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ACS Biochemistry Exam: 270+ Questions and Verified Answers, Exams of Biochemistry

A comprehensive set of 270+ questions and verified answers covering various topics in biochemistry, including protein structure, enzyme kinetics, and carbohydrate chemistry. It is a valuable resource for students preparing for the acs biochemistry exam or seeking to deepen their understanding of the subject.

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2024/2025

Available from 01/14/2025

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ACS BIOCHEMISTRY EXAM / 270 +
QUESTIONS AND CORRECT
VERIFIED ANSWERS BY EXPERTS
2025 LATEST UPDATE GRADED A+.
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Download ACS Biochemistry Exam: 270+ Questions and Verified Answers and more Exams Biochemistry in PDF only on Docsity!

ACS BIOCHEMISTRY EXAM / 270 +

QUESTIONS AND CORRECT

VERIFIED ANSWERS BY EXPERTS

2025 LATEST UPDATE GRADED A+.

ns. 3 main

  1. Henderson-Hasselbach Equation ANSWER pH = pKa + log ([A-] / [HA])
  2. 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.
  3. Salting Out (Purification) ANSWER Changes soluble protein to solid precipitate. Protein precipitates when the charges on the protein match the charges in the solution.
  4. Size-Exclusion Chromatography ANSWER Separates sample based on size with smaller molecules eluting later.
  5. 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.
  6. Hydrophobic/Reverse Phase Chromatography ANSWER Beads are coated with a car- bon chain. Hydrophobic proteins stick better. Elute with non-H-bonding solvent (acetonitrile).
  7. Affinity Chromatography ANSWER Attach a ligand that binds a protein to a bead. Elute with harsh chemicals or similar ligand.
  8. 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.
  9. SDS ANSWER Sodium dodecyl sulfate. Unfolds proteins and gives them uniform negative charge.
  10. Isoelectric Focusing ANSWER Variation of gel electrophoresis where protein charge matters. Involves electrodes and pH gradient. Protein stops at their pI when neutral.
  11. 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.
  12. DTT (dithiothreitol) ANSWER Reduces disulfide bonds.
  13. Iodoacetate ANSWER Adds carboxymethyl group on free -SH groups. Blocks disulfide bonding.
  14. Homologs ANSWER Shares 25% identity with another gene
  15. Orthologs ANSWER Similar genes in different organisms
  16. Paralogs ANSWER Similar "paired" genes in the same organism
  17. Ramachandran Plot ANSWER Shows favorable phi-psi angle combinatio "wells" for ±h- elices, ß-sheets, and left-handed ±h-elices.
  18. Glycine Ramachandran Plot ANSWER Glycine can adopt more angles. (H's for R-group).
  1. 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

  1. ÀR-Ài n g Stacking ANSWER Weird interaction where aromatic rings stack on each other in positive interaction.
  2. Ãh-ole ANSWER Methyl group has area of diminished electron density in center; attracts electronegative groups
  3. 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.
  4. Ka for Binding ANSWER Ka = [PL] / [P][L]
  5. ôv-alue in Binding ANSWER ô = b(ound / total)x100% ô =L[] / ([L] + 1/Ka)
  6. Kd for binding ANSWER Kd = [L] when 50% bound to protein. Kd = 1/Ka
  7. High-Spin Fe ANSWER Electrons are "spread out" and result in larger atom.
  8. Low-Spin Fe ANSWER Electrons are less "spread out" and are compacted by electron rich porphyrin ring.
  9. T-State ANSWER Heme is in high-spin state. H2O is bound to heme.
  10. R-State ANSWER Heme is in low-spin state. O2 is bound to heme.
  11. O2 Binding Event ANSWER O2 binds to T-state and changes the heme to R-state. Causes a 0.4Å movement of the iron.
  12. Hemoglobin Binding Curve ANSWER 4 subunits present in hemoglobin t either T or R - state. Cooperative binding leads to a sigmoidal curve.
  13. 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.
  14. Homotropic Regulation of Binding ANSWER Where a regulatory molecule is also the enzyme's substrate.
  15. Heterotropic Regulation of Binding ANSWER Where an allosteric regulator is present that is not the enzyme's substrate.
  16. Hill Plot ANSWER Turns sigmoid into straight lines. Slope = n (# of binding measurement of binding sites that are cooperative.
  17. 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.
  18. 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.

  1. 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'.
  2. Ionophore ANSWER Hydrophobic molecule that binds to ions and carries them through cell membranes. Disrupts concentration gradients.
  3. Gtransport Equation ANSWER ”Gtransport = RTln([S]out / [S]in) + ZF”¨
  4. Pyranose vs. Furanose ANSWER Pyranose is a 6-membered ring. Furanose is a 5-membered ring.
  5. Mutarotation ANSWER Conversion from ±to ß forms of the sugar at the anomeric carbon.
  6. Anomeric Carbon ANSWER Carbon that is cyclized. Always the same as the aldo or keto carbon in the linear form.
  7. ±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.
  8. Starch ANSWER Found in plants. D-glucose polysaccharide. "Amylose chain". Un- branched. Has reducing and non-reducing end.
  9. Amylose Chain ANSWER Has ±-1 , 4l -inkages that produce a coiled helix similar to an ±h-elix. Has a reducing and non-reducing end.
  10. 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.
  11. Reducing Sugar ANSWER Free aldehydes can reduce FeIII or CuIII. Aldehyde end is the "reducing" end.
  12. Glycogen ANSWER Found in animals. Branched every 8-12 residues and compact. Used as storage of saccharides in animals.
  13. 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.
  14. Chitin ANSWER Homopolymer of N-acetyl-ß-D-glucosamine. Have ß-1,4-linkages. Found in lobsters, squid beaks, beetle shells, etc.
  15. Glycoproteins ANSWER Carbohydrates attached to a protein. Common outside of the cell. Attached at Ser, Thr, or Asn residues.
  16. Membrane Translayer Flip-Flop ANSWER Typically slow, but can be sped up with Flip- pase, Floppase, or Scramblase.
  17. Membrance Fluidity ANSWER Membrane must be fluid. Cis fats increase fluidity, trans fats decrease fluidity.
  18. Type I Integral Membrane Protein ANSWER Membrane protein with C-terminus inside and N-terminus outside
  1. Type IV Integral Membrane Protein ANSWER Membrane protein that contains uncon- nected protein helices
  2. Bacteriorhodopsin ANSWER Type III integral membrane protein with 7 connected he- lices.
  3. ß-Barrel Membrane Protein ANSWER Can act as a large door. Whole proteins can fit inside.
  4. ±h-emolysin ANSWER Secreted as a monomer. Assembles to punch holes in membranes.
  5. Cardiolipin ANSWER "Lipid staple" that ties two proteins (or complexes) together in a membrane. Formed from two phosphoglycerols.
  6. Hydrolysis of Nucleotides ANSWER Base hydrolyzes RNA, but not DNA. DNA is stable in base because of 2' deoxy position.
  7. Chargaff's Rule ANSWER Ratio of A ANSWER T and G ANSWER C are always equal or close to 1
  8. DNA Double-Helix ANSWER Opposite strand direction. 3.4Å distance between comple- mentary bases. 36Å for one complete turn.
  9. A-form DNA ANSWER Condensed form of DNA. Deeper major groove and shallower minor groove.
  10. B-form DNA ANSWER Watson-Crick model DNA. Deep, wide major groove.
  11. Z-form DNA ANSWER Left-handed helical form of DNA
  12. Inverted Repeat in DNA ANSWER Found in double-strands. TTAGCAC|GTGCTAA AATCGTG|CACGATT Forms a cruciform.
  13. Mirror Repeat in DNA/RNA ANSWER Found in single-strands. TTAGCAC|GTGCTAA Forms a hairpin
  14. DNA UV Absorbtion ANSWER Absorbs UV light at 260nm.
  15. Restriction Enzyme ANSWER Cuts DNA at specific restriction sites.
  16. DNA Base-paring ANSWER G-C base pairs have 3 H-bonds A-T base pairs have 2 H-bonds
  17. GPCR (G-protein coupled receptor) ANSWER ±h-elical integral membrane proteins. Is a ±ßc heterotrimer.
  18. ß-adrenergic receptor ANSWER Prototype for all GPCR's. Bind adrenaline/epinephrine to stimulate breakdown of glycogen.
  19. Step 1 of Epinephrine Signal Transduction ANSWER Epinephrine binds to its specific receptor
  20. Step 2 of Epinephrine Signal Transduction ANSWER Hormone complex causes GDP bound to ±s-ubunit to be replaced by GTP, activating ±s-ubunit
  1. Step 3 of Epinephrine Signal Transduction ANSWER Activated ±s-ubunit separates from ßc-complex and moves to adenylyl cyclase, activating it.
  1. FMN ANSWER Single electron transfer.
  1. Step 1 of Glycolysis ANSWER Glucose --> Glucose 6-phosphate. Uses hexokinase enzyme. ATP --> ADP
  2. Step 2 of Glycolysis ANSWER Glucose 6-phosphate <--> Fructose 6-phosphate Uses phosphohexose isomerase enzyme.
  3. Step 3 of Glycolysis ANSWER Fructose 6-phosphate --> Fructose 1,6-bisphosphate Uses PFK- (phosphofructokinase-1) enzyme. ATP --> ADP
  4. First Committed Step of Glycolysis ANSWER Step 3 of Glycolysis. Fructose 6- Phosphate --> Fructose 1,6-bisphosphate. (PFK-1)
  5. Step 4 of Glycolysis ANSWER Fructose 1,6-bisphosphate <--> dihydroxyacetone + glyceraldehyde 3- phosphate. Uses aldolase enzyme.
  6. Step 5 of Glycolysis ANSWER Dihydroxyacetonephosphate <--> glyceraldehyde 3-phosphate Uses triose phosphate isomerase enzyme.
  7. Step 6 of Glycolysis ANSWER Glyceraldehyde 3-Phosphate + Pi <--> 1,3-biphospho- glycerate. Uses G3P dehydrogenase enzyme. NAD+ <--> NADH
  8. First Energy Yielding Step of Glycolysis ANSWER Step 6 of Glycolysis. G3P + Pi <--> 1,3-bisphosphoglycerate
  9. Step 7 of Glycolysis ANSWER 1,3-bisphosphoglycerate + ADP <--> 3-phosphoglycer- ate + ATP Uses phosphoglycerate kinase enzyme.
  10. First ATP Yielding Step of Glycolysis ANSWER Step 7 of Glycolysis. 1,3- bisphosphoglycerate <--> 3-phosphoglycerate
  11. Step 8 of Glycolysis ANSWER 3-phosphoglycerate <--> 2-phosphoglycerate Uses phosphoglycerate mutase enzyme.
  12. Step 9 of Glycolysis ANSWER 2-phosphoglycerate <--> Phosphoenolpyruvate (PEP) Uses enolase enzyme. Dehydration reaction (loss of water).
  • Pi Uses glucose 6-phosphatase.
  1. Cost of Gluconeogenesis ANSWER 4 ATP, 2 GTP, and 2 NADH
  2. Oxidative Pentose Phosphate Pathway ANSWER Uses glucose 6-phosphate to pro- duce 2 NADPH and ribose 5-phosphate used for biosynthesis
  1. Non-Oxidative Pentose Phosphate Pathway ANSWER Regenerates glucose 6-phos- phate from ribose 5- phosphate. Uses transketolase and transaldolase enzymes.
  2. Transketolase ANSWER Transfers a two-carbon keto group
  3. Transaldolase ANSWER Transfers a three-carbon aldo group
  4. Enzyme Km and Substrate Concentration ANSWER Most enzymes have a Km that is near the concentration of the substrate.
  5. 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-
  6. Regulation with fructose 2,6-bisphosphate ANSWER Activates PFK-1 encouraging glycolysis. Inhibits FBPase-1 discouraging gluconeogenesis
  7. Regulation of Pyruvate Kinase ANSWER Inhibited by ATP, Acetyl-Coa, Alanine, long-chain FA's.
  8. 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.
  9. Pyruvate Dehydrogenase ANSWER E1 domain of the PDH complex. Contains TPP cofactor. Releases CO2.
  10. Dihydrolipoyl Transacetylase ANSWER E2 domain of the PDH complex. Catalyzes formation of Acetyl- CoA. Has oxidized, acyl, and reduced lipoyllysine.
  11. Dihydrolipyl Dehydrogenase ANSWER E3 domain of the PDH complex. Catalyzes regeneration of the lipoyllysine using FAD --> FADH
  12. Step 1 of the Citric Acid Cycle ANSWER Acetyl-CoA + Oxaloacetate --> Citrate Uses citrate synthase enzyme H2O --> CoA
  13. Rate-limiting Step of the Citric Acid Cycle ANSWER Step 1 Acetyl- Coa + Oxaloacetate --> Citrate
  14. Step 2 of the Citric Acid Cycle ANSWER Citrate <--> Isocitrate Uses aconitase enzyme H2O <--> H2O
  15. Step 3 of the Citric Acid Cycle ANSWER Isocitrate --> ±k-etoglutarate Uses isocitrate dehydrogenase
  1. Step 5 of the Citric Acid Cycle ANSWER Succinyl-CoA <--> Succinate Uses succinyl- CoA synthetase enzyme GDP + Pi <--> GTP + CoA
  2. Step 6 of the Citric Acid Cycle ANSWER Succinate <--> Fumarate Uses succinate dehydrogenase FAD <--> FADH
  3. Step 7 of the Citric Acid Cycle ANSWER Fumarate <--> L-Malate Uses fumarase enzyme
  1. OH- 2) H+ -->
  1. Step 8 of the Citric Acid Cycle ANSWER L-Malate <--> Oxaloacetate Uses malate dehydrogenase enzyme NAD+ <--> NADH
  2. Net Energy Gain of the Citric Acid Cycle ANSWER 3 NADH, FADH2, and GTP
  3. NADH Producing Steps of the Citric Acid Cycle ANSWER Steps 3, 4, and 8. Isocitrate --> ±k- etoglutarate ±k-etoglutarate --> Succinyl-CoA L-Malate --> Oxaloacetate
  4. FADH2 Producing Steps of the Citric Acid Cycle ANSWER Step 6 Succinate <--

Fumarate Using succinate dehydrogenase enzyme

  1. GTP/ATP Producing Steps of the Citric Acid Cycle ANSWER Step 5 Succinyl-CoA <--> Succinate Using succinyl-Coa synthetase
  2. CO2 Producing Steps of the Citric Acid Cycle ANSWER Steps 3 and 4 Isocitrate --> ±k- etoglutarate ±k-etoglutarate --> Succinyl-CoA
  3. Biotin Structure ANSWER
  4. Biotin Function ANSWER Prosthetic group that serves as a CO2 carrier to separate active sites on an enzyme
  5. 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

  1. 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.