301. Four graphs are described below.
P. Wavelength against pigment absorption
Q. Wavelength against oxygen evolution
R. Light intensity against photosynthetic rate
S. External carbon-dioxide concentration against photosynthetic rate
Which interpretation correctly identifies the principal decision supplied by each graph?
ⓐ. P and Q both measure pigment concentration, while R and S identify chloroplast number
ⓑ. P gives pigment absorption, Q photosynthetic action, and R–S limitation or saturation
ⓒ. P gives a temperature curve, Q gives a water-response curve, and R and S are chromatograms
ⓓ. all four graphs measure the same variable and differ only in axis units
Correct Answer: P gives pigment absorption, Q photosynthetic action, and R–S limitation or saturation
Explanation: Graph P measures how strongly a pigment absorbs different wavelengths and is an absorption spectrum. Graph Q measures the effectiveness of wavelength in producing a photosynthetic response such as oxygen evolution, making it an action spectrum. Graph R varies light intensity and can reveal a low-light linear region, saturation and possible excess-light inhibition. Graph S varies carbon-dioxide concentration and can show carbon limitation, pathway-specific saturation and interaction with light. Correct interpretation depends on both axes and the measured response. Similar curve shapes do not make the graphs equivalent, because wavelength, intensity and gas concentration represent different variables and identify different possible limiting factors. An absorption spectrum records light uptake by a pigment, whereas an action spectrum records photosynthetic effectiveness. Curve separation identifies when the compared factor begins to influence the rate, not merely which curve is higher in the stated light-response comparison.
302. A C4 plant has normal Kranz anatomy, functional PEP carboxylase and normal OAA formation. A mutation prevents four-carbon acids from moving efficiently from mesophyll to bundle-sheath cells. Under high temperature, OAA accumulates in mesophyll cells, bundle-sheath carbon dioxide falls and phosphoglycolate formation increases. Which interpretation best integrates these observations?
ⓐ. RuBisCO has been replaced by PEP carboxylase in the bundle sheath
ⓑ. OAA accumulation has increased carbon-dioxide concentration around RuBisCO
ⓒ. failed acid transport lowers carbon concentration and raises oxygenation
ⓓ. phosphoglycolate formation shows that the Calvin cycle has accelerated
Correct Answer: failed acid transport lowers carbon concentration and raises oxygenation
Explanation: Initial fixation remains functional, as shown by normal OAA formation and its accumulation in mesophyll cells. The mutation blocks the next spatial step: movement of four-carbon acids to bundle-sheath cells. Less acid reaches the decarboxylation site, so less carbon dioxide is released near bundle-sheath RuBisCO. The local \(\mathrm{CO_2:O_2}\) ratio falls, increasing the probability that RuBisCO reacts with oxygen rather than carbon dioxide. Increased phosphoglycolate formation is direct evidence of this oxygenase activity and the initiation of photorespiration. The observations connect transport, carbon concentration and enzyme competition. A functioning initial fixation step cannot preserve the C4 advantage when the fixed carbon fails to reach the compartment containing RuBisCO. A failure of acid transport can interrupt carbon concentration despite normal enzyme activities in the separate cells. Normal OAA formation with poor bundle-sheath delivery localises the defect to transport and predicts reduced carbon-dioxide release around RuBisCO despite intact anatomy.
303. Four C4 leaf treatments are compared under high light and high temperature.
| Treatment | Mesophyll OAA formation | Acid arrival in bundle sheath | Carbon-dioxide retention near RuBisCO | Photorespiration |
|---|
| P | High | High | High | Negligible |
| Q | High | High | Low | High |
| R | Low | Low | Low | High |
| S | High | Low | Low | Moderate |
Which treatment most specifically indicates defective carbon-dioxide retention by unusually gas-permeable bundle-sheath walls?
ⓐ. Treatment Q
ⓑ. Treatment R
ⓒ. Treatment S
ⓓ. Treatment P
Correct Answer: Treatment Q
Explanation: Treatment Q preserves the upstream C4 pathway: mesophyll OAA formation is high and four-carbon acids reach the bundle sheath. Its failure appears after transport, where carbon-dioxide retention near RuBisCO is low and photorespiration is high. This pattern fits bundle-sheath walls that permit released carbon dioxide to diffuse away too readily. Treatment R shows defective initial fixation, while S indicates a transport limitation. P represents the normal coordinated pathway. Effective C4 photosynthesis requires more than carbon capture and transport; carbon dioxide released by decarboxylation must remain concentrated around RuBisCO. Loss of that structural retention lowers the local gaseous advantage and exposes the enzyme to greater oxygen competition despite normal delivery of four-carbon acids. The decisive row combines the relevant observation with the condition needed to interpret it biologically in the stated C4 pathway.
304. Use the cellular arrangement described below. Cell P is a mesophyll cell containing PEP carboxylase. Cell Q is a bundle-sheath cell containing RuBisCO. A four-carbon acid moves from P to Q, where Step R releases carbon dioxide and a three-carbon compound. If Step R is selectively blocked, which paired change is expected first?
ⓐ. increased carbon dioxide near RuBisCO and increased PGA formation
ⓑ. reduced OAA formation in P and increased PEP regeneration in Q
ⓒ. increased oxygen release in Q and reduced light absorption in P
ⓓ. less carbon dioxide near RuBisCO and less three-carbon return product
Correct Answer: less carbon dioxide near RuBisCO and less three-carbon return product
Explanation: Step R is decarboxylation of the transported four-carbon acid inside the bundle-sheath cell. This reaction has two immediate outputs: carbon dioxide and a three-carbon compound. Blocking it lowers the local carbon-dioxide supply available for RuBisCO and simultaneously reduces production of the compound that normally returns to mesophyll cells for PEP regeneration. Initial OAA formation may continue briefly, so the primary defect is not located at PEP carboxylase. Oxygen evolution belongs to the thylakoid light reactions and is not an immediate product of four-carbon-acid breakdown. The described spatial block disrupts both productive Calvin-cycle carbon delivery and closure of the mesophyll–bundle-sheath shuttle. OAA is the first stable four-carbon product, while PEP is the three-carbon initial acceptor. The conclusion applies to the stated biological condition and should not be overgeneralised to every plant or environment in the stated C4 pathway.
305. A C4 leaf receives a \(5\)-second pulse of \(\mathrm{^{14}CO_2}\). Radioactivity first appears in mesophyll OAA. After a short delay, labelled carbon appears in bundle-sheath PGA. Which conclusion is most strongly supported by the time sequence?
ⓐ. PGA moves from bundle-sheath cells to mesophyll cells before OAA forms
ⓑ. Carbon enters a C4 mesophyll product before the bundle-sheath Calvin cycle
ⓒ. RuBisCO performs the first fixation of external carbon dioxide in mesophyll cells
ⓓ. OAA is produced only after labelled carbon has completed the Calvin cycle
Correct Answer: Carbon enters a C4 mesophyll product before the bundle-sheath Calvin cycle
Explanation: The earliest label identifies the first stable destination of newly supplied carbon. Its appearance in mesophyll OAA shows that external inorganic carbon is initially captured through the PEP-carboxylase pathway. The delayed appearance of label in bundle-sheath PGA indicates that the fixed carbon is transported inward as part of a four-carbon acid, released as carbon dioxide and then incorporated by RuBisCO into Calvin-cycle products. Timing distinguishes the preliminary C4 shuttle from the later Calvin cycle rather than merely showing that both compounds eventually contain carbon. A short tracer pulse limits widespread redistribution and makes pathway order easier to infer. The evidence supports mesophyll fixation before bundle-sheath refixation; it does not place RuBisCO at the initial mesophyll step or make PGA the transported four-carbon compound. The sequence demonstrates carbon transfer and refixation across two coordinated cell types.
306. Assertion: C4 plants can maintain greater net carbon gain than C3 plants under warm, bright conditions.
Reason: Decarboxylation of four-carbon acids raises carbon-dioxide concentration around bundle-sheath RuBisCO and suppresses its oxygenase activity.
ⓐ. Both the Assertion and the Reason are true, and the Reason correctly explains the Assertion
ⓑ. Both the Assertion and the Reason are true, but the Reason does not explain the Assertion
ⓒ. The Assertion is true, whereas the Reason is false and cannot explain the Assertion
ⓓ. The Assertion is false, whereas the Reason is true and cannot explain the Assertion
Correct Answer: Both the Assertion and the Reason are true, and the Reason correctly explains the Assertion
Explanation: The Assertion is true. Warm, bright conditions often increase the photorespiratory disadvantage experienced by C3 plants, reducing their net carbon retention. C4 plants initially fix inorganic carbon in mesophyll cells and transport it to bundle-sheath cells as four-carbon acids. The Reason is also true and supplies the relevant mechanism. Decarboxylation releases carbon dioxide close to RuBisCO, favouring carboxylation over oxygenation. Less phosphoglycolate is formed, photorespiratory carbon loss remains negligible and a greater fraction of fixed carbon can contribute to biomass. The Reason directly accounts for the physiological performance described in the Assertion by linking cell-specific transport with altered substrate competition at RuBisCO. C4 carbon concentration suppresses the initiating oxygenase reaction rather than changing the basic catalytic nature of RuBisCO.
307. Two illuminated plant preparations are supplied with isotopes.
| Preparation | Water supplied | Carbon dioxide supplied | Oxygen gas released |
|---|
| P | \(\mathrm{H_2^{18}O}\) | Unlabelled \(\mathrm{CO_2}\) | \(\mathrm{^{18}O_2}\) |
| Q | Unlabelled \(\mathrm{H_2O}\) | \(\mathrm{C^{18}O_2}\) | Predominantly unlabelled \(\mathrm{O_2}\) |
Which conclusion is justified by the combined evidence?
ⓐ. carbon dioxide supplies the oxygen gas only when water is unlabelled
ⓑ. both water and carbon dioxide contribute equally to each oxygen molecule
ⓒ. oxygen gas is formed by reduction of labelled carbon dioxide in the stroma
ⓓ. photosynthetic oxygen originates from water rather than carbon dioxide
Correct Answer: photosynthetic oxygen originates from water rather than carbon dioxide
Explanation: Preparation P transfers the oxygen isotope from labelled water into the released oxygen gas, while Preparation Q does not transfer the carbon-dioxide oxygen label substantially into evolved oxygen. The reciprocal design is important: one treatment follows labelled water, and the other tests whether labelled carbon dioxide could be the alternative source. Together they identify water as the immediate source of photosynthetic oxygen. The result fits photolysis at photosystem II, where water supplies electrons, lumen protons and molecular oxygen. Carbon dioxide instead supplies carbon for carbohydrate synthesis during stromal reactions. Isotope tracing establishes atomic origin rather than merely showing that water and carbon dioxide are both required somewhere in the overall process. The decisive row combines the relevant observation with the condition needed to interpret it biologically in the stated tracer experiment.
308. Two sealed flasks contain comparable destarched leaves. Flask P contains potassium hydroxide and is kept in darkness. Flask Q contains water instead of potassium hydroxide and is illuminated. After several hours, P lacks starch while Q contains starch. Why can this design not isolate the effect of carbon dioxide?
ⓐ. potassium hydroxide cannot absorb carbon dioxide in a sealed flask
ⓑ. light and carbon-dioxide availability changed together
ⓒ. destarching prevents all later starch formation
ⓓ. iodine cannot detect starch formed during illumination
Correct Answer: light and carbon-dioxide availability changed together
Explanation: A valid test of the carbon-dioxide requirement must keep other important conditions, especially light, comparable between the treatment and control. In this design, P lacks carbon dioxide due to potassium hydroxide and also lacks light, while Q retains carbon dioxide and receives illumination. The different starch outcomes could be caused by either variable or by both acting together. The experiment is confounded and cannot assign the result specifically to carbon-dioxide removal. Both flasks should be illuminated equally, with potassium hydroxide present only in the treatment flask and an inert comparison material or water in the control. Destarching is useful since it removes previously stored starch and allows newly formed starch to be interpreted as an experimental product.
309. Match each experimental component with its principal purpose. A Column II entry is used once.
| Column I | Column II |
|---|
| P. Destarching a plant before treatment | 1. Absorb carbon dioxide from a closed space |
| Q. Potassium hydroxide in a flask | 2. Remove pre-existing starch reserves |
| R. Boiling a leaf in alcohol | 3. Remove chlorophyll so the iodine result can be seen clearly |
| S. Leaving part of the same leaf exposed to light | 4. Provide an internal comparison for the covered region |
ⓐ. P-1, Q-2, R-4, S-3
ⓑ. P-3, Q-4, R-1, S-2
ⓒ. P-2, Q-1, R-3, S-4
ⓓ. P-4, Q-3, R-2, S-1
Correct Answer: P-2, Q-1, R-3, S-4
Explanation: Destarching removes starch formed before the experiment, allowing a later iodine-positive result to be attributed to photosynthesis during treatment. Potassium hydroxide absorbs carbon dioxide and creates the intended gas limitation in a closed apparatus. Alcohol extracts chlorophyll after the leaf has been killed, making the blue-black iodine colour easier to observe without green pigment masking it. An uncovered region of the same leaf experiences nearly the same age, internal condition and handling as the covered region, so it provides a strong internal comparison for the effect of light. These components serve different roles: preparation, variable manipulation, result visualisation and experimental control. Correct interpretation depends on keeping their purposes distinct. A positive starch test indicates accumulated carbohydrate after treatment, not the immediate product of the light reaction. The complete mapping must remain internally consistent across all rows, not merely for one familiar pair in the stated starch-test experiment.
310. In an Engelmann-type experiment, a filamentous green alga is illuminated with a spectrum, but the aerobic bacteria used as indicators are altered so that they no longer move toward oxygen. The alga continues photosynthesising normally. If bacteria become uniformly distributed, what is the strongest conclusion?
ⓐ. every wavelength now produces exactly the same amount of oxygen
ⓑ. bacterial distribution no longer reveals the action spectrum
ⓒ. the alga has lost chlorophyll a from its reaction centres
ⓓ. photosynthetic oxygen must now originate from carbon dioxide
Correct Answer: bacterial distribution no longer reveals the action spectrum
Explanation: Engelmann’s inference depends on a biological indicator response. Aerobic bacteria normally accumulate where oxygen concentration is greatest, so their spatial pattern reveals which wavelengths drive the highest oxygen evolution. In the altered experiment, the bacteria have lost the ability to move toward oxygen. Their uniform distribution no longer reports local oxygen production, even though the alga may retain a normal action spectrum. The observation is uninformative rather than evidence for equal photosynthesis at all wavelengths. This distinction separates the process being studied from the behaviour of the measuring system. A failed indicator can erase the expected pattern without changing pigment absorption, reaction-centre function or the water-derived origin of oxygen. The strongest diagnosis comes from locating the first failed step while upstream functions remain intact in the stated spectral comparison.
311. Arrange the steps used to identify an early stable product of carbon fixation with a radioactive-carbon pulse.
P. Separate the cellular compounds.
Q. Expose photosynthesising cells briefly to \(\mathrm{^{14}CO_2}\).
R. Detect which separated compounds contain radioactivity.
S. Stop metabolism rapidly at defined times.
T. Compare early and later labelling patterns.
ⓐ. P → Q → R → S → T
ⓑ. Q → P → S → T → R
ⓒ. S → Q → P → R → T
ⓓ. Q → S → P → R → T
Correct Answer: Q → S → P → R → T
Explanation: Photosynthesising cells must first receive a short pulse of labelled carbon dioxide so that newly fixed carbon enters the pathway. Metabolism is then stopped rapidly at defined times to preserve the distribution of label present at each interval. Cellular compounds are separated, and their radioactivity is detected. Comparing the earliest and later patterns reveals which stable compound receives the label first and how carbon subsequently spreads through the pathway. Separation before exposure would provide no labelled products, while delayed stopping would allow extensive redistribution and obscure early intermediates. The order follows the logic of tracer experimentation: introduce the label, freeze the process, resolve the products, locate the isotope and reconstruct the sequence.
312. A graph plots photosynthetic rate against light intensity for the same plant at \(15^\circ\mathrm{C}\) and \(30^\circ\mathrm{C}\). The curves nearly overlap at very low light. At high light, the \(30^\circ\mathrm{C}\) curve reaches a much higher plateau. Which interpretation is best supported?
ⓐ. light limits both at low intensity; low temperature sets one plateau
ⓑ. temperature is the sole limiting factor at every light intensity
ⓒ. the two temperatures produce identical enzyme activity at high light
ⓓ. high light removes the need for stromal carbon reactions
Correct Answer: light limits both at low intensity; low temperature sets one plateau
Explanation: Near the origin, photon supply is so limited that both temperature treatments operate at similar low rates. Extra thermal capacity cannot be used when insufficient light is available to support ATP and NADPH formation. At high light, the curves separate and the cooler treatment reaches a lower plateau. Light is now adequate, so temperature-sensitive enzyme reactions become more influential. The higher \(30^\circ\mathrm{C}\) plateau indicates faster carbon-reaction capacity for this plant within the tested range. The graph demonstrates a shift in the limiting factor rather than one factor controlling the entire curve. Axis interpretation and comparison of both regions are needed to distinguish low-light limitation from high-light temperature limitation. The axes and the specified curve region determine whether slope, plateau, optimum or decline is being interpreted in the stated light-response comparison.
313. A graph plots photosynthetic rate against external carbon-dioxide concentration for a well-watered leaf and a drought-stressed leaf. Both receive high light and suitable temperature. The well-watered curve rises strongly, while the drought-stressed curve remains low and responds only weakly to increasing external carbon dioxide. Stomata of the stressed leaf are mostly closed. Which explanation fits the graph?
ⓐ. drought has increased carbon-dioxide diffusion into the leaf
ⓑ. external carbon dioxide has become the source of photosynthetic oxygen
ⓒ. stomatal resistance limits enriched carbon dioxide reaching mesophyll cells
ⓓ. high light has stopped all enzyme-controlled reactions in the stressed leaf
Correct Answer: stomatal resistance limits enriched carbon dioxide reaching mesophyll cells
Explanation: The horizontal axis records carbon dioxide outside the leaf, but RuBisCO responds to carbon dioxide that reaches the internal air spaces and photosynthetic cells. In the drought-stressed leaf, mostly closed stomata create a strong diffusion barrier. External enrichment produces only a small rise in internal carbon dioxide and a weak photosynthetic response. The well-watered leaf has more open stomata and can use the added gas, giving the steeper curve. The graph distinguishes atmospheric supply from effective internal availability. Drought may also produce metabolic effects during prolonged stress, but the stated stomatal condition provides the most direct explanation for the limited response under otherwise suitable light and temperature.
314. Photosynthetic rates are measured under four treatments.
| Treatment | Light | Carbon dioxide | Temperature | Relative rate |
|---|
| P | High | High | \(10^\circ\mathrm{C}\) | \(14\) |
| Q | High | High | \(25^\circ\mathrm{C}\) | \(46\) |
| R | High | Low | \(25^\circ\mathrm{C}\) | \(24\) |
| S | Low | High | \(25^\circ\mathrm{C}\) | \(18\) |
Which comparison most directly isolates the effect of temperature?
ⓐ. P and R
ⓑ. Q and S
ⓒ. R and S
ⓓ. P and Q
Correct Answer: P and Q
Explanation: Treatments P and Q have the same high light intensity and high carbon-dioxide availability. Their only stated difference is temperature, which changes from \(10^\circ\mathrm{C}\) to \(25^\circ\mathrm{C}\). The large increase in rate can be assigned most directly to the temperature change under these conditions. The result is consistent with acceleration of enzyme-controlled carbon reactions as temperature approaches a favourable range. Other comparisons alter more than one relevant factor: P and R differ in both temperature and carbon dioxide, Q and S differ in light, and R and S differ in light and carbon dioxide. Isolating one factor requires holding the other major variables constant rather than merely comparing the highest and lowest rates. Reading across rows and then comparing columns isolates the variable responsible for the reported pattern in the stated temperature-response comparison.
315. Under high light and abundant carbon dioxide, a photosynthetic-rate curve reaches a low plateau at \(10^\circ\mathrm{C}\). When temperature is raised to \(25^\circ\mathrm{C}\), the plateau shifts upward. Which factor was most directly limiting the original plateau?
ⓐ. wavelength quality of the light
ⓑ. temperature-dependent biochemical capacity
ⓒ. atmospheric oxygen as the only carbon source
ⓓ. duration of chlorophyll extraction
Correct Answer: temperature-dependent biochemical capacity
Explanation: High light removes ordinary photon limitation, and abundant carbon dioxide supplies the principal external substrate for carbon fixation. The low plateau at \(10^\circ\mathrm{C}\) points to another constraint. Raising only temperature to \(25^\circ\mathrm{C}\) increases the attainable rate, showing that the original restriction involved temperature-sensitive biochemical reactions, especially enzyme-controlled stromal processes. A plateau does not always represent light saturation alone; it identifies the rate ceiling under the complete set of current conditions. Changing the factor that raises the ceiling reveals what was limiting. The result demonstrates Blackman’s principle experimentally and avoids interpreting every flat response region as evidence of inadequate light or carbon dioxide. Very high light can cause photoinhibition, whereas an ordinary plateau usually indicates limitation by another factor. The higher plateau after warming shows that temperature, not light or carbon dioxide, limited enzyme-controlled carbon reactions at the original low temperature.
316. Consider the following statements about interpreting photosynthetic response graphs.
I. A plateau after increasing light means that another factor may be limiting.
II. A decline at extremely high light can indicate photoinhibition rather than ordinary saturation.
III. One temperature optimum must apply universally to all plant species.
IV. External carbon-dioxide enrichment may produce little response when light remains very low.
Which combination is valid?
ⓐ. Statements I and III are correct; statements II and IV are incorrect
ⓑ. Statements II and III are correct; statements I and IV are incorrect
ⓒ. Statements I, II and IV are correct; statement III is incorrect
ⓓ. Statements I, II, III and IV are all correct
Correct Answer: Statements I, II and IV are correct; statement III is incorrect
Explanation: A light-response plateau indicates that additional light no longer raises the rate under the existing conditions, often because carbon dioxide, temperature or internal capacity has become limiting. A decline beyond the plateau is a different region and may reflect damage caused by excessive illumination. Temperature optima are not universal; they vary with pathway type, habitat adaptation and plant physiology, making statement III invalid. Very low light restricts photochemical energy supply, so added carbon dioxide may have little effect until illumination increases. Statements I, II and IV preserve the conditional meaning of graph regions. Interpreting a curve requires attention to axes, environmental context and whether the response rises, plateaus or declines. Correct graph interpretation requires identifying both axes and fixed conditions before assigning a plateau, optimum or decline to a particular limiting factor.
317. Non-cyclic electron flow must supply \(12\) NADPH molecules. Assume each NADPH requires \(2\) electrons and all electrons originate from photolysis according to \(\mathrm{2H_2O\rightarrow4H^++O_2+4e^-}\). How many water molecules are split and oxygen molecules released?
ⓐ. \(12\) water molecules and \(6\) oxygen molecules
ⓑ. \(6\) water molecules and \(3\) oxygen molecules
ⓒ. \(24\) water molecules and \(12\) oxygen molecules
ⓓ. \(12\) water molecules and \(12\) oxygen molecules
Correct Answer: \(12\) water molecules and \(6\) oxygen molecules
Explanation: Formation of \(12\) NADPH molecules requires
\[
12\times2=24
\]
electrons. One complete photolysis relation provides \(4\) electrons by splitting \(2\) water molecules and releasing \(1\) oxygen molecule. The relation must operate
\[
\frac{24}{4}=6
\]
times. Water use is
\[
6\times2=12
\]
molecules, and oxygen production is
\[
6\times1=6
\]
molecules. The calculation links terminal reducing-power formation with the donor-side water reaction. It does not imply that NADPH is produced directly by water splitting; electrons pass through both photosystems and their carriers before reducing \(\mathrm{NADP^+}\). The oxygen count follows from the same electron requirement that supports the stated NADPH output. The electron demand is \(12\times2=24\). Six complete photolysis reactions provide those electrons, and each reaction splits two water molecules while releasing one oxygen molecule. This yields \(12\) water molecules split and \(6\) oxygen molecules without treating water splitting as direct NADPH synthesis.
318. Fixation of \(10\) carbon-dioxide molecules requires \(30\) ATP and \(20\) NADPH. Non-cyclic electron flow supplies exactly \(20\) ATP and \(20\) NADPH. How much additional ATP must cyclic photophosphorylation provide?
ⓐ. \(5\) ATP
ⓑ. \(20\) ATP
ⓒ. \(30\) ATP
ⓓ. \(10\) ATP
Correct Answer: \(10\) ATP
Explanation: The Calvin-cycle requirement for \(10\) carbon-dioxide molecules is already stated as \(30\) ATP and \(20\) NADPH. Non-cyclic flow supplies the full NADPH requirement, so no additional reducing power is needed. Its ATP output is only \(20\), leaving a deficit of
\[
30-20=10
\]
ATP. Cyclic electron flow around photosystem I can supply this missing ATP without producing extra NADPH or oxygen. The calculation explains the regulatory value of cyclic photophosphorylation: it corrects an imbalance between the ATP and NADPH outputs of light reactions and the greater relative ATP demand of carbon fixation. Supplying \(20\) or \(30\) extra ATP would exceed the stated deficit, while \(5\) would leave the pathway energy-limited. The supplied NADPH exactly meets the requirement, but ATP is short by \(30-20=10\). Cyclic photophosphorylation can fill this selective ATP deficit because it increases ATP formation without adding NADPH or oxygen to the non-cyclic output.
319. Six external carbon-dioxide molecules are initially fixed by six PEP molecules in mesophyll cells. Each four-carbon product is later decarboxylated in bundle-sheath cells. Ignoring intermediate names not required here, which symbolic summary correctly represents the carbon flow?
ⓐ. \(\mathrm{6PEP+6CO_2\rightarrow3}\) four-carbon acids \(\rightarrow\mathrm{3CO_2+6}\) three-carbon compounds
ⓑ. \(\mathrm{6PEP+3CO_2\rightarrow6}\) four-carbon acids \(\rightarrow\mathrm{3CO_2+3}\) three-carbon compounds
ⓒ. \(\mathrm{6PEP+6CO_2\rightarrow6}\) four-carbon acids \(\rightarrow\mathrm{6CO_2+6}\) three-carbon compounds
ⓓ. \(\mathrm{6PEP+6O_2\rightarrow6}\) four-carbon acids \(\rightarrow\mathrm{6O_2+6}\) three-carbon compounds
Correct Answer: \(\mathrm{6PEP+6CO_2\rightarrow6}\) four-carbon acids \(\rightarrow\mathrm{6CO_2+6}\) three-carbon compounds
Explanation: Each \(3\)-carbon PEP molecule accepts one external carbon unit and forms one \(4\)-carbon product. Six fixation events produce six four-carbon acids. Decarboxylation of each acid releases one carbon-dioxide molecule and leaves one \(3\)-carbon compound, giving six of each output. Carbon is conserved:
\[
6\times4=24
\]
carbon atoms before decarboxylation, while the products contain
\[
(6\times1)+(6\times3)=24
\]
carbon atoms. The released six carbon-dioxide molecules can enter bundle-sheath Calvin-cycle reactions, while the six three-carbon compounds return toward mesophyll PEP regeneration. The summary captures the shuttle’s carbon transfer without treating the preliminary C4 pathway as a replacement for the Calvin cycle. Six initial fixations produce six four-carbon acids, each carrying one externally acquired carbon. Decarboxylation later releases six carbon-dioxide molecules and leaves six three-carbon compounds. The released carbon is transferred to bundle-sheath RuBisCO; it is not a second set of newly captured atmospheric carbon.
320. Carbon dioxide in a greenhouse is increased from \(0.03\%\) to \(0.05\%\). What is the percentage increase relative to the initial concentration, and what is the correct biological qualifier?
ⓐ. Approximately \(66.7\%\); enrichment may help when other factors are suitable
ⓑ. Approximately \(40\%\); the enrichment guarantees unlimited photosynthesis under all conditions
ⓒ. Approximately \(20\%\); the enrichment can replace the need for light
ⓓ. Approximately \(166.7\%\); the enrichment prevents every form of respiration
Correct Answer: Approximately \(66.7\%\); enrichment may help when other factors are suitable
Explanation: The absolute increase is
\[
0.05\%-0.03\%=0.02\%.
\]
Relative to the initial concentration, the percentage increase is
\[
\frac{0.02}{0.03}\times100\approx66.7\%.
\]
The biological interpretation must remain conditional. Moderate carbon-dioxide enrichment can increase carbon fixation when illumination, temperature, water availability and internal photosynthetic capacity permit the plant to use the added substrate. Under very low light, additional carbon dioxide may produce little response because the photochemical supply of ATP and NADPH remains limiting. Prolonged exposure to concentrations above a favourable range may also become harmful. The calculation measures the proportional increase in carbon-dioxide concentration, while the qualifier applies Blackman’s law by recognising that another factor may limit the photosynthetic response.
