Class 11 Bio Botany · Chapter 11

Samacheer Class 11 Bio Botany - Transport in Plants

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Sections in this chapter
Book Back Questions 10II. Two Mark Questions 26III. 3 Mark Questions 10III. Trans – Membrane Route 7IV. 5 Mark Questions 1Part II 56
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1Book Back Questions10 questions
Q.1In a fully turgid cell:v
  1. (a) DPD = 10 atm; OP = 5 atm; TP = 10 atm
  2. (b) DPD = 0 atm; OP =10 atm; TP = 10 atm
  3. (c) DPD = 0 atm; OP = 5 atm; TP = 10 atm
  4. (d) DPD = 20 atm; OP = 20 atm; TP = 10 atm
Answer:

(b) DPD = 0 atm; OP =10 atm; TP = 10 atm

Q.2Which among the following is correct? i) apoplast is fastest and operate in nonliving part ii) Transmembrane route includes vacuole in) 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, ivv
Answer:

d) i, ii, iii, iv

Q.3What type of transpiration is possible in the xerophyte Opuntia?v
  1. (a) Stomatal
  2. (b) Lenticular
  3. (c) Cuticular
  4. (d) All the above
Answer:

(b) Lenticular

Q.4Stomata of a plant open due to a) Influx of K + b) Effrilx of K + c) Influx of Cl – d) Influx of OH –v
Answer:

a) Influx of K +

Q.5Munch hypothesis is based on:v
  1. (a) translocation of food due to TP gradient and imbibition force
  2. (b) ranslocation of food due to TP
  3. (c) translocation of food due to imbibition force
  4. (d) None of the above
Answer:

(b) ranslocation of food due to TP

Q.6If the concentration of salt in the soil is too high and the plants may wilt even if the field is thoroughy irrigated. Explainv
Answer:

High salt concentration in the soil creates an osmotic potential that is more negative than that of the plant cells, making it difficult for plants to absorb water even when the soil is thoroughly irrigated. When the concentration of salt in the soil is too high, the osmotic potential of the soil solution becomes very low, meaning the soil solution has a lower water potential than the plant cells. Under these conditions, plants must expend more metabolic energy in the form of ATP to actively transport mineral ions and accumulate solutes in their cells in order to lower their own osmotic potential sufficiently to absorb water from the soil. This process is energetically expensive and can deplete the plant's energy reserves. Under extreme salinity conditions, the osmotic potential of the soil solution may become so negative that it exceeds the plant's capacity to generate sufficient osmotic potential through solute accumulation, even with maximum metabolic effort. In such cases, the plant becomes unable to absorb water from the soil and will wilt despite the presence of adequate soil moisture. This phenomenon is referred to as the osmotic or water deficit effect of salinity. The wilting occurs because the plant loses water to the soil through osmosis, as water moves from the plant cells toward the hypertonic soil solution, causing the plant to become dehydrated. This is a significant problem in agricultural areas with saline soils, where crop productivity is severely reduced unless the salt concentration is reduced through irrigation management or soil amendment.

Q.7How phosphorylase enzyme open the stomata in starch sugar interconversion theory?v
Answer:

The discovery of the enzyme phosphorylase in guard cells by Hanes in 1940 provided strong support for the starch-sugar interconversion theory regarding stomatal movement. This theory explains how the interconversion of starch and sugar, influenced by pH changes, leads to the opening and closing of stomata. During the day, photosynthesis occurs in the guard cells, consuming carbon dioxide and leading to an increase in the pH of the guard cell sap. This higher pH activates the enzyme phosphorylase, which then hydrolyses starch into sugar (specifically glucose-1-phosphate, which is then converted to glucose and then sucrose). The accumulation of sugars increases the solute concentration inside the guard cells, lowering their water potential. Consequently, water moves from the adjacent subsidiary cells into the guard cells by osmosis, causing the guard cells to become turgid and bow outwards, leading to the opening of the stomata. Conversely, during the night, photosynthesis ceases, and carbon dioxide accumulates due to respiration, leading to a decrease in the pH of the guard cell sap. This lowered pH inactivates phosphorylase and activates another enzyme, leading to the conversion of sugar back into starch. The decrease in sugar concentration raises the water potential inside the guard cells, causing water to move out of the guard cells and back into the subsidiary cells. As a result, the guard cells become flaccid and collapse, leading to the closure of the stomata.

Q.8List out the non-photosynthetic parts of a plant that need a supply of sucrose?v
Answer:

The non-photosynthetic parts of a plant that require a continuous supply of sucrose, which is the primary form of sugar transported in plants, include various organs that are actively growing, storing food, or performing metabolic functions without producing their own sugars. These essential sink regions include the roots, which are responsible for absorbing water and nutrients and anchoring the plant; tubers, such as potatoes, which are specialized underground stems that serve as storage organs for carbohydrates; developing fruits, which require a significant amount of energy and building blocks for their growth and maturation; and immature leaves, which are still developing and have not yet reached their full photosynthetic capacity, thus relying on sugars transported from mature leaves.

Q.9What are the parameters which control water potential?v
Answer:

1. Slatyer and Taylor (1960) introduced the concept of water potential.
Definition – water potential is the potential energy of water in a system – compared to pure water when temperature and pressure are kept constant.
2. It is also a measure of how freely water molecules can move in a particular environment or system. Water potential is denoted by the Greek symbol Ψ (psi) and measured in Pascal (Pa). At standard temperature, the water potential of pure water is zero
3. Addition of solute to pure water decreases the kinetic energy thereby decreasing the water potential, from zero to negative.
4. So, Comparatively a solution always has low water potential than pure water. In a group of cells with different water potential, a water potential gradient is generated.
5. Water will move from higher water potential to lower water potential.
When potential ( Ψ) can be determined by. Solute concentration or Solute potential ( Ψ s ) Pressure potential ( Ψ p )
By correlating two factors, water potential is written as (Ψ w =Ψ s )+Ψ p
a) Solute potential (Ψ s ) or 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 (Ψ w = Ψ s ).
b) Pressure Potential (Ψ p )
* 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.

2II. Two Mark Questions26 questions
Q.10What is the need for the transport of materials in plants?v
Answer:

The transport of materials in plants is essential for the survival and growth of the plant organism. Water absorbed by the roots from the soil must be transported upward through the xylem tissue to reach the leaves, where it serves as a raw material for photosynthesis and as a medium for the transport of mineral nutrients and other dissolved substances. Without this upward transport of water, the leaves would become desiccated and unable to carry out photosynthesis, which is the fundamental process by which plants produce organic compounds and energy. Conversely, food materials, primarily in the form of sugars and other organic compounds, are synthesized in the leaves during photosynthesis and must be transported to all other parts of the plant, including the roots, through the phloem tissue. This downward and lateral transport of food is necessary to provide energy and building materials for growth, maintenance, and reproduction in non-photosynthetic tissues such as roots, stems, flowers, and fruits. Additionally, various other substances such as hormones, vitamins, and mineral nutrients must be transported throughout the plant to regulate growth and development and to maintain metabolic functions in all cells. Without efficient transport systems, plants would be unable to coordinate their growth, respond to environmental stimuli, or allocate resources appropriately among different organs. Therefore, the transport of materials is a fundamental requirement for plant survival, growth, and reproduction.

Q.11What is osmosisv
Answer:

Osmosis is a special type of diffusion that involves the movement of water molecules across a semipermeable or selectively permeable membrane. Unlike simple diffusion, which can occur for any type of molecule, osmosis is specifically the movement of water molecules in response to differences in solute concentration on either side of a membrane. In osmosis, water molecules move from a region where water is present in higher concentration, which corresponds to a region of lower solute concentration or higher water potential, to a region where water is present in lower concentration, which corresponds to a region of higher solute concentration or lower water potential. This movement continues until the concentration of solutes on both sides of the membrane becomes equal, or until the hydrostatic pressure difference balances the osmotic pressure difference. The semipermeable membrane is crucial to osmosis because it allows water molecules to pass through freely but prevents or restricts the passage of dissolved solute molecules. This selective permeability creates the conditions necessary for osmosis to occur. Osmosis is of fundamental importance in plant physiology because it is the primary mechanism by which plant cells absorb water from the soil and maintain turgor pressure, which is essential for maintaining cell shape, supporting the plant body, and enabling cell growth and expansion. Understanding osmosis is therefore essential for understanding how plants interact with their environment and how water and nutrients are transported within plant tissues.

Q.12Define the term diffusion.v
Answer:

Diffusion is the net movement of molecules from a region of their higher concentration to a region of their lower concentration along a concentration gradient, continuing until an equilibrium is attained. This process occurs spontaneously without the input of energy and is driven by the random thermal motion of molecules. In a region of high concentration, molecules are more densely packed and have a higher probability of moving toward regions of lower concentration simply due to random molecular movement. Over time, as molecules move from high to low concentration regions, the concentration difference decreases until the molecules are uniformly distributed throughout the available space, at which point equilibrium is reached and net diffusion ceases. Diffusion is a fundamental process in plant physiology and occurs across cell membranes, through cell walls, and in intercellular spaces. It is responsible for the movement of various substances including gases such as oxygen and carbon dioxide, dissolved nutrients, and other small molecules. The rate of diffusion is influenced by several factors including the concentration gradient, the temperature of the system, the size of the diffusing molecules, and the nature of the medium through which diffusion occurs. In plants, diffusion plays a crucial role in gas exchange during photosynthesis and respiration, in the movement of nutrients from the soil into root cells, and in the distribution of various substances within plant tissues. Although diffusion is a passive process that does not require metabolic energy, it is essential for maintaining the chemical gradients and concentrations of substances necessary for plant survival and function.

Q.13The touch me plant closes its leaves at the touch – Explain.v
Answer:
  • In the ‘Touch me not’ plant the touching act as stimulus, and it closes the leaves.
  • When we touch the plant, at that time the stem releases some chemicals, which force water to move out of the cell leading to the loss of Turgor pressure and the leaves droop down However after sometime they become normal.
Q.14What is meant by Porin?v
Answer:

Porin is a large transporter protein found in the outer membrane of plastids, mitochondria and bacteria which facilitates the passage of smaller molecules through the membrane. These proteins form channels or pores that allow water and small solutes to cross the membrane while restricting the passage of larger molecules. Porins are essential for maintaining selective permeability and enabling controlled transport of ions and small organic molecules across these organellar membranes.

Q.15Define water potentialv
Answer:
  • The potential energy of water in a system compared to pure water when both temperature and pressure are ketp same.
  • It is a measure of how freely water molecules can move in a given environment
  • Water potential of pure water is = 0
Q.16Define Diffusion Pressure Deficit.v
Answer:
  • Termed by Meyer (1938)
  • The difference between the Diffusion pressure of the solution and its solvent at a particular temperature and atmospheric pressure of the solution and its solvent at a particular temperature and atmospheric pressure is called DPD.
Q.17Differentiate between short distance and Long Distance Transport.v
Answer:

Short distance transport refers to the cell-to-cell movement of substances involving only a few cells in the lateral direction. It occurs through diffusion, osmosis, and active transport across cell membranes and plasmodesmata. This type of transport connects the root hairs to the xylem and phloem tissues. Long distance transport, on the other hand, involves the movement of substances through the extensive network of xylem and phloem tissues over considerable distances within the plant body. It occurs in a direct vertical manner and is the main transport mechanism for moving water and minerals from roots to shoots through xylem, and for translocating organic solutes from source to sink through phloem. Examples of short distance transport include diffusion and osmosis, while long distance transport includes ascent of sap through xylem and translocation of solutes through phloem.

Q.18Differentiate between Passive & Active Transportv
Answer:

Passive transport is a type of movement across a cell membrane that does not require the expenditure of metabolic energy by the cell. It is often referred to as 'downhill' transport because substances move from a region of higher concentration to a region of lower concentration, following their concentration gradient. This process occurs spontaneously and includes mechanisms such as simple diffusion, facilitated diffusion (which involves membrane proteins but no ATP), and osmosis. For example, the movement of oxygen into a cell or water across a semipermeable membrane are instances of passive transport. Active transport, in contrast, is an 'uphill' or 'biological' transport process that requires the cell to expend metabolic energy, typically in the form of ATP, to move substances across a membrane. This movement occurs against the concentration gradient, meaning from a region of lower concentration to a region of higher concentration. Active transport systems often involve specific carrier proteins or pumps embedded in the cell membrane. A classic example is the Na+/K+ ATPase pump, which actively transports sodium ions out of the cell and potassium ions into the cell, maintaining crucial electrochemical gradients essential for nerve impulse transmission and other cellular functions.

Q.19Give two examples of the phenomenon of Imbibition.v
Answer:

Imbibition is a special type of diffusion where water is absorbed by solid colloids, causing them to increase in volume. This process is crucial for various biological phenomena. Two common examples of imbibition include the swelling of dry seeds when placed in water. This swelling is essential for seed germination, as it breaks the seed coat and allows the embryo to grow. Another example is the swelling of wooden furniture, such as windows, tables, and doors, during periods of high humidity, particularly during the rainy season. The wood, being a colloidal substance, absorbs moisture from the atmosphere, leading to its expansion and sometimes causing doors and windows to jam.

Q.20Explain carbonic Acid Exchange theory.v
Answer:
  • Soil solution act as a medium of ion-exchange
  • The CO 2 released by roots combine with water to form carbonic acid (H 2 CO 3 )
  • Carbonic acid dissociates into H + + HCO 3 in the soil solution.
  • H + ions exchange with cations adsorbed on clay particles and cations from micelles get released int c.
Q.21Give Answer in a sentence or two Distinguish between (i) Exomosis & Endomosis (ii) Apoplast & Symplast (iii) Cohesion & Adhesion (v) Influx & Effluxv
Answer:

I) Exomosis
Endomosis
The osmotic outflow of water, when cell placed in hypertonic solution
Osmotic inflow into the cell when placed in hypotonic solution or water
Eg. Preservation of Jam, Jellies, pickles
Eg. Swelling of Dry grapes placed in water
II) Apoplast
Symplast
System of adjacent cell walls – continuous throughout except at the asparian strips of endodermis in the roots
System of interconnected protoplasts of neighbouring cells in plants
III) Cohesion
Adhesion
Attraction between molecules of a similar kind
The attraction between molecules of different kind
IV) Influx
Efflux
The entry of ion into the cell is known as Influx
The exit of ion from the cell into outside is known as Efflux
It can be active or passive
It can be active or passive.

Q.22What is meant by osmotic pressure?v
Answer:

Osmotic pressure is the pressure that develops in a solution when it is separated from its pure solvent by a semipermeable membrane. This pressure arises due to the presence of dissolved solutes in the solution. When a solution and pure water are separated by a semipermeable membrane, water molecules tend to move from the region of higher water potential (pure water) to the region of lower water potential (solution) by osmosis. The accumulation of water molecules in the solution creates a pressure difference, which is termed osmotic pressure. Osmotic pressure is directly proportional to the concentration of solutes and the absolute temperature. It is an important colligative property that depends only on the number of solute particles present, not on their nature or size.

Q.23Define Root Pressure.v
Answer:
  • Stephen Hales – coined the term
  • Stoking (1956) Defined the term.
  • A pressure developing in the tracheary elements of the xylem as a result of metabolic activities of the root.
Q.24Define the term osmosis.v
Answer:

Osmosis, derived from the Latin word 'Osmos' meaning impulse or urge, is a special type of diffusion that represents the spontaneous movement of water or solvent molecules through a selectively permeable membrane. This movement occurs from the region of higher water potential (where solute concentration is lower) to the region of lower water potential (where solute concentration is higher). The selectively permeable membrane allows water molecules to pass through freely but restricts the passage of dissolved solute molecules. Osmosis continues until equilibrium is reached, at which point the water potential on both sides of the membrane becomes equal. This process is fundamental to plant physiology as it drives water uptake by roots, maintains cell turgor pressure, and is essential for various physiological processes including nutrient transport and cell expansion.

Q.25Why plants transport sugars as sucrose and not as starch or Monosaccharide (Glucose & Fructose)v
Answer:

Plants transport sugars primarily as sucrose rather than starch or monosaccharides due to several important advantages. Starch is a polysaccharide and non-reducing sugar that is insoluble in water, making it impossible to transport through the phloem. Glucose and fructose are monosaccharides and reducing sugars that are soluble in water but are less efficient for energy storage and are highly reactive, making them unsuitable for long-distance transport. Sucrose, a disaccharide and non-reducing sugar, possesses ideal properties for transport. It is soluble in water even at high concentrations without increasing viscosity significantly, which allows efficient movement through phloem tissues. Sucrose is more efficient in energy storage compared to monosaccharides and lacks reducing ends, making it chemically inert and stable during transport. These properties make sucrose the ideal transport form of carbohydrates in plants, as it can be easily mobilized from source tissues, transported long distances without causing osmotic stress, and readily converted to other forms at sink tissues for utilization or storage.

Q.26What are the three types of plasmolysis?v
Answer:

Plasmolysis is the process in which plant cells lose water when placed in a hypertonic solution, causing the protoplast to shrink away from the cell wall. This phenomenon is categorized into three distinct types based on the extent of water loss and protoplast shrinkage. These types are incipient plasmolysis, evident plasmolysis, and final plasmolysis. Incipient plasmolysis is the initial stage where the protoplast just begins to pull away from the cell wall at the corners. Evident plasmolysis occurs when the protoplast has significantly shrunk and detached from most of the cell wall. Final plasmolysis represents the complete and irreversible shrinkage of the protoplast, often leading to cell death if prolonged.

Q.27Identify the diagram and Neatly label the partsv
Answer:

The given diagram illustrates the structure of a Hydathode, which is a specialized pore found on the epidermis of leaves, typically at the margins or tips, responsible for guttation. Guttation is the process of exudation of xylem sap from the intact margins of leaves. The labeled parts are A, which represents the Guard cell, although hydathodes are generally characterized by open pores without functional guard cells, sometimes modified epidermal cells are referred to in this context. B points to the Epithem, which is a mass of thin-walled parenchyma cells located beneath the hydathode pore. C indicates the Tracheids, which are the terminal xylem elements that supply water to the epithem cells.

Q.28Identify the Diagram & Label the parts.v
Answer:

The given diagram explains the process of Reverse Osmosis, which is a water purification technology that uses a semi-permeable membrane to remove ions, molecules, and larger particles from drinking water. In this process, pressure is applied to overcome osmotic pressure, forcing water from a region of higher solute concentration through the membrane to a region of lower solute concentration. The labeled parts are A, representing the Pressure applied to the saltwater side, which is necessary to drive the water molecules across the membrane. B indicates the Pure water, which is the filtered water collected on the other side of the membrane after the solutes have been removed. C denotes the Saltwater, which is the solution with a high concentration of dissolved salts that is being treated. D represents the Membrane, specifically a semi-permeable membrane, which allows water molecules to pass through but blocks the passage of most dissolved salts and other impurities.

Q.29Differentiate between Ascent of sap and Translocation of solute.v
Answer:

Ascent of sap and translocation of solutes are two distinct long-distance transport mechanisms in plants. Ascent of sap refers to the upward transport of water along with dissolved mineral nutrients from the roots to the aerial parts of the plant, including stems and leaves. This process occurs through the xylem tissue and is driven by a combination of root pressure, capillarity, and transpiration pull. Translocation of solutes, on the other hand, refers to the transport of organic food materials from the site of synthesis (source) to the site of utilization or storage (sink). This process occurs through the phloem tissue and involves the movement of sugars, amino acids, and other organic compounds. The key differences are that ascent of sap transports water and minerals upward through xylem, while translocation transports organic solutes bidirectionally through phloem. Ascent of sap is primarily driven by physical forces, whereas translocation requires metabolic energy in the form of ATP for active transport processes.

Q.30Give any two objections to starch-sugar interconversion theory.v
Answer:

The starch-sugar interconversion theory, also known as the starch-hydrolysis theory, proposed that the opening and closing of stomata are regulated by the interconversion of starch and sugar within the guard cells. However, this theory faces several significant objections that challenge its universal applicability and accuracy. One major objection is that in many monocotyledonous plants, such as onion and several grasses, the guard cells do not contain starch. If starch is absent, then the proposed interconversion mechanism cannot operate, yet these plants still exhibit stomatal movements. Another critical objection is the lack of conclusive evidence to demonstrate the presence of sugar at the precise time when starch disappears and the stomata open. Experimental observations have not consistently shown a direct correlation between the disappearance of starch and a corresponding increase in sugar concentration sufficient to explain the osmotic changes required for stomatal opening. Furthermore, the theory does not adequately explain the role of potassium ions, which are now widely recognized as key players in stomatal movement.

Q.31Differentiate between cuticular and Lenticular Transpiration.v
Answer:

Cuticular transpiration and lenticular transpiration are two distinct pathways of water loss from plants. Cuticular transpiration is the loss of water through the cuticle, which is the waxy protective layer covering the aerial surfaces of plants. This accounts for only about five to ten percent of the total transpiration in most plants. The rate of cuticular transpiration is inversely related to cuticle thickness; plants with thicker cuticles such as xerophytes experience lesser transpiration. Lenticular transpiration is the loss of water through lenticels, which are small pores present on the woody surfaces of stems and bark. Lenticels allow gaseous exchange and water vapor loss from the underlying tissues. Lenticular transpiration accounts for only about zero point one percent of the total transpiration, making it a minor pathway of water loss. While cuticular transpiration occurs across the entire leaf and stem surface, lenticular transpiration is restricted to specific regions where lenticels are present, primarily on woody stems and older plant parts.

Q.32Mention any two uses of anti – transpirants.v
Answer:

Anti-transpirants are substances applied to plants to reduce the rate of transpiration, which is the loss of water vapor from plant leaves. These compounds are highly beneficial in various agricultural and horticultural practices. One primary use of anti-transpirants is to reduce the enormous loss of water by transpiration in crop plants. By minimizing water loss, these substances help plants conserve water, especially in arid or semi-arid regions, or during drought conditions, thereby improving crop yield and water use efficiency. Another significant application of anti-transpirants is in seedling transplantations in nurseries. When seedlings are transplanted, they often experience transplant shock due to root damage and increased water loss through transpiration. Applying anti-transpirants helps to reduce this water loss, allowing the seedlings to establish more effectively in their new environment and improving their survival rates.

Q.33Give notes an Aquaporin.v
Answer:

Aquaporins are integral membrane proteins that form water channels, facilitating the rapid and selective transport of water across biological membranes. The discovery of aquaporins, specifically Aquaporin-1 (AQP1), by Peter Agre, for which he shared the Nobel Prize in Chemistry in 2003, revolutionized our understanding of water transport in living organisms. These water channel proteins are present in the plasma membrane and other cellular membranes, playing a crucial role in regulating the massive amount of water transport across these barriers. For instance, over 30 types of aquaporins have been identified in maize, highlighting their diversity and importance. Beyond water, some aquaporins can also transport other small, uncharged molecules such as glycerol, urea, carbon dioxide (CO2), ammonia (NH3), metalloids, and reactive oxygen species (ROS). Their primary function is to significantly increase the permeability of the membrane to water, allowing for much faster water movement than simple diffusion across the lipid bilayer. Furthermore, aquaporins are known to confer drought and salt stress tolerance in plants by regulating water uptake and retention under adverse environmental conditions, thereby contributing to plant survival and adaptation.

Q.34Define the term Ion – Exchange.v
Answer:

Ion exchange is a process in which ions of the external soil solution are exchanged with ions of the same charge present in the root cells. Specifically, anions from the soil solution are exchanged with anions from the root cells, and cations from the soil solution are exchanged with cations from the root cells. This process occurs at the root surface and is mediated by the cell membrane. Ion exchange is an important mechanism of mineral nutrient absorption by plants, allowing roots to selectively accumulate essential mineral ions from the soil solution against the concentration gradient.

Q.35A. Differentiate between Cohesion and Adhesion and B. Add a note on their significance.v
Answer:

A. Cohesion and adhesion are two distinct intermolecular forces that operate in the xylem. Cohesion refers to the strong mutual attraction between water molecules themselves, creating a cohesive force that holds water molecules together. Adhesion refers to the attraction between water molecules and the walls of the xylem elements, such as the cellulose and lignin components of vessel walls and tracheid walls. B. Cohesion and adhesion work together synergistically to maintain an unbroken, continuous water column within the xylem tissue. This is crucial for the ascent of sap against gravity. The cohesive force between water molecules is extremely strong, with a magnitude of approximately three hundred and fifty atmospheres, which is far more than sufficient to support the ascent of sap even in the tallest trees. The adhesive forces help anchor this water column to the xylem walls, preventing it from breaking apart. Together, these forces enable water to be pulled upward through the xylem from the roots to the leaves, driven by the transpiration pull created at the leaf surface. Without these cohesive and adhesive properties of water, the ascent of sap would not be possible.

3III. 3 Mark Questions10 questions
Q.36Compare and Contrast Diffusion & Osmosis.v
Answer:

Diffusion
Osmosis
1. 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
It is a special type of diffusion – There is movement of water or solvent molecules through a selectively permeable membrane from a place of its higher concentration to its lower concentration until an equilibrium is attained.
2. it is independent of the living system
It is also independent of the living system
3. Passive process
Passive process
4. Obvious in solids gases & liquids Only in liquid molecules Eg. diffusion of sugar in water
Eg. Dry grapes, when kept in water swells, & becomes turgid.

Q.37Differentiate between osmotic pressure it and osmotic potentialv
Answer:

Osmotic pressure
Osmotic potential
1. The hydrostatic pressure developed in a solution. due to the presence of dissolved solutes when it is separated from a pure solvent by a semi-permeable membrane.
The ratio between the number of solvent particles and the number of solute particles in a solution or (lowering of free energy of water in a system due to the presence of solute particles
2. develops only in a confined system.
develops in confined or an open system
3. The value is positive, though it is numerically equal to osmotic potential
The value is negative though it is numerically opposite to osmotic pressure.

Q.38Do you have an R.O. Purifier ¡n your house? Explain the principle behind it.v
Answer:
  • Yes / No – R.O. is working on the principle of osmosis. but in the reverse direction.
  • In regular osmosis water moves from its higher concentration to its lower concentration through the selectively permeable membrane but here water moves from lower concentration to higher concentration through selectively permeable membrane.
  • Since against concentration gradient, there is the expenditure of energy, to apply pressure, to force water in a reverse direction.
  • Eg- Desalination plants to purify seawater also work like R-O-Purifiers Movement of Water in house hold usage.
Q.39Define Antitranspirant.v
Answer:

An antitranspirant is any material or substance applied to the surface of plants that functions to retard or reduce the rate of transpiration without significantly disturbing the essential processes of gaseous exchange required for respiration and photosynthesis. Antitranspirants work by creating a barrier on the leaf surface that reduces water vapor loss while still allowing carbon dioxide and oxygen to diffuse through for metabolic processes. Common examples of antitranspirants include colorless plastics, silicone oil, and low viscosity waxes. These substances are particularly useful in agriculture and horticulture to reduce water stress in plants during periods of drought or high temperature, to minimize transplant shock, and to protect plants from desiccation. The application of antitranspirants is an important management strategy for conserving water in plants while maintaining their physiological functions.

Q.40What are the inducers of stomatal closure.v
Answer:

Stomatal closure is a critical physiological response in plants, primarily regulated to conserve water, especially under stress conditions. Various factors and substances can induce this closure. Natural anti-transpirants are substances that naturally occur in plants and help reduce water loss by inducing stomatal closure. For example, an increase in carbon dioxide (CO2) concentration within the leaf can inhibit photorespiration and directly induce stomatal closure, as the plant senses a sufficient CO2 supply for photosynthesis. Additionally, certain chemicals, when applied as a foliar spray, can effectively induce stomatal closure for an extended period, typically lasting two to three weeks. Prominent examples of such chemical anti-transpirants include Phenyl Mercuric Acetate (PMA) and Abscisic Acid (ABA). ABA, often referred to as the stress hormone, plays a crucial role in mediating plant responses to drought stress by signaling stomatal closure to minimize water loss through transpiration. These inducers help plants cope with water scarcity and other environmental challenges.

Q.41Fill in the blanks in the tabulations given below The Study Year Scientist associated with it 1. The concept of water potential 1960 …………………………………. 2. Active and Passive absorptions 1949 ………………………………… 3. Pulsation theory 1923 …………………………………….v
Answer:

1) Slatyer & Taylor 2) Kramer 3) J.C. Bose

Q.42Nature of membrane Definition Example 1. Impermeable 1. …………………………….. suberized. cutinizedcell walls 2. ………………………………. Allow diffusion of solvent molecules, do not allow the passage of solute molecules Parched paper 3. Selectively permeable biomembranes allow some solutes to pass in addition to solvent molecules 3. ………………………………….v
Answer:

1) Inhibit the movement of both solvent and solute molecules
2) Semipermeable
3) Tonoplast & plasmalemma

Q.43Explain the capillary theory of Boehm (1809).v
Answer:

The capillary theory of Boehm, proposed in eighteen hundred and nine, suggested that xylem vessels function as capillary tubes. According to this theory, the capillarity of the xylem vessels under normal atmospheric pressure is responsible for the ascent of sap in plants. However, this theory was subsequently rejected by the scientific community for several important reasons. First, the magnitude of capillary force is insufficient to raise water to the heights observed in tall trees; capillary action alone can only raise water to a limited height determined by the vessel diameter and surface tension of water. Second, the xylem vessels are relatively broad and wide, whereas tracheids, which are narrower, actually conduct more water. This observation contradicts the capillary theory, which would predict greater water transport in narrower vessels where capillary forces would be stronger. Third, the theory does not account for the role of transpiration pull and root pressure in driving water transport. These limitations led to the rejection of the capillary theory in favor of more comprehensive explanations involving the cohesion-tension theory and root pressure.

Q.44Explain Phloem loading?v
Answer:

Phloem loading is the process by which the products of photosynthesis, primarily sucrose, are actively transported from the mesophyll cells of leaves into the sieve elements of the phloem for long-distance transport to other parts of the plant. This process is analogous to loading manufactured goods, like cement sacks, onto a vehicle for transportation to their respective destinations. Phloem loading involves several intricate steps. Step I begins in the chloroplasts, where photosynthates are initially produced in the form of starch or triose phosphate. These photosynthates are then transported to the cytoplasm of the mesophyll cells, where triose phosphate is converted into sucrose. Sucrose is the primary form in which sugars are transported in the phloem due to its non-reducing nature, which makes it less reactive during transport. Subsequent steps involve the movement of this sucrose from the mesophyll cells into the companion cells and then into the sieve tube elements, often against a concentration gradient, requiring metabolic energy.

Q.45Explain the theory of photosynthesis in guard cells observed by Von Mohl with its demerits.v
Answer:

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 in the night leads to the closure of stomata.
Demerits:
* The chloroplast of guard cells is poorly developed and incapable of performing photosynthesis.
* The guard cells already possess much amount of stored sugars.
IV. 5 Mark Questions

4III. Trans – Membrane Route7 questions
Q.46Draw & Explain the structure of Stomata.v
Answer:

Stomata are minute pores found predominantly on the epidermis of leaves and, to a lesser extent, on green stems. These pores are crucial for gas exchange, allowing the uptake of carbon dioxide for photosynthesis and the release of oxygen and water vapor during transpiration. The dimensions of stomata typically range from 10 to 40 micrometers in length and 3 to 10 micrometers in breadth. Mature leaves can contain a high density of stomata, ranging from 50 to 500 stomata per square millimeter. Structurally, each stoma is surrounded by a pair of specialized epidermal cells called guard cells. In dicotyledonous plants, these guard cells are typically kidney-shaped or semilunar, while in monocotyledonous plants, they are often dumbbell-shaped. The guard cells enclose a small opening, which is the stoma itself. Surrounding the guard cells are other epidermal cells, sometimes specialized, known as subsidiary cells or accessory cells. These subsidiary cells assist the guard cells in their function. A distinctive feature of guard cells is that their inner wall, facing the stomatal pore, is thicker and less elastic compared to their outer wall. The stomatal pore opens into an interior substomatal cavity, which is an air space within the leaf parenchyma, facilitating efficient gas exchange.

Q.47Explain osmosis by Potato osmoscope Experiment.v
Answer:

Aim: To demonstrate osmosis by Potato osmoscope
Apparatus used: Potato tuber, beaker containing water, sugar solution and pin.
Definition:
Diffusion of water or solvent from the region of higher water potential to a region of lower water potential
is known as osmosis.
Procedure:
Take a peeled potato tuber and make a cavity inside with the help of a knife fill the cavity with concentrated sugar solution and mark the initial level.
Place this set up in a beaker containing pure water After 10 minutes observe the sugar solution level and record your observation.
Observation:
There is rise in the level of the solution. in the cavity of the tuber due to osmosis
Inference: Osmosis has occured, through the potato osmoscope

Q.48Measure Transpiration with Ganong’s Photometerv
Answer:

Aim: To measure the rate of Transpiration with Ganong’s Potometer
Apparatus needed: Ganong’s Potometer, a twig, beaker, water, split rubber cork, and vaseline.
Procedure:
* Ganong’s Potometer is a horizontal graduated tube which is bent in opposite directions at the ends.
* A reservoir is fixed to the horizontal tube hear the wider end Reservoir has stop cock to regulate water flow.
* A twig is fixed to the wider arm through the split cork. The apparatus is filled with water with water from reservoir.
* The apparatus is made air tight by applying vaseline.
* The other bent end of the horizontal tube is dipped into a beaker containing coloured water.
An air bubble is introduced into the graduated tube at the narrow end. Keep the apparatus in bright sunIght
and observe
Observation:
As the twig transpires, the air bubble move towards the twig.
This loss is compensated by water ohsorption from the beaker.
inference:
By the experiment we can study the rate of Transpiration and rate of transpiration is equal to the rate of water absorption.

Q.49Explain Mechanism of Translocation by Munch Mass flow Hypothesisv
Answer:

Munch – Proposed it in 1930 Crafts – elaborated it in 1938
Definition: Organic substances (solute) move from a region of high osmotic pressure (mesophyll) to
region of low OP along TP gradient.
Example – Physical system:
Chamber ‘A’ & chamber ‘B’ made up of semi permeable membrane connected by a tube ‘T’
A – Contain highly concentrated sugar solution (hypertonic)
B – Contain dilute sugar solution (hypotonic)
A – draws water from the reservoir by Endosmosis – TP of chamber ‘A’ increased
* Continuous entry of water in to A – TP increased
* Flow of solute from chamber A to B thro TP gradient.
* The movement continues till both Aand B attain isotonic condition (equilibrium)
(However if new sugar solution added to A system will start to run again)
Example (Biological system)
* Chamber A (Source) – (Equivalent to) – Mesophyll cells of leaves (High concentration of soluble food)
* Chamber B (Sink) – (Equivalent to) – Cells of stem & Roots (Consumption end)
* TubeT – (Analogous to) – Sieve tube to phloem
Steps:
1. Xylem (Reservoir) – Movement of water (Endomosis) – Mesophyll cells (TP increase)
2. Mesophyll cells (High TP) Source – enmass movement of organic solutes through Phloem by TP Gradient – Cells of stem & Root (low TP) (Sink)

Q.50Explain the theory of K + transport – or Explain the mechanism of stomatal movementv
Answer:

Introduction:
Levit (1974) – Proposed it
Raschke (1975) – Elaborated it
Steps:
This process of exchange of ions is called Actie ion exchange ( consume ATP) or Energy
* Increased K + ions in the Guard cells – balanced by CP ions
* Increase in solute concentration (Hypertonic) Decrease in water potential
* Water enters into Guard cells from subsidiary cells
* Wall pressure increase Turgor pressure, Turgid guard cells – fall apart & opens the stoma
* Exit of H +
* Intake of K +
* Exit of K +
* Loss of H 2 O
* Uptake of H 2 O +
* Turgidity of Guard Cells
* Accumulation of CO 2 – Lowering of pH
* Opening of Stoma.
* Activation of ABA
* Closure of Stoma.

Q.51Explain Cytochrome Pump Theory (or) Explain Carrier concept of Active Absorption, through cytochrome Pump theory.v
Answer:

Lunde gardth & Burstom (1933)- Proposed the Cytochrome
Pump theory:
* There is correlation between Respiration & Anion absorption.
* when a plant is transferred from water to salt solution, the rate of respiration increases – known as Anion respiration – or salt respiration
The Assumptions of Cytochrome pump theory:
* The mechanism of anion and cation absorption is different.
* Anion – absorption – through cytochrome pump or chain by Active process
* An oxygen gradient is responsible for oxidation at outer surface of the membrane and reduction at the inner surface.
Explanation:
* On the inner surface, the enzyme dehydrogenase Produces protons (W) and electrons (e)
* Anions are picked up by oxidized cytochrome oxidase and transferred to the other members of the chain.
* Theory assumes the passive movement of cations (C + ) along the electrical gradient created by the accumulation of anions (A – ) at the inner surface of the membrane.
Defects:
*
* Cations also induce respiration
* to fail to explain the selective uptake of ions
* It explains absorption of anions only.

Q.52Explain the opening and closing of stomata by a starch – sugar – Interconversion theory.v
Answer:

i) Lloyd (1908)
According to him, turgidity of Guard cell is due to interconversion of starch → sugar
* Day time:
Guard cells have sugar → so turgid → opening of stomata
* Nighttime:
Guard cells have starch → so loose turgidity (become flaccid) → closure of stomata
ii) Sayre (1920)
According to him, the pH of Guard cell determine opening and closing of stomata
* Day time: Guard cells have high pH →so turgid → opening of stomata
* Nighttime: Guard cells have low pH → become flaccid → closure of stomata to be elaborate
* Day time: Utilisation of CO 2. in photosynthesis → Starch into sugar → high pH → high Turgor pressure→Opening of Stomata
* Night Time: No photosynthesis, so the accumulation of CO 2 → sugar to starch → low pH → decrease in TP → closure of stomata
iii) Hanes (1940)
According to Hanes – Enzyme phosphorylase is responsible for starch sugar conversion in the guard cells.

5IV. 5 Mark Questions1 questions
Q.53Explain ‘routes’ of Water Absorption in the roots.v
Answer:

Introduction
* Root hair & other epidermal cells – By imbibition absorb water from soil –
* By osmosis moves radically & centripetally – across
* cortex
* Endodermis
* Pencycle & Xylem
There are 3 Routes
* Apoplast
* Symplast
* Transmembrane route
I. Apoplast ( GK – Apo – Away) Everything external to PM
1. Cell walls
2. Extra Cellular Space
3. Interior of dead cells (vessel elements Tracheids)
Movement is continuous exclusively through the cell wall or nonliving part of the plant without crossing any membrane.
II. Symplast (GK – Sym = within)
Entire mass of cytosol of all the living cells in a plant + plasmo desmata + inter connecting cytoplasmic channel.
In the movement water has to cross PM, to enter cytoplasm of outer root cell; then move within adjoining
cytoplasm through plasmodesmata around the vacuoles without the necessity to cross more membrane it reaches xylem.
III. Trans – Membrane Route
* Water enters a cell on one side and exits from the other side.
* It crusses 2 membranes for each cell (also through to no plast).

6Part II56 questions
Q.54In plants, cell to cell transport is aided by:v
  1. (a) diffusion alone
  2. (b) osmosis alone
  3. (c) imbibition alone
  4. (d) all the three above
Answer:

(d) all the three above

Q.55The 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) imbibitionv
Answer:

c. Simple diffusion

Q.56Which 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) onlyv
Answer:

(b) (i) and (iii) only

Q.57Solute potential is also known as a) Water potential b) Pressure potential c) Osmotic potential d) Maic potentialv
Answer:

c. Osmotic potential

Q.58The swelling of dry seeds is due to a phenomenon called:v
  1. (a) osmosis
  2. (b) transpiration
  3. (c) imbibition
  4. (d) none of the above
Answer:

(c) imbibition

Q.59Cell 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 barsv
Answer:

a. -14 bars

Q.60The 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 = l0atm, TP = 6atm c) Cell X: Op =-l0atm, TP = 4atm d) CeIlY: OP = -Katm, TP = 4atm The net movement of water shall be from a) A toB and X to Y b) A to B and Y toX c) B to A and X to Y d) B to A and Y to Xv
Answer:

To determine the net movement of water, we need to calculate the Water Potential (Ψ) for each cell using the formula Ψ = OP + TP, where OP is Osmotic Potential (often given as negative, representing solute potential Ψs) and TP is Turgor Potential (Ψp). Water moves from a region of higher water potential to a region of lower water potential. For Cell A: OP = -10 atm, TP = 4 atm. So, ΨA = -10 + 4 = -6 atm. For Cell B: OP = -10 atm, TP = 6 atm. So, ΨB = -10 + 6 = -4 atm. Since ΨB (-4 atm) is higher than ΨA (-6 atm), water will move from B to A. For Cell X: OP = -10 atm, TP = 4 atm. So, ΨX = -10 + 4 = -6 atm. For Cell Y: OP = -8 atm, TP = 4 atm. So, ΨY = -8 + 4 = -4 atm. Since ΨY (-4 atm) is higher than ΨX (-6 atm), water will move from Y to X. Therefore, the net movement of water shall be from B to A and Y to X.

Q.61Water 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 & IVv
Answer:

a) I & II

Q.62………………………. 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 potentialv
Answer:

b. Pressure potential

Q.63Diffusion Pressure Deficit (DPD) was termed by Meyer in:v
  1. (a) 1928
  2. (b) 1828
  3. (c) 1936
  4. (d) 1938
Answer:

(d) 1938

Q.64Imbibants present in plants are generally a) Hydrothermic b) Hydrostatic c) Hydrophilic d) Hydrophobicv
Answer:

c. Hydrophilic

Q.65Kramer (1949) recognised two distinct mechanisms, which independently operate in the absorption of water in plants are:v
  1. (a) osmosis and diffusion
  2. (b) imbibition and diffusion
  3. (c) diffusion and absorption
  4. (d) active absorption and passive absorption
Answer:

(d) active absorption and passive absorption

Q.66Root pressure is totally apsent in Gymnosperms because a) Trachea absent b) Tracheids absent c) Trees are tall d) Trees are comparatively shortv
Answer:

a. Trachea absent

Q.67When 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.v
Answer:

(c) there is a decrease in the rate of respiration and also a decrease in the rate of absorption of water. Respiratory inhibitors such as potassium cyanide and chloroform block the electron transport chain and inhibit aerobic respiration, leading to a significant decrease in the production of ATP. Since the absorption of water and mineral ions by root cells is an active transport process that requires energy in the form of ATP, the inhibition of respiration results in decreased availability of ATP. This causes a corresponding decrease in the rate of water absorption by the roots. Therefore, when respiratory inhibitors are applied, both respiration and water absorption rates decrease simultaneously.

Q.68Find the DPD in a flaccid cell if its OP is 10 a) 20 b) 30 c) 10 d) 40v
Answer:

c.10

Q.69Pulsation theory was proposed by:v
  1. (a) Strasburger
  2. (b) Godsey
  3. (c) J.C. Bose
  4. (d) C.V. Raman
Answer:

(c) J.C. Bose

Q.70When 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 pIasrnoysed d) Will does not show any changev
Answer:

d. Will does not show any change

Q.71Indicate 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)v
Answer:

The correct answer is (d) (i) and (iv). Statement (i) is correct because root pressure is indeed absent in gymnosperms. Statement (iv) is also correct as there is no direct relationship between the ascent of sap and root pressure. Root pressure alone cannot account for the ascent of sap to great heights in tall trees, and sap can rise even in the absence of root pressure. Statement (ii) is incorrect because root pressure does occur in some angiosperms, particularly herbaceous plants and young seedlings. Statement (iii) is incorrect because while root pressure may contribute to sap movement in some cases, it is not the primary mechanism responsible for the ascent of sap in plants.

Q.72I) Water potential – A) Turgor pressure II) Solute potential – B) Osmotic potential + Pressure potential III) Matric potential – C) Osmotic potential IV) Pressure potential – D) Imbibition pressurev
Answer:

b) B C D A

Q.73I) Leaves – A) Antitransport II) Seed – B) Transpiration III) Roots – C) Negative osmotic potential IV) Aspirin – D) Imbibition V) Plasmolyced cell – E. Absorptionv
Answer:

a) C B D E A

Q.74I) 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. Uniportv
Answer:

c) C D A B

Q.75I) Passive transport – A) Uphill transport II) Active transport – B) Short distance transport HI) Cell to cell transport – C) Long-distance transport IV) Ascent of sap – D) Downhill transportv
Answer:

b) D A B C

Q.76The 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 respectivelyv
Answer:

The correct answer is (c) about 10–40μ and 3–10μ respectively. Stomata are microscopic pores found on the leaf surface, and their dimensions are typically measured in micrometers. The length of stomata generally ranges from approximately 10 to 40 micrometers, while the breadth or width ranges from about 3 to 10 micrometers. These measurements can vary slightly depending on the plant species and environmental conditions, but option (c) represents the most accurate general range for stomatal dimensions in most plants.

Q.77A membrane that permits the solvent and not the solute to pass through it is termed is a) Permeable, b) impermeable c) semipermeable d) differentially permeablev
Answer:

c. Semi permeable

Q.78Who did observe that stomata open in light and close in the night:v
  1. (a) Unger
  2. (b) Sachs
  3. (c) Boehm
  4. (d) Von Mohl
Answer:

(d) Von Mohl

Q.79The phosphorylase enzyme in guard cells supports the starch-sugar interconversion theory. The above reaction is:v
  1. (a) oxidation reaction
  2. (b) hydrolyses reaction
  3. (c) reduction reaction
  4. (d) none of the above
Answer:

(b) hydrolyses reaction

Q.80If a cell kept in a solution of unknown concentration gets deplasmolysed the solution is a) hypotonic b) hypertonic c) isotonic d) detonicv
Answer:

a. hypotonic

Q.81A 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 exosmosisv
Answer:

d. water will leave by exosmosis

Q.82Phenyl 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 permanentlyv
Answer:

The correct answer is (a) induces partial stomatal closure for two weeks. Phenyl Mercuric Acetate (PMA) is a chemical compound that acts as a stomatal closure inducer when applied as a foliar spray to plants. When PMA is applied, it causes a partial closure of the stomata, reducing the size of the stomatal aperture. This effect is temporary and lasts for approximately two weeks, after which the stomata gradually return to their normal functioning. This property makes PMA useful in research studies examining stomatal physiology and water loss in plants.

Q.83The osmotic pressure of cell sap is maximum in a) Hydrophytes b) Halophytes c) Xerophytes d) Mesophytesv
Answer:

b. Halophytes

Q.84Say 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, CO 2, NH3 metalloid & ROS IV) The carrier proteins structure does not get modified due to its association with the moleculesv
Answer:

d. False True True False

Q.85I) Hypertonic is a strong solution (low solvent/high solute/ low Ψ ) II) Hypotonic is a weak solution (high solvent/low or zero solutes/ high Ψ) III) Hypertonic is the weak solution (high solvent/low or zero solutes/high Ψ) IV) Hypotonic is a strong solution (low solvent / high solute/low Ψ)v
Answer:

Let's evaluate each statement regarding hypertonic and hypotonic solutions based on their solute concentration, solvent concentration, and water potential (Ψ). I) Hypertonic is a strong solution, meaning it has a low solvent concentration, a high solute concentration, and consequently a low water potential (Ψ). This statement is True. II) Hypotonic is a weak solution, implying it has a high solvent concentration, a low or zero solute concentration, and therefore a high water potential (Ψ). This statement is True. III) Hypertonic is the weak solution, which contradicts the definition of a hypertonic solution. A hypertonic solution is strong, not weak. This statement is False. IV) Hypotonic is a strong solution, which also contradicts the definition of a hypotonic solution. A hypotonic solution is weak, not strong. This statement is False. Therefore, the correct evaluation is True, True, False, False.

Q.86From sieve elements sucrose is translocated into sink organs such as root, tubers etc and this process is termed as:v
  1. (a) Xylem unloading
  2. (b) Xylem uploading
  3. (c) Phloem unloading
  4. (d) Phloem uploading
Answer:

(c) Phloem unloading

Q.87The 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, Ill & IV d) I, II & IVv
Answer:

a. I, II & III

Q.88Hydathodes are generally present in plants that grow in:v
  1. (a) dry places
  2. (b) moist and shady places
  3. (c) sunny places
  4. (d) deserts
Answer:

(b) moist and shady places

Q.89Why 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.v
Answer:

a. It is inactive and highly soluble

Q.90Unloading of pholem at sink includes a) Passive transport b) diffusio c) Osmosis d) Active transportv
Answer:

d. Active transport

Q.91The liquid coming out of the hydathode of grasses is:v
  1. (a) pure water
  2. (b) not pure water
  3. (c) a solution containing a number of dissolved substances
  4. (d) saltwater
Answer:

(c) a solution containing a number of dissolved substances

Q.92In a flaccid cell a) DPD = OP b) DPD = TP c) TP = OP d) OP = Ov
Answer:

a. DPD = OP

Q.93The pathway of water movement involving living part of a cell is a) Apoplast pathway b) symplast pathway c) Transmembrane pathway d) Lateral conductionv
Answer:

b. Symplast pathway

Q.94The 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 abovev
Answer:

a. upward movement of the water plants

Q.95High tensile strength of water is due to a) Adhesion only b) cohesion only c) Both (a) and (b) d) None of thesev
Answer:

c. Both (a) and (b)

Q.96Maximum transpiration occur in a) Mesophytes b) Xerophytes c) Hydrophytes d) Epiphytesv
Answer:

a. Mesophytes

Q.97Supply ends in transport of solutes are a) green leaves b) root and stem c) xylem and phloem d) Hormones and enzymesv
Answer:

c. Xylem and phloem

Q.98For guttation in plants, the process responsible is a) Root pressure b) Atmospheric pressure c) Imbibition d) None of thesev
Answer:

a. Root pressure

Q.99Which 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 theoryv
Answer:

b. Pulsation theory

Q.100Which of the following plant material is an efficient water imbibant? a) Lignin b) Pectin c) Cellulose d) Agarv
Answer:

d. Agar

Q.101Which of the following helps in the Ascent of sap? a) Root pressure b) Transpiration c) Capillarity d) All the abovev
Answer:

d. All the above

Q.102In a girdled plant which of the following dies first? a) Shoot b) root c) Both die simultaneously d) None – the plant survivesv
Answer:

b. root

Q.103Assertion:-A Imbibition is also diffusion Reason -R The movement of water in the above process is along a concentration gradient. a) Both A and Rare 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 Rare falsev
Answer:

a) Both A and R are True and R is correct explanation of A

Q.104Assertion: – 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 onlyv
Answer:

d) Both A and R are false

Q.105Assertion: – 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 imbibatev
Answer:

b) Both A and R are true but R is not the correct explanation of A

Q.106Assertion: – A In phloem loading, food is transported to the sink Reason – R Food is transported from source to sink ‘v
Answer:

d) Both Assertion ‘A’ and Reason ‘R’ are false

Q.107Assertion – A: Xylem a principal water conducting ’ Reason -R: It has been recognised by girdling or ringing experimentsv
Answer:

a) Both A and R are True R is the correct explanation of A

Q.108Assertion: – A In phloem, sugar are translocated in non reducing form Reason – R Non reducing sugars are most reactive sugarsv
Answer:

c) Assertion is true but Reason is false

Q.109Assertion: AIn 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 ’v
Answer:

d) Both A and R are false
II. Two Mark Questions