Students get through the TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants which is useful for their exam preparation.

TN State Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants

Very short answer questions

Question 1.
Define transport in plants. Mention the tissues involved in transportation.
Answer:
Transport is the process of moving water, minerals, and food to all parts of the plant body. Conducting tissues such as xylem and phloem play an important role in transport.

Question 2.
What is the need for transport in plants?
Answer:
Water absorbed from roots must travel up to leaves by xylem for food preparation by photosynthesis. Likewise, food prepared from leaves has to travel to all parts of the plant including roots for growth and other processes.

Question 3.
Compare Active transport with Passive transport.
Answer:

Active Transport Passive Transport
It is an uphill process. It is a downhill process.
Energy is required. Energy is not required.
It is a Biological process. It is a physical process.
(e.g.) Na+ – K+ pump. (e.g.) Osmosis.

TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants

Question 4.
Define the term diffusion.
Answer:
The net movement of molecules from a region of their higher concentration to a region of their lower concentration along a concentration gradient until an equilibrium is attained.

Question 5.
How the diffusing molecules will move?
Answer:
In diffusion, the movement of molecules is continuous and random in order in all directions.

Question 6.
Name the transport proteins of the plasma membrane involved in facilitated diffusion.
Answer:
Channel protein and carrier protein.

Question 7.
Give a brief account of Porin.
Answer:
Porin is a large transporter protein found in the outer membrane of plastids, mitochondria, and bacteria which facilitates smaller molecules to pass through the membrane.

Question 8.
State the role of Channel Protein.
Answer:
Channel protein forms a channel or tunnel in the cell membrane for the easy passage of molecules to enter the cell. The channels are either open or remain closed.

Question 9.
Apart from water, what are the substrates that are transported through aquaporins?
Answer:
Glycerol, Urea, CO2, NH3, Metalloids, and Reactive Oxygen Species (ROS).

TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants

Question 10.
How does a carrier protein function?
Answer:
Carrier protein acts as a vehicle to carry molecules from outside of the membrane to inside the cell and vice versa. Due to the association with molecules to be transported, the structure of carrier protein gets modified until the dissociation of the molecules.

Question 11.
On which basis, the carrier proteins are classified? Mention its types.
Answer:
There are three types of carrier proteins on the basis of handling of molecules and direction of transport. They are,

  1. Uniport,
  2. Symport and
  3. Antiport.

Question 12.
Mention the drawbacks of diffusion.
Answer:
In diffusion, there is a lack of control over the transport of selective molecules. There is a possibility of harmful substances entering the cell by a concentration gradient.

Question 13.
Co-transport and counter transport differ from each other. Justify.
Answer:
In co-transport, two molecules are transported together in the same direction whereas, in counter¬transport two molecules are transported in opposite directions to each other.

Question 14.
State any two vital roles of water in plants.
Answer:

  1. Water maintains the internal temperature of the plant.
  2. It maintains the turgidity of the cell.

TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants

Question 15.
List out a few imbibing.
Answer:
Gum, starch, proteins, cellulose, agar, and gelatin.

Question 16.
Define Imbibition.
Answer:
The phenomenon in which a substance uptake the water and swell up is called Imbibition. The substance is referred to as Imbibant. Example: Swelling of water-soaked seeds.

Question 17.
List out the substances that are transported by facilitated diffusion.
Answer:
Sugars, amino acids, nucleotides, ions, and cell metabolites.

Question 18.
How imbibition, is important for plants?
Answer:
Significance of imbibition:

  1. During the germination of seeds, imbibition increases the volume of the seed enormously and leads to the bursting of the seed coat.
  2. It helps in the absorption of water by roots at the initial level.

Question 19.
Define Water potential.
Answer:
Water potential is the potential energy of water in a system compared to pure water when both temperature and pressure are kept the same. Water potential is denoted by the symbol ¥ (psi).

Question 20.
Define Osmotic Pressure.
Answer:
When a solution and its solvent (pure water) are separated by a semipermeable membrane, pressure is developed in the solution, due to the presence of dissolved solutes. This is called osmotic pressure (OP).

TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants

Question 21.
Mention the symbolic representation of water potential and osmotic pressure.
Answer:
Water potential – ψ (psi) and Osmotic pressure – π (pi).

Question 22.
Mention the alternate terminologies and symbolic representation of solute potential and Matric potential.
Answer:

S. No. Potential Energy Alternate Terminology Symbolic Representation
(a) Solute potential Osmotic potential ψS
(b) Matric potential Imbibition pressure ψM

Question 23.
Expand and Define TP.
Answer:
TP stands for Turgor pressure. When a plant cell is placed in pure water (hypotonic solution) the diffusion of water into the cell takes place by endosmosis. It creates a positive hydrostatic pressure on the rigid cell wall by the cell membrane. Henceforth the pressure exerted by the cell membrane towards the cell wall is Turgor Pressure (TP).

Question 24.
Mention the pressures that act upon the cell to make it a full turgid.
Answer:
Turgor pressure and wall pressure make the cell fully turgid. TP + WP = Turgid.

Question 25.
Define Diffusion Pressure Deficit (DPD).
Answer:
The difference between the diffusion pressure of the solution and its solvent at a particular temperature and atmospheric pressure is called Diffusion Pressure Deficit (DPD).

TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants

Question 26.
Why DPD is also called suction pressure?
Answer:
Increased DPD favors endosmosis or it sucks the water from the hypotonic solution, hence it is called suction pressure.

Question 27.
Define Osmosis.
Answer:
Osmosis represents the movement of water or solvent molecules through a selectively permeable membrane from the place of its higher concentration (high water potential) to the place of its’ lower concentration (low water potential).

Question 28.
State the significance of plasmolysis.
Answer:
Plasmolysis is exhibited only by living cells and so it is used to test whether the cell is living or dead.

Question 29.
What happens if a plant cell is treated with a hypertonic solution?
Answer:
When a plant cell is kept in a hypertonic solution, water leaves the cell due to exosmosis. As a result of water loss, protoplasm shrinks and the cell membrane is pulled away from the cell wall and finally, the cell becomes flaccid. This process is named plasmolysis.

Question 30.
Give an example for plasmolysis and also mention the types of plasmolysis.
Answer:
Wilting of plants noticed under the condition of water scarcity is an indication of plasmolysis. Three types of plasmolysis occur in plants: /) Incipient plasmolysis, ii) Evident plasmolysis, and iii) Final plasmolysis.

Question 31.
Draw a simplified diagram depicting Reverse Osmosis.
Answer:
TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants 1

TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants

Question 32.
If a cell in the cortex with DPD of 5 atm is surrounded by hypodermal cells with DPD of 2 atm, what will be the direction of movement of water?
Answer:
Water will move from low DPD to high DPD (hypodermis 2 atm to cortex 5 atm).

Question 33.
Arrange in the correct sequence in concern with the pathway of water, in roots, (cortex, root hair, xylem, epidermal cell, endodermis, and pericycle).
Answer:
Root hairs → epidermal cell → cortex → endodermis → pericycle → xylem.

Question 34.
Mention the possible routes of water movement across root cells.
Answer:
There are three possible routes of water.
They are i) Apoplast, ii) Symplast and iii) Transmembrane route.

Question 35.
What are the objections to the osmotic active absorption theory?
Answer:
Objections to osmotic theory: (i) The cell sap concentration in the xylem is not always high. (ii) Root pressure is not universal in all plants, especially in trees.

TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants

Question 36.
Name any two respiratory inhibitors.
Answer:
Potassium cyanide (KCN) and chloroform.

Question 37.
Define Ascent of sap.
Answer:
The water within the xylem along with dissolved minerals from roots is called sap and its upward transport is called ascent of sap.

Question 38.
State Relay Pump theory.
Answer:
Relay pump theory of Godlewski (1884)
Periodic changes in osmotic pressure of living cells of the xylem parenchyma and medullary ray act as a pump for the movement of water.

Question 39.
What do you mean by the term ‘Embolism”?
Answer:
Gas bubbles expanding and displacing water within the xylem element is called cavitation or embolism. .

Question 40.
Which is the most widely accepted theory to prove the Ascent of sap? Who proposed it.
Answer:
Cohesion-tension theory or Transpiration pull theory proposed by Dixon and Jolly (1894- 1924).

Question 41.
Define Transpiration.
Answer:
The loss of excess of water in the form of vapor from various aerial parts of the plant is called transpiration.

TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants

Question 42.
State any two theories regarding the mechanism of stomatal movement.
Answer:
Starch-sugar interconversion theory and Active potassium transport ion concept.

Question 43.
Draw and label the structure of stomata.
Answer:
TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants 2

Question 44.
What is an anti transpirant?
Answer:
The term antitranspirant is used to designate any material applied to plants for the purpose of retarding transpiration. An ideal antitranspirant checks the transpiration process without disturbing the process of gaseous exchange.

Question 45.
List out the uses of antitranspirants.
Answer:

  1. Antitranspirants reduce the enormous loss of water by transpiration in crop plants.
  2. Useful for seedling transplantations in nurseries.

TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants

Question 46.
What are hydathodes?
Answer:
Hydathodes are the stomata-like pores present in plants that grow in moist and shady places.

Question 47.
Define the term “Translocation of organic solutes”.
Answer:
The phenomenon of food transportation from the site of synthesis to the site of utilization is known as the translocation of organic solutes. The term solute denotes food material that moves in a solution.

Question 48.
What do you mean by Phloem loading?
Answer:
The movement of photosynthates (products of photosynthesis) from mesophyll cells to phloem sieve elements of mature leaves is known as phloem loading.

Question 49.
List the merits of the Munch-Mass Flow hypothesis.
Answer:

  1. When a woody or herbaceous plant is girdled, the sap contains high sugar-containing exudates from the cut end.
  2. A positive concentration gradient disappears when plants are defoliated.

Question 50.
Define – Flux and its types.
Answer:
The movement of ions into and out of cells or tissues is termed transport or flux. Entry of the ion into the cell is called influx and exit is called efflux.

TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants

Short answer questions

Question 1.
Draw a Flow Chart illustrating various types of cell-to-cell transport.
Answer:
TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants 3

Question 2.
Enumerate the significance of diffusion.
Answer:
Significance of diffusion in Plants

  1. Gaseous exchange of O2 and CO2 between the atmosphere and stomata of leaves takes place by the process of diffusion. O2 is absorbed during respiration and CO2 is absorbed during photosynthesis.
  2. In transpiration, water vapour from intercellular spaces diffuses into the atmosphere through stomata by the process of diffusion.
  3. The transport of ions in mineral salts during passive absorption also takes place by this process.

Question 3.
How semipermeable and selectively permeable membranes differ from each other?
Answer:

Semi-Permeable Selectively Permeable
Semi-permeable allow diffusion of solvent molecules but do not allow the passage of solute molecule. Example: Parchment paper. All biomembranes allow some solutes to pass in addition to the solvent molecules. Example: Plasmalemma, tonoplast, and membranes of cell organelles.

TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants

Question 4.
Give an account on Aquaporin.
Answer:
Aquaporin is a water channel protein embedded in the plasma membrane. It regulates the massive amount of water transport across the membrane. Plants contain a variety of aquaporins. Over 30 types of aquaporins are known from maize. Currently, they are also recognized to transport substrates like glycerol, urea, CO2, NH3, metalloids, and Reactive Oxygen Species (ROS) in addition to water. They increase the permeability of the membrane to water. They confer drought and salt stress tolerance.

Question 5.
Give an account of Matric’s potential.
Answer:
Matric potential represents the attraction between water and the hydrating colloid or gel-like organic molecules in the cell wall which is collectively termed as matric potential. Matric potential is also known as imbibition pressure. The matric potential is maximum (most negative value) in dry material. Example: The swelling of soaked seeds in water.

Question 6.
How DPD differs in various conditions of a cell?
Answer:

  • DPD in normal cell: DPD = OP – TP.
  • DPD in the fully turgid cell: Osmotic pressure is always equal to turgor pressure in a fully turgid cell.
  • OP = TP or OP-TP =0. Hence DPD of fully turgid cell is zero.
  • DPD in the flaccid cell: If the cell is in flaccid condition there is no turgor pressure or TP=0.
    Hence DPD = OP.

Question 7.
Compare Hypertonic, Hypotonic, and Isotonic solutions.
Answer:
TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants 4

TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants

Question 8.
Give an account of Endosmosis and Exosmosis.
Answer:

  1. Endosmosis: Endosmosis is defined as the osmotic entry of solvent into a cell or a system when it is placed in pure water or hypotonic solution. For example, dry raisins placed in the water, it swells up due to turgidity.
  2. Exosmosis: Exosmosis is defined as the osmotic withdrawal of water from a cell or system when it is placed in a hypertonic solution. Exosmosis in a plant cell leads to plasmolysis.

Question 9.
Give an account of Deplasmolysis.
Answer:
The effect of plasmolysis can be reversed, by transferring them back into the water or hypotonic solution. Due to endosmosis, the cell becomes turgid again. It regains its original shape and size. This phenomenon of the revival of the plasmolyzed cell is called deplasmolysis. Example: Immersion of dry raisin in water.

Question 10.
Give an account of Reverse Osmosis and its uses.
Answer:
Reverse Osmosis follows the same principles of osmosis, but in the reverse direction. In this process movement of water is reversed by applying pressure to force the water against a concentration gradient of the solution. In regular osmosis, the water molecules move from the higher concentration (pure water = hypotonic) to the lower concentration (saltwater = hypertonic). But in reverse osmosis, the water molecules move from the lower concentration (saltwater = hypertonic) to higher concentration (pure water = hypotonic) through a selectively permeable membrane.
Uses: Reverse osmosis is used for the purification of drinking water and desalination of seawater.

Question 11.
Differentiate between the types of Plasmolysis.
Answer:
Difference between plasmolysis types.

Incipient Plasmolysis Evident Plasmolysis Final Plasmolysis
No morphological symptoms appear in plants. Wilting of leaves appears. Severe wiping and drooping of leaves appear.
The plasma membrane separates only at the comer from the cell wall of cells. The plasma membrane completely detaches from the cell wall. Plasma membrane completely detaches from cell wall with a maximum shrinkage of volume.
It is reversible. It is reversible. It is irreversible.

TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants

Question 12.
Write in brief about Non-Osmotic active absorption.
Answer:
Bennet-Clark (1936), Thimann (1951), and Kramer (1959) observed absorption of water even if the concentration of cell sap in the root hair is lower than that of the soil water. Such a movement requires an expenditure of energy released by respiration (ATP). Thus, there is a link between water absorption and respiration. It is evident from the fact that when respiratory inhibitors like KCN, Chloroform are applied there is a decrease in the rate of respiration and also the rate of absorption of water.

Question 13.
Explain the pulsation theory of J.C. Bose.
Answer:
Bose invented an instrument called the Crescograph, which consists of an electric probe connected to a galvanometer. When a probe is inserted into the inner cortex of the stem, the galvanometer showed high electrical activity. Bose believed a rhythmic pulsating movement of the inner cortex like a pump (similar to the beating of the heart) is responsible for the ascent of sap. He concluded that cells associated with xylem exhibit pumping action and pumps the sap laterally into xylem cells.

Question 14.
List out the setbacks of Root Pressure Theory.
Answer:
The following objections have been raised against root pressure theory:

  1. Root pressure is totally absent in gymnosperms, which includes some of the tallest plants.
  2. There is no relationship between the ascent of sap and root pressure. For example, in summer, the rate of the ascent of sap is more due to transpiration in spite of the fact that root pressure is very low. On the other hand, in winter when the rate of ascent of sap is low, a.high root pressure is found.
  3. The ascent of sap continues even in the absence of roots.
  4. The magnitude of root pressure is about 2atm, which can raise the water level up to few feet only, whereas the tallest trees are more than 100m high.

Question 15.
Describe the structure of stomata.
Answer:
The epidermis of leaves and green stems possess many small pores called stomata. The length and breadth of stomata are about 10-40 p and 3 – 10 p respectively. Mature leaves contain between 50 and 500 stomata per mm2. Stomata are made up of two guard cells, special semilunar or kidney-shaped living epidermal cells in the epidermis. Guard cells are attached to surrounding epidermal cells known as subsidiary cells or accessory cells. The guard cells are joined together at each end but they are free to separate to form a pore between them. The inner wall of the guard cell is thicker than the outer wall. The stoma opens to the interior into a cavity called the sub-stomatal cavity which remains connected with the intercellular spaces.
TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants 5

TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants

Question 16.
Explain the theory of photosynthesis in guard cells and state its demerits.
Answer:
Theory of Photosynthesis in guard cells:
Von Mohl (1856) observed that stomata open in light and close in the night. According to him, chloroplasts present in the guard cells photosynthesize in the presence of light resulting ‘ in the production of carbohydrate (Sugar) which increases osmotic pressure in guard cells.
It leads to the entry of water from other cells and the stomatal aperture opens. The above process vice versa at the night lead to the closure of stomata.

Demerits

  1. The chloroplast of guard cells is poorly developed and incapable of performing photosynthesis.
  2. The guard cells already possess much amount of stored sugars.

Question 17.
Draw a diagrammatic representation of the steward scheme of stomatal movement.
Answer:
TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants 6

Question 18.
Why wilting occurs? Explain its types.
Answer:
Excessive loss of water through transpiration leads to wilting. In general, there are three types of wilting as follows,
a. Incipient wilting: The water content of plant cells decreases but the symptoms are not visible.
b. Temporary wilting: On hot summer days, the freshness of herbaceous plants reduces turgor pressure in the daytime and regains it at night.
c. Permanent wilting: The absorption of water virtually ceases because the plant cell does not get water from any source and the plant cell passes into a state of permanent wilting.

Question 19.
Give an account of types of structural modifications of leaves for reducing transpirational – loss.
Answer:
Leaf structure: Some anatomical features of leaves like sunken stomata, the presence of hairs, cuticle, the presence of hydrophilic substances like gum, mucilage help to reduce the rate of transpiration. In xerophytes the structural modifications are remarkable. To avoid transpiration, as in Opuntia the stem is flattened to look like leaves called Phylloclade. Cladode or cladophyll in Asparagus is a modified stem capable of limited growth looking like leaves. In some plants, the petioles are flattened and widened, to become phyllodes example Acacia melanoxylon.

TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants

Question 20.
Comment on various chemicals inducing stomatal closure.
Answer:
Carbon-di-oxide induces stomatal closure and acts as a natural antitranspirant. Further, the advantage of using CO2 as an antitranspirant is its inhibition of photorespiration. Phenyl Mercuric Acetate (PMA), when applied as a foliar spray to plants, induces partial stomatal closure for two weeks or more without any toxic effect. The use of abscisic acid highly induces the closing of stomata. Dodecenyl succinic acid also affects on stomatal closure.

Question 21.
Classify translocation based on direction.
Answer:
Phloem translocates the products of photosynthesis from leaves to the area of growth and storage, in the following directions,
Downward direction: From leaves to stem and roots.
Upward direction: From leaves to developing buds, flowers, fruits for consumption and storage. Germination of seeds is also a good example of upward translocation.
Radial direction: From cells of the pith to cortex and epidermis, the food materials are radially translocated.

Question 22.
What do you understand by the term source and sink in plant physiology?
Answer:
The source is defined as any organ in plants which are capable of exporting food materials to the areas of metabolism or to the areas of storage. Examples: Mature leaves and germinating seeds.
TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants 7
A sink is defined as any organ in plants that receives food from a source. Example: Roots, tubers, developing fruits, and immature leaves.

Question 23.
Define Phloem unloading with its steps.
Answer:
From sieve elements, sucrose is translocated into sink organs such as roots, tubers, flowers, and fruits, and this process is termed phloem unloading. It consists of three steps:

  1. sieve element unloading: Sucrose leave from sieve elements.
  2. Short distance transport: Movement of sucrose to sink cells.
  3. Storage and metabolism: The final step when sugars are stored or metabolized in sink cells.

TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants

Question 24.
Enumerate the objections of the Munchi Mass Flow hypothesis.
Answer:

  1. Munchi Mass Flow hypothesis explains the unidirectional movement of solute only. However, bidirectional movement of solute is commonly observed in plants.
  2. The osmotic pressure of mesophyll cells and that of root hair do not confirm the requirements.
  3. This theory gives a passive role to sieve tube and protoplasm, while some workers demonstrated the involvement of ATP.

Question 25.
Donnan equilibrium. Add a note.
Answer:
Within the cell, some of the ions never diffuse out through the membrane. They are trapped within the cell and are called fixed ions. But they must be balanced by the ions of opposite charge. Assuming that a concentration of fixed anions is present inside the membrane, more cations would be absorbed in addition to the normal exchange to maintain the equilibrium. Therefore, the cation concentration would be greater in the internal than in the external solution. This electrical balance or equilibrium controlled by electrical as well as diffusion phenomenon is known as the Donnan equilibrium.

Long answer questions

Question 1.
Write in detail about types of carrier protein.
Answer:
There are three types of carrier proteins classified on the basis of handling of molecules and the direction of transport. They are i. Uniport, ii. Symport, iii. Antiport.

  1. Uniport: In this molecule of a single type move across a membrane independent of other molecules in one direction.
  2. Symport or co-transport: The term symport is used to denote an integral membrane protein that simultaneously transports two types of molecules across the membrane in the same direction.
  3. Antiport or Counter Transport: An antiport is an integral membrane transport protein that simultaneously transports two different molecules, in opposite directions, across the membrane.

TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants

Question 2.
Write a note on Solute potential and Pressure Potential.
Answer:
1. Solute Potential (ψS): Solute potential, otherwise known as osmotic potential denotes the effect of dissolved solute on water potential. In pure water, the addition of solute reduces its free energy and lowers the water potential value from zero to negative. Thus the value of solute potential is always negative. In a solution at standard atmospheric pressure, water potential is always equal to solute potential (ψW = ψS).
2. Pressure Potential (ψP): Pressure potential is a mechanical force working against the effect of solute potential. Increased pressure potential will increase water potential and water enters cells and cells become turgid. This positive hydrostatic pressure within the cell is called Turgor pressure. Likewise, withdrawal of water from the cell decreases the water potential and the cell becomes flaccid.

Question 3.
With the help of a diagram explain the possible route of water across root cells.
Answer:
There are three possible routes of water. They are i) Apoplast, ii) Symplast and iii) Transmembrane route.

  1. Apoplast: The apoplast (Greek: apo = away; plast = cell) consists of everything external to the plasma membrane of the living cell. The apoplast includes cell walls, extracellular spaces, and the interior of dead cells such as vessel elements and tracheids. In the apoplast pathway, water moves exclusively through the cell wall or the non-living part of the plant without crossing any membrane. The apoplast is a continuous system.
    TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants 8
  2. Symplast: The symplast (Greek: sym = within; plast = cell) consists of the entire mass of cytosol of all the living cells in a plant, as well as the plasmodesmata, the cytoplasmic channel that interconnects them. In the symplastic route, water has to cross the plasma membrane to enter the cytoplasm of the outer root cell; then it will move within adjoining cytoplasm through plasmodesmata around the vacuoles without the necessity to cross more membrane, till it reaches xylem.
  3. Transmembrane route: In the transmembrane pathway water sequentially enters a cell on one side and exits from the cell on the other side. In this pathway, water crosses at least two membranes for each cell. Transport across the tonoplast is also involved.

Mechanism of Water Absorption Kramer (1949) recognized two distinct mechanisms which independently operate in the absorption of water in plants. They are, i) active absorption ii) passive absorption.

TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants

Question 4.
Point out the differences between Active absorption and Passive absorption.
Answer:

Active Absorption Passive Absorption
Active absorption takes place by the activity of root and root hairs. The pressure for absorption is not developed in roots and hence roots play a passive role.
Transpiration has no effect on active absorption. Absorption regulated by transpiration.
The root hairs have high DPD as compared to soil solution and therefore water is taken by tension. The absorption occurs due to tension created in xylem sap by transpiration pull, thus water is sucked in by the tension.
Respiratory energy needed. Respiratory energy not required.
It involves the symplastic movement of water. Both symplast and apoplast movement of water involved.                                          ‘

Question 5.
Explain about Cohesion – Tension Theory.
Answer:
Cohesion-tension theory was originally proposed by Dixon and Jolly (1894) and again put forward by Dixon (1914, 1924). This theory is based on the following features:

  1. Strong cohesive force or tensile strength of water: Water molecules have the strong mutual force of attraction called cohesive force due to which they cannot be easily separated from one another. Further, the attraction between a water molecule and the wall of the xylem element is called adhesion. These cohesive and adhesive force works together to form an unbroken continuous water column in the xylem. The magnitude of the cohesive force is much high (350 atm) and is more than enough to ascent sap in the tallest trees.
  2. Continuity of the water column in the plant: An important factor that can break the water column is the introduction of air bubbles in the xylem. Gas bubbles expanding and displacing water within the xylem element is called cavitation or embolism. However, the overall continuity of the water column remains undisturbed since water diffuses into the adjacent xylem elements for continuing ascent of sap.
  3. Transpiration pull or Tension in the unbroken water column: The unbroken water column from leaf to root is just like a rope. If the rope is pulled from the top, the entire rope will move upward. In plants, such a pull is generated by the process of transpiration which is known as transpiration pull. Water vapour evaporates from mesophyll cells to the intercellular spaces near stomata as a result of active transpiration. The water vapours are then transpired through the stomatal pores. Loss of water from mesophyll cells causes a decrease in water potential. So, water moves as a pull from cell to cell along the water potential gradient. This tension, generated at the top (leaf) of the unbroken water column, is transmitted downwards from the petiole, stem and finally reaches the roots. The cohesion theory is the most accepted among plant physiologists today.

TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants

Question 6.
Describe the types of transpiration in plants.
Answer:
Types of Transpiration: Transpiration is of the following three types:

  1. Stomatal transpiration: Stomata are microscopic structures present in high numbers on the lower epidermis of leaves. This is the most dominant form of transpiration and being responsible for most of the water loss (90 – 95%) in plants.
  2. Lenticular transpiration: In stems of woody plants and trees, the epidermis is replaced by periderm because of secondary growth. In order to provide gaseous exchange between the living cells and the outer atmosphere, some pores which look like lens-shaped raised spots are present on the surface of the stem called Lenticels. The loss of water from lenticels is very insignificant as it amounts to only 0.1 % of the total.
  3. Cuticular transpiration: The cuticle is a waxy or resinous layer of cutin, a fatty substance covering the epidermis of leaves and other plant parts. Loss of water through the cuticle is relatively small and it is only about 5 to 10 % of the total transpiration. The thickness of cuticle increases in xerophytes and transpiration is very much reduced or totally absent.

Question 7.
Give an account of Starch – Sugar Interconversion Theory.
Answer:

  1. According to Lloyd (1908), the turgidity of guard cells depends on the interconversion, of starch and sugar. It was supported by Loftfield (1921) as he found guard cells containing sugar during the daytime when they are open and starch during the night when they are closed.
  2. Sayre (1920) observed that the opening and closing of stomata depend upon the change in pH of guard cells. According to him stomata open at high pH during the daytime and become closed at low pH at night. The utilization of CO2 by photosynthesis during the light period causes an increase in pH resulting in the conversion of starch to sugar. Sugar increase in cell favors endosmosis and increases the turgor pressure which leads to opening of stomata. Likewise, the accumulation of CO2 in cells during the night decreases the pH level resulting in the conversion of sugar to starch. Starch decreases the turgor pressure of the guard cell and stomata close.
  3. The discovery of enzyme phosphorylase in guard cells by Hanes (1940) greatly supports the starch-sugar interconversion theory. The enzyme phosphorylase hydrolyses starch into sugar and high pH followed by endosmosis and the opening of stomata during light. The vice versa takes place during the night.
    TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants 9
  4. Steward (1964) proposed a slightly modified scheme of starch-sugar interconversion theory. According to him, Glucose-1-phosphate is osmotically inactive. Removal of phosphate from Glucose-1-phosphate converts to Glucose which is osmotically active and increases the concentration of guard cells leading to the opening of stomata.

Objections to Starch-sugar interconversion theory

  1. In monocots, the guard cell does not have starch.
  2. There is no evidence to show the presence of sugar at a time when starch disappears and stomata open.
  3. It fails to explain the drastic change in pH from 5 to 7 by change of CO2

TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants

Question 8.
Describe the K+ Transport theory on transpiration.
Answer:
The theory of K+ transport theory was proposed by Levit (1974) and elaborated by Raschke (1975). According to this theory, the following steps are involved in the stomatal opening:
TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants 10

In light

  1. In the guard cell, starch is converted into organic acid (malic acid).
  2. .Malic acid in the guard cell dissociates to malate anion and proton (H+).
  3. Protons are transported through the membrane into nearby subsidiary cells with the
    exchange of K+ (Potassium ions) from subsidiary cells to guard cells. This process involves an electrical gradient and is called ion exchange.
  4. This ion exchange is an active process and consumes ATP for energy.
  5. Increased K+ ions in the guard cell are balanced by CL ions. An increase in solute concentration decreases the water potential in the guard cell.
  6. Guard cell becomes hypertonic and favours the entry of water from surrounding cells.
  7. Increased turgor pressure due to the entry of water opens the stomatal pore.

In Dark
TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants 11

  1. In the dark photosynthesis stops and respiration continues with the accumulation of CO2 in the sub-stomatal cavity.
  2. Accumulation of CO2 in the cell lowers the pH level.
  3. Low pH and a shortage of water in the guard cell activate the stress hormone Abscisic acid (ABA).
  4. ABA stops further entry of K+ ions and also induces K+ ions to leak out to subsidiary cells from guard cells.
  5. Loss of water from the guard cell reduces turgor pressure and causes the closure of stomata.

TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants

Question 9.
Draw the structure of hydathode and explain Guttation.
Answer:
TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants 12
During high humidity in the atmosphere, the rate of transpiration is much reduced. When plants absorb water in such a condition root pressure is developed due to excess water within the plant. Thus excess water exudates as a liquid from the edges of the leaves and are called guttation. Example: Grasses, tomato, potato, brinjal, and Alocasia. Guttation occurs through stomata-like pores called hydathodes generally present in plants that grow in moist and shady places. Pores are present over a mass of loosely arranged cells with large intercellular spaces called epithem. This mass of tissue lies near vein endings (xylem and Phloem). The liquid coming out of the hydathode is not pure water but a solution containing a number of dissolved substances.

Question 10.
Explain Ganong’s Potometer experiment and its purpose.
Answer:
TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants 13
Ganongs potometer is used to measure the rate of transpiration indirectly. In this, the amount of water absorbed is measured, and assumed that this amount is equal to the amount of water transpired.
The apparatus consists of a horizontal graduated tube that is bent in opposite directions at the ends. One bent end is wide and the other is narrow. A reservoir is fixed to the horizontal tube near the wider end.
The reservoir has a stopcock to regulate water flow. The apparatus is filled with water from the reservoir. A twig or a small plant is fixed to the wider arm through a split cock. The other bent end of the horizontal tube is dipped into a beaker containing colored water. An air bubble is introduced into the graduated tube at the narrow end keep this apparatus in bright sunlight and observe. As transpiration takes place, the air bubble will move towards the twig. The loss is compensated by water absorption through the xylem portion of the twig. Thus, the rate of water absorption is equal to the rate of transpiration.

Question 11.
Describe the ringing experiment with a diagram.
Answer:
The experiment involves the removal of all the tissue outside to vascular cambium (bark, cortex, and phloem) in woody stems except the xylem. Xylem is the only remaining tissue in the girdled area which connects the upper and lower part of the plant. This setup is placed in a beaker of water. After some time, it is observed that swelling on the upper part of the ring appears as a result of the accumulation of food material. If the experiment continues within days, the roots die first. It is because the supply of food material to the root is cut down by the removal of phloem. The roots cannot synthesize their food and so they die first. As the roots gradually die the upper part (stem), which depends on root for the ascent of sap, will ultimately die.
Ringing experiment.
TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants 14

TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants

Question 12.
Write in detail about Passive Absorption of minerals salts.
Answer:
1. Ion-Exchange: Ions of external soil solution were exchanged with same charged (anion for anion or cation for cation) ions of the root cells. There are two theories explaining this process of ion exchange namely: i. Contact exchange and ii. Carbonic acid exchange.

  1. Contact Exchange Theory: According to this theory, the ions adsorbed on the surface of root cells and clay particles (or clay micelles) are not held tightly but oscillate within a small volume of space called oscillation volume. Due to the small space, both ions overlap each other’s oscillation volume and exchange takes place.
  2. Carbonic Acid Exchange Theory: According to this theory, soil solution plays an important role by acting as a medium for ion exchange. The CO2 released during respiration of root cells combines with water to form carbonic acid (H2CO3). Carbonic acid dissociates into H+ and HCO3 in the soil solution. These H+ ions exchange with cations adsorbed on clay particles and the cations from micelles get released into soil solution and gets adsorbed on root cells.
    TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants 15

Question 13.
Describe Lundegardh’s Cytochrome Pump Theory.
Answer:
Lundegardh’s Cytochrome Pump Theory:
Lundegardh and Burstrom (1933) observed a correlation between respiration and anion absorption. When a plant is transferred from water to a salt solution the rate of respiration increases which is called anion respiration or salt respiration. Based on this observation Lundegardh (1950 and 1954) proposed cytochrome pump theory which is based on the following assumptions:

  1. The mechanism of anion and cation absorption is different.
  2. Anions are absorbed through the cytochrome chain by an active process, cations are absorbed passively.
  3. An oxygen gradient is responsible for oxidation at the outer surface of the membrane and reduction at the inner surface.

According to this theory, the enzyme dehydrogenase on the inner surface is responsible for the formation of protons (H+) and electrons (e). As electrons pass outward through the electron transport chain there is a corresponding inward passage of anions. Anions are picked up by oxidized cytochrome oxidase and are transferred to other members of the chain as they transfer the electron to the next component.
The theory assumes that cations (C+) move passively along the electrical gradient created by the accumulation of anions (A) at the inner surface of the membrane.
Main defects of the above theory are:

  1. Cations also induce respiration.
  2. Fails to explain the selective uptake of ions.
  3. It explains absorption of anions only.
    TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants 16

TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants

Question 14.
Explain Protein-Lecithin Theory.
Answer:
Bennet-Clark’s Protein-Lecithin Theory:
In 1956, Bennet-Clark proposed that the carrier could be a protein associated with phosphatide called lecithin. The carrier is amphoteric (the ability to act either as an acid or a base) and hence both cations and anions combine with it to form the Lecithinion complex in the membrane. Inside the membrane, the Lecithin-ion complex is broken down into phosphatidic acid and choline along with the liberation of ions. Lecithin again gets regenerated from phosphatidic acid and choline in the presence of the enzyme choline acetylase and choline esterase. ATP is required for the regeneration of lecithin.
TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants 17

Higher Order Thinking Skills (HOTs)

Question 1.
Why during rainy seasons, the wooden doors and windows are difficult to close and open? Give the phenomenon behind this and also define the phenomenon.
Answer:
The swelling of wooden windows, doors due to high humidity during the rainy season is due, to imbibition. Imbibition refers to the uptake of water and swelling of substances that are not water-soluble.

Question 2.
While making dry fishes at home high salt concentration is applied. Why? Name the phenomenon.
Answer:
The high salt concentration is applied to extract the excess water from the fishes thereby inhibiting the microbial growth and increase the shelf life. This phenomenon is called exosmosis.

Question 3.
Observe the histogram and answer the following question.
TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants 18
(a) Which type of transpiration does ‘A’ and ‘C’ represent?
(b) Name ‘B’ and also define the plant part. Which is responsible for ‘B’ type of transpiration?
Answer:
(a) A – Stomatal transpiration, C – Cuticular transpiration and B – Lenticular transpiration. Lenticels are the lend-shaped raised spots present on the surface of the stem.

TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants

Question 4.
Plants are highly adaptable to the environment where they survive Opuntia which lives in xeric condition shows phylloclade adaptation. Which part of Opuntia is modified as phylloclade? Why does it modify?
Answer:
In Opuntia, the stem is flattered to look like leaves called phylloclade. It is modified to avoid transpirational loss of water.

Question 5.
Nowadays, water scarcity is becoming a prime problem. To compensate for the need, various strategies are being carried, out by the Governments at national and international levels. One such effective technology is the Desalination of seawater. Which principle is followed in their technology. Define it.
Answer:
Desalination of seawater is being effectively earned out by Reverse Osmosis. In Reverse osmosis, the water molecules are forced by applying pressure from lower concentration to higher concentration through a selectively permeable membrane.

Question 6.
Observe the diagram, it is a plant cell undergoing plasmolysis.
(a) Which stage of plasmolysis does the cell represent?
TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants 19
(b) Whether it is reversible?
(c) What happens if this occurs in the leaf cells?
Answer:
(a) Evident plasmolysis.
(b) Yes, evident plasmolysis is reversible.
(c) Under evident plasmolysis, wilting of leaves appears.

Choose the correct answer.

1. …………. is a downhill process using physical forces.
(a) Short distance transport
(b) Translocation
(c) Active transport
(d) Passive transport
Answer:
(d) Passive transport

2. The smell of the room spray can be felt everywhere inside a closed room. This is because of ………….
(a) Osmosis
(b) Passive transport
(c) Diffusion
(d) Imbibition
Answer:
(c) Diffusion

TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants

3. A mixture of ………… and potassium permanganate is used for fumigation.
(a) Acetaldehyde
(b) Calcium oxide
(c) Formalin
(d) Vinegar
Answer:
(c) Formalin

4. ROS stands for …………
(a) Reduction Oxygen Species
(b) Reactive Oxygen Syndrome
(c) Reducive Oxygen Species
(d) Reactive Oxygen Species
Answer:
(d) Reactive Oxygen Species

5. Peter Agre discovered aquaporin in ………….
(a) RBC
(b) WBC
(c) Platelets
(d) Plasma membrane
Answer:
(a) RBC

6. Protoplasm is made of ……….. of water.
(a) 80-90%
(b) 85-90%
(c) 60-80%
(d) 75-85%
Answer:
(c) 60-80%

TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants

7. ………… does not act as an imbibant.
(a) Protein
(b) Starch
(c) Stone
(d) Gum
Answer:
(c) Stone

8. Which physiological process can be observed in a germinating seed?
(a) Diffusion
(b) Osmosis
(c) Imbibition
(d) Facilitated diffusion
Answer:
(c) Imbibition

9. Water potential is measured in ………….
(a) Watt
(b) Joule
(c) Calories
(d) Pascal
Answer:
(d) Pascal

10. Water potential can be determined by solute potential and ………….
(a) Matric potential
(b) Pressure potential
(c) Osmotic potential
(d) Osmotic potential
Answer:
(b) Pressure potential

TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants

11. Osmotic pressure is represented by the Greek letter ………….
(a) α
(b) π
(c) ψ
(d) θ
Answer:
(b) π

12. If osmotic pressure has a positive value, the osmotic potential has ………… value.
(a) Positive
(b) Negative
(c) Neutral
(d) Zero
Answer:
(b) Negative

13. Which combination of pressures makes a cell full turgid?
(a) TP + WP
(b) OP -TP
(c) SP + OP
(d) WP + SP
Answer:
(a) TP + WP

14. In a hypertonic solution, which of the following substance concentration will be high?
(a) Solute
(b) Solvent
(c) Both a & b
(d) None
Answer:
(a) Solute

15. Dry raisins kept in the water begin to swell. It is a perfect example for ………….
(a) Osmosis
(b) Plasmolysis
(c) Endosmosis
(d) Final plasmolysis
Answer:
(c) Endosmosis

16. Among the following physiological processes, which one occurs only in the living cells?
(a) Diffusion
(b) Osmosis
(c) Exosmosis
(d) Plasmolysis
Answer:
(d) Plasmolysis

TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants

17. To revive a plasmolyzed cell, it should be treated with ………. solution.
(a) Isotonic
(b) Hypertonic
(c) Hypotonic
(d) Neutral
Answer:
(c) Hypotonic

18. Principle behind the desalination of sea water is …………
(a) Endosmosis
(b) Diffusion
(c) Reverse osmosis
(d) Deplasmolysis
Answer:
(c) Reverse osmosis

19. The final destination of water entering the root hair is ………….
(a) Endodermis
(b) Pericycle
(c) Xylem
(d) Cortex
Answer:
(c) Xylem

20. The nature of cell sap is ………….
(a) Hypertonic
(b) Hypotonic
(c) Isotonic
(d) Apoplast
Answer:
(a) Hypertonic

TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants

21. Relay pump theory was put forth by ………….
(a) Atkins and Preistley
(b) Strasburger and Overton
(c) Godlewski
(d) J.C. Bose
Answer:
(c) Godlewski

22. Who is called as father of plant physiology?
(a) J.C. Bose
(b) Stephen Hales
(c) Dixon
(d) Unger
Answer:
(b) Stephen Hales

23. An instrument deviced by J.C. Bose for proving the pulsating movement of cortex is ……….
(a) Seismograph
(b) Galvanometer
(c) Crescograph
(d) Radiograph
Answer:
(c) Crescograph

24. In embolism of xylem, water is displaced by …………
(a) Callose
(b) Photosynthates
(c) Gas bubbles
(d) Porin proteins
Answer:
(c) Gas bubbles

25. Transpiration is a kind of ………. occur through plant body.
(a) Condensation
(b) Sublimation
(c) Evaporation
(d) Percipitation
Answer:
(c) Evaporation

26. Rate of water movement through xylem is …………
(a) 65 cm/second
(b) 75 cm/min
(c) 12 cm/min
(d) 82 cm/sec
Answer:
(b) 75 cm/min

TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants

27. ………… is a fatty substance covering the epidermis of leaves.
(a) Mucin
(b) Cutin
(c) Porin
(d) Suberin
Answer:
(b) Cutin

28. Approximately, a corn plant transpires ……….. litres of H2O per day.
(a) 2
(b) 8
(c) 16
(d) 450
Answer:
(a) 2

29. Starch – sugar interconversion theory was proposed by ………..
(a) Loft field
(b) Lloyd
(c) Von Mohl
(d) Hanes
Answer:
(a) Lloyd

30. ABA stands for ……..
(a) Abscisic acid
(b) Ascorbic acid
(c) Acetyl Butyric Acid
(d) Acetic acid
Answer:
(a) Abscisic acid

TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants

31. Pick out the natural antitranspirant.
(a) O2
(b) SO2
(c) CO2
(d) CO
Answer:
(c) CO2

32. Guttation occurs through …………
(a) Stoma
(b) Epithem
(c) Hydathodes
(d) Epidermis
Answer:
(c) Hydathodes

33. CO2 inhibits …………
(a) Transpiration
(b) Photorespiration
(c) Both a and b
(d) Respiration
Answer:
(c) Both a and b

34. Match the following:

S. No. Particulars S. No. Particulars
1. Theory of K+transport (i) Atkins and Priestley
2. Osmotic active absorption (ii) Unger
3. Capillary theory (iii) Levit
4. Imbibition theory (iv) Boehm

(a) 1 – (iii), 2 – (i), 3 – (iv), 4 – (ii)
(b) 1 – (i), 2 – (ii), 3 – (i), 4 – (iii)
(c) 1 – (ii), 2 – (i), 3 – (iv), 4 – (iii)
(d) 1 – (i), 2 – (iii), 3 – (ii), 4 – (iv)
Answer:
(a) 1 – (iii), 2 – (i), 3 – (iv), 4 – (ii)

TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants

35. …………. is used to measure the rate of transpiration.
(a) Ganongs respiroscope
(b) Ganongs potometer
(c) Arc auxanometer
(d) None of these
Answer:
(b) Ganongs potometer

36. Electro-Osmotic theory was propounded by ……….
(a) Mason and Masked
(b) Fenson and Spanner
(c) Curtis
(d) Levit
Answer:
(b) Fenson and Spanner

37. Translocation of photosynthetic products is a ……… movement.
(a) unidirectional
(b) upward
(c) downward
(d) multi-directional
Answer:
(d) multidirectional