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Photosynthesis in Higher Plants

Biology · Class 11NCERT pp. 131–1502 sets × 25 questions

Key facts and formulas for Photosynthesis in Higher Plants, three worked sample questions, and two full practice sets. Every question has a hint, a reason for every option and a step-by-step solution, with a calculator or animation where one helps. English and Hinglish.

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Key facts for this chapter

IdeaRule or fact
Overall equation6CO₂ + 12H₂O → C₆H₁₂O₆ + 6H₂O + 6O₂; net water used = 6. Oxygen comes from water.
Light reactionZ scheme: H₂O → PS II → acceptor → cytochromes → PS I → acceptor → NADP⁺. P680 in PS II, P700 in PS I.
Cyclic flowOnly PS I, in stroma lamellae (no PS II, no NADP reductase). Makes ATP only.
ChemiosmosisWater splitting, H carrier and NADP reductase build the gradient; flow through CF0 makes ATP at CF1.
Calvin cycleCarboxylation (RuBisCO, 3-PGA), reduction (ATP + NADPH, PGAL), regeneration (ATP). Per CO₂: 3 ATP and 2 NADPH.
Per glucose6 turns, 18 ATP, 12 NADPH in C3. C4 needs 5 ATP per CO₂, so 30 ATP.
C4 plantsPEPcase in mesophyll gives OAA (4C); Calvin cycle only in bundle sheath (Kranz anatomy). Negligible photorespiration, high temperature optimum.
PhotorespirationRuBisCO oxygenase at high O₂, low CO₂; chloroplast → peroxisome → mitochondria. Uses ATP, releases CO₂, no sugar.
Limiting factorsC3 respond to CO₂ above 450 µL/L; C4 saturate near 360. Tomato and bell pepper are grown in CO₂-enriched greenhouses.
ScientistsPriestley (mint), Ingenhousz (light, green parts), Sachs (starch), van Niel (O₂ from water), Engelmann, Arnon, Mitchell, Calvin.

Three sample questions

Q1. In Engelmann's experiment with a prism, Cladophora and aerobic bacteria, the bacteria gathered mostly in
  1. green and yellow regions
  2. the ultraviolet region
  3. the dark gap between colours
  4. blue and red regions of the spectrum
Hint and solution

Hint: Aerobic bacteria go where oxygen is highest. Where does chlorophyll absorb light best, so photosynthesis is fastest?

  1. Aerobic bacteria move to high O₂ — Bacteria that need oxygen swim toward the place where oxygen is released most. So they act as an oxygen detector.
  2. Prism splits light; alga strip placed along it — Engelmann split white light with a prism and laid a filament of Cladophora across the colours.
  3. More photosynthesis → more O₂ → more bacteria — Chlorophyll absorbs mainly blue and red light, so photosynthesis and oxygen release are highest there.
  4. Bacteria gathered at blue and red — So the bacteria clustered in the blue and red regions. This was the first action spectrum. Option D.
Answer: D. Bacteria sense oxygen. They gathered where the alga released most oxygen, which gave the first action spectrum.
Q2. Statement I: In a variegated leaf kept in light, only the green parts test positive for starch. Statement II: This experiment shows that CO2 is needed for photosynthesis.
  1. Both statements are true
  2. Statement I is false and Statement II is true
  3. Both statements are false
  4. Statement I is true and Statement II is false
Hint and solution

Hint: Which part of a variegated leaf lacks chlorophyll? Then ask which factor this leaf tests: chlorophyll or CO2?

  1. Variegated leaf: green parts have chlorophyll, white do not — The leaf has patches with and without chlorophyll, kept in the same light.
  2. Iodine test: only green parts turn blue-black — Starch forms only where chlorophyll is present. So Statement I is true.
  3. This shows need for chlorophyll, not CO2 — To show the need for CO2, a leaf is kept in a split chamber with KOH, which absorbs CO2. Statement II is therefore false.
  4. I true, II false — So the answer is: Statement I is true and Statement II is false. Option D.
Answer: D. The variegated leaf shows the need for chlorophyll. The need for CO2 is shown by the KOH experiment.
Q3. In the overall equation 6CO2 + 12H2O → C6H12O6 + 6H2O + 6O2, the number of water molecules actually consumed in net terms is
  1. 6
  2. 12
  3. 18
  4. 0
Hint and solution

Hint: Some water is used and some is formed again. Subtract the water on the product side from the water on the reactant side.

  1. Water used = 12, water formed = 6 — Read the equation: 12H₂O on the left and 6H₂O on the right.
  2. Net = used − formed — Water formed on the right can be cancelled against water on the left.
  3. Net = 12 − 6 = 6 — Subtract: 12 minus 6 equals 6 molecules. This matches the simple equation 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂.
  4. Answer 6 — So the net number of water molecules consumed is 6. Option A.
Answer: A. 12 are used and 6 are formed, so the net use is 6.

How to use this chapter

Read the NCERT pages first, then attempt Set 1 without the hints. Note every wrong answer in a mistake list, read the reason for each option, and retry Set 2 two days later. A topic is solid when you can say why each wrong option is wrong.

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