Question 1

Pathways to pyruvate: Which of the following pathways can directly catabolize glucose to pyruvate in bacteria?

Accepted Answer

Introduction / Context: Microbes employ several routes to degrade glucose to pyruvate, the central node of carbon metabolism. Two well-known glycolytic routes are the Embden–Meyerhof–Parnas (EMP) pathway (classical glycolysis) and the Entner–Doudoroff (ED) pathway. The tricarboxylic acid (TCA) cycle does not convert glucose to pyruvate; it oxidizes acetyl-CoA derived from pyruvate. Given Data / Assumptions: We focus on direct conversion of glucose to pyruvate. EMP is not listed among the options; ED is. TCA operates downstream of pyruvate formation. Concept / Approach: The ED pathway, present in many Gram-negative bacteria (e.g., Pseudomonas, Zymomonas), converts glucose to pyruvate (and glyceraldehyde-3-phosphate) with a distinct set of enzymes and yields fewer ATP than EMP. The TCA cycle, by contrast, is not a glycolytic route and requires acetyl-CoA input; thus it does not answer the question of converting glucose to pyruvate. Step-by-Step Solution: Identify which listed pathway generates pyruvate directly from glucose. Select ED pathway; exclude TCA as it follows pyruvate decarboxylation to acetyl-CoA. Reject combined option since TCA does not perform the stated step. Verification / Alternative check: Standard pathway maps show glucose → 6-phosphogluconate → KDPG → pyruvate (via ED). TCA begins after pyruvate dehydrogenase forms acetyl-CoA. Why Other Options Are Wrong: TCA cycle / Electron transport: Neither converts glucose to pyruvate. Both / None: Incorrect because ED alone among the options fits the criterion. Common Pitfalls: Assuming all central metabolic cycles start with glucose; TCA is an oxidative hub after glycolysis, not a substitute for it. Final Answer: Entner–Doudoroff pathway directly degrades glucose to pyruvate.

Question 2

ATP yield comparison: Relative to complete aerobic respiration of glucose, the amount of ATP produced by fermentation is generally…

Accepted Answer

Introduction / Context: Cells harvest energy by oxidizing substrates such as glucose. The total ATP yield depends on whether electrons are transferred to oxygen via respiration or to organic acceptors via fermentation. Understanding the difference is fundamental to microbial physiology and industrial bioprocessing. Given Data / Assumptions: Aerobic respiration includes glycolysis, pyruvate oxidation, TCA cycle, and oxidative phosphorylation. Fermentation includes glycolysis followed by reduction of organic molecules to reoxidize NADH to NAD+. Typical bacterial aerobic yields greatly exceed the ~2 ATP per glucose from glycolysis alone. Concept / Approach: Fermentation yields ATP primarily by substrate-level phosphorylation during glycolysis (net 2 ATP per glucose). Aerobic respiration captures additional energy via the electron transport chain and ATP synthase, often yielding an order of magnitude more ATP per glucose depending on organism and conditions. Step-by-Step Solution: Recognize fermentation lacks an external terminal electron acceptor like oxygen. Acknowledge that without oxidative phosphorylation, ATP comes mainly from glycolysis. Conclude fermentation yields less ATP than aerobic respiration. Verification / Alternative check: Empirical measurements of ATP yield or growth yield (g cell per g glucose) consistently show higher values for aerobic conditions compared with fermentative growth. Why Other Options Are Wrong: Greater / Equal: Contradicted by the absence of oxidative phosphorylation in fermentation. Indistinguishable / Zero: Incorrect; mechanisms and yields differ, and fermentation does produce some ATP. Common Pitfalls: Confusing anaerobic respiration (non-oxygen electron acceptors) with fermentation; anaerobic respiration can yield more ATP than fermentation but typically less than aerobic. Final Answer: Lesser than by aerobic metabolism.

Question 3

Fermentation stoichiometry — oxygen requirement: In the classic fermentation of one molecule of glucose to ethanol and carbon dioxide (as in yeast), how many molecular oxygen (O2) molecules are required as reactants?

Accepted Answer

Introduction / Context: Fermentation is an anaerobic energy-yielding process in which an organic molecule acts as both electron donor and electron acceptor. The textbook example is alcoholic fermentation, where glucose is converted to ethanol and carbon dioxide by yeast under oxygen-limited conditions. Given Data / Assumptions: Alcoholic fermentation pathway: glucose → pyruvate → acetaldehyde → ethanol + CO2. Electron acceptor is an organic intermediate (acetaldehyde), not O2. Goal is NADH reoxidation to NAD+ to sustain glycolysis. Concept / Approach: Because no external terminal electron acceptor is used, molecular oxygen is not required as a reactant in fermentation. The pathway operates with internal redox balance: NADH produced in glycolysis is consumed when acetaldehyde is reduced to ethanol, regenerating NAD+. Step-by-Step Solution: Glycolysis: glucose → 2 pyruvate + ATP + NADH. Decarboxylation: pyruvate → acetaldehyde + CO2. Reduction: acetaldehyde + NADH → ethanol + NAD+. Count O2 reactants: none are used; answer is zero. Verification / Alternative check: Balanced equations show conservation of electrons without O2. In the presence of oxygen, yeast preferentially respire (Crabtree effect in high glucose aside), but the question specifies fermentation, which by definition excludes O2 as electron acceptor. Why Other Options Are Wrong: 1, 2, 36, >10: These imply aerobic respiration or oxidative processes, not fermentation. Common Pitfalls: Confusing the need for small amounts of oxygen for sterol or unsaturated fatty acid synthesis in yeasts with the stoichiometric requirement of O2 in the fermentative pathway itself; the latter is zero. Final Answer: 0 molecular oxygen molecules are required for alcoholic fermentation of glucose.

Question 4

Terminal electron acceptor in lactic acid fermentation: During homolactic fermentation of glucose by lactic acid bacteria, which compound serves as the final electron acceptor?

Accepted Answer

Introduction / Context: Lactic acid fermentation is widely used in food biotechnology (yogurt, cheese, pickles) and is central to anaerobic energy metabolism in certain microbes and animal tissues. The key is understanding how cells reoxidize NADH produced in glycolysis when external electron acceptors are absent. Given Data / Assumptions: Glycolysis converts glucose to pyruvate, generating ATP and NADH. Homolactic fermenters (e.g., Lactobacillus, Streptococcus) reduce pyruvate directly to lactate. NADH must be reoxidized to NAD+ to allow continued glycolytic ATP production. Concept / Approach: In homolactic fermentation, pyruvate accepts electrons from NADH, forming lactate and regenerating NAD+. Thus, pyruvate is the final electron acceptor. This internal redox balancing enables ATP yield via substrate-level phosphorylation independent of oxygen or respiratory chains. Step-by-Step Solution: Glycolysis: glucose → 2 pyruvate + 2 NADH + 2 ATP. Reduction step: 2 pyruvate + 2 NADH → 2 lactate + 2 NAD+. Identify the acceptor: pyruvate is reduced to lactate and accepts electrons from NADH. Select “Pyruvate.” Verification / Alternative check: Enzyme lactate dehydrogenase catalyzes the reaction, and measuring lactate accumulation with concurrent oxidation of NADH in vitro confirms the acceptor role of pyruvate. Why Other Options Are Wrong: Fructose / Glucose: Substrates in glycolysis, not final electron acceptors in homolactic fermentation. Acetyl-CoA: Intermediate in respiration; not produced in homolactic fermentation. O2: Not used in fermentation; this is an anaerobic pathway. Common Pitfalls: Confusing heterolactic pathways (which produce multiple products) with homolactic ones; in both cases, an organic molecule derived from glucose, not oxygen, accepts electrons. Final Answer: Pyruvate is the final electron acceptor in lactic acid fermentation.

Question 5

Anaerobic respiration – identifying terminal electron acceptors In microbial energy metabolism under anaerobic respiration (not fermentation), the final electron acceptor can be which of the following?

Accepted Answer

Introduction / Context: Anaerobic respiration is a form of respiration that uses an electron transport chain in the absence of oxygen. Instead of oxygen, microbes reduce alternative, externally supplied terminal electron acceptors. Distinguishing anaerobic respiration from fermentation (which lacks an electron transport chain) is a high-yield concept in microbiology and bioenergetics. Given Data / Assumptions: Anaerobic respiration uses an electron transport system. Terminal electron acceptors are extracellular and inorganic in classic cases. Fermentation uses endogenous organic molecules (e.g., pyruvate) as electron sinks without an electron transport chain. Concept / Approach: Under anaerobic respiration, microbes pass electrons to oxidized inorganic ions such as nitrate (NO₃⁻), nitrite (NO₂⁻), sulphate (SO₄²⁻), elemental sulfur (S⁰), or carbon dioxide (CO₂, yielding methane in methanogenesis). Nitrate reduction to nitrite or further to nitrogenous gases (denitrification) is a classic pathway. Oxygen is the terminal acceptor only in aerobic respiration. Pyruvate and acetyl-CoA are intracellular organic intermediates typically functioning as electron sinks in fermentation or as entry molecules into the Krebs cycle, not as external respiratory acceptors in an electron transport chain. Step-by-Step Solution: Identify the process: anaerobic respiration employs an electron transport chain but lacks O₂.List canonical inorganic acceptors: NO₃⁻, SO₄²⁻, S⁰, CO₂ (in certain archaea).Match options: nitrate (NO₃⁻) fits; oxygen does not; pyruvate and acetyl-CoA are fermentation/central metabolism intermediates.Select nitrate as the correct answer. Verification / Alternative check: Growth yields (ATP per substrate) in nitrate respiration exceed those of fermentation, consistent with energy conservation via an electron transport chain and proton motive force generation. Why Other Options Are Wrong: Oxygen (O₂): defines aerobic respiration, not anaerobic. Pyruvate: an internal organic acceptor in fermentation, not in respiratory chains. Acetyl-CoA: metabolic intermediate; not a terminal acceptor. Sulphate (SO₄²⁻): is indeed an anaerobic acceptor, but the question asks for one correct choice from the list; nitrate is the intended best answer here. Common Pitfalls: Confusing fermentation with anaerobic respiration; assuming any lack of oxygen implies fermentation only. Remember: presence of an electron transport chain distinguishes respiration (aerobic or anaerobic) from fermentation. Final Answer: Nitrate (NO₃⁻).

Question 6

Krebs (citric acid) cycle entry step Before most 2-carbon fragments derived from carbohydrates, fats, or proteins can enter the Krebs citric acid cycle, they must be converted into which common intermediate?

Accepted Answer

Introduction / Context: The Krebs (citric acid) cycle is a central hub of metabolism. Diverse fuels—glucose, fatty acids, and amino acids—are funneled into a common 2-carbon carrier before condensation with oxaloacetate. Recognizing this “gateway” molecule is essential for integrating catabolic pathways. Given Data / Assumptions: Glycolysis yields pyruvate from glucose. Beta-oxidation yields acetyl units from fatty acids. Certain amino acids are catabolized to acetyl units or TCA intermediates. Concept / Approach: Pyruvate is decarboxylated and activated by the pyruvate dehydrogenase complex to form acetyl-CoA (a 2-carbon thioester) plus CO₂ and NADH. Acetyl-CoA then condenses with oxaloacetate to form citrate, initiating the cycle. NADH and FADH₂ are electron carriers generated by prior steps; they are not substrates entering the cycle. Step-by-Step Solution: Trace glucose → pyruvate via glycolysis.Convert pyruvate → acetyl-CoA by oxidative decarboxylation.Feed acetyl-CoA into the Krebs cycle by combining with oxaloacetate to form citrate.Therefore, the required common intermediate is acetyl-CoA. Verification / Alternative check: Fatty acid beta-oxidation and ketogenic amino acid breakdown also produce acetyl-CoA, converging on the same entry route into the TCA cycle, confirming its central role. Why Other Options Are Wrong: Citrate/oxaloacetate: cycle intermediates, not the universal entry product. NADH/FADH₂: electron carriers, not carbon fragments entering the cycle. Pyruvate: must first be converted to acetyl-CoA in aerobic metabolism. Common Pitfalls: Confusing pyruvate with acetyl-CoA; overlooking the regulatory role of the pyruvate dehydrogenase complex in controlling carbohydrate flux into the TCA cycle. Final Answer: Acetyl-CoA.

Question 7

Role of oxygen in cellular metabolism Molecular oxygen (O₂) is used in which direct role within aerobic energy metabolism?

Accepted Answer

Introduction / Context: Understanding precisely where oxygen participates clarifies why aerobic organisms gain more ATP per molecule of glucose than anaerobes. Oxygen’s role is tightly linked to the electron transport chain (ETC) and oxidative phosphorylation in mitochondria or bacterial membranes. Given Data / Assumptions: Glycolysis does not require O₂ directly. The Krebs cycle does not use O₂ directly as a substrate. Photophosphorylation occurs in phototrophs using light-excited electrons. Concept / Approach: In aerobic respiration, electrons from NADH and FADH₂ flow through ETC complexes to oxygen. Oxygen serves as the terminal electron acceptor, being reduced to water (H₂O). This electron flow drives proton pumping and establishes a proton motive force used by ATP synthase to form ATP. Neither glycolysis nor the Krebs cycle consumes molecular oxygen directly; their dependence on oxygen is indirect via the need to reoxidize NADH/FADH₂ through the ETC. Step-by-Step Solution: Identify the pathway that requires O₂ directly: the ETC.Recognize O₂ is reduced to H₂O at cytochrome c oxidase (Complex IV in mitochondria).Link this reduction to proton pumping and ATP synthesis by ATP synthase.Choose “terminal electron acceptor” as the correct role. Verification / Alternative check: Inhibitors of Complex IV (e.g., cyanide) block O₂ reduction, collapse proton motive force, and halt ATP production, confirming oxygen’s role at the chain’s terminus. Why Other Options Are Wrong: Photophosphorylation: driven by light, not oxygen as an electron donor. Glycolysis/Krebs: no direct O₂ substrates in their reactions. Substrate-level phosphorylation: occurs in glycolysis and Krebs independently of O₂. Common Pitfalls: Assuming “aerobic” means oxygen is used in every pathway step; it is specifically the ETC that requires O₂. Final Answer: As the terminal electron acceptor in the electron transport chain.

Question 8

Krebs cycle yields per turn – which product is made in the greatest number? Considering one turn of the Krebs (TCA) cycle starting from one acetyl-CoA, which listed product is generated in the highest number of molecules?

Accepted Answer

Introduction / Context: The Krebs cycle generates reduced electron carriers and a small amount of ATP/GTP per acetyl unit oxidized. Recognizing stoichiometry per cycle helps estimate ATP yield via oxidative phosphorylation and compare substrates. Given Data / Assumptions: Input: one acetyl-CoA condenses with oxaloacetate. Classic stoichiometry per turn: 3 NADH, 1 FADH₂, 1 ATP (or GTP), and 2 CO₂. Counts refer to one cycle turn, not per glucose (which yields two turns). Concept / Approach: NADH is produced at three dehydrogenase steps (isocitrate dehydrogenase, α-ketoglutarate dehydrogenase, and malate dehydrogenase). FADH₂ is produced once (succinate dehydrogenase). Substrate-level phosphorylation yields one ATP (or GTP) at the succinyl-CoA synthetase step. Therefore, the most numerous reduced carrier per turn is NADH. Step-by-Step Solution: List outputs per acetyl-CoA: 3 NADH, 1 FADH₂, 1 ATP/GTP, 2 CO₂.Compare counts: 3 (NADH) > 2 (CO₂) > 1 (FADH₂, ATP/GTP).Select NADH as produced in the greatest number. Verification / Alternative check: Per glucose, the cycle turns twice, giving 6 NADH, 2 FADH₂, and 2 ATP/GTP—again confirming NADH predominance. Why Other Options Are Wrong: Acetyl-CoA: consumed, not produced. FADH₂/ATP: only one each per turn. CO₂: two per turn, fewer than the three NADH. Common Pitfalls: Mixing per-turn counts with per-glucose totals; forgetting that some textbooks denote GTP interchangeably with ATP at the succinyl-CoA step. Final Answer: NADH.

Question 9

Oxidative phosphorylation – required conditions Organisms can synthesize ATP by oxidative phosphorylation under which of the following conditions?

Accepted Answer

Introduction / Context: Oxidative phosphorylation couples the energy released by electron transfer through a respiratory chain to ATP synthesis by ATP synthase. It is distinct from substrate-level phosphorylation and photophosphorylation. Identifying the condition that enables oxidative phosphorylation is critical for understanding energy yields in cells. Given Data / Assumptions: Presence of an electron transport chain (ETC) is essential. A suitable terminal electron acceptor (commonly O₂ in aerobic respiration) is required. Proton motive force across a membrane drives ATP synthase. Concept / Approach: During aerobic respiration, electrons from NADH/FADH₂ flow through ETC complexes (including cytochromes) to O₂. Proton pumping establishes a gradient used by ATP synthase to form ATP—this is oxidative phosphorylation. Fermentation lacks an ETC; glycolysis alone does not constitute oxidative phosphorylation. Photophosphorylation is a light-driven, not oxidative, process in phototrophs. Step-by-Step Solution: Identify options that include an ETC and terminal acceptor.Recognize that electron flow to O₂ via cytochromes defines aerobic oxidative phosphorylation.Select the corresponding option. Verification / Alternative check: Inhibitors of the ETC (e.g., cyanide, antimycin A) or uncouplers (e.g., DNP) block ATP synthesis via oxidative phosphorylation, confirming the dependence on electron transfer to O₂ and proton motive force. Why Other Options Are Wrong: Fermentation/glycolysis: rely on substrate-level phosphorylation; no ETC involvement. Photophosphorylation: light-driven in chloroplasts or photosynthetic membranes, not oxidative phosphorylation. Substrate-level phosphorylation in Krebs: generates GTP/ATP without an ETC. Common Pitfalls: Equating any ATP formation with oxidative phosphorylation; only ETC-driven ATP synthesis qualifies. Final Answer: When electrons pass to oxygen via an electron transport chain containing cytochromes.

Question 10

Glycolysis products – identify what is NOT produced Which of the following molecules is not generated at any step of the glycolytic pathway (from glucose to pyruvate)?

Accepted Answer

Introduction / Context: Glycolysis is a ten-step pathway converting glucose to pyruvate, capturing energy via substrate-level phosphorylation and reducing power in the form of NADH. Knowing exactly which cofactors and nucleotides are consumed versus produced prevents common exam errors. Given Data / Assumptions: NAD⁺ is required as an oxidizing agent at the glyceraldehyde-3-phosphate dehydrogenase step. NADH is produced from the reduction of NAD⁺. ATP is both consumed (investment phase) and produced (payoff phase). ADP is formed when ATP is consumed and serves as a substrate when ATP is produced. Concept / Approach: During glycolysis, NAD⁺ is consumed (reduced) to NADH and must be regenerated elsewhere (e.g., by the ETC in aerobic cells or by fermentation pathways). Thus, NAD⁺ is not a product; it is a reactant. By contrast, NADH, ATP, ADP, and pyruvate all appear as outputs at various steps (net: 2 ATP, 2 NADH, 2 pyruvate per glucose). Step-by-Step Solution: Track cofactors: NAD⁺ + electrons → NADH (produced), meaning NAD⁺ is consumed.Track nucleotides: ATP consumed early → ADP appears; later ADP → ATP produced.End product: pyruvate is formed.Therefore, the molecule not produced in glycolysis is NAD⁺. Verification / Alternative check: Stoichiometry per glucose: glucose + 2 ADP + 2 Pi + 2 NAD⁺ → 2 pyruvate + 2 ATP + 2 NADH + 2 H⁺ + 2 H₂O, confirming NAD⁺ is a substrate, not a product. Why Other Options Are Wrong: NADH: produced at the oxidation of glyceraldehyde-3-phosphate. ADP: produced when ATP is consumed in early steps. ATP: generated by substrate-level phosphorylation at two later steps. Pyruvate: the final carbon product of glycolysis. Common Pitfalls: Confusing cofactor regeneration outside glycolysis (e.g., ETC or fermentation) with glycolytic steps themselves. Final Answer: NAD⁺.

Question 11

P/O ratio (classic value) – NADH to ATP For each pair of electrons transferred from one NADH (generated inside mitochondria) to oxygen via the electron transport chain, how many ATP molecules are formed under classical textbook assumptions?

Accepted Answer

Introduction / Context: The phosphorylation-to-oxygen (P/O) ratio links electron transport to ATP synthesis. While modern bioenergetics often cites approximate values (about 2.5 ATP per mitochondrial NADH), many foundational exam keys use the classic value. Knowing the convention being tested avoids avoidable mistakes. Given Data / Assumptions: Classic textbooks: 3 ATP per NADH, 2 ATP per FADH₂. Modern estimates: ~2.5 ATP per NADH, ~1.5 ATP per FADH₂. The question references “inside the mitochondria,” implying no shuttle losses. Concept / Approach: Using the traditional convention, one NADH donates a pair of electrons to the ETC, driving proton pumping at Complexes I, III, and IV. The resulting proton motive force is sufficient (in the classical model) to generate approximately 3 ATP per NADH oxidized. Many standardized tests retain this rounded value for simplicity. Step-by-Step Solution: Apply the classic P/O ratio for NADH: 3 ATP per NADH.Confirm context: mitochondrial NADH avoids cytosolic shuttle penalties.Select “3” as the answer. Verification / Alternative check: Calculations of total ATP per glucose in older schemas (36 ATP for eukaryotes) derive from 3 ATP per NADH and 2 per FADH₂, validating the testing convention. Why Other Options Are Wrong: 1 or 2: underestimate classical yield. 4: exceeds typical theoretical yields. 2.5: closer to modern estimates but not the traditional exam key. Common Pitfalls: Mixing modern mechanistic values with legacy exam conventions; always follow the convention indicated by companion questions (e.g., total ATP per glucose of 36 in eukaryotes). Final Answer: 3.

Question 12

Total ATP yield in eukaryotic cells (yeast/fungi) A yeast or fungal cell completely oxidizing one molecule of glucose typically nets how many ATP molecules under classical textbook accounting?

Accepted Answer

Introduction / Context: Textbook ATP accounting often assigns standard integer values to ATP yields. For eukaryotes, the classic number for complete oxidation of one glucose is 36 ATP, reflecting mitochondrial respiration and transport costs. Recognizing this convention helps reconcile related questions on P/O ratios and pathway contributions. Given Data / Assumptions: Classic yields: 3 ATP per NADH, 2 ATP per FADH₂. Two ATP are consumed or not realized due to cytosolic NADH shuttling into mitochondria. Prokaryotes are often cited as yielding 38 ATP per glucose under similar assumptions. Concept / Approach: Glycolysis (net 2 ATP + 2 NADH), pyruvate oxidation (2 NADH), TCA cycle (2 ATP/GTP + 6 NADH + 2 FADH₂). Applying classic P/O ratios and accounting for shuttle costs yields 36 ATP in eukaryotes. Modern estimates are lower (about 30–32), but many exams use the traditional figure, especially when paired with a “3 ATP per NADH” assumption elsewhere. Step-by-Step Solution: Sum ATP from substrate-level phosphorylation: 2 (glycolysis) + 2 (TCA) = 4.Sum ATP from NADH/FADH₂ using classic ratios and subtract shuttle costs to reach ~36.Select “36” as the standard textbook answer. Verification / Alternative check: Prokaryotic totals are traditionally given as 38 ATP, reinforcing that the eukaryotic figure is 36 under the classic framework. Why Other Options Are Wrong: 38: classic prokaryotic figure, not eukaryotic with shuttling cost. 34/32/30: align with various modern estimates but not with the conventional exam total tied to a 3/2 P/O scheme. Common Pitfalls: Switching between modern mechanistic yields and legacy numbers within the same problem set; stay consistent with the question set’s conventions. Final Answer: 36.

Question 13

ATP yield of fermentation in bacteria Most bacterial fermentations generate how many net ATP molecules per molecule of glucose?

Accepted Answer

Introduction / Context: Fermentation is a strategy to reoxidize NADH to NAD⁺ in the absence of an electron transport chain, enabling glycolysis to continue. Energy capture is limited to substrate-level phosphorylation. Knowing the net ATP yield is fundamental to comparing metabolic strategies. Given Data / Assumptions: Glycolysis net ATP: 2 ATP per glucose. Fermentation pathways regenerate NAD⁺ but do not add ATP beyond glycolytic substrate-level phosphorylation. No oxidative phosphorylation occurs in fermentation. Concept / Approach: Regardless of the fermentation end product (lactate, ethanol, mixed acids), the ATP yield comes exclusively from glycolysis. The net is 2 ATP per glucose because 2 ATP are invested and 4 ATP are produced at substrate-level phosphorylation steps (net gain 2). Fermentation endpoints serve redox balance, not additional ATP gain. Step-by-Step Solution: Account for ATP in glycolysis: invest 2, harvest 4 → net 2 ATP.Recognize fermentation adds no further ATP via an ETC.Therefore, total net ATP per glucose under fermentation is 2. Verification / Alternative check: Comparisons with aerobic respiration (30–38 ATP depending on convention) underscore how fermentation yields are minimal, explaining slower growth and higher substrate demand under fermentative conditions. Why Other Options Are Wrong: 1 or 0: underestimate yield; glycolysis contributes 2 ATP net. 4 or 6: overestimate; no ETC-driven ATP is added in fermentation. Common Pitfalls: Confusing gross ATP produced (4) with net (2) in glycolysis; forgetting ATP investment steps early in the pathway. Final Answer: 2.

Question 14

Subcellular location of glycolysis enzymes In which cellular compartment are the enzymes of the glycolytic pathway located?

Accepted Answer

Introduction / Context: Glycolysis is a central metabolic pathway present across all domains of life. Its cellular location influences how intermediates are channeled into downstream pathways such as fermentation, the Krebs cycle, or biosynthesis. Given Data / Assumptions: Prokaryotes lack mitochondria and chloroplasts. Eukaryotes compartmentalize many pathways, but glycolysis remains cytosolic. Mitochondrial inner membrane houses the ETC and oxidative phosphorylation, not glycolytic enzymes. Concept / Approach: In both eukaryotic and prokaryotic cells, glycolytic enzymes reside in the cytoplasm (cytosol). In eukaryotes, pyruvate enters mitochondria for oxidative decarboxylation and the Krebs cycle, while NADH and ATP generated in glycolysis remain cytosolic prior to further processing. Membrane surfaces and organelles host other processes (ETC in mitochondria, photosynthesis in chloroplasts), but not the glycolytic sequence. Step-by-Step Solution: Recall pathway compartmentation: glycolysis → cytosol in eukaryotes and cytoplasm in bacteria.Contrast with mitochondrial inner membrane (ETC) and chloroplast membranes (photosynthetic complexes).Select “In the cytoplasm (cytosol)” as correct. Verification / Alternative check: Cell fractionation experiments place glycolytic enzymes (e.g., hexokinase, phosphofructokinase, pyruvate kinase) in cytosolic fractions, corroborating their location. Why Other Options Are Wrong: Plasma membrane inner surface: may host transporters but not the soluble glycolytic enzyme set. Inner mitochondrial membrane: site of ETC and ATP synthase. Outer chloroplast membrane: not where glycolysis resides; chloroplasts host the Calvin cycle in plants. Peroxisomes: house oxidative reactions but not the glycolytic series. Common Pitfalls: Assuming that because respiration occurs in mitochondria, glycolysis must also be mitochondrial; in fact, it is cytosolic. Final Answer: In the cytoplasm (cytosol).

Question 15

The number of ATP molecules produced from one glucose molecule by a bacterium producing lactic acid is

Accepted Answer

approximately equal to producing ethanol

Question 16

In glycolysis, ATP is created by

Accepted Answer

substrate level phosphorylation

Question 17

How many molecules of carbon dioxide will be given off during ten turns of the Krebs cycle?

Accepted Answer

20

Question 18

The enzymes that catalyze the reactions of the Krebs cycle are found in which subcellular organelle of eukaryotes?

Accepted Answer

Mitochondrion

Question 19

Which of the following best explains about the usefulness of the production of ethanol in yeast cells under anaerobic conditions?

Accepted Answer

The process regenerates NAD+, which is required for glycolysis

Question 20

Which of the following is the most complete definition of fermentation?

Accepted Answer

Oxidation of glucose with organic molecules serving as electron acceptors

Energy Release and Conservation Questions