Class 11 Biology – Merit Batch

Animal Kingdom Class 11 Chapter 4 Questions And Answers

February 14, 2025

Class 11 Biology Chapter 4 Animal Kingdom NCERT Solutions

Animal Kingdom NCERT Solutions

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 4 Animal Kingdom:

Animal Kingdom Phylum List

Section Name	Topic Name
4	Animal Kingdom
4.1	Basis of Classification
4.2	Classification of Animals
4.3	Summary

NCERT Solutions Class 11 Biology Biology Sample Papers

NCRT TEXTBOOK QUESTIONS SOLVED

1.What are the difficulties that you would face in classification of animals, if common fundamental features are not taken into account?
soln. The common fundamental features used for classifying animals include body symmetry, arrangement of cells, nature of coelom, level of organisation. Animal classification would be very confusing if fundamental features are not considered.
(i)Animals having different levels of organisation would have been placed in same group. E.g., Sponges and Cnidarians having cellular and tissue level of organisation respectively.
(ii)Animals showing varied types of germinal layers would have been placed together, as diploblastic cnidarians and triploblastic platyhelminthes.
(iii)Animals having different body symmetry would have been placed together, as coelenterates with radial symmetry and platyhelminthes with bilateral symmetry.
(iv)There would have been no classification of animals based on with or without body cavity..
(v)Placing of oviparous and viviparous animals together.

2.If you are given a specimen, what are the steps that you would follow to classify it?
soln. Various steps considered to classify a specimen are:
(i)Mode of nutrition – It can be autotrophic, holozoic, saprophytic or parasitic.
(ii)Complexity of body structure – Whether the specimen is unicellular or multicellular.
(iii)Presence or absence of membrane bound organelles.
(iv)Body symmetry, i.e., the plane by which organism can be divided into two equal halves.
(v)Presence or absence of coelom, it can be acoelomates, pseudocoelomates, eucoelo- mates.
(vi)Phylogenetic relationship.

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3.How useful is the study of the nature of body cavity and coelom in the classification of animals?
soln. Organisms can be classified according to presence or absence of the coelom. The body cavity, which is lined by mesoderm is called coelom. Animals possessing coelom are called coelomates, e.g., annelids, molluscs, arthropods, echinoderms, hemichordates and chordates. In some animals, the body cavity is not lined by mesoderm, instead, the mesoderm is present as scattered pouches in between the ectoderm and endoderm. Such a body cavity is called pseudocoelom and the animals possessing them are called pseudocoelomates, e.g., aschelminthes. In pseudocoelomates, body cavity is derived from blastocoel of the embryo. The animals in which the body cavity is absent are called acoelomates, e.g., platyhelminthes.

4.Distinguish between intracellular and extra-cellular digestion.
soln. Differences between intracellular and extracellular digestion are:
NCERT Solutions For Class 11 Biology Animal Kingdom Q4

NCERT Solutions For Class 11 Biology Animal Kingdom Q4.1

5.What is the difference between direct and indirect development?
soln. Differences between direct development and indirect development are :
NCERT Solutions For Class 11 Biology Animal Kingdom Q5

6.What are the peculiar features that you find in parasitic platyhelminthes?
soln.Following are the peculiar features of parasitic platyhelminthes:
(i) The thick tegument (body covering) resistant to the host’s digestive enzymes and anti-toxins.
(ii)Adhesive organs like suckers in flukes and the hooks and suckers in tapeworms for a firm grip on or in the host’s body.
(iii)Loss of locomotory organs.
(iv)Digestive organs are absent in tapeworms because digested and semidigested food of the host is directly absorbed’ through the body surface.
(v) Reproductive system is best developed in parasitic flatworms.
(vi)Parasitic flatworms, such as liver fluke and tapeworms perform anaerobic respiration.
(vii)They possess a considerable osmotic adaptability, as they can successfully live in different media.

7.What are the reasons that you can think of for the arthropods to constitute the largest group of the animal kingdom?
soln. Arthropods are most successful animals and constitute the largest group of the animal kingdom. They have conquered land, sea and air and make up over three fourth of currently known living and fossil organisms. They range in distribution from deep sea to mountain peaks. Thick, tough, non-living chitinous cuticle forms the exoskeleton which protects the organism from predators, help to withstand temperature upto 100°C or more and prevents water loss. They have ability to reproduce very fast and less time is needed for young ones to hatch from their eggs. Due to metamorphosis, there is less competition among larval and adult forms for food. Cockroaches can even survive nuclear radiations and poisoned earth. All these factors made arthropods the largest phylum among animals.
8.Water vascular system is the characteristic of which group among the following ?
(a) Porifera
(b) Ctenophora
(c) Echinodermata
(d) Chordata
soln. (c) Echinodermata

9.”All vertebrates are chordates but all chordates are not vertebrates”. Justify the statement.
soln. Chordates are the animals that possess notochord (a stiff, supporting rod like structure present on the dorsal side) at some stage of their lives. Phylum Chordata is divided into three Subphyla: Urochordata or tunicata, Cephalochordata and Vertebrata. Subphyla Urochordata and Cephalochordata are often referred to as protochordates and are exclusively marine. In urochordata, notochord is present only in tail of larva and disappears in adults, while in cephalochordata, it extends from head to tail region and persists throughout the life.
The members of Subphylum Vertebrata a possess notochord during the embryonic period and is replaced by a cartilaginous or bony vertebral column in the adult. Thus all vertebrates are chordates but all chordates are not vertebrates.

10.How important is the presence of air bladder in
Pisces?
soln. Bony fishes have a sac-like outgrowth, the swim bladder also called air bladder, that arises as an outgrowth from the dorsal wall of oesophagus. It is hydrostatic in function. It regulates buoyancy and helps them to swim up and down, thus preventing them from sinking. In some species air bladder also helps in respiration. It also serves as resonating chamber to produce or receive sound.

11.What are the modifications that are observed
in birds that help them fly?
soln. Birds have adapted to aerial mode of life through the following modifications:
(i) Body is streamlined and spindle shaped which minimise resistance to the wind.
(ii)Body is covered with feathers. It reduces the friction, prevent loss of heat and help to maintain constant temperature.
(iii)Forelimbs are modified into wings, which help during flight.
(iv)Flight muscles are greatly developed
(v) Most of the bones are pneumatic, hollow and filled vvith air which makes the body lighter and helps in flight.
(vi)Birds are warm-blooded. They maintain a high body temperature (40° – 46°C). This is necessary for flight.
(vii)Heart is four-chambered and functions efficiently with double circulation.
(viiiJAir sacs are present which act as reservoir of air and helps in temperature regulation
(ix)Urinary bladder is absent (except in Rhea) and only one ovary is present which reduces the weight, which is essential for flight.

12. Could the number of eggs or young ones produced by an oviparous and viviparous mother be equal? Why?
soln. No, the number of eggs or young ones produced by an oviparous and viviparous mother respectively cannot be equal. Oviparous mother lays large number of eggs, as the eggs are laid outside the body, so they are not protected from predators and harsh environmental conditions, and therefore destroyed. However in viviparous mother, eggs are not laid outside, but the embryos develop inside the mother and thus are protected from the outside harsh environment, thus, the number of eggs produced are less. Therefore, the number of eggs or young ones produced by an oviparous and viviparous mother respectively cannot be equal.

13.Segmentation in the body is first observed in which of the following?
(a) Platyhelminthes
(b) Aschelminthes
(c) Annelida
(d) Arthropoda
soln. (c) Annelida

14.Match the following:
(a) Operculum (i) Ctenophora
(b) Parapodia (ii)Mollusca
(c) Scales (iii)Porifera
(d) Comb plates (iv)Reptilia
(e) Radula (v) Annelida
(f) Hair (vi)Cyclostomata and
Chondrichthyes
(g) Choanocytes (vii)Mammalia
(h) Gill slits (viii Osteichthyes
soln.(a) – (viii), (b) – (v), (c) – (iv), (d) – (i),
(e) – (ii), (f) – (vii), (g) – (iii), (h) – (vi).

15. Prepare a list of some animals that are found parasitic on human beings.
soln.List of some animals that are found parasitic on human beings :
NCERT Solutions For Class 11 Biology Animal Kingdom Q15

Class 11 Biology NCERT Solutions

Morphology of Flowering Plants Class 11 Chapter 5 Questions And Answers

February 14, 2025

Class 11 Biology Chapter 5 Morphology of Flowering Plants NCERT Solutions

Morphology of Flowering Plants NCERT Solutions

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants:

Section Name	Topic Name
5	Morphology of Flowering Plants
5.1	The Root
5.2	The Stem
5.3	The Leaf
5.4	The Inflorescence
5.5	The Flower
5.6	The Fruit
5.7	The Seed
5.8	Semi-technical Description of a Typical Flowering Plant
5.9	Description of Some Important Families
5.10	Summary

NCRT TEXTBOOK QUESTIONS SOLVED

1.What is meant by modification of root? What type of modification of root is found in the:
(a) Banyan tree
(b) Turnip
(c) Mangrove trees
Soln.Roots of some plants change their shape and structure and become modified to perform certain functions other than absorption and conduction of water and minerals. It is called modification of roots. Roots are modified for support, storage of food and respiration, etc.
(a) Root modification in banyan tree : In banyan tree, the root modifies to form prop roots. Prop roots arise from branches and enter the soil. Thus, they provide mechanical support to densely branched, huge trees.
(b) Root modification in turnip : The
modification of root found in turnip is napiform for food storage. The upper portion of these fleshy roots is inflated or swollen which tapers towards the lower end.
(c) Root modification in mangrove trees : In mangrove plants, i.e., plants growing in saline marshes, the branches of tap root come out of the ground and grow vertically upwards showing negative geotropism. These roots are called pneumatophores. They help to get oxygen for respiration.

2.Justify the following statements on the basis of external features:
(i) Underground parts of a plant are not always roots.
(ii) Flower is a modified shoot.
Soln. (i) Underground parts of plant are not always roots because sometimes the stem also becomes underground and gets modified into various forms to perform different functions of storage, vegetative propagation, perennation, etc. Underground modifications of stems are tuber, rhizome, corm and bulb. The underground stems can be distinguished from roots externally by the presence of nodes and internodes, axillary buds, scale leaves etc. and by absence of root cap and root hairs.
(ii) Flower is the reproductive part of the angiospermic plant and it is defined as the modified shoot because (a) like shoot, flower develops from an axillary or rarely terminal bud. (b) flowers may get modified into fleshy buds or bulbils, (c) A transition from foliage leaves to floral leaves is found in Paeonia. (d) Nymphaea shows transition from sepals to petals and petals to stamens, (e) In Passiflora and Cleome long intemodes occur below gynoecium and stamens.

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3.How is a pinnately compound leaf different from a palmately compound leaf?
Soln.The compound leaves may be of two types, pinnately compound leaf and palmately compound leaf. In pinnately compound leaf, a number of leaflets are present on a common axis, the rachis, which represents the midrib of the leaf as in neem. Pinnately compound leaf may be of different types as unipinnate, bipinna te, tripinna te and decompound. In palmately compound leaf, the leaflets are attached at a common point, i.e., at the tip of petiole, as in silk cotton. Palmately compound leaf may be of different types as unifoliate, bifoliate, trifoliate, quadrifoliate and multifoliate.

4.Explain with suitable examples the different types of phyllotaxy.
Soln.Phyllotaxy is the pattern of arrangement of leaves on the stem or branch. It is usually of three types – alternate, opposite and whorled. In alternate type of phyllotaxy, a single leaf arises at each node in alternate manner, as in china rose, mustard and sunflower plants. In opposite type, a pair of leaves arises at each node and lie opposite to each other as in Calotropis and guava plants.
If more than two leaves arise at a node and form a whorl it is called whorled phyllotaxy as in Alstonia.

5.Define the following terms:
(a) aestivation (b) placentation
(c) actinomorphic (d) zygomorphic
(e) superior ovary (f) perigynous flower (g) epipetalous stamen.
Soln. (a) Aestivation : The mode of arrange¬ment of accessory floral organs (sepals and petals) in relation to one another in floral bud is known as aestivation. The main type of aestivation are valvate, twisted, imbricate, and vexillary.
(b) Placentation : The arrangement of ovules within the ovary is known as placentation. The placentation are of different types namely, marginal, axile, parietal, basal, and free central.
(c) Actinomorphic : When flower can be divided into equal radial halves in any radial plane passing through the centre, it is said to be actinomorphic, e.g., mustard, Datura etc.
(d) Zygomorphic : When a flower can be divided into two similar halves only in one particular vertical plane, it is said to be zygomorphic, e.g., pea, gulmohar, bean, Cassia.
(e) Superior ovary : In hypogynous flower, the gynoecium occupies the highest position while the other parts are situated below it. The ovary in such flowers is said to be superior, e.g., mustard, brinjal.
(f) Perigynous flower: If gynoecium is situated in the centre and other parts of the flower are
located on the rim of the thalamus almost at the same level, it is called perigynous. Here ovary is half superior, e.g., peach, plum.
(g) Epipetalous stamen : When stamens are attached to the petals, they are called epipetalous stamens e.g., brinjal.

6.Differentiate between
(a) Racemose and cymose inflorescence
(b) Fibrous root and adventitious root
(c) Apocarpous and syncarpous ovary
Soln.
(a) Differences between racemose and cymose inflorescence are as follows:
NCERT Solutions For Class 11 Biology Morphology of Flowering Plants Q6
(b) Differences between fibrous and adventitious roots are as follows :

NCERT Solutions For Class 11 Biology Morphology of Flowering Plants Q6.2
(c) Differences between apocarpous and syncarpous ovary are as follows :

7.Draw the labelled diagram of the following:
(i) Gram seed (ii) V. S. of maize seed.
Soln.
(i) Gram seed.
NCERT Solutions For Class 11 Biology Morphology of Flowering Plants Q7
(ii) V.S. of maize seed.

8.Describe modifications of stem with suitable examples.
Soln. Stems are modified to perform different functions. Underground stems of some plants are modified to store food in them. They also act as organs of perennation to tide over conditions unfavourable for growth. Different modifications of stem are :
(i) Underground modifications
(ii)Sub-aerial modifications
(iii)Aerial modifications
(i)Underground modifications of stem are discussed as follows:
(a)Tuber: It is the branch of main stem which accumulates or stores food in it and swells up, e.g., Solarium tuberosum (potato).
(b) Rhizome: It is a branched, prostrate horizontally growing stem having nodes and internodes. On the nodes sessile scale leaves are formed, e.g., Carina, Zingiber officinale (ginger), Curcuma domestica (turmeric) etc.
(c) Corm: This is a spherical,
branched, vertically growing thick underground stem with more diameter than length, e.g., Crocus sativus (saffron), Gladiolus, Colocasia esculenta (arvi) etc.
(d) Bulb: In bulb the stem is highly reduced and can be seen only as a disc-like structure bearing numerous fleshy scaly leaves, e.g., Allium cepa (onion), Allium sativum (garlic) etc.
(ii) Subaerial modifications : Subaerial part of stem grows horizontally on the ground while some part remains underground. Vegetative propagation takes place by means of these. They may be of following kinds.
(a) Runner: It grows prostrate on the surface of soil. It develops at the base of erect shoot called crown. A number of runners arise from one erect shoot which spread in different directions. Each runner has one or more nodes which bear scale leaves and axillary buds, e.g., Cynodon (doob grass).
(b) Stolon: The nodes of horizontally growing underground stem give rise to branches which come out of the soil, e.g., Fragaria (strawberry).
(c) Sucker: Suckers are formed from the node of underground stem. Sucker comes up obliquely in the form of leafy shoot, e.g., Mentha (mint).
(d) Offset: Stem consists of thick and short intemodes. The branches are formed from the main stem and upper portion of each branch bears a group of leaves while the lower portion bears the roots. Each branch is capable of growing as an independent plant after separating from the parent plant, e.g., Eichhornia (water hyacinth), Pistia, etc.
(iii)Aerial modifications : The aerial portion of stem is modified to perform different functions, e.g., climbing, protection, food manufacturing, etc. It may show following types of modifications:
(a) Twinners : The stem is long, flexible and sensitive which can coil around an upright support like a rope, e.g., Ipomoea, Convolvulus.
(b) Climbers : The stem is weak and flexible but is unable to coil around an upright support by itself. It requires the help of clasping or clinging structures. Accordingly, climbers are of four types : root climbers, e.g., Betel; tendril climber, e.g., Passiflora; scramblers, e.g., Bougainvillea and lianas, e.g., Bauhinia.
(c) Phylloclade: The stem performs the function of photosynthesis. The stem modifies into green fleshy leaf-like
structure having distinct nodes and intemodes. Leaves of such plants are reduced into spines in order to prevent loss of water, e.g., Opantia (prickly pear), Euphorbia.
(d) Cladode: It is similar to phylloclade with only one internode, e.g., Asparagus.
(e) Thorn: Stem is modified into stiff, pointed unbranched or branched structures which have lost their growing point and become hard, called as thorns, e.g., Bougainvillea,Pomegranate, Citrus, etc. They perform defensive function.
(f) Tendrils : These are thread like sensitive structures which can coil around a support and help the plant
in climbing, e.g., Cucurbita.
(g) Bulbils: In some plants vegetative buds or floral buds modify into a swollen structure called bulbil. It separates from the parent plant and on approach of favourable condition gives rise to a new plant, i.e., it is an organ of vegetative reproduction, e.g., Agave, Oxalis.

9.Take one flower each of the families Fabaceae and Solanaceae and write their semi-technical description. Also draw their floral diagram after studying them.
Soln.Family Fabaceae (e.g., Pisum sativum) Systematic position:
Class – Dicotyledoneae
Subclass- Polypetalae
Series – Calyciflorae
Order – Rosales
Family – Fabaceae
Vegetative characters:
Habit: herb. Root: tap, branched, with root nodules.
Stem: herbaceous, climbing.
Leaves : pinnately compound, leaf base pulvinate, stipulate, venation reticulate.
Floral characters:
Inflorescence: racemose.
Flower : bisexual, zygomorphic, irregular, hermaphrodite, white or pink, complete, hypogynous to perigynous.
Calyx : sepals five, gamosepalous, ascending, imbricate aestivation, campanulate calyx tube.
Corolla : petals five, polypetalous, vexillary aestivation, papilionaceous, consisting of a posterior standard or vexillum two lateral wings or alae, two anterior ones forming a keel.
Androecium : 10 stamens in two bundles (diadelphous) of (9) + 1, anthers dithecous (bilobed), basifixed, introrse.
Gynoecium : ovary superior, monocarpellary, unilocular with many ovules, marginal placentation, style bent and long, stigma simple and-hairy.
Fruit : legume; seeds one to many, non- endospermic.
Floral formula :
NCERT Solutions For Class 11 Biology Morphology of Flowering Plants Q9
Family Solanaceae (e.g., Solanum nigrum) Systematic position:
Class Subclass Series Order Family
Vegetative characters:
Habit: herbs Stem : herbaceous, aerial, erect, cylindrical, branched.
Leaves: alternate, simple, exstipulate, venation reticulate.
Floral characters:
Inflorescence: cymose.
Flower : ebracteate, ebracteolate, bisexual, actinomorphic, white, hypogynous.
Calyx : sepals five, gamosepalous, persistent, valvate aestivation.
Corolla : petals five, gamopetalous, valvate. aestivation.
Androecium : stamens five, epipetalous, polyandrous, anthers large, bithecous and basifixed.
Gynoecium : bicarpellary, syncarpous,
ovary, obliquely placed carpels in the flower, bilocular, axile placentation, placenta swollen with many ovules.
Fruits : berry with persistent calyx.
Floral formula :
NCERT Solutions For Class 11 Biology Morphology of Flowering Plants Q9.1

10.Describe the various types of placentations found in flowering plants.
Soln.Placenta is a parenchymatous cushion present inside the ovary where ovules are borne. The number, position, arrangement or distribution of placentae inside an ovary is called placentation. The placentation are of different types namely, marginal, axile, parietal, basal and free central.
(i)Marginal placentation : The placenta forms a ridge along the ventral suture of the ovary and the ovules are borne on this ridge forming two rows, e.g., pea.
(ii)Axile placentation : When the placenta is axial and the ovules are attached to it in a multilocular ovary, the placentation is said to be axile, e.g., china rose, tomato and lemon.
(iii)Parietal placentation : The ovules develop on the inner wall of the ovary or on peripheral part. Ovary is one-chambered but it becomes two-chambered due to the formation of the false septum, e.g., mustard and Argemone.
(iv)Free central placentation : When the ovules are borne on central axis and septa are absent, as in Dianthus and primrose the placentation is called free central.
(v)Basal placentation: The placenta develops at the base of ovary and a single ovule is attached to it, as in sunflower, marigold.

11.What is a flower? Describe the parts of a typical angiosperm flower.
Soln.Flower is the reproductive unit in the angiosperms. It is meant for sexual reproduction. A typical flower has four different kinds of whorls arranged successively on the swollen end of the stalk or pedicel, called thalamus or receptacle. These are calyx, corolla, androecium and gynoecium.
Calyx and corolla are accessory organs, while androecium and gynoecium are reproductive organs. In some flowers like lily, the calyx and corolla are not distinct and are termed as perianth. Some flowers have both androecium and gynoecium and are termed hermaphrodite flowers while some flowers have only one of these two whorls.
Calyx : The calyx is the outermost whorl of the flower and its units are called sepals. Generally, sepals are green, leaf like and protect the flower in the bud stage. The calyx may be gamosepalous (sepals united) or polysepalous (sepals free).
Corolla : Corolla is composed of petals. Petals • are usually brightly coloured to attract insects for pollination. Like calyx, corolla may also be free (polypetalous) or united (gamopetalous). The shape and colour of corolla vary greatly in plants. Corolla may be tubular, bell-shaped, funnel-shaped or wheel-shaped.
Androecium : Androecium is the male reproductive part of the flower. It is composed of stamens. Each stamen which represents the male reproductive organ consists of a stalk or a filament and an anther. Each anther is usually bilobed and each lobe has two chambers, the pollen-sacs. The pollen grains are produced in pollen-sacs. A sterile stamen is called staminode.
Gynoecium : Gynoecium is the female reproductive part of the flower and is made up of one or more carpels. A carpel consists of three parts namely stigma, style and ovary. Ovary is the enlarged basal part, on which lies the elongated tube, the style. The style connects the ovary to the stigma. The stigma is usually at the tip of the style and is’ the receptive surface for pollen grains. Each ovary bears one or more ovules attached to a flattened, cushion-like placenta. When more than one carpel is present, they may be free (as in lotus and rose) and are called apocarpous. They are termed syncarpous when carpels are fused, as in mustard and tomato. After fertilisation, the ovules develop into seeds and the ovary matures into a fruit.

12. How do the various leaf modifications help plants?
Soln.Leaves perform various functions besides photosynthesis and thus they are modified into different forms such as –
(i)Leaf tendrils: The different parts of a leaf are modified into tendrils which help the plant in climbing up. Parts of leaf modified into tendrils include stipules e.g., Smiiax ; petiole e.g., Clematis ; leaf apex e.g., Gloriosa ; leaflets e.g., Pisum; whole leaf e.g., Lathyrus.
(ii)Leaf spines: Either for the protection of plant or to lessen the rate of transpiration in xerophytic plants, the leaves modify into sharp, pointed spines. Parts of leaf modified into leaf spines include stipules e.g., Zizyphus; leaf margins e.g., Argemone; leaf apex e.g.r Yucca; entire leaf e.g., Berberis.
(iii)Phyllode: Petioles modify into leaf¬like green, photosynthesising structure e.g., Parkinsonia, Acacia auriculiformis.
(iv)Scale or protective leaves : The leaves modify into hard scaly leaves which protect the vegetative bud by covering them, e.g., Ficus, Artocarpus, Casuarina, etc.
(v) Leaf hooks : They help in climbing e.g., Bignonia.
(vi)Leaf roots : A leaf transforms into roots for balancing on water e.g., Salvinia.
(vii)Leaf pitchers : Leaf is modified into pitcher e.g., Nepenthes (insectivorous), Dischidia (non-insectivorous).
(viii)Leaf bladder: The leaves modify to form bladder like structure which trap insects and then it is closed by a valve present on the mouth of bladder e.g., Utricularia (bladderwort).
(ix) Leaf tentacles: The leaf of sundew plant, Drosera bear minute hairs which have shinning, sticky substance at their tips (tentacles). When any insect sits on the leaf, it is covered by these hairs.

13. Define the term inflorescence. Explain the basis for the different types of inflorescence in flowering plants.
Soln. The arrangement of flowers on the floral axis is termed as inflorescence. A flower is a modified shoot wherein internodes do not elongate and the axis gets condensed. The apex produces different kinds of floral appendages laterally at successive nodes instead of leaves. When a shoot tip transforms into a flower, it is always solitary. Depending on whether the apex gets converted into a flower or continues to grow, two major types of inflorescence are defined – racemose and cymose. In racemose type of inflorescence the main axis continues to grow, the flowers are borne laterally in acropetal succession. In cymose type of inflorescence the main axis terminates in a flower, hence is limited in growth. The flowers are borne in a basipeta! order.

14. Write the floral formula of an actinomorphic, bisexual, hypogynous flower with five united sepals, five free petals, five free stamens and two united carples with superior ovary and axile placentation.
Soln. The floral formula for actinomorphic, bisexual, hypogynous flower with five united sepals, five free petals, five free stamens and two united carples with superior ovary and
axile placentation is:

15.Describe the arrangement of floral members in relation to their insertion on thalamus.
Soln. In a typical flower, the floral members like calyx, corolla, androecium and gynOecium are arranged over the thalamus! Based on the position of calyx, corolla and androecium in respect to ovary on thalamus, the flowers are described as hypogynous, perigynous and epigynous ones. In the hypogynous flower the gynoecium occupies the highest position while the other parts are situated below it. The ovary in such flowers is said to be superior, e.g., mustard, china rose and brinjal. If gynoecium is situated in the centre and other parts of the flower are located on the rim of the thalamus almost at the same level, it is called perigynous. The ovary here is said to be half inferior or sub superior, e.g., plum, rose, peach. In epigynous flowers, the margin of thalamus grows upward enclosing the ovary completely and gets fused with it; the other parts of flower arise above the ovary. Hence, the ovary is said to be inferior as in flowers of guava and cucumber, and the ray florets of sunflower.

Class 11 Biology NCERT Solutions

The Living World Class 11 Chapter 1 Questions And Answers

February 14, 2025

Class 11 Biology Chapter 1 The Living World NCERT Solutions

The Living World NCERT Solutions

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 1 The Living World:

Section Name	Topic Name
1	The Living World
1.1	What is ‘Living’?
1.2	Diversity in the Living World
1.3	Taxonomic Categories
1.4	Taxonomical Aids
1.5	Summary

NCERT Solutions Class 11 Biology Biology Sample Papers

NCERT TEXTBOOK QUESTIONS SOLVED

1. Why are living organisms classified?
Soln. Living organisms are classified because of the following reasons:
(i) Easy identification.
(ii)Study of organisms of other places.
(iii)Study of fossils
(iv)Grouping helps in study of all types of organisms while it is impossible to study individually all of them.
(v) Itbringsoutsimilaritiesanddissimilarities. They help in knowing relationships among different groups.
(vi)Evolution of various taxa can be known.

2. Why are the classification systems changing every now and then?
Soln. From very early days till now biologists use several characters for classification system. These are morphology, anatomy, cytology, physiology, ontogeny, phylogeny, reproduction, biochemistry, etc. But day by day biologists are learning something new about organisms from their fossil records and using” advanced study techniques such as molecular phylogeny, etc. So their point of view about classification keeps changing. Thus the system of classification is modified every now and then.

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3. What different criteria would you choose to classify people that you meet often?
Soln. The various criteria that may be chosen to classify people whom we meet often include behaviour, geographical location, morphology, family members, relatives, friends etc.

4. What do we learn from identification of individuals and populations?
Soln. The knowledge of characteristic of an individual or its whole population helps in identification of similarities and dissimilarities among the individuals of same kind or between different types of organisms. It helps us to classify the organisms in various categories depending upon these similarities and dissimilarities.

5. Given below is the scientific name of mango. Identify the correctly written name.
Mangifera Indica Mangifera indica
Soln. The correctly written scientific name of mango is Mangifera indica.

6. Define a taxon. Give some example of taxa at different hierarchical levels.
Slon. A taxonomic unit in the biological system of classification of organism is called taxon (plural taxa). For example a phylum, order, family, genus or species represents taxon. It represents a rank. For example, all the insects form a taxon. Taxon of class category for birds is Aves and taxon of Phylum category for birds is Chordata. The degree of relationship and degree of similarity varies with the rank of the taxon. Individuals of a higher rank, say Order or Family, are less closely related than those of a lower rank, such as Genus or Species.

7. Can you identify the correct sequence of taxonomical categories?
(a) Species —> Order —> Phylum —> Kingdom
(b) Genus—) Species—> OrderKingdom
(c) Species —> Genus —>Order —> Phylum
Slon. The correct sequence of taxonomical categories is
(c) i.e., Species —>Genus —> Order —> Phylum.

8. Try to collect all the currently accepted meanings for the word ‘species’. Discuss with your teacher the meaning of species in case of higher plants and animals on one hand, and bacteria on the other hand.
Slon. Species occupies a key position in classification. It is the lowest taxonomic category. It is a natural population of individuals or group of populations which resemble one another in all essential morphological and reproductive characters so that they are able to interbreed freely and produce fertile offsprings. Each species is also called genetically distinct and reproductively isolated natural population. Mayr (1964) has defined species as “a group of actually or potentially interbreeding populations that are reproductively isolated from other such groups”.
In higher plants and animals the term ‘species’ refers to a group of individuals that are able to interbreed freely and produce fertile offsprings. But, in case of bacteria interbreeding cannot serve as the best criteria for delimiting species because bacteria usually reproduce asexually. Conjugation, transformation and transduction, which are termed as sexual reproduction methods in bacteria, also do not correspond to true interbreeding. Thus, for bacteria many other characters such as molecular homology, biochemical, physiological, ecological and morphological characters are taken into consideration while classifying them.

9. Define and understand the following terms:
(i) Phylum (ii) Class (iii) Family
(iv) Order (v) Genus
Slon. (i) Phylum – Phylum is a category higher than that of Class. The term Phylum is used for animals. A Phylum is formed of one or more classes, e.g., the Phylum Chordata of animals contains not only the class Mammalia but also Aves (birds), Reptilia (reptiles), Amphibia (amphibians), etc. In plants the term Division is used in place of Phylum.
(ii) Class – A Class is made of one or more related Orders. For example, the Class Dicotyledoneae of flowering plants contains all dicots which are grouped into several orders (e.g., Rosales, Sapindales, Ranales, etc.).
(iii) Family, – It is a taxonomic category which contains one or more related genera. All the genera of a family have some common features or correlated characters. They are separable from genera of a related family by important and characteristic differences in both vegetative and reproductive features. E.g., the genera of cats (Fells) and leopard (Panthera) are included in the Family Felidae. The members of Family Felidae are quite distinct from those of Family Canidae (dogs, foxes, wolves).
Similarly, the family Solanaceae contains a number of genera like Solanum, Datura, Petunia and Nicotiana. They are distinguishable from the genera of the related family Convolvulaceae (Convolvulus, Ipomoea).
(iv) Order – The category includes one or more related families. E.g., the plant Family Solanaceae is placed in the Order Polemoniales alongwith four other related families (Convolvulaceae, Boraginaceae, Hydrophyllaceae and Polemoniaceae). Similarly, the animal families Felidae and Canidae are included under the Order Carnivora alongwith Hyaenidae (hyaenas) and Ursidae (bears).
(v) Genus – It is a group or assemblage of related species which resemble one another in certain correlated characters. Correlated characters are those similar or common features which are used in delimitation of a taxon above the rank of species. All the species of genus are presumed to have evolved from a common ancestor. A genus may have a single living species e.g., Genus Homo. Its species is Homo sapiens – the living or modem man. The Genus Felis has many species, e.g., F. domestica – common cat, F. chaus (jungle cat) etc.

lO.How is a key helpful in the identification and classification of an organism?
Slon.‘Key is an artificial analytic device having a list of statements with dichotomic table of alternate characteristics. Taxonomic
keys are aids for rapid identification of unknown plants and animals based on
the similarities and dissimilarities. Keys are primarily based on stable and reliable characters. The keys are helpful in a faster preliminary identification which can bebacked up by confirmation through comparison with detailed description of the taxon provisionally identified with. Separate taxonomic keys are used for each taxonomic category like Family, Genus and Species.

11.Illustrate the taxonomical hierarchy with suitable examples of a plant and an animal.
Slon. The arrangement of various taxa in a hierarchical order is called taxonomic hierarchy. The hierarchy indicates the various levels of kinship. The number of similar characters of categories decreases from lowest rank to highest rank. The hierarchical system of classification was introduced by Linnaeus.
The hierarchy of major categories is:
Species —►Genus-►Family —► Order—► Class
Kingdom -4— Phylum or Division
Increasing specificity – ► Decreasing specificity
Classification of a plant (Wheat):
Kingdom – Plantae
Division – Angiospermae
Class – Monocotyledonae
Order – Poales
Family – Poaceae
Genus – Triticum
Species – aestivum
Classification of an animal (Housefly):
Kingdom – Animalia
Phylum – Chordata
Class – Insecta
Order – Diptera
Family – Muscidae
Genus – Musca
Species – domestica

Class 11 Biology NCERT Solutions

Cell Cycle and Cell Division Class 11 Chapter 10 Questions And Answers

February 14, 2025

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

Cell Cycle and Cell Division NCERT Solutions

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division:

Section Name	Topic Name
10	Cell Cycle and Cell Division
10.1	Cell Cycle
10.2	M Phase
10.3	Significance of Mitosis
10.4	Meiosis
10.5	Significance of Meiosis
10.6	Summary

NCERT Solutions Class 11 Biology Biology Sample Papers

NCRT TEXTBOOK QUESTIONS SOLVED

1. What is the average cell cycle span for a mammalian cell?
Solution: 24 hours.

2. Distinguish cytokinesis from karyokinesis.
Solution: Differences between cytokinesis and karyokinesis are:
NCERT Solutions For Class 11 Biology Cell Cycle and Cell Division Q2

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3. Describe the events taking place during the interphase.
Solution: The interphase, though called the resting phase, is metabolically quite active. It is the time during which the cell prepares itself for division by undergoing both cell growth and DNA replication in an orderly manner. The interphase is further divided into three phases:
• G1 (Gap 1) phase
• S (Synthesis) phase
• G2 (Gap 2) phase
G1 phase corresponds to the interval between mitosis of previous cell cycle and initiation of DNA replication. During G1 phase the cell is metabolically active and grows continuously 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 doubles per cell. In animal cells, during the S phase, DNA replication occurs in the nucleus, and the centriole duplicates in the cytoplasm. During the G2 phase synthesis of DNA stops while cell growth continues with synthesis of protein and RNA in preparation for mitosis.

4. What is G0 (quiescent phase) of cell cycle?
Solution: G0 phase is the phase of inactivation of cell cycle due to non-availability of mitogens and energy rich compounds. Cells in this stage remain metabolically active but no longer proliferate i.e., do not grow or differentiate unless called on to do so depending on the requirement of the organism. E.g., Nerve and heart cells of chordates are in permanent G0 phase.

5. Why is mitosis called equational division?
Solution: Mitosis is a type of cell division in which chromosomes replicate and become equally distributed in two daughter nuclei so that the daughter cells come to have the same number and type of chromosomes as present in parent cell. So mitosis is called as equational division.

6. Name the stage of cell cycle at which each one of the following events occur:
(i) Chromosomes are moved to 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
(iii) Zygotene of prophase I of meiosis 1
(iv) Pachytene of prophase I of meiosis I

7. Describe the following:
(a) Synapsis
(b) Bivalent
(c) Chiasmata
Draw a diagram to illustrate your answer.
Solution:
(a) Synapsis: During zygotene of prophase I stage homologou s chromosomes start pairing together and this process of association is called synapsis. Electron micrographs of this stage indicate that chromosome synapsis is accompanied by the formation of complex structure called synaptonemal complex.
(b) Bivalent: The complex formed by a pair of synapsed homologous chromosomes is called a bivalent or a tetrad i.e., 4 chromatids or a pair of chromosomes.
NCERT Solutions For Class 11 Biology Cell Cycle and Cell Division Q7
(c) Chiasmata: The beginning of diplotene is recognized by the dissolution of the synaptonemal complex and the tendency of the synapsed homologous chromosomes of the bivalents to separate from each other except at the sites of crossovers. These points of attachment (X-shaped structures) between the homologous chromosomes are called chiasmata.

8. How does cytokinesis in plant cells differ from that in animal cells?
Solution: Plant cytokinesis and animal cytokinesis differ in following respects:
NCERT Solutions For Class 11 Biology Cell Cycle and Cell Division Q8

9. Find examples where the four daughter cells from meiosis are equal in size and where they are found unequal in size.
Solution: During formation of male gametes (i.e., spermatozoa) in a typical mammal (i.e., human being), the four daughter cells formed from meiosis are equal in size. On the other hand, during formation of female gamete (i.e., ovum) in a typical mammal (i.e., human being), the four daughter cells are unequal in size.

10. Can there be DNA replication without cell division?
Solution: Yes. Endomitosis is the multiplication of chromosomes present in a set in nucleus without karyokinesis and cytokinesis result-ing in numerous copies within each cell. It is of 2 types.
Polyteny: Here chromosomes divide and redivide without separation of chromatids so that such chromosomes become multistranded with many copies of DNA. Such polytene (many stranded) chromosomes remain in permanent prophase stage and do not undergo cell cycle e.g., polytene (salivary glands) chromosome of Drosophila has 512- 1024 chromatids. Here number of sets of chromosomes does not change.
Polyploidy (endoduplication) : Here all chromosomes in a set divide and its chromatids separate but nucleus does not divide. This results in an increase in number of sets of chromosomes in the nucleus (4x, 8x….). This increase in sets of chromosomes is called polyploidy. It can be induced by colchicine and granosan. These chromosomes are normal and undergo cell cycle.

11. List the main differences between mitosis and meiosis.
Solution:
NCERT Solutions For Class 11 Biology Cell Cycle and Cell Division Q11

12. Distinguish anaphase of mitosis from anaphase I of meiosis.
Solution: Anaphase of mitosis : It is the phase of shortest duration. APC (anaphase promoting complex) develops. It degenerates proteins -binding the two chromatids in the region of centromere. As a result, the centromere of each chromosome divides. This converts the two chromatids into daughter chromosomes each being attached to the spindle pole of its side by independent chromosomal fibre. The chromosomes move towards the spindle poles with the centromeres projecting towards the poles and the limbs trailing behind. There is corresponding shortening of chromosome fibres. The two pole-ward moving chromosomes of each type remain attached to each other by interzonal fibres. Ultimately, two groups of chromosomes come to lie at the spindle poles.
NCERT Solutions For Class 11 Biology Cell Cycle and Cell Division Q12
Anaphase I of meiosis : Chiasmata disappear completely and the homologous chromosomes separate. The process is called disjunction. The separated chromosomes (univalents) show divergent chromatids and are called dyads. They move towards the spindle poles and ultimately form two groups of haploid chromosomes.
NCERT Solutions For Class 11 Biology Cell Cycle and Cell Division Q12.1

13. What is the significance of meiosis?
Solution: The significance of meiosis is given below:
(i) Formation of gametes – Meiosis forms gametes that are essential for sexual reproduction.
(ii) Genetic information – It switches on the genetic information for the development of gametes or gametophytes and switches off the sporophytic information. ‘
(iii) Maintenance of chromosome number – Meiosis maintains the fixed number of chromosomes in sexually reproducing organisms by halving the same. It is essential since the chromosome number becomes double after fertilisation.
(iv) Assortment of chromosomes – In meiosis paternal and maternal chromosomes assort independently. It causes reshuffling of chromosomes and the traits controlled by them. The variations help the breeders in improving the races of useful plants and animals.
(v) Crossing over – It introduces new combination of traits or variations.
(vi) Mutations – Chromosomal and genomic mutations can take place by irregularities of meiotic divisions. Some of these mutations are useful to the organism and are perpetuated by natural selection.
(vii) Evidence of basic relationship of organisms – Details of meiosis are essentially similar in the majority of organisms showing their basic similarity and relationship.

14. Discuss with your teacher about
(i) haploid insects and lower plants where cell division occurs, and
(ii)some haploid cells in higher plants where cell division does not occur.
Solution:
(i) Cell division occurs in haploid insect, such as drones of honey bee and lower plant like gametophyte of algae, bryophytes, and pteridophytes.
(ii) Synergids and antipodals in embryo sac of ovule are haploid cells where cell division does not occur.

15. Can there be mitosis without DNA replication in S-phase?
Solution: No there cannot be any mitotic division without-DNA replication in ‘S’ phase.

16. Analyse the events during every stage of ceil 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: Number of chromosomes and amount of DNA change during S-phase and anaphase of cell cycle. 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 S phase the number of chromosomes remains the same, i.e., 2N.
In mitotic anaphase, number of chromosomes remains the same. It is only sister chromatids which move towards their respective poles. DNA content remains unchanged. In anaphase I of meiosis, number of chromosomes are reduced to half, i.e., from 2N to IN and also DNA content decrease to one half i.e., from 4C to 2C. In anaphase II of meiosis II DNA content decreases to one half from 2C to 1C but chromosome number remain same.

Class 11 Biology NCERT Solutions

Breathing and Exchange of Gases Class 11 Chapter 17 Questions And Answers

February 14, 2025

Class 11 Biology Chapter 17 Breathing and Exchange of Gases NCERT Solutions

Breathing and Exchange of Gases NCERT Solutions

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 17 Breathing and Exchange of Gases:

Section Name	Topic Name
17	Breathing and Exchange of Gases
17.1	Respiratory Organs
17.2	Mechanism of Breathing
17.3	Exchange of Gases
17.4	Transport of Gases
17.5	Regulation of Respiration
17.6	Disorders of Respiratory System
17.7	Summary

NCERT TEXTBOOK QUESTIONS FROM SOLVED

1. Define vital capacity. What is its significance?
Solution: Vital capacity is defined as the maximum volume of air a person can breathe in after a forced expiration or the maximum volume of air a person can breathe out after a forced inspiration. It represents the maximum amount of air one can renew in the respiratory system in a single respiration. Thus, greater the vital capacity more is the energy available to the body.

2. State the volume of air remaining in the lungs after a normal breathing.
Solution: When a person breathes normally, the amount which remains in the lung after normal expiration, is called functional residual capacity. It is the sum of residual volume and the expiratory reserve volume (FRC = RV + ERV). It is about 2100 – 2300 mL of air.

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3. Diffusion of gases occurs in the alveolar region only and not in the other parts of respiratory system. Why?
Solution: For efficient exchange of gases, respi: atory surface must have certain characteristics such as (i) it must be thin, me ist and permeable to respiratory gases (ii) it must have large surface area, (iii) it must be highly vascular. Only alveolar region has these characteristics. Thus, diffusion of gases occurs in this region only.

4. What are the major transport mechanisms for CO₂? Explain.
Solution: Nearly 20-25 percent of CO₂ is transported by haemoglobin of RBCs, 70 percent of it is carried as bicarbonate ion in
plasma and about 7 percent of CO₂ is carried in a dissolved state through plasma. CO₂ is carried by haemoglobin as carbamino- haemoglobin. This binding is related to the partial pressure of CO₂.

5. What will be the p02 and pCO₂ in the atmospheric air compared to those in the alveolar air?
(i) pO₂ lesser, pCO₂ higher
(ii) pO₂ higher, pCO₂ lesser
(iii) pO₂ higher, pCO₂ higher
(iv) pO₂ lesser, pCO₂ lesser
Solution: (ii) Air that has entered the alveoli through the bronchioles is called alveolar air. It has the same partial pressure of CO₂ and 02 as is in the atmospheric air. Then, there occurs gaseous exchange between the adjacent blood capillaries and the alveoli. CO₂ diffuses from blood into the alveolar air and O₂ diffuses from alveolar air to the blood. As a result, new alveolar air has higher pCO₂and lesser pO₂, than the atmospheric air.

6. Explain the process of inspiration under normal conditions.
Solution: Inspiration is a process by which fresh air enters the lungs. The diaphragm, intercostal muscles and abdominal muscles play an important role. The muscles of the diaphragm and external intercostal muscles are principle muscles of inspiration. Volume of thoracic cavity increases by contraction of diaphragm and external intercostal muscles. During inspiration, relaxation of abdominal muscles also occurs which allows compression of the abdominal organs by diaphragm. Thus, overall volume of the thoracic cavity increases and as a result, there is a decrease of the air pressure in the lungs. The greater pressure outside the body now causes air to flow rapidly into the lungs. The sequence of air flow is.

7. How is respiration regulated?
Solution: Respiration is under both nervous and chemical regulation.
The respiratory centre in brain is composed of groups of neurons located in the medulla oblongata and pons varolii. The respiratory centre regulates the rate and depth of the breathing.
Dorsal respiratory group of neurons are located in the dorsal portion of the medulla oblongata. This group of neurons mainly causes inspiration.
Ventral group of neurons are located in the ventrolateral part of the medulla oblongata. These can cause either inspiration or expiration.
Pneumotaxic centre is located in the dorsal part of pons varolii. It sends signals to all the neurons of dorsal respiratory group and only to inspiratory neurons of ventral respiratory group. Its job is primarily to limit inspiration. Chemically, respiration is regulated by the large numbers of chemoreceptors located in the carotid bodies and in the aortic bodies. Excess carbon dioxide or hydrogen ions mainly stimulate the respiratory centre of the brain and increases the inspiratory and expiratory-signals to the respiratory muscles. Increased C0₂ lowers the pH resulting in acidosis. The role of oxygen in the regulation of respiratory rhythm is quite insignificant.

8. What is the effect of pCO₂on oxygen transport?
Solution: Increase in pCO₂ tension in blood brings rightward shift of the oxygen dissociation curve of haemoglobin thereby decreasing the affinity of haemoglobin for oxygen. This effect is called Bohr’s effect. It plays an important role in the release of oxygen in the tissues.

9. What happens to the respiratory process in a man going up a hill?
Solution: Rate of breathing will increase in order to supply sufficient oxygen to blood because air in mountainous region is deficient in oxygen.

10. What is the site of gaseous exchange in an insect?
Solution: Tracheae (Tracheal respiration) is the site of gaseous exchange in an insect. .

11. Define oxygen dissociation curve. Can you suggest any reason for its sigmoidal pattern?
Solution: The relationship between the partial pressure of oxygen (pO₂) and percentage saturation of the haemoglobin with oxygen (O₂)is graphically illustrated by a curve called oxygen haemoglobin dissociation curve (also called oxygen dissociation curve).
The sigmoidal pattern of oxygen haemoglobin dissociation curve is the result of two properties which play significant role in the transport of oxygen. These two properties are:
(i) Minimal loss of oxygen from haemoglobin occurs above p02 of 70-80 mm Hg despite significant changes in tension of oxygen beyond this. This is depicted by relatively flat portion of the curve.
(ii)Any further decline in p02 from 40 mm Hg causes a disproportionately greater release of oxygen from the haemoglobin. It results in the steeper portion of the curve and causes the curve to be sigmoid.

12. Have you heard about hypoxia? Try to gather information about it, and discuss with your friends.
Solution: Hypoxia is a condition of oxygen shortage in the tissues. It is of two types:
(i) Artificial hypoxia: It results from shortage of oxygen in the air as at high altitude. It causes mountain sickness characterised by breathelessnes, headache, dizziness and bluish tinge on skin.
(ii) Anaemic hypoxia: It results from the reduced oxygen carrying capacity of the blood due to anaemia or carbon monoxide poisoning. In both cases, less haemoglobin is available for carrying 0₂.

13. Distinguish between
(a) IRV and ERV
(b) Inspiratory capacity and expiratory capacity.
(c) Vital capacity and total lung capacity.
Solution:
(a) Differences between IRV and ERV are as follows:
NCERT Solutions For Class 11 Biology Breathing and Exchange of Gases Q13
(b)Differences between inspiratory capacity and expiratory capacity are as follows:

(c) Differences between vital capacity and total lung capacity are as follows:

14. What is tidal volume? Find out the tidal volume (approximate value) for a healthy human in an hour.
Solution: Tidal volume is the volume of air inspired or expired with each normal breath. This is about 500 mL in an adult person. It is composed of about 350 mL of alveolar volume and about 150 mL of dead space volume. The alveolar volume consists of air that reaches the respiratory surfaces of the alveoli and engages in gas exchange. The dead space volume consists of air that does not reach the respiratory surfaces.
A healthy man can inspire or expire approximately 6000 to 8000 mL of air per minute. Therefore, tidal volume for a healthy human in an hour is 360 – 480 mL of air.

Class 11 Biology NCERT Solutions

Body Fluids and Circulation Class 11 Chapter 18 Questions And Answers

February 14, 2025

Class 11 Biology Chapter 18 Body Fluids and Circulation NCERT Solutions

Body Fluids and Circulation NCERT Solutions

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 18 Body Fluids and Circulation:

Section Name	Topic Name
18	Body Fluids and Circulation
18.1	Blood
18.2	Lymph (Tissue Fluid)
18.3	Circulatory Pathways
18.4	Double Circulation
18.5	Regulation of Cardiac Activity
18.6	Disorders of Circulatory System
18.7	Summary

NCERT TEXTBOOK QUESTIONS FROM SOLVED

1. Name the components of the formed elements in the blood and mention one major function of each of them.
Solution: Blood corpuscles are the formed ele-ments in the blood, they constitute 45% of the blood. Formed elements are – (erythrocytes, RBCs or red blood corpuscles), (leucocytes, WBCs or white blood corpuscles) and throm¬bocytes or blood platelets. The major function of RBCs is to transport oxygen from lungs to body tissues and COz from body tissues to the lungs. White blood cells provide immunity to the body. Blood platelets play important role in blood clotting.

2. What is the importance of plasma proteins?
Solution: Plasma proteins constitute about 7 to 8% of plasma. These mainly include albumin, globulin, prothrombin and fibrinogen. Prothrombin and fibrinogen are needed for blood clotting. Albumins and globulins retain water in blood plasma and helps in maintaining osmotic balance. Certain globulins

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3. Match Column I with Column II.
Column I Column II
(a) Eosinophils (i) Coagulation
(b) RBC (ii) Universal recipient
(c) AB Group (iii) Resist infections
(d) Platelets (iv) Contraction of heart
(e) Systol (v) Gas transport
Solutlion.(a) – (iii); (b) – (v); (c) – (ii); (d) – (i); (e) – (iv).

4. Why do we consider blood as a connective tissue?
Solution: A connective tissue connects different tissues or organs of the body. It consists of living cells and extracellular matrix. Blood is vascular connective tissue, it is a mobile tissue consisting of fluid matrix and free cells. Blood transports materials from one place to the other and thereby establishes connectivity between different body parts.

5. What is the difference between lymph and blood?
Solution: The differences between blood and lymph are given below:
NCERT Solutions For Class 11 Biology Body Fluids and Circulation Q5

6. What is meant by double circulation? What is its significance?
Solution: The type of blood circulation in which oxygenated blood and deoxygenated blood do not get mixed is termed double circulation. It includes systemic circulation and pulmonary circulation. The circulatory pathway of double circulation is given in the following flow chart.
NCERT Solutions For Class 11 Biology Body Fluids and Circulation Q6
Flow chart: Double blood circulation Double circulation or separation of systemic and pulmonary circulations provides a higher metabolic rate to the body and also allows the two circulations to have different blood pressures according to the need of the organs they supply.

7. Write the differences between:
(a) Blood and lymph
(b) Open and closed system of circulation
(c) Systole and diastole
(d) P-wave and T-wave
Solution: (a) Refer answer 5.
(b) The differences between open and closed circulatory system are given below:
NCERT Solutions For Class 11 Biology Body Fluids and Circulation Q7

NCERT Solutions For Class 11 Biology Body Fluids and Circulation Q7.1
(c) Systole is contraction of heart chambers in order to pump out blood while diastole is relaxation of heart chambers to receive blood. The contraction of a chamber or systole decreases its volume and forces the blood out of it, whereas its relaxation or diastole brings it back to its original size to receive more blood.
(d) P wave is a small upward wave of elec-trocardiograph that indicates the atrial depolarisation (contraction of atria). It is caused by the activation of SA node. T-wave is a dome shaped wave of electro-cardiograph which represents ventricular repolarisation (ventricular relaxation).

8. Describe the evolutionary change in the pattern of heart among the vertebrates.
Solution: Vertebrates have a single heart. It is a hollow, muscular organ composed of cardiac muscle fibres. Two types of chambers in heart are atria and ventricles. The heart of lower vertebrates have additional chambers, namely sinus venosus and conus arteriosus or bulbus arteriosus or truncus arteriosus. During the course of development, in higher vertebrates, the persistent portions viz, auricles and ventricles are retained. However, these get complicated by incorporating several valves inside them and becoming compartmentali sed.
In fishes, heart is two chambered (1 auricle and 1 ventricle). Both the accessory chambers, sinus venosus and conus arteriosus are present. The heart pumps out deoxygenated blood which is oxygenated by the gills and sent to the body parts from where deoxygenated blood is carried to the heart. It is called single circulation and heart is called venous heart. Lung fish, amphibians and reptiles have three chambered heart, (2 auricles and 1 ventricle). The left atrium gets oxygenated blood from the gills/lungs/skin/buccopharyngeal cavity and the right atrium receives the deoxygenated blood from other body parts. But both oxygenated and deoxygenated blood get mixed up in single ventricle which pumps out mixed blood. This is called incomplete double circulation.
Crocodiles, birds and mammals have a complete four chambered heart (right and left auricles; right and left ventricles). Oxygenated and deoxygenated blood never get mixed. Right parts of the heart receive deoxygenated blood from all other body parts and send it to lungs for oxygenation whereas left parts of heart receive oxygenated blood from lungs and send it to other body parts. This mode of circulation is termed as complete double circulation which includes systemic and pulmonary circulation. There are no accessory chambers in heart of birds and mammals.

9. Why do we call our heart myogenic?
Solution: The heart of molluscs and vertebrates including humans is myogenic. It means heart beat is initiated in heart itself by a patch of modified heart muscle called sino-atrial node or pacemaker which lies in the wall of the right atrium near the opening of the superior vena cava.

10. Sino-atrial node is called the pacemaker of our heart. Why?
Solution: Sino-atrial node (SAN) is a mass of neuromuscular tissue which lies in the wall of right atrium. It is responsible for initiating and maintaining the rhythmic contractile activity of the heart. Therefore, it is called the pacemaker.

11. What is the significance of atrio-ventricular node and atrio-ventricular bundle in the functioning of heart?
Solution: atrio-ventricular node (AVN) is a mass of neuromuscular tissue, which is situated in wall of. right atrium, near the base of inter-atrial septum. AV node is the pacesetter of the heart,- as it transmits the impulses initiated by SA node to all parts of ventricles. Atrio-ventricular bundle (A-V bundle) or bundle of His is a mass of specialised fibres which originates from the AVN. Within the myocardium of the ventricles the branches of bundle of His divide into a network of fine fibres called Purkinje fibres. The bundle of His and the Purkinje fibres convey impulse of contraction from the AVN to the myocardium of the ventricles.

12. Define a cardiac cycle and the cardiac output.
Solution: The sequential events in the heart which are repeated cyclically is called cardiac cycle and it consists of systole (contraction) and diastole (relaxation) of both the atria and ventricles. The duration of a cardiac cycle is 0.8 seconds. Periods of cardiac cycle are atrial systole (0.1 second), ventricular systole (0.3 second) and complete cardiac diastole (0.4 second).
The amount of blood pumped by heart per minute is called cardiac output. It is calculated by multiplying stroke volume (volume of blood pumped by each ventricle per minute) with heart rate (number of beats per minute). The heart of normal person beats 72 times per minute and pumps out about 70 mL of blood per beat. Therefore, cardiac output averages 5000 mL or 5 litres.

13. Explain heart sounds.
Solution: The beating of heart produces characteristic sounds which can be heard by using stethoscope. In a normal person, two sounds are produced per heart beat. The first heart sound Tubb’ is low pitched, not very loud and of long duration. It is caused partly by the closure of the bicuspid and tricuspid valves and partly by the contraction of muscles in the ventricles.
The second heart sound ‘dubb’ is high pitched, louder, sharper and shorter in duration. It is caused by the closure of the semilunar valves and marks the end of ventricular systole.

14. Draw a standard ECG and explain the different segments in it.
Solution: ECG is graphic record of the electric current produced by the excitation of the cardiac muscles. The instrument used to record the changes is an electrocardiograph. A normal electrogram (ECG) is composed of a P wave, a QRS wave (complex) and a T wave. The P Wave is a small upward wave that represents electrical excitation or the atrial depolarisation which leads to contraction of both the atria (atrial contraction). It is caused by the activation of SA node. The impulses of contraction start from the SAnode and spread throughout the artia.
The QRS Wave (complex) represents ventricular depolarisation (ventricular contraction). It is caused by the impulses of the contraction from AV node through the bundle of His and Purkinje fibres and the contraction of the ventricular muscles. Thus this wave is due to the spread of electrical impulse through the ventricles.
The T Wave represents ventricular repolarisation (ventricular relaxation). The potential generated by the recovery of the ventricle from the depolarisation state is called the repolarisation wave. The end of the T-wave marks the end of systole.
ECG gives accurate information about the heart. Therefore, ECG is of great diagnostic value in cardiac diseases.
NCERT Solutions For Class 11 Biology Body Fluids and Circulation Q14

Class 11 Biology NCERT Solutions

Excretory Products and their Elimination Class 11 Chapter 19 Questions And Answers

February 14, 2025

Class 11 Biology Chapter 19 Excretory Products and their Elimination NCERT Solutions

Excretory Products and their Elimination NCERT Solutions

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 19 Excretory Products and their Elimination:

Section Name	Topic Name
19	Excretory Products and their Elimination
19.1	Human Excretory System
19.2	Urine Formation
19.3	Function of the Tubules
19.4	Mechanism of Concentration of the Filtrate
19.5	Regulation of Kidney Function
19.6	Micturition
19.7	Role of other Organs in Excretion
19.8	Disorders of the Excretory System
19.9	Summary

NCERT Solutions Class 11 Biology Biology Sample Papers

NCERT TEXTBOOK QUESTIONS FROM SOLVED

1.Define Glomerular Filtration Rate (GFR).
Solution. The amount of filtrate formed by the kidneys per minute is called glomerular filtration rate (GFR). It is approximately 125 mL/min. in a healthy person.

2.Explain the autoregulatory mechanism of GFR.
Solution. The kidneys have built-in mechanisms for the regulation of glomerular filtration rate. One such efficient mechanism is carried out by juxta glomerular apparatus (JGA). JGA is a special sensitive region formed by cellular modifications in the distal convoluted tubule and the afferent arteriole at the location of their contact. A fall in GFR can activate the JG cells to release renin which can stimulate the glomerular blood flow and thereby the GFR back to normal.

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3.Indicate whether the following statements are true or false.
(a) Micturition is carried out by a reflex.
(b) ADH helps in water elimination, making the urine hypotonic.
(c) Protein-free fluid is filtered from blood plasma into the Bowman’s capsule.
(d) Henle’s loop plays an important role in concentrating the urine.
(e) Glucose is actively reabsorbed in the proximal convoluted tubule.
Solution.(a) True (b) False (c) True (d) True (e) True

4.Give a brief account of the counter current mechanism.
Solution. The kidneys have a special mechanism for concentrating the urine, it is called counter current mechanism. The mechanism is said to be a counter current mechanism because the out flow (in the ascending limb) of Henle’s loop runs parallel to and in the opposite direction of the inflow (in the descending limb) and vasa recta. As the mechanism begins to function, the ascending limb of loop of Henle actively transports chloride and sodium ions out into the vasa recta from where it is secreted into the interstitial fluid. As a result the interstitial fluid around the loop of Henle contains large quantities of NaCl. The filtrate passes from the ascending limb of loop of Henle and enters a collecting duct. The collecting duct passes adjacent to the loop of Henle where the interstitial fluid contains large amounts of NaCl. The high osmotic pressure created by NaCl causes water to diffuse out of the collecting duct in the interstitial fluid and eventually to the blood of vasa recta. The filtrate becomes greatly concentrated and is now called urine. A similar counter current mechanism, operates between the interstitial fluid and blood passing through the vasa recta. As the blood capillary runs along the ascending limb of loop of Henle, NaCl diffuses out of the blood. The direction is reversed as the blood capillary passes along the descending limb of Henle. The blood flows in the vasa recta around the loop of Henle from ascending to the descending side while the fluid passing through the loop of Henle goes in the opposite direction. The arrangement helps to maintain the concentration gradient of NaCl.
The ‘overall function of counter current mechanism is to concentrate sodium chloride in the interstitial fluid and thereby cause water to diffuse out of the collecting ducts and concentrate the urine.

5.Describe the role of liver, lungs and skin in excretion.
Solution. Other than the kidneys, lungs, liver and skin also help in the elimination of excretory wastes. Lungs remove large amounts of C02 (18 litres/day) and also significant quantities of water every day. Liver secretes bile which contains substances like bilirubin, biliverdin, cholesterol, degraded steroid hormones, vitamins and drugs. Most of these substances ultimately pass out along with digestive wastes. The sweat and sebaceous glands in the skin can eliminate certain substances through their secretions. Sweat produced by the sweat glands is a watery fluid containing NaCl, small amounts of urea, lactic acid etc. Sebaceous glands eliminate certain substances like sterols, hydrocarbons and waxes through sebum.

6.Explain micturition.
Solution. The process of passing out urine from the urinary bladder is called micturition. Urine formed by the nephrons is ultimately carried to the urinary bladder where it is stored. This causes stretching of the wall of bladder that leads to the stimulation of stretch receptors on the walls of the bladder. This sends signal to the CNS. The CNS passes on motor messages to initiate the contraction of smooth muscles of the bladder and simultaneous relaxation of the urethral sphincter causing the release of urine.

7.Match the items of column I with those of column II.
Column I Column II
(a) Ammonotelism (i)Birds
(b) Bowman’s capsule (ii)Water reabsorption
(c) Micturition (iii)Bony fish
(d) Uricotelism (iv)Urinary bladder
(e) ADH (v)Renal tubule
Solution. (a) – (iii), (b) – (v), (c) – (iv), (d) – (i), (e) – (ii)

8.What is meant by the term osmoregulation?
Solution. The regulation of water and solute contents of the body fluids by the kidney is called osmoregualtion.

9.Terrestrialanimalsaregenerallyeitherureotelic or uricotelic, not ammonotelic, why?
Solution. Ammonotelic animals are aquatic animals that excrete ammonia which is highly soluble in water, thus large amount of water is also excreted. Terrestrial animals cannot afford to lose such large quantities of water from their bodies as they live in environment having water scarcity. They, therefore, excrete either urea (ureotelic) or uric acid (uricotelic) as these are less soluble in water.

10. What is the significance of juxta glomerular apparatus (JGA) in kidney function?
Solution. Juxta glomerular apparatus (JGA) is a special sensitive region formed by cellular modifications in the distal convoluted tubule and the afferent arteriole at the location of their contact. The JGA plays a complex regulatory role. A fall in glomerular blood flow/ glomerular blood pressure/GFR can activate the JG cells to release renin which converts angiotensinogen in blood to angiotensin I and further to angiotensin II. Angiotensin II, being a powerful vasoconstrictor, increases the glomerular blood pressure and thereby GFR. Angiotensin II also activates the adrenal cortex to release aldosterone. Aldosterone causes reabsorption of Na+ and water from the distal parts of the tubule. This also leads to an increase in blood pressure and GFR.

11 .Name the following.
(a) A chordate animal having flame cells as excretory structures.
(b) Cortical portions projecting between the medullary pyramids in the human kidney.
(c) A loop of capillary running parallel to the Henle’s loop.
Solution. (a) Cephalochordate – Amphioxus
(b) Columns of Bertini
(c) Vasa recta

12.Fill in the gaps.
(a) Ascending limb of Henle’s loop is________to water whereas the descending limb is________to it.
(b) Reabsorption of water from distal parts of the tubules is facilitated by hormone________
(c) Dialysis fluid contains all the constituents as in plasma except________
(d) A healthy adult human excretes (on an average)________gm of urea/day.
Solution.
(a) Ascending limb of Henle’s loop is impermeable to water whereas the descending limb is permeable to it.
(b) Reabsorption of water from distal parts of the tubules is facilitated by hormone ADH.
(c) Dialysis fluid contains all the constituents as in plasma except nitrogenous wastes.
(d) A healthy adult human excretes (on an average) 25 – 30 gm of urea/day.

Class 11 Biology NCERT Solutions

Structural Organisation in Animals Class 11 Chapter 7 Questions And Answers

February 14, 2025

Class 11 Biology Chapter 7 Structural Organisation in Animals NCERT Solutions

Structural Organisation in Animals NCERT Solutions

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 7 Structural Organisation in Animals:

Section Name	Topic Name
7	Structural Organisation in Animals
7.1	Animal Tissues
7.2	Organ and Organ System
7.3	Earthworm
7.4	Cockroach
7.5	Frogs
7.6	Summary

NCRT TEXTBOOK QUESTIONS SOLVED

1. Answer in one word or one line.
(i) Give the common name of Periplaneta americana.
(ii) How many spermathecae are found in earthworm?
(iii) What is the position of ovaries in cockroach?
(iv) How many segments are present in the abdomen of cockroach?
(v) Where do you find Malpighian tubules?
Solution: (i) Cockroach.
(ii) Four pairs.
(iii) In cockroach two large ovaries, lie laterally in the 2nd – 6th abdominal segments’.
(iv) Abdomen of cockroach consists of 10 segments.
(v) Malpighian tubules are present at the junction of midgut and hindgut in cockroach.

2. What are the following and where do you find them in animal body?
(a) Chondrocytes
(b) Axons.
(c) Ciliated epithelium
Solution: (a) Chondrocytes – Chondrocytes are the only cells found in cartilage. They are present in spaces called lacunae and they produce and maintain the matrix of cartilage. Bending ability of cartilage is due to chondrocytes. Cartilage is present at tip of nose, pinna of ear, epiglottis etc.
(b) Axon – Axon is one of the processes of neuron, which is the structural and functional unit of nervous system. The part of cyton – n’here axon arises is axon hillock and axon ends in group of branches called terminal arborizations. It conducts impulses away from the cyton. Neurons (nerve cells)
are present in brain and spinal cord.
(c) Ciliated epithelium – If the columnar or cuboidal cells bear cilia on their free surface they are called ciliated epithelium. 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 tube.

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3. Draw a labelled diagram of the reproductive organs of an earthworm.
Solution:
NCERT Solutions For Class 11 Biology Structural Organisation in Animals Q3

4. Answer the following.
(i) What is the function of nephridia?
(ii) How many types of nephridia are found in earthworm based on their location?
Solution: (i) Nephridia are excretory organs of earthworm, which perform the function of excretion and osmoregulation. Nephridia regulate the volume and composition of the body fluids. A nephridium is a coiled tubular and microscopic structure which starts out as a funnel that collects excess fluid from coelomic chamber. The funnel connects with a tubular wastes through a pore to the surface in the body wall or into the digestive tube.
(ii) In earthworm, nephridia are present in all segments except the first two. There are three types of nephridia on the basis of their location:
(a) Septal nephridia, present on both the sides of intersegmental septa from segment 15 to the last that open into intestine.
(b) Integumentary nephridia, attached to lining of the body wall of segment 3 to the last that open on the body surface and
(c) Pharyngeal nephridia, present as three paired tufts in the 4th, 5th and 6th segments.
NCERT Solutions For Class 11 Biology Structural Organisation in Animals Q4

5. Draw a labelled diagram of alimentary canal of a cockroach.
Solution:
NCERT Solutions For Class 11 Biology Structural Organisation in Animals Q5

6. What are the cellular components of blood?
Solution: Blood is a fluid connective tissue. It is composed of plasma (fluid) and blood cells (corpuscles). Cellular components of blood (blood corpuscles) constitute about 45% of blood volume.
Three types of blood cells are:
(i) Erythrocytes or red blood cells: They are most abundant blood cells. Normal RBC count is 5-5.5 million/mm³ in males and 4.5-5 million/mm³ in females) RBCs help in transport of gases and maintain blood pH.
(ii) Leucocytes or white blood cells: The normal WBC count is 5000-6000/mm³ of blood. They are involved in immune response of body and act as soldiers and scavangers.
(iii) Thrombocytes or blood platelets: There are about 2,50,000 platelets/mm³ of blood. They are involved in blood clotting.

7. Distinguish between the following:
(a) Prostomium and peristomium
(b) Septal nephridium and pharyngeal
Solution: (a) Differences between prostomium and peristomium are
NCERT Solutions For Class 11 Biology Structural Organisation in Animals Q7

NCERT Solutions For Class 11 Biology Structural Organisation in Animals Q7.1
(b) Differences between septal and pharyngeal nephridia are:

8. 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: Areolar tissue, blood and tendon are connective tissues while neuron is a part a nervous tissue.
(b) Cartilage: RBC, WBC and platelets are parts of vascular connective tissue while cartilage is skeletal connective tissue.
(c) Ligament: Ligament is a connective tissue.
(d) Antennae: Maxilla, mandible and labrum are mouth parts of cockroach while antennae are sense organs.
(e) Protonema: Protonema is a filamentous juvenile stage in life cycle of Bryophytes, while mesothorax, metathorax and coxa are appendages of cockroach.

9. Match the terms in column I with those in column II.

Column I	Column II
(a) Compound epithelium (b) Compound eye (c) Septal nephridia (d) Open circulatory system (e) Typhlosole (f) Osteocytes (g) Genitalia	(i) Alimentry canal (ii) Cockroach (iii) Skin (iv) Mosaic vision (v) Earthworm (vi) Phallomere (vii) Bone

Solution: (a) – (iii), (b) – (iv), (c) – (v), (d) – (ii), (e) – (i), (f) – (vii), (g) – (vi)

10. Mention briefly about the circulatory system of earthworm.
Solution: Earthworm possesses a closed type of blood vascular system, as the blood flows through closed blood vessels. Blood is red in colour due to respiratory pigment haemoglobin. Prominent blood vessels in earthworm includes dorsal, ventral, sub- neural, lateral oesophageal and supra- oesophageal blood vessels. There are four pairs of tubular hearts, provided with valves. The anterior two pairs of hearts, known as lateral hearts lie in the 7th and 9th segments and connect the dorsal blood vessel with the ventral blood vessel. They receive blood from the dorsal blood vessel and convey it to the ventral blood vessel. The posterior two pairs of hearts are called latero-oesophageal hearts and are situated in the 12th and 13th segments. The latero-oesophageal hearts apart from connecting the dorsal and ventral blood vessels are also joined with the supra oesophageal blood vessel. Latero-oesophageal hearts carry blood from the dorsal vessel and the supra oesophageal vessel to the ventral blood vessel.Contractions keep blood circulating in one direction. Blood glands are present in the 4th, 5th and 6lh segments which produce blood cells and haemoglobin which is dissolved in blood plasma. Blood cells are phagocytic in nature.

11. Describe various types of epithelial tissues with the help of labelled diagrams.
Solution: Epithelial tissue is a tissue made of one or more layers of compactly arranged cells that covers external surface and internal free surface of body organs and which is underlined by a basement membrane. The various types of epithelial tissue along with the diagram are given below:
(i) Simple epithelium : It is composed of single layer of cells which rest on basement membrane. Simple epithelium generally occurs over secretory and absorptive surfaces and forms lining of body cavities, ducts and tubes. Simple epithelium is of several types.
(a) Squamous epithelium: It consists of single layer of flat cells, tightly linked together and have centrally located oval or spherical nucleus. It is also called pavement epithelium. It is found in walls of blood vessels, air sacs of lungs, and lining of eye lens.
(b) Cuboidal epithelium: Cells of cuboidal epithelium are as tall as wide, with centrally placed nucleus. Its main functions are secretion and absorption. It lines sweat gland, thyroid follicles, salivary glands. Brush bordered cuboidal epithelium, i.e., cells having microvilli on their free surface lines proximal part of uriniferous tubule, pancreatic duct, testis and ovary.
(c) Columnar epithelium: Cells are with basally located nucleus. It helps in secretion and absorption. It occurs in lining of intestine, stomach, gall bladder.
(d) Ciliated epithelium: Free surface of columnar and cuboidal cells are covered with cilia. Cilia help in moving fluids, particles, mucus, etc. in a specific direction. It occurs in the inner surface of Fallopian tubules, nasal passage, bronchioles.
NCERT Solutions For Class 11 Biology Structural Organisation in Animals Q11
(e) Pseudostratified epithelium: It consists of single layer of cells but some cells are shorter than others. Due to difference in size of cells, the epithelium appears 2-3 layered. Pseudostratified columnar epithelium occurs in urethra and parotid salivary gland. Pseudostratified columnar ciliated epithelium (only larger cells ciliated) occurs in lining layer of nasal’ chambers, trachea and large bronchi. It helps in moving mucus and foreign particles.
NCERT Solutions For Class 11 Biology Structural Organisation in Animals Q11.1
(ii) Compoundepithelium/stratifiedepithelium: It is multilayered epithelium where cells of only the lowermost or basal layer are in contact with basement membrane. It provides protection against mechanical and chemical stresses and has limited role in secretion and absorption. It covers dry surface of skin, moist surface of buccal cavity, pharynx, etc. Different types of compound epithelium are:
(a) Stratified squamous epithelium: The cells of outer layer are flattened and squamous while the inner layers are cuboidal cells. It is of two types: Non- keratinised lining oesophagus, pharynx, buccal cavity, cornea, vagina and anal canal and keratinised (comified): forming epidermis of skin, hair, horn and nail.
NCERT Solutions For Class 11 Biology Structural Organisation in Animals Q11.2
(b) Stratified cuboidal epithelium: The outer layer of cuboidal cells and basal layer of columnar cells. It lines ducts of sweat glands, large salivary and pancreatic ducts.
(c) Stratified columnar epithelium: Both upper and basal layers are made of columnar cells, e.g., epiglottis covering, part of urethra.
(d) Stratified ciliated columnar epithelium: Outer layer consists of ciliated columnar cells and basal layer of columnar cells, e.g., larynx.
(iii) Transitional Epithelium: This is stratified epithelium which contains cuboidal or columnar shaped cells, which are thin and stretchable. No basement membrane is present as it would impede stretchability. It lines the inner surface of renal calyces, urinary bladder, ureter. Because of its t distribution, it is also called urothelium.
NCERT Solutions For Class 11 Biology Structural Organisation in Animals Q11.3
(iv) Glandular epithelium: It consists of specialised epithelial cells which synthesise intracellular macromolecules (protein in pancreas, lipids in adrenal glands, glycoprotein in salivary glands and all the three in mammary glands) and pour out the same in the form of a useful fluid secretion which is different from blood or any other extracellular fluid. Glands can be unicellular or multicellular on the basis of number of cells.
(a) Unicellular glands: Single-celled, e.g., goblet (mucous) cells of respiratory tract and alimentary canal.
(b) Multicellular glands: Consist of cluster of cells, e.g., Salivary glands.
NCERT Solutions For Class 11 Biology Structural Organisation in Animals Q11.4
On the basis of presence or absence of duct glands can be:
(a) Exocrine glands : These glands pour their secretion through a duct. They secrete milk, saliva, mucus, earwax. e.g., goblet cells, salivary glands, tear glands, gastric glands, intestinal glands.
(b) Endocrine glands: They are ductless glands, which pour their secretions into blood or lymph for reaching the target region. Their secretion is called hormone e.g., pituitary gland, thyroid gland, parathyroid glands, adrenal glands.
(c) Heterocrine glands: Both exocrine and endocrine, e.g., pancreas.
On basis of mode of secretion glands can be:
(a) Merocrine: Secretion is discharged
through diffusion, e g., goblet cells, sweat glands.
(b) Apocrine glands: Glandular secretion accumulates in the terminal part of the cell which is pinched off, e.g., mammary glands.
(c) Holocrine glands : The cell filled with secretory product disintegrates during discharge of the product, e.g., sebaceous gland.
(v) Modified epithelium : It is of following types:
(a) Germinal epithelium (generally cuboidal, produces gametes), (b) Glandular epithelium (columnar or cuboidal secretes chemicals and mucus), (c) Sensory epithelium or neuroepithelium. Epithelial cells having sensory hair on free surface and connected with nerve fibres on the other surface (generally columnar, receives and conveys stimuli), e.g, nasal epithelium, taste buds, retina, sensory spots of internal ear. (d) Pigmented epithelium – The cells possess melanin granules, e.g, retinal layer in contact with choroid of eye.

12. 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) Differences between simple and compound epithelium are as follows:
NCERT Solutions For Class 11 Biology Structural Organisation in Animals Q12
(b) Differences between cardiac and striated muscles are as follows:

NCERT Solutions For Class 11 Biology Structural Organisation in Animals Q12.2
(c) Differences between dense regular and dense irregular connective tissues are as follows:

(d) Differences between adipose tissue and blood tissue are as follows:

(e) Differences between simple gland and compound gland are as follows:
NCERT Solutions For Class 11 Biology Structural Organisation in Animals Q12.5

13. Draw a neat diagram of digestive system of frog.
Solution:
NCERT Solutions For Class 11 Biology Structural Organisation in Animals Q13

14. Mention the function of the following:
(a) Ureters in frog
(b) Malpighian tubules
(c) Body wall in earthworm.
Solution:
(a) Ureters in frog: Ureter is a transparent duct which arise from outer portion of kidney. In the “male frogs, ureter acts as urinogenital duct which runs backwards from kidneys and opens into the cloaca. It carries both urine and spermatozoa from kidney to the cloaca. In female, ureter conducts only urine from kidneys to the cloaca.
(b) Malpighian tubules: Malpighian tubules are excretory organs present in cockroach. These are present at junction of mid gut and hindgut. These are fine, long, unbranched, yellowish and blind tubules and are 100-150 in number. They help in the removal of excretory products from haemolymph.
(c) Body wall in earthworm: It consists of cuticle, epidermis, muscular layer and parietal peritoneum.
(i) It maintains the characteristic shape of’ the body.
(ii) It protects the internal organs.
(iii) The cuticle prevents excessive evaporation.
(iv) It serves as an ideal respiratory organ.
(v) The receptor cells play a vital sensory function.
(vi) The albumen helps in the formation of cocoon. It also serves as a food for the developing earthworm inside the cocoon.
(vii) Setae and muscles are responsible for locomotion.
(viii) Excretory matter is passed out through nephridiopores.

Class 11 Biology NCERT Solutions

Plant Growth and Development Class 11 Chapter 15 Questions And Answers

February 14, 2025

Class 11 Biology Chapter 15 Plant Growth and Development NCERT Solutions

Plant Growth and Development NCERT Solutions

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 15 Plant Growth and Development:

Section Name	Topic Name
15	Plant Growth and Development
15.1	Growth
15.2	Differentiation, Dedifferentiation and Redifferentiation
15.3	Development
15.4	Plant Growth Regulators
15.5	Photoperiodism
15.6	Vernalisation
15.7	Summary

NCERT Solutions Class 11 Biology Biology Sample Papers

NCERT TEXTBOOK QUESTIONS FROM SOLVED

1. Define growth, differentiation, development, dedifferentiation, redifferentiation, determinate growth, meristem and growth rate.
Solution: Growth is defined as a vital process which brings about an irreversible and permanent change in the shape, size, form, weight and volume of a cell, organ or whole organism, accompanied with increase in dry matter.
Differentiation is a localised qualitative change in size, biochemistry, structure and function of cells, tissues or organs, e.g., fibre, vessel, tracheid, sieve tube, mesophyll, leaf etc. Thus it is a change in form and physiological activity. It results in specialisation for particular functions.
Development may be defined as a process which includes growth, differentiation and maturation in a regular sequence in the life history of a cell, organ or organism viz., seed germination, growth, differentiation, flowering, seed formation and senescence. Dedifferentiation is the process by which the differentiated cells which have lost the ability to divide under certain circumstances, become meristematic and regain the divisibility. Redifferentiation is defined as maturation or differentiation of dedifferentiated cells to form cells which are unable to divide e.g., secondary xylem elements, cork cells etc., are formed by redifferentiation of secondary cambial cells.
Determinate growth is the ability of a cell, tissue or the organism to grow for a limited period of time. Meristem is a tissue consisting of unspecialised immature cells, possessing the power of continuous cell division and adding new cells to the body. Growth rate is defined as the increased growth per unit time.

2. Why is not any one parameter good enough to demonstrate growth throughout the life of a flowering plant?
Solution: A flowering plant consists of a number of organs viz., roots, stem, leaves, flowers, fruits etc. growing differently under different stages of life cycle. These plant organs require different parameters to demonstrate their growth. In plant organs like fruits, bulbs, corms etc. fresh weight is used for measuring their growth. In case of fruits, increase in volume, diameter etc., are also used as other parameters for the measurement of their growth. For flat organs like leaves, increase in surface area is used as the parameter. Stem and roots primarily grow in length and then in girth, thus increase in length and diameter are used for measuring their growth. Consequently, the flowering plants exhibit several parameters to demonstrate growth.

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3. Describe briefly
(a) Arithmetic growth
(b) Geometric growth
(c) Sigmoid growth curve
(d) Absolute and relative growth rates
Solution: (a) Arithmetic growth: If the length of a plant organ is plotted against time it shows a linear curve, the growth is called arithmetic growth. In this growth, the rate of growth is constant and increase in growth occurs in arithmetic progression e.g., length of a plant is measured as 2,4, 6, 8,10,12 cms at a definite interval of 24 hrs. It is found in root or shoot elongating at constant rate. Arithmetic growth is expressed as L_t = L₀ + r_tHere, L_t= length after time t. L₀ = length at the beginning, r = growth rate.
NCERT Solutions For Class 11 Biology Plant Growth and Development Q3
(b) Geometric growth: Geometric growth is the growth where both the progeny cells following mitosis retain the ability to divide and continue to do so. It occurs in many higher plants and in unicellular organisms when grown in nutrient rich medium. Number of cells is initially small so that initial growth is slow which is called lag phase. Later on, there is rapid growth at exponential rate. It is called log or exponential phase.
NCERT Solutions For Class 11 Biology Plant Growth and Development Q3.1
(c) Sigmoid growth curve: Geometric growth cannot be sustained for long. Some cells die. Limited nutrient availability causes slowing down of growth. It leads to stationary phase. There may be actually a decline. Plotting the growth against time will give a typical sigmoid or S-curve.
NCERT Solutions For Class 11 Biology Plant Growth and Development Q3.2
S-curve of growth is typical of most living organisms in their natural environment. It also occurs in cells, tissues and organs of plants.
(d) Absolute growth rate is the measurement of total growth per unit time. Relative growth rate is growth per unit time per unit initial growth.
Growth in given time period/ Measurement at start of time period
Suppose two leaves have grown by 5 cm² in one day. Initial size of leaf A was 5 cm² while that of leaf B was 50 cm². Though their absolute growth is the same (5 cm²/day), relative rate of growth is faster in leaf A(5/5) because of initial small size than in leaf B(5/50).

4. List five main groups of natural plant growth regulators. Write a note on discovery, physiological functions and agricultural/ horticultural applications of any one of them.
Solution: There are five main groups of natural plant growth regulators which are very much recognised as natural hormones in plants. These are:

Auxins
Gibberellins
Cytokinins
Abscisic acid
Ethylene

Discovery of auxin: In 1880, Charles Darwin and Francis Darwin worked with the coleoptile of canary grass (Phalaris sp.) and found the existence of a substance in coleoptile tip, which was able to recognise the light stimulus and leads to the bending of tip towards light. Boysen and Jensen (1910-1913) worked on Avena seedling and explained that the substances secreted in the tip are soluble in water (gelatin).
Paal (1919) reported that the substances secreted in the tip are translocated downwards and caused cell elongation in half portion which was on the dark side and hence bending was observed in opposite direction.
F.W. Went (1928) further refined this experiment and supported the observations of Paal. He was the first person to isolate and name these substances of tip as auxins (Greek Auxein – means ‘to grow’).
In 1931, Kogl and Haagen-Smith isolated
crystalline compounds from human urine.
These were named as auxin-a, auxin-b and
heteroauxin.

Physiological functions of auxins:

Auxins induce cambial cell divisions, shoot cell elongation and early differentiation of xylem and phloem in tissue culture experiments.
In general, auxins initiate rooting but inhibit the growth of roots. IBA is the most potent root initiator.
Auxins inhibit the growth of axillary buds (apical dominance) but enhance the size of carpel and hence earlier fruit formation.
Application of auxins retards the process of senescence (last degradative phase), the abscission of leaves, fruits, branches, etc.
Auxins induce feminisation, i.e., on male plant, female flowers are produced.

Agricultural/horticultural application of auxins:

Application of auxins like IAA, IBA, NAA induce rooting in stem cuttings of many plants. This method is widely used to multiply several economically useful plants.
Normally, auxins inhibit flowering however in litchi and pineapple, application of auxin promotes flowering thus used in orchards.
Auxin induces parthenocarpy in some plants including tomato, pepper, cucumber and Citrus, thus, produces seedless fruits of more economic value.
Auxins like 2, 4-D and 2, 4, 5-T are commercially used as weedicides, due to their low cost and greater chemical stability. They are selective herbicides (killing broad-leaved plants, but not grasses).
For checking premature fruit drop, auxins are applied which prevent the formation of abscission zone in the petiole or just below the fruit. Auxin regulates maturing fruit on the trees of apples, oranges and grape fruit. High doses of auxins can
cause fruit drop. Thus, heavy applications of synthetic auxins are used commercially to promote a coordinated abscission of various fruits to facilitate harvesting.
Auxin, produced in the apical bud, suppresses the development of lateral buds, i.e., apical dominance. Thus practically used in prolonging the dormancy period of potato tubers.
Naphthalene acetamide is used to prevent the lodging (excessive elongation and development of weak plants, specially in gramineae) or falling of crops.
Auxin (2,4-D) promotes callus formation in tissue culture. Complete plantlets are regenerated from callus tissue, using auxins and cytokinin which are then transplanted into the soil. Now-a-days, this is a widely practised method of propagation in the field of agriculture and horticulture.

5. What do you understand by photoperiodism and vernalisation? Describe their significance.
Solution: The physiological mechanism for flower-ing is controlled by two factors: photoperiod or light period, i.e., photoperiodism and low temperature, i.e., vernalisation. Photoperiodism is defined as the flowering response of a plant to relative lengths of light/ dark period. Significance of photoperiodism is as follows:

Photoperiodism determines the season in which a particular plant shall flower. For example, short day plants develop flowers in autumn-spring period (e.g., Dahlia, Xanthium) while long day plants produce flowers in summer (e.g., Amaranthus).
Knowledge of photoperiodic effect is useful in keeping some plants in vegetative growth (many vegetables) to obtain higher yield of tubers, rhizomes etc. or keep the plant in reproductive stage to yield more flowers and fruits.
A plant can be made to flower throughout the year by providing favourable photoperiod.
Helps the plant breeders in effective cross-breeding in plants.
Enable a plant to flower in different seasons.
Vernalisation is promotion or induction of flowering by exposing a plant to low temperature for some time. Significance of vernalisation is as follows :
(i) Crops can be grown earlier.
(ii)Plants can be grown in such regions where normally they do not grow.
(iii)Yield of the plant is increased.
(iv)Resistance to cold and frost is increased.
(v) Resistance to fungal diseases is increased.

6. Why is abscisic acid also known as stress hormone?
Solution: A fairly high concentration of abscisic acid (ABA) is found in leaves of plants growing under stress conditions, such as drought, flooding, injury, mineral deficiency etc. It is accompanied by loss of turgor and closure of stomata. When such plants are transferred to normal conditions, they regain normal turgor and ABA concentration decreases. Since the synthesis of ABA is accelerated under stress condition and the same is destroyed or inactivated when stress is relieved, it is also known as stress hormone.

7. ‘Both growth and differentiation in higher plants are open’. Comment.
Solution: Plant growth is generally indeterminate. Higher plants possess specific areas called meristems which take part in the formation of new cells. The body of plants is built on a modular fashion where structure is never complete because the tips (with apical meristem) “are open ended – always growing and forming new organs to replace the older or senescent ones. Growth is invariably associated with differentiation. The exact trigger for differentiation is also not known. Not only the growth of plants are open- ended, their differentiation is also open. The same apical meristem cells give rise to different types of cells at maturity, e.g., xylem, phloem, parenchyma, sclerenchyma fibres, collenchyma, etc. Thus, both the processes are indeterminate, unlimited and develop into
different structures at maturity i.e., both are open.

8. ‘Both a short day plant and a long day plant can produce flower simultaneously in a given place’. Explain.
Solution: A short day plant (SDP) flowers only when it receives a long dark period and short photoperiod, e.g., Xanthium, Dahlia etc. On the other hand, a long day plant (LDP) will flower only when it receives a long photoperiod and short dark period, e.g., wheat, oat etc. Thus critical photoperiod is that continuous duration of light which must not be exceeded in SDP and should always be exceeded in LDP in order to bring them to flower. Xanthium requires light for less than 15.6 hrs and Henbane requires light for more than 11 hrs. Xanthium (a SDP) and Henbane (DP) will flower simultaneously in light period between 11 to 15.6 hrs.

9. Which one of the plant growth regulators would you use if you are asked to
(a) induce rooting in a twig
(b) quickly ripen a fruit
(c) delay leaf senescence
(d) induce growth in axillary buds
(e) ‘bolt’ a rosette plant
(f) induce immediate stomatal closure in leaves.
Solution: (a) Auxins like IBA, NAA.
(b) Ethylene
(c) Cytokinins
(d) Cytokinins
(e) Gibberellins
(f) Abscisic acid (ABA)

10. Would a defoliated plant respond to photo- periodic cycle? Why?
Solution: No, a defoliated plant would not respond to photoperiodic cycle because photoperiodic stimulus is picked up by the leaves only. Even one leaf or a part of it is sufficient for this purpose. For perception of photoperiodic cycle, there must be the presence of leaves under inductive photoperiod, so that, the hormone responsible for flowering can be produced.

11. What would be expected to happen if:
(a) GA₃ is applied to rice seedlings
(b) dividing cells stop differentiating
(c) a rotten fruit gets mixed with unripe fruits
(d) you forget to add cytokinin to the culture medium.
Solution:
(a) The coleoptile will elongate rapidly, as GA₃ helps in cell growth.
(b) The development of callus (mass of undifferentiated cells) will take place.
(c) The unripe fruits will ripe quickly because of the increased rate of respiration due to emission of ethylene from rotten fruit.
(d) Cell division will retard and shoot will not initiate from the callus.

Class 11 Biology NCERT Solutions

Locomotion and Movement Class 11 Chapter 20 Questions And Answers

February 14, 2025

Class 11 Biology Chapter 20 Locomotion and Movement NCERT Solutions

Locomotion and Movement NCERT Solutions

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 20 Locomotion and Movement:

Section Name	Topic Name
20	Locomotion and Movement
20.1	Types of Movement
20.2	Muscle
20.3	Skeletal System
20.4	Joints
20.5	Disorders of Muscular and Skeletal System
20.1	Summary

NCERT Solutions Class 11 Biology Biology Sample Papers

NCERT TEXTBOOK QUESTIONS FROM SOLVED

1. Draw the diagram of a sarcomere of skeletal muscle showing different regions.
Solution:
NCERT Solutions For Class 11 Biology Locomotion and Movement Q1

2. Define sliding filament theory of muscle contraction.
Solution: According to sliding filament theory of muscle contraction, the actin and myosin filaments slide past each other with the help of cross-bridges to reduce the length of the sarcomeres.

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3. Describe the important steps in muscle contraction.
Solution: Mechanism of muscle contraction is explainei by sliding filament theory which states that contraction of a muscle fibre takes place by the sliding of the thin filaments over the th’ ck filaments. As a nerve impulse reaches the terminal end of the axon, synaptic vesicles fuse with the axon membrane and release a chemical transmitter, acetylcholine and binds to receptor sites of the motor end plate. When depolarization of the motor end plate reaches a certain level, it creates an action potential. An action potential (impulse) passes from the motor end plate over the sarcolemma and then into the T-tubules and sarcoplasmic reticulum and stimulates the sarcoplasmic reticulum to release calcium ions into the sarcoplasm. The calcium ions bind to troponin causing a change in its shape and position. This in turn alters shape and the position of tropomyosin, to which troponin binds. This shift exposes the active sites on the F-actin molecules. Myosin cross-bridges are then able to bind to these active sites. The heads of myosin molecules project laterally from thick myofilaments towards the surrounding thin myofilaments. These heads are called cross bridges. The head of each myosin molecule contains an enzyme mysoin ATPase. In the presence of myosin ATPase,Ca⁺⁺ and Mg⁺⁺ ions, ATP breaks down into ADP and inorganic phosphate, releasing energy in the head.
Energy from ATP causes energized myosin cross bridges to bind to actin.
NCERT Solutions For Class 11 Biology Locomotion and Movement Q3
The energized cross-bridges move, causing thin myofilaments to slide along the thick myofilaments.

4. Write true or false. If false change the statement so that it is true.
(a) Actin is present in thin filament.
(b) H-zone of striated muscle fibre represents both thick and thin filaments.
(c) Human skeleton has 206 bones.
(d) There are 11 pairs of ribs in man.
(e) Sternum is present on the ventral side of the body.
Solution: (a) True
(b) False – H-Zone of striated muscle fibres represents only thick filaments.
(c) True
(d) False – There are 12 pairs of ribs in man.
(e) True

5. Write the differences between:
(a) Actin and Myosin
(b) Red and White muscles
(c) Pectoral and Pelvic girdle
Solution: (a) Actin filaments and myosin filaments can be differentiated as follows:
NCERT Solutions For Class 11 Biology Locomotion and Movement Q5
(b) Differences between red muscle fibres and white muscle fibres are given in the following table:

NCERT Solutions For Class 11 Biology Locomotion and Movement Q5.2

NCERT Solutions For Class 11 Biology Locomotion and Movement Q5.3
(c) Differences between pectoral and pelvic girdles are given in the following table:

6. Match Column I with Column II:
Column I Column II
(a) Smooth muscle (i) Myoglobin
(b) Tropomyosin (ii) Thin filament
(c) Red muscle (iii) Sutures
(d) Skull (iv) Involuntary
Solution.(a) – (iv), (b)-(ii), (c)-(i), (d)-(iii)

7. What are the different types of movements exhibited by the cells of human body?
Solution: The cells of human body show three types of movements: amoeboid, ciliary and muscular.
Amoeboid movements: These are found in leucocytes of blood and phagocytes of certain body organs. In such cells, movements are brought with the help of temporary finger-like cytoplasmic projections, called pseudopodia or false feet. So it is also called pseudopodial movement. These pseudopodia are formed by flow of cytoplasm, called cyclosis (simplest form of movement), and cytoskeletal structures like microfilaments.
Ciliary movements: Large number of our internal tubular organs are lined by ciliated epithelium. For instance, the cilia of the cells lining the trachea, oviducts and vasa efferentia propel dust particles, eggs and sperms respectively by their coordinated movements in specific directions in these organs. Muscular movements: These are brought about by the action of skeleton, joints and muscles. These are of two types: movements of body parts and locomotion.

8. How do you distinguish between a skeletal muscle and a cardiac muscle?
Solution: We can distinguish between a skeletal muscle and a cardiac muscle on the basis of the features discussed in the following table:
NCERT Solutions For Class 11 Biology Locomotion and Movement Q8

9. Name the type of joint between the following:
(a) atlas/axis
(b) carpal/metacarpal of thumb
(c) between phalanges
(d) femur/acetabulum
(e) between cranial bones
(f) between pubic bones in the pelvic girdle
Solution: (a) Pivot joint
(b) Saddle joint
(c) Hinge joint
(d) Ball and socket joint
(e) Fibrous joint
(f) Cartilaginous joint

10. Fill in the blank spaces:
(a) All mammals (except a few) have……. cervical vertebra.
(b) The number of phalanges in each limb of human is…….
(c) Thin filament of myofibril contains two ‘F’ actins and two other proteins namely…….and…….
(d) In a muscle fibre Ca⁺⁺ is stored in …….
(e)…….and…….pairs of ribs are called floating ribs.
(f) The human cranium is made of……. bones.
Solution: (a) 7
(b) 14
(c) tropomyosin, troponin
(d) sarcoplasmic reticulum
(e) 11th and 12th
(f) 8

Class 11 Biology NCERT Solutions

Chemical Coordination and Integration Class 11 Chapter 22 Questions And Answers

February 14, 2025

Class 11 Biology Chapter 22 Chemical Coordination and Integration NCERT Solutions

Chemical Coordination and Integration NCERT Solutions

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 22 Chemical Coordination and Integration:

Section Name	Topic Name
22	Chemical Coordination and Integration
22.1	Endocrine Glands and Hormones
22.2	Human Endocrine System
22.3	Hormones of Heart, Kidney and Gastrointestinal Tract
22.4	Mechanism of Hormone Action
22.5	Summary

NCERT Solutions Class 11 Biology Biology Sample Papers

NCERT TEXTBOOK QUESTIONS FROM SOLVED

1. Define the following:
(a) Exocrine gland,
(b) Endocrine gland,
(c) Hormone.
Solution:
(a) Exocrine gland is a gland that pours its secretion on the surface or into a particular region by means of ducts for performing a metabolic activity, e.g., sebaceous glands, sweat glands, salivary glands and intestinal glands.
(b) Endocrine gland is an isolated gland (separates even from epithelium forming it) which secretes informational molecules or hormones that are poured into venous blood or lymph for reaching the target organ because the gland is not connected with the target organ by any duct. Therefore endocrine gland is also called ductless gland e.g. thyroid gland.
(c) Hormone is a substance that is manu-factured and secreted in very small quantities into the blood stream by an endocrine gland or a specialized nerve cell and regulates the growth or functioning of a specific tissue organ in a distant part of the body e.g insulin.

2. Diagrammatically indicate the location of the various endocrine glands in our body.
Solution:
NCERT Solutions For Class 11 Biology Chemical Coordination and Integration Q2

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3. List the hormones secreted by the following:
(a) Hypothalamus
(b) Pituitary
(c) Thyroid
(d) Parathyroid
(e) Adrenal
(f) Pancreas
(g) Testis
(h) Ovary
(i) Thymus
(j) Atrium
(k) Kidney
(l) G-l Tract.
Solution:
(a) Two types of hormones are produced by hypothalamus : releasing hormones (that stimulate secretion of pituitary hormones) and inhibiting hormones (that inhibit secretion of pituitary hormones).
These hormones are:

Thyrotrophin-releasing hormone Adreno-
corticotrophin-releasing hormone
Follicle-stimulating hormone-releasing hormone
Luteinizing hormone-releasing hormone
Growth hormone-releasing hormone
Growth inhibiting hormone
Prolactin releasing hormone
Prolactin inhibiting hormone
Melanocyte stimulating hormone¬releasing hormone
Melanocyte stimulating hormone- inhibiting hormone.

(b) Different parts of pituitary secrete different hormones.
Hormones secreted by anterior lobe of pituitary are:

Follicle stimulating hormone
Luteinizing hormone
Thyroid stimulating hormone
Adrenocorticotrophic hormone
Somatotrophic or Growth hormone
Prolactin hormone or Luteotrophic hormone.
Middle (intermediate) lobe of pituitary : Melanocyte stimulating hormone.
Posterior lobe of pituitary:
(i) Oxytocin
(ii) Vasopressin or antidiuretic hormone.

Thyroxine or tetraiodothyronine
Triiodothyronine
Calcitonin.

(d) Parathyroid gland secretes a single hormone called parathormone (PTH) or Collip’s hormone.

(e) Adrenal glands have two regions, namely, outer adrenal cortex and inner adrenal medulla. Both these regions secrete different hormones.
Hormones of adrenal cortex are grouped into three categories:

Glucocorticoids, e.g., cortisol
Mineralocorticoids, e.g., aldosterone
Sexcorticoids e.g testosterone. Adrenal medulla secretes two hormones
(i) Epinephrine (adrenaline)
(ii)Nor-epinephrine (nor-adrenaline).

(f) Pancreas secretes following hormones:

Insulin
Glucagon
Somatostatin.

(g) Testis secretes androgens such as testosterone.

(h) Ovary secretes:

Estrogens such as estradiol
Progesterone
Relaxin.

(i) Thymus secretes thymosin hormone.

(j) Atrium secretes atrial natriuretic factor (ANF).

(k) Kidney secretes:
(i) Renin (ii) Erythropoetin

(l) G.I. tract secretes :

Gastrin
Secretin
Cholecystokinin
Enterocrinin
Duocrinin
Villikinin.

4. Fill in the blanks:
Hormones Target gland
(a) Hypothalamic hormones ………………..
(b) Thyrotrophin (TSH) ………………..
(c) Corticotrophin (ACH) ………………..
(d) Gonadotrophins (LH, FSH) ………………..
(e) Melanotrophin (MSH) ………………..
Solution:
(a) Pituitary
(b) Thyroid
(c) Adrenal cortex
(d) Gonads -Testes in male and ovaries in female
(e) Skin.

5. Write short notes on the functions of the following hormones:
(a) Parathyroid hormones (PTH)
(b) Thyroid hormones
(c) Thymosin
(d) Androgens
(e) Estrogens
(f) Insulin and Glucagon.
Solution:
(a) Parathyroid hormone increases the level of calcium and decreases the level of phosphate in the blood.
(b) Thyroid gland secretes three hormones: thyroxine, triiodothyronin and calcitonin. Thyroxine and triiodothyronin control the general metabolism of the body, promote growth of body tissues and stimulates tissue differentiation. Calcitonin regulates the concentration of calcium in the blood.
(c) Thymosin is secreted by thymus. It accelerates cell division, stimulates the development and differentiation of T-lymphocytes and also hastens attainment of sexual maturity.
(d) Androgens are secreted by testis. They stimulate the development of male reproductive system, formation of sperms, development of male accessory sex characters and also determines the male sexual behaviour and the sex urge.
(e) Estrogens are secreted by ovaries. They stimulate the female reproductive tract to grow to full size and become functional, differentiation of ova and development of accessory sex characters.
(f) Insulin is secreted by the |3-cells of the pancreas. It lowers blood glucose level, and promotes synthesis of proteins and fats. Glucagon is secreted by the a-cells of the pancreas. It increases the level of glucose in the blood.

6. Give example(s) of
(a) Hyperglycemic hormone and hypoglyce-mic hormone
(b) Hypercalcemic hormone
(c) Gonadotrophic hormones
(d) Progestational hormone
(e) Blood pressure lowering hormone
(f) Androgens and estrogens.
Solution:
(a)Glucagon, Insulin
(b) Parathormone (PTH)
(c) Follicle stimulating hormone (FSH) and Luteinizing hormone (LH)
(d) Progesterone
(e) Atrial natriuretic factor
(f) Testosterone and Estradiol.

7. Which hormonal deficiency is responsible for the following:
(a) Diabetes meilitus
(b) Goitre
(c) Cretinism.
Solution:
(a) Insulin
(b) Thyroxine and Triiodothyronine
(c) Thyroxine and Triiodothyronine.

8. Briefly mention the mechanism of action of FSH.
Solution: (Folliclestimulatinghormone)being glycoprotein is insoluble in lipids, therefore,
cannot enter the target cells. It binds to the specific receptor molecules located on the surface of the cell membrane to form hormone – receptor complex. This complex causes the release of an enzyme adenylate cyclase from the receptor site. This enzyme forms the cell cyclic adenosine monophosphate (cAMP) from ATP. The cAMP activates the existing enzyme system of the cell. This accelerates the metabolic reactions in the cell. The hormone is called the first messenger and the cAMP is termed the second messenger. The hormone- receptor complex changes the permeability of the cell membrane to facilitate the passage of materials through it. This increases the activities of the cell as it receives the desired materials.

9. Match the following :
Column I Column II
(a) T₄ (i) Hypothalamus
(b) PTH (ii)Thyroid
(c) GnRH (iii)Pituitary
(d) LH (iv) Parathyroid.
Solution:
(a) – (ii); (b) – (iv); (c) – (i); (d) – (iii)

Class 11 Biology NCERT Solutions

Category: Class 11 Biology

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