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Intro Bio Lecture 8

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Intro Bio Lecture 8 Intro Bio Lecture 8 1 Enzyme inhibition • Competitive inhibition (Km affected) • Non-competitive inhibition (Vmax affected) Missing: • Allosteric inhibition 2 Allosteric inhibition Inhibitor binding site , + Active Active Inactive = allosteric inhibitor = 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 not be discussed here. 3 Allosteric inhibitors are used by the cell for feedback inhibition of metabolic pathways Feedback inhibition of enzyme activity, or “End product inhibition” P → Q → R → S → T → U → V End product End product End 4 First committed step is usually inhibited product protein Thr deaminase glucose ...... --> --> threonine -----------------> alpha-ketobutyric acid A Substrate B C isoleucine protein (and no other aa) Allosteric inhibitor Also here: acts as a Feedback inhibitor (inhibitor is dissimilar from substrate) 5 60 minutes, in a minimal medium (glucose + salts) Compare: Rich medium = provide glucose + all 20 amino acids and all vitamins plus other small molecules. Grow any faster? 20 minutes ! in a rich medium Regulation pays off 6 Direction Flowof reactions of glucose in metabolism in E. coli Macromolecules Polysaccharides Lipids Nucleic Acids Proteins monomers intermediates biosynthetic pathway glucose 7 Each arrow = a specific chemical reaction In order for E. coli to double in 1 hour, 3 giant problems 1) Chemical reactions must be catalyzed to go fast. Very fast. Solution: enzymes 2) Chemical reactions must go in the right direction Solution: metabolism to produce and harness energy 3) Design enzymes and metabolism and remember this information Solution: DNA 8 Major concerns about the cell’s essential chemical reactions 1)Speed • We need it to go fast enough to have the cell double in one generation 2) Direction • We need it to go in the direction we want Proteins are used to solve both of these problems. 9 Example: Biosynthesis of a fatty acid 3 glucose’s One 18-carbon fatty acid (18 C’s) 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 (but to be discussed second) 10 Free energy difference determines the direction of a chemical reaction atoms transition state energy catalyzed transition state ree ree f A + B free } energy C + D difference E + F 11 Change in free energy (DG) Energy that can do work For: A + B C + D, left-to-right direction as indicated: Consider the quantity called the change in free energy associated with a chemical reaction, or: Δ G • Δ G < 0 (negative): A AND B WILL TEND TO PRODUCE C AND D (i.e., tends to the right). • Δ G > 0 (positive): C AND D WILL TEND TO PRODUCE A AND B. (i.e., tends to the left) • Δ G = 0: THE REACTION WILL BE AT EQUILIBRIUM: NOT TENDING TO GO IN EITHER DIRECTION IN A NET WAY 12 The change in free energy (DG) A + B C + D products DG = DGo+ RTln [C][D] [A][B] reactants • A, B, C and D = 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 / °K MOLE ( so R =~2) • T = ABSOLUTE TEMP ( oK ) 0oC = 273oK; Room temp = 25o C = 298o K or ~ 300o K • ln = NATURAL LOG • D Go = a CONSTANT: a quantity related to the INTRINSIC properties of A, B, C, and D, . to be explained 13 The change in free energy change (DG) Also abbreviated form: DG = DGo+ RTlnQ (Q for “quotient”) Where Q = ([C][D]/[A][B]) Qualitative term: Quantitative term: Josiah Willard Gibbs What molecules How much of each is (1839 - 1903) are in play. This depends present, at the moment on what A, B, C, and D under consideration are. 14 Standard free energy change (DGo) STANDARD FREE ENERGY CHANGE of a reaction. If all the reactants and all the products are present at 1 unit concentration, then: DG = DGo + RTln(Q) = DGo + RTln([1][1] / [1][1]) = DGo + RTln(1) = DGo + RT x 0, or DG = DGo, when all components are at 1 ….. a special case (when all components are at 1) “1” usually means 1 M 15 Standard free energy change (DGo) So DG and DGo are quite different, and not to be confused with each other. DGo 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. 16 DG and DGo at equilibrium • Because at equilibrium, DG = DGo + RTln(Q) = 0 [C]eq [D]eq and only at equilibrium: Q = Keq = [A]eq [B]eq (a second special case). • So: at equilibrium, o DG = DG + RTln(Keq) = 0 o • And so: DG = - RTln(Keq) • So just measure the Keq, • Plug in R and T • Get: DGo, the standard free energy change • Publish it. It gets written in a book. 17 E.g., let’s say for the reaction A + B C + D, Keq happens to be: [C]eq[D]eq = 2.5 x 10-3 so, not very much product [A]eq[B]eq o -3 Then DG = -RTlnKeq = -2 x 300 x ln(2.5 x 10 ) = -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, +3.6)) So energy is required for the reaction in the left-to-right direction And indeed, very little product accumulates at equilibrium (Keq = 0.0025) 18 Note: If for the reaction A + B C + D, ΔGo = +3.6 Then for the reaction C + D A + B, ΔGo = - 3.6 (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 And RT = 2*300 = 600 cal/deg-mole at room temp., or RT= 0.002*300 = 0.6 kcal/deg-mole at room temp, more usefully 19 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 calculating DGo, 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 o agree to this: 55 M H20 is unit (“1”) concentration for the purpose of defining DG . 20 Some exceptions to the 1M standard condition: 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, we will not bother. But it should be understood we are always talking about ΔGo’ in this course. 21 Summary DG = DGo + RTln(Q) This combination of one qualitative and one quantitative (driving) term tells the direction of a chemical reaction in any particular circumstance o DG = - RTln(Keq) The ΔGo for any reaction is a constant that can be looked up in a book. 22 Energy in a cell: adenosine triphosphate (ATP) -- Hydrolysis of ATP: ATP + HOH → ADP + HPO4 ATP, a cellular small molecule that helps in the transfer of energy from a place where it is made to a place where it is needed. Acid anhydride (new functional group: 2 acids joined) adenine “base” ribose OH OH adenosine phosphate | | HO - P - OH HO - P - OH A-R: a nucleoside ||| ||| O O ATP: a nucleotide (a nucleoside triphosphate) Dehydration between 2 acids 23 The hydrolysis of ATP ATP + HOH → ADP + Pi _ _ _ _ _ _ _ (ADP) The DGo of this reaction is about -7 kcal/mole. Energy is released in this reaction. This is an exergonic reaction under standard conditions Strongly to the right, towards hydrolysis, towards ADP 24 “High energy” bonds • DGo of a least ~ -7 kcal/mole is released upon hydrolysis • Designated with a squiggle (~) often • ATP = A-P-P~P • Rationalized here by the relief of electrical repulsion upon hydrolysis: _ _ _ _ DGo = -7 kcal/mole 25 Hydrolysis of ADP _ AMP AMP _ _ _ _ ADP D o (ADP) G = -7 kcal/mole _ A-R-P~P~P _ Not a high energy bond 26 What happens when we dissolve ATP in water? ATP + HOH → ? + ATPase ATP + HOH ADP + Pi + heat (→7o) 1M ATP, 1 ml water, 1 mmole ATP, @7 kcal/mole, 7000 cal/mole, 7 cal / mmole, 7 cal, @1 cal:1 ml water: 1o C→7o 27 How does the cell harness this energy? • 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 ?? 28 An example: an endergonic reaction Suppose: glucose + Pi glucose-6-phosphate 2- CH2OH CH2OPO3 + Pi o Glucose + Pi --> glucose-6-P + H2O; Δ G = +3.6 kcal/mole. -3 Keq= 2.5 x 10 o 1) ATP + H2O --> ADP + Pi Δ G = -7 kcal/mole o 2) Glucose + Pi --> G6P + H2O Δ G = +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 29 An endergonic reaction, continued Our test tube of 1 M ATP in water: Add glucose + PO4: → nothing Enzymes needed … ATPase? Glucose phosphorylase? Add these enzymes: ATP → ADP + Pi, Glucose + Pi → G6P But just get 7 kcal/mole as heat again.
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