NCERT Class 11 Biology Solutions

NCERT Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division

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NCERT Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division.

Question 1.
What is the average cell cycle span for a mammalian cell?
Solution:
It is significant to note that in the 24-hour average duration of the cell cycle of a human cell or mammalian cells, cell division proper lasts for only about an hour.

Question 2.
Distinguish cytokinesis from karyokinesis.
Solution:

  1. Karyokinesis is a division of the nucleus (mitosis or meiosis) while cytokinesis is a division of the cytoplasm.
  2. Cytokinesis in an animal cells is achieved by the appearance of a furrow in the plasma membrane.
  3. The furrow gradually deepens and ultimately joins in the center dividing the cell cytoplasm into two.
  4. In-plant cell cell wall formation starts in the center of the cell and grows outward to meet the existing lateral walls.
  5. The formation of a new cell wall begins with the formation of a simple precursor, called the cell plate that represents the middle lamella between the walls of two adjacent cells.
  6. It is the time of cytoplasmic division, organelles like mitochondria and plastids get distributed between the two daughter cells.
  7. In some organisms, karyokinesis is not followed by ”cytokinesis as a result of which multinucleate condition arises leading to the formation of syncytium (liquid endosperm of coconut).

Question 3.
Describe the events taking place during the interphase.
Solution:
The interphase though called the resting phase is the time during which the cell is preparing for division by undergoing both cell growth and DNA replication in an orderly manner. The interphase is divided into three further phases :

  • G1 phase (Gap)
  • S phase (Synthesis)
  • G2 phase (Gap2)

G1 phase corresponds to the interval between mitosis and initiation of DNA replication. During G1 phase the cell is metabolically active and continuously grows but does not replicate its DNA. S or synthesis phase marks the period during which DNA synthesis or replication takes place. During this time the amount of DNA per cell doubles. If the initial amount of DNA is denoted as 2C then it increases to 4C. However, there is no increase in the chromosome number; if the cell had diploid or 2n number of chromosomes at G1 even after the S phase the number of chromosomes remains the same i.e. 2n.

In animal cells, during the S phase, as DNA replication begins in the nucleus, the centrioles, initiate replication in the cytoplasm. Dumping the G2 phase proteins are synthesized in preparation for mitosis while cell growth continues.

Question 4.
What is the G0 (quiescent phase) of the cell cycle?
Solution:
Some cells in adult animals do not appear to exhibit division (e.g., heart cell, nerve cell). These cells become inactive and become specialized by differentiating and do not further exit the G1 phase, and enter a stage but divide occasionally called the quiescent stage (G0) of the cell cycle.
NCERT Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division 1
A cell in this stage remains metabolically active but no longer proliferate unless called to do so depending on the requirement of the organism.

Question 5.
Why is mitosis called equational division?
Solution:
M phase is the most dramatic period of the cell cycle involving a major reorganization of virtually all components of the cell. Since the number of chromosomes in the parent and progeny cells is the same it is also called equational division. Though for convenience mitosis has been divided into four stages of nuclear division, it is very essential to understand that cell division is a progressive process and very clear-cut lines cannot be drawn between various stages.
Mitosis is divided into the following four stages:

  • Prophase
  • Meta phase
  • Anaphase
  • Telophase

Question 6.
Name the stage of the cell cycle at which one of the following events occur:
i. Chromosomes are moved to the spindle equator.
ii. Centromere splits and chromatids separate.
iii. Pairing between homologous chromosomes takes place.
iv. Crossing over between homologous chromosomes takes place.
Solution:
i. Metaphase
ii. Anaphase II
iii. Zygotene
iv. Pachytene

Question 7.
Describe the following:
(a) synapsis
(b) bivalent
(c) chiasmata
Draw a diagram to illustrate your answer.
Solution:
(a) During zygotene of prophase I of meiosis homologous chromosomes pair together. This pairing is called synapsis.
NCERT Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division 2
(b) Bivalent: The complex formed by homologous chromosomes during zygotene is called a bivalent.
(c) Chiasmata: During diplotene, the paired chromosomes make an X-shaped structure. This is called chiasmata.

Question 8.
How does cytokinesis in plant cells differ from that in animal cells?
Solution:
In animal cells, cytokinesis is achieved by the appearance of a furrow in the plasma membrane. The furrow gradually deepens and ultimately joins in the centre dividing the cell cytoplasm into two.

Due to the presence of cell walls, cytokinesis is different in plants. In plants, wall formation starts in the centre of the cell and grows outward to meet the existing lateral walls. The formation of the new cell wall begins with the formation of a simple precursor, called the cell plate that represents the middle lamella between the walls of two adjacent cells. Organelles like mitochondria and plastids get distributed between the two daughter cells during this process.

Question 9.
Find examples where the four daughter cells from meiosis are equal in size and where they are found unequal in size.
Solution:
The four daughter cells produced may be equal in size in the sperm of animals. They may be unequal in size as gametes in plants-pollen grains and egg in ovules.

Question 10.
Distinguish anaphase of mitosis from anaphase I of meiosis.
Solution:
The main difference between anaphase of mitosis and anaphase I of meiosis are as follows:
NCERT Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division 3

Question 11.
List the main differences between mitosis and meiosis.
Solution:
The main differences between mitosis and meiosis are as follows :
NCERT Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division 4

Question 12.
What is the significance of meiosis?
Solution:
1. Meiosis ensures the production of the haploid phase in the life cycle of sexually reproducing organisms whereas fertilization restores the diploid phase. We come across meiosis during gametogenesis in plants and animals. This lead to the formation of haploid gemotes.

2. Crossing over takes place during the Pachytene stage of meiosis, in which an exchange of genetic matter occurs. This produces variation, the raw material for evolution.

3. Meiosis has an impact on the genetic consequences due to pairing, crossing over, recombination, and segregation of homologous chromosomes.

Question 13.
Discuss with your teacher about

  1. haploid insects and lower plants where cell-division occurs and
  2. some haploid cells in higher plants where cell-division does not occur.

Solution:

  1. Some insects like honey bee drones are haploid. They are not fertile.
  2. In lower plant, main plant body is haploid produces haploid microspores by mitosis. The gametes of Chlamydomonas fuse to form diploid zygotes. Meiosis take place at this stage in lower plants forming haploid spous which give rise to the new plant.

Question 14.
Can there be mitosis without DNA replication in S-phase?
Solution:
S or synthesis phase marks the period during which DNA synthesis or replication takes place. During this time the amount of DNA per cell doubles. If the initial amount of DNA is denoted as 2C then it increases to 4C. However, there is no increase in the chromosome number; if the cell had diploid or 2n number of chromosomes at G, even after the S phase the number of chromosomes remains the same i.e. 2n.

Question 15.
Can there be DNA replication without cell division?
Solution:
DNA replication takes place in order to prepare cells for division. Cell division is the next logical step after DNA replication.

Question 16.
Analyze the events during every stage of the cell cycle and notice how the following two parameters change
i. Number of chromosomes (N) per cell
ii. Amount of DNA content (C) per cell
Solution:
Cell division is a very important process in all living organisms. During the division of a cell, DNA replication and cell growth also take place. Although cell growth is a continuous process, DNA synthesis occurs only during one specific stage in the cell cycle. The replicated chromosomes (DNA) are then distributed to daughter nuclei by a complex series of events during cell division. These events are themselves under genetic central.

The cell cycle is divided .nto two basic phases: The M phase starts with the nuclear division, corresponding to the separation of daughter chromosomes (Kaiyokinesis), and usually ends with the division of cytoplasm (Cytokinesis’). The interphase though called the resting phase is the time during which the cell is preparing for division by undergoing both cell growth and DNA replication in an orderly manner. The interphase is divided into three further phases:

  • G1 phase (Gap)
  • S phase (Synthesis)
  • G2 phase (Gap2)

VERY SHORT ANSWER QUESTIONS

Question 1.
What is karyokinesis?
Solution:
Division of nucleus is Known as karyokinesis.

Question 2.
Name the phase in which chromatids move apart.
Solution:
Anaphase.

Question 3.
Name the synthetic phase of interphase.
Solution:
S phase

Question 4.
Name the cell divisions which help recombination of genes.
Solution:
Meiosis.

Question 5.
Which type of cell division occurs in the meristematic cell of the root apex?
Solution:
Mitosis

Question 6.
Name the stage during which astral and spindle fibres disappear and nuclear membrane and nucleoli reappear.
Solution:
Telophase.

Question 7.
Name the sub-phases of prophase-I of Meiosis.
Solution:
Leptotene, Zygotene, Pachytene, Diplotene and Diakinesis.

Question 8.
In which stage, the actual reduction of chromosome number occurs in meiosis?
Solution:
Anaphase I

Question 9.
What is Synapsis? (Bijapur, Belgaum, Shimoga 2004)
Solution:
The process of pairing homologous chromosomes is called Synapsis.

Question 10.
What is the peculiarity of zygotene?
Solution:
In the zygotene phase pairing of homologous chromosomes or synapsis takes place.

Question 11.
Define crossing over. Give its significance.
Solution:
It is an exchange of genes between non-sister chromatids of homologous chromosomes. It produces new combination of genes and variation.
Question 12.
Why mitosis is an equational division called?
Solution:
Mitosis is a process of cell division, in which chromosomes are equally distributed into two daughter cells so it is equational division.

SHORT ANSWER QUESTIONS

Question 1.
Differentiate between:
(a) S-phase and G2 phase.
(b) G1 and G2 phase
Solution:
(a) Difference between S-phase and G2 phase:
NCERT Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division 5
(b) Difference between Gj and G2 phase
NCERT Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division 6

Question 2.
Write the significance of Mitosis / Meiosis. (Udupi 2006, D.Kannada 2010)
Solution:
Significance of Mitosis:

  • The distribution of an equal number of chromosomes to the daughter cells maintains a constant chromosome number.
  • Mitosis increases the number of cells so it contributes to growth.

Significance of Meiosis:

  • Meiosis brings genetic crossing over and random distribution of paternal and maternal chromosomes to daughter cells.
  • Recombination produces variations and variations are the sources of organic evolution.

Question 3.
Mention the significance of mitosis.
Solution:

  • In multicellular organisms, the growth of the body is due to the mitotic division of cells.
  • Replacement of worn-out cells and repair of the damaged cells is by mitosis.
  • In unicellular organisms, mitosis results in the asexual reproduction of cells.
  • In plants, vegetative propagation involves only mitotic divisions.

Question 4.
When and why does a reduction in the number of chromosomes take place in meiosis?
Solution:
The actual reduction in the number of chromosomes takes place in anaphase I.
This is because in anaphase I, one member from each homologous pair moves to one pole; the two chromatids of the chromosomes do not separate as the centromeres do not divide at this stage.

Question 5.
Differentiate between prophase I and prophase of mitosis.
Solution:
Difference between prophase I and prophase of mitosis are :
NCERT Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division 7

Question 6.
Imagine a situation if there was no meiosis. Then what would have happened to the next generation?
Solution:
In the absence of meiosis, the next generation would have double the number of chromosomes after the fusion of gametes. This would have resulted in the birth of an altogether new species. The maintenance of characters’ sets would have been possible only through asexual reproduction.

Question 7.
Which stage in meiosis is marked with genetic recombination? How
Solution:
In the pachytene stage of meiosis I the chromosomes appear as tetrads of homologous chromosomes and cross as the genetic material exchange takes place. Thus this stage is marked with genetic recombination.

Question 8.
What is Synapsis and Synaptonemal Complex? In which stage do these occur?
Solution:
The pairing of chromosomes is called Synapsis.
In the chromosomal synapsis, a complex structure is made known as the Synaptonemal complex These occur in the zygotene stage of meiosis.

Question 9.
Describe the events in the prophase of animal cells.
Solution:

  • Chromosomal material condenses to form compact mitotic chromosomes.
  • Chromosomes are seen to be composed of two chromatids attached together at the centromere.
  • The centrioles formed in interphase start moving to the opposite poles of the cell.
  • Initiation of the assembly of the mitotic spindle, the microtubules, the proteinaceous components of the cell cytoplasm help in the process.
  • By the end of the prophase, the nuclear membrane and nucleolus disappear.

Question 10.
Diagrammatically shows the cell cycle and answers the following questions.
NCERT Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division 8
(a) Which is the resting stage.
(b) Which is the stage where replication takes palce.
(c) Which is the stage where mitosis takes place.
(d) Which is the post cell synthetic but pre-cell division stage.
Solution:
(a) Go Phase
(b) S Phase
(c) M Phase
(d) G2 Phase

Question 11.
Define cytokinesis. How is it accomplished in animal and plant cells?
Solution:
The process by which the cytoplasm of the cell divides resulting in the formation of two daughter nuclei is called cytokinesis.

In an animal cell, this is achieved by the appearance of a furrow in the plasma membrane. The furrow gradually deepens and ultimately joins in the centre dividing the cell cytoplasm into two.

In-plant cells, wall formation starts in the centre of the cell and grows outward to meet the existing lateral walls. The formation of a new cell wall begins with the formation of a simple precursor, called the cell plate that represents the middle lamella between the walls to two adjacent cells.

LONG ANSWER QUESTIONS

Question 1.
Describe briefly cytokinesis in animal cells and plant cells.
Solution:
(a) Cytokinesis in animal cells: In the animal cells, the cytoplasm divides by constriction. It appears on the equator and slowly deepens. The constriction converges on all the sides and pinches off the parent cell into 2 daughter cells. Constriction is the result of a peripheral band of microfilaments. This constriction divide the cytoplasm finally,

(b) Cytokinesis in plant cells: Plant cells have a rigid cell wall and this cannot undergo cytokinesis by invaginating cleavage furrow. Therefore, in them, the cytokinesis is accomplished by the formation of phragmoplast from carbohydrate and lipid-containing vesicles of Golgi apparatus and endoplasmic reticulum vesicles. A cell plate at the equator of the dividing cell is formed and divides the cytoplasm.

Question 2.
Mention the significance of mitosis.
Solution:
Significance of mitosis: Mitosis or the equational division is usually restricted to the diploid cells only.

  • However, in some lower plants and in some social insects haploid cells also divide by mitosis.
  • It is very essential to understand the significance of this division in the life of an organism.
  • Mitosis results in the production of diploid daughter cells with identical genetic complement usually.
  • The growth of multicellular organisms is due to mitosis. Cell growth results in disturbing the ratio between the nucleus and the cytoplasm.
  • It, therefore, becomes essential for the cell to divide to restore the nucleo- cytoplasmic ratio. A very significant contribution to mitosis is cell repair.
  • The cells of the upper layer of the epidermis, cells of the lining of the gut, and blood cells are being constantly replaced.
  • Mitotic divisions in the meristematic tissues – the apical and the lateral cambium, result in the continuous growth of plants throughout their life.

Question 3.
Describe meiosis II with the help of suitable diagrams.
Solution:
Meiosis II is divided into four phases.

Prophase II. Meiosis II is initiated immediately after cytokinesis, usually, before the chromosomes have fully elongated. In contrast to meiosis I, meiosis II resembles a normal mitosis. The nuclear membrane disappears by the end of prophase II. The chromosomes again become compact.

Metaphase II- At this stage, the chromosomes align at the equator and the microtubules from opposite poles of the spindle get attached to the kinetochores of sister chromatids. Anaphase II- It begins with the simultaneous splitting of the centromere of each chromosome (which was holding the sister chromatids together), allowing them to move toward opposite poles of the cell.

Telophase II- Meiosis ends with telophase II, in which the two groups of chromosomes once again get enclosed by a nuclear envelope; cytokinesis follows resulting in the formation of a tetrad of cells i.e., four haploid daughter cells.
NCERT Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division 9

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NCERT Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals

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NCERT Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals.

Question 1.
Answer in one word or one line.
(i) Give the common name of Periplaneta americana.
(ii) How many spermathecae are found in earthworms?
(iii) What is the position of ovaries in cockroach?
(iv) How many segments are present in the abdomen of the cockroach?
(v) Where do you find malpighian tubules?
Solution:
(i) Cockroach
(ii) Four pairs of spermathecae are found in the 6“ to 9th segments (one pair in each segment).
(iii) Cockroach includes a pair of ovaries, that lies laterally in the 2nd to 6th abdominal segments of the abdomen.
(iv) Ten
(v) At the junction of midgut and hindgut. 100-150 yellow coloured thin filamentous ring is present in earthworm, which is called malpighian tubules.

Question 2.
Answer the following :
(i) What is the function of nephridia?
(ii) How many types of nephridia are found in earthworms based on their location?
Solution:
(i) Nephridia is the excretory organ of the earthworm or pheretima.
(ii) There are three types of nephridia –
(i) Septal nephridia – Present on both the sides of intersegmental septa of segment 15 to the last that open into the intestine.
(ii) Integumentary nephridia – Attached to the lining of the body wall of segment 3 to the last that opens on the body surface.
(iii) Pharyngeal nephridia – Present as three paired tufts in the 4th, 5th, and 6th segments.
These three different types of nephridia are almost similar in structure.

Question 3.
Draw a labelled diagram of the reproductive organs of an earthworm.
Solution:
NCERT Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals 1

Question 4.
Draw a labelled diagram of the alimentary canal of a cockroach
Solution:
NCERT Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals 2

Question 5.
Distinguish between the followings :
(a) Prostomium and peristomium
(b) Septal nephridium and pharyngeal nephridium
Solution:
(a) The main difference between prostomium and peristomium are :
NCERT Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals 3
(b) The main difference between septal nephridium and pharyngeal nephridium are:
NCERT Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals 4
Question 6.
What are the cellular components of blood?
Solution:
Red blood cells and white blood cells are the cellular components of blood.

Question 7.
What are the following and where do you find them in an animal body.
1. Chondrocytes
2. Axons
3. Ciliated epithelium
Solution:
1. Chondriocytes: The intercellular material of cartilage is solid and pliable and resists compression. Cells of this tissue (Chondriocytes) are enclosed in small cavities within the matrix secreted by them. Most of the cartilages in vertebrate embryos are replaced by bones in adults. Cartilage is present in the tip of the nose, outer ear joints, between adjacent bones of the. vertebral column, limbs, and hands in adults.

2. Axons: It is found in the nervous system neuron. It is a long fiber, the distal end of which is branched. The main function of axons is the transmission of impulses by means of neurotransmitters.

3. Ciliated epithelium: If the columnar or cuboidal cells bear cilia on their free surface they are called the ciliated epithelium. They are found in the lining of the stomach and intestine and help in secretion and absorption, their function is to move particles or mucus in a specific direction over the epithelium. They are mainly present in the inner surface of hollow organs like bronchioles and fallopian tubes.

Question 8.
Describe various types of epithelial tissues with the help of labelled diagrams.
Solution:
Epithelial tissues
Epithelial tissues provide covering to the inner and outer lining of various organs. The cells of epithelial tissues are compactly packed with a little intercellular matrix.
There are two types of epithelial tissues:
(i) Simple epithelium
(ii) Compound epithelium
(i) Simple epithelium :
Composed of a single layer of cells and functions as a lining for body cavities ducts and tubes. It is further divided into three types on the basis of structure modifications-
(a) Squamous epithelium – It is made of a single layer of flattened cells with irregular boundaries.
It is found as a lining for body cavities, ducts, and tubes such as in the walls of blood vessels and air sacs of lungs.
NCERT Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals 5
Functions – It helps in forming a diffusion boundary.
(b) Cuboidal epithelium – It is made of a single, layer of cube-like cells. It is commonly found in the ducts of glands and tubular parts of nephrons in kidneys. Specialized cuboidal cells are capable of producing gametes found in gonads called the germinal epithelium.
Functions- It helps in secretion and absorption and also in moving particles or mucus in a specific direction over the epithelium.
(c) Columnar epithelium – It is composed of a single layer of tall and slender cells. Nuclei are located at the base.
Its free surface may have microvilli.
It is found inlining of the stomach and intestine.
Functions – It helps in secretion and absorption.
NCERT Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals 6

Question 9.
Distinguish between
(a) Simple epithelium and compound epithelium
(b) Cardiac muscle and striated muscle
(c) Dense regular and dense irregular connective tissues
(d) Adipose and blood tissue
(e) Simple gland and compound gland
Solution:
(a) The main difference between simple epithelium and compound epithelium are as following.
NCERT Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals 7
The main difference between Cardiac muscles and striated muscle are as following.
NCERT Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals 8
The main difference between dense regular connective tissues and dense irregular connective tissues are as following.
NCERT Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals 9
The main difference between adipose tissue and blood tissue are as following.
NCERT Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals 10
The main difference between simple gland and compound gland are as following
NCERT Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals 11

Question 10.
Mark the odd one in each series:
(a) Areolar tissue; blood; neuron; tendon
(b) RBC; WBC; platelets; cartilage
(c) Exocrine: endocrine; salivary gland;ligament
(d) Maxilla; mandible; labrum; antennae
(e) Protonema; mesothorax; metathorax; coxa
Solution:
(a) Neuron
(b) Cartilage
(c) Ligament
(d) Antennae
(e) Protonema

Question 11.
Match the terms in column I with those in column II:
Column I Column II
(a) Compound epithelium – (i) Alimentary canal
(b) Compound eye – (ii) Cockroach
(c) Septal nephridia – ( iii) Skin
(d) Open circulatorysystem – (iv) Mosaic vision
(e) Typhlosole – (v) Earthworm
(f) Osteocytes – (vi) Phallomere
(g) Genitalia – (vii) Bone
Solution:
(a) Compound epithelium ( iii) Skin
(b) Compound eye – (iv) Mosaic vision
(c) Septal nephridia – (v) Earthworm
(d) Open circulatory system – (ii) Cockroach
(e) Typhlosole – (i) Alimentarycanal
(f) Osteocytes – (vii) Bone
(g) Genitalia – (vi) Phallomere

Question 12.
Mention briefly the circulatory system of earthworms.
Solution:
Pheretima exhibits a closed type of blood vascular system, consisting of blood vessels capillaries, and heart. Due to the closed circulatory system, blood is confined to the heart and blood vessels. Contractions keep blood circulating in one direction. Smaller blood vessels supply the gut, nerve cord, and body wall. Blood glands are present on the 4th, 5th, and 6th segments. They produce blood cells and hemoglobin which is dissolved in blood plasma. Blood cells are phagocytic in nature. Earthworms lack specialized breathing devices. Respiratory exchange occurs through moist body surfaces into their bloodstream.

Question 13.
Draw a neat diagram of the digestive system of the frog.
Solution:
The digestive system of frog:
NCERT Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals 12

Question 14.
Mention the function of the following:
(a) Ureters in frog
(b) Malpighian tubules
(c) Body wall in earthworm
Solution:
Functions:
(1) Ureters in frog:

  • They carry the urine from the kidneys to the cloaca.
  • In males, it also conducts the sperm as it is the urinogenital duct.
  • In females, the ureters and oviduct open- separately in the cloaca.

(2) Malpighian tubules:

  1. They are the excretory organs of a cockroach.
  2. They collect the nitrogenous wastes from the haeomolymph and send them into the intestine.
  3. Each tubule is lined by glandular and ciliated cells. They absorb nitrogenous waste products and convert them into uric acid which is excreted out through the hindgut.

(3) Body wall of earthworm:
The body wall of the earthworm is covered externally by a thin non-cellular cuticle

VERY SHORT ANSWER QUESTIONS

Question 1.
Name the protein found in white fibres.
Solution:
Collagen.

Question 2.
State the function of setae.
Solution:
It helps in locomotion by gripping the earth.

Question 3.
What is the functional unit of the cockroach eye?
Solution:
Ommatidium.

Question 4.
Name the unit of neural or nervous system.
Solution:
Neurons.

Question 5.
What is worm casting ?
Solution:
It is the insoluble and undigested food that is given out along with soil through anus.

Question 6.
Which cell covers the exposed surfaces of the body (skin) and internal passage ways (digestive tract and glands) ?
Solution:
Squamous epithelium

Question 7.
Name the type of epithelium that lines the inner surface of stomach.
Solution:
Columnar epithelium

Question 8.
Name the type of epithelium that lines the buccal cavity.
Solution:
Stratified squamous epithelium.

Question 9.
N ame of type of tissue that is the most abundant in animal body.
Solution:
Connective tissue.

Question 10.
What is the other name given to the gizzard of cockroach?
Solution:
Proventriculus.

Question 11.
Name the larva of a frog.
Solution:
Tadpole.

Question 12.
What is the scientific name of Indian (bull) frog?
Solution:
Rana tigrina.

SHORT ANSWER QUESTIONS

Question 1.
What are the neuroglia cells?
Solution:
Cells which holds the neuron together are known as neuroglia cells.

Question 2.
What is vermicompositing?
Solution:
The process of increasing soil fertility by earthworms is known as vermicomposting.

Question 3.
What are exocrine glands? Name any two secretions of them.
Solution:
Exocrine glands : Those glands which have ducts to pour their secretion(s) into the respective site of action, are called exocrine.
(i) Salivary glands secrete saliva into the buccal cavity.
(ii) Liver secretes bile into the duodenum.

Question 4.
Write four functions of bones.
Solution:
Functions of bones:
(i) They provide place for attachment of muscles and help in movement and locomotion.
(ii) Bone marrow is the site of manufacture of blood cells.
(iii) Bones provide protection to the internal organs.
(iv) The long bones of the limbs serve the weight-bearing function.
(v) They act as the depot of calcium and phosphorus.

Question 5.
Describe the three types of cell junctions present in the epithelium and other tissues.
Solution:
Cell junctions
(i) Tight junctions – They check leaking of substances across a tissue.
(ii) Adhering junctions – They help in cementing the neighbouring cells together.
(iii) Gap junctions – They facilitate the cells to communicate with each other by cytoplasmic connections, for rapid transfer of ions, small molecules, etc.

Question 6.
How is the gizzard in the alimentary canal of a cockroach suitable for grinding the food?
Solution:
Gizzard of cockroach has following characteristics:
(i) The gizzard has an outer layer of thick circular muscles.
(ii) The inner thick layer of cuticle forms six plate like teeth.
(iii) The movement with the help of muscles and the teeth-like structures help in grinding the food.

Question 7.
What is a typhlosole in an earthworm? Where is it found? What is its function?
Solution:
Typhlosole: It is an internal median fold of the dorsal wall of the intestine. It is found in the intestine between 26th and 35th segments of the body.
It increases the effective area of absorption.

LONG ANSWER QUESTIONS

Question 1.
Describe the female reproductive organs of frog.
Solution:
The female reproductive organs include a pair of ovaries. The ovaries are situated near kidneys and there is no functional connection with kidneys. A pair of oviduct arising from the ovaries opens into the cloaca separately. A mature female can lay 2500 to 3000 ova at a time.
Fertilisation is external and takes place in water. Development involves a larval stage called tadpole. Tadpole undergoes metamorphosis to form the adult.

NCERT Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals 13

Question 2.
Describe with examples, various types of connective tissues.
Solution:
Connective tissues are most abundant and widely distributed in the body of complex animals. They are named connective tissues because of their special function of linking and supporting other tissues/C ••gans of the body. Connective tissues are classified into three types:
(i) Loose connective tissue,
(ii) Dense connective tissue and
(iii) Specialised connective tissue.
(i) Loose connective tissue:

  • It has cells and fibres loosely arranged in a semi-fluid ground substance, for example, areolar tissue present beneath the skin.
  • Often it serves as a support framework for epithelium. It contains fibroblasts (cells that produce and secrete fibres), macrophages and mast cells.
  • Adipose tissue is another type of loose connective tissue located mainly beneath the skin. The cells of this tissue are specialised to store fats.

(ii) Dense connective tissue :

  • Fibres and fibroblasts are compactly packed in the dense connective tissues.
  • Orientation of fibres show a regular or irregular pattern and are called dense regular and dense irregular tissues.
  • In the dense regular connective tissues, the collagen fibres are present in rows between many parallel bundles of fibres.
  • Tendons, which attach skeletal muscles to bones and ligaments which attach one bone to another are examples of this tissue.
  • Dense irregular connective tissue has fibroblasts and many fibres (mostly collagen) that are oriented differently. This tissue is present in the skin.

(iii) Specialised connective tissue :

  • Cartilage, bones and blood are various types of specialised connective tissues.
  • The intercellular material of cartilage is solid and pliable and resists compression.
  • Cells of this tissue (chondrocytes) are enclosed in small cavities within the matrix secreted by them.
  • Cartilage is present in the tip of nose, outer ear joints, between adjacent bones of the vertebral column, limbs and hands in adults.

Bones : It has a hard and non-pliable ground substance rich in calcium salts and collagen fibres which give bone its strength. It is the main tissue that provides structural frame to the body.
Blood : It is a fluid connective tissue containing plasma, red blood cells (RBC), white bloo4 cells (WBC) and platelets. It is the main circulating fluid that helps in the transport of various substances.

Question 3.
Describe the alimentary canal of earthworm.
Solution:
The alimentary canal is a straight tube and mns between first to last segment of the boay. It has following parts.
Mouth : A terminal mouth opens into the buccal cavity (1-3 segments) which leads into muscular pharynx.
oesophagus: A small narrow tube, oesophagus (5-7 segments), continues into a muscular gizzard (8-9 segments). It helps in grinding the soil particles and decaying leaves, etc.
Stomach : The stomach extends from 9-14 segments. The food of the earthworm is decaying leaves and organic matter mixed with soil. Calciferous glands, present in the stomach, neutralise the humic acid present in humus.
• Intestine starts from the 15th segment onwards and continues till the last segment. A pair of short and conical intestinal caecae project from the intestine on the 26th segment.
Typhosole : The characteristic feature of the intestine between 26-35 segments is the presence of internal median fold of dorsal wall called typhlosole. This increases the effective area of absorption in the intestine.
Anus : The alimentary canal opens to the exterior by a small rounded aperture called anus. The ingested organic rich soil passes through the digestive tract where digestive enzymes breakdown complex food into smaller absorbable units. These simpler molecules are absorbed through intestinal membranes and are utilised.

Question 4.
Draw a labelled diagram of external features of cockroach.
Solution:
NCERT Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals 14

We hope the NCERT Solutions for Class 11 Biology at Work Chapter 7 Structural Organization in Animals, help you. If you have any query regarding NCERT Solutions for Class 11 Biology at Work Chapter 7 Structural Organization in Animals, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Biology Chapter 12 Mineral Nutrition

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NCERT Solutions for Class 11 Biology Chapter 11 Transport in Plants

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 11 Transport in Plants.

Question 1.
“All elements that are present in a plant need not be essential to its survival”. Comment.
Solution:
Plants obtain their inorganic nutrients from the air, water, and soil. Plants absorb a wide variety of mineral elements. Not all the mineral elements that they absorb are required by plants. Out of the more than 105 elements discovered so far, less than 21 are essential and beneficial for normal plant growth and development. The elements required in large quantities are called macronutrients. While those required in fewer quantities or in the trace are termed micronutrients. These elements are either essential constituents of proteins, carbohydrates, fats, nucleic acid, etc. and/or take part in various metabolic processes.

Question 2.
Why is the purification of water and nutrient salts so important in studies involving mineral nutrition using hydroponics?
Solution:
The technique of growing plants in a nutrient solution is known as hydroponics. Since a number of improvised methods have been employed to try and determine the mineral nutrients essential for plants. The essence on all these methods involves the culture of plants in a soil-free, defined mineral solution. These method require purified water and mineral nutrients salts. Purification of water and nutrient salt is important to find out other influencing factors

Question 3.
Explain with examples:
Macronutrients, micronutrients, beneficial nutrients, toxic elements, and essential elements.
Solution:
Based upon the criteria only a few elements have been found to be absolutely essential for plant growth and metabolism. These elements are further divided into two broad categories based on their quantitative requirements,

  1. Macronutrients
  2. Micronutrients

Macronutrients must generally be present in plant tissues in the concentration of 1 to 10 mg/L of dry matter. The macronutrients include carbon, hydrogen, oxygen, nitrogen, phosphorous, sulfur, potassium, calcium, and magnesium. Of these, carbon, hydrogen, and oxygen are mainly obtained from CO2, and H20, while the others are absorbed from the soil as mineral nutrition.

Micronutrients or trace elements are needed in very small amounts (equal to or less than 0.1 mg/L of dry matter). These include iron, manganese, copper, molybdenum, zinc, boron, chlorine, and nickel. In addition to the 17 essential elements named above, there are some beneficial elements such as sodium, silicon, cobalt, and selenium. They are required by higher plants.

Essential elements can also be grouped into four broad categories on the basis of their diverse functions. These categories are:
(1) Essentia] elements as components of biomolecules and hence structural elements of cells, (eg: carbon, hydrogen, oxygen, and nitrogen).

(2) Essential elements that are components of energy-related chemical compounds in plants, for example, magnesium in chlorophyll and phosphorous in ATP.

(3) Essential elements that activate or inhibit enzymes, for example, Mg2+ is an activator for both ribulose bisphosphate carboxylase-oxygenase and phosphoenolpyruvate carboxylase, both of which are critical enzymes in photosynthetic carbon fixation; Zn2+ is an activator of alcohol dehydrogenase and Mo of nitrogenase during nitrogen metabolism.

(4) Some essential elements can alter the osmotic potential of a cell. Potassium plays an important role in the opening and closing of stomata. Any mineral ion concentration in tissues that reduces the dry weight of tissues by about 10 percent is considered toxic. Such critical concentrations vary widely among different micronutrients. The toxicity symptom! are difficult to identify.

Toxicity levels for any element also vary for different plants. Many times excess of an element may inhibit the uptake of another element. For example, the prominent symptoms of manganese toxicity are the appearance of brown spots surrounded by chlorotic veins.

Question 4.
Name at least five different deficiency symptoms in plants. Describe them and correlate them with the concerned mineral deficiency.
Solution:

  1. Chlorosis: Chlorosis is the loss of chlorophyll leading to yellowing in leaves. It is caused by a deficiency of N, K, Mg, S, Fe, Mn, Zn, and Mo.
  2. Necrosis: It is the death of tissue. It occurs due to deficiency of Ca, Mg, Cu, K.
  3. Inhibition of cell division: It occurs due to deficiency of N, K, S, Mo.
  4. Stunted plant growth: It occurs due to deficiency of Ca, N, etc.
  5. Premature fall of leaf and buds: It occurs due to deficiency of calcium, magnesium.

Question 5.
If a plant shows a symptom which could develop due to deficiency of more than one nutrient, how would you find out experimentally, the real deficient mineral element?
Solution:

  1. The deficiency symptoms can be distinguished on the basis pf the region of occurrence, presence or absence of dead spots, and chlorosis of entire leaf or interveinal chlorosis.
  2. The region of the appearance of deficiency symptoms depends on the mobility of nutrients in plants. The nutrient deficiency symptoms of N, P, K, Mg, and Mo appear in lower leaves.
  3. Zinc is moderately mobile in plants and deficiency symptoms, therefore, appear in middle leaves.
  4. The deficiency symptoms of less mobile elements (S, Fe, Mn, and Cu) appear on new leaves.
  5. Ca and B are immobile in plants, deficiency symptoms appear on terminal buds.
  6. Chlorine deficiency is less common in crops.
    NCERT Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 1

Question 6.
Why is it that in certain plants deficiency symptoms appear first in younger parts of the plant while in other they do so in mature organs?
Solution:
The deficiency symptoms tend to appear first in the young tissues whenever the elements are relatively immobile and are not transported out of the mature organs, for example, elements like sulphur and calcium are a part of the structural component of the cell and hence are not easily released.

Question 7.
How are the minerals absorbed by the plants?
Solution:
Uptake of mineral ions, by plants, occurs through
two main phases.
Passive Absorption: It is the process of absorption of minerals through it’s outer space(Intercellular space and cell wall) by physical process. Direct expenditure of metabolic energy is not involved. A substance moves from a region of higher chemical potential to lower chemical potential. It occurs through ion channels (transmembrane protein). The theories to explain the movement of ions:
(a) Ion exchange: Both cation and anion gets absorbed on the surface of cell wall. The absorbed ions are exchanged with ions present in soil solution.
(b) Mass flow hypothesis: According to this hypothesis mass flow of ions occur along with absorption of water as a result of transpirational pull.
Active Absorption: It is the process of movement of ions against a concentration gradient, by utilizing ATP as energy. Both influx and efflux of ions are carried out by carrier mechanism. The activated ions combine with carrier proteins and form ion carrier complex. This complex moves
across all the membrane and reaches inner surface, where it breaks and releases ions into the cytoplasm.

Question 8.
What are the conditions necessary for the fixation of atmospheric nitrogen by Rhizobium? What is their role in nitrogen fixation?
Solution:
Rhizobia are unique because they live in a symbiotic relationship with legumes. Necessary conditions:

  • Requires a strong reducing agent and energy in the form of ATP.
  • The enzyme nitrogenase which is very sensitive to oxygen is required.
  • The processes take place in an anaerobic environment
  • The energy is provided by the respiration of host cells.

The reduction of nitrogen to ammonia by living organisms is called biological nitrogen fixation. The enzyme, nitrogenase which is capable of nitrogen reduction is present exclusively in prokaryotes. Several types of symbiotic biological nitrogen-fixing associations are known. The most common association on roots is nodules. Their role in N2– fixation is to supply the plants with nitrogenase that converts nitrogen to amino acids.

Question 9.
What are the steps involved in the formation of root nodule?
Solution:
Nodule formation involves a sequence of multiple interactions between Rhizobium and the roots of the host plant. Stages in the nodule formation are summarised as follows:
Steps in the development of root nodules:
(a) When a root hair of a leguminous plant comes in contact with Rhizobium, it is deformed due to the secretion from the bacterium.
(b) At the site of curling Rhizobia invades the root tissue and proliferate within root hairs.
(c) Some bacteria enlarge to form membrane-bound structures, bacteroids which cannot divide.
(d) The plants form the infection thread, made up of plasma membrane that grows inward, separating the infected tissue from the rest of the plant.
(e) Cell division is stimulated in the infected tissue and more bacteria invade the newly formed tissues.
(f) It is believed that a combination of cytokinin produced by invading bacteria and auxins produced by plant cells, promotes cell division and extension, leading to nodule formation.
NCERT Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 2

Question 10.
Which of the following statements are true? If false correct them:
(a) Boron deficiency leads to the stout axis.
(b) Every mineral element that is present in a cell is needed by the cell.
(c) Nitrogen as a nutrient element, is highly immobile in plants.
(d) It is very easy to establish the essentiality of micronutrients because they are required only in trace quantities.
Solution:
(a) True
(b) False
Correct sentence: Every mineral element that is present in a cell is not needed by the cell.
(c) False
Correct sentence: Nitrogen as a nutrient element is highly mobile in the plants.
(d) False
Correct sentence: It is very difficult to establish the essentiality of micronutrients because they are required only in trace quantities.

VERY SHORT ANSWER QUESTIONS

Question 1.
Name the enzymes that reduce nitrogen in the root nodules of a bean plant.
Solution:
Nitrogenase.

Question 2.
Name the enzymes used in biologically nitrogen fixation. What are the mineral elements needed for the activity of the enzyme?
Solution:
Nitrogenase enzyme.

Question 3.
Name the enzyme that can reduce nitrogen to ammonia.
Solution:
Nitrogenase.

Question 4.
What is the importance of phosphorus for plants?
Solution:
Phosphorus is a constituent of cell membranes, certain proteins, all nucleic acids, and nucleotides, and is required for all phosphorylation reactions.

Question 5.
Name two crops that are commonly produced by hydroponics.
Solution:
Tomato, Lettuce

Question 6.
Name the group of enzymes activated by zinc.
Solution:
Carboxylases.

Question 7.
Define critical concentration of elements with reference to plant nutrition.
Solution:
Critical concentration refers to the concentration of the essential element, below which the plant growth is retarded.

Question 8.
Name the element which is a limiting nutrient for both natural and agricultural ecosystems.
Solution:
Nitrogen.

Question 9.
Name two bacteria that oxidise ammonia into nitrite.
Solution:
Nitrosomonas, Nitrococcus

Question 10.
Name two symbiotic nitrogen-fixing bacteria.
Solution:
Rhizobium, Frankia.

Question 11.
What is hydroponics (Tank farming)?
Solution:
It is plant growth in the liquid culture medium.

Question 12.
What are the framework elements of a plant?
Solution:
Carbon, hydrogen, and oxygen are called framework elements.

Question 13.
What type of condition is created by leghaemoglobin in the root nodules of legumes?
Solution:
Anaerobic condition.

Question 14.
What is meant by active absorption?
Solution:
Active absorption: The uptake of mineral ions against the concentration gradient is called active absorption.

Question 15.
How are amides transported In plants?
Solution:
Amides are transported along with water through the xylem.

Question 16.
What is the function of the enzyme nitrite reductase?
Solution:
It reduces nitrate ions to ammonia.

Question 17.
Name two free-living micro-organisms which can fix nitrogen.
Solution:
Azotobacter, Beijemickia.

SHORT ANSWER QUESTIONS

Question 1.
What is nitrification?
Solution:
Questions It is the conversion of ammonium ion to nitrite and then to nitrate. Nitrosomonas converts ammonium into nitrites, Nitrobacter converts nitrites into nitrates.

Question 2.
Prior to sowing rice, a legume crop was cultivated and ploughed back in this field. Why? Explain.
Solution:
Leguminous plants possess root nodules in which the symbiotic bacteria Rhizobium fixes nitrogen. The fixed nitrogen makes the soil rich in nitrogen fertilizer where the leguminous plant is ploughed back into the field.

Question 3.
Name the organism that fixes nitrogen in symbiotic association with a legume. Where does it live in such plants?
Solution:
Rhizobium. It lives in the root nodules of leguminous plants.

Question 4.
Bring out at similarity and difference between leghaemoglobin and hemoglobin.
Solution:
Both leghaemoglobin and hemoglobin are iron-containing molecules but leghaemoglobin is present in the root nodules in the plants belonging to the family Fabaceae while hemoglobin in human blood pigment.

Question 5.
Bring out at similarity and difference between leghaemoglobin and hemoglobin.
Solution:
NCERT Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 3

Question 6.
In what form is boron absorbed by plants from the soil? Mention its two uses in plants and give two deficiency symptoms of boron in them.
Solution:
Boron is absorbed as \({ H }_{ 2 }{ Po }_{ 4- }\quad and\quad { HPo }_{ 4 }^{ 2- }\)

  • Translocation of carbohydrates
  • Pollen germination

Deficiency symptoms

  • Death of root and shoot tips
  • Abscission of flowers

Question 7.
List the macronutrients and mention three major function, (any three)
Solution:
NCERT Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 4

Question 8.
In what form is boron absorbed by plants from the soil? Mention its two uses in the plants and give two deficiency symptoms of boron in them.
Solution:
NCERT Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 4a
(i) Translocation of carbohydrates.
(ii) Pollen germination.
(iii) Absorption and utilisation of calcium.
(iv) Cell elongation and differentiation.

Question 9.
Distinguish between micronutrients and macronutrients.
Solution:
Differences between micronutrients and macronutrients are as following:
NCERT Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 5

LONG ANSWER QUESTIONS

Question 1.
With the help of a suitable diagram describe the nitrogen cycle.
Solution:
Nitrogen cycle: Plants compete with microbes for the limited nitrogen that is available in the soil. Thus, nitrogen is a limiting nutrient for both natural and agricultural ecosystems.

  • In nature, lightning and ultraviolet radiation provide enough energy to convert nitrogen to nitrogen oxides (NO, NO2, N2O).
  • Industrial combustions, forest fires, automobile exhausts, and power-generating stations are also sources of atmospheric nitrogen oxides.
  • The decomposition of organic nitrogen of dead plants and animals into ammonia is called ammonification. Some of this ammonia

volatilises and re-enters the atmosphere but most of it is converted into nitrate by soil bacteria in the following steps
NCERT Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 6
NCERT Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 7

Question 2.
Name at least five different deficiency symptoms in plants. Describe them and correlate them with concerned mineral deficiency.
Solution:

  • Chlorosis – yellowing of leaves due to loss of chlorophyll caused by the deficiency of N, S, mg, Fe.
  • Necrosis – Death of tissues, especially in leaves caused by the deficiency of Ca, Mg, Ca & K.
  • Delay in flowering caused by the deficiency of molybdenum, nitrogen, and sulphur.
  • Dieback of roots caused by a deficiency of copper.
  • Inhibition of cell division caused by a deficiency of potassium, calcium and nitrogen.

Question 3.
Write notes on :
(a) Reductive animation
(b) Transamination
Solution:
(a) Reduction animation – In this process ammonia (formed by nitrogen assimilation) reacts with a ketoglutaric acid to form the amino acid – glutamic acid. Here ∝ – ketoglutaric acid comes from Kreb’s cycle and hydrogen is donated by co-enzyme NADH or NADPH.
NCERT Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 8

(b) Transamination – Once the glutamic acid is synthesized by reductive amination, other amino acids are synthesized by the transfer of an amino groups to other carbon skeletons. Glutamic acid is the starting material from which 17 other amino acids are formed by the transfer of the amino group of an amino donor compound to the carboxyl position of an amino acceptor compound. Transaminase is the enzyme responsible for such a reaction.

Question 4.
Write an account of the role of mineral elements in a plant.
Solution:
Role of the mineral elements: Plants require mineral elements for various metabolic activities of their body. The following are some of the important functions which the mineral elements perform:

(i) Constituents of the plant body: Elements form the constitution of the plant body. For example carbon, hydrogen and oxygen are essential for the production of carbohydrates hence they are termed framework elements. Nitrogen, sulphur, and phosphorus are required for the synthesis of proteins. Magnesium is an important part of the chlorophyll molecule.

(ii) Influence on the pH of the cell sap: They also influence the pH of the cell sap.

(iii) Maintenance of osmotic pressure in the plant cells: The mineral salts and organic compounds of the cell sap produce necessary osmotic pressure.

(iv) They influence the permeability of cytoplasmic membrane: Different minerals decrease or increase the permeability of plasma membrane.

(v) They have balancing function reactions: Some of the minerals balance the effects of the other.

(vi) They take part in enzymatic reactions: Some elements work as activators while others work as inhibitors in various enzymatic reactions.

We hope the NCERT Solutions for Class 11 Biology at Work Chapter 12 Mineral Nutrition, helps you. If you have any query regarding NCERT Solutions for Class 11 Biology at Work Chapter 12 Mineral Nutrition, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis

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NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis.

Question 1.
By looking at a plant externally can you tell whether a plant is C3 or C4 ? Why and how?
Solution:
Plants that are adapted to dry tropical regions have the C4 pathway. They have a special type of leaf anatomy, they tolerate higher temperatures, they show a response to highlight intensities. Study vertical sections of leaves, one of a C3 plant and the other of a C4 plant.

Question 2.
By looking at which internal structure of a plant can you tell whether a plant is C3 or C4? Explain.
Solution:
In C4 plant internal structure of the leaf possess a special type of anatomy called ‘Kranz’ anatomy. ‘Kranz’ means ‘wreath’ and is a reflection of the arrangement of cells.

The bundle sheath cells may form several layers around the vascular bundles; they are characterised by having large number of chloroplasts, thick walls impervious to gaseous exchange and no intercellular spaces.

While in C3 plants, there is no special type of leaf anatomy. There is only a single type of chloroplast inC3 i.e. granal, while in C4 chloroplasts are dimorphic, i.e, granite in the mesophyll cells and agranal in the bundle sheath cells.

Question 3.
Even though very few cells in a C4 plant carry out the biosynthetic-Calvin pathway, yet they are highly productive, can you discuss why?
Solution:
Though these plants have the C4 oxalacetic acid as the first CO2 fixation product they use the C3 pathway or the Calvin cycle as the main biosynthetic pathway.
In C4 plants photorespiration does not occur. This is because they have a mechanism that increases the concentration of CO2 at the enzyme site.
This takes place when the C4 acid from the mesophyll is broken down in the bundle cells to release CO2 this results in increasing the intracellular concentration of CO2 In turn, this ensures that the Rubisco functions as a carboxylase minimizing the oxygenase activity.

Now that you know that the C4 plants lack photorespiration, you probably can understand why productivity and yields are better in these plants. In addition, these plants show tolerance to higher temperatures.

Question 4.
RuBisCO is an enzyme that acts both as carboxylase and oxygenase. Why do you think RuBisCO carries out more carboxylation in C4 plants.
Solution:
RuBisCO or Ribulose bisphosphate carboxylase – oxygenase enzyme can bind to both C02 and O2. This binding is competitive. The relative concentration of C02 and 02 determines which one of the two will bind to the enzyme.

In C4 plants photorespiration does not occur. This is because they have a mechanism that increases the concentration of C02 at the enzyme site.

This takes place when oxaloacetic acid is broken down in the bundle sheath cells to release C02.

It results in increased intracellular concentration of C02. This ensures that the RuBisCO functions as a carboxylase and minimising the oxygenase activity.

Question 5.
Suppose there were plants that had a high concentration of chlorophyll b, but lacked chlorophyll a, would it carry out photosynthesis? Then why do plants have chlorophyll b and other accessory pigments?
Solution:
Though chlorophyll is the major pigment responsible for trapping light, other thylakoid pigments like chlorophyll b, xanthophylls, and carotenoids, which are called accessory pigments, also absorb light and transfer the energy to ‘chlorophyll a’.

Indeed, they not only enable a wider range of wavelengths of incoming light to be utilized for photosynthesis but also protect ‘chlorophyll a’ from photo-oxidation. Reaction centre chlorophyll-protein complexes are capable of directly absorbing light and performing charge separation events without other chlorophyll pigments but the absorption cross-section is small.

Question 6.
Why is the colour of a leaf kept in the dark frequently yellow, or pale green? Which pigment do you think is more stable?
Solution:
Chlorophyll is unable to absorb energy in the absence of light and loses its stability, giving the leaf a yellowish colour. This shows that xanthophyll is more stable.

Question 7.
Look at leaves of the same plant on the shady side and compare it with the leaves on the sunny side. Or, compare the potted plants kept in the sunlight with those in the shade. Which of them has leaves that are darker green? Why?
Solution:
Light is a limiting factor for photosynthesis Leaves get lesser light for photosynthesis when they are in shade. Therefore, the leaves or plants in shade perform lesser photosynthesis as compared to the leaves or plants kept in sunlight. In order to increase the rate of photosynthesis, the leaves present in shade have more chlorophyll pigments.

This increase in chlorophyll content increases the amount of light absorbed by the leaves, which in turn increases the rate of photosynthesis. Therefore, the leaves or plants in shade are greener than the leaves or plants kept in the sun.

Question 8.
The figure shows the effect of light on the rate of photosynthesis. Based on the graph, answer the following questions.
(a) At which point/s (A, B, or C) in the curve is light a limiting factor?
(b) What could be the Jimiting factor/s in region A?
(c) What do C and D represent on the curve?
Solution:
NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis 1
(a) In the region ‘A’ and half of ‘BTight is limiting factor because rate of photosynthesis is increasing with the intensity of light.
(b) All the other factors except light.
(c) C represents a region where a factor other than light is limiting, e.g., CO2. D represents the light intensity at which rate of photosynthesis is maximum under existing conditions (e.g., CO2).

Question 9.
Give a comparison between the following:
(a) C3 and C4 pathways
(b) Cyclic and non-cyclic photophosphorylation
(c) Anatomy of leaf in C3 and C4.
Solution:
(a) Differences between C3 and C4 pathway
NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis 2
(b) Differences between cyclic and non-cyclic photophosphorylation are
NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis 3
(c) Differences between the anatomy of leaf in C3 plants and anatomy of leaf in C4 plants are
NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis 4

VERY SHORT ANSWER QUESTIONS

Question 1.
Write one anatomical feature of C4 plants.
Solution:
Kranz anatomy in leaf.

Question 2.
Which of the following is not a useful function of the light reaction in photosynthesis?
(a) splitting water
(b) synthesis of NADPH
(c) converting light energy into chemical energy
(d) releasing oxygen for photorespiration
Solution:
(d) Releasing oxygen for photorespiration.

Question 3.
What is the starting substance in the CO2 fixation cycle? (Apr. 91)
Solution:
RuMP.

Question 4.
Where is PS II located in a chloroplast?
Solution:
PS II is located in the appressed regions of grana thylakoid

Question 5.
Name the reaction centre of PS I and PS II.
Solution:
P700 & P680‘

Question 6.
What type of light causes maximum photo-synthesis? (Oct. 1995)
Solution:
Red light

Question 7.
How many molecules of ATP and how many molecules of NADPH are spent to fix three molecules of CO2 in the Calvin cycle?
Solution:
9 ATP and 6 NADPH

Question 8.
Why do the stomata of CAM plants open during the night?
Solution:
As these plants grow in dry areas, they keep stomata close during the day to conserve water and open their stomata during the night for the diffusion of gases.

Question 9.
Mention one useful role of photorespiration in plants.
Solution:
It protects the plants from photooxidative damage.

Question 10.
Cyanobacteria and some other photosynthetic bacteria don’t have chloroplasts. How do they conduct photosynthesis?
Solution:
Cyanobacteria have bluish pigment phycocyanin, which they use to capture light for photosynthesis. Some green bacteria (cyanobacteria) are red or pink due to pigment phycoerythrin. Whatever the colour of cyanobacteria, they are photosynthetic and so can manufacture food.

Question 11.
What is phosphorylation? (M.Q.P.)
Solution:
Synthesis of ATP either with the help of light (during photosynthesis) or in presence of oxygen (during respiration) is called phosphorylation.

Question 12.
Name the organism Englemann used in his experiment.
Solution:
Cladophora.

Question 13.
Write the currently accepted equation of photosynthesis in plants.
Solution:
NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis 5

Question 14.
What is a pigment?
Solution:
A pigment is a substance that absorbs light of certain wavelength(s).

Question 15.
Write the full form of NADP
Solution:
NADP – Nicotinamide adenine dinucleotide Phosphate.

Question 16.
Expand RuBP
Solution:
Ribulose 1, 5 bisphosphates.

Question 17.
Give a reason for the following:
Some bacteria exhibit photosynthesis but they do not produce oxygen. (July 2006)
Solution:
Some photosynthetic bacteria do not use water as their source of hydrogen, hence do not liberate oxygen.

Question 18.
Mention two conditions where light can become a limiting factor.
Solution:
Conditions in which light can become a limiting factor:
(i) Plants in the shade.
(ii) Plants growing under the canopy in a dense forest.

Question 19.
What are antenna molecules?
Solution:
Antenna molecules are light-harvesting pigment molecules that occur on the outer side of a photosynthetic unit.

Question 20.
What is a quantasome? Where is it present?
Solution:
Quantasome means photosynthetic units. It is equivalent is 230 chlorophyll molecules. These are present in the grana lamellae.

SHORT ANSWER QUESTIONS

Question 1.
Specify how C4 photosynthetic pathway increases carbon dioxide concentration in bundle sheath cells of sugarcane?
Solution:
In C4 pathway of sugarcane, C02 from atmosphere enters through the stomata in the mesophyll cell and combines with phosphoenol pyruvate to form a 4-C compound oxaloacetic acid. The OAA is then transported to the bundle sheath where it is decarboxylatedto release C02 in bundle sheath.

Question 2.
Differentiate between absorption spectrum and action spectrum.
Solution:
The main differences between absorption spectrum and action spectrum are as follows.
NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis 6

Question 3.
What are quantasomes? (Oct. 94)
Solution:
Quantasome is a functional unit (Photo-synthetic unit) made of a group of pigment molecules required for carrying out a photochemical reaction. The Pigment molecules are embedded in the grana and differentiated as pigment system I (with chi 670, chi 680, P 700) and pigment system ll(with chi 670, chi 680, P 680, and Xanthophylls)

Question 4.
Distinguish between cyclic and non-cyclic photophosphorylation. (M.Q.P., March 2011)
Solution:
Non cyclic photophosphorylation (a) Cyclic photophosphorylation

  1. The path traversed by an electron is non-cyclic.
    (a) Path traversed by electron is cyclic.
  2. Both PSI and PSII are active.
    (b) Only PSI is active.
  3. It is accompanied by photolysis.
    (c) No photolysis.
  4. The major pathway that takes place.
    (d) Secondary pathway when additional ATP is needed.

Question 5.
What is Blackman’s law of limiting factors?
Solution:
F.F. Blackman (1905) extended a law to formulate the principle of limiting factors. “When a process is conditioned as to its rapidity by a number of separate factors, the rate of the process is limited by the pace of slowest factors.”

Question 6.
In the condition of water stress why the rate of photosynthesis declines?
Solution:
Due to water stress, stomata remain closed and so there is a decrease in CO2concentration and the leaf water potential is also reduced, decline the rate of photosynthesis.

Question 7.
What is a reaction centre? Give the reaction centres of PSI and PSII.
Solution:
Reaction centre is a chlorophyll component of the photosystem and it absorbs as well as accepts energy from other pigments and ejects an electron. The reaction centre of PSII is Chla680 or P680 and PSI is Chla700 or P700

Question 8.
Why is photorespiration considered a wasteful process?
Solution:
Photorespiration considered a wasteful process because
(i) 25% of photosynthetically fixed carbon is lost in the form of C02.
(ii) There is no energy-rich useful compound produced during this process.

Question 9.
Give two reasons as to why photosynthesis is important for sustaining life on earth.
Solution:
Photosynthesis is the most important process because;
(i) it is the only natural process by which oxygen is liberated into the atmosphere.
(ii) it is the process by which food is manufactured for all living organisms.

Question 10.
Why does the rate of photosynthesis decrease at higher light intensities? What plays a protective role in such situations?
Solution:
Rate of photosynthesis decreases for two reasons :
(i) Other factors required for photosynthesis become limiting.
(ii) Destruction of chlorophyll by photo-oxidation.
Carotenoids play a protective role by:
(i) absorbing the excess light and
(ii) acting as an antioxidant to detoxify the effect of activated oxygen species.

Question 11.
What is C4 -pathway? Give an example. (March 2008)
Solution:
CA -pathway is an alternative photosynthetic pathway seen in plants like sugarcane/sorghum/ maize in which the stable compound is oxaloacetate a 4-C compound. It is called the Hatch-slack pathway.

Question 12.
What is kranz anatomy in plants?
Solution:
In Kranz Anatomy vascular bundles are surrounded by a layer of bundle sheath that contains a large number of chloroplasts in mesophyll cells and it is present in C4 plants e.g, Maize, Sugarcane, etc.

LONG ANSWER QUESTIONS

Question 1.
How is photosystem I different from photosystem II?
Solution:
The main differences between photosystem I and photosystem II are
NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis 7

Question 2.
Describe the factors that influence the rate of Photosynthesis. (Oct. 1989)
Solution:
The factors that affect photosynthesis may be both internal & External.

Internal factors:

  • Chlorophyll: it is the light-absorbing pigment and only portions of the plant having chlorophyll can help in photosynthesis.
  • Protoplasmic factor: young seedlings when transferred from darkness to light show the presence of some factors which is believed to be enzymatic initiates photosynthesis and is called the protoplasmic factor.

External factors:

  • Light: It is one of the most important factors which affects the process in 3 ways i.e. quantity, quality, and intensity. Quantity of light is otherwise duration and depends upon the photoperiod that is required by the plant quality refers to the wavelength, maximum photosynthesis occurs in red and blue light while minimum in green light. Intensity favours the process and low intensity decreases the rate of photosynthesis. Very high intensity brings about photooxidation of pigments which is called solarization.
  • CO2: An increase in CO2 concentration favours the process provided other factors are not limiting but very high concentrations are toxic and inhibit photosynthesis.
  • Temperature: Increase in temperature favour photosynthesis but above the optimum range the process decreases due to the denaturation of enzymes.

Question 3.
Explain the process of the biosynthetic phase of photosynthesis occurring in the chloroplasts.
Solution:
The biosynthetic phase of photosynthesis :

  • It occurs in the stroma of chloroplasts.
  • These reactions reduce the carbon dioxide into carbohydrates, making use of the ATP and NADPH2 produced in the photochemical reactions.
  • The reactions are also called as Calvin cycle.
  • The three phases of the Calvin cycle are as follows:

(i) Carboxylation
Six molecules of Ribulose 1,5 bisphosphate
react with six molecules of carbon dioxide to form six molecules of a short-lived 6C- compound.
The reaction is catalysed by RuBP carboxylase (RuBisCo).
The six molecules of the 6C-intermediate break into 12 molecules of 3- phosphoglyceric acid (3-PGA), an SC- compound.
It is through this step that carbon dioxide is fixed in the plant.
NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis 8
(ii) Reduction
12 molecules of 3-phosphoglyceric acid are converted into 12 molecules of 1, 3 diphosphate-glyceric acid, utilising 12 molecules of ATP and then reduced to 3- phosphoglyceraldehyde making use of 12 molecules of NADPH. Two molecules of phosphoglyceraldehyde react to form one molecule of glucose. It is in this step that there is an actual reduction of carbon dioxide leading to sugar formation.

(iii) Regeneration of RuBP
10 molecules of phosphoglyceraldehyde, by a series of complex enzyme-catalyzed reactions, are converted into six molecules of ribulose 1,5-bisphosphate; six molecules of ATP are needed for this step. This step of ‘ regeneration of RuBP is important for the cycle to continue

We hope the NCERT Solutions for Class 11 Biology at Work Chapter 13 Photosynthesis, helps you. If you have any query regarding NCERT Solutions for Class 11 Biology at Work Chapter 13 Photosynthesis, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants

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NCERT Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants.

Question 1.
State the location and function of different types ofmeristems.
Solution:
NCERT Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 1

Question 2.
Cork cambium forms tissues that form the cork. Do you agree with this statement? Explain.
Solution:
Sooner or later, another meristematic tissue called cork cambium or phellogen develops, usually in the cortex region. Cork cambium is a couple of layers thick. It is made of narrow, thin-walled, and nearly rectangular cells. Cork cambium cuts off cells on both sides. The outer cells differentiate into cork or phellem while the inner cells differentiate into secondary cortex or phelloderm. The cork is impervious to water due to suberin deposition in the cell wall. The cells of the secondary cortex are parenchymatous. Phellogen, phellem, and phelloderm are collectively known as periderm.

Question 3.
Explain the process of secondary growth in the stems of woody angiosperms with the help of schematic diagrams. What is its significance?
Solution:
NCERT Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 2
showing secondary growth.
Secondary growth in dicot stem:

  1. It is a “permanent increase in thickness due to the activity of vascular cambium and cork cambium in stellar and extrasolar regions”. In dicot stem intra fascicular cambium is present.
  2. The cells of the medullary ray become meristematic and form interfascicular cambium.
  3. These two cambiums unite and make a complete cambial ring.
  4. The cells of it divide and produce new cells both on its outer and inner sides.
  5. The cells formed on the outer side differentiate into secondary phloem while the cells of the inner side form secondary xylem.
  6. The epidermis is replaced by a secondary protective tissue by an increase in the growth of the stem of the plant. It is made of phellogen (cork cambium).
  7. It arises from the peripheral cells of the cortex. The phellogen forms new cells on the outer side which make phellem (cork) and phelloderm on its inner side also.
  8. Significance: Secondary growth increases the girth or thickness of the plant.
  9. Annual rings of woody angiosperms are very distinct and thus helps in determining the age of the plant.

Question 4.
Draw illustrations to bring out the anatomical difference between
(a) Monocot root and dicot root
(b) Monocot stem and dicot stem
Solution:
NCERT Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 3
NCERT Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 4

Question 5.
Cut a transverse section of the young stem of a plant from your school garden and observe it under a microscope. How would you ascertain whether it is a monocot stem or a dicot stem? Give reasons.
Solution:
After observing the transverse section of the stem we can differentiate that stem is monocot or dicot on the basis of the following characters:
NCERT Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 5
Question 6.
The transverse section of a plant material shows the following anatomical features –

  1. the vascular bundles are conjoint, scattered, and surrounded by a sclerenchymatous bundle sheath,
  2. phloem parenchyma is absent What will you identify it as?

Solution:
The transverse section of a typical young monocotyledonous stem shows that

  1. The vascular bundles are conjoint, scattered, and surrounded by sclerenchymatous bundle sheaths
  2. Phloem parenchyma is absent, and water containing cavities are present within the vascular bundles.

Question 7.
Why xylem and phloem are called complex tissues?
Solution:
Xylem and phloem a composed of several types of cells and they work as a unit. Hence they are called complex tissues.

Question 8.
What is the stomatal apparatus? Explain the structure of stomata with a labelled diagram.
Solution:

  1. Several minute openings or stomata are found on the epidermis of all the green aerial parts of plants but are abundant on the lower surface on the leaves as they regulate the process of transpiration.
  2. A large number of stomata occur on the upper surface of leaves of aquatic plants.
  3. Each stomata is surrounded by two cells known as the guard cells. In the dicotyledons plants these are bean-shaped, but in sedges and grasses these are dumb-bell-shaped.
  4. The guard cell is living. Their outer walls are thin where as the inner ones surrounding the aperture are highly thickened.
  5. Due to this variation in the thickening, the guard cell may become turgid and flaccid, depending upon the supply of water in them, which makes the opening and closing of stomata possible.
  6. Some times a few neighbouring epidermal cells in the vicinity of guard cells become specialized in their shape and size and contents. These are known as subsidiary cells.
  7. The stomatal aperture, guard cells and the surrounding subsidiary cell are together called stomatal apparatus.

Question 9.
Name the three basic tissue systems in the flowering plants. Give the tissue names under each system.
Solution:
On the basis of their structure and location, there are three types of tissue systems. These are the

  1. Epidermal tissue system,
  2. The ground or fundamental tissue system and
  3. The vascular or conducting tissue system.

1. Epidermal tissue system The epidermal tissue system forms the outer-most covering of the whole plant body and comprises epidermal cells, stomata, and the epidermal appendages the trichomes, and hairs.
2. All tissues except epidermis and vascular bundles constitute the ground tissue. It consists of simple tissues such as parenchyma, collenchyma, and sclerenchyma.
3. The vascular system consists of complex tissues, the phloem, and the xylem. The xylem and phloem together constitute vascular bundles.

Question 10.
How is the plant anatomy useful to us?
Solution:
‘The study of plant anatomy is useful in many ways. First of all the study helps us understand the way a plant functions carrying out its routine activities like transpiration, photosynthesis, and growth and repair. Second, it helps botanists and agriculture scientists to understand the disease and cure for plants. Plants are important to maintain the ecological balance of the earth, so understanding plant anatomy is a way to understand the large system of the ecology on this planet.

Question 11.
What is periderm? How does periderm formation take place in the dicot stems?
Solution:
Phellogen, phellem, and phelloderm are collectively known as periderm. Phellogen develops, usually in the cortex region. Phellogen is a couple of layers thick. It is made of narrow, thin-walled, and nearly rectangular cells. Phellogen cuts off cells on both sides. The outer cells differentiate into cork or phellem while the inner cells differentiate into secondary cortex or phelloderm. The cork is impervious to water due to suberin deposition in the cell wall. The cells of the secondary cortex are parenchymatous.

Question 12.
Describe the internal structure of a dorsiventral leaf with the help of labelled diagrams.
Solution:
Dorsiventral (dicotyledonous) leaf: The vertical section of a dorsiventral leaf through the lamina shows three main parts, namely, epidermis, mesophyll, and vascular system.
Epidermis: The epidermis which covers both the upper surface (adaxial epidermis) and lower surface (abaxial epidermis) of the leaf has a conspicuous cuticle. The abaxial epidermis generally bears more stomata than the adaxial epidermis. The latter may even lack stomata.
Mesophyll:

  1. The tissue between the upper and the lower epidermis is called the mesophyll.
  2. It possesses chloroplasts and carries out photosynthesis, is made up of parenchyma.
  3. It has two types of cells – the palisade parenchyma and the spongy parenchyma.
  4. The adaxially placed palisade parenchyma is made up of elongated cells, which are arranged vertically and parallel to each other.
  5. The oval or round and loosely arranged spongy parenchyma is situated below the palisade cells and extends to the lower epidermis.
    NCERT Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 11
    6. There are numerous large spaces and air cavities between these cells.

Vascular system:

  • This includes vascular bundles, which can be seen in the veins and the midrib.
  • The size of the vascular bundles is dependent on the size of the veins.
  • The veins vary in thickness in the reticulate venation of the dicot leaves. The vascular bundles are surrounded by a layer of thick-walled bundle sheath cells.

VERY SHORT ANSWER QUESTIONS

Question 1.
Vascular bundles having cambium are known as.
Solution:
Open, Vascular bundle

Question 2.
Name the two types of sclerenchyma.
Solution:
Sclerenchyma fibers and stone cells.

Question 3.
From where do the secondary meristems originate?
Solution:
Permanent tissue.

Question 4.
What does make the root apical meristem subterminal?
Solution:
The presence of the root cap makes the root apical meristem subterminal.

Question 5.
Where are companion cells located in flowering plants? What are their functions?
Solution:
Companion cells are located in phloem cells of vascular tissues, they support the sieve tubes in water conduction.

Question 6.
What is the advantage of lignocellulose in the wall of the xylem?
Solution:
It provides rigidity, thickness, and resistance

Question 7.
A cross-section of a plant material shows the following features under the microscope: vascular bundles are radially arranged. These are found xylem strands showing exarch condition. What type of plant part of is this?
Solution:
Dicot root.

Question 8.
Based on position, classify various types of meristems
Solution:
Apical, intercalary and lateral meristems.

Question 9.
Name the various component cells of xylem. Which of them does not have a nucleus?
Solution:
Tracheids, vessels, xylem parenchyma andxylem fibres. Only xylem parenchyma have nucleus and living.

Question 10.
Give an example of a secondary meristem.
Solution:
Examples of secondary meristem are cork cambium and interfascicular cambium.

Question 11.
Name the tissue involved in linear and lateral growth in plants.
Solution:
Linear growth is caused by apical meristem and lateral growth is caused by lateral meristem.

Question 12.
Heartwood is more durable than springwood. Why?
Solution:
Heartwood is more durable than spring wood due to its little susceptibility to the attack of pathogens and insects.

Question 13.
Where these present:

  1. Hypodermis layer
  2. Mesophyll tissue
  3. Stomata
  4. Cambium

Solution:

  1. Hypodermis layer – is found in stems
  2. Mesophyll tissue – in leaves
  3. Stomata – lower epidermis in leaves
  4. CambiumIn vascular bundles which are open

SHORT ANSWER QUESTIONS

Question 1.
What are the differences between root hairs and stem hairs?
Solution:
The main difference between stem hairs and root hairs are :
.NCERT Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 9

Question 2.
Draw well labelled diagrams of the T.S. of dicotyledonous leaf.
Solution:
NCERT Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 10

Question 3.
Why is cambium considered to be a lateral meristem?
Solution:
Cambium is responsible for the increase in the thickness of stems and roots as a result of the addition of secondary tissues (secondary cortex, secondary phloem and secondary xylem). They are located at the lateral position so-known as lateral meristems.

Question 4.
Name the plant part in which the endodermis is absent. Give one basic difference between the endodermis and epidermis.
Solution:
The endodermis is absent in leaves. Cells of endodermis possess Casparian strips or bands in their radial and transverse walls which are not found in the epidermis.

Question 5.
What are Casparian strips?
Solution:
These are thickenings of lignin and suberin formed around the lateral walls of the endodermis to prevent plasmolysis.

Question 6.
Which tissue is most abundantly found in plants? Where all is it present in plants?
Solution:
The tissue most abundantly found in plants is parenchyma. It is found in pith, cortex, and in entire mesophyll of the leaves.

Question 7.
What is present in the phloem of leaves besides sieve elements and is it living or dead? How are these functional & used?
Solution:
Besides sieve elements, in phloem parenchyma, living cells are present. These store food other cells are phloem fibres that are dead and provide mechanical strength. These are also used in making ropes and coarse textiles.

LONG ANSWER QUESTIONS

Question 1.
Describe the structure and functions of xylem tissues in angiosperm plants.
Solution:

  • Xylem is a complex tissue. It forms a part of the vascular bundle.
  • It is mainly concerned with the conduction of water and minerals. It also provides mechanical support to the plant.
  • As a conducting strand, xylem forms a continuous channel through the roots, stem, leaves and other aerial parts.
  • It consists of four different types of cells—xylem vessels, trachieds, xylem fibres and xylem parenchyma.
  • Xylem vessels and tracheids are concerned with the conduction of water and minerals from roots to aerial parts of the plant.
  • Xylem fibres provide mechanical strength to the plant body. Xylem parenchyma are the only living components of xylem.
  • These are concerned with the storage of food and other vital functions.

Question 2.
What is collenchyma? Explain its structure and function in the plant body of a herbaceous angiosperm.
Solution:

  • The cells of collenchyma are somewhat elongated with cellulose thickening, found as longitudinal strips.
  • These are usually confined to the comers of the cells.
  • Collenchyma cells appear circular, oval or angular in the transverse section. Internally, each cell possesses a large 4 central vacuole, peripheral cytoplasm and a nucleus.
  • Collenchyma is usually found beneath the epidermis in stem, petiole and leaves of herbaceous dicot plants. It is usually absent in monocot stems and monocot roots.

Functions:

  • It provides tensile strength and rigidity to the plants due to thickening.
  • Chloroplasts containing collenchyma cells are responsible for photosynthesis.
  • Collenchyma also provides elasticity to the plant organs.
  • Collenchyma are alive and also stores food.

Question 3.
Explain sclerenchyma with a well labelled diagram.
Solution:
Sclerenchyma is a simple permanent tissue. It consists of two types of cells. They are sclerenchyma fibres and sclereids.
(a) Sclerenchyma fibres –

  • These are much elongated fibers with tapering ends.
  • On ipaturity, they lose their protoplasm and become dead. Their cell wall is made up of cellulose or lignin, or both.
  • Central cavity of the cell is greatly reduced due to the formation of secondary thickening. Sclerenchyma provides mechanical strength to the plants.
  • They help in conduction when present in the secondary xylem.

(b) Sclereids –

  • They develop from ordinary parenchyma cells by the deposition of lignin.
  • These cells are thick-walled and highly lignified and become dead on maturity.
  • They are broader as compared to fibers and their cell lumen is veiy narrow.
  • Sclereids protect the plant from environmental forces like a strong wind.
  • They provide mechanical strength and rigidity to the plant.
    NCERT Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 7

Question 4.
Describe the structure of a monocotyledonous leaf.
Solution:
Anatomy of Monocot/isobilateral leaf: The upper and lower surfaces are covered by a single-layered epidermis.

  • The upper epidermis has some cells larger than the others; such large cells are known as bulliform/motor cells.
  • Stomata are found on both upper and lower epidermal layers. The mesophyll is not differentiated into palisade and spongy parenchyma.
  • Mesophyll cells are isodiametric and are arranged compactly; they contain a number of chloroplasts. Since monocot leaves have parallel veins, a number of vascular bundles can be seen in a row in the section.
  • Each vascular bundle has sclerenchyma cells (caps) on its upper and lower edges.
  • The xylem is on the upper side and the phloem on the lower side. There is a parenchymatous bundle sheath, which often contains chloroplasts and performs the function of photosynthesis.
    NCERT Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 8

Question 5.
Give two examples & salient features of
(1) Simple Tissue
(2) Complex Tissue
Solution:
(1) Simple Tissue:
(i) Parenchyma
(ii) Collenchyma
(2) Complex Tissue:
(i) Xylem
(ii) Phloem
(1) Simple Tissue:
(i) Parenchyma: These are living, thin-walled cells. It is used for storage of food, induction of substances, provides turgidity to softer parts of the plants
(ii) Collenchyma: These are longer than parenchyma. These are living mechanical tissue, it provides mechanical strength to organs and is present in peripheral position in plants to resist bending my the mind.
(2) Complex Tissue:
(i) Xylem: This is also called Hadrome, which is a water-conducting tissue. It is made up of cells like tracheids, xylem fibers, and xylem parenchyma only xylem parenchyma is living and all others are dead.
(ii) Phloem: This is also called Bast, which is a conducting tissue of food from leaves to all parts of the body. The parts of phloem are sieve elements, companion cells, phloem fibres, and phloem parenchyma. Phloem fibres are dead while parenchyma is living. Together these perform their function.

We hope the NCERT Solutions for Class 11 Biology at Work Chapter 6 Anatomy of Flowering Plants, help you. If you have any query regarding NCERT Solutions for Class 11 Biology at Work Chapter 6 Anatomy of Flowering Plants, drop a comment below and we will get back to you at the earliest.