Section 10/5/06
Junaid Malek, M.D.
Equilibrium
• Occurs when two interconverting states are stable
• This is represented using stacked arrows: State A
State B, where State A
and State B can represent any number of separate entities
• Important to note that equilibrium means stable rates of interconversion, NOT
equal amounts of starting and finishing product
Equilibrium Constant
• To describe relative concentrations of two states in equilibrium, we use the
equilibrium constant, Keq
• Brackets indicate concentration expressed as moles/liter
• If Keq>1, equilibrium favors states to the right of the arrows
• If Keq<1, equilibrium favors states to the left of the arrows
Acidity and pKa
• One specific type of equilibrium encountered in chemistry is that of acids in
aqueous solution
• Acids give up protons; an acid that has given up a proton is said to be
“deprotonated”
• General formula: H-A
A- + H +
• Note that H+ and proton are used interchangeably
• A- is referred to as the “conjugate base”
• The right side of the arrow represents the de-protonated state, while the left
side represents the protonated state
Acidity and pKa
• As the molecules of life exist in an aqueous environment, we can describe the
equilibrium between protonated and deprotonated states as:
A-H + H2O
A- + H3O+
• Keq=[A-][H3O+]/[A-H][H2O]
Acidity and pKa
• Since the equilibrium exists in water, there are many more molecules of water
than acid
• As [H2O] is very large and doesn’t change, we can simplify our Ka as:
Ka =[A-][H3O+]/[A-H]
• Like Keq , Ka indicates which side of the equation equilibrium favors
• The larger the Ka , the more likely that the acid will be in a deprotonated state
(and hence the stronger the acid)
Acidity and pKa
• Since Ka can be very large or very small, it is easier to work on a logarithmic
scale
• Therefore, we use pKa , which is defined as pKa =-log(Ka )
• Thus, the stronger the acid, the lower the pKa
• When Ka changes by one order of magnitue, pKa changes by one unit
Acidity and pKa
• Key Concept: the stronger the acid, the more stable its conjugate base
• When put into water, stronger acids transfer more protons to water
• pH=-log[H3O+]
• The lower the pH, the higher the proton concentration and the stronger the
acid
Henderson-Hasselbalch Equation
• pH and pKa are related via the Henderson-Hasselbalch equation:
pKa = pH + log([A-H]/[A-])
• How is this derived?
Henderson-Hasselbalch Equation: The Derivation
• First consider the acid dissociation constant:
• This equation can be rearranged to isolate the hydrogen ion concentration on
the left. The new equation can be written:
• By definition, log 1/[H3O+]=pH, and log 1/Ka = pKa. Therefore, we can rewrite
the equation to read:
Henderson-Hasselbach Equation
• If you put an acid into a solution where there is a higher concentration of
protons, you protonate more molecules of acid
• If you put an acid into a solution where there is a lower concentration of
protons, more molecules of acid will release their protons to water molecules
• Remember that pKa is a value inherent to a particular acid
• pKa of an acid does not change, regardless of the pH of the solution you put it
into
Henderson-Hasselbalch Equation
• pKa describes the equilibrium between A-H, A- and H3+
• If you change any of these concentrations, you must compensate
by changing the other two in order to maintain equilibrium
• Changing the pH changes the [H3O+]
• To maintain equilibrium the values for [A-H] and [A-] must change
to compensate
Problem #1: Salicylic Acid
O
pKa = 13.4
OH
O
O
pKa = 3.0
OH
O
OH
Aspirin
(acetylsalicylic acid)
Salicylic acid
What is the dominant species at pH=1?
OH
O
OH
What is the dominant species at pH=4?
OH
O
O
What is the dominant species at pH=7?
OH
O
O
What is the dominant species at pH=14?
O
O
O
Problem #2: Saccharin
Saccharin
O O
S
N H
O O
S
N
O
A
+
O
B
Write out the formula for Keq
Keq=[B][H+]/[A]
H+
Problem #2: Saccharin
The pKa for saccharin = ~2(!)
What is the ratio of protonated to deprotonated saccharin
in each of the following solutions?
Diet Soda (pH=3)
1:10
Gastric Juices (pH=1)
10:1
Human Blood (pH~7)
1:100,000
Acidity Trends and Resonance
• Strength of acid related to stability of conjugate base
• What contributes to stability of conjugate base?
• Electronegativity - stable conjugate bases have negative charge borne by
electronegative atom
• Resonance!
Resonance
• Refers to de-localization of electrons throughout several electron orbitals in a
molecule
• This has a stabilizing effect on molecule by distributing negative charge to
several atoms
O-
O
O-
O
• Remember that in reality acetate exists as neither one of these states exist but
rather an average of the two
• The negative charge is split amongst two oxygen atoms, so each bears a
partially negative charge
• The double-bond is also shared between oxygen atoms, resulting in a “1.5”
bond
Acidity: One Final Consideration
• Local environment can affect acidity
• For example, a positively-charged binding pocket of a protein can
stabilize a negatively-charged conjugate base by forming
electrostatic interactions
protein
O
R
O-
Problem #3: Acidity
Which is the stronger acid?
CH3COOH
CH3CH2OH
Which is the stronger acid?
CH3COOH
CCl3COOH
Central Dogma
DNA
RNA
Protein
DNA and RNA: Structre
• DNA and RNA are polymers whose monomers are called nucleotides
• Each nucleotide consists of a sugar, a phosphate and a “base”
• In DNA, sugar is deoxyribose
• Asymmetry enforces the antiparallel directionality
• Chirality defines “handedness” of the helix
DNA and RNA: Structre
• In RNA, the sugar is ribose
• The extra -OH group is thought to contribute to its instability as compared
to DNA
• The phosphate group of DNA and RNA is negatively charged at neutral pH
• This negative charge helps to prevent hydrolysis by repelling lone pairs from
oxygen atom in water
Base Pairing
• Bases of nucleic acids responsible for storing information
• Sequence of base pairs encodes this information
• In DNA, bases on opposite strands of the helix “pair” with complementary
bases through hydrogen bonds
• A pairs with T, and C with G
DNA
RNA
Function
Information
storage
Information
transfer, etc.
Sugar
Deoxyribose
Ribose
Phosphate
Yes
Yes
Bases
A,T,C,G
A,U,C,G
Macromolecular
Structure
Single strand,
Double
many
stranded helix
structures
Entropy and the 2nd Law of Thermodynamics
• Equilibria favor lower energy states
• What does that mean?
• Don’t need to know exact mathematical formula, but know that energy has
two components: energy associated with bonding interactions and energy
associated with disorder
Entropy and the 2nd Law of Thermodynamics
• Bonding interactions are energetically favorable
• More or stronger bonding interactions are energetically favorable
(hence at a lower energy state)
• Energy associated with disorder is called Entropy
• Measure of the disorder of a system
• The higher the disorder, the greater the entropy
Entropy and the 2nd Law of Thermodynamics
• Key concept: It takes energy to keep components of a system in order
• Thus, increasing the amount of disorder decreases the amount of energy
required to maintain the state
• The 2nd Law of Thermodynamics states that systems tend to more towards
states of higher entropy
Entropy and the 2nd Law of Thermodynamics
• Bond energy and entropy define the energy of a state in a complex fashion
• Often, these to components will favor opposite sides
• You may not be able to figure out which component “wins”, but you should
be able to appreciate the effects of each individually
DNA Hybridization
• The pairing of complementary single strands of DNA (hybridization) is highly
favorable energetically
• Why?
• Hydrophobic effect
• Hydrogen bonding between complementary base pairs (also helps ensure
correct base pair matching)
DNA Replication
• Proceeds in semiconservative
manner
DNA Replication
• Requires 4 main components
• A template
• A primer
• dNTP’s
• DNA polymerase
DNA Replication
• First step is hybridization of primer to the DNA template
• Next, DNA polymerase binds to the DNA-primer duplex at the 3’ end of the
primer
• Then, a dNTP must position itself at the 3’ end of the primer and form a
Watson-Crick pair with the opposite base pair on the template
DNA Replication
• If the base pairs match, the polymerase facilitates a reaction which
forms a new bond between the 3’ hydroxyl oxygen of the last
nucleotide on the primer with the alpha phosphorus atom of the
dNTP
• This releases an inorganic phosphate as a product
• PRODUCTION OCCURS IN A 5’->3’ DIRECTION
• Thus, monomers can only be added to the 3’ end of a primer
DNA Polymerase
• This reaction does not occur at any significant speed without DNA polymerase
• Fast and accurate
• Has ability to “proofread”
Polymerase Chain Reaction (PCR)
• Powerful technique used to replicate and amplify DNA in a test tube
• Allows for exponential amplification of a desired region of DNA
Problem #4: Adenine
H
NH2
N
N
N
N
N
N
N
DNA
H
N
N
DNA
Identify all the hydrogen bond donors and acceptors and
indicate which direction they point
H
NH2
N
N
N
N
N
DNA
H
H
N
N
N
N
DNA
N
N H
N
Problem #4: Adenine
• Q: The exocyclic amine and the nitrogen directly attached to the DNA
backbone have lone pairs, but these lone pairs do not serve as hydrogen bond
donors. Why?
• A: These lone pairs are delocalized via resonance, so they no longer have the
electron density necessary to form a hydrogen bond
• Q: Given the direction in which the hydrogen bond donors/acceptors point,
why is it energetically favorable for the DNA bases to stack on top of one
another
• A: All H-bond donors/acceptors are in the plane of the base; the top and
bottom surfaces are therefore hydrophobic. This effect drives bases to stack in
order to consolidate these hydrophobic surfaces
Problem 5: DNA Hybridization
• Q: DNA hybridization is energetically favorable. If you put DNA into
ultra-pure water that contains no dissolved ions, DNA hybridization
becomes unfavorable. Why?
• A: Under normal conditions, the negatively charged backbone will
form electrostatic interactions with cations in solution. This stabilizes
the structure by giving a net neutral charge. Without these cations,
the negative backbones will repel.
Problem 6: DNA Replication
• Each DNA strand has a 5’ and 3’ end. These numbers refer to the numbering of
the hydroxyl groups of the deoxyribose. Two DNA strands hybridize in an
antiparallel fashion to form dsDNA
Problem 6: DNA Replication
Which of these strands are identical? Which are not?
5’-ATGGCCAT-3’ 3’-ATGGCCAT-5’
5’-TACCGGTA-3’ 3’-TACCGGTA-5’
Draw the complementary strand of DNA for the sequence
5’-ATTCAAGC-3’
3’-TAAGTTCG-5’
Draw the complementary strand of DNA for the sequence
5’-GCTTGAAT-3’
3’-CGAACTTA-5’
Are these two pieces of DNA the same or different?
Problem #7: DNA Polymerization
• Q: Why does DNA polymerization occur in the 5’ -> 3’ direction?
• A: Think about the primer and the dNTP. The phosphate on the dNTP is on the
3’ hydroxyl of the primer, so a new bond can only be formed with the 5’ end of
the dNTP