Biochem Review: Proteins, Electrophoresis, Membrane Proteins, Signal Transduction, Exams of Biochemistry

Download Biochem Review: Proteins, Electrophoresis, Membrane Proteins, Signal Transduction and more Exams Biochemistry in PDF only on Docsity! ACS BIOCHEMISTRY EXAM STUDY MATERIAL. Henderson-Hasselbach Equation - Correct answer pH = pKa + log ([A-] / [HA]) FMOC Chemical Synthesis - Correct answer Used in synthesis of a growing amino acid chain to a polystyrene bead. FMOC is used as a protecting group on the N-terminus. Salting Out (Purification) - Correct answer Changes soluble protein to solid precipitate. Protein precipitates when the charges on the protein match the charges in the solution. Size-Exclusion Chromatography - Correct answer Separates sample based on size with smaller molecules eluting later. Ion-Exchange Chromatography - Correct answer Separates sample based on charge. CM attracts +, DEAE attracts -. May have repulsion effect on like charges. Salt or acid used to remove stuck proteins. Hydrophobic/Reverse Phase Chromatography - Correct answer Beads are coated with a carbon chain. Hydrophobic proteins stick better. Elute with non-H-bonding solvent (acetonitrile). Affinity Chromatography - Correct answer Attach a ligand that binds a protein to a bead. Elute with harsh chemicals or similar ligand. SDS-PAGE - Correct answer Uses SDS. Gel is made from cross-linked polyacrylamide. Separates based off of mass with smaller molecules moving faster. Visualized with Coomassie blue. SDS - Correct answer Sodium dodecyl sulfate. Unfolds proteins and gives them uniform negative charge. Isoelectric Focusing - Correct answer Variation of gel electrophoresis where protein charge matters. Involves electrodes and pH gradient. Protein stops at their pI when neutral. FDNB (1-fluoro-2,3-dinitrobenzene) - Correct answer FDNB reacts with the N-terminus of the protein to produce a 2,4-dinitrophenol derivative that labels the first residue. Can repeat hydrolysis to determine sequential amino acids. DTT (dithiothreitol) - Correct answer Reduces disulfide bonds. Iodoacetate - Correct answer Adds carboxymethyl group on free -SH groups. Blocks disulfide bonding. Homologs - Correct answer Shares 25% identity with another gene Orthologs - Correct answer Similar genes in different organisms Paralogs - Correct answer Similar "paired" genes in the same organism Ramachandran Plot - Correct answer Shows favorable phi-psi angle combinations. 3 main "wells" for α-helices, ß-sheets, and left-handed α- helices. Glycine Ramachandran Plot - Correct answer Glycine can adopt more angles. (H's for R-group). Proline Ramachandran Plot - Correct answer Proline adopts fewer angles. Amino group is incorporated into a ring. α-helices - Correct answer Ala is common, Gly & Pro are not very common. Side-chain interactions every 3 or 4 residues. Turns once every 3.6 residues. Distance between backbones is 5.4Å. Helix Dipole - Correct answer Formed from added dipole moments of all hydrogen bonds in an α-helix. N-terminus is δ+ and C-terminus is δ-. ß-sheet - Correct answer Either parallel or anti-parallel. Often twisted to increase strength. T-State - Correct answer Heme is in high-spin state. H2O is bound to heme. R-State - Correct answer Heme is in low-spin state. O2 is bound to heme. O2 Binding Event - Correct answer O2 binds to T-state and changes the heme to R-state. Causes a 0.4Å movement of the iron. Hemoglobin Binding Curve - Correct answer 4 subunits present in hemoglobin that can be either T or R -state. Cooperative binding leads to a sigmoidal curve. Binding Cooperativity - Correct answer When one subunit of hemoglobin changes from T to R-state the other sites are more likely to change to R- state as well. Leads to sigmoidal graph. Homotropic Regulation of Binding - Correct answer Where a regulatory molecule is also the enzyme's substrate. Heterotropic Regulation of Binding - Correct answer Where an allosteric regulator is present that is not the enzyme's substrate. Hill Plot - Correct answer Turns sigmoid into straight lines. Slope = n (# of binding sites). Allows measurement of binding sites that are cooperative. pH and Binding Affinity (Bohr Affect) - Correct answer As [H+] increases, Histidine group in hemoglobin becomes more protonated and protein shifts to T-state. O2 binding affinity decreases. CO2 binding in Hemoglobin - Correct answer Forms carbonic acid that shifts hemoglobin to T-state. O2 binding affinity decreases. Used in the peripheral tissues. BPG (2,3-bisphosphoglycerate) - Correct answer Greatly reduces hemoglobin's affinity for O2 by binding allosterically. Stabilizes T-state. Transfer of O2 can improve because increased delivery in tissues can outweigh decreased binding in the lungs. Michaelis-Menton Equation - Correct answer V0 = (Vmax[S]) / (Km + [S]) Km in Michaelis-Menton - Correct answer Km = [S] when V0 = 0.5(Vmax) Michaelis-Menton Graph - Correct answer Lineweaver-Burke Graph - Correct answer Slope = Km/Vmax Y-intercept = 1/Vmax X-intercept = - 1/Km Lineweaver-Burke Equation - Correct answer Found by taking the reciprocal of the Michaelis-Menton Equation. Kcat - Correct answer Rate-limiting step in any enzyme-catalyzed reaction at saturation. Known as the "turn-over number". Kcat = Vmax/Et Chymotripsin - Correct answer Cleaves proteins on C-terminal endof Phe, Trp, and Tyr Competitive Inhibition Graph - Correct answer Slope changes by factor of α. Slope becomes αKm/Vmax. X-intercept becomes 1/αKm Y-intercept does not change. Vmax does not change. Uncompetitive Inhibition Graph - Correct answer Does not change slope. Changes Km and Vmax. Results in vertical shift up and down. Y-intercept becomes α'/Vmax X-intercept becomes -α'/Km Mixed Inhibition Graph - Correct answer Allosteric inhibitor that binds either E or ES. Pivot point is between X-intercept and Y-intercept. Non-Competitive Inhibition Graph - Correct answer Form of mixed inhibition where the pivot point is on the x-axis. Only happens when K1 is equal to K1'. Ionophore - Correct answer Hydrophobic molecule that binds to ions and carries them through cell membranes. Disrupts concentration gradients. ΔGtransport Equation - Correct answer ΔGtransport = RTln([S]out / [S]in) + ZFΔΨ Pyranose vs. Furanose - Correct answer Pyranose is a 6-membered ring. Furanose is a 5-membered ring. Mutarotation - Correct answer Conversion from α to ß forms of the sugar at the anomeric carbon. Anomeric Carbon - Correct answer Carbon that is cyclized. Always the same as the aldo or keto carbon in the linear form. α vs. ß sugars - Correct answer α form has -OR/OH group opposite from the -CH2OH group. ß form has -OR/OH group on the same side as the -CH2OH group. Starch - Correct answer Found in plants. D-glucose polysaccharide. "Amylose chain". Unbranched. Has reducing and non-reducing end. Amylose Chain - Correct answer Has α-1,4-linkages that produce a coiled helix similar to an α-helix. Has a reducing and non-reducing end. Amylopectin - Correct answer Has α-1,4-linkages. Has periodic α-1,6- linkages that cause branching. Branched every 24-30 residues. Has reducing and non-reducing end. Reducing Sugar - Correct answer Free aldehydes can reduce FeIII or CuIII. Aldehyde end is the "reducing" end. Glycogen - Correct answer Found in animals. Branched every 8-12 residues and compact. Used as storage of saccharides in animals. Cellulose - Correct answer Comes from plants. Poly D-glucose. Formed from ß-1,4-linkage. Form sheets due to equatorial -OH groups that H-bond with other chains. Chitin - Correct answer Homopolymer of N-acetyl-ß-D-glucosamine. Have ß-1,4-linkages. Found in lobsters, squid beaks, beetle shells, etc. Step 3 of Epinephrine Signal Transduction - Correct answer Activated α- subunit separates from ßɣ-complex and moves to adenylyl cyclase, activating it. Step 4 of Epinephrine Signal Transduction - Correct answer Adenylyl cyclase catalyzes the formation of cAMP from ATP Step 5 of Epinephrine Signal Transduction - Correct answer cAMP phosphorylates PKA, activating it Step 6 of Epinephrine Signal Transduction - Correct answer Phosphorylated PKA causes an enzyme cascade causing response to epinephrine Step 7 of Epinephrine Signal Transduction - Correct answer cAMP is degraded, reversing activation of PKA. α-subunit hydrolyzes GTP to GDP and becomes inactivated. cAMP - Correct answer Secondary messenger in GPCR signalling. Formed from ATP by adenylyl cyclase. Activates PKA (protein kinase A). AKAP - Correct answer Anchoring protein that binds to PKA, GPCR, and adenylyl cyclase. GAPs (GTPase activator proteins) - Correct answer Increase activity of GTPase activity in α-subunit of GPCR. ßARK and ßarr - Correct answer Used in desensitization. ßARK phosphorylates receptors and ßarr draws receptor into the cell via endocytosis RTKs (Receptor Tyrosine Kinases) - Correct answer Have tyrosine kinase activity that phosphorylates a tyrosine residue in target proteins INSR (Insulin Receptor Protein) - Correct answer Form of RTK. Catalytic domains undergo auto-phosphorylation. INSR signalling cascade - Correct answer INSR phosphorlates IRS-1 that causes a kinase cascade. INSR cross-talk - Correct answer INSR causes a kinase cascade that alters gene expression and phosphorlates ß-adrenergic receptor causing its endocytosis. NADH - Correct answer FADH2 - Correct answer Single-electron transfer NADPH - Correct answer FMN - Correct answer Single electron transfer. Step 1 of Glycolysis - Correct answer Glucose --> Glucose 6-phosphate. Uses hexokinase enzyme. ATP --> ADP Step 2 of Glycolysis - Correct answer Glucose 6-phosphate <--> Fructose 6-phosphate Uses phosphohexose isomerase enzyme. Step 3 of Glycolysis - Correct answer Fructose 6-phosphate --> Fructose 1,6-bisphosphate Uses PFK-1 (phosphofructokinase-1) enzyme. ATP --> ADP First Committed Step of Glycolysis - Correct answer Step 3 of Glycolysis. Fructose 6-Phosphate --> Fructose 1,6-bisphosphate. (PFK-1) Step 4 of Glycolysis - Correct answer Fructose 1,6-bisphosphate <--> dihydroxyacetone + glyceraldehyde 3-phosphate. Uses aldolase enzyme. Step 5 of Glycolysis - Correct answer Dihydroxyacetonephosphate <--> glyceraldehyde 3-phosphate Uses triose phosphate isomerase enzyme. Step 6 of Glycolysis - Correct answer Glyceraldehyde 3-Phosphate + Pi <--> 1,3-biphosphoglycerate. Uses G3P dehydrogenase enzyme. NAD+ <--> NADH First Energy Yielding Step of Glycolysis - Correct answer Step 6 of Glycolysis. G3P + Pi <--> 1,3-bisphosphoglycerate Step 7 of Glycolysis - Correct answer 1,3-bisphosphoglycerate + ADP <-- > 3-phosphoglycerate + ATP Uses phosphoglycerate kinase enzyme. First ATP Yielding Step of Glycolysis - Correct answer Step 7 of Glycolysis. 1,3-bisphosphoglycerate <--> 3-phosphoglycerate Step 8 of Glycolysis - Correct answer 3-phosphoglycerate <--> 2- phosphoglycerate Uses phosphoglycerate mutase enzyme. Step 9 of Glycolysis - Correct answer 2-phosphoglycerate <--> Phosphoenolpyruvate (PEP) Uses enolase enzyme. Dehydration reaction (loss of water). Step 10 of Glycolysis - Correct answer PEP + ADP --> Pyruvate + ATP Uses pyruvate kinase enzyme. ATP Consuming Steps of Glycolysis - Correct answer Step 1 and 3. Glucose --> Glucose 6-phosphate Fructose 6-phosphate --> Fructose 1,6-bisphosphate ATP Producing Steps of Glycolysis - Correct answer Steps 7 and 10. 1,3-bisphosphoglycerate <--> 3-phosphoglycerate PEP --> Pyruvate NADH Producing Step of Glycolysis - Correct answer Step 6 G3P <--> 1,3-bisphosphoglycerate Total Energy Produced by Glycolysis - Correct answer 2NADH + 4 ATP Lactic Acid Fermentation - Correct answer Pyruvate --> L-Lactate Uses citrate synthase enzyme H2O --> CoA Rate-limiting Step of the Citric Acid Cycle - Correct answer Step 1 Acetyl-Coa + Oxaloacetate --> Citrate Step 2 of the Citric Acid Cycle - Correct answer Citrate <--> Isocitrate Uses aconitase enzyme H2O <--> H2O Step 3 of the Citric Acid Cycle - Correct answer Isocitrate --> α- ketoglutarate Uses isocitrate dehydrogenase NAD(P)+ --> NAD(P)H + CO2 Step 4 of the Citric Acid Cycle - Correct answer α-ketoglutarate --> succinyl-CoA Uses α-ketoglutarate dehydrogenase complex CoA + NAD+ --> NADH + CO2 Step 5 of the Citric Acid Cycle - Correct answer Succinyl-CoA <--> Succinate Uses succinyl-CoA synthetase enzyme GDP + Pi <--> GTP + CoA Step 6 of the Citric Acid Cycle - Correct answer Succinate <--> Fumarate Uses succinate dehydrogenase FAD <--> FADH2 Step 7 of the Citric Acid Cycle - Correct answer Fumarate <--> L-Malate Uses fumarase enzyme 1) OH- 2) H+ --> Step 8 of the Citric Acid Cycle - Correct answer L-Malate <--> Oxaloacetate Uses malate dehydrogenase enzyme NAD+ <--> NADH Net Energy Gain of the Citric Acid Cycle - Correct answer 3 NADH, FADH2, and GTP NADH Producing Steps of the Citric Acid Cycle - Correct answer Steps 3, 4, and 8. Isocitrate --> α-ketoglutarate α-ketoglutarate --> Succinyl-CoA L-Malate --> Oxaloacetate FADH2 Producing Steps of the Citric Acid Cycle - Correct answer Step 6 Succinate <--> Fumarate Using succinate dehydrogenase enzyme GTP/ATP Producing Steps of the Citric Acid Cycle - Correct answer Step 5 Succinyl-CoA <--> Succinate Using succinyl-Coa synthetase CO2 Producing Steps of the Citric Acid Cycle - Correct answer Steps 3 and 4 Isocitrate --> α-ketoglutarate α-ketoglutarate --> Succinyl-CoA Biotin Structure - Correct answer Biotin Function - Correct answer Prosthetic group that serves as a CO2 carrier to separate active sites on an enzyme Regulation of the Citric Acid Cycle - Correct answer Regulation occurs at Steps 1, 2, 4, and 5. High energy molecules (ATP, Acetyl-CoA, NADH) inhibit while low-energy molecules (ADP, AMP, CoA, NAD+) activate these steps Glyoxylate Cycle - Correct answer Found in plants. Produces succinate from 2 acetyl-CoA. Allows oxaloacetate in the CAC to be used in gluconeogenesis. Uses 3 steps from the CAC. Different Steps in the Glyoxylate Cycle - Correct answer Isocitrate --> Glyoxylate (+ succinate) Uses isocitrate lyase Glyoxylate (+ acetyl-coA) --> Malate Uses malate synthase Step 1 of ß-oxidation - Correct answer Fatty acyl-CoA --> trans-Δ2-enoyl- CoA Uses acyl-CoA dehydrogenase FAD --> FADH2 Results in trans double-bond Step 2 of ß-oxidation - Correct answer trans-Δ2-enoyl-CoA (+ H2O) --> L- ß-hydroxy-acyl-CoA Uses enoyl-CoA hydratase TFP (Trifunctional Protein) - Correct answer Protein complex that catalyzes the last three reactions of ß-oxidation. Hetero-octamer (α4ß4) Step 3 of ß-oxidation - Correct answer L-ß-hydroxy-acyl-CoA --> ß- ketoacyl-CoA Uses ß-ketoactyl-CoA dehydrogenase NAD+ --> NAD+ Oxidation of Odd-numbered FA's - Correct answer Results in propionyl- CoA formation. Propionyl-CoA can be converted to succinyl-CoA and used in the CAC Step 4 of ß-oxidation - Correct answer ß-ketoacyl-CoA (+ CoA) --> Fatty acyl-Coa (shorter) Uses thiolase enzyme ß-oxidation in plants - Correct answer Electrons are passed directly to molecular oxygen releasing heat and H2O2 instead of the respiratory chain. ω-oxidation - Correct answer Similar to ß-oxidation but occurs simultaneously on both ends of the molecule. α-oxidation - Correct answer Form of oxidation of branched FA's. Produced propionyl-CoA that must be converted to succinyl-CoA for use in the CAC Complex I in the ETC - Correct answer Accepts two electrons from NADH via an FMN cofactor. Transfers 4H+ to Pside and 2H+ to Q Complex II in the ETC - Correct answer Succinate dehydrogenase. Accepts two electrons electrons from succinate via an FAD group. Q --> QH2 Complex III in the ETC - Correct answer Transfers two electrons from QH2 to cytochrome c via the Q-cycle. Transfers 4H+ to Pside. Complex IV in the ETC - Correct answer Transfers electrons from cytochrome c to O2. Four electrons are used to reduce on O2 into two H2O molecules. Transfers 4H+ to Pside Mitochondrial ATP Synthase - Correct answer Consists of F1 and F0 domains F1 Domain of Mitochondrial ATP Synthase - Correct answer Hexamer of 3 αß dimers. Catalyze ADP + Pi --> ATP via binding-change model F0 Domain of Mitochondrial ATP Synthase - Correct answer Causes rotation of γ-subunit via a half channel and H+ gradient Malate-Aspartate Shuttle - Correct answer Used to maintain gradient of NADH inside of the mitochondria. Involves transport of malate or aspartate; aspartate aminotransferase; and malate dehydrogenase. RuBisCo (Ribulose 1,5-bisphosphate carboxylase/oxygenase) - Correct answer Incorporates CO2 into ribulose 1,5-bisphosphate and cleaves the 6C intermediate into 2 3-phosphoglycerate. Stage 1 of the Calvin Cycle - Correct answer 3 ribulose 1-5-bisphosphate + 3 CO2 --> 6 3-phosphoglycerate. Catalyzed by rubisco Mg2+ in Rubisco - Correct answer Stabilizes negative charge in intermediate and held by Glu, Asp, and carbamoylated Lysine residue Rubisco Activase - Correct answer Triggers removal of ribulose 1,5- bisphosphate or 2-carboxyaarabinitol 1-phosphate so Lys can be carbamoylated. 2-carboxyarabinitol 1-phosphate - Correct answer inhibits carbamoylated rubisco. Synthesized in the dark and is broken down by rubisco activase or light. Stage 2 of the Calvin Cycle - Correct answer 3-phosphoglycerate --> glyceraldehyde 3-phosphate Requires ATP and NADPH Goes through 1,3-bisphosphoglycerate intermediate Stage 3 of the Calvin Cycle - Correct answer Glyceraldehyde 3-phosphate --> Ribulose 1,5-bisphosphate Requires 3 ATP and uses transketolase (TPP). Only uses 8 of the 9 G3P's produced. One G3P is used to make starch/sucrose. Energy Consumption of the Calvin Cycle - Correct answer 9 ATP molecules and 6 NADPH molecules for every 3 CO2 molecules that are fixated. Pi-Triose Phosphate Anti-porter - Correct answer Maintains Pi balance in cytosol/chloroplast due to G3P export to the cytosol. Also exports ATP and NADH to the cytosol. Oxygenase Activity in Rubisco - Correct answer O2 competes with CO2 and reacts to form 2-phosphoglycerate Glycolate Cycle - Correct answer Process of converting 2- phosphoglycerate to 3-phosphoglycerate in chloroplast, peroxisome, and mitochondria. C4 Plants - Correct answer Fix CO2 into PEP to form oxaloacetate (via PEP carboxykinase) that is then converted to malate that carries CO2 to rubisco. Bypasses O2 binding. CAM plants - Correct answer Fix CO2 into PEP to form oxaloacetate (via PEP carboxykinase) that is converted to malate at night and stored until the day time. Malonyl-CoA - Correct answer Formed from Acetyl-CoA and HCO3 via the Acetyl-CoA carboxylase (ACC). Serves as a regulator of FA catabolism and precursor in FA synthesis. ACC (acetyl-CoA carboxylase) Regulation - Correct answer Inhibited by PKA in glucagon chain and activated by pjhosphatase in INSR chain. FAS (Fatty-acid Synthetase) - Correct answer Catalyzes condensation, reduction, dehydration, and reduction of growing fatty acid chain. Requires activation by acetyl-CoA or malonyl-CoA Additional Cost of FAS in Eukaryotes - Correct answer Acetyl-CoA for lipid synthesis is made in mitochondria and must be transferred into the cytosol via citrate transporter. Costs 2 ATP. Cost of FAS in Eukaryotes - Correct answer 3 ATP's per 2 carbon unit added. Phosphatidic Acid - Correct answer Common precursor to TAGs and phospholipids. Consists of a glycerol 3-phosphate with two acyl groups that are attached via acyl transferases. TAGs (Triacylglycerols) - Correct answer Made from phosphatidic acid by removing phosphate with phosphatase and adding an acyl group with acyl transferase. Cholesterol Synthesis - Correct answer Synthesized from 15 acetyl-CoA through a number of intermediates. HMG-CoA Reductase - Correct answer Enzyme that converts ß-hydroxy- ß-methyl glutaryl-CoA to mevalonate in cholesterol metabolism. Regulation of HMG-CoA Reductase - Correct answer Inhibited by AMPK (AMP dependent kinase), glucagon, and oxysterol. Activated by insulin.