NEET Photosynthesis in Higher Plants Previous Year Questions with Solutions

RuBisCO (Ribulose bisphosphate carboxylase-oxygenase) is the most abundant enzyme in the biosphere. Its active site can bind both CO₂ and O₂ competitively: - When CO₂/O₂ ratio is high, it acts as a carboxylase, fixing CO₂ into two 3C molecules (3-PGA) in the Calvin cycle. - When O₂ concentration is high and CO₂ is low, it acts as an oxygenase, initiating the photorespiratory C₂ cycle.

• • Photorespiration does not produce ATP or sugars and wastes fixed carbon.

In the chloroplast stroma, RuBisCO oxygenates RuBP (5-carbon) with molecular oxygen to cleave it into one 3-carbon molecule and one 2-carbon molecule:

• • 3-Phosphoglycerate (PGA) enters the Calvin cycle.
• • 2-Phosphoglycolate is dephosphorylated to glycolate and transported to peroxisomes and mitochondria for carbon salvage.

1. Phosphoenolpyruvate (PEP): 3-carbon CO₂ acceptor in C₄ pathway. 2. 2-Phosphoglycolate: 2-carbon photorespiratory product (X). 3. Oxaloacetate (OAA): 4-carbon organic acid in C₄ pathway / TCA cycle. 4. Malate: 4-carbon dicarboxylic acid in C₄ pathway.

• • Matches Option (2).

Photosystems are composed of two functional parts: 1. Antenna complex (LHC): Hundreds of accessory pigment molecules (Chlorophyll b, xanthophylls, carotenoids) bound to proteins that harvest light of various wavelengths and transfer energy by resonance. 2. Reaction Centre: A unique single molecule of Chlorophyll a that traps excitation energy and ejects high-energy electrons into the electron transport chain.

• • Chlorophyll a is the essential primary photosynthetic pigment.

The protein environment alters the absorption maxima of the reaction centre chlorophyll a in the two photosystems: - In PS I: Reaction centre chlorophyll a has an absorption maximum at 700 nm (designated P700). - In PS II: Reaction centre chlorophyll a has an absorption maximum at 680 nm (designated P680).

• • Chlorophyll a is the unique pigment forming P700.

Chlorophyll a has the absorption peak at 700 nm in the reaction centre PS I (P700).

• • Matches Option (2).

In 1770, Joseph Priestley performed bell jar experiments with a burning candle, a mouse, and a mint plant. He demonstrated that plants restore to the air whatever breathing animals and burning candles remove, establishing the fundamental role of air.

• • Priestley later discovered oxygen in 1774.

Jan Ingenhousz placed an aquatic plant (Hydrilla) in bright sunlight and darkness. He observed that in sunlight, small gas bubbles formed around green parts, but not in the dark. He identified these bubbles as oxygen, proving that sunlight and green tissues are required for oxygen release.

• • Confirmed that green parts release oxygen gas.

In 1854, Julius von Sachs provided evidence that when plants grow, they synthesize glucose, which is stored as starch. He also showed that chlorophyll is housed inside special bodies (later termed chloroplasts).

• • Established the primary organic product of photosynthesis.

Using a glass prism to split white light into spectral components, T.W. Engelmann illuminated a green alga (Cladophora) placed in a suspension of aerobic bacteria. The bacteria accumulated in blue and red light zones, plotting the first action spectrum resembling the absorption spectra of chlorophyll a and b.

• • First experimental action spectrum of photosynthesis.

Combining the milestones sequentially: Role of air (1770) -> Release of oxygen (1779) -> Production of glucose (1854) -> Absorption spectra of chlorophyll a and b (1888). This matches Option (2).

• • Matches official NTA final answer key Option B / (2).

The Calvin cycle consists of three primary stages: Carboxylation (catalyzed by RuBisCO), Reduction, and Regeneration of RuBP. Each single cycle incorporates and fixes 1 molecule of inorganic carbon dioxide (CO₂).

• • Glucose has the molecular formula C₆H₁₂O₆, containing 6 carbon atoms.

Because each turn brings in only 1 carbon atom, 6 separate turns of the cycle are mandatory to fix 6 carbon atoms and yield 1 net molecule of glucose.

• • Requires 18 ATP and 12 NADPH for 1 glucose.

For 3 complete molecules of glucose, the number of required turns scales proportionately:

• • Total CO₂ fixed = 18 molecules.
• • Total ATP consumed = 18 × 3 = 54 ATP.
• • Total NADPH consumed = 18 × 2 = 36 NADPH.

Option (1) is 6 (turns for 1 glucose); Option (2) is 3 (number of glucose molecules requested); Option (3) is 1; Option (4) is 18. The correct choice is Option (4).

• • Matches official NTA final answer key Option D / (4).

Photorespiration occurs prominently in C₃ plants when RuBisCO binds oxygen instead of carbon dioxide under high temperature, high O₂ concentration, and low CO₂ concentration. In contrast, C₄ plants have evolved a CO₂-concentrating mechanism (Kranz anatomy) which prevents photorespiration. Thus, statement (a) is definitively false.

• • Eliminates Option (1) and Option (4).

In photorespiration, RuBisCO functions as an oxygenase, consuming oxygen (O₂) and releasing carbon dioxide (CO₂). Statement (b) claims that "CO₂ is consumed and O₂ is generated", which is the exact opposite (normal photosynthesis). Hence, statement (b) is false.

• • Eliminates Option (3).

When RuBisCO oxygenates ribulose-1,5-bisphosphate (RuBP), the 5-carbon substrate is cleaved into one molecule of 3-carbon phosphoglycerate (3-PGA) and one molecule of 2-carbon 2-phosphoglycolate.

• • Statement (c) is factually correct.

NCERT states verbatim: "In the photorespiratory pathway, there is neither synthesis of sugars, nor of ATP. Rather it results in the release of CO₂ with the utilization of ATP. There is also no synthesis of NADPH." Therefore, statement (d) is factually correct.

• • Photorespiration is a biologically wasteful process.

Statements (c) and (d) are correct, while (a) and (b) are false. The correct option is Option (2).

• • Matches official NTA final answer key Option B / (2).

When leaf pigments are separated via paper chromatography, four distinct pigment bands appear based on their polarity and absorption properties:

• • Chlorophyll a: The primary photosynthetic reaction center pigment; appears bright or blue-green in the chromatogram. Hence, A matches III.
• • Chlorophyll b: An accessory antenna pigment; appears yellow-green in the chromatogram. Hence, B matches I.
• • Xanthophylls: Oxygenated carotenoid accessory pigments; appear distinct yellow in the chromatogram. Hence, C matches II.
• • Carotenoids: Pure hydrocarbon carotenes and related pigments; appear yellow to yellow-orange in the chromatogram. Hence, D matches IV.

Consolidating the four pigment-color associations:

• • A (Chlorophyll a) -> III (Blue-green)
• • B (Chlorophyll b) -> I (Yellow-green)
• • C (Xanthophylls) -> II (Yellow)
• • D (Carotenoids) -> IV (Yellow to Yellow-orange)

The code A-III, B-I, C-II, D-IV uniquely corresponds to Option (2).

Ribulose-1,5-bisphosphate carboxylase-oxygenase (RuBisCO) is the most abundant enzyme on Earth, located in the chloroplast stroma of autotrophs:

• • It possesses bifunctional catalytic capacity: it can act as both a carboxylase and an oxygenase.
• • Under normal physiological conditions, RuBisCO exhibits a significantly higher affinity for $\text{CO}_2$ than for $\text{O}_2$.

The primary physiological reaction catalyzed by RuBisCO:

• • Carboxylation is the most crucial step of the Calvin ($C_3$) cycle.
• • RuBisCO catalyzes the addition of one molecule of $\text{CO}_2$ to the 5-carbon acceptor ribulose-1,5-bisphosphate (RuBP).
• • The unstable 6-carbon intermediate immediately cleaves into two molecules of 3-phosphoglyceric acid (3-PGA), the first stable product of $C_3$ photosynthesis.
• • Hence, statement (4) is undeniably TRUE.

Analyzing why options 1, 2, and 3 are scientifically incorrect:

• • Statement (1): RuBisCO is inactive in the dark; it requires light-driven stromal alkalinization and $Mg^{2+}$ influx.
• • Statement (2): NCERT explicitly clarifies that RuBisCO has a much greater affinity for $\text{CO}_2$ than for $\text{O}_2$; oxygenase activity (photorespiration) occurs only when $\text{CO}_2:\text{O}_2$ ratio decreases.
• • Statement (3): Photolysis of water ($2\text{H}_2\text{O} \rightarrow 4\text{H}^+ + 4e^- + \text{O}_2$) is catalyzed by the water-splitting complex associated with Photosystem II on the luminal side of the thylakoid membrane, completely independent of RuBisCO.

The only true statement is that RuBisCO catalyzes the carboxylation of RuBP, matching Option (4).

Photosynthesis consists of two phases: light reactions (photochemical phase) occurring in thylakoids, producing ATP, NADPH, and O₂; and dark reactions (biosynthetic phase) occurring in the stroma, where sugars are synthesized.

The dark reaction does not directly require light. It requires: 1. Atmospheric carbon dioxide ($CO_2$) as the carbon source (C). 2. $ATP$ to supply energy for reduction and regeneration (D). 3. $NADPH$ to supply reducing equivalents ($H^+$ and electrons) for reduction of 3-PGA to G3P (E).

C, D, and E are the indispensable chemical requirements, corresponding to Option (1).

The Calvin cycle proceeds in three stages: 1. Carboxylation: RuBP ($5C$) binds $CO_2$ via RuBisCO to form two molecules of 3-PGA ($3C$). No ATP or NADPH is consumed.

2. Reduction: Two molecules of ATP for phosphorylation and two of NADPH for reduction are consumed per $CO_2$ molecule fixed. 3. Regeneration: Regeneration of the $CO_2$ acceptor molecule RuBP requires one phosphorylation step using one ATP. Total Energy Consumed per $CO_2$: $$\text{Total ATP} = 2 + 1 = 3\;\text{molecules}$$ $$\text{Total NADPH} = 2\;\text{molecules}$$

To produce 1 molecule of hexose glucose ($C_6H_{12}O_6$), six turns of the cycle are required, consuming $6 \times 3 = 18\;ATP$ and $6 \times 2 = 12\;NADPH$, perfectly matching NCERT Table 13.1.

RuBisCO is the most abundant enzyme in the world and has an active site that can bind both $CO_2$ and $O_2$. In $C_3$ plants, some $O_2$ binds to RuBisCO, and hence $CO_2$ fixation is decreased. The RuBP instead of being converted to 2 molecules of PGA binds with $O_2$ to form one molecule of phosphoglycerate and one molecule of phosphoglycolate. Statement I is true.

In $C_4$ plants photorespiration does not occur. This is because they have a mechanism that increases the concentration of $CO_2$ at the enzyme site. In $C_4$ plants, mesophyll cells completely lack RuBisCO (they possess PEPcase); therefore, photorespiration cannot happen in mesophyll cells. When $C_4$ acids from mesophyll cells are broken down in bundle sheath cells, $CO_2$ is released, ensuring that RuBisCO functions exclusively as a carboxylase. Thus, photorespiration is zero in both cell types. Statement II is false.

Statement I is true but Statement II is false, matching Option (1).

For each single molecule of $CO_2$ entering the cycle, energy is consumed in two specific steps:

Synthesis of one net molecule of glucose ($C_6H_{12}O_6$) requires 6 complete turns of the Calvin cycle fixing 6 $CO_2$ molecules:

The net input/output equation for glucose synthesis in the dark reaction is:

Photosystems comprise hundreds of pigment molecules bound to proteins forming antennae that funnel excitation energy to a unique single chlorophyll $a$ reaction centre:

The two photosystems are designated based on the red absorption peak wavelength of their respective reaction centre chlorophyll $a$ molecules:

NCERT states verbatim: "In PS II the reaction centre chlorophyll a has an absorption peak at 680 nm, and is called P680."

Peter Mitchell's chemiosmotic mechanism explains phosphorylation in both chloroplasts and mitochondria through four cooperative structures:

NCERT explicitly states: "Chemiosmosis requires a membrane, a proton pump, a proton gradient and ATP synthase. Energy is used to pump protons across a membrane, to create a gradient or a high concentration of protons within the thylakoid lumen."

NADP reductase (not "NADP synthase") is a peripheral stroma-facing enzyme facilitating NADP+ reduction, but is not the motor of chemiosmotic ATP synthesis. An "electron gradient" does not drive ATP synthase; the proton-motive force drives it.

Plants exhibiting the Kranz anatomy and the Hatch-Slack pathway are termed $C_4$ plants. Classical examples include maize, sugarcane, and sorghum.

The primary acceptor of $CO_2$ is phosphoenolpyruvate ($PEP$), a 3-carbon compound present in the mesophyll cells. The enzyme responsible is $PEP$ carboxylase ($PEPcase$):

OAA (oxaloacetate) is rapidly reduced to other 4-carbon dicarboxylic acids like malic acid or aspartic acid in the mesophyll cells before transport to bundle sheath cells.

When light of wavelength beyond $680\text{ nm}$ is available, or when stroma lamellae lack PS II and NADP reductase, only Photosystem I ($P_{700}$) operates in a closed cycle:

- Non-cyclic photophosphorylation (Z-scheme): Involves both Photosystems II ($P_{680}$) and I ($P_{700}$) operating in series, generating both ATP and $NADPH + H^+$, with photolysis of water. - Cyclic photophosphorylation: Involves only PS I. Electrons cycle back to the reaction center, synthesizing solely ATP without $NADPH$ production or water splitting.

Statement (4) asserts that "cyclic photophosphorylation involves both PS I and PS II". This is incorrect because PS II does not participate in cyclic electron flow.

RuBisCO (Ribulose-1,5-bisphosphate carboxylase-oxygenase) has an active site that can bind both CO2 and O2 competitively. In C3 plants, when O2 concentration is high and CO2 is low, RuBisCO acts as an oxygenase.

In the oxygenase reaction: $$\text{RuBP (5C)} + \text{O}_2 \xrightarrow{\text{RuBisCO}} \text{3-PGA (3C)} + \text{Phosphoglycolate (2C)}$$ Instead of forming two molecules of 3-PGA as in normal carboxylation, it yields exactly ONE molecule of a 3-C compound (3-PGA) and ONE molecule of a 2-C compound (2-phosphoglycolate).

Option (1) describes carboxylation (2 molecules of 3-C compound). Option (2) correctly states "1 molecule of 3-C compound". Options (3) and (4) propose stoichiometries incompatible with the 5-carbon skeleton of RuBP.

Therefore, the oxygenation activity of RuBisCO enzyme directly results in the formation of 1 molecule of a 3-carbon compound.

When Photosystem II absorbs red light at 680 nm, electrons in chlorophyll a become excited and jump into an orbit farther from the atomic nucleus. These electrons are picked up by a primary electron acceptor (pheophytin).

From pheophytin, electrons are transferred to plastoquinone (PQ), a mobile lipid-soluble quinone embedded within the thylakoid lipid bilayer. Plastoquinone picks up protons from the stroma to form PQH2.

PQH2 transfers its electrons to the multi-subunit Cytochrome b6f complex while translocating protons across the thylakoid membrane into the lumen, generating a proton gradient.

From Cyt b6f, electrons are carried by Plastocyanin (a copper-containing mobile protein) to PS-I. Thus, Plastoquinone uniquely facilitates electron transfer from PS-II to Cyt b6f complex.