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CBSE · Class 11 · Biology

Photosynthesis in Higher Plants

Standard difficulty · 20 questions · 40 marks

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks
1.
The first stable product of the Calvin (C3) cycle is:
  1. glucose
  2. RuBP
  3. oxaloacetic acid
  4. 3-phosphoglyceric acid (PGA)
2.
How many molecules of NADPH does the Calvin cycle use to make 4 molecules of glucose?
  1. 48
  2. 72
  3. 24
  4. 12
3.
Where does the build-up of protons for ATP synthesis take place?
  1. The inner (lumen) side of the thylakoid membrane
  2. The stroma of mesophyll chloroplasts
  3. The bundle sheath cells
  4. The thylakoid lumen
4.
Rice is a:
  1. CAM plant only
  2. non-photosynthetic plant
  3. C4 plant
  4. C3 plant
5.
Hatch and Slack is known for showing that:
  1. the C4 pathway
  2. glucose is made in chloroplasts and stored as starch
  3. the first action spectrum, using Cladophora and aerobic bacteria
  4. only the green parts of plants, in sunlight, restore air
6.
Which leaf pigment appears yellow to yellow-orange on a paper chromatogram?
  1. Xanthophylls
  2. Chlorophyll a
  3. Chlorophyll b
  4. Carotenoids
7.
When 4 molecules of water are split in the light reactions, the number of electrons released is:
  1. 16
  2. 8
  3. 2
  4. 4
8.
Assertion (A): Fixing 30 molecules of CO2 in the Calvin cycle uses 90 molecules of ATP.
Reason (R): Each CO2 fixed uses 2 ATP and 3 NADPH.
  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
Section B — Short answer8 × 2 = 16 marks
9.
What is the difference between cyclic and non-cyclic photophosphorylation?
10.
To make 1 molecule of glucose, how many CO2 molecules must be fixed, how many turns of the Calvin cycle are needed, and how many ATP and NADPH are used?
11.
How many ATP does a C4 plant use to fix 24 molecules of CO2, compared with a C3 plant? Why the difference?
12.
Where does the first fixation of CO2 by PEP carboxylase in C4 plants take place?
13.
What did T. W. Engelmann show about photosynthesis?
14.
In the light reactions, 20 molecules of oxygen are released. How many water molecules were split, and how many electrons were passed to the electron transport chain?
15.
Is sugarcane a C3 or a C4 plant? Name its CO2 acceptor and first stable product.
16.
Which leaf pigment appears yellow to yellow-orange on a chromatogram, and is it the main or an accessory pigment?
Section C — Word problems4 × 4 = 16 marks
17.
Describe the light reactions (the Z scheme).
18.
A C3 plant and a C4 plant each make 10 molecules of glucose. For each, find (a) the CO2 fixed, (b) the ATP used, (c) the NADPH used. (d) Why do C4 plants still do better in hot, dry conditions?
19.
During the light reactions 18 molecules of O2 are released. Find (a) the water molecules split, (b) the electrons released, (c) the protons released into the lumen by photolysis, (d) the NADPH that these electrons can make. Name the photosystem where water is split.
20.
Compare mustard (C3) and pearl millet (bajra) (C4): CO2 acceptor, first product, cells where the Calvin cycle runs, leaf anatomy and photorespiration.
AmbifulWorksheets
Practice that never repeats
CBSE · Class 11 · Biology

Answer key

Photosynthesis in Higher Plants · AW-11B-PHS-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(d) 3-phosphoglyceric acid (PGA)—
2(a) 486 CO2 per glucose; 3 ATP and 2 NADPH per CO2
3(d) The thylakoid lumen—
4(d) C3 plantCalvin cycle only, first product PGA
5(a) the C4 pathway—
6(d) Carotenoids—
7(b) 82H2O → 4H+ + O2 + 4e−
8(c) A is true, but R is false.A is true: 3 × 30 = 90. R is false.
9Non-cyclic uses both photosystems; electrons from water pass to NADP+, so it makes ATP and NADPH and releases oxygen. Cyclic uses only PS I; electrons return to PS I, so it makes only ATP and no NADPH or oxygen.—
106 CO2; 6 turns; 18 ATP and 12 NADPHone turn fixes one CO2, using 3 ATP and 2 NADPH
11C4: 120 ATP; C3: 72 ATP. C4 plants spend 2 more ATP per CO2 to regenerate PEP for the first fixation.5 ATP per CO2 in C4; 3 in C3
12The mesophyll cells—
13The first action spectrum, using Cladophora and aerobic bacteria—
1440 water molecules; 80 electrons2H2O → 4H+ + O2 + 4e−
15C4: PEP (phosphoenol pyruvate); oxaloacetic acid—
16Carotenoids; an accessory pigment—
17Light absorbed by PS II excites the P680 electron, which passes along an electron transport chain (with cytochromes) to PS I; water is split to replace PS II electrons, releasing O2. PS I (P700) is excited again and its electrons pass down a second chain to NADP+, reducing it to NADPH. The electron path looks like a Z. Protons pumped into the lumen drive ATP synthase.—
18(a) 60 CO2 each; (b) C3 180 ATP, C4 300 ATP; (c) 120 NADPH each; (d) they avoid photorespiration by concentrating CO2 round RuBisCO, so they fix carbon efficiently even with stomata partly closedC3: 3 ATP + 2 NADPH per CO2; C4: 5 ATP + 2 NADPH per CO2
19(a) 36; (b) 72; (c) 72; (d) 36 (2 electrons per NADPH); water is split at photosystem II2H2O → 4H+ + O2 + 4e−; NADP+ + 2e− + H+ → NADPH
20Mustard: RuBP; 3-PGA; mesophyll; no Kranz anatomy; photorespiration present. Pearl millet (bajra): PEP; oxaloacetic acid; bundle sheath (after fixation in mesophyll); Kranz anatomy; little or no photorespiration.—