Class 11 Bio Botany · Chapter 11

Samacheer Class 11 Bio Botany - Transport in Plants

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Chapter-wise textbook exercise answers for Transport in Plants with validation-aware solutions.

Answers marked verified were checked during generation against the chapter context and source question text.
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
Solution

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

d) i, ii, iii, iv

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
Solution

(b) Lenticular

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
Solution

a) Influx of K +

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
Solution

(b) ranslocation of food due to TP

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
Solution

High salt concentration results in high be osmotic potential of the soil solution, so the plant has to use more energy to absorb water. Under extreme salinity conditions, plants may be unable to absorb water and will wilt even if the surrounding soil is thoroughly irrigated. This is also referred to as the osmotic or water deficit effect of salinity.

Answer:

High salt concentration results in high be osmotic potential of the soil solution, so the plant has to use more energy to absorb water. Under extreme salinity conditions, plants may be unable to absorb water and will wilt even if the surrounding soil is thoroughly irrigated. This is also referred to as the osmotic or water deficit effect of salinity.

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

* 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 and the opening takes place during the night.
Day:
1.
2. Photosynthesis occur
3. pH – increased
4. Movement of water from
5. subsidiary cells to guard cells
6. Guard cells become turgid
7. Opening of stomata
Night:
1.
2. No Photosynthesis
3. pH – lowered
4. Movement of water from guard cells
5. Guard cells become flaccid
6. Closure of stomata

Answer:

* 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 and the opening takes place during the night.
Day:
1.
2. Photosynthesis occur
3. pH – increased
4. Movement of water from
5. subsidiary cells to guard cells
6. Guard cells become turgid
7. Opening of stomata
Night:
1.
2. No Photosynthesis
3. pH – lowered
4. Movement of water from guard cells
5. Guard cells become flaccid
6. Closure of stomata

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

The non-photosynthetic parts of a plant that need a supply of sucrose:
* Roots
* Tubers
* Developing fruits and
* Immature leaves.

Answer:

The non-photosynthetic parts of a plant that need a supply of sucrose:
* Roots
* Tubers
* Developing fruits and
* Immature leaves.

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

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.

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
Solution

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.

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.

Q.11What is osmosisv
Solution

It is a special type of diffusion almost same like simple diffusion but has a selectively permeable membrane is here, through which osmosis occur.
(OR)
It is the movement of water molecules from a place of its higher concentration, to the place of its lower concentration through a semipermeable membrane.

Answer:

It is a special type of diffusion almost same like simple diffusion but has a selectively permeable membrane is here, through which osmosis occur.
(OR)
It is the movement of water molecules from a place of its higher concentration, to the place of its lower concentration through a semipermeable membrane.

Q.12Define the term diffusion.v
Solution

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.

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.

Q.13The touch me plant closes its leaves at the touch – Explain.v
Solution
  • 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.
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
Solution

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.

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.

Q.15Define water potentialv
Solution
  • 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
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
Solution
  • 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.
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
Solution

SDT
LDT
1. Cell to cell Transport
Involve few cells ni lateral direction
Transport with in the network of xylem and
phloem
2. Connecting link to xylem bind phloem from root hairs to leaf tissues
Direct vertical – main Transport
3. Eg. Diffusion, Osmosis etc
4. Eg. Ascent of sap & Translocation of solutes.

Answer:

SDT
LDT
1. Cell to cell Transport
Involve few cells ni lateral direction
Transport with in the network of xylem and
phloem
2. Connecting link to xylem bind phloem from root hairs to leaf tissues
Direct vertical – main Transport
3. Eg. Diffusion, Osmosis etc
4. Eg. Ascent of sap & Translocation of solutes.

Q.18Differentiate between Passive & Active Transportv
Solution

PT
AT
1. Down hill Transport (Phyical)
Up hill Transport (Biological)
2. Occur According to concentration gradient
Occur against concentration gradient
3. No expenditure of energy
There is expenditure of energy obtained from Respiration
4. Eg. Diffusion – Facilitated Diffusion osmosis etc.
Eg. Na + K + ATP are pump.

Answer:

PT
AT
1. Down hill Transport (Phyical)
Up hill Transport (Biological)
2. Occur According to concentration gradient
Occur against concentration gradient
3. No expenditure of energy
There is expenditure of energy obtained from Respiration
4. Eg. Diffusion – Facilitated Diffusion osmosis etc.
Eg. Na + K + ATP are pump.

Q.19Give two examples of the phenomenon of Imbibition.v
Solution

two examples for the phenomenon of Imbibition:
* The swelling of dry seeds.
* The swelling of wooden windows, tables, doors due to high humidity during the rainy season.

Answer:

two examples for the phenomenon of Imbibition:
* The swelling of dry seeds.
* The swelling of wooden windows, tables, doors due to high humidity during the rainy season.

Q.20Explain carbonic Acid Exchange theory.v
Solution
  • 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.
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
Solution

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.

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
Solution

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

Answer:

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

Q.23Define Root Pressure.v
Solution
  • 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.
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
Solution

Osmosis (Latin: Osmos – impulse, urge) is a special type of diffusion. It 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).

Answer:

Osmosis (Latin: Osmos – impulse, urge) is a special type of diffusion. It 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).

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

Name
Type
Properties
1.Starch
Polysaccharide (non reducing sugar)
Insoluble in water cann’t be transport
2. Glucose & Fructose
Monosaccharides (reducing sugar)
Soluble in water but less efficient in energy storage & reactive
3. Sucrose
Disaccharide(non reducing sugar)
Soluble in water, even at high concentration, low viscosity, more efficient in energy storage no reducing ends make it inert than glucose & Fructose.

Answer:

Name
Type
Properties
1.Starch
Polysaccharide (non reducing sugar)
Insoluble in water cann’t be transport
2. Glucose & Fructose
Monosaccharides (reducing sugar)
Soluble in water but less efficient in energy storage & reactive
3. Sucrose
Disaccharide(non reducing sugar)
Soluble in water, even at high concentration, low viscosity, more efficient in energy storage no reducing ends make it inert than glucose & Fructose.

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

Three types of plasmolysis occur in plants:
* Incipient plasmolysis
* Evident plasmolysis
* Final plasmolysis.

Answer:

Three types of plasmolysis occur in plants:
* Incipient plasmolysis
* Evident plasmolysis
* Final plasmolysis.

Q.27Identify the diagram and Neatly label the partsv
Solution

The given diagram is the structure of Hydathode
A-Guard cell
B-Epithem
C-Tracheids

Answer:

The given diagram is the structure of Hydathode
A-Guard cell
B-Epithem
C-Tracheids

Q.28Identify the Diagram & Label the parts.v
Solution

The given diagram explain Reverse osmosis
A – Pressure
B – Pure water
C – Saltwater
D – Membrane

Answer:

The given diagram explain Reverse osmosis
A – Pressure
B – Pure water
C – Saltwater
D – Membrane

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

Ascent of sap
Translocation of solute
The upward transport of water along with dissolved minerals from roots to the aerial parts is called as Ascent of sap.
The transport of food from the site of synthesis to the site of utilization or from source to sink is known as Translocation of organic solutes (a dissolved substance)
Occur through Xylem
Occur through Phloem

Answer:

Ascent of sap
Translocation of solute
The upward transport of water along with dissolved minerals from roots to the aerial parts is called as Ascent of sap.
The transport of food from the site of synthesis to the site of utilization or from source to sink is known as Translocation of organic solutes (a dissolved substance)
Occur through Xylem
Occur through Phloem

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

Two objections to starch – sugar interconversion theory:
* In monocots, the guard cell does not have starch.
* There is no evidence to show the presence of sugar at a time when starch disappears and stomata open.

Answer:

Two objections to starch – sugar interconversion theory:
* In monocots, the guard cell does not have starch.
* There is no evidence to show the presence of sugar at a time when starch disappears and stomata open.

Q.31Differentiate between cuticular and Lenticular Transpiration.v
Solution

Cuticular Transpiration
Lenticular Transpiration
Loss of water through cuticle is known as cuticular Transpiration
Some pores
It is only about 5 to 10% of the total Transpiration
Present on the woody surface of stem (bark) are known as Lenticels
The thicker the cuticle, the lesser will be the Transpiration. Eg. xerophytes
The loss of water from the lenticel is Lenticular Transpiration – It is only about 0.1 % of the total.

Answer:

Cuticular Transpiration
Lenticular Transpiration
Loss of water through cuticle is known as cuticular Transpiration
Some pores
It is only about 5 to 10% of the total Transpiration
Present on the woody surface of stem (bark) are known as Lenticels
The thicker the cuticle, the lesser will be the Transpiration. Eg. xerophytes
The loss of water from the lenticel is Lenticular Transpiration – It is only about 0.1 % of the total.

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

Two uses of anti – transpirants:
* Anti – transpirants reduce the enormous loss of water by transpiration in crop plants.
* Useful for seedling transplantations in nurseries.

Answer:

Two uses of anti – transpirants:
* Anti – transpirants reduce the enormous loss of water by transpiration in crop plants.
* Useful for seedling transplantations in nurseries.

Q.33Give notes an Aquaporin.v
Solution

* Water pore – Aquaporin in KBC was discovered by Peter Agre (Nobel Prize for chemistry – 2003)
* Water channel protein is present in PM.
* Regulate the massive amount of water transport across PM
* 30 types of Aquaporins are known from maize
They also transporter
* glycerol
* urea
* CO 2
* NH
* metalloids & Reactive oxygen species (ROS)
Function:
* They increase the permeability of the membrane of water
* They confer drought and salt, stress tolerance.

Answer:

* Water pore – Aquaporin in KBC was discovered by Peter Agre (Nobel Prize for chemistry – 2003)
* Water channel protein is present in PM.
* Regulate the massive amount of water transport across PM
* 30 types of Aquaporins are known from maize
They also transporter
* glycerol
* urea
* CO 2
* NH
* metalloids & Reactive oxygen species (ROS)
Function:
* They increase the permeability of the membrane of water
* They confer drought and salt, stress tolerance.

Q.34Define the term Ion – Exchange.v
Solution

Ions of external soil solution are exchanged with the same charged (anion for anion or cation for cation) ions of the root cells.

Answer:

Ions of external soil solution are exchanged with the same charged (anion for anion or cation for cation) ions of the root cells.

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

A.
Cohesion
Adhesion
The strong mutual attraction between water molecules is called cohesion or cohesive force.
The Attraction between a water molecule and the wall of the xylem element is called adhesion.
B. The cohesive and Adhesive forces work together to form an unbroken continuous water column in xylem.
The magnitude of cohesive force is much high (350 atm) and is more than enough to ascent sap in the tallest
trees.
III. 3 Mark Questions

Answer:

A.
Cohesion
Adhesion
The strong mutual attraction between water molecules is called cohesion or cohesive force.
The Attraction between a water molecule and the wall of the xylem element is called adhesion.
B. The cohesive and Adhesive forces work together to form an unbroken continuous water column in xylem.
The magnitude of cohesive force is much high (350 atm) and is more than enough to ascent sap in the tallest
trees.
III. 3 Mark Questions

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

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.

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
Solution

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.

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
Solution
  • 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.
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
Solution

Antitranspirant is any material applied to plants to retard or reduce the rate of transpiration – without disturbing the process of gaseous exchange, for respiration and photosynthesis.
Eg. Colourless plastics silicone oil and low viscosity waxes.

Answer:

Antitranspirant is any material applied to plants to retard or reduce the rate of transpiration – without disturbing the process of gaseous exchange, for respiration and photosynthesis.
Eg. Colourless plastics silicone oil and low viscosity waxes.

Q.40What are the inducers of stomatal closure.v
Solution

* Natural antitranspirants usually induce stomatal closure
Eg. CO 2 – inhibits photorespiration – thereby induces stomata! closure
* Some chemicals, when applied as a foliar spray can induce stomatal closure for 2 – 3 weeks.
Eg. (PMA) Phenyl Mercuric Acetate & (ABA) Abscisic Acid.

Answer:

* Natural antitranspirants usually induce stomatal closure
Eg. CO 2 – inhibits photorespiration – thereby induces stomata! closure
* Some chemicals, when applied as a foliar spray can induce stomatal closure for 2 – 3 weeks.
Eg. (PMA) Phenyl Mercuric Acetate & (ABA) Abscisic Acid.

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
Solution

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

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
Solution

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

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
Solution

Capillary theory: Boehm (1809) suggested that the xylem vessels work like a capillary tube. This capillarity of the vessels under normal atmospheric pressure is responsible for the ascent of sap. This theory was rejected because the magnitude of the capillary force can raise water level only up to a certain height. Further, the xylem vessels are broader than the tracheid which actually conducts more water and against the capillary theory.

Answer:

Capillary theory: Boehm (1809) suggested that the xylem vessels work like a capillary tube. This capillarity of the vessels under normal atmospheric pressure is responsible for the ascent of sap. This theory was rejected because the magnitude of the capillary force can raise water level only up to a certain height. Further, the xylem vessels are broader than the tracheid which actually conducts more water and against the capillary theory.

Q.44Explain Phloem loading?v
Solution

Definition:
The products of Photosynthesis from the Mesophyll of leaves to sieve elements of phloem is known as phloem loading. (Just like the cement sack manufactured in a factory being loaded in a vehicle to be transported the respective site)
It involves 3 steps.
Step I:
* The chloroplast has photosynthate in the form of starch or Trlose phosphate
* It is transported to the cytoplasm, where it is converted into Sucrose.

Answer:

Definition:
The products of Photosynthesis from the Mesophyll of leaves to sieve elements of phloem is known as phloem loading. (Just like the cement sack manufactured in a factory being loaded in a vehicle to be transported the respective site)
It involves 3 steps.
Step I:
* The chloroplast has photosynthate in the form of starch or Trlose phosphate
* It is transported to the cytoplasm, where it is converted into Sucrose.

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

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

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
Solution

1. Definition:
The epidermis of leaves and green stems possess many small pores called – Stomata
2. Length & Breadth
The length – 10- 40μ The Breadth – 3 – 10μ
Number Mature leaves contain 50- 500 stomata / mm 2
3. Structure
a. Guard Cells – A pair of Kidney shaped cells (semilunar) surrounded a small opening called stoma
b. Subsidiary Cells – Guard cells attached to surrounding epidermal cells known as subsidiary cells or accessory cells.
* The inner wall of guard cell is thicker
* The stoma open into an interior substomatal cavity.

Answer:

1. Definition:
The epidermis of leaves and green stems possess many small pores called – Stomata
2. Length & Breadth
The length – 10- 40μ The Breadth – 3 – 10μ
Number Mature leaves contain 50- 500 stomata / mm 2
3. Structure
a. Guard Cells – A pair of Kidney shaped cells (semilunar) surrounded a small opening called stoma
b. Subsidiary Cells – Guard cells attached to surrounding epidermal cells known as subsidiary cells or accessory cells.
* The inner wall of guard cell is thicker
* The stoma open into an interior substomatal cavity.

Q.47Explain osmosis by Potato osmoscope Experiment.v
Solution

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

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
Solution

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.

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
Solution

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)

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
Solution

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.

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
Solution

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.

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
Solution

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.

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
Solution

* 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).

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
Solution

(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
Solution

c. Simple diffusion

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
Solution

(b) (i) and (iii) only

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
Solution

c. Osmotic potential

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
Solution

(c) imbibition

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
Solution

a. -14 bars

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
Solution

d. B to A and Y to X

Answer:

d. 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
Solution

a) I & II

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
Solution

b. Pressure potential

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
Solution

(d) 1938

Answer:

(d) 1938

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

c. Hydrophilic

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
Solution

(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
Solution

a. Trachea absent

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
Solution

(c) there is a decrease in the rate of respiration and also a decrease in the rate of absorption of water.

Answer:

(c) there is a decrease in the rate of respiration and also a decrease in the rate of absorption of water.

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

c.10

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
Solution

(c) J.C. Bose

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
Solution

d. Will does not show any change

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
Solution

(d) (i) and (iv)
Match The Following & Find Out The Correct Order

Answer:

(d) (i) and (iv)
Match The Following & Find Out The Correct Order

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
Solution

b) B C D A

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
Solution

a) C B D E A

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
Solution

c) C D A B

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
Solution

b) D A B C

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
Solution

(c) about 10 – 40μ and 3 – 10μ respectively

Answer:

(c) about 10 – 40μ and 3 – 10μ respectively

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
Solution

c. Semi permeable

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
Solution

(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
Solution

(b) hydrolyses reaction

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
Solution

a. hypotonic

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
Solution

d. water will leave by exosmosis

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
Solution

(a) induces partial stomatal closure for two weeks.

Answer:

(a) induces partial stomatal closure for two weeks.

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

b. Halophytes

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
Solution

d. False True True False

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
Solution

b. True True False False

Answer:

b. 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
Solution

(c) Phloem unloading

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
Solution

a. I, II & III

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
Solution

(b) moist and shady places

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
Solution

a. It is inactive and highly soluble

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
Solution

d. Active transport

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
Solution

(c) a solution containing a number of dissolved substances

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
Solution

a. DPD = OP

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
Solution

b. Symplast pathway

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
Solution

a. upward movement of the water plants

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
Solution

c. Both (a) and (b)

Answer:

c. Both (a) and (b)

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

a. Mesophytes

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
Solution

c. Xylem and phloem

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
Solution

a. Root pressure

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
Solution

b. Pulsation theory

Answer:

b. Pulsation theory

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

d. Agar

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
Solution

d. All the above

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
Solution

b. root

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
Solution

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

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
Solution

d) Both A and R are false

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
Solution

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

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
Solution

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

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
Solution

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

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
Solution

c) Assertion is true but Reason is false

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
Solution

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

Answer:

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