Transcript introacidbase

Biochemistry
Study of chemistry in biological
organisms
Understand how the chemical
structure of a molecule is
determining its function
Focus on important biochemical
macromolecules
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amino acids ----->proteins
fatty acids----->lipids
nucleotides---> nucleic acids
monosaccharides---> carbohydrates
Focus on important processes
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Protein Function
Compartmentalization/regulation
MetabolismDNA 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
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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:
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Intermolecular forces
Properties of water
Equilibrium
Acid/Base Theory
• Definitions
• Buffers
• Relation of structure to pH
Connecting structure and
function requires chemistry
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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
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 - 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
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
conjugate base
Base
conjugate acid
• 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 + AFor every mole of HA titrated, we form a mole of ASo, if we add enough OH- to use up half the HA (it is half-titrated) we end
up with equimolar HA and ALooking 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-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] < [Aand deprotonated form dominates.