biochem review, part i: protein structure and function lecture notes pp. 1-38
TRANSCRIPT
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Biochem Review, Part I:Protein Structure and Function
Lecture Notes pp. 1-38
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An Amino Acid
different for each AA
common to all amino acids*
*except proline, in which the R group forms a ring structure by binding to the amino group
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The TwentyAmino Acids Types of AAs
SMALL: hydrogen or methyl R group.GLYCINEAlanine
NONPOLAR/HYDROPHOBIC: R groups contain largely C, H atoms.
POLAR/HYDROPHILIC: R groups typically contain O, N atoms.
ALIPHATIC: no aromatic rings.
VALINELeucineIsoleucineMethonine*Proline
AROMATIC: contains aromatic rings.
PHENYLALANINETyrosineTryptophan
ACIDIC: acid in R group.
ASPARTATEGlutamate
BASIC: base in R group.
LYSINEArginineHistidine
CHARGEDUNCHARGED: no ionizable group in R group.
ASPARAGINEGlutamineSerineThreonine*Cysteine
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Types of AAs
SMALL: hydrogen or methyl R group.GLYCINEAlanine
NONPOLAR/HYDROPHOBIC: R groups contain largely C, H atoms.
POLAR/HYDROPHILIC: R groups typically contain O, N atoms.
ALIPHATIC: no aromatic rings.
VALINELeucineIsoleucineMethonine*Proline
CHARGEDUNCHARGED: no ionizable group in R group.
ASPARAGINEGlutamineSerineThreonine*Cysteine
an imino acid, with R group bound to amino group
thiol group can participate in disulfide bonding
SpecialAmino Acids
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Peptide Bond Formation
amino terminal residue“N terminal”
carboxyl terminal residue“C terminal”
1 2
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Acid-Base Behavior of AAs
Each amino acid has at least TWO groups that display acid-base behavior (gain or accept H+) – the carboxyl group and amino group.
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Acid-Base Behavior of AAs
Equilibrium constant: Ka = [A-][H+]/[HA]
pH = -log[H+] …convenient shorthand for writing widelyvariable [H+] concentrations
pKa = -log(Ka) …similar shorthand for writing variable Ka values
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The Henderson-Hasselbalch Equation
Relates three terms: pH, pKa, and [A-]/[HA]. If you know two of these values, you can determine the third.
pH = pKa + log([A-]/[HA])
When [A-] = [HA]:pH = pKa + log(1)
pH = pKa + 0
pH = pKa
pka is the pH at which a functional group exists 50% in its protonated form (HA) and 50% in its deprotonated form (A-).
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Isoelectric PointIsoelectric point: the pH at which an AA or
polypeptide has no net charge.
• For a dibasic amino acid:
• For a tribasic amino acid:
pka = 2.4
pka = 9.8
Isoelectric point = average of amino and carboxyl pka values
pka = 2.2
pka = 9.7
pka = 4.3
Isoelectric point = average of the two numerically closest pka values
= (2.2 + 4.3)/2 = 3.25
= (2.4 + 9.8)/2 = 6.1 GLYCINE
GLUTAMATEFormulas can be used for polypeptides as long as they have no more than one ionizable side chain.
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Titration Curves This example shows the curve for a dibasic AA
pka of carboxyl group
pka of amino group
Isoelectric point
pka values always occur at the flattest parts of the curve (around 0.5, 1.5, 2.5 base equivs.)
The isoelectric point will always occur at at 1.0 or 2.0 base equivs.
# of equivs. OH- tells whether AA is dibasic (2 equivs) or tribasic (3 equivs.)
buffering works best at pka
buffering works best at pka
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Practice Problem #1: Titration of Aspartate
1
3
2
Posn.
pH Major form(s) present
Average charge on α -carboxyl (pka = 2.1)
Average charge on R carboxyl (pka = 3.9)
Average charge on amino group (pka = 9.8)
Net charge
1 ~ 0
2 9.8
3 3.0
Curve and structures from: http://www.cem.msu.edu/~reusch/VirtualText/proteins.htm
Rule of thumb: if pH is >2 pH units away from a group’s pka, that group will effectively exist only in a single form. (Below pka – protonated form; above pka – deprotonated form.)
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Practice Problem #1: Titration of AspartatePosn pH Major form(s)
presentAverage charge on α -carboxyl (pka = 2.1)
Average charge on R carboxyl (pka = 3.9)
Average charge on amino group (pka = 9.8)
Net charge
3 3.0
For each relevant functional group , we will use the Henderson-Hasselbalch equation to calculate [A-]/[HA] at pH 3.0.
Henderson-Hasselbalch equation: pH = pka + log([A-]/[HA])
Then, we will convert this ratio into % of groups protonated:
% groups protonated = [HA]/([HA]+[A-]) = 1/(1+[A-]/[HA]) x 100%
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Practice Problem #1: Titration of Aspartate
For the alpha carboxyl:
pH = pka + log([A-]/[HA])pH - pka = log([A-]/[HA])[A-]/[HA] = 10pH-pka
% protonated = 1/(1+10pH-pka) x 100%% protonated = 1/(1+103.0-2.1) x 100%% protonated = 11.18% alpha-COOH100%-11.18% = 88.82% alpha-COO-
Average charge on alpha carboxyl:(0)*(0.1118) + (-1)*(0.8882) = -0.8882
For the R group carboxyl
% protonated = 1/(1+103.0-3.9) x 100%% protonated = 88.82% R-COOH100%-11.18% = 11.18% R-COO-
Average charge on R carboxyl:(0)*(0.8882) + (-1)*(0.1118) = -0.1118
Posn
pH Major form(s) present
Average charge on α -carboxyl (pka = 2.1)
Average charge on R carboxyl (pka = 3.9)
Average charge on amino group (pka = 9.8)
Net charge
3 3.0
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Practice Problem #1: Titration of Aspartate
Finally, we calculate the fraction of molecules in each form by calculating the probability of finding all three functional groups in the protonated/ deprotonated states present in that form.
Posn pH Major form(s) present
Average charge on α -carboxyl (pka = 2.1)
Average charge on R carboxyl (pka = 3.9)
Average charge on amino group (pka = 9.8)
Net charge
3 3.0 -0.89 -0.11 +1 0
P(D) =P(alpha-COO-) *P(R-COOH) * P(NH3+) =(0.8882)(0.8882)(1.0) =0.7889
P(C) = P(alpha-COO-) * P(R-COO-) * P(NH3+) =(0.8882)(0.1118)(1.0) = 0.0993
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Orders of Protein Structure
Primary Structure: the sequence of a protein (the AAs it contains and the order they are in.
Image from: http://www.umass.edu/molvis/workshop/imgs/protein-structure2.png
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Primary Structure
• Genetically determined: specified by sequence of gene that encodes protein
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Primary Structure
Denature Renature
The 3D structure a protein spontaneously assumes is its most stable structure.
• Contains all information necessary to specify higher orders of structure (3D shape)
break disulfide bonds
break disulfide bonds
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Orders of Protein Structure
Primary Structure: the sequence of a protein (the AAs it contains and the order they are in.
Secondary Structure: regular 3D motifs formed by H-bond interactions between N-H and C=O groups of the peptide backbone.
Image from: http://www.umass.edu/molvis/workshop/imgs/protein-structure2.png
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Secondary Structure
• Two major motifs: α helix and β sheet• α helix is “default” structure for an AA chain
(energetically favorable)
The carbonyl of each AA H-bonds with the amino group of an AA 4 residues down.
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Secondary Structure
• Two major motifs: α helix and β sheet• β sheet is composed of parallel strands of AAs
connected to one another by H-bonds
Like the α helix, the β sheet is formed by interactions WITHIN a single polypeptide chain.
N terminus
C terminus
turn in chain
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Secondary Structure
Amino acid identity affects secondary structure.
Helix is destabilized by:– Bulky R groups (e.g., Try)– Adjacent R groups with like charges– Proline, which cannot H-bond
Proline, in particular, tends to appear in unstructured regions (turns).
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Orders of Protein Structure
Primary Structure: the sequence of a protein (the AAs it contains and the order they are in.
Secondary Structure: regular 3D motifs formed by H-bond interactions between N-H and C=O groups of the peptide backbone.
Tertiary Structure: the overall structure of a single polypeptide chain.
Image from: http://www.umass.edu/molvis/workshop/imgs/protein-structure2.png
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Tertiary Structure
• Tertiary structure can be complex and does not typically consist of repeating units.
• A polypeptide will adopt the most stable tertiary structure
= charged
= hydrophobic
= neither
In aqueous environment, this occurs when hydropobic residues are internal and hydrophilic residues are external.
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Orders of Protein Structure
Primary Structure: the sequence of a protein (the AAs it contains and the order they are in.
Secondary Structure: regular 3D motifs formed by H-bond interactions between N-H and C=O groups of the peptide backbone.
Tertiary Structure: the overall structure of a single polypeptide chain.
Quaternary Structure: interaction of multiple polypeptides to form one functional protein.
Image from: http://www.umass.edu/molvis/workshop/imgs/protein-structure2.png
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Quaternary Structure
• Proteins with quaternary structure have multiple subunits and are oligomeric
• Subunits may be identical or different• Oligomeric proteins have the potential for
cooperativity
All cooperative proteins must be oligomeric, but NOT all oligomeric proteins are cooperative.
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Methods to Analyze 1o Structure
• Ion exchange chromatography: tells you composition but NOT sequence
• Edman degradation: tells you sequence of a short (<75 AA) fragment
• Proteolytic cleavage: generates short fragments suitable for Edman degradation
• Electrophoresis: separates proteins according to net charge
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Ion Exchange Chromatography
• Tells you what AAs a protein contains and the relative abundance of each
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Ion Exchange Chromatography
1. Digest protein in strong acid2. Load onto column of negatively
charged (sulfonate coated) beads; AAs bind because positively charged
3. Wash column with solutions of increasing pH, collecting eluate in small fractions
4. For each fraction, determine the presence/absence and abundance of the corresponding AA
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Ion Exchange Chromatography
• An AA will elute approximately when the pH of the solution equals its isoelectric point
AAs with low isoelectric points (such as acidic AAs) come off earlier
AAs with high isoelectric points (such as basic AAs) come off later
Peak height indicates relative abundance
Ion exchange chromatography can also be used on small polypeptides. These too will elute in an order determined by isoelectric point.
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Edman Degradation
• Tells sequence of a short peptide fragment (no greater than 75 AAs)
• N-terminal AA is labeled, cleaved, and its identity determined
• Process is repeated in successive rounds
determine identity
determine identity
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Proteolytic Agents
• Used to cleave a protein into smaller polypeptides (which can then be analyzed)
• Each agent has unique specificity:– Trypsin: cuts after Lys, Arg– Chymotrypsin: cuts after Phe, Tyr, Trp, Leu, Met– Cyanogen Bromide: cuts after Met
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Practice Problem #2: Proteolytic Agents
You have digested a polypeptide with two different agents and obtained these fragments:
Trypsin: [Met-Phe-Val-Arg] [Ala] [Glu-Lys]Chymotrypsin: [Val-Arg-Ala] [Glu-Lys-Met-Phe]
What is the sequence of the polypeptide?
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Practice Problem #2: Proteolytic Agents
You have digested a polypeptide with two different agents and obtained these fragments:
Trypsin: [Met-Phe-Val-Arg] [Ala] [Glu-Lys]Chymotrypsin: [Val-Arg-Ala] [Glu-Lys-Met-Phe]What is the sequence of the polypeptide?
[Glu-Lys][Met-Phe-Val-Arg][Ala] [Glu-Lys-Met-Phe] [Val-Arg-Ala]
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Electrophoresis
• Used to separate proteins into bands according to their net charge.
1. Load protein samples into wells at one end of a gel.2. Apply current; proteins will move through gel matrix
towards the pole opposite their net charge.
wells
(+) pole
(-) pole
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Electrophoresis
• Used to separate proteins into bands according to their net charge.
1. Load protein samples into wells at one end of a gel.2. Apply current; proteins will move through gel matrix
towards the pole opposite their net charge.
wells
(+) pole
(-) pole
Peptide with large net negative charge
Peptide with smaller net negative charge
Peptide with net positive charge
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Methods to Analyze Higher Order Structure
• Nuclear Magnetic Resonance (NMR): can be used for small proteins
• Electron Microscopy: gives overall shape but not atomic resolution
• X-Ray Diffraction: the “gold standard,” determines what atoms are in the protein and the distances between them
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X-Ray Diffraction
1. Crystallize protein of interest2. Expose crystallized protein to X-
ray source (wavelength 1.5 angstroms)
3. Record diffraction pattern 4. Use intensities and positions of
spots to determine atom identity and position
Larger atoms deflect X-rays more than smaller atoms due to greater electron density.
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Homologous Proteins
Homologous proteins are proteins from different organisms that are very similar in structure and function.
Ex: insulin, cytochrome C
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Homologous ProteinsHomologous proteins from different organisms
are similar but (usually) not identical.• Differences arise via mutation• Differences that survive must adequately preserve
function (natural selection)• Differences can only survive at certain residues• AAs at a given position tend to be chemically similar
8 9
Thr Gly Ile
Insulin
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Homologous ProteinsHomologous proteins from different organisms
are similar but not identical.• Differences arise via mutation• Differences that survive must adequately preserve
function (natural selection)• Differences can only survive at certain residues• AAs at a given position tend to be chemically similar• Overall structure must be preserved
(structure function)
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The “Molecular Clock”
Molecular clock theory: AA differences accumulate over time, such that # of differences between homologous proteins can be used to calculate evolutionary distance (time of divergence) for two species.
more AA differences more distantly related
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The “Molecular Clock”
Assume that differences accumulate at a constant rate (# of differences is directly proportional to time of divergence).
Example: Species A and Species B diverged 100 mya. Protein X from Species A and B differs at 10 positions.
Protein X from species C differs from that of Species B at 20 positions. How long ago did Species B and C diverge?
The rate of difference accumulation is unique to each protein (i.e., mutations accumulate at different rates in insulin and CytC).
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Hemoglobin: Intraspecific AA Changes
• Hemoglobin (Hb): found in RBCs, transports oxygen from lungs to tissues
• Tetrameric (4 subunits: 2α, 2β)
• Each subunit has a heme group where O2 binds
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Hemoglobin: Intraspecific AA Changes
• Sickle cell anemia: RBCs sickle and form filaments under low O2 conditions, get stuck in capillaries
• Caused by a single Glu Val mutation at position 6 of the Hb β subunit
low O2
+-
Electrophoresis of HbA and HbS
Glu to Val substitution results in a less negative net charge on HbS and slower migration towards the + pole.
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Hemoglobin: Kinetics & Regulation
Hemoglobin can exist in one of two states:
These states exist in equilibrium.In the presence of certain regulators, one state
will be preferentially stabilized, shifting the equilibrium towards T or R.
binds O2 tightlybinds O2 weakly
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Hemoglobin: Kinetics & Regulation
O2 is a homotropic activator of Hb.
A homotropically activated protein displays cooperativity.
Binding of target molecule (O2) at one active site enhances the affinity of other active sites for target molecule.
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Hemoglobin: Kinetics & Regulation
H+, CO2, and BPG are heterotropic inhibitors of Hb.
They do not resemble O2 and do not bind at the active site.
Binding of heterotropic inhibitors at non-active sites decreases affinity for O2 at the active sites.
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Hemoglobin: Kinetics & RegulationLow O2
Low pHHigh CO2
High O2
High pHLow CO2
O2 Transport
Regulation of Hb is optimized to promote uptake (high saturation) of O2 in lungs and deposition (low saturation) of O2 in tissues.
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Practice Problem 3: Greenglobins
You are studying a strain of mice that have green eyes. You believe that the green color is due to a molecule called protogreen, which is produced in the intestine and transported to the eye.
You hypothesize that a family of proteins called greenglobins are the transporters, and that their affinity for protogreen is affected by retinoin, a small organic molecule present only in the eye.
Image from: http://thebluerepublic.com/Gallery/albums/album03/250px_Mus_Musculus_huismuis.jpg
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Practice Problem 3: GreenglobinsThere are four different greenglobins, A-D. You conduct
binding experiments to learn more about them:
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Practice Problem 3: Greenglobins
1. For each greenglobin, determine whether it is oligomeric or monomeric or if you can’t tell:
Greenglobin A: oligomeric monomeric can’t tellGreenglobin B: oligomeric monomeric can’t tellGreenglobin C: oligomeric monomeric can’t tellGreenglobin D: oligomeric monomeric can’t tell
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Practice Problem 3: Greenglobins
2. For each greenglobin, which type(s) of regulation are illustrated in the graphs above? (Circle all that apply)
Greenglobin A: homotropic activation heterotropic activation heterotropic inhibition Greenglobin B: homotropic activation heterotropic activation heterotropic inhibition Greenglobin C: homotropic activation heterotropic activation heterotropic inhibition Greenglobin D: homotropic activation heterotropic activation heterotropic inhibition
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Practice Problem 3: Greenglobins
3. Rank the different greenglobins in their ability to transport progreen from the intestine to the eye.
1. _______ (best) 2. __________ 3.__________ 4.________ (worst)
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