Question 1

Gluconeogenesis steps: which of the following does NOT occur as a single enzymatic reaction (it requires multiple intermediate steps) during conversion of lower-glycolytic intermediates back to hexose phosphates?

Accepted Answer

Introduction / Context: Gluconeogenesis reverses glycolysis with strategic bypasses. This question targets which transformation is not a single enzymatic step but a pathway segment requiring several enzymes and intermediates. Given Data / Assumptions: Physiological context: hepatic/renal gluconeogenesis. Candidate steps include pyruvate↔oxaloacetate↔PEP and upstream reactions toward hexose phosphates. We are asked to spot a multi-step conversion disguised as if it were a single reaction. Concept / Approach: Some gluconeogenic changes are single enzymes (e.g., pyruvate carboxylase makes oxaloacetate; PEP carboxykinase makes PEP from OAA with decarboxylation). By contrast, PEP does not become fructose 1,6-bisphosphate in one step; it proceeds through several reversible glycolytic reactions. Step-by-Step Solution: 1) Pyruvate → oxaloacetate: a single carboxylation by pyruvate carboxylase (biotin-dependent).2) Oxaloacetate → PEP: a single decarboxylation/phosphorylation by PEP carboxykinase.3) PEP → F1,6BP: requires multiple enzymes (enolase, phosphoglycerate mutase, phosphoglycerate kinase, glyceraldehyde-3-phosphate dehydrogenase, triose phosphate isomerase, aldolase) to reach fructose 1,6-bisphosphate.4) Therefore, the statement that PEP is converted to fructose 1,6-bisphosphate is not a single reaction step and is the correct choice. Verification / Alternative check: Pathway diagrams show distinct nodes: PEP → 2-phosphoglycerate → 3-phosphoglycerate → 1,3-bisphosphoglycerate → glyceraldehyde-3-phosphate/dihydroxyacetone phosphate → fructose 1,6-bisphosphate via aldolase, confirming multi-step nature. Why Other Options Are Wrong: Pyruvate carboxylation: one step catalyzed by pyruvate carboxylase.Pyruvate → PEP: effectively accomplished via two defined, single-enzyme reactions (pyruvate carboxylase, then PEPCK), each discrete.Oxaloacetate decarboxylation to PEP: a single PEPCK reaction. Common Pitfalls: Assuming pathway segments are single reactions or overlooking the aldolase step that forms F1,6BP from triose phosphates. Final Answer: The conversion of PEP to fructose 1,6-bisphosphate

Question 2

Physiology of metabolism: what is the overall function of gluconeogenesis in human biochemistry under fasting or carbohydrate-restricted states?

Accepted Answer

Introduction / Context: Gluconeogenesis synthesizes glucose from non-carbohydrate precursors, sustaining obligate glucose users (e.g., red blood cells, parts of the kidney medulla) when dietary carbohydrate is scarce. Given Data / Assumptions: Physiological settings: fasting, prolonged exercise, low-carbohydrate diet. Key substrates: lactate, glycerol, glucogenic amino acids. Organs: liver (primary), kidney cortex (significant in prolonged fasting). Concept / Approach: Multiple inputs feed gluconeogenesis: lactate via the Cori cycle, glycerol from lipolysis, and amino acids (especially alanine and glutamine) from proteolysis. The goal is to maintain euglycemia when glycogen is depleted. Step-by-Step Solution: 1) Lactate recycling: lactate → pyruvate → glucose in the liver reduces lactic acidosis risk and supports working muscle (Cori cycle).2) Glycerol usage: glycerol → glycerol-3-phosphate → dihydroxyacetone phosphate enters gluconeogenesis after lipolysis.3) Amino acids: transamination/deamination provide carbon skeletons (e.g., alanine → pyruvate; glutamine → oxaloacetate).4) Outcome: maintain plasma glucose for tissues with limited fatty acid oxidation, such as brain (partially) and RBCs (completely dependent). Verification / Alternative check: During overnight fasts, hepatic glycogen supports glucose; as fasting prolongs, gluconeogenesis predominates with renal contribution increasing, consistent with stable glucose levels in healthy adults. Why Other Options Are Wrong: Each single statement is true but incomplete; taken together they capture the full function, hence “All of the above.” Common Pitfalls: Equating gluconeogenesis with only one substrate (e.g., lactate) or forgetting renal gluconeogenesis during prolonged fasting. Final Answer: All of the above

Question 3

In cytosolic glycolysis, what is the NET number of ATP molecules produced per molecule of glucose converted to two molecules of pyruvate (ignoring mitochondrial shuttle costs)?

Accepted Answer

Introduction / Context: ATP accounting in glycolysis is essential for understanding cellular energy balance. This question asks for the net ATP yield per glucose in the cytosol before any mitochondrial oxidative phosphorylation or shuttle adjustments. Given Data / Assumptions: Pathway: glucose → 2 pyruvate. No consideration of NADH oxidation in mitochondria (malate-aspartate or glycerol phosphate shuttles not counted). Standard cytosolic reactions in eukaryotic cells. Concept / Approach: Two ATP are invested in the preparatory phase (hexokinase and phosphofructokinase-1). Four ATP are generated in the payoff phase (phosphoglycerate kinase and pyruvate kinase, each twice per glucose). Net ATP = 4 produced − 2 consumed = 2 ATP. Step-by-Step Solution: 1) Investment: -1 ATP at hexokinase (glucose → G6P) and -1 ATP at PFK-1 (F6P → F1,6BP).2) Payoff: +2 ATP at phosphoglycerate kinase (1,3-BPG → 3-PG, twice).3) Payoff: +2 ATP at pyruvate kinase (PEP → pyruvate, twice).4) Net: 4 − 2 = 2 ATP per glucose. Verification / Alternative check: Two cytosolic NADH are also formed; their ATP equivalence depends on the shuttle used, but that does not change the glycolytic net ATP count of 2. Why Other Options Are Wrong: 4: gross production, ignores the 2 ATP investment.1 or 3: do not match standard stoichiometry. Common Pitfalls: Confusing gross vs net ATP, or double-counting the investment phase. Final Answer: 2

Question 4

Inborn error of metabolism: the classic form of galactosemia is caused by deficiency of which enzyme in the Leloir pathway of galactose utilization?

Accepted Answer

Introduction / Context: Galactosemia is an inborn error of carbohydrate metabolism leading to accumulation of toxic intermediates after milk ingestion. Identifying the defective enzyme guides diagnosis and dietary therapy. Given Data / Assumptions: Clinical setting: neonatal jaundice, vomiting, hepatomegaly after lactose (milk) exposure. Pathway: Leloir pathway converts galactose to glucose-1-phosphate. We focus on the classic (most common) form. Concept / Approach: Classic galactosemia results from GALT deficiency, blocking conversion of galactose-1-phosphate and UDP-glucose to glucose-1-phosphate and UDP-galactose, causing accumulation of galactose-1-phosphate and galactitol (via aldose reductase), with risk of cataracts and liver failure. Step-by-Step Solution: 1) Recognize key enzyme: GALT is central to Leloir pathway flux.2) Predict metabolic consequences: galactose-1-phosphate builds up; galactose is reduced to galactitol.3) Link to clinical picture: toxicity explains hepatic dysfunction and cataracts; dietary galactose restriction improves outcomes. Verification / Alternative check: Other variants exist (galactokinase or UDP-galactose-4-epimerase deficiencies), but the classic severe neonatal presentation is most commonly due to GALT deficiency. Why Other Options Are Wrong: Lactase deficiency: causes lactose intolerance (malabsorption), not systemic neonatal toxicity.Lactose synthetase absence: relates to lactose synthesis in mammary glands, not neonatal catabolism.“Semnase”: not an enzyme in this pathway. Common Pitfalls: Confusing digestive brush-border enzymes (lactase) with hepatic metabolic enzymes (GALT). Final Answer: Absence of galactose-1-phosphate uridyltransferase (GALT)

Question 5

Fructose 1,6-bisphosphate accumulation would have what regulatory effect on carbohydrate metabolism pathways operating in the cytosol?

Accepted Answer

Introduction / Context: Reciprocal regulation prevents futile cycles between glycolysis and gluconeogenesis. Fructose 1,6-bisphosphate (F1,6BP) is a central allosteric signal that coordinates pathway flux, especially in the liver. Given Data / Assumptions: Metabolite of interest: F1,6BP. Target enzymes: pyruvate kinase (glycolysis) and fructose 1,6-bisphosphatase (FBPase-1; gluconeogenesis). Physiological context: cytosolic carbohydrate metabolism. Concept / Approach: F1,6BP exerts feed-forward activation of pyruvate kinase, enhancing glycolytic flux. Conversely, high F1,6BP indicates that FBPase-1 should be less active to avoid simultaneous opposing flux, effectively suppressing gluconeogenesis (while phosphofructokinase-1 is upstream regulated by ATP/citrate/AMP). Step-by-Step Solution: 1) Identify regulatory node: F1,6BP positively regulates pyruvate kinase → stimulates glycolysis.2) Consider gluconeogenesis: elevated F1,6BP reduces the need for FBPase-1 activity (reciprocal control), limiting gluconeogenic flux.3) Outcome: glycolysis stimulated; gluconeogenesis inhibited. Verification / Alternative check: Liver pyruvate kinase is activated by F1,6BP (feed-forward). Meanwhile, gluconeogenesis is favored when F2,6BP is low and ATP/acetyl-CoA are high; high F1,6BP signals glycolytic progress and opposes gluconeogenic pull. Why Other Options Are Wrong: Simultaneous inhibition or stimulation of both pathways violates reciprocal regulation.Inhibit glycolysis and stimulate gluconeogenesis contradicts known feed-forward activation of pyruvate kinase by F1,6BP. Common Pitfalls: Confusing F1,6BP with F2,6BP (a powerful modulator of PFK-1/FBPase-1). Both are regulators but have distinct targets and effects. Final Answer: Stimulate glycolysis and inhibit gluconeogenesis

Question 6

Cellular location check: in which compartment do the reactions of glycolysis predominantly occur in typical eukaryotic cells?

Accepted Answer

Introduction / Context: Knowing the subcellular localization of metabolic pathways is essential for integrating regulation and transport. Glycolysis is the initial stage of carbohydrate catabolism feeding into mitochondrial oxidation. Given Data / Assumptions: Eukaryotic cell with intact organelles. We refer to the enzymatic pathway from glucose to pyruvate. Concept / Approach: All glycolytic enzymes reside in the cytosol (cytoplasm). Pyruvate, NADH, and ATP produced here interface with mitochondrial metabolism via transporters and shuttles; none of the glycolytic steps occur in membranes, the cell wall, or plasmids. Step-by-Step Solution: 1) Enumerate: hexokinase, phosphofructokinase-1, aldolase, GAPDH, phosphoglycerate kinase, phosphoglycerate mutase, enolase, pyruvate kinase—all cytosolic.2) Identify outputs: ATP formed in the cytosol; NADH reoxidized via shuttles or fermentation.3) Conclude: the cytoplasm is the compartment for glycolysis. Verification / Alternative check: Subcellular fractionation and proteomics consistently localize glycolytic enzymes to the cytosolic fraction; mitochondrial metabolism begins with pyruvate conversion to acetyl-CoA by pyruvate dehydrogenase in the matrix. Why Other Options Are Wrong: Cell membrane: not the catalytic site of glycolysis (though glucose transporters reside here).Cell wall: eukaryotic animal cells lack a wall; in any case, not a metabolic enzyme location.Plasmids: DNA elements in bacteria; not a cellular compartment for enzymatic pathways. Common Pitfalls: Confusing transport (membranes) with metabolism, or conflating bacterial plasmids with eukaryotic organelles. Final Answer: Cytoplasm

Question 7

Enzyme activity at low body temperature: why do metabolic reactions in living cells malfunction when an organism is cooled below a critical temperature?

Accepted Answer

Introduction / Context: Temperature strongly influences biochemical reaction rates in living organisms. This question examines why cooling below a critical body temperature leads to widespread metabolic malfunction, focusing on enzyme kinetics and molecular motion rather than permanent structural damage. Given Data / Assumptions: Organism is cooled to a temperature lower than its normal physiological range. Enzymes are proteins that catalyze metabolic reactions by binding substrates at active sites. We consider acute cooling, not prolonged freezing or denaturation by heat. Concept / Approach: Reaction rates depend on the frequency of productive collisions between enzyme and substrate and on the fraction of molecules with sufficient energy to reach the transition state. At lower temperatures, average kinetic energy decreases, diffusion slows, and conformational dynamics within proteins and substrates diminish, reducing catalytic turnover even when enzymes remain correctly folded. Step-by-Step Solution: 1) Cooling reduces molecular kinetic energy, lowering collision frequency between enzymes and substrates.2) Diffusion in the cytosol slows, decreasing encounter rates with the active site.3) Enzyme conformational flexibility (needed for induced fit and transition-state stabilization) is reduced.4) As a result, catalytic rates drop dramatically, and pathways fail to sustain homeostasis. Verification / Alternative check: Q10 rules in physiology show that many biochemical rates approximately double for each 10 °C rise within a physiological range and conversely fall with cooling, consistent with diffusion-limited and activation-barrier processes. Why Other Options Are Wrong: Option A: Loss of 3D shape (denaturation) is typical of high heat or extreme conditions, not mild cooling. Option B: Cooling does not permanently lock substrates; binding is reversible. Option C: Activation energy (Ea) is intrinsic to the reaction and enzyme; temperature changes the fraction of molecules crossing Ea, not Ea itself. Option E: Membranes may become more rigid, but enzymes are not universally “permanently” inactivated first. Common Pitfalls: Confusing cold-induced slowing (reversible kinetic effect) with heat denaturation (irreversible structural loss). Also assuming enzymes stop solely due to membrane phase transitions when the primary immediate effect is reduced molecular motion and dynamics. Final Answer: There is insufficient molecular motion for substrates to interact

Question 8

Metabolic pathways and enzyme specificity: in a typical pathway, how is each individual reaction controlled?

Accepted Answer

Introduction / Context: Cells organize chemistry into metabolic pathways, where each step converts a substrate into a product. This item probes the principle of enzyme specificity and control at the level of individual steps within a pathway. Given Data / Assumptions: Pathways consist of multiple discrete reactions. Each step has its own chemical transformation and kinetic profile. Enzymes accelerate reactions under physiological conditions. Concept / Approach: Enzyme specificity means a particular protein catalyst binds specific substrates and carries out a defined reaction with high selectivity. Although some steps are near-equilibrium and others are far-from-equilibrium (effectively irreversible in vivo), the unifying feature is that each step is catalyzed by its own enzyme, enabling regulation and flux control across the pathway. Step-by-Step Solution: 1) Identify the common characteristic across all steps: enzymatic catalysis.2) Recognize that reversibility varies by step; it is not universal.3) End-product control (feedback inhibition) is one mechanism that targets specific regulatory enzymes, not every step.4) Therefore, the correct general statement is that each reaction is catalyzed by a specific enzyme. Verification / Alternative check: Biochemical maps (e.g., glycolysis, TCA cycle, urea cycle) list a distinct named enzyme for each reaction, confirming universality of step-specific catalysis. Why Other Options Are Wrong: Option A/B: Some steps are reversible, others effectively irreversible; neither property applies to all reactions. Option D: End-product inhibition regulates key nodes, not every reaction. Option E: Physiological rates and specificity require enzymes; spontaneous rates are insufficient. Common Pitfalls: Equating “pathways are regulated” with “every step is controlled by the end product,” and assuming uniform reversibility across all steps. Final Answer: Catalyzed by a specific enzyme

Question 9

Cori cycle physiology: which enzyme is central to interconverting lactate and pyruvate between muscle and liver during the Cori cycle?

Accepted Answer

Introduction / Context: The Cori cycle coordinates energy metabolism between muscle and liver. Muscle produces lactate during intense activity; the liver converts lactate back to glucose that can be returned to muscle. This question targets the key enzyme enabling the lactate–pyruvate interconversion that underpins the cycle. Given Data / Assumptions: Muscle operating anaerobically generates lactate from pyruvate. Liver performs gluconeogenesis using lactate as a substrate. Redox balance between NADH and NAD+ is critical in both tissues. Concept / Approach: Lactate dehydrogenase (LDH) catalyzes the reversible reaction: pyruvate + NADH + H+ ↔ lactate + NAD+. In muscle, LDH reduces pyruvate to lactate to regenerate NAD+ for glycolysis; in liver, LDH oxidizes lactate to pyruvate, which then enters gluconeogenesis. Step-by-Step Solution: 1) Identify the transformation required by the cycle: lactate ↔ pyruvate.2) The enzyme that catalyzes this interconversion is LDH.3) In the liver, pyruvate proceeds to glucose via gluconeogenic enzymes; glucose returns to muscle through the bloodstream. Verification / Alternative check: Physiology texts consistently depict LDH at both ends of the cycle: lactate formation in muscle and lactate utilization in liver. Why Other Options Are Wrong: Glucokinase phosphorylates glucose in liver; G6PD functions in the pentose phosphate pathway; glucose isomerase interconverts glucose/fructose in industrial contexts; pyruvate kinase converts phosphoenolpyruvate to pyruvate in glycolysis but does not handle lactate. Common Pitfalls: Confusing LDH with pyruvate dehydrogenase; PDH converts pyruvate to acetyl-CoA, not lactate. Final Answer: Lactate dehydrogenase

Question 10

Gluconeogenesis coverage check: which of the following processes is NOT excluded from gluconeogenesis (i.e., all listed processes actually participate in glucose formation)?

Accepted Answer

Introduction / Context: Gluconeogenesis is the anabolic pathway that synthesizes glucose from non-carbohydrate precursors such as lactate, glycerol, and glucogenic amino acids, primarily in the liver. This question tests recognition of processes that legitimately belong to gluconeogenesis. Given Data / Assumptions: Options list common steps or on-ramps associated with hepatic glucose production. We interpret the stem under the Recovery-First Policy to clarify intent: identify whether any listed process is “not involved.” If all listed processes are valid components, the appropriate answer is that none are excluded. Concept / Approach: Evaluate each option for its role in gluconeogenesis. Oxaloacetate conversion through phosphoenolpyruvate and up the pathway is central; lactate to pyruvate (via LDH) supplies carbon; glucose-6-phosphatase dephosphorylation is required to release free glucose into blood. Step-by-Step Solution: 1) Option A: Oxaloacetate is converted to phosphoenolpyruvate (via PEPCK) and eventually to glucose—clearly involved.2) Option B: Lactate enters as pyruvate, a major Cori cycle substrate—clearly involved.3) Option C: Glucose-6-phosphatase removes phosphate from G6P in the endoplasmic reticulum, producing exportable glucose—clearly involved.4) Therefore, none of the listed processes is excluded; select “None of the above.” Verification / Alternative check: Standard hepatic gluconeogenesis outlines include LDH (lactate to pyruvate), PEPCK/OAA routing, and glucose-6-phosphatase as essential steps for completing glucose output. Why Other Options Are Wrong: Each named process is a textbook step in gluconeogenesis; declaring any single one as “not involved” would contradict pathway biochemistry. Option E (“all not involved”) is the opposite of reality. Common Pitfalls: Misreading the stem and selecting “all of the above” when the logic demands “none of the above,” or confusing glycolytic enzymes with the bypass steps specific to gluconeogenesis. Final Answer: None of the above (all are involved in gluconeogenesis)

Question 11

Hormonal control of glycogen breakdown: how do glucagon and epinephrine stimulate glycogen phosphorylase to generate G-1-P (and hence G-6-P) from glycogen?

Accepted Answer

Introduction / Context: Glycogenolysis is hormonally regulated to mobilize glucose during fasting or acute stress. This question probes the signaling mechanism by which glucagon (liver-focused) and epinephrine (liver and muscle) activate glycogen breakdown. Given Data / Assumptions: Hormones bind cell-surface G protein–coupled receptors. Glycogen phosphorylase releases glucose-1-phosphate (G-1-P) from glycogen using inorganic phosphate (Pi). Signal amplification cascades are typical in these pathways. Concept / Approach: Glucagon and epinephrine activate Gs-coupled receptors, which stimulate adenylate cyclase to produce cAMP. cAMP activates protein kinase A (PKA), which phosphorylates phosphorylase kinase; phosphorylase kinase then activates glycogen phosphorylase by phosphorylation, increasing glycogen breakdown. G-1-P is converted to G-6-P by phosphoglucomutase. Step-by-Step Solution: 1) Hormone binds GPCR → Gs activates adenylate cyclase.2) Adenylate cyclase converts ATP to cAMP.3) cAMP activates PKA, which phosphorylates phosphorylase kinase.4) Phosphorylase kinase phosphorylates glycogen phosphorylase (b → a), stimulating glycogenolysis.5) G-1-P is produced and then isomerized to G-6-P. Verification / Alternative check: Classic experiments show that cAMP analogs mimic hormonal effects and that PKA inhibitors block glycogen phosphorylase activation, confirming the indirect cAMP-mediated mechanism. Why Other Options Are Wrong: Option A: Epinephrine acts in muscle as well as liver. Option B: The phosphoryl donor at glycogen phosphorylase is inorganic phosphate (Pi), not ATP. Option C: Hormones do not bind the enzyme directly; activation is via signaling. Option E: Flux is not driven by inhibiting phosphoglucomutase; its role is downstream isomerization. Common Pitfalls: Assuming ATP donates phosphate directly in the catalytic step of glycogen phosphorylase, or overlooking the layered kinase cascade initiated by cAMP. Final Answer: Indirectly, by first stimulating adenylate cyclase to make cAMP

Question 12

Features of regulated metabolic pathways: which statements are generally true about how regulated pathways operate in eukaryotic cells?

Accepted Answer

Introduction / Context: Cells must coordinate thousands of reactions so that resources are used efficiently. This question asks about common organizational and regulatory features of metabolic pathways in eukaryotes, especially how compartmentalization and the first committed step guide control of flux. Given Data / Assumptions: We consider typical anabolic/catabolic routes (e.g., glycolysis vs. gluconeogenesis, fatty acid synthesis vs. oxidation). Eukaryotic cells contain membrane-bound organelles (mitochondria, ER, peroxisomes, cytosol). Pathways often include an early “first committed step” that channels intermediates down a defined route. Concept / Approach: Evaluate each statement against standard biochemical principles. Opposing pathways are spatially separated to avoid futile cycles; early irreversible steps are typical regulatory nodes; once the first committed step is traversed, intermediates are effectively destined for the pathway’s end products unless regulation intervenes downstream. Step-by-Step Solution: 1) Compartmentalization: Eukaryotes separate processes (e.g., fatty acid oxidation in mitochondria; fatty acid synthesis in cytosol) enabling differential regulation and metabolite channeling.2) First committed step: Regulatory enzymes (often allosteric and hormonally controlled) reside at the first committed step, ensuring that flux control occurs before significant resource investment.3) Commitment of flux: After the first committed step, subsequent intermediates are typically locked into the pathway, barring special branch regulation.4) Since each statement reflects standard features, the inclusive option “All of these” is correct. Verification / Alternative check: Canonical examples include phosphofructokinase-1 as a control point in glycolysis, acetyl-CoA carboxylase in fatty acid synthesis, and carbamoyl phosphate synthetase I in the urea cycle, each representing an early committed/regulatory step. Why Other Options Are Wrong: Choosing any single statement ignores the fact that all are simultaneously true in most regulated pathways; “None” contradicts well-established organization principles. Common Pitfalls: Confusing “first enzyme” with “first committed enzyme” (the first reaction may be reversible, while control is placed at the first irreversible, committed step). Final Answer: All of these

Question 13

Gluconeogenesis localization: which characteristic enzyme of gluconeogenesis is not found in the cytosol (it is mitochondrial)?

Accepted Answer

Introduction / Context: Gluconeogenesis uses enzymes distributed across cellular compartments. Understanding where each step occurs is crucial for predicting regulation and metabolite transport. This item focuses on identifying a core gluconeogenic enzyme that is not cytosolic. Given Data / Assumptions: Several gluconeogenic reactions occur in the cytosol. Certain steps are localized to organelles to facilitate regulation and substrate channeling. We apply Recovery-First Policy to clarify the intent as a single-best answer question. Concept / Approach: Pyruvate carboxylase catalyzes pyruvate → oxaloacetate in the mitochondrial matrix and requires biotin and acetyl-CoA as an allosteric activator. Fructose-1,6-bisphosphatase is a cytosolic bypass enzyme. Glucose-6-phosphatase is associated with the endoplasmic reticulum membrane with its catalytic face in the ER lumen, functionally non-cytosolic but not mitochondrial. The question asks for a characteristic enzyme that is not in cytosol and highlights the mitochondrial resident as the single-best answer. Step-by-Step Solution: 1) List enzyme localizations: PC (mitochondrial), F-1,6-BPase (cytosolic), G-6-Pase (ER membrane/lumen).2) Identify the one explicitly mitochondrial enzyme among the options: pyruvate carboxylase.3) Select pyruvate carboxylase as the single-best answer satisfying “not cytosolic (mitochondrial).” Verification / Alternative check: Mitochondrial biotin enzymes, including pyruvate carboxylase, are well characterized; deficiency syndromes and isotope tracing support its matrix localization and role in anaplerosis and gluconeogenesis. Why Other Options Are Wrong: Fructose-1,6-bisphosphatase is primarily cytosolic. “Both (a) and (b)” fails because (b) is cytosolic. Glucose-6-phosphatase is non-cytosolic but ER-associated rather than mitochondrial; the prompt seeks a single-best selection. Cytosolic PEPCK isoform, when present, is cytosolic by definition. Common Pitfalls: Conflating “not cytosolic” with “mitochondrial”; ER localization is also non-cytosolic but is a separate compartment. The question aims for the mitochondrial step specifically. Final Answer: Pyruvate carboxylase

Question 14

Terminology check: metabolic reactions that break down complex molecules into smaller units, releasing usable energy, are best described as

Accepted Answer

Introduction / Context: Biochemistry distinguishes pathways that build molecules from those that degrade them. Correctly naming these processes is fundamental to understanding energy flow and metabolic regulation. Given Data / Assumptions: The reactions in question convert large, complex molecules into smaller ones. Energy is released and captured in forms such as ATP, NADH, and FADH2. The question asks for the best single term describing this direction of metabolism. Concept / Approach: Catabolism refers to degradative processes that release free energy, often by oxidizing substrates and coupling to ATP synthesis. Anabolism (biosynthesis) is the opposite, consuming energy to build complexity. The correct term for energy-yielding breakdown is “catabolic.” Step-by-Step Solution: 1) Identify the direction: complex → simple, with energy release.2) Match to terminology: catabolism fits this definition.3) Therefore, select “Catabolic.” Verification / Alternative check: Standard pathways such as glycolysis, beta-oxidation, and the TCA cycle are universally classified as catabolic and are linked to ATP production via oxidative phosphorylation. Why Other Options Are Wrong: “Biosynthetic” denotes anabolism (building up). “Catalytic” describes the general action of enzymes and is not a metabolic direction. “Photosynthetic” is a specialized anabolic process harnessing light energy. “Homeostatic” refers to physiological balance, not a pathway class. Common Pitfalls: Confusing “catabolic” with “catatonic” (a neurological term) or assuming photosynthesis is catabolic because it involves energy; it actually synthesizes carbohydrates. Final Answer: Catabolic

Question 15

Biopolymer stoichiometry: what is the molecular formula of a polysaccharide formed by linking 10 glucose monomers via dehydration synthesis? (Glucose formula: C6H12O6.)

Accepted Answer

Introduction / Context: Condensation (dehydration) reactions form polymers by removing water as monomers join. Determining the empirical formula of the resulting polymer reinforces conservation-of-mass bookkeeping in biochemistry. Given Data / Assumptions: Monomer: glucose, C6H12O6. Number of monomers: n = 10. Each glycosidic linkage eliminates one H2O. Concept / Approach: For a linear polymer from n hexose monomers, total atoms from monomers are C: 6n, H: 12n, O: 6n. The number of water molecules removed is (n - 1). Subtracting those atoms yields the polymer formula: C = 6n; H = 12n - 2(n - 1) = 10n + 2; O = 6n - (n - 1) = 5n + 1. Step-by-Step Solution: 1) Start: for n = 10, C = 6*10 = 60.2) Hydrogen: 12*10 - 2*(10 - 1) = 120 - 18 = 102.3) Oxygen: 6*10 - (10 - 1) = 60 - 9 = 51.4) Combine to get polymer formula: C60H102O51. Verification / Alternative check: General formula for a homopolymer of hexoses: C6nH10n+2O5n+1. Plugging n = 10 gives C60H102O51, matching the calculation. Why Other Options Are Wrong: Option A undercounts hydrogen and oxygen; Option B ignores dehydration (simply 10 monomers summed); Option D is not an empirical molecular formula; Option E uses incorrect water-loss arithmetic. Common Pitfalls: Forgetting that the number of water losses is (n - 1), not n, for a single unbranched chain. Final Answer: C60H102O51

Question 16

Regulation cross-check: an intermediate that inhibits fructose-1,6-bisphosphatase would primarily inhibit which pathway?

Accepted Answer

Introduction / Context: Control points in metabolism are often targeted by feedback signals. Fructose-1,6-bisphosphatase (F-1,6-BPase) is a key gluconeogenic enzyme that reverses the phosphofructokinase-1 step of glycolysis. Understanding which pathway is affected by its inhibition highlights reciprocal regulation between glycolysis and gluconeogenesis. Given Data / Assumptions: F-1,6-BPase converts fructose-1,6-bisphosphate to fructose-6-phosphate. It is a rate-controlling step unique to gluconeogenesis (not present in glycolysis). Allosteric regulators (e.g., AMP, fructose-2,6-bisphosphate) inhibit F-1,6-BPase. Concept / Approach: If an intermediate or regulator inhibits F-1,6-BPase, the glucose-synthesizing pathway is suppressed. In many tissues, increased fructose-2,6-bisphosphate inhibits F-1,6-BPase while stimulating phosphofructokinase-1, thereby favoring glycolysis over gluconeogenesis. Step-by-Step Solution: 1) Identify the enzyme's pathway: F-1,6-BPase is a gluconeogenic bypass enzyme.2) Inhibition reduces conversion of F-1,6-BP to F-6-P.3) This slows glucose production from precursors, i.e., inhibits gluconeogenesis. Verification / Alternative check: Physiological conditions that raise AMP or fructose-2,6-bisphosphate (e.g., fed state, high insulin) suppress hepatic gluconeogenesis, consistent with F-1,6-BPase inhibition. Why Other Options Are Wrong: Glycolysis is enhanced, not inhibited, when F-1,6-BPase is suppressed; “pyrolysis” is a non-biological thermal decomposition process; “none” contradicts known regulation; pentose phosphate pathway is controlled elsewhere (e.g., glucose-6-phosphate dehydrogenase). Common Pitfalls: Assuming every fructose phosphate enzyme belongs to glycolysis; F-1,6-BPase is exclusive to gluconeogenesis. Final Answer: Gluconeogenesis

Question 17

In protein post-translational modification, what is the consensus sequence motif for N-linked glycosylation (attachment of an N-linked oligosaccharide to asparagine) within a polypeptide chain?

Accepted Answer

Introduction / Context: N-linked glycosylation is a major co-/post-translational modification in eukaryotes where an oligosaccharide is attached to the side chain amide nitrogen of asparagine. Recognizing the canonical sequence motif helps in predicting glycoprotein sites and understanding folding, trafficking, and function of membrane and secretory proteins. Given Data / Assumptions: The question asks for the consensus amino-acid sequence around the modified residue. “Consensus” indicates a motif frequently recognized by the oligosaccharyltransferase complex in the endoplasmic reticulum. Standard single-letter position notation: Asn at position 1, any residue (Xaa) at position 2, and Ser or Thr at position 3. Concept / Approach: The well-established motif is Asn-Xaa-Ser/Thr, often written as N-X-S/T. The central position (Xaa) can be almost any residue except proline because proline imposes conformational constraints that reduce enzyme recognition. While not every NXS/T site is glycosylated, most verified N-glycans occur within this sequence context. Step-by-Step Solution: 1) Identify candidate motifs that place Asn (N) at the glycosylation site.2) Check whether the third position is Ser or Thr, which supplies the required hydroxyl for recognition (though the glycan is on Asn).3) Ensure the middle residue is permissive (any residue except Pro preferred).4) Conclude the consensus is Asn-Xaa-(Ser/Thr). Verification / Alternative check: Biochemical mapping and mass spectrometry of glycopeptides repeatedly validate N-X-S/T as the dominant ER cotranslational consensus motif; mutation of any position often abolishes glycosylation. Why Other Options Are Wrong: (Ser or Thr)-Asn-Ala: places Asn in the wrong position. Thr-(Asn or Gln)-Ala: incorrect order and includes Gln, which is not used for N-linkage. None of these: a correct consensus exists. Asn-Pro-(Ser or Thr): proline at X disrupts the motif and typically prevents glycosylation. Common Pitfalls: Confusing O-glycosylation (on Ser/Thr) with N-glycosylation (on Asn), or assuming every N-X-S/T site is necessarily occupied—context and structure also matter. Final Answer: Asn-Xaa-(Ser or Thr)

Question 18

In human metabolism, patients with Von Gierke's disease (Glycogen Storage Disease Type I) cannot perform effective gluconeogenesis to maintain fasting blood glucose because which enzyme is defective?

Accepted Answer

Introduction / Context: Von Gierke's disease (GSD I) is a classic inborn error of carbohydrate metabolism. A key clinical hallmark is severe fasting hypoglycemia because the liver cannot release free glucose into the bloodstream despite active glycogenolysis and gluconeogenesis. Understanding the blocked step pinpoints the responsible enzyme. Given Data / Assumptions: GSD I involves the terminal step of hepatic glucose release. Gluconeogenesis generates glucose-6-phosphate (G6P) in the cytosol. Free glucose export to blood requires dephosphorylation of G6P in the endoplasmic reticulum lumen. Concept / Approach: Glucose-6-phosphatase hydrolyzes G6P to glucose + inorganic phosphate. Without this enzyme (or its transporter G6PT in Type Ib), both glycogenolysis and gluconeogenesis fail to deliver free glucose, causing hypoglycemia and lactic acidosis due to shunting of G6P into glycolysis and lactate production. Step-by-Step Solution: 1) Trace gluconeogenesis to G6P formation.2) Recognize the requirement to convert G6P to glucose for export.3) Identify glucose-6-phosphatase as the essential ER enzyme for this step.4) Defect → G6P cannot be dephosphorylated → no hepatic glucose output. Verification / Alternative check: Biochemical profiles show elevated lactate, triglycerides, and uric acid; hepatomegaly results from glycogen and fat accumulation—findings consistent with a G6Pase pathway defect. Why Other Options Are Wrong: Lactate dehydrogenase: affects lactate/pyruvate interconversion, not final glucose release. Pyruvate carboxylase: its deficiency impairs gluconeogenesis upstream but does not define Von Gierke's disease. Fructose-1,6-bisphosphatase: another gluconeogenic defect (FBPase deficiency) with different presentation. None of these: incorrect because G6Pase is the known culprit in GSD I. Common Pitfalls: Assuming glycogen breakdown alone raises blood glucose; the terminal dephosphorylation step is indispensable for hepatic glucose export. Final Answer: Glucose-6-phosphatase

Question 19

Metabolic pathway design: multistep pathways in cells are considered efficient primarily for which organizational reason?

Accepted Answer

Introduction / Context: Cellular metabolism is organized into multistep pathways rather than single giant reactions. A key efficiency gain comes from the spatial organization of enzymes, which can form complexes (metabolons) or colocalize within specific compartments. This architecture reduces diffusion time, protects labile intermediates, and enables coordinated regulation. Given Data / Assumptions: Pathways involve sequential enzymes. Cells can cluster enzymes or place them within microenvironments (e.g., mitochondria, ER). Efficiency relates to speed, control, and fidelity of intermediate transfer. Concept / Approach: When enzymes are localized together, intermediates are passed directly from one active site to the next (substrate channeling). This minimizes loss to competing reactions, decreases effective transit time, and can couple thermodynamically unfavorable steps to favorable ones through proximity and rapid consumption of intermediates. Step-by-Step Solution: 1) Consider how distance and diffusion limit reaction rates in dilute cytosol.2) Recognize that enzyme clustering reduces intermediate escape and degradation.3) Conclude that colocalization/compartmentation is a central efficiency mechanism.4) Note that many pathways (e.g., TCA cycle, fatty acid synthesis) show physical or compartmental organization. Verification / Alternative check: Experimental reconstitution of multienzyme complexes often shows increased flux compared to dispersed enzymes, consistent with channeling and reduced transit times. Why Other Options Are Wrong: Same substrate for all enzymes: false; each step transforms the substrate into a new intermediate. One enzyme for every step: contradicts the definition of a pathway. Dispersing into different organelles: typically reduces efficiency for a single pathway. No regulation: pathways are highly regulated; localization often enhances regulation. Common Pitfalls: Equating pathway efficiency only with thermodynamics; spatial and temporal organization are equally crucial for flux control and robustness. Final Answer: They localize the pathway’s enzymes together in the same cellular region to facilitate substrate channeling

Question 20

In glycolysis, the aldolase reaction cleaves fructose-1,6-bisphosphate. What are the two triose-phosphate products generated by this split?

Accepted Answer

Introduction / Context: In the payoff phase of glycolysis, the six-carbon sugar fructose-1,6-bisphosphate (F1,6BP) is split by aldolase into two three-carbon intermediates. Correctly naming these products is foundational for understanding how glycolysis doubles the number of triose substrates feeding subsequent ATP-generating steps. Given Data / Assumptions: Substrate: F1,6BP. Enzyme: aldolase (class I in animals). Products are triose phosphates that rapidly interconvert. Concept / Approach: Aldolase catalyzes a reversible aldol cleavage producing dihydroxyacetone phosphate (DHAP, a ketotriose phosphate) and glyceraldehyde-3-phosphate (GAP, an aldotriose phosphate). Triose phosphate isomerase then interconverts DHAP and GAP, funneling both carbons through GAP for oxidation and ATP generation. Step-by-Step Solution: 1) Identify F1,6BP as the six-carbon bisphosphate substrate.2) Apply the aldol cleavage: C3–C4 bond splits to yield DHAP and GAP.3) Recognize that GAP proceeds directly to oxidation; DHAP is converted to GAP.4) Therefore, the pair of products is GAP + DHAP. Verification / Alternative check: Isotopic labeling experiments show distribution of carbon atoms into GAP and DHAP consistent with the aldolase mechanism. Why Other Options Are Wrong: GAP only or DHAP only: aldolase yields both simultaneously. Fructose-6-phosphate + GAP: F6P is upstream, not an aldolase product. None of the above: a standard product pair exists. Common Pitfalls: Confusing aldolase products with phosphofructokinase substrates, or forgetting the rapid isomerization that equalizes triose pools. Final Answer: Glyceraldehyde-3-phosphate and dihydroxyacetone phosphate

Carbohydrate Questions