Biochemistry Exam#1: Lecture #1-#8
Biochemistry Exam#1: Lecture #1-#8 0280
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This 31 page Study Guide was uploaded by Denise Croote on Saturday January 23, 2016. The Study Guide belongs to 0280 at Brown University taught by Arthur Salomon in Spring 2016. Since its upload, it has received 89 views. For similar materials see Introductory Biochemistry in Biology at Brown University.
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Date Created: 01/23/16
Denise Croote Lecture One: Introduction to Amino Acids and Peptides ● we are composed of mostly water (70%), but also protein, nucleic acids, polysaccharides, and lipids ● amino acids: contain an amino group, carboxyl group, and R side chain ● most amino acids are chiral molecules. The LEnantiomer has the amino group on the left of the alpha carbon and the DEnantiomer has the amino group NOT on left ● chiral molecules rotate plane polarized light, mixtures containing equal amounts of L and D enantiomer are optically inactive ● All amino acids that occur in proteins have the L configurations ● categorizing website: http://www.sigmaaldrich.com/lifescience/metabolomics/learningcenter/aminoacid referencechart.html ● aromatic compounds absorb light, absorbance =log(incident light/light transmitted) or = log (1/transmittance) ○ example.) If 90% of incident light is absorbed after 1cm then log(100/10) = 1.0 Absorbance ● a zwitterion can act as a good acid or a good base, free amino acids are zwitterionic at neutral pH ● point 0 = charge of +1 ● pK1 = the COOH loses its proton. At this point half of the ions present have COOH and half have COO ● pI = no net electric charge on the molecule, here the removal of the first proton is complete and the removal of the second proton is just starting ● pI= ½ ( pK1 + pK2) ● pK2 = NH3 loses its proton. At this point half of the ions present have NH3+ and half have NH2 ● point 2= charge of 1 PEPTIDE BOND FORMATION (hydrolysis) ● oligopeptides strings of amino acids. When notated, always begin reading from the NH3 terminal ● peptide bonds are planar because the C=0 and N bond has partial double bond character (look at the resonance) the carbonyl oxygen has a partial negative charge and the N a partial positive charge ● proteases digest proteins to isolate peptides ● Mass Spectrometry tandem MS ○ protein is treated with a protease to hydrolyze it to shorter peptides ○ put into mass spec machine and the fragments are ionized and manipulated so that one of the several peptides produced by the cleavage emerges at the other end ○ peptide collided with a gas to make the molecules fragment at one of the peptide bonds ○ this break generates a charge and charges create peaks, each successive beak has one less amino acid than the peak ahead of it ○ the difference in mass from peak to peak identifies the amino acid that was lost Lecture Two: Protein Structure, Folding, and Stability ● primary structure → amino acid residues ● secondary structure → alpha helix, beta pleated sheet, beta turn ● tertiary structure → polypeptide chain ● quaternary structure → assembled subunits ● peptide bonds occur in the trans formation and have phi and psi angles. Phi φ is the angle between the alpha carbon and the alpha amino nitrogen and psi ψ is the angle of the bond between the alpha carbon and the carboxyl carbon ● when Phi φ and psi ψ are both 0 there is steric clash between the carboxyl of one peptide and the amino H of the other ● when Phi φ and psi ψ are 180 degrees the two peptide planes are parallel and in the fully extended trans formation ● The Alpha Helix (secondary structure) a. linear hydrogen bonds are the strongest b. 3.6 residues per turn, 13 atoms form the ring closed by a H bond c. Phi φ angles are 60 degrees and psi ψ angles are 45 to 50 degrees d. R groups project away from the core, R groups with H bond acceptors and donors are not found in helix b/c they compete with main chain H bonds and destabilize the helix (ex. Asp) e. entropic cost to put glycine in the helix is too high f. proline has a phi angle of 65 degrees instead of 60 degrees. This makes the backbone rigid and pro cannot form H bonds because it lacks the amide H atom. g. Ala, Met, Glu commonly found in the helix h. helix has a dipole moment so in order to neutralize the helix positive residues are found at the C terminus and negative residues are found at the N terminus ● The Antiparallel & Parallel B Sheet (secondary structure) a. Antiparallel have larger Phi φ and psi ψ angle (7A) vs parallel of (6A) b. Sheets are pleated so linear H bonds can be formed c. The alpha helix is continuous while the beta pleated sheet is not gray strands in between the green are other amino acids that make the sheets noncontinuous d. parallel sheets are weaker b/c they have weaker lateral bonds and H bond that are not exactly linear ● The Beta Turns (secondary structure) a. reverse the direction of the backbone, come in one direction and go out the same direction b. proline uses the cis conformation with a phi angle of 65 c. Type I has proline (more common) and type II has glycine ● Ramachandran Plot plots the frequency of angles at alpha carbons, greatest density on plot is usually around 180/ 180 for antiparallel sheets, 60 for alpha helices and 120 for parallel sheets. Very little density at 0 b/c those phi,psi angles are sterically unfavorable ● Supersecondary structures: elements such as helices, sheets, and turns that are arranged into motifs or folds with distinct functions, only about one hundred exist in nature even though there are an infinite number of protein motifs a. larger polypeptides are composed of somewhat independent domains linked by connecting bridges b. ex.) Coiled Coils myosin and keratin, two right handed helices are twisted into a left handed superhelix in a parallel orientation, only 3.5 residues per turn c. ex.) Collagen Triple Helix three parallel left handed helices wrapped around in a right hand sense with 3.3 residues per turn, every 3rd residue in the helix is glycine b/c anything else would be too bulky. Unlike a typical alpha helix, collagen has a lot of proline to stabilize its structure ● water molecules forcefully condense proteins by forming a cage around them, from there electrostatic forces, noncovalent forces, and covalent forces take over ● the native or most thermodynamically favorable conformation is not always the most stable conformation. If the protein is a catalyst you need flexibility, the most thermodynamically favorable conformation freezes the protein in one spot (no flexibility) ● Christian Anfinsen Experiment takes an active enzyme, denatures it w/ urea, waits to see if the primary structure can cause the protein to reform and regain function. Concluded that primary structure IS sufficient to determine the native conformation ● Levinthal paradox too many amino acid conformations for a protein to sample every one and find the most favorable. A protein folds stepwise, transitioning to progressively more stable and lower in energy conformations ● Folding Pathways: a. local secondary structure formation and collapse b. condensation of supersecondary structures c. compaction, molten globule formation ( another name for condensed form) d. conformational adjustment proline isomerization and disulfide bond formation e. subunit assembly ● Theories for protein folding: thermodynamic control says that the outcome of the folding is independent of the path and all proteins reach the global energy minimum vs. theory of kinetic control which says that the pathway determines the outcome and the outcome is a local energy minimum ● Chaperonins are barrel like proteins that provide an environment for other proteins to expose their hydrophobic sides while folding ● You can denature a protein with a high or low pH, a detergent to disrupt the hydrophobic interactions, high temperatures, and chaptropic agents that make more favorable interactions with all residues than water does ex. urea ● column chromatography separates proteins based on properties like size, charge ● SDS gel electrophoresis breaks proteins into monomers and units travel down a gel based on molecular weight, they are stained for visualization Lecture Three: Myoglobin, Hemoglobin, and Immunoglobulins ● some examples of secondary structures are B haripins, Psi loops ● domains are units of compact structure, folding, and function ● most domains are continuous along the structure ● Myoglobin consists of all alpha helices and has a prosthetic group attached (a group that is stably associated with a protein and contributes to its function) ○ heme = prophyrin + iron has a iron atom bound in the Fe2+ state. Fe2+ can simultaneously react with two O2 to make Fe3+ (which is bad because Fe3+ does not bind oxygen) ○ central iron atom has six bonds, 4 to nitrogen, 1 to oxygen, and 1 to histidine ○ when Fe2+ is nestled into the protein a nitrogen from a His residue occupies one of the coordination bonds so that Fe2+ cannot turn into Fe3+ ● free heme binds CO 25,000 times more tightly than 02 while myoglobin binds CO only 200 times more tightly than O2. ● B/c O2 is in a much higher concentration in the body it binds to myoglobin more frequently and only 1% of myoglobin is coordinated with CO ● Theta θ is the fraction of occupied binding sites in a protein = ( occupied sites/ occupied + unoccupied sites) ● a plot of theta vs [L] ligand gives a hyperbolic function (can be defined in partial pressures and concentrations) ● Kd (which is the dissociation constant) is the molar concentration of the ligand at which half of the available ligand binding sites are occupied ● Because of the high concentration of oxygen in the atmosphere, myoglobin is half saturated with oxygen with only 1.2% of atmospheric oxygen ● myoglobin binds too tightly to oxygen to transport it (higher affinity for oxygen and steeper curve than hemoglobin, hemoglobin is better for transport ● Hemoglobin has 2 alpha and 2 beta subunits, can bind 4 oxygens, is found in RBC, and has a similar tertiary structure to myoglobin (even though the aa sequences differ) ● Hb is a heterotetramer(protein containing four noncovalently bound subunits, wherein the subunits are not all identical) ● Have a T state (tense) where the oxygen is not bound, the Fe is pulled back out of the plane by the proximal His ● Have a R state (relaxed) where the O2 is bound to the heme causing it to be centered and pushed forward ● Affinity for oxygen is higher in the R state. Once one O2 binds the conformation of the entire hemoglobin subunit changes to make it bind O2 more tightly in the R state ● Referred to as cooperative binding, binding of first oxygen enhances binding of the remaining three ● During the transition to the R state the iron units in the unoccupied sites are pushed into the central plane and iron has a high affinity for oxygen ● do not want a high affinity R locked state in the tissues because the hemoglobin would not release the oxygen like it is supposed to, would not want too low of a T locked affinity in the lungs because then it would not pick up enough oxygen to bring to the tissues ● myoglobin has no cooperativity b/c it only has one subunit (nH=1). The binding of the last O2 to the Hb is no longer cooperative because all of the remaining few unoccupied sites are on different molecules. ● concerted theory says that all are either in the R state or the T state, sequential theory says the subunits change independently of each other ● Venous blood is more acidic than arterial blood, the more acidic the blood the more the T state is favored and the lower the affinity for oxygen (need a higher concentration of oxygen to achieve the same proportion of binding) ● BGP is a negative allosteric effector. BPG is negatively charged and is attracted to the positively charged His. BPG can only fit with the T tetramer and promotes the T state (reduces O2 affinity) ● When climbing mountains there is less oxygen available, we hyperventilate to get more oxygen, pH of blood goes up, start making more BPG , binds to His, induces T conformation, affinity for oxygen goes down, pick up less oxygen in the lungs, but you are better at dropping it off in the tissues, body’s method maintain efficiency ● Sickle Cell Anemia puts two sticky patches (valines) of the outside of the hemoglobin causing the RBCs to clump ● Heterozygotes for sickle cell has resistance against malaria ● Immunoglobulin IgG has conformation specific antigen binding sites (CDRs) ● SDS Gel Electrophoresis separates a protein into monomers, migration in the gel is a function of molecular weight ● Western Blot coat surface with antigens, incubate with a primary antibody, incubate with a secondary antibodyenzyme complex, add a substrate, look at a color specific antigen Lecture Four: Kinetics ● use kinetics to figure out more about a reaction mechanism, use kinetics to study the effects of substrates and inhibitors on enzymes to gain greater insight into their physiological function, quantitatively measure enzymatic activity to gain information about diseases ● ΔG´0 measures the standard free energy change at a pH of 7 when all concentrations of the solutes are 1M ● ΔG´0 = –RT ln K ´eq where Keq = [P]/[S]. When ΔG´0 is negative, the products have a lower free energy than the reactants and are present in a higher concentration ● a spontaneous (exergonic) reaction tells you the reaction will go but is does not tell you how fast ● The transition state is a short lived unstable state where the the species is equally as likely to go back to the reactants or forward to the products ● The activation energy for the reverse reaction is greater than the activation energy for the forward reaction if ΔG´0 is less than zero. More reactants will reach the transition state than products so P will accumulate ● transition states have high free energy because they have high entropy (substrate has to adopt an ordered conformation to change to product, more ordered means more entropy) and high enthalpy (may have to break covalent bonds, H bonds, and electrostatic interactions in the substrate to reach the product) ● The absolute height to the transition state determines the rate of the reaction whereas the difference in ΔG´0 for the reactants and products determines the thermodynamic outcome of the reaction. ● an enzyme uses an alternative, lower energy pathway and usually operates through a different reaction mechanism than the uncatalyzed pathway. ● an enzyme increases the rate but does NOT change the equilibrium of the reaction ● enzyme active sites are complementary to the transition state conformations of the substrates. They bind to the transition state intermediate and lower the activation energy ● very stable enzyme substrate complexes are not productive because they result in less species reaching the transition state and a slowed reaction rate ● The initial velocity, V0, is the rate of the reaction at the beginning of the steady state, after the equilibrium between E, S, and ES has been reached, but before a significant amount of substrate is consumed or product has been formed. ● V0 = k2[ES] ● the V0 formula is very similar to the formula for ligand binding ● called the MichaelisMenten Equation and it is used to describe the Km, Vmax, and Kcat for each characteristic enzymatic reaction ● Define: Km ≡ (k2 + k–1)/k1 Km is an apparent dissociation constant that correlates with the dissociation constant Kd. Because Kd = k–1/k1 when k2 is really small Km ≈ Kd, a low concentration of substrate is required to achieve the half Vmax if Km is really small ● [ES] =[Et][S] / Km + [S] ● Substitute V0/k2 for [ES] ● V0/k2 = [Et][S]/ Km + [S] ● rearrange to get V0 = k2[Et][S] / Km + [S] ● Km = [S] when V0 = ½ Vmax (derivation shown in text) ● When substrate concentration is much greater than Km, V0 = Vmax ● When substrate concentration is much smaller than Km, there is a linear dependence ● V max is the maximum velocity that can be achieved with a given total enzyme concentration. ● If all enzymes are bound to substrate than the enzyme is working at its maximal rate ● It measures how fast the reaction catalyzed by the enzyme can proceed once [ES] is formed ● Why is the michaelismenten formula useful? ● Why is it beneficial to use an equation that only applies to a small part of the reaction before a significant portion of the substrate is depleted? ○ A.) because when we define these conditions we can view them as a standard and compare different enzymes, or compare the activity of an enzyme in different experimental conditions (pH, temperature, inhibitor presence) ○ B.) because in vivo the product usually does not accumulate in many reactions (because it is used as the substrate for another reaction) and the substrate of the reaction is being constantly replenished so the conditions are actually similar to the initial conditions ● use this formula to determine Km instead of the hyperbolic curve . This is a linear curve with the slope being Km/Vmax and the y intercept being 1/Vmax ● If the values for an enzyme are being held constant, than the rate of formation = the rate of breakdown ● Catalytic constant or turnover number Kcat: number of molecules of substrate converted into product per molecule of enzyme at saturating substrate concentration (Vmax conditions) ● “measure of processitivity” ● Kcat is often thousands of magnitudes larger than Km for certain enzymes (km gives you the best estimate for real world performance) ● making the affinity for E and S larger doesn’t exactly speed up a reaction because when ES becomes more stable, the activation energy rises and the rate of the reaction slows ● When [S] is much less than Km, Vo depends on both the turnover number and the fraction of active sites occupied. (Vo is proportional to both Kcat and 1/Km) ● Catalytic efficiency: Kcat/Km is the rate constant for the conversion of E+S to E+P when [S] is much less than Km ● kcat/Km = V0/([S][Et]) so the higher the catalytic efficiency the greater the initial velocity ● Kcat/Km is limited by the rate at which E and S can diffuse together, catalytic perfection is more easily reached in solution ● an enzyme inactivator reacts irreversibly with an enzyme while an inhibitor reversibly binds ● competitive inhibitors binds only to E and competes with S for the active site, this has an effect of lowering the affinity for E to S (thus raising apparent Km remember Km parallels the dissociation constant) ○ a smaller fraction of active sites are occupied by substrates ○ the Km increases ○ Vmax is not affected ● uncompetitive inhibition binds only to ES and distorts the active site to prevent conversion of ES to E+P. SInce the inhibitor only binds ES it stimulates the formation of ES which decreases Km. ○ decreases Km ○ ESI complex is ineffective so Vmax is lowered ● mixed inhibition inhibitor binds to both E and ES ○ Vmax is lowered because concentration of [ES] is lowered at every substrate concentration ○ raising or lowering Km depends on the relative concentrations of the competitive and uncompetitive inhibitors ● a noncompetitive inhibitor is when the inhibitor binds equally well to E and ES so only Vmax is effected and not Km Lecture Five: Enzyme Catalysis and Chymotrypsin Bisubstrate Reactions ● bisubstrate reactions involved two substrates ● some reactions involve a tertiary complex. For an ordered reaction E1 will form with S1 and then E1S1 will form with E2 to make the tertiary complex E1S1S2. For a random reaction E1 will form with either S1 or S2 first ● If you increase the concentration of S2 than you increase Vmax for these reactions with tertiary complexes because you increase the rate of production because both substrates must bind to the enzyme in order to make the product. ● With tertiary complexes increasing the concentration of S2 decreases the amount of S1 necessary to reach the half maximum rate, this then decreases Km ● A ping pong reaction does not involve a tertiary complex, in this type of reaction E reacts with S1 to give P1, P1 is released and then E' reacts with S2 to make P2 ● in a pinpong reaction increasing the concentration of S2 also increases Vmax but it increases Km because it increases the presence of the enzyme that binds S1, meaning more S1 in needed to fill the binding sites Enzyme Activity ● enzyme binding sites are complementary to substrates ● enzyme binding grooves often shield substrate from the outside environment when bound ● enzymes are usually not complementary to the reactant substrate because that would make them more stable and would increase the energy it would take to reach the transition state ● ● enzymes are complementary to the transition state to make it more stable and decrease its activation energy ● catalytic antibodies have animals produce antibodies that will bind and stabilize the transition state and increase the rate of the reaction by decreasing delta G ● we classify enzymes based on what they catalyze ○ ex.) oxidoreductases are involved in the transfer or electrons, hydrolases are involved in the transfer of functional groups to water ● Modes of catalyzing a reaction include ○ release of binding energy: forming H bonds, electrostatic interactions, and hydrophobic interactions with the transition state ○ dissolving: replacing H bonds between substrate and water ○ positioning of substrate (especially important in bisubstrate reactions) ○ proton transfer ○ covalent catalysis: bond cleavage and formation with the enzyme ○ metal ion catalysis: ionic interaction, or oxidation/reduction reactions ● enzymes need coenzymes and cofactors to function ○ coenzymes can transfer gases (CO2), amino groups, electrons, and hydride ions ○ cofactors (metal ions like Cu2+, Fe2+ ect) are involved in oxidation/reduction reactions, electrostatic interactions, and acidbase catalysis ● an apoenzyme (the protein or polypeptide part of the enzyme) plus a cofactor or a coenzyme make up a holoenzyme ● enzymes entropically enhance the reaction by bringing the reactants closer together in the correct geometry ● if you are trying to condense a reaction, it is best to hide the reaction in a grove where the reactants are not exposed to water. if you are doing a hydrolysis reaction then you will want to expose the reaction to water ● Acid Base Catalysis: sometimes you need to stabalize the transition state and move a proton so that the intermediates don't immediately break down ○ a specific acids base reaction passes to and from H20 molecules ○ a general acid base reaction passes protons between any weak acid and base ○ CHYMOTRYPSIN REACTION MECHANISM ● chymotrypsin is a peptidase meaning it cleaves a reaction at a certain point ● Steps of the Reaction: 1.) when the substrate binds, the side chain of the residue located next to the peptide bond that is going to be cleaved nestles into the hydrophobic pocket of the enzyme. The hydrophobic pocket only accepts aromatic residues. ○ 2.) Asp in the picture above tugs on the His, positioning it so that the lone pairs on the N on Histidine grab and H from the OH group on serine. Once the N rips the H off the O is electronegative and attacks the carbonyl group of the substrate. This forms the ester bond (acyl intermediate) This pushes the electrons from the double bonded carbonyl up onto the oxygen. This oxygen is stabilized by hydrogens in the oxyanion hole ○ 3.) The current tetrahedral is unstable, the oxygen with the negative charge reforms the carbonyl group and the nitrogen that is apart of the amino group gets kicked out, breaking the peptide bond. The amino leaving group gets protonated by His, to make it leave faster. (Product One leaves) ○ 4.) next a water molecule gets deprotonated to make a OH anion. The anion attacks the acyl enzyme linkage to generate a second tetrahedral intermediate with the carbonyl oxygen again in the oxyanion hole. This tetrahedral is unstable and collapses kicking out the O with the serine attacked. The His donates a proton to the serine to stabilize it. The product now has a carboxylic acid. ○ this product then leaves the active site to regenerate the free enzyme ● the Catalytic Triad that makes this reaction possible is the Ser, His, and Asp residues ● OVERALL Review: ○ substrate nestles in hydrophobic pocket and covalently bonds to enzyme ○ enzyme substrate complex formed ○ oxyanion hole forms H bonds with the transition state ○ break the peptide bond ○ break the ester linkage ○ reform the enzyme ● Chymotrypsin speeds up the reaction during the...... ○ acylation phase (cleavage of the peptide bond involves a covalent ester linkage between Ser and the substrate) ○ oxyanion hole (a transient negative oxygen is stabilized by hydrogen bonds from peptide bonds of the enzyme's active site) ○ Hydrolysis of the ester linkage is accompanied by general acid base catalysis of the His to the Ser (His giving H to Ser) ● The pH of the reaction can affect Km (productive substrate binding) which in turn can effect Kcat (the efficiency of the enzyme binding) ○ for ex.) if the pH of the reaction is 7 then His gets protonated, which makes it less likely to steal the hydrogen from the OH on the serine in step 2 ● the mechanism was determined by experimental evidence. Used DIFP (an irreversible inhibitor) that binds to Ser. Used TPCK to irreversibly alkylate (add an alkyl group) the reactive N on His (blocking these affected function and showed that they were important) ● Used the pNitophenylacetate molecule to figure out which part of the reaction was the rate limiting step. Found that the first conversion to break the peptide bond occurs quickly (quick increase in colored molecule) but the second step to break the acyl unit was characterized by a slow steady rate of formation, indicating it was the rate limiting ● proteases are defined by the attacking residue, this reaction would be considered a serine protease ● you do not want proteases to be active all of the time because they would just cut up the proteins before they left the ribosome, you want the to only be active at certain times ● Zymogens are highly active an require a specific protein inhibitor, their activation is irreversible ● in an enzyme cascade, each enzyme will activate the next, which will lead to a signal amplification ex.) could happen rapidly in the case where you need your blood to clot ● Allosteric enzymes are enzymes that once bound by an modulator at its allosteric site (site other than the active site), will change their binding affinity at another location ○ homotropic allosteric regulators are when the substrate is the modulator as well as the target for the enzyme ■ they show sigmoid kinetic behavior which indicates cooperative binding (non Michaelis Menton kinetics) ○ heterotropic allosteric enzymes are those that have different substrate and modulating molecules ● feedback inhibition usually involved enzymes that are heterotropically regulated Lecture Six: Lipids ● saturated fatty acids pack nicely together (strong hydrophobic bonds), unsaturated fatty acids have double bonds that make kinks and do not pack nicely ● most unsaturated fatty acids have double bonds in the cis formation, this increases the fluidity of the membrane by introducing space ● in the notation 12:0 the 12 represents the number of carbon bonds and 0 represents the number of double bonds. ○ something like 16:1 delta 9 means there are 16 carbons, one double bond, and the double bond is on the 9th carbon (start counting from the carboxylic acid end) ● saturated fatty acids have higher melting points than unsaturated fatty acids ● WAXES: esters of long chain fatty acids with long chain alcohols (Fatty acid +Fatty Alcohol) ● ● TRIACYLGLYCEROLS: three fatty acids in ester linkages with a glycerol, these are storage lipids ● lipids are involved in membranes and membranes separate the inside and the outside of the cell, they also compartmentalize cells ● fatty acids have hydrophillic heads and hydrophobic tails, when you put fatty acids in H20 their tails are hydrophpbic and do not like the water environment. When you cluster lipids together, fewer fatty acids are exposed to the water environment (favorable) ● GLYCEROPHOSPHOLIPIDS: (phosphoglycerides) fatty acid tails with a glycerol 3 phosphate head ● GLYCEROGLYCOLIPIDS: attached a sugar molecule to the head. Still has fatty acid tails. specialized for energy transduction ● ● phospholipases cleave specific phospholipid bonds. They cut between the c=o and the o. A lysophospholipid has one fatty acid removed ● ETHER LIPIDS ● ● the Archaea contain mostly ether lipids and they lack fatty acids. One end is a phospholipid and the other is a glycolipid. They are capable of withstanding extreme temperatures. A typical bilayer has hydrophobic tails that oppose and are noncovalently bonded to each other. The Archaea have diphytanyl groups , these groups have stronger covalent bonds to each other an account for its ability to withstand extreme temperatures. ● SPHINGOLIPIDS have an L shaped sphongosine with a fatty acid tail, they do not have an ester bond like most lipids, they have trans double bonds and are commonly found in the brain ● ● STEROLS a series of fused rings, they nestle into the kinks of the membrane, they increase fluidity, they are derived from cholesterol, they lubricate the membrane ● sterols are membrane plasticizers. They interfere with the motion of the fatty acid side chains and prevent highly ordered packing (to increase fluidity and lessen the abruptness of the transition) ● in the paracrystalline state the fatty acids are ordered (unfavorable) ● ● Vitamins are lipids as well. The fat soluble vitamins ADEK are made from lipids. They are isoprenoid compounds ● Lipids diffuse rapidly within the plane of a membrane, but very slowly across membranes. A lipid can circumnavigate an ecoli cell in 2 seconds. ● It takes a long time for a lipid to make it from the outer section of the bilayer to the inner section of the bilayer ○ Flippase uses ATP to speed up a lipid going from outside to inside ○ Floppase uses ATP to speed up a lipid going from inside to outside ○ Scramblase is used to interchange two lipids (one goes inside the other goes outside) ● lipids are distributed asymmetrically within membranes. Sphingomyelin tends to be on the outside while phosphatidylinositol tends to be on the inside ● we want a mixture of lipid types in the membrane so we do not get a catastrophic change ● the composition of membranes surrounding different structures differs, plasma membranes have more cholsterol while nuclear membranes have very little ● membrane fluidity is physiologically regulated. As temperature increases the ratio of unsaturated to saturated fatty acids decreases (meaning more saturated compounds are put into the membrane) this is so that you have less kinks and stronger hydrophobic bonds ● TRANSPORT across the membrane ○ transport molecules shuttle ions and nutrients across the membrane ○ only certain small and nonpolar molecules (like CO2) do not require a transporter ○ to cross the membrane particles can use simple diffusion (nonpolar compounds down a concentration gradient), facilitated diffusion (down an electrochemical gradient voltage difference), or through active transport (against the electrochemical gradient and driven by ATP) ○ tranporters form noncovalent interavtions with the dehydrated solute, replacing the H bonds of the water shell, they maintain this interaction through the transmembrane passage ○ receptor enzymes function when ligands bind extracellulary, an intracellular substrate is then cued to make its product ○ gated ion channels allow for transport of specific ions ○ aquiporins facilitate the transport of water molecules only ● uniport transporters transport one molecule down its concentration gradient ● symport transporters move two molecules in the same direction across the membrane, one moves by facilitated diffusion down the electrochemical gradient and the other moves against the concentration gradient by active transport ● antiport transporters two molecules move in opposite directions in secondary active transport, one moves along its electrochemical gradient and the other against its electrochemical gradient ● ● peripheral membrane proteins are more loosely attached to the membrane, they can be removed by changes in pH, they are associated with membrane proteins through electrostatic interactions ● integral proteins are ingrained in the membrane or covalently attached to a lipid, they are harder to remove and require detergents to break up the hydrophobic interactions ● integral proteins have helices that cross the membrane (typically 6 to 7 turns) ● During glucose transport glucose binds to a receptor, is engulfed, and pumped out the other side to regenerate the opening for more binding (seesaw motion) ● glucose transport uses and amphipathic helix (has hydrophobic and hydrophillic components) ○ these transporters are specific and can be saturated and therefore have a hyperbolic curve with a Vmax and a Km where half of the binding sites are occupied by a substrate ○ hydrophobic plots can predict the membrane spanning regions ● The membrane is a fluid mosaic model. The proteins are like buoys bobbing up and down in the bilayer. You can visualize them by tagging them with fluorescent labeled antibodies ● an erythrocyte chloride bicarbonate exchanger is used to convert CO2 into another form that is more soluble in the plasma ● Active Transport: ○ primary active transport: solute accumulation (uphill) is coupled to an exergonic chemical reaction ■ Example: Na/K transport enzyme transporter binds 3Na+ inside the cell, shuttles them to the outside, releases them, picks of 2K+ and shuttle them to the inside ○ secondary active transport: solute accumulation (uphill) is couple to the exergonic flow of a different solute ■ Example: glucose transport of driven by ion gradients into intestinal epithelial cells ○ when the solute is an ion and its movement is not accompanied by a counter ion, an electrical potential is produced. This electrical potential affects the energetics of the electrogenic transport. ○ normally Δ Gt = RT ln(C 2 / C1 ) but when the solute is charged we need to incorporate the electrical potential difference so.... Δ Gt = RT ln( C2 /C1 ) + ZF Δψ this means that at equilibrium C2 doesnt necessarily equal C1 because we need to incorporate charge Lecture Seven: Carbohydrates and Introduction to Metabolism Active transport requires energy to push a solute against its concentration gradient, primary active transport couples the uphill battle to an exergonic ATP hydrolysis while secondary active transport couples the uphill battle with an exergonic downhill flow of a different solute The sodium/potassium pump is an example of primary active transport, 3Na+ move out and 2K+ move in setting up a negative voltage across the membrane (requires ATP because each are pushed against their concentration gradient) Glucose uses secondary active transport to relocate into the epithelial cells Glucose is polar, but uncharged so its transport does not affect the voltage of the membrane. When an ion is transported without the presence of a counter ion, an electric potential is produced. Electrochemical potential gradients set up by ion moving change the equilibrium so that ΔGt = RT ln(C2 /C1) + ZFΔ Ψ. (z= charge, F= faraday constant, psi = electrical potential). Here at equilibrium C1 does not have to equal C2 Monosaccharides have a ketone on C1 or C2 Monosaccharides are chiral and L/D notation is determined by the carbon farthest from the carbonyl group All sugars are D sugars D-aldoses have the carbonyl as an aldehyde (on the end ex.= glucose) D ketoses have a ketone instead of an aldehyde (ex.= fructose) Epimers: two sugars that only differ in configuration around one carbon atom Hemiacetal = Aldehyde + Alcohol Hemiketals = Ketone + Alcohol These have a hydroxyl and an ether group on the same carbon Most monosaccharides have an aldehyde or ketone and a hydroxyl group on their structure so they are able to react to form hemiacetals or hemiketals (leading to cyclic monosaccharides) The anomeric carbon is bonded to two oxygens Most prefer the Beta D Glucopyranose over the alpha D glucopyranose formation because the beta maximizes equatorial side chains If the hydroxyl group on the anomeric carbon is trans to the terminal CH2OH the sugar is alpha, if it is cis the sugar is beta Alpha and Beta anomers have different physical properties and rotate plane polarized light differently. The rotation between them is called mutarotation and they need not end up in a 1:1 ratio Sugars can cyclize to make either a five member or a six member ring Sugars are easily oxidized to acids, oxidation of the aldehyde group transfers two electrons elsewhere (therefore sugars are reducing agents) Oxidizing glucose gives you D-Gluconate and D-Gluconate has a tendency to form the lactone to lower the energy of the product Hemiacetals and hemiketals can be esterified (add another ROH group) so that an acetal or ketal is formed, these have carbons bonded to two OR groups and are much more stable You can form a disaccharide when you react a hemiacetal from one sugar with an alcohol group from another. This is a condensation reaction. You originally had two hemiacetals exposed but after the disaccharide has formed you only have one, so it is more stable Reducing sugars have a free anomeric carbon while non-reducing sugars do not have a free anomeric carbon. Reducing sugars are named ending with the sugar that has the reducing carbon (the hemiacetal last) Non-reducing sugars can be names from either end Polysaccharies: o Starch: repeating D –glucose units Connected by alpha 1,4 linked D glucose units, they have an angle to their structure that causes the start molecule to form a loose helix o Cellulose: repeating D –glucose units Connected by a beta 1,4 link, are flipped 180° o o Glycogen, cellulose, chitin, peptidoglycans Better to store as polysaccharides because they are easier to dissolve, solubility is parts per whole and you would rather have one unit than a lot of monomers, decreasing the parts decreases the osmolarity Advantage to having only one reducing end is that you have replaced your unstable hemiacetals with stable acetals Advantage to having many non-reducing ends Lecture Eight: Glycolysis ATP hydrolysis can output energy, but since the Pi that comes off is the only product with a resonance, it does not output as much energy as a hydrolysis of a reactant that forms several products with resonance. Hydrolysis of phosphocreatine can output energy as well, the phosphoryl group transfer can drive the formation of ATP from ADP Oxidoreductases are special enzymes that transfer electrons (hydride ions or H atoms) Electrons are transferred in redox reactions, there will always be an electron acceptor and an electron donor Reduction potentials for the redox half reactions can be calculated (∆G’ º) Oxygen is a strong oxidant, meaning that it is good at stealing/gaining electrons) The standard reduction potentials E are relative to the E for the hydrogen atom You measure the standard reduction potential of a redox pair by using the set up to the right. One of the beakers has H in it. If what is in the other beaker has a stronger affinity for electrons, the voltage will be positive, if what is in the other beaker has a lower affinity for electrons the voltage will be negative In redox reactions in the body no electrons are allowed to be left over (all electrons must be transferred somewhere) ∆E rxn = ∆E (molecule gaining electrons) - ∆E (molecule losing electrons) The oxidation state around a carbon atom is the number of electrons around that atom, if connected to other atoms, the electrons will reside on the atom that is more electronegative ∆G = –nF∆E ... this shows that a positive ∆E is associated with an exergonic negative ∆G Electrons are transferred by cofactors acting as electron carriers, pyridine nucleotides and flavin nucleotides are important, their reduced forms are strong reducing agents (electron donators) Need cofactors because you do not want to just release the energy you created as heat, you need to harness it with electron transfer When you reduce NADP NADPH it is not favorable because the addition of the electrons breaks the aromaticity of the compound FMN, FAD FMNH FADH 2, 2 the reduction also disrupts aromaticity, making these compounds have a low affinity for electrons There are several ways to transfer electrons in reactions: o Directly as electrons (Fe2+ + Cu2+ ↔ Fe3+ + Cu+) o As hydrogen atoms (AH2 + B ↔ A + BH2) o As a hydride ion :H— (CH3CH3 + NAD+ ↔ CH2=CH2 + NADH + H+) o Through direct combination with oxygen (RCH3 + 1/2 O2 ↔ RCH2OH) Catabolism = breaking down, take energy containing nutrients like carbs, fats, and proteins and break them down into depleted CO , H O, and NH Give electrons to 2 2 3. anabolic pathways to build things. Cellular Respiration: 1.) Glycolysis splits the sugars and partially oxidizes the products, generates ATP, and reduces cofactors 2.) Citric Acid Cycle completely oxidizes the products of glycolysis to 2O and generates reduced cofactors as well as ATP 3.) Respiratory re-oxidation of the reduced cofactors is coupled to the synthesis of a large amount of ATP Glycolysis the point of glycolysis is to break glucose into two 3 carbon units the average oxidation state for each of the carbons is 4 electrons oxygen has a higher affinity for electrons than glucose and the electron carriers (NADH), therefore the transfer of electrons to these molecules is energetically favorable PART ONE: Preparatory Phase 1. Formation of glucose-6 phosphate…….enzyme: hexokinase gives the glucose molecule a (-) charge, (-) charge prevents it from diffusing across the membrane Addition of the phosphate destabilizes the glucose, further promoting catabolism. ** Addition REQUIRES ATP 2. Isomerization from Glucose 6-Phosphate to Fructose 6- phosphate Use the enzyme phosophohexose isomerase to open the 6 member glucose ring (an aldose) and change it to a 5 membered fructose ring ( a ketose) 3. Second phosphorylation, phosphorylate Fructose 6-phosphate Enzyme: phosphofructokinase (PFK) Reaction speed limited by the presence of this enzyme ** Addition REQUIRES ATP STEPS 1&3 – are allosterically regulated. The product (glucose-6-phosphate) binds to hexokinase to shut it off (step 1) and ATP binds to phosphofructokinase to shut that off (step 3) 4. Adol cleavage ….. enzyme: aldolase Splitting fructose 1,6 biphosphate into dihydroxyacetone phosphate and glyceraldehydes-3 phosphate This reaction is highly endergonic but is driven forward because a fast acting enzyme continually removes the product 5. Isomerization of dihydroxyacetone phosphate to glyceraldehydes-3 phosphate Enzyme: triose phosphate isomerase Going from a ketose to an aldose Also endergonic but later steps pull the reaction forward TiM is a good enzyme because it is isomerizes very quickly and substrate and enzyme bind very quickly upon dissolution of the substrate ** END OF PREPRATORY PHASE PART TWO: Payoff Phase 6. Phosphorylation of glyceraldehydes 3- phosphate Uses inorganic phosphate and enzyme glyceraldehydes 3-phosphate dehydrogenase 7. Formation of ATP from 1,3 Bisphosphoglycerate Enzyme: phosphoglycerate kinase Large negative ∆G (- ∆G) used to convert a lot of ADP to ATP This is substrate level phosphorylation because the 1,3 bisphosphoglycerate is a substrate with high phosphoryl transfer potential The reaction is so exothermic because 3-Phosphoglycerate is a strong acid, quickly loses the H from its OH group and then leave two oxygens to resonate a negative charge 8. 3-Phosphoglycerate 2-Phosphoglycerate 9. Phosphoglycerate Phosphoenolypyruvate 10. Second substrate level phosphorylation Enzyme: Pyruvate kinase Loss of phosphate from Phosoenolpyruvate (PEP) is highly exothermic because PEP is a high energy compound the product can tautomerize from enol for to keto form (keto very stable) Releases ΔG' º = –61.9 kJ/mol, half is captured in the formation of ATP and the other half drives the reaction forward SUMMARY: Glucose is converted to two triose phosphate molecules, at the expense of the hydrolysis of two ATP molecules. Two triose phosphate molecules are converted to pyruvate, regenerating the initial 2 ATP used and generating 2 more ATP plus 2 NADH for each molecule of glucose Even though some of the steps are endergonic (especially when you break the C-C bond) the net reaction is exergonic Glycolysis will completely halt if NADH is not re-oxidized to NAD+, you need to constantly regenerate these NAD molecules After pyruvate is generated you can put the electrons back on pyruvate to regenerate the NAD and produce a lactate. (this is lactic acid buildup in your muscles – done in the absence of oxygen) This does not produce a lot of energy compared to aerobic respiration so the rate of anaerobic respiration is much faster to keep up with the cellular energy demands Yeast take the electrons and put them back on pyruvate to make ethanol Glucose can be utilized in many ways…. o As glycogen, starch, and sucrose for storage o Used for oxidation via glycolysis to make pyruvate o Used to synthesize structural polymers (cell wall, polysaccharides) o Used for oxidation via the pentose phosphate pathway to make Ribose 5- phosphate and NADH Pentose Phosphate Pathway: o Starting material: glucose 6 phosphate o Products: ribose 5 phosphate (which are used to make nucleotides, DNA, and RNA), NADH, CO 2 o This reaction can recycle and take five glucose 6 phosphates and turn them into six ribose 5 phosphates Feeder Pathway o Takes fructose and catabolize it to either a lipid product or a glucose product o Turns galactose (inherently less stable than glucose because the side groups are not all equatorial), attaches a UDP, detaches the UDP and ends with glucose 6 phosphate
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