Thermodynamics II; Acid-base properties Andy Howard Biochemistry Lectures, Spring 2019 22 January 2019, IIT

Thermodynamics; Acid-base reactions Thinking carefully about entropy helps us make sense of biochemical thermodynamics Protonation and deprotonation matter Amino acids exemplify a lot of basic biochemical concepts, and they’re inherently important too 01/22/2019 Thermo II; Acid-base

What we’ll discuss Thermodynamics II Acid-base concepts Amino Acids Free energy Le Chatelier ATP Energy examples Protein folding Acid-base concepts Weak acids Henderson- Hasselbalch Amino Acids What they are a-amino acids 01/22/2019 Thermo II; Acid-base

Free energy and equilibrium Go = -RT ln keq, or keq = exp(-Go/RT) keq is equilibrium constant; formula depends on reaction type For aA + bB  cC + dD, keq = ([C]c[D]d)/([A]a[B]b) If all proportions are equal, keq = ([C][D])/([A][B]) These values ([C], [D] …) denote the concentrations at equilibrium 01/22/2019 Thermo II; Acid-base

Spontaneity and free energy Thus if reaction is just spontaneous, i.e. Go = 0, then keq = 1 If Go < 0, then keq > 1: Exergonic If Go > 0, then keq < 1: Endergonic Distinguishable from exothermic and endothermic, which are concerned only with enthalpy, not the whole free-energy package. 01/22/2019 Thermo II; Acid-base

Free energy as a source of work Change in free energy indicates that the reaction could be used to perform useful work If Go < 0, the system supplies work If Go > 0, we need to do work on the system to make the reaction occur 01/22/2019 Thermo II; Acid-base

What kind of work? Movement (flagella, muscles) Chemical work: Transport molecules against concentration gradients Transport ions against potential gradients Driving otherwise endergonic reactions by direct coupling of reactions by depletion of products 01/22/2019 Thermo II; Acid-base

Coupled reactions Often a single enzyme catalyzes 2 reactions, shoving them together: reaction 1, A  B: Go1 < 0 reaction 2, C  D: Go2 > 0 Coupled reaction: A + C  B + D: GoC = Go1 + Go2 If GoC < 0, reaction 1 is driving reaction 2! 01/22/2019 Thermo II; Acid-base

How else can we win? Concentration of products & reactants may play a role As we’ll discuss in a moment, the actual free energy depends on Go and on concentration of products and reactants So if the first reaction withdraws product of reaction B away, that drives the equilibrium of reaction 2 to the right 01/22/2019 Thermo II; Acid-base

Le Chatelier’s Principle & ΔGo Le Chatelier’s Principle says that a reaction can be driven to the right if one of the products is being removed from availability This can happen in various ways: A gaseous product can evaporate An insoluble product can precipitate A reactive product can get converted to something else. 01/22/2019 Thermo II; Acid-base

Quantitation of Le Chatelier’s Principle Relationship between G (actual free energy experienced in a reaction under real conditions) & standard free energy Go : G = Go + RT ln([products]/[reactants]) So for a typical bimolecular reaction A + B  C + D, G = Go + RT ln{[C][D]/([A][B])} 01/22/2019 Thermo II; Acid-base

Why does this embody Le Chatelier’s Principle? Suppose product D is being taken out of the reaction vessel (e.g. the mitochondrion) either by being converted to something else or by being transported away. Then [D] will be small, so Q=[C][D]/([A][B]) will be small 01/22/2019 Thermo II; Acid-base

Consequence of that! Therefore lnQ will be negative, and G = Go + RT lnQ can become negative even if Go is positive (especially at high T) 01/22/2019 Thermo II; Acid-base

How hard is it to measure Go? Not as bad as you might think. It’s true that it’s hard to set up [C]=[D]=[A]=[B] = 1M; but we don’t actually have to do that. Note that since G =Go +RTln{[C][D]/([A][B])}, we can get to G =Go simply by ensuring that [C][D]/([A][B]) = 1, so that ln{[C][D]/([A][B]) = 0. 01/22/2019 Thermo II; Acid-base

Adenosine Triphosphate ATP readily available in cells Derived from catabolic reactions Contains two high-energy phosphate bonds that can be hydrolyzed to release energy: O O- || | (AMP)-O~P-O~P-O- | || O- O 01/22/2019 Thermo II; Acid-base

Hydrolysis of ATP Hydrolysis at the rightmost high-energy bond: ATP + H2O  ADP + Pi , Go = -33kJ/mol Hydrolysis of middle bond: ATP + H2O  AMP + PPi Go ~ -40kJ/mol BUT PPi + H2O 2 Pi, Go = -31 kJ/mol So, appropriately coupled, we get roughly twice as much! 01/22/2019 Thermo II; Acid-base

ATP as energy currency Any time we wish to drive a reaction that has Go < +30 kJ/mol, we can couple it to ATP hydrolysis and come out ahead If the reaction we want has Go < +60 kJ/mol, we can couple it to ATP  AMP and come out ahead So ATP is a convenient source of energy — an energy currency for the cell 01/22/2019 Thermo II; Acid-base

Coin analogy Think of store of ATP as a roll of quarters Vendors don’t give change Use one quarter for some reactions, two for others Inefficient for buying $0.35 items 01/22/2019 Thermo II; Acid-base

Other high-energy compounds Creatine phosphate: ~ $0.35 Phosphoenolpyruvate: ~ $0.40 So for some reactions, they’re more efficient than ATP 01/22/2019 Thermo II; Acid-base

Why not use those always? There’s no such thing as a free lunch! In order to store a compound, you have to create it in the first place So an intermediate-energy currency is the most appropriate 01/22/2019 Thermo II; Acid-base

Le Chatelier’s principle in ATP-dependent reactions G = Go + RT ln{[C][D]/([A][B]) In an ATP-coupled reaction, D=ADP, B=ATP; C and A are the other reactants But note that G = Go + RT ln{[C][D]/([A][B]) = Go + RT { ln([C]/[A])+ ln([ADP]/[ATP])} 01/22/2019 Thermo II; Acid-base

Why that matters in cells… The fact is that often [ADP]/[ATP] ~ 0.1, so at T=300.6K, RTln[ADP]/[ATP] = -5.8 kJ /mol, so that provides even more available energy to drive the reaction! 01/22/2019 Thermo II; Acid-base

How else can Le Chatelier’s principle help? Often coupled reactions involve withdrawal of a product from availability If that happens, [product] / [reactant] shrinks, the second term becomes negative, and G < 0 even if Go > 0 01/22/2019 Thermo II; Acid-base

Example: glycolysis Later this semester we’ll spend at least one lecture looking at glycolysis, one of the fundamental pathways Some of the glycolytic reactions have Go’ or Go > 0 But all have G values that are negative or zero because of this concentration effect 01/22/2019 Thermo II; Acid-base

How to solve energy problems involving coupled equations General principles: If two equations are added, their energetics add An item that appears on the left and right side of the combined equation can be cancelled Reversing a reaction reverses the sign of G. “Hydrolysis” means “reacting with water” 01/22/2019 Thermo II; Acid-base

A bit more detail Suppose we couple two equations: A + B  C + D, DGo’ = x C + F  B + G, DGo’ = y The result is: A + B + C + F  B + C + D + G, or A + F  D + G, DGo’ = x + y … since B & C appear on both sides 01/22/2019 Thermo II; Acid-base

Slightly more complex… Suppose we couple two equations: A + B  C + D, DGo’ = x H + A  J + C, DGo’ = z Reverse the second equation: J + C  A + H, DGo’ = -z Add this to 1st eqn. & simplify: B + J  D + H, DGo’ = x - z … since A & C appear on both sides 01/22/2019 Thermo II; Acid-base

What do we mean by hydrolysis? It simply means a reaction with water Typically involves cleaving a bond: U + H2O  V + W is described as hydrolysis of U to yield V and W 01/22/2019 Thermo II; Acid-base

Protein Folding Proteins (about which we’ll say a lot soon) are typically folded into a definite conformation at room temp in solution They unfold into a state where they have no definite conformation, at higher temperature We describe this as melting, and it’s frequently a first-order phase transition 01/22/2019 Thermo II; Acid-base

How does that work thermodynamically? Consider the reaction (Protein+Solvent)folded  (Protein+Solvent)unfolded This reaction has a free energy: G = H - TS System includes both protein and solvent: G = Gprotein + Gsolvent = Hprotein + Hsolvent - TSprotein - TSsolvent 01/22/2019 Thermo II; Acid-base

Unfolding: effect on enthalpy Typically Hprotein > 0 because we lose some hydrogen bonds and van der Waals interactions when the protein unfolds Typically Hsolvent ~ 0 because we make extra protein-solvent H-bonds and break some protein-protein H-bonds 01/22/2019 Thermo II; Acid-base

Unfolding: entropy Sprotein > 0 because the protein becomes more disordered Therefore -TSprotein < 0. Ssolvent < 0 because some water molecules become aggregated around the protein; therefore -TSsolvent > 0. 01/22/2019 Thermo II; Acid-base

Putting the entropic terms together Overall entropic term in free energy is -T(Sprotein + Ssolvent) These 2 terms have opposite signs but typically the protein term dominates, just barely Therefore the overall term is slightly negative, and it becomes more negative at higher temperature 01/22/2019 Thermo II; Acid-base

Real melting temperatures % folded Melting temperature is typically around 55ºC for proteins from mesophilic organisms; more like 75ºC for thermophilic organisms 100 Temperature, ºC 01/22/2019 Thermo II; Acid-base

Let’s begin, chemically! Amino acids are important on their own and as building blocks We need to start somewhere: Proteins are made up of amino acids Free amino acids and peptides play significant roles in cells, even though their resting concentrations are low We’ll build from small to large 01/22/2019 Thermo II; Acid-base

Will we really start with amino acids immediately? Not quite yet. We need to say a few things about acid-base reactions We’ll use amino acids to exemplify those in a few minutes. 01/22/2019 Thermo II; Acid-base

Acid-Base Equilibrium In aqueous solution, [H+] ≠ 0, [OH-] ≠ 0. Define: pH  -log10[H+] pOH  -log10[OH-] Product [H+][OH-] = 10-14 M2 (+/-) fact Take common (base-10) log of both sides: log10([H+][OH-]) = log10(10-14) = -14 01/22/2019 Thermo II; Acid-base

Acid-base equilibrium, continued Remember: log(AB) = log(A) + log(B) So log10([H+][OH-]) = log10([H+]) + log10([OH-]) = -14 So -log10([H+]) -log10([OH-]) = 14 So pH + pOH = 14 derived formula 01/22/2019 Thermo II; Acid-base

That’s true everywhere… That equation is true in general. What happens specifically at pH = 7? There, pH = 7, pH + pOH = 14, so pOH = 14 – 7 = 7, so pH = pOH That’s not true at any other pH. 01/22/2019 Thermo II; Acid-base

So what’s the equilibrium constant for this reaction? Note that the chemical reaction is H2O  H+ + OH- Therefore keq = [H+][OH-] / [H2O] But we just said that [H+][OH-] = 10-14M2 We also know that [H2O] = 55.5M (= (1000 g / L )/(18 g/mole)) keq = (10-14M2) /55.5M = 1.8 * 10-16M 01/22/2019 Thermo II; Acid-base

Keep clear in your mind… Often students make mistakes that are easy to avoid if you remember a simple rule: At low pH, there are many protons available and few hydroxide ions available At high pH, there are few protons available and many hydroxide ions available 01/22/2019 Thermo II; Acid-base

Henderson-Hasselbalch Equation If ionizable solutes are present, their ionization will depend on pH Assume a weak acid HA  H+ + A- such that the ionization equilibrium constant is Ka = [A-][H+] / [HA] Define pKa  -log10Ka Then pH = pKa + log10([A-]/[HA]) 01/22/2019 Thermo II; Acid-base

The Derivation is Trivial! Ho hum: Ka = [A-][H+]/[HA] pKa = -log10([A-][H+]/[HA]) = -log10([A-]/[HA]) - log10([H+]) = -log10([A-]/[HA]) + pH Therefore pH = pKa + log10([A-]/[HA]) Often written pH = pKa + log([base]/[acid]) 01/22/2019 Thermo II; Acid-base

How do we use this? Often we’re interested in calculating [base]/[acid] for a dilute solute Clearly if we can calculate log10([base]/[acid]) = pH - pKa then you can determine [base]/[acid] = 10(pH - pKa) 01/22/2019 Thermo II; Acid-base

At, below, above the pKa If pH = pKa, the concentration of the “acid” (more protonated) form and the “base” (less protonated) form are equal If pH < pKa, the acid form predominates If pH > pKa, the base form predominates 01/22/2019 Thermo II; Acid-base

What if a molecule has more than one ionizable group? That’s okay: there will be a pKa characteristic of each group’s ionization Typically the two pKa values will be far apart, so that at any given pH, only at most two forms will be plentiful 01/22/2019 Thermo II; Acid-base

Is the “acid” form always positively charged? No; it simply differs from the “base” form by exactly one charge So they could be: (+1 and 0), or (0 and -1), or even (+2 and +1), or (-1 and -2). 01/22/2019 Thermo II; Acid-base

Using this notion Many amino acid properties are expressed in these terms; but it’s relevant to other biological acids and bases too, like lactate and oleate 01/22/2019 Thermo II; Acid-base

Reading recommendations If the material on ionization of weak acids isn’t pure review for you, I strongly encourage you to read sections 2.3-2.5 of CF&M. in detail. We won’t go over this material in detail in class because it should be review, but you do need to know it! 01/22/2019 Thermo II; Acid-base

pKa values for weak acids Acid Base pKa Acid Base pKa Form Form Form Form Formic acid Formate 3.8 Acetic acid Acetate 4.8 Lactic acid Lactate 3.9 H3PO4 H2PO4- 2.2 H2PO4- HPO4-2 7.2 HPO4-2 PO4-3 12.7 H2CO3 HCO3- 10.2 NH4+ NH3 9.2 CH3NH3+ CH3NH2 10.7 Alanine+ Alanine0 2.4 Alanine0 Alanine- 9.9 01/22/2019 Thermo II; Acid-base

Calculating the pH of a solution of a weak acid Add a weak acid to water at a molarity B, with dissociation constant ka= [H+][A-]/[HA] Then let x = [H+]. We assume that almost all protons in the solution come from solute, not the water, so to a good approximation [A-] = x also. 01/22/2019 Thermo II; Acid-base

pH of a solution of a weak acid, continued Therefore ka = x*x/(B-x) = x2/(B-x) We know B and ka: we want to solve this for x. ka (B-x) = x2 x2 + kax –Bka = 0 x = {-ka ± √[ka2 – 4(1)(-Bka)]} / 2 x = (-ka/2) + √[ka2/4 + Bka] This is an exact result and general. 01/22/2019 Thermo II; Acid-base

Appropriate simplifications In most cases B >> x, so the initial equation is roughly ka = x2/B Solution is then x = (Bka)1/2 Typical values: B = 0.1M, ka= 1.76*10-5M, so x = 1.33*10-3M pH = -log10x = 2.88 01/22/2019 Thermo II; Acid-base

Now: let’s look at amino acids Building blocks of proteins are of the form H3N+-CHR-COO-; these are -amino acids. But there are others, e.g. beta-alanine: H3N+-CH2-CH2-COO- 01/22/2019 Thermo II; Acid-base

These are zwitterions Over a broad range of pH: amino end is protonated and is + charged carboxyl end is unprotonated and is - charged Therefore both ends are charged anywhere near neutral pH Free -amino acids are therefore highly soluble, even if the side chain is apolar 01/22/2019 Thermo II; Acid-base

At low and high pH: At low pH, the carboxyl end is protonated At high pH, the amino end is deprotonated These are molecules with net charges 01/22/2019 Thermo II; Acid-base

Identities of the R groups Nineteen of the twenty ribosomally encoded amino acids fit this form Only variation is in identity of R group (side chain extending off alpha carbon) Complexity ranging from glycine (R=H) to tryptophan (R=-CH2-indole) Sometimes we care about non-ribosomal -amino acids—like ornithine ornithine 01/22/2019 Thermo II; Acid-base

Why “ribosomal”? It’s my slightly pompous way of emphasizing that these 20 amino acids are actually used in the ribosome to build proteins. Other a-amino acids (like ornithine and citrulline) matter, but they don’t get used in making proteins. 01/22/2019 Thermo II; Acid-base

Let’s learn the ribosomal amino acids. We’ll walk through the list of 20, one or two at a time We’ll begin with proline because it’s weird Then we’ll go through the others sequentially You do need to memorize these names and structures, both actively and passively 01/22/2019 Thermo II; Acid-base

Special case: proline Proline isn’t an amino acid: it’s an imino acid Hindered rotation around bond between amine N and alpha carbon is important to its properties Tends to abolish a-helicity because of that hindered rotation Main-chain pKa’s: 2.0, 10.6 01/22/2019 Thermo II; Acid-base

The simplest amino acids Glycine Alanine Main-chain pKa’s: 2.4, 9.8 Somewhat nonpolar methyl Main-chain pKa’s: 2.4, 9.9 01/22/2019 Thermo II; Acid-base

Branched-chain aliphatic aas Main-chain pKa’s: 2.3, 9.7 Valine Isoleucine isopropyl Leucine Main-chain pKa’s: 2.3, 9.7 Main-chain pKa’s: 2.3, 9.8 01/22/2019 Thermo II; Acid-base

Hydroxylated, polar amino acids Serine Threonine hydroxyl Main-chain pKa’s: 2.1, 9.1 Main-chain pKa’s: 2.2, 9.2 01/22/2019 Thermo II; Acid-base

Amino acids with carboxylate side chains Aspartate Glutamate methylene carboxylate Main-chain pKa’s: 2.0, 9.9 Side-chain pKaa: 3.9 Main-chain pKa’s: 2.1, 9.5 Side-chain pKa: 4.1 01/22/2019 Thermo II; Acid-base

Amino Acids with amide side chains Asparagine Glutamine Note: these are uncharged! Don’t fall into the trap of thinking that the side-chain can be ionized! amide Main-chain pKa’s: 2.1, 8.7 Main-chain pKa’s: 2.2, 9.1 01/22/2019 Thermo II; Acid-base

Sulfur-containing amino acids Methionine Cysteine sulfhydryl Main-chain pKa’s: 2.1, 9.3 Main-chain pKa’s: 1.9, 10.7 Side-chain pKa: 8.4 01/22/2019 Thermo II; Acid-base

Positively charged side chains Lysine Arginine Guanidinium Main-chain pKa’s: 2.2, 9.2 Side-chain pKa: 10.5 Main-chain pKa’s: 1.8, 9.0 Side-chain pKa: 12.5 01/22/2019 Thermo II; Acid-base

Aromatic Amino Acids Phenylalanine Tyrosine phenyl Main-chain pKa’s: 2.2, 9.3 Main-chain pKa’s: 2.2, 9.2 Side-chain pKa: 10.5 phenyl 01/22/2019 Thermo II; Acid-base

Histidine: a special case Main-chain pKa’s: 1.8, 9.3 Side-chain pKa: 6.0 imidazole 01/22/2019 Thermo II; Acid-base

Tryptophan: the biggest of all Main-chain pKa’s: 2.5, 9.4 indole 01/22/2019 Thermo II; Acid-base

Common mistakes Thinking that asn and gln sidechains can be ionized Thinking that trp’s side-chain can be ionized Misremembering the two differences between cys and met Misremembering that his is usually neutral 01/22/2019 Thermo II; Acid-base