Selina ICSE Class 10 Biology Solutions Chapter 3 Absorption By Roots The Process Involved

Multiple Choice Questions (MCQs)

Question 1: Absorption of water by the plant cells by surface attraction is called:
(a) Diffusion
(b) Osmosis
(c) Imbibition
(d) Endosmosis

Question 2: A plant cell placed in a certain solution got plasmolyzed. What was the kind of solution?
(a) Isotonic sugar solution
(b) Hypotonic salt solution
(c) Hypertonic salt solution
(d) Isotonic salt solution

Question 3: The state of a cell in which the cell wall is rigid and stretched by the increase in volume due to the absorption of water is called.
(a) Flaccidity
(b) Turgidity
(c) Capillarity
(d) Tonicity

Question 4: Which one of the following is a characteristic NOT related with the suitability of the roots for absorbing water?
(a) Tremendous surface area
(b) contain cell sap at a higher concentration than the surrounding soil water
(c) Root hairs have thin cell walls
(d) Grow downward into the soil

Question 5: Movement of molecules of a substance from the region of their higher concentration to the region of their lower concentration without the involvement a separating membrane, is called
(a) Osmosis
(b) Diffusion
(c) active transport
(d) Capillarity

Question 6: Osmosis and diffusion are the same except that osmosis there is:
(a) a freely permeable membrane
(b) a cell wall in between
(c) a selectively permeable membrane in between
(d) an endless inflow of water into a cell

Question 7: The highest water potential (capacity to move out higher concentrated solution) is that of
(a) Pure water
(b) 10% salt solution
(c) Honey
(d) 50% sugar solution

Question 8: The space between the cell wall and plasma membrane in a plasmolysed cell is filled with.
(a) isotonic solution
(b) hypotonic solution
(c) Hypertonic solution
(d) water

Question 9: What is responsible for guttation?
(a) Osmotic pressure
(b) Root pressure
(c) suction pressure
(d) Capillarity

Question 10: The most appropriate characteristic of a semipermeable membrane is that
(a) it has minute pores
(b) it has no pores
(c) it allows the solute to pass through but not the solvent
(d) it allows a solvent to pass through freely but prevents the passage of the solute.

Very Short Answer Type Questions:

Question 1: Name the following:
(a) The condition of a cell placed in a hypotonic solution.
(b) The process by which intact plants lose water in the form of droplets from leaf margins
(c) The process by which water enters root hairs.
(d) The term for the inward movement of solvent molecules through the plasma membrane of a cell.
(e) The process by which molecules distributes themselves evenly within the space they occupy
(f) The pressure which is responsible for the movement of water molecules across the cortical cells of the root.
(g) The pressure which is responsible for the movement of water molecules across the cortical cells of the root.
Solution 1:
(a) Turgidity
(b) Guttation
(c) Osmosis
(d) Xylem
(e) Endosmosis
(f) Diffusion
(g) Root pressure

Question 2: Give the equivalent terms for the following:
(a) Pressure of the cell contents on the cell wall
(b) The condition in which the cell contents are shrunken
(c) Loss of water through a cut stem
Solution 2:
(a) Turgor pressure
(b) Flaccidity
(c) Bleeding

Question 3: Complete the following statements:
(a) Hypotonic solution is one in which the solution kept outside the cell has lower solute concentration than the cell.
(b) Active transport is one in which the ions outside the roots are ……………
(c) The bending movements of certain flowers towards the sun and the sleep movements of certain plants at night are examples of…………
Solution 3:
(a) the fluids inside
(b) transported inside against their concentration gradient
(c) turgor movements

Question 4: (a) When placed in a more concentrated solution, the cell contents will………… (shrink / swell up)
(b) The pressure by which the Molecules tend to cross the semi-permeable membrane is called osmotic pressure. (salt / water)
(c) Active transport is in a direction to that of diffusion. (opposite / Same)
Solution 4:
(a) Shrink
(b) Water
(c) opposite

Question 5: Match the items in column I with those in column II

Column I	Column II
Xylem	(i) semi – permeable
Phloem	(ii) permeable
Cell membrane	(iii) downward flow of sap
Root pressure	(iv) upward flow of water
Cell wall	(v) guttation

Solution 5:

Column I	Column II
Xylem	(iv) upward flow of water
Phloem	(iii) downward flow of sap
Cell membrane	(i) semi-permeable
Root pressure	(v) guttation
Cell wall	(ii) permeable

Short Answer Type Questions:

Question 1: Differentiate between the following:
(a) Plasmolysis and deplasmolysis
(b) Turgor pressure and wall pressure
(c) Guttation and bleeding
(d) Turgidity and Flaccidity
Solution 1:
(a)

Plasmolysis	Deplasmolysis
When a cell is placed in a hypertonic solution, it shrinks and the plasma membrane pulls away from the cell wall, causing the cytoplasm to shrink and the plasma membrane to pull away from the cell wall.	Deplasmolysis occurs when a plasmolyzed cell is put in water and the cell’s protoplasm swells up again as a result of the re-entry of water.
The cell becomes flaccid during Plasmolysis.	The cell becomes turgid as a result of deplasmolysis.

(b)

Turgor pressure	Wall pressure
The pressure of the cell contents on the cell wall is known as turgor pressure.	The cell wall exerts pressure on the cell content, which is known as wall pressure.

(c)

Guttation	Bleeding
Guttation is a condition in which water drops appear along the leaf margins as a result of excessive root pressure.	The loss of cell sap through a cut stem is known as bleeding.

(d)

Turgidity	Flaccidity
It is the condition of a cell when it can no longer hold any more water and is totally distended.	It is a condition in which the cell content has diminished and the cell is no longer tightly closed.

Question 2: (a) Mention whether the following statements are true (T) or false (F)
(i) A plant cell placed in hypotonic solution gets Plasmolysed.
(ii) Addition of salt to pickles prevents growth of bacteria because they turn turgid.
(iii) Cells that have lost their water content are said to be deplasmolysed.
(iv) Xylem is the water conducting tissue in plants.
(v) The shrinkage of protoplasm, when a cell is kept in hypotonic solution.
(vi) The cell wall of the root cell is a differentially permeable membrane.
(b) Correct the false statements by altering the last word only.
Solution 2:
(a) (i) False
(ii) False
(iii) False
(iv) True
(v) False
(vi) False
(b) (i) A plant cell placed in hypotonic solution gets turgid.
(ii) Addition of salt to pickles prevents growth of bacteria because they turn flaccid.
(iii) Cells that have lost their water content are said to be plasmolysed.
(iv) The shrinkage of protoplasm, when a cell is kept in hypertonic solution.

Question 3: What is the difference between ‘flaccid’ and ‘turgid’? Give one example of flaccid condition in plants.
Solution 3:
When the plant cell wall becomes hard and stretched by an increase in the amount of vacuoles owing to water absorption when placed in hypotonic solution, the cell is said to be turgid.
The cell, on the other hand, is said to be flaccid when the contents of the cell shrink when it is immersed in a hypertonic solution and the cell is no longer tight. Turgidity is the polar opposite of flaccidity.
In a playground, for example, weeds can be eliminated by pouring excessive salts around their base.
                                                                                                                   Or
When a plant cell is submerged for 30 minutes in a hypertonic solution such as salt solution, it becomes flaccid or limp.

Question 4: Give reasons for the following:
(a) If you sprinkle some common salt on grass growing on a lawn, it is killed at that spot.
(b) If you uproot a plant from the soil, its leaves soon wilt.
(c) It is better to transplant seedlings in a flower-bed in the evening and not in the morning.
(d) A plant cell when kept in a hypertonic salt solution for about 30 minutes turns flaccid.
(e) Potato cubes when placed in water become firm and increase in size.
Solution 4:
(a) When common salt is sprinkled on grass, it induces Plasmolysis, which causes the grass cells to die. As a result, sprinkling ordinary salt on grass growing on a lawn kills it instantly.
(b) When a plant is uprooted, the leaves continue to lose water through transpiration, but the roots absorb no more water. As a result, there is no way to compensate for the loss of water through transpiration, and the uprooted plant’s leaves quickly wilt.
(c) Seedlings are stressed during transplantation. If the seedlings are transplanted in the morning, they will be subjected to the extra stress of the hot afternoon’s heavy transpiration. Because the amount of water absorbed exceeds the amount of water lost through transpiration, transplanting in the evening allows seedlings to acclimate for a longer period of time during the night (lower temperatures). As a result, planting seedlings in a flower bed in the evening rather than the morning is preferable.
(d) In a hypertonic solution, the solute concentration outside the cell is higher than the solute concentration inside the cell. As a result, exosmosis causes water to flow out of the plant cell. The cytoplasm shrinks as the plasma membrane pulls away from the cell wall, resulting in a flaccid cell. As a result, after 30 minutes in a hypertonic salt solution, a plant cell becomes flaccid.
(e) When compared to the water in which the cubes are placed, potato cubes have an excess of salts and sugars. As a result of endosmosis, water from the environment penetrates the potato cubes, making them firmer and larger.

Question 5: Mentions whether the following statements are true (T) or false (F) and give explanation in support of your answer.
Solution 5:
(a) Plasmolysis occurs when a cell is placed in a hypertonic solution, resulting in the outflow of water from the cell, causing the cytoplasm to shrink away from the cell wall. Deplasmolysis, on the other hand, occurs when water is reintroduced into a plasmolyzed cell after it has been placed in a hypotonic solution, causing the protoplasm to inflate and press against the cell wall.

(b) Guttation is the leakage of cell sap through a severed stem, whereas bleeding is the appearance of drops of water along leaf margins due to high root pressure.

(c) In a seed, there is only one seed coat.

(d) Because the twig’s xylem remains intact, the leaves stay turgid. Xylem is important for water conduction in plants.

(e) Excessive root pressure causes guttation.
It is at its peak when root pressure is at its highest, which occurs early in the morning or late at night.
This is because transpiration is very low and water absorption is quite high at these times.

(f) Imbibition cause dry seeds to expand when submerged in water. Dry seeds absorb water and swell when they come into touch with it.

Long Answer Type Questions:

Question 1: Give two examples of turgor movements in plants.
Solution 1:
Examples of turgor movements in plants:
(i) In Mimosa pudica, a sensitive plant, the stimulus of touch causes a decrease of turgor called pulvinus at the base of the leaflets and the base of the petioles. The plant’s leaves fold and droop as a result of this.
(ii) Insectivorous plants’ leaves shut up to entrap a living prey. When an insect comes into contact with a leaf, it loses its turgor, which causes the plant’s leaves to close.
(iii) The bending of certain flowers in the direction of the sun.

Question 2: Explain the mechanism of closing and opening of the stomata.
Solution 2:
The turgidity of the guard cells determines when the stomata close and open. On the side facing the stoma, each guard cell has a thicker wall, whereas the opposing side has a thin wall. Chloroplasts are found in guard cells. The osmotic pressure of the contents of the guard cells rises as a result of glucose synthesis during photosynthesis and other chemical changes, and they take more water from neighboring cells, becoming turgid. The guard cells grow more arched outwards as a result of turgor, and the aperture between them widens, opening the stoma

The guard cells become flaccid at night or when there is a lack of water in the leaf, and their inner stiff walls become straight, sealing the stomatal hole.

Question 3: Concentration of mineral nutrient elements is higher inside the root hairs than in the surrounding soil. How do roots take them in from the soil?
Solution 3:
If the concentration of mineral nutrient elements inside the root-hairs is higher than in the surrounding soil, roots use ‘active transport’ to take them in from the soil. Mineral ions are forcibly transported from the surrounding soil, where they have a lower concentration, into the roots, where they have a higher concentration, across the cell membrane by expending energy. The cell provides this energy in the form of ATP.

Question 4: Explain how soaked seeds swell up and burst their seed coats.
Solution 4:
The seeds expand up when immersed in water due to imbibition and endosmosis. Water enters the cell via these two steps. The seed coat can no longer withstand the turgor pressure due to endosmosis, and the seed coat bursts.

Question 5: Leaves of the sensitive plant wilt and droop down on a slight touch. What mechanism brings about this change?
Solution 5:
Due to turgor movement, the sensitive plant’s leaves wilt and droop down when touched. Turgid pulvinus tissue supports the delicate plant’s petiole. The decrease of turgor at the base of the leaflets and the base of the petioles, referred to as pulvinus, is caused by the stimulus of touch. The bottom half of the pulvinus loses water, causing the petiole to collapse. The leaves wilt and droop as a result of this.

Question 6: What is transpiration pull? How is it caused?
Solution 6: Due to the tendency of water molecules to remain linked, more water molecules are drawn up as water is lost from the leaf surface through transpiration. The resulting ‘transpiration pull’ creates a continuous column of water throughout the stem. The xylem creates a negative pressure or tension that pushes water away from the roots and soil. Transpirational pull is a powerful force that causes sap to rise

Structured/ Application/ Skill Type Questions:

Question 1: The following diagram represents a plant cell after being placed in a strong sugar solution.

Guidelines 1 to 5 indicate the following:
(1) Cell wall, (2) Strong sugar solution, (3) Protoplasm (4) Large vacuole, (5) Nucleus
(a) What is the state of the cell shown in the cell shown in the diagram?
(b) Name the structure which acts as a selectively permeable membrane.
(c) If the cell had been placed in distilled water instead of strong sugar solution, which feature would not have been seen?
(d) Name any one feature of this plant cell which is not present in an animal cell.
Solution 1:
(a) The cell is flaccid, which means it has been plasmolyzed.
(b) Plasma Membrane
(c) There would be no plasmolysis and no flaccidity, i.e. the protoplasm would not have shrunk away from the cell wall.
(d) Animal cells lack a cell wall.

Question 2: A leaf cell of a water plant was placed in a liquid other than pond water. After sometime, it assumed a shape as shown below.

(a) Give the term for the state of the cell it has acquired.
(b) Comment on the nature (tonicity) of the liquid surrounding the cell.
(c) Redraw in the space provided, the diagram of the cell is it is soon placed in ordinary water for some time.
Solution 2:
(a) Flaccid Cell
(b) The liquid is hypertonic solution. The solute concentration outside the cell is higher than the solute concentration inside the cell.
(c)

Question 3: The diagram given below represents an experimental set – up to demonstrate a certain process. Study the same and answer the questions that follow:

(a) Name the process.
(b) Define the above named process.
(c) what would you observe in the experimental set-up after an hour or so?
what control experiment can be set up for comparison?
(e) Keeping in mind the root- hair, cell and its surroundings, name the parts that correspond to (1) concentrated sugar solution (2) parchment paper and (3) water in the beaker.
(f) Name any other substance that can be used instead of parchment paper in the above experiment.
Mention two advantages of the process to the plants.
Solution 3:
(a) Osmosis
(b) Osmosis is the passage of water molecules from a more dilute solution (with a lower solute concentration) to a less dilute solution over a semi-permeable membrane (with a higher solute concentration).
(c) After about an hour, the sugar solution level in the thistle funnel will rise and the water level in the beaker will drop significantly.
(d) The water will be contained in the beaker for the control experiment. At the same time, the thistle funnel with the cellophane paper tied on its mouth and inverted in the beaker will contain water instead of the sugar solution.
(e)
(1) Concentrated sugar solution – Cell sap (of higher concentration than that of the surrounding water) within the root hair.
(2) Parchment paper- cell membrane of root hair.
(3) Water in the beaker – water in soil.
(f) cellophane paper, egg membrane, animal bladder (any one)
(i) Osmosis allows plant roots to absorb water and nutrients from the surrounding soil.
(ii) Osmosis is a process that allows plants to absorb water from the earth, allowing them to keep cells alive in their roots, stems, and leaves.
Osmosis also has a role in the opening and closing of stomata, which is crucial for processes like transpiration and photosynthesis. (any two)

Question 4: The diagram below represents a layer of epidermal cells showing a fully grown root hair. Study the diagram and answer the questions that follow:

(a) Name the parts labelled A, B, C and D.
(b) The root hair cell is in a turgid state. Name and explain the process that caused this state.
(c) Mention one distinct difference between the parts labelled A and B.
(d) Draw a diagram of the above root hair cell as it would appear when a concentrated solution of fertilizers is added near it.
Solution 4:
(a) A – Cell wall
B – Cell membrane
C – Cytoplasm
D – Nucleus
(b) A root hair gets turgid because of the absorption of water from the surrounding. Absorption of water by root hair is achieved by the process of osmosis. The concentration of water in the surrounding is more than that of the interior of the cell; this causes the water from the surrounding to move in because of endosmosis.
(c)

Cell wall	Cell membrane
A root hair’s cell wall is porous, allowing both salt and water to pass through.	A root hair’s cell membrane is semi-permeable, allowing only small dissolved salt molecules to pass through.

Question 5: Two potatoes cubes each 1 cm3 in size, were placed separately in two containers (A&B), the container (A) having water and the other (B) containing concentrated sugar solution. After 24 hours when the cubes were examined, those placed in water were found to be firm and had increased slightly in size and those placed in concentrated sugar solution were found to be soft and had somewhat decreased in size. Use the above information to answer the questions that follow:
(a) Account for the firmness and increase in the size of the potato cubes placed in water.
(b) Account for the softness and decrease in size of the potato cubes which were places in sugar solution.
(c) Name and define the physical process being investigated in this experiment.
Solution 5:
(a) Because water is hypotonic to potato cells, endosmosis occurs and water enters the cells. The protoplasm swells to the point that it presses on the cell wall. The cells are turgid, or totally distended. When the potato cubes are placed in water, this allows them to firm up and grow in size.
(b) The sugar solution is hypertonic to the potato cells, causing exosmosis and water to leak out of them. The cytoplasm shrinks and the plasma membrane pulls away from the cell wall, giving the potato cell a less bloated appearance. The cells become flaccid or floppy. When the potato cubes are placed in sugar solution, they become soft and shrink in size.
(c) The procedure being researched is osmosis. Osmosis is the process of moving water molecules from a dilute (lower solute concentration) solution to a less dilute solution through a semi-permeable membrane (with a higher solute concentration).

Question 6: Given below is the diagrammatic representation of the transverse section of a part of a plant. Study it and answer the questions that follow:

(a) Name the part of the plant that is shown
(b) Label the parts 1 to 6
(c) Write the functions of parts 3 and 5
Solution 6:
(a) It’s a diagrammatic cross-section of a root segment.
(b) 1 – Root hair
2 – Epidermis
3 – Cortex
4 – Endodermis
5 – Phloem
6 – Xylem
(c) The ground tissue, cortex (label 3), is involved in the intake of water and minerals. It aids in the storage of photosynthetic products as well.
Phloem (label 5) aids in the transportation of prepared food from the leaves to various areas of the plant.

Question 7: Study the diagram given below and answer the questions that follow:

(a) Name the process being studied in the above experiment
(b) Explain the process mentioned in (a) above
(c) why is oil placed over water?
Solution 7:
a) The osmosis process of water absorption by plant roots is being investigated.
b) A root-hair includes cell sap, which has a higher salt concentration than outside soil water. Osmosis is triggered as a result of this differential, and outside water diffuses into the root hair. Water travels from the root-hair-bearing cell to the next cells and then to the xylem vessels.
c) To avoid water loss due to evaporation, the water’s surface was coated with oil.

Question 8:
Show by a series of diagrams, the change which a plant cell will undergo when placed in
(a) Hypertonic salt solution and
(b) Hypotonic salt solution
Solution 8:

(a)

(b)