401. Preparation P is a soluble extract supplied with PEP and ADP; it forms ATP without a membrane or oxygen. Preparation Q is an inner-membrane vesicle supplied with ADP and inorganic phosphate; it forms ATP only when a proton gradient is imposed. Which classification is correct?
ⓐ. P and Q both demonstrate substrate-level phosphorylation
ⓑ. P and Q both demonstrate chemiosmotic ATP synthesis
ⓒ. P demonstrates chemiosmotic ATP synthesis, while Q demonstrates fermentation
ⓓ. P shows substrate-level phosphorylation; Q shows chemiosmotic ATP synthesis
Correct Answer: P shows substrate-level phosphorylation; Q shows chemiosmotic ATP synthesis
Explanation: In Preparation P, phosphoenolpyruvate transfers a phosphate group directly to ADP while pyruvate is formed. This membrane-independent reaction is substrate-level phosphorylation. Preparation Q behaves differently: ATP appears only when a transmembrane proton gradient is supplied, so proton flow through ATP synthase is the immediate energy source for phosphorylation of ADP. The vesicle demonstrates chemiosmotic ATP synthesis. It isolates the gradient-use step of the mitochondrial mechanism but does not by itself demonstrate complete oxidative phosphorylation, since no respiratory oxidation or electron-transfer chain establishes the gradient in the preparation. In intact mitochondria, electron transfer forms the proton gradient and ATP synthase uses that gradient, linking oxidation with phosphorylation. This distinction prevents ATP-synthase activity alone from being labelled complete oxidative phosphorylation when the oxidation-linked formation of the gradient is absent. The comparison separates direct phosphate transfer from ATP formation driven by an experimentally imposed proton-motive difference.
402. Oxygenated cells show normal glycolytic ATP formation but accumulate pyruvate and lactate. Their isolated mitochondria consume oxygen normally when supplied with succinate. Supplied pyruvate fails to produce acetyl CoA, carbon dioxide or NADH, whereas supplied acetyl CoA restores citrate formation and TCA activity. Which diagnosis best integrates the observations?
ⓐ. Complex IV is defective, forcing all mitochondrial carriers to remain reduced
ⓑ. the link reaction is defective, diverting cytosolic pyruvate toward lactate
ⓒ. Glycolysis is blocked before glucose phosphorylation
ⓓ. The cells are obligate anaerobes lacking a functional respiratory chain
Correct Answer: the link reaction is defective, diverting cytosolic pyruvate toward lactate
Explanation: Normal glycolytic ATP confirms that cytoplasmic glucose breakdown remains functional. Failure of pyruvate to form acetyl CoA, carbon dioxide and NADH localises the primary defect to oxidative decarboxylation of pyruvate. Direct acetyl CoA addition bypasses this step and restores citrate formation and TCA activity, showing that the downstream matrix pathway is functional. Succinate-supported oxygen consumption further demonstrates that complex II and the later electron-transport chain can operate, excluding a general respiratory-chain failure. Pyruvate accumulates in the cytoplasm and is reduced to lactate, allowing regeneration of \(\mathrm{NAD^+}\) and continued glycolysis even though oxygen is present. The cells are not obligate anaerobes; they possess functional mitochondrial electron transport but cannot connect glycolytic pyruvate efficiently to acetyl-CoA formation.
403. Growth and metabolic records of three microorganisms are shown below.
| Organism | Oxygen present | Oxygen absent |
|---|
| P | Rapid growth, high ATP yield, little ethanol | Slower growth, ethanol and \(\mathrm{CO_2}\) formed |
| Q | No growth | Growth with lactate formation |
| R | Growth | No growth or fermentation product |
If oxygen is suddenly removed from actively growing cultures of all three organisms, which prediction is best supported?
ⓐ. P stops after oxygen removal, Q continues with lactate, and R switches to ethanol formation
ⓑ. P and R retain their oxygenated growth rates, while Q stops growing
ⓒ. P continues more slowly with ethanol and \(\mathrm{CO_2}\), Q can grow with lactate, and R stops
ⓓ. P forms lactate, Q switches to alcoholic fermentation, and R continues by fermentation
Correct Answer: P continues more slowly with ethanol and \(\mathrm{CO_2}\), Q can grow with lactate, and R stops
Explanation: Organism P can use oxygen for a high-yield pathway but also grows without oxygen while forming ethanol and carbon dioxide. Oxygen removal should therefore shift P toward slower alcoholic fermentation rather than stop it immediately. Organism Q grows only in the oxygen-free treatment and forms lactate, so the same condition supports its anaerobic metabolism. Organism R grows when oxygen is present but shows neither growth nor a fermentation product when oxygen is absent. Its record provides no evidence of a usable anaerobic alternative, so removal of oxygen should stop sustained growth. The prediction integrates metabolic flexibility, product identity and energy yield across all three records: P is facultatively anaerobic, Q requires the oxygen-free condition, and R depends on oxygen under the stated observations. Biologically, the common glycolytic stage precedes the organism- and condition-dependent pyruvate fate.
404. Leaf discs are placed in a sealed transparent chamber containing carbon dioxide but no external molecular oxygen. In light, oxygen concentration inside the chamber rises and mitochondrial ATP production continues. When the chamber is transferred to darkness, oxygen concentration and mitochondrial ATP production both decline. The strongest inference is:
ⓐ. photosynthetic oxygen briefly supports respiration in illuminated leaf tissue
ⓑ. Mitochondria produce oxygen whenever external oxygen is absent
ⓒ. Carbon dioxide replaces oxygen as the terminal electron acceptor in light
ⓓ. Respiration occurs only when photosynthesis is actively producing carbohydrate
Correct Answer: photosynthetic oxygen briefly supports respiration in illuminated leaf tissue
Explanation: The chamber initially contains no external oxygen, yet oxygen accumulates during illumination. The leaf discs must be producing oxygen through photosynthesis. Continued mitochondrial ATP formation shows that this internally generated oxygen can reach respiratory sites and serve as the terminal electron acceptor. Transfer to darkness stops photosynthetic oxygen production, while respiration continues consuming the remaining oxygen. The subsequent decline in both oxygen concentration and mitochondrial ATP production follows naturally. Mitochondria do not manufacture molecular oxygen, and carbon dioxide cannot substitute for oxygen at complex IV. The experiment demonstrates a local interaction between two cellular processes without merging their functions: chloroplasts generate oxygen in light, while mitochondria use oxygen to maintain electron transport and oxidative phosphorylation.
405. Two sucrose molecules are hydrolysed completely, and every released monosaccharide completes glycolysis. What is the combined glycolytic output?
ⓐ. \(\mathrm{4\ pyruvate,\ 4\ net\ ATP,\ 4(NADH+H^+)}\)
ⓑ. \(\mathrm{8\ pyruvate,\ 4\ net\ ATP,\ 8(NADH+H^+)}\)
ⓒ. \(\mathrm{4\ pyruvate,\ 8\ net\ ATP,\ 8(NADH+H^+)}\)
ⓓ. \(\mathrm{8\ pyruvate,\ 8\ net\ ATP,\ 8(NADH+H^+)}\)
Correct Answer: \(\mathrm{8\ pyruvate,\ 8\ net\ ATP,\ 8(NADH+H^+)}\)
Explanation: Invertase hydrolyses each sucrose molecule into one glucose and one fructose. Two sucrose molecules provide four monosaccharide molecules. Both glucose and fructose can enter the common glycolytic route and are treated here as hexose substrates completing the standard pathway. Each hexose forms two pyruvates, two net ATP and two \(\mathrm{NADH+H^+}\). Multiplying each output by four gives eight pyruvates, eight net ATP and eight reduced nicotinamide coenzymes. The calculation concerns glycolysis only, so products of pyruvate oxidation, the TCA cycle and oxidative phosphorylation are excluded. The result also distinguishes the number of sucrose molecules from the number of glycolytically processed hexose units produced after hydrolysis. Sucrose hydrolysis itself produces no glycolytic ATP or reduced coenzyme; it only supplies the four hexose units that enter the pathway. All listed products arise after those monosaccharides are processed through glycolysis. Carbon conservation also gives eight three-carbon pyruvates from the four hexoses.
406. Both ATP-consuming hexose-phosphorylation reactions of glycolysis are disabled in a cell-free extract. The downstream triose-level enzymes remain active. Glucose or a compound already converted from glycerol into PGAL is supplied separately. Which outcome is expected?
ⓐ. Both substrates form pyruvate normally since ATP investment is optional
ⓑ. the PGAL-derived substrate forms NADH, ATP and pyruvate; glucose cannot reach the payoff phase
ⓒ. Glucose forms pyruvate, whereas the PGAL-derived substrate cannot enter glycolysis
ⓓ. Neither substrate can form ATP since all glycolytic ATP is produced during the investment phase
Correct Answer: the PGAL-derived substrate forms NADH, ATP and pyruvate; glucose cannot reach the payoff phase
Explanation: Glucose must pass through the ATP-investment reactions before the six-carbon pathway can be cleaved into triose phosphates. Disabling both phosphorylation events prevents glucose from reaching the PGAL stage and prevents its normal payoff reactions. The glycerol-derived compound is already supplied as PGAL, placing it downstream of the blocked hexose steps. It can undergo oxidation to BPGA, reduce \(\mathrm{NAD^+}\), form ATP during the two substrate-level phosphorylation events and reach pyruvate. Glycolytic ATP is not produced during investment; those early reactions consume ATP. The experiment illustrates how alternative respiratory substrates can enter downstream of an upstream lesion and how pathway position determines whether a substrate can bypass the block. Supplying a downstream intermediate therefore bypasses the disabled investment phase, whereas glucose remains trapped upstream of both blocked phosphorylation reactions.
407. Five glucose molecules undergo anaerobic metabolism. Three follow alcoholic fermentation, while two follow lactic acid fermentation. Which combined record is correct?
ⓐ. \(\mathrm{6\ ethanol,\ 4\ lactate,\ 10CO_2,\ 5\ net\ ATP,\ 5NAD^+\ regenerated}\)
ⓑ. \(\mathrm{3\ ethanol,\ 2\ lactate,\ 6CO_2,\ 10\ net\ ATP,\ 10NAD^+\ regenerated}\)
ⓒ. \(\mathrm{6\ ethanol,\ 4\ lactate,\ 4CO_2,\ 10\ net\ ATP,\ 8NAD^+\ regenerated}\)
ⓓ. \(\mathrm{6\ ethanol,\ 4\ lactate,\ 6CO_2,\ 10\ net\ ATP,\ 10NAD^+\ regenerated}\)
Correct Answer: \(\mathrm{6\ ethanol,\ 4\ lactate,\ 6CO_2,\ 10\ net\ ATP,\ 10NAD^+\ regenerated}\)
Explanation: Each glucose produces two pyruvates, two net ATP and two glycolytic NADH. In alcoholic fermentation, the two pyruvates form two ethanol and two carbon-dioxide molecules, while two NADH are oxidised to regenerate two \(\mathrm{NAD^+}\). Three glucose molecules form six ethanol, six carbon dioxide and regenerate six \(\mathrm{NAD^+}\). Each glucose following lactic fermentation forms two lactate without carbon-dioxide release and regenerates two \(\mathrm{NAD^+}\). Two such glucose molecules add four lactate and regenerate four more \(\mathrm{NAD^+}\). All five glucose molecules contribute \(\mathrm{5\times2=10}\) net ATP through glycolysis. Adding both branches gives the stated combined record.
408. A graph follows a facultative anaerobe before and after oxygen is introduced at time \(t_1\). After \(t_1\), the slope of ethanol accumulation decreases, oxygen consumption begins, the slope of ATP accumulation increases and glucose consumption per unit ATP falls. Which interpretation best fits the graph?
ⓐ. Oxygen inhibits energy production and forces faster glucose use
ⓑ. aerobic respiration raises ATP yield and lowers glucose demand
ⓒ. Ethanol becomes the terminal electron acceptor of the aerobic pathway
ⓓ. Glycolysis stops completely once oxygen becomes available
Correct Answer: aerobic respiration raises ATP yield and lowers glucose demand
Explanation: Before oxygen is supplied, the organism relies heavily on alcoholic fermentation. This pathway regenerates \(\mathrm{NAD^+}\) but yields only the two net ATP produced during glycolysis, so substantial glucose must be consumed to meet energy demand. After \(t_1\), oxygen permits terminal electron transfer, reoxidation of reduced coenzymes and oxidative phosphorylation. ATP accumulates more rapidly, while less glucose is needed per unit ATP. The reduced slope of ethanol production indicates that fewer pyruvate molecules are being diverted into fermentation. The presence of oxygen does not stop glycolysis; it remains the common upstream pathway. The graph identifies a metabolic shift through coordinated changes in product formation, oxygen use, energy yield and substrate economy. The four slope changes together identify a shift in pathway use, not merely an effect of oxygen on growth rate.
409. Match each entry compound with the products formed from that entry point until oxaloacetate is regenerated. A Column II entry is used once.
| Column I | Column II |
|---|
| P. Acetyl CoA | 1. \(\mathrm{0CO_2,\ 1(NADH+H^+),\ 0FADH_2,\ 0ATP\ equivalent}\) |
| Q. Alpha-ketoglutarate | 2. \(\mathrm{1CO_2,\ 2(NADH+H^+),\ 1FADH_2,\ 1ATP\ equivalent}\) |
| R. Succinyl CoA | 3. \(\mathrm{0CO_2,\ 1(NADH+H^+),\ 1FADH_2,\ 1ATP\ equivalent}\) |
| S. Malate | 4. \(\mathrm{2CO_2,\ 3(NADH+H^+),\ 1FADH_2,\ 1ATP\ equivalent}\) |
ⓐ. P-2, Q-4, R-1, S-3
ⓑ. P-4, Q-3, R-2, S-1
ⓒ. P-4, Q-2, R-3, S-1
ⓓ. P-3, Q-2, R-4, S-1
Correct Answer: P-4, Q-2, R-3, S-1
Explanation: Acetyl CoA enters a complete TCA turn and produces two carbon dioxide, three NADH, one FADH\(_2\) and one ATP equivalent. Alpha-ketoglutarate enters after the first decarboxylation and first cycle NADH-forming reaction, so its remaining route produces one carbon dioxide, two NADH, one FADH\(_2\) and one ATP equivalent. Succinyl CoA enters after both decarboxylations and forms one ATP equivalent, one FADH\(_2\) and one NADH before oxaloacetate is restored. Malate has only its final oxidation remaining, producing one NADH and oxaloacetate. The mapping shows how pathway entry position determines the remaining carbon loss and energy capture available from a respiratory carbon skeleton. Oxidised coenzymes connect mitochondrial carbon oxidation with downstream electron transport. In terms of carbon flow, entry at a downstream intermediate changes the remaining product set compared with entry as acetyl CoA.
410. Both acetyl CoA molecules produced from one glucose enter the TCA cycle but are withdrawn as alpha-ketoglutarate immediately after the first oxidative decarboxylation. Which consequence applies to both interrupted turns?
ⓐ. each forms NADH and \(\mathrm{CO_2}\), but no FADH\(_2\), GTP or oxaloacetate regeneration
ⓑ. Each completes all reduced-coenzyme-forming reactions but releases no carbon dioxide
ⓒ. Each forms GTP and FADH\(_2\) before withdrawal as alpha-ketoglutarate
ⓓ. Each regenerates oxaloacetate before the five-carbon intermediate leaves the pathway
Correct Answer: each forms NADH and \(\mathrm{CO_2}\), but no FADH\(_2\), GTP or oxaloacetate regeneration
Explanation: Acetyl CoA first combines with oxaloacetate to form citrate, which is rearranged to isocitrate. Oxidative decarboxylation of isocitrate produces alpha-ketoglutarate, one carbon dioxide and one \(\mathrm{NADH+H^+}\). Withdrawal at this point prevents every later part of that turn. The second decarboxylation and its NADH do not occur, succinyl CoA is not formed, and the GTP-producing conversion to succinate is lost. Succinate oxidation cannot form FADH\(_2\), and the later four-carbon sequence cannot regenerate oxaloacetate. The case demonstrates that biosynthetic withdrawal can preserve some early respiratory output while eliminating the remainder of a cycle turn. Merely entering the TCA pathway does not guarantee completion of its normal product balance.
411. A reconstituted system contains three fractions.
Region P contains cytoplasmic glycolytic enzymes.
Region Q contains soluble mitochondrial-matrix enzymes.
Boundary R contains correctly oriented inner-membrane respiratory complexes and ATP synthase.
Glucose, ADP, inorganic phosphate and oxygen are supplied, and necessary metabolites can move between included fractions. What is the smallest combination capable of converting glucose completely to carbon dioxide and water while producing the major oxidative ATP yield?
ⓐ. Region P only
ⓑ. Regions P and Q only
ⓒ. Regions Q and R only
ⓓ. Regions P, Q and R
Correct Answer: Regions P, Q and R
Explanation: Region P is required to convert glucose into pyruvate through glycolysis. Region Q is then needed for oxidative decarboxylation of pyruvate and for the TCA cycle, which releases carbon dioxide and generates most of the reduced coenzymes. Boundary R contains the electron-transfer system that oxidises NADH and FADH\(_2\), transfers electrons to oxygen and forms water. Its ATP synthase uses the proton gradient to produce the major oxidative ATP yield. P alone stops at pyruvate. P and Q can release carbon dioxide and form reduced coenzymes but cannot obtain the large membrane-linked ATP contribution. Q and R lack the glycolytic machinery needed to process supplied glucose. Complete aerobic respiration from glucose requires coordination of all three fractions.
412. A tissue is stated to oxidise only protein with \(\mathrm{RQ=0.9}\) and fat with \(\mathrm{RQ=0.7}\) under aerobic conditions. Repeated measurements give an overall \(\mathrm{RQ=1.0}\). Which conclusion is justified?
ⓐ. equal oxygen use by protein and fat must produce an RQ of one
ⓑ. the two-substrate assumption or the gas measurements must be re-examined
ⓒ. fat oxidation alone can raise the quotient above the protein value
ⓓ. the result proves that protein has an RQ greater than one
Correct Answer: the two-substrate assumption or the gas measurements must be re-examined
Explanation: For a mixture, total carbon-dioxide evolution is the sum of the contributions from each substrate, and overall RQ is obtained by dividing that total by total oxygen consumption. When the only component values are \(0.9\) and \(0.7\), any oxygen-weighted mixture must lie between \(0.7\) and \(0.9\). It cannot reach \(1.0\), regardless of the relative amounts of protein and fat being oxidised. The observation therefore conflicts with at least one stated condition. Another substrate such as carbohydrate may be contributing, a non-respiratory gas flux may be included, or the measurements may require correction. The evidence does not justify changing the characteristic substrate values. A weighted biological ratio must remain within the range of its contributing component ratios when all weights are non-negative.
413. Three molecules enter respiration at different points: glucose at the beginning of glycolysis, PGAL at the triose-payoff stage and acetyl CoA at the TCA-cycle entry point. Assuming complete oxidation from each stated entry, which ordering applies both to the number of carbon-dioxide molecules ultimately released and to the number of ATP or GTP molecules formed directly by substrate-level phosphorylation?
ⓐ. Acetyl CoA \(\gt\) PGAL \(\gt\) glucose
ⓑ. Glucose \(\gt\) acetyl CoA \(\gt\) PGAL
ⓒ. Glucose \(\gt\) PGAL \(\gt\) acetyl CoA
ⓓ. PGAL \(\gt\) glucose \(\gt\) acetyl CoA
Correct Answer: Glucose \(\gt\) PGAL \(\gt\) acetyl CoA
Explanation: Complete oxidation of one glucose releases six carbon-dioxide molecules. Its two PGAL molecules each support two ATP-forming substrate-level reactions during the glycolytic payoff phase, giving four directly formed ATP, and its two TCA turns add two GTP or ATP equivalents. The direct total is six, not the net glycolytic value after subtracting the earlier ATP investment. The two ATP consumed before hexose cleavage affect net yield but do not reduce the count of ATP molecules formed later by substrate-level phosphorylation. One supplied PGAL forms one pyruvate, releases three carbon-dioxide molecules through the link reaction and TCA cycle, forms two ATP directly in the remaining glycolytic steps and forms one GTP in the TCA cycle, giving three direct nucleotide triphosphates. One acetyl CoA releases two carbon-dioxide molecules and forms one GTP directly. The carbon counts also match the sizes of the supplied carbon skeletons: six carbons in glucose, three in PGAL and two in acetyl CoA. Both rankings are glucose \(\gt\) PGAL \(\gt\) acetyl CoA.
414. Conversion of alpha-ketoglutarate into succinyl CoA is blocked. Which alternative respiratory input can bypass the blocked step and still permit formation of GTP, FADH\(_2\), NADH and regenerated oxaloacetate downstream?
ⓐ. deaminated amino-acid carbon entering as succinyl CoA
ⓑ. Fatty-acid-derived acetyl CoA entering citrate formation
ⓒ. Glycerol-derived carbon entering near PGAL
ⓓ. Pyruvate entering the mitochondrial matrix
Correct Answer: deaminated amino-acid carbon entering as succinyl CoA
Explanation: The blocked reaction lies between alpha-ketoglutarate and succinyl CoA. Supplying a carbon skeleton that enters directly as succinyl CoA places the substrate immediately downstream of the lesion. It can undergo the GTP-forming conversion to succinate, produce FADH\(_2\) during succinate oxidation and produce NADH during malate oxidation before oxaloacetate is regenerated. Acetyl CoA, glycerol-derived carbon and pyruvate all enter upstream of the block. Their carbon would eventually reach alpha-ketoglutarate but could not cross the inhibited conversion. This bypass illustrates how different amino-acid carbon skeletons can join at specific TCA intermediates and preserve only the reactions lying downstream of a lesion. Entry farther downstream bypasses earlier ATP, redox and carbon-dioxide-forming reactions.
415. From a single glucose, glycolysis forms two pyruvates. Pyruvate P undergoes alcoholic fermentation, whereas Pyruvate Q undergoes complete aerobic oxidation. Which final profile is expected?
ⓐ. Two ethanol, two lactate and no carbon dioxide
ⓑ. One ethanol, three carbon dioxide and no oxidative phosphorylation
ⓒ. one ethanol, four carbon dioxide and ATP by both phosphorylation modes
ⓓ. No ethanol, six carbon dioxide and the full theoretical \(\mathrm{38\ ATP}\)
Correct Answer: one ethanol, four carbon dioxide and ATP by both phosphorylation modes
Explanation: Alcoholic processing of one pyruvate forms one acetaldehyde, releases one carbon dioxide and then produces one ethanol while oxidising one glycolytic NADH. Complete oxidation of the other pyruvate releases three carbon dioxide: one during the link reaction and two during its TCA turn. Total carbon-dioxide output is four, and one ethanol molecule remains as the reduced fermentative product. Both pyruvates arose through glycolysis, so substrate-level ATP has formed. The aerobically processed pyruvate also generates matrix NADH and FADH\(_2\), allowing oxidative phosphorylation. The cell cannot obtain the full theoretical glucose yield since half of the pyruvate carbon remains in ethanol and one glycolytic NADH is consumed during fermentative redox regeneration.
416. Two treatments are applied separately to inner-mitochondrial-membrane preparations.
Treatment P removes a small mobile protein from the membrane surface and interrupts transfer from complex III to complex IV.
Treatment Q removes a lipid-soluble mobile carrier and interrupts transfer from both complexes I and II to complex III.
The removed carriers in P and Q are, respectively:
ⓐ. Ubiquinone and cytochrome c
ⓑ. Cytochrome c and ubiquinone
ⓒ. Complex IV and complex I
ⓓ. \(F_1\) and \(F_0\)
Correct Answer: Cytochrome c and ubiquinone
Explanation: Cytochrome c is a small mobile protein associated with the outer surface of the inner mitochondrial membrane. It accepts electrons from complex III and transfers them to complex IV, so its removal produces the defect described for Treatment P. Ubiquinone is a lipid-soluble carrier that moves within the membrane. It accepts electrons from both complex I and complex II and carries them toward complex III. Removing it disrupts both donor routes before complex III, matching Treatment Q. The comparison distinguishes the two mobile carriers by physical location as well as pathway position. ATP-synthase components conduct protons and catalyse ATP formation rather than shuttle electrons between respiratory complexes. The two treatments are distinguished by both carrier chemistry and position: a surface protein links complexes III and IV, while a membrane-soluble carrier links the entry complexes with complex III.
417. A cell has two simultaneous defects: glucose cannot pass through the early ATP-investment portion of glycolysis, and cytosolic pyruvate cannot enter the mitochondrial matrix. The TCA-cycle and electron-transport systems remain functional. Which pair of alternative substrates can still supply mitochondrial respiration downstream of both defects?
ⓐ. fatty-acid-derived acetyl CoA plus a deaminated TCA intermediate
ⓑ. Glycerol converted to PGAL and glucose released from stored carbohydrate
ⓒ. Sucrose hydrolysed into glucose and fructose
ⓓ. Glycerol converted to PGAL and pyruvate produced from fructose
Correct Answer: fatty-acid-derived acetyl CoA plus a deaminated TCA intermediate
Explanation: Fatty acids can be converted into acetyl CoA, entering respiration downstream of glycolysis and pyruvate transport. A deaminated amino-acid carbon skeleton that enters directly as a TCA intermediate also bypasses both defective stages. Glycerol-derived carbon enters near PGAL and can proceed to pyruvate, but the second defect prevents that pyruvate from reaching the matrix. Glucose, fructose and sucrose-derived monosaccharides require the blocked early glycolytic sequence. The correct pair contains two inputs that reach the mitochondrial carbon pathway without depending on either hexose processing or cytosolic pyruvate transport. This integration shows how alternative substrates enter a shared respiratory network at different positions and respond differently to combined upstream lesions. The electron streams enter the chain at unequal positions, which permits different ATP assignments despite their shared terminal acceptor.
418. Consider the following statements about responses to changing oxygen availability.
I. A facultative anaerobe may shift from aerobic respiration to fermentation when oxygen is removed.
II. Its ATP yield per glucose generally falls after this shift.
III. An obligate anaerobe necessarily grows faster when oxygen is introduced.
IV. Fermentation products may accumulate in the oxygen-free condition.
ⓐ. I and III only
ⓑ. II and III only
ⓒ. I, III and IV only
ⓓ. I, II and IV only
Correct Answer: I, II and IV only
Explanation: Facultative anaerobes can alter their respiratory mode according to oxygen availability. When oxygen is removed, they may use fermentation to regenerate \(\mathrm{NAD^+}\) and keep glycolysis operating. The ATP yield per glucose then falls markedly since oxidative phosphorylation is lost and the net gain remains largely glycolytic. Ethanol, lactate or other fermentative products may accumulate according to the organism. An obligate anaerobe requires oxygen-free conditions and may fail to grow when oxygen is introduced, so statement III is false. The valid combination distinguishes metabolic flexibility from a compulsory positive response to oxygen. Oxygen improves energy yield only in organisms possessing and tolerating the aerobic machinery needed to use it. After the common glycolytic stage, organism-specific enzymes determine the organic end product.
419. Green and non-green sectors of a variegated leaf are enclosed together in a transparent chamber containing carbon dioxide but no external oxygen. In light, oxygen around the non-green sector remains sufficient for mitochondrial respiration. Covering only the green sector with an opaque screen causes oxygen around the non-green sector to fall. The strongest inference is:
ⓐ. Non-green cells begin photosynthesis when oxygen is scarce
ⓑ. The green sector transports ATP directly into the non-green sector
ⓒ. oxygen from the green sector diffuses locally to adjacent non-green tissue
ⓓ. Carbon dioxide becomes the terminal respiratory acceptor in the covered leaf
Correct Answer: oxygen from the green sector diffuses locally to adjacent non-green tissue
Explanation: The non-green sector cannot produce substantial photosynthetic oxygen, yet it maintains respiration while the neighbouring green sector is illuminated. Selective covering of the green region removes its light supply without directly covering the non-green tissue. The resulting oxygen decline around the non-green sector links its earlier oxygen availability with photosynthesis in the green cells. Oxygen can diffuse through interconnected internal air spaces and support mitochondrial respiration in nearby non-photosynthetic cells. ATP itself is not supplied as a long-distance product of the green sector, and carbon dioxide cannot replace oxygen in terminal electron transfer. The experiment provides a local tissue-level example of photosynthetic oxygen supplementing respiratory gas supply within an illuminated plant organ.
420. Begin with the theoretical \(\mathrm{38\ ATP}\) balance for one glucose. One of the two glycolytic NADH molecules does not contribute to mitochondrial ATP formation, and one acetyl CoA is diverted to fatty-acid synthesis after its link reaction but before entering the TCA cycle. All other counted events remain unchanged. What revised ATP total is obtained?
ⓐ. \(\mathrm{20\ ATP}\)
ⓑ. \(\mathrm{26\ ATP}\)
ⓒ. \(\mathrm{35\ ATP}\)
ⓓ. \(\mathrm{23\ ATP}\)
Correct Answer: \(\mathrm{23\ ATP}\)
Explanation: Loss of the oxidative contribution from one glycolytic NADH removes \(\mathrm{3\ ATP}\) under the stated ATP assignment. The diverted acetyl CoA has already been formed, so its link-reaction NADH remains counted. What is lost is one complete TCA turn: three NADH contribute \(\mathrm{9\ ATP}\), one FADH\(_2\) contributes \(\mathrm{2\ ATP}\), and one GTP or ATP equivalent contributes \(\mathrm{1\ ATP}\). The missing TCA contribution is \(\mathrm{12\ ATP}\). Total loss is \(\mathrm{3+12=15\ ATP}\), giving \(\mathrm{38-15=23\ ATP}\). The calculation combines failure of the glycolytic-NADH transfer assumption with anabolic withdrawal of an acetyl-CoA intermediate. The two changes act at different accounting stages. Failure to use a cytosolic NADH removes only its oxidative contribution, whereas carbon diversion occurs after the link reaction and removes only the products of one TCA turn. Glycolysis, both link reactions and the other cycle turn remain counted.
