Biochemistry

1. Biochemistry Study of chemistry in biological organisms Understand how the chemical structure of a molecule is determining its function

2. Focus on important biochemical macromolecules • amino acids ----->proteins • fatty acids----->lipids • nucleotides---> nucleic acids • monosaccharides---> carbohydrates

3. Focus on important processes • Protein Function • Compartmentalization/regulation • Metabolism- • DNA synthesis/replication

4. Protein Function • What is a protein’s structure and what role does it play in the body? • What are some important proteins in the body? • What are some key principles behind protein’s functions?

5. Enzymes • What are enzymes? • What is the role of enzymes in an organism? • How do they work?

6. Lipids • What are lipids and their structures • What are roles of lipids

7. Membranes and Transport • What is the structure of a membrane? • What is compartmentalization and why is it important? • How can molecules and information get across a membrane?

8. Carbohydrates • What the structures of carbohydrates and what is their role?

9. Metabolism • Glycolysis, Krebs cycle, Oxidative Phosphorylation, beta oxidation • How does a cell convert glucose to energy? • How does a cell convert fat to energy? • Roles of ATP, NAD and FAD • vitamins

10. Nucleic Acids • What are their structures? • What their functions? • How do they replicate? • What is the relationship between nucleic acids and proteins?

11. Connecting structure and function requires chemistry • Chemistry knowledge needed: • Intermolecular forces • Properties of water • Equilibrium • Acid/Base Theory • Definitions • Buffers • Relation of structure to pH

12. Connecting structure and function requires chemistry • Oxidation-Reductions • Thermodynamics: study of energy flow • Organic functional groups • Important organic reactions

13. Intermolecular forces • Hydrogen bonds • Dipole/dipole interactions • Nonpolar forces

14. Dipole/Dipole interactions • Polarity in molecules • Polar bonds • Asymmetry • Positive side of one polar molecule sticks to negative side of another

15. Dipole-Dipole interactions

16. Hydrogen Bonding • Special case of dipole dipole interaction • Hydrogen covalently attached to O, N, F, or Cl sticks to an unshared pair of electrons on another molecule • H-bond donors • Have the hydrogen • H-bond acceptors • Have the unshared pair • Strongest of intermolecular forces

17. Hydrogen bonding

18. Hydrogen bonding • Affect the properties of water • Water has a higher boiling point than expected • Water will dissolve only substances that can interact with its partially negative and partially positive ends

19. Nonpolar forces • Nonpolar molecules stick together weakly • Use London dispersion forces • Examples are carbon based molecules like hydrocarbons • Velcro effect • Many weak interactions can work together to be strong

20. Dissolving process • Solute—solute + solvent—solvent - 2 solute---solvent • Have to break solute—solute interactions as well as solvent—solvent interactions • Replace with solute-solvent interactions

21. Like dissolves like • Hydrophobic = nonpolar • Hydrophilic = polar • Overall, like dissolves like means that polar molecules dissolve in polar solvents and nonpolar solutes dissolve in nonpolar solvents

22. Like dissolves like • Salt dissolving in water

23. Amphipathicity • Some molecules have both a hydrophilic and hydrophobic part • soap is an example

24. Amphipathicity

25. Equilibrium Two opposing processes occurring at the same rate: walking up the down escalator treadmill

26. Equilibrium For chemical equilibrium, It is when two opposing reactions occur at the same rate. mA + nB <= pC + q D • Two reactions: • Forward: mA + nB - pC + qD • Reverse: pC + qD - mA + nB • Equilibrium when rates are equal

27. Reaction Rates Rate of reaction depends on concentration of reactants For the reaction: mA + nB => pC + qD Forward rate (Rf) = kf[A]m[B]n Reverse rate (Rr) = kr[C]p[D]q (rate constants kf and kr as well as superscripts have to be determined experimentally)

28. Equilibrium • When rates are equal: • Rf = Rr so (from previous slide) • kf[A]m[B]n = kr[C]p[D]q • Putting constants together: (Law of Mass Action) • kf = [C]p[D]q = Keq kr [A]m[B]n • Keq is the equilibrium constant • Solids and liquids don’t appear…they have constant concentration

29. Equilibrium in quantitative terms • The equilibrium state is quantified in terms of a constant called the Equilibrium Constant Keq. It is the ratio of products/reactants • It is determined by Law of Mass Action

30. Possible Situations at Equilibrium • 1. There are equal amounts of products and reactants. K=1 or close to it • 2. There are more products than reactants due to strong forward reaction • equilibrium lies right) • K >>1 • 3. There are more reactants than products due to strong reverse reaction • equlibrium lies left • K <<1

31. Keq Constant Expression • Given the following reactions, write out the equilibrium expression for the reaction • CaCO3(s) + 2HCl(aq) ---> CaCl2(aq) + H2O(l) + CO2(g) • 2SO2(g) + O2(g) --->2SO3(g)

32. Answers [CaCl2][CO2] [HCl]2 [SO3]2 SO2]2 [O2]

33. Le Chatelier’s Principle • When a system at equilibrium is stressed out of equilibrium, it shifts away from the stress to reestablish equilibrium. • Shifts away from what is added • Shifts towards what is removed

34. Le Chatelier’s Examples • N2 + 3 H2=> 2 NH3 • If we add nitrogen or hydrogen, it shifts to the right, making more ammonia • Removal of ammonia accomplishes the same thing • Shifts to the left if add ammonia

35. Le Chatelier’ and Regulation of Metabolism • What the diet industry doesn’t want you to know! • Food - ABCDenergy • A  fat • What happens if energy is used up? • What happens if eat a big meal and don’t use energy

36. Acid/Base Theory • Definitions • Acid is a proton (H+) donor • Produces H3O+ in water • HCl + H2O - H3O+ + Cl- • Base is a proton (H+) acceptor • Produces OH- in water • NH3 + H2O > NH4+ + OH-

37. Strong acids v weak acids • Strong 100 percent ionized • No Equilibrium or equilibrium lies to the right • K eq >>> 1 and is too large to measure • Weak acids not completely ionized • Equilibrium reactions • Have Keq • For acids, Keq called a Ka

38. Acetic Acid as Example of a Weak Acid • HC2H3O2(aq) <---> H+(aq) + C2H3O2-(aq) • K = [H+] [C2H3O2-] [HC2H3O2] • value is 1.8 x 10-5 • 1.8 x 10-5 <<< 1

39. Weak acids, Ka and pKa • pKa = - log Ka • For weak acids, weaker will be less dissociated • Make less H3O+ • Eq lies further to left • Lower Ka • Since pKa and Ka inversely related: the lower the Ka, the higher the pKa, the weaker the acid

40. pH • pH= -log [H+] • increasing the amount of H+ (in an acidic solution), decreases the pH • increasing the amount of OH-decreases the amount of H+ (in a basic solution), therefore, the pH increases • pH< 7 acidic • pH>7 basic

41. Conjugate Base Pairs • Whatever is produced when the acid (HA) donates a proton (H+) is called its conjugate base (A-). • Whatever is produced when the base (B) accepts a proton is called a conjugate acid (HB+).

42. Conjugate Base Pairs • HA(aq)+ H2O(l) H3O+(aq)+ A–(aq) • Acid Base conjugate acid conjugate base • differ by one H+ for acids/bases • Example: HC2H3O2 and C2H3O2- • acid conj. base

43. Buffers • A bufferis a solution that resists a change in pH upon addition of small amounts of acid or base. • It is a mixture of a weak acid/weak base conjugate pair • Ex: HA/ A-

44. Buffer with added acid • Weak base component of the buffer neutralizes added acid • A- + H+ -- HA

45. Buffers with added base • Weak acid component of the buffer neutralizes added base • Equation: OH- + HA --> H2O + A-

46. Relationship of pH to structure • We can think of a weak acid, HA, as existing in two forms. • Protonated = HA • Deprotonated = A- • Protonated is the acid • Deprotonated is the conjugate base • Titrated form

47. Henderson-Hasselbach Equation • pH = pKa + log ([A-] / [HA]) • Can be used quantitatively to make buffers • Ka is the equilibrium constant for the acid • HA(aq) + H2O(l) < H3O+(aq) + A-(aq) • Ka = [H3O+][A-] [HA] • Higher Ka = more acidic acid

48. Henderson Hasselbach continued • pH = pKa + log ([A-] / [HA]) • pKa = -logKa Since negative, lower pKa = more acidic

49. Henderson Hasselbach and structure In a titration if we add base to the acid: HA + OH- - H2O + A- For every mole of HA titrated, we form a mole of A- So, if we add enough OH- to use up half the HA (it is half-titrated) we end up with equimolar HA and A- Looking at the equation: pH = pKa + log ([A-] / [HA]) If [A-] = [HA] then [A-] / [HA] = 1 and log ([A-] / [HA]) = log (1) – 0 So pH = pKa

50. We can now relate the pH of the solution to the structure of weak acid using Henderson-Hasselbach pH = pKa + log ([A-] / [HA]) If pH = pKa, we have equal amounts of protonated and deprotonated forms If, pH < pKa, means log term is negative so [HA]>[A-] and protonated form dominates If pH > pKa, means log term is postive so [HA] < [A- and deprotonated form dominates. So what?