Chapter 1 The Living World

Transcription

1 Class XI Chapter 1 The Living World Biology Question 1: Why are living organisms classified? A large variety of plants, animals, and microbes are found on earth. All these living organisms differ in size, shape, colour, habitat, and many other characteristics. As there are millions of living organisms on earth, studying each of them is impossible. Therefore, scientists have devised mechanisms to classify all living organisms. These methods of classification are based on rules and principles that allow identification, nomenclature, and finally classification of an organism. For example, based on certain principles, once an organism is identified as an insect, it will be given a scientific name and then grouped with other similar organisms. Thus, various groups or taxon include organisms based on their similarity and differences. Therefore, the biological classification helps in revealing the relationship between various organisms. It also helps in making study of organisms easy and organized. Question 2: Why are the classification systems changing every now and then? Millions of plants, animals, and microorganisms are found on earth. Many of these have been identified by the scientists while many new species are still being discovered around the world. Therefore, to classify these newly discovered species, new systems of classification have to be devised every now and then. This creates the requirement to change the existing systems of classification. Question 3: What different criteria would you choose to classify people that you meet often? Page 1 of 7

2 Class XI Chapter 1 The Living World Biology To classify a class of forty students, let us start the classification on the basis of sexes of the students. This classification will result in the formation of two major groups- boys and girls. Each of these two groups can be further classified on the basis of the names of the students falling in these groups. Since it is possible that more than one student can have a particular name, these names can be further divided based on the surnames. Since there is still some chance that more than one student can have the same surname, the final level of classification will be based on the roll numbers of each student. Question 4: What do we learn from identification of individuals and populations? The knowledge of characteristics of an individual or its entire population helps in the identification of similarities and dissimilarities among the individuals of same kind or between different types of organisms. It helps the scientists to classify organisms in various categories. Question 5: Given below is the scientific name of Mango. Identify the correctly written name. Mangifera Indica Mangifera indica In binomial system of nomenclature, the generic name of a species always starts with a capital letter whereas the specific name starts with a small letter. Therefore, the correct scientific name of Mango is Mangifera indica. Page 2 of 7

3 Class XI Chapter 1 The Living World Biology Question 6: Define a taxon. Give some examples of taxa at different hierarchical levels. Each unit or category of classification is termed as a taxon. It represents a rank. For example, the basic level of classification is species, followed by genus, family, order, class, phylum or division, in ascending order. The highest level of classification is known as kingdom. Question 7: Can you identify the correct sequence of taxonomical categories? (a) Species Order Phylum Kingdom (b) Genus Species Order Kingdom (c) Species Genus Order Phylum The correct hierarchical arrangement of taxonomic categories in ascending order is Species Genus Family Order Class Phylum Kingdom Therefore, both (a) and (c) represent correct sequences of taxonomic categories. In sequence (b), species should be followed by genus. Therefore, it does not represent the correct sequence. Question 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. In biological terms, species is the basic taxonomical rank. It can be defined as a group of similar organisms that are capable of interbreeding under natural conditions to produce fertile offsprings. Therefore, a group of similar individuals that are respectively isolated form a species. Species can also be defined as group of individuals that share the same gene pool. Page 3 of 7

4 Class XI Chapter 1 The Living World Biology Question 9: Define and understand the following terms: (i) Phylum (ii) Class (iii) Family (iv) Order (v) Genus (i) Phylum Phylum is the primary division of kingdom. It includes one or more related classes of animals. In plants, instead of phylum, the term division is used. (ii) Class Class is a taxonomic group consisting of one or more related orders. For example, the class, Mammalia, includes many orders. (iii) Family Family is a taxonomic group containing one or more related genera. In plants, families are categorized on the basis of vegetative and reproductive features. (iv) Order Order is a taxonomic group containing one or more families. For example, the order, carnivore, includes many families. (v) Genus Genus is a taxonomic group including closely related species. For example, the genus, Solanum, includes many species such as nigrum, melongena, tuberosum,etc. Question 10: How is a key helpful in the identification and classification of an organism? Key is another taxonomical aid that helps in identification of plant and animal species. These keys are based on similarities and dissimilarities in characters, generally in a pair called couplet. Each statement in a taxonomic key is referred to as a lead. For categorizing each taxonomic rank, such as family, genus, species, etc., different keys are used. It is also useful in identification of unknown organisms. Page 4 of 7

5 Class XI Chapter 1 The Living World Biology Keys are of two types- indented and bracketed keys. Indented key provides a sequence of choices between two or more statements while in bracketed key, a pair of contrasting characters are used. (i) Indented key to identify different species of Rhododendron. 1. Leaves evergreen 2. leaves densely hairy below, orange or white hair; flower appears to have separate petals. Rhododendron groenlandicum 2. hair absent on leaves, flower has five petals fused in a shallow tube. Rhododendron maximus 1. Leaves deciduous 3. pink flowers with two free petals and three fused petals. Rhododendron canadense 3. white to pink flowers with all petals fused together (ii) Bracketed key to identify different species of Rhododendron. 1. Leaves evergreen Leaves deciduous Leaves densely hairy below, orange or white hair; flower appears to have separate petals.. Rhododendron groenlandicum 2. Hair absent on leaves, flower has five petals fused in shallow tube..rhododendron maximus 3. Pink flowers with two free petals and three fused petals Rhododendron canadense 3. White to pink flowers with all petals fused together Page 5 of 7

6 Class XI Chapter 1 The Living World Biology Question 11: Illustrate the taxonomical hierarchy with suitable examples of a plant and an animal. The arrangement of various taxa in a hierarchical order is called taxonomic hierarchy. In this hierarchy, species is present at the lowest level whereas kingdom is present at the highest level. Kingdom Phylum or division Class Order Family Genus Species A Taxonomic hierarchy Classification of a plant As an example, let us classify Solanum melongena (Brinjal). Kingdom Plantae Division Angiospermae Class Dicotyledonae Order Solanales Family Solanaceae Genus Solanum Species melongena Page 6 of 7

7 Class XI Chapter 1 The Living World Biology Classification of an animal As an example, let us classify Columba livia (Blue rock Dove). Kingdom Animalia Phylum Chordata Class Aves Order Columbiformes Family Columbidae Genus Columba Species livia Page 7 of 7

8 Class XI Chapter 2 Biological Classification Biology Question 1: Discuss how classification systems have undergone several changes over a period of time? The classification systems have undergone several changes with time. The first attempt of classification was made by Aristotle. He classified plants as herbs, shrubs, and trees. Animals, on the other hand, were classified on the basis of presence or absence of red blood cells. This system of classification failed to classify all the known organisms. Therefore, Linnaeus gave a two kingdom system of classification. It consists of kingdom Plantae and kingdom Animalia. However, this system did not differentiate between unicellular and multicellular organisms and between eukaryotes and prokaryotes. Therefore, there were large numbers of organisms that could not be classified under the two kingdoms. To solve these problems, a five kingdom system of classification was proposed by R.H Whittaker in On the basis of characteristics, such as cell structure, mode of nutrition, presence of cell wall, etc., five kingdoms, Monera, Protista, Fungi, Plantae, and Animalia were formed. Question 2: State two economically important uses of: (a) Heterotrophic bacteria (b) Archaebacteria (a) Heterotrophic bacteria (1) They act as decomposers and help in the formation of humus. (2) They help in the production of curd from milk. (3) Many antibiotics are obtained from some species of bacteria. (4) Many soil bacteria help in fixation of atmospheric nitrogen. Page 1 of 10

9 Class XI Chapter 2- Biological Classification Biology (b) Archaebacteria (1) Methane gas is produced from the dung of ruminants by the methanogens. (2) Methanogens are also involved in the formation of biogas and sewage treatment. Question 3: What is the nature of cell-walls in diatoms? The cell walls of diatoms are made of silica. Their cell wall construction is known as frustule. It consists of two thin overlapping shells that fit into each other such as a soap box. When the diatoms die, the silica in their cell walls gets deposited in the form of diatomaceous earth. This diatomaceous earth is very soft and quite inert. It is used in filtration of oils, sugars, and for other industrial purposes. Question 4: Find out what do the terms algal bloom and red-tides signify. Algal bloom Algal bloom refers to an increase in the population of algae or blue-green algae in water, resulting in discoloration of the water body. This causes an increase in the biological oxygen demand (BOD), resulting in the death of fishes and other aquatic animals. Red-tides Red tides are caused by red dinoflagellates (Gonyaulax) that multiply rapidly. Due to their large numbers, the sea appears red in colour. They release large amounts of toxins in water that can cause death of a large number of fishes. Question 5: How are viroids different from viruses? Page 2 of 10

10 Class XI Chapter 2- Biological Classification Biology Viroids were discovered in 1917 by T.O. Denier. They cause potato spindle tuber disease. They are smaller in size than viruses. They also lack the protein coat and contain free RNA of low molecular weight. Question 6: Describe briefly the four major groups of Protozoa. Protozoa are microscopic unicellular protists with heterotrophic mode of nutrition. They may be holozoic, saprobic, or parasitic. These are divided into four major groups. (1) Amoeboid protozoa or sarcodines They are unicellular, jelly-like protozoa found in fresh or sea water and in moist soil. Their body lacks a periplast. Therefore, they may be naked or covered by a calcareous shell. They usually lack flagella and have temporary protoplasmic outgrowths called pseudopodia. These pseudopodia or false feet help in movement and capturing prey. They include free living forms such as Amoeba or parasitic forms such as Entamoeba. (2) Flagellated protozoa or zooflagellates They are free living, non-photosynthetic flagellates without a cell wall. They possess flagella for locomotion and capturing prey. They include parasitic forms such as Trypanosoma, which causes sleeping sickness in human beings. (3) Ciliated protozoa or ciliates They are aquatic individuals that form a large group of protozoa. Their characteristic features are the presence of numerous cilia on the entire body surface and the presence of two types of nuclei. All the cilia beat in the same direction to move the water laden food inside a cavity called gullet. They include organisms such as Paramoecium, Vorticella,etc. (4) Sporozoans Page 3 of 10

11 Class XI Chapter 2- Biological Classification Biology They include disease causing endoparasites and other pathogens. They are uninucleate and their body is covered by a pellicle. They do not possess cilia or flagella. They include the malaria causing parasite Plasmodium. Question 7: Plants are autotrophic. Can you think of some plants that are partially heterotrophic? Plants have autotrophic mode of nutrition as they contain chlorophyll pigment. Thus, they have the ability to prepare their own food by the process of photosynthesis. However, some insectivorous plants are partially heterotrophic. They have various means of capturing insects so as to supplement their diet with required nutrients derived from insects, causing proliferation of growth. The examples include pitcher plant (Nepenthes), Venus fly trap, bladderwort, and sundew plant. Question 8: What do the terms phycobiont and mycobiont signify? Phycobiont refers to the algal component of the lichens and mycobiont refers to the fungal component. Algae contain chlorophyll and prepare food for fungi whereas the fungus provides shelter to algae and absorbs water and nutrients from the soil. This type of relationship is referred to as symbiotic. Question 9: Give a comparative account of the classes of Kingdom Fungi under the following: (i) Mode of nutrition (ii) Mode of reproduction (A) Phycomycetes- This group of fungi includes members such as Rhizopus, Albugo, etc. Page 4 of 10

12 Class XI Chapter 2- Biological Classification Biology (i) Mode of nutrition They are obligate parasites on plants or are found on decaying matter such as wood. (ii) Mode of reproduction Asexual reproduction takes place through motile zoospores or non-motile aplanospores that are produced endogenously in sporangium. Sexual reproduction may be of isogamous, anisogamous, or oogamous type. It results in the formation of thick-walled zygospore. (B) Ascomycetes- This group of fungi includes members such as Penicillium, Aspergillus, Claviceps, and Neurospora. (i) Mode of nutrition They are sporophytic, decomposers, parasitic or coprophilous (growing on dung). (ii) Mode of reproduction Asexual reproduction occurs through asexual spores produced exogenously, such as conidia produced on conidiophores. Sexual reproduction takes place through ascospores produced endogenously in saclike asci and arranged inside ascocarps. (C) Basidiomycetes- This group of fungi includes members such as Ustilago, Agaricus and Puccinia. (i) Mode of nutrition They grow as decomposers in soil or on logs and tree stumps. They also occur as parasites in plants causing diseases such as rusts and smuts. (ii) Mode of reproduction Asexual reproduction takes place commonly through fragmentation. Asexual spores are absent. Sex organs are absent but sexual reproduction takes place through plasmogamy. It involves fusion of two different strains of hyphae. The resulting dikaryon gives rise to a basidium. Four basidiospores are produced inside a basidium. (D) Deuteromycetes This group of fungi includes members such as Alternaria, Trichoderma, and Colletotrichum. Page 5 of 10

13 Class XI Chapter 2- Biological Classification Biology (i) Mode of nutrition Some members are saprophytes while others are parasites. However, a large number act as decomposers of leaf litter. (ii) Mode of reproduction Asexual reproduction is the only way of reproduction in deuteromycetes. It occurs through asexual spores called conidia. Sexual reproduction is absent in deuteromycetes. Question 10: What are the characteristic features of Euglenoids? Some characteristic features of Euglenoids are as follows. Euglenoids (such as Euglena) are unicellular protists commonly found in fresh water. Instead of cell wall, a protein-rich cell membrane known as pellicle is present. They bear two flagella on the anterior end of the body. A small light sensitive eye spot is present. They contain photosynthetic pigments such as chlorophyll and can thus prepare their own food. However, in absence of light, they behave similar to heterotrophs by capturing other small aquatic organisms. They have both plant and animal-like features, which makes them difficult to classify. Question 11: Give a brief account of viruses with respect to their structure and nature of genetic material. Also name four common viral diseases. Page 6 of 10

14 Viruses Class are XIsub-microscopic Chapter infectious 2- Biological agents that Classification can infect all living organisms. Biology A virus consists of genetic material surrounded by a protein coat. The genetic material may be present in the form of DNA or RNA. Most of the viruses, infecting plants, have single stranded RNA as genetic material. On the other hand, the viruses infecting animals have single or double stranded RNA or double stranded DNA. Bacteriophages or viruses infecting bacteria mostly have double stranded DNA. Their protein coat called capsid is made up of capsomere subunits. These capsomeres are arranged in helical or polyhedral geometric forms. A.I.D.S, small pox, mumps, and influenza are some common examples of viral diseases. Question 12: Organise a discussion in your class on the topic- Are viruses living or non-living? Viruses are microscopic organisms that have characteristics of both living and nonliving. A virus consists of a strand of DNA or RNA covered by a protein coat. This presence of nucleic acid (DNA or RNA) suggests that viruses are alive. In addition, they can also respond to their environment (inside the host cell) in a limited manner. However, some other characters, such as their inability to reproduce without using the host cell machinery and their acellular nature, indicate that viruses are nonliving. Therefore, classifying viruses has remained a mystery for modern systematics. Page 7 of 10

15 Class XI Chapter 2- Biological Classification Biology Question 1: Name the parts of an angiosperm flower in which development of male and female gametophyte take place. The male gametophyte or the pollen grain develops inside the pollen chamber of the anther, whereas the female gametophyte (also known as the embryo sac) develops inside the nucellus of the ovule from the functional megaspore. Question 2: Differentiate between microsporogenesis and megasporogenesis. Which type of cell division occurs during these events? Name the structures formed at the end of these two events. (a) Microsporogenesis Megasporogenesis 1. It is the process of the formation of microspore tetrads from a microspore mother cell through meiosis. It is the process of the formation of the four megaspores from a megaspore mother cell in the region of the nucellus through meiosis 2. It occurs inside the pollen sac of the anther. It occurs inside the ovule. (b) Both events (microsporogenesis and megasporogenesis) involve the process of meiosis or reduction division which results in the formation of haploid gametes from the microspore and megaspore mother cells. (c) Microsporogenesis results in the formation of haploid microspores from a diploid microspore mother cell. On the other hand, megasporogenesis results in the formation of haploid megaspores from a diploid megaspore mother cell. Page 8 of 10

16 Class XI Chapter 2- Biological Classification Biology Question 3: Arrange the following terms in the correct developmental sequence: Pollen grain, sporogenous tissue, microspore tetrad, pollen mother cell, male gametes The correct development sequence is as follows: Sporogenous tissue pollen mother cell microspore tetrad Pollen grain male gamete During the development of microsporangium, each cell of the sporogenous tissue acts as a pollen mother cell and gives rise to a microspore tetrad, containing four haploid microspores by the process of meiosis (microsporogenesis). As the anther matures, these microspores dissociate and develop into pollen grains. The pollen grains mature and give rise to male gametes. Question 4: With a neat, labelled diagram, describe the parts of a typical angiosperm ovule. An ovule is a female megasporangium where the formation of megaspores takes place. Page 9 of 10

17 Chapter 2- Biological Classification Biology The various parts of an ovule are (1) Funiculus It is a stalk-like structure which represents the point of attachment of the ovule to the placenta of the ovary. (2) Hilum It is the point where the body of the ovule is attached to the funiculus. (3) Integuments They are the outer layers surrounding the ovule that provide protection to the developing embryo. (4) Micropyle It is a narrow pore formed by the projection of integuments. It marks the point where the pollen tube enters the ovule at the time of fertilization. (5) Nucellus It is a mass of the parenchymatous tissue surrounded by the integuments from the outside. The nucellus provides nutrition to the developing embryo. The embryo sac is located inside the nucellus. (6) Chalazal It is the based swollen part of the nucellus from where the integuments originate. Question 5: What is meant by monosporic development of female gametophyte? The female gametophyte or the embryo sac develops from a single functional megaspore. This is known as monosporic development of the female gametophyte. In most flowering plants, a single megaspore mother cell present at the micropylar pole of the nucellus region of the ovule undergoes meiosis to produce four haploid megaspores. Later, out of these four megaspores, only one functional megaspore develops into the female gametophyte, while the remaining three degenerate. Page 10 of 10

18 Class XI Chapter 3 Plant Kingdom Biology Question 1: What is the basis of classification of algae? Algae are classified into three main classes Chlorophyceae, Phaeophyceae, and Rhodophyceae. These divisions are based on the following factors: (a) Major photosynthetic pigments present (b) Form of stored food (c) Cell wall composition (d) Number of flagella and position of insertion Class I Chlorophyceae Common name Green algae Major pigments Chlorophylls a and b Stored food Starch Cell wall composition Cellulose Flagella number and position 28; equal and apical Class II Phaeophyceae Common name Brown algae Major pigments Chlorophylls a and c, and fucoxanthin Stored food Mannitol and laminarin Cell wall composition Cellulose and algin Flagella number and position 2; unequal and lateral Class III Rhodophyceae Common name Red algae Major pigments Chlorophylls a and b, and phycoerythrin Stored food Floridean starch Cell wall Cellulose, pectin, and polysulphate esters Flagella number Absent Page 1 of 12

19 Class XI Chapter 3 Plant Kingdom Biology Question 2: When and where does reduction division take place in the life cycle of a liverwort, a moss, a fern, a gymnosperm and an angiosperm? Liverwort In liverworts, the main plant-body is haploid (gametophytic). It bears the male and female sex organs which produce gametes. These gametes fuse to form a zygote. The zygote develops on the gametophytic plant-body to form a sporophyte. The sporophyte is differentiated into the foot, seta, and capsule. Many haploid spores are produced as a result of the reduction division taking place inside the capsule. Moss In mosses, the primary protonema (developed in the first stage) develops into the secondary protonema. Both these stages are haploid or gametophytic. The secondary protonema bears the sex organs which produce gametes. These gametes fuse to form a zygote. The zygote develops into a sporophyte. Many spores are formed as a result of the reduction division taking place in the capsule of this sporophyte. Fern In ferns, the main plant-body is sporophytic. Its leaves are known as sporophylls and these bear the sporangia. Reduction division takes place in these sporangia, thereby producing many spores. Gymnosperm In gymnosperms, the main plant-body is sporophytic. They bear two types of leaves microsporophylls and megasporophylls. Reduction division takes place in the microsporangia present on the microsporophylls (producing pollen grains) and on the megasporangia present on the megasporophylls (producing megaspores). Angiosperm In angiosperms, the main plant-body is sporophytic and bears flowers. The male sex organ in the flower is the stamen, while the female sex organ is the pistil. Reduction division takes place in the anthers of the stamen (producing haploid pollen grains) and in the ovary of the pistil (producing eggs). Page 2 of 12

20 Class XI Chapter 3 Plant Kingdom Biology Question 3: Name three groups of plants that bear archegonia. Briefly describe the life cycle of any one of them. Archegonium is the female sex organ that produces the female gamete or egg. It is present in the life cycles of bryophytes, pteridophytes, and gymnosperms. Life cycle of a fern (Dryopteris) Dryopteris is a common fern with pinnately-compound leaves. The main plant-body is sporophytic. Many sporangia are borne on the lower surfaces of its mature leaves. Each sporangium has spore mother cells which undergo meiosis to produce haploid spores. On maturing, these spores dehisce and germinate to give rise to a heartshaped gametophyte called prothallus. The prothallus bears the male and female sex organs called antheridia and archegonia respectively. The antheridia produce sperms that swim in water to reach the archegonia. The egg is produced by the archegonia. As a result of fertilisation, a zygote is formed. The zygote forms an embryo, which in turn develops into a new sporophyte. The young plant comes out of the archegonium of the parent gametophyte. Page 3 of 12

21 Page 4 of 12 Class XI Chapter 3 Plant Kingdom Biology Question 4: Mention the ploidy of the following: protonemal cell of a moss; primary endosperm nucleus in dicot, leaf cell of a moss; prothallus cell of a fern; gemma cell in Marchantia; meristem cell of monocot, ovum of a liverwort, and zygote of a fern. (a) Protonemal cell of a moss Haploid (b) Primary endosperm nucleus in a dicot Triploid (c) Leaf cell of a moss Haploid (d) Prothallus of a fern Haploid (e) Gemma cell in Marchantia Haploid (f) Meristem cell of a monocot Diploid (g) Ovum of a liverwort Haploid (h) Zygote of a fern Diploid Question 5: Write a note on economic importance of algae and gymnosperms. Economic importance of algae Algae have diverse economic uses. They perform half of the total carbon dioxidefixation on earth by photosynthesis, acting as the primary producers in aquatic habitats. (a) Food source: Many species of marine algae such as Porphyra, Sargassum, and Laminaria are edible. Chlorella and Spirulina are rich in proteins. Thus, they are used as food supplements. (b) Commercial importance: Agar is used in the preparation of jellies and icecream. It is obtained from Gelidium and Gracilaria. Carrageenin is used as an emulsifier in chocolates, paints, and toothpastes. It is obtained from the red algae. (c) Medicines: Many red algae such as Corallina are used in treating worm infections. Economic importance of gymnosperms

22 Class XI Chapter 3 Plant Kingdom Biology (a) Construction purposes: Many conifers such as pine, cedar, etc., are sources of the soft wood used in construction and packing. (b) Medicinal uses: An anticancer drug Taxol is obtained from Taxus. Many species of Ephedra produce ephedrine, which can be used in the treatment of asthma and bronchitis. (c) Food source: The seeds of Pinus gerardiana (known as chilgoza) are edible. (d) Source of resins: Resins are used commercially for manufacturing sealing waxes and water-proof paints. A type of resin known as turpentine is obtained from various species of Pinus. Question 6: Both gymnosperms and angiosperms bear seeds, then why are they classified separately? Gymnosperms and angiosperms are seed-producing plants with diplontic life cycles. In gymnosperms, the sporophylls are aggregated to form compact cones. The microsporophylls are broad and are not distinguished into filaments and anthers. The megasporophylls are woody and lack the ovary, style, and stigma, because of which the ovules lie exposed. The female gametophyte consists of archegonia. The fertilisation process involves the fusion of a male gamete with the female gamete. Their endosperm is haploid. The produced seeds are naked as there is no fruit formation. Angiosperms are also known as flowering plants. They have sporophylls that aggregate to form flowers with the perianth. The microsporophylls consist of stamens containing pollen sacs. These sacs bear the male gametes called pollen grains. The megasporophylls are delicate and rolled, forming carpels that contain the ovary, style, and stigma. The ovules are present inside the ovary. The archegonium is replaced by an egg apparatus. Two male gametes enter the egg apparatus at the time of fertilisation. One male gamete fertilises the egg and the other fuses with the diploid secondary nucleus to form an endosperm. The resulting endosperm is thus Page 5 of 12

23 Class XI Chapter 3 Plant Kingdom Biology triploid. In addition, in angiosperms, the development of seeds takes place inside the fruits. Question 7: What is heterospory? Briefly comment on its significance. Give two examples. Heterospory is a phenomenon in which two kinds of spores are borne by the same plant. These spores differ in size. The smaller one is known as microspore and the larger one is known as megaspore. The microspore germinates to form the male gametophyte and the megaspore germinates to form the female gametophyte. The male gametophyte releases the male gametes and these reach the female gametophyte to fuse with the egg. The development of the zygote takes place inside the female gametophyte. This retention and germination of the megaspore within the megasporangium ensures proper development of the zygote. The zygote develops into the future sporophyte. The evolution of the seed habit is related to the retention of the megaspore. Heterospory is thus considered an important step in evolution as it is a precursor to the seed habit. Heterospory evolved first in pteridophytes such as Selaginella and Salvinia. Question 8: Explain briefly the following terms with suitable examples:- (i) protonema (ii) antheridium (iii) archegonium (iv) diplontic (v) sporophyll (vi) isogamy Page 6 of 12

24 Class XI Chapter 3 Plant Kingdom Biology (i) Protonema It is the first stage in the life cycle of a moss, developing directly from the spore. It consists of creeping, green, branched, and often filamentous structures. (ii) Antheridium It is the male sex organ present in bryophytes and pteridophytes and is surrounded by a jacket of sterile cells. It encloses the sperm mother cells, which give rise to the male gametes. (iii) Archegonium It is the female sex organ present in bryophytes, pteridophytes, and gymnosperms. In bryophytes and pteridophytes, it generally has a swollen venter and a tubular neck, and contains the female gamete called the egg. (iv) Diplontic It is the term used for the life cycles of seed-bearing plants (gymnosperms and angiosperms). In these plants, the diploid sporophyte is dominant, photosynthetic, and independent. The gametophyte is represented by a single-celled (or a few-celled) structure. (v) Sporophyll In pteridophytes, the sporophytic plant body bears sporangia. These sporangia are subtended by leaf-like appendages known as sporophylls. In gymnosperms, microsporophylls and megasporophylls are found. These bear microspores and megaspores respectively. (vi) Isogamy It is a type of sexual reproduction involving the fusion of morphologically-similar gametes. This means that the gametes are of the same size, but perform different functions. This type of reproduction is commonly observed in Spirogyra. Question 9: Differentiate between the following:- (i)red algae and brown algae (ii) liverworts and moss (iii) homosporous and heterosporous pteridophyte (iv) syngamy and triple fusion Page 7 of 12

25 Class XI Chapter 3 Plant Kingdom Biology (i) Red algae and brown algae Red algae Brown algae 1. Red algae are grouped under the class Rhodophyceae. 1. Brown algae are grouped under the class Phaeophyceae. 2. They contain floridean starch as stored food. 2. They contain mannitol or laminarin as stored food. They contain the photosynthetic They contain the photosynthetic 3. pigments chlorophylls a and d, 3. pigments chlorophylls a and c, and phycoerythrin. and fucoxanthin. 4. Their cell walls are composed of cellulose, pectin, and phycocolloids. 4. Their cell walls are composed of cellulose and algin. 5. Flagella are absent 5. Two flagella are present (ii) Liverworts and moss Liverworts Moss 1. They have unicellular rhizoids. 1. They have multicellular rhizoids. 2. Scales are present very often 2. Scales are absent 3. They are generally thalloid, with dichotomous branching. 3. They are foliage, with lateral branching. 4. Gemma cups are present 4. Gemma cups are absent Page 8 of 12

26 Class XI Chapter 3 Plant Kingdom Biology 5. Sporophyte has very little photosynthetic tissue 5. Sporophyte has abundant photosynthetic tissue (iii) Homosporous and heterosporous pteridophyte Homosporous pteridophytes Heterosporous pteridophytes 1. They bear spores that are of the same type. 1. They bear two kinds of spores microspores and megaspores. 2. They produce bisexual gametophytes. 2. They produce unisexual gametophytes. (iv) Syngamy and triple fusion Syngamy Triple fusion 1. It is the process of fusion of the male gamete with the egg in an angiosperm. 1. It is the process of fusion of the male gamete with the diploid secondary nucleus in an angiosperm. 2. A diploid zygote is formed as a result of syngamy. 2. A triploid primary endosperm is formed as a result of triple fusion. Page 9 of 12

27 Class XI Chapter 3 Plant Kingdom Biology Question 10: How would you distinguish monocots from dicots? Monocots and dicots can be differentiated through their morphological and anatomical characteristics. Characteristic Monocot Dicot Morphology Roots Fibrous roots Tap roots Venation Generally venation parallel Generally venation reticulate Flowers Trimerous flowers Pentamerous flowers Cotyledons in seeds One Two Anatomy No. of vascular bundles in stem Numerous Generally 2 6 Cambium Absent Present Leaves Isobilateral Dorsiventral Page 10 of 12

28 Page 11 of 12 Class XI Chapter 3 Plant Kingdom Biology Question 11: Match the followings (column I with column II) Column I Column II (a) Chlamydomonas (i) Moss (b) Cycas (ii) Pteridophyte (c) Selaginella (iii) Algae (d) Sphagnum (iv) Gymnosperm Column I Column II (a) Chlamydomonas (iii) Algae (b) Cycas (iv) Gymnosperm (c) Selaginella (ii) Pteridophyte (d) Sphagnum (i) Moss Question 12: Describe the important characteristics of gymnosperms. Important features of gymnosperms: 1. The term gymnosperm refers to plants with naked seeds (gymnos naked, sperma seeds), i.e., the seeds of these plants are not enclosed in fruits. 2. The plant-body ranges from medium to tall trees and shrubs. The giant redwood tree Sequoia is one of the tallest trees in the world. 3. The root system consists of tap roots. The coralloid roots present in Cycas are associated with nitrogen-fixing cyanobacteria.

29 Class XI Chapter 3 Plant Kingdom Biology 4. The stem can be branched (as in Pinus and Cedrus) or un-branched (as in Cycas). 5. The leaves can be simple (as in Pinus)or compound (pinnate in Cycas). The leaves are needle-like, with a thick cuticle and sunken stomata. These help in preventing water loss. 6. Gymnosperms are heterosporous. They bear two kinds of spores microspores and megaspores. 7. Flowers are absent. The microsporophylls and megasporophylls are arranged to form compact male and female cones. 8. Pollination occurs mostly through wind and pollen grains reach the pollen chamber of the ovule through the micropyle. 9. The male and female gametophytes are dependent on the sporophyte. 10. The seeds contain haploid endosperms and remain uncovered. Page 12 of 12

30 Class XI Chapter 4 Animal Kingdom Biology Question 1: What are the difficulties that you would face in classification of animals, if common fundamental features are not taken into account? For the classification of living organisms, common fundamental characteristics are considered. If we consider specific characteristics, then each organism will be placed in a separate group and the entire objective of classification would not be achieved. Classification of animals is also important in comparing different organisms and judging their individual evolutionary significance. If only a single characteristic is considered, then this objective would not be achieved. Question 2: If you are given a specimen, what are the steps that you would follow to classify it? There is a certain common fundamental feature that helps in classification of living organisms. The features that can be used in classification are as follows. Page 1 of 8

31 Class XI Chapter 4- Animal Kingdom Biology On the basis of above features, we can easily classify a specimen into its respective category. Question 3: How useful is the study of the nature of body cavity and coelom in the classification of animals? Coelom is a fluid filled space between the body wall and digestive tract. The presence or absence of body cavity or coelom plays a very important role in the classification of animals. Animals that possess a fluid filled cavity between body wall and digestive tract are known as coelomates. Annelids, mollusks, arthropods, echinodermates, and chordates are examples of coelomates. On the other hand, the animals in which the body cavity is not lined by mesoderm are known as pseudocoelomates. In such animals, mesoderm is scattered in between ectoderm and endoderm. Aschelminthes is an example of pseudocoelomates. In certain animals, the body cavity is absent. They are known as acoelomates. An example of acoelomates is platyhelminthes. Question 4: Distinguish between intracellular and extracellular digestion? Page 2 of 8

32 Class XI Chapter 4- Animal Kingdom Biology Intracellular digestion Extracellular digestion 1. The digestion of food occurs within the cell. 1. The digestion occurs in the cavity of alimentary canal. Digestive enzymes are secreted by Digestive enzymes are secreted by 2. the surrounding cytoplasm into the 2. special cells into the cavity of food vacuole. alimentary canal. 3. Digestive products are diffused into the cytoplasm. 3. Digestive products diffuse across the intestinal wall into various parts of the body. 4. It is a less efficient method. 4. It is a more efficient method of digestion. 5. It occurs in unicellular organisms. 5. It occurs in multicellular organisms. Question 5: What is the difference between direct and indirect development? Page 3 of 8

33 Class XI Chapter 4- Animal Kingdom Biology Direct development Indirect development 1. It is a type of development in which an embryo develops into a mature individual without involving a larval stage. 1. It is a type of development that involves a sexually-immature larval stage, having different food requirements than adults. 2. Metamorphosis is absent. 2. Metamorphosis involving development of larva to a sexuallymature adult is present. 3. It occurs in fishes, reptiles, birds, and mammals. 3. It occurs in most of the invertebrates and amphibians. Question 6: What are the peculiar features that you find in parasitic platyhelminthes? Taenia (Tapeworm) and Fasciola (liver fluke) are examples of parasitic platyhelminthes. Peculiar features in parasitic platyhelminthes are as follows. 1. They have dorsiventrally flattened body and bear hooks and suckers to get attached inside the body of the host. 2. Their body is covered with thick tegument, which protects them from the action of digestive juices of the host. 3. The tegument also helps in absorbing nutrients from the host s body. Question 7: What are the reasons that you can think of for the arthropods to constitute the largest group of the animal kingdom? Page 4 of 8

34 Class XI Chapter 4- Animal Kingdom Biology The phylum, Arthropoda, consists of more than two-thirds of the animal species on earth. The reasons for the success of arthropods are as follows. i. Jointed legs that allow more mobility on land ii. Hard exoskeleton made of chitin that protects the body iii. The hard exoskeleton also reduces water loss from the body of arthropods making them more adapted to terrestrial conditions. Question 8: Water vascular system is the characteristic of which group of the following: (a) Porifera (b) Ctenophora (c) Echinodermata (d) Chordata Water vascular system is a characteristic feature of the phylum, Echinodermata. It consists of an array of radiating channels, tube feet, and madreporite. The water vascular system helps in locomotion, food capturing, and respiration. Question 9: All vertebrates are chordates but all chordates are not vertebrates. Justify the statement. The characteristic features of the phylum, Chordata, include the presence of a notochord and paired pharyngeal gill slits. In sub-phylum Vertebrata, the notochord present in embryos gets replaced by a cartilaginous or bony vertebral column in adults. Thus, it can be said that all vertebrates are chordates but all chordates are not vertebrates. Question 10: How important is the presence of air bladder in Pisces? Gas bladder or air bladder is a gas filled sac present in fishes. It helps in maintaining buoyancy. Thus, it helps fishes to ascend or descend and stay in the water current. Page 5 of 8

35 Class XI Chapter 4- Animal Kingdom Biology Question 11: What are the modifications that are observed in birds that help them fly? Birds have undergone many structural adaptations to suit their aerial life. Some of these adaptations are as follows. (i) Streamlined body for rapid and smooth movement (ii) Covering of feathers for insulation (iii) Forelimbs modified into wings and hind limbs used for walking, perching, and swimming (iv) Presence of pneumatic bones to reduce weight (v) Presence of additional air sacs to supplement respiration Question 12: Could the number of eggs or young ones produced by an oviparous and viviparous mother be equal? Why? The numbers of eggs produced by an oviparous mother will be more than the young ones produced by a viviparous mother. This is because in oviparous animals, the development of young ones takes place outside the mother s body. Their eggs are more prone to environmental conditions and predators. Therefore, to overcome the loss, more eggs are produced by mothers so that even under harsh environmental conditions, some eggs might be able to survive and produce young ones. On the other hand, in viviparous organisms, the development of young ones takes place in safe conditions inside the body of the mother. They are less exposed to environmental conditions and predators. Therefore, there are more chances of their survival and hence, less number of young ones is produced compared to the number of eggs. Question 13: Page 6 of 8

36 Class XI Chapter 4- Animal Kingdom Biology Segmentation in the body is first observed in which of the following: (a) Platyhelminthes (b) Aschelminthes (c) Annelida (d) Arthropoda The body segmentation first appeared in the phylum, Annelida (annulus meaning little ring). Question 14: Match the following: (a) Operculum (b) Parapodia (c) Scales (d) Comb plates (e) Radula (f) Hairs (g) Choanocytes (h) Gill slits (i) Ctenophora (ii) Mollusca (iii) Porifera (iv) Reptilia (v) Annelida (vi) Cyclostomata and Chondrichthyes (vii) Mammalia (viii) Osteichthyes Column I (a) Osteichthyes Parapodia Page 7 of 8

37 Class XI Chapter 4- Animal Kingdom Biology (c) Scales (iv) Reptilia (d) Comb plates (i) Ctenophora (e) Radula (ii) Mollusca (f) Hairs (vii) Mammalia (g) Choanocytes (iii) Porifera (h) Gill slits (vi) Cyclostomata and Chondrichthyes Question 15: Prepare a list of some animals that are found parasitic on human beings. S. No. Name of organism Phylum 1 Taenia solium Platyhelminthes 2 Fasciola hepatica Platyhelminthes 3 Ascaris lumbricoides Aschelminthes 4 Wuchereria bancrofti Aschelminthes 5 Ancyclostoma Aschelminthes Page 8 of 8

38 Class XI Chapter 5 Morphology of Flowering Plants Biology Question 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 Primarily, there are two types of root systems found in plants, namely the tap root system and fibrous root system. The main function of the roots is to absorb water and minerals from the soil. However, roots are also modified to perform various other functions. The roots of some plants act as storage sites for food, some provide support to massive plant structures, while others absorb oxygen from the atmosphere. Roots and its modifications in various plants: (a) Banyan tree The banyan tree (Ficus benghalensis) has massive pillar-like adventitious roots arising from the aerial part of the stem. These roots grow towards the ground and provide support to the tree. Such roots are called prop roots. (b) Turnip The roots of turnip (Brassica rape) help in the storage of food. Similar food-storing roots are found in radishes, carrots, and sweet potatoes. (c) Mangrove tree The roots of mangrove plants grow vertically upwards from the soil for the absorption of oxygen from the atmosphere as the soil is poorly aerated. These types of roots are called pneumatophores. Question 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 Page 1 of 14

39 Class XI Chapter 5 Morphology of Flowering Plants Biology (i) Various parts of plants are modified into underground structures to perform various functions such as stems, leaves, and even fruits. The stems in ginger and banana are underground and swollen due to storage of food. They are called rhizomes. Similarly, corm is an underground stem in Colocasia and Zamin-khand. The tips of the underground stem in potato become swollen due to the accumulation of food and forms tuber. Tubers bear eyes, which are subtended by a leaf scar. Basal leaves in onions become fleshy because of the accumulation of food. In peanuts, the flower after fertilization gets pushed inside the soil by growing a flower stalk. The formation of fruits and seeds takes place inside the soil. (ii) During the flowering season, the apical meristem gives rise to the floral meristem. The axis of the stem gets condensed, while the internodes lie near each other. Instead of leaves, various floral appendages arise from the node. Therefore, it can be said that the flower is a modified shoot. Question 3: How is pinnately compound leaf different from palmately compound leaf? vpinnately compound leaf Palmately compound leaf The leaflets are attached to the common axis, called rachis. The leaflets are attached at a common point on the leaf stalk. Examples include neem and Cassia fistula( also called golden shower plant) Examples include silk cotton (Bombax ) and Cannabis. Question 4: Explain with suitable examples the different types of phyllotaxy? Page 2 of 14

40 Class XI Chapter 5 Morphology of Flowering Plants Biology Phyllotaxy refers to the pattern or arrangement of leaves on the stem or branch of a plant. It is of three types, alternate, opposite, and whorled phyllotaxy. In alternate phyllotaxy, a single leaf arises from the node of a branch. This type of phyllotaxy is observed in the sunflower, mustard, and peepal. Plants with opposite phyllotaxy have two leaves arising from the node in opposite directions. It is found in guava and jamun plants. Plants with whorled phyllotaxy have three or more leaves arising from the node. It is found in Alstonia. Question 5: Define the following terms: (a) Aestivation (b) Placentation (c) Actinomorphic (d) Zygomorphic (e) Superior ovary (f) Perigynous flower (g) Epipetalous Stamen (a) Aestivation The term aestivation refers to the mode in which sepals or petals are arranged in a floral bud with respect to other floral members. There are four types of aestivation in plants i.e., valvate, twisted, imbricate, and vexillary. (b) Placentation The term placentation refers to the arrangement of ovules within the ovary of a flower. It is primarily of five types, namely marginal, basal, parietal, axile, and free central. (c) Actinomorphic Actinomorphic flowers can be divided into two radial halves by any radial plane passing through its centre. Examples of these flowers include chilly and mustard. Page 3 of 14

41 Class XI Chapter 5 Morphology of Flowering Plants Biology (d)zygomorphic Zygomorphic flowers are those flowers which can be divided into two similar halves by a single vertical plane. Examples of these flowers include pea and beans. (e) Superior ovary Superior ovary flowers are those flowers in which the gynoecium is present at the highest position, while other floral parts are arranged below it. A flower with this arrangement is described as hypogynous. Examples include brinjal and mustard. (f) Perigynous flower In perigynous flowers, the gynoecium is present in the centre and the rest of the floral parts are arranged at the rim of the thalamus at the same level. Examples include plum and rose. (g) Epipetalous Stamen Epipetalous stamens are stamens attached to the petals. They are found in brinjal. Question 6: Differentiate between (a) Racemose and cymose inflorescence (b) Fibrous roots and adventitious roots (c) Apocarpous and syncarpous ovary Racemose inflorescence Cymose inflorescence Page 4 of 14

42 Class XI Chapter 5 Morphology of Flowering Plants Biology 1) Younger flowers are present at the tip while older flowers are arranged at the base of this inflorescence. Such an arrangement is called acropetal succession. 2) The main axis in racemose inflorescence continues to grow and produce flowers laterally. 1) Younger flowers are present at the base of the inflorescence, while older flowers are present at the top. Such an arrangement is called basipetal succession. 2) The main axis in cymose inflorescence has limited growth, which later terminates into a flower. Fibrous root Adventitious root 1) In monocots, the primary root which develops from the radicle of the seed is short-lived and is replaced by a large number of roots arising from the base of the stem. 2) It is found in wheat and other cereals. 1) These roots arise from any part of the plant other than the radicle of seeds. 2) It is found in banyan, Monstera, and other plants. Apocarpous ovary Syncarpous ovary 1) The flowers with apocarpus ovary have more than one carpel. These carpels are free. 2) It is found in lotus and rose flowers. 1) The flowers with syncarpous ovary have more than one carpel. However, these carpels are fused. 2) It is found in the flowers of tomato and mustard. Page 5 of 14

43 Class XI Chapter 5 Morphology of Flowering Plants Biology Question 7: Draw the labelled diagram of the following: (i) Gram seed (ii) V.S. of maize seed (i) (ii) Question 8: Describe modifications of stem with suitable examples Stems of various plants have undergone modifications to perform different functions. Underground stems or storage stems: Examples: Rhizomes, Corms, tubers In ginger and banana, the underground stem is called a rhizome. The underground stem in Colocasia (arvi) is known as corm. Rhizomes and corms are underground stems, modified for the storage of food. Also, these stems help in vegetative Page 6 of 14

44 Class XI Chapter 5 Morphology of Flowering Plants Biology reproduction of these plants. The tips of the underground stem in potato plants become swollen due to the accumulation of food. The potato is a tuber that helps in the storage of food and bears eyes on it. Subtended by a leaf scar, these eyes bear buds that give rise to new plants. Supportive stems Example: tendril The stem in some weak plants bear thin, slender, and spirally-coiled structures called tendrils that help the plant get attached to nearby structures for support. Tendrils are found in cucumbers, melons, and other members of the family Cucurbitaceae. Protective stems Example: Thorns The stem in bougainvillea and citrus plants (like lemon and orange) bear sharp, pointed structures called thorns, which provide protection to the plant from herbivores. Photosynthetic stems Example: Opuntia The stem in the Opuntia is green. It carries out the process of photosynthesis in the absence of leaves. Others stem modifications In some plants, underground stems such as grasses spread in the soil and help in perenation. These stems are called runners. The short lateral stem called the offset in some aquatic plants (such as Eichhornia) bears leaves and tufts of roots at the node and gives rise to new plants. Question 9: Take one flower each of families Fabaceae and Solanaceae and write its semitechnical description. Also draw their floral diagrams after studying them. Page 7 of 14

45 Class XI Chapter 5 Morphology of Flowering Plants Biology (1) Family Fabaceae/Papilionaceae (pea plant) Fabaceae/Papilionaceae is a sub-family of the Leguminoseae family. Vegetative features: Habit: Pinnately compound, alternately arranged with leaf tendrils with the pulvinus present at the leaf base along folacious stipules. Root: Tap root system with root nodules. Floral features: Inflorescence: Racemose, generally axial than terminal Flower: Zygomorphic and bisexual flowers are found Calyx: It contains five sepals which are gamosepalous while aestivation is imbricate. Corolla: It contains five petals (polypetalous) with vexillary aestivation. Androecium: It consists of ten anthers that are diadelphous with dithecous anthers. Gynoecium: Monocarpellary superior ovary which is unilocular with marginal placentation. Fruit: Legume pod with non-endospermic seeds Floral formula: Economic importance: Peas are used as vegetables for making various culinary preparations. (2) Flowers of Solanum nigrum Family Solanaceae Vegetative features: Habit: Erect, herbaceous plant Page 8 of 14

46 Class XI Chapter 5 Morphology of Flowering Plants Biology Leaves: Simple, exstipulate leaves with reticulate venation Stem: Erect stem with numerous branches. Floral features: Inflorescence: Solitary and axillary Flowers: Actinomorphic, bisexual flowers Calyx: Calyx is composed of five sepals that are united and persistent. Aestivation is valvate. Corolla: Corolla consists of five united petals with valvate aestivation. Androecium: It consists of five epipetalous stamens. Gynoecium: It consists of bicarpellary syncarpous superior ovary with axile placentation. Fruits: Berry Seeds: Numerous, endospermous Floral formula: Economic importance: Used for medicinal purposes. Question 10: Describe the various types of placentations found in flowering plants. Page 9 of 14

47 Class XI Chapter 5 Morphology of Flowering Plants Biology Placentation refers to the arrangement of ovules inside the ovary. It is of five basic types. (A) Marginal placentation: The ovary in which the placenta forms a ridge along the ventral suture of the ovary and the ovules develop on two separate rows is known to have marginal placentation. This type of placentation is found in peas. (B) Parietal placentation When the ovules develop on the inner walls of the ovary, the ovary is said to have parietal placentation. C) Axile placentation In axile placentation, the placenta is axial and ovules are attached to it. Examples include China rose, lemon, and tomato. (D) Basal placentation Page 10 of 14

48 Class XI Chapter 5 Morphology of Flowering Plants Biology The ovary in which the placenta develops from its base and a single ovule is found attached to the base is said to have basal placentation. It is found in marigold and sunflower. (E) Free central placentation In free central placentation, the ovules develop on the central axis while the septa are absent. This type of placentation is found in Dianthus and primrose. Question 11: What is a flower? Describe the parts of a typical angiosperm flower? A flower can be defined as the reproductive unit of any flowering plant (angiosperms). Flowers carry out sexual reproduction in angiosperms. A typical flower is a modified stem with a condensed axis. A flower has four different parts i.e., the calyx, corolla, androecium, and gynoecium. Androecium and gynoecium represent the male and female reproductive organs of a flower (respectively). Bisexual flowers are those which contain both androecium and gynoecium, while unisexual flowers contain either gynoecium or androecium. The corolla and the calyx are generally distinct, but may sometimes be fused (called perianth). A flower that contains all four floral parts is called a complete flower. Parts of flowers Page 11 of 14

49 Class XI Chapter 5 Morphology of Flowering Plants Biology (A) The calyx forms the outermost whorl of a flower, which contains sepals. They are green, leaf like structures that cover and protect the flowers during the bud stage. When the sepals of a flower are free, they are called polysepalous, while fused sepals of a flower are called gamosepalous. (B) The corolla of a flower is a layer that lies inside the calyx. It contains beautifully coloured petals, which help in attracting insects for pollination. When the petals are free, they are called polypetalous, while fused petals are called gamopetalous. (C) The androecium or the stamen is the male reproductive part of a flower. It consists of two parts, the filament and the bilobed anther. The bilobed anther is the site for meiosis and the generation of pollen grains. (D) Gynoecium represents the female reproductive part of a flower. It consists of an ovary. The ovary is connected by a long tube (called style) to the stigma. The ovary bears numerous ovules attached to the placenta. Question 12: How do the various leaf modifications help plants? The main function of the leaves is to carry out the process of photosynthesis. However, in a few plants, leaves are modified to perform different functions. (a) Tendrils: The leaves of a pea plant are modified into tendrils that help the plant in climbing. (b) Spines: The leaves in cactus are modified into sharp spines that act as an organ of defense. (c) Phyllode: The leaves of some Australian acacia are short-lived and soon replaced by flattened, green structures called phyllodes that arise from the petiole of the leaves. The petioles in these plants synthesize food. (d) Pitcher: The leaves of the pitcher plant are modified into pitcher-like structures, which contain digestive juices and help in trapping and digesting insects. Page 12 of 14

50 Class XI Chapter 5 Morphology of Flowering Plants Biology Question 13: Define the term inflorescence. Explain the basis for the different types of inflorescence in flowering plants. Inflorescence is the manner in which the flowers are arranged on the flowering axis. During the flowering season, the vegetative apex of the stem gets converted into a floral meristem. Based on whether the floral axis continues to grow or end in a flower, inflorescence is classified into racemose and cymose. In racemose inflorescence, the floral axis continues to grow and produces flowers laterally. On the other hand, in cymose inflorescence, the main axis terminates into a flower. Hence, it is limited in growth. Question 14: Write the floral formula of an actinomorphic bisexual, hypogynous flower with five united sepals, five free petals. Five free stamens and two united carpals with superior ovary and axile placentation. The floral formula of the described flower is represented as: Actinomorphic flowers are represented by the symbol. A bisexual flower is indicated by. The calyx contains five united sepals which can be represented as K (5). The corolla consists of five free petals and it represented as C 5. The androecium consists of five free stamens and is represented by A 5. The gynoecium consists of a superior ovary with two united carpels and axile placentations, which can be represented as Question 15: Page 13 of 14

51 Class XI Chapter 5 Morphology of Flowering Plants Biology Describe the arrangement of floral members in relation to their insertion on thalamus? Based on the position of the calyx, corolla, and androecium (with respect to the ovary on the thalamus), the flowers are described as hypognous, perigynous, and epigynous. In hypogynous flowers, the ovary occupies the highest position on the thalamus while other floral parts are situated below it. In such flowers, the ovary is superior e.g., China rose, mustard etc. In perigynous flowers, the ovary is situated at the centre and other floral parts are arranged on the rim of the thalamus. The ovary here is said to be half inferior e.g., plum, rose, peach In epigynous flowers, the thalamus grows around the ovary fusing with its wall. The other floral parts are present above the ovary. Hence, the ovary is said to be inferior e.g., flowers of guava and cucumber. Page 14 of 14

52 Class XI Chapter 6 Anatomy of Flowering Plants Biology Question 1: State the location and function of different types of meristem. Meristems are specialised regions of plant growth. The meristems mark the regions where active cell division and rapid division of cells take place. Meristems are of three types depending on their location. Apical meristem It is present at the root apex and the shoot apex. The shoot apical meristem is present at the tip of the shoots and its active division results in the elongation of the stem and formation of new leaves. The root apical meristem helps in root elongation. Intercalary meristem It is present between the masses of mature tissues present at the bases of the leaves of grasses. It helps in the regeneration of grasses after they have been grazed by herbivores. Since the intercalary meristem and the apical meristem appear early in a plant s life, they constitute the primary meristem. Lateral meristem It appears in the mature tissues of roots and shoots. It is called the secondary meristem as it appears later in a plant s life. It helps in adding secondary tissues to the plant body and in increasing the girth of plants. Examples include fascicular cambium, interfascicular cambium, and cork cambium Question 2: Cork cambium forms tissues that form the cork. Do you agree with this statement? Explain. When secondary growth occurs in the dicot stem and root, the epidermal layer gets broken. There is a need to replace the outer epidermal cells for providing protection to the stem and root from infections. Therefore, the cork cambium develops from the cortical region. It is also known as phellogen and is composed of thin-walled rectangular cells. It cuts off cells toward both sides. The cells on the outer side get

53 Class XI Chapter 6 Anatomy of Flowering Plants Biology differentiated into the cork or phellem, while the cells on the inside give rise to the secondary cortex or phelloderm. The cork is impervious to water, but allows gaseous exchange through the lenticels. Phellogen, phellem, and phelloderm together constitute the periderm. Question 3: Explain the process of secondary growth in stems of woody angiosperm with help of schematic diagrams. What is the significance? In woody dicots, the strip of cambium present between the primary xylem and phloem is called the interfascicular cambium. The interfascicular cambium is formed from the cells of the medullary rays adjoining the interfascicular cambium. This results in the formation of a continuous cambium ring. The cambium cuts off new cells toward its either sides. The cells present toward the outside differentiate into the secondary phloem, while the cells cut off toward the pith give rise to the secondary xylem. The amount of the secondary xylem produced is more than that of the secondary phloem. Page 2 of 11

54 Class XI Chapter 6 Anatomy of Flowering Plants Biology The secondary growth in plants increases the girth of plants, increases the amount of water and nutrients to support the growing number of leaves, and also provides support to plants. Question 4: Draw illustrations to bring out anatomical difference between (a) Monocot root and dicot root (b) Monocot stem and dicot stem (a)monocot root and dicot root Page 3 of 11

55 Page 4 of 11 Class XI Chapter 6 Anatomy of Flowering Plants Biology (b)monocot stem and dicot stem

56 Class XI Chapter 6 Anatomy of Flowering Plants Biology Question 5: Cut a transverse section of young stem of a plant from your school garden and observe it under the microscope. How would you ascertain whether it is a monocot stem or dicot stem? Give reasons. The dicot stem is characterised by the presence of conjoint, collateral, and open vascular bundles, with a strip of cambium between the xylem and phloem. The vascular bundles are arranged in the form of a ring, around the centrally-located pith. The ground tissue is differentiated into the collenchyma, parenchyma, endodermis, pericycle, and pith. Medullary rays are present between the vascular bundles. Page 5 of 11

57 Class XI Chapter 6 Anatomy of Flowering Plants Biology The monocot stem is characterised by conjoint, collateral, and closed vascular bundles, scattered in the ground tissue containing the parenchyma. Each vascular bundle is surrounded by sclerenchymatous bundle-sheath cells. Phloem parenchyma is absent and water-containing cavities are present. Page 6 of 11

58 Page 7 of 11 Class XI Chapter 6 Anatomy of Flowering Plants Biology Question 6: The transverse section of a plant material shows the following anatomical features, (a) the vascular bundles are conjoint, scattered and surrounded by sclerenchymatous bundle sheaths (b) phloem parenchyma is absent. What will you identify it as? The monocot stem is characterised by conjoint, collateral, and closed vascular bundles, scattered in the ground tissue containing the parenchyma. Each vascular bundle is surrounded by sclerenchymatous bundle-sheath cells. Phloem parenchyma and medullary rays are absent in monocot stems. Question 7: Why are xylem and phloem called complex tissues? Xylem and phloem are known as complex tissues as they are made up of more than one type of cells. These cells work in a coordinated manner, as a unit, to perform the various functions of the xylem and phloem. Xylem helps in conducting water and minerals. It also provides mechanical support to plants. It is made up of the following components: Tracheids (xylem vessels and xylem tracheids) Xylem parenchyma Xylem fibres Tracheids are elongated, thick-walled dead cells with tapering ends. Vessels are long, tubular, and cylindrical structures formed from the vessel members, with each having lignified walls and large central cavities. Both tracheids and vessels lack protoplasm. Xylem fibres consist of thick walls with an almost insignificant lumen. They help in providing mechanical support to the plant. Xylem parenchyma is made up of thin-walled parenchymatous cells that help in the storage of food materials and in the radial conduction of water. Phloem helps in conducting food materials. It is composed of:

59 Class XI Chapter 6 Anatomy of Flowering Plants Biology Sieve tube elements Companion cells Phloem parenchyma Phloem fibres Sieve tube elements are tube-like elongated structures associated with companion cells. The end walls of sieve tube elements are perforated to form the sieve plate. Sieve tube elements are living cells containing cytoplasm and nucleus. Companion cells are parenchymatous in nature. They help in maintaining the pressure gradient in the sieve tube elements. Phloem parenchyma helps in the storage of food and is made up of long tapering cells, with a dense cytoplasm. Phloem fibres are made up of elongated sclerenchymatous cells with thick cell walls. Question 8: What is stomatal apparatus? Explain the structure of stomata with a labelled diagram. Stomata are small pores present in the epidermis of leaves. They regulate the process of transpiration and gaseous exchange. The stomatal pore is enclosed between two bean-shaped guard cells. The inner walls of guard cells are thick, while the outer walls are thin. The guard cells are surrounded by subsidiary cells. These are the specialised epidermal cells present around the guard cells. The pores, the guard cells, and the subsidiary cells together constitute the stomatal apparatus. Page 8 of 11

60 Page 9 of 11 Class XI Chapter 6 Anatomy of Flowering Plants Biology Question 9: Name the three basic tissue systems in the flowering plants. Give the tissue names under each system. No. Tissue system Tissues present 1. Epidermal tissue system Epidermis, trichomes, hairs, stomata 2. Ground tissue system Parenchyma, collenchyma, sclerenchyma, mesophyll 3. Vascular tissue system Xylem, phloem, cambium Question 10: How is the study of plant anatomy useful to us? The study of plant anatomy helps us to understand the structural adaptations of plants with respect to diverse environmental conditions. It also helps us to distinguish between monocots, dicots, and gymnosperms. Such a study is linked to plant physiology. Hence, it helps in the improvement of food crops. The study of plant-structure allows us to predict the strength of wood. This is useful in utilising it to its potential. The study of various plant fibres such as jute, flax, etc., helps in their commercial exploitation. Question 11: What is periderm? How does periderm formation take place in dicot stem? Periderm is composed of the phellogen, phellem, and phelloderm. During secondary growth, the outer epidermal layer and the cortical layer are broken because of the cambium. To replace them, the cells of the cortex turn meristematic,

61 Class XI Chapter 6 Anatomy of Flowering Plants Biology giving rise to cork cambium or phellogen. It is composed of thin-walled, narrow and rectangular cells. Phellogen cuts off cells on its either side. The cells cut off toward the outside give rise to the phellem or cork. The suberin deposits in its cell wall make it impervious to water. The inner cells give rise to the secondary cortex or phelloderm. The secondary cortex is parenchymatous. Question 12: Describe the internal structure of a dorsiventral leaf with the help of labelled diagrams. Dorsiventral leaves are found in dicots. The vertical section of a dorsiventral leaf contains three distinct parts. [1] Epidermis: Epidermis is present on both the upper surface (adaxial epidermis) and the lower surface (abaxial epidermis). The epidermis on the outside is covered with a thick cuticle. Abaxial epidermis bears more stomata than the adaxial epidermis. [2] Mesophyll: Mesophyll is a tissue of the leaf present between the adaxial and abaxial epidermises. It is differentiated into the palisade parenchyma (composed of tall, compactly-placed cells) and the spongy parenchyma (comprising oval or round, loosely-arranged cells with inter cellular spaces). Mesophyll contains the chloroplasts which perform the function of photosynthesis. [3] Vascular system: The vascular bundles present in leaves are conjoint and closed. They are surrounded by thick layers of bundle-sheath cells. Page 10 of 11

62 Class XI Chapter 6 Anatomy of Flowering Plants Biology Page 11 of 11

63 Class XI Chapter 7 Structural Organisation in Animals Biology Question 1: 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 the cockroach? (iv) How many segments are present in the abdomen of cockroach? (v) Where do you find malphigian tubules? (i) The common name of Periplaneta americana is the American cockroach. (ii) Four pairs of spermathecae are present in earthworms. They are located between sixth and the ninth segments. They help in receiving and storing the spermatozoa during copulation. (iii) In a cockroach, the pair of ovaries is located between 12 th and 13 th abdominal segments. (iv) In both sexes, the abdomen of a cockroach consists of ten segments. (v) Malphigian tubules are main excretory organs of cockroaches. They form a part of the alimentary canal. Question 2: the following: (i) What is the function of nephridia? (ii) How many types of nephridia are found in earthworm based on their location? (i) Nephridia are segmentally arranged excretory organs present in earthworms. (ii) On the basis of their location, three types of nephridia are found in earthworms. They are: (a) Septal nephridia: These are present on both sides of the inter-segmental septa behind the 15 th segment. They open into the intestines. (b) Integumentary nephridia: These lie attached to the body wall from the third segment to the last segment, which opens on the body surface. Page 1 of 11

64 Class XI Chapter 7 Structural Organisation in Animals Biology (c) Pharyngeal nephridia: These are present as three paired tufts in fourth, fifth, and sixth segments. Question 3: Draw a labelled diagram of the reproductive organs of an earthworm. Question 4: Draw a labelled diagram of alimentary canal of a cockroach. Page 2 of 11

65 Class XI Chapter 7 Structural Organisation in Animals Biology Question 5: Distinguish between the following (a) Prostomium and peristomium (b) Septal nephridium and pharyngeal nephridium (a) Prostomium Peristomium Prostomium is a small fleshy lobe, which overhangs the mouth of an earthworm. It helps the organism push into the soil and is sensory in function. The first body segment in the earthworm is called the peristomium. It surrounds the mouth opening. (b) Septal nephridium Pharyngeal nephridium Page 3 of 11

66 Class XI Chapter 7 Structural Organisation in Animals Biology They are present on both sides of intersegmental septa behind the 15 th segment. They open into the intestines. They are present as three paired tufts in the fourth, fifth, and sixth segments. Question 6: What are the cellular components of blood? Components of blood include erythrocytes (RBCs), leucocytes (WBCs), and thrombocytes (platelets). These components form 45% of blood. They are suspended in the remaining fluid portion, called plasma. Mammalian erythrocytes are biconcave, coloured cells devoid of a nucleus. They help in transporting respiratory gases. Leucocytes or white blood cells are nucleated cells. They can be divided into two types, granulocytes (neutrophils, eosinophils, and basophils) and agranulocytes (lymphocytes and monocytes). They help fight against various disease-causing germs entering the body. Thrombocytes are cell fragments produced from megarkaryocytes of the bone. They play a major role during blood coagulation. Question 7: What are the following and where do you find them in animal body (a) Chondriocytes (b) Axons (c) Ciliated epithelium Chondriocytes: They are cells of cartilages, and are present in small cavities within the matrix secreted by them. Axons: Page 4 of 11

67 Class XI Chapter 7 Structural Organisation in Animals Biology They are long, slender projections of neurons that help in carrying nerve impulses from the neuron body. Axons aggregate in bundles which make up the nerves. Ciliated epithelium: It consists of simple columnar or cuboidal epithelium with cilia on their free surfaces. It is present on the inner surface of the oviducts and bronchioles. It helps in the movement of eggs or mucus in specific directions. Question 8: Describe various types of epithelial tissues with the help of labelled diagrams. Epithelial tissue lines the surface of a body and forms a protective covering. Epithelium cells are packed tightly together with little intercellular matrix. Epithelial tissue in the body is of two types. (a) Simple epithelium: It consists of a single layer of cells where cells are in direct contact with the basement membrane. It is further sub-divided into the following types: (i) Simple squamous epithelium: It consists of a single layer of flat cells with irregular boundaries. It is found in the walls of the blood vessels and in the lining of alveoli. (ii) Simple cuboidal epithelium: It consists of a single layer of cube-like cells. It is present in regions where secretion and absorption of substances takes place such as the proximal convoluted tubule region of the nephron. (iii) Simple columnar epithelium: It consists of a single layer of tall, slender cells with their nuclei present at the base of the cells. They may bear micro-villi on the free surfaces. Columnar epithelium forms the lining of the stomach and intestines, and is involved in the function of secretion and absorption. (iv) Ciliated epithelium: It consists of columnar or cuboidal cells with cilia on their free surfaces. They are present in bronchioles and oviducts from where they direct mucus and eggs in specific directions. Page 5 of 11

68 Class XI Chapter 7 Structural Organisation in Animals Biology (v) Glandular epithelium:it consists of columnar or cuboidal cells involved in the secretion of substances. Glands are of two types, unicellular glands (goblet cells of the alimentary canal) and multicellular glands (salivary glands). They can be classified as exocrine (ductless glands) and endocrine glands (duct glands) by the method through which they release enzymes. (b) Compound epithelium: It consists of many layers of cells. It is involved mainly in the function of providing protection and has a limited role in secretion and absorption. Examples of compound epithelium include the dry surface of the skin or moist inner lining of the buccal cavity, pharynx, pancreatic ducts, and the inner lining of ducts of salivary glands. Question 9: Distinguish between (a) Simple epithelium and compound epithelium. (b) Cardiac muscle and striated muscle Page 6 of 11

69 Class XI Chapter 7 Structural Organisation in Animals Biology (c) Dense regular and dense irregular connective tissues (d) Adipose and blood tissue (e) Simple gland and compound gland (a) Simple epithelium Compound epithelium 1. It is composed of only one layer of cells. 2. It is mainly involved in the function of absorption and secretion. 3. It is present in the lining of the stomach, intestine. 1. It is composed of many layers of cells. 2. It is mainly involved in the function of protection and has a limited role in absorption and secretion. 3. It is present in the lining of the pharynx and buccal cavity. (b) Cardiac muscles Striated muscles 1. They are involuntary in function. 2. They are multi-nucleate and branched. 3. They are found only in the heart. 1. They are voluntary in function. 2. They are multi-nucleate and unbranched. 3. They are found only in triceps, biceps, and limbs. (c) Dense regular connective tissues Dense irregular connective tissues Page 7 of 11

70 Class XI Chapter 7 Structural Organisation in Animals Biology 1. In dense regular connective tissues, collagen fibres are present in rows between parallel boundless fibres. 2. They are present in tendons and ligaments. 1. In dense irregular connective tissues, fibres are arranged irregularly. 2. They are present in the skin. (d) Adipose tissue Blood tissue 1. It is composed of collagen fibres, elastin fibres, fibroblasts, macrophages, and adipociytes. 2. It helps in the synthesis, storage, and metabolism of fats. 3. It is present beneath the skin. 1. It is composed of RBCs, WBCs, platelets, and plasma. 2. It helps in the transportation of food, wastes, gases, and hormones. 3. It is present in the blood vessels. (e) Simple glands Compound glands 1. They contain isolated glandular cells. 2. They are unicellular. 3. Examples include goblet cells of the alimentary canal. 1. They contain a cluster of secretory cells. 2. They are multicellular. 3. Examples include salivary glands. Question 10: Mark the odd one in each series: (a) Areolar tissue; blood; neuron; tendon Page 8 of 11

71 Class XI Chapter 7 Structural Organisation in Animals Biology (b) RBC; WBC; platelets; cartilage (c) Exocrine; endocrine; salivary gland; ligament (d) Maxilla; mandible; labrum; athorax; coxa (a) Areolar tissue, blood, and tendons are examples of connective tissues. Neuron is an example of neural tissue. (b) RBCs, WBCs, and platelets are the three most important components of blood. Cartilage is therefore, the odd one out. (c) Exocrine, endocrine, and salivary glands are examples of simple glandular epithelium. Ligament is a connective tissue. (d) Maxilla, mandible, and labrum are mouthparts of a cockroach. Antennae, on the other hand, are present in the head region of cockroaches. (e) Protonema forms the developmental stage in the life cycle of a moss. Mesothorax, metathorax, and coxa are parts or segments present in the legs of a cockroach. Question 11: 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 Column I Column II Page 9 of 11

72 Class XI Chapter 7 Structural Organisation in Animals Biology (a) Compound epithelium (b) Compound eye (c) Septal nephridia (d) Open circulatory system (e) Typhlosole (f) Osteocytes (g) Genitalia (iii) Skin (iv) Mosaic vision (v) Earthworm (ii) Cockroach (i) Alimentary canal (vii) Bone (vi) Phallomere Question 12: Mention briefly about the circulatory system of earthworm Earthworms (Pheretima) have closed blood vascular systems, which consists of the heart, blood vessels, and capillaries. The heart pumps blood for circulating it in one direction. Blood is supplied by smaller blood cells to the gut nerve cord and the body wall. Blood glands are present in the 4 th, 5 th, and 6 th segments, which produce blood cells and haemoglobin dissolved in blood plasma. Blood cells in earthworms are phagocytic in nature. Question 13: Draw a neat diagram of digestive system of frog. Page 10 of 11

73 Class XI Chapter 7 Structural Organisation in Animals Biology Question 14: Mention the function of the following (a) Ureters in frog (b) Malpighian tubules (c) Body wall in earthworm (a) Ureters in frogs: A ureter acts as a urinogenital duct, which carries sperms along with urine in male frogs. (b) Malphigian tubules: Malphigian tubules are excretory organs in cockroaches. (c) Body wall in earthworms: In earthworms, the body wall consists of muscle layers. It helps in movement and burrowing. Page 11 of 11

74 Class XI Chapter 8 Cell The Unit of Life Biology Question 1: Which of the following is not correct? (a) Robert Brown discovered the cell. (b) Schleiden and Schwann formulated the cell theory. (c) Virchow explained that cells are formed from pre-existing cells. (d) A unicellular organism carries out its life activities within a single cell. (a) Robert Brown did not discover the cell. The cell was discovered by Robert Hook. Question 2: New cells generate from (a) bacterial fermentation (b) regeneration of old cells (c) pre-existing cells (d) abiotic materials (c) According to the biogenic theory, new cells can only arise from pre-existing cells. Only complete cells, in favourable conditions, can give rise to new cells. Question 3: Match the following (a) Cristae (i) Flat membranous sacs in stroma (b) Cisternae (ii) Infoldings in mitochondria (c) Thylakoids (iii) Disc-shaped sacs in Golgi apparatus Column I Column II (a) Cristae (ii) Infoldings in mitochondria (b) Cisternae (iii) Disc-shaped sacs in Golgi apparatus (c) Thylakoids (i) Flat membranous sacs in stroma Page 1 of 10

75 Class XI Chapter 8 Cell The Unit of Life Biology Question 4: Which of the following is correct: (a) Cells of all living organisms have a nucleus. (b) Both animal and plant cells have a well defined cell wall. (c) In prokaryotes, there are no membrane bound organelles. (d) Cells are formed de novo from abiotic materials. (c) Membrane-bound organelles are organelles surrounded by a double membrane. Nucleus, mitochondria, chloroplasts, etc., are examples of such organelles. These cell organelles are absent from prokaryotes. (a) Only eukaryotic cells have nuclei. They are absent from prokaryotes. (b) Cell walls are only present in plant cells. They are absent from all animal cells. (d) All cells arise from pre-existing cells. Question 5: What is a mesosome in a prokaryotic cell? Mention the functions that it performs. Mesosome is a convoluted membranous structure formed in a prokaryotic cell by the invagination of the plasma membrane. Its functions are as follows: (1) These extensions help in the synthesis of the cell wall, replication of DNA. They also help in the equal distribution of chromosomes into the daughter cells. (2) It also increases the surface area of the plasma membrane to carry out various enzymatic activities. (3) It helps in secretion processes as well as in bacterial respiration. Page 2 of 10

76 Class XI Chapter 8 Cell The Unit of Life Biology Question 6: How do neutral solutes move across the plasma membrane? Can the polar molecules also move across it in the same way? If not, then how are these transported across the membrane? Plasma membrane is the outermost covering of the cell that separates it from the environment. It regulates the movement of substances into the cell and out from it. It allows the entry of only some substances and prevents the movement of other materials. Hence, the membrane is selectively-permeable. Movement of neutral solutes across the cell membrane Neutral molecules move across the plasma membrane by simple passive diffusion. Diffusion is the movement of molecules from a region of higher concentration to a region of lower concentration. Movement of polar molecules across the cell membrane The cell membrane is made up of a phospholipid bilayer and proteins. The movement of polar molecules across the non-polar lipid bilayer requires carrier-proteins. Carrier-proteins are integral protein particles having certain affinity for specific solutes. As a result, they facilitate the transport of molecules across the membrane. Question 7: Name two cell-organelles that are double membrane bound. What are the characteristics of these two organelles? State their functions and draw labelled diagrams of both. Mitochondria and chloroplasts are the two organelles that are double-membranebound. Page 3 of 10

77 Class XI Chapter 8 Cell The Unit of Life Biology Characteristics of the mitochondria Mitochondria are double- membrane-bound structures. The membrane of a mitochondrion is divided into the inner and outer membranes, distinctly divided into two aqueous compartments outer and inner compartments. The outer membrane is very porous (containing the organelle), while the inner membrane is deeply-folded. These folds are known as cristae. Cristae increase the surface area inside the cell. They are the sites for ATP-generating chemical reactions. The membrane of a mitochondrion contains specific enzymes meant for specific mitochondrial functions. Hence, the mitochondria are the sites for aerobic respiration. They have their own DNA and ribosomes. Thus, they are able to make their own proteins. This is why they are considered as semi-autonomous organelles Characteristics of chloroplasts Chloroplasts are double-membrane-bound structures. They are divided into outer and inner membranes, further divided into two distinct regions: (i) Grana are stacks of flattened discs containing chlorophyll molecules. The flattened membranous sacs are called thylakoids. The thylakoids of adjacent grana are connected by membranous tubules called stroma lamellae. (ii) Stroma is a homogenous mixture in which grana are embedded. It contains several enzymes that are used for the synthesis of carbohydrates and proteins. It also contains its own DNA and ribosomes. Page 4 of 10

78 Class XI Chapter 8 Cell The Unit of Life Biology Functions of the mitochondria: (i) They are the sites for cellular respiration. (ii) They provide energy in the form of ATP for all vital activities of living cells. (iii) They have their own DNA and ribosomes. Hence, they are regarded as semiautonomous organelles. (iv) They have several enzymes, intermediately required for the synthesis of various chemicals such as fatty acids, steroids, and amino acids. Functions of chloroplasts: (i) They trap solar energy and utilise it for manufacturing food for plants. Hence, they are involved in the process of photosynthesis. (ii) They contain the enzymes required for the synthesis of carbohydrates and proteins. Question 8: What are the characteristics of prokaryotic cells? Prokaryotic cell is a unicellular organism lacking membrane-bound organelles. The characteristics of prokaryotic cells are as follows: (i) Most of them are unicellular. Page 5 of 10

79 Class XI Chapter 8 Cell The Unit of Life Biology (ii) They are generally small in size. The size of a prokaryotic cell varies from µm. (iii) The nuclear region of a prokaryotic cell is poorly-defined because of the absence of a nuclear membrane. Hence, a prokaryotic cell lacks a true nucleus. (iv) The genetic materials of prokaryotic cells are naked. They contain single, circular chromosomes. In addition to the genomic DNA, they have a small, circular plasmid DNA. (v) They have specialised membranous structures called mesosomes. Mesosomes are formed by the invagination of the cell membrane. These extensions help in the synthesis of the cell wall, replication of DNA. They also help in the equal distribution of chromosomes into the daughter cells. (vi) Membrane-bound cell organelles such as mitochondria, plastids, and endoplasmic reticulum are absent from a prokaryotic cell. (vii) Most prokaryotic cells contain a three-layered structure outermost glycocalyx, middle cell wall, and the innermost plasma membrane. This structure acts as a protective unit. Examples of prokaryotic cells include blue green algae, bacteria, etc. Question 9: Multicellular organisms have division of labour. Explain. Multicellular organisms are made up of millions and trillions of cells. All these cells perform specific functions. All the cells specialised for performing similar functions are grouped together as tissues in the body. Hence, a particular function is carried out by a group of cells at a definite place in the body. Similarly, different functions are carried out by different groups of cells in an organism. This is known as division of labour in multicellular organisms. Page 6 of 10

80 Class XI Chapter 8 Cell The Unit of Life Biology Question 10: Cell is the basic unit of life. Discuss in brief. Cells are the basic units of life capable of doing all the required biochemical processes that a normal cell has to do in order to live. The basic needs for the survival of all living organisms are the same. All living organisms need to respire, digest food for obtaining energy, and get rid of metabolic wastes. Cells are capable of performing all the metabolic functions of the body. Hence, cells are called the functional units of life. Question 11: What are nuclear pores? State their function. Nuclear pores are tiny holes present in the nuclear membrane of the nucleus. They are formed by the fusion of two nuclear membranes. These holes allow specific substances to be transferred into a cell and out from it. They allow molecules such as RNA and proteins to move in both directions, between the nucleus and the cytoplasm. Question 12: Both lysosomes and vacuoles are endomembrane structures, yet they differ in terms of their functions. Comment. Lysosomes are membrane-bound vesicular structures holding a variety of enzymes such as lipases, proteases, and carbohydrases. The purpose of lysosomes is to digest worn out cells. They are involved in the intracellular digestion of foreign food particles and microbes. Sometimes, they also act as suicidal bags. They are involved in the self digestion of cells. They are a kind of waste disposal systems of a cell. On the other hand, vacuoles are storage sacs found in cells. They might store the waste products of cells. In unicellular organisms, the food vacuole contains the consumed Page 7 of 10

81 Class XI Chapter 8 Cell The Unit of Life Biology food particles. It also plays a role in expelling excess water and some wastes from the cell. Question 13: Describe the structure of the following with the help of labelled diagrams. (i) Nucleus (ii) Centrosome (i) Nucleus Nucleus controls all the cellular activities of the cell. It is spherical in shape. It is composed of the following structures: Nuclear membrane: It is a double membrane separating the contents of the nucleus from the cytoplasm. The narrow space between the two membranes is called the perinuclear space. Nuclear membrane has tiny holes called nuclear pores. These holes allow specific substances to be transferred into a cell and out from it. Nucleoplasm/Nuclear matrix: It is a homogenous granular fluid present inside the nucleus. It contains the nucleolus and chromatin. Nucleolus is a spherical structure that is not bound by any membrane. It is rich in protein and RNA molecules, and is the site for ribosome formation. Chromatin is an entangled mass of thread-like structures. It contains DNA and some basic proteins called histones. (ii) Centrosome Centrosome consists of two cylindrical structures called centrioles. Centrioles lie perpendicular to each other. Each has a cartwheel-like organisation. A centriole is made up of microtubule triplets that are evenly spaced in a ring. The adjacent triplets are linked together. There is a proteinaceous hub in the central part Page 8 of 10

82 Class XI Chapter 8 Cell The Unit of Life Biology of a centriole. The hub is connected to the triplets via radial spokes. These centrioles help in organising the spindle fibres and astral rays during cell division. They form the basal body of cilia and flagella. Question 14: What is a centromere? How does the position of centromere form the basis of classification of chromosomes. Support your answer with a diagram showing the position of centromere on different types of chromosomes. Centromere is a constriction present on the chromosomes where the chromatids are held together. Chromosomes are divided into four types based on the position of the centromere. (i) Metacentric chromosome The chromosomes in which the centromere is present in the middle and divides the chromosome into two equal arms is known as a metacentric chromosome. (ii) Sub-metacentric chromosome Page 9 of 10

83 Class XI Chapter 8 Cell The Unit of Life Biology The chromosome in which the centromere is slightly away from the middle region is known as a sub-metacentric chromosome. In this, one arm is slightly longer than the other. (iii) Acrocentric chromosome The chromosome in which the centromere is located close to one of the terminal ends is known as an acrocentric chromosome. In this, one arm is extremely long and the other is extremely short. (iv) Telocentric chromosome The chromosome in which the centromere is located at one of the terminal ends is known as a telocentric chromosome. Page 10 of 10

84 Class XI Chapter 9 Biomolecules Biology Question 1: What are macromolecules? Give examples. Macromolecules are large complex molecules that occur in colloidal state in intercellular fluid. They are formed by the polymerization of low molecular weight micromolecules. Polysaccharides, proteins, and nucleic acids are common examples of macromolecules. Question 2: Illustrate a glycosidic, peptide and a phospho-diester bond. (a) Glycosidic bond is formed normally between carbon atoms, 1 and 4, of neighbouring monosaccharide units. (b) Peptide bond is a covalent bond that joins the two amino acids by NH CO linkage. Page 1 of 13

85 Class XI Chapter 9 Biomolecules Biology (c) Phosphodiester bond is a strong covalent bond between phosphate and two sugar groups. Such bonds form the sugar phosphate backbone of nucleic acids. Question 3: What is meant by tertiary structure of proteins? The helical polypeptide chain undergoes coiling and folding to form a complex threedimensional shape referred to as tertiary structure of proteins. These coils and folds are arranged to hide the non-polar amino acid chains and to expose the polar side chains. The tertiary structure is held together by the weak bonds formed between various parts of the polypeptide chain. Question 4: Find and write down structures of 10 interesting small molecular weight biomolecules. Find if there is any industry which manufactures the compounds by isolation. Find out who are the buyers. (a) Page 2 of 13

86 Class XI Chapter 9 Biomolecules Biology Molecule Structure 1. Adenosine 2. Thymidine 3. Sucrose 4. Maltose Page 3 of 13

87 Class XI Chapter 9 Biomolecules Biology 5. Lactose 6. Ribose 7. DNA Page 4 of 13

88 Class XI Chapter 9 Biomolecules Biology 8. RNA 9. Glycerol 10. Insulin (b) Compound Manufacturer Buyer 1. Starch products 2. Liquid glucose Kosha Impex (P) Ltd. Marudhar apparels Research laboratories, educational institutes, and other industries, which use biomolecules as a precursor for making other products. Page 5 of 13

89 Class XI Chapter 9 Biomolecules Biology 3. Various enzymes such as amylase, protease, cellulase Map (India) Ltd Question 5: Proteins have primary structure. If you are given a method to know which amino acid is at either of the two termini (ends) of a protein, can you connect this information to purity or homogeneity of a protein? Yes, if we are given a method to know the sequence of proteins, we can connect this information to the purity of a protein. It is known that an accurate sequence of a certain amino acid is very important for the functioning of a protein. If there is any change in the sequence, it would alter its structure, thereby altering the function. If we are provided with a method to know the sequence of an unknown protein, then using this information, we can determine its structure and compare it with any of the known correct protein sequence. Any change in the sequence can be linked to the purity or homogeneity of a protein. For example, any one change in the sequence of haemoglobin can alter the normal haemoglobin structure to an abnormal structure that can cause sickle cell anaemia. Question 6: Find out and make a list of proteins used as therapeutic agents. Find other applications of proteins (e.g., cosmetics, etc.) Proteins used as therapeutic agents are as follows: 1. Thrombin and fibrinogen They help in blood clotting. 2. Antigen (antibody) It helps in blood transfusion. 3. Insulin It helps in maintaining blood glucose level in the body. 4. Renin It helps in osmoregulation. Page 6 of 13

90 Class XI Chapter 9 Biomolecules Biology Proteins are also commonly used in the manufacture of cosmetics, toxins, and as biological buffers. Question 7: Explain the composition of triglyceride. Triglyceride is a glyceride, which is formed from a single molecule of glycerol, esterified with three fatty acids. It is mainly present in vegetable oils and animal fat. Structure of triglyceride The general chemical formula of triglyceride is These three fatty acids can be same or different., where R 1, R 2, and R 3 are fatty acids. Question 8: Can you describe what happens when milk is converted into curd or yoghurt from your understanding of proteins. Proteins are macromolecules formed by the polymerization of amino acids. Structurally, proteins are divided into four levels. (a) Primary structure It is the linear sequence of amino acids in a polypeptide chain. (b) Secondary structure The polypeptide chain is coiled to form a threedimensional structure. (c) Tertiary structure The helical polypeptide chain is further coiled and folded to form a complex structure. Page 7 of 13

91 Class XI Chapter 9 Biomolecules Biology (d) Quaternary structure More than one polypeptide chains assemble to form the quaternary structure. Milk has many globular proteins. When milk is converted into curd or yoghurt, these complex proteins get denatured, thus converting globular proteins into fibrous proteins. Therefore, by the process of denaturation, the secondary and tertiary structures of proteins are destroyed. Question 9: Can you attempt building models of biomolecules using commercially available atomic models (Ball and Stick models). Ball and stick models are 3-D molecular models that can be used to describe the structure of biomolecules. In ball and stick model, the atoms are represented as balls whereas the bonds that hold the atoms are represented by the sticks. Double and triple bonds are represented by springs that form curved connections between the balls. The size and colour of various atoms are different and are depicted by the relative size of the balls. It is the most fundamental and common model of representing biomolecular structures. Page 8 of 13

92 Class XI Chapter 9 Biomolecules Biology In the above ball and stick model of D-glucose, the oxygen atoms are represented by red balls, hydrogen atoms by blue balls, while carbon atoms are represented by grey balls. Question 10: Attempt titrating an amino acid against a weak base and discover the number of dissociating ( ionizable ) functional groups in the amino acid. Titrating a neutral or basic amino acid against a weak base will dissociate only one functional group, whereas titration between acidic amino acid and a weak acid will dissociate two or more functional groups. Question 11: Draw the structure of the amino acid, alanine. Structure of alanine Question 12: What are gums made of? Is Fevicol different? Gums are hetero-polysaccharides. They are made from two or more different types of monosaccharides. On the other hand, fevicol is polyvinyl alcohol (PVA) glue. It is not a polysaccharide. Page 9 of 13

93 Class XI Chapter 9 Biomolecules Biology Question 13: Find out a qualitative test for proteins, fats and oils, amino acids and test any fruit juice, saliva, sweat and urine for them. (a) Test for protein Biuret s test If Biuret s reagent is added to protein, then the colour of the reagent changes from light blue to purple. (b) Test for fats and oils Grease or solubility test (c) Test for amino acid Ninhydrin test If Ninhydrin reagent is added to the solution, then the colourless solution changes to pink, blue, or purple, depending on the amino acid. Item Name of the test Procedure Result Inference 1. Fruit juice Biuret s test Fruit juice + Biuret s reagent Colour changes from light blue to purple Protein present. is Grease test To a brown paper, add a few drops of fruit juice. No translucent spot Fats and oils are absent or are in negligible amounts. Ninhydrin test Fruit juice + Ninhydrin reagent + boil for 5 minutes Colourless solution changes to pink, blue, or purple colour Amino are present. acids 2. Saliva Biuret s Saliva + Biuret s Colour changes from Proteins are Page 10 of 13

94 Class XI Chapter 9 Biomolecules Biology test reagent light blue to purple present. Grease test On a brown paper, add a drop of saliva. No translucent spot Fats/oils absent. are Ninhydrin test Saliva + Ninhydrin reagent + boil for 5 minutes Colourless solution changes to pink, blue, or purple colour Amino are present. acids 3. Sweat Biuret s test Sweat + Biuret s reagent No colour change Proteins absent. are Solubility test Sweat + Water Oily appearance Fats/oil may be present. Ninhydrin test Sweat + Ninhydrin reagent + boil for 5 minutes No colour change, solution remains colourless Amino are absent. acids 4. Urine Biuret s test Few drops of urine + Biuret s reagent Colour changes from light blue to purple Proteins present. are Solubility test Few drops of urine + Water Little bit of oily appearance Fats may or may not be present. Ninhydrin Few drops of Colourless solution Amino acids test urine + changes to pink, are present. Page 11 of 13

95 Class XI Chapter 9 Biomolecules Biology Ninhydrin reagent + boil for 5 minutes blue, or purple colour depending on the type of amino acid Question 14: Find out how much cellulose is made by all the plants in the biosphere and compare it with how much of paper is manufactured by man and hence what is the consumption of plant material by man annually. What a loss of vegetation! Approximately, 100 billion tonnes of cellulose are made per year by all the plants in the biosphere and it takes 17 full grown trees to make one ton of paper. Trees are also used to fulfil the other requirements of man such as for timber, food, medicines, etc. Hence, it is difficult to calculate the annual consumption of plant material by man. Question 15: Describe the important properties of enzymes. Properties of enzymes (1) Enzymes are complex macromolecules with high molecular weight. (2) They catalyze biochemical reactions in a cell. They help in the breakdown of large molecules into smaller molecules or bring together two smaller molecules to form a larger molecule. (3) Enzymes do not start a reaction. However, they help in accelerating it. (4) Enzymes affect the rate of biochemical reaction and not the direction. (5) Most of the enzymes have high turnover number. Turnover number of an enzyme is the number of molecules of a substance that is acted upon by an enzyme per minute. High turnover number of enzymes increases the efficiency of reaction. Page 12 of 13

96 Class XI Chapter 9 Biomolecules Biology (6) Enzymes are specific in action. (7) Enzymatic activity decreases with increase in temperature. (8) They show maximum activity at an optimum ph of 6 8. (9) The velocity of enzyme increases with increase in substrate concentration and then, ultimately reaches maximum velocity. Page 13 of 13

97 Class XI Chapter 10 Cell Cycle and Cell Division Biology Question 1: What is the average cell cycle span for a mammalian cell? The average cell cycle span for a mammalian cell is approximately 24 hours. Question 2: Distinguish cytokinesis from karyokinesis. Cytokinesis Karyokinesis Cytokinesis is the biological Karyokinesis is the biological (i) process involving the division of a cell s cytoplasm during mitosis or (i) process involving the division of a cell s nucleus during mitosis or meiosis. meiosis. (ii) It is divided into four stages prophase, metaphase, anaphase, and telophase. (ii) Stages such as prophase, metaphase, anaphase, and telophase are not present in karyokinesis. Question 3: Describe the events taking place during interphase. Interphase involves a series of changes that prepare a cell for division. It is the period during which the cell experiences growth and DNA replication in an orderly manner. Interphase is divided into three phases. (i) G 1 phase (ii) S phase (iii) G 2 phase Page 1 of 10

98 Class XI Chapter 10- Cell Cycle and Cell Division Biology G 1 phase It is the stage during which the cell grows and prepares its DNA for replication. In this phase, the cell is metabolically active. S phase It is the stage during which DNA synthesis occurs. In this phase, the amount of DNA (per cell) doubles, but the chromosome number remains the same. G 2 phase In this phase, the cell continues to grow and prepares itself for division. The proteins and RNA required for mitosis are synthesised during this stage. Question 4: What is G 0 (quiescent phase) of cell cycle? G 0 or quiescent phase is the stage wherein cells remain metabolically active, but do not proliferate unless called to do so. Such cells are used for replacing the cells lost during injury. Question 5: Why is mitosis called equational division? Mitosis is the process of cell division wherein the chromosomes replicate and get equally distributed into two daughter cells. The chromosome number in each daughter cell is equal to that in the parent cell, i.e., diploid. Hence, mitosis is known as equational division. Page 2 of 10

99 Class XI Chapter 10- Cell Cycle and Cell Division Biology Question 6: Name the stage of cell cycle at which 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 (i) Metaphase (ii) Anaphase (iii) Zygotene of meiosis I (iv) Pachytene of meiosis I Question 7: Describe the following: (a) synapsis (b) bivalent (c) chiasmata Draw a diagram to illustrate your answer. (a) Synapsis The pairing of homologous chromosomes is called synapsis. This occurs during the second stage of prophase I or zygotene. (b) Bivalent Bivalent or tetrad is a pair of synapsed homologous chromosomes. They are formed during the zygotene stage of prophase I of meiosis. Page 3 of 10

100 Class XI Chapter 10 Cell Cycle and Cell Division Biology (c) Chiasmata Chiasmata is the site where two sister chromatids have crossed over. It represents the site of cross-over. It is formed during the diplotene stage of prophase I of meiosis. Question 8: How does cytokinesis in plant cells differ from that in animal cells? Cytokinesis in plant cells Cytokinesis is animal cells (i) The division of the cytoplasm takes place by cell plate formation. (i) The division of the cytoplasm takes place by cleavage. Cell plate formation starts at the Cleavage starts at the periphery (ii) centre of the cell and grows (ii) and then moves inward, dividing outward, toward the lateral walls. the cell into two parts. Page 4 of 10

101 Class XI Chapter 10 Cell Cycle and Cell Division Biology Question 9: Find examples where the four daughter cells from meiosis are equal in size and where they are found unequal in size. (a) Spermatogenesis or the formation of sperms in human beings occurs by the process of meiosis. It results in the formation of four equal-sized daughter cells. (b) Oogenesis or the formation of ovum in human beings occurs by the process of meiosis. It results in the formation of four daughter cells which are unequal in size. Question 10: Distinguish anaphase of mitosis from anaphase I of meiosis. Anaphase of mitosis Anaphase I of meiosis Anaphase is the stage during which the centromere splits and the chromatids separate. The chromosomes move apart, toward the opposite poles. These chromosomes are genetically identical. During anaphase I, the homologous chromosomes separate, while the chromatids remain attached at their centromeres. Hence, in anaphase I, the chromosomes of each bivalent pair separate, while the sister chromatids remain together. Page 5 of 10

102 Class XI Chapter 10 Cell Cycle and Cell Division Biology Question 11: List the main differences between mitosis and meiosis. Mitosis Meiosis 1. In mitotic division, a single division results in two daughter cells. 1. Meiotic division involves two successive divisions meiosis I and meiosis II. These divisions result in four daughter cells. Meiosis I is known as reductional 2. Mitosis is known as equational division. This is because the daughter cells have the same diploid number of chromosomes as the parent. 2. division. This is because the chromosome number is reduced to half. Meiosis II is known as equational division. This is because the sister chromatids separate and the chromosome number remains the Page 6 of 10

103 Class XI Chapter 10 Cell Cycle and Cell Division Biology same. 3. Prophase is short and does not comprise any phase. 3. Prophase I is very long and comprises 5 phases leptotene, zygotene, pachytene, diplotene, and diakinesis. 4. There is no pairing of chromosomes, crossing-over, or chiasmata-formation during prophase. 4. In the zygotene stage of prophase, the pairing of chromosomes occurs. During pachytene, the crossing-over occurs. The chiasmata are formed in the diplotene stage. 5. Synaptonemal complex is not formed. 5. Synaptonemal complex is formed during the zygotene stage of prophase I. During anaphase I, the homologous chromosomes separate, while the Anaphase involves the separation chromatids remain attached at their 6. of the chromatids of each 6. centromeres. chromosome. During anaphase II, the chromatids separate as a result of the splitting of the centromere. Mitosis plays a significant role in Meiosis brings about variation and 7. the healing, repair, and growth of 7. maintains the chromosome number a cell. from generation to generation. Question 12: Page 7 of 10

104 Class XI Chapter 10 Cell Cycle and Cell Division Biology What is the significance of meiosis? Meiosis is the process involving the reduction in the amount of genetic material. It comprises two successive nuclear and cell divisions, with a single cycle of DNA replication. As a result, at the end of meiosis II, four haploid cells are formed. Significance of meiosis 1. Meiosis maintains the chromosome number from generation to generation. It reduces the chromosome number to half so that the process of fertilisation restores the original number in the zygote. 2. Variations are caused by the cross-over and the random distribution of homologous chromosomes between daughter cells. Variations play an important role in evolution. 3. Chromosomal mutations are brought about by the introduction of certain abnormalities. These chromosomal mutations may be advantageous for an individual. Question 13: 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. (i) In some insects and lower plants, fertilization is immediately followed by zygotic meiosis, which leads to the production of haploid organisms. This type of life cycle is known as haplontic life cycle. (ii) The phenomenon of polyploidy can be observed in some haploid cells in higher plants in which cell division does not occur. Polyploidy is a state in which cells contain multiple pairs of chromosomes than the basic set. Polyploidy can be artificially induced in plants by applying colichine to cell culture. Question 14: Page 8 of 10

105 Class XI Chapter 10 Cell Cycle and Cell Division Biology Can there be mitosis without DNA replication in S phase? Mitotic cell division cannot take place without DNA replication in S phase. Two important events take place during S phase one is the synthesis or duplication of DNA and the other is the duplication of the centriole. DNA duplication is important as it maintains the chromosome number in the daughter cells. Mitosis is an equational division. Therefore, the duplication of DNA is an important step. Question 15: Can there be DNA replication without cell division? There can be DNA replication without cell division. During cell division, the parent cell gets divided into two daughter cells. However, if there is a repeated replication of DNA without any cell division, then this DNA will keep accumulating inside the cell. This would increase the volume of the cell nucleus, thereby causing cell expansion. An example of DNA duplication without cell division is commonly observed in the salivary glands of Drosophila. The chromosome undergoing repeated DNA duplication is known as polytene chromosome. Question 16: Analyse the events during every stage of cell cycle and notice how the following two parameters change (i) Number of chromosomes (N) per cell (ii) Amount of DNA content (C) per cell During meiosis, the number of chromosomes and the amount of DNA in a cell change. (i) Number of chromosomes (N) per cell During anaphase I of the meiotic cycle, the homologous chromosomes separate and start moving toward their respective poles. As a result, the bivalents get divided into Page 9 of 10

106 Class XI Chapter 10 Cell Cycle and Cell Division Biology two sister chromatids and receive half the chromosomes present in the parent cell. Therefore, the number of chromosomes reduces in anaphase I. (ii) Amount of DNA content (C) per cell During anaphase II of the meiotic cycle, the chromatids separate as a result of the splitting of the centromere. It is the centromere that holds together the sister chromatids of each chromosome. As a result, the chromatids move toward their respective poles. Therefore, at each pole, a haploid number of chromosomes and a haploid amount of DNA are present. During mitosis, the number of chromosomes remains the same. The DNA duplicated in the S phase gets separated in the two daughter cells during anaphase. As a result, the DNA content (C) of the two newly-formed daughter cells remains the same. Page 10 of 10

107 Class XII Chapter 12 Biotechnology and its Applications Biology Crystals of Bt toxin produced by some bacteria do not kill the bacteria themselves because bacteria are resistant to the toxin toxin is immature: toxin is inactive: bacteria encloses toxin in a special sac. toxin is inactive: In bacteria, the toxin is present in an inactive form, called prototoxin, which gets converted into active form when it enters the body of an insect. What are transgenic bacteria? Illustrate using any one example. Transgenic bacteria contain foreign gene that is intentionally introduced into its genome. They are manipulated to express the desirable gene for the production of various commercially important products. An example of transgenic bacteria is. In the plasmid of, the two DNA sequences corresponding to A and B chain of human insulin are inserted, so as to produce the respective human insulin chains. Hence, after the insertion of insulin gene into the bacterium, it becomes transgenic and starts producing chains of human insulin. Later on, these chains are extracted from and combined to form human insulin. Page 1 of 6

108 Class XII Chapter 12 Biotechnology and its Applications Biology Compare and contrast the advantages and disadvantages of production of genetically modified crops. The production of genetically modified (GM) or transgenic plants has several advantages. Most of the GM crops have been developed for pest resistance, which increases the crop productivity and therefore, reduces the reliance on chemical pesticides. Many varieties of GM food crops have been developed, which have enhanced nutritional quality. For example, golden rice is a transgenic variety in rice, which is rich in vitamin A. These plants prevent the loss of fertility of soil by increasing the efficiency of mineral usage. They are highly tolerant to unfavourable abiotic conditions. The use of GM crops decreases the post harvesting loss of crops. However, there are certain controversies regarding the use of genetically modified crops around the world. The use of these crops can affect the native biodiversity in an area. For example, the use of Bt toxin to decrease the amount of pesticide is posing a threat for beneficial insect pollinators such as honey bee. If the gene expressed for Bt toxin gets expressed in the pollen, then the honey bee might be affected. As a result, the process of pollination by honey bees would be affected. Also, genetically modified crops are affecting human health. They supply allergens and certain antibiotic resistance markers in the body. Also, they can cause genetic pollution in the wild relatives of the crop plants. Hence, it is affecting our natural environment. What are Cry proteins? Name an organism that produces it. How has man exploited this protein to his benefit? Cry proteins are encoded by cry genes. These proteins are toxins, which are produced by bacteria. This bacterium contains these proteins in their inactive from. When the inactive toxin protein is ingested by the insect, it gets activated by the alkaline ph of the gut. This results in the lysis of epithelial cell and eventually the death of the insect. Therefore, man has exploited this protein to develop certain transgenic crops with insect resistance such as Bt cotton, Bt corn, etc. Page 2 of 6

109 Class XII Chapter 12 Biotechnology and its Applications Biology What is gene therapy? Illustrate using the example of adenosine deaminase (ADA) deficiency. Gene therapy is a technique for correcting a defective gene through gene manipulation. It involves the delivery of a normal gene into the individual to replace the defective gene, for example, the introduction of gene for adenosine deaminase (ADA) in ADA deficient individual. The adenosine deaminase enzyme is important for the normal functioning of the immune system. The individual suffering from this disorder can be cured by transplantation of bone marrow cells. The first step involves the extraction of lymphocyte from the patient s bone marrow. Then, a functional gene for ADA is introduced into lymphocytes with the help of retrovirus. These treated lymphocytes containing ADA gene are then introduced into the patient s bone marrow. Thus, the gene gets activated producing functional T4 lymphocytes and activating the patient s immune system. Diagrammatically represent the experimental steps in cloning and expressing an human gene (say the gene for growth hormone) into a bacterium like? DNA cloning is a method of producing multiple identical copies of specific template DNA. It involves the use of a vector to carry the specific foreign DNA fragment into the host cell. The mechanism of cloning and transfer of gene for growth hormone into is represented below. Page 3 of 6

110 Class XII Chapter 12 Biotechnology and its Applications Biology Page 4 of 6

111 Class XII Chapter 12 Biotechnology and its Applications Biology Can you suggest a method to remove oil (hydrocarbon) from seeds based on your understanding of rdna technology and chemistry of oil? Recombinant DNA technology (rdna) is a technique used for manipulating the genetic material of an organism to obtain the desired result. For example, this technology is used for removing oil from seeds. The constituents of oil are glycerol and fatty acids. Using rdna, one can obtain oilless seeds by preventing the synthesis of either glycerol or fatty acids. This is done by removing the specific gene responsible for the synthesis. Find out from internet what is golden rice. Golden rice is a genetically modified variety of rice,,which has been developedas a fortified food for areas where there is a shortage of dietary vitamin A. It contains a precursor of pro4vitamin A, called beta4carotene, which has been introduced into the rice through genetic engineering. The rice plant naturally produces beta4 carotene pigment in its leaves. However, it is absent in the endosperm of the seed. This is because beta4carotene pigment helps in the process of photosynthesis while photosynthesis does not occur in endosperm. Since beta4carotene is a precursor of pro4 vitamin A, it is introduced into the rice variety to fulfill the shortage of dietary vitamin A. It is simple and a less expensive alternative to vitamin supplements. However, this variety of rice has faced a significant opposition from environment activists. Therefore, they are still not available in market for human consumption. Does our blood have proteases and nucleases? No, human blood does not include the enzymes, nucleases and proteases. In human beings, blood serum contains different types of protease inhibitors, which protect the blood proteins from being broken down by the action of proteases. The enzyme, nucleases, catalyses the hydrolysis of nucleic acids that is absent in blood. Page 5 of 6

112 Class XII Chapter 12 Biotechnology and its Applications Biology Consult internet and find out how to make orally active protein pharmaceutical. What is the major problem to be encountered? Orally active protein pharmaceuticals contain biologically active materials such as peptides or proteins, antibodies, and polymeric beads. It is administrated orally into the body through various formulations. It involves the encapsulation of protein or peptide in liposomes or formulations using penetration enhancers. These proteins or peptides are used for treatment of various diseases and are also used as vaccines. However, the oral administration of these peptides or proteins has some problems related to it. Once these proteins are ingested, the proteases present in the stomach juices denature the protein. As a result, their effect will be nullified. Hence, it is necessary to protect the therapeutic protein from digestive enzymes, if taken orally. This is the reason for the proteins to be injected directly into the target site. Page 6 of 6

113 Class XI Chapter 11 Transport in Plants Biology Question 1: What are the factors affecting the rate of diffusion? Diffusion is the passive movement of substances from a region of higher concentration to a region of lower concentration. Diffusion of substances plays an important role in cellular transport in plants. Rate of diffusion is affected by concentration gradient, membrane permeability, temperature, and pressure. Diffusion takes place as long as there is a difference between the concentrations of a substance across a barrier. However, diffusion stops, when the concentrations of the substance on either side of the barrier become equal. The permeability of a membrane affects the rate of diffusion. Diffusion rate increases as membrane permeability increases. Changes in temperature and pressure values also affect the diffusion of substances. Pressure plays an important role in the diffusion of gases as gases diffuse from a region of higher partial pressure to a region of lower partial pressure. Question 2: What are porins? What role do they play in diffusion? Porins are types of proteins which form pores of large sizes in the outer membranes of plastids such as chloroplast, mitochondria and the membranes in bacteria. They help in facilitating the passive transport of small-sized protein molecules. Question 3: Describe the role played by protein pumps during active transport in plants. In plant cells, active transport occurs against the concentration gradient, i.e., from a region of lower concentration to a region of higher concentration. The process of active transport involves specific protein pumps. The protein pumps are made up of specific proteins called trans-membrane proteins. These pumps first make a complex Page 1 of 11

114 Class XI Chapter 11 Transport in Plants Biology with the substance to be transported across the membrane, using the energy derived from ATP. The substance finally gets liberated into the cytoplasm as a result of the dissociation of the protein substance complex. Question 4: Explain why pure water has the maximum water potential. Water potential quantifies the tendency of water to move from one part to the other during various cellular processes. It is denoted by the Greek letter Psi or Ψ. The water potential of pure water is always taken as zero at standard temperature and pressure. It can be explained in terms of the kinetic energy possessed by water molecules. When water is in liquid form, the movement of its molecules is rapid and constant. Pure water has the highest concentration of water molecules. Therefore, it has the highest water potential. When some solute is dissolved in water, the water potential of pure water decreases. Question 5: Differentiate between the following: (a) Diffusion and Osmosis (b) Transpiration and Evaporation (c) Osmotic Pressure and Osmotic Potential (d) Imbibition and Diffusion (e) Apoplast and Symplast pathways of movement of water in plants. (f) Guttation and Transpiration. (a) Diffusion and osmosis Diffusion Osmosis 1. Diffusion is the passive 1. Osmosis is the process in which Page 2 of 11

115 Class XI Chapter 11 Transport in Plants Biology movement of particles, ions, and molecules along the concentration gradient. the diffusion of a solvent (water) occurs across a semi-permeable membrane. 2. It can occur in solids, liquids, and gases. 2. It occurs in the liquid medium. 3. It does not require a semipermeable membrane. 3. It requires a semi-permeable membrane. (b) Transpiration and evaporation Transpiration Evaporation 1. It occurs in plants. 1. It occurs from any free surface and involves living and non-living surfaces. 2. It is a physiological process. 2. It is a physical process. 3. It occurs mainly through the stomatal pores on plant leaves. 3. It is occurs through any free surface. 4. It is controlled by environmental factors as well as physiological factors of plants such as root-shoot ratio and number of stomata. 4. It is entirely driven by environmental factors. (c) Osmotic pressure and osmotic potential Osmotic pressure Osmotic potential 1. It is expressed in bars with a positive sign. 1. It is expressed in bars with a negative sign. Page 3 of 11

116 Class XI Chapter 11 Transport in Plants Biology 2. It is a positive pressure. 2. It is a negative pressure. 3. Its value increases with an increase in the concentration of solute particles. 3. Its value decreases with an increase in the concentration of solute particles. (d) Imbibition and diffusion Imbibition Diffusion 1. Imbibition is a special type of diffusion. In this process, water is absorbed by solids and colloids, causing an enormous increase in volume. 1. Diffusion is the passive movement of particles, ions, and molecules along the concentration gradient. 2. It usually involves water. 2. It involves solids, liquids, and gases. (e) Apoplast and symplast pathways of movement of water in plants Apoplast pathway Symplast pathway 1. The apoplast pathway involves the movement of water through the adjacent cell walls of the epidermis and cortex. The movement of water is restricted at the casparian strips of the root endodermis. 1. The symplast pathway involves the movement of water through the interconnected protoplasts of the epidermis, cortex, endodermis, and root pericycle. 2. It is a faster process of water movement and water moves through mass flow. 2. It is a slower process of water movement. (f) Guttation and transpiration Page 4 of 11

117 Class XI Chapter 11 Transport in Plants Biology Guttation Transpiration 1. It occurs usually at night. 1. It occurs usually during the day. 2. Water is lost from the leaves in the form of liquid droplets. 2. Water is lost from the leaves in the form of water vapour. 3. It occurs through the vein endings of leaves. 3. It occurs through the stomata. 4. It is an uncontrolled process. 4. It is a controlled process. Question 6: Briefly describe water potential. What are the factors affecting it? Water potential quantifies the tendency of water to move from one part to the other during various cellular processes such as diffusion, osmosis, etc. It is denoted by the Greek letter Psi or Ψ and is expressed in Pascals (Pa). The water potential of pure water is always taken as zero at standard temperature and pressure. Water potential (Ψw) is expressed as the sum of solute potential (Ψs) and pressure potential (Ψp). Ψw = Ψs + Ψp When some solute is dissolved in water, the water potential of pure water decreases. This is termed as solute potential (Ψs), which is always negative. For a solution at atmospheric pressure, Ψw = Ψs. The water potential of pure water or a solution increases on the application of pressure values more than atmospheric pressure. It is termed as pressure potential. It is denoted by Ψp and has a positive value, although a negative pressure potential is present in the xylem. This pressure potential plays a major role in the ascent of water through the stem. Question 7: Page 5 of 11

118 Class XI Chapter 11 Transport in Plants Biology What happens when a pressure greater than the atmospheric pressure is applied to pure water or a solution? The water potential of pure water or a solution increases on the application of pressure values more than atmospheric pressure. For example: when water diffuses into a plant cell, it causes pressure to build up against the cell wall. This makes the cell wall turgid. This pressure is termed as pressure potential and has a positive value. Question 8: (a) With the help of well-labelled diagrams, describe the process of plasmolysis in plants, giving appropriate examples. (b) Explain what will happen to a plant cell if it is kept in a solution having higher water potential. (a) Plasmolysis can be defined as the shrinkage of the cytoplasm of a plant cell, away from its cell wall and toward the centre. It occurs because of the movement of water from the intracellular space to the outer-cellular space. This happens when the plant cell is placed in a hypertonic solution (i.e., a solution having more solute concentration than the cell cytoplasm). This causes the water to move out of the cell and toward the solution. The cytoplasm of the cell shrinks and the cell is said to be plasmolysed. This process can be observed in an onion peel kept in a highly concentrated salt solution. (b) When a plant cell is placed in a hypertonic solution or a solution having higher water potential, the water diffuses into the cell (i.e., movement is observed from Page 6 of 11

119 Class XI Chapter 11 Transport in Plants Biology higher to lower water pressure region). The entry of water in the plant cell exerts pressure on the rigid cell wall. This is called turgor pressure. As a result of its rigid cell wall, the plant cell does not burst. Question 9: How is the mycorrhizal association helpful in absorption of water and minerals in plants? Mycorrhiza is a symbiotic association of fungi with the root systems of some plants. The fungal hyphae either form a dense network around the young roots or they penetrate the cells of the roots. The large surface area of the fungal hyphae is helpful in increasing the absorption of water and minerals from the soil. In return, they get sugar and nitrogenous compounds from the host plants. The mycorrhizal association is obligate in some plants. For example, Pinus seeds do not germinate and establish in the absence of mycorrhizal. Question 10: What role does root pressure play in water movement in plants? Root pressure is the positive pressure that develops in the roots of plants by the active absorption of nutrients from the soil. When the nutrients are actively absorbed by root hairs, water (along with minerals) increases the pressure in the xylem. This pressure pushes the water up to small heights. Root pressure can be observed experimentally by cutting the stem of a well-watered plant on a humid day. When the stem is cut, the solution oozes from the cut end. Root pressure is also linked to the phenomenon of guttation, i.e., the loss of water in the form of liquid droplets from the vein endings of certain herbaceous plants. Root pressure is only able to transport water up to small heights. However, it helps in re-establishing the continuous chains of water molecules in the xylem. Page 7 of 11

120 Class XI Chapter 11 Transport in Plants Biology Transpirational pull maintains the flow of water molecules from the roots to the shoots. Question 11: Describe transpiration pull model of water transport in plants. What are the factors influencing transpiration? How is it useful to plants? In tall trees, water rises with the help of the transpirational pull generated by transpiration or loss of water from the stomatal pores of leaves. This is called the cohesion-tension model of water transport. During daytime, the water lost through transpiration (by the leaves to the surroundings) causes the guard cells and other epidermal cells to become flaccid. They in turn take water from the xylem. This creates a negative pressure or tension in the xylem vessels, from the surfaces of the leaves to the tips of the roots, through the stem. As a result, the water present in the xylem is pulled as a single column from the stem. The cohesion and adhesion forces of the water molecules and the cell walls of the xylem vessels prevent the water column from splitting. In plants, transpiration is driven by several environmental and physiological factors. The external factors affecting transpiration are wind, speed, light, humidity, and temperature. The plant factors affecting transpiration are canopy structure, number and distribution of stomata, water status of plants, and number of open stomata. Although transpiration causes water loss, the transpirational pull helps water rise in the stems of plants. This helps in the absorption and transport of minerals from the soil to the various plant parts. Transpiration has a cooling effect on plants. It helps maintain plant shape and structure by keeping the cells turgid. Transpiration also provides water for photosynthesis. Question 12: Discuss the factors responsible for ascent of xylem sap in plants. Page 8 of 11

121 Class XI Chapter 11 Transport in Plants Biology Transpirational pull is responsible for the ascent of water in the xylem. This ascent of water is dependent on the following physical factors: Cohesion Mutual attraction between water molecules Surface tension Responsible for the greater attraction between water molecules in liquid phase than in gaseous phase Adhesion Attraction of water molecules to polar surfaces Capillarity Ability of water to rise in thin tubes These physical properties of water allow it to move against gravity in the xylem. Question 13: What essential role does the root endodermis play during mineral absorption in plants? In plants, nutrients are absorbed through the active and passive transports. The endodermal cells of the roots containing suberin allow only selected minerals to pass through them. The transport proteins present in the membranes of these cells act as check points for the various solutes reaching the xylem. Question 14: Explain why xylem transport is unidirectional and phloem transport bi-directional. During the growth of a plant, its leaves act as the source of food as they carry out photosynthesis. The phloem conducts the food from the source to the sink (the part of the plant requiring or storing food). During spring, this process is reversed as the food stored in the sink is mobilised toward the growing buds of the plant, through the phloem. Thus, the movement of food in the phloem is bidirectional (i.e., upward and downward). The transport of water in the xylem takes place only from the roots to the leaves. Therefore, the movement of water and nutrients in the xylem is unidirectional. Page 9 of 11

122 Class XI Chapter 11 Transport in Plants Biology Question 15: Explain pressure flow hypothesis of translocation of sugars in plants. According to the pressure flow hypothesis, food is prepared in the plant leaves in the form of glucose. Before moving into the source cells present in the phloem, the prepared food is converted into sucrose. Water moves from the xylem vessels into the adjacent phloem, thereby increasing the hydrostatic pressure in the phloem. Consequently, the sucrose moves through the sieve cells of the phloem. The sucrose already present in the sink region is converted into starch or cellulose, thereby reducing the hydrostatic pressure in the sink cells. Hence, the pressure difference created between the source and the sink cells allows sugars to be translocated from the former to the latter. This starch or cellulose is finally removed from the sink cells through active transport. Question 16: What causes the opening and closing of guard cells of stomata during transpiration? The tiny pores present on the surfaces of leaves, called stomata, help in the exchange of gases. Each stoma consists of bean-shaped or dumbbell-shaped guard cells. The epidermal cells surrounding the guard cells are modified to form subsidiary cells. The opening and closing of the guard cells is caused by a change in their turgidity. The inner walls of the guard cells are thick and elastic, while the outer walls are thin. The numerous microfibrils present in the guard cells facilitate the opening and closing of the guard cells. At the time of the opening of the stomata, the turgidity of the guard cells increases. As a result, the outer walls bulge and the inner walls become crescent-shaped. The stomatal opening is facilitated by the radial arrangement of the microfibrils. Page 10 of 11

123 Class XI Chapter 11 Transport in Plants Biology At the time of the closing of the stomata, the guard cells lose their turgidity, the outer and inner walls retain their original shapes, and the microfibrils get arranged longitudinally. Page 11 of 11

124 Class XI Chapter 12 Mineral Nutrition Biology Question 1: All elements that are present in a plant need not be essential to its survival. Comment. Plants tend to absorb different kinds of nutrients from soil. However, a nutrient is inessential for a plant if it is not involved in the plant s physiology and metabolism. For example, plants growing near radioactive sites tend to accumulate radioactive metals. Similarly, gold and selenium get accumulated in plants growing near mining sites. However, this does not mean that radioactive metals, gold, or selenium are essential nutrients for the survival of these plants. Question 2: Why is purification of water and nutrient salts so important in studies involving mineral nutrition using hydroponics? Hydroponics is the art of growing plants in a nutrient solution in the absence of soil. Since the plant roots are exposed to a limited amount of the solution, there are chances that the concentrations of oxygen and other minerals in the plant roots would reduce. Therefore, in studies involving mineral nutrition using hydroponics, purification of water and nutrient salts is essential so as to maintain an optimum growth of the plants. Question 3: Explain with examples: macronutrients, micronutrients, beneficial nutrients, toxic elements and essential elements. Macronutrients: They are the nutrients required by plants in large amounts. They are present in plant tissues in amounts more than 10 mmole kg 1 of dry matter. Examples include hydrogen, oxygen, and nitrogen. Page 1 of 6

125 Class XI Chapter 12 Mineral Nutrition Biology Micronutrients: They are also called trace elements and are present in plant bodies in very small amounts, i.e., amounts less than 10 mmole kg 1 of dry matter. Examples include cobalt, manganese, zinc, etc. Beneficial nutrients: They are plant nutrients that may not be essential, but are beneficial to plants. Sodium, silicon, cobalt and selenium are beneficial to higher plants. Toxic elements: Micronutrients are required by plants in small quantities. An excess of these nutrients may induce toxicity in plants. For example, when manganese is present in large amounts, it induces deficiencies of iron, magnesium, and calcium by interfering with their metabolism. Essential elements: These elements are absolutely necessary for plant growth and reproduction. The requirement of these elements is specific and non-replaceable. They are further classified as macro and micro-nutrients. Question 4: Name at least five different deficiency symptoms in plants. Describe them and correlate them with the concerned mineral deficiency. The five main deficiency symptoms arising in plants are: Chlorosis Necrosis Inhibition of cell division Delayed flowering Stunted plant growth Chlorosis or loss of chlorophyll leads to the yellowing of leaves. It is caused by the deficiencies of nitrogen, potassium, magnesium, sulphur, iron, manganese, zinc, and molybdenum. Necrosis is the death of plant tissues as a result of the deficiencies of calcium, magnesium, copper, and potassium. Page 2 of 6

126 Class XI Chapter 12 Mineral Nutrition Biology Inhibition of cell division is caused by the deficiencies of nitrogen, potassium, sulphur, and molybdenum. Delayed flowering is caused by the deficiencies of nitrogen, sulphur, and molybdenum. Stunted plant growth is a result of the deficiencies of copper and sulphur. Question 5: If a plant shows a symptom which could develop due to deficiency of more than one nutrient, how would you find out experimentally, the real deficient mineral element? In plants, the deficiency of a nutrient can cause multiple symptoms. For example, the deficiency of nitrogen causes chlorosis and delayed flowering. In a similar way, the deficiency of a nutrient can cause the same symptom as that caused by the deficiency of another nutrient. For example, necrosis is caused by the deficiency of calcium, magnesium, copper, and potassium. Another point to be considered is that different plants respond in different ways to the deficiency of the same nutrient. Hence, to identify the nutrient deficient in a plant, all the symptoms developed in its different parts must be studied and compared with the available standard tables. Question 6: Why is that in certain plants deficiency symptoms appear first in younger parts of the plant while in others they do so in mature organs? Deficiency symptoms are morphological changes in plants, indicating nutrient deficiency. Deficiency symptoms vary from one element to another. The plant part in which a deficiency symptom occurs depends on the mobility of the deficient element in the plant. Elements such as nitrogen, potassium, and magnesium are highly mobile. These elements move from the mature organs to the younger parts of a plant. Therefore, the symptoms for the deficiencies of these elements first appear in Page 3 of 6

127 Class XI Chapter 12 Mineral Nutrition Biology the older parts of the plant. Elements such as calcium and sulphur are relatively immobile. These elements are not transported out of the older parts of a plant. Therefore, the symptoms for the deficiencies of these elements first appear in the younger parts of the plant. Question 7: How are the minerals absorbed by the plants? The absorption of soil nutrients by the roots of plants occurs in two main phases apoplast and symplast. During the initial phase or apoplast, there is a rapid uptake of nutrients from the soil into the free spaces of plant cells. This process is passive and it usually occurs through trans-membrane proteins and ion-channels. In the second phase or symplast, the ions are taken slowly into the inner spaces of the cells. This pathway generally involves the expenditure of energy in the form of ATP. Question 8: What are the conditions necessary for fixation of atmospheric nitrogen by Rhizobium. What is their role in N 2 -fixation? Rhizobium is a symbiotic bacteria present in the root nodules of leguminous plants. The basic requirements for Rhizobium to carry out nitrogen fixation are as follows: (a) Presence of the enzyme nitrogenase (b) Presence of leg-haemoglobin (c) Non-haem iron protein, ferrodoxin as the electron-carrier (d) Constant supply of ATP (e) Mg 2+ ions as co-factors Rhizobium contains the enzyme nitrogenase a Mo-Fe protein that helps in the conversion of atmospheric free nitrogen into ammonia. Page 4 of 6

128 Class XI Chapter 12 Mineral Nutrition Biology The reaction is as follows: N 2 + 8e + 8H ATP 2 NH 3 + H ADP + 16Pi The Rhizobium bacteria live as aerobes under free-living conditions, but require anaerobic conditions during nitrogen fixation. This is because the enzyme nitrogenase is highly sensitive to molecular oxygen. The nodules contain leghaemoglobin, which protects nitrogenase from oxygen. Question 9: What are the steps involved in formation of a root nodule? Multiple interactions are involved in the formation of root nodules. The Rhizobium bacteria divide and form colonies. These get attached to the root hairs and epidermal cells. The root hairs get curled and are invaded by the bacteria. This invasion is followed by the formation of an infection thread that carries the bacteria into the cortex of the root. The bacteria get modified into rod-shaped bacteroides. As a result, the cells in the cortex and pericycle undergo division, leading to the formation of root nodules. The nodules finally get connected with the vascular tissues of the roots for nutrient exchange. increases in region C. Question 10: Which of the following statements are true? If false, correct them: (a) Boron deficiency leads to stout axis. (b) Every mineral element that is present in a cell is needed by the cell. (c) Nitrogen as a nutrient element, is highly immobile in the plants. (d) It is very easy to establish the essentiality of micronutrients because they are required only in trace quantities. (a) True Page 5 of 6

129 Class XI Chapter 12 Mineral Nutrition Biology (b) All the mineral elements present in a cell are not needed by the cell. For example, plants growing near radioactive mining sites tend to accumulate large amounts of radioactive compounds. These compounds are not essential for the plants. (c) Nitrogen as a nutrient element is highly mobile in plants. It can be mobilised from the old and mature parts of a plant to its younger parts. (d) True Page 6 of 6

130 Class XI Chapter 13 Photosynthesis in Higher Plants Biology Question 1: By looking at a plant externally can you tell whether a plant is C 3 or C 4? Why and how? One cannot distinguish whether a plant is C 3 or C 4 by observing its leaves and other morphological features externally. Unlike C 3 plants, the leaves of C 4 plants have a special anatomy called Kranz anatomy and this difference can only be observed at the cellular level. For example, although wheat and maize are grasses, wheat is a C 3 plant, while maize is a C 4 plant. Question 2: By looking at which internal structure of a plant can you tell whether a plant is C 3 or C 4? Explain. The leaves of C 4 plants have a special anatomy called Kranz anatomy. This makes them different from C 3 plants. Special cells, known as bundle-sheath cells, surround the vascular bundles. These cells have a large number of chloroplasts. They are thick-walled and have no intercellular spaces. They are also impervious to gaseous exchange. All these anatomical features help prevent photorespiration in C 4 plants, thereby increasing their ability to photosynthesise. Question 3: Even though a very few cells in a C 4 plant carry out the biosynthetic Calvin pathway, yet they are highly productive. Can you discuss why? The productivity of a plant is measured by the rate at which it photosynthesises. The amount of carbon dioxide present in a plant is directly proportional to the rate of photosynthesis. C 4 plants have a mechanism for increasing the concentration of carbon dioxide. In C 4 plants, the Calvin cycle occurs in the bundle-sheath cells. The C 4 compound (malic acid) from the mesophyll cells is broken down in the bundle- Page 1 of 6

131 Class XI Chapter 13 Photosynthesis in Higher Plants Biology sheath cells. As a result, CO 2 is released. The increase in CO 2 ensures that the enzyme RuBisCo does not act as an oxygenase, but as a carboxylase. This prevents photorespiration and increases the rate of photosynthesis. Thus, C 4 plants are highly productive. Question 4: RuBisCo is an enzyme that acts both as a carboxylase and oxygenase. Why do you think RuBisCo carries out more carboxylation in C 4 plants? The enzyme RuBisCo is absent from the mesophyll cells of C 4 plants. It is present in the bundle-sheath cells surrounding the vascular bundles. In C 4 plants, the Calvin cycle occurs in the bundle-sheath cells. The primary CO 2 acceptor in the mesophyll cells is phosphoenol pyruvate a three-carbon compound. It is converted into the four-carbon compound oxaloacetic acid (OAA). OAA is further converted into malic acid. Malic acid is transported to the bundle-sheath cells, where it undergoes decarboxylation and CO 2 fixation occurs by the Calvin cycle. This prevents the enzyme RuBisCo from acting as an oxygenase. Question 5: Suppose there were plants that had a high concentration of Chlorophyll-b, but lacked chlorophyll-a, would it carry out photosynthesis? Then why do plants have chlorophyll-b and other accessory pigments? Chlorophyll-a molecules act as antenna molecules. They get excited by absorbing light and emit electrons during cyclic and non-cyclic photophosphorylations. They form the reaction centres for both photosystems I and II. Chlorophyll-b and other photosynthetic pigments such as carotenoids and xanthophylls act as accessory pigments. Their role is to absorb energy and transfer it to chlorophyll-a. Carotenoids and xanthophylls also protect the chlorophyll molecule from photo-oxidation. Therefore, chlorophyll-a is essential for photosynthesis. Page 2 of 6

132 Class XI Chapter 13 Photosynthesis in Higher Plants Biology If any plant were to lack chlorophyll-a and contain a high concentration of chlorophyll-b, then this plant would not undergo photosynthesis. Question 6: Why is the colour of a leaf kept in the dark frequently yellow, or pale green? Which pigment do you think is more stable? Since leaves require light to perform photosynthesis, the colour of a leaf kept in the dark changes from a darker to a lighter shade of green. Sometimes, it also turns yellow. The production of the chlorophyll pigment essential for photosynthesis is directly proportional to the amount of light available. In the absence of light, the production of chlorophyll-a molecules stops and they get broken slowly. This changes the colour of the leaf gradually to light green. During this process, the xanthophyll and carotenoid pigments become predominant, causing the leaf to become yellow. These pigments are more stable as light is not essential for their production. They are always present in plants. Question 7: Look at leaves of the same plant on the shady side and compare it with the leaves on the sunny side. Or, compare the potted plants kept in the sunlight with those in the shade. Which of them has leaves that are darker green? Why? Light is a limiting factor for photosynthesis. Leaves get lesser light for photosynthesis when they are in shade. Therefore, the leaves or plants in shade perform lesser photosynthesis as compared to the leaves or plants kept in sunlight. In order to increase the rate of photosynthesis, the leaves present in shade have more chlorophyll pigments. This increase in chlorophyll content increases the amount of light absorbed by the leaves, which in turn increases the rate of photosynthesis. Therefore, the leaves or plants in shade are greener than the leaves or plants kept in the sun. Page 3 of 6

133 Class XI Chapter 13 Photosynthesis in Higher Plants Biology Question 8: Figure shows the effect of light on the rate of photosynthesis. Based on the graph, answer the following questions: (a) At which point/s (A, B or C) in the curve is light a limiting factor? (b) What could be the limiting factor/s in region A? (c) What do C and D represent on the curve? (a) Generally, light is not a limiting factor. It becomes a limiting factor for plants growing in shade or under tree canopies. In the given graph, light is a limiting factor at the point where photosynthesis is the minimum. The least value for photosynthesis is in region A. Hence, light is a limiting factor in this region. Page 4 of 6

134 Class XI Chapter 13 Photosynthesis in Higher Plants Biology (b) Light is a limiting factor in region A. Water, temperature, and the concentration of carbon dioxide could also be limiting factors in this region. (c) Point D represents the optimum point and gives the light intensity at which the maximum photosynthesis is recorded. The rate of photosynthesis remains constant after this point, even though the intensity of light Question 9: Give comparison between the following: (a) C 3 and C 4 pathways (b) Cyclic and non-cyclic photophosphorylation (c) Anatomy of leaf in C 3 and C 4 plants (a) C 3 and C 4 pathways C 3 pathways C 4 pathways 1. The primary acceptor of CO 2 is RUBP a six-carbon compound. 1. The primary acceptor of CO 2 is phosphoenol pyruvate a threecarbon compound. 2. The first stable product is 3- phosphoglycerate. 2. The first stable product is oxaloacetic acid. 3. It occurs only in the mesophyll cells of the leaves. 3. It occurs in the mesophyll and bundle-sheath cells of the leaves. It is a slower process of carbon It is a faster process of carbon 4. fixation and photo-respiratory losses 4. fixation and photo-respiratory are high. losses are low. Page 5 of 6

135 Class XI Chapter 13 Photosynthesis in Higher Plants Biology (b) Cyclic and non-cyclic photophosphorylations Cyclic photophosphorylation Non-cyclic photophosphorylation 1. It occurs only in photosystem I. 1. It occurs in photosystems I and II. 2. It involves only the synthesis of ATP. 2. It involves the synthesis of ATP and NADPH 2. In this process, photolysis of water In this process, photolysis of 3. does not occur. Therefore, oxygen is 3. water takes place and oxygen is not produced. liberated. 4. In this process, electrons move in a closed circle. 4. In this process, electrons do not move in a closed circle. (c) Anatomy of the leaves in C 3 and C 4 plants C 3 leaves C 4 leaves 1. Bundle-sheath cells are absent 1. Bundle-sheath cells are present 2. RuBisCo is present in the mesophyll cells. 2. RuBisCo is present in the bundlesheath cells. The first stable compound produced is The first stable compound produced 3. 3-phosphoglycerate a three-carbon 3. is oxaloacetic acid a four-carbon compound. compound. 4. Photorespiration occurs 4. Photorespiration does not occur Page 6 of 6

136 Class XI Chapter 14 Respiration in Plants Biology Question 1: Differentiate between (a) Respiration and Combustion (b) Glycolysis and Krebs cycle (c) Aerobic respiration and Fermentation (a) Respiration and combustion Respiration Combustion 1. It is a biochemical process. 1. It is a physiochemical process. 2. It occurs in the living cells. 2. It does not occur in the living cells. 3. ATP is generated 3. ATP is not generated 4. Enzymes are required 4. Enzymes are not required 5. It is a biologically-controlled process. 5. It is an uncontrolled process. (b) Glycolysis and Krebs cycle Glycolysis Krebs cycle 1. It is a linear pathway. 1. It is a cyclic pathway. 2. It occurs in the cell cytoplasm. 2. It occurs in the mitochondrial matrix. 3. It occurs in both aerobic and anaerobic respiration. 3. It occurs in aerobic respiration. It generates 2 NADH 2 and 2 ATP It produces 6 NADH 2, 2FADH 2, and 2 4. molecules on the breakdown of 4. ATP molecules on the breakdown of one glucose molecule. two acetyl-coa molecules. Page 1 of 10

137 Class XI Chapter 14 Respiration in Plants Biology (c) Aerobic respiration and fermentation Aerobic respiration Fermentation 1. Oxygen is used for deriving energy 1. Occurs in the absence of oxygen 2. Occurs in the cytoplasm and mitochondria 2. Occurs in the cytoplasm 3. End products are carbon dioxide and water 3. End products are ethyl alcohol and carbon dioxide 4. Complete oxidation of the respiratory substrate takes place 4. Incomplete oxidation of the respiratory substrate takes place 5. About 36 ATP molecules are produced 5. Only 2 ATP molecules are produced Question 2: What are respiratory substrates? Name the most common respiratory substrate. The compounds oxidised during the process of respiration are called respiratory substrates. Carbohydrates, especially glucose, act as respiratory substrates. Fats, proteins, and organic acids also act as respiratory substrates. Question 3: Give the schematic representation of glycolysis? Page 2 of 10

138 Class XI Chapter 14 Respiration in Plants Biology Page 3 of 10

139 Class XI Chapter 14 Respiration in Plants Biology Question 4: What are the main steps in aerobic respiration? Where does it take place? The major steps in aerobic respiration and the sites where they occur are listed in the given table. Step Site of occurrence 1. Glycolysis 1. Cytoplasm 2. Krebs cycle 2. Matrix of mitochondria 3. Electron system transport 3. Inner mitochondrial membrane 4. Oxidative phosphorylation 4. F 0 -F 1 particles in the inner mitochondrial membrane Question 5: Give the schematic representation of an overall view of Krebs cycle. Page 4 of 10

140 Class XI Chapter 14 Respiration in Plants Biology Question 6: Explain ETS. ETS or electron transport system is located in the inner mitochondrial membrane. It helps in releasing and utilizing the energy stored in NADH+H + and FADH 2. NADH + H +, which is formed during glycolysis and citric acid cycle, gets oxidized by NADH dehydrogenase (complex I). The electrons so generated get transferred to Page 5 of 10

141 Class XI Chapter 14 Respiration in Plants Biology ubiquinone through FMN. In a similar manner, FADH 2 (complex II) generated during citric acid cycle gets transferred to ubiquinone. The electrons from ubiquinone are received by cytochrome bc 1 (complex III) and further get transferred to cytochrome c. The cytochrome c acts as a mobile carrier between complex III and cytochrome c oxidase complex, containing cytochrome a and a 3, along with copper centres (complex IV). During the transfer of electrons from each complex, the process is accompanied by the production of ATP from ADP and inorganic phosphate by the action ATP synthase (complex V). The amount of ATP produced depends on the molecule, which has been oxidized. 2 ATP molecules are produced by the oxidation of one molecule of NADH. One molecule of FADH 2, on oxidation, gives 3 ATP molecules. Page 6 of 10

142 Class XI Chapter 14 Respiration in Plants Biology Question 7: Distinguish between the following: (a) Aerobic respiration and Anaerobic respiration (b) Glycolysis and Fermentation (c) Glycolysis and Citric acid Cycle (a) Aerobic respiration and Anaerobic respiration Aerobic respiration Anaerobic respiration 1. It uses oxygen for deriving energy. 1. It occurs in the absence of oxygen. 2. It occurs in cytoplasm and mitochondria. 2. It occurs in cytoplasm. 3. The end products of aerobic respiration are carbon dioxide and water. 3. The end products of fermentation are ethyl alcohol and carbon-dioxide. 4. Complete oxidation of respiratory substrate takes place. 4. Incomplete oxidation of respiratory substrate takes place ATP molecules are produced. 5. Only 2 ATP molecules are produced. (b) Glycolysis and Fermentation Glycolysis Fermentation 1. Glycolysis occurs during aerobic and anaerobic respiration. 1. Fermentation is a type of anaerobic respiration. 2. Pyruvic acid is produced as its end product. 2. Ethanol or lactic acid is produced as its end product. Page 7 of 10

143 Class XI Chapter 14 Respiration in Plants Biology (c) Glycolysis and citric acid cycle Glycolysis Citric acid cycle (Krebs cycle) 1. It is a linear pathway. 1. It is a cyclic pathway. 2. It occurs in the cell cytoplasm. 2. It occurs in the mitochondrial matrix. 3. It occurs in both aerobic and anaerobic respiration. 3. It occurs in aerobic respiration. One glucose molecule breaks It produces 6 NADH 2, 2 FADH 2, and 2 4. down to generate 2 NADH 2 and 2 4. ATP molecules on breakdown of two ATP molecules. acetyl-coa molecules. Question 8: What are the assumptions made during the calculation of net gain of ATP? For theoretical calculation of ATP molecules, various assumptions are made, which are as follows. (a) It is assumed that various parts of aerobic respiration such as glycolysis, TCA cycle, and ETS occur in a sequential and orderly pathway. (b) NADH produced during the process of glycolysis enters into mitochondria to undergo oxidative phosphorylation. (c) Glucose molecule is assumed to be the only substrate while it is assumed that no other molecule enters the pathway at intermediate stages. (d) The intermediates produced during respiration are not utilized in any other process. Page 8 of 10

144 Class XI Chapter 14 Respiration in Plants Biology Question 9: Discuss The respiratory pathway is an amphibolic pathway. Respiration is generally assumed to be a catabolic process because during respiration, various substrates are broken down for deriving energy. Carbohydrates are broken down to glucose before entering respiratory pathways. Fats get converted into fatty acids and glycerol whereas fatty acids get converted into acetyl CoA before entering the respiration. In a similar manner, proteins are converted into amino acids, which enter respiration after deamination. During synthesis of fatty acids, acetyl CoA is withdrawn from respiratory pathway. Also, in the synthesis of proteins, respiratory substrates get withdrawn. Thus, respiration is also involved in anabolism. Therefore, respiration can be termed as amphibolic pathway as it involves both anabolism and catabolism. Question 10: Define RQ. What is its value for fats? Respiratory quotient (RQ) or respiratory ratio can be defined as the ratio of the volume of CO 2 evolved to the volume of O 2 consumed during respiration. The value of respiratory quotient depends on the type of respiratory substrate. Its value is one for carbohydrates. However, it is always less than one for fats as fats consume more oxygen for respiration than carbohydrates. It can be illustrated through the example of tripalmitin fatty acid, which consumes 145 molecules of O 2 for respiration while 102 molecules of CO 2 are evolved. The RQ value for tripalmitin is 0.7. Question 11: What is oxidative phosphorylation? Page 9 of 10

145 Class XI Chapter 14 Respiration in Plants Biology Oxidative phosphorylation is a process in which electrons are transferred from electron donors to oxygen, which acts as electron acceptor. The oxidation-reduction reactions are involved in the formation of proton gradient. The main role in oxidative phosphorylation is played by the enzyme ATP synthase (complex V). This enzyme complex consists of F 0 and F 1 components. The F 1 headpiece is a peripheral membrane protein complex and contains the site for ATP synthesis from ADP and inorganic phosphate. F 0 component is a part of membrane protein complex, which acts as a channel for crossing of the protons from inner mitochondrial membrane to the mitochondrial matrix. For every two protons passing through F 0 F 1 complex, synthesis of one ATP molecule takes place. Question 12: What is the significance of step-wise release of energy in respiration? The process of aerobic respiration is divided into four phases glycolysis, TCA cycle, ETS, and oxidative phosphorylation. It is generally assumed that the process of respiration and production of ATP in each phase takes place in a step-wise manner. The product of one pathway forms the substrate of the other pathway. Various molecules produced during respiration are involved in other biochemical processes. The respiratory substrates enter and withdraw from pathway on necessity. ATP gets utilized wherever required and enzymatic rates are generally controlled. Thus, the step-wise release of energy makes the system more efficient in extracting and storing energy. Page 10 of 10

146 Class XI Chapter 15 Plant Growth and Development Biology Question 1: Define growth, differentiation, development, dedifferentiation, redifferentiation, determinate growth, meristem and growth rate. (a) Growth It is an irreversible and permanent process, accomplished by an increase in the size of an organ or organ parts or even of an individual cell. (b) Differentiation It is a process in which the cells derived from the apical meristem (root and shoot apex) and the cambium undergo structural changes in the cell wall and the protoplasm, becoming mature to perform specific functions. (c) Development It refers to the various changes occurring in an organism during its life cycle from the germination of seeds to senescence. (d) De-differentiation It is the process in which permanent plant cells regain the power to divide under certain conditions. (e) Re-differentiation It is the process in which de-differentiated cells become mature again and lose their capacity to divide. (f) Determinate growth It refers to limited growth. For example, animals and plant leaves stop growing after having reached maturity. (g) Meristem In plants, growth is restricted to specialised regions where active cell divisions take place. Such a region is called meristem. There are three types of meristems apical meristem, lateral meristem, and intercalary meristem. (h) Growth rate It can be defined as the increased growth in plants per unit time. Page 3 of 9

147 Class XI Chapter 15 Plant Growth and Development Biology Question 2: Why is not any one parameter good enough to demonstrate growth throughout the life of a flowering plant? In plants, growth is said to have taken place when the amount of protoplasm increases. Measuring the growth of protoplasm involves many parameters such as the weight of the fresh tissue sample, the weight of the dry tissue sample, the differences in length, area, volume, and cell number measured during the growth period. Measuring the growth of plants using only one parameter does not provide enough information and hence, is insufficient for demonstrating growth. Question 3: Describe briefly: (a) Arithmetic growth (b) Geometric growth (c) Sigmoid growth curve (d) Absolute and relative growth rates (a) Arithmetic growth In arithmetic growth, one of the daughter cells continues to divide, while the other differentiates into maturity. The elongation of roots at a constant rate is an example of arithmetic growth. (b) Geometric growth Geometric growth is characterised by a slow growth in the initial stages and a rapid growth during the later stages. The daughter cells derived from mitosis retain the ability to divide, but slow down because of a limited nutrient supply. (c) Sigmoid growth curve The growth of living organisms in their natural environment is characterised by an S- shaped curve called sigmoid growth curve. This curve is divided into three phases lag phase, log phase or exponential phase of rapid growth, and stationary phase. Page 3 of 9

148 Class XI Chapter 15 Plant Growth and Development Biology Exponential growth can be expressed as: Where, W 1 = Final size W 0 = Initial size r = Growth rate t= Time of growth e = Base of natural logarithms (d) Absolute and relative growth rates Absolute growth rate refers to the measurement and comparison of total growth per unit time. Relative growth rate refers to the growth of a particular system per unit time, expressed on a common basis. Question 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. Plant growth regulators are the chemical molecules secreted by plants affecting the physiological attributes of a plant. There are five main plant growth regulators. These are: (i) Auxins (ii) Gibberellic acid (iii) Cytokinins (iv) Ethylene (v) Abscisic acid (i) Auxins Discovery: Page 3 of 9

149 Class XI Chapter 15 Plant Growth and Development Biology The first observations regarding the effects of auxins were made by Charles Darwin and Francis Darwin wherein they saw the coleoptiles of canary gross bending toward a unilateral source of light. It was concluded after a series of experiments that some substance produced at the tip of coleoptiles was responsible for the bending. Finally, this substance was extracted as auxins from the tips of coleoptiles in oat seedlings. Physiological functions: 1. They control plant cell-growth. 2. They cause the phenomenon of apical dominance. 3. They control division in the vascular cambium and xylem differentiation. 4. They induce parthenocarpy and prevent abscission of leaves and fruits. Horticultural applications: 1. They are used as the rooting hormones in stem cuttings D is used weedicide to kill broadleaf, dicotyledonous weeds. 3. They induce parthenocarpy in tomatoes. 4. They promote flowering in pineapple and litchi. (ii) Gibberellic acid Discovery: Bakane or the foolish rice seedling disease was first observed by Japanese farmers. In this disease, rice seedlings appear to grow taller than natural plants, and become slender and pale green. Later, after several experiments, it was found that this condition was caused by the infection from a certain fungus Gibberella fujikuroi. The active substance was isolated and identified as gibberellic acid. Physiological functions: 1. It causes elongation of internodes. 2. It promotes bolting in rosette plants. 3. It helps in inducing seed germination by breaking seed dormancy and initiating the synthesis of hydrolases enzymes for digesting reserve food. Horticultural applications: Page 4 of 9

150 Class XI Chapter 15 Plant Growth and Development Biology 1. It helps in increasing the sugar content in sugarcane by increasing the length of the internodes. 2. It increases the length of grape stalks. 3. It improves the shape of apple. 4. It delays senescence. 5. It hastens maturity and induces seed-production in juvenile conifers. (iii) Cytokinins Discovery: Through their experimental observations, F. Skoog and his co-workers found that the tobacco callus differentiated when extracts of vascular tissues, yeast extract, coconut milk, or DNA were added to the culture medium. This led to the discovery of cytokinins. Physiological functions: 1. They promote the growth of lateral branches by inhibiting apical dominance. 2. They help in the production of new leaves, chloroplasts, and adventitious shoots. 3. They help in delaying senescence by promoting nutrient mobilisation. Horticultural applications: 1. They are used for preventing apical dominance. 2. They are used for delaying senescence in leaves. (iv) Ethylene Discovery: It was observed that unripe bananas ripened faster when stored with ripe bananas. Later, the substance promoting the ripening was found to be ethylene. Physiological functions: 1. It helps in breaking seed and bud dormancy. 2. It promotes rapid internode-elongation in deep-water rice plants. 3. It promotes root-growth and formation of root hairs. 4. It promotes senescence and abscission of leaves and flowers. 5. It hastens the respiration rate in fruits and enhances fruit ripening. Horticultural applications: Page 5 of 9

151 Class XI Chapter 15 Plant Growth and Development Biology 1. It is used to initiate flowering and synchronising the fruit set in pineapples. 2. It induces flowering in mango. 3. Ethephon is used to ripen the fruits in tomatoes and apples, and accelerate the abscission of flowers and leaves in cotton, cherry, and walnut. 4. It promotes the number of female flowers in cucumbers. (v) Abscisic acid Discovery: During the mid 1960s, inhibitor-b, abscission II, and dormin were discovered by three independent researchers. These were later on found to be chemically similar and were thereafter called ABA (Abscisic acid). Physiological functions: 1. It acts as an inhibitor to plant metabolism. 2. It stimulates stomatal closure during water stress. 3. It induces seed dormancy. 4. It induces abscission of leaves, fruits, and flowers. Horticultural application: It induces seed dormancy in stored seeds. Question 5: What do you understand by photoperiodism and vernalisation? Describe their significance. Photoperiodism refers to the response of plants with respect to the duration of light (i.e., period of day and night). On the basis of its response to the duration of light, a plant is classified as a short-day plant, a long-day plant, or a day-neutral plant. Short-day plants flower when they are exposed to light for a period less than the critical day-length (for example: Chrysanthemum). Long-day plants flower when they are exposed to light for a period more than the critical day-length (for example: radish). When no marked correlation is observed between the duration of exposure Page 6 of 9

152 Class XI Chapter 15 Plant Growth and Development Biology to light and the flowering response, plants are termed as day-neutral plants (for example: tomato). It is hypothesised that the hormonal substance responsible for flowering is formed in the leaves, subsequently migrating to the shoot apices and modifying them into flowering apices. Photoperiodism helps in studying the response of flowering in various crop plants with respect to the duration of exposure to light. Vernalisation is the cold-induced flowering in plants. In some plants (such as the winter varieties of wheat and rye and biennials such as carrot and cabbage), exposure to low temperature is necessary for flowering to be induced. The winter varieties of rye and wheat are planted in autumn. They remain in the seedling stage during winters and flower during summers. However, when these varieties are sown in spring, they fail to flower. Similar response is seen in cabbage and radish. Question 6: Why is Abscisic acid also known as stress hormone? Abscisic acid is called stress hormones as it induces various responses in plants against stress conditions. It increases the tolerance of plants toward various stresses. It induces the closure of the stomata during water stress. It promotes seed dormancy and ensures seed germination during favourable conditions. It helps seeds withstand desiccation. It also helps in inducing dormancy in plants at the end of the growing season and promotes abscission of leaves, fruits, and flowers. Question 7: Both growth and differentiation in higher plants are open. Comment. Growth and development in higher plants is referred to as being open. This is because various meristems, having the capacity for continuously dividing and producing new cells, are present at different locations in these plant bodies. Page 7 of 9

153 Class XI Chapter 15 Plant Growth and Development Biology Question 8: Both a short day plant and a long day plant can flower simultaneously in a given place. Explain. The flowering response in short-day plants and long-day plants is dependent on the durations for which these plants are exposed to light. The short-day plant and longday plant can flower at the same place, provided they have been given an adequate photoperiod. Question 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. (a) Induce rooting in a twig Auxins (b) Quickly ripen a fruit Ethylene (c) Delay leaf senescence Cytokinins (d) Induce growth in axillary buds Cytokinins (e) Bolt a rosette plant Gibberellic acid (f) Induce immediate stomatal closure in leaves Abscisic acid Question 10: Would a defoliated plant respond to photoperiodic cycle? Why? A defoliated plant will not respond to the photoperiodic cycle. Page 8 of 9

154 Class XI Chapter 15 Plant Growth and Development Biology It is hypothesised that the hormonal substance responsible for flowering is formed in the leaves, subsequently migrating to the shoot apices and modifying them into flowering apices. Therefore, in the absence of leaves, light perception would not occur, i.e., the plant would not respond to light. Question 11: What would be expected to happen if: (a) GA 3 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. (a) If GA 3 is applied to rice seedlings, then the rice seedlings will exhibit internodeelongation and increase in height. (b) If dividing cells stop differentiating, then the plant organs such as leaves and stem will not be formed. The mass of undifferentiated cell is called callus. (c) If a rotten fruit gets mixed with unripe fruits, then the ethylene produced from the rotten fruits will hasten the ripening of the unripe fruits. (d) If you forget to add cytokinin to the culture medium, then cell division, growth, and differentiation will not be observed. Page 9 of 9

155 Class XI Chapter 16 Digestion and Absorption Biology Question 1: Choose the correct answer among the following: (a) Gastric juice contains (i) pepsin, lipase and rennin (ii) trypsin lipase and rennin (iii) trypsin, pepsin and lipase (iv) trypsin, pepsin and renin (b) Succus entericus is the name given to (i) a junction between ileum and large intestine (ii) intestinal juice (iii) swelling in the gut (iv) appendix (a): (i) Pepsin, lipase, and rennin Gastric juice contains pepsin, lipase, and rennin. Pepsin is secreted in an inactive form as pepsinogen, which is activated by HCl. Pepsin digests proteins into peptones. Lipase breaks down fats into fatty acids. Rennin is a photolytic enzyme present in the gastric juice. It helps in the coagulation of milk. (b): (ii) Intestinal juice Succus entericus is another name for intestinal juice. It is secreted by the intestinal gland. Intestinal juice contains a variety of enzymes such as maltase, lipases, nucleosidases, dipeptidases, etc. Question 2: Match column I with column II Column I Column II (a) Bilirubin and biliverdin (i) Parotid (b) Hydrolysis of starch (ii) Bile Page 1 of 12

156 Class XI Chapter 16 Digestion and Absorption Biology (c) Digestion of fat (iii) Lipases (d) Salivary gland (iv) Amylases Column I Column II (a) Bilirubin and biliverdin (ii) Bile (b) Hydrolysis of starch (iv) Amylases (c) Digestion of fat (iii) Lipases (d) Salivary gland (i) Parotid Question 3: briefly: (a) Why are villi present in the intestine and not in the stomach? (b) How does pepsinogen change into its active form? (c) What are the basic layers of the wall of alimentary canal? (d) How does bile help in the digestion of fats? Page 2 of 12

157 Class XI Chapter 16 Digestion and Absorption Biology (a) The mucosal wall of the small intestine forms millions of tiny finger-like projections known as villi. These villi increase the surface area for more efficient food absorption. Within these villi, there are numerous blood vessels that absorb the digested products of proteins and carbohydrates, carrying them to the blood stream. The villi also contain lymph vessels for absorbing the products of fat-digestion. From the blood stream, the absorbed food is finally delivered to each and every cell of the body. The mucosal walls of the stomach form irregular folds known as rugae. These help increase the surface area to volume ratio of the expanding stomach. (b) Pepsinogen is a precursor of pepsin stored in the stomach walls. It is converted into pepsin by hydrochloric acid. Pepsin is the activated in the form of pepsinogen. Pepsinogen Pepsin + Inactive peptide (Inactive) (Active) (c) The walls of the alimentary canal are made up of four layers. These are as follows: (i) Serosa is the outermost layer of the human alimentary canal. It is made up of a thin layer of secretory epithelial cells, with some connective tissues underneath. (ii) Muscularis is a thin layer of smooth muscles arranged into an outer longitudinal layer and an inner circular layer. (iii) Sub-mucosa is a layer of loose connective tissues, containing nerves, blood, and lymph vessels. It supports the mucosa. Page 3 of 12

158 Class XI Chapter 16 Digestion and Absorption Biology iv. Mucosa is the innermost lining of the lumen of the alimentary canal. It is mainly involved in absorption and secretion. (d) Bile is a digestive juice secreted by the liver and stored in the gall bladder. Bile juice has bile salts such as bilirubin and biliverdin. These break down large fat globules into smaller globules so that the pancreatic enzymes can easily act on them. This process is known as emulsification of fats. Bile juice also makes the medium alkaline and activates lipase. Question 4: State the role of pancreatic juice in digestion of proteins. Pancreatic juice contains a variety of inactive enzymes such as trypsinogen, chymotrypsinogen, and carboxypeptidases. These enzymes play an important role in the digestion of proteins. Physiology of protein-digestion The enzyme enterokinase is secreted by the intestinal mucosa. It activates trypsinogen into trypsin. Trypsinogen Trypsin + Inactive peptide Trypsin then activates the other enzymes of pancreatic juice such as chymotrypsinogen and carboxypeptidase. Chymotrypsinogen is a milk-coagulating enzyme that converts proteins into peptides. Chymotrypsinogen (Inactive) Chymotrypsin (Active) Proteins Peptides Carboxypeptidase acts on the carboxyl end of the peptide chain and helps release the last amino acids. Hence, it helps in the digestion of proteins. Peptides Smaller peptide chain + Amino acids Page 4 of 12

159 Class XI Chapter 16 Digestion and Absorption Biology Thus, in short, we can say that the partially-hydrolysed proteins present in the chyme are acted upon by various proteolytic enzymes of the pancreatic juice for their complete digestion. Proteins, peptones Dipeptides and proteases Question 5: Describe the process of digestion of protein in stomach. The digestion of proteins begins in the stomach and is completed in the small intestine. The digestive juice secreted in the gastric glands present on the stomach walls is called gastric juice. The food that enters the stomach becomes acidic on mixing with this gastric juice. The main components of gastric juice are hydrochloric acid, pepsinogen, mucus, and rennin. Hydrochloric acid dissolves the bits of food and creates an acidic medium so that pepsinogen is converted into pepsin. Pepsin is a protein- digesting enzyme. It is secreted in its inactive form called pepsinogen, which then gets activated by hydrochloric acid. The activated pepsin then converts proteins into proteases and peptides. Proteins Proteases + Peptides Rennin is a proteolytic enzyme, released in an inactive form called prorennin. Rennin plays an important role in the coagulation of milk. Page 5 of 12

160 Class XI Chapter 16 Digestion and Absorption Biology Question 6: Given the dental formula of human beings The dental formula expresses the arrangement of teeth in each half of the upper jaw and the lower jaw. The entire formula is multiplied by two to express the total number of teeth. The dental formula for milk teeth in humans is: Each half of the upper jaw and the lower jaw has 2 incisors, 1 canine, and 2 molars. Premolars are absent in milk teeth. The dental formula for permanent teeth in humans is: Each half of the upper jaw and the lower jaw has 2 incisors, 1 canine, 2 premolars, and 3 molars. An adult human has 32 permanent teeth. Question 7: Bile juice contains no digestive enzymes, yet it is important for digestion. Why? Bile is a digestive juice secreted by the liver. Although it does not contain any digestive enzymes, it plays an important role in the digestion of fats. Bile juice has bile salts such as bilirubin and biliverdin. These break down large fat globules into smaller globules so that the pancreatic enzymes can easily act on them. This process is known as emulsification of fats. Bile juice also makes the medium alkaline and activates lipase. Question 8: Describe the digestive role of chymotrypsin. What two other digestive enzymes of the same category are secreted by its source gland? Page 6 of 12

161 Class XI Chapter 16 Digestion and Absorption Biology The enzyme trypsin (present in the pancreatic juice) activates the inactive enzyme chymotrypsinogen into chymotrypsin. Chymotrypsinogen Chymotrypsin (Inactive) (Active) The activated chymotrypsin plays an important role in the further breakdown of the partially-hydrolysed proteins. Proteins Peptides The other digestive enzymes of the same category are trypsinogen and carboxypeptidase. These are secreted by the same source-gland, pancreas. Trypsinogen is present in an inactive form in the pancreatic juice. The enzyme enterokinase secreted by the intestinal mucosa activates trypsinogen into trypsin. Trypsinogen Trypsin + Inactive peptide The activated trypsin then further hydrolyses the remaining trypsinogen and activates other pancreatic enzymes such as chymotrypsinogen and carboxypeptidase. Trypsin also helps in breaking down proteins into peptides. Proteins Peptides Carboxypeptidases act on the carboxyl end of the peptide chain and help in releasing the last amino acids. Peptides Small peptide chain + Amino acids Question 9: How are polysaccharides and disaccharides digested? The digestion of carbohydrates takes place in the mouth and the small intestine region of the alimentary canal. The enzymes that act on carbohydrates are collectively known as carbohydrases. Digestion in the mouth: Page 7 of 12

162 Class XI Chapter 16 Digestion and Absorption Biology As food enters the mouth, it gets mixed with saliva. Saliva secreted by the salivary glands contains a digestive enzyme called salivary amylase. This enzyme breaks down starch into sugar at ph 6.8. Starch Maltose + Isomaltose + Limit dextrins Salivary amylase continues to act in the oesophagus, but its action stops in the stomach as the contents become acidic. Hence, carbohydrate-digestion stops in the stomach. Digestion in the small intestine: Carbohydrate-digestion is resumed in the small intestine. Here, the food gets mixed with the pancreatic juice and the intestinal juice. Pancreatic juice contains the pancreatic amylase that hydrolyses the polysaccharides into disaccharides. Starch Disaccharides (Polysaccharides) Similarly, the intestinal juice contains a variety of enzymes (disaccharidases such as maltase, lactase, sucrase, etc.). These disaccharidases help in the digestion of disaccharides. The digestion of carbohydrates is completed in the small intestine. Maltose Lactose Sucrose 2Glucose Glucose + Galactose Glucose + Fructose Question 10: What would happen if HCl were not secreted in the stomach? Hydrochloric acid is secreted by the glands present on the stomach walls. It dissolves bits of food and creates an acidic medium. The acidic medium allows pepsinogen to be converted into pepsin. Pepsin plays an important role in the digestion of proteins. Therefore, if HCl were not secreted in the stomach, then pepsin would not be Page 8 of 12

163 Class XI Chapter 16 Digestion and Absorption Biology activated. This would affect protein digestion. A ph of about 1.8 is necessary for proteins to be digested. This ph is achieved by HCl. Question 11: How does butter in your food gets digested and absorbed in the body? Digestion of fats: Butter is a fat product and gets digested in the small intestine. The bile juice secreted by the liver contains bile salts that break down large fat globules into smaller globules, so as to increase their surface area for the action of lipase. This process is referred to as emulsification of fats. After this, the pancreatic lipase present in the pancreatic juice and the intestinal lipase present in the intestinal juice hydrolyse the fat molecules into triglycerides, diglycerides, monoglycerides, and ultimately into glycerol. Fats Triglycerides + Diglycerides Diglycerides and monoglycerides Fatty acids + Glycerol Absorption of fats: Fat absorption is an active process. During fat digestion, fats are hydrolysed into fatty acids and glycerol. However, since these are water insoluble, they cannot be directly absorbed by the blood. Hence, they are first incorporated into small droplets called micelles and then transported into the villi of the intestinal mucosa. They are then reformed into small microscopic particles called chylomicrons, which are small, protein-coated fat globules. These chylomicrons are transported to the lymph vessels in the villi. From the lymph vessels, the absorbed food is finally released into the blood stream and from the blood stream, to each and every cell of the body. Question 12: Page 9 of 12

164 Class XI Chapter 16 Digestion and Absorption Biology Discuss the main steps in the digestion of proteins as the food passes through different parts of the alimentary canal. The digestion of proteins begins in the stomach and is completed in the small intestine. The enzymes that act on proteins are known as proteases. Digestion in the stomach: The digestive juice secreted in the gastric glands present on the stomach walls is called gastric juice. The main components of gastric juice are HCl, pepsinogen, and rennin. The food that enters the stomach becomes acidic on mixing with this gastric juice. The acidic medium converts inactive pepsinogen into active pepsin. The active pepsin then converts proteins into proteases and peptides. Proteins Proteases + Peptides The enzyme rennin plays an important role in the coagulation of milk. Digestion in the small intestine: The food from the stomach is acted upon by three enzymes present in the small intestine pancreatic juice, intestinal juice (known as succus entericus), and bile juice. Action of pancreatic juice Pancreatic juice contains a variety of inactive enzymes such as trypsinogen, chymotrypsinogen, and carboxypeptidases. The enzymes are present in an inactivated state. The enzyme enterokinase secreted by the intestinal mucosa activates trypsinogen into trypsin. Trypsinogen Trypsin + Inactive peptide The activated trypsin then activates the other enzymes of pancreatic juice. Chymotrypsinogen is a proteolytic enzyme that breaks down proteins into peptides. Chymotrypsinogen Proteins Peptides Chymotrypsin Page 10 of 12

165 Class XI Chapter 16 Digestion and Absorption Biology Carboxypeptidases act on the carboxyl end of the peptide chain and help in releasing the last amino acids. Peptides Smaller peptide chain + Amino acids Action of bile juice Bile juice has bile salts such as bilirubin and biliverdin which break down large, fat globules into smaller globules so that pancreatic enzymes can easily act on them. This process is known as emulsification of fats. Bile juice also makes the medium alkaline and activates lipase. Lipase then breaks down fats into diglycerides and monoglycerides. Action of intestinal juice Intestinal juice contains a variety of enzymes. Pancreatic amylase digests polysaccharides into disaccharides. Disaccharidases such as maltase, lactase, sucrase, etc., further digest the disaccharides. The proteases hydrolyse peptides into dipeptides and finally into amino acids. Dipeptides Amino acids Pancreatic lipase breaks down fats into diglycerides and monoglycerides. The nucleases break down nucleic acids into nucleotides and nucleosides. Question 13: Explain the term thecodont and diphyodont. Thecodont is a type of dentition in which the teeth are embedded in the deep sockets of the jaw bone. Ankylosis is absent and the roots are cylindrical. Examples include living crocodilians and mammals. Diphyodont is a type of dentition in which two successive sets of teeth are developed during the lifetime of the organism. The first set of teeth is deciduous and the other set is permanent. The deciduous set of teeth is replaced by the permanent adult teeth. Page 11 of 12

166 Class XI Chapter 16 Digestion and Absorption Biology This type of dentition can be seen in humans. Question 14: Name different types of teeth and their number in an adult human. There are four different types of teeth in an adult human. They are as follows: (i) Incisors The eight teeth in the front are incisors. There are four incisors each in the upper jaw and the lower jaw. They are meant for cutting. (ii) Canines The pointy teeth on either side of the incisors are canines. They are four in number, two each placed in the upper jaw and the lower jaw. They are meant for tearing. (iii) Premolars They are present next to the canines. They are eight in number, four each placed in the upper jaw and the lower jaw. They are meant for grinding. (iv) Molars They are present at the end of the jaw, next to the premolars. There are twelve molars, six each placed in the upper jaw and the lower jaw. Hence, the dental formula in humans is This means each half of the upper jaw and the lower jaw has 2 incisors, 1 canine, 2 premolars, and 3 molars. Hence, an adult human has 32 permanent teeth. Question 15: What are the functions of liver? Liver is the largest and heaviest internal organ of the body. It is not directly involved in digestion, but secretes digestive juices. It secretes bile which plays a major role in the emulsification of fats. Page 12 of 12

167 Class XI Chapter 17 Breathing and Exchange of Gases Biology Question 1: Define vital capacity. What is its significance? Vital capacity is the maximum volume of air that can be exhaled after a maximum inspiration. It is about litres in the human body. It promotes the act of supplying fresh air and getting rid of foul air, thereby increasing the gaseous exchange between the tissues and the environment. Question 2: State the volume of air remaining in the lungs after a normal breathing. The volume of air remaining in the lungs after a normal expiration is known as functional residual capacity (FRC). It includes expiratory reserve volume (ERV) and residual volume (RV). ERV is the maximum volume of air that can be exhaled after a normal expiration. It is about 1000 ml to 1500 ml. RV is the volume of air remaining in the lungs after maximum expiration. It is about 1100 ml to 1500 ml. FRC = ERV + RV ml Functional residual capacity of the human lungs is about ml. Question 3: Diffusion of gases occurs in the alveolar region only and not in the other parts of respiratory system. Why? Each alveolus is made up of highly-permeable and thin layers of squamous epithelial cells. Similarly, the blood capillaries have layers of squamous epithelial cells. Oxygen-rich air enters the body through the nose and reaches the alveoli. The deoxygenated (carbon dioxide-rich) blood from the body is brought to the heart by the veins. The heart pumps it to the lungs for oxygenation. The exchange of O 2 and Page 1 of 9

168 Class XI Chapter 17 Breathing and Exchange of Gases Biology CO 2 takes place between the blood capillaries surrounding the alveoli and the gases present in the alveoli. Thus, the alveoli are the sites for gaseous exchange. The exchange of gases takes place by simple diffusion because of pressure or concentration differences. The barrier between the alveoli and the capillaries is thin and the diffusion of gases takes place from higher partial pressure to lower partial pressure. The venous blood that reaches the alveoli has lower partial pressure of O 2 and higher partial pressure of CO 2 as compared to alveolar air. Hence, oxygen diffuses into blood. Simultaneously, carbon dioxide diffuses out of blood and into the alveoli. Question 4: What are the major transport mechanisms for CO 2? Explain. Plasma and red blood cells transport carbon dioxide. This is because they are readily soluble in water. (1) Through plasma: About 7% of CO 2 is carried in a dissolved state through plasma. Carbon dioxide combines with water and forms carbonic acid. Since the process of forming carbonic acid is slow, only a small amount of carbon dioxide is carried this way. (2) Through RBCs: About 20 25% of CO 2 is transported by the red blood cells as carbaminohaemoglobin. Carbon dioxide binds to the amino groups on the polypeptide chains of haemoglobin and forms a compound known as carbaminohaemoglobin. (3) Through sodium bicarbonate: About 70% of carbon dioxide is transported as sodium bicarbonate. As CO 2 diffuses into the blood plasma, a large part of it combines with water to form carbonic acid in Page 2 of 9

169 Class XI Chapter 17 Breathing and Exchange of Gases Biology the presence of the enzyme carbonic anhydrase. Carbonic anhydrase is a zinc enzyme that speeds up the formation of carbonic acid. This carbonic acid dissociates into bicarbonate (HCO 3 ) and hydrogen ions (H + ). Question 5: What will be the po 2 and pco 2 in the atmospheric air compared to those in the alveolar air? (i) po 2 lesser, pco 2 higher (ii) po 2 higher, pco 2 lesser (iii) po 2 higher, pco 2 higher (iv) po 2 lesser, pco 2 lesser : (ii) po 2 higher, pco 2 lesser The partial pressure of oxygen in atmospheric air is higher than that of oxygen in alveolar air. In atmospheric air, po 2 is about 159 mm Hg. In alveolar air, it is about 104 mm Hg. The partial pressure of carbon dioxide in atmospheric air is lesser than that of carbon dioxide in alveolar air. In atmospheric air, pco 2 is about 0.3 mmhg. In alveolar air, it is about 40 mm Hg. Question 6: Explain the process of inspiration under normal conditions. Page 3 of 9

170 Class XI Chapter 17 Breathing and Exchange of Gases Biology Inspiration or inhalation is the process of bringing air from outside the body into the lungs. It is carried out by creating a pressure gradient between the lungs and the atmosphere. When air enters the lungs, the diaphragm expands toward the abdominal cavity, thereby increasing the space in the thoracic cavity for accommodating the inhaled air. The volume of the thoracic chamber in the anteroposterior axis increases with the simultaneous contraction of the external intercostal muscles. This causes the ribs and the sternum to move out, thereby increasing the volume of the thoracic chamber in the dorsoventral axis. The overall increase in the thoracic volume leads to a similar increase in the pulmonary volume. Now, as a result of this increase, the intra-pulmonary pressure becomes lesser than the atmospheric pressure. This causes the air from outside the body to move into the lungs. Question 7: How is respiration regulated? The respiratory rhythm centre present in the medulla region of the brain is primarily responsible for the regulation of respiration. The pneumotaxic centre can alter the function performed by the respiratory rhythm centre by signalling to reduce the inspiration rate. The chemosensitive region present near the respiratory centre is sensitive to carbon dioxide and hydrogen ions. This region then signals to change the rate of expiration for eliminating the compounds. The receptors present in the carotid artery and aorta detect the levels of carbon dioxide and hydrogen ions in blood. As the level of carbon dioxide increases, the respiratory centre sends nerve impulses for the necessary changes. Page 4 of 9

171 Class XI Chapter 17 Breathing and Exchange of Gases Biology Question 8: What is the effect of pco 2 on oxygen transport? pco 2 plays an important role in the transportation of oxygen. At the alveolus, the low pco 2 and high po 2 favours the formation of haemoglobin. At the tissues, the high pco 2 and low po 2 favours the dissociation of oxygen from oxyhaemoglobin. Hence, the affinity of haemoglobin for oxygen is enhanced by the decrease of pco 2 in blood. Therefore, oxygen is transported in blood as oxyhaemoglobin and oxygen dissociates from it at the tissues. Question 9: What happens to the respiratory process in a man going up a hill? As altitude increases, the oxygen level in the atmosphere decreases. Therefore, as a man goes uphill, he gets less oxygen with each breath. This causes the amount of oxygen in the blood to decline. The respiratory rate increases in response to the decrease in the oxygen content of blood. Simultaneously, the rate of heart beat increases to increase the supply of oxygen to blood. Question 10: What is the site of gaseous exchange in an insect? In insects, gaseous exchange occurs through a network of tubes collectively known as the tracheal system. The small openings on the sides of an insect s body are known as spiracles. Oxygen-rich air enters through the spiracles. The spiracles are connected to the network of tubes. From the spiracles, oxygen enters the tracheae. From here, oxygen diffuses into the cells of the body. The movement of carbon dioxide follows the reverse path. The CO 2 from the cells of the body first enters the tracheae and then leaves the body through the spiracles. Page 5 of 9

172 Class XI Chapter 17 Breathing and Exchange of Gases Biology Question 11: Define oxygen dissociation curve. Can you suggest any reason for its sigmoidal pattern? The oxygen dissociation curve is a graph showing the percentage saturation of oxyhaemoglobin at various partial pressures of oxygen. The curve shows the equilibrium of oxyhaemoglobin and haemoglobin at various partial pressures. In the lungs, the partial pressure of oxygen is high. Hence, haemoglobin binds to oxygen and forms oxyhaemoglobin. Tissues have a low oxygen concentration. Therefore, at the tissues, oxyhaemoglobin releases oxygen to form haemoglobin. The sigmoid shape of the dissociation curve is because of the binding of oxygen to haemoglobin. As the first oxygen molecule binds to haemoglobin, it increases the affinity for the second molecule of oxygen to bind. Subsequently, haemoglobin attracts more oxygen. Question 12: Have you heard about hypoxia? Try to gather information about it, and discuss with your friends. Hypoxia is a condition characterised by an inadequate or decreased supply of oxygen to the lungs. It is caused by several extrinsic factors such as reduction in po 2, inadequate oxygen, etc. The different types of hypoxia are discussed below. Hypoxemic hypoxia In this condition, there is a reduction in the oxygen content of blood as a result of the low partial pressure of oxygen in the arterial blood. Anaemic hypoxia In this condition, there is a reduction in the concentration of haemoglobin. Stagnant or ischemic hypoxia Page 6 of 9

173 Class XI Chapter 17 Breathing and Exchange of Gases Biology In this condition, there is a deficiency in the oxygen content of blood because of poor blood circulation. It occurs when a person is exposed to cold temperature for a prolonged period of time. Histotoxic hypoxia In this condition, tissues are unable to use oxygen. This occurs during carbon monoxide or cyanide poisoning. Question 13: Distinguish between (a) IRV and ERV (b) Inspiratory capacity and Expiratory capacity (c) Vital capacity and Total lung capacity (a) Inspiratory reserve volume (IRV) Expiratory reserve volume (ERV) 1. It is the maximum volume of air that can be inhaled after a normal inspiration. 2. It is about ml in the human lungs. 1. It is the maximum volume of air that can be exhaled after a normal expiration. 2. It is about ml in the human lungs. (b) Inspiratory capacity (IC) Expiratory capacity (EC) Page 7 of 9

174 Class XI Chapter 17 Breathing and Exchange of Gases Biology 1. It is the volume of air that can be inhaled after a normal expiration. 2. It includes tidal volume and inspiratory reserve volume. IC = TV + IRV 1. It is the volume of air that can be exhaled after a normal inspiration. 2. It includes tidal volume and expiratory reserve volume. EC = TV + ERV (c) Vital capacity (VC) Total lung capacity (TLC) 1. It is the maximum volume of air that can be exhaled after a maximum inspiration. It includes IC and ERV. 2. It is about 4000 ml in the human lungs. 1. It is the volume of air in the lungs after maximum inspiration. It includes IC, ERV, and residual volume. 2. It is about ml in the human lungs. Question 14: What is Tidal volume? Find out the Tidal volume (approximate value) for a healthy human in an hour. Tidal volume is the volume of air inspired or expired during normal respiration. It is about 6000 to 8000 ml of air per minute. The hourly tidal volume for a healthy human can be calculated as: Tidal volume = 6000 to 8000 ml/minute Tidal volume in an hour = 6000 to 8000 ml (60 min) = ml to ml Page 8 of 9

175 Class XI Chapter 17 Breathing and Exchange of Gases Biology Therefore, the hourly tidal volume for a healthy human is approximately ml to ml. Page 9 of 9

176 Class XI Chapter 18 Body Fluids and Circulation Biology Question 1: Name the components of the formed elements in the blood and mention one major function of each of them. The component elements in the blood are: (1) Erythrocytes: They are the most abundant cells and contain the red pigment called haemoglobin. They carry oxygen to all parts of the body. Red blood cells are produced continuously in some parts of the body such as the marrow of long bones, ribs, etc. There are about 4 6 million RBCs per cubic millimetre of blood. (2) Leukocytes Leucocytes are colourless cells. These cells do not contain haemoglobin. They are the largest cells of the body and are divided into two main categories. (a) Granulocytes These leucocytes have granules in their cytoplasm and include neutrophils, eosinophils, and basophiles. Neutrophils are phagocytic cells that protect the body against various infecting agents. Eosinophils are associated with allergic reactions, while basophiles are involved in inflammatory responses. (b) Agranulocytes Lymphocytes and monocytes are agranulocytes. Lymphocytes generate immune responses against infecting agents, while monocytes are phagocytic in nature. (3) Platelets Platelets are small irregular bodies present in blood. They contain essential chemicals that help in clotting. The main function of platelets is to promote clotting. Question 2: What is the importance of plasma proteins? Page 1 of 12

177 Class XI Chapter 18 Body Fluids and Circulation Biology Plasma is the colourless fluid of blood which helps in the transport of food, CO 2, waste products, and salts. It constitutes about 55% of blood. About 6.8% of the plasma is constituted by proteins such as fibrinogens, globulins, and albumins. Fibrinogen is a plasma glycoprotein synthesised by the liver. It plays a role in the clotting of blood. Globulin is a major protein of the plasma. It protects the body against infecting agents. Albumin is a major protein of the plasma. It helps in maintaining the fluid volume within the vascular space. Question 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) Systole (v) Gas transport Column I Column II (a) Eosinophils (iii) Resist infections (b) RBC (v) Gas transport (c) AB Group (ii) Universal Recipient Page 2 of 12

178 Class XI Chapter 18 Body Fluids and Circulation Biology (d) Platelets (i) Coagulation (e) Systole (iv) Contraction of heart Question 4: Why do we consider blood as a connective tissue? Connective tissues have cells scattered throughout an extra-cellular matrix. They connect different body systems. Blood is considered as a type of connective tissue because of two reasons. (i) Like the other connective tissues, blood is mesodermal in origin. (ii) It connects the body systems, transports oxygen and nutrients to all the parts of the body, and removes the waste products. Blood has an extra-cellular matrix called plasma, with red blood cells, white blood cells, and platelets floating in it. Question 5: What is the difference between lymph and blood? Lymph Blood 1. It is a colourless fluid that does not contain RBCs. 1. It is a red-coloured fluid that contains RBCs. 2. It contains plasma and lesser number of WBCs and platelets. 2. It contains plasma, RBCs, WBCs, and platelets. 3. It helps in body defence and is a part of the immune system. 3. It is associated with the circulation of oxygen and carbon dioxide. 4. Its plasma lacks proteins. 4. Its plasma has proteins, calcium, Page 3 of 12

179 Class XI Chapter 18 Body Fluids and Circulation Biology and phosphorus. It transports nutrients from the tissue It transports nutrients and 5. cells to the blood, through lymphatic 5. oxygen from one organ to vessels. another. 6. The flow of lymph is slow. 6. The flow of blood in the blood vessels is fast. Question 6: What is meant by double circulation? What is its significance? Double circulation is a process during which blood passes twice through the heart during one complete cycle. This type of circulation is found in amphibians, reptiles, birds, and mammals. However, it is more prominent in birds and mammals as in them the heart is completely divided into four chambers the right atrium, the right ventricle, the left atrium, and the left ventricle. The movement of blood in an organism is divided into two parts: (i) Systemic circulation (ii) Pulmonary circulation Systemic circulation involves the movement of oxygenated blood from the left ventricle of the heart to the aorta. It is then carried by blood through a network of arteries, arterioles, and capillaries to the tissues. From the tissues, the deoxygenated blood is collected by the venules, veins, and vena cava, and is emptied into the left auricle. Pulmonary circulation involves the movement of deoxygenated blood from the right ventricle to the pulmonary artery, which then carries blood to the lungs for oxygenation. From the lungs, the oxygenated blood is carried by the pulmonary veins into the left atrium.

180 Class XI Chapter 18 Body Fluids and Circulation Biology Hence, in double circulation, blood has to pass alternately through the lungs and the tissues. Significance of double circulation: The separation of oxygenated and deoxygenated blood allows a more efficient supply of oxygen to the body cells. Blood is circulated to the body tissues through systemic circulation and to the lungs through pulmonary circulation. Question 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 (a) Blood and lymph Blood Lymph 1. Blood is a red-coloured fluid that contains RBCs. 1. Lymph is a colourless fluid that lacks RBCs. 2. It contains plasma, RBCs, WBCs, and platelets. It also contains proteins. 2. It contains plasma and lesser number of WBCs and platelets. It lacks proteins. Blood transports nutrients and Lymph plays a role in the defensive 3. oxygen from one organ to 3. system of the body. It is a part of the another. immune system. Page 5 of 12

181 Class XI Chapter 18 Body Fluids and Circulation Biology (b) Open and closed systems of circulation Open system of circulation Closed system of circulation In this system, blood is pumped by In this system, blood is pumped by 1. the heart, through large vessels, into 1. the heart, through a closed body cavities called sinuses. network of vessels. 2. The body tissues are in direct contact with blood. 2. The body tissues are not in direct contact with blood. Blood flows at low pressure. Hence, it Blood flows at high pressure. 3. is a slower and less efficient system 3. Hence, it is a faster and more of circulation. efficient system of circulation. 4. The flow of blood is not regulated through the tissues and organs. 4. The flow of blood can be regulated by valves. 5. This system is present in arthropods and molluscs. 5. This system is present in annelids, echinoderms, and vertebrates. (c) Systole and diastole Systole Diastole 1. It is the contraction of the heart chambers to drive blood into the aorta and the pulmonary artery. 1. It is the relaxation of the heart chambers between two contractions. During diastole, the chambers are filled with blood. Systole decreases the volume of Diastole brings the heart chambers back 2. the heart chambers and forces 2. into their original sizes to receive more the blood out of them. blood. Page 6 of 12

182 Class XI Chapter 18 Body Fluids and Circulation Biology (d) P-wave and T-wave P-wave T-wave In an electrocardiogram (ECG), the P- In an electrocardiogram (ECG), 1. wave indicates the activation of the SA 1. the T-wave represents node. ventricular relaxation. During this phase, the impulse of During this phase, the ventricles 2. contraction is generated by the SA 2. relax and return to their normal node, causing atrial depolarisation. state. 3. It is of atrial origin. 3. It is of ventricular origin. Question 8: Describe the evolutionary change in the pattern of heart among the vertebrates. All vertebrates possess a heart a hollow muscular organ composed of cardiac muscle fibres. The function of the heart is to pump oxygen to all parts of the body. The evolution of the heart is based on the separation of oxygenated blood from deoxygenated blood for efficient oxygen transport. In fishes, the heart was like a hollow tube. This evolved into the four-chambered heart in mammals. Piscean heart Fish has only two chambers in its heart one auricle and one ventricle. Since both the auricle and the ventricle remain undivided, only deoxygenated blood passes through it. The deoxygenated blood enters the gills for oxygenation from the ventricle. It has additional chambers such as sinus venosus and conus arteriosus. Page 7 of 12

183 Class XI Chapter 18 Body Fluids and Circulation Biology Amphibian heart Amphibians, such as frogs, have three-chambered hearts, with two auricles and one ventricle. The auricle is divided into a right and a left chamber by an inter-auricular septum, while the ventricle remains undivided. Additional chambers such as sinus venosus and conus arteriosus are also present. The oxygenated blood from the lungs enters the left auricle and simultaneously, the deoxygenated blood from the body enters the right auricle. Both these auricles empty into the ventricle, wherein the oxygenated and deoxygenated blood get mixed to some extent. Reptilian heart Reptiles have incomplete four-chambered hearts, except for crocodiles, alligators, and gharials. They have only one accessory chamber called sinus venosus. The reptilian heart also shows mixed blood circulation. Page 8 of 12

184 Class XI Chapter 18 Body Fluids and Circulation Biology Avian and mammalian hearts They have two pairs of chambers for separating oxygenated and deoxygenated bloods. The heart is divided into four chambers. The upper two chambers are called atria and the lower two chambers are called ventricles. The chambers are separated by a muscular wall that prevents the mixing of the blood rich in oxygen with the blood rich in carbon dioxide. Question 9: Why do we call our heart myogenic? In the human heart, contraction is initiated by a special modified heart muscle known as sinoatrial node. It is located in the right atrium. The SA node has the inherent Page 9 of 12

185 Class XI Chapter 18 Body Fluids and Circulation Biology power of generating a wave of contraction and controlling the heart beat. Hence, it is known as the pacemaker. Since the heart beat is initiated by the SA node and the impulse of contraction originates in the heart itself, the human heart is termed myogenic. The hearts of vertebrates and molluscs are also myogenic. Question 10: Sino-atrial node is called the pacemaker of our heart. Why? The sino-atrial (SA) node is a specialised bundle of neurons located in the upper part of the right atrium of the heart. The cardiac impulse originating from the SA node triggers a sequence of electrical events in the heart, thereby controlling the sequence of muscle contraction that pumps blood out of the heart. Since the SA node initiates and maintains the rhythmicity of the heart, it is known as the natural pacemaker of the human body. Question 11: What is the significance of atrio-ventricular node and atrio-ventricular bundle in the functioning of heart? The atrioventricular (AV) node is present in the right atrium, near the base of the inter-auricular septum that separates the right auricle from the ventricle. It gives rise to the bundle of His that conducts the cardiac impulses from the auricles to the ventricles. As the bundle of His passes the ventricle along the inter-ventricular septum, it divides into two branches the right ventricle and the left ventricle. The end branches of this conducting system then forms a network of Purkinje fibres that penetrate into the myocardium. The auricular contraction initiated by the wave of excitation from the sino-atrial node (SA node) stimulates the atrio-ventricular node, thereby leading to the contraction of ventricles through the bundle of His and Purkinje fibres. Hence, the atrio-ventricular node and the atrioventricular bundle play a role in the contraction of ventricles. Page 10 of 12

186 Class XI Chapter 18 Body Fluids and Circulation Biology Question 12: Define a cardiac cycle and the cardiac output. Cardiac cycle is defined as the complete cycle of events in the heart from the beginning of one heart beat to the beginning of the next. It comprises three stages atrial systole, ventricular systole, and complete cardiac diastole. Cardiac output is defined as the amount of blood pumped out by the ventricles in a minute. Question 13: Explain heart sounds. Heart sounds are noises generated by the closing and opening of the heart valves. In a healthy individual, there are two normal heart sounds called lub and dub. Lub is the first heart sound. It is associated with the closure of the tricuspid and bicuspid valves at the beginning of systole. The second heart sound dub is associated with the closure of the semilunar valves at the beginning of diastole. These sounds provide important information about the condition and working of the heart. Question 14: Draw a standard ECG and explain the different segments in it. Electrocardiogram is a graphical representation of the cardiac cycle produced by an electrograph. The diagrammatic representation of a standard ECG is shown below. Page 11 of 12

187 Class XI Chapter 18 Body Fluids and Circulation Biology A typical human electrocardiogram has five waves P, Q, R, S, and T. The P, R, and T-waves are above the base line and are known as positive waves. The Q and S- waves are below the base line and are known as negative waves. The P-wave is of atrial origin, while the Q, R, S, and T-waves are of ventricular origin. (a) The P-wave indicates atrial depolarisation. During this wave, the impulse of contraction is generated by the SA node. The PQ-wave represents atrial contraction. (b) The QR-wave is preceded by ventricular contraction. It represents the spread of the impulse of contraction from the AV node to the wall of the ventricle. It leads to ventricular depolarisation. (c) The RS-wave represents ventricular contraction of about 0.3 sec. (d) The ST-wave represents ventricular relaxation of about 0.4 sec. During this phase, the ventricles relax and return to their normal state. (e) The T-wave represents ventricular relaxation. Page 12 of 12

188 Class XI Chapter 19 Excretory Products and their Elimination Biology Question 1: Define Glomerular Filtration Rate (GFR) Glomerular filtration rate is the amount of glomerular filtrate formed in all the nephrons of both the kidneys per minute. In a healthy individual, it is about 125 ml/minute. Glomerular filtrate contains glucose, amino acids, sodium, potassium, urea, uric acid, ketone bodies, and large amounts of water. Question 2: Explain the autoregulatory mechanism of GFR. The mechanism by which the kidney regulates the glomerular filtration rate is autoregulative. It is carried out by the juxtaglomerular apparatus. Juxtaglomerular apparatus is a microscopic structure located between the vascular pole of the renal corpuscle and the returning distal convoluted tubule of the same nephron. It plays a role in regulating the renal blood flow and glomerular filtration rate. When there is a fall in the glomerular filtration rate, it activates the juxtaglomerular cells to release renin. This stimulates the glomerular blood flow, thereby bringing the GFR back to normal. Renin brings the GFR back to normal by the activation of the reninangiotensin mechanism. Question 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. (a) True Page 1 of 7

189 Class XI Chapter 19 Excretory Products and their Elimination Biology (b) False (c) True (d) True (e) True Question 4: Give a brief account of the counter current mechanism. The counter current mechanism operating inside the kidney is the main adaptation for the conservation of water. There are two counter current mechanisms inside the kidneys. They are Henle s loop and vasa rectae. Henle s loop is a U-shaped part of the nephron. Blood flows in the two limbs of the tube in opposite directions and this gives rise to counter currents. The Vasa recta is an efferent arteriole, which forms a capillary network around the tubules inside the renal medulla. It runs parallel to Henley s loop and is U-shaped. Blood flows in opposite directions in the two limbs of vasa recta. As a result, blood entering the renal medulla in the descending limb comes in close contact with the outgoing blood in the ascending limb. Page 2 of 7

190 Class XI Chapter 19 Excretory Products and their Elimination Biology The osmolarity increases from 300 mosmoll -1 in the cortex to 1200 mosmoll -1 in the inner medulla by counter current mechanism. It helps in maintaining the concentration gradient, which in turn helps in easy movement of water from collecting tubules. The gradient is a result of the movement of NaCl and urea. Question 5: Describe the role of liver, lungs and skin in excretion. Liver, lungs, and skin also play an important role in the process of excretion. Role of the liver: Liver is the largest gland in vertebrates. It helps in the excretion of cholesterol, steroid hormones, vitamins, drugs, and other waste materials through bile. Urea is formed in the liver by the ornithine cycle. Ammonia a toxic substance is quickly changed into urea in the liver and thence eliminated from the body. Liver also changes the decomposed haemoglobin pigment into bile pigments called bilirubin and biliverdin. Role of the lungs: Lungs help in the removing waste materials such as carbon dioxide from the body. Role of the skin: Skin has many glands which help in excreting waste products through pores. It has two types of glands sweat and sebaceous glands. Sweat glands are highly vascular and tubular glands that separate the waste products from the blood and excrete them in the form of sweat. Sweat excretes excess salt and water from the body. Sebaceous glands are branched glands that secrete an oily secretion called sebum. Question 6: Explain micturition. Page 3 of 7

191 Class XI Chapter 19 Excretory Products and their Elimination Biology Micturition is the process by which the urine from the urinary bladder is excreted. As the urine accumulates, the muscular walls of the bladder expand. The walls stimulate the sensory nerves in the bladder, setting up a reflex action. This reflex stimulates the urge to pass out urine. To discharge urine, the urethral sphincter relaxes and the smooth muscles of the bladder contract. This forces the urine out from the bladder. An adult human excretes about litres of urine per day. Question 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 (d) ADH (v) Renal tubule Column I Column II (a) Ammonotelism (iii) Bony fish (b) Bowman s capsule (v) Renal tubule (c) Micturition (iv) Urinary bladder (d) Uricotelism (i) Birds (d) ADH (ii) Water reabsorption Page 4 of 7

192 Class XI Chapter 19 Excretory Products and their Elimination Biology Question 8: What is meant by the term osmoregulation? Osmoregulation is a homeostatic mechanism that regulates the optimum temperature of water and salts in the tissues and body fluids. It maintains the internal environment of the body by water and ionic concentration. Question 9: Terrestrial animals are generally either ureotelic or uricotelic, not ammonotelic, why? Terrestrial animals are either ureotelic or uricotelic, and not ammonotelic. This is because of the following two main reasons: (a) Ammonia is highly toxic in nature. Therefore, it needs to be converted into a less toxic form such as urea or uric acid. (b) Terrestrial animals need to conserve water. Since ammonia is soluble in water, it cannot be eliminated continuously. Hence, it is converted into urea or uric acid. These forms are less toxic and also insoluble in water. This helps terrestrial animals conserve water. Question 10: What is the significance of juxtaglomerular apparatus (JGA) in kidney function? Juxtaglomerular apparatus (JGA) is a complex structure made up of a few cells of glomerulus, distal tubule, and afferent and efferent arterioles. It is located in a specialised region of a nephron, wherein the afferent arteriole and the distal convoluted tubule (DLT) come in direct contact with each other. The juxtaglomerular apparatus contains specialised cells of the afferent arteriole known as juxtaglomerular cells. These cells contain the enzyme renin that can sense blood pressure. When glomerular blood flow (or glomerular blood pressure or Page 5 of 7

193 Class XI Chapter 19 Excretory Products and their Elimination Biology glomerular filtration rate) decreases, it activates juxtaglomerular cells to release renin. Renin converts the angiotensinogen in blood into angiotensin I and further into angiotensin II. Angiotensin II is a powerful vasoconstrictor that increases the glomerular blood pressure and filtration rate. Angiotensin II also stimulates the adrenal cortex of the adrenal gland to produce aldosterone. Aldosterone increases the rate of absorption of sodium ions and water from the distal convoluted tubule and the collecting duct. This also leads to an increase in blood pressure and glomerular filtration rate. This mechanism, known as renin-angiotensin mechanism, ultimately leads to an increased blood pressure. Question 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. Page 6 of 7

194 Class XI Chapter 19 Excretory Products and their Elimination Biology (a) Amphioxus is an example of a chordate that has flame cells as excretory structures. Flame cell is a type of excretory and osmoregulatory system. (b) The cortical portions projecting between the medullary pyramids in the human kidney are the columns of Bertini. They represent the cortical tissues present within the medulla. (c) A loop of capillary that runs parallel to Henle s loop is known as vasa rectae. Vasa rectae, along with Henle s loop, helps in maintaining a concentration gradient in the medullary interstitium. Question 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. (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 the 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. Page 7 of 7

195 Class XI Chapter 20 Locomotion and Movement Biology Question 1: Draw the diagram of a sarcomere of skeletal muscle showing different regions. The diagrammatic representation of a sarcomere is as follows: Question 2: Define sliding filament theory of muscle contraction. The sliding filament theory explains the process of muscle contraction during which the thin filaments slide over the thick filaments, which shortens the myofibril. Each muscle fibre has an alternate light and dark band, which contains a special contractile protein, called actin and myosin respectively. Actin is a thin contractile protein present in the light band and is known as the I-band, whereas myosin is a thick contractile protein present in the dark band and is known as the A-band. There is an elastic fibre called z line that bisects each I-band. The thin filament is firmly anchored to the z line. The central part of the thick filament that is not overlapped by the thin filament is known as the H-zone. During muscle contraction, the myosin heads or cross bridges come in close contact with the thin filaments. As a result, the thin filaments are pulled towards the middle of the sarcomere. The Z line attached to the actin filaments is also pulled leading to the shortening of the sarcomere. Hence, the length of the band remains constant as its original length and the I-band shortens and the H-zone disappears. Page 1 of 8

196 Class XI Chapter 20 Locomotion and Movement Biology Question 3: Describe the important steps in muscle contraction. During skeletal muscle contraction, the thick filament slides over the thin filament by a repeated binding and releases myosin along the filament. This whole process occurs in a sequential manner. Step 1: Muscle contraction is initiated by signals that travel along the axon and reach the neuromuscular junction or motor end plate. Neuromuscular junction is a junction between a neuron and the sarcolemma of the muscle fibre. As a result, Acetylcholine (a neurotransmitter) is released into the synaptic cleft by generating an action potential in sarcolemma. Step 2: The generation of this action potential releases calcium ions from the sarcoplasmic reticulum in the sarcoplasm. Page 2 of 8

197 Class XI Chapter 20 Locomotion and Movement Biology Step 3: The increased calcium ions in the sarcoplasm leads to the activation of actin sites. Calcium ions bind to the troponin on actin filaments and remove the tropomyosin, wrapped around actin filaments. Hence, active actin sites are exposed and this allows myosin heads to attach to this site. Step 4: In this stage, the myosin head attaches to the exposed site of actin and forms cross bridges by utilizing energy from ATP hydrolysis. The actin filaments are pulled. As a result, the H-zone reduces. It is at this stage that the contraction of the muscle occurs. Step 5: After muscle contraction, the myosin head pulls the actin filament and releases ADP along with inorganic phosphate. ATP molecules bind and detach myosin and the cross bridges are broken. Stage 6: This process of formation and breaking down of cross bridges continues until there is a drop in the stimulus, which causes an increase in calcium. As a result, the concentration of calcium ions decreases, thereby masking the actin filaments and leading to muscle relaxation. Question 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. (a) : True (b) : False H -zone of striated muscle fibre is the central part of the thick filament that is not overlapped by the thin filament. (c) : True (d) : False Page 3 of 8

198 Class XI Chapter 20 Locomotion and Movement Biology There are 12 pairs of ribs in a man. (e) : True Question 5: Write the difference between: (a) Actin and Myosin (b) Red and White muscles (c) Pectoral and Pelvic girdle (a) Actin and Myosin Actin Myosin 1 Actin is a thin contractile protein. 1 Myosin is a thick contractile protein. 2. It is present in light bands and is called an isotropic band. 2 It is present in dark bands and is called an anisotropic band. (b) Red and White muscles Red muscle fibre White muscle fibre 1 Red muscle fibres are thin and smaller in size. 1 White muscle fibres are thick and larger in size. 2 They are red in colour as they contain large amounts of myoglobin. 2 They are white in colour as they contain small amounts of myoglobin 3 They contain numerous mitochondria. 3 They contain less number of mitochondria. 4 They carry out slow and sustained contractions for a long period. 4 They carry out fast work for short duration. Page 4 of 8

199 Class XI Chapter 20 Locomotion and Movement Biology 5 They provide energy by aerobic respiration. 5 They provide energy by anaerobic respiration. (c) Pectoral and Pelvic girdle Pectoral girdle Pelvic girdle 1 It is a skeletal support from where the forelimbs of vertebrates are attached. 1 It is a skeletal support form where the hind limbs of vertebrates are attached. 2 It is composed of two Bones namely, clavicle or collar bones and scapula or shoulder bone. 2 It is composed of three bones, upper ileum, inner pubic, and ischium. Question 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 Column I Column II (a) Smooth muscle (iv) Involuntary Page 5 of 8

200 Class XI Chapter 20 Locomotion and Movement Biology (b) Tropomyosin (ii) Thin filament (c) Red muscle (i) Myoglobin (d) Skull (iii) Sutures Question 7: What are the different types of movements exhibited by the cells of human body? Movement is a characteristic feature of living organisms. The different types of movement exhibited by cells of the human body are: Amoeboid movement: Leucocytes present in the blood show amoeboid movement. During tissue damage, these blood cells move from the circulatory system towards the injury site to initiate an immune response. Ciliary movement: Reproductive cells such as sperms and ova show ciliary movement. The passage of ova through the fallopian tube towards the uterus is facilitated by this movement. Muscular movement: Muscle cells show muscular movement. Question 8: How do you distinguish between a skeletal muscle and a cardiac muscle? Skeletal muscle Cardiac muscle 1. The cells of skeletal muscles are unbranched. 1. The cells of cardiac muscles are branched. 2. Intercalated disks are absent. 2. The cells are joined with one another by intercalated disks that help in coordination or Page 6 of 8

201 Class XI Chapter 20 Locomotion and Movement Biology synchronization of the heart beat. 3. Alternate light and dark bands are present. 3. Faint bands are present. 4. They are voluntary muscles. 4. They are involuntary muscles. 5. They contract rapidly and get fatigued in a short span of time. 5. They contract rapidly but do not get fatigued easily. 6. They are present in body parts such as the legs, tongue, hands, etc. 6. These muscles are present in the heart and control the contraction and relaxation of the heart. Question 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 Page 7 of 8

202 Class XI Chapter 20 Locomotion and Movement Biology (a) atlas/axis: Pivotal joint (b) carpal/metacarpal of thumb: Saddle joint (c) between phalanges: Hinge joint (d) femur/acetabulum: Ball and socket joint (e) between cranial bones: Fibrous joint (f) between pubic bones in the pelvic girdle: Ball and socket joint Question 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 2 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. (a) All mammals (except a few) have cervical vertebra. (b) The number of phalanges in each limb of a human is. (c) Thin filament of myofibril contains 2 F actins and two other proteins, namely and. (d) In a muscle fibre, Ca ++ is stored in the. (e) And pairs of ribs are called floating ribs. (f) The human cranium is made up of bones. Page 8 of 8

203 Class XI Chapter 21- Neural Control and Coordination Biology Question 1: Briefly describe the structure of the following: (a) Brain (b) Eye (c) Ear (A) Brain: Brain is the main coordinating centre of the body. It is a part of nervous system that controls and monitors every organ of the body. It is well protected by cranial meninges that are made up of an outer layer called dura mater, a thin middle layer called arachnoid, and an inner layer called pia mater. It is divided into three regions forebrain, midbrain, and hindbrain. Forebrain: It is the main thinking part of the brain. It consists of cerebrum, thalamus, and hypothalamus. (a) Cerebrum: Cerebrum is the largest part of the brain and constitutes about four fifth of its weight. Cerebrum is divided into two cerebral hemispheres by a deep longitudinal cerebral fissure. These hemispheres are joined by a tract of nerve fibre known as corpus callosum. The cerebral hemispheres are covered by a layer of cells known as cerebral cortex or grey matter. Cerebrum has sensory regions known as association areas that receive sensory impulses from various receptors as well as from motor regions that control the movement of various muscles. The innermost part of cerebrum gives an opaque white appearance to the layer and is known as the white matter. (b) Thalamus: Thalamus is the main centre of coordination for sensory and motor signalling. It is wrapped by cerebrum. (c) Hypothalamus: It lies at the base of thalamus and contains a number of centres that regulate body temperature and the urge for eating and drinking. Some regions of cerebrum, along with hypothalamus, are involved in the regulation of sexual behaviour and expression of emotional reactions such as excitement, pleasure, fear, etc. Page 1 of 22

204 Class XI Chapter 21- Neural Control and Coordination Biology Midbrain: It is located between the thalamus region of the forebrain and pons region of hindbrain. The dorsal surface of midbrain consists of superior and inferior corpora bigemina and four round lobes called corpora quadrigemina. A canal known as cerebral aqueduct passes through the midbrain. Midbrain is concerned with the sense of sight and hearing. Hindbrain: It consists of three regions pons, cerebellum, and medulla oblongata. (a) Pons is a band of nerve fibre that lies between medulla oblongata and midbrain. It connects the lateral parts of cerebellar hemisphere together. (b) Cerebellum is a large and well developed part of hindbrain. It is located below the posterior sides of cerebral hemispheres and above medulla oblongata. It is responsible for maintaining posture and equilibrium of the body. (c) Medulla oblongata is the posterior and simplest part of the brain. It is located beneath the cerebellum. Its lower end extends in the form of spinal cord and leaves the skull through foramen magnum. (B) Eye: Eyes are spherical structures that consist of three layers. (a) The outer layer is composed of sclera and cornea. (i) Sclera is an opaque tissue that is usually known as white of the eye. It is composed of a dense connective tissue. (ii) Cornea is a transparent anterior portion of eye that lacks blood vessels and is nourished by lymph from the nearby area. It is slightly bulged forward and helps in focusing light rays with the help of lens. (b) The middle layer of eye is vascular in nature and contains choroid, ciliary body, and iris. (i) Choroid lies next to the sclera and contains numerous blood vessels that provide nutrients and oxygen to the retina and other tissues. (ii) Ciliary body: The choroid layer is thin over posterior region and gets thickened in the anterior portion to form ciliary body. It contains blood vessels, ciliary muscles, and ciliary processes. Page 2 of 22

205 Class XI Chapter 21- Neural Control and Coordination Biology (iii) Iris: At the junction of sclera and cornea, the ciliary body continues forward to form thin coloured partition called iris. It is the visible coloured portion of eye. The eye contains a transparent, biconvex, and elastic structure just behind the iris. It is known as lens. The lens is held in position by suspensory ligaments attached to the ciliary body. The lens divides the eye ball into two chambers an anterior aqueous and posterior vitreous chamber. (c) The innermost nervous coat of eye contains retina. Retina is the innermost layer. It contains three layers of cells inner ganglion cells, middle bipolar cells, and outermost photoreceptor cells. The receptor cells present in the retina are of two types rod cells and cone cells. (a) Rod cells The rods contain the rhodopsin pigment (visual purple) that is highly sensitive to dim light. It is responsible for twilight vision. (b) Cone cells The cones contain the iodopsin pigment (visual violet) and are highly sensitive to high intensity light. They are responsible for daylight and colour visions. The innermost ganglionic cells give rise to optic nerve fibre that forms optic nerve in each eye and is connected with the brain. (C) Ear: Ear is the sense organ for hearing and equilibrium. It consists of three portions external ear, middle ear, and internal ear. 1. External ear: It consists of pinna, external auditory meatus, and a tympanic membrane. (a) Pinna is a sensitive structure that collects and directs the vibrations into the ear to produce sound. (b) External auditory meatus is a tubular passage supported by cartilage in external ear. (c) Tympanic membrane is a thin membrane that lies close to the auditory canal. It separates the middle ear from external ear. 2. Middle ear: It is an air filled tympanic cavity that is connected with pharynx through eustachian tube. Eustachian tube helps to equalize air pressure in both sides of tympanic membrane. The middle ear contains a flexible chain of three middle bones called ear Page 3 of 22

206 Class XI Chapter 21- Neural Control and Coordination Biology ossicles. The three ear ossicles are malleus, incus, and stapes that are attached to each other. 3. Internal ear: It is also known as labyrinth. Labyrinth is divided into bony labyrinth and a membranous labyrinth. Bony labyrinth is filled with perilymph while membranous labyrinth is filled with endolymph. Membranous labyrinth is divided into 2 parts. (a) Vestibular apparatus Vestibular apparatus is a central sac like part that is divided into utriculus and sacculus. A special group of sensory cells called macula are present in sacculus and utriculus. Vestibular apparatus also contains three semi circular canals. The lower end of each semi circular canal contains a projecting ridge called crista ampularis. Each ampulla has a group of sensory cells called crista. Crista and macula are responsible for maintaining the balance of body and posture. (b) Cochlea: Cochlea is a long and coiled outgrowth of sacculus. It is the main hearing organ. Cochlea consists of three membranes. The organ of corti, a hearing organ, is located on the basilar membrane that has hair cells. Question 2: Compare the following: (a) Central neural system (CNS) and Peripheral neural system (PNS) (b) Resting potential and action potential (c) Choroid and retina (a) Central neural system (CNS) and Peripheral neural system (PNS) Central neural system Peripheral neural system 1. It is the main 1. It is not the main coordinating centre of the Page 4 of 22

207 Class XI Chapter 21- Neural Control and Coordination Biology coordinating centre of the body. body. 2. It includes brain and spinal cord. 2. It includes cranial and spinal nerves that connect central nervous system to different parts of the body. (b) Resting potential and action potential Resting potential Action potential 1. It is the potential difference across the nerve fibre when there is no conduction of nerve impulse. 1. It is the potential difference across nerve fibre when there is conduction of nerve impulse. 2. The membrane is more permeable to K + ions than to Na + ions. 2. The membrane is more permeable to Na + ions than to K + ions. (c) Choroid and retina Choroid Retina 1. Choroid is the middle vascular layer of eye. 1. Retina is the innermost nervous coat of eye. 2. It contains numerous blood vessels that provide nutrients and oxygen to retina and other tissues. 2. It contains photoreceptor cells, rods and cones that are associated with twilight and colour vision respectively. Question 3: Explain the following processes: (a) Polarisation of the membrane of a nerve fibre Page 5 of 22

208 Class XI Chapter 21- Neural Control and Coordination Biology (b) Depolarisation of the membrane of a nerve fibre (c) Conduction of a nerve impulse along a nerve fibre (d) Transmission of a nerve impulse across a chemical synapse (a) Polarisation of the membrane of a nerve fibre During resting condition, the concentration of K + ions is more inside the axoplasm while the concentration of Na + ions is more outside the axoplasm. As a result, the potassium ions move faster from inside to outside as compared to sodium ions. Therefore, the membrane becomes positively charged outside and negatively charged inside. This is known as polarization of membrane or polarized nerve. (b) Depolarisation of the membrane of a nerve fibre When an electrical stimulus is given to a nerve fibre, an action potential is generated. The membrane becomes permeable to sodium ions than to potassium ions. This results into positive charge inside and negative charge outside the nerve fibre. Hence, the membrane is said to be depolarized. (c) Conduction of a nerve impulse along a nerve fibre There are two types of nerve fibres myelinated and non myelinated. In myelinated nerve fibre, the action potential is conducted from node to node in jumping manner. This is because the myelinated nerve fibre is coated with myelin sheath. The myelin sheath is impermeable to ions. As a result, the ionic exchange and depolarisation of nerve fibre is not possible along the whole length of nerve fibre. It takes place only Page 6 of 22

209 Class XI Chapter 21- Neural Control and Coordination Biology at some point, known as nodes of Ranvier, whereas in non myelinated nerve fibre, the ionic exchange and depolarization of nerve fibre takes place along the whole length of the nerve fibre. Because of this ionic exchange, the depolarized area becomes repolarised and the next polarized area becomes depolarized. (d) Transmission of a nerve impulse across a chemical synapse Synapse is a small gap that occurs between the last portion of the axon of one neuron and the dendrite of next neuron. When an impulse reaches at the end plate of axon, vesicles consisting of chemical substance or neurotransmitter, such as acetylcholine, fuse with the plasma membrane. This chemical moves across the cleft and attaches to chemo receptors present on the membrane of the dendrite of next neuron. This binding of chemical with chemo receptors leads to the depolarization of membrane and generates a nerve impulse across nerve fibre. The chemical, acetylcholine, is inactivated by enzyme acetylcholinestrase. The enzyme is present in the post synaptic membrane of the dendrite. It hydrolyses acetylcholine and this allows the membrane to repolarise. Question 4: Draw labelled diagrams of the following: (a) Neuron (b) Brain (c) Eye (d) Ear (a) Neuron Page 7 of 22

210 Class XI Chapter 21- Neural Control and Coordination Biology (b) Brain (c) Eye Page 8 of 22

211 Class XI Chapter 21- Neural Control and Coordination Biology (d) Ear Question 5: Write short notes on the following: (a) Neural coordination (b) Forebrain (c) Midbrain (d) Hindbrain (e) Retina (f) Ear ossicles (g) Cochlea (h) Organ of Corti (i) Synapse (a) Neural coordination The neural system provides rapid coordination among the organs of the body. This coordination is in the form of electric impulses and is quick and short lived. All the physiological processes in the body are closed linked and dependent upon each other. For example, during exercise, our body requires more oxygen and food. Hence, the breathing rate increases automatically and the heart beats faster. This leads to a faster supply of oxygenated blood to the muscles. Moreover, the cellular functions require regulation continuously. These functions are carried out by the hormones. Hence, the neural system along with the endocrine system control and coordinate the physiological processes. (b) Forebrain It is the main thinking part of the brain. It consists of cerebrum, thalamus, and hypothalamus. Page 9 of 22

212 Class XI Chapter 21- Neural Control and Coordination Biology (i) Cerebrum: Cerebrum is the largest part of the brain and constitutes about four fifth of its weight. Cerebrum is divided into two cerebral hemispheres by a deep longitudinal cerebral fissure. These hemispheres are joined by a tract of nerve fibres known as corpus callosum. The cerebral hemispheres are covered by a layer of cells known as cerebral cortex or grey matter. Cerebrum has sensory regions known as association areas that receive sensory impulses from various receptors as well as from motor regions that control the movement of various muscles. The innermost part of cerebrum gives an opaque white appearance to the layer and is known as the white matter. (ii) Thalamus: Thalamus is the main centre of coordination for sensory and motor signalling. It is wrapped by cerebrum. (iii) Hypothalamus: It lies at the base of thalamus and contains a number of centres that regulate body temperature and the urge for eating and drinking. Some regions of cerebrum, along with hypothalamus, are involved in the regulation of sexual behaviour and expression of emotional reactions such as excitement, pleasure, fear, etc. (c) Midbrain It is located between the thalamus region of the forebrain and pons region of hindbrain. The dorsal surface of midbrain consists of superior and inferior corpora bigemina and four round lobes called corpora quadrigemina. A canal known as cerebral aqueduct passes through the midbrain. Midbrain is concerned with the sense of sight and hearing. (d) Hindbrain It consists of three regions pons, cerebellum, and medulla oblongata. (i) Pons is a band of nerve fibres that lies between medulla oblongata and midbrain. It connects the lateral parts of cerebellar hemisphere together. Page 10 of 22

213 Class XI Chapter 21- Neural Control and Coordination Biology (ii) Cerebellum is a large and well developed part of hindbrain. It is located below the posterior sides of cerebral hemispheres and above the medulla oblongata. It is responsible for maintaining posture and equilibrium of the body. (iii) Medulla oblongata is the posterior and simplest part of the brain. It is located beneath the cerebellum. Its lower end extends in the form of spinal cord and leaves the skull through foramen magnum. (e) Retina Retina is the innermost layer. It contains three layers of cells inner ganglion cells, middle bipolar cells, and outermost photoreceptor cells. The receptor cells present in the retina are of two types rod cells and cone cells. (i) Rod cells The rods contain rhodopsin pigment (visual purple), which is highly sensitive to dim light. It is responsible for twilight vision. (ii) Cone cells The cones contain iodopsin pigment (visual violet) and are highly sensitive to high intensity light. They are responsible for daylight and colour visions. The innermost ganglionic cells give rise to optic nerve fibre that forms optic nerve in each eye and is connected with the brain. In this region, the photoreceptor cells are absent. Hence, it is known as the blind spot. At the posterior part, lateral to blind spot, there is a pigmented spot called macula lutea. This spot has a shallow depression at its middle known as fovea. Fovea has only cone cells. They are devoid of rod cells. Hence, it is the place of most distinct vision. (f) Ear ossicles The middle ear contains a flexible chain of three middle bones called ear ossicles. The three ear ossicles are as follows. (i) Malleus (ii) Incus (iii) Stapes The malleus is attached to tympanic membrane on one side and to incus on the other side. The incus is connected with stapes. Stapes, in turn, are attached with an oval membrane, fenestra ovalis, of internal ear. The ear ossicles act as a lever that transmits sound waves from external ear to internal ear. Page 11 of 22

214 Class XI Chapter 21- Neural Control and Coordination Biology (g) Cochlea Cochlea is a long, coiled outgrowth of sacculus. It is the main hearing organ. The cochlea forms three chambers. (i) Upper scala vestibule (ii) Middle scala media (iii) Lower scale tympani The floor of the scala media is basilar membrane while its roof is Reissner s membrane. Reissner s membrane gives out a projection called tectorial membrane. The organ of corti, a hearing organ, is located on the basilar membrane. Organ of corti contains receptor hair cells. The upper scala vestibule and lower scala tympani contain perilymph. (h) Organ of corti Organ of corti is the hearing organ. It is located on the basilar membrane that contains hair cells. Hair cells act as auditory receptors. They are present on the internal side of organ of corti. (i) Synapse Synapse is a junction between the axon terminal of one neuron and the dendrite of next neuron. It is separated by a small gap known as synaptic cleft. There are two types of synapses. (a) Electrical synapse (b) Chemical synapse Page 12 of 22

215 Class XI Chapter 21- Neural Control and Coordination Biology In electrical synapses, the pre and post synaptic neurons lie in close proximity to each other. Hence, the impulse can move directly from one neuron to another across the synapse. This represents a faster method of impulse transmission. In chemical synapses, the pre and post synaptic neurons are not in close proximity. They are separated by a synaptic cleft. The transmission of nerve impulses is carried out by chemicals such as neurotransmitters. Question 6: Give a brief account of: (a) Mechanism of synaptic transmission (b) Mechanism of vision (c) Mechanism of hearing (a) Mechanism of synaptic transmission Synapse is a junction between two neurons. It is present between the axon terminal of one neuron and the dendrite of next neuron separated by a cleft. There are two ways of synaptic transmission. (1) Chemical transmission (2) Electrical transmission 1. Chemical transmission When a nerve impulse reaches the end plate of axon, it releases a neurotransmitter (acetylcholine) across the synaptic cleft. This chemical is synthesized in cell body of the neuron and is transported to the axon terminal. The acetylcholine diffuses across the cleft and binds to the receptors present on the membrane of next neuron. This causes depolarization of membrane and initiates an action potential. 2. Electrical transmission In this type of transmission, an electric current is formed in the neuron. This electric current generates an action potential and leads to transmission of nerve impulse across the nerve fibre. This represents a faster method of nerve conduction than the chemical method of transmission. Page 13 of 22

216 Class XI Chapter 21- Neural Control and Coordination Biology (b) Mechanism of vision Retina is the innermost layer of eye. It contains three layers of cells inner ganglion cells, middle bipolar cells, and outermost photoreceptor cells. A photoreceptor cell is composed of a protein called opsin and an aldehyde of vitamin A called retinal. When light rays are focused on the retina through cornea, it leads to the dissociation of retinal from opsin protein. This changes the structure of opsin. As the structure of opsin changes, the permeability of membrane changes, generating a potential difference in the cells. This generates an action potential in the ganglionic cells and is transmitted to the visual cortex of the brain via optic nerves. In the cortex region of brain, the impulses are analysed and image is formed on the retina. (c) Mechanism of hearing The pinna of the external region collects the sound waves and directs it towards ear drum or external auditory canal. These waves strike the tympanic membrane and vibrations are created. Then, these vibrations are transmitted to the oval window, fenestra ovalis, through three ear ossicles, named as malleus, incus, and stapes. These ear ossicles act as lever and transmit the sound waves to internal ear. These vibrations from fenestra ovalis are transmitted into cochlear fluid. This generates sound waves in the lymph. The formation of waves generates a ripple in the basilar membrane. This movement bends the sensory hair cells present on the organ of corti against tectorial membrane. As a result of this, sound waves are converted into nerve impulses. These impulses are then carried to auditory cortex of brain via auditory nerves. In cerebral cortex of brain, the impulses are analysed and sound is recognized. Page 14 of 22

217 Class XI Chapter 21- Neural Control and Coordination Biology Question 7: briefly: (a) How do you perceive the colour of an object? (b) Which part of our body helps us in maintaining the body balance? (c) How does the eye regulate the amount of light that falls on the retina? (a) Photoreceptors are cells that are sensitive to light. They are of two types rods and cones. These are present in the retina. Cones help in distinguishing colours. There are three types of cone cells those responding to green light, those responding to blue light, and those responding to red light. These cells are stimulated by different lights, from different sources. The combinations of the signals generated help us see the different colours. (b) Vestibular apparatus is located in the internal ear, above the cochlea and helps in maintaining body balance. Crista and macula are the sensory spots of the vestibular apparatus controlling dynamic equilibrium. (c) Pupil is the small aperture in the iris that regulates the amount of light entering the eye. Cornea, aqueous humour, lens, and vitreous humour act together and refract light rays, focussing them onto the photoreceptor cells of the retina. Page 15 of 22