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Biochem Review, Part I: Protein Structure and Function Lecture Notes pp. 1-38 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 The Twenty Amino Acids SMALL: hydrogen or methyl R group. GLYCINE Alanine Types of AAs NONPOLAR/HYDROPHOBIC: R groups contain largely C, H atoms. ALIPHATIC: no aromatic rings. AROMATIC: contains aromatic rings. VALINE Leucine Isoleucine Methonine *Proline PHENYLALANINE Tyrosine Tryptophan POLAR/HYDROPHILIC: R groups typically contain O, N atoms. CHARGED ACIDIC: acid in R group. BASIC: base in R group. UNCHARGED: no ionizable group in R group. LYSINE Arginine Histidine ASPARAGINE Glutamine Serine Threonine *Cysteine ASPARTATE Glutamate Special Amino Acids SMALL: hydrogen or methyl R group. GLYCINE Alanine Types of AAs NONPOLAR/HYDROPHOBIC: R groups contain largely C, H atoms. ALIPHATIC: no aromatic rings. VALINE Leucine Isoleucine Methonine *Proline POLAR/HYDROPHILIC: R groups typically contain O, N atoms. CHARGED UNCHARGED: no ionizable group in R group. an imino acid, with R group bound to amino group thiol group can participate in disulfide bonding ASPARAGINE Glutamine Serine Threonine *Cysteine Peptide Bond Formation 1 amino terminal residue “N terminal” 2 carboxyl terminal residue “C terminal” 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. Acid-Base Behavior of AAs Equilibrium constant: Ka = [A-][H+]/[HA] pH = -log[H+] pKa = -log(Ka) …convenient shorthand for writing widely variable [H+] concentrations …similar shorthand for writing variable Ka values 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) pk is the pH at which a functional group exists 50% pH = pKa + 0 in its protonated form (HA) and 50% in its deprotonated pH = pKa form (A-). a Isoelectric Point Isoelectric point: the pH at which an AA or polypeptide has no net charge. • For a dibasic amino acid: Isoelectric point = average of amino and carboxyl pka values = (2.4 + 9.8)/2 = 6.1 pka = 2.4 pka = 9.8 GLYCINE • For a tribasic amino acid: Isoelectric point = average of the two numerically closest pka values pka = 2.2 pka = 9.7 = (2.2 + 4.3)/2 = 3.25 Formulas can be used for polypeptides as long as they have no more than one ionizable side chain. pka = 4.3 GLUTAMATE Titration Curves This example shows the curve for a dibasic AA pka of amino group pka values always occur at the flattest parts of the curve (around 0.5, 1.5, 2.5 base equivs.) pka of carboxyl group buffering works best at pka Isoelectric point The isoelectric point will always occur at at 1.0 or 2.0 base equivs. buffering works best at pka # of equivs. OH- tells whether AA is dibasic (2 equivs) or tribasic (3 equivs.) Practice Problem #1: Titration of Aspartate 2 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.) 3 1 Posn . pH 1 ~0 2 9.8 3 3.0 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 Curve and structures from: http://www.cem.msu.edu/~reusch/VirtualText/proteins.htm Practice Problem #1: Titration of Aspartate Posn pH 3 3.0 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 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% Practice Problem #1: Titration of Aspartate Pos n pH 3 3.0 Major form(s) present Average charge on α -carboxyl (pka = 2.1) For the alpha carboxyl: Average charge on R carboxyl (pka = 3.9) Average charge on amino group (pka = 9.8) Net charge For the R group 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-COOH 100%-11.18% = 88.82% alpha-COO- % protonated = 1/(1+103.0-3.9) x 100% % protonated = 88.82% R-COOH 100%-11.18% = 11.18% R-COO- Average charge on alpha carboxyl: (0)*(0.1118) + (-1)*(0.8882) = -0.8882 Average charge on R carboxyl: (0)*(0.8882) + (-1)*(0.1118) = -0.1118 Practice Problem #1: Titration of Aspartate Posn pH 3 3.0 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 -0.89 -0.11 +1 0 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. 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 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 Primary Structure • Genetically determined: specified by sequence of gene that encodes protein Primary Structure • Contains all information necessary to specify higher orders of structure (3D shape) Denature Renature break disulfide bonds break disulfide bonds The 3D structure a protein spontaneously assumes is its most stable structure. 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 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 Hbonds with the amino group of an AA 4 residues down. 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 N terminus turn in chain C terminus Like the α helix, the β sheet is formed by interactions WITHIN a single polypeptide chain. 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). 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 Tertiary Structure • Tertiary structure can be complex and does not typically consist of repeating units. • A polypeptide will adopt the most stable tertiary structure In aqueous environment, this occurs when hydropobic residues are internal and hydrophilic residues are external. = charged = hydrophobic = neither 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 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. 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 Ion Exchange Chromatography • Tells you what AAs a protein contains and the relative abundance of each Ion Exchange Chromatography 1. Digest protein in strong acid 2. 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 Ion Exchange Chromatography • An AA will elute approximately when the pH of the solution equals its isoelectric point Ion exchange chromatography can also be used on small polypeptides. These too will elute in an order determined by isoelectric point. AAs with low isoelectric points (such as acidic AAs) come off earlier Peak height indicates relative abundance AAs with high isoelectric points (such as basic AAs) come off later Edman Degradation determine identity determine identity • 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 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 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? 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] 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. (+) pole wells (-) pole 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. Peptide with large net negative charge Peptide with smaller net negative charge Peptide with net positive charge (+) pole wells (-) pole 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 X-Ray Diffraction 1. Crystallize protein of interest 2. Expose crystallized protein to Xray 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. Homologous Proteins Homologous proteins are proteins from different organisms that are very similar in structure and function. Ex: insulin, cytochrome C Homologous Proteins Homologous 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 Thr Gly Ile 8 9 Insulin Homologous Proteins Homologous 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) 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 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). 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 Hemoglobin: Intraspecific AA Changes low O2 - + Electrophoresis of HbA and HbS • 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 Glu to Val substitution results in a less negative net charge on HbS and slower migration towards the + pole. Hemoglobin: Kinetics & Regulation Hemoglobin can exist in one of two states: binds O2 weakly binds O2 tightly These states exist in equilibrium. In the presence of certain regulators, one state will be preferentially stabilized, shifting the equilibrium towards T or R. 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. 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. Hemoglobin: Kinetics & Regulation Low O2 Low pH High CO2 O2 Transport High O2 High pH Low CO2 Regulation of Hb is optimized to promote uptake (high saturation) of O2 in lungs and deposition (low saturation) of O2 in tissues. 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 Practice Problem 3: Greenglobins % of progreen binding sites occupied There are four different greenglobins, A-D. You conduct binding experiments to learn more about them: % of progreen binding sites occupied Practice Problem 3: Greenglobins 1. For each greenglobin, determine whether it is oligomeric or monomeric or if you can’t tell: Greenglobin A: Greenglobin B: Greenglobin C: Greenglobin D: oligomeric oligomeric oligomeric oligomeric monomeric monomeric monomeric monomeric can’t tell can’t tell can’t tell can’t tell % of progreen binding sites occupied 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: Greenglobin B: Greenglobin C: Greenglobin D: homotropic activation homotropic activation homotropic activation homotropic activation heterotropic activation heterotropic activation heterotropic activation heterotropic activation heterotropic inhibition heterotropic inhibition heterotropic inhibition heterotropic inhibition % of progreen binding sites occupied 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)