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Henderson-Hasselbach Equation - ANSWER pH = pKa + log ([A-] / [HA]) FMOC Chemical Synthesis - ANSWER Used in the 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) - ANSWER Changes soluble protein to solid precipitate. Protein precipitates when the charges on the protein match the charges in the solution. Size-Exclusion Chromatography - ANSWER Separates sample based on size with smaller molecules eluting later. Ion-Exchange Chromatography - ANSWER Separates sample based on charge. CM attracts +, DEAE attracts -. May have a repulsion effect on like charges. Salt or acid is used to remove stuck proteins. Hydrophobic/Reverse Phase Chromatography - ANSWER Beads are coated with a carbon chain. Hydrophobic proteins stick better. Elute with non-H- bonding solvent (acetonitrile). Affinity Chromatography - ANSWER Attach a ligand that binds a protein to a bead. Elute with harsh chemicals or similar ligand. SDS-PAGE - 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 - ANSWER Sodium dodecyl sulfate. Unfolds proteins and gives them uniform negative charge.
Isoelectric Focusing - 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) - 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) - ANSWER Reduces disulfide bonds. Iodoacetate - ANSWER Adds carboxymethyl group on free - SH groups. Blocks disulfide bonding. Homologs - ANSWER Shares 25% identity with another gene Orthologs - ANSWER Similar genes in different organisms Paralogs - ANSWER Similar "paired" genes in the same organism Ramachandran Plot - ANSWER Shows favorable phi-psi angle combinations. 3 main "wells" for α-helices, ß-sheets, and left-handed α-helices. Glycine Ramachandran Plot - ANSWER Glycine can adopt more angles. (H's for R-group). Proline Ramachandran Plot - ANSWER Proline adopts fewer angles. Amino group is incorporated into a ring. α-helices - 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 - 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 - ANSWER Either parallel or anti-parallel. Often twisted to increase strength.
Temperature Denaturation of Protein - ANSWER Midpoint of reaction is Tm. Cooperative Protein Folding - ANSWER Folding transition is sharp. More reversible. Folding Funnel - ANSWER Shows 3D version of 2D energy states. Lowest energy is stable protein. Rough funnel is less cooperative. Protein-Protein Interfaces - ANSWER "Core" and "fringe" of the interfaces. Core is more hydrophobic and is on the inside when interfaced. Fringe is more hydrophilic. π-π Ring Stacking - ANSWER Weird interaction where aromatic rings stack on each other in positive interaction. σ-hole - ANSWER Methyl group has area of diminished electron density in center; attracts electronegative groups Fe Binding of O2 - ANSWER Fe2+ binds to O2 reversible. Fe3+ has an additional + charge and binds to O2 irreversibly. Fe3+ rusts in O2 rich environments. Ka for Binding - ANSWER Ka = [PL] / [P][L] ϴ-value in Binding - ANSWER ϴ = (bound / total)x100% ϴ = [L] / ([L] + 1/Ka) Kd for binding - ANSWER Kd = [L] when 50% bound to protein. Kd = 1/Ka High-Spin Fe - ANSWER Electrons are "spread out" and result in larger atom. Low-Spin Fe - ANSWER Electrons are less "spread out" and are compacted by electron rich porphyrin ring. T-State - ANSWER Heme is in high-spin state. H2O is bound to heme.
R-State - ANSWER Heme is in low-spin state. O2 is bound to heme. O2 Binding Event - ANSWER O2 binds to T-state and changes the heme to R- state. Causes a 0.4Å movement of the iron. Hemoglobin Binding Curve - ANSWER 4 subunits present in hemoglobin that can be either T or R - state. Cooperative binding leads to a sigmoidal curve. Binding Cooperativity - 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 - ANSWER Where a regulatory molecule is also the enzyme's substrate. Heterotropic Regulation of Binding - ANSWER Where an allosteric regulator is present that is not the enzyme's substrate. Hill Plot - 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) - 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 - ANSWER Forms carbonic acid that shifts hemoglobin to T-state. O2 binding affinity decreases. Used in the peripheral tissues. BPG (2,3-bisphosphoglycerate) - 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 - ANSWER V0 = (Vmax[S]) / (Km + [S]) Km in Michaelis-Menton - ANSWER Km = [S] when V0 = 0.5(Vmax)
Pyranose vs. Furanose - ANSWER Pyranose is a 6-membered ring. Furanose is a 5-membered ring. Mutarotation - ANSWER Conversion from α to ß forms of the sugar at the anomeric carbon. Anomeric Carbon - ANSWER Carbon that is cyclized. Always the same as the aldo or keto carbon in the linear form. α vs. ß sugars - 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 - ANSWER Found in plants. D-glucose polysaccharide. "Amylose chain". Unbranched. Has reducing and non-reducing end. Amylose Chain - ANSWER Has α-1,4-linkages that produce a coiled helix similar to an α-helix. Has a reducing and non-reducing end. Amylopectin - 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 - ANSWER Free aldehydes can reduce FeIII or CuIII. Aldehyde end is the "reducing" end. Glycogen - ANSWER Found in animals. Branched every 8-12 residues and compact. Used as storage of saccharides in animals. Cellulose - 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 - ANSWER Homopolymer of N-acetyl-ß-D-glucosamine. Have ß-1,4- linkages. Found in lobsters, squid beaks, beetle shells, etc.
Glycoproteins - ANSWER Carbohydrates attached to a protein. Common outside of the cell. Attached at Ser, Thr, or Asn residues. Membrane Translayer Flip-Flop - ANSWER Typically slow, but can be sped up with Flippase, Floppase, or Scramblase. Membrance Fluidity - ANSWER Membrane must be fluid. Cis fats increase fluidity, trans fats decrease fluidity. Type I Integral Membrane Protein - ANSWER Membrane protein with C- terminus inside and N-terminus outside Type II Integral Membrane Protein - ANSWER Membrane protein with N- terminus inside and C-terminus outside Type III Integral Membrane Protein - ANSWER Membrane protein that contains connected protein helices Type IV Integral Membrane Protein - ANSWER Membrane protein that contains unconnected protein helices Bacteriorhodopsin - ANSWER Type III integral membrane protein with 7 connected helices. ß-Barrel Membrane Protein - ANSWER Can act as a large door. Whole proteins can fit inside. α-hemolysin - ANSWER Secreted as a monomer. Assembles to punch holes in membranes. Cardiolipin - ANSWER "Lipid staple" that ties two proteins (or complexes) together in a membrane. Formed from two phosphoglycerols. Hydrolysis of Nucleotides - ANSWER Base hydrolyzes RNA, but not DNA. DNA is stable in base because of 2' deoxy position. Chargaff's Rule - ANSWER Ratio of A:T and G:C are always equal or close to 1
Step 2 of Epinephrine Signal Transduction - ANSWER Hormone complex causes GDP bound to α-subunit to be replaced by GTP, activating α-subunit Step 3 of Epinephrine Signal Transduction - ANSWER Activated α-subunit separates from ßɣ-complex and moves to adenylyl cyclase, activating it. Step 4 of Epinephrine Signal Transduction - ANSWER Adenylyl cyclase catalyzes the formation of cAMP from ATP Step 5 of Epinephrine Signal Transduction - ANSWER cAMP phosphorylates PKA, activating it Step 6 of Epinephrine Signal Transduction - ANSWER Phosphorylated PKA causes an enzyme cascade causing response to epinephrine Step 7 of Epinephrine Signal Transduction - ANSWER cAMP is degraded, reversing activation of PKA. α-subunit hydrolyzes GTP to GDP and becomes inactivated. cAMP - ANSWER Secondary messenger in GPCR signalling. Formed from ATP by adenylyl cyclase. Activates PKA (protein kinase A). AKAP - ANSWER Anchoring protein that binds to PKA, GPCR, and adenylyl cyclase. GAPs (GTPase activator proteins) - ANSWER Increase activity of GTPase activity in α-subunit of GPCR. ßARK and ßarr - ANSWER Used in desensitization. ßARK phosphorylates receptors and ßarr draws receptor into the cell via endocytosis RTKs (Receptor Tyrosine Kinases) - ANSWER Have tyrosine kinase activity that phosphorylates a tyrosine residue in target proteins INSR (Insulin Receptor Protein) - ANSWER Form of RTK. Catalytic domains undergo auto-phosphorylation.
INSR signalling cascade - ANSWER INSR phosphorlates IRS-1 that causes a kinase cascade. INSR cross-talk - ANSWER INSR causes a kinase cascade that alters gene expression and phosphorlates ß-adrenergic receptor causing its endocytosis. NADH - ANSWER FADH2 - ANSWER Single-electron transfer NADPH - ANSWER FMN - ANSWER Single electron transfer. Step 1 of Glycolysis - ANSWER Glucose --> Glucose 6-phosphate. Uses hexokinase enzyme. ATP --> ADP Step 2 of Glycolysis - ANSWER Glucose 6-phosphate <--> Fructose 6- phosphate Uses phosphohexose isomerase enzyme. Step 3 of Glycolysis - ANSWER Fructose 6-phosphate --> Fructose 1,6- bisphosphate Uses PFK-1 (phosphofructokinase-1) enzyme. ATP --> ADP First Committed Step of Glycolysis - ANSWER Step 3 of Glycolysis. Fructose 6-Phosphate --> Fructose 1,6-bisphosphate. (PFK-1) Step 4 of Glycolysis - ANSWER Fructose 1,6-bisphosphate <--> dihydroxyacetone + glyceraldehyde 3-phosphate. Uses aldolase enzyme. Step 5 of Glycolysis - ANSWER Dihydroxyacetonephosphate <--> glyceraldehyde 3-phosphate Uses triose phosphate isomerase enzyme.
Lactic Acid Fermentation - ANSWER Pyruvate --> L-Lactate NADH --> NAD+ Regenerates NAD+ for use in glycolysis Ethanol Fermentation - ANSWER Pyruvate --> Acetalaldehyde --> Ethanol Uses pyruvate decarboxylase (TPP) and alcohol dehydrogenase. NADH --> CO2(TPP) + NAD+ TPP Cofactor Structure - ANSWER TPP Cofactor - ANSWER Common acetaldehyde carrier. Used in pyruvate decarboxylase, pyruvate dehydrogenase, α-ketoglutarate dehydrogenase, and transketolase Bypass Reactions in Gluconeogenesis - ANSWER Steps 1,3, and 10 must be bypassed. Gluconeogenic Bypass of Step 10 - ANSWER Bicarbonate + Pyruvate --> Oxaloacetate Pyruvate decarboxylate (biotin) ATP --> ADP Oxaloacetate --> PEP PEP carboxykinase GTP --> GDP + CO Gluconeogenic Bypass of Step 3 - ANSWER Fructose 1,6-bisphosphate + H2O --> Fructose 6-phosphate + Pi Uses FBPase-1 (coordinated with PFK-1) Gluconeogenic Bypass of Step 1 - ANSWER Glucose 6-phosphate + H2O --> Glucose + Pi Uses glucose 6-phosphatase. Cost of Gluconeogenesis - ANSWER 4 ATP, 2 GTP, and 2 NADH
Oxidative Pentose Phosphate Pathway - ANSWER Uses glucose 6-phosphate to produce 2 NADPH and ribose 5-phosphate used for biosynthesis Non-Oxidative Pentose Phosphate Pathway - ANSWER Regenerates glucose 6 - phosphate from ribose 5-phosphate. Uses transketolase and transaldolase enzymes. Transketolase - ANSWER Transfers a two-carbon keto group Transaldolase - ANSWER Transfers a three-carbon aldo group Enzyme Km and Substrate Concentration - ANSWER Most enzymes have a Km that is near the concentration of the substrate. Fructose 2,6-bisphosphate - ANSWER Not a glycolytic intermediate. Interconverts between fructose 2,6-bisphosphate and fructose 6-phosphate using PFK-2 and FBPase- 2 Regulation with fructose 2,6-bisphosphate - ANSWER Activates PFK- 1 encouraging glycolysis. Inhibits FBPase-1 discouraging gluconeogenesis Regulation of Pyruvate Kinase - ANSWER Inhibited by ATP, Acetyl-Coa, Alanine, long-chain FA's. PDH (Pyruvate Dehydrogenase Complex) - ANSWER Large complex that converts pyruvate + Coa --> Acetyl-Coa + CO Uses pyruvate dehydrogenase, dihydolipoyl transacetylase, and dihydrolipoyl dehydrogenase. Inhibited by phosphorylation by ATP. Pyruvate Dehydrogenase - ANSWER E1 domain of the PDH complex. Contains TPP cofactor. Releases CO2. Dihydrolipoyl Transacetylase - ANSWER E2 domain of the PDH complex. Catalyzes formation of Acetyl-CoA. Has oxidized, acyl, and reduced lipoyllysine.
Net Energy Gain of the Citric Acid Cycle - ANSWER 3 NADH, FADH2, and GTP NADH Producing Steps of the Citric Acid Cycle - ANSWER Steps 3, 4, and 8. Isocitrate --> α-ketoglutarate α-ketoglutarate --> Succinyl-CoA L-Malate --> Oxaloacetate FADH2 Producing Steps of the Citric Acid Cycle - ANSWER Step 6 Succinate <--> Fumarate Using succinate dehydrogenase enzyme GTP/ATP Producing Steps of the Citric Acid Cycle - ANSWER Step 5 Succinyl-CoA <--> Succinate Using succinyl-Coa synthetase CO2 Producing Steps of the Citric Acid Cycle - ANSWER Steps 3 and 4 Isocitrate --> α-ketoglutarate α-ketoglutarate --> Succinyl-CoA Biotin Structure - ANSWER Biotin Function - ANSWER Prosthetic group that serves as a CO2 carrier to separate active sites on an enzyme Regulation of the Citric Acid Cycle - 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 - 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 - ANSWER Isocitrate --> Glyoxylate (+ succinate) Uses isocitrate lyase
Glyoxylate (+ acetyl-coA) --> Malate Uses malate synthase Step 1 of ß-oxidation - ANSWER Fatty acyl-CoA --> trans-Δ2-enoyl-CoA Uses acyl-CoA dehydrogenase FAD --> FADH Results in trans double-bond Step 2 of ß-oxidation - ANSWER trans-Δ2-enoyl-CoA (+ H2O) --> L-ß- hydroxy-acyl-CoA Uses enoyl-CoA hydratase TFP (Trifunctional Protein) - ANSWER Protein complex that catalyzes the last three reactions of ß-oxidation. Hetero-octamer (α4ß4) Step 3 of ß-oxidation - ANSWER L-ß-hydroxy-acyl-CoA --> ß-ketoacyl-CoA Uses ß-ketoactyl-CoA dehydrogenase NAD+ --> NAD+ Oxidation of Odd-numbered FA's - ANSWER Results in propionyl-CoA formation. Propionyl-CoA can be converted to succinyl-CoA and used in the CAC Step 4 of ß-oxidation - ANSWER ß-ketoacyl-CoA (+ CoA) --> Fatty acyl-Coa (shorter) Uses thiolase enzyme ß-oxidation in plants - ANSWER Electrons are passed directly to molecular oxygen releasing heat and H2O2 instead of the respiratory chain. ω-oxidation - ANSWER Similar to ß-oxidation but occurs simultaneously on both ends of the molecule. α-oxidation - ANSWER Form of oxidation of branched FA's. Produced propionyl-CoA that must be converted to succinyl-CoA for use in the CAC
1) ATP --> ADP
- Aspartate --> AMP Step 3 of the Urea Cycle - ANSWER Arginosuccinate --> Argininine Uses arginosuccinase Produces fumarate byproduct Step 4 of the Urea Cycle - ANSWER Arginine --> Ornithine Uses arginase enzyme H2O --> Urea N-acetylglutamate - ANSWER Upregulates the production of carbamoyl phosphate and the urea cycle. Formed from acetyl-CoA and glutamate. PCR (Protein Chain Reaction) - ANSWER Process by which DNA is replicated. Has melting step, annealing step, replication step. pKa of Arginine R-group - ANSWER 12. pKa of Aspartate R-group - ANSWER 3. pKa of Cysteine R-group - ANSWER 8 pKa of Glutamate R-group - ANSWER 4 pKa of Histidine R-group - ANSWER 6. pKa of Lysine R-group - ANSWER 10. pKa of Tyrosne R-group - ANSWER 10 FAD Structure - ANSWER Q (Ubiquinone/Coenzyme Q) Structure - ANSWER Q (Ubiquinone/Coenzyme Q) Function - ANSWER Lipid soluble electron carrier. Carries 2 electrons with 2 H+.
ETC (Electron Transport Chain) - ANSWER Consists of 4 functional protein complexes. Complex I in the ETC - ANSWER Accepts two electrons from NADH via an FMN cofactor. Transfers 4H+ to Pside and 2H+ to Q Complex II in the ETC - ANSWER Succinate dehydrogenase. Accepts two electrons electrons from succinate via an FAD group. Q --> QH Complex III in the ETC - ANSWER Transfers two electrons from QH2 to cytochrome c via the Q-cycle. Transfers 4H+ to Pside. Complex IV in the ETC - 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 - ANSWER Consists of F1 and F0 domains F1 Domain of Mitochondrial ATP Synthase - ANSWER Hexamer of 3 αß dimers. Catalyze ADP + Pi --> ATP via binding-change model F0 Domain of Mitochondrial ATP Synthase - ANSWER Causes rotation of γ- subunit via a half channel and H+ gradient Malate-Aspartate Shuttle - 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) - ANSWER Incorporates CO2 into ribulose 1,5-bisphosphate and cleaves the 6C intermediate into 2 3-phosphoglycerate. Stage 1 of the Calvin Cycle - ANSWER 3 ribulose 1- 5 - bisphosphate + 3 CO2 -
6 3-phosphoglycerate. Catalyzed by rubisco Mg2+ in Rubisco - ANSWER Stabilizes negative charge in intermediate and held by Glu, Asp, and carbamoylated Lysine residue
