Samacheer Kalvi Class 11 Bio Botany Solutions Chapter 11 Transport in Plants

Part - I

 

Question 1. In a fully turgid cell:
(a) DPD = 10 atm; OP = 5 atm; TP = 10 atm
(b) DPD = 0 atm; OP = 10 atm; TP = 10 atm
(c) DPD = 0 atm; OP = 5 atm; TP = 10 atm
(d) DPD = 20 atm; OP = 20 atm; TP = 10 atm
Answer: (b) DPD = 0 atm; OP = 10 atm; TP = 10 atm
In simple words: When a plant cell is fully turgid, it means it is completely full of water and cannot take in any more. At this point, the DPD (Diffusion Pressure Deficit) becomes zero, indicating no more water can enter. The osmotic potential (OP) and turgor potential (TP) are equal but opposite in sign to balance each other.

🎯 Exam Tip: Remember the relationship: DPD = OP - TP. For a fully turgid cell, water potential is maximum (zero), meaning DPD is zero.

 

Question 2. Which among the following is correct?
i) apoplast is fastest and operate in nonliving part
ii) Transmembrane route includes vacuole
iii) Symplast interconnect the nearby cell through plasma desmata
iv) Symplast and the transmembrane route is in the living part of the cell
a) i and ii
b) ii and iii
c) iii and iv
d) i, ii, iii, iv
Answer: (d) i, ii, iii, iv
In simple words: All the given statements about how water moves in plants are correct. The apoplast pathway is quick through non-living parts, the transmembrane route involves crossing cell membranes and the vacuole, and the symplast pathway connects living cells via plasmodesmata, which are small channels. Both symplast and transmembrane routes work through the living parts of the cell.

🎯 Exam Tip: Understand each water transport pathway (apoplast, symplast, transmembrane) by listing its key features, speed, and the parts of the cell it uses.

 

Question 3. What type of transpiration is possible in the xerophyte Opuntia?
(a) Stomatal
(b) Lenticular
(c) Cuticular
(d) All of the options
Answer: (b) Lenticular
In simple words: In Opuntia, a desert plant, most water loss happens through lenticels, which are small pores on the plant's surface. Stomatal transpiration is reduced to save water, and cuticular transpiration through the waxy layer is also minimal. Lenticular transpiration allows for some gas exchange without major water loss.

🎯 Exam Tip: Xerophytes are plants adapted to dry conditions; their features (like reduced stomata or thick cuticles) aim to minimize water loss, so lenticular transpiration becomes more prominent.

 

Question 4. Stomata of a plant open due to
a) Influx of K+
b) Efflux of K+
c) Influx of Cl-
d) Influx of OH¯
Answer: (a) Influx of K+
In simple words: Stomata open when potassium ions (K+) move into the guard cells. This movement causes water to follow, making the guard cells swell up. When they swell, they bend, and this bending opens the stomatal pore, allowing gases to pass through.

🎯 Exam Tip: The influx of K+ ions into guard cells is a key mechanism for stomatal opening, leading to increased turgor pressure.

 

Question 5. Munch hypothesis is based on:
(a) translocation of food due to TP gradient and imbibition force
(b) ranslocation of food due to TP
(c) translocation of food due to imbibition force
(d) None of the options
Answer: (b) ranslocation of food due to TP
In simple words: The Munch hypothesis, also known as the pressure flow hypothesis, explains how food moves through plants. It states that food (sugars) move from areas where there's high turgor pressure (like leaves) to areas with low turgor pressure (like roots or fruits). This pressure difference pushes the sugars along.

🎯 Exam Tip: The turgor pressure gradient is the driving force for the mass flow of food in plants, according to the Munch hypothesis.

 

Question 6. If the concentration of salt in the soil is too high and the plants may wilt even if the field is thoroughly irrigated. Explain
Answer: A high salt concentration in the soil makes the soil solution have a low osmotic potential. This means the water in the soil is held more tightly by the salt, making it harder for plant roots to absorb it. Even if there is plenty of water, the plants have to use more energy to pull the water in. Under very salty conditions, plants might not be able to absorb enough water and will wilt, even when the soil is wet. This effect is also known as osmotic or water deficit effect of salinity, where the plant experiences 'physiological drought'.
In simple words: Too much salt in the soil makes it hard for plants to take up water, even when there is enough water available. The salty water pulls water away from the plant roots, causing the plant to wilt.

🎯 Exam Tip: Explain that high salt concentration lowers the soil's water potential, making it osmotically challenging for plants to absorb water, leading to wilting.

 

Question 7. How phosphorylase enzyme open the stomata in starch sugar interconversion theory?
Answer: The starch-sugar interconversion theory helps explain how stomata open and close. Hanes (1940) discovered that the enzyme phosphorylase in guard cells plays a key role. During the day, this enzyme helps change starch into sugar, which increases the pH and osmotic pressure within the guard cells. This process is crucial for the plant's survival as it regulates water loss and gas exchange.

Day:
1. Starch \( \xrightarrow{\text{Phosphorylase}} \) Sugar
2. Photosynthesis occurs
3. pH increases
4. Water moves from subsidiary cells into guard cells
5. Guard cells become turgid (swollen)
6. Stomata open

Night:
1. Sugar \( \xrightarrow{\text{Phosphorylase}} \) Starch
2. No Photosynthesis
3. pH lowers
4. Water moves out of guard cells
5. Guard cells become flaccid (limp)
6. Stomata close
In simple words: During the day, the phosphorylase enzyme changes starch into sugar inside the guard cells. This makes the cells absorb water and swell, opening the stomata. At night, sugar changes back to starch, and the cells lose water and shrink, closing the stomata.

🎯 Exam Tip: Highlight the role of phosphorylase in converting starch to sugar (day) and sugar to starch (night), linking it directly to osmotic potential changes and stomatal movement.

 

Question 8. List out the non-photosynthetic parts of a plant that need a supply of sucrose?
Answer: The non-photosynthetic parts of a plant that need a supply of sucrose for energy and growth are:
1. Roots
2. Tubers
3. Developing fruits
4. Immature leaves.
These parts rely on sugars produced elsewhere, like mature leaves, to fuel their cellular activities. Sucrose is the primary form of sugar transported in plants because it is less reactive and more stable than glucose.
In simple words: Parts of the plant that do not make their own food through photosynthesis, such as roots, tubers, growing fruits, and young leaves, need sugar (sucrose) supplied from other parts of the plant.

🎯 Exam Tip: Remember that sucrose is the main transport sugar in plants, supplying energy to all non-photosynthetic "sink" organs for growth and storage.

 

Question 9. What are the parameters which control water potential?
Answer: Water potential is a critical concept for understanding water movement in plants, and it is controlled by several parameters. Slatyer and Taylor (1960) defined water potential as the potential energy of water in a system compared to pure water, assuming constant temperature and pressure. This potential drives water movement across cells and tissues.
1. Water potential is denoted by the Greek symbol \( \Psi \) (psi) and measured in Pascals (Pa). At standard temperature and pressure, the water potential of pure water is zero.
2. The addition of solute to pure water lowers its kinetic energy, making the water potential decrease from zero to a negative value.
3. Therefore, a solution always has a lower water potential than pure water.
4. In a group of cells with differing water potentials, a water potential gradient is established, causing water to move from a region of higher water potential to a region of lower water potential.
Water potential (\( \Psi \)) is determined by two main factors: Solute potential (\( \Psi_s \)) and Pressure potential (\( \Psi_p \)). The overall water potential is given by the equation: \( \Psi_w = \Psi_s + \Psi_p \).

a) Solute potential (\( \Psi_s \)) or Osmotic potential
• This denotes the effect of dissolved solutes on water potential.
• In pure water, adding solutes reduces the free energy of water, lowering its water potential from zero to negative.
• Thus, the value of solute potential is always negative. In a solution at standard atmospheric pressure, water potential is equal to solute potential (\( \Psi_w = \Psi_s \)).

b) Pressure Potential (\( \Psi_p \))
• Pressure potential is a mechanical force that counteracts the effect of solute potential. It's the pressure exerted by the cell wall on the protoplast.
• Increased pressure potential increases water potential, causing cells to become turgid.
• This positive hydrostatic pressure within the cell is called turgor. Conversely, water withdrawal from the cell decreases water potential, making the cell flaccid.
In simple words: Water potential is affected by how much dissolved stuff (solute potential) is in the water and how much pressure is pushing on the water (pressure potential). More dissolved stuff makes water potential lower (more negative), while more pressure makes it higher (more positive). These two factors together decide which way water will move.

🎯 Exam Tip: Clearly define water potential as the sum of solute and pressure potentials, explaining how each component affects the overall value and direction of water movement.

 

Question 10. An artificial cell made of selectively permeable membrane immersed in a beaker (in the figure) Read the values and answer the following questions?
a) Draw an arrow to indicate the direction of water movement
b) Is the solution outside the cell isotonic, hypotonic or hypertonic?
c) Is the cell isotonic, hypotonic, or hypertonic?
d) Will the cell become more flaccid, more turgid or stay in original size?
e) With reference to artifical cell state, the process is endomosis or exomosis? Give reasons
Answer:
a)

b) The outside solution is **hypotonic** (pure water).
c) The cell is **hypertonic** (the problem description implies the cell has solutes, making its water potential lower than pure water, although initial diagram shows \( \Psi_w = 0, \Psi_s = 0, \Psi_p = 0 \)). Assuming the cell has a lower water potential than pure water for the question to make sense, it is hypertonic relative to the pure water outside.
d) The cell will become **more turgid**.
e) The process is **endo-osmosis** because the solvent (water) moves inside the cell.
Reason: Endosmosis is defined as the osmotic entry of a solvent into a cell when it is placed in pure water or a hypotonic solution. In this setup, the solution in the beaker outside the cell is pure water, where \( \Psi_w = 0 \). Water will enter the artificial cell, moving from the hypotonic (pure water) environment to the hypertonic (cell) environment.
In simple words: When the artificial cell is put into pure water, water will move from the outside (pure water) into the cell. This makes the cell swell up (become turgid). This process is called endo-osmosis because water enters the cell. The pure water outside is hypotonic compared to the cell inside, which is hypertonic.

🎯 Exam Tip: Always remember that water moves from a region of higher water potential (hypotonic) to a region of lower water potential (hypertonic) across a semi-permeable membrane, causing endosmosis or exosmosis.

I - Choose The Correct Answers

 

Question 1. In plants, cell to cell transport is aided by:
(a) diffusion alone
(b) osmosis alone
(c) imbibition alone
(d) all the three above
Answer: (d) all the three above
In simple words: In plants, water and substances move from one cell to another through a combination of diffusion (spreading out), osmosis (water moving across a membrane), and imbibition (water being absorbed by solid particles). All these processes help transport things within the plant.

🎯 Exam Tip: Recognize that cell to cell transport is a complex process involving multiple physical phenomena, not just one. Diffusion, osmosis, and imbibition all contribute.

 

Question 2. The smell from a lightened incense stick or mosquito coil or open perfume bottle in a closed room is due to
a) Osmosis
b) Facilitated diffusion
c) Simple diffusion
d) imbibition
Answer: (c) Simple diffusion
In simple words: The smell spreads because the tiny particles of the incense, coil, or perfume move from where there are many of them (the source) to where there are fewer (the rest of the room). This natural spreading out of particles from a high concentration area to a low concentration area is called simple diffusion.

🎯 Exam Tip: Diffusion is the random movement of particles from a region of higher concentration to a region of lower concentration, often observed with gases and liquids.

 

Question 3. Which of the following statements are correct?
(i) Cell membranes allow water and non-polar molecules to permeate by simple diffusion.
(ii) Polar molecules like amino acids can also diffuse through the membrane.
(iii) Smaller molecules diffuse faster than larger molecules.
(iv) Larger molecules diffuse faster than smaller molecules.
(a) (i) and (iv) only
(b) (i) and (iii) only
(c) (i) and (ii) only
(d) (ii) and (iv) only
Answer: (b) (i) and (iii) only
In simple words: Cell membranes let water and non-polar molecules pass through easily by simple diffusion. Also, smaller molecules generally move across membranes much faster than larger ones. Polar molecules usually need special help (like protein channels) to cross the membrane.

🎯 Exam Tip: Remember that simple diffusion favors small, non-polar molecules. Polar and larger molecules often require facilitated diffusion or active transport to cross cell membranes.

 

Question 4. Solute potential is also known as
a) Water potential
b) Pressure potential
c) Osmotic potential
d) Maic potential
Answer: (c) Osmotic potential
In simple words: Solute potential is another name for osmotic potential. It's a measure of how much dissolved substances (solutes) reduce the water potential in a solution. The more solutes present, the lower (more negative) the solute potential.

🎯 Exam Tip: Understand that "solute potential" and "osmotic potential" are interchangeable terms, both referring to the effect of dissolved solutes on water potential.

 

Question 5. The swelling of dry seeds is due to a phenomenon called:
(a) osmosis
(b) transpiration
(c) imbibition
(d) none of the options
Answer: (c) imbibition
In simple words: When dry seeds get wet, they soak up water and swell a lot. This happens because the solid materials in the seed (like proteins) attract water molecules and absorb them. This process of absorbing water by solid particles, causing them to increase in volume, is called imbibition.

🎯 Exam Tip: Imbibition is a specific type of absorption where water is taken up by solid substances (colloids), causing them to swell, essential for seed germination.

 

Question 6. Cell A has an osmotic potential of -20 bars and a pressure potential of +6 bars. What will be its water potential?
a) -14 bars
b) +14 bars
c) -20 bars
d) +20 bars
Answer: (a) -14 bars
In simple words: To find the water potential, you add the osmotic potential and the pressure potential. So, -20 bars plus +6 bars equals -14 bars. This tells us the overall tendency of water to move into or out of the cell.

🎯 Exam Tip: Use the formula \( \Psi_w = \Psi_s + \Psi_p \) (Water Potential = Solute Potential + Pressure Potential) for calculations involving water potential.

 

Question 7. The OP and TP of two pairs of cells A – B, and X-Y are under
a) Cell A: OP=-I0atm, TP=4atm
b) Cell B : OP = I0atm, TP = 6atm
c) Cell X: Op =-l0atm, TP = 4atm
d) CellY: OP = -Katm, TP = 4atm
The net movement of water shall be from
a) A to B and X to Y
b) A to B and Y to X
c) B to A and X to Y
d) B to A and Y to X
Answer: (d) B to A and Y to X
In simple words: Water always moves from a region of higher water potential to a region of lower water potential. We need to calculate the water potential (WP = OP + TP) for each cell. Water will move from the cell with the higher water potential to the cell with the lower water potential. So, water moves from B to A, and from Y to X.

🎯 Exam Tip: Calculate the water potential for each cell using DPD = OP - TP (or Water Potential = OP - TP if considering osmotic pressure as positive) and determine the direction of water flow from higher to lower water potential.

 

Question 8. Water potential is influenced by which of the two factors among the given four
I) Concentration
II) Pressure
III) Temperature
IV) gravity
a) I & II
b) II & III
c) III & IV
d) I & IV
Answer: (a) I & II
In simple words: The water potential of a solution is mainly affected by two things: how much dissolved material (concentration) is in it, and how much pressure is pushing on it. Temperature and gravity usually have less significant direct influence on water potential in typical plant physiological contexts.

🎯 Exam Tip: Remember the primary components of water potential are solute potential (related to concentration) and pressure potential. While temperature and gravity have minor effects, they are usually less critical than concentration and pressure in calculations.

 

Question 9. __________ is equal to TP and is positive except plasmolysed cell and in xylem vessel where it is negative
a) Water potential
b) Pressure potential
c) Solute potential
d) Hydrostatic potential
Answer: (b) Pressure potential
In simple words: Pressure potential is usually positive and equals the turgor pressure (TP) inside a plant cell. However, in cells that have lost a lot of water (plasmolysed cells), or in the narrow tubes of the xylem, this pressure can become negative. This negative pressure is important for water to be pulled upwards in plants.

🎯 Exam Tip: Pressure potential reflects the physical pressure exerted on water, which is typically positive as turgor pressure, but can become negative in conditions like plasmolysis or within xylem under tension.

 

Question 10. Diffusion Pressure Deficit (DPD) was termed by Meyer in:
(a) 1928
(b) 1828
(c) 1936
(d) 1938
Answer: (d) 1938
In simple words: The term "Diffusion Pressure Deficit" (DPD), which helps describe how much water a cell wants to absorb, was first introduced by a scientist named Meyer in the year 1938. It's a key concept in understanding water movement in plants.

🎯 Exam Tip: Knowing the historical context and the scientists associated with key biological terms like DPD can earn extra marks in descriptive answers.

 

Question 11. Imbibants present in plants are generally
a) Hydrothermic
b) Hydrostatic
c) Hydrophilic
d) Hydrophobic
Answer: (c) Hydrophilic
In simple words: Imbibants are substances that absorb water and swell up, like dry seeds or wood. In plants, these substances are usually hydrophilic, meaning they "love water." They have parts that are attracted to water molecules, which allows them to take in a lot of water.

🎯 Exam Tip: Imbibition relies on the hydrophilic nature of plant components (e.g., cellulose, proteins) to attract and absorb water molecules, a crucial step in seed germination.

 

Question 12. Kramer (1949) recognised two distinct mechanisms, which independently operate in the absorption of water in plants are:
(a) osmosis and diffusion
(b) imbibition and diffusion
(c) diffusion and absorption
(d) active absorption and passive absorption
Answer: (d) active absorption and passive absorption
In simple words: Kramer identified that plants absorb water in two main ways. One is active absorption, where the plant uses energy to pull water in. The other is passive absorption, which happens without the plant using its own energy, often driven by transpiration pull from the leaves. Both processes work together to ensure the plant gets enough water.

🎯 Exam Tip: Differentiate between active (energy-dependent) and passive (energy-independent, driven by transpiration) water absorption to explain how plants manage their water uptake.

 

Question 13. Root pressure is totally absent in Gymnosperms because
a) Trachea absent
b) Tracheids absent
c) Trees are tall
d) Trees are comparatively short
Answer: (a) Trachea absent
In simple words: Gymnosperms, which are ancient plants like conifers, do not have trachea (also called vessels), which are wide tubes for water transport. Instead, they primarily use tracheids, which are narrower. The absence of these wider vessels means they don't develop significant root pressure, which is a pushing force from the roots that helps move water up. Their transport system primarily relies on other forces.

🎯 Exam Tip: Connect the absence of root pressure in gymnosperms to their specific xylem anatomy, particularly the lack of true vessels (trachea).

 

Question 14. When respiratory inhibitors like KCN, chloroform are applied:
(a) there is a decrease in the rate of respiration and an increase in the rate of absorption of water.
(b) there is an increase in the rate of respiration and a decrease in the rate of absorption of water.
(c) there is a decrease in the rate of respiration and also a decrease in the rate of absorption of water.
(d) there is an increase in the rate of respiration and also in the rate of absorption of water.
Answer: (c) there is a decrease in the rate of respiration and also a decrease in the rate of absorption of water.
In simple words: Respiratory inhibitors stop the process of respiration, which is how cells get energy. Since active water absorption (where plants use energy to take in water) needs energy from respiration, blocking respiration also slows down water absorption. So, both respiration and water absorption rates go down.

🎯 Exam Tip: Understand that active water absorption is an energy-dependent process, directly linked to respiration; thus, inhibiting respiration will reduce active water uptake.

 

Question 15. Find the DPD in a flaccid cell if its OP is 10
a) 20
b) 30
c) 10
d) 40
Answer: (c) 10
In simple words: In a flaccid cell (a cell that is limp and has lost water), there is no turgor pressure (TP = 0). Since DPD = OP - TP, if OP is 10 and TP is 0, then DPD is 10 - 0 = 10. This means the cell still wants to absorb water due to its osmotic potential.

🎯 Exam Tip: For a flaccid cell, the turgor pressure (TP) is zero, so DPD (Diffusion Pressure Deficit) is simply equal to the Osmotic Potential (OP).

 

Question 16. Pulsation theory was proposed by:
(a) Strasburger
(b) Godsey
(c) J.C. Bose
(d) C.V. Raman
Answer: (c) J.C. Bose
In simple words: The Pulsation theory, which suggests that rhythmic contractions of living cells in the inner bark help in the ascent of sap, was put forward by the famous Indian scientist J.C. Bose. He proposed that living cells might contribute to water movement, though this theory is not widely accepted today as the primary mechanism.

🎯 Exam Tip: Associate J.C. Bose with the Pulsation theory, even though other theories like the cohesion-tension theory are more widely accepted for water transport in tall plants.

 

Question 17. When a cell is kept in 0.5m solution of sucrose it's volume does not alter. If the same cell is placed in 0.5M solution of sodium chloride, the volume of the cell
a) Increase
b) Decrease
c) cell will be plasrnoysed
d) Will does not show any change
Answer: (d) Will does not show any change
In simple words: If a cell's volume doesn't change in a 0.5M sucrose solution, it means that solution is isotonic to the cell. When the same cell is moved to a 0.5M sodium chloride solution, its volume still won't change. This is because both solutions have the same osmotic effect on the cell, maintaining a balance of water movement.

🎯 Exam Tip: An isotonic solution has the same solute concentration as the cell, leading to no net water movement and thus no change in cell volume. This principle applies regardless of the specific solute, as long as the molarity is equivalent in its osmotic effect.

 

Question 18. Indicate the correct statements:
(i) Root pressure is absent in gymnosperms.
(ii) Root pressure is totally absent in angiosperms.
(iii) There is a relationship between the ascent of sap and root pressure.
(iv) There is no relationship between the ascent of sap and root pressure.
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (ii) and (iv)
(d) (i) and (iv)
Answer: (d) (i) and (iv)
In simple words: Root pressure, which is a pushing force from the roots, is not found in gymnosperms (like pine trees). Also, root pressure does not play a major role in pulling water up tall trees (ascent of sap). The main force for water ascent is transpiration pull. So, there is no direct relationship between root pressure and the main way water moves up tall plants.

🎯 Exam Tip: Remember that root pressure is generally low and absent in gymnosperms, and it contributes minimally to the ascent of sap, especially in tall plants where transpiration pull is the primary mechanism.

Match The Following & Find Out The Correct Order

 

Question 19. I) Water potential – A) Turgor pressure
II) Solute potential – B) Osmotic potential + Pressure potential
III) Matric potential – C) Osmotic potential
IV) Pressure potential – D) Imbibition pressure
I II III IV
(a) B C A D
(b) B C D A
(c) C B D A
(d) C D B A
Answer: (b) B C D A
In simple words: This question asks to match plant-related terms with their correct definitions or associated concepts. Water potential relates to both osmotic and pressure potential. Solute potential is the same as osmotic potential. Matric potential is linked to imbibition pressure, and pressure potential corresponds to turgor pressure.

🎯 Exam Tip: When matching concepts, carefully read each pair and consider their scientific definitions to ensure you link them correctly.

 

Question 20. I) Leaves – A) Antitransport
II) Seed – B) Transpiration
III) Roots – C) Negative osmotic potential
IV) Aspirin – D) Imbibition
V) Plasmolyced cell – E. Absorption
I II III IV V
(a) B D E A C
(b) D B A E C
(c) B D A E C
(d) A B D C E
(e) B E A D C
Answer: (a) B D E A C
In simple words: This question matches plant parts or substances with their functions or properties. Seeds typically show imbibition. Transpiration happens through leaves. Roots are involved in absorption. A plasmolyzed cell would have negative osmotic potential. Aspirin, as a chemical, can influence antitransport processes.

🎯 Exam Tip: For complex matching questions, first identify the most obvious pairs. This can help narrow down the options for the remaining terms.

 

Question 21. I) Transport of substance from a region of lower concentration to a region of higher concentration is with the expenditure of energy – A. Antiport
II) The movement of two types of molecules across the membrane in opposite direction – B. Symport The movement of a molecule across III) a membrane independent of other molecules – C. Active port
IV) The movement of two types of molecules across the membrane in the same direction – D. Uniport
I II III IV
(a) C A D B
(b) A C B D
(c) C D A B
(d) B C A D
Answer: (c) C D A B
In simple words: This question tests knowledge of different types of transport across cell membranes. Active transport moves substances against a concentration gradient using energy. Antiport moves two molecules in opposite directions. Uniport moves one molecule. Symport moves two molecules in the same direction.

🎯 Exam Tip: Understanding the prefixes "uni-", "sym-", and "anti-" can help you remember the definitions of uniport, symport, and antiport, respectively.

 

Question 22. I) Passive transport – A) Uphill transport
II) Active transport – B) Short distance transport
III) Cell to cell transport – C) Long-distance transport
IV) Ascent of sap – D) Downhill transport
I II III IV
(a) A D C B
(b) D A B C
(c) D A C B
(d) D C B A
Answer: (b) D A B C
In simple words: This question is about different transport mechanisms in plants. Passive transport moves substances down a gradient (downhill), while active transport moves them uphill, requiring energy. Cell-to-cell transport is short distance. Ascent of sap is a long-distance process.

🎯 Exam Tip: Differentiate between "uphill" (against gradient, active) and "downhill" (with gradient, passive) transport for clarity.

 

Question 23. The length and breadth of stomata is:
(a) about 10 – 30μ and 2 – 10μ respectively
(b) about 10 – 14μ and 3 – 10μ respectively
(c) about 10 – 40μ and 3 – 10μ respectively
(d) about 5 – 30μ and 5 – 10μ respectively
Answer: (c) about 10 – 40μ and 3 – 10μ respectively
In simple words: Stomata are tiny pores on plant surfaces. Their length typically ranges from 10 to 40 micrometers, and their width ranges from 3 to 10 micrometers.

🎯 Exam Tip: Remember specific numerical ranges for plant structures like stomata, as these details are often tested.

 

Question 24. A membrane that permits the solvent and not the solute to pass through it is termed is
(a) Permeable,
(b) impermeable
(c) semipermeable
(d) differentially permeable
Answer: (c) semipermeable
In simple words: A semipermeable membrane is like a sieve that only lets tiny solvent molecules (like water) pass through, but blocks larger solute molecules. It is a key feature in processes like osmosis.

🎯 Exam Tip: Distinguish between permeable (lets everything through), impermeable (lets nothing through), semipermeable (lets solvent but not solute through), and differentially permeable (selectively lets some solutes through too).

 

Question 25. Who did observe that stomata open in light and close in the night:
(a) Unger
(b) Sachs
(c) Boehm
(d) Von Mohl
Answer: (d) Von Mohl
In simple words: The scientist Von Mohl was the first to notice that plant stomata open when it's light and close when it's dark. This observation helps us understand how plants control water loss and gas exchange.

🎯 Exam Tip: Knowing the key scientists associated with important discoveries can help you answer historical questions in biology.

 

Question 26. The phosphorylase enzyme in guard cells supports the starch-sugar interconversion theory. The above reaction is:
(a) oxidation reaction
(b) hydrolyses reaction
(c) reduction reaction
(d) None of the options
Answer: (b) hydrolyses reaction
In simple words: The phosphorylase enzyme helps change starch into sugar and vice versa in guard cells. This process involves adding water to break down starch, which is a hydrolysis reaction.

🎯 Exam Tip: Remember that hydrolysis reactions involve the breaking down of a compound using water, while oxidation and reduction involve changes in electron states.

 

Question 27. If a cell kept in a solution of unknown concentration gets deplasmolysed the solution is
(a) hypotonic
(b) hypertonic
(c) isotonic
(d) detonic
Answer: (a) hypotonic
In simple words: Deplasmolysis means a shrunken plant cell regains water and becomes turgid again. This happens when the cell is placed in a hypotonic solution, which has a lower solute concentration than the cell, causing water to move into the cell.

🎯 Exam Tip: A hypotonic solution causes a cell to swell (deplasmolysis), an isotonic solution causes no change, and a hypertonic solution causes a cell to shrink (plasmolysis).

 

Question 28. A cell placed in a strong salt solution will shrink because
(a) the cytoplasm will decompose
(b) mineral salts will break the cell wall
(c) salt will leave the cell
(d) water will leave by exosmosis
Answer: (d) water will leave by exosmosis
In simple words: When a plant cell is put into a strong salt solution, the water inside the cell moves out into the salt solution. This movement is called exosmosis, and it makes the cell shrink.

🎯 Exam Tip: Remember that water always moves from an area of higher water concentration to an area of lower water concentration (or higher solute concentration) across a semipermeable membrane.

 

Question 29. Phenyl Mercuric Acetate (PMA), when applied as a foliar spray to plants:
(a) induces partial stomatal closure for two weeks.
(b) induces partial stomatal opening for two weeks.
(c) induces partial stomatal closure for four weeks.
(d) induces stomatal closure permanently
Answer: (a) induces partial stomatal closure for two weeks.
In simple words: Phenyl Mercuric Acetate (PMA) is a chemical that helps plants save water. When sprayed on leaves, it makes the stomata (tiny pores) close a bit, reducing water loss through transpiration for about two weeks.

🎯 Exam Tip: PMA acts as an anti-transpirant by reducing stomatal opening, which is useful in drought conditions to conserve water in plants.

 

Question 30. The osmotic pressure of cell sap is maximum in
(a) Hydrophytes
(b) Halophytes
(c) Xerophytes
(d) Mesophytes
Answer: (b) Halophytes
In simple words: Halophytes are plants that live in salty environments. They have a very high osmotic pressure in their cell sap to help them absorb water from the salty soil, which naturally makes water want to move out of the plant.

🎯 Exam Tip: Halophytes need high osmotic pressure to counter the low water potential of their salty surroundings and draw water into their cells.

 

Question 31. Say true or false and on that basis choose the right answer.
I) In facilitated diffusion, molecules move across the cell membrane with the help of special proteins, with the expenditure of energy
II) Porin is a larger transport protein, facilitates smaller molecules to pass through.
III) Aquaporins are recognized to transport urea, CO2, NH3 metalloid & ROS
IV) The carrier proteins structure does not get modified due to its association with the molecules
I II III IV
(a) False False True True
(b) True False True False
(c) False False True False
(d) False True True False
Answer: (d) False True True False
In simple words: Let's check each statement: facilitated diffusion does not use energy, so I is false. Porins are large proteins that help small molecules pass, so II is true. Aquaporins transport many small substances like urea and CO2, so III is true. Carrier proteins *do* change their shape when binding to molecules, so IV is false.

🎯 Exam Tip: Understand the key differences between passive transport (diffusion, facilitated diffusion - no energy) and active transport (uses energy) when evaluating such statements.

 

Question 32. I) Hypertonic is a strong solution (low solvent/high solute/low \( \Psi \) )
II) Hypotonic is a weak solution (high solvent/low or zero solutes/ high \( \Psi \) )
III) Hypertonic is the weak solution (high solvent/low or zero solutes/high \( \Psi \) )
IV) Hypotonic is a strong solution (low solvent / high solute/low \( \Psi \) )
I II III IV
(a) False True False True
(b) True True False False
(c) True False True False
(d) False False True True
Answer: (b) True True False False
In simple words: This question asks you to identify true or false statements about hypertonic and hypotonic solutions. A hypertonic solution is indeed strong with low water potential. A hypotonic solution is weak with high water potential. Statements III and IV contradict these definitions, making them false.

🎯 Exam Tip: Clearly define hypertonic (more solute, less water, low water potential) and hypotonic (less solute, more water, high water potential) solutions to avoid confusion.

 

Question 33. From sieve elements sucrose is translocated into sink organs such as root, tubers etc and this process is termed as:
(a) Xylem unloading
(b) Xylem uploading
(c) Phloem unloading
(d) Phloem uploading
Answer: (c) Phloem unloading
In simple words: Sucrose, a type of sugar, moves from the sieve elements of the phloem into storage parts like roots and tubers. This process where sugar is taken out of the phloem into the sink organs is called phloem unloading.

🎯 Exam Tip: Remember that phloem is responsible for transporting sugars, and "unloading" refers to the movement of these sugars into storage or utilization areas (sinks).

 

Question 34. The value of pure water is zero in which three aspects of the given options
I) Osmotic pressure
II) Osmotic potential
III) Water potential
IV) Pressure potential
(a) I, II, & III
(b) II, III & IV
(c) I, III & IV
(d) I, II & IV
Answer: (a) I, II & III
In simple words: For pure water, its osmotic pressure is zero, its osmotic potential is zero, and its water potential is also zero. This is because there are no solutes to create osmotic effects, and pure water is used as the reference point for water potential.

🎯 Exam Tip: Pure water is the standard reference for water potential, with a value of zero. Any dissolved solutes will lower this potential.

 

Question 35. Gases such as oxygen and carbon dioxide cross the cell membrane by
(a) Passive diffusion through the lipid bilayer
(b) Primary active transport
(c) Specific gas transport proteins
(d) Secondary active transport
Answer: (a) Passive diffusion through the lipid bilayer
In simple words: Oxygen and carbon dioxide are small, non-polar molecules. They can easily move directly through the cell's lipid bilayer from an area of high concentration to an area of low concentration without needing energy or special proteins. This process is called passive diffusion.

🎯 Exam Tip: Small, non-polar molecules like O2 and CO2 typically cross cell membranes via simple diffusion, following their concentration gradients.

 

Question 36. Hydathodes are generally present in plants that grow in:
(a) dry places
(b) moist and shady places
(c) sunny places
(d) deserts
Answer: (b) moist and shady places
In simple words: Hydathodes are special pores that release excess water from plants, a process called guttation. They are usually found in plants that live in moist and shady areas where transpiration (water loss as vapor) is low, but water absorption is high, leading to a build-up of water pressure.

🎯 Exam Tip: Hydathodes help release water as liquid droplets (guttation) when transpiration is insufficient, which is common in high-humidity environments.

 

Question 37. Why sugars are transported in the form of su-crose in phloem?
(a) It is inactive and highly soluble
(b) It is active
(c) It yields high ATP
(d) It is lighter in weight.
Answer: (a) It is inactive and highly soluble
In simple words: Plants transport sugars as sucrose because it is very soluble in water, making it easy to move through the phloem. Also, sucrose is non-reducing and less reactive, so it doesn't get used up or changed chemically during transport to other parts of the plant.

🎯 Exam Tip: Sucrose's stability and solubility are key advantages for long-distance sugar transport in plants, ensuring efficient delivery to growth and storage areas.

 

Question 38. Unloading of pholem at sink includes
(a) Passive transport
(b) diffusio
(c) Osmosis
(d) Active transport
Answer: (d) Active transport
In simple words: Phloem unloading, which is the movement of sugars out of the phloem into storage areas (sinks), usually requires energy. This process is often driven by active transport mechanisms to move sugars against their concentration gradient into the sink cells.

🎯 Exam Tip: Phloem loading and unloading are generally active processes that require metabolic energy (ATP) to move sugars efficiently against concentration gradients.

 

Question 39. The liquid coming out of the hydathode of grasses is:
(a) pure water
(b) not pure water
(c) a solution containing a number of dissolved substances
(d) saltwater
Answer: (c) a solution containing a number of dissolved substances
In simple words: The liquid released by hydathodes in guttation is not pure water. It is actually a dilute solution that contains various dissolved substances, like mineral salts and organic compounds, from the plant's xylem sap.

🎯 Exam Tip: Guttation fluid differs from transpiration, as it contains dissolved solutes, indicating it's expelled directly from the xylem without evaporation.

 

Question 40. In a flaccid cell
(a) DPD = OP
(b) DPD = TP
(c) TP = OP
(d) OP = O
Answer: (a) DPD = OP
In simple words: A flaccid cell is one that has lost water and is not firm. In such a cell, the turgor pressure (TP) is zero. Since DPD = OP - TP, when TP is zero, the Diffusion Pressure Deficit (DPD) becomes equal to the Osmotic Pressure (OP). This means the cell's demand for water is equal to its osmotic potential.

🎯 Exam Tip: Remember the relationship DPD = OP - TP. For a flaccid cell, TP is 0, making DPD = OP. For a fully turgid cell, DPD is 0, meaning OP = TP.

 

Question 41. The pathway of water movement involving living part of a cell is
(a) Apoplast pathway
(b) symplast pathway
(c) Transmembrane pathway
(d) Lateral conduction
Answer: (b) symplast pathway
In simple words: The symplast pathway is how water moves from one plant cell to another by going through the living parts of the cells. This means water travels through the cytoplasm of cells, connected by tiny channels called plasmodesmata, rather than through cell walls.

🎯 Exam Tip: Distinguish between symplast (through living cytoplasm via plasmodesmata) and apoplast (through non-living cell walls and intercellular spaces) pathways for water movement in plants.

 

Question 42. The ascent of sap is
(a) Upward movement of water in plants
(b) downward movement of water in plants
(c) upward and downward movement of water plants
(d) None of the options
Answer: (a) Upward movement of water in plants
In simple words: The ascent of sap is the process where water, along with dissolved minerals, moves upwards from the roots to the leaves and other aerial parts of a plant. This upward flow is crucial for photosynthesis and maintaining plant structure.

🎯 Exam Tip: Ascent of sap specifically refers to the upward transport of water and minerals through the xylem against gravity.

 

Question 43. High tensile strength of water is due to
(a) Adhesion only
(b) cohesion only
(c) Both (a) and (b)
(d) None of the options
Answer: (c) Both (a) and (b)
In simple words: Water has high tensile strength, meaning it can resist being pulled apart, because of both adhesion and cohesion. Cohesion is when water molecules stick to each other, forming a continuous column. Adhesion is when water molecules stick to the walls of the xylem vessels, helping to counteract gravity.

🎯 Exam Tip: Remember that both cohesive (water-water attraction) and adhesive (water-surface attraction) forces are essential for maintaining the continuous water column in the xylem during the ascent of sap.

 

Question 44. Maximum transpiration occur in
(a) Mesophytes
(b) Xerophytes
(c) Hydrophytes
(d) Epiphytes
Answer: (a) Mesophytes
In simple words: Mesophytes are plants that grow in moderate conditions, neither extremely dry nor extremely wet. They typically have a normal amount of stomata and well-developed root systems, allowing them to transpire (lose water vapor) at a higher rate compared to plants in very dry or very wet environments.

🎯 Exam Tip: Mesophytes are adapted to average moisture, unlike xerophytes (dry, less transpiration) or hydrophytes (water, very little transpiration).

 

Question 45. Supply ends in transport of solutes are
(a) green leaves
(b) root and stem
(c) xylem and phloem
(d) Hormones and enzymes
Answer: (c) xylem and phloem
In simple words: The main transport systems for solutes in plants are the xylem and phloem. Xylem carries water and minerals from the roots upwards. Phloem carries sugars (food) from the leaves to other parts of the plant. Together, they form the plant's vascular system.

🎯 Exam Tip: Xylem and phloem are the primary vascular tissues responsible for long-distance transport of water, minerals, and organic solutes in plants.

 

Question 46. For guttation in plants, the process responsible is
(a) Root pressure
(b) Atmospheric pressure
(c) Imbibition
(d) None of the options
Answer: (a) Root pressure
In simple words: Guttation is the process where plants release water droplets from their leaves, usually at night or in high humidity. This happens because water absorbed by the roots builds up positive pressure, called root pressure, pushing water out of the hydathodes.

🎯 Exam Tip: Guttation is driven by positive root pressure, which forces liquid water out, unlike transpiration, which is driven by negative pressure (pull) and releases water vapor.

 

Question 47. Which of the following theories for Ascent of sap was proposed by famous Indian scientist. J.C. Bose.
(a) Transpiration pull theory
(b) Pulsation theory
(c) Root pressure theory
(d) Atmospheric pressure theory
Answer: (b) Pulsation theory
In simple words: The Indian scientist J.C. Bose proposed the Pulsation theory to explain how sap rises in plants. This theory suggests that living cells in the inner cortex of the root actively pump water upwards in rhythmic pulsations.

🎯 Exam Tip: Recognize key historical theories in plant physiology and the scientists associated with them, like J.C. Bose's Pulsation Theory.

 

Question 48. Which of the following plant material is an efficient water imbibant?
(a) Lignin
(b) Pectin
(c) Cellulose
(d) Agar
Answer: (d) Agar
In simple words: Among the options, agar is a highly effective imbibant, meaning it can absorb large amounts of water and swell significantly. While pectin and cellulose also imbibe water, agar demonstrates a stronger capacity for this process due to its unique molecular structure.

🎯 Exam Tip: Imbibition is the absorption of water by solid particles. Materials with strong hydrophilic (water-attracting) properties, like agar, are excellent imbibants.

 

Question 49. Which of the following helps in the Ascent of sap?
(a) Root pressure
(b) Transpiration
(c) Capillarity
(d) All of the options
Answer: (d) All of the options
In simple words: The upward movement of water (ascent of sap) in plants is a complex process. It is helped by root pressure pushing water from below, transpiration pulling water from the leaves, and capillarity (water moving up narrow tubes) in the xylem vessels. All these factors work together.

🎯 Exam Tip: The Cohesion-Tension-Transpiration Pull model is the most widely accepted explanation, integrating transpiration, cohesion, and adhesion, while root pressure plays a smaller role.

 

Question 50. In a girdled plant which of the following dies first?
(a) Shoot
(b) root
(c) Both die simultaneously
(d) None – the plant survives
Answer: (b) root
In simple words: Girdling means removing a ring of bark, which includes the phloem, from around a tree trunk. This stops food (sugars) from reaching the roots, so the roots starve and die first. After the roots die, the shoot will eventually die too.

🎯 Exam Tip: Girdling specifically disrupts phloem transport, highlighting the importance of phloem in supplying roots with food produced by the leaves.

 

Question 51. Assertion:-A Imbibition is also diffusion
Reason -R The movement of water in the above process is along a concentration gradient.
(a) Both A and R are true and R is correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A true but R false
(d) Both A and R are false
Answer: (a) Both A and R are true and R is correct explanation of A
In simple words: Imbibition is a type of diffusion where water moves along a concentration gradient into solid particles. The movement of water from a higher to lower concentration explains why imbibition happens.

🎯 Exam Tip: For assertion-reason questions, first determine if each statement is true, then if the reason correctly explains the assertion. Imbibition is a special case of diffusion.

 

Question 52. Assertion: – A In rooted plant, the transport of water and minerals in xylem is essentially multi-directional
Reason – R Organic compound and nuitrient undergoes undirectional transport only
(a) Both A and R are true and R is correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A true but R false
(d) Both A and R are false
Answer: (d) Both A and R are false
In simple words: The transport of water and minerals in xylem is primarily unidirectional (upwards), not multi-directional. Also, organic compounds (food) transported in phloem can move in multiple directions, depending on the plant's needs, not just unidirectionally.

🎯 Exam Tip: Xylem transport is generally unidirectional (roots to shoot), while phloem transport is bidirectional (source to sink) for organic solutes.

 

Question 53. Assertion: – A The adsorption of water by solid particles of an adsorbant with out forming a solution is known as imbibition
Reason: – R The liquid which is imbided is known as imbibate
(a) Both A and R are true and R is correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A true but R false
(d) Both A and R are false
Answer: (b) Both A and R are true but R is not the correct explanation of A
In simple words: Imbibition is when solid particles absorb water without making a true solution. The liquid absorbed during imbibition is indeed called imbibate. Both statements are true. However, the reason only defines "imbibate" and doesn't explain *why* solids adsorb water without forming a solution, so it's not the correct explanation.

🎯 Exam Tip: Understand that imbibition involves physical adsorption, not chemical dissolution, and that the imbibate is the liquid that gets absorbed by the imbibant (the solid).

 

Question 54. Assertion: – A In phloem loading, food is transported to the sink
Reason - R Food is transported from source to sink
(a) Both A and R are true and R is correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A true but R false
(d) Both A and R are false
Answer: (d) Both Assertion 'A' and Reason 'R' are false
In simple words: In phloem loading, food is transported *into* the phloem from the source (like leaves), not directly to the sink. Also, food is transported *from* a source (where it's made) *to* a sink (where it's used or stored), so the reason statement is correct in its general principle, but the given answer states both are false. Let's re-evaluate. Phloem loading moves food *into* phloem *from* source. So assertion 'A' is false. Reason 'R' that food is transported from source to sink is generally true. Therefore, A is false and R is true. This would mean option (c) is the correct one, however, following the provided answer key, both are indicated as false.

🎯 Exam Tip: Phloem loading involves moving sugars from source cells into sieve tubes, and transport then occurs from source to sink. This distinction is crucial for understanding nutrient distribution in plants.

 

Question 55. Assertion – A: Xylem a principal water conducting '
Reason -R: It has been recognised by girdling or ringing experiments
(a) Both A and R are True R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A true but R false
(d) Both A and R are false
Answer: (a) Both A and R are True R is the correct explanation of A
In simple words: Xylem is indeed the main tissue that conducts water in plants. Girdling experiments, where the phloem is removed but the xylem is left intact, show that water transport continues, proving xylem's role in water conduction. Therefore, both statements are true, and the reason correctly explains the assertion.

🎯 Exam Tip: Girdling experiments are classic demonstrations in botany to differentiate the functions of xylem and phloem by selectively removing one tissue.

 

Question 56. Assertion: – A In phloem, sugar are translocated in non reducing form
Reason – R Non reducing sugars are most reactive sugars
(a) Both A and R are true and R is correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A true but R false
(d) Both A and R are false
Answer: (c) A true but R false
In simple words: In phloem, sugars are transported as sucrose, which is a non-reducing sugar, making it stable for transport. So, the assertion is true. However, non-reducing sugars are actually *less* reactive than reducing sugars, not more. Therefore, the reason is false.

🎯 Exam Tip: Sucrose is a disaccharide formed from glucose and fructose, and its non-reducing nature prevents it from reacting with other molecules during transport, ensuring efficient delivery.

 

Question 57. Assertion: A In ringing experiment a narrow continuous band of tissues external to the phloem is removed
Reason: R Ringing experiment proves that phloem is involved in water transport '
(a) Both A and R are true and R is correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A true but R false
(d) Both A and R are false
Answer: (d) Both A and R are false
In simple words: In a ringing experiment, the tissues *external* to the xylem are removed, which includes the phloem. So, the assertion is false. The ringing experiment actually shows that phloem is involved in *food* (sugar) transport, not water transport, so the reason is also false.

🎯 Exam Tip: The ringing experiment primarily demonstrates the role of phloem in transporting organic nutrients downwards, as the accumulation of sugars above the ring inhibits root growth.

II. Two Mark Questions

 

Question 1. What is the need for the transport of materials in plants?
Answer: Plants need to transport materials to ensure all their parts get what they need to survive and grow. Water absorbed by the roots must travel to the leaves for photosynthesis. Similarly, the food produced in the leaves needs to be transported to all other parts of the plant, including the roots, for energy and growth. Without this transport, different parts of the plant would not receive the necessary resources.
In simple words: Plants transport water, minerals, and food so every part can get what it needs to live and grow.

🎯 Exam Tip: Focus on the two main substances transported—water/minerals (xylem) and food/sugars (phloem)—and their destination/purpose in different plant parts.

 

Question 2. What is osmosis
Answer: Osmosis is a unique type of diffusion involving the movement of water molecules across a selectively permeable membrane. Water moves from a region where it is in higher concentration (or has higher water potential) to a region where it is in lower concentration (or has lower water potential). This movement happens without the plant using its own energy.
In simple words: Osmosis is when water moves from a place with more water to a place with less water, through a special skin that only lets water pass.

🎯 Exam Tip: Key terms for osmosis are "selectively permeable membrane" and "movement of water molecules down their water potential gradient."

 

Question 3. Define the term diffusion.
Answer: Diffusion is the net movement of molecules from an area where they are highly concentrated to an area of lower concentration. This movement continues along a concentration gradient until the molecules are evenly spread out and a balance is reached. Diffusion is a key process for gases and small particles moving in and out of cells without using energy.
In simple words: Molecules move from a crowded area to a less crowded area until they are spread equally.

🎯 Exam Tip: Remember that diffusion does not require any energy from the cell, as it relies on the natural random motion of molecules.

 

Question 4. The touch me plant closes its leaves at the touch - Explain.
Answer:
• In the 'Touch me not' plant, the physical touch acts as a signal or trigger, causing its leaves to close.
• When the plant is touched, specific chemicals are released in the stem. These chemicals cause water to move out of the cells.
• This movement of water leads to the loss of stiffness (turgor pressure) in the cells, which makes the leaves droop and fold.
• After some time, when the chemicals are no longer present, water moves back into the cells, and the leaves return to their normal open state. This quick movement is a defense mechanism to protect the plant from harm, like grazing animals.
In simple words: When touched, the plant releases chemicals that make water leave its cells, causing the leaves to fold up. Later, the leaves open again.

🎯 Exam Tip: This response to touch is called thigmonasty, a rapid reversible movement triggered by mechanical stimulation.

 

Question 5. What is meant by Porin?
Answer: Porins are large transport proteins found in the outer membrane of certain cell organelles like plastids and mitochondria, as well as in bacteria. These proteins create channels that make it easier for smaller molecules to pass through the membrane. These proteins act like gatekeepers, allowing specific small substances to enter or exit.
In simple words: Porins are big proteins that make tunnels in cell membranes, helping small molecules get through.

🎯 Exam Tip: Think of porins as non-selective channels for small molecules, unlike highly specific carrier proteins.

 

Question 6. Define water potential
Answer:
• Water potential is the potential energy of water within a system when compared to pure water. This comparison is made when both temperature and pressure are kept constant.
• It is a way to measure how freely water molecules can move in a given environment or system.
• Pure water has a water potential of zero, which is its highest possible value. Water always moves from an area of higher water potential to an area of lower water potential, which is crucial for plant water uptake.
In simple words: Water potential is a measure of how much energy water has to move. Pure water has the highest water potential.

🎯 Exam Tip: Remember that water always moves down a water potential gradient, from higher to lower potential.

 

Question 7. Define Diffusion Pressure Deficit.
Answer:
• The term Diffusion Pressure Deficit (DPD) was first introduced by Meyer in 1938.
• DPD represents the difference between the diffusion pressure of a solution and the diffusion pressure of its pure solvent. This difference is measured at a particular temperature and atmospheric pressure. Essentially, DPD indicates the water-absorbing power of a cell or solution.
In simple words: DPD is the pressure difference that makes water move into a solution, first named by Meyer.

🎯 Exam Tip: DPD is a critical concept for understanding water uptake by plant cells; a higher DPD means a greater tendency to absorb water.

 

Question 8. Differentiate between short distance and Long Distance Transport.
Answer:

Short-distance transport handles local movement, while long-distance transport connects distant parts of the plant efficiently.
In simple words: Short distance transport moves things cell by cell, while long distance transport moves things through tubes like xylem and phloem over long paths.

🎯 Exam Tip: Remember that short-distance transport is often local and less organized, while long-distance transport relies on specialized vascular tissues.

 

Question 9. Differentiate between Passive & Active Transport
Answer:

Passive transport is like floating downstream, while active transport is like swimming upstream, needing effort.
In simple words: Passive transport moves substances without using energy, usually from high to low concentration. Active transport uses energy to move substances, often from low to high concentration.

🎯 Exam Tip: A key difference is the use of energy: passive transport relies on concentration gradients, while active transport requires ATP.

 

Question 10. Give two examples of the phenomenon of Imbibition.
Answer: Here are two examples of imbibition:
1. Dry seeds swell up when they soak in water. This is why seeds get bigger before they sprout.
2. Wooden windows, tables, and doors can swell during the rainy season because they absorb moisture from the air, making them harder to open or close. Imbibition is a type of absorption where a solid material takes in liquid, causing it to increase in volume.
In simple words: Dry seeds getting bigger in water and wooden things swelling in humid weather are examples of imbibition.

🎯 Exam Tip: Remember that imbibition is specific to solids absorbing liquids, leading to an increase in volume and pressure.

 

Question 11. Explain carbonic Acid Exchange theory.
Answer:
• The soil solution acts as a medium for the exchange of ions.
• Carbon dioxide (\( \text{CO}_2 \)), which is released by plant roots, combines with water to form carbonic acid (\( \text{H}_2\text{CO}_3 \)).
• This carbonic acid then breaks down (dissociates) into hydrogen ions (\( \text{H}^+ \)) and bicarbonate ions (\( \text{HCO}_3^- \)) in the soil solution.
• The \( \text{H}^+ \) ions then exchange with other positive ions (cations) that are attached to clay particles and tiny soil particles called micelles. This process releases these cations, making them available for the plant to absorb. This exchange helps plants absorb essential minerals from the soil.
In simple words: Roots release CO2, which makes carbonic acid in the soil. This acid then helps plants swap hydrogen for other good nutrients.

🎯 Exam Tip: Remember that this theory highlights the role of root respiration in making nutrients available for uptake.

 

Question 12. Give Answer in a sentence or two Distinguish between (i) Exomosis & Endomosis (ii) Apoplast & Symplast (iii) Cohesion & Adhesion (v) Influx & Efflux
Answer:

These different transport mechanisms allow plants to efficiently move water, nutrients, and signals throughout their structure.
In simple words: Exosmosis is water leaving a cell, endosmosis is water entering. Apoplast is water moving through cell walls, symplast is through cell cytoplasm. Cohesion is water sticking to water, adhesion is water sticking to other things. Influx is ions entering, efflux is ions leaving.

🎯 Exam Tip: For differentiation questions, focus on one key distinguishing feature for each pair and provide an example if possible.

 

Question 13. What is meant by osmotic pressure?
Answer: Osmotic pressure is the force or pressure that develops in a solution when it is separated from its pure solvent (like pure water) by a semipermeable membrane. This pressure is created due to the presence of dissolved substances (solutes) in the solution. This pressure helps water move into plant cells, keeping them firm.
In simple words: Osmotic pressure is the push that happens in a liquid when it's trying to get water through a special filter because it has stuff dissolved in it.

🎯 Exam Tip: Remember that osmotic pressure is directly proportional to the concentration of solutes in the solution.

 

Question 14. Define Root Pressure.
Answer:
• Stephen Hales was the scientist who first used the term "root pressure."
• Stoking (1956) provided a clear definition of root pressure.
• Root pressure is the positive pressure that develops in the water-carrying vessels (tracheary elements) of the xylem in plant roots. This pressure is a result of the metabolic activities happening within the root cells. Root pressure is strongest when transpiration rates are low, such as during the night.
In simple words: Root pressure is a push created in plant roots that helps move water up, especially when the plant isn't losing much water from its leaves.

🎯 Exam Tip: Root pressure helps push water up the stem in small plants and contributes to guttation (water droplets on leaf edges).

 

Question 15. Define the term osmosis.
Answer: Osmosis comes from the Latin word "Osmos," meaning impulse or urge. It is a special type of diffusion. Osmosis describes the movement of water or other solvent molecules through a selectively permeable membrane. This movement occurs from an area of higher water concentration (high water potential) to an area of lower water concentration (low water potential). This process is vital for cells to maintain their water balance.
In simple words: Osmosis is when water moves through a special filter from where there's lots of it to where there's less.

🎯 Exam Tip: Always emphasize the role of a selectively permeable membrane in osmosis, distinguishing it from simple diffusion.

 

Question 16. Why plants transport sugars as sucrose and not as starch or Monosaccharide (Glucose & Fructose)
Answer:

Sucrose is an ideal transport sugar because its non-reducing nature prevents unwanted reactions during transport.
In simple words: Plants move sugar as sucrose because it dissolves well, doesn't react easily, and is good for storing energy during its journey. Starch is too big, and glucose/fructose are too reactive.

🎯 Exam Tip: Highlight that sucrose's non-reducing nature makes it chemically stable, reducing the risk of unwanted reactions during transport.

 

Question 17. What are the three types of plasmolysis?
Answer: The three main types of plasmolysis observed in plants are:
1. **Incipient plasmolysis:** This is the earliest stage where the cell membrane just begins to pull away from the cell wall, usually at the corners.
2. **Evident plasmolysis:** In this stage, the cell membrane has clearly shrunk and significantly separated from the cell wall.
3. **Final plasmolysis:** The cell has lost almost all its water, and the protoplast (living part of the cell) is completely detached and shrunken in the center of the cell. Observing these stages helps us understand how plant cells react to different water environments.
In simple words: Plasmolysis has three steps: the cell membrane just starts to pull away, then it clearly shrinks, and finally, it shrinks completely, leaving the cell wall empty.

🎯 Exam Tip: Understanding the progressive stages of plasmolysis is crucial for recognizing cell responses to hypertonic solutions.

 

Question 18. Identify the diagram and Neatly label the parts
Answer: The diagram shows the structure of a Hydathode.
**A - Guard cell**
**B - Epithem**
**C - Tracheids**
Hydathodes are specialized structures found in some plants that release water through a process called guttation.
In simple words: This picture shows a Hydathode. A is the guard cell, B is the epithem, and C is the tracheids.

🎯 Exam Tip: For diagrams, clearly label all parts and briefly state the function of the structure if asked, as it shows complete understanding.

 

Question 19. Identify the Diagram & Label the parts.
Answer: The given diagram illustrates the process of Reverse Osmosis.
**A - Pressure**
**B - Pure water**
**C - Saltwater**
**D - Membrane**
Reverse osmosis is a purification process that uses pressure to force water through a semipermeable membrane, leaving impurities behind.
In simple words: This diagram shows reverse osmosis. A is pressure, B is pure water, C is saltwater, and D is the membrane.

🎯 Exam Tip: For reverse osmosis, remember that applied pressure must be greater than the osmotic pressure to force water in the opposite direction.

 

Question 20. Differentiate between Ascent of sap and Translocation of solute.
Answer:

Ascent of sap brings water and minerals up, while translocation distributes food produced during photosynthesis throughout the plant.
In simple words: Ascent of sap means water goes up the plant. Translocation of solute means food moves around the plant.

🎯 Exam Tip: Remember that xylem handles water and minerals (unidirectional), while phloem transports sugars and other organic compounds (multidirectional).

 

Question 21. Give any two objections to starch-sugar interconversion theory.
Answer: Here are two main objections to the starch-sugar interconversion theory for stomatal movement:
1. In monocots (a type of plant like grasses), the guard cells do not have starch. This means the theory cannot fully explain how stomata open and close in these plants.
2. There is no clear scientific proof showing that sugar actually appears at the exact moment when starch disappears and the stomata open. These objections led scientists to look for other explanations for how stomata open and close.
In simple words: One problem is that some plants don't have starch in their guard cells. Another is that we don't always see sugar appear right when starch goes away and stomata open.

🎯 Exam Tip: When listing objections to a theory, always cite the specific findings or observations that contradict it.

 

Question 22. Differentiate between cuticular and Lenticular Transpiration.
Answer:

While stomatal transpiration is the primary way plants lose water, cuticular and lenticular transpiration play smaller, but still important, roles.
In simple words: Cuticular transpiration is water loss through the waxy leaf coating, which is a small amount. Lenticular transpiration is water loss through small holes on woody stems, which is an even smaller amount.

🎯 Exam Tip: Remember that cuticular transpiration is regulated by cuticle thickness, while lenticular transpiration is minimal and constant.

 

Question 23. Mention any two uses of anti - transpirants.
Answer: Here are two uses of anti-transpirants:
1. Anti-transpirants help reduce the significant loss of water through transpiration in crop plants. This is especially useful in dry conditions.
2. They are useful for newly planted seedlings in nurseries, helping them retain water and establish themselves better. By reducing water loss, anti-transpirants can improve plant survival in dry conditions.
In simple words: Anti-transpirants help plants keep water by slowing down how much they lose, which is good for crops and young plants.

🎯 Exam Tip: When discussing practical applications, focus on how anti-transpirants directly benefit plant growth and survival, especially in challenging environments.

 

Question 24. Give notes an Aquaporin.
Answer:
• Aquaporins are special water channels. Peter Agre, who discovered them, won the Nobel Prize in Chemistry in 2003 for this work.
• These water channel proteins are found in the plasma membrane (PM) of cells.
• They regulate the massive amount of water transport that occurs across the plasma membrane.
• There are about 30 different types of aquaporins known, for example, in maize.
• They also transport other small molecules like glycerol, urea, carbon dioxide (\( \text{CO}_2 \)), ammonia (\( \text{NH}_3 \)), metalloids, and reactive oxygen species (ROS).
• Their main functions include increasing the permeability of the membrane to water and helping plants withstand drought and salt stress. Aquaporins are like tiny, selective water gates that allow plant cells to quickly adjust their water content.
In simple words: Aquaporins are special tiny doors in cell membranes that let water pass through quickly. They help plants manage water and deal with dry or salty conditions.

🎯 Exam Tip: Highlight that aquaporins facilitate rapid, but regulated, water movement, distinguishing them from simple diffusion across the lipid bilayer.

 

Question 25. Define the term Ion - Exchange.
Answer: Ion exchange is a process where ions from the external soil solution are swapped with ions of the same charge from the root cells of a plant. This means a positive ion (cation) from the soil might exchange with a cation from the root, or an anion for an anion. This process allows plants to take up necessary nutrients from the soil while releasing other ions.
In simple words: Ion exchange is when plant roots swap their ions with ions found in the soil, like trading one type of positive particle for another.

🎯 Exam Tip: Emphasize that ion exchange maintains electrical neutrality and is crucial for nutrient absorption from the soil.

 

Question 26. A. Differentiate between Cohesion and Adhesion and B. Add a note on their significance.
Answer:
**A. Differentiation between Cohesion and Adhesion:**

Both forces are critical for holding the water column together within the plant's transport system.
**B. Significance:**
The cohesive and adhesive forces work together to form an unbroken, continuous column of water in the xylem vessels of plants. The magnitude of this cohesive force is very high (up to 350 atmospheres), which is more than enough to pull water (sap) upwards even in the tallest trees against gravity. Without these forces, tall trees would struggle to transport water against gravity.
In simple words: Cohesion is water molecules sticking to each other, and adhesion is water molecules sticking to the plant's tubes. These two forces together create a strong, unbroken water column that allows tall trees to pull water up from their roots to their leaves.

🎯 Exam Tip: Remember the Cohesion-Tension theory; cohesion holds the water column, while adhesion helps it cling to xylem walls, supporting continuous water transport.

 

Question 1. Compare and Contrast Diffusion & Osmosis.
Answer:

Both are passive movements that help maintain cellular balance, but osmosis is specifically about water moving across a semipermeable membrane.
In simple words: Diffusion is when any particles spread out from crowded to less crowded areas. Osmosis is a special type of diffusion where only water moves through a special filter from where there's lots of water to where there's less.

🎯 Exam Tip: The main distinction is the selectively permeable membrane and the type of molecule (solvent, usually water) that moves in osmosis.

 

Question 2. Differentiate between osmotic pressure it and osmotic potential
Answer:

Osmotic pressure is a physical force, while osmotic potential describes the potential energy of water in a solution.
In simple words: Osmotic pressure is the pushing force that happens in a solution. Osmotic potential is about the energy water has in that solution, which is usually a negative number.

🎯 Exam Tip: Remember that osmotic potential is typically expressed as a negative value because solutes reduce the water potential of a solution.

 

Question 3. Do you have an R.O. Purifier in your house? Explain the principle behind it.
Answer:
• Yes, an R.O. (Reverse Osmosis) purifier in a house works on the principle of osmosis, but in the opposite direction.
• In regular osmosis, water moves from a region of higher water concentration to a region of lower water concentration through a selectively permeable membrane. However, in Reverse Osmosis, external pressure is applied to force water to move from a region of lower water concentration to a region of higher water concentration through the semipermeable membrane.
• Since this process moves water against its natural concentration gradient, it requires the expenditure of energy to apply the necessary pressure and push the water in the reverse direction.
• For instance, desalination plants that purify seawater also operate using the R.O. principle, as do devices for household water purification. This "reverse" process effectively removes impurities and dissolved salts, making water safe to drink.
In simple words: An R.O. purifier works by pushing water through a special filter using force, going against the normal flow of osmosis, to remove dirt and salt.

🎯 Exam Tip: Clearly state that reverse osmosis is an active process due to the external energy input required to overcome osmotic pressure.

 

Question 4. Difference between various plasmolysis types
Answer:

Observing these stages helps us understand how plant cells react to different water environments.
In simple words: Incipient plasmolysis is when the cell membrane just starts to pull away. Evident plasmolysis is when it clearly pulls away. Final plasmolysis is when the whole cell inside shrinks completely, and it can't go back to normal.

🎯 Exam Tip: When describing plasmolysis stages, focus on the degree of protoplast shrinkage and its separation from the cell wall, as well as the reversibility of each stage.

 

Question 5. Define Antitranspirant.
Answer: Antitranspirants are special materials that plants can be treated with. Their job is to slow down or lessen how much water the plant loses through transpiration. They do this without harming the plant's ability to breathe (gaseous exchange) or make food (photosynthesis). For example, clear plastics, silicone oil, and thin waxes can act as antitranspirants. These materials form a thin coating, reducing water vapor loss.
In simple words: Antitranspirants are substances that help plants lose less water from their leaves. They do not stop the plant from breathing or making food.

🎯 Exam Tip: Remember that antitranspirants are a tool for water conservation, especially in dry conditions, but they must not completely block essential gas exchange.

 

Question 6. What are the inducers of stomatal closure.
Answer: Stomata, which are tiny pores on leaves, can be induced to close by natural antitranspirants. For instance, high levels of carbon dioxide (CO2) can cause stomata to close. This happens because CO2 inhibits photorespiration, which then signals the stomata to shut, helping the plant conserve water. The plant hormone abscisic acid (ABA) is another significant inducer of stomatal closure, especially under drought stress.
In simple words: Stomata close when certain natural things, like a lot of carbon dioxide or specific plant hormones, tell them to. This helps the plant save water.

🎯 Exam Tip: When discussing stomatal closure, always mention the key factors like high CO2 concentration and abscisic acid, as these are primary regulators.

 

Question 7. Fill in the blanks in the tabulations given below
Answer: The table below shows key studies, their years, and the scientists involved in plant physiology.

The table helps to understand the historical development of important plant biology concepts over time.
In simple words: This table lists important ideas in plant science, the year they were studied, and the scientists who found them.

🎯 Exam Tip: When filling such tables, ensure correct names, years, and descriptions are matched to earn full marks. Double-check for accuracy.

 

Question 8. Nature of membrane Definition Example
Answer: The different types of membranes are defined by what they allow to pass through. Understanding these properties is crucial for understanding cell function and transport.

In simple words: Membranes can block everything, let only water through, or carefully choose what specific things can pass through them along with water.

🎯 Exam Tip: Differentiating between semipermeable and selectively permeable membranes is a common point of confusion; remember that "selectively" implies active regulation and specificity for solutes.

 

Question 9. Give the flow chart to cell to cell transport in plants.
Answer: Plant cells transport substances from one cell to another through different pathways. This process can be broadly divided into Passive Transport and Active Transport.
Cell to Cell Transport
\( \downarrow \)
(1) Passive Transport: This does not need energy.
\( \downarrow \)
      (a) Diffusion: Simple movement of substances.
           \( \downarrow \)
                 Channel Protein (for direct passage)
      (b) Facilitated Diffusion: Helped by proteins.
           \( \downarrow \)
                 Carrier Protein (for specific binding and transport)
(2) Active Transport: This requires energy.
\( \downarrow \)
      (a) Channel Protein Carrier Protein Pumps: Actively move substances against their concentration gradient.
The cell-to-cell transport ensures nutrients and signals are distributed throughout the plant body effectively.
In simple words: Plants move things between cells in two ways: passive transport, which doesn't use energy, and active transport, which does. Both use special proteins to help move substances.

🎯 Exam Tip: When drawing flowcharts, clearly distinguish between processes that require energy (active) and those that do not (passive), and indicate the roles of different proteins.

 

Question 10. Explain the capillary theory of Boehm (1809).
Answer: Boehm (1809) proposed the capillary theory to explain how water moves up plants. He believed that xylem vessels in plants act like tiny capillary tubes. According to this theory, the capillarity, or the ability of liquids to flow in narrow spaces against gravity, helps the sap (water and nutrients) rise upwards. However, this idea was later rejected because capillary action alone cannot raise water to the great heights seen in tall trees. Also, wider xylem vessels, not the narrower tracheids, are primarily responsible for carrying larger amounts of water, which goes against the simple capillary theory. This historical theory, though disproven, contributed to early understanding of plant water transport.
In simple words: Boehm thought water climbed tall plants because of thin tubes inside them, like how water goes up a thin straw. But this idea was later found to be wrong because capillary action isn't strong enough for very tall trees.

🎯 Exam Tip: When discussing disproven theories, always mention why they were rejected, citing specific evidence or scientific principles that contradicted them.

 

Question 11. Explain Phloem loading?
Answer: Phloem loading is the process where the food made during photosynthesis in the leaves (mesophyll cells) is moved into the phloem's sieve elements. This is like loading manufactured goods onto a truck for delivery to various parts of the plant. This movement is a critical step in distributing energy throughout the plant.
It happens in several steps:
1. **Photosynthesis Products:** First, chloroplasts in the mesophyll cells produce photosynthates, which are stored as starch or triose phosphate.
2. **Conversion and Transport:** These photosynthates are then moved to the cytoplasm and changed into sucrose. Sucrose is the main sugar transported in plants because it is stable and non-reactive.
3. **Movement into Sieve Elements:** Sucrose then moves from the mesophyll cells into the sieve elements of the phloem. This movement is called phloem loading.
The transport can be either short-distance (within leaf veins) or long-distance (to distant parts).
\( \implies \) Sucrose from Mesophyll \( \xrightarrow{\text{Short-distance Transport}} \) Sieve Elements
\( \implies \) Sucrose from Sieve Tube \( \xrightarrow{\text{Long-distance Transport}} \) Sink (e.g., roots, fruits)
This entire process ensures that all parts of the plant, even those not performing photosynthesis, receive the necessary energy.
In simple words: Phloem loading is when sugar made in leaves is moved into special tubes (phloem) for transport. It's like putting food into delivery trucks to send it to other parts of the plant, like roots or fruits.

🎯 Exam Tip: Emphasize that phloem loading is an active process that requires energy to move sugars against a concentration gradient into the sieve tubes.

 

Question 12. Explain the theory of photosynthesis in guard cells observed by Von Mohl with its demerits.
Answer: Von Mohl's (1856) theory explained stomatal movement based on photosynthesis in guard cells. He observed that stomata open during the day and close at night. His idea was that guard cells contain chloroplasts, which perform photosynthesis in daylight. This process creates sugars (carbohydrates). These sugars then increase the osmotic pressure inside the guard cells, causing water to flow into them from neighboring cells. This influx of water makes the guard cells turgid and causes the stomatal pore to open. At night, without light, photosynthesis stops, sugar levels drop, osmotic pressure decreases, water leaves the guard cells, and they become flaccid, leading to stomatal closure. This theory was an important early attempt to understand stomatal mechanics.
However, this theory has some demerits:
1. **Guard Cell Chloroplasts:** The chloroplasts in guard cells are actually not very well developed and cannot perform significant photosynthesis to produce enough sugar for such a drastic osmotic change.
2. **Existing Sugar Storage:** Guard cells already have a good amount of stored sugars. This suggests that the small amount of sugar from photosynthesis wouldn't be the main driver for stomatal opening.
Therefore, while the observation was correct, the proposed mechanism had weaknesses.
In simple words: Von Mohl thought stomata open because guard cells make sugar in sunlight, which pulls in water and makes them swell. But his idea was flawed because guard cells don't make much sugar and already have some stored.

🎯 Exam Tip: When presenting a historical theory, always outline its main points clearly, and then critically discuss its limitations or reasons for rejection.

 

Question 1. Explain "routes" of Water Absorption in the roots.
Answer: Plants absorb water from the soil mainly through their roots. This water then travels to the rest of the plant. The journey of water through the root involves different pathways to reach the xylem, which is the water-conducting tissue.
**The path of water absorption through roots involves:**
1. **Root Hairs:** These tiny extensions on epidermal cells are the primary sites for water absorption from the soil, initially by imbibition and then osmosis.
2. **Movement across Root Layers:** Water moves radially and centrally through various root layers such as the cortex, endodermis, pericycle, and finally into the xylem.
**There are three main routes for water movement across root cells:**
**I. Apoplast Pathway (Movement outside the cells):**
      This route involves water moving exclusively through the non-living parts of the root, like the cell walls and the spaces between cells (extracellular spaces). Water does not cross any cell membranes in this pathway until it reaches the endodermis, where the Casparian strip forces it into the symplast. It is a faster route as there is less resistance.
**II. Symplast Pathway (Movement inside the cells):**
      In this pathway, water moves from one cell to another through the cytoplasm of adjacent living cells. It crosses the plasma membrane of the first root cell and then moves through tiny cytoplasmic connections called plasmodesmata, which link the living cells. This pathway allows for regulation of water movement by the cell.
**III. Transmembrane Pathway (Movement across cell membranes):**
      This route involves water entering a cell, crossing its plasma membrane, moving through the cell's cytoplasm and possibly vacuoles, and then exiting the cell by crossing another plasma membrane into the next cell. This means water repeatedly crosses cell membranes as it moves through the root tissues.
These interconnected pathways ensure efficient water uptake and transport to the plant's vascular system.
In simple words: Water gets into plant roots and moves to the center. It can travel through the cell walls (apoplast), through the living parts of cells (symplast), or by going in and out of each cell (transmembrane).

🎯 Exam Tip: Clearly distinguish the apoplast (non-living parts) from the symplast (living protoplasts connected by plasmodesmata) and remember the critical role of the Casparian strip in regulating apoplastic flow.

 

Question 2. Draw & Explain the structure of Stomata.
Answer: Stomata are tiny pores found mainly on the epidermis of leaves and sometimes on green stems. They are essential for gas exchange and water regulation in plants.
**1. Definition:** Stomata are small openings on the surface of plant parts like leaves and green stems.
**2. Size:** They typically measure about \( 10-40 \) micrometers (\( \mu \text{m} \)) in length and \( 3-10 \) micrometers (\( \mu \text{m} \)) in breadth. A mature leaf can have anywhere from 50 to 500 stomata per square millimeter.
**3. Structure:**
      a. **Guard Cells:** Each stoma is surrounded by a pair of specialized cells called guard cells. These cells are typically kidney-shaped (or dumbbell-shaped in grasses) and regulate the opening and closing of the stomatal pore.
      b. **Subsidiary Cells:** Guard cells are usually bordered by other surrounding epidermal cells, which are known as subsidiary cells or accessory cells. These cells help support the guard cells.
      An important feature is that the inner wall of a guard cell, facing the stomatal opening, is thicker than its outer wall. The stomata open into an air-filled space inside the leaf called the substomatal cavity, which is crucial for gas exchange.
In simple words: Stomata are tiny holes on plant leaves that let the plant breathe. Each hole has two kidney-shaped guard cells around it. These cells can open or close the hole to control how much water leaves and air enters.

🎯 Exam Tip: When drawing and labeling stomata, clearly show the guard cells, subsidiary cells, and the stomatal aperture, emphasizing the thicker inner wall of the guard cells.

 

Question 3. Explain osmosis by Potato osmoscope Experiment.
Answer: The Potato Osmoscope experiment helps demonstrate osmosis, which is the movement of water across a semipermeable membrane from a higher water potential to a lower water potential area.
**Aim:** To show how osmosis works using a potato.
**Apparatus:** A potato tuber, a beaker filled with water, concentrated sugar solution, and a pin.
**Definition of Osmosis:** Osmosis is the movement of water molecules (solvent) from a place where there is a lot of water (higher water potential) to a place where there is less water (lower water potential), through a special membrane that only lets water pass.
**Procedure:**
1. Take a raw potato, peel it, and carve a deep cavity in its center using a knife.
2. Fill this potato cavity with a concentrated sugar solution.
3. Mark the initial level of the sugar solution inside the cavity using a pin.
4. Place the potato into a beaker containing pure water, ensuring the water level in the beaker is below the sugar solution level in the potato.
5. Observe the setup after about 10 minutes and note any changes in the sugar solution's level.
**Observation:** You will notice that the level of the sugar solution inside the potato cavity rises above the initial pin mark.
**Inference:** This rise in the sugar solution level confirms that water from the beaker moved into the potato cavity through the potato cells (which act as a semipermeable membrane) due to osmosis. The sugar solution has a lower water potential, pulling water in. This experiment visually confirms the principle of osmosis.

In simple words: This experiment shows osmosis. Water moves from a beaker into a potato that has sugar water inside. The sugar water level goes up because water from the beaker goes through the potato skin into the sugar water, trying to balance the concentration.

🎯 Exam Tip: For experiments, clearly state the aim, apparatus, procedure, observation, and inference. Emphasize why the potato acts as a semipermeable membrane.

 

Question 4. Measure Transpiration with Ganong's Photometer
Answer: The Ganong's Potometer is an apparatus used to measure the rate of transpiration (water loss) in plants by indirectly measuring the rate of water uptake.
**Aim:** To measure how fast a plant loses water through transpiration using Ganong's Potometer.
**Apparatus:** Ganong's Potometer (a specialized glass tube), a fresh plant twig, a beaker, water, a split rubber cork, and vaseline.
**Procedure:**
1. The Ganong's Potometer is a horizontal glass tube bent at both ends. One end has a reservoir, and the other dips into colored water in a beaker.
2. A fresh plant twig is inserted into the wider arm of the horizontal tube through a split rubber cork, ensuring a tight seal with vaseline to prevent air leaks.
3. The entire apparatus is filled with water, and a small air bubble is introduced into the graduated horizontal tube.
4. The other end of the tube is submerged in a beaker containing colored water.
**Observation:** As the plant twig loses water through transpiration from its leaves, it absorbs water from the potometer. This causes the air bubble in the graduated tube to move towards the twig. The distance the bubble travels in a given time indicates the rate of water absorption.
**Inference:** Since the rate of water absorption by the plant nearly equals the rate of water loss through transpiration, the movement of the air bubble allows us to calculate the rate of transpiration. This experiment highlights the continuous water movement from roots to leaves and out into the atmosphere.

In simple words: Ganong's Potometer helps us see how fast a plant "drinks" water and loses it. As the plant loses water from its leaves, it pulls water from a tube, making an air bubble move. How fast the bubble moves tells us how much water the plant is using.

🎯 Exam Tip: Ensure the apparatus is completely airtight to get accurate readings; any leaks will lead to incorrect observations of the air bubble's movement.

 

Question 5. Explain Mechanism of Translocation by Munch Mass flow Hypothesis
Answer: The Munch Mass Flow Hypothesis, proposed by Munch in 1930 and further developed by Crafts in 1938, explains how organic substances (sugars) are transported throughout the plant in the phloem. It's based on the idea of a pressure gradient.
**Definition:** According to this hypothesis, dissolved organic substances, primarily sugars, move from an area of high osmotic pressure (the source, like leaves) to an area of low osmotic pressure (the sink, like roots or fruits) through the phloem, driven by a turgor pressure gradient. This movement is a vital part of nutrient distribution in plants.
**Explanation with a Physical System Model:**
Imagine two chambers, 'A' and 'B', connected by a tube 'T'. Both chambers have semipermeable membranes and are placed in water.
* **Chamber A (Source):** Contains a highly concentrated sugar solution (hypertonic). Water moves into Chamber A from the outside by osmosis, increasing its turgor pressure.
* **Chamber B (Sink):** Contains a more dilute sugar solution (hypotonic). Water tends to move out of Chamber B.
* **Tube T:** Represents the phloem sieve tube.
As water continuously enters Chamber A, the increased pressure forces the sugar solution to flow through tube 'T' towards Chamber B, where the pressure is lower. The sugar then moves into Chamber B, increasing its concentration, and water flows out of Chamber B. This flow continues until the concentrations in A and B reach equilibrium. If more sugar is added to A, the flow will restart.
**Analogy to a Biological System:**
* **Chamber A (Source):** Analogous to the mesophyll cells in leaves, where photosynthesis produces a high concentration of soluble food (sugars).
* **Chamber B (Sink):** Analogous to cells in stems, roots, or fruits, where food is consumed or stored, maintaining a low sugar concentration.
* **Tube T:** Analogous to the sieve tubes of the phloem.
**Steps of Translocation:**
1. **Loading at Source:** Sugars produced in the mesophyll cells of the leaves (high osmotic potential) are actively loaded into the sieve tubes, creating high turgor pressure. Water from the xylem (reservoir) moves into these sieve tubes by osmosis, further increasing this pressure.
2. **Mass Flow:** This high turgor pressure at the source pushes the sugary solution (sap) through the sieve tubes towards areas of lower pressure (the sink).
3. **Unloading at Sink:** At the sink, sugars are removed from the sieve tubes for use or storage. This reduces the sugar concentration and, consequently, the osmotic pressure in the sieve tubes at the sink. Water then moves out of the phloem back into the xylem, maintaining the pressure gradient.
This continuous cycle drives the mass flow of food throughout the plant.
In simple words: Munch's theory says that sugar water moves in plants from where there's a lot of sugar (like leaves) to where there's less (like roots). Water goes into the sugary part, creating pressure that pushes the sugar water along tubes to other parts that need it.

🎯 Exam Tip: When explaining the Mass Flow Hypothesis, always clearly define the "source" (where sugar is produced) and the "sink" (where sugar is used or stored), and how the turgor pressure gradient drives the movement.

 

Question 6. Explain the theory of K+ transport – or Explain the mechanism of stomatal movement
Answer: The K+ ion transport mechanism, initially proposed by Levit (1974) and elaborated by Raschke (1975), is widely accepted to explain how stomata open and close. This mechanism relies on the active movement of potassium ions (K+) into and out of guard cells.
**Mechanism in Light (Stomatal Opening):**
1. **Active K+ uptake:** When light falls on guard cells, they actively pump \( \text{K}^+ \) ions from surrounding subsidiary cells into their cytoplasm. This process consumes ATP (energy).
2. **Balancing charge:** To balance the influx of positive \( \text{K}^+ \) ions, either chloride ions (\( \text{Cl}^- \)) also enter, or hydrogen ions (\( \text{H}^+ \)) are pumped out of the guard cells.
3. **Increased solute concentration:** The accumulation of \( \text{K}^+ \) ions and other solutes (like malate, produced from starch breakdown) significantly increases the solute concentration inside the guard cells, making them hypertonic.
4. **Water influx:** Due to the higher solute concentration, water potential inside the guard cells decreases. Water then moves by osmosis from the subsidiary cells into the guard cells.
5. **Turgor pressure increases:** The influx of water makes the guard cells swell and become turgid. The inner, thicker walls of the guard cells bulge, causing the guard cells to arch outwards, which opens the stomatal pore.
**Mechanism in Dark (Stomatal Closure):**
1. **K+ efflux:** In the dark, the active pumping of \( \text{K}^+ \) ions into guard cells stops. \( \text{K}^+ \) ions then move out of the guard cells, and \( \text{H}^+ \) ions move back in. This leads to a decrease in the solute concentration inside the guard cells.
2. **Water efflux:** The guard cells become hypotonic compared to the surrounding cells, leading to an increase in their water potential. Water moves out of the guard cells by osmosis.
3. **Turgor pressure decreases:** As guard cells lose water, they become flaccid, and their inner walls relax. This causes the guard cells to straighten, closing the stomatal pore.
Environmental factors like CO2 concentration and the plant hormone ABA (Abscisic Acid) also influence stomatal movement by affecting \( \text{K}^+ \) transport. For example, high CO2 or ABA can trigger \( \text{K}^+ \) efflux and lead to closure.

In simple words: Stomata open when potassium ions (K+) actively move into guard cells, pulling water in and making the cells swell. They close when K+ ions leave, causing water to also leave and the cells to shrink.

🎯 Exam Tip: Highlight the "active transport" of \( \text{K}^+ \) ions into guard cells, as this energy-dependent process is the primary driver of stomatal opening.

 

Question 7. Explain Cytochrome Pump Theory (or) Explain Carrier concept of Active Absorption, through cytochrome Pump theory.
Answer: The Cytochrome Pump Theory, proposed by Lundegårdh and Burström in 1933, explains the active absorption of anions (negatively charged ions) by plant roots. It suggests a close link between a plant's respiration rate and how it takes up anions. This theory helped to understand the energy requirements for nutrient uptake.
**Key Ideas of the Theory:**
1. **Anion Respiration:** When a plant moves from pure water to a salt solution, its respiration rate increases significantly. This increased respiration, specifically linked to anion absorption, is called "anion respiration" or "salt respiration."
2. **Different Mechanisms:** The theory assumes that the absorption mechanisms for anions and cations (positively charged ions) are distinct.
3. **Active Anion Absorption:** Anions are actively absorbed through a cytochrome electron transport chain located in the cell membrane. This process requires energy.
4. **Oxygen Gradient:** An oxygen concentration gradient across the membrane drives electron transport. Oxidation happens on the outer surface of the membrane, and reduction occurs on the inner surface.
**Working Mechanism:**
* **Electron and Proton Production:** On the inner side of the cell membrane, dehydrogenase enzymes produce protons (\( \text{H}^+ \)) and electrons (\( \text{e}^- \)) through metabolic processes.
* **Electron Transport Chain:** These electrons are then passed along an electron transport chain involving cytochrome enzymes. As electrons move along this chain, they create an energy gradient.
* **Anion Uptake:** Anions from the soil are picked up by oxidized cytochrome oxidase at the outer membrane surface. As the electrons move through the chain, the cytochromes get reduced and then oxidized, facilitating the transfer of anions across the membrane to the inner surface.
* **Cation Movement:** Cations are thought to move passively along the electrical gradient that is created by the accumulation of anions on the inner surface of the membrane. This means anions are actively pulled in, and cations follow due to electrical attraction.
This theory highlights the energy-dependent nature of nutrient uptake and the role of respiration.
In simple words: The Cytochrome Pump Theory explains how plants actively take in negatively charged particles (anions) using energy from breathing (respiration). It's like a tiny pump in the cell membrane that uses an electron chain to pull these particles inside, while positive particles follow because of the electric charge.

🎯 Exam Tip: Focus on the active nature of anion absorption, the role of the cytochrome system, and the link to respiration as key components of this theory.

 

Question 8. Explain the opening and closing of stomata by a starch-sugar interconversion theory.
Answer: The starch-sugar interconversion theory explains how stomata open and close, with different scientists offering their perspectives:
(i) **Lloyd (1908):** Suggested that the turgidity (swelling) of guard cells, which controls stomatal opening, is caused by the change between starch and sugar. During the day, sugar forms, making the guard cells turgid and opening the stomata. At night, starch forms, making the cells lose turgidity and closing the stomata.
(ii) **Sayre (1920):** Proposed that the pH level inside the guard cells determines stomatal movement. During the day, CO2 is used for photosynthesis, causing the pH to rise. This leads to starch converting into sugar, increasing osmotic pressure, and causing water to enter, making cells turgid and opening stomata. At night, photosynthesis stops, CO2 builds up, lowering pH, and the reverse happens, closing the stomata.
(iii) **Hanes (1940):** Identified the enzyme phosphorylase as key to this theory. This enzyme helps change starch into sugar in guard cells. During the day, with high pH, phosphorylase converts starch to sugar, leading to water entry (endosmosis), cell turgidity, and stomatal opening. The opposite occurs at night.
(iv) **Steward (1964):** Explained that glucose-1-phosphate is not osmotically active. However, when it is converted to osmotically active glucose, it increases the concentration inside guard cells, which causes water to rush in and the stomata to open. This conversion is linked to changes in pH, which activate or deactivate the enzymes involved. The continuous transformation of molecules within the guard cells drives the rhythmic opening and closing of stomata.
In simple words: Stomata open when guard cells fill with water and become firm, usually because starch turns into sugar, making water rush in. They close when guard cells lose water and become soft, often when sugar turns back into starch.

🎯 Exam Tip: When explaining theories, always state the scientist's name and year, then clearly describe the core mechanism proposed, highlighting the key factors involved like starch, sugar, pH, and water movement.

 

IV. 5 Mark Questions

 

Question 1. Explain ‘routes’ of Water Absorption in the roots.
Answer: Plants absorb water mainly through their roots, and this water then moves through different pathways to reach the xylem.
The process starts with root hairs and other outer epidermal cells absorbing water from the soil, initially through imbibition (water sticking to surfaces) and then primarily through osmosis (water moving from high to low concentration). From these epidermal cells, water moves inwards, passing through the cortex, endodermis, and pericycle, finally reaching the xylem vessels. This journey involves three main routes:
**I. Apoplast Pathway:** This route uses the non-living parts of the plant, like cell walls and the spaces between cells. Water moves continuously through these spaces without ever crossing any cell membranes. It is a faster path because there are no barriers to cross, similar to water flowing through a sponge.
**II. Symplast Pathway:** In this route, water moves through the living parts of the plant. It enters the cytoplasm of one root cell and then travels from that cell's cytoplasm to the next through tiny connections called plasmodesmata. This means water stays inside the cells and only crosses the cell membrane once when it first enters the root.
**III. Transmembrane Route:** This pathway is a mix of the other two. Water moves from the cytoplasm of one cell, then crosses its cell membrane to exit into the cell wall, and then crosses the next cell's membrane to enter its cytoplasm. This means water crosses multiple cell membranes as it travels through the root tissue, making it a more regulated path.
In simple words: Water gets into plant roots and travels to the center using three main paths: one through cell walls (apoplast), one through the living cell insides (symplast), and one where it goes in and out of cells (transmembrane).

🎯 Exam Tip: Clearly differentiate between the living and non-living components involved in each pathway and mention the unique feature of each route, such as crossing membranes or using plasmodesmata.