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Chapter 2

Plant Physiology (Botany): water relations, photosynthesis, respiration, plant hormones, growth and movement

NEB Class 12 past exam questions

Past questions and answers

27 questions set from this chapter, 3 of them more than once. Most repeated first.

  • Asked 2 times
  • 2082 Supplementary · 4 marks
  • 2079 · Model question · 3+1 marks

You are given seeds for germination. Briefly describe the key factors that influence normal seed germination. How would exposing the seeds to low temperatures affect the germination process?

Answer

Key factors influencing normal seed germination

  1. Water (imbibition): The seed must absorb water so the seed coat softens and the embryo cells become metabolically active. Water activates enzymes and hydrolyses stored food (starch, protein, fat) into soluble forms.
  2. Temperature: An optimum temperature (usually 25–35 °C for most crop seeds) is needed for enzyme activity. Very low or very high temperatures slow or stop germination.
  3. Oxygen (air): Germination is an aerobic process. Oxygen is needed for respiration to release energy for the growing embryo. Waterlogged or compact soil reduces oxygen and prevents germination.
  4. Light: Most seeds germinate in dark or light, but some are light-sensitive. Light is essential only for certain seeds (e.g., lettuce); for most crop seeds it is not a limiting factor.
  5. Viability and dormancy: Only viable (living) seeds with a healthy embryo can germinate. Some seeds have hard seed coats or internal dormancy that must be broken first.
  6. Soil/medium conditions: Proper moisture, aeration and a suitable pH of the medium support normal germination.

Effect of exposing seeds to low temperature

Low temperature slows down enzyme activity and metabolic reactions, so germination is delayed or inhibited. However, in some seeds low temperature is beneficial: it provides stratification (chilling treatment), which breaks seed dormancy by overcoming the inhibitory effect of abscisic acid and promoting gibberellin activity. Thus, for dormant seeds of temperate plants (e.g., apple, peach), a period of low temperature is required for normal germination, while for non-dormant seeds it simply delays germination.

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  • Asked 2 times
  • 2082 HISSAN · 8 marks
  • 2080 Baishakh · 5+1+2 marks

Define photosynthesis. Explain non-cyclic electron flow during the light reaction of photosynthesis.

Answer

Photosynthesis is the process by which green plants, algae and some bacteria use light energy to convert carbon dioxide and water into carbohydrates (glucose), releasing oxygen as a by-product. It occurs in the chloroplast and is summarised as:

6CO2+12H2O→chlorophylllightC6H12O6+6O2+6H2O6CO_2 + 12H_2O \xrightarrow[\text{chlorophyll}]{\text{light}} C_6H_{12}O_6 + 6O_2 + 6H_2O

Photosynthesis has two phases: the light reaction (photochemical phase) in the thylakoid membranes, and the dark reaction (Calvin cycle) in the stroma. Non-cyclic electron flow (non-cyclic photophosphorylation) is the main light-driven electron transport of the light reaction.

Non-cyclic electron flow

It is called "non-cyclic" because the electron does not return to its original source; instead it is passed to NADP⁺, and the electron lost by chlorophyll is replaced by electrons obtained from the photolysis of water. It involves both photosystems (PS I and PS II) and produces ATP, NADPH and O₂.

Steps:

  1. Excitation of PS II. Light of wavelength 680 nm is absorbed by the chlorophyll a molecules of PS II (P680). The electrons become excited and are raised to a higher energy level.

  2. Electron transport from PS II to PS I. The excited electrons are picked up by the primary electron acceptor (pheophytin) and passed down an electron transport chain — pheophytin → plastoquinone (PQ) → cytochrome b₆f complex → plastocyanin (PC) — to PS I. As electrons move down this chain, energy is released and used to pump protons into the thylakoid lumen, creating a proton gradient.

  3. Photolysis of water (Hill reaction). To replace the electrons lost by P680, water is split in the presence of light and Mn²⁺/Cl⁻ ions:

2H2O→light, PS II4H++4e−+O22H_2O \xrightarrow{\text{light, PS II}} 4H^+ + 4e^- + O_2

The electrons replace those lost by P680, the H⁺ ions accumulate in the lumen, and O₂ is released as a by-product. Thus water is the electron donor in non-cyclic flow.

  1. Excitation of PS I. Light of wavelength 700 nm is absorbed by PS I (P700). Its electrons are excited and accepted by ferredoxin (Fd) via the primary acceptor.

  2. Reduction of NADP⁺. The enzyme NADP reductase transfers electrons from ferredoxin to NADP⁺, which combines with H⁺ (from the lumen) to form NADPH:

NADP++2e−+H+→NADPHNADP^+ + 2e^- + H^+ \rightarrow NADPH

  1. Synthesis of ATP (chemiosmosis). The protons pumped into the lumen during electron transport create a proton gradient. These protons flow back to the stroma through the ATP synthase (CF₀–CF₁) channel, and the energy released drives the synthesis of ATP from ADP and inorganic phosphate. This is called non-cyclic photophosphorylation.

Net result: 2H₂O are split, giving 1 O₂, and the electron pair passing through the two photosystems yields about 2 ATP and 2 NADPH (per 2 electrons).

The ATP and NADPH formed are used in the stroma to reduce CO₂ to carbohydrate in the Calvin cycle.

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  • Asked 2 times
  • 2080 Baishakh · 1+3 marks
  • 2081 Baishakh · 4 marks

What is ascent of sap? Describe how transpiration pull, cohesive and adhesive properties cause ascent of sap in tall trees (cohesion-tension theory), explaining the processes involved for the maximum uptake of water in plants.

Answer

Ascent of sap is the upward movement of water and dissolved minerals from the roots through the xylem to the leaves and other aerial parts of the plant, against gravity.

Cohesion-tension (transpiration pull) theory (Dixon and Joly)

This is the most widely accepted explanation of how water reaches the top of tall trees. It rests on three properties:

  1. Transpiration pull (tension): Water evaporates from the mesophyll cells through the stomata. The mesophyll cells then have a lower water potential and draw water from neighbouring cells and finally from the xylem vessels of the leaf veins. This creates a strong negative pressure (tension) in the xylem that pulls the water column upward.
  2. Cohesion: Water molecules attract each other by hydrogen bonds. This holds them together as a continuous, unbroken column from leaf to root, and gives it high tensile strength.
  3. Adhesion: Water molecules also stick to the hydrophilic walls of the narrow xylem vessels and tracheids. This, together with capillarity, supports the column and stops it from breaking or slipping down.

Steps in the process

  • Transpiration from the leaf lowers the water potential of the leaf cells.
  • Water is pulled out of the xylem of the leaf, and the tension is transmitted down the continuous water column to the root.
  • Water enters the root from the soil by osmosis, passes across the cortex to the xylem, and replaces the water lost.

Maximum uptake of water

Because the pull is generated at the leaf and passes through the unbroken water column, the plant takes up water at a rate equal to the transpiration rate. Uptake is greatest when the continuity of the water column is maintained, the xylem is narrow, many stomata are open, and humidity is low and the air is warm and moving. The root works passively, acting as an absorbing surface. In this way the cohesion-tension mechanism can raise water to the top of trees that are over 100 m tall.

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  • 2083 Baishakh · 1 mark

What is called the process in which the shrinkage of protoplasm of fresh plant cell kept in hypertonic solution? a. Deplasmolysis b. Endosmosis c. Imbibition d. Plasmolysis

Answer

(d) Plasmolysis

When a fresh plant cell is kept in a hypertonic solution, water leaves the protoplasm by exosmosis and the protoplasm shrinks away from the cell wall — this process is called plasmolysis.

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  • 2083 Baishakh · 1 mark

A gardener removes the apical bud of a coffee plant to promote lateral branching. Which hormonal change best explains this effect? a. Reduced auxins allows lateral buds to grow b. Increased gibberellin causes stem elongation c. Increased auxin induces dormancy d. Increased auxins stimulates lateral buds

Answer

(a) Reduced auxins allows lateral buds to grow

The apical bud produces auxin, which travels downward and suppresses the growth of lateral (axillary) buds — a phenomenon called apical dominance. Removing the apical bud removes this source of auxin, so the lateral buds are released from inhibition and grow out into branches.

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  • 2083 Baishakh · 3+1 marks

Glycolysis involves partial oxidation of glucose to pyruvate with release of energy. Write the flow chart of major phases of EMP pathway and state the net gain of ATP and NADH per molecule of glucose in it.

Answer

Flow chart of the EMP pathway (glycolysis)

Glycolysis occurs in the cytoplasm and is divided into two phases: the preparatory (investment) phase and the pay-off (oxidative) phase.

A. Preparatory phase (glucose → 2 glyceraldehyde-3-phosphate)

Glucose (6C)
   │  ATP → ADP   (hexokinase)
   ▼
Glucose-6-phosphate (6C)
   │            (phosphoglucose isomerase)
   ▼
Fructose-6-phosphate (6C)
   │  ATP → ADP   (phosphofructokinase)
   ▼
Fructose-1,6-bisphosphate (6C)
   │            (aldolase)
   ▼
Dihydroxyacetone phosphate ⇌ Glyceraldehyde-3-phosphate
        (3C)                    (3C)

B. Pay-off phase (2 glyceraldehyde-3-phosphate → 2 pyruvate)

2 Glyceraldehyde-3-phosphate (3C)
   │  + 2 NAD⁺ → 2 NADH   (glyceraldehyde-3-phosphate dehydrogenase)
   ▼
2 1,3-bisphosphoglycerate (3C)
   │  2 ADP → 2 ATP   (phosphoglycerate kinase)
   ▼
2 3-phosphoglycerate (3C)
   │            (phosphoglycerate mutase)
   ▼
2 2-phosphoglycerate (3C)
   │            (enolase)
   ▼
2 Phosphoenolpyruvate (3C)
   │  2 ADP → 2 ATP   (pyruvate kinase)
   ▼
2 Pyruvate (3C)

Net gain per molecule of glucose

ProductProducedUsedNet gain
ATP4 (2 from phosphoglycerate kinase + 2 from pyruvate kinase)2 (in preparatory phase)2 ATP
NADH2 (from oxidation of 2 glyceraldehyde-3-phosphate)02 NADH

So, one molecule of glucose yields a net gain of 2 ATP and 2 NADH (plus 2 molecules of pyruvate) through the EMP pathway.

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  • 2083 Baishakh · 4+3+1 marks

Dark reaction or Black man's reaction of photosynthesis occurs in absence of light inside stroma region of chloroplast. It uses ATP and NADP and reduce to to form sugar. Provide the flow chart for Cycle highlighting 3 major steps, and explain it. Write the chemical equation of photosynthesis.

Answer

Dark Reaction (Blackman's Reaction) — the Calvin Cycle

The dark reaction (also called the Calvin cycle or C₃ cycle) takes place in the stroma of the chloroplast. It does not need light directly, but it uses the ATP and NADPH produced during the light reaction to fix CO₂ into sugar. It has three major steps:

  1. Carboxylation — CO₂ is accepted by ribulose-1,5-bisphosphate (RuBP) with the help of the enzyme RuBisCO, forming an unstable 6-carbon compound that splits into two molecules of 3-phosphoglyceric acid (3-PGA).
  2. Reduction — Each 3-PGA is phosphorylated by ATP and reduced by NADPH to form glyceraldehyde-3-phosphate (G3P).
  3. Regeneration — Most of the G3P is used to regenerate RuBP (with more ATP) so the cycle can continue; the rest is used to build glucose.

Flow chart of the Calvin cycle

                 CO₂
                  │
        ┌─────────┴──────────┐
        │  1. CARBOXYLATION  │
        │  RuBP + CO₂        │
        │  (RuBisCO enzyme)  │
        └─────────┬──────────┘
                  │
         2 × 3-PGA (3-phosphoglyceric acid)
                  │
        ┌─────────┴──────────┐
        │  2. REDUCTION      │
        │  ATP → ADP + Pi    │
        │  NADPH → NADP⁺     │
        └─────────┬──────────┘
                  │
         2 × G3P (glyceraldehyde-3-phosphate)
                  │
        ┌─────────┴──────────┐
        │  3. REGENERATION   │
        │  ATP → ADP + Pi    │
        │  G3P → RuBP        │
        └─────────┬──────────┘
                  │
              RuBP (reused)
                  │
        (6 turns → 1 glucose)

Explanation of the three steps

Step 1 — Carboxylation (fixation of CO₂): RuBP (a 5-carbon compound) combines with one molecule of CO₂. The enzyme RuBisCO catalyses this reaction. The product is an unstable 6-carbon compound which immediately breaks into two molecules of 3-PGA (3-carbon each). This is the actual fixation of carbon.

RuBP (C5)+CO2→2×3-PGA (C3)\text{RuBP (C}_5\text{)} + \text{CO}_2 \rightarrow 2 \times \text{3-PGA (C}_3\text{)}

Step 2 — Reduction: Each 3-PGA molecule is first phosphorylated by ATP to 1,3-bisphosphoglyceric acid, then reduced by NADPH (giving up H) to G3P. For every CO₂ fixed, 2 ATP and 2 NADPH are used here.

3-PGA→ATP1,3-bisphosphoglycerate→NADPHG3P\text{3-PGA} \xrightarrow{\text{ATP}} \text{1,3-bisphosphoglycerate} \xrightarrow{\text{NADPH}} \text{G3P}

Step 3 — Regeneration of RuBP: Of the G3P formed, five out of every six molecules are used (using 1 ATP each) to rebuild RuBP, so the cycle keeps running. The remaining G3P molecules leave the cycle and are joined to make glucose.

Overall balance for one glucose (6 turns of the cycle):

6 CO2+18 ATP+12 NADPH→C6H12O6+18 ADP+18 Pi+12 NADP+6\,\text{CO}_2 + 18\,\text{ATP} + 12\,\text{NADPH} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 18\,\text{ADP} + 18\,\text{P}_i + 12\,\text{NADP}^+

Chemical equation of photosynthesis

6 CO2+12 H2O→chlorophylllightC6H12O6+6 O2+6 H2O6\,\text{CO}_2 + 12\,\text{H}_2\text{O} \xrightarrow[\text{chlorophyll}]{\text{light}} \text{C}_6\text{H}_{12}\text{O}_6 + 6\,\text{O}_2 + 6\,\text{H}_2\text{O}

(For the dark reaction alone, the equation is simply: 6 CO2+12 NADPH+18 ATP→C6H12O6+12 NADP++18 ADP+18 Pi6\,\text{CO}_2 + 12\,\text{NADPH} + 18\,\text{ATP} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 12\,\text{NADP}^+ + 18\,\text{ADP} + 18\,\text{P}_i.)

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  • 2082 Baishakh · 1 mark

Which of the following structures in a plant is primarily responsible for regulating transpiration? (a) Xylem (b) Stomata (c) Root hairs (d) Phloem

Answer

(b) Stomata

Stomata are the small pores on the leaf epidermis whose opening and closing (by guard cells) controls the loss of water vapour, and hence regulates transpiration. Xylem conducts water, root hairs absorb it, and phloem translocates food — none of these regulate transpiration.

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  • 2082 Baishakh · 1 mark

If a plant is placed in a sealed bell jar with limited oxygen, how will its energy production be affected over time? (a) ATP production will increase due to higher glycolysis. (b) The plant will rely on anaerobic respiration producing less ATP. (c) The citric acid cycle will become faster to make up for lack of oxygen. (d) The electron transport chain will continue functioning normally.

Answer

(b) The plant will rely on anaerobic respiration producing less ATP.

In a sealed jar the oxygen is used up and not replenished, so aerobic respiration (Krebs cycle and electron transport chain) cannot continue. The plant then carries out anaerobic respiration (fermentation), which yields only 2 ATP per glucose molecule instead of about 36–38 ATP in aerobic respiration, so energy production falls over time.

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  • 2082 Baishakh · 2+2 marks

How does diffusion differ from osmosis based on the types of molecules involved across a cell membrane? Explain how environmental factors influence the rate of diffusion.

Answer

Diffusion vs osmosis (based on types of molecules)

  • Diffusion involves the movement of any type of molecule or ion (e.g., O₂, CO₂, glucose, ions) from a region of higher concentration to a region of lower concentration, directly across the membrane or in the surrounding medium. It does not require a semipermeable membrane.
  • Osmosis involves the movement of only solvent molecules, usually water, across a semipermeable membrane from a region of higher water potential (lower solute concentration) to a region of lower water potential (higher solute concentration).

So the key difference is that diffusion can involve solute particles as well as gases, while osmosis is restricted to water (solvent) molecules moving through a selectively permeable membrane.

Environmental factors influencing the rate of diffusion

  1. Concentration gradient — The steeper the difference in concentration between the two regions, the faster the rate of diffusion. A greater gradient drives more net movement.
  2. Temperature — Higher temperature increases the kinetic energy of molecules, so they move faster and diffusion occurs more rapidly. Lower temperature slows it down.
  3. Surface area — A larger surface area available for exchange (e.g., alveolar surface, root hair surface) increases the rate of diffusion, as more molecules can cross at a time.
  4. Distance (thickness of the medium/membrane) — The shorter the distance over which diffusion must occur, the faster the rate. Thicker barriers slow diffusion.
  5. Size and nature of molecules — Smaller and lighter molecules (e.g., CO₂) diffuse faster than larger, heavier ones (e.g., proteins).
  6. Medium/density — Diffusion is faster in gases than in liquids because gas molecules are farther apart and move more freely.

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  • 2082 Baishakh · 2+3+3 marks

How do different photosynthetic pigments contribute to capturing light energy? Explain how cyclic and non-cyclic photo-phosphorylation generate ATP and NADPH, and illustrate these with appropriate charts.

Answer

Photosynthetic pigments and capture of light energy (2 marks)

Photosynthetic pigments are coloured compounds in the thylakoid membranes. Each absorbs different wavelengths, so together they capture a wide part of the solar spectrum.

PigmentColourAbsorption peaks (nm)Role
Chlorophyll aBlue-greenabout 430 and 662Primary pigment; forms the reaction centre (P680 in PS II, P700 in PS I) and converts light energy into chemical energy
Chlorophyll bYellow-greenabout 453 and 642Accessory (antenna) pigment; passes energy to chlorophyll a
XanthophyllYellowabout 450Accessory pigment; widens the range absorbed and protects chlorophyll from photo-oxidation
CaroteneOrangeabout 450 to 480Accessory pigment; absorbs blue light and transfers energy to chlorophyll a
  • Accessory pigments absorb wavelengths that chlorophyll a absorbs poorly and pass the energy by resonance to chlorophyll a at the reaction centre.
  • Hence the action spectrum of photosynthesis follows the combined absorption spectrum of all pigments.

Non-cyclic photophosphorylation (3 marks)

It involves both PS II and PS I. Electron flow is one-way, water is the electron donor, and ATP, NADPH and O₂ are produced.

  1. Light excites P680 of PS II; the electron goes to the primary acceptor (pheophytin).
  2. Photolysis of water replaces the lost electron: 2H2O→light4H++4e−+O22H_2O \xrightarrow{\text{light}} 4H^+ + 4e^- + O_2
  3. The electron passes through plastoquinone, cytochrome b₆f and plastocyanin to PS I. Protons are pumped into the thylakoid lumen, forming a proton gradient.
  4. Light excites P700 of PS I; the electron goes to ferredoxin.
  5. NADP reductase transfers it to NADP⁺: NADP++2e−+H+→NADPHNADP^+ + 2e^- + H^+ \rightarrow NADPH
  6. Protons flow back through ATP synthase (CF₀-CF₁) and ADP + Pi becomes ATP (chemiosmosis).

Cyclic photophosphorylation (included in the 3 marks)

It involves PS I only. The electron from P700 goes to ferredoxin, then back through cytochrome b₆f and plastocyanin to P700. Protons pumped during this return drive ATP synthase, so only ATP is made (no NADPH, no O₂). It supplies extra ATP when the Calvin cycle needs more ATP than NADPH.

Charts (3 marks)

Non-cyclic (Z-scheme):

H2O --> PS II (P680) --> Pheophytin --> PQ --> Cyt b6f --> PC --> PS I (P700) --> Fd --> NADP reductase --> NADPH
 |  (photolysis: O2 + H+ + e-)  light       |                  light
 |                                          +--> H+ pumped into lumen --> ATP synthase --> ATP

Cyclic:

        light
          |
   PS I (P700) --> Fd --> Cyt b6f --> PC --+
        ^                |                 |
        +----------------|-----------------+
                         +--> H+ gradient --> ATP synthase --> ATP
FeatureNon-cyclicCyclic
PhotosystemsPS II and PS IPS I only
Electron pathOpen (one-way)Closed
Electron donorWaterNone (same electron returns)
ATPYesYes
NADPHYesNo
O₂ releasedYesNo
SiteGrana thylakoidsStroma lamellae

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  • 2082 Supplementary · 1 mark

Which of the following is the end product of glycolysis? (a) Ethanol (b) Lactic acid (c) Pyruvic acid (d) Acetyl CoA

Answer

(c) Pyruvic acid

Glycolysis breaks one molecule of glucose into two molecules of pyruvic acid (pyruvate), along with 2 ATP and 2 NADH. Ethanol and lactic acid are formed later by fermentation, and acetyl CoA is formed when pyruvate enters the mitochondrion.

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  • 2082 Supplementary · 1 mark

A farmer wants to grow larger seedless fruits. Which hormone should he/she use? (a) Auxin (b) Cytokinin (c) Gibberellin (d) Abscisic acid

Answer

(c) Gibberellin

Gibberellin induces parthenocarpy (development of fruit without fertilisation) and also promotes cell elongation, so it gives larger seedless fruits — as commonly done in grapes.

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  • 2082 Supplementary · 1+2+5 marks

Under certain environmental conditions, plants need to maximize their ability to capture light energy. Describe the process of photosynthesis and explain the role of chlorophyll pigments in this process. Using your understanding of the non-cyclic electron transport system, explain how ATP and NADPH₂ are generated to support the dark reaction.

Answer

Process of photosynthesis (1 mark)

Photosynthesis is the process by which green plants use light energy, chlorophyll, CO₂ and water to make carbohydrate and release oxygen:

6CO2+12H2O→chlorophylllightC6H12O6+6O2+6H2O6CO_2 + 12H_2O \xrightarrow[\text{chlorophyll}]{\text{light}} C_6H_{12}O_6 + 6O_2 + 6H_2O

It occurs in the chloroplast in two phases: the light reaction in the thylakoid membranes (light is trapped; ATP and NADPH₂ are made and water is split) and the dark reaction (Calvin cycle) in the stroma (CO₂ is reduced to sugar using ATP and NADPH₂).

Role of chlorophyll pigments (2 marks)

  • Chlorophyll a is the primary pigment and the reaction-centre pigment (P680 in PS II, P700 in PS I). Only it can donate the excited electron to the electron transport chain. Absorption peaks are near 430 nm (blue) and 662 nm (red).
  • Chlorophyll b and the carotenoids (carotene, xanthophyll) are accessory pigments. They absorb other wavelengths and pass the energy to chlorophyll a, so more of the spectrum is used. This helps plants capture more light when it is limiting, for example in shade.
  • The Mg²⁺ porphyrin head absorbs photons and the phytol tail anchors the molecule in the thylakoid membrane. An absorbed photon raises an electron to a higher energy level, which starts the light reaction.

Non-cyclic electron transport: ATP and NADPH₂ (5 marks)

It involves both PS II and PS I. Electron flow is one-way, with water as the electron donor, and it gives ATP, NADPH₂ and O₂.

  1. Excitation of PS II: Light of about 680 nm excites P680; the electron is taken up by the primary acceptor (pheophytin).
  2. Photolysis of water: The lost electrons are replaced from water, and O₂ is released: 2H2O→light, PS II4H++4e−+O22H_2O \xrightarrow{\text{light, PS II}} 4H^+ + 4e^- + O_2
  3. Transport to PS I: The electron passes along pheophytin → plastoquinone → cytochrome b₆f → plastocyanin. Protons are pumped into the lumen, creating a proton gradient.
  4. Excitation of PS I: Light of about 700 nm excites P700; its electron goes to ferredoxin.
  5. Formation of NADPH₂: NADP reductase passes the electrons to NADP⁺: NADP++2e−+H+→NADPH2NADP^+ + 2e^- + H^+ \rightarrow NADPH_2
  6. Formation of ATP: Protons flow back through ATP synthase (CF₀-CF₁), making ATP from ADP and Pi (photophosphorylation).

Support to the dark reaction: In the stroma, NADPH₂ reduces 3-PGA to glyceraldehyde-3-phosphate, and ATP phosphorylates 3-PGA and regenerates RuBP, so CO₂ is fixed into carbohydrate.

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  • 2081 Baishakh · 8 marks

The light-independent reactions of photosynthesis involve the conversion of carbon dioxide into glucose. Explain the steps involved in this process and how the products of the light dependent reactions (ATP and NADPH) are utilized. [5+3]

Answer

The light-independent reactions (also called the Calvin cycle or C₃ cycle) occur in the stroma of the chloroplast. They do not require light directly, but they depend on the ATP and NADPH produced by the light-dependent reactions. The cycle fixes CO₂ into carbohydrate through three phases: carboxylation, reduction, and regeneration.

1. Carboxylation (CO₂ fixation) CO₂ from the atmosphere enters the stroma and combines with a 5-carbon acceptor molecule, ribulose-1,5-bisphosphate (RuBP). This reaction is catalysed by the enzyme RuBisCO (ribulose-1,5-bisphosphate carboxylase/oxygenase), the most abundant enzyme on Earth. The unstable 6-carbon intermediate immediately splits into two molecules of 3-phosphoglycerate (3-PGA), each with 3 carbon atoms. This is the actual step where inorganic carbon becomes organic.

2. Reduction Each molecule of 3-PGA is phosphorylated by ATP to form 1,3-bisphosphoglycerate, and then reduced by NADPH to form glyceraldehyde-3-phosphate (G3P). For every CO₂ fixed, 2 ATP and 2 NADPH are consumed in this phase. G3P is a 3-carbon sugar phosphate.

3. Regeneration of RuBP Most of the G3P (5 out of every 6 molecules) is used to regenerate RuBP through a series of reactions so that the cycle can continue. This regeneration step requires 1 ATP per CO₂ fixed. The remaining 1 molecule of G3P is the net product.

Overall balance To produce one molecule of glucose (6C), the cycle must turn 6 times: 6 CO2+18 ATP+12 NADPH→C6H12O6+18 ADP+18 Pi+12 NADP+6\,\text{CO}_2 + 18\,\text{ATP} + 12\,\text{NADPH} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + 18\,\text{ADP} + 18\,\text{P}_i + 12\,\text{NADP}^+


Utilization of ATP and NADPH (3 marks)

The ATP and NADPH formed in the light-dependent reactions are used in the Calvin cycle as follows:

  • NADPH acts as the reducing agent. It donates hydrogen (electrons) to reduce 3-PGA into G3P during the reduction phase. Without NADPH, the fixed carbon cannot be reduced to sugar.
  • ATP provides chemical energy and phosphate groups. It is used in two places:
    • In the reduction phase, ATP phosphorylates 3-PGA to 1,3-bisphosphoglycerate, making it reactive enough to accept hydrogen from NADPH.
    • In the regeneration phase, ATP is used to convert G3P back into RuBP, allowing the cycle to continue.
  • Thus, ATP supplies energy while NADPH supplies reducing power. Together they drive the endergonic conversion of CO₂ into glucose. Once used, they form ADP, Pi, and NADP⁺, which return to the thylakoid membrane to be recharged by the light-dependent reactions.

In short, the Calvin cycle is the biosynthetic (dark) phase of photosynthesis, where ATP and NADPH from the light reactions drive the reduction of CO₂ into glucose.

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  • 2082 HISSAN · 1 mark

The major force for the ascent of sap is generated in the: (a) Roots (b) Stem (c) Leaves (d) Flowers

Answer

(c) Leaves

The transpiration pull generated in the leaves (due to water loss from stomata) is the major force that pulls water up the xylem.

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  • 2082 HISSAN · 4 marks

Identify the type of germination shown in the given figure and explain its mechanism.
Figure: Labeled diagram of seed germination showing cotyledons lifted above the soil surface.

Answer

Type of germination: Epigeal germination (the cotyledons are lifted above the soil surface).

Mechanism:

  1. Imbibition and swelling: The seed absorbs water, swells, and the seed coat ruptures. Enzymes become active and digest the stored food.

  2. Emergence of radicle: The radicle emerges first and grows downward into the soil, anchoring the seed and absorbing water and minerals.

  3. Rapid elongation of the hypocotyl: The hypocotyl (the region between the radicle and the cotyledons) elongates rapidly and grows upward. It is curved in the form of an arch (the hypocotyl arch), which protects the delicate plumule tip from injury as it pushes through the soil.

  4. Lifting of cotyledons: As the hypocotyl elongates, it carries the cotyledons above the soil surface. The cotyledons thus become aerial and turn green (in many plants) to carry out photosynthesis for a short period.

  5. Straightening and plumule growth: Once above the ground, the hypocotyl straightens, the seed coat is shed, and the plumule (embryonic shoot) develops into the shoot with the first foliage leaves.

  6. Food supply: The cotyledons supply stored food to the growing seedling until the young plant becomes green and self-sufficient, after which they shrivel and fall off.

Examples: Bean (Phaseolus), castor, papaya, onion, tomato.

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  • 2081 Baishakh · 1 mark

What causes cell elongation in the internodal regions of plants? (a) Indole acetic acid (b) Gibberellins (c) Abscisic acid (d) Cytokinins

Answer

(b) Gibberellins

Gibberellins stimulate cell elongation and division in the subapical meristem of the stem, which lengthens the internodes (for example, they reverse dwarfism and cause bolting in rosette plants). IAA mainly promotes elongation in coleoptiles and apical dominance, ABA inhibits growth, and cytokinins promote cell division.

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  • 2081 Baishakh · 1 mark

How does plasmolysis alter the osmotic process across the cell membrane? (a) Alters solute equilibrium, promoting water withdrawal from the cell (b) Blocks water exit, causing water accumulation within the cell (c) Disturbs solute balance, causing water to enter the cell (d) Increases solute diffusion across the membrane, leading to osmotic equilibrium

Answer

(a) Alters solute equilibrium, promoting water withdrawal from the cell

Plasmolysis occurs when a cell is placed in a hypertonic solution. The outside solute concentration is higher than inside, so water leaves the cell by exosmosis and the protoplasm shrinks away from the cell wall. Options (b) and (c) describe water entering or being trapped, which is the opposite of what happens.

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  • 2081 Baishakh · 3+5 marks

How does aerobic respiration differ from anaerobic respiration? Describe the cycle that involves breakdown of two carbon compound Acetyl CoA in aerobic respiration with necessary diagram. [3+5]

Answer

Differences between aerobic and anaerobic respiration

Aerobic respirationAnaerobic respiration
Needs free oxygenOccurs without oxygen
Glucose is oxidised completely to CO2CO_2 and H2OH_2OGlucose is oxidised incompletely to ethanol and CO2CO_2 (or lactic acid)
Occurs in cytoplasm (glycolysis) and mitochondria (Krebs cycle, ETS)Occurs only in cytoplasm
Net gain of about 30-32 ATP per glucoseNet gain of only 2 ATP per glucose
Final electron acceptor is oxygenFinal electron acceptor is an organic compound such as pyruvate

Krebs cycle (citric acid cycle)

It takes place in the mitochondrial matrix. Pyruvate from glycolysis is converted to two-carbon acetyl CoA (oxidative decarboxylation), which enters the cycle.

  1. Condensation: acetyl CoA (2C) combines with oxaloacetate (4C) to form citrate (6C).
  2. Citrate is isomerised to isocitrate (6C).
  3. Isocitrate is oxidised and decarboxylated to α\alpha-ketoglutarate (5C), releasing CO2CO_2 and forming NADH.
  4. α\alpha-Ketoglutarate is oxidised and decarboxylated to succinyl CoA (4C), releasing CO2CO_2 and forming NADH.
  5. Succinyl CoA is converted to succinate, forming one GTP (ATP) by substrate-level phosphorylation.
  6. Succinate is oxidised to fumarate, forming FADH2FADH_2.
  7. Fumarate is hydrated to malate.
  8. Malate is oxidised to oxaloacetate, forming NADH, and the cycle is ready to start again.

Per acetyl CoA: 3 NADH, 1 FADH2FADH_2, 1 ATP (GTP) and 2 CO2CO_2. For one glucose (2 turns): 6 NADH, 2 FADH2FADH_2, 2 ATP and 4 CO2CO_2.

Diagram: draw a circle with oxaloacetate at the top, acetyl CoA entering, and the intermediates citrate, isocitrate, α\alpha-ketoglutarate, succinyl CoA, succinate, fumarate and malate arranged clockwise, with arrows showing where CO2CO_2, NADH, FADH2FADH_2 and ATP are produced.

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  • 2080 Baishakh · 1 mark

Two cells A and B are contiguous as shown in the diagram. Cell A has osmotic pressure 12 atm, turgor pressure 9 atm, and diffusion pressure deficit 5 atm. Cell B has osmotic pressure 10 atm, turgor pressure 5 atm, and diffusion pressure deficit 7 atm. Which one of the following results will happen? (a) No movement of water (b) Equilibrium between the two cells A and B (c) Movement of water from cell A to B (d) Movement of water from cell B to A

Answer

(c) Movement of water from cell A to B

Water moves from the cell with the lower diffusion pressure deficit (DPD, higher water potential) to the cell with the higher DPD. Cell A has a DPD of 5 atm and cell B has 7 atm, so water moves from A to B until the two are in equilibrium.

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  • 2079 Jestha · 1 mark

When a plant cell is placed in a hypertonic solution, it gets plasmolysed as shown in the diagram. Which of the following occupies the space between the cell wall and the shrunken protoplast in such a plasmolysed cell? (a) Water (b) Hypertonic solution (c) Isotonic solution (d) Hypotonic solution

Answer

(b) Hypertonic solution

The cell wall is freely permeable, so the external hypertonic solution passes through it. When the protoplast shrinks away from the wall by exosmosis, the gap is filled by this external hypertonic solution.

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  • 2079 Jestha · 4 marks

What are plant growth hormones? Write the physiological functions of auxin. Mention its shortcoming of hyper use in crops.

Answer

Plant growth hormones: organic chemicals produced in small amounts in one part of the plant that are transported to other parts, where they promote, inhibit or modify growth and development. The main ones are auxins, gibberellins, cytokinins, abscisic acid and ethylene.

Physiological functions of auxin (e.g. IAA)

  • Promotes cell elongation, particularly in stems and coleoptiles.
  • Causes apical dominance (the apical bud suppresses lateral buds).
  • Stimulates root formation in cuttings and fruit setting and growth (parthenocarpy).
  • Causes phototropism and geotropism responses.
  • Delays leaf and fruit abscission, and promotes cambial activity.
  • Used as selective weed killer (2,4-D).

Shortcomings of excessive use in crops

  • High concentration inhibits growth instead of promoting it and can kill the plant (2,4-D is a herbicide).
  • Causes abnormal growth, deformed leaves and fruits, and may harm non-target broad-leaved plants.
  • Residues may contaminate soil and water and affect the environment.

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  • 2079 · Model question · 1 mark

Which one of the following parts is not responsible for transpiration in plants? (a) Root (b) Stem (c) Leaves (d) All aerial parts

Answer

(a) Root

Transpiration is the loss of water vapour from the aerial parts of a plant: through stomata of leaves, lenticels of stems and cuticle. Roots are underground and absorb water, so they do not transpire.

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  • 2079 · Model question · 1 mark

Some of the C4C_4 plants possess diagnostic features of Kranz anatomy. Select which of the following plant contains such features? (a) Wheat (b) Maize (c) Barley (d) Oats

Answer

(b) Maize

Maize is a C4C_4 plant. Its leaf shows Kranz anatomy, in which bundle sheath cells containing large chloroplasts form a wreath around the vascular bundles and mesophyll cells surround them. Wheat, barley and oats are C3C_3 plants.

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  • 2079 · Model question · 1+1+5+1 marks

Three German biochemists Embden, Meyerhof and Parnas (EMP) worked out a pathway which involved the formation of pyruvic acid from a glucose molecule. Identify this pathway. Name the site where this pathway takes place in a cell. Construct a step by step pathway developed by them. List the net gain of ATP in this pathway. [1+1+5+1]

Answer

Pathway: Glycolysis (EMP pathway), the conversion of one glucose molecule into two molecules of pyruvic acid.

Site: The cytoplasm (cytosol) of the cell. It does not need oxygen.

Steps of the pathway

Preparatory (energy investment) phase

  1. Glucose (6C) is phosphorylated by hexokinase, using 1 ATP, to give glucose-6-phosphate.
  2. Glucose-6-phosphate is isomerized by phosphohexose isomerase to fructose-6-phosphate.
  3. Phosphofructokinase adds a second phosphate using 1 ATP, forming fructose-1,6-bisphosphate.
  4. Aldolase splits it into two 3C sugars, dihydroxyacetone phosphate (DHAP) and glyceraldehyde-3-phosphate (G3P). DHAP is converted to G3P by triose phosphate isomerase, so two G3P are available.

Pay-off (energy generation) phase, occurring twice per glucose

  1. G3P is oxidized and phosphorylated by G3P dehydrogenase with inorganic phosphate, forming 1,3-bisphosphoglycerate and reducing NAD+NAD^+ to NADH (2 NADH in total).
  2. 1,3-bisphosphoglycerate gives its phosphate to ADP by phosphoglycerate kinase, forming 3-phosphoglycerate and ATP (substrate-level phosphorylation).
  3. 3-phosphoglycerate is converted to 2-phosphoglycerate by phosphoglyceromutase.
  4. Enolase removes water, forming phosphoenolpyruvate (PEP).
  5. PEP gives its phosphate to ADP by pyruvate kinase, forming pyruvic acid and ATP.

Net gain of ATP

ATP used =2= 2; ATP produced =2×2=4= 2\times 2 = 4.

Net ATP=4−2=2 ATP\text{Net ATP} = 4 - 2 = 2 \text{ ATP}

along with 2 NADH and 2 pyruvate per glucose.

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Questions from NEB Class 12 2083 Biology (Board exam, Baishakh 2083; typed text), NEB Class 12 2082 Biology (Board exam, Baishakh 2082; typed text), NEB Class 12 2082 Supplementary Biology Set A (Grade-increment/supplementary exam, Set A; typed text), NEB Class 12 2081 Biology (Board exam, Baishakh 2081; html text plus PDF), NEB Class 12 2080 Biology (Board exam, 2080; scanned PDF), NEB Class 12 2079 Biology (First new-course board exam, Jestha 2079; scanned PDF), HISSAN Class 12 2082 Biology Set D1 (HISSAN pre-board, Set D1), NEB Class 12 2081 Biology 2021 (second paper, nebexam text) (NEB board exam, 2081; a different paper from the dhanraj 2081 paper already on the site; typed text on nebexam), NEB Class 12 2080 Biology 2021 (second paper, nebexam text) (NEB board exam, 2080; a different paper from the scanned 2080 paper already on the site; typed text on nebexam), NEB Class 12 2079 Biology 2021 (second set, nebexam text) (NEB board exam, 2079 (first new-course exam); a different set from the scanned 2079 paper already on the site; typed text on nebexam) and NEB Class 12 Biology Model Question (2079 (2023)) (NEB model question paper (subject code 2021) issued for the 2080 exam; text also at https://www.nebexam.com/class-12/biology/question-bank/biology-model-question). Answers are written for this site; check them against your class notes.

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