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Photosynthesis occurs in chloroplasts; grana house light reaction, stroma hosts dark reaction. Example: grana contain stacked thylakoids with chlorophyll a.
Light reaction converts light energy to chemical energy (ATP and NADPH); occurs in thylakoid membranes. Example: non-cyclic photophosphorylation produces both ATP and NADPH.
Z-scheme describes electron flow in photosystems II and I; PSII absorbs 680 nm (P680), PSI absorbs 700 nm (P700). Verify from NCERT.
Photolysis of water occurs in PSII, releasing O?, electrons, and protons; catalyzed by Mn?² and Cl?. Example: 2H?O-4H? + 4e? + O?.
Chemiosmotic hypothesis explains ATP synthesis: proton gradient across thylakoid membrane drives ATP synthase. Example: pH drops in lumen, rises in stroma.
Dark reaction (Calvin cycle) fixes CO? into glucose using ATP and NADPH; occurs in stroma. Example: RuBP (5C) + CO?-2 molecules of 3-PGA (3C).
Calvin cycle has three phases: carboxylation (RuBisCO), reduction (uses ATP and NADPH), regeneration (of RuBP). Example: RuBisCO is the most abundant enzyme on Earth.
One turn of Calvin cycle fixes 1 CO?; 3 turns produce 1 molecule of G3P; 6 turns produce 1 glucose. Verify from NCERT.
RuBisCO catalyzes both carboxylation and oxygenation; oxygenation leads to photorespiration. Example: high O? and low CO? favor photorespiration.
C3 plants fix CO? directly into 3-PGA via RuBisCO; first stable product is 3-phosphoglycerate (3C). Example: rice, wheat.
C4 plants minimize photorespiration via spatial separation: mesophyll cells fix CO? into 4C oxaloacetate, bundle sheath cells release CO? for Calvin cycle. Example: maize, sugarcane.
C4 pathway uses PEP carboxylase (in mesophyll), which has high affinity for CO? and no oxygenase activity. Example: PEP + CO?-oxaloacetate (catalyzed by PEPcase).
Kranz anatomy is characteristic of C4 plants: large bundle sheath cells with numerous chloroplasts surrounding veins. Example: absence in C3 plants like sunflower.
CAM plants fix CO? at night via PEPcase into malic acid; decarboxylation occurs during day for Calvin cycle. Example: Bryophyllum, pineapple.
CAM stands for Crassulacean Acid Metabolism; temporal separation of CO? fixation and Calvin cycle. Example: stomata open at night, closed during day.
Photorespiration does not produce ATP or sugar; it consumes energy and releases CO?. Example: occurs in C3 plants under high O?, low CO?.
C4 plants are more efficient in hot, dry climates due to reduced photorespiration. Example: maize yields higher than wheat in tropics.
ATP required per CO? fixed: 3 ATP and 2 NADPH in C3 cycle; C4 cycle requires 5 ATP and 2 NADPH per CO?. Verify from NCERT.
Emerson enhancement effect shows increased photosynthesis when both 680 nm and 700 nm light are provided together. Example: proves two photosystems.
Hill reaction: isolated chloroplasts produce O? in presence of light and electron acceptor (e.g., ferric salts), but not glucose. Example: proves light reaction independence.
Intermediate — requires understanding of biochemical pathways, spatial/temporal distinctions, and energy calculations; diagrams frequently tested.
Trap: Assuming photorespiration produces energy like respiration. Avoid: Photorespiration consumes ATP and releases CO? without producing sugar; it is a wasteful process.
Trap: Confusing C4 and CAM pathways as structurally similar. Avoid: C4 uses spatial separation (mesophyll vs. bundle sheath), CAM uses temporal separation (night vs. day).
Trap: Thinking RuBisCO only fixes CO?. Avoid: RuBisCO has dual activity: carboxylase (in C3 cycle) and oxygenase (triggers photorespiration under high O?).
Q1. In which part of the chloroplast does the light reaction of photosynthesis occur? A. Stroma B. Outer membrane C. Thylakoid membrane D. Inner membrane
Answer: C Explanation: Light reaction occurs in thylakoid membranes where photosystems and electron transport chain are located. Why others fail: Stroma (A) is for dark reaction; membranes (B, D) are structural but not sites of photochemical reactions.
Q2. Which of the following is the primary CO? acceptor in C4 plants? A. RuBP B. PEP C. PGA D. Malate
Answer: B Explanation: Phosphoenolpyruvate (PEP) in mesophyll cells accepts CO? to form oxaloacetate in C4 plants. Why others fail: RuBP (A) is CO? acceptor in C3 plants; malate (D) is a product, not initial acceptor.
Q3. What is the first stable product of carbon fixation in C3 plants? A. RuBP B. Glucose C. Phosphoglyceraldehyde (PGAL) D. 3-Phosphoglycerate (PGA)
Answer: D Explanation: CO? combines with RuBP to form two molecules of 3-PGA, the first stable compound in C3 cycle. Why others fail: PGAL (C) appears later; glucose (B) is final product; RuBP (A) is acceptor, not product.
Q4. Which of the following correctly represents the ATP and NADPH requirement for the synthesis of one molecule of glucose in the Calvin cycle? A. 6 ATP and 6 NADPH B. 9 ATP and 6 NADPH C. 12 ATP and 12 NADPH D. 18 ATP and 12 NADPH
Answer: D Explanation: 6 turns of Calvin cycle produce one glucose, requiring 18 ATP and 12 NADPH (3 ATP + 2 NADPH per CO?). Why others fail: 9 ATP and 6 NADPH (B) is for 3 CO? molecules (one G3P), not full glucose.
Q5. Which of the following plants exhibits Kranz anatomy? A. Rice B. Wheat C. Sugarcane D. Potato
Answer: C Explanation: Sugarcane is a C4 plant with Kranz anatomy—large bundle sheath cells with chloroplasts. Why others fail: Rice, wheat, potato (A, B, D) are C3 plants lacking Kranz anatomy.
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