer (and not always present, actually) 2) Pyruvate, our end-product of glycolysis In E. coli, humans: Pyruvate lactate, NADH2 NAD, coupled In Yeast: Pyruvate ethanol + CO2 53 54 Glucose GAL-3-P ATP 1,3-Di-PGA ATP Glucose NAD NADH2 Biosynthetic pathway to NAD Lactate Pyruvate excreted Handout 7-1b yeast humans E. coli 55 H C=O | CH3 Acetaldehyde detail 56 Fermentation: anaerobiosis (no oxygen) Lactate fermentation Ethanolic fermentation Mutually exclusive, depends on organism Other types, less common fermentations, exist – (e.g., propionic acid fermentation, going on in Swiss cheese) 57 The efficiency of fermentation glucose--> 2 lactates, without considering the couplings for the formation of ATP's (no energy harnessing): Δ Go = -45 kcal/mole So 45 kcal/mole to work with. Out of this comes 2 ATPs, worth 14 kcal/mol. So the efficiency is about 14/45 = ~30% Where did the other 31/45 kcal/mole go? Wasted as HEAT. Fermentation goes all the way to the right 58 glucose--> 2 lactates, without considering the couplings for the formation of ATP's (no energy harnessing): Δ Go = -45 kcal/mole kcal/mole Out of this comes 2 ATPs, worth 14 kcal/mol. So the efficiency is about 14/45 = ~30% Since 2 ATPs ARE produced, taking them into account, for the reaction: Glucose + 2 ADP + 2 Pi 2 lactate + 2 ATP ΔGo = -31 kcal/mole (45-14) Very favorable. All the way to the right. Keep bringing in glucose, keep spewing out lactate, Make all the ATP you want. That’s fermentation. Gl;ycerol as an alternative sole carbon and energy source for E. coli ATP glycerol +NAD glycerol phosphate - O2 ? 59 + NADH2 DHAP (dihydroxy acetone phosphate) glycolysis +O2 CO2 + H2O and ADP + Pi ATP 60 Glycerol + ATP → glycerol phosphate → DHAP NAD → NADH2 61 +NAD ATP glycerol glycerol phosphate - O2 + NADH2 DHAP (dihydroxy acetone phosphate) glycolysis +O2 Glycerol cannot be fermented. CO2 + H2O E. coli CANNOT grow on glycerol in the absence of air These pathways are real, and they set the rules. Stoichiometry of chemical reactions must be obeyed. No magic is involved 62 Energy yield But all this spewing of lactate turns out to be wasteful. Using oxygen as an oxidizing agent glucose could be completely oxidized, to: … CO2 That is, burned. How much energy released then? Glucose + 6 O2 6 CO2 + 6 H2O ΔGo = -686 kcal/mole ! Compared to -45 to lactate (both w/o ATP production considered) Complete oxidation of glucose, Much more ATP But nature’s solution is a bit complicated. The fate of pyruvate is now different