Figure 24-24 gives the electric j! potential V as a function of x. (a) Rank the five regions according to the magnitude of the x component of the electric field within them, greatest first. What is the direction @ x -4q " x -2q 5 -- x of the field along the x axis in Fig. 24-24 Question 4. (b) region 2 and (c) region 4?
Unit 1: Proteins Section 1.1: Introduction: pH & Charge of Amino Acids Prokaryote cell Eukaryotic cell Robert Hooke coined the biological term cell so called because his observations of plant cells reminded him of monks' cells which were called "cellula". Cytoplasm: the fluid contained within the cell that holds and surround the cell's organelles in a liquid environment which is necessary for many of the cell's vital functions to occur. The materials that are found within the cytoplasm are also typically found within the cell membrane. Typically, the cytoplasm contains materials that are known as cytoplasmic inclusions. o These inclusions are typically starch granules, mineral crystals, or lipid droplets that are floating around within the cytoplasm. Cytosol: the gellike translucent fluid that is part of the cytoplasm, consisting mostly of water. Suspends inclusions and organelles where most of the cells chemical reactions occur [EX: transportation and communication] Unit 1: Proteins Ionic Bonds: salt bonds Covalent bonds: stronger that hydrogen bonds Hydrogen bond: When a H atom is shared between two atoms of N and/or O (and sometimes S) Directionality of the hydrogen bond affects its strength o The attraction between the partial electric charges is greatest when the three atoms involved in the bond (in this case O, H, and O) lie in a straight line. Linear = stronger Bent = weaker + + pH = log (H ) = log[1/(H )] log x = log (1/x) Acid dissociation: - OH H O Unit 1: Proteins Acetic acid Acetate (Ac )- K = = 1.74 x + a H 10-5M Buffering action is simply the consequence of two reversible reactions taking place simultaneously and reaching their points of equilibrium as governed by tweiraequilibrium constants, K and K . HendersonHasselbalch Equation: (A-) (HA) pH = pK + log a Titration of acetic acid: pK1 pH = Buffering Region: where the change in pH is minimal. Its absolute minimum is when pH=pKa Unit 1: Proteins When pH=pKa the acid concentration equals the conjugate base concentration and the net charge changes by 1, (loses/gains a proton depending on which way the equation is being driven) Change on Charge as pH increases: pK1 - 0 1 0 -0.5 -1.0 The net charge of acetic acid before OH is added is 0 and the net charge after OH is added is 1 because 1 mole of acetic acid loses a proton for every 1 mole of OH added . (net charge is in red under the equation) As pH decreases, the net charge of any biological system has to become more positive or LESS negative. Titration comparison between 3 different acids + + NH4 ↔ NH 3 + H pKa = 9.25 H PO - ↔ HPO 2- + H+ pK = 6.86 2 4 4 a Acetic acid ↔ acetate + H pK = 4.76 a Unit 1: Proteins Structure of Amino acids: All of the amino acids normally found in proteins (except glycine) are chiral and have the L configuration. The R group, or side chain, attached to the αcarbon is different in each amino acid. required to recognize the 20 amino acids (tests are multiple choice) and know their abbreviation Major ionic form at pH 7 & nonpolar aliphatic R group Glycine Alanine Proline Valine Gly, G Ala, A Pro, P Val, V Leucine Isoleucine Methionine Leu, L Ile, I Met, M Unit 1: Proteins Major Ionic form at pH 7 & aromatic R group Phenylalanine Tyrosine Tryptophane(e) Phe, F Y W Major ionic form at pH 7 & polar & uncharged R group Serine Threonine Cysteine Ser, S Thr, T Cys, S Unit 1: Proteins Asparagine Glutamine Asn, N Gln, Q Major ionic form at pH 7 & positively charged & R groups Lysine Arginine Histidine Lys, K Arg, R His H Major ionic form at pH 7 & negatively charged & R group Aspartate Glutamate Asp, D Glu, E Unit 1: Proteins Example: Titration of Glycine pH = 9.60 pH = 2.34 pK1 pK2 +1 0 -1 For every OH added, Structure of: glycine loses an H, Group Acid Base ~pK a range reducing its charge by 1. The carboxyl group loses a Carboxyli proton before the c 1.5 to 4 ammonium group because it is a stronger acid (stronger acids lose Imidazole 6 to 7 protons easier/faster than weaker acids/bases) Sulfhydry 7 to 9 l Amino 8 to 10 Phenolic 8 to 10 Guanido 11 to 13 Unit 1: Proteins Example: Titration of Glutamic Acid pH = 9.67 pH = 4.25 pH = 2.34 +1.0 +0.5 0 -0.5 -1.0 -1.5 -2.0 Example: Titration of Histidine +1 02 pK1 pH = 9.17 D E H Y K R Amin Unit 1: Proteins o acid: pH: pH = 6.0 0 1 pH = 1.82 2 3 4 pKpK2 5 6 +2 01 7 8 9 10 11 12 13 Unit 1: Proteins Section 1.2: Proteins Unit 1: Proteins Proteins are all polypeptides, but not all polypeptides are proteins Peptide bonds are amides and formed by a condensation reaction - H O + 2 + H O 2 The elimination of water between amino acid 1 and amino acid 2 form the peptide bond between the two amino acids. The amino acids join between the ammonia and the carboxyl groups. Peptides are named beginning with the aminoterminal residue, which by convention is placed at the left. The peptide bonds are shown here in red. Structure of a protein: the primary structure = the primary sequence = the amino acid sequence. Amino acids are numbered form the amino (N) terminal to the carboxyl (C) terminal. Amino (N) terminal Carboxyl (C) terminal Unit 1: Proteins Oxidation of 2silfhydryl groups to a disulfide bond + - + + 2 H + 2 e Insulins Amino acids chains bind together by inter or intra disulfide bonds. In amino acids chains, specific Cysteine’s will bind together. Cysteine’s that are part of the same chain bond together via interdisulfide bonds while Cysteine’s that are part of two different amino acid chains bond via interdisulfide bonds. Huma T I N T n Porcin e " " " A Bovin A V " A Difeerences between amino acid sequence of insulin between humans and other animals Feline A V H A Human insulin is made thru genetic engineering by inserting the human gene into a bacterium. The bacterium will make human insulin. Before this method was developed, pig pancreases were used to harvest insulin for use by humans. (Unfortunate for people whose body’s recognized the insulin to be foreign.) The greater the difference in structure between the host protein and the foreign protein, the greater chance the host sees the foreign protein as "notoneofus" and thus turns on the immune response against the foreign protein. Porcine and bovine insulins are not on the market anymore because they are not very marketable. Now, people have to purchase human insulin which sucks for animals with diabetes. Unit 1: Proteins Conformation Proteins are polypeptides with molecular weight of at least 10,000 On average, 1 amino acid weighs 110D Example: How many amino acids are in a protein with molecular weight of 10,000 40,000 10,000 / 110D = 90 amino acids 40,000 / 110D = 360 amino acids Example: What is the molecule weight of a protein containing 512 amino acids 512 x 110D = 55,000 Constraints are what give proteins their shape. o Each peptide bond has some doublebond character due to resonance and cannot rotate. o As with all amides, the carbonyl O has a partial charge and the amide N a partial + charge. o Almost (but not all) peptide bonds in proteins have this trans configuration. (Resonance) A peptide bond (The 6 atoms forming the plane are in red) Models of the αhelix Nterminal DO NOT confuse and alpha and double helix C atoms = green; O atoms = red; N atoms = blue; side groups = orange; H atoms are not shown. Unit 1: Proteins Cterminal Pitch is the distance along the alpha helix for one turn of the helix = 5.4 A Rise is the distance along the alpha helix for one amino acid residue = 1.5 A or 3.6 amino acid residues per turn The dimensions are fixed and known. Hydrogen bonds between atoms in the polypeptide backbone (not in the side groups) give the protein its helical structure o Hydrogen bonds are weak but a bunch of them together are strong. o Hydrogen bonds occur at between atoms that are four amino acids apart [Ex: amino acid 100 bonds to amino acid 104, 101 bonds to 105, 102 binds to 106, etc] (n and n+4) R 1 R2 R 3 R 4 R5 R6 R 7 R8 N C C N C C N C C N C C N C C N C C N C C N C C H H O H H O H H O H H O H H O H H O H H O H H O Placement of the hydrogen bonds. The red atoms share hydrogen bonds with the next red atom and etc for the rest of the colors The conformation of each and every protein must be experimentally determined. Only 70% of hemoglobin or myoglobin is found to be in the helical structure. (100% keratin is in alphahelix and 0% chymotrypsinogen is in the alphahelical structure) Hydrophobic Interactions When molecules combined, the surface are to volume ratio decreases (this increases the stability of hydrophobic molecules) M The same concept can be applied to children and adults. Children have a smaller MM surface area to volume ratio so they M loses heat faster than adults. Food will also cook differently depending on the amount you cook at once because the surface area to volume ration is different. Less stable: More stable: More Less interaction interaction Hydrophilic Interactions NaCl dissolves in water. Na will interact with O while Cl will interact with H . “Like dissolves like” Unit 1: Proteins Proteins have hydrophobic and hydrophilic regions Hydrophobicity is the main driving force behind protein conformation General rule: Hydrophobic amino acids in the interior of a folded protein; hydrophilic amino acids on the outside where they can interact with the aqueous solvent. The black spheres represent hydrophobic amino acids while the white spheres represent hydrophilic amino acids. Some amino acids are amphipathic (neither hating nor loving of water/ having a passion for both) Hydrophobic amino acids: F, M, I, L, V (in decreasing order of hydrophobicity) Hydrophilic amino acids: H, Q, N, E, K, D The rest are amphipathic (W and Y can act as either hydrophobic or hydrophilic) Cellular Membrane ~30 Å An amino acid is 1.5A long and there are 20 amino acids per 30A in a lipid bilayer 30A / (1.5A/aa) = 20 amino acids Conversion: 30A = 20 amino acids Unit 1: Proteins 1.5A = 1 amino acid Reminder: 1Å = 10 10m The tails (hydrophobic region) in the phospholipid bilayer of the membrane often form an alphahelical structure the more bonds formed, the more stable the compound (if a carbonyl O and an amino N cannot hydrogen bond with water, they tend to hydrogen bond with themselves) Most proteins are stabilized to a large extent by forming lots of H bonds But amino acids in the lipid portion of a membrane cannot form H bonds with water because water is excluded from that region. But these amino acids can form H bonds with themselves Since alphahelices are an especially common and stable conformation that depends upon H bonding between amino acids alphahelices are common in hydrophobic environments Oxygen and nitrogen readily form hydrogen bonds because they are so polar Bacteriorhodopsin: A 7transmembrane protein The single polypeptide chain folds into seven hydrophobic α helices, each of which traverses the lipid bilayer roughly perpendicular to the plane of the membrane. Transmembrane proteins transverse the membrane once Nterminal Outside APTActivated Channel Inside Schematic representation showing the membrane topology of a typical P2X receptor subunit. First and second transmembrane domains are labeled TM1 and TM2. Unit 1: Proteins alphahelix is within the dynamite. PX2 receptors are involved in a variety of physiological processes. [Ex: Apoptosis, contraction of the vas deferens during ejaculation, modulation of vascular tone, macrophage activation] Integral and Peripheral Membrane Protein An alphahelical structure generally (but not always) form in the membranespanning region of a protein o The more bonds formed, the more stable is a compound. Most proteins are stabilized to a large extent by forming lots of H bonds, especially with the polypeptide backbone. But amino acids in the lipid portion of a membrane cannot form H bonds with water because water is excluded from that region. However, these amino acids can form H bonds with themselves. Since alphahelices are an especially common and stable conformation that depends upon H bonding between atoms (N and O) in the polypeptide backbone alpha helices are common in hydrophobic environments. Note that these H bonds are in the polypeptide backbone, not in the side groups o Beta sheets can also form, since they are also stabilized by H bonds between amino acids, but they are much less common in membranespanning proteins than are alphahelices. Antiparallel BetaPleated Sheets There can be several layers of amino acid chains (only two layers are shown for simplicity) N C C N The hydrogen bonds in antiparallel betasheets are linear (and hence stronger than the bent H bonds in parallel betasheets). Unit 1: Proteins N C Parallel BetaPleated Sheet N C The hydrogen bonds in parallel betasheets are bent (instead of being linear). AntiParallel Beta Pleated Sheets More stable than parallel Bsheets More common than parallel Parallel Beta Pleated Sheets Unit 1: Proteins R groups alternate up and down Beta sheets are often twisted, not flat The arrow points to the C terminal BetaBarrel Structure Found in Porins nonhelical structure Tertiary structure of a protein folding of the 2° structural elements plus the spatial arrangements of the aminoacid side chains Denaturation is the disruption of conformation without breaking peptide bonds. Proteins unfold (results in loss of biological activity which is usually irreversible) [Ex: heating, alcohol, high/low pH] Unit 1: Proteins egg whites will lose solubility when it is cooked (eggs have to be denatured before being eaten because the enzymes have to get to the peptide bonds to digest the egg) fevers develop in order to denature the virus Alcohol is used to sterilize by denaturing bacteria (the bacteria lose biological activity) Changing pH also changes charge. (repel or attract proteins) Folding proteins Proteins are synthesized in an unfolded state. Folding is directed by auxiliary proteins. The primary sequence also helps a protein fold correctly. o Many problems in the biotech industry is incorrect folding Incorrect folding results in cystic fibrosis, Alzheimer’s disease, mad cow disease, etc Quaternary structure Many proteins have more than one polypeptide (subunits) a human cell contains about 20,000 different proteins. Contains about 100,000,000 proteins in total Units 1: Proteins Section 1.3: Hemoglobin & Myoglobin Myo means muscle and globin means blood: Myoglobin is a protein found in the muscle while hemoglobin is found in blood. Many require cofactors and a prosthetic group Cofactor: a small molecule required for the biological activity of a protein Prosthetic group: a cofactor tightly bound to the protein via covalent bond(s) Myoglobin and hemoglobin: Mb + O Mb-O 2 2 The Porphyrin Ring Hb + O Hb-O System 2 2 The structure of the cofactor, heme Porphyrins consist of 4 pyrrole rings linked by methene bridges Structure is more or less planar There is continuous resonance over the entire protoporphyrin ring system. The iron must be in +2 redox state to function 2 to bind to O and F8 histidine), must have 6 bonds 2+ 3+ o Fe is red while Fe is brown Myoglobin showing heme cofactor 8 alpha helical structures bound by primary sequence structures 153 amino acids in human myoglobin C-terminal molecular weight is 17,199D N-terminal Units 1: Proteins A coordinate covalent bond is a covalent bond (i.e., shared electrons between 2 atoms) where one atom alone provides both electrons. Hemoglobin in red blood cells Found only in erythrocytes (red blood cells) Has quaternary structure (alpha 2 beta 2) (unlike myoglobin) and 4 subunits Each subunit of hemoglobin molecule has 1 heme; 4 hemes per 1 hemoglobin; 4 O 2can bind to a hemoglobin Structure of human deoxyhemoglobin: The alpha subunits are in blue; The beta subunits in purple; Alpha chain of hemoglobin: molecular weight is 15,126D 141 amino acids Beta chain of hemoglobin: molecular weight is 15, 867D 146 amino acids Respiratory system in invertebrates Takes up oxygen in lugs/gills and the hemoglobin binds oxygen. The hemoglobin transports oxygen through the arteries to the peripheral tissue. The oxygen is released and can be used for metabolism producing CO 2 Some of the oxygen is bound to myoglobin in muscle. The CO 2nd bicarbonate is transported through blood (in veins) (by hemoglobin) and out of body. Oxygen concentration in lungs is higher than in tissue Erythrocytes Red blood cells (erythro=red, cyte=cell) Adults have ~25 trillion erythrocytes ~280 million hemoglobin molecules in 1 erythrocyte Allosteric Proteins is a protein whose functional site can be altered by the binding of a small molecule at a non overlapping site. Units 1: Proteins It is triggered to alter its function in response to the binding of a target compound (allosteric ligand) at a site that is distal from the active site of the protein Allosteric ligands bind at the allosteric site, resulting in conformational changes that alter the functional site(s) Allosteric ligand = allosteric effector (may be + or ) Functional site Allosteric site Allosteric ligand (+ or Cooperativity is a - effector) type of an allosteric regulation (Allosteric ligand or effector) Binding of the ligand to one site on the protein affects the binding properties of a different site on the same protein Can be positive or negative modulation of biological activity (↑ or ↓ functionality) In cooperativity, the allosteric ligand is both a functional ligand for the pro2ein (e.g., O for hemoglobin; a substrate for an enzyme), as well as an allosteric modulator of the protein's activity It is generally an activator of the protein o E.g., O 2is a positive allosteric effector of hemoglobin as well as the biological ligand at the active site Oxygen binding curves for myoglobin and hemoglobin % DeoxyMb or % OxyMb Or DeoxyHb DexoyHb Units 1: Proteins Mb Lungs HbA (sea level) Mb + O ↔ Mb(O ) 2 2 Tissues Hb + n O ↔ 2 Partial Pressure of Oxygen (torr) Myoglobin has a hyperbolic curve while hemoglobin has a sigmoid shaped curve Myoglobin contains 1 polypeptide while hemoglobin contains 4 polypeptides making hemoglobin it a quaternary structure. This is why the curves are different Hemoglobin does not release all of the oxygen it picks up, some of it returns to the lungs Fraction saturation = Bound2O / total heme groups % Saturation = fraction saturation x 100 Torr = 1mm Hg at 0 C = 0.133 kPa 760 torr = 1 atm 750 torr = 100 kPa Deoxyhemoglobin has several salt bonds (formed because of positive and negatively charged atoms) that stabilize the “T state” shape. Oxyhemoglobin does not have salt bonds, it is in the “R state” shape. o T (deoxyHb) = tight, tense, taut o R (oxyHb) = relaxed The binding of oxygen to heme iron in myoglobin or hemoglobin The conformation of deoxy is more curved than the oxy conformation His (F8) His (F8) Units 1: Proteins Deoxy (T state) Oxy (R state) When a molecule of oxygen binds to a heme, the heme iron ion is drawn into the plane of the porphyrin ring in its respective subunit. This causes the same conformational change in the other three subunits, increasing the affinity of their heme groups for oxygen. Deoxyhemoglobin resists oxygenation because it is stabilized by hydrogen bonds and salt bridges in its tense state (oxygenation will break these stabilizing interactions) Color of blood Color of blood in arteries is bright red while the color of blood in veins is dull red The iron gives blood its red color In the arteries, iron binds 6 oxygen. Iron binds 5 oxygen’s in veins. Conformation of Oxy and Deoxyhemoglobin Oxy (R) Deoxy (T) Sequential Model for binding of O2 by hemoglobin 2,3bisPhosphoglycerate (BPG) Units 1: Proteins = glycerate % Sa Effect of BPG on binding of oxygen to hemoglobin tur No BPG ati on 5 mM BPG 8 mM BPG Partial Pressure of Oxygen Without BPG, hemoglobin is not functional because it is shifted so far to the left, it would not deliver oxygen to the tissue. (0xy form has been stabilized) • Shift of the oxygen-binding curve for hemoglobin to the right = shift of the oxygen-binding curve for hemoglobin down = stabilizing the T state (deoxy form) of hemoglobin • Shift of the oxygen-binding curve for hemoglobin to the left = shift of the oxygen-binding curve for hemoglobin up = stailOne adaptation to higher altitudes is to increase BPG production stabilizing the deoxy form of hemoglobin and delivering more oxygen to tissue one BPG binds to the Beta subunits of deoxyhemoglobin in the middle of the molecule by its negative charges binding to the positive charges on the side groups of the hemoglobin as oxygen binds, less BPG binds stabilizing the oxy (R) form and vice versa DeoxyHb (T) Units 1: Proteins Another way of showing this balancing act: Hb-O intermediate deoxyHb + O 2 2 Oxygen is picked up in the lungs/gills and transported through the arteries by hemoglobin the tissue or stores (for aquatic invertebrates). Once the oxygen has been utilized carbon dioxide is transported back to the lungs through veins. Effect of pH on binding of oxygen to hemoglobin pH 7.6 pH 7.4 pH 7.2 When CO 2 dissolves in water in tissue, pHartial Pressure of Oxygen (torr) decreases and the oxygen binding curve shifts to the right/down, stabilizing the Deoxy form. As CO2 is removed the pH increases and the oxy form is stabilized Effect of pH on the conformation of hemoglobin When pH decreases (increases in ionic bonds/hydrogen concentration increases) the Deoxy form is favored and vise versa Comparative binding of oxygen to HbA and HbF HbA = alpha beta HbF = alpha gamma The fetal hemoglobin curve sits to the left of the adult hemoglobin. The fetal hemoglobin favors the oxy form (R) form. Fetal oxygen binds more oxygen than adult hemoglobin o When adult and fetal hemoglobin pass each other in capillaries, the fetal hemoglobin “steals” oxygen form the adult hemoglobin and binds oxygen more tightly. Units 1: Proteins HbF HbA Hb + n O ↔ % 2 Hb(O ) 2 n Partial Pressure of Oxygen (torr) Difference between the HbABeta chain and the HbFGamma chain: ENFRLLGNVL VCVLAHHFGK EFTPPVQAAY QKVVAGVANA LAHKYH ENFKLLGNVL VTVLAIHFGK EFTPEVQASW QKMVTAVASA LSSRYH (Ser instead of HisB143) Ser has no charge (unlike His which is positively charged) so there are reactions in HbF that are missing Binding of 2,3bisphosphoglycerate to beta chain of deoxyhemoglobin BPG binds less tightly in HbF, therefore there is less shift toward the Deoxy form. Because the oxy form is more favored in HbF than in HbA, HbF can steal oxygen form HbA In HbF, the amino acid at position beta 143 is Ser, which does not have an ionic side group and thus does not ionically bond the negativelycharged BPG, thus lowering the affinity of the hemoglobin for BPG. This shifts the oxygenbinding curve toward the right (or up or toward the R form of hemoglobin). Thus HbF can "steal" oxygen from maternal HbA. Ser in gam Deoxy Ser in gam Units 1: Proteins Adults retain some of their fetal hemoglobin Persistent fetal hemoglobinemia: disorder that causes adult to produce fetal hemoglobin instead of normal hemoglobin. There are no major or minor symptoms and people live normal lives. Hemoglobin’s (and all globins) have invariant positions that require certain amino acids (like His at the F8 position or Gly in the B6 position). o The His binds to the iron in heme which is required for life and cannot be changed. o Gly has a the smallest side groups so the B helix can pass by (fit by) the E helix Section 1.4: Enzymes Almost all enzymes are proteins. They have a binding site and a catalytic site. They speed up the rate of a reaction. o Enzymes do not alter the equilibrium constant or the concentrations. Enzymes shorten the time it takes to reach equilibrium o Enzymes are specific as to which reactions they catalyze and substrates they bind to Substrates must collide a the correct orientation and have more energy than the product to react Enzymes only bind substrates at substratebinding sites Units 1: Proteins E + S ES EP E + P E = enzyme (specifically the substrate-binding site) S = substrate ES = the enzyme-substrate complex EP = the enzyme-product complex P = product Substratebinding site = those amino acids in the enzyme involved in binding the substrate(s) Active site = those amino acids in the enzyme directly involved in catalysis o The substratebinding site and the active site always overlap (where the substrate binds is where catalysis occurs) Substrate binding and induced fit Induced fit: when a substrate binds to the substratebinding site on an enzyme, the conformation of the enzyme changes. Units 1: Proteins The rate of reaction slows with time. As a reaction approaches equilibrium the rate slows As substrate concentration increases the rate of reaction increases but the initial rate of reaction does not change with substrate concentration V0 = how fast P forms (or S disappears) when there is no reverse rx The maximum velocity, Vmaxis extrapolated from the plot, becau0e V approaches but never quite reaches Vmax The substrate concentration at whic0 V is half maximam is K , the Michaelis constant 1 0.75 Vmax Vo 0.5 0.25 0 0 25 50 75 100 (Substrate) V /2 max K m [S] >> [E] MichaelisMenten plot (the Curve) is obtained from multiple experiments Km = the substrate concentration that results 0n a V of 1/2 of the maximal initial velocity (Vmax) Vmax[S] Km + [S] Units 1: Proteins Michaelis-Menten equation: V = ------------ 0 If [S] is low, Km tends to be low As the substrate concentration increases, the concentration of the enzymesubstrate complex increases until all the binding sites on the enzyme are occupied (= saturating concentration of This is three differnet experiments substrate) Km values differ greatly. They cannot be predicted. They must be determined experimentally o Km is independent of enzyme concentration If [S] << m , rate is relatively slow but have "fine" control of rate (V ) by varying [S] If [S] >> m , rate is close to max but have less control oo rate (V ) by varying [S] If V is high, the rate of a reaction tends to be high and vice versa max V max = 1.0, 0.7, 0.3 mM per min Km is the same for all three enzymes V 0 (mM per min) Vmaxepends on 2 major parameters Substrate Concentration (mM) 1) Vmax is proportional to [enzyme] Decrease [enzyme] by 50%, Vmax will decrease by 50% 2) Total Vmax is proportional to the Vmax for a single enzyme molecule Competitive Inhibition Competitive inhibition can be overcome by increasing the concentration of substrate so that there are so many substrate molecules that they out compete the inhibitor Units 1: Proteins Competitive inhibitors typically have structures similar enough to the structure of the substrate that they "fool" the enzyme active site into binding them. However, the inhibitor structure does not allow the reaction to proceed to produce product. In competitive I, the competitive inhibitors bind to the enzyme's active site. In the presence of a competitive inhibitor: Vmax is not altered; Km is increase. (it takes more substrate for Vo to reach 1/2 Vmax) V0 (I) = "Km " = (mM per min) 0 mM 1 mM 10 mM 6 mM Substrate Concentration (mM) 50 mM 26 mM Antifreeze ingestion (ethylene glycol poisoning) is countered with alcohol. o The ethanol competes with ethylene glycol for the same binding site on alcohol dehydrogenase. As the ethanol concentration increases, less ethylene glycol can bind to the active site and thus formation of toxic glycolaldehyde is slowed. After about a day, the very watersoluble ethylene glycol is excreted by the kidneys. Carbon monoxide can bind to the iron in hemoglobin causing carbon monoxide poisoning o Counteract by increasing the oxygen concentration where oxygen can compete with the carbon monoxide. (carbon monoxide binds more tightly to iron than oxygen) Units 1: Proteins pH optimum for the activity of an enzyme These curves are constructed from measurements of initial velocities when the reaction is carried out in buffers of different pH. The pH optimum for the activity of an enzyme is generally close to the pH of the environment in which the enzyme is normally found. Pepsin Chymotrypsin Relative reaction velocity 1 3 5 7 9 11 pH Optimal temperature of an enzyme The rate of almost all chemical reactions increase with increased temperature Since enzymecatalyzed reactions depend on the native conformation of the enzyme, thermal denaturation of the catalyst is a competing process Thermal denaturation of the enzyme No Relative Velocity Yes Temperature (K) Units 1: Proteins Optimal temperature and pH is an adaptation of the cell to exist in its environment Unit 1: protein Section 1.5: Enzyme Mechanisms 1 Chymotrypsin is a protease in the small intestine that helps digest proteins to amino acids. It is synthesized and stored in the pancreases until it is needed. Has three polypeptide chains held together by disulfide bonds His57, Asp102, and Ser195 are involved in the mechanism H2O + The amino acid on the Nterminal (R1) is usually, F, Y, or W Chymotrypsin is an endopeptidase o an endopeptidase catalyzes the hydrolysis of a peptide bond that is NOT at the Nterminal or C terminal site (catalyzes a peptide bond that is on the "inside" of the peptide) Basecatalyzed hydrolysis of a peptide bond HOH + + 7 But the biological reaction must proceed at pH ~7 where (OH ) is far too low (~10 M) o Solution: use a RO oxyanion as the base o Problem: alcohols are not acidic in water; they aave a pK ~16. The only way to create RO at pK ~7 is to ↓ pa of the alcohol (ROH from ~16 to ~8 ) Outline of Mechanism of serine proteases 1st product + + + Unit 1: protein (Must ↓ pK af the serine alcohol from ~16 to ~8 so can form RO to act as base to attack the carbonyl C) + + + 2nd product First (as with any enzyme) the substrate binds to the substratebinding site on chymotrypsin. Now the SerO atom must attain enough of a charge to attack the electrophilic carbonyl C of the substrate peptide bond (shown as a thick black bond in the slide). Since an alcohol is normally not an aciddissociable group in water (remember that pure vodka is not sour; pKa for water ≈ 14; pKa for ethanol ≈ 16), the Ser hydroxyl needs some help to lower its aK sufficiently so that the O takes on basic properties (i.e., has a charge) so it can attack the carbonyl C in the substrate. o This help comes from the adjacent base form of imidazole which can attract the hydroxyl H while the hydroxyl O is attacking the carbonyl C of the substrate. Mechanism of serine protease: The catalytic triad Substrate The adjacent imidazole group (in the base form)lowers the pKa of the serine hydroxyl from ~16 to a low enough value so that the O atom can serve as a base (i.e., have a charge) to attack the carbonyl C atom of the substrate peptide bond. But the imidazole group by itself cannolower the pKa of the OH group sufficiently; it needs the adjacent negative charge on the Asp side chain traise the pKa of the imidazole group (which is normally ~6 to 7) so it can interact with the Ser to make the hyrdroxyl O more electronegative (= nucleophilic). Unit 1: protein When a pK aecreases, the base form is stabilized; when a aK increases, the acid form is stabilized. The tetrahedral C intermediate is stabilized by the oxyanion hole, which we do not discuss here (even though it is very important). But remember that a tetrahedral C with 3 O or N atoms attached is inherently unstable; one of the CO or CN bonds will likely break. Another way of stating this: The charge on the adjacent Asp carboxylate stabilizes the acid form (with a + charge) of the imidazole group; i.e., tha pK for the imidazole group significantly from the "usual" value of 6 to 7. This allows the uncharged imidazole group to act as a stronger base (since the acid form is more stable now); i.e., extract a proton from the serine alcohol group. The unstable tetrahedral intermediate breaks the CN bond (the original peptide bond in the substrate) to split the original peptide into 2 smaller peptides: o The product peptide with the new Nterminal is transiently bonded to the imidazole group; the peptide product with the new Cterminal group is still covalently attached to the enzyme via an ester bond with the Ser hyrdroxyl group. o The imidazole group returns to the base (uncharged) form. H HOH R' N H + 1st product (shorter peptide with The base fonew N-terminal amino acidcts)a proton from water allowing a hydroxide (with charge) to attack the electrophilic C of the ester. Net effecta ↓ pK of water = ↑ the (HO). Unit 1: protein Now water (which is actually a substrate in this reaction) comes into play. It displaces the peptide product with the new Nterminal; i.e., one of the two products of the reaction is now free. The other peptide product is still covalently attached to the Ser hydroxyl group. The base form of the imidazole extracts a proton from the water to allow the resulting hydroxide ion (with a charge) to attack the carbonyl C (with a partial + charge) of the ester to form again a relatively unstable tetrahedral intermediate. 2nd product (shorter peptide with new The tetrahedral C is unstable resulting in cleavage of the ester bond between the Ser hydroxyl and the product carboxylic acid, which is the Cterminal of the C-terminal shorter peptide product. The second product is releases and the enzyme (molecule with serine) is now amino acid) ready to accept the next substrate Unit 1: protein Summary: The Binding site or chymotrypsin, Trypsin, and Elastase (serine proteases) Unit 1: protein Chymotrypsin Trypsin Elastase F, Y, W K, R G, A The H atoms shown for chymotrypsin and trypsin are the side groups of glycine residues. The binding pocket of chymotrypsin is lined with hydrophobic amino acids that interact with a hydrophobic sidechain of Trp, Tyr, Phe or Met in the substrate. The binding pocket of trypsin has an aspartate group at the end of the binding pocket (Asp189 instead of Ser189 in chymotrypsin). The negative charge on the aspartate can form a salt bond with the postitive charge on the side group of lysine or arginine in the substrate molecule. Note that the aspartate in the binding pocket is not the same aspartate that is in the catalytic triad The binding pocket of elastase is blocked by protruding aminoacid side groups. The pocket thus becomes shallow and only the small aliphatic sidegroups of Gly, Ala and Val of the substrate can bind there Catalytic triad inhibitor Unit 1: protein Enzymes with the catalytic triad (SHD) are inhibited by organofluorophosphates The organofluorophosphate binds to the serine of the catalytic triad, irreversibly inhibiting the enzyme Acetylecholinesterase Acetylcholine Acetate Choline Organofluorophosphates inhibit cholinesterase which inactivates the neurotransmitter acetylcholine This esterase has the same catalytic mechanism as serine proteases Irreversible inhibition by sarin Unit 1: protein Sarin is an organofluorophosphate Inhibits cholinesterase which inactivates the neuro transmitter acetylcholine Ser of the Acetylcholine + H2O → Acetate + Choline catalytic This esterase has the same catalytic mechanism triad as serine proteases Nerve poison used by Sadam Hussein, Aum Shinrikyo and Bashar alAssad Unit 1: protein Section 1.6: Enzyme Mechanisms 2 Thymidylate synthesis is a key reaction in the synthesis of deoxythymidylate (dTMP), a precursor of DNA synthesis. o The enzyme replaces a H atom with a methyl group (not easily complicated mechanism) o Key role in cell division important target for chemotherapy o THFA: tetrahydrofolic acid (a reduced form of the B vitamin folic acid) O O H H C H C CH 3 N C N C C dUMC C C dTMP N N O H O H dRibose5'P dRibose5'P dRibose5'P dRibose5'P 5 6 1 5-F-uracil 5-FdUMP 5Furacil is converted to 5FdUMP by a series of 4 enzymecatalyzed reactions (shown immediately above) that are in an existing pathway for deoxynucleotide synthesis. For each of these 4 enzymes, the F atom at the 5 position of uracil is perceived as a H atom; reaction proceeds the same whether uracil or 5 Furacil is the pyrimidine. This is because F is a very small atom, as is H. In addition, none of the reactions catalyzed by the first 4 enzymes involve the C5. It is only when the 5fluoro compound reaches thymidylate synthase that the presence of a F instead of a H at position 5 makes a difference They must be converted into the active compound before achieving the desired pharmacological effect. Unit 1: protein Mechanism for Thymidylate synthase 146 1) A base (S) in Cys of the enzyme attacks the C6 of uracil. A carbanion is formed around the C5 of uracil 2) The carbanion at C5 of uracil attacks the methylene group of folic acid, breaking a CN bond. 3) A second proton is extracted from uracil by base attack, breaking the other CNbond. The H comes off as H . F cannot form F , only F. Why Answer: F is very electronegative. 4) A hydride shift from C6 of tetrahydrofolate to the CH2 group on the C5 of uracil completes the reaction. H 5FdUMP as a suicide inhibitor mechanismbased inhibitor/Trojan horse inhibitor Highly specific and very effective in inhibiting thymidylate synthesis Binding of a substrate analogue to the substratebinding site of an enzyme with the formation of an irreversible complex via a covalent bind during the normal catalytic reaction