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

2. 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
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3. 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
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4. 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
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5. 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.
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6. 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
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7. 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
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8. 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!
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9. 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
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10. 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.
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11. 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])}
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12. 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
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13. 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)
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14. 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.
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15. 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
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16. 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!
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17. 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
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18. 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
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19. Other high-energy compounds
Creatine phosphate: ~ $0.35
Phosphoenolpyruvate: ~ $0.40
So for some reactions, they’re more efficient than ATP
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20. 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
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21. 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])}
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22. 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!
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23. 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
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24. 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
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25. 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”
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26. 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
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27. 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
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28. 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
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29. 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
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30. 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
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31. 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
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32. 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.
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33. 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
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34. Real melting temperatures
% folded
Melting temperature is typically around 55ºC for proteins from mesophilic organisms; more like 75ºC for thermophilic organisms
100
Temperature, ºC
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35. 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
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36. 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.
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37. Acid-Base Equilibrium
In aqueous solution, [H+] ≠ 0, [OH-] ≠ 0.
Define:
pH  -log10[H+]
pOH  -log10[OH-]
Product [H+][OH-] = M2 (+/-) fact
Take common (base-10) log of both sides: log10([H+][OH-]) = log10(10-14) = -14
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38. 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
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39. 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.
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40. 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
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41. 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
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42. 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])
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43. 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])
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44. 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)
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45. 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
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46. 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
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47. 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).
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48. 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
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49. Reading recommendations
If the material on ionization of weak acids isn’t pure review for you, I strongly encourage you to read sections 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!
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50. 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 H2PO
H2PO4- HPO HPO4-2 PO
H2CO3 HCO NH4+ NH3 9.2
CH3NH3+ CH3NH Alanine+ Alanine0 2.4
Alanine0 Alanine- 9.9
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51. 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.
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52. 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.
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53. 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
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54. 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-
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55. 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
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56. 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
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57. 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
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58. 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.
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59. 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
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60. 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
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61. 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
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62. 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
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63. Hydroxylated, polar amino acids
Serine
Threonine
hydroxyl
Main-chain pKa’s: 2.1, 9.1
Main-chain pKa’s: 2.2, 9.2
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64. 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
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65. 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
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66. Sulfur-containing amino acids
Methionine
Cysteine
sulfhydryl
Main-chain pKa’s: 2.1, 9.3
Main-chain pKa’s: 1.9, Side-chain pKa: 8.4
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67. 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
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68. 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
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69. Histidine: a special case
Main-chain pKa’s: 1.8, 9.3 Side-chain pKa: 6.0
imidazole
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70. Tryptophan: the biggest of all
Main-chain pKa’s: 2.5, 9.4
indole
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71. 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
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