Doc-1103B-B.P.S.-XI-Biology-Chapterwise-Topicwise-Worksheets-with-Solution-2014-15.pdf

BRILLIANT PUBLIC SCHOOL , SITAMARHI (Affiliated up to +2 level to C.B.S.E., New Delhi) Affiliation No. - 330419

XI - Biology Chapterwise Topicwise Worksheets with Solution

Session : 2014-15

Office: Rajopatti, Dumra Road, Sitamarhi(Bihar), Pin-843301 Website: www.brilliantpublicschool.com; E-mail: [email protected]

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BIOLOGY (Class 11) Index Chapter

page

1.

The Living World

01

2.

Biological Classification

11

3.

Plant Kingdom

24

4.

Animal Kingdom

41

5.

Morphology of Flowering Plants

56

6.

Anatomy of Flowering Plants

74

7.

Structural Organisation in Animals

85

8.

Cell : The Unit of Life

97

9.

Biomolecules

109

10.

Cell Cycle and Cell Division

120

11.

Transport in Plants

128

12.

Mineral Nutrition

137

13.

Photosynthesis in Higher Plants

146

14.

Respiration in Plants

155

15.

Plant Growth and Development

166

16.

Digestion and Absorption

175

17.

Breathing and Exchange of Gases

184

18.

Body Fluids and Circulation

193

19.

Excretory Products and their Elimination

202

20.

Locomotion and Movement

212

21.

Neural Control and Coordination

221

22.

Chemical Coordination and Integration

230

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (The Living World)

1.

Name the three fields of systematics.

[1]

2.

Give the two name system of organisms?

[1]

3.

Write the correct order of sequence of taxonomical categories?

[1]

4.

What are the advantages of giving scientific names of the organisms?

[2]

5.

Give the role of botanical gardens?

[2]

6.

Differentiate between species & taxon?

[2]

7.

Name the guidelines for naming of organisms?

[3]

8.

What is Biological classification? What is the need of classification?

[3]

9.

What is Binomial system of nomenclature? Who proposed this system? Why is

[5]

binomial nomenclature the most acceptable mode of naming organism?

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (The Living World) [ANSWERS] Ans. 1 Nomenclature, classification & taxonomy. Ans.2 Binomial Nomenclature Ans.3 Species

genus

family

order

class

Phylum

kingdom.

Ans.4 i) Scientific names are universally accepted in the world because they are based on same principles that are universal. ii) The advantage of a technical term is the relationship & comparison too the others. Ans.5 i) Botanical gardens provide plant materials for taxonomic studies. ii) Plant species are grown for identification. iii) Plants are grown for research.

iv) To maintain records of local flora.

Ans.6 SPECIES

TAXON

i) It is the basic taxonomic category

i) It is a level of taxonomic category

ii) It is a rank

ii) It is a group of concrete biological aspects

iii) It is monophylectic

iii) It may be mono or polyphylectic.

Ans.7 Guidelines for naming of organisms include :i)

A scientific name generally has two words in Latin or derived from latin irrespective of their origin.

ii) First word denotes the genus where as second word for species. iii)

Names are printed in italics or are separately underlined to indicate the Latin origin.

iv)

Each taxonomic group has only one correct name.

v) The name must be short, precise and easy to pronounce. vi)

Generic name begins with a capital letter & the specific name with small letter eg. Homo sapiens.

vii)

The name of author is written in abbreviated form after species name & it is printed in Roman.

Ans.8 Biological classification is the naming of organisms by two words. One is generic name & other is specific name for eg. Man is called Home sapiens classification becomes essential for the following reasons:i)

It is very essential for the systematic study of living beings. Without this study of different organisms would be in confusion.

ii) It is impossible to study each & every organism. iii) All the types of organisms do not occur in a given locality. iv) Without a proper system so classification, it is impossible to recognize or identify different types of organism. v) Classification helps in knowing the relationships among different groups of animals & plants. vi) Classification makes the study of organisms easier & gives a comparative account of them. Ans.9 Naming of plants & animals with two words one generic & other specific name is called binomial system of nomenclature. Carolus Linnaeus introduced this scientific system to name a species. He gave two names to a species eg. Mangifere is generic name and indica is the specific name. Binomial nomenclature is universally accepted all over the world because it is written according to universal rules of nomenclature framed by ICBN, ICZN, & ICNPC etc. It has two parts generic name & specific name followed by name if scientist who discovered it at last in abbreviated form. It must be in Latin or derived from Latin. It must be binomial. The genus starts with capital letter while species by small letter. Handwritten name is underlined it indicates relationship with other species present in same genus. The rules & regulations present of binomial nomenclature must be observed before a taxonomist names a new found organism. This maintains stability in taxa, avoids the use of names that may cause error ambiguity & confusion.

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (The Living World)

1.

Give the unit of classification?

[1]

2.

Who gave binomial name of classification?

[1]

3.

What is meant by identification of a species?

[1]

4.

Why are classification systems changing every now & then?

[2]

5.

Differentiate between taxon & category?

[2]

6.

Describe the role of museum in studying systematic?

[2]

7.

“Botanical gardens are living herbaria”. Comment ?

[2]

8.

State any five objectives of classification.

[3]

9.

Explain the utility of systematic & mention the characterstics of new.

[3]

10.

What are the major divisions of classification, classify man.

[3]

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (The Living World) [ANSWERS] Ans. 1 Rank / category Ans.2 Carolus Linnaeus Ans.3 The identification is aimed at finding correct name & proper position of a species in established scheme of classification. Ans.4 The organisms are classified on the basis of characteristics. Earliest classification were based on the uses of various organisms but now the humans are interested in knowing more about different kinds of organisms & their diversities & their relationship also. Ans.5 TAXON

CATEGORY

i) Taxon represents to a group of i) Category refers to a rank of status of taxon.

organisms. ii) It is only of one eg. Dicots, ii) Category is of two types i.e.

Monocots,

a) Major rank – kingdom, division, class. b) minor rank – Genus & species

Ans.6 i) Museums have collections of plants & animals ii) These are used to deposit type specimens. iii) Important centres for taxonomic studies. iv) Provide information about local flora & fauna as well as of other areas.

Ans.7

Botanical gardens are repositories of information useful for taxonomic studies. Herbaria are most permanent records of plant specimens. Living plants are maintained in botanical gardens. They play key roles in conservation, research, ecology, library & herbaria etc.

Ans.8

Objectives of classification:i)

Development of a system for easily identifying a species if known or unknown

ii)

The description of various species.

iii)

Recognition of different species.

iv)

To bring circulated characteristics at various levels in hierarchy.

v)

The grouping of species in taxonomic classification.

vi)

To establish natural relationship board on phylogeny on the basis of resemblances of the organisms of the organisms.

Ans.9

Systematics is defined as “the study of classification of organisms based on evolutionary relationships”. i)

It provides useful information about organism, its evolution & adaptation name & classification etc.

ii) Systematics helps us in the identification of useful & harmful animals or plants in applied field of biology. iii) It plays economical role. New systematics has the following features:a) Species are regarded as dynamic unit & not as static unit of classical systematic. b) The importance of species as such is reduced since most of the work is done with sub-divisions of species. c) The morphological species definition has been replaced by a biological one which takes ecology, genetics, geography, cytology & behaviour into consideration. Ans.10 i)

Kingdom:- It is the highest category of classification. There are 2 kingdom – Animal & plant kingdom.

ii)

Phylum:- A group of closely related classes having certain common characters.

iii)

Class:- A group of closely related orders having certain common characters.

iv)

Order:- A group of closely related families having certain common characters.

v)

Family:- A group of closely related genera having certain common characters.

vi)

Genus:- A group of closely related species having certain common characters.

vii)

Species:- Individuals having certain common characters. Classification of man:Kingdom

Animalia

Phylum

Chordate

Class

Mammalia

Order

Primates

Family

Hominidae

Genus

Homo

Species

Sapiens

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (The Living World)

1.

Name the highest categories of classification?

[1]

2.

What are the three codes of nomenclatures?

[1]

3.

What do you mean by “chemotaxonomy?

[1]

4.

Why are living organisms classified?

[2]

5.

What is Taxonomic key? How is it helpful in the identification & classification

[2]

of an organism? 6.

Differentiate between taxonomy & systematic.

[2]

7.

What is a taxon? Illustrate the taxonomical hierarchy with a suitable example?

[2]

8.

What are taxonomic aids? Mention some of the taxonomic aids for

[3]

identification 9.

How would you set up a herbarium?

[3]

10.

Differentiate between classical taxonomy & Modern taxonomy.

[3]

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (The Living World) [ANSWERS] Ans. 1

Kingdom.

Ans.2

International code of botanical, zoological & bacteriological nomenclature.

Ans.3

Understanding taxonomic relationships based on the distribution of certain characteristic chemical constituents is called chemotaxonomy.

Ans.4

There are various kinds of life that differ in shape, size & colour etc. The biological diversity is the range of life occurring in biological world. The diversity develops due to the evolution and development of adaptations to overcome competitions among life forms due to limited resources.

Ans.5

Key is a taxonomic aid for identification of unknown organisms based on similarities & dissimilarities. It is a taxonomic literature based on couplet. These are analytical in nature & separate keys are needed for every taxonomic category eg. genus, order, family etc & species for identification of organisms.

Ans.6 TAXONOMY i) The science of identification, nomenclature & classification is called taxonomy. ii) It deals with the rules & the principles of classification.

SYSTEMATICS i) It refers to the science of identification description, nomenclature & classification. ii) It deals with unique characteristics at every level of classification.

Ans.7

Taxon is “a unit of classification of organisms which can be recognized & assigned a definite category at any level of classification” eg. order primates & carnivores are included in mammala. Various classes eg. Pisces, animals, reptilia aves & mammalia form phylum- chordata. All phyla are included in kingdom animalia .

Ans.8

Taxonomic aids are devices used to study, Identification & classify organism, some of these are:i) Herbarium :- collections of present /preserved or mounted plant specimens. arranged systematically to provide information on sheets ii) Botanical gardens :- specialized gardens for collection of living plants, it is maintained for references & identification purposes in which each plant is labelled showing its biological name.

iii)

iv)

Ans.9

Zoological parks:- places with live animals are called zoos or zoological parks. The animal live in their natural habitat there are separate places for birds, tigers, lions, reptiles etc. Museums :- These are mostly set up in institutions where collection of preserved plants & animals for reference & taxonomic studies are placed in preservatives eg. Alcohol & formalin.

SETTING UP OF HERBARIUM involves the following steps:i) Visit to a specific area to get intact part or plant, seeds or flowers. ii) Information about habitat, season & time of collection as well topography etc. iii) For collection, some tools are needed, notebook, digger, scissor, knife polyethene, newspaper etc. iv) Spreading of specimens & drying, change the paper sheets after 3-4 days, plant press may be used for it. The dried specimens are pasted on herbarium sheets & pesticides like CS2, naphthalene Hgcl2 etc. v) Put label on specimen & mention its place of collection, time of collection, common name scientific name etc.

Ans.10 CLASSICAL TAXONOMY i) It is called old taxonomy or systematic ii) The species was considered a basic, concrete & separate unit that was fixed or static entity & the work of creator.

MODERN TAXONOMY i) It is called Neo- systematic or Biosystematic. ii) The species is considered related to one another, mutable & the work of gradual modification as wall as dynamic & ever- changing. iii) In it, classification was based on the iii) In it, classification was based on morphological features only phylogenetic relationships of the organisms iv) Few individuals were studied. iv) large number of individuals are studied v) The species was delimited on v) Emphasis in population instead of morphological characters. species. Morphological delimitation was replaced by biological delimitation.

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (Biological Classification)

1.

Who introduced the five kingdom classification of organisms?

[1]

2.

To which kingdom the multicellular decomposers belong?

[1]

3.

Expand PPLO.

[1]

4.

What is the basis of modern classification?

[2]

5.

Give one example of a fungus as a soure of antibiotics?

[2]

6.

How are viroids different from viruses?

[2]

7.

Explain sexual reproduction in bacteria?

[3]

8.

Discuss the salient features of viruses with the help of diagram?

[3]

9.

Write the distinct characters of fungi & explain using a diagram.

[5]

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (Biological Classification) [ANSWERS] Ans.1

R.H. Whittaker (1969)

Ans.2

Kingdom fungi

Ans.3

Pleuropneumonia like organisms.

Ans.4

Modern taxonomy is based on :-

i)

Evolutionary relationship.

ii)

The similarities in the genetic codes of species.

iii)

Ecological characters.

Ans.5

Penicillium is the genus which is the source of an antibiotic penicillin. Penicillium is known as green & blue moulds. Penicillium chryosogenum is utilized for production of antibiotic penicillin.

Ans.6 VIRIODS

VIRUSES

i) Viriods are smaller than viruses & lack i) Viruses are non- cellular organisms

Ans.7

protein coat.

having protein coat.

ii) Genetic material is free RNA

ii) Genetic material is RNA or DNA.

Bacteria do not have true sexual reproduction but they show genetic recombination by three ways:-

i)

Conjugation:- It was discovered by Lederberg & Tatum. The donor or male call is identified by the presence of plasmid called F- factor in cells. Donor cell bears cylindrical hollow sex Pilli for attachment to recipient bacterium. Donor & recipient come in physical contact with the help of sex pilli. Plasmid or plant of donor DNA is transferred into recipient cell.

ii) Transformation :- It was discovered by Griffith. It includes death of bacterial donor cell resulting in release of its DNA into external medium DNA gets fragmented & gets incorporated into metabolically active cells. Recipient cell after incorporation of donor DNA is known as recombination that expresses all its character together with character of donor cell. iii) Transduction :- It was discovered by Zinder & Lederberg. Donor genes are transferred into recipient all by a virus. A phage causes lysis of bacterium & incorporates bacterial genes into phages then is liberated & they infect new bacterial genes. Ans.8

Features of viruses:-

i)

They are smaller then bacteria.

ii) They can be filtered iii) They are able to reproduce in host cell by using enzymes & metabolic machinery of host cell iv) DNA/RNA is their genetic material v) These are obligate parasites, self replicating & non – cellular organisms. vi) They have protein coat called capsid that protects nucleic acid. vii) They cause disease in plants like mosaic, leaf curling, leaf role, vein clearing etc.

Ans.9 i) Cell type – eukaryotic except yeast. ii) Cell wall – present but made up of chitin. iii) Chloroplast – absent. iv) Mitochondria – present

v) Nuclear envelope – present vi) Tissues – present but limited, yeast is a unicellular fungi, hyphae mycelium coenocytic, septate vii) Motility – Cilia, flagella in some treat absent in most forms. viii) Nutrition – Heterotrophic, saprophytes, parasites, absorb food or as symbionts in lichens ix) Reproduction

–

fertilization

or

meiosis

in

sexual

reproduction & fission, budding fragmentation, conidia formation etc in asexual reproduction. x) Nervous system – absent xi) Occurrence – air, water, soil, animals or plants xii) Examples

–

yeast,

Penicillium,

Agaricus,

Rhizopus,

phytophthora, Asperigillus claviceps, Rust, smut.

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Biological Classification)

1.

Name the five kingdoms in which the organisms are grouped together?

[1]

2.

Which organisms are known as “Jokers of plant kingdom”

[1]

3.

In which class of fungi sexual reproduction does not occur?

[1]

4.

Explain phylogenetic system of classification?

[2]

5.

What is the basis of Whittaker’s system of classification?

[2]

6.

Find out what do the terms “algal bloom”& “red tides” signify?

[2]

7.

Distinguish between bacteria & cyanobacteria?

[3]

8.

Describe the salient features of protists?

[3]

9.

Explain the various methods of asexual & sexual reproduction in fungi?

[5]

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Biological Classification) [ANSWERS] Ans. 1

Monera, protista, fungi, plantae & animalia.

Ans.2

Mycoplasma

Ans.3

Deuteromycetes

Ans.4

phylogenetic system of classification is based on evolutionary relationships of organisms. It reflects true relationship between organisms. It is not static but not dynamic. Its sources are fossils records that are never complete due to difficulty in formation, exposure, discovery & study.

Ans.5

Whittaker based his classification on following three criteria :i) Structure of cell i-e. prokaryotic Vs. Eukaryotic organization. ii) Unicellular Vs. multi cellular organisms iii) Different modes of nutrition – parasitic, autotrophic or heterotrophic.

Ans.6

i)

Algal bloom refers to the excessive growth of algae in water body due to enrichment of excessive nutrients in it.

ii)

The red dinoflagellates undergo rapid multiplication eg. Gonyaulux which make the sea appear red. It is called red tide.

Ans.7 BACTERIA i) cells are comparatively smaller ii) They have lesser structural elaboration iii) Most bacteria have flagella iv) Are autotrophic heterotrophic both v) Possess bacteriochlorophyll vi) Reserve food is glycogen

CYANOBACTERIA i) Cells are comparatively larger. ii) They exhibit high degree of morphological complexity & structural elaboration. iii) Do not have flagella. & iv) Are autotrophic. v) Possess chlorophyll. vi) reserve food is cyanophycean starch

starch

like

Ans.8

i)

They

are

single

celled

colonial

filamentous eukaryotes. ii)

They

grow

in

humid

&

moist

environment. iii)

Some are photosynthetic some are not.

iv)

Some forms are like plants & some like animals.

v)

Contain membrane bound organelles.

vi)

Protozoan’s are unicellular heterotrophic

vii) Examples- protozoan’s, slime moulds, Euglenoid, diatoms, dinoflagellate

Ans.9

1)

ASEXUAL REPRODUCTION:- Special types of reproductive cells are formed in asexual reproduction in fungi. They are known as spores. a) Zoospores :- Zoospores are motile eg. phycomycetes. They may have one or more flagella. On germination zoospores produces new mycelium. b) Oidia :- Some oval or spherical spores are found in mucor. They are formed by small segment of hyphae. c)

Conidia:- conidia are formed in some fungi as a means of asexual reproduction. They are borne on conidiospores eg. penicillium.

d) Chlamydospores :- Thick walled resting spores are produced in some fungi. They may be terminal or intercalary.

2)

SEXUAL REPRODUCTION:- There are three kinds of sexual reproduction in fungi:i)

Isogamy:- It is the fusion of morphologically similar gametes.

ii) Anisogamy:- It is the fusion of two morphologically & physiologically dissimilar gametes. iii) Oogamy:- It is the fusion of female egg with that of male antherozoids. These gametes are produced in oogonium & anthridium respectively.

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (Biological Classification)

1.

Who is known as “Father of classification”?

[1]

2.

Name the fungus from which LSD drug is obtained?

[1]

3.

It is advised to grow one pulse crop in between two main crops in the same

[1]

field why? 4.

Why are some fungi grouped under “fungi imperfecti”?

[2]

5.

Explain “Numerical taxonomy”.

[2]

6.

What are the demerits of five kingdom classification?

[2]

7.

Give scientific name of species of fungus:-

[2]

a) Produces a plant disease. b) Is edible c) A source of antibiotic d) Used in manufacture of ethanol. 8.

Why is natural system of classification better then artificial system of

[3]

classification? 9.

Give a comparative account of classes of kingdom fungi on the basis of mode

[3]

of nutrition & mode of reproduction. 10.

Write the diagnostic characters of kingdom monera.

[5]

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (Biological Classification) [ANSWERS] Ans. 1

Carolus Linnaeus

Ans.2

claviceps purpurea.

Ans.3

To increase the fertility of soil

Ans.4

The fungi commonly reproduce asexually. A part of mycelium is detached by fragmentation into small pieces which grow into new mycelium but in some fungi sexual reproduction also takes place. There is a group of fungi which reproduces completely by asexual spores & not by sexual spores so they are grouped under fungi imperfecti.

Ans.5

Numerical taxonomy refers to use of technological methods in taxonomy. The observable characters are studied. The number & codes are assigned to them for computer like (+) & (-). The date processed by computer scores the taxa as per number of unit characters possessed by them.

Ans.6

Ans.7

Ans.8

i)

Kingdom monera & protista include autotrophic & heterotrophic organisms.

ii)

Phylogenetic relationships in lower organisms are not specific & clear.

iii)

Multicellular groups have evolved from the protists.

i)

Phytophthora infestans – causes late blight of potato

ii)

Agaricus campestris – Edible mushroom.

iii)

Penicillium notatum – Produces antibiotic Penicillin

iv)

Sacchromyces cerevisae – used in production of ethanol.

Natural selection not only brings out natural relationships but also studies evolutionary tendencies & phylogeny with help of all available data including fossils. It is better than artificial system of classification due to following reasons:-

i)

This system brings out natural relationships amongst organisms.

ii)

This places only related organizations of group.

iii)

It avoids coming together of unrelated organisms.

iv)

It shows phylogenetic relationships & origin of different taxa.

Ans.9 PHYCOMYCETES i) They are obligate parasites on plants ii) The spores are produced in sporangia. Asexual spores are oospores or zygospores formed by union of gametes. (iii) Sexual spores are zoospores or aplanospores

Ans.10

ASCOMYCETES They are saprophytic or parasitic Asexual spores are ascospores. Asci are arranged in ascocarps

BASIDIOMYCETES DEUTEROMYCETES They are pasites They are saprophytes or parasites Basidia are Asexual spores are arranged in conidia basidiocarp.

Sexual spores are ascospores produced in ascus.

Plasogamy occur by fusion of somatic & vegetative cells

Monera i)

It is a kingdom of prokaryotes.

ii)

It

includes-

bacteria,

Cyanobacteria

&

actinomycetes. iii)

Microscopic organisms without nucleus but having a cell wall in some

iv)

Nutrition

is

either

heterotrophic

autotrophic. v)

They are decomposers & mineralizers.

vi)

Some monerans are archaebacteria .

vii)

eg. spirulina, nostoc, oscillotoria, bacillus.

or

Sexual reproduction is absent in them.

CBSE TEST PAPER-04 CLASS - XI BIOLOGY (Biological Classification)

1.

Define experimental taxonomy?

[1]

2.

Name the fungus causes the rust of wheat?

[1]

3.

What are distributed organisms which have not been included under any

[1]

kingdom? 4.

Compare salient features of monera & protista.

[2]

5.

State an economically important use of

[2]

i) Heterotrophic bacteria. ii) Archaebacteria. 6.

Write the importance of classification of organism.

[2]

7.

What are insectivorous plants? Give an example.

[2]

8.

Discuss different systems of classification briefly.

[3]

9.

What are the different groups of fungi?

[3]

10.

Compare the kingdoms under five kingdom classification in terms of cell type

[5]

cell organelles Nucleus, motility, cellularity.

CBSE TEST PAPER-04 CLASS - XI BIOLOGY (Biological Classification)

[ANSWERS] Ans.1

It is the identification of evolutionary units within species by experimentally determining their genetical origin

Ans.2

Peccinia graminis tritici.

Ans.3

Virus & Viriods

Ans.4 MONERA i)

It

includes

PROTISTA unicellular

achaebacteria, cyanobacteria

bacteria, i) It includes photosynthetic algae,

slime moulds, protozoan etc.

ii) They are prokaryotic, photosynthetic ii) These are eukaryotic unicellular,

&some heterotrophs

Ans.5

autotrophy or saprophytes or parasites

i) Heterotrophic bacteria are decomposers mostly. Some are helpful to make curd milk, fixing nitrogen etc while some are pathogens & cause diseases. ii) Archaebacteria, bacteria include methanogens that produce biogas from cow dung etc.

Ans.6

i)

It is essential for systematic study of living beings to classify them as more than millions of plants are known today

Ans.7

ii)

All types of organisms do not occur on same locality

iii)

It is not possible to study all organisms at one time.

iv)

It helps in knowing evolutionary relationships between different groups

v)

It makes easier to recognize & identify each organism.

Insectivorous plants are carnivorous plants. They trap insects to supplement nutritional requirement of nitrogen. These are green plants & their leaves are modified to trap insects to overcome shortage of nitrogen eg. in pitcher plant (Nepenthes) leaf blade is modified into a pitcher.

Ans.8

Three different groups of fungi are i)

ii)

iii)

Ans.9

Phycomycetes :- They have multinucleated, aseptate mycelium. Asexual reproduction occurs by aplanospores & sexual reproduction occurs by isogamy or oogamy. These are found in water or damp places eg. mucor Albugo etc. Ascomycetes:- They are unicellular or multicellular mycelium which is septate. Asexual spores formed in chains are called conidia. Sexual reproduction occurs by ascospores beared in cup shaped structure called asci eg. yeast penicillium, Aspergillus. Basidiomycetes :- They are called club fungi due to club- shaped end of mycelium called basidium. They have septate mycelium and bears asexual spores basidiospores. Eg mushroom smut rust.

Different systems of classification are:i) ii) iii)

iv)

Artificial classification- It takes into account easily observable few characteristics only & not anatomical relationships. Natural classification- It relies on natural affinities among organisms. It employs external & internal both features. Phylogenetic classification:- It is based upon evolutionary relationships among the organisms i-e. Organism belonging to same group have common ancestory. Phenotypic classification :-additional criteria & methodologies are employed to classify organisms to avoid problem establishing evolutionary relationship.

Ans.10

i) Cell type ii)cell organelle iii) Nucleus iv) Motility

Kingdom Monera Prokaryotic Absent

Kingdom protista Eukaryotic present

Absent Bacterial flagella

present present Cilia, flagella or Cilia or amoeboid flagella movement Absent Present but limited

v) Tissue or Absent Multi cellularity

Kingdom fungi Eukaryotic present

Kingdom plantae Eukaryotic present

Kingdom animalia Eukaryotic present

present Parts shows movement not plant Present in all plants

present Contractile fibres Present in all animals

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (Plant Kingdom)

1.

Name seedless vascular plants.

[1]

2.

Which pigment is responsible for red colour of red algae?

[1]

3.

What is a cone?

[1]

4.

What features led to dominance of vascular plants?

[2]

5.

Differentiate between Red algae & brown algae?

[2]

6.

Give some important features of dicots?

[2]

7.

Explain in brief the structure of prothallus of fern?

[3]

8.

Point out differences in sexual reproduction of moss & fern?

[3]

9.

Explain the life cycle in green algae?

[5]

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (Plant Kingdom) [ANSWERS] 1.

Pteridophytes.

2.

Phycobilin, phycoerythrin & phycocyanin.

3.

The fruiting body of gymnosperms which consists of micro & megasporophyll are called as cone.

4.

Three important features have to dominance of Angiosperm:i) Development of deep roots capable of penetrating the soil. ii) Development of water- proofing material eg. cutin on aerial surfaces, to reduce water loss through evaporation. iii) Development of strong woody material anchor & support above ground structures.

5. RED ALGAE i) mostly unicellular & microscopic ii) Phycoerythrin, phycocyanin & phycobilin pigments are present. Ii iii) Reserve food material is Floridian starch iv) chlorophyll ‘a’ present v) eg. Gelidium, polysiphonia

BROWN ALGAE i) filamentous & heterotrichous. ii) fucoxanthin pigment is present. iii) Reserve food material is Laminarian starch. iv) chlorophyll ‘a’ absent v) Laminaria, focus & sargassum

6.

The dicotyledons are characterized by either woody or herbaceous habit, their flower parts usually are in four or five their leaves are net-veined, vascular bundles are arranged in a circle or ring within the stem. The dicots have two cotyledons in their seeds.

7.

Prothallus of Fern:i) It is a heart-shaped structure. ii) The sex organs are present on the lower surface of the prothallus below the apical notch. iii) Sex organs are antheridia & archaegonia. iv) Prothallus is produced from the meiospore as gametophyte of fern. v) Below the sex organs are rhizoids

vi) Archegonia are flask shaped but antheridia are globose. vii) Male & female gametes are produced in antheridia & archegonia. NCERT page no. 37 fig no. 3.3 (c) 8. MOSS (Bryophyte) Fern (pteridophytes) i) Sex organs are borne on the i) Sex organs are borne on an inconspicuous gametophytic plant body. gametophyte or prothallus which represents an alternate phase to sporophytic plant body. ii). Antheridia are well developed ii). Antheridia are less developed & mostly & often possess a stalk. devoid of a stalk. iii). Antheridial jacket made up of iii). Antheridial jacket mostly made up of only 3 several cells – cells. iv). Sperms biflagellate iv). Multiflagellate sperms v). Archegonia often have stalk v). Archegonia do not have stalk vi). Neck is 6 - rowed vi). Neck is 4 – rowed.

9.

There are three types of life cycle are found in green algae:a) HAPLONTIC LIFE CYCLE:- The dominant phase is haploid. Diploid state is found only in the form of zygote or zygospores. Meiosis takes place at time of its germination. Eg. ulothrix, spirogyra. b) DIPLONTIC LIFE CYCLE:- The dominant phase alga is diploid. It gives rise to haploid gametes through meiosis. Gametes unit & the zygote regenerates diploid phase. c) DIPLOHAPLONTIC LIFE CYCLE:- It has well developed multicellular haploid & diploid phase. These are respectively called gametophyte & sporophyte. Haploid gametophyte produce haploid gametes. Fusion product of gametes grows directly into diploid sporophytes. Sporophytes produce haploid spores by meiosis. The meiospores germinate into new gametophyte.

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Plant Kingdom)

1.

Name a unicellular algae.

[1]

2.

Why some bryophytes are called liverworts.

[1]

3.

What are rhizoids?

[1]

4.

List four classes of plants belonging to fern group.

[2]

5.

How will you differentiate between red algae & green algae.

[2]

6.

Write two important characteristics of gymnosperms?

[2]

7.

Describe the main features of pteridophytes?

[3]

8.

“Algae & Bryophytes are different from each other.” Point out the main

[3]

differences between them? 9.

Explain briefly the alternation of generation in bryophytes?

[5]

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Plant Kingdom) [ANSWERS] 1.

Chlamydomonas

2.

Some bryophytes are called liverworts as they are liver shaped eg. marchantia.

3.

Rhizoids are slender, unicellular or multicellular hair like structure, which penetrate in the moist soil & absorb the water for plants.

4.

Pteridophytes or fern group are divided into four classes:i)

Class 1: psilopsida eg. psilotum nudum.

ii)

Class 2: Lycopsida eg. Lycopodium phlegmaria.

iii)

Class 3: Sphenopsida eg. Equisetum

iv)

Class 4: pteropsida eg. Adiantum & pteridium

5. RED ALGAE

GREEN ALGAE

i) It belongs to rhodophyta

i) It belongs to chlorophyta

ii)

Phycoerythrin,

phycocyanin, ii) Chlorophyll ‘a’ & ‘b’ with β -carotene &

phycobilin & chlorophyll ‘a’ is present.

carotinoids are present.

iii) Reserve food material is Floridian iii) Reserve food material is starch.

starch iv) Unicellular & microscopic but few are iv) Unicellular or multicellular may be

6.

filamentous & heterotrichous

motile & flagellated

v) eg. geladuim, porphyra.

v) eg. spirogyra, Volvox

i)

It is a group of vascular plant which possess naked seeds attached to surface of megasporophyll

ii)

Megasporophyll is not folded to form on ovary so there is no fruit formation

7.

i)

They are small sized & occur in humid & tropic climate mostly growing as epiphytes.

ii)

The plant body is divided into root, stem & leaves.

iii)

Some ferns appears like small trees.

iv)

Lycopoduim,

selaginella

&

equistem are some members of pteridophytes. v)

The leaves are of two types compound leaves & sporophylls.

vi)

Plant body is sporophytic.

vii)

They are vascular cryptograms.

viii) Alternation of generation is present. ix)

Prothallus represent gametophytic phase.

x)

Pteridophyta is divided into 4- classes :- psilopsida, lycopsida, sphenopsida & pteropsida NCERT Page no 37 fig No. 3.3 (a) and (b)

8. ELGAE i) Mostly aquatic

BRYOPHYTES i) mostly terrestrial, found in damp, shady places. ii) Thallus single celled to branched ii) Thallus made of parenchymatous cells.

filaments iii) No tissue differentiation

iii) Tissue differentiation well marked

iv) Stomata absent

iv) Stomata present

v) Rhizoids absent

v) Rhizoids present

vi) Asexual reproduction by aplanospores vi) Asexual reproduction absent

or zoospores. vii)

Sexual

reproduction

isogamous vii) Sexual reproduction is of oogamous

anisogamoes or oogamous.

type

viii) No embryo formed after fertilisation

viii) Embryo formed after fertilization

ix) Eg. ulothrix, volvox, ulva, chladophora

ix) Eg. Ruccia, marchantia, funaria, porella

9.

The life cycle of moss represesents two distinct generations GAMETOPHYTIC & SPOROPHYTIC. Moss plant is a gametophyte. Spore is the beginning of gametophytic generation. It develops into protenema which give rise to male & female gametophytes. Gametophyte consists of green thallus having archegoniophores & antheridiophores which bear sex organs & the gametes are produced in them either monoecious or diecious. Club shaped antheriduim bears biflagellate sperms or antherozoids. Flask shaped archegonium encloses the female egg. Zygote is formed after the fertilization of male & female gametes with the help of water. Repeated divisions of the zygote give rise to the embryo (2N) which soon develops into sporophyte. The sporophyte of moss gets differenliated into three parts foot seta & capsule. Inside the capsule single celled spores are produced. After the dehiscence, they begin to germinate & give rise to protonema to start the cycle again. Gametophytic Generation alternates the sporophytic generation.

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (Plant Kingdom)

1.

Name the algae which is used for fodder to poultry birds.

[1]

2.

Which groups of plants is called vascular cryptogam ?

[1]

3.

What is a cone ?

[1]

4.

How do fungi differs from algae ?

[2]

5.

Both gymnosperms & angiosperm bear seeds but then why are they

[2]

classified separately?

6.

List any three characterstic features of Bryophytes.

[2]

7.

What are the identifying features of Angiosperms flowering plants.

[3]

8.

Describe the similarities in sexual reproduction of moss & fern.

[3]

9.

Describe the common mode of reproduction in Angiosperms.

[5]

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (Plant Kingdom) [ANSWERS] 1.

Laminaria.

2.

Pteridophyta.

3.

The fruiting body of gymnosperm which consists of micro & megasporophyll are called as clone.

4.

5.

Algae

Fungi

i) Chlorophyll present so they are green

i) chlorophyll absent so they are non- green.

ii) Autotrophic nutrition

ii) Saprophytic or parasitic nutrition

iii) Absorbs inorganic & mineral salts

iii) Absorbs organic or mineral salts

iv) Eg chlamydomonas, ulothrix

iv) Eg. albugo & yeast

The gymnosperms are plants that bears ovules which are not covered by any ovary wall & remain exposed. The seeds of gymnosperms are not covered that is they are naked but in the Angiosperm, the seeds remains closed inside the fruit so these are classified seperalely.

6.

i) They are small, erect plant growing in moist shady places ii) They have no leaf like, stem like or root like structure. iii) Most plants are gametophytes. They develop from haploid spores.

7.

i)

Majority of the plants around us are Angiosperms.

ii) flowering plants show great number of diversities in habitat, habits, forms, duration of life, mode of nutrition etc. iii) The plants with stem varying from a few mm to metre or so in height are termed as herbs, medium sized plants with woody stem are termed as shrubs & tall woody plants are known as trees.

iv) Plants which live for a year or part of year are termed annual, which live for two year are termed as biennials & which live more than two years are termed as perennials. v) Plants which live in extremely dry conditions are termed as Xerophytes; plants living in water are termed as hydrophytes; those living in moderate conditions are termed as mesophytes. vi) All flowering plants have roots, stem & leaves. They produce flowers, seeds & fruits.. vii) The economic uses of plants are varied. Plants provide us with materials for our food, clothing & shelter. 8.

i)

Oogamous mode of sexual reproduction which involves fertilization of nonmotile female gametes ar egg by means of a motile male gamete or sperm.

ii)

In both the male sex organ consists of a jacket of sterile cells that enclose a spermatogenous tissue.

iii)

Sperms are flagellate.

iv)

Female sex organ or archaegonia are flask shaped with tubular neck & a swollen basal venter. Venter encloses a single egg or oosphere & sterile venter canal cell. Neck has one or more neck canal cell.

v)

Are external source of water is needed for the swimming of the sperms so as to reach the open archegonia.

9.

vi)

Dependence of the embryo upon gametophytic phase.

vii)

Occurrence of hetromorphic or heterologous alternation of generation.

Stamens & pistils are the two reproductive parts of a flower. The stamen consist of a slender filament with anthers at the tip. Each pistil is made of three parts- ovary, style & stigma. Ovary contains one to many ovules. Each ovule contains megaspore mother cell it produces four haploid megaspores after meiosis of them three degenerate & remaining

one

is

functional

megaspore.

It

divides

by

meiosis

forming

megagametophyte. It consists of 8 haploid nuclei embedded in cytoplasm of which 3 cells lie at the micropylar end & 3 antipodal lie at chalazal end. The two remaining nuclei move to centre to make a diploid nucleus.

The anthers have pollen sac & contains many microspore mother cells. Each of them produces

four

haploid

microspores

after

meiosis

&

each

becomes

a

microgametophyte. It contains two nuclei generative nucleus & tube nucleus. The pouen iscarried away by air & other agencies & reaches stigma of pistil of same or difference plants. This process is called pollination. Pollen grains germinates & produces a pollen tube it grows within style & reaches ovule of ovary. The generative nucleus divides pollen tube producing two male gametes. On reaching ovule, pollen tube burst to release male gamtes. One of the two gametes fertilise egg & forms a diploid zygote. Other male gamete fertilizes with polar nuclei to form triploid endosperm. This is known as double fertilization.

CBSE TEST PAPER-04 CLASS - XI BIOLOGY (Plant Kingdom)

1.

What are cone bearing plants called?

[1]

2.

Name any red algae which is used as vegetables?

[1]

3.

What do you mean by thallus?

[1]

4.

List any two differences’ between gymnosperms & angiosperms?

[2]

5.

What is the role of capsule in life history of moss?

[2]

6.

What is the difference between syngamy & triple fusion?

[2]

7.

Mention some of the uses of ferns?

[2]

8.

Why are Bryophytes regarded as “the amphibians of plant kingdom”?

[3]

9.

Describe the important characterslics of gymnosperms?

[3]

10.

Classify plant kingdom?

[5]

CBSE TEST PAPER-04 CLASS - XI BIOLOGY (Plant Kingdom) [ANSWERS] 1.

Conifers

2.

Porphyra

3.

It is a plant body which is not differentiated into root, stem & leaves.

4. Gymnosperm i) Seed plants without flowers & with naked seeds. ii) There are about 9000 species of gymnosperm iii) Eg. cycas, Pinus

Angiosperm i) Angiosperms are known as flowering plants which have covered seeds ii) There are about 250,000 sp. Of angiosperm. iii) Eg. delonix, Rosa.

5.

Capsule is an important structure of sporophyte of moss plant. In the capsule, spores are produced. When the capsule ripens its dehiscence takes place & spores are liberated by winds. The spores develop favorable conditions into protonema.

6.

Syngamy is fusion of male gamete (sperm) to the female gamete (egg) to form a zygote while triple fusion is fusion of another male gamete to the diploid secondary nucleus to form primary endosperm nucleus.

7.

i) ii) iii) iv) v)

8.

Ferns are much used by florists for decoration. They are also grown as ornamental plants. Wood from tropical tree fern are used as building material because it resists termite decay. Ferns are used as astringent during childbirth to stop bleeding. Maiden hair fern is a source of expectorant.

Amphibians live on land & water with equal case but they must come to water during the breeding season to lay their egg. Water is therefore, essential for amphibians for breeding. In the same way, bryophytes live on land but they must get water for completing their life history because only through the medium of water antherozoides reaches the archegonia & fertilise the egg. If therefore, water is not available to bryophytes during

the period they shall not survive so on account of this similarity the bryophytes are called as “Amphibians of plant kingdom” 9.

i) ii) iii) iv) v) vi) vii) viii) ix) x) xi)

They grow in cool & warm climate in hills & in plains. Gymnosperms are evergreen woody & perennial plants They have well developed vascular system but compared to seed plants their xylem has no vessel & phloem is without companion cell. Plants are heterosporous. Conifers are cone bearing trees eg. pines, cedrus fir. They usually have evergreen needle like leaves which are well adapted to withstand extremes of temperature, humidity & wind. Reduction of gametophytic generation. The leaves have a reduced surface area thick cuticle & sunken stomata to conserve moisture & reduce the water loss by transpiration. Ovules are exposed to receive pollen grains. Gymnosperms possess exposed or naked seeds. Polyembryony is common occurrence.

10. Kingdom Plantae

Cryptogamae Thallophyta i) ii) iii)

iv) v)

Bryopnyta

Phanerogamae Pteridophyta

Gymnosperm

Angiosperm

Thallophytic:- plant body is thallus i.e. not differentiated into root, stem & leaves eg. chlorella, ulothrix, spirogyra etc. Bryophyta:- Amphibious in habit, water is necessary for fertilization, Vascular tissues are absent eg. Riccia, Marchantia, funaria. Pteridophyta:- plant body is differentiated into distinct underground stem like rhizome bearing roots & aerial shoots with leaves. They are called “primitive vascular plants “ eg. equistem, Adiantum, pteris Gymnosperm:- seeds are naked eg. cycas, pinus, cedars Angiosperm:- seeds are protected inside the friuts eg. i) monocotyledones eg. grass, maize, rice & ii). Dicotyledons eg gram, pea, sunflower.

CBSE TEST PAPER-05 CLASS - XI BIOLOGY (Plant Kingdom)

1.

Name the vascular plants which produces only spores but no flowers or

[1]

seeds?

2.

Where are the antheridia & archaegonia located in ferns?

[1]

3.

What are the two main classes of bryophytes?

[1]

4.

Tabulate differences between Gymnosperm & pteridophytes

[2]

5.

What is heterospory? What is its significance?

[2]

6.

What are gymnosperms? What are its four classes?

[2]

7.

How would you distinguish between monocots & dicots?

[2]

8.

List common modes of reproduction in Algae?

[3]

9.

What are ferns? Describe its salient features.

[3]

10.

Differentiate between Red, Brown & Green algae.

[3]

CBSE TEST PAPER-05 CLASS - XI BIOLOGY (Plant Kingdom) [ANSWERS] 1. 2. 3. 4.

Pteridophytes. Prothallus Liverwort & Mosses. Gymnosperm i) found in temperate climatic region ii) cambium present iii) pollentube is formed iv) Neck canal cells are absent

Pteridophytes i) found is shady & moist places ii) cambium lacking iii) pollen tube is not formed iv) Neck canal cells are present

5.

Heterospory refers to the production of two kinds of spores in pteridophytes eg. salivinia and selaginella produces two kinds of spores macrospores & microspores. These mega & microspores germinate & then give rise to male & female gametophytes. The female gametophyte is retained on parent sporophyte for variable period of time. So, this event is a precursor to seed habit.

6.

Gymnosperms are vascular plants with naked seeds. The seeds are exposed on surface of sporophyll. The reproductive organs are usually borne in cones on which spores are spirally arranged. Gymnosperms are classified into four groups Conifers, Cycads, Ginkgo, Gnetophytes.

7. Root Stem Leaf Floral parts Cotyledons. seeds

8.

Monocots Aaventitious Soft & herbaceous Parallel Trimerous One cotyledon Endospermic seeds

Dicots Tap roots Woody & herbaceous Reticulate Venation Tetra – or pentamerous Two cotyledon Non – endospermic seeds.

Reproduction in algae occurs by the following method:i)

Vegetative reproduction:- It occurs by fragmentation, zoospores, aplanospares palmella stages etc. akinites are also formed in asexual reproduction.

ii)

iii)

iv)

9.

Sexual Reproduotion in chlamydomonas:- In chlamydomonas, the flagellated & motile gametes which are isogamous unite to form a quadriflagellate zygote. It is converted into zygospore. When the flagella are lost & a cyst wall is formed around it zygospore germinate by meiosis to form four haploid meiospores. Palmella stage:- If the conditions are unfavourable, the daughter cells instead of forming zoospore divided repeatedly into numerous cells. Their walls become gelatinous & cells remains together. This stage is called palmella stage. On return of favourable conditions, the cells inside the gelatinous mass & develop cilia. Asexual Reproduction in chlamydomonas:- It takes place by formation of zoospores In the formation of zoospores, the cilia from chlamydomonas are withdrawn. The cell content divide into 4 & 8 daughter cells. In this way, they become motile & called as zoospores.

Ferns are found in warm moist tropical region & dry rocky places. The plant body is distinguished into three parts- i) underground stem rhizome ii) it bear roots & iii) it sends caerial shoots with leaves. Leaves of ferns are of two types- a) simple leaves with single vein & b) compound leaves with several leaflets. The sporophyte phase is dominant in ferns. On underside of leaflets are borne sori which contains sporangia. Where the spores are produced after meiosis division, the sporangium has an annulus. It is made of band of thickened cells that dry out pulling it open. So spores are released. These spores germinate into a porthallus the gametophyte. The gametophyte bears antheridia & archaegmia on underside. The antheridia bear flagellated sperms & egg lies at the base of archaegonia. The process of fertilization occurs when water is available for flagellated sperms to swim to reach the egg.

10. RED ALGAE i) Mainly marine ii) Only few are unicellular

BROWN ALGAE i) Marine form ii) Unicellular forms almost exist iii) Occurs in group of three a iv) chlorophyll a & c present

iii) Thylakoid unstacked iv) Only chlorophyll present v) fucoxanthin present vi) Phycobilin present vii) Reserve food is starch viii) Motile stages are not observed

v) Fucoxanthin present vi) Phycobilin absent vii) Reserve food is laminarin viii) Present

GREEN ALGAE i) Freshwater mostly ii) Unicellular species are more iii) stacked in groups of 2-20 iv) Chlorophyll a & b present v) Fucoxanthin absent vi) Phycobilin absent vii) Reserve food is starch viii) Present.

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (Animal Kingdom)

1.

Name an example of egg- laying mammals

[1]

2.

What is polymorphism

[1]

3.

Which animal is popularly called ploughman of nature & why ?

[1]

4.

List any four identifying features of arthopoda & give examples.

[2]

5.

Distinguish between diploblastic & triploblastic animals

[2]

6.

What is protochordates? How is it classified.

[2]

7.

“All vertebrates are chordates but all chordates are not vertebrates” justify

[3]

the statement. 8.

“Mammals are the most successful & dominant animals today” Give evidence

[3]

9.

How are non chordates different from chordates. Write the major phyla of

[5]

non- chordate & give examples.

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (Animal Kingdom) [ANSWERS] 1.

Duck- bill platyhelminthes

2.

The phenomena when an organism have different kinds of zooids for different function is called polymorphism.

3.

Earthworms are popularly known as Natures ploughman because it brings subsoil over the surface & create fine burrows for aeration.

4.

i)

Animals having jointed appendages

ii)

Triploblastic, coelomate, & bilaterally symmetrical

iii)

Body is covered by chitinous cuticle & segments are not separated by septa

iv)

Arthropods are unisexual animals

v)

eg. crab, Apis, spider, Anopheles

5. Diploblastic animals

Triploblastic animals

Diploblastic animals have two germ Triploblastic animals have three germ layer layers outer ectoderm & inner endoderm outer ectoderm middle mesoderm & inner

6.

in their embryo gastrula stage

endoderm in their embryo gastrula

Eg. Hydra, Obelia, Porpcta

Eg. all animals except porefera & coelentrata

Protochordates are the primitive non vertebrate ehordates. There are three subphyla a) b) c)

7.

Hemichordata eg. Belanoglossus. Urochordata eg. salpa & Herdmania. Cephalochordate eg. Amphioxus.

All vertebrates are chordates because they possess three basic chordate features as:i)

All chordates posses a notochord

ii)

All chordates have a dorsal hollow nerve cord.

iii)

All chordates have pharyngeal gill cleft in some stages of lift cycle

All chordates are not vertebrates. Vertebrates have vertebral column but protochordates & agnatha have notochord that is not replaced by vertebral column.

8.

Mammals are the most successful & dominant animals today. They thrive very well in most environment of world & The unique characteristics of mammals are:i) ii) iii) iv) v) vi) vii) viii) ix)

Body covered with hair Presence of sebaceous & sweat glands in skin Presence of mammary glands in females Presence of a pair of external ears & three ear osciscles Heart is four chambered RBCS are biconcave & enucleated Corpus callosum unites two cerebral hemisphere Testis are extra abdominal Mostly viviparous & embryo attached to uterine wall by placenta.

9. Non - Chordates

Chordates

i) Notochord is present

i) Notochord is absent

ii) Central Nervous system dorsal, hollow & single.

ii) Central nervous system is ventral solid & double

iii) Pharynx is perforated by slits

iii) Gill slits absent

iv) Heart Ventral

iv) Heart dorsal

v) A post anal metamerically segmented tail present

v) Terminal part unsegmented

Major phyla of non – chordates are:i)

Phylum - porifera:- adults sessile having cellular grade of organization & body is porous eg. Spongilla.

ii)

Phylum – coelentrata:- Radially symmetrical & tentacles present in polyps & medusa eg. Aurelia.

iii)

Phylum – Platyhelminthes:- Dorsoventrally flattened & organ of excretion is protonephridia eg. Taenia.

iv)

Phylum - Nematoda:- Parasitic forms with elongated round body eg. Enterobius.

v)

Phylum - Annlida:- Body metamerically segmented eg. Hiduneria.

vi)

Phylumn - Arthopoda:- Exoskeleton of chitin, Jointed appendages eg. Bombax mori

vii)

Phylum – Mollusc:- soft bodies shelled animals having foot, mantle & visceral mass eg. chiton

viii) Phylum – Echinodermata:- Exclusively marine having spiny skin & water vascular system with tube feet eg. ophiothrix.

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Animal Kingdom)

1.

What are the organs of excretion in annelids & insects

[1]

2.

Name a free living & a parasitic Platyhelminths

[1]

3.

Name two adaptations for an aerial mode of life

[1]

4.

Mention the unique features of nematodes

[2]

5.

Point out differences between dog fish & cat fish

[2]

6.

Outline the role of coelom in animals

[2]

7.

Enlist the main characteristics & examples of plylum porifera

[3]

8.

What are the basis of classification of animalia ?

[3]

9.

Enlist the main features of Aschelminthes & give examples

[5]

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Animal Kingdom) [ANSWERS] 1.

Annelida – nephridia & insect – malphigian tubule.

2.

Freeliving – planaria & parasitic – Taenia.

3.

i) ii)

Forlimbs modified into wings Uricotelic excretion & pneumatic bones.

4.

i) ii) iv) v)

Syncytical without mesodermal lining Intestine non – muscular but formed endoderm alone Body wall musculature & made of special types of muscles. Sexual dimorphism is quite clear.

5. Dog fish i) It belongs to phylum – chordate Class – Pisces Subclass - Chondrichthyes ii) It is a cartilaginous fish iii) Body streamlined & divisible into head, trunk & tail

Cat fish It belongs to phylum – chordate Class – Pisces Subclass – Osteichthyes It is a bony fish Its endoskeleton is made up of bones.

6.

Coelom is the space between body wall & alimentary canal of organisms it is lined by mesoderm. Visceral organs lie in the coelom. Flatworm does not have coelom. Hence they are called acoelamata. Pseudocoelom is found in the round worm. Annelids are coelomate animals.

7.

i) ii) iii) iv)

v)

They are commonly called as sponges They are generally marine, diploblastic, bilerally symmetrical They have water transport mechanism They are very primitive multi-cellular animals with cellular level of organization. Water can enter by pores Ostia in body wall directly or through canal into spongocoel. From it goes out by means of osculum. It is called canal system

vi) vii) viii) ix) x) 8.

The digestion is intracellular Body is supported by a skeleton which consists of spicules Sexes are not separate They reproduce asexually by fragmentation or sexually by formation of gametes Fertilization is internal example- euplectella, Sycon, Spongilla, Euspongia

Animals are classified on the basis of following characteristics:i)

Notochord:- It is a rod – like structure found on in the chordates. Non – chordates do not have it

ii)

Symmetry:- It is the plan of arrangement of body parts.” There are three types – asymmetric, radially symmetrical & bilaterally symmetrical.

iii)

Organisation:- Animals have cellular grade of organization. Their bodies are made up of cell others have tissues organs & organ system.

iv)

Embryonic layers:- Ectoderm, mesoderm& endoderm give rise to different organs in the body. These are called germinal layers. Some animals are diploblastic eg. sponges but others are triploblastic having three germinal layers.

9.

i)

They are called Round worm as they appear circular in C.S.

ii)

Free living, aquatic, terrestrial or parasitic

iii)

Organization of body is organ level

iv)

Bilaterally symmetrical animals

v)

They are triploblastic & pseudocoelomate

vi)

Alimentation complete with muscular or pharynx

vii) Sexes are Separate viii) Body is covered by cuticle ix)

Fertilization is internal

x)

Examples are filarial worm (wuchereria), Ascaris, Pinworm (Enterobuis) Hookworm (Ancyclostoma)

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (Animal Kingdom)

1.

Name the organs of defense in paramecium

[1]

2.

Name the second largest animal phylum

[1]

3.

What are acoelomate animals ?

[1]

4.

Mention the unique features of phylum mollusca.

[2]

5.

Distinguish between insect & arachnida.

[2]

6.

Why are echinoderms considered closer to chordates than any other

[2]

phylum ? 7.

Give important characters of phylum Nemathelmintnes

[3]

8.

Members of which phylum are known as “segmented worm” Write about

[3]

their body symmetry, mode of excretion & respiration. 9.

Enlist the main salient features of phylum ctenophora.

[5]

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (Animal Kingdom) [ANSWERS] 1.

Trichocysts are organ of defense in paramaecium

2.

Molluscs

3.

The animals which do not have a coelom or body cavity are called acoelomate eg. porifera, coelenterates, flatworms.

4.

i)

Body soft as well as unsegmented

ii)

The body is covered by calcareous shell & mantle.

iii)

Body is divided into – head, visceral mass &foot.

iv)

Buccal mass possesses radula.

5. Insect

Arachnida

i) Body is divided into three parts i) Body is divided into two parts –

head, thorax & abdomen ii). Wings are found

cephalthorax & abdomen ii) wings are lacking

iii).Appendages on head are antennae, iii) Appendages on head are pair of

mandibles & maxiliae iv) Walking legs are three pairs

chellcerae & a pair of pedipalpa. iv) walking legs are four pairs

6.

Echinoderms are considered closer to chordates because like chordates, they are deuterostomes where the anal region develops earlier than mouth region. Their larve are also closer to protochordata.

7.

i) ii) iii) iv)

They are commonly known as roundworms or nematodes & are covered by cuticle. Body is bilaterally Symmetrical Animals with elongated cylindrical & spindle shaped body with pointed ends The body cavity is a false coelom called pseudocoelom

v) vi) vii) 8.

The alimentary canal lacks a muscle layer. Respiratory organ & blood vascular system are absent Example – Ascaris, Ancylostoma, Rhabditis.

The members of the phylum Annelida are known as “the segmented worms” Their body is metamerecally segmented eg. Neiris, pheretima & Hirudinaria. i)

Body Symmetry:- Segmented worms have typical metameric segmentation. Their body consists of segments called somites or metamere & ring like grooves known as annuli

ii)

Excretion:- the excretory unit of these invertebrates are coiled tubules called nephridia.

iii)

Respiration:- Respiration occurs by gills or by skin. The skin is richly supplied with blood vessels. It is permeable. The exchanges of gases take place there.

9.

i)

Ctenophares are marine animals with transparent & flat ar oval body shape.

ii)

Polyp phase is absent in life cycle.

iii)

These are bilaterally symmetrical & devoid of cnidoblast cells.

iv)

When the tentacles are present they are two in number & contain colloblast cells.

v)

They move by cilia which join together to from comb plates, they are eight median comb plates.

vi)

They gastrovascular cavity is branched & open to the exterior by stomodaeum. Example of Ctenophora (Pleurobrachia )

vii)

They are diploblastic animals but the mesoglea is different from that of cnidaria.

viii)

The presence of special sense organs at the opposite end of the mouth is the characteristic of this phylum.

ix)

They reproduce only by sexual means

x)

They do not have larval phase in their life cycle

xi).

Eg. ctenophore, ctenoplana, Beroe, & Hormiphora.

CBSE TEST PAPER-04 CLASS - XI BIOLOGY (Animal Kingdom)

1.

Name the larva found in mollusca & annelid.

[1]

2.

Name two viviparous fishes.

[1]

3.

What are flame cells ?

[1]

4.

Distinguish between bony fish & cartilaginous fish.

[2]

5.

Give reason why a snail & an octopus are classified under the same

[2]

phylum?

6.

List three basic chordate characters

[2]

7.

Give any four characteristics of hemichordate.

[2]

8.

Differentiate between Annelida & Arthropada.

[3]

9.

What are basic plans of body design in animals?

[3]

10.

Mention the important characters of phylum echinodermata & give

[3]

examples.

CBSE TEST PAPER-04 CLASS - XI BIOLOGY (Animal Kingdom) [ANSWERS] 1.

Trochophore larvae

2.

Pristis & scoliodon

3.

Flame cells are excretory organs of platyhelminthes which possesses flickering cilia or flagella for driving the absorbed excretory product into system of ducts

4. Bony fishes Cartilaginous fishes i) They are called osteichthyes i) They are called chondrichthyes. ii) Their endoskeleton is bony. ii) Their endoskeleton is cartilaginous. iii) They are found in sea & fresh water iii) All are marine forms. both. iv) They have swim bladder. iv) They have five pairs of gills v) Gills are covered by opercula. v) operaculum absent vi) Eg. Salmon, catla Rohu. v) Rays, scoliodon, electric ray.

5.

Snails & octopus are classified under the phylum mollusca because they have following three characters:i) Presence of mantle in both ii) Presence of foot in both iii) Presence of shell in both

6.

i) ii) iii)

7.

i) ii) iii) iv)

Notochord :- a dorsal solid notochord is present throughout life or in larval stage. Nerve cord :- a dorsal hollow nerve cord is present Pharyngeal gill slits :- a perforated pharynx is present in young condition or throughout life. These are worm like marine animals that have organ- system level of organization. They are bilaterally symmetrical, triploblastic & eucoelomate. Body is cylindrical & is divided into anterior proboscis, collar & long trunk. Respiration occurs through gills.

8. Annelida i) Elongated & metamerically segmented body ii) Appendages borne on body segments iii) Setae present iv) Body wall dermomuscular v) Body cavity is coelom vi) Respiratory pigment is haemoglobin vii) Blood is red viii) Blood vascular system is close type ix) Cilia & nephridia present x) No exoskeleton

Arthropoda i) Body segmented & differentiated into cephalic, thoracic & abdominal region ii) Appendages may be segmented or jointed. iii) Setae absent iv) Body wall is not dermomuscular v) Body canal is haemocoel vi) Respiratory pigment is absent vii) Blood is colourless or bluish viii) Blood vascular system is open type ix) Cilia & nephridia absent x) Exoskeleton is chitinous

9.

Animals can be divided into three basic plans: i) Cell Aggregate plan:- It is found is simple animals eg. sponges in which clusters of cells with rudimentary division of labour is found in them. ii) Blind sac plans:- It is found in coelenterates & flat wors. They have a digestive cavity with only one opening to the outside. Through this opening the mouth food is ingested & undigested waste is thrown out. The cells are more specialized & have division of labour. iii) Tube- within a tube plan:- It is found in more complex forms In this plan body cavity forms one tube within which is situated another tube alimentary canal, opening on one side by mouth & other side by anus.

10.

i)

The word Echinodermate means “ spiny skin” which is optly used for group of animals represented by such common forms e. starfish, Sea urchin. ii) The skin forms a hard spiny protective skeletal covering iii) They are sluggish marine forms. iv) Forms usually show a pentamerous radial symmetry v) The radial symmetry is superficial & body in fact can be divided only in two halves. vi) They have a coelom & water vascular system. vii) Locomotion takes place by numerous hollow tube feet viii) Excretion by diffusion through body ix) Fertilization in open sea. x) Development includes free swimming diploneural larva. Eg. Asterias, searerchin, sea cucumber.

CBSE TEST PAPER-05 CLASS - XI BIOLOGY (Animal Kingdom)

1.

Name a vertebrate in which jaws are absent

[1]

2.

Assign the phylum to which following animals belongs – pheretima &

[1]

sponge 3.

What is metamerism ?

[1]

4.

Distinguish between centipede & millipede

[2]

5.

Give reason why arthopda constitute the largest group of animal kingdom

[2]

6.

Differentiate between male & female ascaris.

[2]

7.

List three adaptations that help the birds (Aves) in flying.

[2]

8.

Give three important distinguishing characters of arthropodo, reptiles &

[3]

mammals

9.

Mention the important characteristics of coelentrata & give examples

[3]

10.

Differentiate between flightless & flying birds.

[3]

CBSE TEST PAPER-05 CLASS - XI BIOLOGY (Animal Kingdom) [ANSWERS] 1.

Petromyzon

2.

Pheretima – Annelida & sponge – porifera.

3.

In some bilateria, the body consists of many segments & shows repetition of parts. This type of segmentation is called metamerism.

4. centipede millipede i) Dirsoventrally flattened body i) cylindrical body ii) There are two parts of body – head & ii) There are three parts of body – head, trunk thorax & abdomen. iii) Maxillae are 2 pairs iii) Maxillae are only one pair.

5.

Arthopods constitutes the largest group of animal kingdom:i) Have organ level of organization. ii) bilaterally symmetrical, segmented, triploblastic, encoelomate animals. iii) Body enclosed by chitinous cuticle. iv) They have jointed appendages. v). Trachea or book gills for respiration.

6. Male Ascaris i) 15-30 cm long ii) Posterior end curved. iii) Vulva absent iv) There are 2 pineal spicules from cloacal pore. v) Pre- anal or post- anal papillae present

7.

i) ii) iii)

8.

i)

Female Ascaris i) 20-40 cm long ii) Posterior end straight iii) Vulva present iv) No pineal spicules. v) There are no such structures.

The avian flight muscles are used for fast short fly. Flight muscles contain white fibres which are poorer in mitochondria & free of myoglobin. The long bones are hollow & connected by air passages. Arthopods:- jointed appendages, segmented body divisible into head, thorax & abdomen, presence of hard non- living exoskeleton of chitin, eyes compound eg. insects, centipede etc.

ii) iii)

9.

i) ii) iii) iv) v)

vi) vii) viii)

Reptiles:- cold blooded, Body covered by scales, Two pairs of limbs, lay eggs eg. lizard, snake etc. Mammals:- warm blooded, body covered by hairs, an external ear is present, give birth to young ones, They have small pointed teeth & long snout insectivores are primitive mammals. They are marine animals which may be silitary or colonial The body is two layered or diploblastic The body possesses a radial symmetry They are acoelomate animals i.e. true coelom. They exhibit blind sac body plan. The body encloses a large central cavity known as coelenterons which has a single opening to the exterior. Coelenterons is called gastro vascular cavity. They commonly show polymorphism. Two kinds of individuals presenthybroid & medusoid They possess tentacles which are usually thread- like out growths. Stinging cells or nematoeytes are present. Eg. Obelia. Aurelia.

10. i) Classification ii) Wings iii) Feathers iv) Sternum v) Ribs vi) Tail vertebrate vii) Flying viii) Distribution ix) Example

Flightless Birds Belong to suborder Retitae Wings vestigial No interlocking mechanism Sternum raft like No uncinate process Pygostyle may be small or absent. Cannot fly Restricted in distribution Rheo, cassowary emu ostrich.

Flying Birds Belong to carinatae Wings are well developed Possess interlocking mechanism Sternum boat shaped Ribs uncinate process Pygostyle found Can fly Found all over the world House sparrow cuckoo hornbill quail peacock fowl parrot crow.

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (Morphology of Flowering Plants)

1.

Name one monocot & one dicot in which endosperm is present?

[1]

2.

Why are date palm referred to as dioecious?

[1]

3.

What is placentation?

[1]

4.

What is Rhizome? Give its two examples.

[2]

5.

Differentiate between epigynous & perigynous flowers.

[2]

6.

Give reason to justify that onion bulb is a modified stem?

[2]

7.

Describe that parts of a typical angiospermic leaf?

[3]

8.

Differentiate between a maiz grain & a bean seed?

[3]

9.

Describe the various types of placentations found in flowering plants &

[5]

represent diagrammatically.

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (Morphology of Flowering Plants) [ANSWERS] 1.

Monocot: Maize grain & Dicot: castir oil seed.

2.

Because male & female flowers are borne on different plants.

3.

The arrangement of ovule in the ovary is called placentation.

4.

Rhizome is a prostrate & a thickened underground stem having distinct nodes, internodes scales, leaves as well as buds. It creeps horizontally under the ground eg. Ginger, turmeric.

5. Epigynous flowers i) The thalamus is cup shaped & is fused with the ovary so that floral parts rise on the top of ovary. ii) Ovary is inferior eg. Apple, cucumber

Perigynous flowers i) The thalamus is cup-shaped structure around the ovary but is not fused & bears sepals, petals & stamens. ii) Ovary is half inferior eg. Rose.

6.

Onion bulb is a modified, highly condensed & disc like. It has a large number of fleshy scale leaves. Terminal & auxiliary buds are present. On the lower posterior side a cluster of adventitious roots are present.

7.

A typical angiospermic foliage leaf possesses the following parts. i) LEAF BASE:- It is the region in the stem, from which leaf arises. Its main function is to attach the leaf with the stem are a branch. ii) PETIOLE:- The stalk of a leaf is called petiole. The leaves having petiole is called petiolate. As in banyan leaf, some leaves may lack petioles. iii) LAMINA:- The green, flattened part of the leaf attached with petiole is known as “lamina”. It is the part which performs photosynthesis, respiration & transpiration. There is a “midrib” in the middle of the lamina. The midrib in compound leaf is called rachis. The lamina may be of different shapes in different kinds of leaves.

8. MAIZE GRAIN i) It is single seeded fruit called the caryopsis ii) The fruit wall or the pericarp is fused with testa. iii) There is one seed coat which inseperably fused with pericarp iv) The grain is endospermic v) The grain has no hilum, micropyle & chalaza on its surface. vi) There is no ridge like raphe

BEAN SEED i) It is a true seed formed inside a fruit called the pod or legume. ii) The pericarp is free from testa.

iii) There are two seed coats called testa & tegmen They are fused with each other. iv) The seed is non endospermic. v) The chalaza, hilum & micropyle are clearly visible. vi) The raphe is clearly risible. vii) The plumale & radical are not covered vii) The plumule & radical are protected by any such protective sheath. by distinct sheath called the coleoptinct sheath called the coleoptiles & coleorhizae respectively viii) The cotyledon acts as the absorbing viii) The cotyledons are merely food structure that absorbs food from storage organs. endosperm & transfers it to embryo.

9. i) ii) iii) iv) v) vi)

The various types of placentations found in flowering plants are:MARGINAL:- Ovary one chambered and ovules lies along the margin of the ovary eg. pea & gram PARIETAL:- Ovary one chambered and ovules lies at the level of the fusion of the fusion of the carpels. Eg. mustard. AXILE:- Ovary many chambered and ovules are attached to the central column eg. onion & lemon. FREE CENTRAL:- Ovary one chambered and at the centre it bears many ovules eg. Dianthus, Primula. BASAL:- Ovary one chambered and ovules develop on the thalamus eg. sunflower. SUPERFICIAL:- Ovary is multilocular & syncarpous. Ovules develop on the minor surface of the ovary. Eg. Nymphea.

Types of placentation :(a) Marginal (b) Axile (c) Parietal (d) Free central (e) Basal

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Morphology of Flowering Plants)

1.

Write floral formula of Brassica Campestris.

[1]

2.

Why are flowers of cucumber referred to as epigynous ?

[1]

3.

What is false fruit?

[1]

4.

What is the difference between alternate & whorled phylotaxy.

[2]

5.

Define venation? What are two types of venation?

[2]

6.

Why is leaf of Bombax categorized as palmately compound multifoliate leaf?

[2]

7.

Describe the arrangement of floral members in relation to their insertion on

[3]

thalamus. 8.

How is herbaceous stem different from a woody stem?

[3]

9.

What is a flower? Describe the parts of typical angiospermic plants with the

[5]

help of a diagram.

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Morphology of Flowering Plants) [ANSWERS]

1.

2.

Because the floral parts lie above the ovary & the ovary is inferior.

3.

When floral parts other than ovary takes part in formation of fruit & become edible, it is called false fruit.

4. Alternate phyllotaxy

Whorled phyllotaxy

i) Only one leaf arises at each node.

i) More than two leaves arises at each node

ii) Leaves arises alternately on left ii) Leaves arise in whorl from one point

& right sides of the stem iii) Eg. chinarose, mango

5.

6.

iii) Eg. Neruim

Veins arrangement in leaf lamina is called venation. There are two types of venation:a)

Parallel:- when veins are arranged parallel to each other on lamina

b)

Reticulate:- when veins forms a network on leaf lamina.

Leaf of Bombax is categorized as palmately compound multifoliate leaf because the petiole bears leaflets in its lips in pinnately compound leaf. Five or more leaflets are articulated on a long axis & the shape of leaf is like the palm of a hand in Bombax. This type of leaf is called digitate.

7.

Based on the position of calyx, corolla & the androecium in respect of ovary on the thalamus, flowers may be explained into 3 kinds. a).

HYPOGYNOUS FLOWERS:- Gynoecium located at highest position & rest whorls of flower lies below it. eg. mustard, chinarose.

b).

PERIGYNOUS FLOWERS:- The gynoecium is situated in center, other parts of flower lie on the rim of thalamus almost at same level, Ovary is half inferior eg. plum, rose.

C)

EPIGYNOUS FLOWERS:- The margins of thalamus grows upwards enclosing the ovary fully & getting to it, rest parts of the flower arises above the ovary. i.e. Ovary is inferior in these flowers. Eg. Guava, sunflower, cucumber.

8. HERBACEOUS STEM

WOODY STEM

i) Annual or biennial & short- lived

i) Always perennial & long. Lived.

ii) Green, soft and fleshy and on bending ii) Brown or grey & hard and break on

does not break. iii)

The

protective

bending superficial

epidermis forms the outer covering.

layer iii) Thy epidermis is replaced by corky

layer or bark.

iv) Stomata are present throughout its iv) It develops dot- like pores called

length for gaseous exchange

lenticels for gaseous exchange

v) Buds often naked

v) buds are often protected by scales

vi) They consist of primary permanent vi) They consist of secondary permanent

tissues. 9.

tissues.

The flower can be defined as a modified shoot bearing nodes & modified floral leaves. It consists of following parts:-

i).

CALYX:- It is the outermost whorl of flower. It is green. Leaf- like structure it may be polysepalous (sepals free) or gamosepalous (sepals united) calyx may be regular or irregular.

ii).

COROLLA:- It is the second whorl of the flower inside the sepals. The petals are usually brightly coloured. The insects are attracted due to colour of the petals so they help in pollination. The narrow stalk like lower portion of petal is called a claw & the upper extended portion is known as limb.

iii).

ANDROECIUM:- It represents as male reproductive parts. It consists of stamens in each stamen there are three parts:a)

Anther:- Knob like bilobed structure containing pollen grains. Each lobe contains two chambers called pollen sac.

b)

Connective:- A strip of tissue, which connects the anther lobe is called connective.

c)

iv).

Filament:- a slender stalk by which anther lobes are attached is called filament.

GYNOECIUM:- It is the female part of the flower it is made up of three parts a)

Stigma:- upper part of pistil which receives pollen grains

b)

Style: - The stalk between stigma & ovary.

c).

Ovary:- basal part containing ovules.

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (Morphology of Flowering Plants)

1.

What is the term used for a plant bearing both male & female flowers.

[1]

2.

What are runners?

[1]

3.

Why are flowers of mustard referred to as hypogynous.

[1]

4.

Explain with suitable examples of different types phyllotaxy

[2]

5.

Draw a well labeled diagram of V.S. of maize seed.

[2]

6.

Write differences between phyllode & phylloclade.

[2]

7.

How do various leaf modifications help plants?

[3]

8.

Differentiate between Tuber & Bulb.

[3]

9.

Describe the aerial modifications of stem.

[5]

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (Morphology of Flowering Plants) [ANSWERS] 1. Monoecious flowers 2. A long creeping stem with long internodes running horizontally on the surface of the soil is called a runner. 3. Because ovary is situated at the top & other three whorls are inserted below the pistil. 4. Phyllotaxy is the arrangement of leaves on the stem or branch. It can be of two types:i).

OPPOSITE PHYLLOTAXY:- Two leaves at each node opposite to each other. Eg. calotropis Guava.

ii).

WHORLED PHYLLOTAXY:- Where more than two leaves arise at each mode eg. nerium

5. .

6. PHYLLODE

PHYLLOCLADE

i) Modification of petiole

i) Modification of stem

ii) Bears an bud in its axil

ii) Developed in axial of leaf

iii) Nodes internodes are not borne

iii) Nodes internodes are found.

iv) Does not have leaves & flowers

iv) Has reduced bristles spiny leaves &

flowers.

7. The normal functions of leaves are photosynthesis, respiration & transpiration. Besides these function the leaves have to perform other functions. Hence, they modify themselves in different ways as follows:i)

TENDRIL:- In some plants the entire leaf or part of it gets modified to coiled thread like structure called tendrils . Tendrils help the plants to climb up eg. pea, clematis.

ii)

SPINES:- In many plants the leaves or their apices are modified into thin sharp & pointed structure known as spines. They help in defence eg. opuntia, yucea. Etc.

iii)

SCALE LEAVES:- In onion mostly all the leaves are present in the form of fleshy scale leaves.

iv)

PITCHER:- It is the modification of leaf in insectivorous plant in which the lamina takes the form of a pitcher, apex in the form of a lid to trap the insects. There are number of digestive glands in the inner walls of the pitcher. These glands secrete a fluid which digests insects eg. Nepenthes.

v)

PHYLLODE :- The petiole becomes green, flattened & leaf like & is called phyllode eg. Australian Acacia.

8. TUBER (POTATO)

BULB (ONION)

i) Stem is very well developed

i) Stem is reduced to a disc.

ii) Adventitious roots absent

ii) Adventitious roots are present.

iii) Potato plant can bear numerous tubers

iii) Only one bulb develops in one onion

plant. iv) Food is stored in stem.

iv) Food is stored in fleshy scale leaves.

v) Food stored in the form of starch.

v) Food not stored in the form of starch.

vi) Buds external

vi) Buds internal

vii) Distinct nodes & internodes are present vii) Nodes & internodes are indistinct viii) Scale leaves found in the nodal region viii) Scale leaves are fleshy & conspicuous

are very small. ix) The tuber is a total stem.

ix) The bulb is a shoot.

9. AERIAL MODIFICATIONS OF STEM INCLUDES:1. STEM TENDRIL:- Stem tendrils are thin leafless slender & spirally coiled structures which develop from auxiliary buds. They help the plant such as cucumber, water melon, grape vine etc. to climb. 2. STEM THORN:- sometimes the auxiliary buds grows into hard, woody straight & pointed structures called thorns. It arises in the axil of leaf or at the tip of branch. Sometimes thorn bears leaves also. They are commonly found on plants eg. citrus durantha, Bougainvillea etc. 3. PHYLLOCLADE:- It is the green flattened or cylindrical stem which takes the form and function of leaf. They contain chlorophyll & carry photosynthesis. They have many nodes & internodes. Their true leaves are reduced, spines or scales. It is commonly found in xerophytic plants eg. opuntia, epiphyllum etc. 4. CLADODE:- This is a phylloclade of limited growth which develops, from the node of the stem or branch & in the axil of a scale leaf eg. asparagus, Ruscus, asculentus etc. cladodes are green flat & leaf like structures which carry on photosynthesis. 5. BULBILS:- This is a modified vegetative or floral bud meant for the production of a new plant. It detaches itself from mother plant & grows into an independent plant. Bulbils are found in oxalis, Agava american, Lilium etc.

CBSE TEST PAPER-04 CLASS - XI BIOLOGY (Morphology of Flowering Plants)

1.

Name the two layers of seed coat.

[1]

2.

Which family has characteristically a swollen axile placenta.

[1]

3.

Why root system is poorly developed in aquatic plants.

[1]

4.

“Flower is a modified shoot.” justify the statement.

[2]

5.

Distinguish between prop root & stilt roots

[2]

6.

What is inflorescence? What are its two types?

[2]

7.

Draw the floral formula & floral diagram of family solanaceae.

[2]

8.

Give four types of underground stem & give examples for each.

[3]

9.

Compare Trailer, runner & sucker.

[3]

10.

What do you mean by “modification of roots”. Describe some of the

[3]

modifications of tap roots giving suitable example.

CBSE TEST PAPER-04 CLASS - XI BIOLOGY (Morphology of Flowering Plants) [ANSWERS] 1.

Testa & Tegmen.

2.

Solanaceae.

3.

Because in aquatic plants there no soil to anchor firmly rather, absorption of water occurs through diffusion hence root system is not completely developed.

4.

“Flower is considered as modified shoot” because the internodes in a flower are highly condense & the appendages such as sepals, petals, stamens & carpels are generally large in number.

5. PROP ROOTS

STILT ROOTS

i) arises from horizontal aerial i) Arises from basal nodes of stem.

branches of a free stem ii) Long & provide support to plant ii) Short roots and grows downward obliquely

6.

like pillars

to provide support to stem like rope of tent.

iii) Eg. banyan tree

iii) Maize, Jowar.

The arrangement of flowers an the floral axis is called inflorescence. Inflorescence are of two major types:a)

Racemose inflorescence:- main axis continues to grow & flowers are borne laterally in acropetal succession.

b)

Cymose inflorescence: - main axis terminates in a flower hence, is limited in growth, flowers are borne in basipetal order.

7.

Floral formula:-

Floral Diagram :-

8.

FOUR TYPES OF UNDERGROUND STEMS:i)

RHIZOME:- The stem is prostate, thickened & grows horizontally under the soil. Stem is much branched & each branch ends in terminal bud. Adventitious roots arise in profusion eg. fern, water lily, turmeric.

ii)

BULB :- Highly condensed & discoidal stem. Terminal bud in the centre produces aerial root that produces flowers. From base of stem adventitious roots develop. Leaves store food material. Terminal bud & scale leaves are present eg. onion garlic.

iii)

CORM:- Condensed form of rhizome with auxiliary buds & scale leaves. It is swollen base of underground stem axis. Nodes & internodes are present eg. zimikand, saffron, colocasia.

iv)

TUBER:- It grows horizontally & swells at the apex. Adventitious roots arise during sprouting. It has many buds that grow into new plants eg. potato, Halianthus.

9. TRAILER

RUNNER

SUCKER

i) Semi aerial creeping stem i) Prostate, sub-aerial stem. It i)

it does not roots at intervals

is green & root at intervals.

Underground

non

green stem.

ii) Does not participate in ii) Does not participate in ii) Helps in perennation.

perennation

perennation.

iii) No help in vegetative iii)

propagation.

10.

Helps

in

vegetative iii) Helps in vegetative

propagation.

propagation.

The functions other than normal functions of roots eg. fixation, absorption & conduction are to be carried out by roots. These are called modifications of roots. The modifications of top roots includes:-

a).

FUSIFORM:- This roots is swollen in the middle & tapers at both the ends gradually eg. Raddish.

b).

NAPIFORM:- The shape of this root becomes almost spherical but tapers abruptly downward eg. turnip.

c).

CONICAL:- The shape becomes cone like eg. carrot.

d).

TUBEROUS:- It is a swollen root having no specific shape eg. mirablis, Trichosanthes.

CBSE TEST PAPER-05 CLASS - XI BIOLOGY (Morphology of Flowering Plants)

1.

Name two plants where seeds do not have endosperm?

[1]

2.

Which plants part has given rise to following modifications:-

[1]

a) Spines of opuntia

b) Pitcher of Nepenthes.

3.

Why is leaf of Neem called unipinnately compound.

[1]

4.

Differentiate between true fruit & false fruit.

[2]

5.

Write the floral formula & draw the floral diagram of family Liliaceae.

[2]

6.

“Underground parts of a plant are not always roots” justify the statement.

[2]

7.

How would you differentiate leaflets of a compound leaf from simple leaves

[2]

on a branch? 8.

What is aestivation? What are its different types give examples.

[3]

9.

Describe the sub- aerial modifications of stem.

[3]

10.

Explain with examples. What are the different modifications of adventitious

[3]

roots?

CBSE TEST PAPER-05 CLASS - XI BIOLOGY (Morphology of Flowering Plants) [ANSWERS] 1.

Bean, gram, pea.

2.

a)

3.

Leaf of Neem is called unipinnately compound because leaflets are found in pairs on

modified stem

b)

modified leaf.

either side of rachis. 4. TRUE FRUIT

FALSE FRUIT

i) it develops from the ovary

i) it develops from other parts along with the ovary

ii) No other part is involved in ii) Thalamus and perianth takes part in fruit

fruit formation

formation.

iii) Eg. pea.

iii) Eg. apple.

5.

Floral formula: -

Floral diagram: -

6.

Usually roots develop below the ground. But in potato, the stem gets modified into “tuber” like structure for the storage of reserve food material. These tubers develop & grow under the ground. Potato is a stem because it bears scale leaves, buds, nodes etc.

7. SIMPLE LEAF

COMPOUND LEAF

i) Lamina is not divided into distinct lobes i) Lamina is incised into two or more

or leaflets.

distinct leaflets.

ii) Axilliary bud is present in the axil of ii) Bud is present in the axil of whole leaf.

simple leaf. iii) Simple leaves are in acropetal iii) Leaflets of compound leaf are not in

succession on stem

acropetal succession.

iv) Base of leaf may have stipules

iv) Stipules may be present base of

compound leaf v) Simple leaves appear in one or more v) Leaflets in a compound leaf lie in one

plane.

8.

plane only.

The mode of arrangement of sepals or petals in floral bud with respect to the other members of same whorl is known as AESTIVATION. The main types of aestivation are:a)

VALVATE:- when sepals or petals in a whorl just touch one another at the margin without overlapping eg. calotropis.

b)

TWISTED:- if one margin of appendage overlaps that of the next one & so on & is called twisted eg. chinarose, ladyfinger.

c)

IMBRICATE:- If margins of sepals or petals overlaps one another but not in a particular direction eg. cassia & gulmohar.

d)

VEXILLARY:- There are five petals the largest overlaps the two lateral petals which in turn overlaps the two smallest anterior petals eg. bean, pea.

9.

The main function of sub-aerial modification of stems is vegetative propagation. They are of following types:i)

RUNNERS:- These stems are long & thin with branches which creep along the ground & develop root at the nodes. Many such branches are produced by mother plant & they spread out in all direction. They may break off & start living as independent plants eg. oxalis, doob grass.

ii)

STOLON:- This is also a thin lateral branch which arises from the base of stem. It grows upward & bent down again developing roots at the tip & producing a bud. The bud grows into a new plant eg. mint, strawberry.

iii)

OFFSET:- This is a thickened horizontal branch arising in the axil of a lower leaf. It is a short branch which produces a cluster of leaves above & tufts of roots below. Offset can break off from mother plant & start living independent life. Eg. Water lettuce, water hyacinth.

iv)

SUCKER:- The sucker is a lateral branch which develops from underground part of stem. It grows upward in obliquely manner & directly give rise to new plant eg. banana, pineapple.

10.

MODIFICATIONS OF ADVENTITIOUS ROOTS:i)

TUBEROUS:- It is swollen root & shapeless occurring singly eg. sweet potato.

ii)

FASCICULATED:- Several tuberous roots arise from the same place in a cluster eg. dahlia, Asparagus.

iii)

BEADED ROOTS:- These roots have swollen parts at frequent intervals eg. portulaca, vitis.

iv)

PROP ROOTS:- These are pillars like roots hanging vertically downward from aerial branch of plant eg. Banyan tree.

v)

STILT ROOTS:- The roots are short which grow obliquely from near the base of the main stem & they provide anchorage & support to the stem eg. sugarcane, maize, sorghum.

vi)

PARASITIC ROOTS:- These roots penetrate into the host cells & absorb nutrients from host tree eg. cuscutta.

vii)

ASSIMTLATORY ROOTS:- Adventitious roots in certain plants become green to carry out photosynthesis & are called assimilatory roots eg. tinospora, trapa.

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (Anatomy of Flowering Plants)

1.

Name two specialized kinds of parenchyma.

[1]

2.

What is the function of companion cells in phloem ?

[1]

3.

Define meristem.

[1]

4.

Why is cambium considered to be a lateral meristem ?

[2]

5.

Mention four characteristics of sunflower’s vascular bundles.

[2]

6.

Differentiate between tracheids & vessels.

[2]

7.

Explain the structure & function of collenchyma.

[3]

8.

What are sieve elements? Explain their types & functions.

[3]

9.

Describe the internal structure of a monocot root with the help of a labeled

[5]

diagram.

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (Anotomy of Flowering Plants) [ANSWERS] 1.

i) aerenchyma

2.

Companion cells help the sieve tube members in translocation of food material

3.

All the cells of an embryo of the plant are capable of division but, in a localized region cell division occur continuously. It is called meristem.

4.

These meristems are present along the lateral sides of stem & roots therefore these are called lateral meristem. Interstealer cambium ring formed by intrafasicular & inter fascicular are two examples of lateral meristem.

5.

i) ii) iii) iv)

ii) Chlorenchyma.

Xylem & phloem occurs as alternate separate patches on different radii. Xylem is exarch. The number of rays is equivalent to the number of xylem bundles & accordingly xylem condition in the root may be called as monarch. Diarch, triarch, tetrarch, pentarch, hexarch & poly arch.

6.

7.

TRACHEIDS

VESSELS

i) found in all vascular plants

i) Found in angiosperms only

ii) They are shorter & dead at maturity

ii) They are very big & dead at maturity.

iii) Lumen is narrow

iii) Lumen is wider.

iv) Tracheids have pointed ends.

iv) End walls mostly absent.

Collenchymas has polygonal cells & has unevenly thickened walls which are prominent at the corners. It is an example of simple tissue. Cells are more or less elongated with primary, non-lignified cell wall. The wall thickening is primary in nature & is composed of cellulose, hemicelluloses & pectin materials with high percentage of water. The thickening may be primarily at the corners or angles of the cells. They are found mostly in the hypodermis of herbaceous dicots in the form of homogenous layers or in the patches.

FUNCTION:- The main function of this tissue is to give strength to the plant parts. They also provide elasticity & support to the growing organs 8.

Sieve elements are the parts of phloem. They are meant for translocation & conduction of food material. Sieve elements are of two types:a)

Sieve

cells:-

sieve

cells

are

present

in

pteridophytes and gymnosperms. The cell wall is perforated. There are sieve plates throughout end walls & lateral walls. b)

Sieve tubes:- sieve tubes are present in angiosperms. Many sieve cells are connected to each-other to form a channel. There are sieve plates of the walls.

9.

A T.S. of monocot root shows the following tissues:a)

EPIDERMIS:- It is the outermost layer of root having no intercellular spaces stomata & cuticle. It bears unicellular root hairs.

b)

CORTEX:- It is present beneath the epidermis. It consists of many layers of parenchymatous

cells

with

large

intercellular spaces. c)

ENDODERMIS:- It is the innermost layer of cortex. Its cells are barrel shaped with casparian strips on their antinunal walls. The passage cells are seen just opposite the protoxylem ends.

d)

PERICYCLE:- It consists of single layer of thin walled parenchymatous cells.

e)

VASCULAR BUNDLE:- The vascular bundles are radial, alternating xylem & phloem. The xylem & phloem bundles are always more than six. The xylem is exarch in condition. The central portion is occupied by large pith of parechyomatous cells. The conjuctive tissue is found between the xylem & phloem strand.

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Anatomy of Flowering Plants)

1.

When does vascular bundle refer to as closed bundles.

[1]

2.

Name the aerating pores in the bark of stems.

[1]

3.

What are sclereids?

[1]

4.

What are tracheary elements? Of what use are these to plants ?

[2]

5.

Distinguish between collenchymas& sclerenchyma.

[2]

6.

Why large number of stomata are seen on lower surface of dicot leaves in

[2]

terrestrial plants. 7.

State the location & function of different types of meristems.

[3]

8.

Describe the internal structure of a dicot root.

[3]

9.

What is wood? What are its different types?

[5]

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Anotomy of Flowering Plants) [ANSWERS] 1.

When cambium is absent.

2.

Lenticels

3.

Sclerieds are thick walled, hard & strongly lignified selerenchyma cells.

4.

Tracheary elements are vessels & tracheids. They are conducting cells of the xylem. The xylem vessels have perforations in their end walls while perforations are absent in tracheids, they form a continuous channel through root, stem & leaves for conduction of water & minerals.

5. COUENCHYMA

SCLERENCHYMA

i) Living mechanical tissue contains i) Mechanical tissue is dead.

protoplasm ii) Thickening in cell wall due to cellulose, ii) Thickening on cell walls due to

6.

hemicelluloses & pectin

deposition of lignin cellulose or both.

iii) High water content in cells

iii) Low water content in cells

iv) Cell lumen is wide.

iv) Cell lumen is narrow.

Stomata are found on the epidermis of green aerial parts of plants but they are abundant on lower surface of leaves of dicot plants as they are helpful in regulation of the process of transpiration.

7.

A meristem is a group of cells that are in a continuous state of division and thus continuously produce new cells on the basis of location & function, the meristem are of following types:-

a)

APICAL MERISTEM:- These are present at the apices of stems, roots & branches the activity of apices of stem adds to length of plant or its parts.

b)

INTERCALARY MERISTEM:- These meristems are intercalated in between the permanent tissues. They may be present cither at the base of internode as in stem of various grasses & wheat, the activities of these meristems also add, to length of plant or its organ.

c)

LATERAL MERISTEMS:- These meristems are present along the side of the stem these include cambium & cork cambium. The activity of lateral meristem adds to thickness of plan.

8.

A T.S. of dicot root shows the following structures:a)

EPIBLEMA:- It is called piliferous layer. Unicellular root hairs extend to outside from the epiblema.

b)

CORTEX:- It is the main part of root having many layers of rounded parenchymatous cells contain starch grains. Intercellular spaces are present in between

them.

It

stores

formed

substances. c)

ENDODERMIS:- It lies inner to cortex & contain barrel shaped cells having no intercellular spaces. Radial walls of its cells may have lignified casparian strip water & minerals pass through passage cells to phloem.

d)

STELE:- It is the central part of dicot root. Inner to endodermis lays pericycle which is single layered thick only. Phloem & xylem are present in different radii to form separate bundles.

9.

Botanically, a secondary xylem is called as wood. It is formed by the metabolism of the plant i.e. secondary growth by cambium & constitutes the bulk of plant body in dicot stem & dicot root. Wood can be classified into following categories. i)

Hardwood:- It is wood produced by angiosperms. It consists mainly of xylem vessels & hence called porous wood.

ii)

Soft wood:- It is wood produced by gymnosperm. It consists mainly of xylem trachieds & hence called non-porous wood.

iii)

Heart wood:_ It is the central core of wood formed during secondary growth. It consists of dead cells. The cells are dark in color due to the presence of extractives like gums, resins, tannins, etc.

iv)

Sap wood:_ It is the peripneral part of wood formed during secondary growth. It consists of living cells. The cells are lighter in colour as extractives are absent.

v)

Early wood:- It is the wood formed during favorable season. Vessels & tracheids formed are larger in dimensions.

vi)

Late wood:- It is the wood formed during unfavorable seasons. The vessels & tracheids formed are smaller in dimensions.

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (Anatomy of Flowering Plants)

1.

Name the tissue represented by jute fibres used for making ropes?

[1]

2.

Why xylem & phloem are called complex tissues?

[1]

3.

Name the types of wood in which vessels are absent.

[1]

4.

What are the functions of tracheids.

[1]

5.

What is stomatal apparatus ? Draw a well labelled diagram of stomata.

[2]

6.

How can you identify a monocot stem and a dicot stem? Give reasons.

[2]

7.

Differentiate between xylem & phloem

[2]

8.

Draw a well labeled diagram of T.S. of monocot stem.

[2]

9.

What is phellogen? What does it produce?

[2]

10.

Describe the elements of xylem with the help of suitable diagram.

[3]

11.

Distinguish between dicot root & monocot root.

[3]

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (Anatomy of Flowering Plants) [ANSWERS] 1.

Sclerenchyma.

2.

Because they are made up of more than one type of cells that work together as a unit.

3.

Soft wood eg. pinus.

4.

Tracheids transports water & give mechanical support to the tree.

5.

The stomata occurs on the surface of leaves. They regulate transpiration in plant & exchange of gases. Each stomata is made of 2 bean shaped cells called guard cell. The guard cells possess chloroplast & regulate opening & closing of stomata. The stomatal aperture, guard cells & surrounding subsidiary cells make the stomatal apparatus.

6.

In monocot stem, the vascular bundles are scattered. No distinction between pitch & cortex. Cambium is not present. Vascular bundles are closed whereas, dicot stem shows epidermis, cortex & stele. Epidermis bears appendages-trichomes. The vascular bundles are open & are arranged in rings. Cortex & pith are distinct cambium present.

7. PHLOEM

XYLEM

i) conduction of food

i) conduction of water & minerals

ii) Phloem fibres are dead, sieve tube, ii) Tracheids, vessels & sclerenchyma are

companion.

Cells

and

phyoem dead. Xylem parenchyma are living.

pareuchyma are living iii) It occurs in small quantity

iii) It occurs in large quantity.

8.

9.

Phellogen is called cork cambium. It is developed to protect the inner tissues in dicot stems it develops from hypodermal cells which are collenchymatous or even from epidermal cells near to cortex. Phellogen or cork cambium produce secondary tissue more on outer side then inner side.

10.

Xylem being a complex tissue is made up of different types of cells as follows:a)

TRACHEIDS:- They are elongated tube like structures. They do not have perforation or openings at their ends. They are dead. They help in conduction of water & minerals.

b)

VESSELS:- They are narrow tube like structures having annular & spiral thickening in protoxylem. They are wider & have spiral, reticulate & pitted thickening in metaxylem. They are dead. They help to conduct water & mineral from roots to upper parts of plant.

c)

XYLEM PARENCHYMA:-They are living cells. They are called as wood parenchyma they help in storage of food & lateral transport of substances.

d)

XYLEM FIBRES:- They are long, slender, pointed, dead sclerenchymatous cells. They are called wood fibres. They have small pits & thickened walls they give strength & support to plants.

11. DICOT ROOT

MONOCOT ROOT

i) diarch/ triarch/ telrarch/ pentarch or i) always polyarch

hexarch ii) Cortex narrow

ii) Cortex very wide.

iii) The casparian strips are more iii) The casparian strips are not very

prominent in endodermal cells.

prominent in endodermal cells.

iv) Pericycle gives rise to primordial of iv) Pericycle give rise to lateral roots only

lateral roots, cork cambium as well as part of vascular cambium v) Vessels & tracheids polygonal in T.S

v) vessels & tracheiols oval in T.S

vi) Secondary growth is present

vi) Secondary growth is absent

vii)

Conjuctive

parenchyma

makes vii) Conjuctive parenchyma do not make

vascular cambium.

vascular cambium.

viii) Pith very small or absent.

viii) Pith is very large

ix)

Passage

cells

are

absent

in ix)

Passage

endodermis

endodermis

x) conjuctive tissue is parenchymatous

x)

cells

conjuctive

are

tissue

present

in

can

be

parenehymatous or sclerenchymatous.

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (STRUCTURE ORGANISATION IN ANIMALS)

1.

Name the kind of tissue which forms the lining of blood vessels?

[1]

2.

Name the chemical which helps in transmitting nerve impulse at the

[1]

synapse? 3.

What is the main function of WBCs?

[1]

4.

What are the organs of excretion in insects?

[1]

5.

Give the characteristic of epithelial tissues?

[2]

6.

How many types of nephridia are found in earthworm based on their

[2]

location? 7.

What do you mean by haemopoiesis?

[2]

8.

Differentiate between blood & lymph?

[2]

9.

What are nissl’s granules? Where are they found?

[2]

10.

i) Give three differences between frogs & toads?

[3]

ii) What do you understand by open type of circulatory system? 11.

What are the cellular components of blood?

[3]

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (STRUCTURE ORGANISATION IN ANIMALS) [ANSWERS] 1.

Squamous epithelium.

2.

Acetylcholine.

3.

They protect the body from infection & diseases.

4.

Malphigian tubules.

5.

Epithelial tissue forms a layer on the free surface i.e. external surface of animal body & internal surface of visceral organs, body cavity & blood vessels. Cells of epithelium are set closely, separated by very thin film of extracellular material. Adjacent cells are held together by cell junctions.

6.

Nephridia are of three types:i) SEPTAL NEPHRIDIA:- Those present on both the sides of intersegmental septa & open into intestine ii) INTEGUMENTAL NEPHRIDIA:- Those found attached to the lining of the body wall & open on the body surface. iii) PHARYNGEAL NEPHRIDIA:- Those found on the 4th, 5th, & 6th segment in the form of three paired tufts are pharyngeal nephridia.

7.

Haelmopoeisis is the formation of new erythrocytes from the haemopoietic tissue. The haemopoietic tissues in the young foetus in liver & spleen whereas in the adults, it is the bone marrow of long bone. The haemopoitic tissue synthesizes millions of RBC’s every minute & its excess lot is stored in the spleen.

8. BLOOD i) It is vascular tissue ii) It is found in blood vessels. iii) It is made of plasma, erythrocytes, leucocytes & platelets. Neutrophils are most abundant. iv) It has haemoglobin v) It helps in transport of materials inside the body.

LYMPH i) It is white (straw coloured) vascular tissue ii) It is found in lymph vessels. iii) It is made of plasma, leucocytes, erythrocytes & platelets are absent. Lymphocytes are most abundant. iv) Haemoglobin is absent v) It functions as middle man between blood & body cells.

9.

Nissl’s granules are the small basophilic bodies found in the cytoplasm of soma & dendrites. They are found in nervous tissues.

10

a) FROGS TOADS i) Scientific name of frog is Rana i) Scientific name of toad is Bufo tigrina melanostictus ii) Frogs are diurnal ii) Toads are nocturnal iii) Parotid glands absent iii) Parotid glands present iv) Skin moist & slipper iv) Skin dry & rough. V) amphibious animals v) Terrestrial for egg laying.

b)

11.

In open circulatory system the blood vessels are poorly developed & open into spaces rather capillaries. All the vesceral organs are bathed in blood (haemolymph) e.g. cockroach.

The blood consists of two parts- plasma (liquid part) & corpuscles (solid parts). The blood corpuscles float in the plasma & are of three major types:a) Red Blood cells or Erythrocytes:- They are circular disc–shaped biconcave cells without nucleus. They contains a pigment haemoglobin which has great affinity towards oxygen. In normal healthy individuals, the number of RBCs / mm3 ranges between 4.5-5 millions in females & 5.5-6.0 millions in males. b)

White Blood cells or Leucocytes :- These are colourless nucleated corpuscles & can pass through the capillary walls into lymph & tissue fluid. Their count is 6000 to 10,000 per mm3 of blood. Their main function is protection against any foreign substance. Leucocytes are further of two types:i)

AGRANULOCYTES:- They have clear cytoplasm without granules & a bilobed nucleus they are further divided into two types lymphocytes & monocytes.

ii)

GRANULOCYTES:- They are of three main types Basophils :- which are stained with basic dye (methylene blue) & have a bilobed nucleus. Neutrophils:- which are stained with neutral dye & have multilobed nucleus. Eosinophils :- which are stained with acidic dye (eosin) & have bilobed nucleus.

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (STRUCTURE ORGANISATION IN ANIMALS)

1.

Which tissue has fat globules?

[1]

2.

Name two anticoagulants of blood of man?

[1]

3.

Name the type of epithelium that lines the inner surface of stomach?

[1]

4.

What causes fatigue of the muscle fibres?

[1]

5.

Discuss the structure of haversian system in the histology of bone?

[2]

6.

Distinguish between myosin & actin filament?

[2]

7.

What are chondriocytes? Where are they found?

[2]

8.

Name the major class of plasma protein & mention their functions.

[2]

9.

What is the function of nephridia ?

[2]

10.

How do erytherocytes transport oxygen & carbon dioxide in the blood?

[3]

11.

Describe the different types of connective tissues & give examples?

[3]

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (STRUCTURE ORGANISATION IN ANIMALS) [ANSWERS] 1.

Adipose tissue.

2.

Heparin & antiprothrombin.

3.

Columnar epithelium.

4.

Due to accumulation of lactic acid or due to prolonged contraction

5.

Bone consists of connective tissue having matrix surrounded by periosteum. In mammalian bone haversian canal which carry blood vessels & nerves of the bone are surrounded by a number of concentric lamellae of intercellular matrix & bone.

6. MYOSIN FILAMENT i) It is found in only A- band

ACTIN FILAMENT i) It is found in I band & also projects in Aband. ii) It is thicker (100A) ii) It is thinner (50A) iii) Cross bridges are present iii) Cross bridges are absent iv) About 1500 myosin filaments are iv) About 3000 actin filament are found found per myofibril per myofibril.

7.

In the matrix of cartilage, in chondrin there lay some large, bluntly angular cartilage cells called chondriocytes. They lie scattered in chondrin. Chondriocytes occurs in clusters of 2 or 3 cells in small spaces called the lacunae.

8.

Three major classes of plasma proteins are:a) Serum b) Serum globulin c) Fibrinogen FUNCTIONS OF PLASMA PROTEINS:i) Providing body immunity ii) Prevention of blood loss. iii) Retention of fluids in the blood. iv) Transport of material v) Maintaining PH of blood. vi) Conducting heat to skin for dissipation.

9.

Annelids have long & coiled excretory tubes called nephridia. They lie in body cavity & collect excretory wastes like NH3, uric acid, urea etc, from body fluids. Nephrostome is called ciliated funnel. It then passes them into looped, coiled ducts. They are removed out from the body through small apertures called nephridiopores.

10.

During development of erythrocytes there is formation of special substance called haemoglobin which is capable of transporting oxygen. It is a complex protein & is composed of two components a) a protein called globin & b) a Fe2+ porphyrin ring called heme. This haemoglobin when exposed to high partial pressure of oxygen combines with oxygem to form oxy-haemoglobin which carries 4 molecules of oxygen loosely bound to four Fe2+ ions. When this oxy haemoglobin reaches the tissues where there is low oxygen pressure oxyhaemoglobin dissociates into oxygen & deoxyhaemoglobin. In this way, erythrocytes transports oxygen from lungs to tissue. Similarly haemoglobin also transports carbon dioxide from tissues to lungs.

11.

On the Basis of matrix, connective tissues are of two main types:I) Connective tissue proper:- It connects & supports many tissues & organs. Its matrix is dense. Eg. i) Areolar tissue :- It consists of three types of cells & types of fibres, all distributed in the matrix. Fiborblasts are irregularly shaped flat cells with long protoplasmic processes, they secrete collagen & elastin proteins for the fibres. ii) Adipose tissue:- It consists of collagen fibres, elastin fibres, fibroblast, maerophages, & adipocytes which stores fat. It prevents heat loss by forming one insulating layer beneath skin. II) Supportive connective tissues :- It consists of following types of connective tissue:i) Cartilage:- It is the endoskeletal material of the vertebrates, it is in the form of solid matrix formed of chondrin with few collagen fibres & chondrioblast cells. ii) Bone:- The matrix consists of bone cells, osteocytes, fibres & a ground substance impregnated by calcium phosphate, calcium carbonate, magnesium phosphate & calcium fluoride. Due to these salts, it becomes very hard & forms skeletal support of body.

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (STRUCTURE ORGANISATION IN ANIMALS)

1.

Name the type of epithelium lines the buccal cavity?

[1]

2.

Why muscle cells are usually called muscle fibres?

[1]

3.

Define glands.

[1]

4.

How many spermathecae are present in earthworms.

[1]

5.

Distinguish between smooth & striated muscles.

[2]

6.

What are the functions of mast cells?

[2]

7.

How can a male frog be distinguished from a female frog ?

[2]

8.

Give reason why earthworms are known as friends of farmers.

[2]

9.

Write short note on adipose tissues.

[2]

10.

Describe briefly the structure of voluntary muscles

[3]

11.

How does blood gets coagulated on coming out from an injured vessel. How

[3]

coagulation is normally prevented uninjured vessels.

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (STRUCTURE ORGANISATION IN ANIMALS) [ANSWERS] 1.

Stratified squamous epithelium

2.

Muscle cells are usually called muscle fibres because muscle cells are thin & elongated into thread like structures.

3.

Glands are secretary structures formed of epithelial tissues.

4.

Four pairs of spermathecae are found in each of 6 to 9 segments of earthworm

5. SMOOTH MUSCLES

STRIATED MUSCLES

i) They are called involuntary muscles.

i) they are called voluntary muscles

ii) They are found in hollow organs

ii) They are mostly attached to bones by

tendons iii) They are uninucleate.

iii) They are multinucleate.

iv) They do not show any striation

iv) They show striated appearance i.e.

alternate light & dark bands.

6.

Mast cells are granular large irregularly shaped cells present in areolar connective tissue a)

They store inflammation producing substance histamine. When histamine is released inflammation is caused due to some reason.

b)

They also release heparin which prevent activation of prothrombin this preventing coagulation of blood.

7.

The male frogs may be distinguished by presence of sound producing vocal sacs. They also has a copulatory pad on the first digit of the forelimbs. Vocal sacs & copulatory pads are absent in female frogs.

8.

Earthworms are known as “friends of farmers” since it makes burrows in the soil. Due to it, soil becomes porous. It facilitates respiration as well as penetration for the developing roots of the plants, the earthworm eats decaying vegetation & in the burrows, it enriches the soil.

9.

It is a specialized form of aerolar tissue where it contains mainly fat cells or adipocytes. The matrix contains fibroblasts, macrophages, collagen fibres & elastin fibres. This tissue lies beneath the skin, around kidney & in mesentery & bone marrow. It synthesizes, stores & mebabolises fat & forms insulating layer beneath the skin.

10.

A voluntary muscle is a bundle of numerous striated muscle fibres. Each fibres is long, unbranched measuring 40 mm in length & 20ym in thickness. Each fibre is enclosed in a membrane called sarcolemma & its cytoplasm is called sarcoplasm. The sarcoplasm contains many myofibrils that are long, thin, unbranched & cross striated. Each myofibril consists of alternating thick A & light I-band. A band is formed of protein myosin & I-band with actin protein. The thick filament bands lie parallel to one another. The thin filament extends between them upto a considerable distance in an orderly manner. At the center of the I-band is a fine, dense, dark, Z-line. Each segment of myofibril from one Z-band to the next functions as a contractile unit & is called sarcomere.

11.

When a blood vessel is injured & blood comes out of it, the thrombocytes clump together, break & release the coagulation promoting substances called thromboplastin. Thromboplastin helps in the formation of enzyme thrombokinase. This enzyme thrombokinase hydrolyses prothrombin in the plasma into thrombin ca2+ ions are needed for both activation & functioning of thrombin. Thrombin catalyses the hydrolysis of soluble fibrinogen in the plasma into insoluble fibrin. The fibrin precipitates as a network of fibres & traps many blood cells to form a red solid mass called blood clot. The clot seals the wound in the blood vessel to stop bleeding. However in uninjured tissues & blood vessels don’t release thromboplastin. That’s why coagulation is prevented in an uninjured vessel.

CBSE TEST PAPER-04 CLASS - XI BIOLOGY (Structural Organization in Animals)

1.

Name the proteins which constitute muscle fibres.

[1]

2.

Which type of epithelium is found in urinary bladder

[1]

3.

From which germ layers do the following organs originate

[1]

a) kidney b) urinary bladder. 4.

What are neuroglia cells?

[2]

5.

How does saltatory conduction takes place along a nerve fibre

[2]

6.

What is mucosa?

[2]

7.

Write short note on gaseous exchange in cockroach.

[2]

8.

Draw a well labelled diagram of a nerve cell.

[2]

9.

Distinguish between tendon & ligament

[2]

10.

Name the various fibres of connective tissue & compare them

[3]

11.

Give an account of alimentary canal of frog.

[3]

CBSE TEST PAPER-04 CLASS - XI BIOLOGY (Structural Organization in Animals)

[ANSWERS] 1.

Actin & myosin

2.

Transitional epithelium.

3.

a) Mesoderm b) Endoderm.

4.

Neuroglia cells are cells that hold the neurons together.

5.

Along a myelinated nerve fibre, the conduction of impulse is called salutatory conduction. This is so because the ionic changes & consequent depolarization taking place only.

6.

Mucosa is the mucous secreting epithelial tissue alongwith the supporting connective tissue beneath it. It lines some hollow organs or cavities of the body eg. alimentary canal, nose, trachea & lungs etc.

7.

Cockroach is an insect & has a system of trachea. Hence tracheal respiration occurs in these animals. It is a complicated system of air tubes. They divide & form tracheoles. Tracheoles are connected to the spiracles located in the segments of thorax & abdomen. The body cells or fluid come in direct communication of air.

8.

9. TENDON

LIGAMENT

i) It is formed of white fibrous connective tissue

i) It is formed of yellow fibrous tissues

ii) Fibroblasts are arranged in rows between the ii) Fibroblasts are scattered in matrix

bundles of white fibres iii) It is tough & non flexible

iii) It is elastic & flexible.

iv) It joins muscles to bones

iv) It joins bones together.

10. Nature i) colour ii) Protein iii) Occurrence iv) Nature

Collagen fibres White Formed of protein, tropho- collagen Found in bundles unbranched

v) fibres vi) elasticity vii) Location

Thick, long wavy Tough, non elastic Abundant in tendon

11.

Elastin fibres Yellow Elastin protein

Reticular fibres White Reticulin protein

Singly Branched & anastomosing Thin, long straight Elastic Abundant in ligament

Singly Branched but form a network Short Delicate Abundant in ambryo in lymphoid as well as blood forming tissues.

ALIMENTARY CANAL OF FROG:It is a short tube starting from mouth to cloaca. Mouth opens into buccopharyngeal cavity. It has many maxillary teeth as

the

margin

of upper jaw.

Vomerine teeth lie at the floor of this cavity. The tongue is bilobed & muscular. It is used to capture the prey.

Gullet

opens

into

the

oesophagus which is distended into stomach. Stomach follows small & large intestine. The rectum opens into the cloaca. Liver & pancreas are digestive canals.

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (Cell the Unit of Life)

1.

Define totipotency?

[1]

2.

Name two cell organelles which contain their own DNA?

[1]

3.

Which cell organelle functions as “seggregation apparatus”?

[1]

4.

Give two examples of gram positive bacteria?

[2]

5.

What is the significance of plasma membrane?

[2]

6.

Differentiate between gram positive and gram negative bacteria?

[2]

7.

Why lysosomes are called “suicidal bags”?

[2]

8.

Explain the functions of centrosome?

[2]

9.

What is meant by active transport across a cell membrane?

[2]

10.

Describe the ultrastructure of a cillium or flagellum?

[3]

11.

Distinguish between prokaryotic & eukaryotic cell?

[3]

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (Cell the Unit of Life)

[ANSWERS] 1.

Each vegetative plant cell has capacity to develop into a full plant. This characteristic of plant is called totipotency.

2.

Mitochondria & chloroplast.

3.

Endoplasmic Reticulum (ER)

4.

Mycobacterium & clostridium tetani.

5.

Significance of plasma Membrane:i)

It forms the outer boundary of cell thus giving cell a definite shape

ii)

It protects inner contents of the cell.

iii)

It forms a molecular boundary between cell & its environment.

6. GRAM-POSITIVE BACTERIA

GRAM-NEGATIVE BACTERIA

i) Their cell wall is only single layered & i) Their cell wall consists of two layers &

7.

100-200 A0 thick.

is 70-120 A0 in thickness.

ii) They are stained by gram stain

ii) They are not stained by gram stain

iii) They do not have pilli.

iii) They have pilli

iv) Mesosomes present

iv.) Mesosomes absent

Lysosomes are sac-like structures bounded by a single membrane which contains several digestive enzymes. These enzymes when released from lysosomes bring about breaks down of various cytoplasmic structures. It helps in digestion of food particles, other foreign bodies, old worn out organelles of cell often resulting in death of cell hence are referred as suicidal bags of cell.

8.

Function of Centrosomes :a)

Centrioles form basal bodies.

b)

At the time of cell division, they organize spindle and form asters.

c)

They give rise to cilia and flagella.

d)

Out of the two centrioles, the distal centrioles of sperms forms the axial filament or axoneme of sperm tail.

9.

When molecules moves from a region of lower concentration to a region of higher concentration i.e. against concentration gradient, the process is known as active transport. The energy is required for the movement of molecules or ions in opposite direction. The enzyme responsible for the pumping of compounds into or out of cell believed to be a component of the membrane eg. Na+- K+ pump.

10.

Cilia & flagella have fundamentally the same structures. Each cilium or flagellum consists of eleven microtubules. These microtubules are arranged in two radii. Of these, nine are doublets. These are situated at the periphery & the remaining

two

are

single

microtubules situated in the centre. The microtubules are enclosed in a cytoplasmic matrix to form an axial filament. The outer tubules are 360 A0 in diameter & are composed of two sub- units. The smaller of these have two arms in A- tubule & the smaller is B- tubule. These are found around the cylinder. The central microtubules are enclosed in a common sheath. From the centre arise nine secondary filaments. These are connected with tubules of the outer doublets.

11. PROKARYOTIC CELL

EUKARYOTIC CELL

i) It lacks well organized nucleus. The i) Nucleus is well developed.

genetic material is present in the form of nucleoid. ii) DNA is in circular form & is not packed ii) Linear DNA packed into chromosomes

into chromosomes. iii) Nuclear membrane is lacking

iii) Nuclear membrane is present.

iv) Mitochondria absent

iv) Mitochondria present.

v) Chloroplast absent

v) Chloroplast is present in plant cell only.

vi) Membrane bound organelles are vi)

Membrane

bound

organelle

are

absent

present.

vii) The ribosomes are of 70stype

vii).The ribosomes are of 80s type

viii) Cell wall consist of mucoptides

viii) Cell wall is absent in animal cells in

plant cell, cell wall is made up of cellulose, hemicelluloses, lignin etc. ix) Flagella are simple

ix) Flagella are specialized.

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Cell the Unit of Life)

1.

Which structure is called little nucleus?

[1]

2.

What is the function of contractile vacuole?

[1]

3.

Name the enzymes present in peroxysomes?

[1]

4.

Who gave the statement “Omnis cellular cellula”?

[1]

5.

“Both lysosomes & vacuoles are endomembrane structures yet they differ

[2]

in terms of their functions” comment. 6.

Who proposed cell theory? Give its postulates?

[2]

7.

Which cell organelle is known as powerhouse of cell & why?

[2]

8.

What are the main functions of cell wall?

[2]

9.

State differences between SER & RER?

[2]

10.

Explain the fluid mosaic model of plasma membrane

[3]

11.

Describe the structure of a typical eukaryotic chloroplast.

[3]

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Cell the Unit of Life)

[ANSWERS] 1.

Nucleolus.

2.

Water balance or osmoregulation.

3.

Catalase & B- hydroxyoxidase.

4.

Rudolf Virchow.

5.

Lysosomes & the vacuoles are endomembranous structures yet these differ in terms of their functions:i)

Lysosomes contains hydrolytic enzymes eg. lipase, protease which are able to digest lipids, proteins, nucleic acid & carbohydrate.

ii)

Vacoules are membrane bound spaces which facilitates transport of many ions & other materials against the concentration gradient.

6.

M. J. Scheilden & Theodore Schwann gave the famous cell theory which states as follows:i)

All living things are made of cells & cell products.

ii)

The cell is the structural & functional unit of all living organisms.

iii)

All metabolic reactions in the living things take place with in the cell

The cell theory was later modified by Rudolf Virchow who stated that “all new cells arise from the pre- existing cells”.

7.

The double membrane mitochondria are actively associated with aerobic respiration & the release of energy for cellular activity. The biological oxidation of the fats & carbohydrates release much amount of energy which is utilized by mitochondria for ATP synthesis. When required energy is released form ATP molecules for various cell processes in cells so they are termed as “Power house of the cell”

8.

FUNCTIONS OF CELL WALL:i)

It provides a definite shape to the cell.

ii)

It protects inner contents of cells

iii)

It protects delicate plasma membrane present below it.

iv)

It allows transport of various substances to & from the cell.

v)

It prevents cell contents from drying up.

9. SER

RER

i) SER do not have ribosomes & is i) RER have ribosomes on its outer

composed of vesicles & tubules

surface & is composed of cisternal

ii) It synthesizes steroids & lipids eg. fat ii) Its main function is protein synthesis

cell lipid secretory cells of liver

due to the presence of ribosomes.

iii) Gives rise to sphaerosomes

iii) Gives rise to Golgi bodes, vacuoles as

well as lysosomes. iv) Free of ribosomes.

10.

iv) Bears ribosomes.

The fluid mosaic model was proposed by G.Nicholson & s. singer. According to this each phospholipids

layer

is

bimolecular & their hydrophilic ends are pointed towards top & bottom respectively.

In this, proteins are of two categories- peripheral (extrinsic) & integral (intrinsic). The integral proteins are tightly held in place by strong hydrophilic or hydrophobic interactions or both and are difficult to remove from the membranes. Two peripheral proteins are superficially arranged on either side membrane selectively permeable thus this model explains cell membrane is quasifluid & is made up of “protein icebergs in the sea of lipids”.

11.

Chloroplasts are bounded by two membranes, about 3000 A0 in total thicknesses. Each membrane is 40-60 A0 thick. The inner membrane is very intricately elaborated to form a system of lamellae. Internally the chloroplasts is divisible into two parts

(a)

stroma- colourless, ground substance

(b)

Membrane system- made of closed flattened sacs called thylakoids. These thylakoids are closely packed & appears as piles of coins. These structures are

called

Grana.

The

arrangement can be in the form of simple

parallel

sacs

running

lengthwise, or may be in a complex interconnecting network of the sacs. The chloroplasts invariably have some starch granules which often accumulate near a special region known as pyrenoid in algae

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (Cell the Unit of Life)

1.

Which organelle is called the engine of the cell?

[1]

2.

What is mycoplasma ?

[1]

3.

Why is karyotype done at metaphase?

[1]

4.

Expand PPLO

[1]

5.

What are nuclear pores? State their functions?

[2]

6.

Give differences between cell wall & cell membrane?

[2]

7.

Which organelle is responsible for increasing the surface area of absorption

[2]

in a cell? How? 8.

What is mesosome in a prokaryotic cell? Mention the function that it

[2]

performs? 9.

“plasma membrane is described as” protein iceberg in sea of lipids”. why ?

[2]

10.

Mention three similarities & three differences between mitochondria &

[3]

chloroplasts? 11.

“multicellular

organisms

counterpart” why?

have

better

survival

than

their

cellular

[3]

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (Cell the Unit of Life) [ANSWERS] 1.

Ribosomes where protein synthesis occurs

2.

Mycoplasma is aerobic prokaryote. Cell wall is absent in them & they have a nucleoid.

3.

Because metaphase chromosomes with two chromatids strands of each double chromosome held together at the centromere are clearly seen.

4.

Pleuropneumonia like organisms.

5.

Nuclear envelope contains two parallel membranes & the thickness is 10-50 nm. Outer membrane has small pores called the nuclear pores formed by fusion of two membranes. These pores are the passages through which movement of RNA & protein molecules occurs in both directions between nucleus & cytoplasm.

6.

7.

CELL WALL

CELL MEMBRANE

i) present in plant cell exclusively

i) present predominantly in animal cells

ii) Made up of cellulose

ii) Made up of proteins fats & water

iii) Thick & tough in nature

iii) Extremely thin & elastic in nature

iv) Thickening of various kinds present

iv) No thickenings

v) it is not selectively permeable

v) selectively permeable membrane

The endoplasmic reticulum is responsible for increasing the surface area for absorption. It remains in the form of convulated tubule in the cytoplasm in the form of network. This provides more area for chemical reactions and increases the surface area of absorption.

8.

Mesosome in a prokaryotic cell is formed by extensions of plasma membrane into the cell it may be in form of vesicle, tubule or lamella. They help in cell wall formation. They help in replication of DNA & distribution of it to daughter cells. They help in secretion respiration, & increase plasma membrane surface area.

9.

The plasma membrane as described by singer & Nicolson is of fluid mosaic model type. The lipid & proteins are arranged in a mosaic fashion. The matrix is highly viscous fluid of two layers of phospholipids molecules having two types of globular proteins i) peripheral or extrinsic proteins & ii) integral or intrinsic proteins. The proteins present superficially or tightly with the membrane are enzymatic can move across the matrix & help in the active & passive transport of ions through the membrane.

10.

SIMILARITIES BETWEEN MITOCHONDRIA & CHLOROPLAST i)

Mitochondria & chloroplasts are semi-automous organelle & they possess their own DNA, RNA as well as ribosomes.

ii)

They both develop & originate in the same way, formed by division of preexisting organelle

iii)

Both of them contain circular DNA.

DIFFERENCES BETWEEN MITOCHONDRIA & CHLOROPLAST i)

Mitochondria occurs in all eukaryotic cells while chloroplast are present only in plant cells.

ii)

Pigments are absent in mitochondria but always present in chloroplast.

iii)

The inner membrane of mitochondria are folded into cristae where as cristae are absent in chloroplast.

11.

In unicellular organisms, there is no division of labour. The single cell of the organism is capable of performing all the vital activities of life respiration, movement, digestion & reproduction etc. Respiration, nutrition & excretion generally occur through general body surface no special organs for these are present in them because they are too small to need them. In multicellular organisms all the body cells do not perform all the vital activities of life rather these cells play more specialized role in life activities eg. some cells of the body perform the function of movement some perform the function of digestion or respiration or removal of wastes from the body some cells perform the function of

transport. These cells would perform no other function except for which they are specialized. The group of similar cells performing similar function is termed as tissues.

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (Biomolecules)

1.

Which is the important energy carrier in the cell?

[1]

2.

Name the monomer subunits which form Nucleic acids?

[1]

3.

What are macromolecules? Give example.

[1]

4.

Differentiate between nucleotide & nucleoside?

[2]

5.

How are glycosidic bonds formed?

[2]

6.

What do you mean by steady state?

[2]

7.

Enumerate the functions of lipids?

[3]

8.

Describe the lock & key hypothesis of enzyme action?

[3]

9.

Explain briefly four levels of protein structure?

[5]

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (Biomolecules)

[ANSWERS] 1.

Adenosine tri phosphate (ATP)

2.

Nucleotide.

3.

Macromolecules are large complex molecules formed by polymerization of micromolecules & have high molecular weight.

4. NUCLEOTIDE

NUCLEOSIDE

i) Nucleotide is made up of base, sugar & i) Nitrogenous base & sugar form a

phosphoric acid. ii)

Nucleotide

nucleoside of

RNA

is

called ii)

Nucleoside

of

RNA

is

called

ribonucleotide & nucleotide of DNA is ribonucleoside & nucleoside of DNA is called deoxyribonucleotid

called deoxyribonucleoside

iii) E.g adenylic acid, guanylic acid, iii) Eg. Adenosine, fuanosine, cytidine,

thymidylic acid, uridylic acid

5.

thymidine, uridine

The glycosidic or ketone group of a monosaccharide can react & bind with an alcoholic group of another organic compound to join the two compounds together. This bond is known as glycosidic bond.

6.

An open system always remains in steady state i.e. the rate of in put of energy & matter is always equal to the output of energy & matter.

7.

i)

Most of the plants & animals fats constitute storage compound. Fat is stored mainly in adipose cells in the animals.

ii)

In oil seed plants, oil provides nourishment to developing embryo during seed germination. Oil extracted from these seeds is used in cooking.

iii)

Fats provide energy to the body.

iv)

Fats serve as insulators & protect body from cold. It gets deposited underneath skin.

v)

Phospholipid form an structural component of all bio- membranes in cell.

vi)

Cholesterol acts as precursor for synthesis of various hormones, vitamins & bile salts.

vii)

The lipid form the white matter, grey matter of brain & myelin sheath of neurons.

8.

According to Fischer’s lock & key hypothesis of enzyme action:- if the right key fits in the right lock, the lock can be opened otherwise not. To explain the above in context with enzyme action it is bedewed that molecules have specific geometric shapes. Proteins are able to act as enzyme because their shape provides space configuration into which other molecules can fit. The molecules which are acted upon by the enzymes are called substrates of the enzymes. Under the above assumption only those substrate molecule with proper geometric shape can fit into the active site of the enzymes. However, under special circumstances some other molecules which are similar to the substrate can also combine with active site of enzyme. In such cases molecules may compete with substrate & the reaction may either slow down or stop. This is called competitive inhibition.

9.

FOUR LEVELS OF PROTEIN STRUCTURE:a)

PRIMARY STRUCTURE:- The protein exists as a long chain of amino acids arranged in a particular sequence such a polypeptide is non- functional

b)

SECONDARY STRUCTURE:-first amino acid is N-terminal amino acid & last is known as c-terminal amino acid. There is interaction between every fourth amino acid by formation of hydrogen bond the polypeptide is folded in a helical shape eg. keratin. When two or more polypeptide chains are held together by intermolecular hydrogen bonds the structure is known as pleated sheet.

c)

TERTIARY STRUCTURE:- The polypeptide becomes stabilized by folding & coating by the formation of ionic bonds or hydrophobic bonds or disulfide

bridges. It is called tertiary structure. It gives a three dimensional view of proteins. Biological activity of protein depends on its tertiary structure. d)

QUATERNARY STRUCTURE:- Such proteins are farmed of more than one polypeptide or subunits each one having primary secondary & tertiary structure. This is called quaternary structure. Each polypeptide chain functions as subunit of the proteins.

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Biomolecules)

1.

Identify the polymer which makes exoskeleton of insects.

[1]

2.

Name the following:- i) sugar present is DNA ii) Base not found in DNA

[1]

3.

Why proteins are called biological polymer?

[1]

4.

Which molecule has the capacity to duplicate?

[1]

5.

What is metabolism? Mention the role of enzymes is metabolism ?

[2]

6.

Why are enzymes called as biocatalyst?

[2]

7.

Give the functions of carbohydrates?

[2]

8.

What do you meant by activation energy?

[2]

9.

List the different types of lipids

[2]

10.

Describe the structure & function of ATP?

[3]

11.

Differentiate between cofactors, coenzymes & prosthetic group.

[3]

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Biomolecules) [ANSWERS] 1.

Chitin a polymer of glucosamine that forms exoskeleton of insects,

2.

i) Deoxyribose sugar ii) Uracil

3.

As proteins are able to perform multiple functions eg. Protection mechanical support, transportation, movement etc, they are called as biological polymers.

4.

Deoxyribonucleic acid (DNA)

5.

Metabolism is defined as the sum total of the living processes in the body. Enzymes direct metabolic pathways. Enzymes act as catalysts. Enzymes are highly specialized organic catalysts produced by living cell. Biochemical pathways refer to the reactions occurring in the cells in sequences. Enzymes guide the biochemical pathways along desired directions. They have active site. The substrate binds at active site of enzyme & form enzyme substrate complex.

6.

The substances which changes the rate of chemical reaction without altering the equilibrium point of reaction is called catalyst. The catalysts of the organism are called enzymes & they are synthesized in the living cell. Hence called as Biocatalysts.

7.

i)

Carbohydrates play role in all metabolic reactions of body & formed as intermediate compounds in pathways of the processes.

8.

ii)

Ribose & deoxyribose sugar are found in nucleic acids.

iii)

Glucose is oxidised in respiration to yield energy.

iv)

Glucose is used in synthesis of fats as well as proteins.

Activation energy is the energy required to initiate a chemical or biochemical reaction. Activation energy overcomes the energy barriers of the reactants which occurs amongst the reactants due to i) presence of electrons over their surface ii) Absence of precise & forceful collisions essential for bringing the reactive sites of the chemical together.

9.

Lipids are of three types:i)

Simple lipids:- they are of alcohols or triglycerides containing fatty acid & glycerol.

ii)

Compound lipids:- They are simple lipids with a biologically active compound in them eg. glycolipids ( carbohydrate lipid) lipoprotein ( protein + lipids)

iii) Derived lipids:- They are hydrolysed products of simple lipids such as fatty acids & alcohol. 10.

ATP is primary & universal carrier of chemical energy in the cell living cell capture store & transport energy in a chemical form, largely ATP & it is the ATP which is the carrier & intermediate source of chemical energy to those reactions in the cell which do not occur simultaneously. These reactions can take place only if chemical energy is released. The ATP molecule consists of a nitrogenous base adenine a pentose sugar of ribose type & three inorganic phosphate molecules two phosphate bonds are high energy bonds & one is relatively poor in energy. Energy released in living cell is thus stored in the chemical bonds of the ATP molecule which then serve as major energy yielding & energy requiring substance in the cell. ATP is broken down into ADP whenever energy is needed. ATP –> ADP + ip + energy.

11. COFACTORS

COENZYMES

PROSTHETIC GROUP

i) It is a non protein i) it is a non protein group i) it is a non protein part or

substance or group that which is loosely attached to group which gets attached gets

attached

to

an the open enzyme in a to open enzyme.

enzyme. ii)

It

functional enzyme is

functioning

essential

it

may

for ii) NAD is coenzyme for ii) Some prosthetic group

be dehydrogenase

organic or inorganic or metallic factor

have

porphyrin

cytochrome.

of

the

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (Biomolecules)

1.

Name the abundant proteins in biosphere?

[1]

2.

Lipids are not biomacromolecules why?

[1]

3.

Which lipid can cause heart ailment?

[1]

4.

What are micro- nutrients?

[1]

5.

Enlist three properties of enzymes?

[2]

6.

Enumerate differences between DNA & RNA?

[2]

7.

Why are monosaccharide’s sugars are are known as reducing sugars?

[2]

8.

How does temperature affects enzyme catalysed reaction?

[2]

9.

What is enzymatic competitive inhibition? Give one example?

[2]

10.

How does enzymes brings about high rate of chemical conversions?

[3]

11.

What are nucleic acids? Describe the structure of DNA.

[3]

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (Biomolecules) [ANSWERS] 1.

RUBISCO

2.

Lipids are not biomacromolecule because their molecular weight does not exceed 800.

3.

Cholesterol.

4

Minerals required by plants in trace quantity eg. Mn, Co, Zn, B, etc. are called micronutrients.

5.

i)

An enzyme is specific for a substrate & catalyses only a particular reaction. because of the specific shape of active site & substrate.

ii)

Every enzyme requires an optimum temperature for its functioning.

iii)

The enzymes are sensitive to PH & each enzyme shows its maximum activity at a specific PH called optimum PH.

6. DNA

RNA

i) it consists of a double helical of two i) It consists of only one helical of single

polynucleotide chains

polynucleotide chain.

ii) Deoxyribose sugar is present in the ii) Ribose sugar is present in nucleotide

nucleotides. iii) Pyrimidine bases are thymidine & iii) Pyrmidine bases are uracil & cytosine

cytosine. iv)

DNA

contains

all

the

genetic iv) RNA helps in protein synthesis.

information

7.

Monosaccharides sugars are called reducing sugars because they have a free aldehyde or ketone group & can reduce Cu2+ to Cu+. Disaccharides like sucrose does not reduce Cu2+ to Cu+ so, it not a reducing sugar.

8.

The temperature affects the velocity of enzyme action. When the temperature is high, there is a sudden decrease in enzyme action due to denaturation. Mostly enzymatic reactions occur below 450c

9.

Some chemicals prevent the enzyme to function, are known as inhibitors. Enzymatic competitive inhibition is done by the substrate which very closely resembles the substrate in its molecular structure. Enzyme + Inhibitor

Enzyme inhibitor complex.

Eg. malonate inhibits the action of succinate dehydrogenase because it shows close resemblance with succinate substrate. 10.

A chemical that is converted into a product is known as the substrate. Therefore the enzymes with tertiary structures including an active site convert a substrate into a product. The substrate ‘S’ must bind enzymes at its active site within a given cleft. So an obligatory formation of an ES substrate complex occurs. At a state when the substrate is bound to an enzyme active site, a new structure of substrate is formed. In the graph, if ‘P’ is at lower level than ‘S’ reaction is exothermic i-e energy is supplied to make product ‘P’. The ‘S’ has to go through much

higher

energy

state

known

as

“transition state. The enzymes brings down energy barrier making transition of ‘S’ to ‘P’ more easy. The difference in average energy content between that of ‘S’ & this transition state is termed as activation energy. 11.

Nucleic acids are found in acid soluble fraction of living tissue. They are linear polymers of deoxyribonucleotides or ribonucleotides A nucleotide has 3 distinct components. DNA is a double stranded structure & each strand is a polymer of deoxyribonucleotide. The backbone of the nucleic acid is uniformly consisting of alternating pentose sugar & phosphate group

i)

The steps composed of nitrogenous bases adenine guanine cytosine & thymine & hydrogen bonds hold two strands together.

ii)

Two strands are complementary to each other.

iii)

They run in an antiparallel manner.

iv)

It is genetic material in all organisms.

v)

It has the property to replicate

vi)

At one end of strand, 5-c of pentose sugar is free on other end; third carbon of pentose is free.

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (CEEL CYCLE & CELL DIVISION)

1.

Name the stage of cell division in which paired homologous chromosomes

[1]

get shortened & thickened?

2.

Which structure of animal cell forms the asters of spindle?

[1]

3.

Name the cells in which meiosis occurs?

[1]

4.

What is the importance of chromosomes replication during interphase?

[2]

5.

Distinguish between metaphase of mitosis & metaphase I of meiosis?

[2]

6.

How does duration affect the cell cycle in organism?

[2]

7.

What is the significance of meiosis?

[3]

8.

Differentiate between animal cell mitosis & plant cell mitosis?

[3]

9.

Explain the various phases of meiosis II division?

[5]

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (CEEL CYCLE & CELL DIVISION) [ANSWERS] 1.

Pachytene.

2.

Centrosome

3.

Reproductive cells or germ cells.

4.

Interphase is a stage between the successive cell divisions. It is considered as the resting stage of nucleus as it does not show any morphological changes. But physiologically it is a very active stage in the life of a cell as the cell prepares itself for division & many biochemical changes occur during this stage.

5. METAPHASE OF MITOSIS

METAPHASE OF MEIOSIS: I

i) Each chromosome consist of two i)

Homologous

chromosomes

form

chromatids which are held together by bivalent each bivalent consists of four centromere

chromatids & two centromeres

ii) The chromosomes line up in one plane ii) Bivalents become arranged in the

to make up the equatorial plate.

plane of the equator forming equatorial plate.

6.

The duration depends on type of cell & external factors like temperature, food & oxygen. Time period for G, S, G, & M-phase is species under specific environmental conditions like 20 min. for bacterial cell, 10 hrs for intestinal epithetial cell 20 hrs for onion root tip cell. It shows that time required for every step have been pre-set within cell of organisms.

7.

Significance of Meiosis:a)

It reduces number of chromosomes to half in daughter cells.

b)

It is very essential phenomenon in life cycle of sexually reproducing animals as it restores the fixed number of chromosomes.

c)

Gametes are formed as a result of meiosis. Each gamete possesses half the number of chromosomes present in somatic cells.

d)

It avoids the multiplication of chromosomes & thus maintains the stability & constant number of chromosomes of the species.

e)

During the crossing over, exchange of nuclear material, genetic variations within the species takes place with the result that new combinations of genetic material are formed.

8. ANIMAL CELL MITOSIS

PLANT CELL MITOSIS

i) occurs in bone marrow & many i) occurs in meristems

epithelia ii) Animal cell becomes spherical before ii) Cell shape does not change before

9.

cell division

division

iii) Several hormones induce cell division

iii) Induced by plant hormone cytokine

iv) Centro some present

iv) Centrosome absent

v) mitotic apparatus contains asters

v) mitotic apparatus has no asters

vi) Mid body is formed

vi) Mid body is not formed.

vii) occurs through cleavage

Vii) Occurs by cell-plate formation

viii) Microfilaments are involved in it

viii) Microfilaments are not formed

ix) Cleavage proceeds centripetally in it

ix) Cell grows centrifugally in it.

STAGES OF MELOSIS – II:i)

PROPHASE II:- Meiosis II is initiated immediately after cytokinesis usually before chromosomes have fully elongated. The nuclear membrane disappears by the end of prophase-II. The chromosomes again become compact.

ii)

METAPHASE-II:- At this stage the chromosomes align at the equator & the microtubules form opposite poles of the spindle get attached to the kinetochores of sister chromatids.

iii)

ANAPHASE-II:- It begins with the simultaneous splitting of the centromere of each chromosome allowing then to move towards opposite poles of the cell.

iv)

TELOPHASE-II:- Meiosis ends with telophase-II, in which two groups of chromosomes once again get enclosed by nuclear envelope, cytokinesis follows resulting in the formation of tetrad of cell i.e. four haploid daughter cells.

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Cell Cycle and Cell Division)

1.

At which stage of meiosis crossing over of genetic material takes place?

[1]

2.

What is Go phase?

[1]

3.

Name the cell division concerned with cancer?

[1]

4.

Why is meiosis called reductional division & mitosis called equational

[2]

division? 5.

Write three processes which take place in interphase?

[2]

6.

Enumerate the significance of mitosis?

[2]

7.

Write six differences between mitosis & meiosis?

[3]

8.

What are homologous chromosomes? What happens to homologous

[3]

chromosomes during meiosis ?

9.

What is mitosis? Give a brief account of mitosis in animal cell?

[5]

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Cell Cycle and Cell Division) [ANSWERS] 1.

pachytene.

2.

A stage when cell cycle is arrested during interphase is called Go phase

3.

Mitosis

4.

In meiosis, the number of chromosomes is reduced to half so, it is called, reductional division. The gametes are formed in sexually reproducing organisms in germ cell. While in mitosis, number of chromosomes remains constant after division hence, it is called equational division.

5.

Three processes in interphase:-

i)

The replication of DNA with the synthesis of histones & nuclear proteins.

ii)

Division of centriole to new centriole which lie at right to each other.

iii)

The synthesis of energy- rich compound to provide energy for mitosis.

6. i)

The number of chromosomes in mitosis cell division remains constant in daughter cells

ii)

Asexual reproduction occurs with the help of mitosis.

iii)

Size of cell is controlled by mitosis.

iv)

Growth & development of the zygote is maintained through mitosis

7. MITOSIS

MELOSIS

i) Chromosome doubling is followed by i) There is doubling of chromosomes once

separation of daughter chromosomes the but it is followed by two nuclear divisions. cell divides only once.

The cell divides twice.

ii) Mitosis occurs in all the somatic cells

ii) It occurs in reproductive or germ cells

iii) It is completed in one sequence of iii) The whole process completes into two

stages

successive divisions

iv) Synopsis is absent

iv) Synopsis is present

v) No crossing over & chaisnata formation

v) crossing over & chaismata formation

occurs vi) A cell produces two diploid cells.

8.

vi) A cell produces four haploid cells.

Homologous chromosomes are pairs of similar chromosomes having corresponding genes governing the same set of traits. During the heterotypic division of meiosis in leptotene, chromosomes are thread shaped & coiled. During zygotene, the homologous chromosomes start pairing. In pachytene, the chromosomes show thickening & shortening. Diplotene, is marked by cessation of attraction force between two homologous chromoses uncoiling of homologous chromosomes tends to separate them from each other but remain attached at chiasmata. During diakinesis, the separation of homologous chromosome is complete. Exchange of parts between chromatids of homologous chromosomes may take place. During Anaphase I the centromere of homologous compounds of bivalents repel each other After separation of centromere, the homologous chromosomes begin to move apart. In telophase-I the chromosomes reach poles & become shortened.

9.

Mitosis is an equational cell division in which number of chromosomes in parent & progeny cell remains same. STAGES OF MITOSIS:1)

PROPHASE:-

a)

chromosome material condenses to form compact mitotic chromosomes. Chromosomes are seen to be composed of two chromatids attached together at centromere.

b)

Initiation of assembly of mitotic spindle, the microtubules the protein components of the cell cytoplasm help in the process.

2)

METAPHASE:- a) b)

Spindle fibres attach to kinetochores of chromosomes Chromosomes are moved to spindle equator & get aligned along metaphase plate through spindle fibres to both poles.

3)

ANAPHASE:-

a)

centromere splits and chromatids separate

b)

Chromatids move to opposite poles.

4)

TELOPHASE:- a)

Chromosomes cluster at opposite spindle poles & their identity is lost as discrete elements

b)

Nuclear envelope assembles around the chromosome clusters.

c)

Nucleolus, Golgi complex & ER reform.

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (TRANSPORT IN PLANTS)

1.

Name two anti-transpirants?

[1]

2.

Define translocation?

[1]

3.

When does wilting occur?

[1]

4.

Why cell is called an osmotic system?

[2]

5.

Distinguish between active transport & passive transport

[2]

6.

Give the main purposes of transpiration?

[2]

7.

Explain pressure flow hypothesis for translocation of sugars in plants?

[3]

8.

Explain why pure water has maximum water potential?

[3]

9.

Describe the cohesive force theory of ascent of sap in plants?

[5]

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (TRANSPORT IN PLANTS) [ANSWERS] 1.

Phenyl mercuric acetate, Absissic acid.

2.

Transport of food through phloem vascular system is known as translocation.

3.

Wilting occurs when the loss of water by evaporation exceeds the rate of uptake by roots.

4.

A cell is called as an osmotic system because:i) It has all the components of an efficient osmotic system i.e. a semi permeable Membrane & different concentration of sap on the two sides of membranes. ii) Flow of water occurs from higher water potential to lower water potential.

5. Active Transport i) This is a rapid process. ii) Energy is needed iii) It is a vital process. iv) Movement is one direction only v) Requires carrier proteins vi) Movement of metabolite against conc. gradient

Passive Transport i) this is a slow process ii) No need of energy iii) It is a physical process. iv) Movement is in both directions v) does not require carrier proteins vi) Movement of metabolite along the conc. gradient.

6.

i) Supplies water for the process of photosynthesis. ii) Transports minerals from soil to all parts of a plant. iii) Cools leaf surface by some 10-15 c by evaporative cooling iv) Maintains shape of plant & structure by keeping cells turgid.

7.

The most accepted mechanism for translocation of sugars source to sink is known as pressure flow hypothesis. The glucose is prepared at the source, it is converted into sucrose. The sugars is then moved in the form of sucrose into the companion cell, then into the living sieve tube cells by active transport loading at the source creates a hypertonic condition in vascular tissue- the phloem.

8.

Pure water has maximum water potential because:i) Water molecules have kinetic energy, in liquid as well as the gaseous state they are in constant motion. ii) The greater the concentration of water in a system, the greater its kinetic energy of its water potential. iii) The pure water will have the greatest water potential. iv) It two systems having water are in contact eg. soil & air or cell & solution, random movement of water molecules will occur from the system with higher energy to one with lower energy. At equilibrium, the water will move from the system containing water at higher water potential to one having low water potential. v) Water potential is represented by Psi or Ψ . It is expressed in pressure units such as Pascal. Water potential of pure water at defined temperature & pressure is taken to be zero. vi) If on pure water a solute is dissolved, its concentration decreases thereby reducing its water potentials so, all solutions have lower water potential than pure water.

9.

Cohesive force theory or transpirational pull theory was proposed by Dixon & jolly. Main aspects of this theory are:i) Strong cohesive force or tensile strength of water:- The water molecules attract each other by mutual force which is called the “cohesive force”. The attraction between the walls of xylem elements & the water molecules is called adhesion the cohesive force helps to maintain a long column of water under tension. ii) Continuity of water column in plant:- air breaks do not hinders the total cohesive system & the tree are able to maintain a rapid flow of sap. The water or the sap can be pulled by the transpiration force. iii) Transpiration pull or tension of unbroken column of water:- As a result of transpiration water is drawn in the intercellular spaces from the mesophyll cells which consequently draw water osmotically from nearby cells & thus a diffusion pressure deficit or suction force is developed. Due to suction force, the adjacent cells take water from xylem of these veins of the leaves. As the xylem of these veins is connected with xylem of roots through stem xylem a tension is set up in the water column of xylem & whole column is physically pulled up.

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (TRANSPORT IN PLANTS)

1.

Name two factors that affect water potential?

[1]

2.

Define plasmodesmata?

[1]

3.

Why is salt added in excess to pickles?

[1]

4.

State the significance of plasmolysis?

[2]

5.

Why is c4 photosynthetic system more beneficial than c3 photosynthetic

[2]

system? 6.

Distinguish between transpiration & evaporation?

[2]

7.

Explain facilitated diffusion?

[3]

8.

Describe water potential? What are the factors influencing it?

[3]

9.

What forces are involved in absorption of water from soil by root hairs.

[5]

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (TRANSPORT IN PLANTS) [ANSWERS] 1.

Amount of solute & external pressures

2.

Plasmodesmata are protoplasmic connections between adjacent cells.

3.

High concentration of salts prevents the growth of microorganisms & thus it prevents spoilage of pickles.

4.

Significance of plasmolysis :i) It explains osmosis ii) Essential to know the biotic nature of the cell. iii) Essential in killing of weeds. iv) To determine the osmotic pressure of the cell. v) As a proof of cell wall permeability.

5.

Evolution of c4 photosynthetic system is perhaps one of the strategies for maximizing availability of co2 while minimizing water loss. C4 plants are twice as efficient as C3 varieties in terms of carbon fixation eg. C4 plants will lose only 300g of water by evaporation for every gram of co2 fixed whereas C3 plants loses 600g of water for same grams of CO2 fixed.

6. Transpiration i) It is a physiological process where loss of water occurs through aerial parts of plants. ii) It takes place during the daytime iii) It is regulated by activity of guard cells around stomata

7.

Facilitated diffusion is also called carrier mediated diffusion or transport by carrier proteins. It occurs along the concentration gradient assisted by carrier proteins eg. diffusion of glucose in RBC’s. The membrane provides sites at which some molecules cross the membrane.

Evaporation i) It is not physiological process but simply a physical process. ii) It takes place at all the times iii) It is not so.

They do not set up a concentration gradient. A concentration gradient is present for molecules to diffuse even if facilitated by proteins membrane without expenditure of ATP energy. It cannot cause net transport of molecules from a low to high concentration as it requires input of energy. Transport rate reaches maximum when all of the protein transporters are being used. It is very specific & permits a cell to select with protein side chains. Facilitated diffusion is stereo specific the carrier proteins are permeases.

8.

Water potential is represented by Ψ = Ψm + Ψs + Ψp. The water mover from the point where water potential is greater to the other where it is less. It is denoted by psi or Ψ . Water potential is the chemical energy of water. It is measured in terms of pressure & is also influenced by concentration gravity as well as pressure. The chemical potential of water is called water potential. It indicates free energy rotated to water. Water potential of pure water is zero. Water moves into the cell from outside & hydrostatic pressure is increased. It increases water potential of a solute ( Ψs ) but the difference between inside & outside is reduced. Water moves from outside into the cell due to water potential gradient. Three factors affect water potential. They are matric potential ( Ψm ) , solute potential

( Ψs ) &

pressure potential ( Ψp ) . Solute potential is always negative. Pressure

potential is usually positive & is denoted by Ψp . Thus, water potential is affected by both solute & pressure potential. The relationship between them is 9.

Ψ = Ψs + Ψp.

a)

A negative tension is exerted down the roots due to transpiration on pull by the aerial parts of the plants this causes a decrease in water potential of roots which favours the uptake of water from the soil.

b)

The decrease of water potential in the root cells than the soil favours the absorption of water from the soil.

c)

The cohesive forces among the water molecules & adhesive forces between the water & xylem vessels maintain an unbroken column of water in capillaries of xylem vessels. The gradient of water potential exists in the xylem vessel starting from leaf to roots which favour uptake of water from the soil.

d)

The water from the soil enters into the root hairs & from there it reaches the xylem vessel with lower water potential. It results in formation of root pressure. This root pressure pushes water to aerial parts of plant body.

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (TRANSPORT IN PLANTS)

1.

What determines the direction of flow of water from one cell to another

[1]

cell? 2.

Define guttation.

[1]

3.

What is the water potential of pure water?

[1]

4.

Mention two conditions necessary for imbibitions to occur?

[2]

5.

What are the factors affecting the rate of diffusion.

[2]

6.

What is the role of osmotic potential in regulating water potential of plant

[2]

cells. 7.

Distinguishes between imbibitions & diffusion?

[3]

8.

Describe the plant cell as an osmotic system?

[3]

9.

How is opening & closing of stomata controlled?

[3]

10.

Define transpiration? Why is it useful? Mention any three environmental

[5]

factors that affect the transpiration?

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (TRANSPORT IN PLANTS) [ANSWERS] 1.

Water potential of cell (Ψ).

2.

Water loss in liquid phase from plants is called guttation

3.

Zero bars.

4.

i) ii)

5.

The rate of diffusion is affected by the following:i) Gradient of concentration ii) Permeability of cell membrane separating them. iii) Temperature iv) Pressure

6.

Osmotic potential refers to the pressure which is needed to stop the movement of water from outside into the solution. An isolated solution which is not bounded by any membrane has no osmotic pressure. It has only the potential to result into a pressure when placed in an osmometer

Water potential gradient between the surface of absorbent & the liquid imbibed The affinity between the absorbent & the imbibed liquid.

7. Imbibition i) it occurs in living & dead both ii) It refers “to the absorption of water by general surface”

Diffusion i) it occurs in solids, liquids & gases. ii) It refers “to the movement of molecules, ions of gases, solids, liquids from the region of higher concentration to lower concentration. iii) An absorbent is involved but iii) No need of semi- permeable membrane no membrane in it. iv) It is a reversible process. iv) It is not a reversible process.

8.

The plasma membrane in plant cell with the vacuolar membrane & cytoplasmic film or alone is more or less semi permeable in mature. This membrane allows the water molecules to get through it freely whereas it allows certain molecules to enter & prevent others. The cytoplasm is surrounded by the cell wall. It possesses very much higher concentration than the solutions entering the plant cells via osmosis. So the plant cell functions as an osmotic system if it fulfills

the following two conditions:i) It has a semi-permeable membrane. ii) It possesses a liquid substance having much higher concentration therefore, plant cell acts as osmotic system. 9.

The factors affecting stomata opening & closing are:a) Light:- Light intensity needed for stomatal opening is low the stomata open in light but close in dark. In CAM plants, stomata open in dark & closed during daytime. b) Temperature:- If temperature is increased, then the stomata open but when there is decrease in temperature the stomata close. c) Availability of water:- The stomata are closed due to water stress or moisture deficit. d) Concentration of CO2 :- If there is an increase in CO2 concentration inside the leaf the stomatal openings are closed even in light. When CO2 is used up by plant in photosynthesis the stomata open.

10.

Transpiration is a phenomenon naturally occurring in plants through which water is lost from plants in the form of water vapours through their aerial parts. It is useful to plants because i) It helps in movement of xylem sap. ii) It causes cooling of leaf surface & thereby protects leaf from heat injury by intense sunlight Transpiration is affected by 3 environmental factors:a) humidity:- water is lost slowly in the atmosphere, if the humidity is high or increased. b) temperature:- the rate of transpiration is doubled with rise in temperature by 10.c c) wind speed:- high wind speed or a dry breeze greatly increases the transpiration

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (Mineral Nutrition)

1.

Name a soil bacteria which is capable of converting ammonia to nitrates.

[1]

2.

Which macronutrient is essential for synthesis of auxin.

[1]

3.

What do you mean by “chlorosis”?

[1]

4.

A farmer adds azotobacter culture to the soil before sowing maize. How does

[2]

it increase the yield of maize? 5.

Name the pigment found in root nodules of legumes. What is its function?

[2]

6.

What is hydroponics? Mention its uses?

[2]

7.

What do you understand by “Donnan Equilibrium?

[3]

8.

What are essential mineral elements?

[3]

9.

Describe the process of development of root nodules in leguminous plant.

[3]

Name the oxygen scavenger molecule present in root nodules?

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (Mineral Nutrition) [ANSWERS] Ans 01.

Nitrosomonas.

Ans 02.

Zinc

Ans 03.

Lack of development of chlorophyll in the leaves.

Ans 04.

Azotobacter provides nitrogen fixing bacteria which converts free nitrogen into nitrous and nitrites. It increases soil fertility. So it increases yield of maize.

Ans 05.

Root nodules of leguminons plants contain pigment leghaemglobin. Its function is to protect nitrogenase from oxygen. Hence called ‘oxygen scavenger’.

Ans 06.

Hydroponics is the cultivation of plants in the nutrient solution by placing their rooted part in nutrient solution. By hydroponics or water culture experiment, essentiality of an element for plant growth can be determined by exceeding a particular element in culture solution and by observing the symptoms caused by its deficiency.

Ans 07.

Ans 08.

This theory explains that the passive accumulation of ion that are fixed on nondiffusible, against an ecp gradient. A membrane that separates a cell from the external medium and allows exchange of some ions and not the other. On the inner side of this membrane are anions (fixed & non-diffusible). The membrane becomes impermeable to these anions. In such condition (for equilibrium) mobile cations are needed to balance the negative charges of the anions. According to it Donnan equilibrium is reached, if the product of anions and cations is the internal solution becomes equal to the product of anions and cations in the external solution. [Ci+] [Ai-] = [Co+] [Ao-] Where Ci+ = cations inside Ai- = Anions in side Co+ = Cations outside Ao- = Anions outside. Mineral elements found in soil which may enter plants through the roots. More than 60 elements of 105 discovered so far occur in different plants. Some

accumulate selenium but some others gold. Some plants growing near nuclear test sites takes up radioactive strontium. Ans 09.

Formation of root nodules in a leguminous plant: 1) When a root hair of a leguminous plant comes in contact with Rhizobium, the root hair becomes curled or deformed, due to chemicals secreted by bacterium. 2) At the site of curling or deformation, the bacteria invade the root and multiply within the root hair. 3) Some of the bacteria enlarge to become membrane – bound structures known as bacteroids, which help in spreading infection. 4) An infection thread made of plasma membrane is formed by the host that separates the infected cell from rest of the tissue. 5) Cell division is stimulated in the infected tissue and more bacteria enter the newly formed cells. Leghaemoglobin (Lb) is the oxygen scavenger found in root nodules of legume plants.

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Mineral Nutrition)

1.

Name any two elements having toxic effect on protoplasm?

[1]

2.

What is hydroponics?

[1]

3.

Give the function of enzyme nitrate reductase.

[1]

4.

What is balanced nutrient solution?

[2]

5.

What is nitrification? Name any two nitrifying bacteria in soil?

[2]

6.

In what form is magnesium absorbed by plants from the soil. Given two

[2]

functions of magnesium in plants & its deficiency symptoms. 7.

Differentiate between active & passive absorption.

[3]

8.

List the criteria for essentiality of elements as nutrient in plants.

[3]

9.

Write role of different elements in a plant?

[5]

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Mineral Nutrition) [ANSWERS] Ans 01.

Lead, mercury and arsenic.

Ans 02.

Plant growth in nutrient rich liquid culture medium.

Ans 03.

It reduces nitrate ions to ammonia.

Ans 04.

Balanced nutrient solution or balanced salt solution is prepared by dissolving definite proportions of salts in distilled water needed for normal plant growth. The plants growing in these solutions survive munch longer as they get all essential as well as the trace elements from medium.

Ans 05.

Nitrification is the process of conversion of ammonia into nitrites. It involves two steps:(i) Ammonium ions are oxidized into nitrates by the bacteria like Nitrosomonas and Nitrosococcus. (ii) Nitrites are converted into nitrates by the bacteria like Nitrobacter.

Ans 06.

Magnesium is absorbed by the plants from the soil in the form of divalent Mg2+. Functions – 1) Synthesis of DNA and RNA. 2) It activates enzymes in respiration and photosynthesis. Deficiency symptoms – 1) Chlorosis between the leaf veins. 2) Premature leaf abscission.

Ans 07. 1.

2. 3. 4. 5.

Active Absorption Passive Absorption Absorption of minerals is against Absorption of minerals is along the the concentration gradient. concentration gradient by simple diffusion. Energy is utilized for absorption. Energy is not utilized for absorption It is fast. It is slow. It is unidirectional. It may be bidirectional. It is a biochemical process. It is a physical process.

Ans 08.

Criteria for essentiality of element are fallowing1) The element must be absolutely essential for supporting normal growth and reproduction. 2) The requirement are need of the element must be specific and not replaceable by another element. 3) An element should be directly involved in the metabolism of the plant.

Ans 09.

Some important functions of mineral elements are – (1) Maintenance of the osmotic pressure in the plant cells – The mineral salts and organic compounds of the cell sap produce necessary osmotic pressure. (2) Constituents of the plant body – Elements form constitution of the plant body. For ex – Carbon, Hydrogen and oxygen are essential constituents of carbohydrates. Hence, called framework elements. Nitrogen, sulphur and phosphorous are required for synthesis of proteins. Magnesium is important constituent of chlorophyll. (3) Influence on the PH of the cell sap – They also influence the PH of the cell sap. (4) They influence the permeability of cytoplasmic membrane – They increase or decrease the permeability of the plasma membrane. (5) They take part in enzymatic reactions – some elements work as activators while the others works as inhibitors in various enzymatic reactions. (6) They have balancing functions reactions – some of the minerals balance the effects of the other.

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (Mineral Nutrition)

1.

Name essential components of biomolecules.

[1]

2.

Name the enzyme that can reduce nitrogen to ammonia.

[1]

3.

What are micronutrients?

[1]

4.

List the four broad groups of essential elements.

[2]

5.

How is hydroponics useful?

[2]

6.

What is mineral nutrition? Name one essential element that is a component of

[2]

energy – related chemical compounds. 7.

Describe the process of nitrogen fixation in plants. Mention the site where this

[3]

process actually occur is such plants. 8.

Differentiate between apoplast and symplast.

[3]

9.

Mention the role of micronutrients in plants life?

[5]

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (Mineral Nutrition) [ANSWERS] Ans 01.

C, H, O and N.

Ans 02.

Nitrogenase enzyme.

Ans 03.

Elements which are required by the plants in very small or trace quantities are termed as micronutrients for eg; zinc copper, etc.

Ans 04. 1. 2. 3. 4.

Type Components of bimolecules

Examples and its Role C,H,O,N is a component of nucleic acids and proteins. Components of energy related Mg. P. P is a constituent of ATP chemical compounds. molecule. Activate or Inhibit enzymes. Mg2+, Zn2+, Mo. Mo is activator of nitrogenase in N2 metabolism. Alter osmotic potential of a cell. Potassium. It is a essential. In opening of closing of stomata leaves.

Ans 05.

Hydroponics is essential to know the following(i) Essentiality of mineral element. (ii) Deficiency symptom due to non – availability of specific nutrient. (iii) Toxicity to plant if element is present is excess. (iv) Role of essential elements is metabolism of a plant.

Ans 06.

Plants require mineral nutrients for their growth and development which do not occur in the plant body in Free State. The utilization of there elements by the plants for its growth and development is called mineral nutrition. Mg in chlorophyll is essential component of energy-related chemical compound.

Ans 07.

Nitrogen fixation occurs in nodules in legume plants is gram and arhar. They act as sites for it. The legume plants like pea, gram show symbiosis or mutualism with bacterium, Rhizobium leguminasauram. Nitrogen fixation occurs with the help of enzymes nitrogenase & leghaemoglobin. Leghaemoglobin act as O2 scavenges and nitrogenase catalyse the conversion of N2 into NH3 N2 + 8e- + 16ATP + 8H+ → H2 + 2NH3 + 16ADP + 16 Pi

Ans 08. 1. 2.

3.

Apoplast (outer space) It includes cell wall and intercellular space. Uptake of ions into space is a passive process and involves no expenditure of energy. Initially, the ions are taken up quickly into outer space into medium.

Symplast (Inner space) It includes the cytoplasm and vacuole of the cell. Uptake of ions is an active process and involves expenditure of energy. Uptake of ions occurs slowly from the outer space.

Ans 09. Micronutrient 1. Boron (B) 2. Molybdenum (Mo) 3. Manganese (Mn) 4. Copper (Ca) 5. Chloride (Cl) 6. Zinc (Zn)

Role in plants life Pectin formation in cell wall, Translocation of sugar, Absorption of water. Reduction of nitrates constituent of nitrate reductases. Activation for photophosphorylation. Nitrogen metabolism, chlorophyll synthesis, Activation of enzymes. Component of enzymes. Component of plastocyanin. Transfer of electron. Synthesis of auxins. Acts as an activator.

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (Photosynthesis in higher plants)

1.

Expand NADP.

[1]

2.

Name one plant that carries out photosynthesis at night?

[1]

3.

Name the cell – organelles involved in photorespiration.

[1]

4.

What is red Drop?

[2]

5.

What are the enzymes that catalyze the dark reaction of carbon fixation

[2]

located? 6.

What are the two main functions of pigments other than chlorophyll in green

[2]

leaves? 7.

Explain -There is no oxygen evolution in bacterial photosynthesis.

[3]

8.

What is the advantage of using chlorella rather than a higher plant?

[3]

9.

(a) suggest some habitats or natural circumstances in which

[5]

(i) Light intensity (ii) CO2 concentration (iii) temperature might be limiting factors in photosynthesis. (b) In C4 plants which type of chloroplast is specialized for light reactions and which for dark reactions? (c) Why is it an advantage that bundle sheath chloroplast lack grana?

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (Photosynthesis in higher plants) [ANSWERS] Ans 01.

Nicotinamide adenine dinucleotide phosphate.

Ans 02.

Opuntia, chenopodium.

Ans 03.

Mitcohondria, chloroplast and peroxisomes.

Ans 04.

It is the occasional fall in photosynthetic yield beyond red region of spectrum. This is also called Emerson effect.

Ans 05.

The stroma contains enzymes which are capable of utilizing ATP and NADPH2 to produce carbohydrate during dark reaction. The carbon fixation occurs in the stroma by a series of enzymes catalysed steps which are located outside the thylakoids.

Ans 06.

Ans 07.

(i)

To absorb light energy and transfer it to chlorophyll for photosynthesis.

(ii)

To protect the chlorophyll molecule from photo oxidation.

In bacterial photosynthesis, the raw material for the supply of proton is H2S than H2O Thus, these is production of S than O2 during splitting in light reaction. 2H2S → 2HS + + 2H+ HS + HS → H2S + S

Ans 08.

Photosynthesis in chlorella and higher plants is biochemically similar but chlorella was used by Melvin Calvin (1954) due to following reasons – (i)

Chlorella culture is a chloroplast culture as a large volume of every cell is occupied by a single chloroplast.

(ii)

A synchroneous culture may easily be obtained in a short span of time.

(iii)

Cells are very quickly exposed to radioactive carbondioxcide and are quickly killed; thus handling chlorella for experiments is easier.

Ans 09.

(a)

Some situations are –

(i)

In a shaded community; dawn and twilight in a warm climate.

(ii)

Carbon dioxide is normally limiting, but it may be more so in a crowded stand of plants under sunny, warm conditions.

(iii)

(b)

On a bright day winter.

Mesophyll chloroplast for light reaction. Bundle sheath chloroplast for dark reaction.

(c)

Oxygen production is related to grana and oxygen would compete with CO2 for RuBP carboxylase and stimulate photorespiration. Grana occupy a large area of the chloroplast. In their absence in the bundle sheath there is more stroma, and so more RuBP carboxylase and more storage space for starch.

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Photosynthesis in higher plants)

1.

Why photosynthesis important?

[1]

2.

Define photosynthesis?

[1]

3.

What is the site for photosynthesis in Opuntia?

[1]

4.

Differentiate between respiration and photorespiration.

[2]

5.

Explain the role of water in photosynthesis.

[2]

6.

What is the law of limiting factor?

[2]

7.

What is the advantage of having more than one pigment molecule in a photo

[3]

centre? 8.

Why are C4 plant preferred in the tropical region?

[3]

9.

Briefly explain the chemisomotic hypothesis?

[5]

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Photosynthesis in higher plants) [ANSWERS] Ans 01.

(a) (b)

Primary source of all food on earth. O2 is released by green plants into the atmosphere.

Ans 02.

Photosynthesis is an anabolic endergonic as well as oxidation reduction process in which green plants manufacture food by raw materials in sunlight.

Ans 03.

Stem

Ans 04. 1. 2.

Respiration It occurs in all plants (C3 and C4) Glucose is the substrate of this reaction.

Photorespiration It occurs in C3 plants only. Glycolate is the substrate which is connected into Glycine, NH3 and CO2 in peroxisomes.

Ans 05

(i) (ii) (iii) (iv)

It is a reactant in light reaction. Water stress causes closure of stomata. It reduces the availability of CO2 Reduces surface area of leaves.

Ans 06.

This law states that “if a chemical process is affected by more than one factor which is nearest to its minimal value; then the rate will be determined by the factor which is nearest to its minimal value; it is the factor that directly affects process if its quantity is changed”.

Ans 07.

Light reaction depends upon the amount of solar energy trapped by the pigment. Energy trapped by a single pigment molecule is not enough to start the initial reaction which may occur in light. Hence, a number of pigment molecules provide protection to the chlorophyll molecule against photo oxidation.

Ans 08.

C4 plants utilize 30 ATP’s to produce one molecule of glucose favoured in tropical region. In these plants photorespiration is the mechanism not to lose the photosynthetic carbon. In the process of photorespiration RuBP is catabolised to a 3-carbon atom compound instead of combining with CO2. More than 50% CO2 fixed by photosynthesis is lost in photorespiration. Photorespiration acts to undo the work of photosynthesis as no energy rich compound is produced during this

process. Thus C4 plants are better photosynthesizes than C3 plants and C4 pathway is of adaptive advantage in tropical region and thus these plants are preferred. Ans 09.

Chemiosmotic hypothesis explained the mechanism of ATP synthesis in chloroplast. In photosynthesis, ATP synthesis is linked to development of proton gradient across a membrane. These are membrane of thylakoids. The proton accumulation is towards the inside of the membrane (in the lumen). The processes which occur during activation of electrons and their transport to determine the steps that causes a proton gradient to develop. ATP synthesis is linked to development of proton gradient.

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (Photosynthesis in higher plants)

1.

Explain chlorophyll is an essential photosynthetic pigment?

[1]

2.

What is the end product of light reaction?

[1]

3.

Give examples of photosynthetic micro – organisms which also fixes

[1]

atmospheric nitrogen? 4.

Compare between chlorophyll ‘a’ and chlorophyll ‘b’?

[2]

5.

What is kranz anatomy?

[2]

6.

Give advantages of C4 cycle over C3 Cycle.

[2]

7.

Distinguish between photo system – I and Photo system – II

[3]

8.

How does temperature affect photosynthesis?

[3]

9.

Explain the process of bio-synthetic phase of photosynthesis occurring in

[5]

chloroplast.

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (Photosynthesis in higher plants) [ANSWERS] Ans 01.

Chlorophyll – b and other pigments of a reaction centre or photosystem absorb solar energy and transfer it to chlorophyll–a. Ultimately it is chlorophyll–a that initiates photosynthesis process.

Ans 02.

ATP, NADPH2 and O2

Ans 03.

Anabaena, Nostoc.

Ans 04. 1.

2.

3. 4.

Ans 05.

Chlorophyll a Chlorophyll a has methyl group at the 3rd carbon position of II pyrrole ring of porphyrin head. It shows maximum absorption at 429 nm (blue) and 660 nm (red) wavelength. It is highly soluble in petroleum, ether. It is blue – green in colors.

Chlorophyll b It has an aldhyde group at the 3rd carbon position of II pyrrole ring or porphyrin head. It shaves maximum absorption at 453 nm (blue) and 642 nm (red) wavelengths. It is highly soluble in methyl alcohol. It is yellow green in colors.

Kranz Anatomy – The anatomy in which, the vein of the leaf is surrounded by the bundle sheath containing a member of chloroplasts, having a bull form cells in upper epidermis is known as “Kranz anatomy”.

Ans 06.

(i)

C4 cycle is more efficient than C3 cycle.

(ii)

The photorespiration is lacking in C4 plants.

(iii)

C4 cycle can use CO2 at very low concentrations in comparison to C3 plants.

(iv)

C4 cycle operates in plants adapted to high integrity of light, high temperature and low water availability, C3 cycle cannot operate under these conditions at all.

Ans 07. Photosystem – I 1. It is the cluster of pigment molecules which absorb light wavelengths at or below 700nm. 2. The light absorbed by any pigment molecule of the cluster is transferred to P700, which is the reaction centre. 3. It has a high ratio of chlorophyll – a to chlorophyll – b.

Ans 08.

Ans 09.

Photosystem – II It is the cluster of pigment molecules which absorb light wavelength at or below 680nm. The light absorbed by any pigment molecule of the cluster is transferred to P680 which is the reaction centre. It contains relatively more chlorophyll – b than chlorophyll – a.

The dark reactions are temperature controlled. The C4 plants respond to higher temperatures, C4 plants exhibit high rate of photosynthesis. C3 have much low temperature optimum. Tropical plants have higher temperature for photosynthesis.

Biosynthetic phase (Dark Reaction) : The process by which carbon – dioxide is reduced to carbohydrates is known as carbon fixation in plants. The fixation of carbon takes place in the stroma of chloroplasts, by a series of enzyme – catalyzed reactions. C3 pathway: It is known as Calvin cycle. The path of carbon in the dark reaction was traced by Melvin Calvin through a technique called autoradiography, using 14C, hence this pathway is called Calvin cycle. Calvin cycle consist of three phases: (i) Carboxylation (ii) Glycolytic reversal (iii) Regeneration of RuBP. (i) Carboxylation – Six molecules of Ribulose 1, 5 biphoshate react with six molecules of carbon-dioxide to form 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 – compound break into 12 molecules of 3-phosphoglyceric acid (PGA), a 3C – compound PGA is the first stable compound in this pathway. (ii) Reduction – 12 molecules of phosphoglyceric acid are converted into 12 molecules of 1,3 diphosphoglycerate and then reduced to phosphogly acetaldehyde (PGAL) using ATP and NADPH molecules respectively. Two molecules of PGAL are diverted for the synthesis of sugar and then into the starch. (iii) Regeneration of RuBP – For the cycle to continue, the primary acceptor of carbon-dioxide, i.e, RuBP has to regenerated 10 molecules of PGAL, by a series of complex reaction, are converted into 6 molecules of 5C – compound, RuBP. Formation of 6 molecules of RuBP requires six ATP molecules.

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (Respiration in Plants)

1.

Define fermentation and aerobic respiration.

[1]

2.

What are the different types of respiration occurs in plants?

[1]

3.

Name the energy currency of the cells.

[1]

4.

Define RQ. What is its value for fats?

[2]

5.

What is the importance of F0-F1 particles in ATP production during aerobic

[2]

Respiration?

6.

What is oxidative decarboxylation? What happens to pyruvate immediately

[2]

after this reaction? 7.

Describe the mechanism of Respiration.

[3]

8.

What are the various steps involved in glycolysis?

[3]

9.

Describe the process and role of citric acid cycle in living organisms.

[5]

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (Respiration in Plants) [ANSWERS] Ans 01.

Fermentation is partial breakdown of glucose. Aerobic respiration glucose is completely degraded into CO2 and H2O.

Ans 02.

Aerobic respiration and Anaerobic respiration.

Ans 03.

ATP.

Ans 04.

Respiratory Quotient (RQ) : The ratio of the volume of CO2 evolved to the volume of O2 consumed in respiration is termed as the respiratory quotient or respiratory ratio.

Its value for fats is less than one. Ans 05.

F1 head piece contains the site for ATP synthesis from ADP and phosphate. F0 forms the channel through which protons cross the inner membrane.

Ans 06.

Oxidative decarboxylation – It is the process in which carbon is removed from a compound as carbon-dioxide and the compound is oxidized. Pyruvate is oxidatively decarboxylated into 2C acetate unit, which joins coenzyme A (COA) to form acetyl CO – A.

Ans 07.

Mechanism of respiration – Glucose molecule is broken down into an intermediate molecule, Pyruvic acid. a) Breakdown of pyruvic acid in anaerobic respiration – In this process in absence of oxygen the pyruvic acid is incompletely reduced to ethyl alcohol. Glucose → Ethyl alcohol + CO2 + 2ATP b) Breakdown of pyruvic acid in aerobic respiration – In this process the pyruvic acid is completely oxidized into CO2 and H2O is the presence of oxygen. This process occurs in the mitochondria of the cell and is khown as kreb’s cycle.

Glucose (C6H12O6) Cytosol Glycolysis (in Cytoplasm) O2 absent (Anaerobic) Pyruvic acid

(C2H5OH+CO2) Fermentation

TcA Cycle

O2 present (aerobic respiration)

CO2 H 2O

ATP NADH Mitochondria The broad schene of respiration.

Ans 08.

Steps of Glycolysis –

Ans 09.

It is called “tricarboxylic acid cycle”. Following steps are present for completing this cycle(i) In this step, CO2 is removed from pyruvic acid and resulting 2- carbon unit with the sulphur containing compound coenzyme A forming Acetyl CoA. During this process the hydrogen released is accepted by NAD and NADH2 is produced. Pyruvic acid _ CoA + NAD → Acetyl CoA + NADH2 + CO2 (ii) Acetyl coenzyme A reacts with a 4 – carbon compound oxaloacetic acid to form citric acid. (iii) The citrate remains in equilibrium with cisaconitic acid and isocitric acid in the presence of the enzyme aconitase. Citric acid → Isocitric acid. (iv) Isocitrate is dehydrogenated in the presence of isocitrate dehydrogenase enzyme to form oxalosuccinate. The hydrogen released is accepted by NAD to form NADH2. Isocitric acid + NAD → Oxalosuccinate + NADH2. (v) A molecule of CO2 is lost from oxalosuccinate and a 5 – carbon compound ∝ –ketoglutaric acid is formed in the presence of decarboxylate enzyme. (vi) ∝ – ketoglutarate loses a molecule of CO2 and 4 – carbon compound succinyl CoA is formed. ∝ – keloglutarate + CoA + NAD → succinyl + CoA + NADH2 + CO2 (vii) Succinyl CoA forms succinate, and ATP is found by linking ADP and inorganic phosphate (Pi) Succinate CoA + ADP + Pi → Succinylate + CoA + ATP (viii) Succinate is oxidized into fumarate in the presence of succinate dehydrogenase enzyme. The hydrogen liberated is accepted by FAD and FADH2 is formed. Succinate + FAD → Fumerate + FADH2 (ix) In this step the fumarate is converted into malate in the presence of enzyme fumarate hydrase (fumarase) Fumarate → Malate.

(x)

Malate is changed into oxaloacetate in the presence of the enzyme malate denydrogenase. NAD is reduced to NADH2 by the liberated hydrogen. Thus oxaloacetic acid produced is ready to combine with the fresh acetyl CoA obtained from pyruvic acid for completing one cycle.

Net yield kreb’s cycle :- 1 Pyruvic acid + 1ADP + 4NAD + 1FAD
3CO2 + 1FADH2 + 4NADH2 + 1ATP Thus total yield of energy 1ATP = 1 ATP 3×4NADH2 = 12ATP 2×1FADH2 = 2ATP Total = 15 ATP Thus 2 Pyruvic acid in ghycolysis yield, 15×2=30ATP.

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Respiration in Plants)

1.

What are the other two names for kreb’s cycle?

[1]

2.

In which organelle does kreb’s cycle occur in living cells?

[1]

3.

Mention the conditions under which

[1]

(i) RQ is 1 (ii) R.Q is less than 1 4.

What is respiration?

[2]

5.

Why less energy is produced during anaerobic respiration?

[2]

6.

What is the function of phosphofructokinase in glycolysis?

[2]

7.

Explain Respiratory Balance sheet.

[3]

8.

What is the significance of stepwise release of energy in respiration?

[3]

9.

Explain ETS.

[5]

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Respiration in Plants) [ANSWERS] Ans 01.

Citric acid cycle (CAC), Tricarboxylic acid cycle (TCA)

Ans 02.

Mitochondria.

Ans 03.

(i) (ii)

If carbohydrates are used as substrate and are fully oxidized the R.Q will be 1. If fats are used in respiration, the R.Q well be less than 1.

Ans 04.

A process of physicochemical change by which environmental oxygen is taken into, to oxidize the stored food, for release of CO2, water and energy. The energy released is used for doing various life activities, where as CO2 is used by the plaints.

Ans 05.

i) ii)

iii) iv)

Incomplete breakdown of respiratory substrate takes place. Some of the products of anaerobic respiration can be oxidized further to release energy which shows that anaerobic respiration does not liberate the whole of energy contained on the respiratory substrate. O2 is not utilized for securing electrons & protons. NADH2 does not produce ATP as electron transport is absent.

Ans 06.

It catalyses the formation of fructose -1, 6- biphosphate from fructose-6phosphate and adenosine –tri- phosphate (ATP) Fructose -1,6- biophosphate is splited into 2 molecules of triose phosphate – 3 phosphoglyceraldhyde and dihydroxyacetone phosphate.

Ans 07.

a)

b) c) d)

As equetial, orderly pathway functioning, with one substrate forming next one with glycolysis, TCA cycle and ETS pathway following one after another. NADH synthesized in glycolysis. It is transferred into mitochondria and undergoes oxidative phosphorylation. None of intermediates in pathway are used to form any other compound. Only glucose is being respired; no other alternative substrates enter in pathway at any of intermediary stages.

Ans 8.

Advantages of step wise oxidation during respirationsa) It facilitates the utilization of a relatively higher proportion of that energy in ATP synthesis. b) Activities of enzymes for the different steps may be enhanced or inhibited by specific compounds. This provides a means of controlling the rate of the pathway and the energy output according to the need of the cell. c) The same pathway may be utilized for forming intermediates used in the synthesis of other bimolecular like amino acids.

Ans 09.

Mechanism of Electron transport system – Glucose molecule is completely oxidized by the end of the citric acid cycle. The energy is not released unless NADH and FADH are oxidized through the ETS. The oxidation means ‘removal of electrons from it’. Metabolic pathway through which the electron passes from one carrier to another is called “Electron transport system” It is operative in the inner mitochondria membrane. Electrons from NADH produced in mitochondrial matrix are oxidized by NADH dehydrogenase (complex I) and electrons are then transferred to ubiqinone located within the inner membrane also receives reducing equivalents via FADH; that is generated during oxidation of succinate, through activity of enzyme named succinate dehydrogenase (complex II). Reduced ubiquinone is then oxidized with the transfer of electrons to cytochrome complex (complex III). Cytothrome is small protein attached to outer surface of inner membrane and acts as a mobile carries for transfer of electrons between complex III and complex IV. (complex IV) is cytochromes ‘c’ oxidize complex having cytochromes ‘a’ and a3. When electron pass from one carries to another via complex I to IV in ETS, they are coupled to ATP synthase (complex V) for production ATP from ADP and inorganic phosphate. Oxidation of one molecule of NADH, gives rise to 3 molecules of ATP, while that of FADH, produces 2 molecules of ATP. Electrons are curried by cytochromes and recombine with their protons before the final stage when hydrogen atom is accepted by oxygen to form water. O2 acts as final hydrogen acceptor. Whole process by which oxygen allows the production of ATP by phosphorylation of ADP is called ‘oxidative phosphorylation’.

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (Respiration in Plants)

1.

What is respiration?

[1]

2.

Give two types of cellular respiration.

[1]

3.

How many carbon atoms are present in the molecule of each of :

[1]

(i) Glucose and (ii) Pyruvate? 4.

Give difference between Breathing and Respiration?

[2]

5.

Define aerobic respiration?

[2]

6.

What is compensation point?

[2]

7.

Write the significance of citric acid cycle.

[3]

8.

Explain fermentation.

[3]

9.

Give the various steps involved in Glycolysis.

[5]

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (Respiration in Plants) [ANSWERS] Ans 01.

A process of physiochemical change by which environmental oxygen is taken into, to oxidize the stored food, for release of CO2, water and energy. The energy released is used for doing various life activities, whereas CO2 is used by the plants.

Ans 02.

(a) Aerobic (b) Anaerobic

Ans 03.

(i) 6 carbon in glucose (ii) 3 carbon in pyruvate.

Ans 04. Breathing a. It is a biophysical process. b. Oxygen is taken in and CO2 is given out.

Respiration It is a biochemical process Water, carbon dioxide and energy is released by the oxidation of carbohydrates.

Ans 05.

The process of release of energy through intake of molecular oxygen and release of CO2 is known as aerobic respiration.

Ans 06.

At low concentration of CO2 and non-limiting light intensity, photosynthetic rate of a given plant will be equal to the total amount of respiration. Atmospheric concentration of CO2 at which photosynthesis just compensates for respiration is referred to as the CO2 compensation point.

Ans 07.

i) (ii) (iii)

Ans 08.

It explains the process of breaking of pyruvate into CO2 and water. It is major pathway of generation of ATP. More energy is released (30 ATP) in this process as compared to glycolysis. Many intermediates compounds are formed. They are used in the synthesis of other bimolecules like amino-acids, nucleotides, Chlorophyll, cytochromes and fats.

It occurs in some organisms like some bacteria that produce lactic acid from pyruvic acid. In animal cells, such as muscles during exercise, when O2 is inadequate for cellular exercise, the pyruvic acid is reduced to lactic acid by lactate

dehydrogrogenase. Reducing agent is Glucose C6H12O6 NADH + H+ that is reoxidised to NAD+ in Glyceraldehyde 3 - phosphate + both processes. Lactic acid NAD In both lactic acid and alcohol 3-phosphoglyceric fermentation not much energy is acid Pyreevic acid + relased; less than seven per cent of the NADH+H + energy in glucose is released and not all NAD Phosphoenol it is trapped as high energy bonds of Pyruvic acid Ethanol+CO2 ATP. The processes are hazardous either the acid or alcohol is produced. Yeasts Major pathway of anaerobic respiration poison themselves to death when the concentration of alcohol reaches approximately 13%. Ans 09.

The steps are as follows1) Glucose is phosphorylated in the presence of ATP, catalyzed by the enzyme hexokinase. 2) Glucose – 6 – phosphate is changed into its isomer fructose – 6 – phosphate catalyzed by phosphohexose isomers. 3) Fructose – 6 – phosphate is phosphorylated in the presence of ATP to form Fructose 1, 6 biphosphate. 4) Fructose 1, 6 biposphate is split into two molecules of triose phosphate one of 3 – phosphoglyceral dehyde and one of dihydroxyacetose phosphate, which are interconvertible. This reaction is catalysed by phosphofructokinase. 5) 3 – phosphoglyceraldehyde is oxieised to 1,3 biphosphoglycerate, with the reduction of NAD to NADH. 6) Phosphoglycerate kinase catalyses the formation of 3-phosphoglycerate to 1,3 biposphoglycerate and 1 molecule of ATP is produced directly (substrate phosphation). 7) 3-phosphoglycerate is converted into 2-phosphoglycerate and then into phosphoenlpysuvate (PEP) 8) PEP is converted into pyrucate long with the formation of one molecule of ATP directly. The enzyme pyruvate kinase catalyses this step. The end products of glycolysis are 2 molecules of pyruvc acid + 2 NADH + 2ATP.

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (Plant Growth and Development)

1.

What is growth?

[1]

2.

Which hormone act as “stress hormone”?

[1]

3.

Name a non-acidic growth substance.

[1]

4.

What would happen to tissue culture of parenchyma if-

[2]

a) Auxin and cytokinin were present in equal quantities. b) More cytokinin than auxin was present. c) More auxin than cytokinin was present 5.

Define vernalization?

[2]

6.

What is heterophylly?

[2]

7.

What are the conditions necessary for growth?

[3]

8.

What does the sigmoid growth curve of a population mean?

[3]

9.

What is photoperiodism? How do you categories the angiosperms on the basis

[5]

of their flowering response.

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (Plant Growth and Development)

[ANSWERS]

Ans 01.

Irreversal permanent increase in size of an organism.

Ans 02.

Abscisic acid (ABA)

Ans 03.

Kinetin.

Ans 04.

i) cells divide but do not differentiate ii) shoot buds develop from the callus. iii) Root develop.

Ans 05.

The term vernalization is promoter of flowering by a previous cold treatment. In flowering plant, plants requiring cold treatment usually behave as biennials. They germinate and grow vegetative in first season and produce flower in second season.

Ans 06.

The plants follow various pathways in response to environment / phase of life to form different kind of structures. It is known as plasticity. Some examples are heterophylly in cotton, coriander and larkspur. The leaves of juvenile plants are different in shape from those in mature plant. Difference in the shapes of leaves produced in air and those produced in water in buttercup also represent the heterophyllous development due to environment. The phenomenon of heterophylly is an example of plasticity in plants.

Ans 07.

The necessary conditions for growth are water, oxygen and nutrients. The plant cell grows in size by cell enlargement that needs water. The plant growth and further development are intimately liked to water status of plants. Water provides medium for enzymatic activities needed for growth O2 helps in releasing metabolic energy for growth. Nutrients both macro and micro essential elements are needed by plants for synthesis of the protoplasm. Moreover, they act as source of energy.

Every plant has an optimum temperature range best suited for its growth. Any deviation from it may be detrimental to its survival. Environmental signal like light and gravity also influence various phases or stages of growth in plaints. In biological organization growth occurs at many levels, from the molecular level upto the ecosystem level. It can be measured at different levels such as the growth of cell organism or population. If it is measured in length, area, volume, mass or number of cells or individuals and plotted against time, and s-shaped curve is obtained. This is known as sigmoid curve. An Time analysis of this curve shows a lag phase during which slow growth occurs. This gradually attains a rapid growth, followed by a period of slow growth and ultimately a decline called stationary phase. Since the same patterns of growth is sigmoid curve observed at all levels of organization it is said to be universal.

Ans 09.

Photoperiodism – The phenomenon of inducing flowering in plants with response to length of daily period of light or relative day & night length is call photoperiodism. The angiosperms are classified into the following three categories on the basis of photoperiodism. (i) short day plants (SDP) – They require a relating short day length than critical period for flowers e.g chrysanthemum, Nicotiana, soyabean. (ii) Long day plants (LDP) – They require a relating longer day length than critical period for flowing e.g. wheat, Maize, Radish. (iii) Day neutral Plants (DNP) – The flowering response in their plants remain unaffected by the length of day. These plants are also called as photoneutrals or indifferent plants e.g. Cotton, Pea, Tomato & sunflower.

Growth rate

Ans 08.

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Plant Growth and Development)

1.

What is aleurone layer?

[1]

2.

Name the growth regulator which was first isolated from corn kernel and

[1]

coconut milk? 3.

What is the full form of IAA?

[1]

4.

‘Both growth and differentiation in higher plants are open’ comment.

[2]

5.

What is bolting? Which hormone is responsible for it?

[2]

6.

Why is the term short plants a misnomer?

[2]

7.

Differentiate between photoperiodism and vernilisation?

[3]

8.

Discuss the statement : ‘The growth is measurable’

[3]

9.

i)

What do you understand by the tem Development?

[5]

ii)

Explain the sequence of development process in a plant cell.

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Plant Growth and Development)

[ANSWERS] Ans 01.

It is special tissue layer which surrounds the endosperm in maize grain.

Ans02.

Zeatin (Cytokinin) was first isolated from corn kernel and coconut milk.

Ans 03.

Indole Acetic Acid.

Ans 04.

Growth and differentiation in plants are open as all the cells as well as the tissues arising from the same meristem may possess different structures at maturity. The maturity is determined by the location of cells or tissues e.g. it may be at shoot apex, root apex. Cambium etc.

Ans 05.

Enormous elongation of inter nodes resulting increase in stem height. Gibberellins cause the plants to bolt and flower.

Ans 06.

These plants require a relatively short day light period usually 8-10 hours and a continous dark period of about 14-16 hours for flowering. In short day plants dark period is critical and must be continuous. They are known as long Night plants and the term short day plant is a misnomer with long night plants.

Ans 07. a

b c d

e

Photoperiodism Photoperiodism is the flowering response of the plants to the duration of light and dark period in the diurnal cycle. In this stimulus perceived by green leaves only. In this florigen is produced under photoinductive conditions. Photoperiodic induction cannot be reversed by exposing to noninductive conditions. GA3 has the capability to replace the requirements of photo-inductive conditions in long day plants only.

Vernilization Vernalisation prepares the plants for perceiving stimulus for flower induction by chilling treatment. Stimulus is perceived by young embryos, meristems and even leaves. Vernalin is produced by chilling treatment. Vernalisation can be reversed when maintained higher temperature GA3 can replace cold treatment to induce vernalisation.

Ans 08.

The growth (at a cellular level) is basically a consequence of increase in the amount of protoplasm. Since we cannot measure growth directly it is measured by some quantity that is more or less proportional to it so the growth is measured by a variety of parameters like increase in fresh weight; dry weight; length; area; volume and cell number etc. One single maize root apical meristem may give rise to more than 17,500 new calls per hour. The cells in a watermelon can increase in size by upto 3,50,000 times. Therefore, growth may be expressed as increase in cell number or as increase in size of cell. The growth of a pollen tube is measured in terms of length. An increase in surface area measures growth in a dorsiventral leaf or dicot leaf.

Ans 09.

It is a term “that includes all changes that an organism goes through during its life cycle from germination of the seed to senescence.” Diagrammatic representation of the sequence of processes in development of a cell of a higher plant. D eath

ce ll D ivison Th e ce lls divide A M e ristem atic cell

Plasm atic gro w th

Expan sio n or Elon gation

D ifferen tiatio n

M atu ratio n

S en e scen ce

A M atu re cell

Sequence of the development process in a plant cell.

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (Plant Growth and Development)

1.

Name the hormone which is responsible for elongation of intermodal regions of

[1]

green plants. 2.

Would a defoliated plant respond to photoperiodic cycle? Why?

[1]

3.

Mention the names of two such substances that cause seed dormancy?

[1]

4.

Explain the role played by phytohormone in seed germination.

[2]

5.

What is ‘Bioassay’?

[2]

6.

Name any two synthetic auxins. How are they used in agriculture.

[2]

7.

What is apical dominance name the hormone that controls it.

[3]

8.

Write the principal characteristics of PGR’s.

[3]

9.

How is growth measured in plants?

[5]

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (Plant Growth and Development)

[ANSWERS]

Ans 01.

Ethylene

Ans 02.

No as the site of perception is the leaves so it will not respond to photoperiodic cycle.

Ans 03.

Abscissic acid and phenolic acid.

Ans 04.

It is a regulatory pigment which controls several light dependent developmental processes in plants besides seed germination. Phytohormones exist in two inter convertible forms : Pr and Pfr. On absorbing red light Pr becomes Pfr and Pfr becomes Pr either rapidly by absorbing far red light or slowly in darkness. Germination is promoted by Pfr and red light is needed to promote this. Darkness (far-red) promotes Pr formation which induces dormancy and inhibits germination.

Ans 05.

A bioassay is the evaluation of the effect of a substance on living organism under controlled conditions.

Ans 06.

Two synthetic auxins are (i) Napthalene Acetic Acid (NAA) (ii) Indole Bytyric Acid (IBA) They are used as weedicides

Ans 07.

“The inhibition of growth of lateral buds into the branches in the presence of an apical bud.” Apical dominance is under the control of auxins IAA (indole acctic acid) is the principal auxin found in all the plants. Lateral buds start their development when apical bud is removed. The process is again reversed if you apply IAA to decapitulate apex of plant.

Ans 08.

Characteristics – PGRs are small, simple molecules of diverse chemical structure occurring in plants. They are indole compounds (indole 3 acetic acid, IAA); adenine derivatives CN6 of (arotenoids) and the fatty acids (abscisic acid, ABA); terpenes (gibberllic acid, GA3) or gases (ethylene, C2H4). PGR’s are called plant growth substances or plant hormones.

They are broadly divided into two groups based on their function in a living plant body:a) On group of PGR’s involved in growth promoting activities, e.g, cell division, cell enlargement, pattern formation, tropic growth, flowering, fruiting and seed formation. They also termed plant growth promoter e.g. auxons, gibberellins, cytokinins. b) PGR’s of other group are in plant responses to wounds and stresses of biotic and abiotic origin. These may be involved in different growth inhibiting activities like dormancy and abscission; e.g, abscissic acid (ABA). The gaseous PGR is ethylene. It is inhibitor of growth activities mostly. Ans 09.

Growth in plants is measured by are indicator arc auxanometer. Pfeffer’s auxanometer have two wheels attached to a stem. The lip of the potted plant is connected to small pulley and its other end is strengthened to a weight. A pointer is attached to a big pulley by weight and also to a cylinder having smoothed paper.

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (Digestion and Absorption)

1.

Name the secretions of Goblet cell & parietal cells.

[1]

2.

Name the three parts of small intestine of man.

[1]

3.

Which is the largest gland in our body?

[1]

4.

What is the role of micelles in the fat absorption?

[2]

5.

Give two functions of trypsin?

[2]

6.

What are the specific functions of food?

[2]

7.

How is DNA content in our food digested in the body?

[3]

8.

How would it affect the digestion of proteins if there is blockade in the

[3]

pancreatic duct?

9.

Draw a labeled diagram of human alimentary canal & Describe its different parts.

[5]

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (Digestion and Absorption) [ANSWERS] Ans 01.

Goblet cells secrete mucus. Parietal cells secrete hydrochloric acid (HCl) and intrinsic factor.

Ans 02.

Duodenum, Jejunum and ileum.

Ans 03.

Liver.

Ans 04.

During digestion, the fat in the intestine is converted to monoglycerides diglycerides and fatty acids which are insoluble in water. They cannot be directly absorbed from the intestinal contents. They are first incorporated into small, spherical and water soluble droplets called micelles by bile salts. It is from these micelles that fatty acids, glycosides, sterols and fat soluble vitamins are absorbed into the intestinal cells.

Ans 05.

1)

Trypsin converts chymotrypsinogen into chymotrypsin.

2)

Trypsin acts on proteoses and peptones and convert them into peptides. Trypsin + Peptones + Proteoses → Peptides.

Ans 06.

Specific functions of food are – (i) Food on oxidation inside the body supplies energy to perform various functions. (ii) It serves to supply the material for growth and development of the body. (iii) It also serves as a reserve material mainly as fat and glycogen. These can be utilized at the time of emergency. (iv) It protects the body from diseases.

Ans 07.

DNA content is digested in the intestinal part of our alimentary canal by the enzymes present in pancreatic juice & sucous entericus. DNA content Deoxyribonuclease Enzyme

Deoxyribonucleotides.

Deoxyribonucleotides.

Nucleosidase Enzyme

PO4 +Deoxyribonucleotides.

Deoxyribonucleotides.

Nucleosidase Enzyme

Deoxyribose + Purine + Pyrimidine.

Ans 08.

Ans 09.

Pancreatic duct in addition to pancreatic juice brings bile juice also. The pancreatic juice contains many enzymes which are as fallowsa) Trypsin – It acts on protein, proteases and peptones and converts them into amino acids. b) Amylopsin – It acts on starch and converts it into soluble sugars. c) steapsin or lipase – It emulsify the fats and convert them into fatty acids and glycerol. Hence, if there is a blockade in the pancreatic duct then there will be no digestion of proteins, fats and starch because the digestive enzymes will be absent. The alimentary canal of man is a long coiled tube of varying diameter. It measures from 8 to 10 meters in length. It is divisible into the following parts – a) Oral cavity – It is the initial enlarged part of the alimentary canal. It opens by mouth and consists of lips, cheeks, gums, teeth and the palate and its muscles. The salivary glands open into the oral cavity. b) Pharynx – The oral cavity passes into pharynx. c) Oesophagus – It is a muscular tube about 10 inches long through which food passes into the stomach where it joins the cardiac stomach. d) The stomach is a sac – like structure and situated below the diaphragm. The wall of the stomach contains many small gastric pits into which ducts of gastric glands open. e) Small intestine – It is a long tube – like structure measuring about 5-7 meters. It is divisible into 3 parts – duodenum, the jejunum and the ileum. The duodenum is the first part and u – shaped. In this open the opening of pancreatic duct and bile duct. f) Large intestine – The large intestine is about 1.5m long. It consists of caecum with vermiform appendix, colon and rectum. The rectum opens to the exterior by anus.

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Digestion and Absorption)

1.

What is the main function of bile salt?

[1]

2.

Name the watery fluid secreted from Bruner’s gland in duodenum.

[1]

3.

What is atheroma?

[1]

4.

How does fat absorption takes place?

[2]

5.

How is food absorbed?

[2]

6.

What are enzymes?

[2]

7.

What is the action of salivary amylase? Differentiate between lipases and

[3]

peptidases? 8.

It is absolutely not necessary to produce amylase in an active form in our

[3]

body. But it is not in the case of trypsin. Given reasons.

9.

Name the enzymes for protein digestion in the gastric, pancreatic and intestinal, the substrate they digest and products of their action.

[5]

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Digestion and Absorption) [ANSWERS] Ans 01.

It reduces the surface tension of fat droplets causing their break down into many small ones.

Ans 02.

Mucoid fluid is secreted from Bruner’s gland in duodenum.

Ans 03.

Deposition of cholesterol on the walls of arteries.

Ans 04.

Fat absorption – It occurs as monoglycerides and fatty acids. These are resynthesized into triglycerides which in turn, combine with cholesterol. They form chylonicrons chylomicrons pass into the lymphatic system for circulation.

Ans 05.

The food eaten up by individuals is in complex form. The digestive glands secrete enzymes in different parts of alimentary canal and digest it into simpler form, mainly soluble form. The digested food consist of fatty acid and glycerol are absorbed through intestinal wall through lacteals. The sugars, amino acids, salts and water passed into blood circulation, water absorption takes place in colon (large intestine).

Ans 06.

Enzymes are defined as “an organic catalyst found in a living organism, which alters the fate unchanged at the end of the reaction; but itself remain unchanged at the end of the reaction; and is produced by the living organism but is not itself alive.

Ans 07.

Salivary amylase digest starch into sugars. Difference between lipases and Peptidases. Lipases

Ans 08.

Peptidases

1. They are insoluble in water.

They are soluble in water.

2. These hydrolyse fats & oils.

These hydrolyse proteins,

Salivary amylase is secreted in buccal cavity and it digests starch and sugar (carbohydrates). Since amylase does not act on protein of which animal tissues (buccal cavity) is made from, it is secreted in its original form.

Trypis – It acts on proteins. The wall of the alimentary canal is also made of protein. Hence it is very essential that it is secreted in an inactive form and it should become active when food protein is available in the alimentary canal. Thus to prevent damage (digestion of body) it is secreted in an inactive form. Ans 09. Juices 1 Gastric

Enzymes

Substrates

Products

Pepsin Renin

Proteins, casein (milk) casein

Peptones, Paracasein (curd) Para

Juice 2 Pancreatic

casein Trypsin

i) Protein

Peptides

ii) Chymotrypsinogen (inactive)

Chymotrypsin (active)

iii)Procarboxypeptidases (inactive)

Carboxy peptidases (active)

iv)Protelactase (inactive)

Elactase (active)

v) Fibrinogen (blood)

Fibrin (clot)

Chymotrypsin

casein

Paracasein

Carboxypeptidase

Peptides

Small peptides, amino acids

Enterokinase

Trypsinogen (inactive)

Trysin (active)

Amino peptidases

Peptides

small peptides, amino acid

Dipeptidases

Dipeptides

Amino acids.

Juice

3 Intestinal Juice

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (Digestion and Absorption)

1.

What is egestion?

[1]

2.

What are micelles?

[1]

3.

What are crypts of lieberkuhn?

[1]

4.

If a major part of the small intestine of a mammal be removed, will this affect

[2]

absorption of food? 5.

What is the role of micelles in the fat absorption?

[2]

6.

Differentiate chylomicron & micelles on the basis of their structural

[2]

components. 7.

Describe coagulation of milk in alimentary canal.

[3]

8.

Name three enzymes secreted by pancreas specify the substance and the

[3]

product of each. 9.

Explain the absorption of digested products.

[5]

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (Digestion and Absorption) [ANSWERS] Ans 01.

Passing out of undigested food from the body.

Ans 02.

Monoglycerids, long chain fatty acids and digested fats unite with bile salts and form small spherical droplets known as micelles.

Ans 03.

Pits into the sub mucosa of gastrointestinal tract wall.

Ans 04.

The major part of the food is absorbed only in the small intestine, only some part of water is absorbed in the stomach. So, if the major part of the small intestine is removed it would seriously affect the absorption of digested food.

Ans 05.

Fats are digested into monoglycerides, diglycerides and fatty acids, which are insoluble in water. These are first incorporated into small, spherical and water soluble droplets called micelles. Micelles help in the absorption of fatty acids, glycerols, sterols and fat soluble vitamins into the intestinal cells.

Ans 06. Chylomicrons 1. Protein coated water soluble fat droplets of about 150 mm released into the lymph. 2. In this form fats / lipids are put into circulation.

Micelles It is formed by combination of fatty acids, mono acylglycerols and the bile salts. In this form, digested fats are absorbed in intestinal cells in alimentary canal.

Ans 07.

When the food or milk reaches the stomach, the protein digestion starts. Pepsin stimulates the digestion of proteins in milk (casein) HCl activates pepsinogn into pepsin. It hydrolyses soluble casein into paracasein which precipitated as calcium paracaseinate to make solid curd i.e., coagulation of milk. There is a milk – coagulating enzyme called rennin which is found in calf gastric juice. Rennin is secreted as pro-rennin (inactive form) but in the presence of HCl, it is hydrolyses casein into paracasein leading to milk coagulation.

Ans 08.

Pancreas is a composite gland. It has exocrine and endocrine parts. The exocrine parts secretes pancreatic juice. It contain trypsin, amylopsin and steapsin. a) Trypin – It converts proteins, peptones and proteoses into amino acids.

b) Amylopsin – It acts upon starch and converts them into soluble sugars. c) Steapsin or lipase – It emulsifies fats and converts them into fatty acid & glycerol. Ans 09.

Absorption of Digested products – The absorption is defined “as the process by which end products of digestion pass through the intestinal mucosa onto the blood or lymph” . The process of absorption is carried out by 3 mechanisms: by passive, active or transport mechanisms. The monosaceharides such as glucose, amino acids and certain electrolytes e.g. chloride are mostly absorbed by the process of simple diffusion against the concentration gradient some substances e.g., fructose and some amino acids are absorbed the help of carries ions like Na+. It is known as facilitated transport. The transport of water – It depends upon osmotic gradient. Active transport takes place against the concentration gradient and it needs energy. The amino acids, monosaccharides like Glucose, electrolytes like Na+ are absorbed into the blood by active transport. The fatty acids and glycerol – These are insoluble and so cannot be absorbed into blood. They are incorporated into small droplets termed as micelles. They move into the intestinal mucosa. They again form very small protein – coated fat globules or the chylomicrons. The chylomicrons are transported into lymph vessel or lacteals found in the villi. They ultimately release absorbed substances into the blood. The absorption of various substances occurs in various parts of alimentary canal, mouth, stomach, small intestine and large intestine. Maximum absorption takes place in small intestine. The small intestine contains villi for it.

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (Breathing and Exchange of gases)

1.

Define partial pressure of a gas.

[1]

2.

Name the other pigments which are present in animals besides haemoglobin.

[1]

3.

What is the difference between alveolar air and inspired air?

[1]

4.

Give role of intercoastals muscles in respiration.

[2]

5.

Explain Erythrocytes can carry out anaerobic metabolism only.

[2]

6.

Describe how our brain gets a continuous supply of oxygen form the

[2]

atmosphere. 7.

What is hypoxia, artificial hypoxia & Anaemic hypoxia?

[3]

8.

How is respiration regulated?

[3]

9.

Describe transport mechanism of CO2.

[5]

CBSE TEST PAPER-1 CLASS - XI BIOLOGY (Breathing and Exchange of gases) [ANSWERS] Ans 01.

Pressure contributed by an individual gas in a mixture of gases is called partial pressure of gas and it is represented as PO2 for O2 and PCO2 for CO2.

Ans 02.

Haemocyanin and haemoerythrin.

Ans 03.

Alveolar air – The air present in the alveoli. Inspired air – The amount of air inspired at a time.

Ans 04.

The contraction of the external intercoastals muscles & diaphragm increases the volume of the thoracic cavity lowers the pressure in the lungs. To fill up the gap, the fresh air reaches to the lungs resulting in the inspiration. The relaxation of the inspiratory muscles decreases the volume of the thoracic cavity and subsequently, pressure in the lungs increase. To equalize this pressure, the air from the lungs rushes out through the respiratory passage to bring about expiration.

Ans 05.

Erythrocytes lack mitochondria and respiratory enzymes to perform the process of aerobic respiration. Therefore, they undergo aerobic respiration to carry out anaerobic metabolism only.

Ans 06.

Passage of air which contains oxygen: Inhalation of fresh air → trachea → bronchi → lungs → alveoli → diffusion of O2 into blood (RBC) → formation of oxygemoglobin → some in plasma → pulmonary vein → carry it to heart → left auricle → to ventricle → Dorsal aorta → Carotid artery to the brain dissociation of oxyhaemoglobin, O2 supplied to the tissue, partial pressure of O2 facilitates diffusion.

Ans 07.

Hypoxia – It is a condition of oxygen shortage in the tissues. It is 2 kinds – artificial hypoxia and anaemic hypoxia. Artificial hypoxia – It is the result of shortage of air over 2400m altitudes. The mountain sickness is caused by headache, dizziness, nausea, vomiting, mental fatigue and breathless mess etc. Anaemic hypoxia – results due to reduced O2 capacity of blood due to less content of Hb or carbon monoxide poisoning.

Ans 08.

Respiratory centre located in floor of the medulla oblongata of the brain controls respiration. The centre is bilateral and its two halves which are connected together by commissural neurons. The sides of this centre are connected with motor respiratory neuron. The nerve cells of the centre are connected with the breathing apparatus forming a reflex arc. These nerve cells are secretive to chemical composition of blood. Half of the respiratory centre is an inspiratory centre and expiratory centre. It is believed that the inspiratory centre work in normal breathing and expiratory centre during other conditions like coughing, sneezing and laughing. These two centers control the entire breathing in man with his knowing about it. Dorsal respiratory group, ventral respiratory group and pneumoptaxic groups act as respiration centers in the brain. Neumotaxic centre is located dorsally in upper pons. It transmits signals to inspiratory area. It controls the switch off point of inspiration.

Ans 09.

Transport of CO2 in the blood. (i) In the dissolved form About 5 – 7 % of carbon – dioxide is transported in dissolved form in the plasma of blood. (ii) In the form of bicarbonate. The remaining part of carbon dioxide enters the erythrocytes, where it reacts with the water to form carbonic acid (H2 CO3) ; this reaction is catalysed by carbonic anhydrase.. o Carbonic acid immediately dissociates into hydrogen ions (H+) and bicarbonate (HCO3-) o These H+ ions combine with haemoglobin, after the oxyglobin (KHbO2) dissociates to liberate the oxygen ; as a result haemoglobinic acid (H.Hb) is formed. o The majority of bicarbonate ions (HCO-3) diffuses out of the erythrocytes into the plasma, following a concentration gradient. o In response, chloride ions (Cl-) diffuse from the plasma into the erythrocytes (Chloride shift) and electrical neutrality is maintained. o The chloride ions combine with potassium to form potassium chloride. o In the plasma the bicarbonate ions combine with sodium and transported as sodium bicarbonate (NaHCO3) o Nearly 70% of the carbon dioxide is transported in this form. (iii) In the form of carbaminohaemoglobin o Same amount of CO2 reacts with the amine radicals (NH2+) of haenoglobin and form carbaminohaemoglobin (HbCO2) molecule. o About 23% of carbon – dioxide is transported in this form.

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Breathing and Exchange of Gases)

1.

Define vital capacity.

[1]

2.

What is the role of carbonic anhydrase in RBC’s?

[1]

3.

What is carbamino haemoglobin?

[1]

4.

What is chloride shift? Explain.

[2]

5.

Explain briefly the first step is respiration?

[2]

6.

Write a note on bronchitis and its prevention.

[2]

7.

Differentiate between vital lung capacity and total lung capacity.

[3]

8.

Explain the mechanism of breathing in humans.

[3]

9.

Describe in brief the respiratory organs of man.

[5]

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Body fluids and circulation) [ANSWERS] Ans 01.

Vital Capacity is the volume of air breathed out by a maximum forceful expiration.

Ans 02.

About 70% of CO2 reacts with water to form carbonic acid in RBCs in the presence of enzyme carbonic anhydrase. CO2 + H2O ⇌ H2CO3

Ans 03.

Carbaminohaemoglobin is formed when CO2 combines with globin is reduced haemoglobin.

Ans 04.

The diffusion of chloride ions from blood plasma into RBS’s is known as chloride shift. a) Occurs from plasma to RBC’s in human body. b) It maintains ionic balance and electrochemical neutrality.

Ans 05.

First step in respiration is called breathing. In breathing atmospheric air is taken in by inspiration and alveolar air is released out by expiration. The exchange of O2 and CO2 between deoxygenated blood and alveoli, transport of gases throughout body by blood, exchange of O2 and CO2 between the oxygenated blood and tissues and utilization of O2 by the cells are the other steps involved in it.

Ans 06.

It is “inflammation of the bronchi and is characterized by hypertrophy hyperplasia seromucous glands and goblet cells lining the bronchi” Symptoms are coughing with thick greenish-yellow sputum. It shows infection, that excessive secretion of mucus. It is caused by pollutants as well as the cigarette smoking. Prevention of Bronchitis – 1) Avoiding exposure to allergens. 2) Treatment involves antibiotic theory & bronchodilator dugs, etc.

Ans.07. Vital Capacity (VC)

Total lung Capacity. (CT2C)

1.

Sum total of tidal volume, expiratory reserve Sum total of vital capacity and residual and inspiratory reserve volume. volume.

2.

VC = Vt + ERV + IRV

TLC = VC + RV

3.

Value is 3500-4500ml.

Value is 5000 – 6000ml

4.

Represents maximum amount of air that a Represents maximum total amount of air person can expel after filling the lungs to the which can be present in lungs after maximum. maximum inspiratory effort.

Ans 08.

The mechanism of breathing in human involves breathing in of air in the lungs and breathing out of air from lungs thoracic cavity. The form is called inspiration and later expiration. The lungs are located in the closed thoracic cavity. A muscular partition called diaphragm separates the thoracic cavity from the abdominal cavity. During inspiration the diaphragm is lowered due to contraction intercostals muscle. This result into the increase of volume of thorax causing fall of air pressure in the thoracic cavity lowers the pressure in the lungs and the air rushes from outside into lungs through external nares, trachea & bronchi. During expiration the diaphragm move upward and the lateral thoracic walls move inwards due to the relaxation of muscles of diaphragm and the intercoastals muscles. This decrease the volume of thorax and the pressure inside the thorax and lungs is increased which results in the expulsion of some of air from the lungs to the atmosphere outside the body.

Ans 09. The following are the main respiratory organs:1) Nostrils – These are the paired openings situated at the anterior and posterior ends of the nasal chambers. They are lined up with ciliated epithelium and mucous cells. These prevent the entrance of dust into lungs and help in warming and moistening the air. The nasal chamber opens interiorly by external nostril and posterior internal nostril into the pharynx. 2) Larynx – It is situated at the anterior part of trachea and communicates with the pharynx. The glottis is protected by a stiff cartilage called epiglottis. The larynx contains pairs of vocal cords which set into vibrations when the air enters into it and produces the sound. 3) Trachea – It is a long ringed tube. It is supported c – shaped elastic cartilaginou rings to prevent its collapsing. It is lined internally with mucous membrane to hold the dust particles, bacteria and other foreign bodies. It also warms the air. 4) Bronchi – Inside the thorax, the trachea bifurcates into two branchy and each of which enters into one lung. In each lung, the bronchus again redivides into numerous small branches known as bronchioles. These bronchioles further divide at its ends to form respiratory bronchioles. 5) Lungs – There are two large bag-like spongy structures which are the main respiratory organs. These are enclosed by double pleural membranes. The lungs are divided externally by lobes. The right lung consists of four lobes and left by two lobes. Inside the lungs, the respiratory bronchioles give rise to alveolar ducts, alveolar sac and finally alveoli. Each lung contains millions of alveoli. Each alveolus is exceptionally thin walled. Its walls are highly permeable and richly supplied with blood capillaries. The blood is supplied to the lungs by a pair of pulmonary arteries. These bring blood which poor in oxygen & rich in CO2. The exchange of gases occurs in the alveoli of the lungs. The oxygenated blood from alveolar capillaries is called by pair of pulmonary vein to be conveyed to the heart.

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (Breathing and Exchange of Gases)

1.

Name the place where actual exchange of gases takes place in insects.

[1]

2.

What is the percentage of O2 in inspired & expired air?

[1]

3.

What is the utility of chloride shift?

[1]

4.

What is the difference between carbaminohaemoglobin and oxyhaemoglobin.

[2]

5.

What is functional residual capacity?

[2]

6.

Describe the transport of O2 and CO2?

[2]

7.

Define oxygen dissociation curve? Why it has sigmoidal pattern?

[3]

8.

What is the role of carbonic anhydrase? Show by series of reactions how

[3]

carbonic anhydrase starts the reactions leading to the formation of hemoglobinic acid?

9.

Explain how our heart muscles get a continuous supply of atmospheric oxygen.

[5]

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (Body fluids and circulation) [ANSWERS] Ans 01.

Tracheoles.

Ans 02.

Inspired air has 21% O2 and expired air has 16% O2.

Ans 03.

It maintains the ionic balance and electrochemical neutrality.

Ans 04. Oxyhaemoglobin 1. It is formed by the combination of oxygen with the Fe2+ part of haemoglobin. 2. It formation occurs on the alveolar surface.

Carbominohaemoglobin It is formed by the combination of carbon dioxide with the amine radical of haemoglobin. Its formation occurs in the tissues.

When a person inhales and exhales in a normal way, the volume air that remains in the lungs is known as functional residual capacity (FRC). It includes the residual volume and expiratory reserve volume, i. e, FRC = RV + ERV.

Ans 06.

O2 is transported as oxyhaemoglobin. In the alveoli of lungs (PO2 is higher), O2 gets bound to hemoglobin that dissociates at tissues where PO2 and H+ concentration are high. Approx 70% CO2 transported as bicarbonate (HCO3-) with the help of the enzyme carbonic anhydrase, 20-25% CO2 is carried by haemoglobin as carbaminohamoglobin. In tissues PCO2 is high its gets bound to blood but in alveoli where PCO2 is low and PO2 is high, this removed from blood.

Ans 07.

The relationship between O2 tension and its absorption by haemoglobin produces a graph called oxygen dissociation curve (O2 equilibrium curve). At about 100 mm Hg O2 tension Hb is 98% saturated (complete formation of haemoglobin). As it falls, the saturation of Hb decreases slowly. When O2 tension is about 40mm Hg, oxyhaemoglobin dissociates and O2 is available to the tissues. The O2 gets bound to Hb in lung surface and it gets dissociated at tissues.

2

O

A

m m C

120 110 100 90 80 70 60 50 40 30 20 10 0

0

% saturation wita O

2

Ans 05.

40

mm

CO

2

10 20 30 40 50 60 70 80 O xygen pressure in m ercury

Ans 08.

Carbonic Anhydrate : CO2 reacts with water in presence of carbonic anhydrase in erythrocytes, Carbonic acid (H2CO3) is dissociated into hydrogen (H+) and bicarbonate (HCO3-) ions). Oxyhaemoglobin (HbO2) of RBC’s is weakly acidic and remain in association with K+ ions as KHbO2. H+ ions combine with haemoglobin. Bricarbonate ions diffuse out into plasma and combine with haemoglobin to from haemoglobinic acid (H. Hb)

Ans 09.

When inspiration occurs, the O2 is taken into lungs. O2 mixes with air already present in alveoli and becomes alveolar air, whose PO2 is 100 m Hg. As PO2 of blood in the vessels is 40 mmHg oxygen differs into blood vessels from alveoli and the oxyhaemoglobin is formed when oxygen combines loosely with the Fe++ ions of haemoglobin. Oxygenated blood from the lungs reaches the left auricle through pulmonary vein; to left ventricle is pumped at through aorta also. The branch supplying blood to heart muscles is coronary artery. In heart muscles, as the PO2 is lower than that of the blood in the branches of coronary artery, oxyhaemoglobin dissociates and releases O2 to cardiac muscles.

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (Body fluids and circulation)

1.

Which of the four chambers of the human heart has the thickest muscular

[1]

wall?

2.

Where are RBCs formed from in an adult human?

[1]

3.

What is ECG technique?

[1]

4.

Distinguish between mitral and tricuspid value?

[2]

5.

Why does the fish heart pump only deoxygenated blood?

[2]

6.

How is heart failure different from heart attack?

[2]

7.

What is cardiac cycle?

[3]

8.

Differentiate between right ventricle and left ventricle.

[3]

9.

Describe the structure of human heart.

[5]

CBSE TEST PAPER-1 CLASS - XI BIOLOGY (Body fluids and circulation) [ANSWERS] Ans 01.

Left ventricle.

Ans 02.

RBCS are formed from the bone marrow.

Ans 03.

It is a technique to record and photograph the various electric cal changes in the working of the heart.

Ans 04. 1. 2. 3. 4.

Ans 05.

1) 2) 3)

Mitral Value It is called bicuspid value

Tricuspid value It lies in the region of right atrioventricular aperture. All the two flaps are of almost All the three flaps are different in size. equal size. There are two flaps in this flap. There are three flaps in this flap. Check back flow of oxygenated Check back flow of the deoxygenated blood into left auricle. blood into right auricle.

Atrium receives deoxygenated blood from all parts of the body. It is pumped into the ventricle from where it is pumped to the gills. The oxygenated blood flows from the gills to various parts.

Ans 06. 1.

2.

Ans 07.

Heart failure It refers to the state of the heart when the heart is not pumping blood sufficient to meet the need of the body. It is often due to congestion of lungs.

Heart attack It refers to the state where the heart stops beating. It is due to inadequate blood supply to the heart.

Cardiac cycle – The rhythmic contraction and relaxation of cardiac muscles is known as cardiac cycle or heart beat. It is involuntary (automatic). The contraction and relaxation of heart muscles are called systole and diastole respectively. One complete cardiac cycle occurs in 0.8 sec. Three stages of cardiac cycle are1) arterial systole 2) ventricular systole 3) Joint diastole.

Ans 08. 1. 2. 3. 4. 5. 6.

Ans 09.

Right ventricle Right ventricle is smaller than the left ventricle. Moderator band present in it. Columnae carneae thicker but less intricate. Receives and pushes deoxygenated blood. Crescent shaped. The wall of right ventricle is thinner than left ventricle.

Heft ventricle Left ventricle is comparatively larger than right ventricle. Moderator band is lacking in it. Columnae carneae narrower but more intricate. Receives and pumps oxygenated blood. Biconvex in shape. The wall of it is thicker than right ventricle.

The heart is a muscular organ situated in thoracic cavity which lies above the diaphragm between the two lungs. It is situated almost in the middle of the chest tilted at its apex to the left. It is enclosed in a double walled membranous sac, the pericardium fitted with pericardial fluid. The heart continuous working without stopping throughout the life of an individual. The heart of an average person at rest under normal circumstances beats. 70 to 80 times in a minute when it contracts its forces and pumps the blood into arteries which supply the blood to body organs. In man and other mammals heart is four chambered structure divisible into two halves right and left. The right & left halves of the heart are completely separated by septa. Each half has an upper chamber called the auricle and the lower chamber called the ventricle. Each auricle opens into the ventricle of its one side through an auricuo–ventricular aperature. The two apertures are guarded by valves which open only into the ventricle and prevent the back flow of the blood. The mitral valve or bicuspid valve having two flaps is present at the AV opening on the left side and the tricuspid valve (with three flaps) on the right side of the heart. The left ventricle is provided with tendinous cords called chordae tendinae and papillary muscle which prevent the valves from being pushed into auricles when the ventricles contract. The starting point of the aorta at left ventricle there is another set of semilunar valves.

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Body fluids and circulation)

1.

In which mammal, the RBC are nucleated?

[1]

2.

Name any two substances which prevent blood coagulation in uninjured blood

[1]

vessels. 3.

Name the type of granulocytes that play an important role in detoxification?

[1]

4.

Name the different types of granulocytes. Give the function of the one which

[2]

constitutes maximum percentage of total leucocytes. 5.

Why is closed circulatory system considered advantageous?

[2]

6.

What it is the name of the straw coloured fluid left after clotting of blood? How

[2]

is it different from blood? 7.

Write a note on “Regulation of cardiac activity”?

[3]

8.

Why does lymph contain much less proteins than the blood plasma? Name the

[3]

two principal lymph vessels in humans. 9.

What is lymphatic system? Discuss its importance.

[5]

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Body fluids and circulation) [ANSWERS] Ans 01.

Camel.

Ans 02.

Heparin, Antithrombin.

Ans 03.

Eosinophils.

Ans 04.

Different types of granulocytes are: (i) Neutrophils – 62% (ii) Acidophils (eosinophils) – 3% (iii) Basophils - 0.5% to 1% Neutrophils are phagocytic i.e, responsible for protection against infection.

Ans 05.

Closed circulatory system is considered advantageous for the following reasonsa) It maintains sufficient high blood pressure, blood flows at a high velocity; this quickens the supply of needed material and removal of wastes from the tissues. b) The volume of blood flowing to a particular organ / tissues can be regulated to the need of the tissues.

Ans 06.

It is called serum Plasma without coagulation factors is called serum. It differs from plasma in having much less quantity of proteins; it is outside the blood vessels.

Ans 07.

(i)

(ii)

(iii) (iv)

Ans 08.

The special neural centre located in medulla oblongata of brain can moderate cardiac function through autonomic nervous systems. Therefore help in controlling heart regulation. The parasympathetic neural signals, (component of ANS) decrease rate of heart beat, speed of conduction of action potential and also the cardiac output. The adrenal medullary hormones enhance cardiac output (C.O). The neural signals through sympathetic nerves may increase heart beat rate and the strength of ventricular contraction and also cardiac output.

Lymph contains mush less protein than plasma, because the capillary wall is impermeable to larger molecules like proteins.

The two principle lymph vessels are – Right lymphatic duct and thoracic duct. Ans 09.

Lymph is a colourlese tissue fluid resembling the blood except that it has no haemoglobin and RBCs. In comparison to blood, lymph contains less blood proteins, more of waste matter, increased amount of food material and a large number of WBC’s The tissue fluid is filtered from the blood plasma through the walls of capillaries some WBC also come out from there capillaries Now this tissue fluid enters into lymphatic capillaries as is known as lymph so the tissue fluid is converted into lymph. Circulation of lymph: Lymph vessels : Almost all of the body organs have a large number of lymph vessels and lymph capillaries. The walls of lymph vessels have valves (like veins). They form the network in the organs – one is superficial and other is deep seated. The flow of lymph in these vessels is only one side i.e., from the organs but never to the organs. In human body the following two large lymph vessels are present. Ductus Thoracious – It start from the abdominal cavity with a dilation called receptaculum chyli. Then it passes into the thoracic cavity then to the left of the neck region. It receives the lymph from the following organs – lower extremities, region of the true pelvis, abdominal region, left upper extremities the left half of the thorax, head, face & neck. Lymph nodes – These are small globular masses of lymphatic tissue and these arranged in groups from each region organs of the body the lymph flows into definite lymph nodes. The nodes are called regional nodes. Function of lymph: (i) It serves to return interstitial fluid into blood. (ii) It allows plasma proteins macromoleclues to diffuse through the lymph vessels. (iii) It transport digested fat through lacteals in villi of intestine.

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (Body fluids and circulation)

1.

A cardiologist observed an enlarged QR wave in the ECG of a patient. What

[1]

does it indicate?

2.

Name the double layered membranous covering of the heart.

[1]

3.

Why lymphatic circulation takes place very slowly?

[1]

4.

Why is swelling of feet of leg caused when a person stands immobile for a long

[2]

time? 5.

How are the two heart sounds produced during cardiac cycle? Which one of

[2]

these is of longer duration?

6.

What is average number of thrombocytes in blood? What is their function?

[2]

7.

Differentiate between arteries and veins.

[3]

8.

Explain the chemical events that take place to form a blood clot to seal the

[3]

wound?

9.

Explain double circulation with the help of diagram.

[5]

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (Body fluids and circulation) [ANSWERS] Ans 01.

Enlarged Q and R waves are the indication of myocardial infraction.

Ans 02.

Pericardium.

Ans 03.

Lymphatic circulation occurs due to squeezing action of surrounding muscles and not heart.

Ans 04.

When a person stands immobile for a long time, the flow of blood to the leg and feet is reduced temporarily. This leads to an accumulation of fluid in the leg and feet tissues resulting in swelling. But this swelling is subsided when he moves for short time because blood starts circulating again the veins normally.

Ans 05.

The two heart sounds are ‘lubb’ and ‘dupp’ - The first heart sound ‘lubb’ is produced by the closure of AV – valves at the start of ventricular systole. - The second heart sound ‘dupp’ is produced by the closure of semi lunar values at the start of ventricular diastole.

Ans 06.

1,50,000 to 3,00,000 / mm3 of blood The release substances that are concerned with the clotting of blood.

Ans 07. 1. 2. 3. 4. 5. 6. 7.

Arteries These are vessels containing blood flowing away form the heart. In these blood flows under great pressure. Their walls are elastic, thick and muscular. They are non-collapsible Their cavity is small. Valves are not present in them. Blood flows with jerks.

Veins These are vessels containing blood flowing towards heart. Blood flows under less pressure. Walls are thin, non-elastic, fibrous, Collapsible. Cavity is large. Valves present. Blood flows without jerks.

Ans 08.

Coagulation of blood – 1) When blood comes out of a blood vessel, the platelets clump together, break and release platelet factors, thromboplastin. 2) The prothrombin initiates the conversion of prothrombin into thrombin. 3) Thrombin catalyse the conversion of fibrinogen into fibrin which forms a mesh / network in which blood cells get entangled. 4) Ca++ ions are necessary for both the above steps.

Ans 09.

The heart is the pumping organ. It pumps blood to the various body organs, through closed vessels. Form the left ventricle blood goes with aorta which send it to arteries for supplying the body organs. From the body tissues blood is returned to the right atrium through two veins superior and inferior vena cava. This type of circulation is known as systemic circulation. From the right ventricle blood is pumped into the pulmonary trunk which divides into the pulmonary arteries each of which goes to the lung. Here the blood is oxygenated. The oxygenated blood is returned to left atrium through pulmonary veins. This is called pulmonary circulation.

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (Excretory Products and their elimination)

1.

In which part of nephron filtration takes place?

[1]

2.

What difference is observed in the ascending and descending limb of Henle’s

[1]

loop witch reference to permeability of water? 3.

What is the PH of urine.

[1]

4.

Differentiate between Rennin and Renin?

[2]

5.

What are the two intrinsic mechanisms that provide auto regulation of

[2]

glomerular filtrate? Explain any one of these. 6.

How is the permeability of the distal convoluted tubule and the collecting

[2]

tubule controlled for regulating the water content inside the body? 7.

Person suffering from very low blood pressure pass no urine why? What

[3]

suggestion would you offer for the removal of waste products from the blood in such a situation. 8.

Explain briefly how micturation is a reflex process; but is also under some

[3]

voluntary control. 9.

Describe briefly the structure and function of renal corpuscle.

[5]

CBSE TEST PAPER-1 CLASS - XI BIOLOGY (Excretory Products and their elimination) [ANSWERS] Ans 01.

Glomerulus.

Ans 02.

Ascending limp of Henle’s loop is impermeable to water. Descending limb of Henle’s loop is permeable to water.

Ans 03.

It is slightly acidic, PH – 6.0

Ans 04. 1.

2. 3. 4.

Rennin Renin Secreted from the peptic cell or Secreted from the juxtaglomerular cells gastric glands in stomach. of afferent renal artery in the renal cortex of kidney. The rennin is proteolysis enzyme. Renin is a hormone but it acts as enzyme also. Helpful in the digestion of milk of Converts angiotensinogen into protein. angiotensin – II Released in inactive form. Released in active form.

Ans 05.

Two intrinsic mechanisms that provide autoregulation, Myogenic mechanism and Juxtaglomerular apparatus (JGA) JGA – A special cellular apparatus is located in kidney where DCT passes close to Bowman’s capsule between afferent and efferent arterioles. The JGA cells secrete rennin that modulate blood pressure there by regulating renal flow and GFR.

Ans 06.

1) 2) 3) 4)

Ans 07.

When the water content inside the body is low, the osmorecepters stimulates theadenohypophysis to releases vasopressin / ADH. Vasopressin / ADH render the DCT and collecting tubule permeable to water. So, water is reabsorbed. When the water content in the body is normal, there is no release of ADH. The tubule is impermeable to water and water is eliminated in the urine.

It is because, the blood to pass through the glomerulus of the nephron must have required amount of pressure in it. If the pressure is not sufficient it will not flow through glomerulus and filtration would not the take place, hence no urine would be formed. This is quite harmful to the person as waste products go on accumulating in the body.

To avoid this, a person should be advised take sufficient amount of water and medicine to keep the blood pressure at optimum level. Ans 08.

1) 2)

3)

4)

Ans 09.

Micturation is act of voiding urine It accomplished by the simultaneous contraction of the smooth muscles of urinary bladder wall and the relaxation of the skeletal muscle sphincters around the opening of the bladder. As the bladder wall becomes stretched due to accumulation of urine, the stretch receptors in the wall of the bladder generate nerve impulse that are carried to sense neurons to the spinal cord and brain produce the sensation of fullness. But the sphincter muscles can also be relaxed voluntarily and there by the smooth muscles of the bladder are allowed to contract under autonomic control and the content of the bladder can be emptied.

Renal Corpuscle – It is the main excretory organ in the kidney. Nephrons are the functional units of kidney of renal corpuscle. There are about 102 million nephrons in each kidney in man. Structure of nephron – The nephron is a thin, long, twisted tubular structure. The tubule of each nephron starts as a up – shaped called Bowman’s capsule. There is a globular tuft of capillaries in the hollow of the cup. The Bowman’s capsule and the glomerulus together form a globular body called the renal corpuscle. Blood enters the glomerular capillaries through an afferent arteriole and leaves the glomerulus through as efferent arteriole. Urine is formed by the filtration of a protein free fluid from the glomerulus into lumen of the Bowman’s capsule. There are 3 parts of a nephron – (i) proximal nephron (ii) hoop of Henle and (iii) distal nephron. A long highly coiled and tubule trusted starts form the neck of the Bowman’s capsule. It is called the PCT (Proximal Convoluted tubule). It continues into a thin–walled straight tubule, then loops like segment of the tubule is called the Henle’s loop. It has thin descending limb and thick ascending limb. The Henle’s loop continuous into another segment of coiled and twisted tubule called DCT (Distal convoluted tubule). The terminal part of DCT is a straight short tubule called the collecting duct. The collecting duct runs down to the medulla again conducting the collected urine towards the medulla.

The collecting ducts unite with each other in the medulla to form the larger ducts called Ducts of Bellini. These ducts rue through the renal pyramids and open into renal pelvis. The efferent arteriole gives a capillary network around the tubule in the cortex. It also fives rise to some parallel wise, thin walled straight capillaries called vasa rectae. They help to maintain reabsorbed ions and urea in the intestinal fluid and maintain osmotic pressure in the kidney. Functions – Various part of nephron perform deferent function but the main function is liberation of metabolic waste from the body and maintain osmotic pressure of fluid in the body.

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Excretory Products and their elimination)

1.

Name the three kinds of nitrogen excretion.

[1]

2.

What are podocytes?

[1]

3.

Besides water, name any two constituents of human sweat.

[1]

4.

Kidneys do not play a major role in excretion in ammonotelic animals Justify.

[2]

5.

Define glomerular filtration rate. What is its value in a healthy human?

[2]

6.

What is the significance of frog’s tadpole being ammonetelic and the adult

[2]

frog being ureotelic? 7.

Describe urea cycle.

[3]

8.

What is a dialysis machine? When is it needed?

[3]

9.

Describe the mechanism of urine formation.

[5]

CBSE TEST PAPER-2 CLASS - XI BIOLOGY (Excretory Products and their elimination) [ANSWERS] Ans 01.

(a) Ammonotelism (b) Ureotelism (c) Uricotelism

Ans 02.

Epithelial cells of Bowman’s capsule are called podocytes.

Ans 03.

Sodium chloride and urea.

Ans 04.

Kidneys do not play any significant role in elimination of ammonia – (i) Ammonia is readily soluble in water and diffuses across the body surfaces. (ii) Ammonia is excreted as ammonium ions through gill surface.

Ans 05.

The amount of filtrate formed by kidneys per minute is known as Glomerular filtration rate. (GFR). In a healthy individual, GFR is approx. 125ml / minute i.e. 180 liters per day.

Ans 06.

Tadpole is ammonotelic, because excretion of ammonia requires a large volume of water, which the tadpole has in its surrounding. An adult frog is ureotelic because elimination of urea requires a moderate volume of water that is much less compared to ammonia.

Ans 07.

It is called kreb’s Ornitnine cycle.

Ans 08.

Dialysis machine is also known as artificial kidney. It filters blood when the kidneys fail. In dialysis small solute moleclues diffuse through a semi permeable membrane as a substitute for glomerulus. It has a cellophane membrane where the blood of a patient is made to flow on one side of the membrane and the surrounding fluid on other side. The wastes from the blood move into the surrounding fluid though cellophane membrane. It is needed when kidney fails to work and urine is not formed.

Ans 09.

Urine formation involves three main processes – (i) Glomerular filtration – A protein – free fluid is filtered from blood of glomerular capillaries into the lumen of Bowman’s capsule. The filtration occurs through three layers, which form the filtration membrane; they are : (i) Endothelium of glomerular capillaries (ii) Epithelium of Bowman’s capsule. (iii) Basement membrane between the two layers. The epithelial cells or podocytes of the Bowman’s capsule are arranged in an intricate manner, to leave some filtration slits. The blood is filtered so finely that the composition of filtrate is very similar to plasma except for the plasma proteins. The glomerular filtrate rate is about (25ml / min) (ii) Reabsorption Nearly 90% of the filtrate is reabsorbed in the renal tubule of the epithelial cells of the lining of renal tubule. Certain substances like glucose, amino acids, Na+ ions, K+ ions and Ca2+ ions are reabsorbed actively. Water is reabsorbed passively by osmosis. Certain other substance like Cl- ions are absorbed passively. (iii) Tubular secretion – It is the process by which certain substances / ions like K+ and ammonia are directly secreted into the lumen of the nephron. The step is important in urine formation, as it helps to maintain the ionic balance and PH of the body fluids.

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (Excretory Products and their elimination)

1.

What happens is glomerulonephritis?

[1]

2.

Name the excretory organ of cockroach.

[1]

3.

Name the hormone which controls the concentration of sodium in the body.

[1]

4.

Describe the blood vessels called vasa rectae found in relation to uriniferous

[2]

tubules. What is their function? 5.

What is chief nitrogenaus waste product in birds? Give two advantages of this

[2]

mode of excretion. 6.

Terrestrial animals are generally either ureotelic ar uric telic, Not

[2]

ammonotelic. Why? 7.

Suppose the kidneys of a person are damaged, can you predict what is going to

[3]

happen to him? 8.

How does liver both as a digestive as well as an excretory organ?

[3]

9.

Describe the renal excretory system of man.

[5]

CBSE TEST PAPER-3 CLASS - XI BIOLOGY (Excretory Products and their elimination) [ANSWERS] Ans 01.

Inflammation of glomeruli of kidney.

Ans 02.

Malphigian tubules.

Ans 03.

Aldosterene is a hormone which controls concentration of sodium in the body.

Ans 04.

Vasa recta are u–shaped, thin walled capillaries that arise from the efferent arteriole; they run parallel to Henle’s loop. They retain the reabsorbed ions in the medullary tissue fluid and maintain its high osmolarity.

Ans 05.

Chief mitrogenous waste product of bird is uric acid. It is advantageous them as – (i) Uric acid requires very little or no water for its elimination. (ii) Uric acid is far less toxic and can be eliminated slowly.

Ans 06.

Ammonia is highly toxic and it has to eliminate as rapidly as it is formed. - Land animals have an integument that is impervious to gas exchange. - It requires a large volume of water needed for elimination - They do not access to such a large volume of water needed for elimination of ammonia. - So they are ureotelic or uricotelic.

Ans 07.

The kidneys are considered as master chemists of the body. If they are damaged, it would disturb the normal functioning of the life processes. Due to the effect of toxins produced by some bacteria the fillers of tiny uriniferous tubules are damaged. They become perforated with larger holes and allow blood cells, proteins also to pass through them along with the urea and water during filtration of blood in formation of urine. Thus urine contains the blood proteins etc. It is a serious disease.

Ans 08.

Liver serve as digestive organ – It secretes bile; bile helps in the digestion of fats. Liver serve as excretory organ - It secretes following waste products in the bile; bilirubin, biliverdin (products of degradation of haemoglobin), cholesterol, inactivated steroid hormones, drugs, etc in the bile; these wastes are eliminated along with the digestive wastes or faecal matter.

Ans 09. The urinary system consist of following organs. (a) A pair of kidneys (b) A pair of ureters (c) Urinary bladder (d) Urethra kidney – The kidneys are located in the abdomen, one on each side of the vertebral column just below the diaphragm, which remains protected by last two pairs of ribs.. The left kidney is usually placed higher than right one. Each kidney is 10cm in length 5 cm in breadth and 4cm in thickness. Each kidney is somewhat bean shaped with concavity along the inner border. Blood vessels, nerves, lymph ducts and ureters enter the kidney at this point. In the gross anatomy of kidney, two regions can be clearly marked out. There are outer cortex and inner medulla. Internal structure of kidney – Each kidney composed of several tiny units called nephrons or uriniferous tubule, all similar in structure and function. Each nephron is made up of vascular component, the glomerulus and tubular component, and surrounded by a network of capillaries. The tubule is composed of single layer of epithelial cells which differ in structure and function in different parts of the tubule. The tubule originates as a blind sac, which is known as Bowman’s capsule which is lined by a single layer of thin epithelial cells – Bowman’s capsule is ultimately associated with vascular glomerulus which protrudes into Bowman’s capsule and is completely covered by the linning of the capsule. The glomerulus is formed by afferent and efferent arteries. The afferent arterioles bring blood to the tubular and efferent arterioles takes blood away from the tubule. Due to this, the blood in the glomerulus is separated from the space within the capsule only by (a) a thin layer of a tissue composed of the single – celled capillary lining (b) a layer material called basement membrane and (c) the single celled lining of Bowman’s capsule. This extremely thin barrier permits the filtration of the fluid from the capillaries into Bowman’s capsule. The glomerular capillaries combine together to form efferent arterioles which further divide into many capillaries distributed all over the surface of the tubule. These capillaries are termed as particular capillaries. These capillaries join together to form the venous channels which take blood away from the kidney. The tubule is divided into three pates – Proximal convoluted part in which Bowman’s capsule opens. The next part is Henle’s loop and last part is distal convoluted part which finally runs as a collecting duct. Ureters – These are two tubes or about 30 cm long – one coming out from each kidney. They run downwards and open into urinary bladder. Urinary bladder- It is a bag like structure where urine is stared. The bladder has three openings two of the ureters and one to urethra. Urethra – the urethra is the passage through which urine is passed out.

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (Locomotion and Movement)

1.

What is a tendon?

[1]

2.

Which type of movable joint makes the hip joint?

[1]

3.

Name the heaviest and longest bone in the human body?

[1]

4.

Why can a red muscle fiber work for a prolonged period, while a white muscle

[2]

fibre suffers from fatigue soon? 5.

What is the function of girdles?

[2]

6.

What makes the synovial joints freely movable? List any four types of

[2]

synovial joints.

7.

Differentiate between Endoskeleton and Exoskeleton.

[3]

8.

Explain the following –

[3]

a) Antagonistic muscles b) Tetanus c) Threshed stimulus 9.

Explain sliding filament theory of muscle contraction.

[5]

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (Locomotion and Movement) [ANSWERS] Ans 01.

Sarcomere.

Ans 02.

Inflammation of the joints.

Ans 03.

Left pectoral girdle – 2 ; Left arm – 30.

Ans 04.

i) ii) iii) iv)

Skeletal system forms the framework for the body. The bone of skeletal system protects delicate internal organs of the body. Skeleton provides attachment surface for the body muscles, tendons and other similar things and thus helps in movement It gives shape and posture to the body.

Ans 05.

The place of articulation of two or more bones are called as joints. At the joints, the surface of the two bones is in opposite to each other. These joints facilitate the movement of bones in different ways.

Ans 06.

Osteoporosis is an age – dependent systemic disorder, characterized by low bone–mass and micro–architectural deterioration of the bones. Factors – (1) Deficiency of calcium & vitamin D. (2) Imbalance of hormones like parathyroid hormone, thyrocalcitonin and sex hormones.

Ans 07.

Sarcoplasmic Reticulum – It releases calcium ions, that bind to the troponing and bring about conformational changes; so the active site on F – actin for myosin becomes exposed. Myosin head – It provides the specific binding site for F – actin to from cross – bridges, it also shows ATP are active site F – actin – F – actin are specific to myosin head for cross bridge formation.

Ans 08.

Types of muscle tissue – It is of 3 types (i) Striated / Skeletal muscles –
The muscle fibers are cylindrical, unbranched and show prominent striations.
They are innervated by voluntary nervous system.

(ii)

(iii)

Ans 09.


They are under the control of conscious mind and be moved at well.
Attached to skeletal system. Smooth muscles –
They are innervated by autonomic nervous system.
They are not under voluntary control. Cardiac muscles –
These muscles are found exclusively in the heart.
They are involuntary in function.

A nerve impulse arriving at neuromuscular junction stimulates contractile response. Due to the depolarization of the surface of sarcomeres it spreads quickly. Neurotransmitter is releases at the enormous collar junction. It enters into the sarcomere through membrane channel. Na+ moves inside the sarcomere. It is called inflow of Na+. Action potential is generated in the sarcomere. Action potential travels to the full length of muscle fiber. The sarcoplasmic reticulum, then release the Ca2+ which is stored here. It binds to the specific sites found in the hooping of the thin filament. Their filaments are called actins, change fakes in troponing active sites of F – actin are exposed then to myosin head. Myosin head shows Mg2+ dependent ATPase activity. During relaxation of muscle Ca2+ is pumped back into sarcoplasmic reticulum. Consequently the troponing component is treed to inhibit the active sites for myosin head. Cross bridges are broken. Their filaments assume their normal position. The muscle fiber is then in relaxed state.

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (Locomotion and Movement)

1.

Name the functional contractile unit of muscle.

[1]

2.

What is arthritis?

[1]

3.

What is the total member of bones present in the left pectoral girdle and the

[1]

left arm respectively in a normal human?

4.

List functions of skeleton in higher animals?

[2]

5.

Define a joint.

[2]

6.

What is osteoporosis? Name two factors which are responsible for

[2]

osteoporosis. 7.

Explain the initiation of muscle contraction. What is the role of sarcoplasmic

[3]

reticulum, Myosin head and F – actin during contraction in striated muscles?

8.

What are the three types of muscle tissue? Write two characteristic points

[3]

about the structure of each of them? 9.

What is the role of Ca++ and ATP in muscle contraction?

[5]

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Locomotion and Movement)

1.

Name the tissue which connects muscles to the bone?

[1]

2.

What is the function of myoglobin?

[1]

3.

What causes fatigue of muscle fibers?

[1]

4.

Which kinds of muscle fibers are richly found in the extensor muscles present on

[2]

the bact of human body? What characteristics enable those fibers to serve their purpose? 5.

Give differences between red and white muscle fibers, other than color.

[2]

6.

What are floating ribs? How many of them are there?

[2]

7.

Represent diagrammatically a sarcomere and label its parts. Which of these

[3]

parts shorten during muscle contraction?

8.

Describe any three disorders of the muscular system.

[3]

9.

Describe the various kinds of joint in human body. According to mobility

[5]

giving one example of each.

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Locomotion and Movement) [ANSWERS] Ans 01.

Tendon.

Ans 02.

Myoglobin stores oxygen in the muscles.

Ans 03.

Lactic acid.

Ans 04.

Red muscle fibers. Characteristics:- They are rich in mitochondria, myoglobin, slow acting, no lactic acid. Aerobic respiration takes place.

Ans 05.

Red muscles – Have more mitochondria. - Can contract for longer period. White muscles – Have less mitochondria. - Can contract for short period.

Ans 06.

Floating ribs – The last two pairs (11 and 12 pairs) ribs are called floating ribs. They are dorsally attached to the respective thoracic vertebrae and are free ventrally.

Ans 07.

z band (sarcomere)

cross bridge

Sarcomere contracts during muscle contraction

A myofibril Actin filament

Ans 08.

Myosin Filament

Disorders of muscular system (i) Myasthenia gravi’s It is an auto – minimum disorder, that affects the neuron muscular junction, leading to progressive weakening and paralysis of skeletal muscles.

(ii) Muscular dystrophy – It is a genetic disorder resulting in progressive degeneration of skeletal muscles. (iii) Tatany – It refers to the rapid spasm (wild contraction) or the continued state of contraction due to low Ca+ in the body fluids. Ans 09.

Various types of joints – When two or more bones articular with each other at a particular point it forms a joint. The joints help in performing various types of movements. In the body of vertebrates three types of joints are found – 1) Perfect joints – The joints have synovial joints are capable of performing movements in more than one plane. These joints may be of the following categories(a) Ball and socket joints – As the name suggests in this type of joint one bone forms a cup like depression of socket in which ball like structure fits. The head or ball can move freely can the joint in any direction. Ex – shoulder and hip joints. (b) Hinge joint – In this type of joint movement is performed only in one direction. Ex elbow, knee joint, joints of phalanges of fingers & toes. (c) Gliding joints – such are the joints in which one surface glides over another such joints are found in the vertebral column. (d) Pivot joint – One bones of the joint is always fixed and other is able to move freely over the former. Such joints can be seen in the skull of man which moves freely over the adontoid process the 2nd neck vertebra. (e) Saddle joints : It resembles ball & socket joint except that ball and socket are not fully developed. 2) Imperfect joint – the joints which do not possesses synovial capsule or connecting ligaments are called imperfect joints. Ex – the joint between the ilium of pelvic girdle and transverse process by sacral vertebra. 3)

Immovable joints:- Joints which are permanently fixed and cannot perform any movement are termed as immovable joints, these also do not posses synovial capsule of ligaments. Various bones of the skull are joined with such joints. These joints do not allow any kind of movements.

CBSE TEST PAPER-3 CLASS - XI BIOLOGY (Locomotion and Movement) [ANSWERS] Ans 01.

The tough non – elastic connective tissue that joins muscle to a bone.

Ans 02.

Ball & socket joint.

Ans 03.

Femur.

Ans 04.

Red muscle fibers contain myoglobin that stores oxygen in the form of oxymyoglobin since there is a continuous supply of oxygen; for oxidation of good materials to release energy, the red muscle fibers energy, and the red muscles fibres do not become fatigued and work for long periods. White muscle fibers lock myoglobin. They carry out anaerobic respiration and become fatigued.

Ans 05.

There are two girdles in the body, pectoral girdle & pelvic girdle. 1) Pectoral girdle – It provides surface to the soft organs of the body of the pectoral region. It also provides surface (glenoid cavity) for the articulation of forelimbs. In the glenoid cavity fits the head of the humerous bone. 2) Pelvic girdle – It protects the organs of pelvic region and provides surface (acetabulum) for the articulation of the hind limbs. (the femur bone of the thigh fits in the acetabulum)

Ans 06.

Synovial Joints – In the synovial joints not only, a space called synovial cavity is present between articulating bones. This cavity is filled with synovial fluid, that reduces the friction on the articulating surface of bones; so the synovial joints are freely movable. Synovial joints are of the following types: (i) Ball and socket joint (ii) Hinge joint (iii) Pivot joint.

Ans 07. 1. 2. 3.

Exoskeleton Made up of hard parts on the surface of body. Examples scales, feathers, hair, claws, hooves, nails and horns in vertebrates Formed by ectoderm.

Endoskeleton Made up of hard parts found inside the body. Cartilage and bones form endoskeleton in vertebrates. Formed by mesoderm.

Ans 08.

a)

b)

c)

Ans 09.

Antagonistic muscles – Contraction of muscles which results in the opposite movements at the same joint are called antagonistic muscle e. g biceps is a flexor for the elbow joint and triceps is its antagonistic. And an extension for that point. During flexion at the elbow biceps contracts and triceps relaxes; during extension triceps contracts and biceps relaxes. Tetnus – If a muscle fiber is stimulated by many nerve impulses or electric shocks it will remain in the state of contraction till the stimulation continued of contraction is known as tetnus. Threshold stimulus – Each skeletal muscle is made of many muscle fibers and each muscle fiber is supplied by a nerve. These nerves sends nerve impulse to the muscle fibers. As a result of this the muscle is stimulated and contraction of the muscle takes place. But for contraction muscle fibres requires a minimum strength of the nerve impulse. This is called threshold stimulus.

Sarcomeres are small units of myofibrils: the sarcomere consists of A – band in the centre with halves of two I – bands on its two sides i. e. the distance between two z – membranes. When stimulus is given to muscles, the thin (actin or I – band) filaments slide in the space between the thick (myosin or A – bond) but neither of them change its length. Due to sliding of I filaments, there is breakage and rearrangement of the cross – linkage between actin and myosin filaments. The ATP is broken by the enzyme ATPase myosin which provides energy for interaction between actin and myosin filaments. Consequently, the thin action filaments slide deeper into the A bands and z – lines are drawn closer with each other by the disappearance of H – zone finally the sarcomeres becomes shortened due to shortening of its I–band. By relaxation, the cross linkage between the filaments are rearranged. The cross–bridges disappear due to the pulling of the filaments by the active sites on the actin filaments. The actin filaments are slide out from the A – band. Consequently, this elongates I – band, pushing the z – line away form each other. Thus contrition and relaxation of muscles occurs due to repetitive formation and breakage of cross bridges between thick filament of A – band and thin filament of I– band.

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (Neural Control and Coordination)

1.

How does an impulse travel across a synapse?

[1]

2.

How many pairs of cranial nerves are present in man?

[1]

3.

What is saltatory conduction?

[1]

4.

What is a reflex?

[2]

5.

What happens when the membrane of a nerve cell carries out a sodium pump?

[2]

6.

What are the events that take place at the point of stimulation of axon?

[2]

7.

Differentiate between dorsal spinal roots and ventral spinal roots.

[3]

8.

Describe human neural system.

[3]

9.

Draw a labeled diagram to show the structural view of human ear in the

[5]

sectional view.

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (Neural Control and Coordination) [ANSWERS] Ans 01.

The impulse travel across a synapse from the axons to the cell body and dendrites to the next neuron.

Ans 02.

12 pairs.

Ans 03.

Saltatory conduction refers to the type of conduction of nerve impulse by a myelinated nerve fibre, where the action potential jumps form one node of Ranvier to the other.

Ans 04.

Reflex is an involuntary action performed by muscle under the direction of spinal cord in response to the stimulus. It is an automatic response to a stimulus which is not under conscious control. A large number of activities of animals are conducted by reflexes e. g. Respiration, peristalsis, watering of the mouth, secretion of saliva in the mouth, etc.

Ans 05.

When the membrane carries a sodium pump, it carries three sodium ions from the axoplasm to the cell exterior: - It transfers two potassium ions exchange from the ECF to the cell interior. - The exterior is positively charged.

Ans 06.

At the point of stimulation the membrane permeability changes; it becomes freely permeable to Na+ ions. There is a rapid inflow of Na+ ions and the interior / axoplasm becomes positively charged and the exterior becomes negatively charged. This condition is known as depolarized state and the potential difference across the membrane is known as action potential. Now the current flows through the axoplasm from the depolarized region to due next polarised region and through the ECF from the polarised region to the depolarised region.

Ans 07. Dorsal spinal Roots 1. They are made of sensory (afferent) nerves. 2. They have dorsal root ganglia. 3. Their cell bodies are located in dorsal root ganglia.

Ventral spinal Roots They are made of motor (efferent) nerves. They have no ganglia. The cell bodies of ventral spinal nerve root is located in ventrolateral horn of grey matter.

Ans 08.

Ans 09.

It is divided into two parts1) Central Neural system (CNS) – CNS includes brain and spinal cord. This is the site of information processing and control. 2) Peripheral neural system (PNS) – PNS consists of all nerves of the body associated with the CNS. Nerve fibers of PNS are of two types i.e. afferent fibers and efferent fiber. (a) Afferent nerve fibers transmit impulses from tissues / organs to CNS. (b) Efferent nerve fibers transmit impulses from CNS to concerned peripherel tissues / organs. PNS is further divided into – (1) Somatic neural system – It relays impulse from CNS to skeletal muscles. (2) Autonomic neural system – ANS transmits impulses from CNS to involuntary organs as well as the smooth muscles of body It is again divided into two parts a) sympathetic neural system b) Para sympathetic neural system.

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Neural Control and Coordination)

1.

Name the band of nerve fibers that joins the two cerebral hemisphere in

[1]

mammals.

2.

What is threshold stimulus for nerve cell?

[1]

3.

What is a compound eye?

[1]

4.

Give parts of neuron.

[2]

5.

Describe the role & location of ciliary body in human eye.

[2]

6.

What is mosaic vision?

[2]

7.

Why do giant squids have very thick nerve fiber?

[3]

8.

Where are synaptic vesicles found? Name their chemical contents? What is

[3]

the function of these contents?

9.

What is meant by the resting membrane potential of neuron. How do ion channels & sodium – potassium pumps contribute to the resting potential?

[5]

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Neural Control and Coordination) [ANSWERS] Ans 01.

Corpus callosum.

Ans 02.

The minimum intensity / strength of a stimulus required to initiate depolarization of neuron is called threshold stimulus.

Ans 03.

In insects the eye is composed of many independent visual elements called commatidia such an eye is called compound eye.

Ans 04.

Ans 05.

Neuron is a microscopic structure made up of 3 partsa) Cell body – In contains cytoplasm with typical cell organelles and some granular bodies called Nissl’s granules. b) Dendrites – The short fibers that branch repeatedly and project out of the cell body. They transmit impulse towards the cell body or cyton. c) Axon – It is a long fibre. Its distal end is branched. Each branch terminates into bulblike structure called as synoptic knob. Location:Function:

The choroid becomes thick where the cornea & sclera meet; It is called ciliary body. The ciliary body continues in front of the lens to form an opaque structure called iris.

Ans 06.

This type of vision is found insects due to compound eye. A complete image of the object as seen by the compound eye is formed by a number of small lineages each of which is contributed by an ommatidium. Such an image formed by many bits of images is called a mosaic image and the vision as the mosaic image vision.

Ans 07.

The velocity of a nerve impulse in a nerve fiber depends on two factors i. e. on its myelinated and also on the thickness of the fibers. The impulses travel faster in thicker nerve fibers since giant squids are very large sized aquatic animals they have thick nerve fibers.

Ans 08.

Synaptic vesicles are found in the bulbous expansion called synaptic knob, at the nerve terminalEach synaptic vesicle contains as many as 10,000molecules of a neurotransmitter substance that is responsive for transmission of nerve impulse across the synapse. When a wave of depolarization reaches the presynaptic membrane, the voltage gated calcium channels concentrated at the synapse open & Ca++ ions diffuse into the terminal form the surrounding fluid. - The Ca++ = ions stimulate the synaptic vesicles to move to the terminal membrane, fuse with it and then and then rupture by exocytosis into the cleft. - This neurotrarmitter diffuses across the synapse and stimulates the membrane of the next neuron.

Ans 09.

Resting membrane potential. - The electrical potential difference across the membranes of a resting neuron is called resting membrane potential. - The membrane is polarized, with a negative interior and positively charged exterior. - The permeability of membrane to K+ ions is greater than its permeability to Na+ ions. - The negatively charged protein molecules can cross the membrane. - The sodium pump transports 3 Na+ ions to the exterior, while in exchange only 2K+ ions comes inside. - Hence the surface carries a positively charge, which the interior negatively charged.

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (Neural Control and Coordination)

1.

What types of neurons are found in dorsal root of spinal nerve?

[1]

2.

What is the basic unit of neural system?

[1]

3.

Why is blind spot devoid of the ability for vision?

[1]

4.

Where does cerebrospinal fluid occur in our body? Mention two if its function.

[2]

5.

What is the chemical and difference between rods & cones?

[2]

6.

Why are gray matter and white matter contained in human nervous system

[2]

named so? 7.

Give the location and function in the human eye, of the following –

[3]

(i) cornea (ii) Iris (iii) Vitreous humor 8.

Why are nerve impulses conducted more rapidly in myelinated nerve fiber than

[3]

in a non – myelinated one? Explain. 9.

Taking one example, describe the functioning of the various components of a spinal reflex arc.

[5]

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (Neural Control and Coordination) [ANSWERS] Ans 01.

Sensory neurons.

Ans 02.

Neuron.

Ans 03.

Blind spot has no photoreceptor cells – rods or cones.

Ans 04.

Cerebrospinal fluid is found in the subarachnoid space between arachnoids and parameter of the menings around the brain and spinal cord and also in the cavities of the brain. Functions – 1) It protects brain and spinal card by acting as a cushion to absorb shocks. 2) It helps in removing harmful metabolites drugs etc. away from the brain.

Ans 05. 1. 2. 3.

Rods These are more secretive to light and are meant for vision in dim light. They do not have the ability to make colored image They contain the visual pigment rhodopsin.

Cones These are meant for vision in bright light. They have ability to make colored image. These contain the pigment iodopsin.

Ans 06.

Gray matter contains spindle, pyramidal, cell bodies with grayish brown appearance and hence called as gray matter. White matter contains millions of myelinated axons; the large amount of myelin gives this tissue an opaque white appearance and hence called white matter.

Ans 07.

1)

2)

3)

Cornea – It is the dome – shaped part of sclera that is transparent and more curved. Function – It refract light towards retina. Iris – It is the colored (pigmented) at front and formed by choroid. Functions :- (i) It encloses pupil. (ii) Iris contains cilliary muscles which regulate the size of pupil and controls the amount of light. Vitreous humor – It is present in posterior chamber of eye. Functions:- (i) It helps in shape to the eye & supports retina & lens. (ii) It refracts the light rays.

Ans 08.

In a myelinated nerve fiber, the lipid rich myelin acts as an insulator and depolarization occurs in the nodes of Ranvier where myelin sheath is absent. Since the action potential jumps from one node of Ravines to another, the conduction becomes faster and such a type of conduction is called saltatory conduction. In a non–myelinated fiber, the depolarization occurs all along its length and hence conduction becomes slower.

Ans 09.

A Reflex arc is the specific neural pathway from stimulus to reflex. Components of Reflex arc are – (1) Receptors – These are the organs / tissues which receive the stimulus and send it as an impulse. (2) Sensory or afferent nerves – These are neurons which conduct the impulse from the receptor to the central Nervous system (spinal cord) (3) Relay or intermediate neurons – They are neurons which conduct the impulse from the afferent neurons to the efferent neurons. (4) Effectors / motor neurons – These neurons conduct the impulse from the spinal cord/ relay neurons to the effectors organ concerned. (5) Effectors – It is the organ / tissue or gland that functions accordingly.

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (Chemical Co – Ordination and Integration)

1.

What are hormones.

[1]

2.

Name the gland of emergency.

[1]

3.

Which gland secrete glucagon?

[1]

4.

Differentiate hormone & neurohormone?

[2]

5.

What are gonadotropics?

[2]

6.

Why oxytocin is called as ‘birth hormone’?

[2]

7.

Describe the physiological functions & disorders of thyroid gland.

[3]

8.

Write full form of ADH and describe how it affects the functioning of kidney

[3]

tubules. 9.

Name the hormone that regulates each of the following and mention the source of it. 1) urinary elimination of water. 2) storage of glucose as glycogen. 3) Na+ and K+ metabolism. 4) Basal metabolic rate 5) Descent of testes iota scrotum

[5]

CBSE TEST PAPER-01 CLASS - XI BIOLOGY (Chemical Co – Ordination and Integration) [ANSWERS] Ans 01.

Hormones (endocrines) are the secretions of endocrine glands.

Ans 02.

Adrenals.

Ans 03.

Pancreas.

Ans 04. 1. 2.

Hormone Secreted by endocrine glands. It stimulates the growth & metabolism of endocrine glands and body cells.

1. 2.

Neurohormone Secreted by neuro – secretary cells. It stimulates the secretion of hormones of pituitary.

Ans 05.

These are the gonad stimulating hormones secreted by the anterior lobe of pituitary e.g.- follicle stimulating Hormone and prolactin.

Ans 06.

Oxytocin causes the contraction of smooth muscles of uterus during child birth. So it is called ‘birth hormone’.

Ans 07.

Thyroid gland consists of a two lobed structure in the region of larynx. It secretes the hormone thyroxin which contains Iodine the thyroxin stimulates the rate of cellular oxidation and control the basal metabolic rate. It also maintains balance of the Ca++ in blood thyroid hormones also promotes growth of body tissues both physical growth and mental development are stimulated. They stimulate tissue differentiation because of this action they promote metamorphosis of tadpoles into adult frogs. Thyroid secretes the hormones which stimulate all metabolic actions. They are controlled by hormones secreted by anterior pituitary gland. Disorders – 1) Hyperthyroidism :- It reduces the basic metabolic rate 2) Cretinism:- The delayed growth (mental, bodily and sexual). The patient is pot – bellied and pigeon – cheated and has a protruding tongue. 3) Myxodema (Adult) :- Dry coarse skin, loss of hair, reduced cerebration, temperature and pulse rate, slowed speech. The patient gains weight, reproductive failure and has a puffy appearance and lacks alertness.

Ans 08.

ADH – Anti diuretic hormone It affects kidney tubules in following ways – 1) It renders the distal convoluted tubule, collecting tubule and collecting duct of the nephrons permeable water so that water is reabsorbed from filtrate in these segments and urine becomes hypertonic. 2) It also regulates the arterial blood pressure.

Ans 09. 1. 2. 3. 4. 5.

Activity Urinary elimination of water. Storage of glucose as glycogen Na+ & K+ metabolism Basal Metabolic Rate Descends of testis into scrotum

Hormone ADH

Source Posterior pituitary

Insulin Glucagon Aldosterone Thyroxin, Triidothyroxin FSH

Islets of langerhans Adrenal cortex Thyroid gland Anterior pituitary

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Chemical Co – Ordination and Integration)

1.

Distinguish between diabetes mellitus and diabetes insipidus.

[1]

2.

Name the hormones of fight or flight.

[1]

3.

Name the hormone secreted from outermost cellular layer of adrenal cortex?

[1]

4.

What usually can cause over secretion of parathormone in human body? List

[2]

any two effects on the body because of this hormone. 5.

What is the function of pineal gland?

[2]

6.

Explain the hormones of kidney and GI tract.

[2]

7.

Differentiate between exocrine, endocrine & heterocrine glands.

[3]

8.

Name the T3 and T4 components of thyroid hormone. Explain their specific

[3]

function.

9.

Explain the Hormones of adrenal gland and their action on target tissue in a tabular from.

[5]

CBSE TEST PAPER-02 CLASS - XI BIOLOGY (Chemical Co – Ordination and Integration) [ANSWERS] Ans 01.

Diabetes mellitus is caused due to less secretion of Insulin by β cells of Islet of langerhans in pancreas. Diabetes insipidus is caused due to less secretion of ADH (vasopressin) by posterior pituitary gland.

Ans 02.

Adrenaline and nor – adrenaline.

Ans 03.

Aldosterone, a mineral ocorticoid.

Ans 04.

A tumors in parathyroid glands causes the over secretion / hypersecretion of parthormone. Due to demineralization, the bones become deformed and are early fractured. If untreated, it can lead to osteitis fibrosa cystica disease in human beings.

Ans 05.

It secretes a hormone the melatonin. It reduces the reproductive activity and may also delay the sexual development in an individual.

Ans 06.

Kidney – Juxtaglimerular cells of kidney secrete a peptide hormone called erythropoietin. It stimulates erythropoiesis or formation of RBC’s of bloodG – I tract – The endocrine cells found in various parts of gastro-intestine tract secrete 4 peptide hormones – Gastrin, secretin, cholecystokinin (CCK) as well as gastric Inhibitory peptide or GIP.

Ans 07. 1. 2.

Ans 08.

Exocrine glands It has a duct

Endocrine glands It is ductless gland.

Their secretions are carried by the ducts to the internal parts or body surface e.g salivary gland in mouth.

Their secretions are carried by blood to the target organs e.g. Parathyroid, pituitary and adrenals.

Heterorine glands It is partly endocrine & partly exocrine Endocrine part releases hormones into blood stream while exocrine part into ducts associated with it e.g. pancreas, ovary’s, testis.

T3 = Thyroxin. It contains 4 atoms of iodine. T4 = Triiodothyroxine. It has 3 atoms of iodine T3 and T4 have identical effects on target cells. They are called together as TH (Thyroid hormone) They : 1) regulate metabolic rate 2) regulate metabolism 3) help in metamorphosis of frog.

Ans 09. Endocrine glands & Hormones 1. Mineral corticoids (Aldosterone)

2. Glucocorticoids (corticosol corticosterone & cortisone)

3. Gonadocorticoids (Androgens and estrogens)

4. Adrenaline

5. Nor adrenaline Hormone

Principal Action They control electrolyte and water metabolism. The increase blood level of Na+ and water. They decrease blood levels of K+ by stimulating kidney tubules to reabsorb more Na+ Cl- and water and less K+. They raise blood glucose level. They promote gluconeogenesis and also promote liver glycogen formation and breakdown of plasma proteins. They increase availability of amino acids for enzymes synthesis by liver general resistance to long term. Stress counter inflammatory and allergic responses, and decreases antibody production. Concentrations secreted by adults are low. Their effects are usually insignificant. They stimulate development of secondary sexual characteristics specially in males. Stimulates elevation of blood glucose by converting liver glycogen to glucose, hormone. Rise in blood pressure acceleration of rate and force of heart beat, constriction of skin and visceral smooth muscle capillaries muscles, dilation of arterioles of heart and skeleton increase in breakdown of lipids Increase in oxygen consumption erection of hairs, dilation of pupils. They initiate stress responses. It stimulates reactions similar to those produced by adrenaline.

Target tissue Kidney tubules

Liver

Gonads

Skeletal muscles fat cells, cardiac muscles, smooth muscles, blood vessel.

-

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (Chemical Co – Ordination and Integration)

1.

What is the function of Leydeig’s cells?

[1]

2.

Name the gland which secrets vasopressin.

[1]

3.

Name one mineralocorticoid.

[1]

4.

In general, how steroid hormones do effects changes in their target cells.

[2]

5.

What is corpus luteum? How does it function as a endocrine gland?

[2]

6.

Name the gland that functions as a biological clock in our body where it is

[2]

located? Name its one secretion. 7.

Differentiate between vitamin, hormone & enzyme.

[3]

8.

A patient was complaining of frequent urination, excursive thirst, hunger and

[3]

tiredness. His fasting glucose level was found higher than 130 mg / dL an two occasions : (i) Name the disease (ii) Give the root cause of this disease (iii) Explain why the blood glucose level is higher than 130 mg / dL. 9.

Explain the mechanism of hormone action.

[5]

CBSE TEST PAPER-03 CLASS - XI BIOLOGY (Chemical Co – Ordination and Integration) [ANSWERS] Ans 01.

Leydig cells or interstitial cells of the testes secretes testosterone hormone. It stimulates the development of external male sex characters such as beards, moustaches and low pitch voice is man & stimulates the formation of sperms in testis.

Ans 02.

It is secreted by posterior part of the pituitary gland.

Ans 03.

Aldosterone.

Ans 04.

Steroid hormones are lipid soluble. These quickly pass through plasma membrane of a target cell into the cytoplasm. There they bind to intercellular receptor proteins and form a complex. This complex enters the nucleus and binds itself to specific regulatory sites on the chromosomes. This binding changes gene expression and stimulates transcription of same genes. It may repress some other genes. Finally in RNA acts for protein synthesis. The lipid soluble hormones are slow in action. They are last longer hormones.

Ans 05.

Corpus luteum is the structure formed by the ruptured ovarian follicle after ovulation. - It secrets the hormone progesterone, which is necessary for pregnancy changes.

Ans 06.

Pineal gland functions as biological clock in our body. Location – It is located on the dorsal side of the forebrain. Secretion – It secretes melatonin.

Ans 07. 1. 2. 3. 4.

Vitamin It is carried in the food.

Hormone It is carried by the blood. It is used up during the It is consumed during process. the metabolic reaction. It is obtained from food. It is produced by an endocrine gland. It may be organic acid, It is glycoprotein, amide, amine, ester, steroid or polypeptide. alcohol or steroid.

Enzyme It is not carried by the blood. It remains unchanged after the reaction. It is produced by exocrine gland. It is always proteinaceous in nature.

5. 6.

It act as coenzyme.

It act as a stimulating It act as a biocatalyst. substance. Its deficiency causes It excess as well as It is required in small deficiency diseases. deficiency causes many amount. hormonal disorders & diseases.

Ans 08.

(i) (ii) (iii)

The disease is diabetes mellitus. It is caused by under secretion of insulin resulting in hyperglycemia. In the absence of insulin, the following functions are impaired. - utilization and uptake of glucose by adipocytes and hepatocytes. - Conversion of glucose into glycogen by the above target cells.

Ans 09.

Upon the target cells, two main kinds of hormone action have been observed i) action at meanbrane level ii) Induction of protein synthesis at gene level. 1)

Hormone acts as first messenger : It is attached to some integral protein at specific receptor site on the surface of cell membrane (of target cell). It stimulates adenyl cylase (enzyme). It catalyses conversion of ATP to cyclic AMP which acts as second messengers). It affects cell metabolism.

2)

Gene activation – The steroid hormone enters into the cytoplasm through cell membrane and binds to protein receptors there of the target cells. This hormone receptors complex stimulates the gene to synthesize a particular enzyme. Fig : Mechanism of hormone action