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Selina Concise Biology Class 10 ICSE Solutions Cell Cycle, Cell Division and Structure of Chromosomes

November 10, 2023May 14, 2022 by Veerendra

Selina Concise Biology Class 10 ICSE Solutions Cell Cycle, Cell Division and Structure of Chromosomes

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 2 Cell Cycle, Cell Division and Structure Of Chromosomes. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Biology Chapter 2 Cell Cycle, Cell Division and Structure of Chromosomes

Exercise 1

Solution A.1.
(b) DNA and Histones

Solution A.2.
(c) Coloured bodies

Solution B.1.
(a) – Nucleotides.
(b) – Nucleosome.
(c) – Hydrogen Bond.
(d) – Phosphate, Sugar and Nitrogenous base.

Solution C.1.
Chromatin fibre is unfolded, uncondensed, extended DNA. It is only visible when cell under goes division whereas chromosomes are condensed DNA and they are visible when the cell is divided.

Solution C.2.
Rungs of DNA ladder is made of nitrogenous bases which includes Adenine (A), Guanine (G), Cytosine (C) and Thymine (T).

Solution C.3.
(a) The four nitrogenous bases in the DNA ladder are Guanine, Thymine, Adenine and Cytosine.
(b) Genes are specific sequences of nucleotides on a chromosome.
(c) A nucleotide is composed of a phosphate, sugar (pentose) and a nitrogenous base.
(d) Nucleosomes are groups of histone molecules surrounded by DNA strands.
(e) If there are 46 chromosomes in a cell there will be 46 chromatin fibres inside the nucleus during interphase.

Solution D.1.
Nucleosome is basic structural unit of DNA. Each strand of DNA winds around a core of eight histone molecules. This core can be imagined like a football, around which a long rope is wound with one or two loops. Each such complex structure is called a nucleosome. A single human chromosome may have about a million nucleosomes.

Solution D.2.
Gene is a structural and functional unit of heredity and variations. Genes are specific sequences of nucleotides on a chromosome that encode particular proteins which express in the form of some particular feature of the body. In other words, gene is the DNA segment of the chromosome and it controls the expression of characteristics.

Solution E.1.
(a) 2
(b) 2 on each strand
(c) 1- Phosphate, 2- Sugar, 3- Bases, 4- Hydrogen Bond, 5 – Base
(d)Nucleotide

Solution E.2.
B, C and A.

Exercise 2

Solution A.1.
(c) both ovary and testis

Solution A.2.
(c) Anaphase, telophase

Solution A.3.
(c) DNA

Solution B.1.
Cell A: 2
Cell B: 4

Solution B.2.
(a) – Metaphase.
(b) – Telophase.
(c) – Prophase.
(d) – Anaphase.

Solution B.3.
(a) Somatic (body).
(b) Four.
(c) Reproductive.
(d) 23 and 23.
(e) Haploid.
(f) Centriole.

Solution C.1.

(a) A chromosome is an organized structure of DNA and protein found in cells. It is a single piece of coiled DNA containing many genes, regulatory elements and other nucleotide sequences whereas a chromatid is one of the two copies of DNA making up a duplicated chromosome, which are joined at their centromeres, for the process of cell division (mitosis or meiosis).

(b) The centrosome is an area in the cell where microtubules are produced. Within an animal cell centrosome, there is a pair of small organelles called the centrioles. During animal cell division, the centrosome divides and the centrioles replicate (make new copies) whereas each chromosome in its condensed form consists of two chromatids joined at some point along the length. This point of attachment is called centromere.

(c) An aster is a cellular structure shaped like a star, formed around each centrosome during mitosis in an animal cell whereas spindle fibers are aggregates of microtubules that move chromosomes during cell division.

(d) A haploid cell is a cell that contains one complete set of chromosomes. Gametes are haploid cells that are produced by meiosis whereas a diploid cell is a cell that contains two sets of chromosomes. One set of chromosomes is donated from each parent.

Solution C.2.
In this statement, reduction means that the number of chromosomes are reduced to half i.e. out of the 23 pairs of chromosomes in humans, only single set of chromosomes are passed on to the sex cells.

Solution C.3.
Gametes must be produced by meiosis for sexual reproduction because the numbers of chromosomes are reduced to half during meiosis and then the normal diploid numbers of chromosomes are regained during the process of fertilization.

Solution C.4.

(a) F; Surface skin cells are continuously lost and replaced by the underlying cells.

(b) T; All types of human cells, have 46 chromosomes. The only type of cell which does not have 46 chromosomes are the sex cells, which have only half of the number, so they have 23 chromosomes. The egg cell is a sex cell (found in female). So it must have 23 chromosomes.

(d) T; Mitotic cell division can be a mode of asexual reproduction in unicellular organisms like amoeba or yeast cell which divides into two daughter cells.

(e) T; While the maternal and paternal chromosomes are separating, the chromatid material gets exchanged between the two members of a homologous pair resulting in genetic recombination.

Solution D.1.
a.

Centromere
Spindle fibres
Chromatids

b. The stage described in the diagram is the late anaphase of mitosis in an animal cell. The stage can be identified by the presence of separated chromatids which are found at the two poles of the cell. The appearance of the furrow in the cell membrane classifies the stage as the late anaphase.
c. The division is mitotic division and this kind of cell division occurs in all the cells of the body except for the reproductive cells.
d. The stage before anaphase is metaphase.
Selina Concise Biology Class 10 ICSE Solutions Cell Cycle, Cell Division and Structure of Chromosomes image - 1

Solution D.2.

Selina Concise Biology Class 10 ICSE Solutions Cell Cycle, Cell Division and Structure of Chromosomes image - 2

Solution D.3.
The exchange of chromatids between homologous chromosomes is called crossing-over. This is the process by which the two chromosomes of a homologous pair exchange equal segments with each other.
Crossing over occurs in the first division of meiosis. At that stage each chromosome has replicated into two strands called sister chromatids. The two homologous chromosomes of a pair synapse, or come together. While the chromosomes are synapsed, breaks occur at corresponding points in two of the non-sister chromatids, i.e., in one chromatid of each chromosome.
Since the chromosomes are homologous, breaks at corresponding points mean that the segments that are broken off contain corresponding genes, i.e., alleles. The broken sections are then exchanged between the chromosomes to form complete new units, and each new recombined chromosome of the pair can go to a different daughter sex cell. It results in recombination of genes found on the same chromosome, called linked genes that would otherwise always be transmitted together.

Solution D.4.

(a) Late prophase. Because the nuclear membrane and nucleolus have disappeared.
(b) Centrioles.
(c)

Centromere
Chromatids.
Spindle fibre.

(d) Metaphase. The centromeres of chromosomes are drawn to the equator by equal pull of two chromosomal spindle fibres that connects each centromere to the opposite poles, forming a metaphasic plate.

(e)

Mitosis	Meiosis
(i) Two daughter cells are produced.	(i) Four daughter cells are produced.
(ii) It is equational division i.e. the number of chromosome in the daughter cells or parent cells remains the same.	(ii) It is reductional division i.e. the number of chromosomes is reduced to half in the daughter cells.

Solution D.5.
(a) Metaphase.
(b) 4.
(c) A – Animal
B – Animal
C – Plant
(d) (iv)

Solution D.6.
(a) This is an animal cell because:

The outline is circular (in plants it would be angular {rectangular or polygonal}) and cell wall is absent.
Centrosomes on centrioles are present. (These are found only in animal cells)

(b) Mitosis.
(c) B, C, D, A.
(d) Interphase.
(e)
Selina Concise Biology Class 10 ICSE Solutions Cell Cycle, Cell Division and Structure of Chromosomes image - 3

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Selina Concise Biology Class 10 ICSE Solutions Genetics Some Basic Fundamentals

November 10, 2023May 14, 2022 by Veerendra

Selina Concise Biology Class 10 ICSE Solutions Genetics- Some Basic Fundamentals

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 3 Genetics Some Basic Fundamentals. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Biology Chapter 3 Genetics – Some Basic Fundamentals

Exercise 1

Solution A.1.
d) Ascaris

Solution A.2.
a) 3 : 1

Solution B.1.
(a) – (iii) Study of laws of inheritance of characters
(b) – (v) Chromosomes other than the pair of sex chromosomes
(c) – (iv) A gene that can express when only in a similar pair
(d) – (ii) The alternative forms of a gene
(e) – (i) Chromosomes similar in size and shape

Solution B.2.
Lion, tiger, domestic cat (Any two)

Solution B.3.
Colour-blindness, Thalassaemia, Sickle cell anaemia and Haemophilia (Any two)

Solution B.4.
Homozygous dominant – RR
Homozygous recessive – rr

Solution C.1.

(a)

Phenotype	Genotype
The observable Characteristic which is genetically controlled is called phenotype.	The set of genes present in the cells of an organism is called its genotype.

(b)

Character	Trait
Any heritable feature is called a character.	The alternative form of a character is called trait.

(c)

Monohybrid cross	Dihybrid cross
It is a cross between two pure breeding parent organisms with different varieties taking into consideration the alternative trait of only one character.	It is a cross between two pure breeding parent organisms with different varieties taking into consideration the alternative trait of two characters.

Solution C.2.
The characteristics of a species such as physical appearance, body functions and behavior are not only the outcome of chromosome number, but these depend on the genotype of every organism. That means the set of genes present in the organisms may very and therefore lion, tiger and domestic cat have the same number of 38 chromosomes, their characteristics (like different appearances) are the result of the genes located on the chromosomes.

Solution C.3.

Character	Dominant trait	Recessive trait
Flower Colour	Purple	White
Seed Colour	Yellow	Green
Seed Shape	Round	Wrinkled
Pod Shape	Inflated	Constricted
Flower Position	Axial	Terminal

(Any 3)

Solution C.4.

Colour-blindness is caused due to recessive genes which occur on the X chromosome.
Males have only one X chromosome. If there is recessive gene present on X chromosome, then the male will suffer from colour-blindness.
Females have two X chromosomes. It is highly impossible that both the X chromosomes carry abnormal gene. Hence, if one gene is abnormal and since it is recessive, its expression will be masked by the normal gene present on the other X chromosome. Females are unlikely to suffer from colour-blindness.

Solution C.5.
Phenotypic Ratio – 3 (Black Fur) :1 (Brown Fur)
Genotypic Ratio – 1(Homozygous Black Fur):2 (Heterozygous Black Fur): 1 (Homozygous Brown Fur)

Solution D.1.

(a) Heterozygous: The condition in which a pair of homologous chromosomes carries dissimilar alleles for a particular character.

For example –

A daughter (XX^o) from a normal homozygous mother for colour vision (XX) and a colour blind father has one normal and one defective allele (X^oY).
Certain tongue rollers are heterozygous with Rr genotype.

(b) Homozygous: The condition in which a pair of homologous chromosomes carries similar alleles for a particular character.

For example –

A colorblind daughter (X^oX^o) will have both the X chromosomes with defective alleles.
A non-roller will have rr (homozygous) genotype.

(c) Pedigree Chart: A pedigree chart is a diagram that shows the occurrence and appearance or phenotypes of a particular gene or organism and its ancestors from one generation to the next. In the pedigree chart, males are shown by squares and females by circles.

Solution D.2.
Mendel’s laws of inheritance are:

Law of Dominance: Out of a pair of contrasting characters present together, only one is able to express itself while the other remains suppressed. The one that expresses is the dominant character and the one that is unexpressed is the recessive one.
Law of Segregation : The two members of a pair of factors separate during the formation of gametes. The gametes combine together by random fusion at the time of zygote formation. This law is also known as ‘law of purity of gametes’.
Law of Independent Assortment: When there are two pairs of contrasting characters, the distribution of the members of one pair into the gametes is independent of the distribution of the other pair.

Solution D.3.

- The sex of the child depends on the father. The egg contains only one X chromosome, but half of the sperms contain X-chromosome whereas the other half contains Y-chromosome. It is simply a matter of chance as to which category of sperm fuses with the ovum and this determines whether the child will be male or female.
- If the egg fuses with X-bearing sperm, the resulting combination is XX and the resulting child is female.
- If the egg fuses with Y-bearing sperm, the resulting combination is XY and the resulting child is male.

Solution E.1.

Selina Concise Biology Class 10 ICSE Solutions Genetics Some Basic Fundamentals image - 2

Solution E.2.
(a) Black
(b) No

Solution E.3.

Selina Concise Biology Class 10 ICSE Solutions Genetics Some Basic Fundamentals image - 3

Solution E.4.
(a) Father
(b) Two sons and three daughters
(c) The child 1 (daughter) is colour blind
(d) X chromosome
(e) Haemophilia

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Selina Concise Biology Class 10 ICSE Solutions Absorption by Roots : The Processes Involved

November 10, 2023May 14, 2022 by Veerendra

Selina Concise Biology Class 10 ICSE Solutions Absorption by Roots : The Processes Involved

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 4 Absorption by Roots : The Processes Involved. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Biology Chapter 4 Absorption by Roots : The Processes Involved

Exercise 1

Solution A.1.
c) Imbibition

Solution A.2.
c) Hypertonic salt solution

Solution A.3.
b) Turgidity

Solution A.4.
d) Grow downward into the soil

Solution A.5.
b) Diffusion

Solution A.6.
c) a selectively permeable membrane in between

Solution A.7.
a) Pure water

Solution A.8.
d) water

Solution A.9.
b) Root pressure

Solution A.10.
d) it allows a solvent to pass through freely but prevents the passage of the solute

Solution B.1.
(a) Turgidity
(b) Guttation
(c) Osmosis
(d) Xylem
(e) Endosmosis
(f) Diffusion
(g) Root pressure

Solution B.2.
(a) Turgor pressure
(b) Flaccidity
(c) Bleeding

Solution B.3.
(a) the fluids inside
(b) transported inside against their concentration gradient
(c) turgor movements

Solution B.4.
(a) shrink
(b) water
(c) opposite

Solution B.5.

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

Solution C.1.
(a)

Plasmolysis

Deplasmolysis

1. It refers to the shrinkage of the cytoplasm and withdrawal of the plasma membrane from the cell wall caused due to the withdrawal of water when placed in a hypertonic solution.

2. In Plasmolysis, the cell becomes flaccid.

1.Deplasmolysis is the recovery of a plasmolysed cell when it is placed in water, wherein the cell’s protoplasm again swells up due to the re-entry of water.

2. In deplasmolysis, the cell becomes turgid.

(b)

Turgor pressure	Wall pressure
Turgor pressure is the pressure of the cell contents on the cell wall.	Wall pressure is the pressure exerted by the cell wall on the cell content.

(c)

Guttation	Bleeding
Guttation is the process by which drops of water appear along leaf margins due to excessive root pressure.	Bleeding is the loss of cell sap through a cut stem.

(d)

Turgidity	Flaccidity
1. It is the state of a cell in which the cell cannot accommodate any more water and it is fully distended.	1. It is the condition in which the cell content is shrunken and the cell is not tight.

Solution C.2.
(a)

False
False
False
True
False
False

(b)

A plant cell placed in hypotonic solution gets turgid.
Addition of salt to pickles prevents growth of bacteria because they turn flaccid.
Cells that have lost their water content are said to be plasmolysed.
The shrinkage of protoplasm, when a cell is kept in hypertonic solution.

Solution C.3.
The cell is said to be turgid when the plant cell wall becomes rigid and stretched by an increase in the volume of vacuoles due to the absorption of water when placed in hypotonic solution. On the other hand, the cell is said to be flaccid when the cell contents get shrunken when the cell is placed in hypertonic solution and the cell is no more tight. Flaccidity is the reverse of turgidity.

Example: Weeds can be killed in a playground by sprinkling excessive salts around their base.
Or
A plant cell when immersed in hypertonic solution like salt solution for about 30 minutes will become flaccid or limp.

Solution C.4.

(a) Common salt when sprinkled on the grass causes the Plasmolysis of grass cell ultimately leading them to death. Hence, if we sprinkle some common salt on grass growing on a lawn, it is killed at the spot.

(b) If a plant is uprooted, the leaves continue losing water by transpiration, but there is no more water absorbed the roots. This does not allow the compensation for the loss of water by transpiration; hence the leaves of the uprooted plant wilt soon.

(c) Transplantation causes stress to the seedlings. If the seedlings are transplanted in the morning, they would have to immediately bear the additional stress of excessive transpiration occurring during the hot afternoon. Transplantation in the evening helps the seedlings to adjust for a longer time during the night (cooler temperatures) because the quantity of water absorbed exceeds the loss of water through transpiration. Therefore, it is better to transplant seedling in a flower bed in the evening and not in the morning.

(d) In a hypertonic solution, the solution outside the cell has higher solute concentration than the fluids inside the cell. Therefore, water flows out from the plant cell due to exosmosis. The cytoplasm shrinks and the plasma membrane withdraws away from the cell wall and this the cell becomes flaccid. Hence a plant cell when kept in a hypertonic salt solution for about 30 minutes turns flaccid.

(e) Potato cubes contain excess of salts and sugars as compared to the water in which the cubes are placed. Hence, due to endosmosis, water from the surrounding enters the potato cubes making them firm and increasing their size.

Solution C.5.

(a) True.
Plasmolysis occurs due to outflow of water from the cell when placed in hypertonic solution due to which the cytoplasm shrinks away from the cell wall. On the other hand, deplasmolysis is the result of the re-entry of water into the plasmolysed cell when placed in hypotonic solution due to which the protoplasm again swells up pressing tight against the cell wall.

(b) False.
Guttation is the process by which drops of water appear along leaf margins due to excessive root pressure whereas bleeding is the loss of cell sap through a cut stem.

(d) False.
The leaves of the twig remain turgid since its xylem is intact and xylem is responsible for water conduction in plants.

(e) False.
Guttation occurs due to excessive root pressure. It is maximum when root pressure is maximum which occurs in the early mornings or at night. This is because during these times, transpiration is very low and water absorption is very high.

(f) False.
Dry seeds when submerged in water swell up due to imbibitions. On contact with water dry seeds imbibe water and swell up.

Solution D.1.
Examples of turgor movements in plants:

In Mimosa pudica, a sensitive plant, the stimulus of touch leads to loss of turgor at the base of the leaflets and at the base of the petioles called pulvinus. This causes the folding and drooping of leaves of the plant.
The leaves of insectivorous plants close up to entrap a living prey. When the insect come in contact with the leaf, it loses it turgor hence closing the leaves of the plant.
The bending movements of certain flowers towards the sun.
(Any two)

Solution D.2.
The closing and opening of the stomata depends on the turgidity of the guard cells. Each guard cell has a thicker wall on the side facing the stoma and a thin wall on the opposite side. Guard cells contain chloroplasts. As a result of the synthesis of glucose during photosynthesis and some other chemical changes, the osmotic pressure of the contents of the guard cells increases and they absorb more water from the neighbouring cells, thus becoming turgid. Due to turgor, the guard cells become more arched outwards and the aperture between them widens, thereby opening the stoma.
Selina Concise Biology Class 10 ICSE Solutions Absorption by Roots The Processes Involved image - 1
At night or when there is shortage of water in the leaf, the guard cells turn flaccid and their inner rigid walls become straight, thus closing the stomatal aperture.

Solution D.3.
If the concentration of mineral nutrient elements is higher inside the root-hairs than in the surrounding soil, then roots take them in from the soil by ‘active transport’. In active transport, the mineral ions are forcibly carried from the surrounding soil i.e. the region of their lower concentration into the roots i.e. the region of their higher concentration through the cell membrane by expenditure of energy. This energy is supplied by the cell in the form of ATP.

Solution D.4.
When soaked in water, the seeds swell up due to imbibition and endosmosis. During these two processes water enters the cell. Due to endosmosis, at some point, the seed coat is unable to bear the turgor pressure and hence, the seed coat bursts.

Solution D.5.
Leaves of the sensitive plant wilt and droop down on a slight touch due to turgor movement. Petiole of sensitive plant is held up by turgid pulvinus tissue. The stimulus of touch leads to loss of turgor at the base of the leaflets and at the base of the petioles i.e. pulvinus. The cells of the lower side of pulvinus lose water and the petiole collapses. This causes the wilting and drooping of the leaves.
Selina Concise Biology Class 10 ICSE Solutions Absorption by Roots The Processes Involved image - 2

Solution D.6.
As water is lost from the leaf surface by transpiration, more water molecules are pulled up due to the tendency of water molecules to remain joined i.e. cohesion. This produces a continuous column of water throughout the stem which is known as ‘transpiration pull’. A negative pressure or tension is produced in the xylem that pulls the water from the roots and soil. Transpirational pull is an important force which causes the ascent of sap.
Selina Concise Biology Class 10 ICSE Solutions Absorption by Roots The Processes Involved image - 3

Solution E.1.
(a) The cell is flaccid i.e. it is plasmolysed.
(b) Plasma Membrane
(c) Plasmolysis would not occur and flaccidity would not be seen i.e. the protoplasm would not have shrunken away from the cell wall.
(d) Cell Wall is absent in animal cell.

Solution E.2.
(a) Flaccid Cell
(b) The liquid is hypertonic solution. It has higher solute concentration outside the cell than the fluids inside the cell.
(c)
Selina Concise Biology Class 10 ICSE Solutions Absorption by Roots The Processes Involved image - 4

Solution E.3.

(a) Osmosis

(b) Osmosis is the diffusion of water molecules across a semi-permeable membrane from a more dilute solution (with a lower solute concentration) to a less dilute solution (with a higher solute concentration).

(c) After an hour or so, the level of sugar solution in the thistle funnel will rise and the level of water in the beaker will drop slightly.

(d) For control experiment, the beaker will contain the water. At the same time, instead of the sugar solution; the thistle funnel with the cellophane paper tied on its mouth and inverted in the beaker will also contain water.

(e)

concentrated sugar solution – Cell sap (of higher concentration than that of the surrounding water) within the root hair.
parchment paper – cell membrane of root hair.
water in the beaker – water in soil.

(f) cellophane paper, egg membrane, animal bladder (any one)

(g)

The roots of plants absorb water and minerals from surrounding soil due to osmosis.
Osmosis allows plants to absorb water from the soil which helps plants to keep cells alive in roots, stems and leaves.
Osmosis is also important in the opening and closing of stomata which is an important feature for the processes like transpiration and photosynthesis. (Any two)

Solution E.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.

Cell wall	Cell membrane
The cell wall of a root hair is freely permeable and allows both salt and water to pass through.	The cell membrane of a root hair is semi-permeable and does not allow large dissolved salt molecules to pass through.

Selina Concise Biology Class 10 ICSE Solutions Absorption by Roots The Processes Involved image - 5

Solution E.5.

(a) Water is hypotonic to the potato cells, due to which endosmosis occurs and water enters the potato cells. The protoplasm swells up pressing tight against the cell wall. The cells are fully distended i.e. turgid. This causes the firmness and increase in the size of the potato cubes when placed in water.

(b) Sugar solution is hypertonic to the potato cells, due to which exosmosis occurs and water flows out of the potato cells. The potato cell loses its distended appearance, the cytoplasm shrinks and the plasma membrane withdraws from the cell wall. The cells become limp or flaccid. This causes the softness and decrease in size of the potato cubes when placed in sugar solution.

(c) The process being investigated is osmosis. Osmosis is the diffusion of water molecules across a semi-permeable membrane from a more dilute solution (with a lower solute concentration) to a less dilute solution (with a higher solute concentration).

Solution E.6.
(a) It is the diagrammatic cross-section of a part of a root.

(b)

Root hair
Epidermis
Cortex
Endodermis
Phloem
Xylem

(c) Cortex (label 3) is the ground tissue and is active in the uptake of water and minerals. It also helps in storage of photosynthetic products.
Phloem (label 5) helps in transporting the prepared food from leaves to different parts of the plant.

Solution E.7.
(a) The process of water absorption by plant roots through osmosis is being studied here.

(b) A root-hair contains cell sap which contains higher concentration of salts as compared to outside soil water. This difference sets off osmosis and outside water diffuses into the root-hair. From the cell bearing root-hair, water passes into adjoining cells one after another to finally the xylem vessels.

Solution E.8.

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Selina Concise Biology Class 10 ICSE Solutions Transpiration

November 10, 2023May 14, 2022 by Veerendra

Selina Concise Biology Class 10 ICSE Solutions Transpiration

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 5 Transpiration. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Biology Chapter 5 Transpiration

Exercise 2

Solution A.1.
(a) Open stomata, dry atmosphere and moist soil

Solution A.2.
(a) increase

Solution A.3.
(b) temperature is high

Solution A.4.
(c) sunken stomata

Solution A.5.
(d) hydathodes

Solution A.6.
(d) transpiration

Solution A.7.
(d) hot, dry and windy

Solution A.8.
(b) Lenticels

Solution A.9.
(b) evaporation of water from the aerial surfaces of a plant

Solution B.1.
(a) Lenticels
(b) Guttation
(c) Potometer
(d) Nerium
(e) Ganong’s photometer
(f) Stomata and cuticle
(g) Hydathodes
(h) Guttation

Solution B.2.
(a) vapour, aerial
(b) stomata, transpiration
(c) suction, water (heat)

Solution C.1.
(a) guttation
(b) protection and reduced transpiration
(c) transpiration
(d) reduced transpiration

Solution C.2.
(i) False
(ii) True
(iii) True
(iv) False
(v) Most transpiration occurs at mid-day.
(vi) Potometer is an instrument used for measuring the rate of transpiration in green plants.

Solution C.3.
(a) Transpiration increases with the velocity of wind. If the wind blows faster, the water vapours released during transpiration are removed faster and the area surrounding the transpiring leaf does not get saturated with water vapour.

(b) When the rate of transpiration far exceeds the rate of absorption of water by roots, the cells lose their turgidity. Hence, excessive transpiration results in wilting of the leaves.

(c) Plants absorb water continuously through their roots, which is then conducted upwards to all the aerial parts of the plant, including the leaves. Only a small quantity of this water i.e. about 0.02% is used for the photosynthesis and other activities. The rest of the water is transpired as water vapour. Hence water transpired is the water absorbed.

(d) There are more stomatal openings on the lower surface of a dorsiventral leaf. More the number of stomata, higher is the rate of transpiration. Hence more transpiration occurs from the lower surface.

(e) Cork and Bark of trees are tissues of old woody stems. Bark is thick with outermost layer made of dead cells and the cork is hydrophobic in nature. These properties make them water-proof and hence they prevent transpiration.

(f) In both perspiration and transpiration, water is lost by evapouration from the body of the organism as water vapour. This evaporation reduces the temperature of the body surface and brings about cooling in the body of the organism.

(g) On a bright sunny day, the rate of transpiration is much higher than any other days. The leaves of certain plants roll up on a bright sunny day to reduce the exposed surface and thus reduce the rate of transpiration.

Solution C.4.
(a) False
Reason: Potometer is used to measure the rate of transpiration in a plant. Demonstration of transpiration occurring from the lower surface of a leaf is done by analyzing the changes in colour of pieces of dry cobalt chloride paper attached (and held in place) to the two surfaces of a leaf.

(b) True
Reason: Transpiration carried out by the large number of trees in a forest. This increases the moisture in the atmosphere and brings rain.

(c) False
Reason: Hydathodes are special pores present on the ends of leaf veins through which guttation occurs and water droplets are given out. Their openings cannot be regulated. Stomata on the other hand are minute openings in the epidermal layer of leaves through which exchange of gases as well as transpiration occurs. Water is given out as water vapour. Stomatal opening is regulated by guard cells.

(d) False
Reason: Transpiration is reduced during high atmospheric humidity. High humidity in the air reduces the rate of outward diffusion of the internal water vapour across stomata, thereby reducing the rate of transpiration.

(e) True
Reason: Desert plants need to reduce transpiration as much as possible so as to survive in the hot and dry environment. Hence some of them have sunken stomata as an adaptation to curtail transpiration.

(f) True
Reason: During the day, the stomata are open to facilitate the inward diffusion of carbon dioxide for photosynthesis. During mid-day, the outside temperature is higher, due to which there is more evaporation of water from the leaves. Therefore more transpiration occurs during mid-day.

Solution C.5.

Guttation	Bleeding
It is the removal of excess of water from the plants because of excess water buildup in the plant.	It is the removal of water from the plant because of injury.
Water escapes from specialisedstructures called hydathodes.	Water escapes in the form of sap from the injured part of the plant.

Solution D.1.
Wilting refers to the loss of cellular turgidity in plants which results in the drooping of leaves or plant as a whole because of lack of water.
During noon the rate of transpiration exceeds the rate of absorption of water by roots. Due to the excessive transpiration, the cells of leaves lose their turgidity and wilt.

Solution D.2.
The lower surface of leaf is sheltered from direct sunlight. If more stomata are on the upper surface of a leaf, then excessive transpiration would occur, resulting in quick wilting of the plant. Hence most plants have more numerous stomata on the lower surface of a leaf to control the rate of transpiration.

Solution D.3.
Take the small potted rose plant and cover it with a transparent polythene bag. Tie its mouth around the base of the stem. Leave the plant in sunlight for an hour or two.
Selina Concise Biology Class 10 ICSE Solutions Transpiration image - 1
Drops of water will soon appear on the inner side of the bag due to the saturation of water vapour given out by the leaves. A similar empty polythene bag with its mouth tied and kept in sunlight will show no drops of water. This is the control to show that plants transpire water in the form of water. If tested with dry cobalt chloride paper, the drops will be confirmed as water only.

Solution D.4.
Selina Concise Biology Class 10 ICSE Solutions Transpiration image - 2
Potometer is a device that measures the rate of water intake by a plant. This water intake is almost equal to the water lost through transpiration. Potometers do not measure the water lost due to transpiration but measure the water uptake by the shoot.

Solution D.5.

Transpiration occurring through lenticels i.e. minute openings on the surface of old stems is called lenticular transpiration.
Stomatal transpiration is controlled by the plant by altering the size of the stoma, where as this does not happen in case of lenticular transpiration. This is because the lenticels never close, but remain open all the time.
The amount of stomatal transpiration is much more than the amount of lenticular transpiration.

Solution D.6.
The factors that accelerate the rate of transpiration are:

High intensity of sunlight
High temperature
Higher wind velocity
Decrease in atmospheric pressure
(Any three)

Solution D.7.
Forests have large number of plants especially trees. Each plant loses water in the form of water vapour everyday into the atmosphere through transpiration. A large apple tree loses as much as 30 litres of water per day. So huge amount of water is escaped into the atmosphere by forests. This increases the moisture in the atmosphere and brings more frequent rains.

Solution D.8.
The advantages of transpiration to the plants are:

Transpiration brings about a cooling effect to the plant body since evaporation of water reduces the temperature of leaf surface.
Transpiration helps in the ascent of sap by producing a suction force acting from the top of the plant.
Transpiration helps in distributing water and mineral salts throughout the plant body.
Transpiration helps in eliminating excess water.

Solution D.9.

If the water content of the leaves decreases due any reason, the guard cells turn flaccid, thereby closing the stomatal opening and transpiration stops.
Some plants have sunken stomata whereas others have reduced number of stomata to reduce transpiration.
In some plants, leaves may be dropped or may be absent or changed into spines as an adaptation to reduce transpiration.
The leaves may be covered by thick cuticle such as in Banyan tree, so as to reduce transpiration.

Solution D.10.
No, they are not dew drops.

This is water given out by the plant body through guttation. Since the banana plant is growing in humid environment, transpiration is hampered. But the roots continue to absorb water from the soil. This builds up a huge hydrostatic pressure within the plant and forces out the excess water from the hydathodes, which are pores present at the tips of veins in the leaf. This is observed especially during the mornings.

Solution D.11.
(a) Intensity of light – During the day, the stomata are open to facilitate the inward diffusion of carbon dioxide for photosynthesis. At night they are closed. Hence more transpiration occurs during the day. During cloudy days, the stomata are partially closed and the transpiration is reduced.
Selina Concise Biology Class 10 ICSE Solutions Transpiration image - 3
(b) Humidity of the atmosphere – When the air is humid; it can receive very less water vapour. Thus, high humidity in the air reduces the rate of outward diffusion of the internal water vapour across stomata, thereby reducing the rate of transpiration.

Solution E.1.

(i) The leaf D would become most limp. This is because water would be lost through transpiration from upper as well as the lower surface of leaf D since it is uncoated.

(ii) The least limping would be shown by leaf C since its upper and lower surfaces have been coated with vaseline. So no water is lost from the leaf through transpiration since the stomatal openings get blocked by vaseline.

Solution E.2.
(a)
Guard Cell
Inner wall of the Guard Cell
Stoma/Stomatal Aperture
(b) Open state
(c) The structure of stoma remains same in monocots as well as in dicots. Hence, the stoma from the diagram can be of a monocot leaf or of a dicot leaf.
Selina Concise Biology Class 10 ICSE Solutions Transpiration image - 4

Solution E.3.
(a) Transpiration
(b) Oil is put on the surface of water to prevent loss of water by evaporation.
(c) Yes, the transpiration rate will increase. Transpiration would occur faster. The observable changes will occur in less time.
(d) The spring balance progressively measures the change in weight of the set-up. This because as the plant transpires, it creates the suction force in plant which allows roots to absorb more water from the test tube. Hence, the water in the test will get reduced. Thus, the weight of the entire set will decrease.

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Selina Concise Biology Class 10 ICSE Solutions Photosynthesis: Provider of Food for All

November 10, 2023May 14, 2022 by Veerendra

Selina Concise Biology Class 10 ICSE Solutions Photosynthesis: Provider of Food for All

APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 10 Biology Chapter 6 Photosynthesis: Provider of Food for All. You can download the Selina Concise Biology ICSE Solutions for Class 10 with Free PDF download option. Selina Publishers Concise Biology for Class 10 ICSE Solutions all questions are solved and explained by expert teachers as per ICSE board guidelines.

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Selina ICSE Solutions for Class 10 Biology Chapter 6 Photosynthesis: Provider of Food for All

Exercise 1

Solution A.1.
b) glucose formed in photosynthesis soon gets converted into starch

Solution A.2.
b) twelve

Solution A.3.
b) humidity

Solution A.4.
c) trapping light energy

Solution A.5.
a) continue to live, but will not be able to store food

Solution A.6.
a) Carbon dioxide is reduced and water is oxidised

Solution A.7.
c) activate chlorophyll

Solution A.8.
d) ensure that the leaves are free from starch

Solution A.9.
a) CO₂

Solution B.1.
(a) Producers / Autotrophs
(b) Chloroplasts
(c) ATP (Adenosine triphosphate)
(d) Glucose
(e) Green plants
(f) Carbon dioxide dissolved in water
(g) Stroma
(h) Phloem

Solution C.1.

(a)

Respiration	Photosynthesis
The gas released during respiration is carbon dioxide.	The gas released during photosynthesis is oxygen.

(b)

Light Reaction	Dark Reaction
Hydrogen and oxygen are produced here, along with release of electrons, which converts ADP into ATP.	Glucose is the main product formed during dark reaction.

(c)

Producers	Consumers
Producers show autotrophic mode of nutrition i.e. they are able to produce their own food from basic raw materials. For example: green plants	Consumers show heterotrophic mode of nutrition i.e. they depend directly or indirectly on the producers for their food. For example: Animals

(d)

Grass	Grasshopper
Green grass being a producer is capable of producing its own food by photosynthesis.	Grasshopper is a primary consumer (herbivore) and directly feeds on producers like grass.

(e)

Chlorophyll	Chloroplast
Chlorophyll is the green pigment present in cell organelles called chloroplasts.	Chloroplasts are cell organelles, situated in the cytoplasm of plant cells. They are present mainly in the mesophyll cells and in the guard cells of stomata.

Solution C.2.
(a) False
Correct Statement: Dark reaction of photosynthesis is independent of light and occurs simultaneously with light reaction.

(b) True

(c) False
Correct Statement: Starch produced in a leaf is stored temporarily in the leaf until the process of photosynthesis. At night it is converted back into soluble sugar and translocated to different part of the body either for the utilization or for the storage.

(d) True

(e) False
Correct Statement: Green plants are producers.

(f) False
Correct Statement: Respiration results in loss of dry weight of the plants.

(g) False
Correct Statement: Photosynthesis stops at a temperature of above 40oC.

(h) True
(i) True
(j) True

Solution C.3.
(a) grana
(b) iodine solution
(c) chloroplast
(d) Calvin cycle
(e) Sucrose

Solution C.4.

(a) False
Photosynthesis increases with the light intensity up to a certain limit only and then it gets stabilized.

(b) False
The atmospheric temperature is an important external factor affecting photosynthesis. The rate of photosynthesis increases up to the temperature 35^oC after which the rate falls and the photosynthesis stops after 40^oC.

(c) False
Ice cold water will hamper the process of photosynthesis in the immersed leaf, even if there is sufficient sunshine because the temperature is an important factor for the rate of photosynthesis.

(d) False
For destarching, the potted plant can kept in a dark room for 24-48 hours.

(e) False
There is no start point or end point in the carbon cycle, the carbon is constantly circulated between the atmosphere and the living organisms.

(f) False
If a plant is kept in bright light all the 24 hours for a few days, the dark reaction (biosynthetic phase) will continue to occur because the dark reaction is independent of light and it occurs simultaneously with the light dependent reaction.

(g) True

Solution C.5.
Photons, grana, water molecules, hydrogen and hydroxyl ions, oxygen

Solution C.6.

Photosynthesis	Respiration
Carbon dioxide is used up and oxygen is released.	Oxygen is used up and carbon dioxide is released.
Photosynthesis occurs in plants and some bacteria.	Respiration occurs in all living organisms.
Photosynthesis results in gain of dry weight of the plants.	Respiration results in loss of dry weight of the plants.
Glucose is produced which is utilized by the plants.	Glucose is broken down to obtain energy.
The raw materials for the photosynthesis are water, carbon dioxide and sunlight.	The raw material for respiration is glucose.

(Any 4)

Solution C.7.
Oxygen is released during photosynthesis. Some of this oxygen may be used in respiration in the leaf cells, but the major portion of it is not required and it diffuses out into the atmosphere through the stomata. However, in a sense, even this oxygen is not a waste because all organisms require it for their existence including the plants.

Solution C.8.
The presence of starch is regarded as evidence of photosynthesis. Hence before starting an experiment on photosynthesis, the plant should be placed in the dark for 24-48 hours to destarch the leaves. During this period, all the starch from the leaves will be sent to the storage organs and the leaves will not show the presence of starch. So the various experiments on photosynthesis can be carried out effectively.

Solution C.9.
Destarching means removal of starch. Destarching is carried out so that all the starch from the leaves will be sent to the storage organs. Hence all the leaves will not show the presence of starch and photosynthesis can be studied. Destarching ensures that any starch present after the experiment has been formed under experimental conditions.

Solution C.10.
If a green plant is kept in bright light, it tends to use up all the CO₂ produced during respiration, for photosynthesis. Thus, the release of CO₂cannot be demonstrated. Hence, it is difficult to demonstrate respiration as these two processes occur simultaneously.

Solution C.11.
The chloroplasts are concentrated in the upper layers of the leaf which helps cells to trap the sunlight quickly. Also the epidermis is covered by a waxy, waterproof layer of cuticle. This layer is thicker on the upper surface than the lower one. Hence most leaves have the upper surface more green and shiny than the lower one.

Solution C.12.

Place hydrilla plant (a water plant) in a beaker containing pond water and cover it by a short-stemmed funnel. (Make sure the level of water in the beaker is above the level of the stem of the funnel)
Invert a test tube full of water over the stem of the funnel.
Place the set up in the sun light for a few hours.

Observation:
Bubbles appear in the stem which rise and are collected in the test tube. When sufficient gas gets collected, a glowing splinter will be introduced in the test tube, which will burst into flames.

Inference:
The splinter glows due the presence of oxygen in the test tube which proves that the gas collected in the test is released by hydrilla during photosynthesis.
Selina Concise Biology Class 10 ICSE Solutions Photosynthesis Provider of Food for All image - 1

Solution C.13.

(i) Light Reaction:
The light reaction occurs in two main steps:

Activation of chlorophyll – On exposure to light energy, chlorophyll becomes activated by absorbing photons.
Splitting of water – The absorbed energy is used in splitting the water molecule into hydrogen and oxygen, releasing energy. This reaction is known as photolysis of water.

The fate of H⁺, e^– and (O) component are as follows:
The hydrogen ions (H⁺) obtained from above are picked up by a compound NADP (Nicotinamide adenine dinucleotide phosphate) to form NADPH.

The oxygen (O) component is given out as molecular oxygen (O₂).
2O → O₂The electrons (e^–) are used in converting ADP into energy rich ATP by adding one inorganic phosphate group P_i.
ADP + P_i → ATP
This process is called photophosphorylation.

(ii) Dark reaction: The reactions in this phase does not require light energy and occur simultaneously with the light reaction. The time gap between the light and dark reaction is less than one thousandth of a second. In the dark reaction, ATP and NADPH molecules (produced during light reaction) are used to produce glucose (C₆H₁₂O₆) from carbon dioxide. Fixation and reduction of carbon dioxide occurs in the stroma of the chloroplast through a series of reactions. The glucose produced is either immediately used up by the cells or stored in the form of starch.
Selina Concise Biology Class 10 ICSE Solutions Photosynthesis Provider of Food for All image - 4

Solution C.14.
Complete the following food chains by writing the names of appropriate organisms in the blanks:
(i) Grass → Rabbit. → Snake → Hawk
(ii) Grass/Corn → Mouse → Snake → Peacock

Solution C.15.
Non-green plants such as fungi and bacteria obtain their nourishment from decaying organic matter in their environment. This matter comes from dead animals and plants. Fungi and bacteria break down the organic matter to obtain the nourishment and they release carbon dioxide back in the atmosphere.

Solution C.16.
Chlorophyll is the foundation site for the photosynthesis in green plants. The initiation of photosynthesis takes place when the chlorophyll molecule traps the light energy. The light energy is then converted into chemical energy in the form of glucose using carbon dioxide (CO₂) from the atmosphere, and water (H₂O) from the soil. All other organisms, directly or indirectly depend on this food for their survival. The starting point of any food chain is always a plant. If green plants were to suddenly disappear, then so would virtually all life on Earth. Thus, we can say that all life owes its existence to chlorophyll.

Solution C.17.
To test the leaf for starch, the leaf is boiled in water to kill the cells. It is next boiled in methylated spirit to remove chlorophyll. The leaf is placed in warm water to soften it. It is then placed in a dish and iodine solution in added. The region, which contains starch, turns blue-black and the region, which does not contain starch, turns brown.

Solution D.1.

a. The student wanted to show that sunlight is necessary for photosynthesis. / The role of sunlight in photosynthesis is being investigated.

b. Yes. The other uncovered leave of the potted plant act as a control.

c. Destarching ensures that any starch present after the experiment has been formed under experimental conditions. Therefore, the plant was kept in the dark before the experiment.

The student dipped the leaf in boiling water for a minute to kill the cells.
Then he boiled the leaf in alcohol/methylated spirit over a water bath to remove chlorophyll. The leaf becomes hard and brittle.
He then places the leaf in hot water to soften it.
Next the student spreads the leaf in a dish and pours iodine solution on it. The presence of starch is indicated by a blue-black colour.
The uncovered portion (exposed to sunlight) turned blue-black colour and the covered portion showed brown colour. The difference in the colours of covered and uncovered part of leaves indicates the importance of sunlight in photosynthesis.

Solution D.2.
(a) Guard cells: They regulate the opening and closing of stomata and thus regulate the entry of carbon dioxide through the stomata.

(b) Cuticle: Cuticle is transparent and water proof due to which light can penetrate this later easily.

(c) Mesophyll cells: Mesophyll cells are the main sites for photosynthesis. Chloroplasts are mainly contained in the mesophyll cells. When sunlight falls on the leaf, the light energy is trapped by the chlorophyll of the upper layers of mesophyll, especially the palisade cells.

(d) Xylem Tissue in the Leaf Veins: Water is essential for photosynthesis to occur. Water is taken up by the roots from the soil, sent up through the stem and finally brought to the leaves (site of photosynthesis) through the xylem tissue. The water is then distributed in the mesophyll tissue.

(e) Phloem Tissue in the Leaf Veins: The prepared food is transported from leaves to all parts of the plant by the phloem tissue. The glucose is converted into insoluble starch and later into soluble sugar i.e. sucrose, which is transported in solution through the phloem in the veins of the leaf and down through the phloem of the stem.

(f) Stoma: The main function of stoma is to let in carbon dioxide from the atmosphere for photosynthesis. Also most of the oxygen produced during photosynthesis diffuses out into the atmosphere through the stomata.

Solution D.3.
a.

Sunlight
Oxygen
Glucose
Xylem

b. A – Transpiration
B – Translocation

Solution D.4.
a. Food chain
b. Hawk, eagle
c. Photosynthesis
d. Carbon

Solution D.5.
Test to determine the presence of starch in a leaf:

Dip a leaf in boiling water for a minute to kill the cells.
Boil the leaf in methylated spirit in a water bath to remove the chlorophyll, till the leaf turns pale blue and becomes hard and brittle.
Now place the leaf in hot water to soften it.
Place the leaf in a Petri dish and pour iodine solution over it.
The appearance of a blue-black colour on the leaf is indicative of the presence of starch.
The absence of starch is indicated by a brown colouration.

Solution D.6.
a. To demonstrate the importance of carbon dioxide in photosynthesis
b. No, the experiment will not work satisfactorily, as the beaker contains lime water and not potassium hydroxide to absorb CO₂.
c. Place potassium hydroxide in the beaker instead of lime water
d. Before starting the experiment, it is necessary to destarch the leaves of the plant by keeping the plant in complete darkness for 48 hours. This is because if the plant is not destarched, then the experiment will give false results because starch stored previously may be detected in the leaf placed in the beaker even if no starch is produced during the experiment.

Solution D.7.
Selina Concise Biology Class 10 ICSE Solutions Photosynthesis Provider of Food for All image - 5

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