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ACS BIOCHEMISTRY CERTIFICATION SCRIPT 2026 QUESTIONS WITH SOLUTIONS GRADED A+, Exams of Biochemistry

Liberty University (LU)Biochemistry

ACS BIOCHEMISTRY CERTIFICATION SCRIPT 2026 QUESTIONS WITH SOLUTIONS GRADED A+

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2025/2026

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ACS BIOCHEMISTRY CERTIFICATIONACS BIOCHEMISTRY CERTIFICATION

SCRIPT 2026 QUESTIONS WITHSCRIPT 2026 QUESTIONS WITH

SOLUTIONS GRADED A+SOLUTIONS GRADED A+

•Henderson-Hasselbach Equation.

Answer: pH = pKa + log ([A-] / [HA])

•pI (isoelectric point).

Answer: pH at which the protein has no net charge.

•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.

•pH.

Answer: Measure of the acidity or basicity of a solution.

•pKa.

Answer: The pH at which half of a given species is deprotonated.

•Net charge of proteins.

Answer: Depends on solution pH relative to pKa values.

•Salting Out (Purification).

Answer: Changes soluble protein to solid precipitate. Protein precipitates

when the charges on the protein match the charges in the solution.

•Primary structure.

Answer: Amino acid sequence of a protein.

•Secondary structure.

Answer: Alpha helices and beta sheets in proteins.

•Tertiary structure.

Answer: 3D folding of a protein, involving hydrophobic core, H-bonds, and

salt bridges.

Discover Exams of Biochemistry Liberty University (LU)

Partial preview of the text

Download ACS BIOCHEMISTRY CERTIFICATION SCRIPT 2026 QUESTIONS WITH SOLUTIONS GRADED A+ and more Exams Biochemistry in PDF only on Docsity!

ACS BIOCHEMISTRY CERTIFICATIONACS BIOCHEMISTRY CERTIFICATION

SCRIPT 2026 QUESTIONS WITHSCRIPT 2026 QUESTIONS WITH

SOLUTIONS GRADED A+SOLUTIONS GRADED A+

Henderson-Hasselbach Equation. Answer: pH = pKa + log ([A-] / [HA])
pI (isoelectric point). Answer: pH at which the protein has no net charge.
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.
pH. Answer: Measure of the acidity or basicity of a solution.
pKa. Answer: The pH at which half of a given species is deprotonated.
Net charge of proteins. Answer: Depends on solution pH relative to pKa values.
Salting Out (Purification). Answer: Changes soluble protein to solid precipitate. Protein precipitates when the charges on the protein match the charges in the solution.
Primary structure. Answer: Amino acid sequence of a protein.
Secondary structure. Answer: Alpha helices and beta sheets in proteins.
Tertiary structure. Answer: 3D folding of a protein, involving hydrophobic core, H-bonds, and salt bridges.

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 repulsion effect on like charges. Salt or acid used to remove stuck proteins.
Quaternary structure. Answer: Assembly of multiple polypeptides into a functional protein.
Peripheral proteins. Answer: Loosely attached to membranes by ionic/H-bond interactions.
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.
Integral proteins. Answer: Proteins that span the membrane, often with hydrophobic helices.
X-ray crystallography. Answer: Determines 3D structure without using electromagnetic radiation directly.
SDS. Answer: Sodium dodecyl sulfate. Unfolds proteins and gives them uniform negative charge.

Ramachandran Plot. Answer: Shows favorable phi-psi angle combinations. 3 main "wells" for α-helices, ß-sheets, and left-handed α-helices.
Dialysis. Answer: Separates small molecules from proteins via a semi-permeable membrane.
Glycine Ramachandran Plot. Answer: Glycine can adopt more angles. (H's for R-group).
Size exclusion chromatography. Answer: Technique where larger proteins elute first.
Ion-exchange chromatography. Answer: Separates proteins by charge; proteins bind based on pI.
SDS-PAGE. Answer: Technique that separates proteins by size, with smaller proteins migrating faster.
Km. Answer: Substrate concentration at which the reaction rate is half of Vmax.
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.
Vmax. Answer: Maximum rate of an enzyme-catalyzed reaction.
Competitive inhibitor. Answer: Increases Km without changing Vmax.

Non-competitive inhibitor. Answer: Decreases Vmax without changing Km.
Uncompetitive inhibitor. Answer: Decreases both Km and Vmax.
Transition State Analog. Answer: Molecules that mimic the high-energy transition state and bind enzymes tightly.
Michaelis-Menten plot. Answer: A hyperbolic curve representing the rate of enzyme reactions.
Lineweaver-Burk plot. Answer: A double reciprocal linear plot (1/V vs. 1/[S]).
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.
Carbonic anhydrase. Answer: Enzyme that uses Zn2+ to activate water for nucleophilic attack.
Anti-parallel ß-sheet. Answer: Alternating sheet directions (C & N-termini don't line-up). Has straight H-bonds.
Parallel ß-sheet. Answer: Same sheet directions (C & N-termini line up). Has angled H-bonds.
Serine protease. Answer: Enzyme with a catalytic triad (Ser-His-Asp) that hydrolyzes peptide bonds.
ß-turns.

Hemoglobin 4° Structure. Answer: Tetramer. Dimer of dimers. α2ß2 tetramer.
O2 Saturation Curves. Answer: Myoglobin: hyperbolic curve; Hemoglobin: sigmoidal curve (cooperativity).
Kinases. Answer: Add phosphate groups.
Phosphatases. Answer: Remove phosphate groups.
α/ß Protein Folding. Answer: Less distinct areas of α and ß folding.
Muscle Contraction. Answer: Myosin uses ATP hydrolysis to 'walk' along actin filaments.
α+ß Protein Folding. Answer: Two distinct areas of α and ß folding.
Mechanism of Denaturants. Answer: Highly soluble, H-binding molecules. Stabilize protein backbone in water. Allows denatured state to be stabilized.
Temperature Denaturation of Protein. Answer: Midpoint of reaction is Tm.
Coupled Reactions. Answer: Unfavorable reactions driven by coupling to favorable ones (ATP hydrolysis).
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
ΔG Calculations. Answer: ΔG = ΔG°' + RT ln([products]/[reactants]). Coupled ΔGs are additive.
Glycosidic Linkages. Answer: α(1→4) in starch; β(1→4) in cellulose.
Reducing Sugars. Answer: Sugars with free anomeric carbon.
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.
Glycoproteins. Answer: Carbohydrates linked to Asn (N-linkage) or Ser/Thr/Tyr (O-linkage).
Lactate. Answer: Produced by anaerobic glycolysis (regenerates NAD+).
Ka for Binding. Answer: Ka = [PL] / [P][L]
Pyruvate Dehydrogenase Complex. Answer: Converts pyruvate → acetyl-CoA (needs TPP, lipoate, FAD,

Answer: Heme is in high-spin state. H2O is bound to heme.

Unsaturated Fats. Answer: More fluid, lower melting points.
Na+/K+ Pump. Answer: Active transporter: pumps 3 Na+ out, 2 K+ in, against gradient.
LDL. Answer: Delivers cholesterol to tissues ('bad cholesterol').
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.
HDL. Answer: Removes cholesterol from tissues ('good cholesterol').
Chylomicrons. Answer: Transport dietary fats.
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.

B-Oxidation. Answer: Takes place in mitochondria; fatty acids → acetyl-CoA + NADH + FADH2.
Transaminases. Answer: Transfer amino groups to α-ketoglutarate to form glutamate.
Hill Plot. Answer: Turns sigmoid into straight lines. Slope = n (# of binding sites). Allows measurement of binding sites that are cooperative.
Electron Transport Chain. Answer: NADH → Complex I → CoQ → Complex III → Cyt C → Complex IV → O2.
FADH2 Entry. Answer: Enters at Complex II.
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.
Overall Reaction. Answer: O2 → H2O, generates proton gradient.
Uncouplers. Answer: Dissipate proton gradient, halt ATP production.
Chargaff's Rules. Answer: %A = %T, %G = %C.
DNA Forms. Answer: A-DNA, B-DNA (common form), Z-DNA (zigzag shape).
CO2 binding in Hemoglobin. Answer: Forms carbonic acid that shifts hemoglobin to T-state. O2 binding affinity decreases. Used in the peripheral tissues.
Melting Temperature (Tm).

Answer: Slope changes by factor of α. Slope becomes αKm/Vmax.

Sanger Method. Answer: Chain-termination with ddNTPs.
Uncompetitive Inhibition Graph Changes Km and Vmax. Results in vertical shift up and down. Y-intercept becomes α'/Vmax X-intercept becomes -α'/Km. Answer: Does not change slope.
Mixed Inhibition Graph Pivot point is between X-intercept and Y-intercept.. Answer: Allosteric inhibitor that binds either E or ES.
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.
PCR. Answer: Denaturation → Annealing → Extension.
Restriction Enzymes. Answer: Cut DNA at specific sequences.
Cloning. Answer: Inserting DNA fragments into vectors.
DNA Replication. Answer: DNA polymerase synthesizes 5'→3', needs primer.
Reverse Transcriptase. Answer: Retroviruses use it to synthesize DNA from RNA.
Transcription. Answer: RNA polymerase makes mRNA from DNA.
Translation.

Answer: mRNA codons translated into protein at ribosome.

Fatty Acid Synthesis. Answer: Occurs in cytosol; requires ATP, NADPH, and acetyl-CoA.
ΔGtransport Equation. Answer: ΔGtransport = RTln([S]out / [S]in) + ZFΔΨ
Photosynthesis. Answer: Light energy → ATP + NADPH.
Pyranose vs. Furanose Furanose is a 5-membered ring.. Answer: Pyranose is a 6-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 ß form has -OR/OH group on the same side as the -CH2OH group.. Answer: α form has -OR/OH group opposite from the -CH2OH group.
Starch. Answer: Found in plants. D-glucose polysaccharide. "Amylose chain". Unbranched. Has reducing and non-reducing end.
Pentose Phosphate Pathway. Answer: Generates NADPH and ribose-5-phosphate.
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.
Primary protein structure. Answer: The level of protein structure that is stabilized primarily by hydrogen bonds between backbone atoms.

Iron atom in hemoglobin. Answer: The iron atom in hemoglobin binds oxygen reversibly.
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.
Primary regulatory enzyme of glycolysis. Answer: The primary regulatory enzyme of glycolysis is Phosphofructokinase-1 (PFK-1).
Coenzyme used by pyruvate dehydrogenase complex. Answer: TPP is a coenzyme used by the pyruvate dehydrogenase complex.
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.
Fate of lactate under aerobic conditions. Answer: Under aerobic conditions in human cells, lactate is oxidized to pyruvate.

Intermediate in glycolysis with high-energy phosphate bond. Answer: 1,3-Bisphosphoglycerate contains a high-energy phosphate bond.
Type I Integral Membrane Protein. Answer: Membrane protein with C-terminus inside and N-terminus outside
Cofactor carrying electrons in electron transport chain. Answer: NAD+ carries electrons in the electron transport chain.
Type II Integral Membrane Protein. Answer: Membrane protein with N-terminus inside and C-terminus outside
Molecule binding to Fe²⁺ in heme of hemoglobin. Answer: O₂ binds to the Fe²⁺ in heme of hemoglobin.
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
Effect of 2,3-BPG on hemoglobin. Answer: 2,3-BPG decreases oxygen affinity of hemoglobin.
Bacteriorhodopsin. Answer: Type III integral membrane protein with 7 connected helices.
Bond linking two monosaccharides in maltose. Answer: Maltose is linked by an α(1→4) glycosidic bond.
Best method for separating proteins by size under native conditions. Answer: Gel-filtration chromatography is best for separating proteins by size under native conditions.
ß-Barrel Membrane Protein. Answer: Can act as a large door. Whole proteins can fit inside.
Commonly phosphorylated amino acid side chain. Answer: Serine is commonly phosphorylated.

Answer: Condensed form of DNA. Deeper major groove and shallower minor groove.

Type of enzyme for insulin receptor. Answer: The insulin receptor is a kinase.
B-form DNA. Answer: Watson-Crick model DNA. Deep, wide major groove.
β-oxidation. Answer: Produces Acetyl-CoA directly.
Mass spectrometry. Answer: Determines protein mass most precisely.
PCR. Answer: The extension step synthesizes DNA.
Z-form DNA. Answer: Left-handed helical form of DNA
Size-exclusion column. Answer: A protein elutes first if it is very large.
Inverted Repeat in DNA TTAGCAC|GTGCTAA AATCGTG|CACGATT Forms a cruciform.. Answer: Found in double-strands.
Mirror Repeat in DNA/RNA TTAGCAC|GTGCTAA Forms a hairpin. Answer: Found in single-strands.
SDS-PAGE. Answer: A 50 kDa protein migrates further than expected if it has a large carbohydrate group.
Transition state. Answer: The peak energy of the reaction.
DNA UV Absorbtion. Answer: Absorbs UV light at 260nm.

Restriction Enzyme. Answer: Cuts DNA at specific restriction sites.
Cation exchange chromatography. Answer: A protein with pI 6.8 will bind and elute last in a buffer of pH 5.0.
DNA Base-paring A-T base pairs have 2 H-bonds. Answer: G-C base pairs have 3 H-bonds
Transition state analog. Answer: Structurally similar to the transition state.
GPCR (G-protein coupled receptor). Answer: α-helical integral membrane proteins. Is a αßɣ 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 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.
Transferases. Answer: Enzyme class that catalyzes the transfer of phosphate groups.
Step 4 of Epinephrine Signal Transduction. Answer: Adenylyl cyclase catalyzes the formation of cAMP from ATP
Western blot. Answer: Detects protein size and presence.

ACS BIOCHEMISTRY CERTIFICATION SCRIPT 2026 QUESTIONS WITH SOLUTIONS GRADED A+, Exams of Biochemistry

Related documents

Partial preview of the text

Download ACS BIOCHEMISTRY CERTIFICATION SCRIPT 2026 QUESTIONS WITH SOLUTIONS GRADED A+ and more Exams Biochemistry in PDF only on Docsity!

ACS BIOCHEMISTRY CERTIFICATIONACS BIOCHEMISTRY CERTIFICATION

SCRIPT 2026 QUESTIONS WITHSCRIPT 2026 QUESTIONS WITH

SOLUTIONS GRADED A+SOLUTIONS GRADED A+