CBSE class 11 Biology Solutions

NCERT Exemplar Solutions for Class 11 Biology Chapter 9 Biomolecules

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NCERT Exemplar Solutions for Class 11 Biology Chapter 9 Biomolecules

These Solutions are part of NCERT Exemplar Solutions for Class 11 Biology. Here we have given NCERT Exemplar Solutions for Class 11 Biology Chapter 9 Biomolecules.

VERY SHORT ANSWER QUESTIONS

Question 1.
Medicines are either man made (i.e. synthetic) or obtained from living organisms like plants, bacteria, animals, etc., and hence, the latter are called natural products. Sometimes, natural products are chemically altered by man to reduce toxicity or side effects. Write against each of the following whether they were initially obtained as a natural product or as 3 synthetic chemical.
(a) Penicillin
(b) Sulphonamide
(c) Vitamin-C
(d) Growth hormone
Solution:
(a) Penicillin is a group oT antibiotics derived from fungi Penicillium obtained naturally.
(b) Sulphonamide an antimirobial agent is a synthetic chemical.
(c) Vitamin-C or L-ascorbic acid or ascorbate is a natural product and an essential nutrient for humans. It is present in citrus fruits.
(d) Growth hormone also known as somatotropin or somatropin is a peptide hormone occurring naturally in the body it stimulates growth.

Question 2.
Write the name of any one amino acid, sugar, nucleotide and fatty acid.
Solution:
(a) Amino acid — Leucine
(b) Sugar — Lactose
(c) Nucleotide — Adenosine triphosphate
(d) Fatty acid — Palmitic acid

Question 3.
Reaction given below is catalysed by oxidoreductase between two substrates A and
A’, complete the reaction.
A reduced + A’ oxidised —>
Solution:
Oxidoreductase is an enzyme that catalyses oxidation reduction reactions. This enzyme is associated in catalysing the transfer of electron from one molecule (the reduction), also called as electron donor, to another molecule (the oxidant), also called as electron acceptor.
The complete reaction is
NCERT Exemplar Solutions for Class 11 Biology Chapter 9 Biomolecules 1.1

Question 4.
How are prosthetic groups different from co¬factors?
Solution:
Organic compounds that are tightly bound to the apoenzyme, (an enzyme without cofactor) by covalent or non-covalent bonds are prosthetic groups e.g., peroxidase and
catalase catalyse the breakdown of hydrogen peroxide to water and oxygen where haeme is the prosthetic group and it is a part of the active site of the enzyme.
Co-factor is small, heat stable and non-protein part of conjugate enzyme. It may be inorganic or organic in nature. Co-factors when loosely bound to an enzyme is called coenzyme and when tightly bound to apoenzyme is called prosthetic group.

Question 5.
Glycine and alanine are different with respect to one substituent on the a-carbon. What are 4. the other common substituent groups?
Solution:
The common substituted groups in both the amino acids are NH2 COOH and H.
NCERT Exemplar Solutions for Class 11 Biology Chapter 9 Biomolecules 1.2

SHORT ANSWER QUESTIONS

Question 1.
Enzymes are proteins. Proteins are long chains of amino acids linked to each, other by peptide bonds. Amino acids have many functional groups in their structure.
NCERT Exemplar Solutions for Class 11 Biology Chapter 9 Biomolecules 1.3
These functional groups are many of them at least, ionisable. As they are peak acids and bases in chemical nature, this ionisation is influenced by pH of the solution. For many enzymes, activity is influenced by surrounding pH. This is depicted in the curve below, explain briefly.
Solution:
Enzymes, generally function in a narrow range of pH. Most of the enzymes show their highest activity at a particular pH called optimum pH- activity declines below and above this value. Extremely high or low pH values generally results in complete loss of activity for most enzymes. The given graph represents the maximum enzyme activity at the optimum pH.

Question 2.
Explain the association of carbohydrate to the plasma membrane and its significance.
Solution:
Secondary metabolites are chemicals produced by plants which do not play any [role] in growth, photosynthesis reproduction or other primary functions of the plant. Rubber (cis 1,4- polyisopyrene) is a secondary metabolite.
(i) Rubber is extracted from Hevea brasiliensis (rubber tree)
(ii) It is a byproduct of the lactiferous tissue of the vessels that are in the form of latex.
(iii) It contains over 400 isoprene units and thus is the largest of the terpenoids.
(iv) It is elastic, water proof and a good conductor of electricity.

Question 3.
Nucleic acids exhibit secondary structure, justify with example.
Solution:

  1. Nucleic acids are large biological molecules, essential for all known forms of life.
  2. The secondary structure of a nucleic acid molecule refers to the base pairing interactions within a single molecule or set of interacting molecules.
  3. DNA and RNA represent two main nucleic acids, their secondary structures however differ the secondary structure of DNA comprises of two complementary strands of polydeoxyribonucleotide, spirally coiled on a common axis forming a helical structure.
  4. This double helical structure of DNA is stabilized by phosphodiester bonds (between 5’ of sugar of one nucleotide and 3 sugar of another nucleotide), hydrogen bonds (between bases, and ionic interactions.

Question 4.
Comment on the statement ‘living state is a non-equilibrium steady state to be able to perform work’
Solution:

  1. Living organism are not in equilibrium because work cannot be performed by a system at equilibrium.
  2. The living organisms exist in a steady state characterised by concentration of each of the biomoleculSs.
  3. These biomolecules are in a metabolic flux. Any chemical or physical process moves simultaneously to equilibrium.
  4. Living organisms work continuously and they cannot afford to reach equilibrium.
  5. The living state thus is an a non-equilibrium steady-state to be able to perform work. This achieved by energy input provided by metabolism.

LONG ANSWER QUESTIONS

Question 1.
What are different classes of enzymes? Explain any two with the type of reactions they catalyse.
Solution:
Enzymes are divided into six classes each with
4-13 sub-classes and named accordingly by a number comparising of four digits.
(i) Oxidoreductases/dehydrogenases : These enzymes take part in oxidation, reduction or transfer of electrons,
(ii) Transferase : These enzymes transfer a functional group (other than hydrogen).
from one molecule to another. The transfer , chemical group does not occur in free state.
(iii) Hydrolases : These enzymes catalyse the hydrolysis of bonds like ester, ether,
peptide, glycosidic C-C, C-halide, P-N etc.
NCERT Exemplar Solutions for Class 11 Biology Chapter 9 Biomolecules 1.4
(iv) Lyases cause cleavage, removal of groups without hydrolysis and addition of groups to double bonds or removal of groups producing double bonds.
NCERT Exemplar Solutions for Class 11 Biology Chapter 9 Biomolecules 1.5
(v) Isomerases rearrangement of molecular structure to effect isomeric changes. They are of three types isomerases, epimerases and mutases.
NCERT Exemplar Solutions for Class 11 Biology Chapter 9 Biomolecules 1.6
(vi) Ligases catalyse bonding of two chemicals with the help of energy obtained from ATP resulting formation of bonds such as C—O, C—S, C—N and P—O e.g., pyruvate carboxylase
Pyruvric acid + C02 + ATP + H20
NCERT Exemplar Solutions for Class 11 Biology Chapter 9 Biomolecules 1.7

We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 9 Biomolecules, help you. If you have any query regarding NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 9 Biomolecules, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Solutions for Class 11 Biology Chapter 8 Cell: The Unit of Life

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NCERT Exemplar Solutions for Class 11 Biology Chapter 8 Cell: The Unit of Life

These Solutions are part of NCERT Exemplar Solutions for Class 11 Biology. Here we have given NCERT Exemplar Solutions for Class 11 Biology Chapter 8 Cell: The Unit of Life.

VERY SHORT ANSWER QUESTIONS

Question 1.
Mention a single membrane bound organelle which is rich in hydrolytic enzymes.
Solution:
The membrane bourld vesicular structures formed by Golgi apparatus are Lysosomes. These vesicles have been found to be rich in all types of hydrolytic enzymes as hydrolase, lipases, proteases and carbohydrases which digest carbohydrates proteins, lipids and nucleic acid at an acidic pH.

Question 2.
What are gas vacuoles? State their functions.
Solution:
Gas vacuoles also known as pseudovacuoles or air vacuoles are the characteristic feature 1 of prokaryotes. They store metabolic gases and take part in regulation of buoyancy.

Question 3.
What is the function of a polysome? (Gk. Poly – many, Soma = body).
Solution:
A polysome consists a cluster of ribosomes that are held simultaneously by a strand of messenger KNA in rosette or helical group. They contain a portion of the genetic code that each ribosome is translating and are used in formation of multiple copies of same polypeptide. They are found in the cyloplasm during the process of active protein synthesis.

Question 4.
What is the feature of a metacentric chromosome?
Solution:
The centromere is median, in metacentric chromo-some. The centromere lies in the middle portion and forms two equal arms of chromosome.
NCERT Exemplar Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 1.1

Question 5.
What is the feature of a metacentric chromosome?
Solution:
Additional constriction or secondary constriction at the chromosomal ends as distal part of the arm formed by chromatin thread are known satellite chromosomees. These constriction gives appearance of an outgrowth or a small fragment.
These are also known as (sat) chromosomes or marker chromosomes. Chromosomes 13,14, 15, 16, 21 and 21 satellite chromosomes.
NCERT Exemplar Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 1.2

SHORT ANSWER QUESTIONS

Question 1.
Discuss briefly the role of nucleous in the cells activity involved in protein synthesis.
Solution:
The round, naked and a slightly irregular structure, which is attached to the chromatin at a specific region called as Nucleolar Organizer Region (NOR). Nucleous was first discovered by Fontana (1781).
NCERT Exemplar Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 1.3
The role of nucleolus can be described as:
(i) Nucleolus is the chief site for the synthesis of ribosomal RNA.
(ii) It is the centre for the formation of ribosome components.
(iii) It is the colloidal complex that fills the nucleus.
(iv) It combines rRNA with proteins to produce ribosomal sub-units. The ribosomes sub-units after their formation pass out and get established in the cytoplasm.
(v) It also receives and stores ribosomal proteins formed in the cytoplasm.
(vi) These ribosomal proteins formed are the sites for protein synthesis in the cell.
(vii) Nucleolus is essential for spindle formation during nuclear division as well.

Question 2.
Explain the association of carbohydrate to the plasma membrane and its significance.
Solution:

  1. The plasma membrane, surrounds th cell. It consists of lipids, proteins and carbohydrates that are imperative in both structure and function of the cell.
  2. Carbohydrates attach either with proteins or lipids usually making up less than 10% of the membrane weight.
  3. They can give rise to a wide variety of structures in relatively short chains. They give distinguishing features to individual cell types and thus they may be involved.
  4. Cell Recognition like ABC surfaces have carbohydrates arranged in branched chains: difference in the arrangement give rise to different blood group antigens (i.e., A, B and O).
  5. Cell surface differences are also responsible for the specificity of action of cells with hormones, drugs, viruses or bacteria. The cause of difference of cell surface is related to characteristic surface due to carbohydrate component.

Question 3.
Briefly describe the cell theory.
Solution:
Schleiden and Schwann formulated the cell theory, in 1938-39 which stated
(i) All living beings are made up of cells and products formed by the cells.
(ii) Cells are the structural and functional units oflife
The cell theory stated by Schleiden and Schwann failed to explain the question of origin of cells.
A major expansion of the cell theory was expressed by Virchow in his statment ‘Omnis cellula e cellula’ (all cells arise from pre-existing cells) in 1855.
This concept, was the actual idea of Nagelli (1846), which later on was elaborated by Virchow, along with considerable evidences in its support. The work of Nagelli and Virchow established cell division as the central pehnomenon in the continuity oflife.
The modem cell theory is thus based on two facts
(i) All living organisms are composed of cells and products of cells.
(ii) Cells are the basic structural and functional units oflife.
(iii) All cells arise from pre-existing cells. Vimses are exception to cell theory as they are .pot composed of cell. They consist of a nucleic acid (DNA or RNA) surrounded by a protein sheet and are incapable of independant existence, self regulation and self reproduction.

Question 4.
Give the biochemical composition of plasma membrane. How are lipid molecules arranged in the membrane?
Solution:
Chemcial composition of plasma membrane includes
Component                      Composition
Lipids                               (20-79%)
Proteins                           (20-70%)
Carbohydrates                (1-5%)
Water                                20%
Lipids form the continuous structural frame of the cell membrane and hence are the major components of the cell membrane. Lipids such as phqspholipics, glycolipids, and steroids are found
NCERT Exemplar Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 1.4
The lipid molecule possess both polar hydrophilic (water loving) and non-polar hydrophobic (water repelling) ends. The hydrophilic region is in the form of a head, while the hydrophobic part contains fatty acid tails. Hydrophobic tail is present towards the centre of the membrane. This structures results is the formation of lipid bilayer known as unit membrane/biological membrane/cell membrane. Proteins are embedded within the lipid bilayer – Carbohydrates are structure upon proteins.

Question 5.
What are plasmids? Describe their role in bacteria.
Solution:

  1. A plasmid is usually a circular (sometimes linear), double stranted DNA that can autonomously replicate.
  2. These are found in the cytoplam of the bacterial cell. Plasmids normally remain separated from the chromosome, but sometimes may temporarily integrate into it and replicate with it incidentally.
  3. Role and Plasmids in Bacteria Plasmids are the extra chromosomal circular, independently replicating unit besides nucleoid in the bacterial cell.
  4. Plasmids are used to transfer information from one cell to another, i.e., transfer of important genes, enabling to metabolise a nutrient, which normally a bacteria is unable to. It also helps in conjugation of bacteria.
  5. These days plasmids are used in a variety of recombination experiments, as cloning vectors.

LONG ANSWER QUESTIONS

Question 1.
Is there a species specific or region specific type of plastids? How does one distinguish one from the other?
Solution:
Plastids are specific to different species and are found in all plant cells and in euglenoids. They bear certain pigments that impart specific colour^ to the part of the plant possesing them. Plastids ar classified into three main types, based on the type of pigments- leucoplasts, chromoplast and chloroplast.
Leucoplasts are colourless plastids which store food material. They are of three types based on their storage products.
(a) Amyloplasts store starch, e.g., tuber of potato, grain of rice, grain of wheat.
(b) Elaioplasts store fats, e.g., rose
(c) Aleuropiasts are protein storing plastids, e.g., castor endosperm.
Chromoplast are non photosynthetic coloured plastids which synthesise and store carotenoid pigments. They appear orange, red or yellow. These mostly occur in ripe fruits (tomato and chilies) carrot roots, etc.
Chloroplasts are green color plastids which help in synthesising food material by photosyntheis. They contain chrophyll and carotenoid pigments which trap light energy.
Each chloroplast is oval or spherical, double membrane bound cell organelle. The space present inside inner membrane is called stroma Anumberrof oiganised flattenedmembranous sacs called thylakoids are present in the stroma. Thylakoids are arranged in stacks called grana.
The thylakoids of different grana are connected by membranous tubules called the stroma lamellae. The stroma of the lamellae contain the enzymes that are required for the synthesis of carbohydrates and proteins.
NCERT Exemplar Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 1.5

Question 2.
Write the functions of the following
(a) Centromere
(b) Cell wall
(c) Smooth ER
(d) Golgi apparatus
(e) Centrioles
Solution:
(a) Centromere is required for proper chromosome segregation. The centromere consists of two sister chromatids. It is also necessary for attachment of chromosomes to the spindle apparatus during mitosis and meiosis.
NCERT Exemplar Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 1.6
(b) Cell wall gives a definite shape to the cell 1 and protects the cell from mechanical injury
and infections. It also aids in cell to cell interaction and acts as a barrier for undesirable macromolecules.
(c) Smooth ER helps in synthesis of lipids, metabolism of carbohydrates, regulation of calcium concentration, drug detoxification and attachment of receptors on cell membrane proteins.
The smooth ER also contains enzymes- glucose 6 phosphatase, which converts glucose 6 phosphate to glucose essential in glucose metabolism.
NCERT Exemplar Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 1.7
(d) Golgi apparatus is an important site for the formation of glycoprotein and glyco lipids also involved in the synthesis of cell wall materials and plays an important role in formation of cell plate during cell divisionas well.
NCERT Exemplar Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 1.8
(e) Centrioles form the base body of cilia and flagella and spindle fibres that gives rise to spindle apparatus during cell division in ‘animal cells. They help in formation of microtubules and sperm tail. They also help in cell division by forming asters, which acts as spindle pole.

We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 8 Cell: The Unit of Life, help you. If you have any query regarding NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 8 Cell: The Unit of Life, drop a comment below and we will get back to you at the earliest.

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

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

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

VERY SHORT ANSWER QUESTIONS

Question 1.
State the number of segments in earthworm which are covered by a prominent dark band or clitellum.
Solution:
Segments 14-16 bare covered by a prominent dark band of glandular tissue called clitellum in a mature earthworm. It secretes mucus and albumen that help in formation of cocoon.
NCERT Exemplar Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals 1.1

Question 2.
Where are sclerites present in cockroach?
Solution:
In all the body segments of cockroach sclerites are present. They are of two types dorsal sclerites often known as tergites, and ventral sclerites which are referred to as sternites.

Question 3.
How many times do nymphs moult to reach the adult form of cockroach?
Solution:
The nymph grows by moulting about 13 times to reach. In cockroach’, the development is indirect and paurometabous adult form has three stages, i.e., egg, nymph and adult. The nymph resembles adult except/or undeveloped wings and genitalia.

Question 4.
Identify the sex of a frog in which sound producing vocal sacs are present.
Solution:
Sex of frogs can be distinguished on the basis of presence of sound producing vocal sacs. These organs are present in males which make them crock lauder than females, so as to attract females for mating.

Question 5.
A muscle fibre tapers at both ends and does not show striations. Name the muscle fibre.
Solution:
Muscle fibres that taper at both the ends (fusiform) and do not show striations are smooth muscle fibres. They are also called involuntary muscles.

Question 6.
Name the different cell junctions found in tissues.
Solution:
The different cell junctions found in tissue include:
(i) Tight junctions are regions where plasma membrane of adjacent epithelial cells are held close together. They check the movement of material between then.
(ii) Gap junctions are meant for chemical exchange between adjacent cells.
(iii) Adhering junctions function to keep neighbouring cells together.

Question 7.
Give two identifying features of an adult male frog.
Solution:
The two identifying features of an adult male frog include
(a) Nuptial Pad is a copulatory pad present on the first digit of the forelimb of male frog and helps in closing female during amphelexus.
(b) Vocal Sacs are loose skin folds on throat of male frogs for producing louder croak to attract females for mating purposes.

Question 8.
Which mouth part of cockroach is comparable to our tongue?
Solution:
In cockroach, hypopharynx acts as a tongue and lies within cavity enclosed by the mouth parts.

Question 9.
The digestive system of frog is made of the following parts. Arrange them in an order beginning from mouth. Mouth, oesophagus, buccal cavity, stomach, intestine, cloaca, rectum, cloacal aperture.
Solution:
The correct arrangement of the part of digestive system in frog is
Mouth —> Buccal cavity —> Oesophagus —> Stomach —>Intestine —> Rectum —> Cloaca —> Cloacal aperture.

Question 10.
What is the difference between cutaneous and pulmonary respiration?
Solution:
In frog respiration takes place via the skin as well lungs.
Pulmonary respiration and occurs outside the water through lungs. Cutaneous respiration takes place in water as well as land, occurs through highly vascularised moist skin.

Question 11.
Special venous connection between liver and intestine and between kidney and intestine is found in frog, what are the called?
Solution:
In frog, venous connection between liver and intestine is called hepatic portal system and venus connection between the kidney and the lower parts of the frog is called renal portal system.

SHORT ANSWER QUESTIONS

Question 1.
Stratified epithelial cells have limited role in secretion. Justify their role in our skin.
The edible part of the peach or pear pome fruit for the fleshy thalamus.
Solution:

  1. Stratified epithelium consists of epithelial cells in which the innermost layer is made up of columnar or cuboidal cells.
  2. It is a type of compound epithelium and a waterproof protein called keratin is present few outer layers.
  3. These layers of dead cells is called homy layer which is shed at intervals due to frictions and thus has a limited role in secretions and absorption.
  4. The main function of stratified epithelium is to provide protection to the body against mechanical and chemical stresses.

Question 2.
How does a gap junctions facilitate intercellular communication ?
Solution:
Intercellular communication is facilitated by gap Junction allowing small signaling molecules to pass from cell to cell.
These are fine hydrophilic channels, between two adjacent animal cells that are formed with the help of two protein cylinders called connexus.
Each connexus consists of six proteins subunits that surround a hydrophilic channel. Opening or closing of channel is controlled by pH and Ca2+ ion concentration.

Question 3.
Why are blood, bone and cartilage called connective tissue?
Solution:

  1. Connective tissue pt’ovides the structural framework and support to different organs forming tissue.
  2. Blood is a fluid or vascular connective tissue, which connects various organs and transports substances from one place to another.
  3. Bone is a solid, rigid and strong skeletal connective tissue, which supports the body and helps in locomotion.
  4. Cartilage is also a skeletal connective tissue, not as rigid bone but piable and resists compression.
  5. It plays role in support and protection and present in tip of nose, outer ear joints etc.

Question 4.
How do you distinguish between dorsal and ventral surface of the body of earthworm?
Solution:
The body of an earthworm can be distinguished into dorsal and ventral sides due to the presence of certain peculiar feature in it which include the following.
(i) The dorsal surface is darker than ventral surface because it is marked by a dark median mid dorsal line along the longitudinal axis of body. This is due to dorsal blood vessel, seen through integument.
(ii) Genital openings (pores), are present in the ventral surface of both male and female.
(iii) On vental surface genital papilla is located and helps in copulation.
NCERT Exemplar Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals 1.2

Question 5.
Complete the following statement.
(a) In cockroach grinding of food particle is performed by ……..
(b) malpighian tubules help in removal of …….
(c) Hind gut of cockroach is differentiated into……
(d) In cockroach blood vessels open into spaces called ……
Solution:
(a) Gizzard is a muscular and greatly folded structure which marks the end of foregut in cockroach and bears six plates with teeth for crushing and grinding the food.
(b) Malpighian tubules are excretory in ‘ function as they help in the removal of
nitrogenous wastes in arthropods.
(c) Ileum, colon and rectum and rectum opens and through anus.
(d) Haemocoel is the body cavity of cockroach divided into sinuses and contains visceral organs of cockroach floating in haemolymph.

Question 6.
Mention special features of eye in cockroach. Discuss compound eye in arthropods and mention its structural features.
Solution:

  1. In cockroach the eyes are large sessile, paired bean-shaped and present on either side of head.
  2. The are compound in nature. Each compound eye consists of a large number of visual elements called ommatidia.
  3. Each ommatidium is composed of a diopteric region and reticular (receptor) region. It is capable of producing a separate image of a small part of object seen.
  4. Thus, the image of the object viewed consists of several pieces and is known as mosaic image.
  5. Fine nerve fibres arise from the inner end of each ommatidium all of which combine to form one optic nerve connected to the brain.
    NCERT Exemplar Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals 1.3

Question 7.
Frog is a poikilotherm, exhibits camouflage and undergoes aestivation and hibernation, how are all these benficial to it?
Solution:

  1. A trait with a current functional role in the life history of an organism that is maintained and evolved by means of natural selection and evolution and help organism in its survival is an adapture triat.
  2. Frog is a poikilotherm or a (cold blooded animal). It regulates its body temperature according to its environment.
  3. It undergoes winter sleep (hibernation) for withstanding very cold temperatures and sujnmer sleep in hot temperatures (aestivation).
  4. During this period, it lives in a dormant stage with very minimal vital body activities.
  5. Frog is capable of changing its body colour as well, though gradually, with the change in its surrounding and climatic conditions.
  6. This capability in frog is called as camouflage which lets it escape from the predators, an essential survival parameter for living.

Question 8.
Write the functions in brief in Column II, appropriate to the structures given in column I.
NCERT Exemplar Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals 1.4
Solution:
(a) Nictitating Membrane in frog protects the eye from water and any damage by covering the eye ball of frog.
(b) Tympanum is present on each side of the frog head and is involved in the hearing process.
(c) Copulatory Pad present in the limbs of the male frog and helps in copulation by holding the female during its sexual activity.

Question 9.
Using appropriate examples, differentiate between false and true body segmentation.
Solution:

  1. The serial repetition of similar body parts along the length of an animal is segmentation. The body of animals can be truely segmented or pseudo/false segmented.
  2. True segmentation is found in annelids, arthropods and some chordates. In these organisms there is a linear repetition of body parts and each repeated unit is called somite (metamere).
  3. In earthworms, the successive somites are externally and intemaly. ‘
  4. Pseudosegmentation is seen when body is divided into number of false segments which are independent of each other.
  5. Each segment is able to perform all the vital function of body. Growth occurs by the addition of new segments from the anterior end, e.g., tapeworm.

Question 10.
What is special about tissue present in the heart?
Solution:
Special tissue present in heart is called cardiac muscle, which has the following features
(i) Cardiac muscle fibres are supplied with both central and autonomic nervous system and are not under the control of will of animal.
(ii) These muscles show rhythmicity and are immune to fatigue.
(iii) They have a rich supply of blood.
(iv) They are myogenicas. They possess the property of contraction even if completely isolated from the body.
NCERT Exemplar Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals 1.5

LONG ANSWER QUESTIONS

Question 1.
Comment upon the gametic exchange in earthworm during mating. Discuss the physiology in reproduction of earthworm.
Solution:
In earthworm mating is a unique process. Earthworm is a hermaphrodite. In which breeding takes place during rainy season and copulation begins soon after maturation of the sperms.
The gametic exchange and the physiology of reproduction during mating can described in the following manner.
(i) Earthworms are protandrous animal (i.e., maturation of sperm takes place much earlier then that of ova).
(ii) Mating process in earthworm occurs through process of cross-fertilisation.
(iii) The mating process involves exchange of gametic materials between the two worms.
(iv) Two individuals from adjacent burrows emerge half but and lie in contact with each other, and exchange packets of sperms called spermatophores opposite gonadal opening.
NCERT Exemplar Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals 1.6
(v) The skin encircling male pore, elevates a little during the process to form a temporary papilla that fits like a penis into the opposite spermathecal pore to keep it open.
(vi) The copulating worm after filling of spermatheca moves a bit to adjust another pair of spermathecae to face the other male pores. This is accomplished in about an hour’s copulation.
(vii) The sperms mostly remain in their diverticula within the spermathecae and the ampulla is associated with the secretion of nutritive substances for the sperms.
(viii) The sperm and egg are passed into cocoon, secreted by clitellar gland.
(ix) Fertilisation is therefore external.

Question 2.
Explain the digestive system of cockroach with the help of labelled sketch.
Solution:
The alimentary canal of cockroach is divided
into three regions foregut, midgut and hindgut.
(i) Mouthy cavity, pharynx, oesophagus, crop and gizzard are included in foregut.
(ii) Mouth cavity is a small space, surrounded by mouth parts. Food is crushed and acted upon by the salivary secretion in mouth.
(iii) The mouth opens into a short tubular pharynx, leading towards the narrow tubular passage called oesophagus and then into a sac-like structure called crop which acts as a storage organ.
NCERT Exemplar Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals 1.7
(iv) The crop is followed further by gizzard ‘ (proventriculus). Gizzard is composed of thick circular muscles and thick inner cuticle forming six highly chitinous plates called as teeth. It associated with the grinding and crushing of food particles. A thick cuticle lines the entire foregut.
(v) About one-third middle part of alimentry canal comprises of midgut or mesentron. The internal lining of midgut is an endodermal epithelium of columnal cells raised into several small villi like folds.
(vi) Anterior most part of midgut surrounding the stomadaeal valve is called cardia. Finger like blind processes called as enteric or hepatic caeca are present at the junction of foregut and midgut.
(vii) A ring of yellow filamentous structures is formed between the midgut and hindgut that aid in the removal of excretory products from haemolymph.
(viii) The remaining one-third posterior part of alimentary canal is Hindgut. It is relatively thicker than the midgut lined by cuticle and ectodermal epithelium.
(ix) Hindgut is diffrentiated into three parts anterior Ileum, middle colon and posterior rectum. Ileum is short and relatively narrower and its cuticie bears minute spines. Colon is the longest, relatively thicker and a coiled part of hindgut. Rectum is a small and oval chamber that opens out through anus.

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

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

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

VERY SHORT ANSWER QUESTIONS

Question 1.
Product of photosynthesis is transported from the leaves to various parts of the plants and stored in some cell before being utilised. What are the cells/tissues that store them?
Solution:
The first product of photosynthesis is glucose. It is highly reactive molecule and gets converted into a disaccharide-sucrose for storage.
The food gets stored in specialised prarenchymatous cells present either in roots and stems or in their modifications in the form of a polysaccharide called starch.

Question 2.
Protoxylem is the first formed xylem. If the protoxylem lies next to pholem what kind of arrangement of xylem would you call it?
Solution:
The condition of the xylem arrangement if protoxylem lies next to phloem is called as exarch. It is found in roots.

Question 3.
What is the function of phloem parenchyma?
Solution:
The main function of phloem parenchyma is to store food and other substances like resins, latex and mucilage. They help in transport of food as well.

Question 4.
What is present on the surface of the leaves which helps the plant prevent loss of water but is absent in roots?
Solution:
Cuticle is a waxy coating covering the entire surface of the plant body. It is absent in roots, it prevents the loss of water through the surface of the plant.

Question 5.
What is the epidermal cell modification in plants which prevents water loss?
Solution:
Bulliform or motor cells are modified epidermal cells meant for checking water loss present in monocots or grasses. They help in shutting down stomata and thus reduce water loss through transpiration under stressed conditions.

Question 6.
What constitutes the cambial ring?
Solution:
The cambium present in between the xylem and phloem is called fasicular or intrafasicular cambium and the newly formed cambium between the two vascular bundle is known as interfascular cambium. Both type of cambium combine to form the cambial ring.

Question 7.
Give one basic functional difference between phellogen and phelloderm.
Solution:
Phelloderm is a permanent tissue while phellogen is a meristematic tissue. Phellogen (cork cambium) develops from the cortical cells, sometimes from pericycle cells. These cells actively divide and forms phellem on outerside and phelloderm (cortex cells) innerside so phelloderm originates from phellogen.
NCERT Exemplar Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 1.1

Question 8.
Arrange the following in the sequence you would find them in a plant starting from the periphery-phellem, phellogen, phelloderm.
Solution:
The outer most layer is phellem or cork followed by phellogen (cork cambium) which in turn is followed by phelloderm (secondary cortex.

Question 9.
If one debarks a tree, what parts of the plant is being removed?
Solution:
Debarking refers to removal of bark, i.e., all tissues exterior to the vascular cambium, including secondary phloem. Bark includes periderm (phellogen, phellem and phelloderm) and secondary phloem.

SHORT ANSWER QUESTIONS

Question 1.
While eating peach or pear it is usually seen that some stone like structures get entangled in the teeth, what are these stone like structures called?
The edible part of the peach or pear pome fruit for the fleshy thalamus.
Solution:
The stone cells are present in the pulpy part of fruit of peach and pear. These are sclerenchymatous cells and which are dead in nature. They provide mechanical support to the soft tissue.
NCERT Exemplar Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 1.2

Question 2.
What is the commercial source of cork? How frs it formed in the plant?
Solution:
The source of commercial cork is the cork tissue of Quercus suber, which yields bottle cork. Cork is formed by cork cambium or phellogen cell, cells of cork cambium divide periclinally, cutting cells towards the inside and outside. The cells that cut off towards the outside become suberised and dead.
These are compactly packed in radial rows without intercellular spaces and form cork of phellem. Cork is impervious to water due to presence suberin and provides protection to the underlying tissues.

Question 3.
Below is a list of plant fibres. From which part of the plant these are obtained.
(a) Coir
(b) Hemp
(c) Cotton
(d) Jute
Solution:
(a) Coir is a natural fibre obtained from coconut husk. It is the fibrous mesoderm of the fruit of Cocos nucifera (coconut).
(b) Hemp fibre is obtained from the stems of Cannabis sativa. It is the bast fibre (soft or stem fibre) obtained from secondary phloem.
(c) Cotton fibre is the epidermal growth in cotton (Gossypium hirsutum) seed. It is an elongated structure made up of cellulose.
(d) Jute is a natural bast fibre made up of cellulose and lignin obtained from Corchorus capsularis.

Question 4.
Epidermal cells are often modified to perform specialised functions in plants. Name some of them and function they perform.
Solution:
The epidermal tissue system comprises of one cell thick layer of epidermal tissue and forms the outer most covering of the whole plant body. ,
Modification of Epidermal Cells
Following are the modifications of the epidermal tissue (i) root hair
Structure
These unicellular hairs are the extensions of epidermal cell of roots in the root hair zone.
NCERT Exemplar Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 1.3
Function
They increase the surface area for absorption of water and minerals.
(ii)Epidermal Appendages
Structure
These are called trichomes and are epidermal cell modifications. There any be unicellular or multicellular.
NCERT Exemplar Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 1.4
Appendages of epidermis of leaves
A-Stellate hair of a Alyssum
B-Glandular hair of Pelargonium
C-Short glandular hair of Lavandula
D-Floccose hair of Malva
E-Glandularhair of solanum
F-Urtivating hair of Verbascum
Function
They produce some glandular secretions.

Question 5.
The lawn grass (Cyandon dactyl on) needs to be mowed frequently to prevent its overgrowth. Which tissue is responsible for its rapid growth?
Solution:
The rapid growth of mowed lawn grass is due to meristematic tissue. When the apex of grass is cut frequently, it leads to the growth of the lateral branches, that makes it more bushy.

Question 6.
Plants require water for their survival. But when watered excessively, plants die. Discuss.
Solution:
Plants use water for several metabolic process as photosynthesis, transpiration and respiration. Plants when watered in excess die because excess water removes the air trapped between the soil particles.
The plant roots do not get 02 for respiration. Once cells of root die, water and mineral absorption is stopped and this leads to gradual death of a plant.

Question 7.
A transverse section of the trunk of a tree shows concentric rings which are known as growth rings. How are these rings formed? What is the significance of these rings?
Solution:

  1. The concentric growth rings are called annual rings. These rings are formed due to the secondary growth.
  2. Secondary growth occurs due to the activity of cambium which is a meristermatic tissue in dicot trees.
  3. The rate of activity of cambium is more in spring so wood formed has larger wider xylem cells, whereas wood formed in autumn has narrower and smaller xylem elements.
  4. This results in the formation of two rings called growth rings.
  5. By counting these rings, age of the tree can be determined. This branch of science is known as dendrochronology or growth ring analysis.
    NCERT Exemplar Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 1.5

Question 8.
Trunks of some of the aged tree species appear to be composed of several fused trunks. Is it a physiological or anatomical abnormality? Explain in detail.
Solution:
The appearance of several fused trunks is anatomical abnormality. It is an abnormal type of secondary growth where a regular vascular cambium or cork cambium is not formed in its normal position. Anomalous secondary growth produces cortical and medullary vascular bundles in case of old tree trunks. Thus, the additional or accessory vascular bundles given appurtenance of several fused trunks.

Question 9.
What is the difference between lenticels and stomata? The gaseous exchange in all plants. Occurs by means of several openings present in the plant body.
Solution:
NCERT Exemplar Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 1.6

Question 10.
Write the precise function of
(a) sieve tube
(b) interfascicular cambium
(c) collenchyma
(d) aerenchyma
Solution:
Sieve tube It’s function is to transport of synthesised food throughout the plant. It is present in the phleom tissue.
NCERT Exemplar Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 1.7
Interfascicular Cambium It is a kind of  secondary meristermatic tissue present in between two vascular bundles. It is function is to bring about secondary growth in the dicot stem and root.
NCERT Exemplar Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 1.8
Collenchyma cells have angular thickening at corners. There function is to provide mechanical support to young growing herbaceous stem.
NCERT Exemplar Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 1.9
Aeronchyma is a specilised parenchyma having large air spaces. It provides buyoncy to the hydrophytic plants.
NCERT Exemplar Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 1.10

Question 11.
The stomatal pore is guarded by two kidney shaped guard cells. Name the epidermal cells surrounding the guard cells. How does a guard cell differ from an epidermal cell? Use a diagram to illustrate your answer.
Solution:
Stematal apparatus is a special modification of epidermal tissue present over leaf area. The epidermal cells surrounding the guard cells of stomata are called subsidiary cells include.
Differences between guard cells and epidermal cells include:
NCERT Exemplar Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 1.11

LONG ANSWER QUESTIONS

Question 1.
Is Pirns an evergreen Tree? Comment.
Solution:

  • The plants which have persistent leaves in all the four seasons are evergreen. Deciduous plants in contrast completely loose their foliage during winter or dry season.
  • Pinus belonging to gymnosperms is an evergreen tree. Under conditions of extreme cold the flowering plants shed their leaves and become dormant.
  • In Pinus due to the presence of a thick bark thick needle-like leaves and sunken stomata to reduce the rate of transpiration the leaves we not shed.
  • The cold areas are both physiologically and physically dry due to scanty rainfall, precipitation as snow, decreased root absorption at low temperature and exposed habitats.
  • Pinus however is well adapted to such conditions. It continues to manufacture food during this period and grows to domiante other plants. This show that Pinus is an evergreen tree.
  • It does not shed its leaves or needles under any condition.

Question 2.
Assume that a pencil box held in your hand, represents a plant cell. In how many possible planes can it be cut? Indicate these cuts with the help of line drawings.
Solution:
A. If a plant cell is cut in different plane if result, in radial symmetry.
NCERT Exemplar Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 1.12
B. If a plant cell is cut in two equal halves it result in bilateral symmetry.
NCERT Exemplar Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 1.13

Question 3.
Each of thefollowing terms has some anatomical significance. What do these terms mean? Explain with the help of line diagrams.
(a) Plasmodesmata
(b) Middle Lamella
(c) Secondary wall
Solution:
NCERT Exemplar Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 1.14

We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 6 Anatomy of Flowering Plants, help you. If you have any query regarding NCERT Exemplar 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.

NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants

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NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants

These Solutions are part of NCERT Exemplar Solutions for Class 11 Biology. Here we have given NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants.

VERY SHORT ANSWER QUESTIONS

Question 1.
Roots obtain oxygen from air in the soil for respiration. In the absence or deficiency of 02, root growth is restricted or completely stopped. How do, the plants growing in marsh lands or swamps obtain their 02 required for root respiration?
Solution:
The roots of the plants as Rhizophora that grow in marsh/swamp areas become negatively geotropic. They grow vertically upwards in air, above the soil level and respire. They are thus called respiratory roots or pneumatophores.

Question 2.
In Opuntia, the stem is modified into a flattened green structure to perform the function of leaves, (i.e., photosynthesis). Cite some other examples of modifications of plant parts for the purpose of photosytnthesis.
Solution:
In Opuntia a xerophytic plant leaves are modified into spine to reduce the rate of transpiration and they do not perform the photosynthesis at all.
The function of photosynthesis in Opuntia plant is performed by stem which is thick fleshy and flattened structure containing chlorophyll and stores food and known as phylloclade.
In some plants similarly roots become assimilatory e.g., case of Trapa and Tinospom. These roots grow outside the soil, develop chlorophyll in them and perform photosynthesis.

Question 3.
In swampy areas like the sunderbans in West Bengal, plants bear special kind of roots called……
Solution:
Pneumatophores Roots are meant for the absorption of water and minerals from the soil. Cells of roots require 02 to respire. In swampy areas, soil does not have air, so no 07 is available to them.
In such cases, roots come out of the soil showing negative geotropism and breathe after coming in contact with air, e.g., Rhizophora. Such roots are called pneumatophores or respiratory roots.

Question 4.
In aquatic plants like Pistia and Eichhornia, leaves and roots are found near…..
Solution:
In Pistia and Eichhonia, the stem is like a runner where it branches to form leaves at the
v apex and roots below. Both the plants are hydrophytes and thus the roots are found near the surface of water.

Question 5.
Which parts in ginger and onion are edible?
Solution:
The edible part of ginger is rhizome the modified stem which stores food material whreas the edible part in onion is fleshy leaves, where the internode becomes shortened, leaves get condensed to form a tunic and store food material.

Question 6.
In epigynous flower, ovary is situated below the……….. .
Solution:
Ovary is situated below the thalamus (inferior) in epigynous flower while the other whorls of flower like sepals, petals and androecium grows above the ovary (superior), e.g., carrot, guava, Cucurbit a, sunflower, etc.
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.1

Question 7.
Add the missing floral organs of the given floral formula of Fabaceae.
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.14
Solution:
The floral formula of fabaceae family is
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.2
The flower of fabaceae is bisexual, zygomorphic, pentameros, gamosepalous, corolla-petals 5, androecium is ten diadelphous, gynoecium- superior, ovary monocarpellay.

SHORT ANSWER QUESTIONS

Question 1.
Give two examples of roots that develop from different parts of the angiospermic plant other than the radicle.
Solution:
Prop roots are meant for support. Prop roots develop from the lower nodes of stem of banyan tree. They grow downwards and touch the soil.
Stilt roots arise from the lower nodes of stem in sugarcane and enter the soil to provide strength to the plant. These protect the plant against winds.
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.3

Question 2.
The essential functions of roots are anchorage and absorption of water and minerals in the terrestrial plant. What functions are associated with the roots of aquatic plants. How are roots of aquatic plants and terrestrial plants different?
Solution:
Usually the terrestrial roots show a branched network that helps in anchorage and absorption of water and minerals from soil to the plant. While in aquatic plants, roots show modification and deviation from their normal function.
Ex – in plants like Trapu, Tinospora the roots are green and highly branched to increase the photosynthetic area, whereas in plants like Jussiaecci they get inflated due to air project out of water so a to help the plant in floating and exchange of gases.
Difference between roots of aquatic plants and terrestrial plants are as:
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.4

Question 3.
Draw diagrams of a typical monocot and dicot leaves to show their venation pattern.
Solution:
The pattern of distribution of veins and veinlets in the lamina of leaf is called Venation. It’s pattern is different in monocot and dicot leaf.
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.5

Question 4.
A typical angiosperm flower consists of four floral parts. Give the names of the floral parts and their arrangements sequentially.
Solution:
Following are the four floral parts of typical angiospermic flower.

  1. Calyx is the outermost whorl of the flower and comprised of sepals. These are usually green and (in bud stage) are protective in function.
  2. Corolla is composed of petals, usually bright coloured to attract insects for pollination.
  3. Androecium is composed of stamens, the male reproductive organ. Each stamen consists of stalk or filament and anther (containing pollen sac and pollen grains).
  4. Gynoecium is the female reproductive part and comprised of one or more carpels. Each
    NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.6

Question 5.
Reticulate venation is found in dicot leaves while in monocot leaves venation is of parallel type. Biology being a ‘Science of exceptions’, find out any exception to this generalisation.
Solution:
Reticulate venation is a characteristic of dicots and parallel venation is of monocots. But few exceptions are also seen in this generalisation, parallel venation is also found in dicot plants, e.g, Calophyllum, Corymbium, etc and reticulate venation is also found in monocot plants such as Alocasia, smilax, etc.

Question 6.
You have heard about several insectivorous plants that fee on insects. Nepenthes or the pitcher plant is one such example, which usually grows in shallow water or in march lands. What part of the plant is modified into a pitcher? How does this modification help the plant for food even though it can photosynthesise like any other green plant?
Solution:

  1. In insectivorous plant like Nepenthes, the leaf lamin is modified to form a pitcher and anterior part of petiole coils like tendril which keeps the pitcher in a vertical direction.
  2. Posterior part of the petiole remains flattened like a leaf. The apex of lamina forms a lid.
  3. Pitcher contains digestive enzyme for digesting trapped insects.
  4. All these modifications and adaptations are developed to make up for the nitrogen deficiency in the plant because these plants are found in N2 deficient soil, (marshy/swamp soils).
    NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.7

Question 7.
How can you differentiate between free central and axile placentation?
Solution:
The arrangement of ovules on the walls of ovary with the help of special kind of tissue called placenta is placentation. Plants show different types of placentation.
Difference between free central placentation and axile placentation include:
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.8
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.9

Question 8.
Why is maize grain usually called as a fruit and not a seed?
Solution:
The maize grain is usually known as fruit because it is a ripened ovary which contains a ripened ovule, e.g., a single seed. This fruit is known as caryopsis in which the pericarp is fused with the seed coat. The maize grain occurs attached to a thick cob or peduncle.

Question 9.
Tendrils of grapevins are homologous to the tendril of pumpkins, but are analogous to that of pea. Justify the above statement.
Solution:
Homologous organs are organs that have similar origin but they differ functionally. Axillary bud of stem gives rise to tendril of both grapevine and pumpkins so they have same origin, i.e., homologous, whereas analogous organs are organs having different origin, but perform same function. The tendril of pea arises from the leaf and helps the plant to climb.

Question 10.
Rhizome of ginger is like the roots of other plants that grows underground. Despite this fact ginger is a stem and not a root. Justify.
Solution:
Rhizome of Ginger is a type of modified underground stem which grows horizontally underground and bears nodes, intemodes and scaly leaves and buds, which gives rise to aerial shoots.
The adventitious root arises from the lower surface of nodes. It is not a true root because root does not have nodes and intemodes. The rhizome does not perform the function of anchorage and absorption, rather serve as reservoir for food storage. All these characteristics support the fact that ginger is a stem and not a root.
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.10

LONG ANSWER QUESTIONS

Question 1.
Distinguish between families – Fabaceae, solanaceae, Liliaceae on the basis of gynoecium characteristics (with figures). Also write economic importance of any one of the above family.
Solution:
The families in plant kingdom mainly differ from each other in their reproductive structures.
Based on characteristics of gynoecium the difference between the three families include the following:
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.11
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.12

Question 2.
Describe various stem modifications associated with food storage climbing and protection.
Solution:
The aerial part of plant bearing nodes, intemodes, buds, flowers, fruits and seeds is stem. Besides these functions and forms, it gets modified and perform under spfccial conditions.
The various stem modifications include:
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.13

We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 5 Morphology of Flowering Plants, help you. If you have any query regarding NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 5 Morphology of Flowering Plants, drop a comment below and we will get back to you at the earliest.