Chap 3-5 / Exam 1 Flashcards
1
Q
What is the zwitterionic form of an amino acid
A
- Have both positively and negatively charged groups (hybrid).
- Aa’s usually exist as zwitterions
- During a pH titration, these compounds go from a net positive charge, through 0, to a net negative charge
- The isoelectric point (pI) is the pH where the compound has a zero net charge
2
Q
What is the L- type of aa used for
A
Used by the ribosome to make peptides and proteins. Natural proteins produced by translation on the ribosome only use L- type amino acids.
3
Q
What is the R- type of aa used for
A
used by specialized non-ribosomal systems to make certain peptides:
- bacterial cell wall components
- antibiotics (actinomycin, gramacidin)
- other compounds (cyclosporin)
4
Q
Chiral (alpha “α”)
A
Object is not identical to its reflection. Non-superimposable. Enantiomers.
5
Q
Achiral
A
Object is identical to its reflection. Superimposable. Some Diastereomers.
6
Q
Which molecule is an alpha amino acid?
A
7
Q
What are the designations of carbon in an aa
A
The chiral carbon is designated α, and carbons in the R group are designated β, γ, δ, ε. The carboxylate carbon is numbered 1 and remaining carbons accordingly
8
Q
Which classification of aa are non-polar? And what are the associated aa’s?
A
Aliphatic (R-chain=alkyl group,
G - Glycine (achiral, can be on surface loop bc small size)
A - Alanine (can be on surface loop bc small size)
P - Proline (an imino acid – imines defined as secondary amines (N bonded to two C inside of group, not one C)
V - Valine
L - Leucine
I - Isoleucine
M - Methionine
Aromatic (delocalized electrons inside of rings, ):
F - Phenylalanine
Y - Tyrosine
W - Tryptophan
Note: polar atoms in tyrosine and tryptophan, but bulk of sidechains are hydrophobic
9
Q
Lambert-Beer Law
A
- aromatic aa absorb UV
light around 280nm - W absorbs most
strongly, then Y, then F
Lambert-Beer Law: A=εBC - A is Absorbance (experimentally measured)
- ε is extinction coefficient (constant, can be measured/estimated)
- B is path length (known parameter of spectrophotometer)
- C is concentration (unknown)
10
Q
Spectrophotometer
A
Used to determine UV absorption in aromatic rings
11
Q
Which classification of aa are polar? And what are the associated aa’s?
A
Neutral:
S - Serine
T- Threonine
C - Cysteine (sidechain can form disulfide bonds within a protein or between chains)
N - Asparagine
Q - Glutamine
Positive (0 or +1 charge):
K - Lysine
R - Arginine
H - Histidine
Negative (0 or -1 charge, acids in their protonated form):
D - Aspartate
E - Glutamate
12
Q
Disulfide bridge
A
- Two residues of cysteine covalently bound by oxidation of the thiol groups (-SH) make a disulfide bridge
- Different parts of a protein (even far
away) can be stabilized with disulfide
bounds - Happens on exterior of cell (extracellular space)
13
Q
Isoelectric point (pI) and formula
A
- The isoelectric point (pI) is the pH where the compound has a zero net charge.
- iso=equal hence 0
- pI determined by α-amino (H+ accepter basic), α-carboxyl (H+ donor acidic), and ionizable R-groups.
- only terminal α-amino and α-carboxyl groups in peptide ionize, not
internal ones in peptide bonds - the pI of a peptide is more complicated to calculate than that of amino acids bc of interactions btwn side chains
- the pI of proteins can tremendously vary from the calculated value - often by 2-3 pH units
Formula: 1/2 (pKa1+pKa2)
14
Q
Buffering capacity
A
Each pKa near each charge
15
Q
Based on this titration plot for His, when
does His have strong buffering capacity?
A
When pH is near 1.82, 6.0, or 9.17
16
Q
How to find pI (isoelectric point)
A
- Draw structure of amino acid completely protonated (i.e. imagine being at ultra-low pH)
- Imagine titrating with base like NaOH – begin removing protons from functional groups with lowest pKa.
- Main chain carboxyl has lowest pKa - first to lose its proton
- Side chain carboxyl (if any) has second lowest pKa - next to lose its proton
- Side chain amine (if any) may have a higher or lower pKa than the main chain
amine - Draw the structure of the amino acid after each removal of proton and calculate the amino acid’s net charge
- Identify the structure with a net charge of 0
- Calculate pI by averaging the pKa values that ionize before & after this structure using the pI formula (1/2(pKa1+pKa2))
17
Q
Determine the pI of the tripeptide, Asp-Gly-Glu
A
Step 1: Identify the ionizable groups.
The peptide Asp-Gly-Glu has four ionizable groups:
1.Aspartate N-terminal amino group (NH₃⁺) → pKa ≈ 9.6
2.Aspartate side chain (COOH) → pKa ≈ 3.65
3.Glutamate side chain (COOH) → pKa ≈ 4.25
4.Glutamate C-terminal carboxyl group (COOH) → pKa ≈ 2.19
No contribution of Gly because it is in the middle of the polipeptide and has no ionizable R group
Step 2: Determine the charge at very low pH (pH <2.19) At pH = 1 (that is lower than 2.19), all groups are fully protonated:
*N-terminal (NH₃⁺): +1
*Aspartic acid side chain (COOH): 0
*Glutamic acid side chain (COOH): 0
*C-terminal (COOH): 0
Total charge = +1
Step 3: Increase the pH and see when groups lose protons
As the pH rises, groups will deprotonate (lose H⁺) in order of their pKa values:
1.pH ≈ 2.19 (C-terminal pKa):
C-terminal COOH → COO⁻ (-1 charge)
Total charge = 0
2. pH ≈ 3.65 (Aspartic acid pKa):
Aspartic acid COOH → COO⁻ (-1 charge
Total charge = -1
3. pH ≈ 4.25 (Glutamic acid pKa):
Glutamic acid COOH → COO⁻ (-1 charge)
Total charge = -2
4.pH ≈ 9.6 (N-terminal pKa):
N-terminal NH₃⁺ → NH₂ (0 charge)
Total charge = -3
Step 4: Find the pI (where net charge = 0) REMEMBER: The isoelectric point (pI) is the pH where the net charge is zero.
*At pH 2.19, the charge goes from +1 → 0 after the C-terminal deprotonates.
*At pH 3.65, it goes from 0 → -1 after the Asp r group deprotonates.
So, the pI is between pH 2.19 and 3.65 → ( 2.19+3.65)/2=2.92
18
Q
pI for an aa without an ionizable side chain
A
Calculating the pI for the carbonyl and amine group
19
Q
pI for an aa with an ionizable side chain
A
20
Q
pI example with different titratable side chain
A
21
Q
What forms during Primary Structure of an aa?
A
Peptide links.
Ribosomes link amino
acids by peptide
bonds in a condensation (or dehydration) reaction
during translation forming amino acid residues (loss of H2O)
22
Q
Oligopeptide
A
Few AA (like pentapeptide = 5mer below)
23
Q
Polypeptide
A
Many AA (Molecular Weight (MW) < 10,000 Daltons)
24
Q
Protein
A
Big polypeptide (Molecular Weight (MW) > 10,000 Daltons)
25
How is amino acid sequence written
always written N -> C
(amino -> carboxyl)
26
Peptide bond characteristics
The planar nature of the peptide bond restricts the possible conformations that a
polypeptide chain can assume
27
Dihedral Angles φ (phi) and ψ (psi)
They define a Peptide
* A dihedral angle (or torsion angle) is defined between three successive chemical bonds and is the angle at the intersection of two planes.
* Two dihedral angles are sufficient to define the conformation of the polypeptide chain: phi (φ) and psi (ψ)
* Omega (ω) is between C-N = peptide bond
* Å (Angstrom) = 0.1 nm (nanometer); 1 nm = 10-9 meter
* Conformation in which both phi (φ) and psi (ψ) are zero degrees is
prohibited due to steric clash
28
Forces and Interactions that Stabilize Protein Structures
* Protein folding is dominated by weak, noncovalent interactions or forces including ionic interactions, van der Waals interactions, and the hydrophobic effect.
* Protein structure is determined in part by the planar structure of the peptide bond and can be described in part by the dihedral angles that describe the orientations of adjacent peptide bonds relative to one another
29
Which of the following is FALSE about the peptide bond formation
reaction?
A) Generates one molecule of ATP
B) Releases a water molecule
C) Is a condensation or dehydration reaction
D) Links the carboxyl group of one amino acid to the amino group of
another amino acid
E) Happens in the ribosome
Generates one molecule of ATP
30
Centrifugation: Cellular components
start: Lyse/homogenize/break cells then: centrifuge
High speed centrifugation of lysed cells produces two components: the supernatant (cytosol) and the pellet (organelles and plasma membrane)
31
Purification of proteins and complexes
What is Recombinant expression?
* Genetically engineer bacteria, yeast or
human cell lines to express target protein(s)
* pro: can obtain high levels of target, mutants
* con: may not have all native partners present
32
Purification of proteins and complexes
What is "Purification from native tissue"?
* Purify complexes from native sources
* pro: native complexes present
* con: yields often limiting, difficult to engineer
33
Types of Protein chromatography
* Affinity chromatography
* Ion exchange chromatography
* Size exclusion chromatography
* each of these methods separates proteins and their complexes from
each other depending on some physical characteristic
* each method determined by the types of resin being used
34
Affinity chromatography
* based on protein’s affinity to ligand covalently attached to solid resin
* highly selective — proteins are very selective with ligand they bind to
* can allow for single step purification
* generality greatly enhanced by the ability to generate fusion proteins of
your target with a protein with known affinity
35
Ion exchange chromatography
* based on protein’s charge at a particular pH
* not so selective, since many proteins can
have the same charge at a given pH — but very general
* resin in figure has negative charge = binds
positive proteins (cation exchange chromatography)
* one can also use positively-charged resins
to bind negatively-charged proteins (anion exchange chromatography)
36
Size exclusion chromatography
* based on protein’s MW
* not so selective, since many proteins have similar MW
* porous resin with lots of different sized channels; smaller proteins enter resin particles while larger
proteins cannot = longer path for smaller proteins than larger ones, takes more time to go
through column
37
SDS-PAGE: sodium dodecyl sulfate polyacrylamide gel electrophoresis
* proteins are coated with SDS (negative charge)
* SDS mostly removes contributions due to
conformation (denaturant) and inherent charge in
the folded protein (instead, now there is equal charge per MW)
* SDS denatures proteins, so it does not give information about native structure, folding, or activity.
To study native state, Native PAGE (without SDS) is used
* this effectively allows separation by MW
* smaller things move faster through the gel matrix — so at the end, low MW at bottom and high MW at top
* resolution ~1 kDa = ~1000 Da
38
SDS-PAGE Process
* first, electrophoresis to separate proteins by MW
* Then treat with a stain such as Coomassie blue to visualize
* note calibration markers (“ladder”) at left
* example shown here: recombinant protein expression
* start with a complex mixture, become simpler with purification
39
Overview of Protein Structure
* The 3D structure of a protein is determined by the AA sequence
* Protein conformation is stabilized by non-covalent interactions (hydrophobic effect, hydrogen bonds, ionic interactions, and van der Waals forces) and disulfide bonds (cysteine).
* Most isolated proteins adopt a number of stable structures called conformations
* Each protein has a specific function
40
How could you tell the
# of aa in chymotrypsin?
Using Molecular Weight (MW) Estimation
The molecular weight of chymotrypsin is approx 25 kDa (kilodaltons).
The average molecular weight of an amino acid is about 110 Da (Daltons).
So to estimate the number of amino acids:
25,000 Da / 110 Da per/amino acid≈227 amino acids
This rough estimate aligns closely with the known sequence length (~241 aa)
41
Secondary Structure Characteristics
* Secondary structures are defined by particular values of φ and ψ
* established by non-covalent interactions: H-bonds and steric clashes
2 structure types:
* A helical conformation referred to as α-helix
* An extended conformation referred to as β-strand
* Both types of structure satisfy the hydrogen bonding requirements of protein backbone atoms
42
Secondary structure: α-helix
* the α-helix (right-handed) is shown
wrapped around an imaginary axis (a)
* R-groups protrude outward
* the repeating unit is a single turn of
the helix extending ~ 5.4 Å (3.6
AA/turn, 1.5 Å rise/AA = 5.4 Å
rise/turn)
* Note: 3.6 AA/turn, not exactly 4!
* H-bonds between carbonyl of residue n and amide protons of residue n+4
* every peptide bond has potential to
participate in H-bonding
* Q: What is the exception?
PROLIN : lack of free amide proton
43
Pattern of H-bonding in α-helices
44
Why aren't all the aa are suited for an α-helix
The ideal α-helix formers are Alanine (Ala), Leucine (Leu), Methionine (Met), and Glutamate (Glu) because they have side chains that fit well into the helical structure and favor hydrogen bonding patterns. Conversely, Proline, Glycine, bulky, charged, and strongly polar amino acids tend to destabilize α-helices.
“Breaking residues” of α-helix:
* adjacent turns of Glu/Asp (negative charges repel)
* adjacent turns of Lys/Arg (positive charges repel)
* bulky R-groups
* Pro - unable to H-bond and restricted dihedral angles
A bigger number = less propensity to
take up the α-helical
conformation
45
Is this peptide helical?
RDFTKEYP
No, b/c the same-charged residue separated by 4 residues (n,n+4): RxxxK; DxxxE; Y, P
How to analyze:
1. Assess Each Amino Acid's Helix Propensity
Different amino acids have varying tendencies to stabilize or destabilize α-helices. Below is a quick classification:
Strong Helix Formers:
Alanine (Ala, A) → Most favorable
Leucine (Leu, L), Methionine (Met, M), Glutamate (Glu, E), Lysine (Lys, K) → Common in α-helices
Neutral or Moderate Helix Formers:
Aspartate (Asp, D), Arginine (Arg, R), Threonine (Thr, T) → Can be present but not always ideal
Helix Breakers:
Proline (Pro, P) → Major helix breaker
Glycine (Gly, G) → Too flexible, usually destabilizing
2. Analyze the Given Peptide (RDFTKEYP)
3. Prediction: Is It Helical?
The peptide has some strong helix formers (K, E), which help α-helix formation.
However, Proline (P) at the end is a known helix breaker, meaning it can terminate or severely disrupt an α-helix.
The presence of Asp (D) and Thr (T) can also destabilize the helix due to electrostatic repulsion or H-bond competition.
4. Conclusion
❌ This peptide is unlikely to form a stable α-helix because:
Proline at the end disrupts the helix.
Aspartate and Threonine can interfere with hydrogen bonding.
The mix of bulky and charged residues may prevent a stable helix from forming.
46
α-helix and dipole
The electric dipole of a peptide bond is transmitted along the α-helix, resulting in an overall helix dipole.
Negatively charged amino acids are often found near the amino terminus of the helical segment to stabilize interaction with the positive charge of the helix dipole and vice versa
**Positively charged amino group at terminal end is destabilizing**
47
α-helix handedness
The α helix can be “right-handed” or “left-handed,” although the right-handed version is far more common in proteins.
48
Constraints affecting the stability of an α helix
1. the intrinsic propensity of an amino acid residue to form an α helix;
2. the interactions between R groups, particularly those spaced three (or four) residues apart;
3. the bulkiness of adjacent R groups;
4. the occurrence of Pro and Gly residues
5. interactions between amino acid residues at the ends of the helical
segment and the electric dipole inherent to the α helix.
49
Why is Gly often in the left-handed alpha-helix region?
Glycine is found in the left-handed α-helix region because its small size and high flexibility allow it to adopt φ and ψ angles that are sterically prohibited for most other amino acids.
50
Secondary structure: β-strands and β-sheet
* The backbone is extended in a zigzag shape
* H-bonds are formed between adjacent segments of polypeptide chain
* The chains can be parallel or antiparallel (same or different amino-to-carboxyl orientation), and may form with one or more peptide chains
51
β-sheets/strands: β- and γ-turns
* globular proteins have numerous turns where the polypeptide reverses direction
* β-turns type I and II have 4 AA that make a 180° turn
* γ-turn has 3 AA (not shown, rarer)
* Gly (flexible) and Pro (assumes cis
configuration) are well suited for a tight turn
* predominantly found on protein surface as residues 2 and 3 need to H-bond with wate
52
Ramachandran plots
Describe the Distribution of Secondary Structure in a Protein
* describes the dihedral angles phi (φ) and psi (ψ) of a polypeptide
* Ramachandran plot is required to test the modeled geometry of a protein structure
53
Circular dichroism (CD) spectroscopy
* Structural asymmetry in a molecule
gives rise to difference in absorption
(via peptide bond) of left-handed
versus right-handed plane-polarized
light.
* The α-helix and β conformations have characteristic CD spectra
* Using CD spectra, biochemists can
determine whether proteins are
properly folded, estimate the fraction
of the protein that is folded in either of the common secondary structures
and monitor transitions between the
folded and unfolded states.
54
Which of the following peptide sequences would be MOST likely to form a helix?
A) APAINPAINPA
B) YFWAFYHWYG
C) SKIMVENALR
D) RLIMRKANIK
E) none of the above
SKIMVENALR
A) APAINPAINPA — Pro = bad
B) YFWAFYHWYG — bulky Tyr/Phe/Trp = bad
C) SKIMVENALR — nothing bad
D) RLIMRKANIK — Arg/Lys & Arg/Lys on adjacent turns = bad
E) none of the above
55
Tertiary Structure Characteristics
Formation of a polypeptide that are established by non-covalent interactions Hydrogen bonds, van der waals, (exception: disulfide bonds)
* Tertiary structures are well-defined, three-dimensional folds determined by amino acid sequence.
* Structural biologists can define the three-dimensional structures of proteins.
* Two major groups in which proteins can be classified for their tertiary structure: fibrous proteins and globular proteins
56
Tertiary Structure: Fibrous proteins: Keratins
Usually consists largely of a single type of secondary structure, and tertiary structure is simple
* Mainly serve structural roles (strength/flexibility)
* A single secondary structure element makes a tertiary structure
* Insoluble in water
* High hydrophobic AA content. Packed intra- and inter-molecularly
Example: α-keratins:
* Dry weight of hair, wool, nails, claws…
* Right-handed helix with a tighter turn due to hydrophobic AA packing intra- and inter-molecularly
* Helices pack to form left-handed super helix
* Stabilized by disulfide bonds
57
Tertiary Structure: Fibrous proteins: Collagen
* Collagen is the most abundant
protein in mammals.
* Connective tissue such as tendons,
cartilage…
* 3 left-handed helices (α chains) with
3 amino acids per turn pack together
* Requires repeating motif: Gly-X-Y
where X is often Pro and Y is -Pro-
OH — hydroxy-proline
* The final structure is a right-handed
superhelix (3 left-handed folded)
58
What is the secondary structure of collagen?
a) α helix
b) β conformation
c) both
d) neither
neither - 3 left-handed helices (α chains) with 3 amino acids per turn
59
Tertiary Structure: Globular proteins
* Cytoplasmic proteins
* 75% is packed, 25% is cavities
(filled with water or empty)
* Unique tertiary structures (folds/motifs)
* Different combinations of α-helices
and β-sheets can be present in a
protein
* α-helices and β-sheets are found in
different layers because they can
not H-bond to one another using
backbone H-bond donors or acceptors
60
The first protein structure
Myoglobin
* Myoglobin is a globular protein that stores oxygen and facilitates oxygen diffusion in rapidly contracting muscle
tissue.
* Myoglobin contains a single polypeptide chain of 153 amino acid residues of known sequence and a single iron protoporphyrin (heme) group.
61
Multi-domained proteins
* A domain is a compact, independent, and stable ensemble of packaged secondary structure elements
* The different colored regions
correspond to different domains in the same protein
* Domains can have different structures and different stability
* Domains can be inserted within other domains, or be completely independent.
62
Intrinsically disordered proteins
Intrinsically disordered proteins (IDPs) lack stable tertiary structures
* Some proteins have regions that are disordered; some proteins are even entirely disordered
* Example on right: p53 has ordered middle, but disordered N- and C-termini; C-terminus becomes ordered when it binds other proteins — but in a different way for each partner
63
The pH in your stomach is very low (in the 1-4 range). A protein called pepsin is a “protease” enzyme that breaks down other proteins in the stomach to aid digestion. Which of the following is true of pepsin when studied in the lab at pH 7, relative to
when it is in its native environment?
A) Its structure is likely to be more stable because the environment is less acidic
B) Its net charge is higher
C) Its net charge is lower
D) Its pI is higher
E) Its pI is lower
A) Its structure is likely to be more stable because the environment is less acidic wrong b/c proteins have evolved to be stable in their natural environment
B) Its net charge is higher - wrong b/c opposite of the correct answer
C) Its net charge is lower
CORRECT b/c higher pH = fewer protons = some positive groups become neutral, some
neutral groups become negative
D) Its pI is higher - wrong b/c pI doesn’t change
E) Its pI is lower - wrong b/c pI doesn’t change
64
Quaternary structure
Protein complexes that are established by non-covalent interactions (exception: disulfide bonds)
* Multiple polypeptide chains (same
or different) that oligomerize into a
quaternary structure (tetramer)
* Protomers in multimers may be
arranged with rotational or helical
symmetry, meaning that the protomers can be superimposed by
rotation on rotational axis or by helical rotation
65
Hemoglobin is a tetramer of 4
chains = 2 ɑ subunits + 2 β subunits
What is its symmetry?
Hemoglobin (Hb) exhibits D2 (dihedral) symmetry.
Why?
Hemoglobin consists of 4 subunits: 2 α subunits + 2 β subunits.
The α and β subunits are arranged in a tetrameric structure.
The molecule has twofold rotational symmetry around three perpendicular axes.
This D2 symmetry means that the hemoglobin tetramer can be rotated 180° around three different axes and still look the same. It is not perfectly symmetric like an ideal tetrahedral or cubic arrangement but follows a dihedral symmetry pattern.
66
How can Biomolecular Structures be Determined
Common tools: x-ray crystallography or cryo-electron microscopy (cryo-EM)
Note: Structural biology — the study of the three-dimensional structures of
biomolecules — combines biochemical approaches with physical tools and computational methods to obtain structures.
67
X-ray crystallography
* Electrons scatter X-rays and the
amplitude of the wave scattered
by an atom is proportional to its
number of electrons.
* The scattered waves recombine
= constructive interference.
* The ways in which the scattered
waves recombine depends on
the atomic arrangement.
* Diffraction pattern observed &
used to construct a model of the
structure
Strength: High resolution
Weakness: Static structure
68
Nuclear Magnetic Resonance (NMR) spectroscopy
* NMR-active nuclei – most
commonly, 1H, 13C and 15N – will
have a distinct chemical shift
depending on its environment
(covalent and non-covalent
interactions)
* Specific magnetization is
transferred into sample to probe
particular environments
* Nuclei-nuclei distances can be
measured and used to calculate
the structure
Strength: Records dynamics in solution at individual atoms
Weakness: Low sensitivity & signal overlap for large proteins
69
Cryo-electron microscopy (CryoEM)
* Images of protein or complex suspended in vitrified ice are
taken using an electron microscope
* The images are averaged and used
to calculate a 3D model of the protein/complex
Strength: Records dynamics and high-resolution
Weakness: Limited to large macromolecules or complexes
70
The Protein Data Bank (PDB)
* The number of known three-dimensional protein structures is now more than 100,000, doubling every couple of years.
* PDB is an important archive where scientists post and access these
verified structures.
* Each structure in the PDB is assigned an identifying label, a four-
character identifier called the PDB ID.
* The data files in the PDB describe the spatial coordinates of each atom,
which can be used to visualize the structure and create notes on how it
was verified.
71
Alpha helices are stabilized by hydrogen bonds between the carbonyl group of residue i and the amide group of residue i+4 of the next helical turn. But the first and last turns don’t have a turn before or after them, respectively. This means that the first turn of a helix has N-H groups with unsatisfied hydrogen bonds. Which of the following amino acids do you predict is very common near the start of a helix to make up for these missing interactions with its R group and thereby stabilize the first helical turn?
A) Gly
B) Phe
C) Asp
D) Arg
E) Ala
A) Gly - While glycine is flexible, it does not provide a stabilizing hydrogen bond with its side chain.
B) Phe - A bulky hydrophobic side chain, which does not help with hydrogen bonding or electrostatic interactions.
C) Asp – ASP has a negatively charged COO⁻ side chain at physiological pH, which can form favorable electrostatic interactions with the partially positive N-H groups of the helix backbone, stabilizing the structure.
D) Arg - Positively charged, but this would not stabilize the N-terminus (it could help at the C-terminus, interacting with the negatively charged C=O groups).
E) Ala- A small non-polar residue that does not contribute hydrogen bonding from its side chain.
72
Why are most proteins are marginally stable
* As proteins fold, strong forces favor — and oppose — the folding process
* For proteins to fold, the net sum of these forces must be energetically favorable (ΔG < 0).
* Most proteins can be unfolded by:
* slight changes in solution conditions (e.g. temperature, pH, salt), often leading to an irreversible loss of structure and function
* amino acid mutations that remove favorable interactions (or generate unfavorable interactions)
* In either case, this often leads to an irreversible loss of structure and function
73
Protein folding
Local structures such as α-helices
and β-sheets form spontaneously
into supersecondary structures.
This promotes long-range
interactions to form structure.
74
The thermodynamics of protein folding
Spontaneous “collapse” is driven by the release of water molecules forming a cage around hydrophobic AA to a “molten globule” state, then other AA “find” correct conformation by sampling the energy landscape depicted. Native structure has the
lowest energy.
75
Protein misfolding & disease
* Many proteins have transient alternative folds
* Alternative folds may be stabilized by mutations
* Some such folds include exposed β-
sheet edges and may cause assembly
into amyloid fibrils
* Amyloid fibrils are associated with
several diseases including Alzheimer’s,
diabetes
Example: blood coagulation factor IX
* Red = AA’s with a mutation detected
in clinical patients with hemophilia
(impaired blood clotting)
* Each mutation either unfolds the
protein or changes its structure
enough to disrupt its function in the
blood coagulation cascade
76
What does the loss of Protein Structure result in?
Results in Loss of Function
(Structure = Function)
* A loss of three-dimensional structure that is sufficient to cause loss of function is called denaturation.
* Most proteins can be denatured by heat, extremes of pH, organic solvents, solutes such as urea and guanidine hydrochloride, or detergents like sodium dodecyl sulfate (SDS).
* Structural features that increase stability of the folded state can increase resilience to these denaturing effects.
* Denaturation often leads to protein precipitation, for example, the white of a boiled egg is no longer soluble.
77
Denaturation with Increasing Temperature
Proteins are only marginally stable, meaning that the folded state is only slightly more favorable than the unfolded state under physiological conditions. When temperature
increases, the system's entropy (disorder) increases, making the unfolded state more thermodynamically favorable.
Once a critical temperature (Tm) is reached, the protein unfolds cooperatively; that is, once part
of the structure destabilizes, the rest quickly follows.
In summary, temperature-induced denaturation occurs because thermal energy disrupts non-covalent forces that stabilize the folded conformation, leading to an unfolded, non-functional protein.
78
Protein folding Pathways
* The folding pathway of a large polypeptide chain is complicated and is directed by ionic interactions and the hydrophobic effect.
* In general, local secondary structures form first, followed by longer-range interactions between
elements of secondary structure that come together to form stable folded structures.
* Folding often requires chaperone proteins that interact with partially folded or improperly folded
polypeptides to facilitate correct folding pathways or to provide microenvironments in which
folding can occur.
* KEY TERM chaperone protein: A member of any of several classes of proteins or protein complexes that direct the accurate folding of proteins in all cells
79
Molecular Chaperones
* These are sophisticated
multicomponent systems
* GroEL/GroES = “chaperonins” (Hsp60 family) are required by the 10-15% of proteins that do not fold spontaneously; this number increases to approximately 30% after heat shock
* Unfolded proteins go inside large
cavity in GroEL and GroES complexes
(heptamers), mechanism poorly
understood.
* ATP hydrolysis is used to promote
conformational changes of GroEL
80
The famous Anfinsen protein folding experiment was the first to show that the amino acid sequence of a protein can be sufficient to determine its structure. This experiment used the enzyme ribonuclease A, which contains four disulfide bonds. First, urea and beta-mercaptoethanol (a reducing agent) were added to a solution of RNase A, with the result that protein activity was lost. Next, urea was slowly removed by dialysis. Finally, beta-mercaptoethanol was removed by dialysis, with the result that almost all of the original protein activity was regained. What happens when the order of the last two
steps is swapped: beta-mercaptoethanol is removed first, then urea is removed?
A) Non-native disulfide bonds form after beta-mercaptoethanol is removed, so the protein cannot refold correctly
B) Half the original activity is regained
C) The original disulfide bonds reform, but the protein still cannot refold
D) The disulfide bonds do not reform, but activity is regained anyway
E) The protein still refolds the same way
A) Non-native disulfide bonds form after beta-mercaptoethanol is removed, so the protein cannot refold correctly - CORRECT b/c SS reform once BME removed, but scrambled bc urea still present at that point, so protein still unfolded
81
Protein and ligand characteristics
* Proteins breathe and are dynamic
* Ligand is a molecule (e.g. protein, DNA, peptide, hormone…) that binds to another molecule
* Ligands can regulate the activity of a protein (e.g. ON/OFF)
* Binding can be reversible or irreversible
* Binding site is usually complementary to ligand, with compatible shapes, charges, etc. to facilitate favorable non-covalent interactions
* The binding site may form through an induced fit (conformational change) of protein and/or ligand
* Multisubunit proteins are composed of multiple polypeptide chains
* Conformational change in one subunit of a multisubunit proteins may affect the conformation of
another subunit to regulate the function of the protein (regulation could be positive or negative)
* A ligand that is chemically modified by a protein (i.e. an enzyme) is called a substrate
* Drugs are ligands/substrates that can reversibly or irreversibly bind to proteins
82
The vertebrate oxygen transport and storage system
* Animals require a steady supply of O2 to their cells and a steady removal of waste products such as CO2 and CO
* With the exception of insects, diffusion of O2 through blood is not fast enough
* Therefore nearly all animals pump O2 from their lungs/gills through their arteries to the tissues and return CO2 via venous blood to their lungs/gills
83
Red blood cells
~6-8μm
No organelles and full of hemoglobin
84
Heme-containing proteins: O2 storage and transport
For O2 storage and transport, mammals use:
* storage = myoglobin (Mb): stores oxygen in tissues like muscle where there is a high demand for oxygen
* transport = hemoglobin (Hb): transports oxygen from lungs/gills to tissues, and transports carbon dioxide and protons back to the lungs/gills
85
Myoglobin
is a monomer protein composed of a
single 153 residue polypeptide chain
liganded to a prosthetic group called a heme
86
Hemoglobin (Hb)
Hemoglobin is a hetero-tetramer
protein composed of 2 copies of an α- and 2 copies of a β-polypeptide chain – two gene products that are homologs (evolutionary related)
* 4 protein chains, each of which bind heme
* arrows are pointing at hemes
87
Heme structure
* Composed of two parts: a porphyrin ring (protoporphyrin IX) + iron
* Synthesized in immature red blood cells
* Iron (Fe2+) binds O2 reversibly; Fe3+ does not bind O2
* Mb/Hb prevent oxidation from Fe2+ (ferrous) to Fe3+ (ferric) by bound O2
88
Myoglobin (Mb): Oxygen storage
* 16.7 kDa MW monomer
* 8 α-helical segments, ~80% of AA
are in helices
* helices named A, B, C, … H
* non-helical connecting bends
named AB, BC, …
* stores O2, mostly in muscle
* high concentration in diving
mammals
* first protein to have its 3D structure
determined, by X-ray diffraction
89
Myoglobin (Mb): O2 binding site
* Fe2+ is normally octahedrally
coordinated – attached to 6 ligands/binding groups
* Nitrogen atoms of porphyrin ring
account for 4 ligands
* The proximal His of hemoglobin/myoglobin accounts for
1 ligand
* O2 accounts for last ligand. This O2
interacts with the distal His
90
Binding of O2 to myoglobin
(protein/ligand binding equilibria)
P + L ⇌ PL
Keq = Ka = [PL] / [P][L]
Kd = 1/Ka = [P][L] / [PL
P: protein (Mb), L: ligand (O2)
Ka = association constant (not acid dissociation constant Ka)
Kd = dissociation constan
Kd is [L] at which half of the available ligand-binding sites are occupied
theta (θ aka Y) is fraction of binding sites occupied. Formula: binding sites occupied / total binding sites
θ = [PL] / ([PL] + [P]) (goes from 0 to 1)
θ = [L] / ([L]+Kd)
θ = 0.5 when [L] = Kd
Note: [L] is for the unbound ligand
91
Dissociation constants
Here, a ligand L is added
to two different proteins A
and B
92
Mb: Quantitatively thinking about O2 binding
Myoglobin (muscle) accepts oxygen released from hemoglobin, and delivers the oxygen to the mitochondria when oxygen needs are sufficiently great (e.g. during exercise)
* pO2 high (100 mm Hg, 13.3 kPa) in lungs/gills — Hb present
* pO2 low (30 mm Hg, 4.0 kPa) in tissues — Mb present
* P50 = Kd = [O2] that results in 50% (half saturation) of Mb
* The strong affinity of myoglobin for O2 is needed in order for myoglobin to successfully bind the limited amount of O2 present in the capillaries/tissue: ~30 mm Hg = 4.0 kPa
* Myoglobin releases its O2 when the pO2 concentration of the cell drops below myoglobin’s affinity for O2
93
Mb vs. Hb
94
O2-triggered 4° structure changes in Hb
Hb exists in two states:
* T (tense) state without O2
* R (relaxed) state with O2
* Binding of O2 causes a conformational change in Hb from T to R state
* Change occurs because ionic interactions between the α2-β1 and α1-β2 are eliminated
* The T to R transition results in closing of hole between β subunits
95
Deoxygenated-Hb vs Oxygenated-Hb
* Deoxyhemoglobin favors the tense (T) state, which contains a greater number of ion pairs (more stable)
* Oxyhemoglobin favors the relaxed (R) state
* Mechanism for this shift: Binding of O2 to heme creates steric clash that changes the α2-β1 and α1-β2 interfaces, shifting Hb from T to R state
* T state binds O2 more weakly than R state
* Binding of O2 to one chain shifts T->R, increasing Hb’s affinity for more O2 = positive cooperativity
96
Structural effects of O2 binding
* Binding of O2 results in a large number of movements in Hb. These involve movements in the side chains that are propagated into quaternary movements of the subunits
* All protein quaternary changes stem
from small changes that propagate and expand into large scale changes
97
Comparing Mb & Hb O2 binding curves
98
The Monod/Wyman/Changeux (MWC) model
99
Regulation and alteration of Hb function
* Ligand regulation of Hb: pH and CO2
* Poisoning by CO
* Alteration of Hb function by amino acid substitution
100
pH dependence of oxygen binding: Bohr Effec
* pH is lower in tissues; excess H+
must be disposed of somehow
* H+ bind to Hb at specific sites
(including His residues) & form salt
bridges that stabilize T state
* This reduces the affinity of Hb for O2
* Thus, in tissues: low pH -> [H+]
bound, O2 released
101
Interlinking of Bohr effect and CO2 transport
CO2 transport from peripheral tissues is handled:
* ~8% physically dissolved CO2 gas
* ~25% carried by Hb via reversible reactions w/ amino groups
* ~67% transported as bicarbonate (HCO3-)
* For the ~67% as HCO3-: CO2(g) is modestly soluble in H2O, but HCO3- is highly soluble
* RBCs carry carbonic anhydrase to catalyze the following conversion (picture)
In tissues:
* H+ produced by this reaction helps facilitate O2 release by Hb (by Bohr effect)
* That H+ uptake by Hb helps facilitate carbonic anhydrase (by Le Chatelier’s principle)
In lungs:
* Reverse above process
102
CO/CN poisoning
* CO binds Fe2+ in Hb-bound heme
~200x stronger than O2, hence can
block O2 transport = is a toxic gas
* CO binds Fe2+ in free heme even
more tightly than in Hb-bound
* But in Hb, distal histidine lowers CO
affinity by preventing direct linear
interaction between CO and Fe2+
103
HbS: sickle cell anemia
* Sickle cell anemia originates from HbS, a form that contains a mutation replacing Glu 6 with a Val residue on the β subunits (E6V)
* This mutation generates a hydrophobic patch on the surface of Hb that is exposed only when Hb is deoxygenated. Such patches can lead to interactions between different Hb complexes, eventually forming long fibers that distort the shape of RBCs into sickled appearance
* The kinetics of process, like many aggregation-type processes, are highly non-linear with respect to monomer concentration. In this case, the rate is dependent on [HbS]10
* Heterozygotes in HbS are asymptomatic as their [HbS] is half that of homozygotes; 0.510 ~ 0.001
* Fiber formation often occurs in small blood vessels, blocking these vessels
and setting up vicious cycle
104
HbS: Sickle cells
* The sickle cell mutation is more common in certain parts of Africa
* This is because it also confers a mild resistance to malaria
* Thus evolution has struck a balance between the positive effects of fighting malaria vs. the negative effects of sickle cell anemia
105
What is the Bohr Effect?
The Bohr Effect describes how lower pH in tissues leads to increased H+ binding to hemoglobin (Hb), reducing its affinity for O2 and facilitating O2 release.
106
How does pH affect oxygen binding in hemoglobin?
Lower pH increases [H+] bound to Hb, resulting in O2 release.
107
What percentage of CO2 transport is carried by hemoglobin?
~25% carried by Hb via reversible reactions with amino groups.
108
What form of CO2 accounts for the majority of its transport in the body?
~67% transported as bicarbonate (HCO3-).
109
What role does carbonic anhydrase play in CO2 transport?
It catalyzes the conversion of CO2 and H2O to H+ and HCO3- in red blood cells.
110
What happens to H+ produced by the carbonic anhydrase reaction in tissues?
It helps facilitate O2 release by hemoglobin through the Bohr effect.
111
What is CO/CN poisoning?
CO binds to Fe2+ in hemoglobin with ~200x stronger affinity than O2, blocking O2 transport.
112
What mutation causes sickle cell anemia?
A mutation replacing Glu 6 with Val on the β subunits of hemoglobin (HbS).
113
What does the sickle cell mutation create on hemoglobin?
A hydrophobic patch that leads to interactions between different Hb complexes, forming long fibers.
114
Why are heterozygotes for HbS asymptomatic?
Their [HbS] is half that of homozygotes, reducing the risk of sickle cell effects.
115
What is the relationship between sickle cell mutation and malaria?
The mutation confers mild resistance to malaria, leading to a balance between positive and negative effects.
116
What is the function of plant phytoglobin?
It is involved in the maintenance of redox and energy status during hypoxia reducing fermentation.
117
What is the structure of IgG?
IgG consists of two heavy chains and two light chains, with a Fab region for antigen binding and an Fc region for signaling.
118
What does the Fab region of IgG do?
It binds to antigens.
119
What is the molecular weight of IgG?
150 kDa.
120
What is the significance of the Kd value in antibody-antigen binding?
Kd as low as 10-10 M indicates high affinity and specificity.
121
What is the principle behind ELISA?
To exploit antibody specificity for rapid screening and quantitation of antigens.
122
In an ELISA, what indicates a positive result?
More yellow color produced by the enzyme attached to the secondary antibody.
123
What is the function of leghemoglobin in plants?
It protects O2-sensitive nitrogenase and provides O2 to bacteria.
124
Fill in the blank: The sickle cell mutation generates a _______ on the surface of Hb that is exposed when Hb is deoxygenated.
hydrophobic patch
125
True or False: The binding of IgG to antigen is governed by chemical complementarity.
True
