NCERT Solutions for Class 11 Biology Chapter 3 Plant Kingdom

NCERT Solutions for Class 11 Biology Chapter 3 Plant Kingdom

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 3 Plant Kingdom.

Question 1.
What is the basis of the classification of algae?
Solution:
Classification of Algae: The algae are divided into three main classes: Chlorophyceae, Phaeophyceae, and Rhodophyceae.

  1. Chlorophyceae: The members of Chlorophyceae are commonly called green algae.
  2. Phaeophyceae: The members of Phaeophyceae or brown algae, are found primarily in marine habitats.
  3. Rhodophyta is commonly called red algae. (See Table on the opposite page)

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?
Solution:
I. Liverwort – The main part of the body of liverwort is thalloid. Haploid gametes are produced from the male and female sex organs garnets fuses to form zygote. Zygote develops in the form of sporophytes. These sporophytes are further differentiated into foot, seta and capsule. As a result of reduction division many haploid spores are produced in capsule.
II. Moss – In the first stage in moss primary protonema develops into secondary ’ protonema. Both these stages are haploid. Zygote formed by the fusion of gametes further produce sporophytes.
III. Fern – Leaves of sporophyte bear sporangia in which spores are produced by reduction division in meiosis.
IV. Gymnosperm – In microsporophylls and megasporophylls that bear microsporangia and megasporangia respectively, reduction division occurs to produce microspores (pollen grains) and megaspore.
V. Angiosperm – Main part of the body is sporophytic and bears flowers. Reduction division takes place in anthers of stamen i.e. haploid pollen grains and in the ovary of pistil producing eggs.

Question 3.
Name three groups of plants that bear archegonia. Briefly describe the life cycle of any one of them.
Solution:

Archegonia is the female reproductive organ of

  1. Bryophytes
  2. Pteridophytes and
  3. Gymnosperms groups of plants.

The life cycle of gymnosperms: The gymnosperms are heterosporous, they produce haploid microspores and megaspores. The two kinds of spores are produced within sporangia that are borne oil sporophylls which are arranged spirally along an axis to form lax or compact strobili or cones.
The strobili bearing microsporophylls and microsporangia are called microsporangia or male strobili. The microspores develop into a male gametophytic generation which is highly reduced and is confined to only a limited number of cells. This reduced gametophyte is called pollen-grain. The development of pollen- grains takes place within the microsporangia.

The cones Rearing megasporophylls with ovules or megasporangia are called microsporangia or female strobili. The male or female cones or strobili may be borne on the same tree or on different trees. The megaspore mother cell is differentiated from one of the cells of the nucellus. The nucellus is protected by envelopes and the composite structure is called an ovule.

The ovules are borne on megasporophylls which may be clustered to form the female cones. The female cones are borne on the main plant body of the sporophyte. The megaspore mother cell divides meiotically to form four megaspores. One of the megaspores enclosed within the megasporangium (nucleus) develops into a multicellular female gametophyte that bears two or more archegonia or female sex organs. The multicellular female gametophytes are also retained within megasporangium.

Three groups of plant that bear archegonia are
1. Bryophyta,
2. Pteridophyta and
3. Gymnosperm.

  1. Life cycle of gymnosperms: The gymnosperms are heterosporous, they produce haploid microspores and megaspores.
  2. The two kinds of spores are produced within sporangia that are borne on sporophyll which are arranged spirally along an axis to form strobili or cones.
  3. The strobili bearing microsporophylls and microsporongia are called microsporangiate or male strobili.
  4. The microspores develop into a male gametophytic generation which is highly reduced, and confined to only limited number of cells. The reduced gametophyte is called pollen grain.
  5. The development of pollen grain takes place within microsporangia.
    The cones bearing megasporophylls with ovules or megasporangia are called macrosporangiate or female strobili.
  6. The male or female cones or strobili may be borne on the same tree or different trees. The megaspore mother cell is differentiated from one of the cells of the nucellus.
  7. The nucellus is protected by envelopes and the composite structure is called an ovule. The ovules are borne on megasporophylls which may be clustered to form the female cones.
  8. The females cones are borne on the main plant body of the sporophyte. The megaspore mother cell divides meiotically to form four megaspore.
  9. One of the megaspore enclosed within the megasporangium develops into a multicellular female gametophyte that bears 2 or more archegonia or female sex organs.
  10. The multicellular female gametophyte is also remaining within megasporangium.

Question 4.
Mention the ploidy of the following: protonemal cells of a moss; primary endosperm nucleus in dicot, leaf cell of a moss; prothallus cell of a fern; gemma cell in Marchantia; Meristem cells of monocot, ovum of liverwort and zygote of a fern.
Solution:
Protonemal cells of a moss – Haploid (n) Primary endosperm nucleus in dicot – Triploid (3n)
Leaf cell of a moss – Haploid (n)
Prothallus cell of a fem – Haploid (n)
Gemma cell in Marchantia – Haploid (n) Meristem cells of monocot – Diploid (2n) Ovum of liverwort – Haploid (n)
Zygote of a fem – Diploid (2n)

Question 5.
Write a note on economic importance of algae and gymnosperms.
Solution:

  1. Algae like Chlorella, Gelidium (produce agar- agar) are used as food. Many algae like diatoms (used in manufacture of glass, polish, etc), algin (used in vulcanisation, artificial fibres, etc.), are used in industry.
  2. Nos toe, Anabaena, etc., are useful in increasing fertility of soil. Antibiotic chlorellin is extracted from Chlorella.
  3. Many algae have harmful effect also, for example, Microcystis, Chlrococcus, Oscillatoria cause water blooms and Cephaleuros species of algae are parasitic on tea leaves and cause harm to tea industry.
  4. v Alginic acid are extracted from the members Phaeophycea such as laminaria, Macrocystis and carrageenin is extracted from red algae chondris Crispos.

Economic importance of gymnosperms :

  1. Gymnosperms helps in checking soil erosion.
  2. Seeds of Pinns gerardiana, Gnetnm gnemon and Ginkgo biloba are eaten.
  3. Conifers like Pinus longifolia, Cedrus deodara, Picea, Tsngo, etc., produce soft wood. Bark of Tsugo yields tannins for making inks, seeds and bark of Cycas are used as poultica for wounds and sores.
  4. Ephedra, Gnetum, Taxus baccata, Cycles rumphii are used for medicinal purposes.

Question 6.
Both gymnosperms and angiosperms bear seeds, then why are they classified separately?
Solution:
Gymnosperms are ‘naked seeded’ plants because their seeds are not enclosed in fruit wall whereas angiosperm are ‘enclosed seeds’ as seeds (ovules) are found enclosed in the ovary wall.

Question 7.
What is heterospory? Briefly comment on its significance. Give two examples.
Solution:

  1. Genera like Selaginella and Salvinia which produce two kinds of spores macro and microspores are known as heterosporous.
  2. The megaspore and microspores germinate and give rise to female and male gametophytes, respectively.
  3. The female gametophytes in these plants are retained on the parent sporophyte for variable periods.
  4. The development of the zygotes into young embryo takes place within the female gametophytes.
  5. This event is considered as an important step in evolution leading to seed habit. Heterospory is considered as the first step towards seed habit. Selaginella and Marsilea, show seed habit.

Question 8.
Explain briefly
(i) Protonema
(ii) Antheridium
(iii) Archegonium
(iv) Diplontic
(v) Sporophyll
(vi) Isogamy
Solution:
(i) Protonema: The predominant stage of moss gametophyte which directly develops from spore is known as protonema.
(ii) Antheridium: The male sex organ in bryophytes, pteridophytes and gymnosperms is called antheridium. It bears male gamete.
(iii) Archegonium : It is the female sex organ found in bryophytes, pteridophytes and gymnosperms. It bears a female gamete.
(iv) Diplontic: In the life cycle of plants when their diploid stage is dominant for long time then this is called diplontic.
(v) Sporophyll: The sporophyte bears sporangia that are subtended by leaf-like appendages called sporophylls.
(vi) Isogamy: When the gametes involved in sexual reproduction are morphologically similar then this is called isogamy. These gametes are physiologically different.

Question 9.
Differentiate between the following:-
• Red algae and brown algae
• Liverworts and moss
• Homosporous and heterosporous pteridophyte
• Syngamy and triple fusion 
Solution:

Differences between red algae and brown algae are as follows
NCERT Solutions for Class 11 Biology Chapter 3 Plant Kingdom 1
NCERT Solutions for Class 11 Biology Chapter 3 Plant Kingdom 2

Question 10.
How would you distinguish monocots from dicots?
Solution:
Dicots are characterized by having two cotyledons in their seeds while the monocotyledons have only one.

Question 11.
Match the following content of column I with column II
Column I Column II
(a) Chlamydomonas (i) Moss
(b) Cycas (ii) Pteridophyta
(c) Selaginella (iii) Algae
(d) Sphagnum (iv) Gymnosperm
Solution:
(a) Chlamydomonas  (iii) Algae
(b) Cycas                    (iv) Gymnosperm
(c) Selaginella            (ii) Pteridophyta
(d) Sphagnum           (i) Moss

Question 12.
Describe the important characteristics of gymnosperms.
Solution:
The seed forming vascular plants which produce seeds but no fruits are called gymnosperms. General characters of gymnosperm are as follows:

  • Fertilization does not require water.
  • Leaves may be of two kinds: foliage leaves and scale leaves.
  • The ovules are orthotropus.
  • Most primitive seed-bearing plants.
  • Mostly these plants are evergreen.
  • Have no ovary wall, seeds are naked.
  • Exhibit polyembryony. Sexual reproduction oogamous type.
  • Large, tall and woody trees.

VERY SHORT ANSWER QUESTIONS

Question 1.
Which group of plants are commonly called “Amphibians of Plant Kingdom”?
Solution:
Bryophytes

Question 2.
Which pigment does provide red colour to red algae?
Solution:
Phycoerythrin.

Question 3.
Which alga does reproduce sexually by conjugation?
Solution:
Spiro gym

Question 4.
Name the organ that fixes the Plant body of Riccia to the soil.
Solution:
Hepaticopsida.

Question 5.
Which plant group is called vascular cryptograms?
Solution:
Pteridophytes.

Question 6.
Which plant is commonly called as walking fern?
Solution:
Adiantum

Question 7.
Which group of plants produces seed but not fruits?
Solution:
Gymnosperm.

Question 8.
Which part of ovule is haploid in gymnosperm?
Solution:
Endosperm

Question 9.
What is name of megasporophyll bearing ovules of angiosperm?
Solution:
Carpel.

Question 10.
Why do gymnosperms fail to produce fruits?
Solution:
Fruits are formed from ovaries. Since the gymnosperm ovules are not enclosed inside the ovaries, they do not produce fruit.

Question 11.
Why life cycle of angiosperm is called diplontic?
Solution:
Life cycle of angiosperm is called diplontic because the diploid (sporophytic) phase is more prominent and long-lived whereas the haploid (gametophytic) phase is short-lived.

Question 12.
Which algae is known as Rolling alga?
Solution:
Volvox

Question 13.
Which are the specialized structures in selaginella which bear adventitious roots?
Solution:
Rhizophores

Question 14.
What is the number of cells and nuclei present in the embryo sac of angiosperm?
Solution:
7 cells and 8 nuclei

Question 15.
Which plant is known as a living fossil?
Solution:
Cycas

SHORT ANSWER QUESTIONS

Question 1.
What are the salient features of pteridophyte?
Solution:
Pteridophytes are vascular cryptogams i.e., plants of this group possess vascular tissue (i. e., xylem, and phloem) for the conduction of water and minerals and for the translocation of foods. They are flowerless and seedless plants.
The general characters of pteridophytes are as following.
I. Primary root is short-lived. It is replaced by adventitious roots.
II. All vegetative parts possess vascular tissues. A cambium is altogether absent. In xylem trachea are absent and in phloem companion cells are absent.
III. Pteridophytes show origin and evolution of stele {i.e., vascular tissue, pericycle, and if present the pith).
IV. Pteridophytes are characterized by having only tracheids in their xylem and only sieve tube in their phloem.
V. The main plant body is the sporophyte (diploid), usually differentiated into true roots, true stems, and true leaves. A stem is usually an underground rhizome or an erect trunk as in tree ferns. Leaves are large (megaphyllous) and variously shaped.

Question 2.
What is the Alternation of Generation?
Solution:
gametophytic and sporophytic generations alternate in the life cycle, it is called the Alter-nation of generation. Or vice versa gametophyte produces gametes and their fusion product zygote or Oospore. Oospore produces sporophyte. The sporophyte produces spores. Spores germinate and produce gametophyte.

Question 3.
What are gemmae? Name two plants that produce gemmae.
Solution:
Gemmae are green, multicellular asexual buds, which develop in small receptacles, called gemmae cups, on the thallus.
The gemmae become detached from the parent thallus and germinate to form new individuals. e.g., Marchantia, Riccia.

Question 4.
Differentiate between cytotaxonomy and chemotaxonomy.
Solution:
Cytotaxonomy that is based on cytological information like chromosome number, structure, behaviour and chemotaxonomy that uses the chemical constituents of the plant to resolve confusions, are also used by taxonomists these days.

Question 5.
Differentiate between monocotyledon and dicotyledons.
Solution:
NCERT Solutions for Class 11 Biology Chapter 3 Plant Kingdom 3

Question 6.
Mention four types of Placentation with suitable examples.
Solution:

  • Marginal placentation: Example members of the family fabaceae (Crotalaria)
  • Axile placentation: Example members of the family Malvaceae (Hibiscus)
  • Parietal placentation. Example: Cucumber.
  • Basal placentation Example: Sunflower.

Question 7.
What are the features of Gemmae found in bryophytes?
Solution:
Gemmae are the means of asexual reproduction found in many bryophytes (ex-Liverworts). They are one to many-celled, specially produced clonal plant fragments. They are green multicellular, asexual buds which develop in small receptacles called “gemma cups” located on the thalli. Gemmae become detached from the parent body and germinate to form a new individual

Question 8.
What is the basis of the classification of the phylogenetic system?
Solution:
It indicates evolutionary as well as the genetic relationship among organism, it is based on the fossil record, biochemical, anatomical, morphological, embryological, physiological, genetics, Karyotype, and other studies.

Question 9.
What are the main features of Anthocerotopsida?
Solution:
These are also known as hornworts because typical horn-like appearances are present of their sporophyte. These contain thalloid gametophyte, distinctly dorsiventral, rhizoids are present, Thalloid do not possess air chambers and scales. Each cell of thallus has a single large chloroplast with a pyrenoid.

LONG ANSWER QUESTIONS

Question 1.
Enumerate distinguishing characters of Bryophytes.
Solution:

  • Bryophytes are called “Amphibians of the plant kingdom” They grow in moist shady places. They require water for their sperms to swim towards eggs, hence the term ‘amphibians’.
  • The plant body is gametophyte so that it produces garnets.
  • The gametophyte is nonvascular so that xylem and phloem tissues are absent.
  • The gametophyte is thalloid in structure or it may be differentiated into stem-like, root-like and leaf-like organs.
  • Gametophyte reproduces sexually. The male sex organs are antheridia and the female sex organs are archegonia. Antheridia produce flagellated motile sperms. Archegonia produce female gamete or egg.
  • Fertilization results in the formation of diploid oospore. It produces diploid sporophyte Sporophyte reproduces asexually and produces haploid spores. They germinate and produce haploid gametophyte. The sporophyte is dependant on the gametophyte.
  • In the life cycle, there is an alternation of generation between haploid gametophyte and diploid sporophyte.
  • Bryophytes also show vegetative reproduction by fragmentation, gemmae formation etc.

Question 2.
Write short notes on the following :
(a) Peristomial teeth of moss
(b) Protonema of moss
(c) Archegonia of moss
Solution:
(a) Peristomal teeth are located just below the operculum. It helps in the dispersal of spores by the hygroscopic movement of its outer ring while the inner ring does not show the hygroscopic movements.
(b) Each spore produces a filamentous juvenile stage called protonema. Protonema has two types of branches, subterranean nongreen rhizoidal, and green Mediterranean branches. Buds develop on green prostrate branches which grow to form new moss plants.
(c) Female reproductive organ of moss is called archegonium.
NCERT Solutions for Class 11 Biology Chapter 3 Plant Kingdom 4

 

Question 3.
Explain the general characters of the Gymnosperm plant body. (Tumkur 2008)
Solution:

  • Gymnosperm plant body is sporophyte.
  • Sporophyte consists of stem, root and leaves. Stem is unbranched in cycas. the stem bears crown of leaves at the tip.
  • The foliage leaves are pinnately compound in cycas and needle-like in conifers. Brown small non-photosynthetic leaves called scale leaves/cataphylls are present in cycas.
  • In cycas, the cataphylls are thickly covered with brown hairs called ramenta.
  • The foliage leaves are green. The pinnae are tough, leathery and has only midrib without veins.
  • Young leaves show circinate vernation.
  • The roots are inhabited by blue-green algae like Nostoc help the plant in nitrogen fixation
  • The sporophyte produces two types of spores so it is heterosporous.

Question 4.
Give a comparative account of gymnosperms and angiosperms.
Solution:
Comparison of gymnosperms and angiosperms
NCERT Solutions for Class 11 Biology Chapter 3 Plant Kingdom 5

Question 5.
Describe major features of Plant kingdom.
Solution:
Features of Plant Kingdom:

  • Plants are autotrophic, except for some carnivorous, plants. They trap photo energy from sunlight and convert it to chemical energy through I photosynthesis. Because of this plants are the main channel for supplying energy in the food chain on earth.
  • Reproduction in plants can be by any of the following modes: Vegetative or Asexual, and Sexual Reproduction.
  • Plant cell is unique because of presence of cell wall and large vacuoles. Green parts of plant contain chlorophyll, which helps them in trapping the photo energy.
  • Sizes of plants can vary from microscopic to a very large tree. Plants are mainly divided into Algae, Bryophytes, Pteridophytes, Gymnosperms and Angiosperms.
  • Lower plants, like algae and bryophytes, have thalloid structures, while higher plants, like gymnosperms and angiosperms, have clearly defined roots and stems.
  • In higher plants root gives a means to anchor in the soil and helps the plant in taking minerals and water from the soil. Green leaves on the stem help them in photosynthesis.
  • Most of the plant growth as a result of photosynthesis. After photosynthesis, extra food is utilized to facilitate growth.
  • Usually in higher plants growth is unlimited and some taller trees can live a life of more than 1000 years.
  • Being the main carbon fixation agent, plants are very important for the whole ecology.
  • The whole food basket for humans is being filled by the plant kingdom. Even animal products, like milk and poultry, are indirect results of a plant’s carbon fixation.
  • Plants supply raw materials for a majority of economic activities. Wood for furniture and building materials comes from plants. The whole paper industry is dependent on the plant kingdom. Think of life if there was no paper and you may understand the larger impact on human civilization.
  • Angiosperms have special organs, called flowers, to bear sexual parts. Flowers are a helpful tool in facilitating variations and further evolution of the plant kingdom.

NCERT Solutions for Class 11 Biology Chapter 3 Plant Kingdom 6

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

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

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

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 from pre¬existing cells.
(d) A unicellular organism carries out its life activities within single cell.
Solution:
(a) Robert brown discovered the cell.

Question 2.
New cells generate from
(a) bacterial fermentation
(b) regeneration of old cells
(c) pre-existing cells
(d) abiotic material
Solution:
(c) pre-existing cells

Question 3.
Match the following
Column A Column B
(a) Cristae (i) Flat membranous sac in stroma
(b) Cisternae (ii) Infoldings in mitochondria
(c) Thylakoids (iii) Disc-shaped sacs in Golgi apparatus
Solution:
(a) Cristae (ii) Infoldings in mitochondria
(b) Cisternae (iii) Disc-shaped sacs in Golgi apparatus
(c) Thylakoids (i) Flat membranous sac in stroma

Question 4.
Which of the following is correct
(a) Cells of living organisms have a nucleus.
(b) Both animals 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
Solution:
(c) In prokaryotes, there are no membrane-bound organelles.

Question 5.
What are mesosomes in a prokaryotic cells? Mention the function that it performs.
Solution:
A special membranous structure is a mesosome that is formed by the extensions of the plasma membrane into the cell. These extensions are in the form of vesicles, tubules, and lamellae. They help in cell wall formation, DNA replication, and distribution to daughter cells. They also help in respiration, in the secretion process increase plasma membrane surface and enzymatic content.

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?
Solution:
Neutral solutes may move across the membrane by the process of simple diffusion along the concentration gradient i.e. from higher concentration to the lower. Water may also move across this membrane from higher to lower concentrations. The movement of water by diffusion is called osmosis. As the polar molecules cannot pass through the non-polar lipid bilayer, they require a carrier protein of the membrane to facilitate their transport across the membrane.

A few ions or molecules are transported across the membrane against their concentration gradient i.e. from lower to the higher concentration. Such transport is an energy-dependent process, in which ATP is utilized and is called active transport. e,g., Na’/K* Pump.

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.
Solution:
Two double membrane-bound cell organelles:
(a) Mitochondria: It has finger-like folds in the inner membrane called cristae. Mitochondria is the place for aerobic respiration.
NCERT Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 1
(b) Chloroplast: Chloroplast is responsible for converting light energy into chemical energy. Chloroplast contains stacked thylakoid in its matrix.
NCERT Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 2
Functions of mitochondria: The double membrane mitochondria are actively associated with aerobic respiration and the release of energy for cellular activity. The biological oxidation of the fats and carbohydrate release much amount of energy which is utilised by mitochondria for ATP synthesis. The required energy is released from ATP molecules for various cell processes in cells so they are termed as “the powerhouse of the cell”
Functions of chloroplast :
(i) Their main function of the chloroplast is to trap the sun’s energy and to convert it into chemical energy of food by photosynthesis.
(ii) Storage of starch.
(iii) Chloroplast in fruits and flowers changes into chromoplasts.

Question 8.
What are the characteristics of prokaryotic cells?
Solution:
The prokaryotic cells are represented by bacteria, blue-green algae, mycoplasma, and PPLO (Pleuro Pneumonia-like organisms). They are generally smaller and multiply more rapidly than the eukaryotic cells. They may vary greatly in shape and size. The four basic shapes of bacteria are bacillus (rod-like), coccus (spherical), vibrio (comma-shaped), and spirillum (spiral).

The organization of the prokaryotic cell is fundamentally similar even though prokaryotes exhibit a wide variety of shapes and functions. All prokaryotes have a cell wall surrounding the cell membrane. The fluid matrix filling the cell is the cytoplasm. There is no well-defined nucleus. The genetic material is basically naked being not enveloped by a nuclear membrane.

In addition to the genomic DNA (the single chromosome/ DNA circle), many bacteria have small DNA circles outside the genomic DNA. These smaller DNA circles are called plasmids, The plasmid DNA confers certain unique phenotypic to antibiotics. Prokaryotes have something unique in the form of inclusions.

A specialised differentiated form of the cell membrane called mesosome is the characteristic of prokaryotes. They are essentially infoldings of the cell membrane.

Question 9.
Multicellular organisms have a division of labour. Explain.
Solution:
In unicellular organisms, there is no division of labour.

  1. The single cell of the organisms is capable of performing all the vital activities of life i.e., respiration, movement, digestion and reproduction, etc. Respiration, nutrition, and excretion in most of these unicellular organisms takes place through the general body surface.
  2. No special organs for these are present in them because they are too small to need them. Most of these unicellular organisms reproduce by simple binary division, to maintain their continuity.
  3. However, in some, sexual reproduction has also been observed.

Question 10.
Cell is the basic unit of life. Discuss in brief.
Solution:
Cell: The Basic Unit of life: All living organisms are composed of small, tiny structures or compartments called cells. These cells are called the ‘building blocks’ of life.

  1. The cells in true sense are considered as the basic unit of life because all the life processes i.e., metabolism, responsiveness, reproduction are carried out by the cells.
  2. Respiration, nutrition, release of energy for the body are carried out within the cells only.
  3. Even the animals and plants reproduce because the cells reproduce individually.
  4. Growth occurs because cell grow and multiply.
    In Amoeba all the life processes are performed within the boundaries of the single cell.
  5. This is true of all other multicellular organisms. The only difference in the multicellular organisms is that the body of these organisms is made up of many cells.
  6. In these organisms, the cell do not behave independently but get organized into tissues. Each tissue is specialized to perform specific functions. Different tissues then get organised into tissues.
  7. Each tissue is specialized to perform specific functions. Different tissues then get organised into organs which perform certain specific functions.
  8. Different organs are finally organised to form organ systems. Now it must be very clear that the basic structure to tissues, organs and organ system are the cells only.
  9. These tissues, organs and organ system of the organisms work because the cells work.
  10. Thus “the cells are structural and functional unit of the living beings” hence it is the basic unit of life.

Question 11.
What are nuclear pores? State their function.
Solution:
At a number of places, the nuclear envelope is intercepted by minute pores which are called nuclear pores. These are formed by the fusion of two nuclear membranes. These nuclear pores are the passages through which the movement of RNA and protein molecules takes place 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.
Solution:
Lysosomes are filled with hydrolytic enzymes that are capable of digesting carbohydrates, proteins, lipids and nucleic acids whereas vacuoles contain water, sap, excretory product and other materials not useful for cell.

Question 13.
Describe the structure of the following with the help of labelled diagrams.
(i) Nucleus
(ii) Centrosome
Solution:
NCERT Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 3
(i) Nucleus:

    1. The nucleus is a large organelle controlling all the activities of the eukaryotic cells. Some cells have more than one nucleus.
    2. Binucleate cells have 2 nuclei per cell eg. Paramoecium. Multinucleate cells have many nuclei e.g. Ascaris.
    3. Some cells lack nucleus (anucleate) at maturity. Examples: mammalian RBCs and sieve tube cells in vascular plants.
    4. The nucleus is bounded by two membranes, which make the nuclear envelope.
    5. The outer and inner membranes are separated by a narrow space, perinuclear space.
    6. The outer membrane remains in continuation with endoplasmic reticulum (ER) and the inner one surrounds the nuclear contents.
    7. At some points, the nuclear evelope is interrupted by the presence of small structures called nuclear pores.
    8. These pores help in exchange of materials between nucleoplasm and cytoplasm. Nuclear membrane dissappears during cell division. It reappears during nuclear reorganization in stage.
    9. The nucleoplasm contains chromatin and nucleolus. The nucleolus is a rounded structure. It is not separated from the rest of the nucleoplasm by membrane.
    10. It is associated with a specific nucleolar organizing region (NOR) of some chromosomes. Nucleolus is the “site for ribosomal RNA synthesis”.
    11. The cells which remain engaged in protein synthesis have larger and more numerous nuclei in their nucleoplasm.

NCERT Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 4

(ii) Centrosome:

  1. Under the electron microscope, each centriole is seen to be formed of nine sets of tubular structures arranged in a circular fashion.
  2. Each of these sets is a triplet composed of three microtubules. Each microtubule has a diameter of about 250A. The triplets are found in the matrix.
  3. Sometimes delicate strands appear to connect sets of the triplet to each other.
  4. Also can be seen radiating from the central core of the cylinder, delicate strands which connect sets of the triplets to each other giving a cartwheel appearance.
  5. Basal bodies are structures similar to the centrioles. They produce cilia and flagella.

Question 14.
What is a centromere? How does the position of the centromere form the basis of the classification of chromosomes? Support your answer with a diagram showing the position of the centromere on different types of chromosomes.
Solution:
Eukaryotic chromosomes: The chromosomes are uncoiled in a loose, indistinct network called the chromatin that contains DNA, RNA and protein in interphase.

The types of proteins present and associated with DNA are histone and non-histone proteins.

Chromosomes are thread-like structures. They become visible (under light microscope) during cell division.

In higher organisms, the well-developed nucleus contains a definite number of chromosomes of definite size and shape.
NCERT Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 5
The shape of a chromosome is usually observable at metaphase and anaphase when the position of primary constriction {centromere) is clearly seen. Based on the position of the centromere, chromosomes are of 3 types:

  • telocentric – with terminal centromere,
  • the acrocentric – terminal centromere is capped by a telomere
  • submetacentric – the centromere is subterminal in position
  • metacentric – these have median centromere.

VERY SHORT ANSWER QUESTIONS

Question 1.
Who discovered the Golgi body?
Solution:
Camillo Golgi (1898).

Question 2.
Give the location of 70S ribosomes.
Solution:
Prokaryotic cells, plastids, and mitochondria

Question 3.
Name the cell organelle rich in acid hydrolases.
Solution:
Lysosomes

Question 4.
Who proposed the cell theory?
Solution:
Schleiden and Schwann.

Question 5.
Expand PPLO.
Solution:
PPLO (Pleuro Pneumonia Like Organisms)

Question 6.
Name the organelle responsible for protein synthesis in a cell.
Solution:
Ribosome

Question 7.
Give the full form of SER and RER.
Solution:

  • SER – Smooth endoplasmic reticulum.
  • RER – Rough endoplasmic reticulum.

Question 8.
Name the membrane which surrounds the vacuole in the cell.
Solution:
Tonoplast

Question 9.
Name two types of constituents of the plasma membrane.
Solution:
Proteins and lipids.

Question 10.
Name two processes of passive transport.
Solution:

  • Osmosis
  • Diffusion

Question 11.
What is plasmodesmata? What is its function?
Solution:
Plasmodesmata: Adjoining the cells and the linking gap of cytoplasmic protoplasmic presence is called Plasmodesmata. It links the neighbouring cells together.

SHORT ANSWER QUESTIONS

Question 1.
Why fluid-mosaic model is more accepted than other models of the plasma membrane?
Solution:
The fluid mosaic model explains
(i) quasifluid state of the plasma membrane,
(ii) It differentiates two 6. types of proteins
(iii) It explains functional specificity and variability in two surfaces of PM.

Question 2.
Name three types of elements in the Golgi body. List two major functions of the Golgi body.
Solution:
Three types of elements in golgi body are cistemae, vesicles and vacuoles. The main function of golgi bodies are cellular secretion and acrosome formation.

Question 3.
What is peculiar about mitochondrial DNA?
Solution:
Mitochondrial DNA is circular double-stranded and not associated with histone proteins.

Question 4.
How does cytokinesis take place in plant and animal cells?
Solution:
In plant cell cytokinesis take place by cell plate formation and in animal cells it occurs by constriction

Question 5.
Differentiate between prokaryotic and eukaryotic cells.
Solution:
The main differences between prokaryotic cell and eukaryotic cell are
NCERT Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 6

Question 6.
Differentiate between active and passive transport across the membrane.
Solution:
The main differences between active transport and passive transport are
NCERT Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 7

Question 7.
Describe the different methods of transport of nutrients in cell.
Solution:
Different methods of transport of nutrients in the cell are:
(i) Simple diffusion : It is the movement of ions/ molecules of any substance from a region of higher concentration to ‘the region of lower concentration, until equilibrium is reached. Many neutral solutes move by diffusion.
(ii) Osmosis : It is the movement of solvent molecules across a semipermeable membrane, from the region of higher concentration to the region of lower concentration, until equilibrium is reached. Water moves by osmosis from one cell to the other.
(iii) Facilitated diffusion : It refers to the movement of ions/molecules across the membrane with the help of transmembrane proteins.

Question 8.
Why are ribosomes of prokaryotes different from eukaryotes?
Solution:
The type of ribosomes of prokaryotes is different from eukaryotes because the prokaryotes were primitive, simpler and have remained intact during evolution while at the base level eukaryotes have adapted with the environment and are retaining their kind of entities for the complex structure. Prokaryotes have 70S ribosomes with 30S and 50S subunit and eukaryotes have 80S ribosome with 40S subunit and 60S sub unit.

Question 9.
What is the function of
(1) Nuclear Pores
(2) Slimy Capsule in Bacteria
(3) Golgi Bodies:
(4) Centrosome with Centrioles
Solution:
(1) Nuclear Pores: There is exchange of RNA and proteins through the nuclear pores
(2) Slimy Capsule in Bacteria : A slimy capsule is the outer covering of cell wall of bacteria and is an additional protection for the bacteria.
(3) Golgi Bodies :
(i) It takes part in packaging materials delivered either to the intra-cellular targets or secteted outside the cell.
(ii) It is also a important site of formation of glycoproteins and glycolipids.
(4) Centrosome with Centrioles
(i) Centrioles help in organising the spindle fibres and astral rays during cell division.
(ii) It also provides basal bodies which give rise to cilia and flagella.

LONG ANSWER QUESTIONS

Question 1.
Describe the structure of cell wall.
Solution:

  • A non-living rigid structure called the cell wall forms an outer covering for the plasma membrane of fungi and plants.
  • Cell wall does not only give shape to the cell and but protect the cell from mechanical damage and infection.
  • It also helps in cell-to-cell interaction and provides barrier to undesirable macromolecules.
  • Algae have cell wall, made of cellulose, galactans, mannans and minerals like calcium carbonate, while in other plants it consists of cehulose, hemicellulose, pectins and proteins.
  • The cell wall of a young plant cell, the primary wall is capable of growth, which gradually diminishes as the cell matures and the secondary wall is formed on the inner side of the cell.
  • The middle lamella is a layer mainly of calcium pectate which holds or glues the different neighbouring cells together.
  • The cell wall and middle lamellae may be transferred by plasmodesmata which connect the cytoplasm of neighbouring cells.

Question 2.
Give an account of prokaryotic cells.
Solution:

  • The prokaryotic cells are represented by bacteria, blue green algae, mycoplasma and PPLO (Pleuro Pneumonia Like Organisms). They are generally smaller and multiply more rapidly than the eukaryotic cells.
  • They may vary greatly in shape and size. The four basic shapes of bacteria are bacillus (rod like), coccus (spherical), vibrio (comma shaped) and spirillum (Spiral).
    The organisation of the prokaryotic cell is fundamentally similar even though prokaryotes exhibit a wide variety of shapes and functions. All prokaryotes have a cell wall surrounding the cell membrane.
  • The fluid matrix filling the cell is the cytoplasm. There is no well defined nucleus.
  • The genetic material is basically naked, not enveloped by a nuclear membrane.
  • In addition to the genomic DNA (the single chromosome/circular DNA), many bacteria have small circular DNA outside the genomic DNA. These smaller DNA is called plasmids.
  • The plasmid DNA confers certain unique phenotypic characters to such bacteria. One such character is resistance to antibiotics. Nuclear membrane is found in eukaryotes.
  • No organelles, like the ones in eukaryotes, are found in prokaryotic cells except ribosomes.
  • Prokaryotes have something unique in the form of inclusions. A specialised differentiated form of cell membrane called mesosome is the characteristic of prokaryotes which helps in respiration process.
  • They are essentially infoldings of cell membrane.

Question 3.
Give an ultrastructure of mitochondria.
Solution:

  • Mitochondria, unless specifically stained, are not easily visible under the microscope.
  • The number of mitochondria per cell is variable depending on the physiological activity of the cells.
  • In terms of shape and size also, considerable degree of variability is observed.
  • Typically it is sausage shaped or cylindrical having a diameter of 0.2-1.0 ft m (average 0.5 film) and length (1.0 -4.1 ft).
  • Each mitochondrion is a double membrane-bound structure with the outer membrane and the inner membrane dividing its lumen distinctly into two aqueous compartments, i.e. the outer compartment and the inner compartment.
  • The inner compartment is called the matrix. The outer membrane forms the continuous limiting boundary of the organehe.
  • The inner membrane forms a number of infoldings called the cristae. The cristae increase the surface area.
  • The two membranes have their own specific enzymes associated with the mitochondrial function. Mitochondria are the sites of aerobic respiration.
  • They produce cellular energy in the form of ATP, hence they are called. “Power houses” of the cell.
  • The matrix also possesses single circular DNA molecule, a few RNA molecules, ribosomes (70s) and the components required for the synthesis of proteins. The Mitochondria divide by fission.
    NCERT Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 8

Question 4.
Describe the fluid mosaic model of membrane.
Solution:
The characteristic features of fluid mosaic model
• This model was proposed by Singer and Nicholson.
• According to this model, there is a central bilipid layer (of phospholipids) with their polar head group toward the outside and the non-polar tails pointing inwards.
• Some proteins which are embedded in the lipid layer are called integral proteins and they cannot be separated from the membrane easily.
• Some large globular integral proteins which project beyond the lipid layer on both the sides are believed to have channels through which water soluble materials can pass across.
• Those proteins which are superficially attached are called peripheral (extrinsic) proteins and they can be easily removed.
• Some membrane lipids and integral proteins remain bound to oligosaccharides; such oligosaccharides project into the extracellular fluid and they influence the manner in which cells interact with the other cell.
• There are also certain specific proteins called membrane receptors, which mediate the flow of materials and information into the cell.

Question 5.
What are plastids? How are they classified on the basis of the type of pigments? Name them and their pigments and mention their functions.
Solution:
Plastids are double-membrane bound organelles
of different shapes, that are found only in plant
cells and contain pigments and storage products.
They are of three types :
(i) Leucoplasts
These are the oval, spherical, rod-like or filamentous colourless plastids which are found in storage organs. Their main function is to store reserve materials like starch (amyloplasts), proteins (aleuroplasts) and fats (elaioplasts).
(ii) Chromoplasts
• These are coloured plastids containing mainly the yellow, red and orange pigments (carotene and xanthophyll).
• These are found in petals of flowers and skin of fruits.
• They attracts agents for pollination and dispersal of fruits/seeds.
(iii) Chloroplasts
• These are the green plastids containing mainly chlorophylls and very little carotene and xanthophyll.
• Their main function is photosynthesis and the formation of starch.

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

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

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

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

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

Question 2.
Distinguish cytokinesis from karyokinesis.
Solution:

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

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

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

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

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

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

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

  • Prophase
  • Meta phase
  • Anaphase
  • Telophase

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

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

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

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

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

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

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

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

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

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

Question 13.
Discuss with your teacher about

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

Solution:

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

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

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

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

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

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

VERY SHORT ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

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

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

SHORT ANSWER QUESTIONS

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

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

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

Significance of Meiosis:

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

Question 3.
Mention the significance of mitosis.
Solution:

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

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

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

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

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

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

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

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

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

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

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

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

LONG ANSWER QUESTIONS

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

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

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

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

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

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

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

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

We hope the NCERT Solutions for Class 11 Biology at Work Chapter 10 Cell Cycle and Cell Division, help you. If you have any query regarding NCERT Solutions for Class 11 Biology at Work Chapter 10 Cell Cycle and Cell Division, drop a comment below and we will get back to you at the earliest.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

(2) Malpighian tubules:

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

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

VERY SHORT ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

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

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

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

SHORT ANSWER QUESTIONS

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

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

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

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

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

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

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

LONG ANSWER QUESTIONS

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

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

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

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

(ii) Dense connective tissue :

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

(iii) Specialised connective tissue :

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

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

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

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

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NCERT Solutions for Class 11 Biology Chapter 12 Mineral Nutrition

NCERT Solutions for Class 11 Biology Chapter 11 Transport in Plants

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

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

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

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

  1. Macronutrients
  2. Micronutrients

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

VERY SHORT ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

SHORT ANSWER QUESTIONS

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

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

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

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

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

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

  • Translocation of carbohydrates
  • Pollen germination

Deficiency symptoms

  • Death of root and shoot tips
  • Abscission of flowers

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

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

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

LONG ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis

NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

VERY SHORT ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

SHORT ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

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

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

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

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

LONG ANSWER QUESTIONS

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

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

Internal factors:

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

External factors:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Vascular system:

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

VERY SHORT ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

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

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

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

Question 13.
Where these present:

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

Solution:

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

SHORT ANSWER QUESTIONS

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

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

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

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

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

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

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

LONG ANSWER QUESTIONS

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

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

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

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

Functions:

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

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

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

(b) Sclereids –

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

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

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

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

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

NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants

NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants.

Question 1.
Differentiate between
(a) Respiration and Combustion
(b) Glycolysis and Krebs’ cycle
(c) Aerobic respiration and Fermentation
Solution:
(a) Differences between respiration and combustion are as follows :
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 1
(b) Differences between glycolysis and krebs’ cycle are as follows :
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 2
(c) Differences between aerobic respiration and fermentation are as follows :
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 3

Question 2.
What are respiratory substrates? Name the most common respiratory substrate.
Solution:
The compounds that are oxidized during this process are known as respiratory substrates. Usually, carbohydrates are oxidized to release energy, but proteins, fats, and even organic acids can be used as respiratory substances in some plants, under certain conditions.

Question 3.
Give the schematic representation of glycolysis.
Solution:
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 4

Question 4.
What are the main steps in aerobic respiration? Where does it take place?
Solution:
In aerobic respiration which takes place within the mitochondria, the final product of glycolysis, pyruvate is transported from the cytoplasm into the mitochondria.
The crucial events in aerobic respiration are:
• The complete oxidation of pyruvate by the stepwise removal of all the hydrogen atoms, leaving three molecules of CO2.
• The passing on of the electrons removed as part of the hydrogen atoms to molecular O2 with the simultaneous synthesis of ATP.
• The first process takes place in the matrix of the mitochondria while the second process is located on the inner membrane of the mitochondria.
• Pyruvate, which is formed by the glycolytic catabolism of carbohydrates in the cytosol, after it enters mitochondrial matrix undergoes oxidative decarboxylation by a complex set of reactions catalysed by pyruvic dehydrogenase. The reactions catalysed by pyruvic dehydrogenase require the participation of several coenzymes, including NAD+ and Coenzyme A.
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 5
During this process, two molecules of NADH are produced from the metabolism of two molecules of pyruvic acid (produced from one glucose molecule during glycolysis).
The acetyl CoA then enters a cyclic pathway, the tricarboxylic acid cycle, more commonly called as Krebs’ cycle.

Question 5.
Give the schematic representation of an overall view of Krebs’ cycle.
Solution:
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 6

Question 6.
Explain ETS.
Solution:
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 FADH2 NADH+ H+, which is formed during glycolyis and citric acid cycle, gets oxidized by NADH dehydrogenase. The electrons so generated get transferred to ubiquinone through FMN. In a similar manner, FADH2 generated during citric acid cycle gets transferred to ubiquinone. The electrons from ubiquinone are received by cytochrome bc1, 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 a3, along with copper centres.

NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 7
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. The amount of ATP produced depends on the molecule, which has been oxidized.

Question 7.
Distinguish between the following:
(a) Aerobic respiration and anaerobic respiration.
(b) Glycolysis and fermentation.
(c) Glycolysis and citric acid cycle.
Solution:
Differences between aerobic respiration and anaerobic respiration are as follows :
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 8
Differences between glycolysis and fermentation are as follows :
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 9
Differences between glycolysis and citric acid cycle are as follows :
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 10

Question 8.
What are the assumptions made during the calculation of the net gain of ATP?
Solution:

  • There is a sequential, orderly pathway functioning, with one substrate forming the next and with glycolysis, TCA cycle, and ETS pathway following one after another.
  • The NADH synthesized in glycolysis is transferred into the mitochondria and undergoes oxidative phosphorylation.
  • None of the intermediates in the pathway are utilized to synthesize any other compound.
  • Only glucose is being respired -no other alternative substrates are entering the pathway at any of the intermediary stages.

Question 9.
Discuss “The respiratory pathway is an amphibolic pathway”.
Solution:
Respiration is generally assumed to be a catabolic process because, during respiration, various substrates are broken down for deriving energy. Carbohydrates are broken down into glucose before entering respiratory pathways. Fats get converted into fatty acids and glycerol whereas fatty acids get converted into acetyl CoA before entering respiration. In a similar manner, proteins are converted into amino acids, which enter respiration after deamination.

During the synthesis of fatty acids, acetyl CoA is withdrawn from the respiratory pathway. Also, in the synthesis of proteins, respiratory substances 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?
Solution:
The ratio of the volume of CO2 evolved to the volume of O2 consumed in respiration is called respiratory quotient (RQ) or respiratory ratio.
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 11

Question 11.
What is oxidative phosphorylation ?
Solution:

Although the aerobic process of respiration takes place only in the presence of oxygen, the role of oxygen is limited to the terminal stage of the process. Yet, the presence of oxygen is vital, since it drives the whole process by removing hydrogen from the system. Oxygen acts as the final hydrogen acceptor. Unlike photophosphorylation where it is the light energy that is utilized for the production of proton gradient required for phosphorylation, in respiration, it is the energy of oxidation-reduction utilized for the same process. It is for this reason that the process is called oxidative phosphorylation.

Question 12.
What is the significance of the step-wise release of energy in respiration?
Solution:
During oxidation within a cell, all the energy contained in respiratory substrates is not released free into the cell, or in a single step. It is released in a series of slow step-wise reactions controlled by enzymes, and it is trapped as chemical energy in the form of ATP.

Hence, it is important to understand that the energy released by oxidation in respiration is not used directly but is used to synthesise ATP, which is broken down whenever (and wherever) energy needs to be utilised. Hence, ATP acts as the energy currency of the cell.

This energy trapped in ATP is utilised in various energy-requiring processes of the organisms, and the carbon skeleton produced during respiration is used as precursors for the biosynthesis of other molecules in the cell.

VERY SHORT ANSWER QUESTIONS

Question 1.
What is anaerobic respiration? (Oct. 83)
Solution:
Incomplete or partial breakdown of fuel molecules into compounds such as ethyl alcohol, lactic acid in the absence of molecular oxygen.

Question 2.
Name the final acceptor of an electron in ETC.
Solution:
Oxygen is the electron acceptor of ETC.
Question 3.
The function of oxygen in aerobic respiration:
(i) It acts as the final electron acceptor.
(ii) It drives the whole process by removing hydrogen from the system.
Solution:
The function of oxygen in aerobic respiration:
(i) It acts as the final electron acceptor.
(ii) It drives the whole process by removing hydrogen from the system.

Question 4.
What is respiration? (Oct. 86)
Solution:
The oxidative process in which chemically bound energy from complex organic fuel molecules such as carbohydrates, proteins, and fats is captured in the form of ATP.

Question 5.
Where does the electron transport system operate in the mitochondria?
Solution:
Phosphofructokinase catalyses the formation of fructose 1, 6 bisphosphates from fructose 6-phosphate.

Question 6.
Give the function of phosphofructokinase in glycolysis.
Solution:
Hexokinase-helps in the phosphorylation of glucose.

Question 7.
Name the enzyme that catalyses the phosphorylation of glucose.
Solution:
The formation of acetyl CoA takes place in the mitochondrial matrix.

Question 8.
Where does the formation of acetyl CoA take place in a cell?
Solution:
The first step in the Krebs cycle is the condensation of an acetyl group (acetyl CoA) with oxaloacetic acid (OAA) to form citric acid and release the Coenzyme A.

Question 9.
What is the first step of reaction in the TCA cycle?
Solution:
Fatty acids may be converted to acetyl CoA before they from the respiratory substrates.

Question 10.
What is alcoholic fermentation?
Solution:
Alcoholic fermentation is the process by which yeast cells breakdown glucose into ethyl alcohol and carbon-dioxide under anaerobic conditions.

Question 11.
Name the oxidative pathway through which intermediate metabolites of glucose, fatty acids, and amino acids are finally oxidised.
Solution:
36 ATP/38 ATP molecules are obtained in the process of respiration and it is related to the aerobic respiration type.

Question 12.
What is lactic acid fermentation? (Oct. 2001)
Solution:
It is the process of fermentation by which lactose found in milk is converted to lactic acid by the action of lactobacillus.

Question 13.
What are the two molecules obtained by the action of aldolase from fructose -1, -6- biphosphate?
Solution:
ATP is produced.

SHORT ANSWER QUESTIONS

Question 1.
How is the proton gradient established?
Solution:
The proton gradient is established by passing proton (H+) from the matrix across the inner mitochondrial membrane into intermembrane space with the energy released during electron transfers in ETC.

Question 2.
Describe the steps in the formation of lactic acid from pyruvic acid.
Solution:
Pyruvic acid is catalysed by the enzyme lactic dehydrogenase. NADH formed in glycolysis is used up for the reduction.
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 12

Question 3.
How is ATP formed by the energy released during the electron transport system in mitochondria?
Solution:
ATP formations require an enzyme called ATP synthase. It has two components F0– F1. ATP- synthase becomes active in ATP formation when the concentration of H+ on the Fo side is higher than the F1 side. Fligher proton concentration in the outer chamber causes the proton to pass the inner chamber. F1 particle induced by the flow of proton through Fo channel. The energy of the proton gradient attaches the phosphate radicle to ADP. This produces ATP.

Question 4.
Give a detailed account of the net gain of ATP at a different stages of respiration.
Solution:
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 13
In most eukaryotic cells 2 molecules of ATP are required for transporting NADH produced in glycolysis to mitochondria for further oxidation. Hence net gain of ATP is 36 molecules.

Question 5.
Enumerate the functions of ATP.
Solution:
Functions of ATP:-
(i) ATP functions as a universal energy carrier of living systems.
(ii) ATP stores small packets of energy in its molecules.
(iii) It is mobile in the cell. Therefore, it reaches all parts of the cell away from the region of ATP synthesis.
(iv) It activates a number of chemicals by functioning as a phosphorylating agent.
(v) ATP provides energy for muscle contraction.
(vi) It is involved in the transport of substances against a concentration gradient.

Question 6.
Where is cytochrome c located? What is its function?
Solution:
Cytochrome c is located on the outer surface of the inner mitochondrial membrane. It acts as a mobile carrier for the transfer of electrons between complex III and complex IV of the electron transport system.

Question 7.
Define respiratory quotient.
Solution:
The respiratory quotient is defined as the ratio of the volume of carbon dioxide evolved to the volume of oxygen consumed in respiration.

Question 8.
What is oxidative phosphorylation?
Solution:
The whole process by which oxygen effectively allows the production of ATP by phosphorylation of ADP is called oxidative phosphorylation.

Question 9.
The energy yield in terms of ATP is higher in aerobic respiration than during anaerobic respiration. Why is there anaerobic respiration even in organisms that live in aerobic conditions like human beings and angiosperms?
Solution:
Aerobic organisms do face situations where oxygen availability is little. For example, overworked muscles do not receive enough oxygen during strenuous exercise. Similarly, deep-seated tissues of angiosperms do not receive enough oxygen through diffusion from outside. In such situations, only anaerobic respiration can help in the survival of the tissue.

Question 10.
Comment on the statement- “Respiration is an energy-producing process but ATP is used in some steps of the process”.
Solution:
ATP is required in all those reactions where phosphorylative activation of the substrate is required. Therefore, despite producing energy (as ATP), respiration requires ATP in certain steps, e.g., glucose – glucose 6-phosphate, fructose 6-phosphate —fructose 1, 6- bisphosphate.

LONG ANSWER QUESTIONS

Question 1.
Explain the major steps in Krebs’ cycle. Why is this cycle also called the citric acid cycle?
Solution:
Krebs cycle: This process occurs in the mitochondrial matrix.
Major steps of Krebs cycle are as follows :

  • Acetyl Co-A, formed by the oxidative decarboxylation of pyruvic acid enters the Krebs’ cycle.
  • It combines with oxalo acetic acid (OAA), a 4C-compound, to form a 6C-compound, citric acid; the reaction is catalysed by citrate synthase.
  • Citrate is then isomerised into isocitrate.
    NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 14
  • Isocitrate is converted into oxalosuccinic acid in the presence of NAD and isocitrate dehydrogenase.
  •  Oxalosuccinic acid is then decarboxylated into a-ketoglutaric acid (KG), in the presence of a decarboxylase enzyme.
  •  a-ketoglutaric acid is converted into succinyl Co-A in the presence of NAD, Co- A, and enzyme a-ketoglutarate dehydrogenase.
  •  When succinyl Co-A is converted into succinic acid, one molecule of GTP is formed and Co-A is released.
  •  In the remaining part of the cycle, succinic acid is converted into OAA, so that the citric acid cycle can continue to operate.
  •  During this cycle, three molecules of NAD and one molecule of FAD are reduced to NADH and FADH respectively.
  •  This cycle is called as a citric acid cycle because the first product is citric acid which is 3-C compound.

Question 2.
Name the end product of glycolysis. Where is it produced in the cell? Discuss oxidative decarboxylation.
Solution:
Glycolysis results in the formation of two molecules of pyruvic acid, NADH, and ATP. It occurs in the cytosol of the cell.
Aerobic oxidation: One of the three carbons of pyruvic acid is oxidised to carbon dioxide in the reaction called oxidative decarboxylation. Pyruvic acid is first decarboxylated and then oxidised by the enzyme pyruvic dehydrogenase. The two-carbon units are readily accepted by coenzyme-A (Co-A) to form acetyl Co-A. The summary of the reaction is given in the following equation :
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 15
Thus, pyruvic acid enters the Krebs cycle as acetyl Co-A. Krebs’ cycle occurs in the mitochondrial matrix.
Acetyl Co-A, formed by the oxidative decarboxylation of pyruvic acid enters the Krebs’ cycle.

Question 3.
Represent schematically the interrelationship among metabolic pathways in a plant, showing respiration mediated breakdown of different organic compounds.
Solution:
Schematic representation among metabolic pathways showing respiration mediated breakdown of different organic molecules to CO2 and H2O:
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 16
Question 4.
How do plants manage the exchange of gases? Give an overview of respiration in plants.
Solution:
Plants, unlike animals, have no specialized organs for gaseous exchange but they have stomata and lenticels for this purpose. There are several reasons why plants can get along without respiratory organs.

  • Each plant part takes care of its own gas- exchange needs. There is very little transport of gases from one plant part to another.
  • Plants do not present great demands for gas exchange. Roots stem and leave respire at a lower rate than animals do.
  • Only during photosynthesis, large volumes of leases exchanged and, each leaf is well adapted to take care of its own needs during these periods.
  • When cells perform photosynthesis, the availability of O2 is not a problem in these cells since 02 is released
  • The distance that gases must diffuse even in large, bulky plants is not great. Each living cell in a plant is located quite close to the surface of the plant.
  • Even in woody stems, the ‘living’ cells are organised in thin layers inside and beneath the bark. They also have openings called lenticels. The cells in the interior are dead and provide only mechanical support.
  • Thus, most cells of a plant have attested to a part of their surface in contact with air. This is also facilitated by the loose packing of parenchyma cells in leaves, stems, and roots, which provide an interconnected network of air spaces.
  • The complete combustion of glucose, which produces C02 and H20 as end products, yields energy. Most of the energy is given out as heat.
    C6H12O6 + 6O2 → 6C02 + 6H20 + Energy
  • If this energy is to be useful to the cell, it should be able to utilise it to synthesis other molecules that the cell requires.
  • The strategy that the plant cell uses is to catabolize the glucose molecule in such a way that not all the liberated energy goes out as heat.
  • The key is to oxidise glucose not in one step but in several small steps enabling some steps to be just large enough so that the energy released can be coupled to ATP synthesis.

We hope the NCERT Solutions for Class 11 Biology at Work Chapter 14 Respiration in Plants, helps you. If you have any query regarding NCERT Solutions for Class 11 Biology at Work Chapter 14 Respiration in Plants, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Biology Chapter 2 Biological Classification

NCERT Solutions for Class 11 Biology Chapter 2 Biological Classification

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 2 Biological Classification.

Question 1.
Discuss how classification systems have undergone several changes over a period of time?
Solution:
The classification was born instinctively out of a need to all organisms for our own use since the dawn of civilization. Aristotle was the earliest to attempt a more scientific basis for classification. He used simple morphological characteristics to classify plants into trees, herbs and shrubs. He also divided animals into Animals with red blood and those who do not have red blood. Linnaeus proposed a two-kingdom system of classification with Plantae and Animalia including plants and animals respectively.

The above system did not distinguish eukaryotes and prokaryotes, unicellular and multicellular, photosynthetic and non-photosynthetic organisms. A need was felt for including besides gross morphology, other characteristics like cell structure, nature of cell wall, mode of nutrition, habitat, method of reproduction, evolutionary relationships etc., Recently R.H. Whittaker proposed a five-kingdom classification to answer above the Five kingdoms are

  • Monera
  • Protista
  • Fungi
  • Plantae
  • Animalia.

Question 2.
State two economically important uses of
(a) heterotrophic bacteria
(b) archaebacteria
Solution:
(a) Heterotropic bacteria : These bacteria are natural scavengers. The souring of milk into lactic acid and alcohol to vinegar is brought about by some saprophytic bacteria, e.g., Lactic acid bacteria and acetic acid bacteria respectively.
A number of antibiotic are extracted from actinomycetes especially from the genus Streptomyces e.g. Streptomycin, Chloramphenicol, Oilorotetracycline, Erythromycin, Terramycin etc.
(b) Archaebacteria live as symbionts in the rumen of herbivorous animals.
Methanogens are present in the guts of several ruminant animals such as cows and buffaloes and they are responsible for the production of methane (biogas) from the dung of these animals.

Question 3.
What is the nature of the cell wall in diatoms?
Solution:
In diatoms, ail walls form two thin overlapping shells which fit together as in a soapbox. The walls are embedded with silica and thus the walls are indestructible. Thus diatoms have left behind a large amount of cell wall deposits in their habitat.

Question 4.
Find out what do the terms ‘algal bloom’ and ‘red tides’ signify?
Solution:
Algal bloom : When colour of water changes due to profuse growth of coloured phytoplankton, it is called algal bloom.
Red tides : Redness of the red sea is due to luxurient growth of Trichodesrium erythrium, a member of cynobacteria (blue green alage).

Question 5.
How are viroids different from viruses?
Solution:
Viroids are simpler than viruses, consisting of
a single RNA molecule that is not covered by protein capsid. The genetic material of viruses are surrounded by protein coat.

Question 6.
Describe briefly the four major groups of protozoa.
Solution:
The four major group of protozoa are flagellated protozoan, amoeboid protozoan, sporozoan, ciliated protozoan. The main characters of these group are as follows:
NCERT Solutions for Class 11 Biology Chapter 2 Biological Classification 1

Question 7.
Plants are autotrophic. Can you think of some plants that are partly hetrotrophic?
Solution:
Bladderwort and venus fly trap are examples of insectivorous plants and Cuscuta is a parasite. These are plants which are partially heterotrophic.

Question 8.
What do the terms phycobiont and mycobiont signify?
Solution:
Lichens shows symbiotic association between algae and fungi. The fungal component of lichen is called mycobiont and the algal component is called phycobiont.

Question 9.
Give a comparative account of the classes of kingdom fungi under the following:
(a) Mode of nutrition
(b) Mode of reproduction
Solution:
Kingdom fungi has four classes, these are Phycomycetes, ascomycetes, basidiomycetes and Deuteromycetes. The comparison between these classes are as follows :
NCERT Solutions for Class 11 Biology Chapter 2 Biological Classification 2

Question 10.
What are the characteristic features of Euglenoids?
Solution:
Euglenoids show the following characteristic features:

  • They store carbohydrates in the form of paramylon.
  • Since euglenoids are green and holophytic like other plants.
  • Few are non-green and saprophytic, some are holotropic.
  • They bear a red-pigmented eyespot and a gullet near the base of flagellum.
  • All the euglenoids have one or two flagella which help in swimming.
  • Absence of cell-wall but contain flexible pellicle made up of protein.
  • Freshwater, free-living found in ponds and ditches.

Question 11.
Give a brief account of viruses with respect to their structure and the nature of genetic material. Also, name four common viral diseases.
Solution:
Viruses have the following characteristics:
(i) All plant viruses have single-stranded RNA and all animal viruses have either single or double-stranded RNA or double-stranded DNA.
(ii) Protein vims also contain genetic material RNA or DNA. A vims is a nucleoprotein and the genetic material is infectious, These are obligate parasites, self replicating, non-cellular organisms.
(iii) Vimses are smaller than bacteria and their genetic material is surrounded by protein I coat called capsid. Capsid is made up of small subunits called capsomeres.
Four common viral diseases are :
(a) Cough and cold
(b) Mumps
(c) Influenza
(d) Smallpox

Question 12.
Organise a discussion is your class on the topic are viruses living or non-living?
Solution:
Vimses are link between living and non-living. They possess some living characters and some non-living characters. Crystallization is a non-living character but it can reproduce inside living body.
Actually vimses are metabolically inert when outside the host-cell. They reproduce using the metabolic machinery of the host cell.

VERY SHORT ANSWER QUESTIONS

Question 1.
Who wrote the books ‘Species Plantarum’ and ‘Systema Naturae?
Solution:
Carolus Linnaeus.

Question 2.
Name the two kingdoms of the living world proposed by Linnaeus.
Solution:
Plantae and Animalia

Question 3.
What are protists?
Solution:
Protists are unicellular, eukaryotic organisms.

Question 4.
Which organism was earlier placed in the plant as well as animal kingdoms and why?
Solution:
Euglena because it has locomotory organelle, flexible pellicle, contractile vacuole and reproduce by binary fission like animals and chloroplasts and pyrenoids like plants.

Question 5.
Name the 5 kingdoms of organisms in the order of their supposed evolution.
Solution:
Monera, Protista, Fungi, Animalia and Plantae.

Question 6.
Mention 2 traits in which fungi resemble animalia.
Solution:
Heterotrophy and glycogen as reserve food.

Question 7.
Define
(a) Plasmogamy
(b) Karyogamy
Solution:
(a) Plasmogamy – Fusion of protoplasms between two motile or non-motile gametes.
(b) Karyogamy – Fusion of two nuclei.

Question 8.
What is a retrovirus? Give an example
Solution:
Retrovirus is organisms that have RNA s as genetic material. For example HTV

Question 9.
Give two salient features of slime moulds.
Solution:
The two salient features of slime moulds are:

  1. These do not have a cell wall
  2. These have pseudopodia for movement

Question 10.
What is called the jokers of microbiology and why?
Solution:
Jokers of microbiology are mycoplasma as they have no cell wall and no definite shape.

Question 11.
Give the names of two diseases caused by Protozoans
Solution:
Two diseases caused by protozoans are
(1) Amoebiasis
(2) Malaria

SHORT ANSWER QUESTIONS

Question 1.
Cyanobacteria play a major role in our ecology. Discuss.
Solution:
Cyanobacteria, also known as ‘blue-green algae’ help in carbon fixation in a major way on the ocean surface.

They are helpful in nitrogen fixation in paddy fields leading to a better harvest. About 80% of photosynthesis on ocean surface is done by cyanobacteria. So, it can be said that they play a major role in our ecology.

Question 2.
What is the role of methanogens?
Solution:
Methanogens are type of bacteria which live in the gut of ruminating animals.

They assist those animals in digestion and the byproduct of that digestive process is methane.

More number of livestock population results in increased methane level in the environment leading to global warming. So, indirectly methanogens can be responsible for global warming.

Question 3.
What are lichens? What are the roles of lichen in water pollution ?
Solution:
Lichens are symbiotic associations i.e. mutually useful associations, between algae and fungi.

The algal component is known as phycobiont and fungal component as mycobiont, which are autotrophic and heterotrophic, respectively.

Algae prepare food for fungi and fungi provide shelter and absorb mineral nutrients and water for its partner. Lichens are very good pollution indicators as they do not grow in polluted areas.

Question 4.
On what factors is the 5 kingdom classification of Whittaker based?
Solution:
The five kingdom classification is based upon the following factors :
(i) Complexity of cell structure – Prokaryotes or Eukaryotes
(ii) Complexity of organisms body – Unicellular or Multicellular
(iii) Mode of obtaining nutrition – Autotrophs or Heterotrophs
(iv) Phylogenetic relationships

Question 5.
Give the technical terms used for the following:
(a) Remains of an organism of a former geological age.
(b) Science of classification of organisms.
(c) Evolutionary history of a group of organisms.
(d) Organisms which synthesize their own food, using chemical energy.
Solution:
(a) Fossils
(b) Taxonomy
(c) Evolution
(d) Autotrophs

Question 6.
What are the kinds (shapewise) bacteria found in nature. Name the pathogen with the disease caused
Solution:
NCERT Solutions for Class 11 Biology Chapter 2 Biological Classification 3

 

Question 7.
Why is
(i) Basidiomycetes called club fungi?
(ii) Ascomycetes called sac fungi?
Solution:
(i) After sexual reproduction basidium is formed which form the shape a club and this chin these fungi are called Club Fungi.
(ii) In sexual reproduction ascospores are formed in a sac like asci and thus this fungi is called sac fungi.

LONG ANSWER QUESTIONS

Question 1.
Give an account of early work in taxonomy.
Solution:

  • Since the dawn of civilisation, there have been many attempts to classify living organisms.
  • It was done instinctively not using criteria that were scientific but borne out of a need to use organisms for our own use – for food, shelter and clothing.
  • Aristotle was the earliest to attempt a more scientific basis for classification. He used simple morphological characters to classify plants into trees, shrubs and herbs.
  • He also divided animals into two groups, those which had red blood and those that did not.
  • In Linnaeus’ time a Two Kingdom system of classification with Plantae and Animalia kingdoms was developed that included all plants and animals respectively.
  • Classification of organisms into plants and animals was easily done and was easy to understand, inspite, a large number of organisms did not fall into either category.
  • R.H. Whittaker (1969) proposed a Five Kingdom Classification.
  • The kingdoms defined by him were named Monera, Protista, Fungi, Plantae and Animalia.
  • The main criteria for classification used by him include cell structure, thallus organisation, mode of nutrition, reproduction and phylogenetic relationships.

Question 2.
Differentiate briefly characteristics of kingdom Plantae and Animalia.
Solution:

  • Kingdom Plantae includes all eukaryotic chlorophyll-containing organisms commonly called plants.
  • A few members are partially heterotrophic such as the insectivorous plants or parasites.
  • Bladderwort and Venus fly trap are examples of insectivorous plants and Cuscuta is a parasite.
  • The plant cells have an eukaryotic structure with prominent chloroplasts and cell wall mainly made of cellulose.
  • Plantae includes algae, bryophytes, pteridophytes, gymnosperms and angiosperms.
  • The animal kingdom is characterised by v heterotrophic eukaryotic organisms that are multicellular and their cells lack cell walls.
  • They directly or indirectly depend on plants for food. They digest their food in an internal cavity and store food reserves as glycogen or fat.
  • Their mode of nutrition is holozoic – by ingestion of food. They follow a definite growth pattern and grow into adults that have a definite shape and size.
  • Higher forms show elaborate sensory and neuromotor mechanism. Most of them are capable of locomotion.

Question 3.
Give the economic importance of diatoms. Diatoms are used
Solution:
(1) as a cleaning agent in tooth pastes and metal polishes.
(2) Adding to make sound proof rooms.
(3) In Alteration of sugar, alcohol and antibiotics
(4) as put in paints to ad the paint visibility at night
(5) as an insulating material in Refrigerators, fumances etc.

Question 4.
What are the distinguishing characters of kingdom fungi?
Solution:
The distinguishing characters of kingdom fungiare as follows :
(i) Fungi are non-vascular, non-seeded, non-flowering, eukaryotic achlorophyllous (nongreen), heterophic (heterophytic) spore bearing, thalloid, multicellular decomposers and mineralisers of organic wastes and help in recycling of matter in the biosphere.
(ii) In true fungi the plant body is thallus. It may be non-mycelial or mycelial.
a. Non mycelial: The non-mycelial forms are unicellular; however they may form a pseudomycelium by budding,
b. Mycelial: In mycelial form plant body is made up of thread like structures called hyphae. Hyphae are usually branched tube like structure bounded by a cell-wall of chitin. The hyphae may be septate (higher fungi) or aseptate (lower fungi).
Septate hyphae may be of 3 kinds, uninucleate (monokaryotic hyphae), with binucleate cells (dikaryotic hyphae) ormultinucleate. Some fungi are aseptate and known as coenocytic fungi, with hundreds of nuclei in continuous cytoplasmic mass.
(iii) The cell shows eukaryotic organization but lack chloroplast and Golgi bodies. The genetic material is DNA and mitosis is intracellular (karyochorisis).
(iv) Fungi lack chlorophyll, hence, they do not prepare food by photosynthesis. Thus they can grow everywhere, where organic material is available.
(v) Fungi are heterotrophs that acquire their nutrient by absorption. They store their food in the form of glycogen.
(vi) The primitive fungi have oogamous type of sexual reproduction where as most advanced ones do not have sexual reproduction.

Question 5.
Compare the main features of Monera with Protista.
Solution:
NCERT Solutions for Class 11 Biology Chapter 2 Biological Classification 4

 

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NCERT Solutions for Class 11 Biology Chapter 1 The Living World

NCERT Solutions for Class 11 Biology Chapter 1 The Living World

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 1 The Living World.

Question 1.
Why are living organisms classified?
Solution:
Classification of living organisms grouped them in special categories, which is based on observable characters. It makes their study easy and convenient. For example, Mammals are those who possess mammary glands, the hair on the body, external pinnae, etc.

Question 2.
Why are classification systems changing every now and then?
Solution:
The classification system changes when more information becomes available about the organisms. Additional information are updated from time to time about different organisms at this stage there is a need arises to make changes in the classification system.

Question 3.
What different criteria would you choose to classify people that you meet often?
Solution:
Classification means the arrangement of organisms into groups on the basis of their affinities or relationships. The branch of biology that deals with the study of principles and procedures of biological classification are called taxonomy. Some fundamental elements of taxonomy are discussed below.

Nomenclature: It is the science of providing distinct and proper names to organisms. It is the determination of the correct name as per established universal practices and rules.

Classification: It deals with the mode of arranging organisms or group^ of organisms into categories according to a systematic plan or ah order. The categories used in the classification of animals are Class, Order, Family, Genus, and Species. Each category is a unit and is also called a taxon (PI. Taxa).

Identification: It is the determination of the correct name and place of an organism in a system of classification. It determines that the particular organism is similar to some other organism of known identity. This implies assigning an organism to a particular taxonomic group. Suppose there are three plants say x, y, z. AH represent different species. Another plant w resembles y. The recognition of the plant was identical to the already known plant y is its identification.
One of the important features of systematics is the naming of living organisms. The organisms have been given two types of names i.e

  • common or vernacular names
  • Scientific names.

Question 4.
What do we learn from the identification of individuals and populations?
Solution:
Identification of individuals and populations determines their exact place or position in the set plan of classification.

Question 5.
Given below is the scientific name of mango. Identify the correctly written name.
(a) Mangifera Indica
(b) Mangifera indica
Solution:
(b) Mangifera indica

Question 6.
Define a taxon. Give some examples of taxa at different hierarchical levels.
Solution:
“Taxon is a unit of classification or a rank or a level of hierarchy in system of classification. The following chart gives taxonomical categories showing a hierarchical arrangement in ascending order.

Kingdom
↑
phylum or Division
Class
↑
Order
↑
Family
↑
Genus
↑
Species

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
Solution:
(c) Species → Genus → Order → Phylum

Question 8.
Try to collect all the currently accepted meanings for the word ‘species’. Discuss with your teacher the meaning of species in the case of higher plants and animals on one hand and bacteria on the other hand.
Solution:

  1. Species is one of the basic units of biological classification. A species is often defined as a
    group of organisms capable of interbreeding aid in producing fertile offspring.
  2. Sometimes more precise or differing measures such as similarity of DNA, morphology o,^ecological niche are used to define the basis of species.
  3. In case of animals, the name of species is defined by the specific name or the specific epithet. For example, gray wolves belong to the species Canis lupus, golden Jackals to Cam’s aureus etc.
  4. Both of them belong to same genus Canis, but species name varies. But species name of plant is only called species epithet.
  5. The ‘specific name’ in botany is always the combination of genus name and species epithet such as saccharum in Acer saccharum (Sugar maple).
  6. But bacteria are grouped under four categories based on their shape – spherical, rod-shaped, comma and spiral shaped and species of bacteria is according to their shapes. Thus the meaning of species in higher organism and bacteria are different.

Question 9.
Define and understand the following terms:
(i) Phylum (ii) Class (iii) Family (iv) Order (v) Genus
Solution:
(i) Phylum: A phylum is a group of related classes having some common features, e.g., protozoa.
(ii) Class: A class is a group of related orders, for e.g., order Rodentia, Lagomorpha and Carnivora all having hair and milk glands are placed in class Mammalia.
(iii) Family: A family is a group of related genera. The genus Felis of cats and the genus Panthera of lion, tiger and leopard are placed in the family Felidal.
(iv) Order: An order is a group of related families. The family Felidae of cats and the »family Coridal of dogs are assigned to the order Carnivora. Cats and dogs have large canine teeth and are flesh-eaters.
(v) Genus: A genus is a group of species alike in the broad features of their organization but different in detail. As per the rules of binomial nomenclatures, a species can not be named without assigning it to a genus.
NCERT Solutions for Class 11 Biology Chapter 1 The Living World 1

Question 10.
How is key helpful in the identification and classification of organisms?
Solution:
Keys are contrasting pairs of characters (couplet), it represents the choice made between two opposite options. This results in acceptance of only one and rejection of the other. Each statement in the key is called a lead. Separate taxonomic keys are required for each taxonomic category such as family, genus, and species for identification purposes.

Question 11.
Illustrate the taxonomical hierarchy with suitable examples of a plant and an animal.
Solution: 
NCERT Solutions for Class 11 Biology Chapter 1 The Living World 2

VERY SHORT ANSWER QUESTIONS

Question 1.
Name the basic Unit of classification.
Solution:
Species.

Question 2.
Who introduced the hierarchy in taxonomy?
Solution:
Linnaeus

Question 3.
Who is the father of taxonomy?
Solution:
Carolus Linnaeus.

Question 4.
What is meant by cytotaxonomy?
Solution:
Classification based on chromosome number.

Question 5.
Who devised the binomial nomenclature?
Solution:
Carolus Linnaeus

Question 6.
What is a type specimen?
Solution:
Establishment of the name of the new species on the basis of the original specimen is called type specimen

Question 7.
In which language binomial nomenclature is written?
Solution:
Latin

Question 8.
What term is used to describe organisms without a well-developed nucleus?
Solution:
Prokaryote

Question 9.
Is inter-specific breeding possible?
Solution:
Yes, both.

Question 10.
What are DNA viruses / RNA viruses?
Solution:
Viruses that possess DNA as the genetic material are called DNA viruses.

Question 11.
What is speciation?
Solution:
Formation of a new species from an existing one by the appearance of mutation.

Question 12.
What are correlated characters?
Solution:
The common features the species have to qualify for inclusion in a genus are called correlated characters

Question 13.
Why classification of plants and animals is necessary?
Solution:
Classification divides millions of plant and animal species into convenient groups that make their study easier

Question 14.
What is cohort or order?
Solution:
The cohort is a unit of classification higher than the 6. family

Question 15.
Give an example of symbiotic bacteria.
Solution:
Rhizobium leguminosarum

Question 16.
Give botanical and zoological names of the following:
(1) Pea
(2) Wheat
(3) Man
(4) Potato
Solution:
(1) Pea → Pisumsatinum
(2) Wheat → Triticumaextivum
(3) Man → Homo sapiens
(4) Potato → Solanum tuberosum

SHORT ANSWER QUESTIONS

Question 1.
Write a note on bacteriophages. (Dharwar. 2004, Belgaum. 04,2005)
Solution:
The viruses that infect bacteria are called bacteriophages. They were discovered by Twort. They are Tadpole shaped. They have DNA as their genetic material. They are distinguished into T – odd phages as well as T – even phages.

Question 2.
What is a taxonomic aid?
Solution:
A taxonomic aid is storage of record of either live or dead specimens of flora or fauna, which helps scientists in taking reference to study classification

Question 3.
Give the classification of man.
Solution:
Common Name – Human
Scientific Name – Homo sapiens
Genera – Homo
Families – Hominidae
Orders – Primata
Classes – Mammalia
Phyla/Division – Chordate

Question 4.
What is a museum? How many kinds of museums are found?
Solution:
Museum in an institution where artistic and educational materials are exhibited to the public. The material available for observation and study is called a collection.
Kinds of Museums:

  • Art Museum
  • History Museum
  • Applied Science Museum
  • Natural Science Museum

Question 5.
Give a reason for the following.
Bacteria are the Natural Scavengers ‘ (D.Kannada 2006)
Solution:
because they bring about the decomposition of organic debris and clean the earth’s surface.

Question 6.
What is the role of characteristics of living beings in classification?
Solution:
A group of common features of living beings are placed under a common category of classification and when uncommon under a different category. It means more systematic a process for further study, research, protection and recording.

Question 7.
What is the significance of a HERBARIUM?
Solution:
HERBARIUM:- A book, case, or room containing an orderly collection of dried plants is called Herbarium. It develops interest in Nature for the activists in it. It can be used to gain knowledge and be updated about plants and their scientific names and even compare various samples. It is a small scale it can be proactive to do. One can make projects too from it for schools, colleges and research institutions.

Question 8.
Explain the role of blue-green algae in soil fertility.
Solution:
Blue-green algae like Nostoc, Anabaena fix atmospheric nitrogen. Heterocyst contains nitrogens enzyme that helps in nitrogen fixation. Nitrogen-fixing blue-green algae are inoculated in the rice field to increase soil fertility.

LONG ANSWER QUESTIONS

Question 1.
Write a short note on Binomial Nomencia? ture and guidelines for Binomial nomenclature.
Solution:
Binomial Nomenclature was introduced by Carolus Linnaeus. In this method every organism is given a scientific name, which has two parts, the first is the name of the genus (generic name) and the second is the name of the species (specific epithet) e.g.: Homo sapiens In the above examples, Homo is a generic name, while sapiens is the name of the species belonging to Homo.

Guidelines:

  • scientific names are generally in Latin or derived from Latin irrespective of their origin
  • The scientific names are written in italics or underlined (when handwritten)
  • The first word denotes the name of the genus and the second word denotes the specific epithet
  • The generic name starts with a capital letter, while the specific name starts with a small letter (If a specific name starts with a capital letter it denotes the name of a person or place)
  • The name of the author is written in an abbreviated form after the specific name. e.g.: Homo sapiens Linn.

Question 2.
What is the difference between living and nonliving?
Solution:
Question 3.
Explain the binomial system of nomenclature.
Solution:
Binomial nomenclature system was developed by Linnaeus. Binomial nomenclature is the system of providing organisms with appropriate and distinct names consisting of two words, first generic and second specific. The first or 4.

  • generic word is also called genus. It is like a noun and its first letter is written in capital form.The second word or specific epithet represents the species.
  • It is like an adjective. Its first letter is written in small form except occasionally when it denotes a person or place. The two word name is appended with the name of the taxonomist who discovered the organism and provided with its scientific name, e.g., Ficus bengalensis L., Mangifera indica Linn, The name of taxonomist can be written in full or in abbreviated form.
  • There are several technical names which have three words, e.g., Homo sapien sapiens, Acacia nilotica indica, Gerilla gorilla. Here the first word is generic, the second specific while the third word represents variety (mostly in botanical literature) or subspecies (mostly in zoological literature).
  • If the same scientific name is to be written time and again, the name of the genus can be abbreviated, e.g., F. bengalensis.

Question 4.
What is the role of zoological parks in wildlife conservation?
Solution:

  • In the early stages, the zoological parks were considered as places of relaxation and enjoyment for public, however, there has been a change in the objective of the purposefulness of these parks.
  • The establishment of zoological parks help in providing knowledge about different native and exotic wild mammals, birds, reptiles, fish and flora to the public in general and school children in particular.
  • Since the key to wildlife conservation lies in the education of the masses and the involvement of voluntary organisations, zoological parks are very useful in spreading knowledge on the wildlife wealth of the country.
  • These are also important centres for organising seminars, training and researches on the management of wildlife species and for study of their social behaviour, breeding and ecological species.

We hope the NCERT Solutions for Class 11 Biology at Work Chapter 1 The Living World, help you. If you have any query regarding NCERT Solutions for Class 11 Biology at Work Chapter 1 The Living World, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Biology Chapter 22 Chemical Coordination and Integration

NCERT Solutions for Class 11 Biology Chapter 22 Chemical Coordination and Integration

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 22 Chemical Coordination and Integration.

Question 1.
Define the following:
(a) Exocrine glands
(b) Endocrine glands Hormones
Solution:
(a) Glands with duct is known as an exocrine gland. They secrete their secretions through ducts. Eg: Pancreas.
(b) Gland without duct is known as an endocrine gland. They directly secrete their secretions. Eg: Pituitary gland.
(c) Hormones are non-nutrient chemicals which act as intercellular messengers and are produced in trace amounts. Eg: thyroxine.

Question 2.
Diagrammatically indicate the location of the various endocrine glands in our body.
Solution:
NCERT Solutions for Class 11 Biology Chapter 22 Chemical Coordination and Integration 1

Question 3.
List the hormones secreted by the following.
(a) Hypothalamus
(b) Pituitary
(c) Thyroid
(d) Parathyroid
(e) Adrenal
(f) Pancreas
(g) Testis
(h) Ovary
(i) Thymus
(j) Atrium
(k) Kidney
(l) G-ITract
Solution:
(a) Hypothalamus: Gonadotrophic releasing hormone (GnRH), somatostatin.
(b) Pituitary: Growth hormone, prolactin, thyroid-stimulating hormone, luteinizing hormone and follicle stimulating hormone, oxytocin, vasopressin.
(c) Thyroid: triiodothyronine (T3) and tetraiodothyronine (T4).
(d) Parathyroid: Parathyroid hormone.
(e) Adrenal: Adrenaline or epinephrine and noradrenaline or norepinephrine.
(f) Pancreas: Insulin and glucagon.
(g) Testis: A group of androgens mainly testosterone.
(h) Ovary: Estrogens and progesterone.
(i) Thymus: Thymosins.
(j) Atrium: Atrial Natriuretic factor (ANF)
(k) Kidney: Erythropoietin.
(l) G-I tract: Gastrin, secretin, cholecystokinin (CCK).

Question 4.
Fill in the blanks:
Hormones                      Target gland
(a) Hypothalamic hormones ……………….
(b) Thyrotrophin (TSH) ………………
(c) Corticotrophin (ACTH) ………………
(d) GonadobTjpfains(LH,FSH) ……………….
(e) MeIanotrophin(MSH) ………………..
Solution:
(a) Anterior pituitary and posterior pituitaiy.
(b) Thyroid glands
(c) Adrenal cortex
(d) Gonads (testis and ovary)
(e) Pituitary

Question 5.
Write short notes on the functions of the following hormones:
(a) Parathyroid hormone (PTH)
(b) Thyroid hormones
(c) Thymosins
(d) Androgens
(e) Estrogens
(f) Insulin and Glucagon
Solution:
(a) The secretion of PTH is regulated by the circulating levels of calcium ions. Parathyroid hormone (PTH) increases the Ca2+ levels in the blood. PTH acts on bones and stimulates the process of bone resorption. PTH also stimulates reabsorption of Ca2+ by the renal tubules and increases Ca2+ absorption from the digested food. PTH is a hypercalcemic hormone, i.e., it increases the blood Ca2+ levels. Along with TCT, it plays a significant role in calcium balance in the body.

(b) Thyroid hormones play an important role in the regulation of the basal metabolic rate. These hormones also support the process of red blood cell formation. Thyroid hormones control the metabolism of carbohydrates, proteins, and fats. Maintenance of water and electrolyte balance is also influenced by thyroid hormones. The thyroid gland also secretes the protein hormone thyrocalcitonin (TCH) which regulates blood calcium levels.

(c) Thymosins play a major role in the differentiation of T – lymphocytes, which provide cell-mediated immunity. In addition, thymosins also promote the production of antibodies to provide humoral immunity.

(d) Androgens regulate the development, maturation and functions of the male accessory sex organs like epididymis, vas deferens, seminal vesicles, prostate gland, urethra etc. These hormones stimulate muscular growth, growth of facial and axillary hair, aggressiveness, low pitch of voice etc. Androgens play a major stimulatory role in the process of spermatogenesis and act on the central neural system and influence male sexual behaviour. These hormones produce anabolic effects on protein and carbohydrate metabolism.

(e) Estrogens produce wide-ranging actions such as stimulation of growth and activities of female secondary sex organs, development of growing ovarian follicles, appearance of female secondary sex characters, mammary gland development. Estrogen also regulate female sexual behaviour.

(f) Glucagon is a peptide hormone, and plays an important role in maintaining normal blood glucose levels. Glucagon acts mainly on the liver cells and stimulates glycogenolysis resulting in increased blood sugar. In addition, this hormone stimulates the process of gluconeogenesis which also contributes to hyperglycemia. Glucagon reduces cellular glucose uptake and utilisation. Thus, glucagon is a hyperglycemic hormone.

Insulin is a peptide hormone, which plays a major role in the regulation of glucose homeostasis. Insulin acts mainly on hepatocytes and adipocytes and enhances cellular glucose uptake and utilisation. As a result, there is a rapid movement of glucose from blood to hepatocytes and adipocytes resulting in decreased blood glucose levels (hypoglycemia). Insulin also stimulates conversion of glucose to glycogen in the target cells. The glucose homeostasis in the blood is thus maintained jointly by the two insulin and glucagons.

Question 6.
Give example(s) of:
(a Hyperglycemic hormone and hypoglycemic hormone
(b Hypercalcemic hormone
(c) Gonadotrophic hormones
(d Progestational hormone
(e) Blood pressure lowering hormone
(f) Androgens and estrogens
Solution:
(a) Glucagon and insulin
(b) Parathyroid hormone
(c) LH and FSH
(d) Progesterone
(e) Atrial Natriuretic Factor (ANF)
(f) Testosterone and estradiol

Question 7.
Which hormonal deficiency is responsible for the following:
(a) Diabetes mellitus
(b) Goitre
(c) Cretinism
Solution:
Diabetes mellitus: Insulin deficiency.
Goiter: Deficiency of thyroid hormones due to deficiency of iodine.
Cretinism: Deficiency of thyroid hormones during childhood.

Question 8.
Briefly mention the mechanism of action of FSH.
Solution:
Follicle-stimulating hormone (FSH) and LH Stimulate gonadal activity. In males, FSH and androgens regulate spermatogenesis. In females, FSH is responsible for the growth and development of the ovarian follicles, maturation of egg, and secretion of estrogens

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

VERY SHORT ANSWER QUESTIONS

Question 1.
Give two examples of endocrine glands.
Solution:
Examples of endocrine glands are
(i) pituitary gland
(ii) pineal gland.

Question 2.
Which gland secretes glucagon?
Solution:
Pancreas secretes glucagon.

Question 3.
What is the location of the pituitary gland?
Solution:
The pituitary gland is located in a bony cavity called Sella tursica and is attached to the hypothalamus by a stalk.

Question 4.
Which hormone is secreted by pars intermedia?
Solution:
Pars intermedia secretes melanocyte-stimulating hormone (MSH).

Question 5.
Which hormone is secreted by the pineal gland?
Solution:
Pineal gland secretes melatonin hormone.

Question 6.
Name one disorder caused by hyperfunctioning of the pituitary.
Solution:
Gigantism caused by hyperfunctioning of the pituitary.

Question 7.
How many lobes does the thyroid gland have?
Solution:
The thyroid gland is composed of two lobes which are located on either side of the trachea.

Question 8.
Which gland produces the hormone called thyrocalcitonin (TCT) which regulates the blood calcium levels?
Solution:
Thyroid gland secretes a protein hormone called thyrocalcitonin (TCT) which regulates the blood calcium levels.

Question 9.
Name the organ which secretes progesterone.
Solution:
Ovary secretes progesterone.

Question 10.
Name two hormones of pancreas.
Solution:
Glucagon and insulin are two hormones of pancreas.

Question 11.
Which gland secretes somatostatin?
Solution:
Hypothalamus secretes somatostatin.

Question 12.
Which complex is formed during the binding of hormone to a receptor ?
Solution:
Hormone-receptor complex is formed during the binding of a hormone to a receptor.

Question 13.
Name one hormone secreted by gastro-intestinal tract.
Solution:
Gastrin is secreted by gastro-intestinal tract.

Question 14.
Which hormone is secreted by testes and write the function of hormone also.
Solution:
Testosterone is secreted by testes. It controls growth and development of male secondary
v-sexual characters.

Question 15.
What are membrane bound receptors?
Solution:
Hormone receptors that are present on the cell membrane of the target cells are called membrane bound receptors.

Question 16.
How many types of cells are present in Islets of Langerhans?
Solution:
There are two types of cells present in Islets of Langerhans which are a-cells and P-cells.

Question 17.
Why is oxytocin called as birth hormone ?
Solution:
Oxytocin stimulates the contraction of smooth muscles of uterus and facilitates the child birth.

Question 18.
Why is vasopressin known as antidiuretic hormone?
Solution:
Vasopressin stimulates the reabsorption of water and electrolytes and reduces the loss of water through urine, known as diuresis. Hence it is called as anti-diuretic hormone.

Question 19.
What is cretinism ?
Solution:
Cretinism is reduction in body growth as well as underdevelopment of brain resulting in various structural and functional defects mainly due to deficiency of thyroxine (hypothyroidism) in infants and children.

Question 20.
Which hormone interacts with membrane bound receptor and does not normally enter the target cell?
Solution:
Follicle stimulating hormone interacts with membrane bound receptor and does not normally enter the target cell.

Question 21.
Which hormone opposes parathormone?
Solution:
Thyrocalcitonin opposes parathormone.

Question 22
Which hormone is known as anti-aging hormone?
Solution:
Melatonin, secreted by pineal gland is known as anti-ageing hormone.

SHORT ANSWER QUESTIONS

Question 1.
Discuss the location and function of parathyroid hormone.
Solution:
There are four parathyroid glands present on the back side of thyroid glands, two on each of the lobes of the thyroid gland.
The parathyroid glands secrete parathyroid hormone (PTH), a peptide hormone. Parathyroid hormone (PTH) increases the Ca2+ levels in the blood. PTH acts on bones and stimulates the process of bone resorption (dissolution/ demineralization. PTH also stimulates reabsorption of Ca2+ by the renal tubules and increases Ca2+ absorption from the digested food.

Question 2.
What is progesterone? Name two important functions of progesterone.
Solution:
Progesterone is a steroid hormone secreted by ovary.
Functions of progesterone are as follows:
• It supports pregnancy.
• It acts on mammary glands and stimulates the formation of alveoli (sac-like structures which store milk) and milk secretion.

Question 3.
Name the principal mineralocorticoid secreted by adrenal cortex. Give its any two functions
Solution:
Aldosterone is main mineralocorticoid secreted by adrenal cortex.
Functions of aldosterone are :
• It stimulates reabsorption of Na+ and water.
• It stimulates excretion of K+ and phosphate ions.

Question 4.
What are hormone receptors? What are the modes of their action ?
Solution:
Receptors are specific proteins present on the surface of target cell which bind with hormones and produce physiological changes in cell. Their are two types of hormone receptors which are discussed below:
NCERT Solutions for Class 11 Biology Chapter 22 Chemical Coordination and Integration 2

Question 5.
Name the gland that functions as a biological clock in our body. Where is this gland located ? Name its one secretion.
Solution:
Pineal gland functions as a biological clock in our body.

  • It is located on the dorsal side of the forebrain.
  • It secretes melatonin.

Question 6.
Where is thymus gland located in the human body ? Describe its role.
Solution:
The thymus gland is a lobular structure located on the dorsal side of the heart and the aorta.
– It secretes hormone thymosin, which has a stimulating effect on the immune system.
– Thymosin promotes proliferation and maturation of T-lymphocytes.

Question 7.
How does insulin act on high glucose content in the blood in a normal human body ?
Solution:
• Insulin is a peptide hormone, which plays a major role in the regulation of glucose homeostasis. Insulin acts mainly on hepatocytes and adipocytes (cells of adipose tissue), and enhances cellular glucose uptake and utilisation. As a result, there is a rapid movement of glucose from blood to hepatocytes and adipocytes resulting in decreased blood glucose levels (hypoglycemia)
• Insulin also stimulates conversion of glucose to glycogen (glycogenesis) in the target cells.
• The glucose homeostatasis in blood is thus maintained jointly by the two enzyme insulin and glucagon.

Question 8.
What is corpus luteum ? Name its one secretion.
Solution:
Corpus luteum is the structure formed by the ruptured ovarian follicles after ovulation.
– It mainly secretes the hormone progesterone.

Long ANSWER QUESTIONS

Question 1.
Explain briefly the structure and functions of middle ear.
Solution:
Calcitonin (from thyroid) : Essential for maintaining bone strength as it does not allow calcium mobilisation from bones. It also lowers plasma level of calcium ifthe same is high. Parathormone: Low level of parathormone secretion decreases blood plasma level of calcium, prevents reabsorption from bones and causes tetany. Higher levels of parathormone (PTH) increase plasma level of Ca2+ by withdrawal from bones resulting in (i) Inflammation and tenderness in bones due to dissolution of calcium from bones, formation of cavities which get plugged with fibrous nodes and cysts (osteitis fibrosa cystica) making bones soft deformed and prone to fracture, (ii) Excess plasma calcium gets deposited in various parts of body which often leads of obstruction and death.

Question 2.
Draw a well labelled diagram showing the location of parathyroid gland. Discuss the function/(s) of parathormone and hyperparathyroidism.
Solution:
NCERT Solutions for Class 11 Biology Chapter 22 Chemical Coordination and Integration 3
Function : Parathormone (Collip’s hormone) regulates the metabolism of calcium and phosphate between blood and other tissue.
(i) Hyperparathyroidism : It leads to demineralisation resulting in softing and bending of bones. An excess of parathormone cause osteoporosis and kidney stones.

Question 3.
What are the causes for following disorders?
(a) Acromegaly
(b) Cretinism
(c) Gigantism
(d) Myxoedema
Solution:
Disorders and their causes.
(a) Acromegaly (Acro-extremity, megaly – large) – It is caused by excess secretion of growth hormone after adulthood is reached.
(b) Cretinism : It is caused by deficiency of thyroid hormone in infant.
(c) Gigantism : It is caused by excess of growth hormone from early age.
(d) Myxoedema: It is caused by deficiency of thyroid hormone in adults.

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