301. Arrange the following carbon events for complete aerobic oxidation of one glucose molecule.
P. One \(\mathrm{6C}\) glucose forms two \(\mathrm{3C}\) pyruvates.
Q. Two pyruvates form two \(\mathrm{2C}\) acetyl groups and release \(\mathrm{2CO_2}\).
R. Each acetyl group combines with oxaloacetate and enters one TCA turn.
S. The two TCA turns release a further \(\mathrm{4CO_2}\).
ⓐ. \(Q\rightarrow P\rightarrow S\rightarrow R\)
ⓑ. \(P\rightarrow R\rightarrow Q\rightarrow S\)
ⓒ. \(R\rightarrow S\rightarrow P\rightarrow Q\)
ⓓ. \(P\rightarrow Q\rightarrow R\rightarrow S\)
Correct Answer: \(P\rightarrow Q\rightarrow R\rightarrow S\)
Explanation: Glycolysis first divides the six-carbon glucose framework into two three-carbon pyruvate molecules without releasing carbon dioxide. Each pyruvate then undergoes oxidative decarboxylation, losing one carbon as carbon dioxide and producing a two-carbon acetyl group. The two acetyl groups enter separate TCA turns by combining with oxaloacetate. Each turn releases two carbon-dioxide molecules, so the cycle contributes four carbon dioxide per glucose. Adding the two link-reaction molecules gives a total of six carbon dioxide, accounting for all six original glucose carbons. The order must follow substrate formation and entry: acetyl groups cannot enter the cycle before pyruvate oxidation has produced them. Stage-wise totals prevent carbon dioxide from being confused with the later ATP value of electron carriers. The carbon ledger closes at six: two carbons leave during the link reactions and four more leave during the two complete TCA turns.
302. The stage-wise products from one completely oxidised glucose are shown below.
| Stage | Carbon product | Reduced coenzymes | Direct ATP equivalents |
|---|
| Glycolysis | \(\mathrm{2\ pyruvate}\) | \(\mathrm{2(NADH+H^+)}\) | \(\mathrm{2}\) net |
| Two link reactions | \(\mathrm{2CO_2+2\ acetyl\,CoA}\) | \(\mathrm{2(NADH+H^+)}\) | \(\mathrm{0}\) |
| Two TCA turns | \(\mathrm{4CO_2}\) | Missing | \(\mathrm{2}\) |
The missing entry is:
ⓐ. \(\mathrm{4(NADH+H^+)+4FADH_2}\)
ⓑ. \(\mathrm{6(NADH+H^+)+2FADH_2}\)
ⓒ. \(\mathrm{8(NADH+H^+)+2FADH_2}\)
ⓓ. \(\mathrm{6(NADH+H^+)+4FADH_2}\)
Correct Answer: \(\mathrm{6(NADH+H^+)+2FADH_2}\)
Explanation: One TCA turn per acetyl CoA produces three \(\mathrm{NADH+H^+}\) and one \(\mathrm{FADH_2}\). One glucose supplies two acetyl CoA molecules, so the cycle turns twice. Doubling the per-turn reduced-coenzyme output gives \(\mathrm{6(NADH+H^+)}\) and \(\mathrm{2FADH_2}\). These values belong only to the TCA cycle. The two link reactions provide two additional NADH, and glycolysis provides another two, producing ten NADH across the complete pathway. The table also shows that direct ATP formation is much smaller than the energy stored in reduced coenzymes. Most theoretical ATP is generated later when these coenzymes donate electrons to oxidative phosphorylation. Direct ATP, reduced coenzymes and carbon dioxide belong in separate columns until the final summation. Separate ledgers for the link reaction and TCA cycle prevent omissions and double counting. The TCA-cycle row excludes the two NADH formed in the link reactions from its reduced-coenzyme total.
303. Three glucose molecules undergo complete aerobic oxidation. Before oxidative phosphorylation is counted, what totals are produced for carbon dioxide, reduced coenzymes and direct ATP equivalents?
ⓐ. \(\mathrm{18CO_2,\ 30(NADH+H^+),\ 6FADH_2,\ 12\ direct\ ATP\ equivalents}\)
ⓑ. \(\mathrm{12CO_2,\ 24(NADH+H^+),\ 6FADH_2,\ 12\ direct\ ATP\ equivalents}\)
ⓒ. \(\mathrm{18CO_2,\ 30(NADH+H^+),\ 3FADH_2,\ 6\ direct\ ATP\ equivalents}\)
ⓓ. \(\mathrm{18CO_2,\ 24(NADH+H^+),\ 6FADH_2,\ 18\ direct\ ATP\ equivalents}\)
Correct Answer: \(\mathrm{18CO_2,\ 30(NADH+H^+),\ 6FADH_2,\ 12\ direct\ ATP\ equivalents}\)
Explanation: The complete pre-oxidative-phosphorylation output per glucose is \(\mathrm{6CO_2}\), \(\mathrm{10(NADH+H^+)}\), \(\mathrm{2FADH_2}\) and four direct ATP equivalents. The direct ATP total includes two net ATP from glycolysis and two GTP or ATP equivalents from the TCA cycle. For three glucose molecules: \[\mathrm{3\times6CO_2=18CO_2}\] \[\mathrm{3\times10(NADH+H^+)=30(NADH+H^+)}\] \[\mathrm{3\times2FADH_2=6FADH_2}\] \[\mathrm{3\times4\ direct\ ATP=12\ direct\ ATP}\] These twelve ATP equivalents exclude the much larger ATP contribution obtained when the thirty NADH and six FADH\(_2\) molecules are oxidised. The calculation integrates carbon release, electron capture and substrate-level phosphorylation without mixing pathway stages. The threefold scaling must be applied separately to carbon dioxide, each reduced-coenzyme class and direct ATP, preventing oxidative-phosphorylation products from being counted prematurely.
304. Which symbolic relation best summarises the major products formed per glucose before oxidation of reduced coenzymes is converted into ATP?
ⓐ. \(\mathrm{Glucose\rightarrow2CO_2+6(NADH+H^+)+4FADH_2+2\ direct\ ATP}\)
ⓑ. \(\mathrm{Glucose\rightarrow6CO_2+8(NADH+H^+)+4FADH_2+4\ direct\ ATP}\)
ⓒ. \(\mathrm{Glucose\rightarrow6CO_2+10(NADH+H^+)+2FADH_2+4ATP\ equivalents}\)
ⓓ. \(\mathrm{Glucose\rightarrow4CO_2+10(NADH+H^+)+2FADH_2+38\ direct\ ATP}\)
Correct Answer: \(\mathrm{Glucose\rightarrow6CO_2+10(NADH+H^+)+2FADH_2+4ATP\ equivalents}\)
Explanation: Complete oxidation of one glucose releases all six carbon atoms as six carbon-dioxide molecules. Glycolysis forms two NADH, the two link reactions form two more and the two TCA turns form six, giving ten NADH in total. The TCA cycle also produces two FADH\(_2\), one per turn. Direct phosphorylation contributes two net ATP during glycolysis and two GTP or ATP equivalents during the cycle, producing four direct ATP equivalents. The theoretical value of thirty-eight ATP includes later oxidative phosphorylation and cannot be labelled a direct ATP output. The relation deliberately separates reduced-coenzyme production from the ATP that those coenzymes may generate through the inner-membrane respiratory system. This is a pre-oxidative ledger: the ten NADH and two FADH\(_2\) are energy carriers, not ATP molecules already formed. Their later oxidation supplies the difference between four direct equivalents and the theoretical total.
305. All six carbon atoms of one glucose molecule are labelled. Glycolysis and both link reactions occur normally, but entry of acetyl CoA into the TCA cycle is completely blocked. Where will the labelled carbon be found immediately after the link reactions?
ⓐ. All \(\mathrm{6C}\) in carbon dioxide
ⓑ. All \(\mathrm{6C}\) in two acetyl groups
ⓒ. \(\mathrm{4C}\) in carbon dioxide and \(\mathrm{2C}\) in acetyl groups
ⓓ. \(\mathrm{2C\ as\ CO_2,\ 4C\ as\ two\ acetyl\ groups}\)
Correct Answer: \(\mathrm{2C\ as\ CO_2,\ 4C\ as\ two\ acetyl\ groups}\)
Explanation: Glycolysis preserves all six glucose carbons by forming two three-carbon pyruvate molecules. Each pyruvate then undergoes one oxidative decarboxylation. One labelled carbon is removed from each pyruvate as carbon dioxide, producing two labelled carbon-dioxide molecules in total. The remaining two-carbon portion of each pyruvate becomes an acetyl group attached to coenzyme A. The two acetyl groups contain four labelled carbons altogether. Since TCA entry is blocked, those four carbons have not yet entered the cycle or undergone further decarboxylation. Carbon conservation gives \(\mathrm{2C+4C=6C}\), accounting for the complete labelled glucose skeleton at the stated stopping point. The blocked TCA entry prevents any further labelled carbon loss, so the immediate distribution remains two carbons in carbon dioxide and four in acetyl groups. This stopping point is crucial: later cycle reactions cannot redistribute the four acetyl carbons.
306. Equal amounts of glucose are supplied to two cell preparations. Preparation P lacks oxygen and converts pyruvate into lactate. Preparation Q has functional mitochondria and completely oxidises pyruvate. Both initially complete glycolysis. Which comparison is most accurate?
ⓐ. P releases all six glucose carbons as carbon dioxide, while Q releases none
ⓑ. P gains glycolytic ATP only; Q also gains ATP through mitochondrial oxidation
ⓒ. P forms more reduced coenzymes permanently than Q since it lacks oxygen
ⓓ. Q must stop glycolysis as soon as pyruvate enters the mitochondrion
Correct Answer: P gains glycolytic ATP only; Q also gains ATP through mitochondrial oxidation
Explanation: Both preparations begin with glycolysis and obtain a net gain of two ATP through substrate-level phosphorylation. In Preparation P, pyruvate is reduced to lactate. This regenerates \(\mathrm{NAD^+}\) but adds no further net ATP and retains all glucose carbon in reduced organic products. Preparation Q sends pyruvate through the link reaction, TCA cycle and electron-transport system. These stages release carbon dioxide, form reduced coenzymes and produce a much larger ATP yield through oxidative phosphorylation. The fermentative route prevents permanent accumulation of NADH by reoxidising it during lactate formation. The comparison highlights the difference between maintaining glycolysis under oxygen shortage and extracting substantially more energy through complete aerobic oxidation. A cell that cannot reoxidise NADH gradually loses the coenzyme form needed at the PGAL oxidation step. The decisive energetic difference is that Q reoxidises reduced coenzymes through the mitochondrial chain, whereas P obtains no ATP beyond the glycolytic net gain.
307. Consider the following statements about ATP-forming mechanisms in respiration.
I. Substrate-level phosphorylation involves direct phosphate transfer from a metabolic intermediate.
II. Oxidative phosphorylation depends on electron transfer, a proton gradient and ATP synthase.
III. Every substrate-level phosphorylation reaction stops immediately when oxygen is removed.
IV. Formation of GTP during conversion of succinyl CoA into succinate is substrate-level phosphorylation.
ⓐ. I and III only
ⓑ. II, III and IV only
ⓒ. I, II and IV only
ⓓ. I, II, III and IV
Correct Answer: I, II and IV only
Explanation: Substrate-level phosphorylation forms a nucleotide triphosphate through direct transfer or conservation of phosphate energy in a defined metabolic reaction. It occurs during two glycolytic steps and during formation of GTP in the TCA cycle. Oxidative phosphorylation uses a different mechanism: respiratory electron flow establishes a proton gradient, and ATP synthase uses proton return to form ATP. Oxygen removal directly prevents sustained electron transport and oxidative phosphorylation. Cytoplasmic substrate-level phosphorylation may continue under anaerobic conditions when fermentation regenerates the \(\mathrm{NAD^+}\) needed by glycolysis. Statement III is too broad. The valid statements distinguish ATP-forming mechanisms by energy source and coupling process rather than merely by whether they occur somewhere within respiration.
308. Match each blocked respiratory event with the ATP-forming consequence that follows most directly. A Column II entry is used once.
| Column I | Column II |
|---|
| P. BPGA cannot become PGA | 1. Loss of the first glycolytic substrate-level ATP-forming step |
| Q. PEP cannot become pyruvate | 2. Loss of the second glycolytic substrate-level ATP-forming step |
| R. Succinyl CoA cannot become succinate | 3. Loss of TCA-cycle GTP formation |
| S. Proton return through \(F_0\) is blocked | 4. Loss of ATP formation by oxidative phosphorylation |
ⓐ. P-1, Q-2, R-3, S-4
ⓑ. P-2, Q-1, R-4, S-3
ⓒ. P-1, Q-3, R-2, S-4
ⓓ. P-4, Q-2, R-1, S-3
Correct Answer: P-1, Q-2, R-3, S-4
Explanation: Conversion of BPGA into PGA transfers phosphate directly to ADP and supplies the first substrate-level ATP-forming event of the triose payoff sequence, so P-1. Conversion of PEP into pyruvate supplies the second direct glycolytic ATP-forming event, giving Q-2. The succinyl CoA-to-succinate reaction forms GTP by substrate-level phosphorylation in the TCA cycle, so R-3. Proton return through \(F_0\) drives the \(F_1\) catalytic component of ATP synthase; blocking that flow removes oxidative ATP formation, giving S-4. The mapping distinguishes four immediate losses rather than merely naming phosphorylation categories. It also separates direct phosphate-transfer reactions from the membrane-coupled mechanism that depends on a proton gradient. Loss of membrane impermeability breaks the coupling between carrier oxidation and catalytic ATP formation. The mapping separates three direct phosphate-transfer events from proton-gradient-driven ATP formation, so each block removes only its corresponding ATP-producing step.
309. Assertion: All ATP attributed to complete aerobic respiration is formed by oxidative phosphorylation.
Reason: Glycolysis and the TCA cycle also produce ATP or an ATP equivalent through substrate-level phosphorylation.
ⓐ. Both Assertion and Reason are true, and Reason correctly explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Assertion is false, but Reason is true
Explanation: The assertion is false since complete aerobic respiration includes more than one ATP-forming mechanism. Most of the theoretical ATP yield arises through oxidative phosphorylation when NADH and FADH\(_2\) donate electrons to the respiratory chain. A smaller but important portion forms directly. Glycolysis produces a net gain of two ATP through substrate-level phosphorylation, and two TCA turns produce two GTP or ATP equivalents. The reason correctly identifies these direct contributions and explains why the total cannot be assigned entirely to oxidative phosphorylation. Under the stated balance, four ATP equivalents per glucose arise directly, while the remaining theoretical contribution is linked with oxidation of reduced coenzymes at the inner mitochondrial membrane.
310. Complex IV is inhibited for a brief period in a cell that still contains glucose, glycolytic enzymes, ADP and an available route for cytoplasmic \(\mathrm{NAD^+}\) regeneration. Which pattern is expected?
ⓐ. Both glycolytic substrate-level phosphorylation and oxidative phosphorylation increase
ⓑ. Oxidative phosphorylation falls, while glycolytic ATP formation continues briefly
ⓒ. Glycolysis becomes the source of mitochondrial oxygen
ⓓ. TCA-cycle GTP formation increases enough to replace all lost ATP
Correct Answer: Oxidative phosphorylation falls, while glycolytic ATP formation continues briefly
Explanation: Complex IV is required for terminal electron transfer to oxygen. Its inhibition stops normal downstream electron flow, prevents maintenance of the proton gradient and sharply reduces oxidative phosphorylation. Glycolysis is located in the cytoplasm and forms ATP through direct phosphate transfer. An available route for regenerating cytoplasmic \(\mathrm{NAD^+}\) allows the PGAL-oxidation step and subsequent ATP-producing reactions to continue temporarily. This remaining ATP supply is small compared with the lost mitochondrial contribution. TCA activity will eventually decline as mitochondrial NADH accumulates and \(\mathrm{NAD^+}\) becomes limited. Direct glycolytic ATP formation and oxidative phosphorylation can respond differently to the same respiratory-chain inhibitor. Normal oxidative phosphorylation requires both electron transfer and a membrane capable of maintaining unequal proton concentrations.
311. A chemical makes the inner mitochondrial membrane freely permeable to protons without directly inhibiting glycolytic enzymes or TCA-cycle substrate-level phosphorylation. Which response is most likely?
ⓐ. Oxidative ATP formation falls; direct ATP formation initially continues
ⓑ. Every ATP-forming reaction stops before electron transport changes
ⓒ. The proton gradient increases and ATP synthase produces more ATP
ⓓ. PEP and succinyl CoA begin donating electrons directly to oxygen
Correct Answer: Oxidative ATP formation falls; direct ATP formation initially continues
Explanation: Free proton movement collapses the gradient that normally stores energy across the inner mitochondrial membrane. Electron transfer and oxygen consumption may continue, but proton return no longer occurs primarily through ATP synthase, so oxidative ATP formation declines. The treatment does not directly remove the glycolytic enzymes that transfer phosphate from BPGA or PEP, nor does it directly block GTP formation from succinyl CoA. Those substrate-level reactions can continue while their required substrates and coenzymes remain available. Longer-term effects may arise as cellular energy balance and pathway regulation change, but the immediate distinction remains clear. A proton-permeability defect uncouples oxidation from phosphorylation rather than abolishing every direct phosphate-transfer reaction.
312. Four treatments are applied separately to respiring cells.
| Treatment | Primary direct effect |
|---|
| P. Block PEP-to-pyruvate conversion | Loss of one glycolytic substrate-level phosphorylation |
| Q. Block ATP synthase | Loss of oxidative phosphorylation |
| R. Block succinyl CoA-to-succinate conversion | Loss of TCA-cycle GTP formation |
| S. Remove oxygen while glycolytic \(\mathrm{NAD^+}\) regeneration remains possible | Loss of sustained oxidative phosphorylation |
Which pair directly removes oxidative phosphorylation without directly blocking a substrate-level phosphorylation enzyme?
ⓐ. P and R
ⓑ. P and Q
ⓒ. Q and S
ⓓ. R and S
Correct Answer: Q and S
Explanation: Treatment Q directly disables ATP synthase, the enzyme that forms ATP from the proton gradient during oxidative phosphorylation. Treatment S removes the terminal electron acceptor, preventing sustained respiratory electron flow and collapse of oxidative ATP production. Neither treatment directly inhibits the enzymes that transfer phosphate during glycolysis or the enzyme system that forms GTP from succinyl CoA. Treatment P directly removes a glycolytic substrate-level phosphorylation reaction. Treatment R directly removes the TCA-cycle substrate-level event. The comparison separates the site of ATP formation from indirect downstream effects. Oxygen removal may eventually reduce matrix pathway activity, but its primary ATP-forming target is the electron-transport-linked mechanism rather than a direct phosphate-transfer enzyme. Electron flow may continue across a damaged inner membrane, yet loss of proton separation prevents efficient ATP capture. Treatments Q and S remove the membrane-dependent ATP route, whereas P and R directly interrupt enzymes that transfer phosphate from metabolic intermediates.
313. A graph follows an aerobic cell through two changes. At time \(t_1\), oxygen is removed. NADH rises, oxygen consumption becomes zero and mitochondrial ATP production falls, while lactate begins accumulating. At time \(t_2\), oxygen is restored. NADH declines, lactate accumulation slows and mitochondrial ATP production recovers. The best interpretation is:
ⓐ. Oxygen removal directly activates the TCA cycle and inhibits glycolysis
ⓑ. Lactate is the terminal product of complete glucose oxidation
ⓒ. Restoration of oxygen forces all ATP production to become substrate-level
ⓓ. oxygen availability controls carrier oxidation and determines later pyruvate fate
Correct Answer: oxygen availability controls carrier oxidation and determines later pyruvate fate
Explanation: Oxygen removal blocks terminal electron acceptance at complex IV. NADH then accumulates as the respiratory chain loses its ability to reoxidise reduced coenzymes, and mitochondrial ATP production falls with loss of oxidative phosphorylation. Glycolysis can continue if pyruvate is reduced to lactate, a reaction that regenerates cytoplasmic \(\mathrm{NAD^+}\). Lactate accumulation is a response to restricted aerobic processing rather than evidence of complete oxidation. Restoring oxygen reopens terminal electron flow, lowers NADH, permits mitochondrial pyruvate oxidation and reduces dependence on fermentation. The graph integrates redox state, product formation and ATP source into one reversible shift controlled by oxygen availability. The reversal after \(t_2\) shows that the original changes arose from temporary terminal-acceptor limitation rather than permanent damage. Coordinated recovery of carrier oxidation and mitochondrial ATP formation explains the simultaneous slowing of lactate accumulation.
314. A waterlogged germinating seed has abundant stored carbohydrate but limited access to oxygen. Ethanol and carbon dioxide accumulate, the seed gains only a small amount of ATP per glucose and reserve consumption becomes rapid. Which explanation fits all observations?
ⓐ. Complete aerobic oxidation is operating with an RQ of one
ⓑ. alcoholic fermentation regenerates \(\mathrm{NAD^+}\) while producing only a small ATP yield
ⓒ. Fatty acids are the only substrate and enter exclusively through glycolysis
ⓓ. Oxidative phosphorylation is producing the ethanol and carbon dioxide
Correct Answer: alcoholic fermentation regenerates \(\mathrm{NAD^+}\) while producing only a small ATP yield
Explanation: Waterlogging reduces the oxygen available in soil spaces and limits aerobic electron transport in the germinating seed. Glycolysis can still form pyruvate and a net gain of two ATP. Plant cells under oxygen limitation may convert pyruvate through acetaldehyde into ethanol, releasing carbon dioxide and regenerating \(\mathrm{NAD^+}\). This recycling allows glycolysis to continue, but the organic end product retains much of the original glucose energy. The seed must consume reserves rapidly to meet its ATP demand. Ethanol and carbon dioxide identify alcoholic fermentation rather than lactic fermentation. Oxidative phosphorylation is restricted under the stated condition and cannot account for either the low ATP yield or the fermentative products. The product pattern and low yield must be interpreted together.
315. Four respiratory profiles are shown below.
| Profile | Oxygen condition | Major end products | ATP pattern | Principal location beyond glycolysis |
|---|
| P | Present | \(\mathrm{CO_2}\) and \(\mathrm{H_2O}\) | High theoretical yield | Matrix and inner membrane |
| Q | Absent | Ethanol and \(\mathrm{CO_2}\) | \(\mathrm{2}\) net ATP | Cytoplasm |
| R | Limited | Lactic acid | \(\mathrm{2}\) net ATP | Cytoplasm |
| S | Present | Lactic acid only | \(\mathrm{38}\) ATP from fermentation | Inner membrane |
Which profile is biologically inconsistent?
ⓐ. Profile P
ⓑ. Profile Q
ⓒ. Profile S
ⓓ. Profile R
Correct Answer: Profile S
Explanation: Profile S combines features that cannot belong to one respiratory pathway. Lactic acid fermentation occurs in the cytoplasm, where pyruvate is reduced to lactate and \(\mathrm{NAD^+}\) is regenerated for continued glycolysis. It does not use the inner mitochondrial membrane as its principal site. Fermentation also contributes no ATP beyond the two net ATP formed during glycolysis for each glucose molecule. The theoretical yield of \(38\) ATP requires complete aerobic processing through glycolysis, oxidative decarboxylation, the TCA cycle, electron transfer and oxidative phosphorylation under the stated ideal assumptions. Merely supplying oxygen does not convert lactate formation into a high-yield mitochondrial process. The decisive conflict is the simultaneous assignment of a fermentative product, an inner-membrane location and the complete aerobic ATP yield. The combination of lactate as the sole major product, inner-membrane location and \(38\)-ATP fermentative yield makes Profile S biologically inconsistent.
316. Use the pathway arrangement described below. Region P contains glucose, PGAL and pyruvate. An arrow carries pyruvate into Region Q, where acetyl CoA, citrate and oxaloacetate occur. Reduced carriers formed in Q transfer electrons to complexes in Boundary R, where oxygen is reduced and ATP synthase operates. Which interpretation is correct?
ⓐ. P is cytoplasm, Q is matrix and R is inner mitochondrial membrane
ⓑ. P is matrix, Q is cytoplasm and R is plasma membrane
ⓒ. P is intermembrane space, Q is inner membrane and R is cytoplasm
ⓓ. P is cytoplasm, Q is outer mitochondrial membrane and R is matrix
Correct Answer: P is cytoplasm, Q is matrix and R is inner mitochondrial membrane
Explanation: Glucose, PGAL and pyruvate identify the glycolytic pathway, placing Region P in the cytoplasm. Acetyl CoA, citrate and oxaloacetate identify the link reaction and TCA cycle, placing Region Q in the mitochondrial matrix. Boundary R contains respiratory complexes, oxygen reduction and ATP synthase, all characteristic of the inner mitochondrial membrane. The arrows also show the functional connections among compartments. Pyruvate carries carbon from glycolysis into matrix oxidation, while NADH and FADH\(_2\) transfer reducing equivalents from matrix reactions to the membrane carrier system. This organisation permits carbon oxidation and chemiosmotic ATP formation to occur in specialised but coordinated cellular regions. The arrangement demonstrates that complete aerobic respiration requires exchange of metabolites and reducing equivalents across compartment boundaries. The evidence also reflects that integrated accounting keeps glycolytic, link-reaction, TCA and electron-transport contributions separate until the final biological conclusion.
317. Two glucose molecules undergo complete aerobic oxidation under the theoretical \(\mathrm{38\ ATP}\)-per-glucose assumptions. What combined gas and energy values are expected?
ⓐ. \(\mathrm{6O_2\ consumed,\ 12CO_2\ evolved,\ 38ATP}\)
ⓑ. \(\mathrm{12O_2\ consumed,\ 6CO_2\ evolved,\ 76ATP}\)
ⓒ. \(\mathrm{12O_2\ consumed,\ 12CO_2\ evolved,\ 38ATP}\)
ⓓ. \(\mathrm{12O_2\ consumed,\ 12CO_2\ evolved,\ 76ATP}\)
Correct Answer: \(\mathrm{12O_2\ consumed,\ 12CO_2\ evolved,\ 76ATP}\)
Explanation: Complete oxidation of one glucose follows the gas relation \(\mathrm{C_6H_{12}O_6+6O_2\rightarrow6CO_2+6H_2O}\). Two glucose molecules consume \(\mathrm{2\times6=12}\) oxygen molecules or equivalent gas units and release \(\mathrm{2\times6=12}\) carbon-dioxide units. Under the stated idealised balance, each glucose is assigned thirty-eight ATP: \[\mathrm{2\times38\ ATP=76\ ATP}\] Equal oxygen and carbon-dioxide quantities also give an RQ of one. The gas coefficients and ATP values must be scaled by the same number of glucose molecules. The result remains theoretical and assumes complete pathway operation, full stated yield from reduced coenzymes, no intermediate withdrawal and glucose as the sole respiratory substrate. The combined output follows because the link reaction removes one carbon before the two TCA decarboxylations and supplies reduced coenzymes to the membrane system. Accounting remains valid only when products are counted from the stated entry point and under the supplied assumptions. Biological interpretation follows from the dependency between stages rather than from the numerical total alone.
318. Which relationship best classifies the major pathways involved in glucose respiration?
ⓐ. Fermentation is the common pathway that occurs before glycolysis and aerobic respiration
ⓑ. Glycolysis is upstream; pyruvate then enters fermentation or mitochondrial oxidation
ⓒ. Oxidative phosphorylation includes glycolysis, fermentation and the TCA cycle as membrane reactions
ⓓ. The TCA cycle is the common initial pathway for both alcoholic and lactic acid fermentation
Correct Answer: Glycolysis is upstream; pyruvate then enters fermentation or mitochondrial oxidation
Explanation: Glycolysis is the shared cytoplasmic pathway that converts glucose into pyruvate. The pathway does not directly require oxygen and can supply pyruvate under both aerobic and anaerobic conditions. Pyruvate then occupies the branch point. In oxygen-limited cells it may be processed through alcoholic or lactic acid fermentation, depending on the organism or tissue. Under aerobic conditions in eukaryotic cells, pyruvate enters the mitochondrial matrix, forms acetyl CoA and proceeds through the TCA cycle and oxidative phosphorylation. Fermentation is not upstream of glycolysis, and the TCA cycle does not initiate the fermentative routes. The classification places the shared pathway first and the condition-dependent alternatives afterward.
319. Three inhibitors are tested separately in aerobic cells.
P. A hexokinase inhibitor causes glucose accumulation and lowers pyruvate, carbon dioxide and ATP formation.
Q. A pyruvate-transport inhibitor causes cytosolic pyruvate accumulation and lowers matrix carbon-dioxide production.
R. A complex-IV inhibitor causes reduced electron carriers to accumulate while oxygen consumption falls.
Which evaluation is correct?
ⓐ. Only P identifies the inhibited step correctly
ⓑ. Only Q and R identify their inhibited steps correctly
ⓒ. P, Q and R each show the expected upstream and downstream pattern
ⓓ. None of the records is consistent with respiratory compartmentation
Correct Answer: P, Q and R each show the expected upstream and downstream pattern
Explanation: Hexokinase acts near the beginning of glycolysis, so its inhibition leaves glucose unprocessed and reduces every downstream product derived from glycolytic pyruvate. A pyruvate-transport block permits cytoplasmic glycolysis to continue but prevents normal delivery of pyruvate into the matrix, producing cytosolic accumulation and reduced mitochondrial decarboxylation. Complex IV is the terminal electron-transfer complex. Its inhibition prevents electrons from reaching oxygen, so upstream carriers accumulate in reduced form and oxygen consumption decreases. Each record follows the same diagnostic principle: the substrate or reduced carrier before a block accumulates, while products or processes beyond the block decline. Together, the treatments map glycolytic, transport and membrane stages of aerobic respiration. The three records together trace sequential failure at the cytoplasmic pathway, matrix-entry step and terminal membrane complex.
320. An illuminated green leaf has an adequate carbohydrate supply and is exchanging gases normally. At time \(t_1\), it is transferred to darkness. At time \(t_2\), external oxygen is also removed while carbohydrate remains available. Which sequence is most likely?
ⓐ. Photosynthetic oxygen stops at \(t_1\); continuing respiration may favour fermentation after \(t_2\)
ⓑ. Respiration stops at \(t_1\); photosynthesis continues, and the TCA cycle accelerates after \(t_2\)
ⓒ. Glycolysis stops at \(t_2\), while oxidative phosphorylation continues without oxygen
ⓓ. Carbon dioxide production ends at \(t_1\), and glucose becomes the terminal electron acceptor after \(t_2\)
Correct Answer: Photosynthetic oxygen stops at \(t_1\); continuing respiration may favour fermentation after \(t_2\)
Explanation: Darkness stops the light-dependent production of oxygen by photosynthesis, but it does not directly stop cellular respiration. The leaf can continue glycolysis and aerobic mitochondrial oxidation while external or internal oxygen remains available. Removal of external oxygen at \(t_2\) progressively limits terminal electron acceptance, oxidative phosphorylation and regeneration of mitochondrial oxidised coenzymes. Glycolysis itself does not directly require oxygen and may continue if a fermentative pathway regenerates cytoplasmic \(\mathrm{NAD^+}\). The ATP yield then becomes much smaller than under aerobic conditions. This sequence integrates the independence of respiration from light, the terminal role of oxygen and the redox purpose of fermentation in an oxygen-limited plant tissue. After oxygen removal, the key metabolic change is loss of the high-yield aerobic route while glycolysis may persist only if fermentative coenzyme recycling is available.
