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WGU BIOCHEMISTRY OA PRACTICE
EXAMINATION 2026 QUESTIONS
WITH ANSWERS GRADED A+
◍ true. Answer: T/F: The majority of the reactions of the urea cycle occur in the cytosol.
◍ Henderson-Hasselbach Equation. Answer: pH = pKa + log ([A-] / [HA])
◍ false. Answer: T/F: The majority of the reactions of the urea cycle occur in the mitochondrial matrix.
◍ 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.
◍ false. Answer: T/F: FADH2 yields 2.5 ATP molecules after donating electrons and hydrogen atoms to the ETC.
◍ Salting Out (Purification). Answer: Changes soluble protein to solid precipitate. Protein precipitates when the charges on the protein match the charges in the solution.
◍ 2.5 ATP per NADH. Answer: How many ATP molecules does NADH yield after donating electrons and hydrogen atoms to the electron transfer chain (ETC)?
◍ 1.5 ATP per FADH2. Answer: How many ATP molecules does FADH2 yield after donating
electrons and hydrogen atoms to the electron transfer chain (ETC)?
◍ Size-Exclusion Chromatography. Answer: Separates sample based on size with smaller molecules eluting later.
◍ true. Answer: T/F: ATP is invested in the urea cycle.
◍ false. Answer: T/F: There are 5 complexes associated with the ETC.
◍ 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.
◍ Hydrophobic/Reverse Phase Chromatography. Answer: Beads are coated with a carbon chain. Hydrophobic proteins stick better. Elute with non-H-bonding solvent (acetonitrile).
◍ false. Answer: T/F: In the production of ATP via ATP synthase, the H+ ions move from the mitochondrial matrix to the intermembrane space.
◍ Affinity Chromatography. Answer: Attach a ligand that binds a protein to a bead. Elute with harsh chemicals or similar ligand.
◍ true. Answer: T/F: In the production of ATP via ATP synthase, the H+ ions move from the intermembrane space to the mitochondrial matrix.
◍ 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.
◍ urea. Answer: What is the following chemical species?
◍ ammonium (NH4+). Answer: What is the following chemical species?
◍ aspartate. Answer: What amino acid is the following structure?
◍ arginine. Answer: What amino acid is the following structure?
◍ ornithine. Answer: What amino acid is the following structure?
◍ 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).
◍ complex I. Answer: NADH donates electrons and H+ to complex
◍ Proline Ramachandran Plot. Answer: Proline adopts fewer angles. Amino group is incorporated into a ring.
◍ complex II. Answer: FADH2 donates electrons and H+ to complex
◍ 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.
◍ 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.
◍ complexes I, III, IV. Answer: Which of the ETC complexes pump H+?
◍ complex II. Answer: Which of the ETC complexes do not pump H+?
◍ cytochrome c. Answer: What protein shuttles electrons from complex III to complex IV?
◍ ß-turns. Answer: Tight u-turns with specific phi-psi angles. Must have gly at position
- Proline may also be at ß-turn because it can have a cis-omega angle.
◍ coQ. Answer: What molecule transfers electrons and H+ from both complex I and complex II to complex III?
◍ Loops. Answer: Not highly structured. Not necessary highly flexible, but can occasionally move. Very variable in sequence.
◍ α+ß Protein Folding. Answer: Two distinct areas of α and ß folding.
◍ complex II. Answer: At which complex does the following reaction take place?FADH
◍ 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.
◍ from the matrix to the intermembrane space. Answer: During the electron transport chain, protons are pumped from ___ in the mitochondria.
◍ 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.
◍ as protons flow through ATP synthase and back into the mitochondrial matrix. Answer: Energy released ___ is utilized for the synthesis of ATP.
◍ 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.
◍ F1. Answer: At which part of ATP synthase is ATP produced?
◍ Ka for Binding. Answer: Ka = [PL] / [P][L]
◍ muscle twitches, muscles weakness, and degradation of nerve cells. Answer: An individual with the MERRF disease tends to have symptoms of
◍ ϴ-value in Binding. Answer: ϴ = (bound / total)x100%ϴ = [L] / ([L] + 1/Ka)
◍ inner membrane. Answer: In mitochondria, the molecule that produces ATP is located in the
◍ rotation of ATP synthase. Answer: What process results in ATP formation?
◍ Kd for binding. Answer: Kd = [L] when 50% bound to protein. Kd = 1/Ka
◍ F and F2. Answer: Which of the following is not a part of ATP synthase?- F- Fo- F1- F
◍ CAC. Answer: Oxidative phosphorylation uses NADH and FADH2 generated from the ___.
◍ High-Spin Fe. Answer: Electrons are "spread out" and result in larger atom.
◍ synthesis of ATP from ADP + Pi. Answer: Oxidative phosphorylation is the synthesis of ___ from ___.
beta oxidation and the urea cycle occur. This close proximity of reaction cycles is efficient because it allows NADH and FADH2, produced by the citric acid cycle, to also participate in the electron transport chain by donating electrons and hydrogen ions. B. The complexes of the ETC are located in the inner membrane and accept the electrons from the coenzymes.. Answer: Describe the critical structure of the mitochondrion that are important for the production of ATP.A. What structural features of the mitochondrion support ATP formation.B. Where are the complexes of the ETC located.
◍ O2 Binding Event. Answer: O2 binds to T-state and changes the heme to R-state. Causes a 0.4Å movement of the iron.
◍ The chemical reactions of the ETC, through the oxidation of coenzymes produced by the CAC, create energy that pumps H+ ions into the intermembrane space of mitochondria. The accumulation of these H+ ions in the intermembrane space creates an H+ ion gradient. This H+ gradient provides the driving force for the rotation of ATP synthase, which results in the production of AT P. Therefore, the ETC is responsible for the production of ATP via oxidative phosphorylation.. Answer: According to the chemiosmotic theory, how is the ETC connected to ATP production? Explain in at least 4 complete sentences.
◍ 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)
◍ Michaelis-Menton Graph. Answer: nan
◍ Lineweaver-Burke Graph. Answer: Slope = Km/VmaxY-intercept = 1/VmaxX-intercept = - 1/Km
◍ Lineweaver-Burke Equation. Answer: Found by taking the reciprocal of the Michaelis-Menton Equation.
◍ Kcat.
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
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.
◍ 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: nan
◍ FADH2. Answer: Single-electron transfer
◍ NADPH. Answer: nan
◍ 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-phosphateUses phosphohexose isomerase enzyme.
◍ Step 3 of Glycolysis. Answer: Fructose 6-phosphate --> Fructose 1,6-bisphosphateUses PFK- (phosphofructokinase-1) enzyme.ATP --> ADP
◍ First Committed Step of Glycolysis. Answer: Step 3 of Glycolysis.Fructose 6-Phosphate --> Fructose
6-phosphate --> Fructose 1,6-bisphosphate
◍ ATP Producing Steps of Glycolysis. Answer: Steps 7 and 10.1,3-bisphosphoglycerate <--> 3-phosphoglyceratePEP --> Pyruvate
◍ NADH Producing Step of Glycolysis. Answer: Step 6G3P <--> 1,3-bisphosphoglycerate
◍ Total Energy Produced by Glycolysis. Answer: 2NADH + 4 ATP
◍ Lactic Acid Fermentation. Answer: Pyruvate --> L-LactateNADH --> NAD+Regenerates NAD+ for use in glycolysis
◍ Ethanol Fermentation. Answer: Pyruvate --> Acetalaldehyde --> EthanolUses pyruvate decarboxylase (TPP) and alcohol dehydrogenase.NADH --> CO2(TPP) + NAD+
◍ TPP Cofactor Structure. Answer: nan
◍ 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 --> OxaloacetatePyruvate decarboxylate (biotin)ATP --> ADPOxaloacetate --> PEPPEP carboxykinase GTP --> GDP + CO
◍ Gluconeogenic Bypass of Step 3. Answer: Fructose 1,6-bisphosphate + H2O --> Fructose 6-phosphate + PiUses FBPase-1 (coordinated with PFK-1)
◍ Gluconeogenic Bypass of Step 1. Answer: Glucose 6-phosphate + H2O --> Glucose + PiUses 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-
◍ Regulation with fructose 2,6-bisphosphate. Answer: Activates PFK-1 encouraging glycolysis. Inhibits FBPase- 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 +
