401. Priestley’s bell-jar observations, Ingenhousz’s light comparison and Engelmann’s bacterial distribution collectively show that illuminated green tissue restores combustion-supporting air and evolves most oxygen in blue and red light. Which claim still requires additional isotope evidence?
ⓐ. Green tissue can alter enclosed air under illumination
ⓑ. The oxygen atoms released during photosynthesis originate from water
ⓒ. Blue and red wavelengths support high oxygen evolution
ⓓ. Darkness prevents the normal illuminated restoration observed by Ingenhousz
Correct Answer: The oxygen atoms released during photosynthesis originate from water
Explanation: The classical experiments establish several functional relationships. Priestley showed that plants can restore air altered by combustion or animal respiration. Ingenhousz demonstrated that this restoration depends on light and green tissue. Engelmann linked different wavelengths with local oxygen evolution by using aerobic bacteria as indicators. None of these observations traces individual oxygen atoms back to a molecular reactant. Water and carbon dioxide are both present in photosynthesis, so identifying which supplies evolved oxygen requires an atomic tracer. Experiments using oxygen-18 show that labelling water labels the released oxygen gas, whereas labelling carbon dioxide does not produce the same principal result. The distinction separates evidence for oxygen evolution from evidence for oxygen origin. A historical observation may establish that a gas is produced without revealing the precise molecular source of its atoms.
402. Organism P fixes \(\mathrm{^{14}CO_2}\), uses water as its hydrogen donor and releases oxygen. Organism Q also fixes \(\mathrm{^{14}CO_2}\), but uses hydrogen sulphide and releases sulphur. Which conclusion best follows?
ⓐ. Carbon dioxide supplies oxygen gas in P but sulphur in Q
ⓑ. Water and hydrogen sulphide determine which carbon atoms enter carbohydrate
ⓒ. Q does not perform photosynthesis since it fails to evolve oxygen
ⓓ. Carbon dioxide supplies carbon; the donor determines the oxidised product
Correct Answer: Carbon dioxide supplies carbon; the donor determines the oxidised product
Explanation: Carbon-14 from labelled carbon dioxide enters organic compounds in both organisms, identifying carbon dioxide as their carbon source. Their released products differ because their electron and hydrogen donors differ. Water oxidation in oxygenic photosynthesis yields molecular oxygen, while oxidation of hydrogen sulphide in anoxygenic photosynthesis yields sulphur or related sulphur products rather than oxygen. The comparison supports Van Niel’s generalisation that photosynthesis involves reduction of carbon dioxide using hydrogen from an oxidisable donor. Oxygen evolution is not a universal requirement for every photosynthetic pathway, nor does carbon dioxide determine the donor-derived oxidation product. The two organisms share the central function of light-supported carbon reduction but differ in the substance supplying electrons and hydrogen. This evidence also supports the conclusion that green-plant oxygen originates from water rather than carbon dioxide.
403. A chloroplast is separated into three functional fractions. Fraction G contains granal membranes with both photosystems. Fraction L contains stroma lamellae supporting cyclic flow around photosystem I. Fraction S contains stromal enzymes. Under white light with water, ADP, inorganic phosphate, \(\mathrm{NADP^+}\), carbon dioxide and RuBP, which minimum combination can produce both oxygen and carbohydrate?
ⓐ. Fractions G and S
ⓑ. Fractions L and S
ⓒ. Fractions G and L
ⓓ. Fraction S alone
Correct Answer: Fractions G and S
Explanation: Fraction G contains both photosystem II and photosystem I, allowing non-cyclic electron flow. With water, \(\mathrm{NADP^+}\), ADP and inorganic phosphate, it can produce oxygen, NADPH and ATP. Fraction S contains RuBisCO and the remaining Calvin-cycle enzymes, which use ATP and NADPH to fix carbon dioxide and form carbohydrate intermediates. Combining G and S supplies the complete minimum functional partnership for oxygenic photosynthesis under the stated conditions. Fraction L can produce additional ATP through cyclic flow but lacks photosystem II and cannot supply oxygen or normal non-cyclic NADPH. G plus L lacks the stromal carbon-fixation machinery, while S alone lacks photochemical energy and reducing-power production. Granal membranes supply oxygen, ATP and NADPH, while stromal enzymes use ATP and NADPH for carbon fixation; together G and S provide the complementary phases needed for carbohydrate formation.
404. The water-splitting complex of photosystem II is completely inhibited, and the leaf is supplied with a very high external oxygen concentration. Oxygen evolution, ATP formation and NADPH production remain low. Why does external oxygen fail to rescue the pathway?
ⓐ. Oxygen cannot diffuse into green leaves
ⓑ. Photolysis supplies electrons and protons while releasing oxygen
ⓒ. External oxygen converts \(\mathrm{NADP^+}\) directly into NADPH
ⓓ. RuBisCO uses oxygen to regenerate electrons for \(\mathrm{P_{680}}\)
Correct Answer: Photolysis supplies electrons and protons while releasing oxygen
Explanation: The water-splitting complex performs more than oxygen production. It supplies electrons that replace those lost from oxidised \(\mathrm{P_{680}}\) and releases protons into the thylakoid lumen, contributing to the chemiosmotic gradient. External molecular oxygen cannot be converted backward into these required electrons and protons under the photosynthetic pathway described. Without replacement electrons, photosystem-II output declines, electron supply to photosystem I falls and NADPH production becomes limited. Reduced electron-linked proton movement also weakens ATP formation. The overall equation may list oxygen as a product, but adding a final product cannot substitute for the donor-side reaction that generates electron continuity. RuBisCO’s oxygenase activity acts on RuBP and initiates photorespiration; it does not restore photosystem-II reaction-centre electrons.
405. After a pulse of \(\mathrm{^{14}CO_2}\), an illuminated chloroplast accumulates labelled PGA but forms very little labelled triose phosphate. ATP concentration is normal, NADPH concentration is very low, and addition of external NADPH causes labelled triose phosphate to appear rapidly. Which conclusion is best supported?
ⓐ. RuBisCO carboxylation fails, while NADPH supply remains normal
ⓑ. RuBP regeneration fails, while PGA reduction remains normal
ⓒ. Stromal PGA reduction works, but photochemical NADPH supply fails
ⓓ. PGA acts as the primary acceptor, while RuBP remains unused
Correct Answer: Stromal PGA reduction works, but photochemical NADPH supply fails
Explanation: Labelled PGA formation shows that carbon dioxide enters the Calvin cycle through RuBisCO-mediated carboxylation of RuBP. The block lies after PGA formation, where reduction toward triose phosphate requires both ATP and NADPH. ATP is normal, while NADPH is deficient. Rapid appearance of labelled triose phosphate after external NADPH addition demonstrates that the stromal reduction machinery can function when reducing power is supplied. The primary defect is upstream in photochemical NADPH production or delivery, not in the Calvin-cycle enzymes themselves. PGA is a first stable product rather than the primary acceptor, and RuBisCO is evidently active since labelled PGA forms. The rescue experiment is more informative than the initial metabolite pattern alone: supplying one missing product identifies the limiting light-reaction output and confirms that the downstream pathway remains competent. The strongest diagnosis comes from locating the first failed step while upstream functions remain intact in the stated tracer experiment.
406. A mutant C4 leaf retains Kranz anatomy, but PEP carboxylase is present only in bundle-sheath cells and RuBisCO only in mesophyll cells. External carbon dioxide enters through stomata into the mesophyll air spaces. Which consequence is most likely under high temperature?
ⓐ. Disrupted carbon concentration exposes mesophyll RuBisCO to oxygen
ⓑ. Carbon concentration improves since both enzymes have moved closer to the vascular bundle
ⓒ. OAA becomes the primary carbon-dioxide acceptor in mesophyll cells
ⓓ. Photorespiration remains negligible since Kranz anatomy alone is sufficient
Correct Answer: Disrupted carbon concentration exposes mesophyll RuBisCO to oxygen
Explanation: Normal C4 organisation places PEP carboxylase in mesophyll cells, where incoming inorganic carbon is initially fixed into OAA. Four-carbon acids then move inward and release carbon dioxide around bundle-sheath RuBisCO. Swapping the enzymes destroys this functional sequence. Mesophyll cells no longer perform efficient initial PEP carboxylation at the principal entry site of external carbon dioxide, while RuBisCO is relocated outside the carbon-dioxide-retaining bundle sheath. Under high temperature, mesophyll RuBisCO faces stronger competition from oxygen and greater photorespiration. Kranz anatomy provides a structural framework but cannot create carbon concentration when enzyme localisation is reversed. PEP remains the primary acceptor; OAA is the first stable product. The case demonstrates that C4 efficiency depends on coordinated anatomy, enzyme distribution and metabolite transport rather than on any one feature in isolation.
407. Two C3 chloroplast preparations receive identical light-reaction supplies of ATP and NADPH and initially show equal RuBisCO carboxylation. Preparation P is exposed to conditions causing extensive photorespiration, while Q experiences little oxygenation. Even when carbon dioxide is replenished continuously, P shows lower sustained Calvin-cycle turnover. Which explanation is most appropriate?
ⓐ. Photorespiration produces excess sugar that inhibits the Calvin cycle
ⓑ. Oxygenation supplies additional NADPH but removes ATP
ⓒ. Phosphoglycolate directly replaces RuBP as the carbon-dioxide acceptor
ⓓ. Photorespiration releases fixed carbon and consumes useful ATP
Correct Answer: Photorespiration releases fixed carbon and consumes useful ATP
Explanation: Continuous carbon-dioxide replenishment removes one possible substrate shortage, but it does not eliminate the energetic and carbon costs of photorespiration. Oxygenation of RuBP produces phosphoglycolate, which enters a recovery pathway that consumes ATP and releases carbon dioxide. ATP diverted into this pathway is unavailable for Calvin-cycle reduction and RuBP regeneration. Loss of previously fixed carbon also lowers the productive return from the original carboxylation events. Preparation P can show reduced sustained Calvin-cycle turnover despite receiving the same initial light-reaction output and external carbon dioxide as Q. Photorespiration does not generate sugar, ATP or NADPH, and phosphoglycolate cannot substitute for RuBP. The comparison reveals that photorespiration imposes both a carbon penalty and an opportunity cost on the limited ATP supply. The strongest diagnosis comes from locating the first failed step while upstream functions remain intact in the stated RuBisCO comparison.
408. Fifteen four-carbon acids are formed in C4 mesophyll cells. Exactly \(80\%\) reach the bundle sheath and are decarboxylated. RuBisCO fixes \(75\%\) of the carbon dioxide released there. Which accounting is correct?
ⓐ. \(15\) acids are decarboxylated, \(12\) carbon-dioxide molecules are fixed and no acids remain in the mesophyll
ⓑ. \(12\) acids are decarboxylated, \(12\) carbon-dioxide molecules are fixed and \(3\) three-carbon compounds form
ⓒ. \(12\) acids are decarboxylated, \(9\) carbon-dioxide molecules are fixed and \(3\) four-carbon acids remain in the mesophyll
ⓓ. \(9\) acids are decarboxylated, \(12\) carbon-dioxide molecules are fixed and \(6\) acids remain in the mesophyll
Correct Answer: \(12\) acids are decarboxylated, \(9\) carbon-dioxide molecules are fixed and \(3\) four-carbon acids remain in the mesophyll
Explanation: Transport succeeds for \(15\times0.80=12\) four-carbon acids, leaving \(3\) four-carbon acids untransported in the mesophyll. Each transported acid releases one carbon-dioxide molecule during bundle-sheath decarboxylation, so \(12\) carbon-dioxide molecules become locally available. RuBisCO fixes \(75\%\) of this released carbon dioxide, giving \(12\times0.75=9\) fixed molecules. The remaining \(3\) released carbon-dioxide molecules are not fixed under the stated efficiency; they must not be confused with the \(3\) four-carbon acids that failed to reach the bundle sheath. The calculation therefore separates transport success from bundle-sheath fixation success. Acid formation occurs first, transport determines the decarboxylated pool, and RuBisCO efficiency determines the fraction of released carbon dioxide entering productive fixation.
409. Photosynthesis of each leaf increases linearly below \(100\) light units according to \(\text{rate}=0.1\times\text{light intensity}\), and saturates at \(10\) rate units per leaf above that intensity. Plant P has \(10\) leaves receiving \(160\) light units each. Plant Q has \(20\) leaves receiving \(70\) units each. Assume all other factors are non-limiting and leaf rates are additive. Which comparison is correct?
ⓐ. P gains \(100\) units and Q gains \(70\) units, so P is higher by \(30\)
ⓑ. P gains \(100\) units and Q gains \(140\) units, so Q is higher by \(40\)
ⓒ. P gains \(160\) units and Q gains \(140\) units, so P is higher by \(20\)
ⓓ. Both plants gain \(100\) units since only saturated leaves contribute
Correct Answer: P gains \(100\) units and Q gains \(140\) units, so Q is higher by \(40\)
Explanation: Every leaf of Plant P receives light above the saturation point, so each contributes the maximum \(10\) rate units. Its whole-plant rate is \(10\times10=100\). Each leaf of Plant Q receives \(70\) units, which lies in the linear range, giving \(0.1\times70=7\) rate units per leaf. With \(20\) leaves, Q reaches \(20\times7=140\). Plant Q exceeds P by \(140-100=40\) units. The calculation integrates an external factor, light intensity, with an internal factor, leaf number. A plant with lower illumination per leaf can achieve greater whole-plant photosynthesis when it possesses enough productive leaf area. Saturation also means that the excess light received by P cannot raise its per-leaf rate under the stated conditions. Each P leaf is saturated at \(10\) units, giving \(10\times10=100\). Each Q leaf remains linear at \(0.1\times70=7\) units, giving \(20\times7=140\). Whole-plant gain depends on both per-leaf rate and leaf number, not on the brighter treatment alone.
410. A destarched variegated leaf is illuminated inside a sealed chamber containing fresh KOH. After several hours, both green and non-green regions remain yellow-brown with iodine. Which conclusion is most appropriate?
ⓐ. chlorophyll is unnecessary because neither region formed detectable starch
ⓑ. the setup cannot isolate chlorophyll's role because carbon dioxide was removed from both regions
ⓒ. green tissue carried out normal photosynthesis but iodine failed in sealed air
ⓓ. non-green tissue prevented KOH from absorbing carbon dioxide around the leaf
Correct Answer: the setup cannot isolate chlorophyll's role because carbon dioxide was removed from both regions
Explanation: KOH absorbs carbon dioxide from the sealed chamber. Both green and non-green regions lack an essential raw material for photosynthetic carbohydrate formation, even though the leaf is illuminated. A negative iodine result in both regions cannot establish whether chlorophyll-containing tissue is required; carbon-dioxide removal prevents the green region from photosynthesising. The comparison changes two relevant conditions at once: pigment status differs between regions, while carbon dioxide is absent for both. A valid variegated-leaf test of chlorophyll must provide carbon dioxide and light to both regions so that chlorophyll presence is the decisive difference. Iodine can still detect starch in sealed air, and non-green tissue does not stop KOH from absorbing the gas. The result demonstrates how an uncontrolled limiting factor can make both experimental groups negative and obscure the variable being investigated.
411. All light intensities in the following treatments lie within the linear low-light range, and contributions from different wavelengths are additive.
| Wavelength region | Relative action coefficient |
|---|
| Blue | \(0.8\) |
| Green | \(0.2\) |
| Red | \(1.0\) |
Treatment P supplies \(40\) blue-light units and \(60\) green-light units. Treatment Q supplies \(20\) blue-light units and \(80\) red-light units. Which prediction is correct?
ⓐ. P produces \(80\) units, while Q produces \(100\) units
ⓑ. P produces \(44\) units, while Q produces \(84\) units
ⓒ. Treatment P produces more since it contains a larger blue-light component
ⓓ. Treatment Q produces \(96\) units compared with \(44\) units for P
Correct Answer: Treatment Q produces \(96\) units compared with \(44\) units for P
Explanation: Each wavelength contribution equals the supplied light units multiplied by the corresponding action coefficient. For P, blue light contributes \(40\times0.8=32\) units and green light contributes \(60\times0.2=12\), giving a total of \(44\). For Q, blue contributes \(20\times0.8=16\) and red contributes \(80\times1.0=80\), giving \(96\). Q performs better despite receiving less blue light because most of its illumination lies in the highly effective red region. Total light quantity alone cannot predict the output when wavelengths differ in photosynthetic effectiveness. The calculation is valid because all intensities lie in the stated linear range and the contributions are additive. At saturating intensities, simple proportional addition could fail because another factor would restrict the response. The coefficients represent action, not merely pigment colour, so the comparison integrates the usable contribution of each wavelength to the measured photosynthetic process.
412. In an Engelmann-type experiment, aerobic bacteria cluster mainly in the blue and red regions around an illuminated filamentous alga. A concern is raised that the bacteria might move preferentially toward blue and red light even without an oxygen gradient. Which control most directly tests this alternative explanation?
ⓐ. Repeat the bacterial distribution test without the alga while maintaining uniform oxygen
ⓑ. Replace the aerobic bacteria with additional chlorophyll-containing algal cells
ⓒ. Illuminate the original preparation only with white light and ignore bacterial position
ⓓ. Remove carbon dioxide while keeping the alga and bacteria under the spectrum
Correct Answer: Repeat the bacterial distribution test without the alga while maintaining uniform oxygen
Explanation: Engelmann’s inference depends on bacterial movement representing local oxygen concentration rather than a direct response to wavelength. The proposed control retains the spectral light pattern but removes the alga, and hence removes wavelength-specific photosynthetic oxygen production. Uniform oxygen conditions ensure that no oxygen gradient guides movement. If bacteria still cluster in blue and red regions, their distribution cannot be used confidently as an oxygen indicator. If they remain uniform, the original clustering is more reasonably attributed to oxygen released by the alga. Replacing bacteria with algal cells removes the indicator system rather than testing its bias. White light eliminates the wavelength comparison, while carbon-dioxide removal alters algal photosynthesis but does not isolate a possible direct bacterial response to coloured light. The control must preserve the suspected confounding variable while removing the proposed causal signal.
413. Stroma lamellae are selectively destroyed in a chloroplast, while granal membranes containing both photosystems and the stromal Calvin-cycle enzymes remain functional. Which change is most likely under illumination?
ⓐ. Non-cyclic electron flow stops completely, but cyclic photophosphorylation remains normal
ⓑ. Oxygen, NADPH and supplementary cyclic ATP all remain unchanged
ⓒ. Cyclic ATP supplementation falls, while grana still form oxygen and NADPH
ⓓ. RuBisCO is destroyed since it is embedded in the stroma lamellae
Correct Answer: Cyclic ATP supplementation falls, while grana still form oxygen and NADPH
Explanation: Stroma lamellae are associated prominently with photosystem I and can support cyclic electron flow, which forms ATP without producing NADPH or oxygen. Their selective destruction removes an important route for supplementing ATP when stromal carbon reactions require more ATP relative to NADPH. Intact granal membranes still contain the machinery for non-cyclic electron flow involving photosystem II and photosystem I. Water splitting can release oxygen, and terminal electron transfer can form NADPH while photophosphorylation produces some ATP. The main defect is an impaired capacity to adjust the ATP-to-NADPH output through cyclic flow, not total loss of all light reactions. RuBisCO is a soluble stromal enzyme rather than a component embedded in stroma lamellae. The altered chloroplast may become relatively ATP-limited during active carbon fixation even while oxygen and NADPH continue to appear.
414. Destarched isolated stem segments are treated under identical carbon-dioxide and temperature conditions. An illuminated green stem segment forms starch, while an illuminated non-green segment and a green segment kept in darkness remain starch-negative. Which conclusion is best supported?
ⓐ. green stems can photosynthesise without illumination
ⓑ. green organs other than leaves can photosynthesise
ⓒ. non-green stems can photosynthesise under illumination
ⓓ. green stems can photosynthesise only in darkness
Correct Answer: green organs other than leaves can photosynthesise
Explanation: The illuminated green segment supplies all three relevant conditions and forms starch. The illuminated non-green segment tests the need for chlorophyll-containing tissue, while the dark green segment tests the need for light. Their negative results support local photosynthesis in the illuminated green stem rather than passive starch appearance in every stem region. Leaves remain the principal photosynthetic organs in higher plants, largely owing to their broad surface and chloroplast-rich mesophyll, but they are not the only structures capable of the process. Green stems and other chlorophyll-containing plant parts may also assimilate carbon dioxide. The experiment does not compare total organ productivity or chloroplast numbers, so it cannot establish that stems exceed leaves in importance. Its conclusion is limited to photosynthetic competence of the tested green stem tissue under favourable conditions.
415. Arrange the following discoveries from the broadest environmental observation to the most specific evidence about atomic origin.
P. A plant restores enclosed air altered by a candle or animal.
Q. Restoration occurs in light and is associated with green tissue.
R. Blue and red wavelengths produce the greatest local oxygen evolution.
S. Oxygen isotope from labelled water appears in the evolved gas.
ⓐ. P → Q → R → S
ⓑ. Q → P → S → R
ⓒ. P → R → Q → S
ⓓ. S → R → Q → P
Correct Answer: P → Q → R → S
Explanation: Observation P establishes only that a plant can restore a property of altered enclosed air. Q adds two necessary conditions by linking the effect with illumination and green tissue. R increases explanatory resolution further by showing that oxygen production varies with wavelength and is greatest in blue and red regions. S reaches the atomic-source level by tracing labelled oxygen from water into the evolved gas. The sequence is not merely a list of scientist names; each stage answers a question left unresolved by the preceding evidence. Air restoration does not identify light dependence, wavelength effectiveness or molecular origin. Spectral oxygen evidence still cannot reveal whether water or carbon dioxide supplies the released oxygen atoms. Isotope tracing provides that final mechanistic distinction and links the historical observations to photolysis at photosystem II.
416. Four thylakoid preparations are illuminated to establish a proton gradient and then transferred suddenly to darkness.
| Treatment | Gradient after darkness | Brief ATP formation after darkness |
|---|
| P | Collapses rapidly | Present |
| Q | Persists strongly | Absent |
| R | Collapses rapidly | Absent |
| S | Was very low even during illumination | Absent |
Which treatment most specifically indicates that the \(CF_0\) proton channel is blocked while the membrane remains otherwise intact?
ⓐ. Treatment P
ⓑ. Treatment R
ⓒ. Treatment S
ⓓ. Treatment Q
Correct Answer: Treatment Q
Explanation: A blocked \(CF_0\) channel prevents protons accumulated in the lumen from returning to the stroma through ATP synthase. The proton difference should persist after illumination stops, since the controlled route for gradient breakdown is unavailable. ATP formation is absent because proton flow through \(CF_0\) is the event that supplies energy to the \(CF_1\) catalytic head. Treatment P represents an intact system: the stored gradient drives a brief ATP pulse and then collapses. R is more consistent with an inactive \(CF_1\) component or uncoupled proton passage, since the gradient disappears without ATP production. S suggests failure to establish or retain the gradient during illumination, such as a proton leak. The paired observation of persistent gradient and absent ATP distinguishes channel blockage from failed gradient generation or catalytic-head damage. A persistent gradient without ATP therefore identifies blocked proton passage through \(CF_0\) while the membrane still retains protons.
417. Eighteen RuBP molecules react with RuBisCO. Twelve undergo carboxylation and six undergo oxygenation. Count all immediate products, but calculate energy only for the standard Calvin-cycle processing of the twelve fixed carbon-dioxide molecules; exclude photorespiratory costs. Which account is correct?
ⓐ. \(24\) PGA, \(6\) phosphoglycolate, \(18\) ATP and \(12\) NADPH
ⓑ. \(30\) PGA, \(12\) phosphoglycolate, \(36\) ATP and \(24\) NADPH
ⓒ. \(30\) PGA, \(6\) phosphoglycolate, \(36\) ATP and \(24\) NADPH
ⓓ. \(36\) PGA, \(6\) phosphoglycolate, \(54\) ATP and \(36\) NADPH
Correct Answer: \(30\) PGA, \(6\) phosphoglycolate, \(36\) ATP and \(24\) NADPH
Explanation: Each RuBisCO carboxylation produces two PGA molecules, so twelve carboxylation events yield \(12\times2=24\) PGA. Each oxygenation event produces one PGA and one phosphoglycolate, adding \(6\) PGA and \(6\) phosphoglycolate. The immediate product totals are \(30\) PGA and \(6\) phosphoglycolate. Energy accounting includes only the twelve productive carbon-dioxide fixations. At \(3\) ATP and \(2\) NADPH per fixed carbon dioxide, the Calvin-cycle requirement is \(12\times3=36\) ATP and \(12\times2=24\) NADPH. Processing the phosphoglycolate would impose an additional photorespiratory energy cost, but that cost is deliberately excluded. Separating immediate product formation from productive fixation prevents the six oxygenation events from being counted as extra Calvin-cycle carboxylations. Product counting and energy counting use different event sets here. All eighteen RuBisCO reactions contribute immediate products, but only twelve carboxylations fix carbon dioxide productively. Hence \(30\) PGA and \(6\) phosphoglycolate coexist with the restricted Calvin demand of \(36\) ATP and \(24\) NADPH.
418. Photosynthesising algae receive a brief pulse of \(\mathrm{^{14}CO_2}\), after which unlabelled carbon dioxide is supplied. Label appears first in PGA, later in triose phosphate and eventually in RuBP, while the total RuBP concentration remains approximately constant. Which inference is strongest?
ⓐ. Reduced intermediates help regenerate the carbon-dioxide acceptor
ⓑ. PGA is converted directly into oxygen before RuBP forms
ⓒ. RuBP is a terminal carbohydrate product removed from the cycle
ⓓ. The isotope enters RuBP without passing through reduction reactions
Correct Answer: Reduced intermediates help regenerate the carbon-dioxide acceptor
Explanation: The earliest label in PGA identifies it as an early stable product of Calvin-cycle carboxylation. During the unlabelled chase, the original radioactive carbon moves into later products rather than being continually introduced from the atmosphere. Its appearance in triose phosphate shows passage through the reduction phase. Later labelling of RuBP demonstrates that part of the reduced carbon pool is reorganised to regenerate the \(5\)-carbon acceptor. The approximately constant total RuBP concentration indicates cycling: RuBP is consumed during carboxylation and restored through regeneration rather than accumulating as a terminal product. The time sequence rules out direct transfer from carbon dioxide to RuBP and shows that regeneration is connected to earlier reduced intermediates. Pulse–chase evidence can reveal pathway direction even when the concentrations of cycle compounds remain relatively steady. Observation must be separated from conclusion: the measured result supports only the mechanism tested by the design in the stated tracer experiment.
419. Plasmodesmatal connections between C4 mesophyll and bundle-sheath cells are selectively sealed, while PEP carboxylase, decarboxylating enzymes, RuBisCO and the Calvin cycle remain functional within their original cells. Which pattern is expected?
ⓐ. Four-carbon acids move inward normally, but only oxygen fails to leave the bundle sheath
ⓑ. Bundle-sheath RuBisCO relocates to mesophyll cells and restores the pathway
ⓒ. C4 acids accumulate, three-carbon return fails and PEP regeneration declines
ⓓ. OAA is converted directly into glucose without intercellular transport
Correct Answer: C4 acids accumulate, three-carbon return fails and PEP regeneration declines
Explanation: The C4 shuttle requires movement in both directions between the two photosynthetic cell types. Four-carbon acids formed in mesophyll cells must enter bundle-sheath cells for decarboxylation, and the resulting three-carbon compound must return to mesophyll cells for PEP regeneration. Sealing the cellular connections interrupts both transfers. Initially formed C4 acids accumulate on the mesophyll side, while bundle-sheath cells receive less decarboxylation substrate and produce less local carbon dioxide. Any three-carbon compound already generated in the bundle sheath also fails to return efficiently. The mesophyll PEP pool declines after existing acceptor molecules are used. Functional enzymes cannot compensate when their substrates remain trapped in the wrong cell. The two-direction transport defect collapses the carbon-concentrating cycle even though no enzyme has been chemically inhibited. The relation remains valid only under the specified transport and bundle-sheath fixation conditions.
420. Net photosynthetic rate is plotted against external carbon-dioxide concentration for a C3 plant and a C4 plant at \(20^\circ\mathrm{C}\) and \(35^\circ\mathrm{C}\). At \(35^\circ\mathrm{C}\) and ordinary carbon dioxide, the C4 curve lies far above the C3 curve. Carbon-dioxide enrichment raises the C3 curve strongly and narrows the gap, while the C4 curve changes comparatively little. At \(20^\circ\mathrm{C}\), the initial gap is smaller. Which explanation best integrates the curves?
ⓐ. Carbon-dioxide enrichment converts C3 plants into anatomical C4 plants at high temperature
ⓑ. C4 plants lack RuBisCO and cannot respond to external carbon dioxide
ⓒ. Temperature affects only light absorption and not RuBisCO activity
ⓓ. Heat raises C3 photorespiration; enrichment favours carboxylation; C4 changes less
Correct Answer: Heat raises C3 photorespiration; enrichment favours carboxylation; C4 changes less
Explanation: The axes compare external carbon dioxide with net photosynthesis under two temperatures. At \(35^\circ\mathrm{C}\), a C3 leaf experiences substantial competition between oxygen and carbon dioxide at RuBisCO, increasing photorespiration and lowering net carbon gain. External carbon-dioxide enrichment shifts RuBisCO activity toward carboxylation and raises the C3 curve markedly. The C4 plant already releases carbon dioxide around bundle-sheath RuBisCO through its acid shuttle, so additional external enrichment produces a smaller response. At \(20^\circ\mathrm{C}\), the C3 photorespiratory disadvantage is less severe, making the initial pathway difference smaller. The curves reflect interacting gas and temperature effects rather than pathway conversion or absence of RuBisCO in C4 plants. C4 anatomy and enzyme compartmentation remain unchanged by greenhouse enrichment. The axes and the specified curve region determine whether slope, plateau, optimum or decline is being interpreted in the stated carbon-dioxide response.
