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. 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 (R f ) = k f [A] m [B] n Reverse rate (R r ) = k r [C] p [D] q (rate constants k f and k r as well as superscripts have to be determined experimentally)

28. Equilibrium When rates are equal: –R f = R r so (from previous slide) k f [A] m [B] n = k r [C] p [D] q –Putting constants together: (Law of Mass Action) k f = [C] p [D] q = K eq k r [A] m [B] n K eq 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 K eq. 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. K eq Constant Expression Given the following reactions, write out the equilibrium expression for the reaction CaCO 3 (s) + 2HCl(aq) ---> CaCl 2 (aq) + H 2 O(l) + CO 2 (g) 2SO 2 (g) + O 2 (g) --->2SO 3 (g)

32. Answers [CaCl 2 ][CO 2 ] [HCl] 2 [SO 3 ] 2 SO 2 ] 2 [O 2 ]

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 N 2 + 3 H 2  => 2 NH 3 –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 H 3 O + in water HCl + H 2 O -  H 3 O + + Cl - –Base is a proton (H + ) acceptor Produces OH - in water NH 3 + H 2 O  > NH 4 + + 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 K eq –For acids, K eq called a K a

38. Acetic Acid as Example of a Weak Acid HC 2 H 3 O 2 (aq) H + (aq) + C 2 H 3 O 2 - (aq) K = [H + ] [C 2 H 3 O 2 - ] [HC 2 H 3 O 2 ] value is 1.8 x 10 -5 1.8 x 10 -5 <<< 1

39. Weak acids, K a and pK a –pK a = - log K a –For weak acids, weaker will be less dissociated Make less H 3 O + Eq lies further to left Lower K a –Since pKa and Ka inversely related: the lower the K a, the higher the pK a, 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)+ H 2 O(l)  H 3 O + (aq)+ A – (aq) Acid Base conjugate acid conjugate base differ by one H + for acids/bases Example: HC 2 H 3 O 2 and C 2 H 3 O 2 - acid conj. base

43. Buffers A buffer is 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 --> H 2 O + 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 = pK a + log ([A - ] / [HA]) Can be used quantitatively to make buffers K a is the equilibrium constant for the acid –HA (aq) + H 2 O (l) <  H 3 O + (aq) + A - (aq) –K a = [H 3 O + ][A - ] [HA] –Higher K a = more acidic acid

48. Henderson Hasselbach continued pH = pK a + log ([A - ] / [HA]) pK a = -logK a Since negative, lower pK a = more acidic

49. Henderson Hasselbach and structure In a titration if we add base to the acid: HA + OH - -  H 2 O + 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 = pK a + log ([A - ] / [HA]) If [A - ] = [HA] then [A - ] / [HA] = 1 and log ([A - ] / [HA]) = log (1) – 0 So pH = pK a

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