Biochemistry

Study of chemistry in biological organisms

Understand how the chemical structure of a molecule is determining its function

Focus on important biochemical macromolecules

– amino acids ----->proteins – fatty acids----->lipids – nucleotides---> nucleic acids– monosaccharides---> carbohydrates

Focus on important processes

– Protein Function– Compartmentalization/regulation– Metabolism-– DNA synthesis/replication

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?

Enzymes

• What are enzymes?

• What is the role of enzymes in an organism?

• How do they work?

Lipids

• What are lipids and their structures

• What are roles of lipids

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?

Carbohydrates

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

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

Nucleic Acids

• What are their structures?

• What their functions?

• How do they replicate?

• What is the relationship between nucleic acids and proteins?

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

Connecting structure and function requires chemistry

– Oxidation-Reductions– Thermodynamics: study of energy flow – Organic functional groups– Important organic reactions

Intermolecular forces

• Hydrogen bonds

• Dipole/dipole interactions

• Nonpolar forces

Dipole/Dipole interactions

• Polarity in molecules– Polar bonds– Asymmetry

• Positive side of one polar molecule sticks to negative side of another

Dipole-Dipole interactions

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

Hydrogen bonding

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

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

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

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

Like dissolves like

• Salt dissolving in water

Amphipathicity

• Some molecules have both a hydrophilic and hydrophobic part

• soap is an example

Amphipathicity

Equilibrium

Two opposing processes occurring at the same rate:

walking up the down escalator

treadmill

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

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)

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

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

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

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)

Answers

[CaCl2][CO2]

[HCl]2

[SO3]2

SO2]2 [O2]

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

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

Le Chatelier’ and Regulation of Metabolism

• What the diet industry doesn’t want you to know!

– Food - ABCDenergy

• A fat

– What happens if energy is used up?

– What happens if eat a big meal and don’t use energy

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-

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

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

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

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

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+).

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

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-

Buffer with added acid• Weak base component of the buffer

neutralizes added acid

• A- + H+ -- HA

Buffers with added base

• Weak acid component of the buffer neutralizes added base

• Equation: OH- + HA --> H2O + A-

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

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

Henderson Hasselbach continued

• pH = pKa + log ([A-] / [HA])

• pKa = -logKa

Since negative, lower pKa = more acidic

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

So what?We can now relate the pH of the solution to the

structure of weak acid using Henderson-HasselbachpH = 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.