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ACS BIOCHEMISTRY EXAM / 270 +
QUESTIONS AND CORRECT
VERIFIED ANSWERS BY EXPERTS
2025 LATEST UPDATE GRADED A+.
ns. 3 main
- Henderson-Hasselbach Equation ANSWER pH = pKa + log ([A-] / [HA])
- FMOC Chemical Synthesis 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) 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 at- tracts +, DEAE attracts - . May have repulsion effect on like charges. Salt or acid used to remove stuck proteins.
- Hydrophobic/Reverse Phase Chromatography ANSWER Beads are coated with a car- bon 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 combinatio "wells" for ±h- elices, ß-sheets, and left-handed ±h-elices.
- Glycine Ramachandran Plot ANSWER Glycine can adopt more angles. (H's for R-group).
- 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.
hat can be
- ÀR-Ài n g Stacking ANSWER Weird interaction where aromatic rings stack on each other in positive interaction.
- Ãh-ole 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 O irreversibly. Fe3+ rusts in O2 rich environments.
- Ka for Binding ANSWER Ka = [PL] / [P][L]
- ôv-alue in Binding ANSWER ô = b(ound / total)x100% ô =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 t 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 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.
sites). Allows
69. Mixed Inhibition Graph ANSWER Allosteric inhibitor that binds either E o Pivot point is r ES.
between X-intercept and Y-intercept.
- Non-Competitive Inhibition Graph ANSWER Form of mixed inhibition where the pivot point is on the x- axis. Only happens when K1 is equal to K1'.
- Ionophore ANSWER Hydrophobic molecule that binds to ions and carries them through cell membranes. Disrupts concentration gradients.
- ” Gtransport Equation ANSWER ”Gtransport = RTln([S]out / [S]in) + ZF”¨
- Pyranose vs. Furanose ANSWER Pyranose is a 6-membered ring. Furanose is a 5-membered ring.
- Mutarotation ANSWER Conversion from ±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.
- ±vs. ß sugars ANSWER ±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". Un- branched. Has reducing and non-reducing end.
- Amylose Chain ANSWER Has ±-1 , 4l -inkages that produce a coiled helix similar to an ±h-elix. Has a reducing and non-reducing end.
- Amylopectin ANSWER Has ±-1 , 4l -inkages. Has periodic ±-1 , 6l -inkages that cause branch- ing. 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 Flip- pase, 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 IV Integral Membrane Protein ANSWER Membrane protein that contains uncon- nected protein helices
- Bacteriorhodopsin ANSWER Type III integral membrane protein with 7 connected he- lices.
- ß-Barrel Membrane Protein ANSWER Can act as a large door. Whole proteins can fit inside.
- ±h-emolysin 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 ANSWER T and G ANSWER C are always equal or close to 1
- DNA Double-Helix ANSWER Opposite strand direction. 3.4Å distance between comple- mentary bases. 36Å for one complete turn.
- A-form DNA ANSWER Condensed form of DNA. Deeper major groove and shallower minor groove.
- B-form DNA ANSWER Watson-Crick model DNA. Deep, wide major groove.
- Z-form DNA ANSWER Left-handed helical form of DNA
- Inverted Repeat in DNA ANSWER Found in double-strands. TTAGCAC|GTGCTAA AATCGTG|CACGATT Forms a cruciform.
- Mirror Repeat in DNA/RNA ANSWER Found in single-strands. TTAGCAC|GTGCTAA Forms a hairpin
- DNA UV Absorbtion ANSWER Absorbs UV light at 260nm.
- Restriction Enzyme ANSWER Cuts DNA at specific restriction sites.
- DNA Base-paring ANSWER G-C base pairs have 3 H-bonds A-T base pairs have 2 H-bonds
- GPCR (G-protein coupled receptor) ANSWER ±h-elical integral membrane proteins. Is a ±ßc heterotrimer.
- ß-adrenergic receptor ANSWER Prototype for all GPCR's. Bind adrenaline/epinephrine to stimulate breakdown of glycogen.
- Step 1 of Epinephrine Signal Transduction ANSWER Epinephrine binds to its specific receptor
- Step 2 of Epinephrine Signal Transduction ANSWER Hormone complex causes GDP bound to ±s-ubunit to be replaced by GTP, activating ±s-ubunit
- Step 3 of Epinephrine Signal Transduction ANSWER Activated ±s-ubunit separates from ßc-complex and moves to adenylyl cyclase, activating it.
- 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- (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.
- Step 6 of Glycolysis ANSWER Glyceraldehyde 3-Phosphate + Pi <--> 1,3-biphospho- glycerate. Uses G3P dehydrogenase enzyme. NAD+ <--> NADH
- First Energy Yielding Step of Glycolysis ANSWER Step 6 of Glycolysis. G3P + Pi <--> 1,3-bisphosphoglycerate
- Step 7 of Glycolysis ANSWER 1,3-bisphosphoglycerate + ADP <--> 3-phosphoglycer- ate + ATP Uses phosphoglycerate kinase enzyme.
- First ATP Yielding Step of Glycolysis ANSWER Step 7 of Glycolysis. 1,3- bisphosphoglycerate <--> 3-phosphoglycerate
- Step 8 of Glycolysis ANSWER 3-phosphoglycerate <--> 2-phosphoglycerate Uses phosphoglycerate mutase enzyme.
- Step 9 of Glycolysis ANSWER 2-phosphoglycerate <--> Phosphoenolpyruvate (PEP) Uses enolase enzyme. Dehydration reaction (loss of water).
- 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 pro- duce 2 NADPH and ribose 5-phosphate used for biosynthesis
- Non-Oxidative Pentose Phosphate Pathway ANSWER Regenerates glucose 6-phos- phate 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 be- tween fructose 2,6- bisphosphate and fructose 6-phosphate using PFK-2 and FB- Pase-
- 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 dehy- drogenase. 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.
- Dihydrolipyl Dehydrogenase ANSWER E3 domain of the PDH complex. Catalyzes regeneration of the lipoyllysine using FAD --> FADH
- Step 1 of the Citric Acid Cycle ANSWER Acetyl-CoA + Oxaloacetate --> Citrate Uses citrate synthase enzyme H2O --> CoA
- Rate-limiting Step of the Citric Acid Cycle ANSWER Step 1 Acetyl- Coa + Oxaloacetate --> Citrate
- Step 2 of the Citric Acid Cycle ANSWER Citrate <--> Isocitrate Uses aconitase enzyme H2O <--> H2O
- Step 3 of the Citric Acid Cycle ANSWER Isocitrate --> ±k-etoglutarate Uses isocitrate dehydrogenase
- Step 5 of the Citric Acid Cycle ANSWER Succinyl-CoA <--> Succinate Uses succinyl- CoA synthetase enzyme GDP + Pi <--> GTP + CoA
- Step 6 of the Citric Acid Cycle ANSWER Succinate <--> Fumarate Uses succinate dehydrogenase FAD <--> FADH
- Step 7 of the Citric Acid Cycle ANSWER Fumarate <--> L-Malate Uses fumarase enzyme
- OH- 2) H+ -->
- Step 8 of the Citric Acid Cycle ANSWER L-Malate <--> Oxaloacetate Uses malate dehydrogenase enzyme NAD+ <--> NADH
- 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 --> ±k- etoglutarate ±k-etoglutarate --> 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 --> ±k- etoglutarate ±k-etoglutarate --> 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.