Laboratory Manual Biology not to be republished · PDF fileInvestigatory Project Work Project 1 : To study the effect of pH on seed germination Project 2 : ... Biology, like any other

Laboratory Manual

BiologyClass XII

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FOREWORD

The National Council of Educational Research and Training (NCERT) is theapex body concerning all aspects of refinement of School Education. It hasrecently developed textual material in Biology for Higher Secondary stagewhich is based on the National Curriculum Framework (NCF)–2005. The NCFrecommends that children’s experience in school education must be linkedto the life outside school so that learning experience is joyful and fills the gapbetween the experience at home and in community. It recommends to diffusethe sharp boundaries between different subjects and discourages rote learning.The recent development of syllabi and textual material is an attempt toimplement this basic idea. The present Laboratory Manual will becomplementary to the textbook of Biology for Class XII. It is in continuationto the NCERT’s efforts to improve upon comprehension of concepts andpractical skills among students. The purpose of this manual is not only toconvey the approach and philosophy of the practical course to students andteachers but to provide them appropriate guidance for carrying outexperiments in the laboratory. The manual is supposed to encourage childrento reflect on their own learning and to pursue further activities and questions.Of course the success of this effort also depends on the initiatives to be takenby the principals and teachers to encourage children to carry out experimentsin the laboratory and develop their thinking and nurture creativity.

The methods adopted for performing the practicals and theirevaluation will determine how effective this practical book will proveto make the children’s life at school a happy experience, rather thana source of stress and boredom. The practical book attempts to providespace to opportunities for contemplation and wondering, discussionin small groups, and activities requiring hands-on experience. It ishoped that the material provided in this manual will help studentsin carrying out laboratory work effectively and will encourage teachersto introduce some open-ended experiments at the school level.

YASH PAL

Professor and ChairpersonNational Steering Committee

National Council of EducationalResearch and Training

New Delhi21 May 2008

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PREFACE

The development of the present laboratory manual is in continuation tothe NCERT’s efforts to improve upon comprehension of concepts andpractical skills among the students. The present laboratory manual willbe complementary to the textbook of Biology for Class XII.

The expansion of scientific knowledge and consequently the changein the system of education has led to the development of new methods ofinstructions. Today the stress is laid on the enquiry approach anddiscussion method instead of lecture method of teaching. Biology is nowsomething more than observation of living organisms. Study of Biologyincludes microscopic observations to reveal minute internal details of theorganism, biochemical testing to understand complex reactions takingplace inside the organisms, experiments with live organism to understandvarious physiological processes and even much more. In other wordsexperiments in Biology truly represents an interdisciplinary approach oflearning.

The new syllabus of Biology has been designed to cater to the needs ofpupil who are desirous of pursuing science further. The fundamentalobjective of this course is to develop scientific attitude and desiredlaboratory skills required for pursuing Biology as a discipline at this level.A similar approach has been taken while formulating the practical syllabusof Biology for higher secondary stage. The practical syllabus includescontent based experiments, which help in comprehension of the concepts.There are altogether twenty-five exercises in the present manual whichare based on Biology curriculum for Class XII. For each practical work,principle, requirements, procedure, precautions, observations, discussionand the questions are given in the book. The methodology of preparationof any reagent, if required, has been given alongwith the requirements,for the convenience of students and teachers. The questions are aimed todevelop learner’s understanding of the related problems. However, teachermay provide help in case the problem is found to be beyond the capabilityof the learner. Precautions must be well understood by the learners beforeproceeding with the experiments and projects. In addition to the coreexperiments enlisted in the syllabus for Class XII emphasis has also beengiven for pursuing Investigation Project Work. It is expected that theseinformations will motivate the students to take up independent projectwork on topics of their interest.

Appropriate appendices related to the observation and study oforganisms are given along with the experiment. International symbols forunits, hazards and hazard warnings are given at appropriate places inthe book. It is expected that this will make the learners more careful aboutthe environment and make them careful while dealing with the equipmentsand chemicals in the laboratory.

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It gives me a pleasure to express my thanks to all those who have beenassociated at various stages of development of this laboratory manual. It ishoped that this practical book will improve teaching-learning process inBiology to a great extent. The learners will be able to understand the subjectwell and will be able to apply the acquired knowledge in new situations.I acknowledge with thanks the dedicated efforts and valuable contributionof Dr Dinesh Kumar, coordinator of this programme and other team memberswho contributed and finalised the manuscript. I especially thank ProfessorG. Ravindra, Director (Incharge), NCERT for his administrative support andkeen interest in the development of this laboratory manual. I am also gratefulto the participating teachers and subject experts who participated in thereview workshop and provided their comments and suggestions which helpedin the refinement of this manual. We warmly welcome comments andsuggestions from our readers for further improvement of this manual.

HUKUM SINGH

Professor and Head

Department of Education in

Science and Mathematics

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MEMBERS

Animesh K. Mohapatra, Associate Professor, Regional Institute of Education,

NCERT, Ajmer

B.K. Tripathi, Professor, DESM, NCERT, New Delhi

C.V. Shimray, Assistant Professor, DESM, NCERT, New Delhi

N.V.S.R.K. Prasad, Associate Professor in Botany, Sri Venkateshwara College,

New Delhi

P.K. Durani, Professor (Retired), DESM, NCERT, New Delhi

Sunita L. Varte, Assistant Professor, DESM, NCERT, New Delhi

S.P. Sinha, Professor of Zoology (Retired), TM Bhagalpur University, Bhagalpur

V.V. Anand, Associate Professor , Regional Institute of Education, Mysore

MEMBER-COORDINATOR

Dinesh Kumar, Associate Professor, DESM, NCERT, New Delhi

LABORATORY MANUAL DEVELOPMENT TEAM

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The National Council of Educational Research and Training (NCERT) acknowledges the valuable

contribution of the individuals and organisations involved in the development of this laboratory

manual. The council also acknowledges the valuable contribution of the following academics for

reviewing and refining the manuscript of the laboratory manual: A.K. Sharma, Reader in Zoology,

University of Lucknow, Lucknow; Iswant Kaur, D.M. School, RIE, Bhopal; K. Muralidhar,

Professor, Department of Zoology, University of Delhi, Delhi; K.K. Sharma, Professor Department

of Zoology, M.D.S. University, Ajmer; M.M. Chaturvedi, Professor Department of Zoology,University of Delhi, Delhi; Nazir Ahmad Kakpori, Department of Education, Govt of Jammu &

Kashmir, Srinagar; Reena Mohapatra, St. Stephen’s Senior Secondary School, Ajmer; Savita

Sharma, Mount Carmel School, Dwarka, New Delhi; Savithri Singh, Professor and Principal,

Acharya Narendra Dev College, New Delhi; Shalu Dhawan, Amity International School, Saket,

New Delhi; Shivani Goswami, Mother’s International School, New Delhi; V.K. Srivastava, Reader

in Zoology, J.N. College, Pasighat; Vijay Kumar, Delhi State Science Teacher Forum, New Delhi.

We also acknowledge the contributions of Anil Kumar and Binita Kumari, Junior Project

Fellows, DESM, NCERT, New Delhi.

Special thanks are also due to Hukum Singh, Professor and Head, DESM, NCERT for his

interest in the work and administrative support.The Council also acknowledges the efforts of Surender Kumar, Narender Kumar Verma, Monika

Rajput and Girish Goyal, DTP Operators, for helping in shaping this laboratory manual. The

contributions of Publication Department of NCERT in printing out this laboratory manual are

also duly acknowledged.

ACKNOWLEDGEMENT

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FOREWORD

PREFACE

Introduction

Exercise 1 : To study the reproductive parts of commonlyavailable flowers

Exercise 2 : To calculate percentage of pollen germination

Exercise 3 : To study pollen tube growth on stigma

Exercise 4 : To study the discrete stages of gametogenesis inmammalian testis and ovary

Exercise 5 : To study and identify various stages of femalegametophyte development in the ovary of a flower

Exercise 6 : Preparation and study of mitosis in onion roottips

Exercise 7 : Study of stages of meiosis using permanent slides

Exercise 8 : To study the blastula stage of embryonicdevelopment in mammals, with the help ofpermanent slide, chart, model or photograph

Exercise 9 : To verify Mendel's Law of Segregation

Exercise 10 : To verify the Mendel’s Law of IndependentAssortment

Exercise 11 : Preparation and analysis of Pedigree Charts

Exercise 12 : To perform emasculation, bagging and taggingfor controlled pollination

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CONTENTS

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Exercise 13 : Staining of nucleic acid by acetocarmine

Exercise 14 : To identify common disease-causing organismsand the symptoms of the diseases

Exercise 15 : To study the texture of soil samples

Exercise 16 : To determine the water-holding capacity of soils

Exercise 17 : To study the ecological adaptations in plants livingin xeric and hydric conditions

Exercise 18 : To study the adaptations in animals living in xericand hydric conditions

Exercise 19 : To determine the pH of different water and soilsamples

Exercise 20 : To study turbidity of water samples

Exercise 21 : To analyse living organisms in water samples

Exercise 22 : To determine the amount of SuspendedParticulate Matter (SPM) in air at different sites ina city

Exercise 23 : To study plant population density by quadratmethod

Exercise 24 : To study plant population frequency by quadratmethod

Exercise 25 : Study of homologous and analogous organs inplants and animals

Investigatory Project Work

Project 1 : To study the effect of pH on seed germination

Project 2 : Quantitative analysis of phytoplankton in a waterbody

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Laboratory is a place where ideas and concepts can be tested throughexperiments. Biology, like any other discipline of science, is based onexperimental work and therefore practical forms an integral part of learning.Biology laboratory provides a unique learning environment where learnersinculcate scientific temper, develop relevant skills and get exposed to realmsof techniques and methodologies of scientific investigations. Laboratoryinvestigations in Biology increase the reasoning abilities, bring scientificattitude in a learner and also help in acquisition of skills of scientific processes.Also, observation of nature and the living organisms found in it is no lessimportant for the understanding of many aspects of the subject especiallythe diversity of the living organisms, their systematic study, their relationshipsamong themselves and with the environment. Knowledge in the field ofBiology can be acquired or constructed only on the basis of correctobservations and experimentally verifiable processes.

Biology laboratory thus provides the learners an environment where theprocess of learning is facilitated by hands-on experiments. Biology is aunique discipline in the sense that it does not merely deal with the study ofmorphology, anatomy, physiology and reproduction of the living organisms,rather, understanding of the subject requires understanding of a number ofinterdisciplinary areas and approaches. On one hand, a biologist needs tobe sufficiently skilled in handling the enormous diversity of the livingorganisms, be it plants, animals, fungi or even microscopic bacteria, whileon the other hand, a biologist should be able to understand the biochemical,molecular, physiological, behavioural, genetic and many other phenomenapertaining to the living organisms. The study of intricate relationship ofdifferent types of organisms among themselves and also with its environmentis an important concern of a biologist. Thus, experiments and exercises inBiology train a learner about skills of observations, manipulation of theorganisms for the study of internal details, biochemical as well as molecularcomposition and processes, investigation of the abiotic environment and even

analysis of phenomena like inheritance and evolution.As far as the study of the living organism is concerned, correctness of the

method is very important. Such a study may be very simple, e.g., study ofhabit, habitat and external features of the plants or animals, or, it may involvecertain manipulations like dissection and section cutting of the parts of theorganisms to study the minute details. Very often observation and study ofthe magnified image of the minute parts under a microscope provides abetter insight about the features of the organisms. However, microscopicstudy involves certain specific skills depending on type of the organisms/tissues/cells to be studied. It involves specific preparations (peeling, section

Introduction

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LABORATORY MANUAL: BIOLOGY

cutting, fixation, staining, dehydration, mounting, etc.) so that microscopicexamination reveals the expected details. As histological and cytologicalobservations give us only static pictures of the continuous processes, analysisof biochemical, physiological and ecological aspects need certain other kindsof skills such as preparation of chemicals and reagents, designing andperforming an experiment, observation and recording of data and ultimatelyinterpretation and drawing conclusions. While performing experiments,honesty in recording of data and its correct presentation is very importantas it is not only useful in the logical interpretation but also helps in theidentification of errors.

In order to perform experiments successfully, a learner needs to go tothe Biology laboratory well prepared. This includes the following:

1. Laboratory Record Book: For maintaining all the information includingrecording of data and its interpretation.

2. Dissection Box: A dissection box is required in the Biology laboratoryfor various purposes like handling and manipulation of living materials,performing experiments, preparation of slide, etc. A dissection boxshould contain scissors (two pairs, one small with fine tip and onelarger), scalpels (one small and one medium sized), forceps (two, onesmall with sharp fine tips and the other medium sized with blunt tips),dissecting needles (two), razor, hand lens, dropper, fine brush, etc.

3. Laboratory Manual

4. A Laboratory Coat or Apron

5. A Hand Towel

While in the laboratory a student should be very careful and methodical.One should listen carefully to the instructions given by the teacher/instructor before performing an experiment. In the biology laboratory astudent has to handle a number of sharp objects and hence necessaryprecaution and care should always be taken while handling objects likescissors, forceps, needles, scalpel, razor, etc. It is also very important tofollow the safety instructions mentioned on the instruments and/or on thelabel of the reagent/chemical. Students should also be aware about theuse of the First-aid Box so that in case of any accident or injury thepreliminary aid can be provided to the affected person.

While describing the experiment students are expected to follow a patternin which the aim of the experiment, its principle, list of the materials to beused, procedure, observation table (if required), inference and discussionshould be given. Necessary precautions to be taken should also bementioned appropriately in the procedure or at the end. There are a fewexperiments in which field visit is essentially required. For this all thenecessary preparations (materials, equipments, reagents and chemicals)

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EXERCISE 1

should be made in advance. Drawing of illustrations is also an importantcomponent of the practical in Biology. Students are expected to follow certainfundamental rules while drawing the illustrations so that it reflects yourobservations correctly.• Make your illustrations using pencil only and always use white drawing

sheet. Illustration should be in the centre of the page.• Drawing of an object (plant, animal or experimental set-up) should be

proportionate in size.• Draw your illustrations keeping the object before you.• Drawing must be clear with simple outlines.• Appropriately label your drawing. Parts of the drawing should be

indicated by straight horizontal line or arrow. Two lines or arrow shouldnever cross each other. As far as possible, labelling should be done onthe right side of the drawing. An appropriate legend or heading of thedrawing should also be given below it.

About the Manual

The main objective of the manual is to introduce the students of highersecondary stage to the fascinating world of plants, animals and microbesand their complex biological phenomena. The manual covers a completedescription of the experiments and exercises. The suggested experimentscover almost all the units/topics including those on diversity in living world,plant, animal and human physiology, genetics, bio-technology and humanwelfare and environment. A standard format has been used to describe eachexperiment which includes

• Aim: It gives a brief title of the experiment under investigation.

• Principle: It is a very brief introduction of the experiment underinvestigation and explains the biological phenomenon involved. It givesbrief but comprehensive ideas about the design of the experiment andexplains the significance of the phenomenon being studied.

• Materials required: This includes the names of plants/animals to beused as 'samples', the type of apparatus, the type and quantity ofglasswares required, reagents, chemicals and solutions needed, theirconcentration and other specifications, method of preparations ofsolutions and reagents. If a particular material/chemical/glassware isnot available, sufficient alternatives have been suggested.

• Procedure: This section includes full details of experimental procedureexplained stepwise, including special precautions necessary to be takenwhile the experiment is being conducted. Drawings of the samples,apparatus and the experimental setup, wherever found necessary, havebeen included to facilitate the students to perform the experiment asaccurately as possible.

INTRODUCTION

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• Observation and Results: This section deals with the recording of allobservations made during the experiment. Students are advised toconsider the entire data. Data can be represented in the form of tables,graphs and histograms wherever possible. Use of units in which variousquantities are measured has been indicated in the manual.

• Discussion: Included in this heading is a statement of the conclusions

drawn from the experimental results and compared thesis (whereverpossible) with any comparable data from other sources. The relevanceof the conclusions drawn from the experimental results to the variousprocesses under investigation and to the life of plant, animal and microbeshas been prompted out.

• Precaution: This section contains all the necessary precautions to be

taken during experimentation to obtain results free of errors. However,attempts have been made to mention required precautions along withthe procedure also.

A great emphasis has been laid on a student getting valid results andinterpreting them. It is essential that the teacher should properly explaineach experiment so that inexperienced students will be able to obtain accurateresults within a reasonable time. Teachers are also expected to help studentsin identifying errors and mistakes committed during experiments and waysfor correcting them. It is possible that some of the students may undoubtedlybe capable of doing more sophisticated work than that represented in themanual. But introductory course of this sort has been designed to help allstudents for some useful and joyful experience by conducting theexperiments of their own. The manual also aims that students and teachersnot be discouraged by either incomplete experiments or experiments whichyield apparently meaningless results.

With the objectives of inculcating scientific temper among learners andproviding them an opportunity to undertake independent scientificinvestigation, Investigatory Project Work has been included as an integralpart of the practical curriculum of Class XII. Such investigatory projectsare expected to provide thrill in the learning process. It is also expected toserve the real purpose of practicals, i.e., developing an ability to hypothesiseand design experiments to address certain problems, to make observationsmethodically and to draw conclusions out of the experimental data. Acomprehensive idea about undertaking investigatory project has been givenin the book with a list of a few problems on which investigatory projectwork can be undertaken. However, the list is only suggestive and consideringthe wider scope students can undertake any kind of investigatory projectwork depending on their region, its climatic condition, availability ofresources, etc.

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Aim: To study the reproductive parts of commonly available f lowers

Principle : The male reproductive parts of a flower are the stamens collectively calledandroecium and the female reproductive parts are the carpels/pistils collectively calledgynoecium. The individual units of stamen consist of a filament, which supports the antherlobes. Gynoecium consists of stigma, style and ovary. Many variations are found in differentcharacteristics of both the stamens and carpels. We shall try to study these variations in thereproductive parts of flowers in the exercise.

Requirement: Commonly available flowers, needles, forceps, razor/scalpel blade, brush, slides,cover slip, watch glass, magnifying lens, dissecting microscope, compound microscope, etc.

Procedure(i) Familiarise with the terms to describe the reproductive parts of flowers

given in annexures of Exercise No. 11 of Laboratory Manual: Biology(Class XI) and at the end of this experiment.

(ii) Observe the flower with the naked eye, hand lens or under a dissectingmicroscope. Study their reproductive parts and count the number ofstamens and record their cohesive and adhesive features.

(iii) Cut L.S. of the flower and place it on a slide to observe the followingcharacters:

(a) Placement of anthers

(b) Position of the ovary: epigynous/perigynous/hypogynous.

(iv) Mount one stamen on a slide and study the following characters:

(a) Attachment of filament to anther

(b) Dehiscence pattern of the anther lobes for discharge of pollen.

(v) Cut T.S. of anther lobe to observe the number of pollen sacs.

(vi) Mount the pistil on a slide and study style, stigma and ovary. Recordthe number of stigma and nature of pistil.

(vii) Cut T.S. of ovary, mount it on a slide and observe

(a) Number of locules in the ovary

Exercise 1

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(b) Type of placentation

(c) Number of ovules per locule

(viii) Draw labelled figures of your preparation and observations.

Questions

1. Name the most common type of placentation observed.

2. What is the most common type of dehisence pattern in anthers?

3. Name a few unisexual flower-bearing plants studied by you.

4. “Flower is a modified shoot.” Justify the statement based on your observation.

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EXERCISE 1

Annexure 1

Description of reproductive parts of flowers

Androecium

Number of stamens The number of stamens may vary from a few to many in dif-ferent flowers

Stamens may be free or united. If united they can be of thefollowing type:

( i ) Syngenesious : Filaments free and anthers united,e.g., Sunflower.

( i i ) Synandrous : Stamens fused all through their length,e.g., Cucurbita

( i i i ) Adelphous: Anthers remain free and filaments are united.Adelphous condition can be

(a) Monoadelphous—United to form 1 bundle, e.g.,China rose

(b) Diadelphous—United to form 2 bundles, e.g., Pea

(c) Polyadelphous—United into more than twobundles, e.g., Lemon

Fusion of stamens with other parts of the flower

( i ) Epipetalous : Stamens fused with petals,e.g., Sunflower, Datura

( i i ) Epiphyllous : Stamens fused with perianth,e.g., Lily

( i ) Basifixed: Filament attached to the base of anther,

e.g., Mustard

( i i ) Adnate : Filament attached along the whole length ofanther, e.g., Michelia, Magnolia

(i i i ) Dorsifixed : Filament attached to the back of anther,e.g., Passion flower

(iv) Versatile : Anther lobes attached with filament in themiddle portion with both ends free, e.g., Gramineaefamily

( i ) Porous : Pollens released through pores, e.g., Brinjal,Potato

( i i ) Longitudinal: Pollens released through the longitudinalslit of another lobes, e.g., China rose, Cotton

Cohesion(Fig. 1.1 a–e)

Adhesion(Fig. 1.2)

Attachment of filament toanther(Fig.1.3 a–d)

Dehiscence pattern

(Fig. 1.4 a,b)

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Gynoecium

Number of stamens The number of stamens may vary from a few to many indifferent flowers

(i) Epigynous: Position of ovary inferior to other floral parts,

e.g., Mustard, China rose

( i i ) Perigynous : Other floral parts are attached around theovary, e.g., Apple, Guava

(ii i ) Hypogynous: Position of ovary superior to other floralparts, e.g., Sunflower

If number of carpels is more than one, they may be

(i ) Apocarpous : Carpels are free. Each carpel has its ownstyle and stigma, e.g., Rose

( i i ) Syncarpous: Carpels are united, e.g., Lady’s finger, Tomato

Vary from one to many

(i ) Unilocular : One locule, e.g., Rose, Pea

( i i ) Bilocular: Two locules, e.g., Datura

(ii i ) Multilocular : Many locules, e.g., Lady’s finger, China rose

(i ) Marginal : The placenta forms a ridge along the ventralsuture of the ovary and the ovules are borne on this ridge,e.g., Pea

( i i ) Axile: The ovary is partitioned into several chambers orlocules and the placentae are borne along the septa ofthe ovary, e.g., Tomato, China rose

(ii i ) Parietal: The ovules develop on the inner wall of theovary or on peripheral part. Ovary unilocular but in somecases becomes two chambered due to formation of a falseseptum, e.g., Mustard

(iv) Free central : Ovules are borne on the central axis andsepta are absent, e.g., Carnation, Chilly

(v) Basal: Placenta develops at the base of the ovary,e.g. ,Sunflower.

Position of ovary(Fig. 1.5 a–d)

Cohesion(Fig. 1.6 a–c)

Number of locules in ovary

Placentation(Fig. 1.7 a–e)

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EXERCISE 1

Fig.1.1 Cohesion of stamens: (a) Syngenesious (b) Synandrous(c) Monoadelphous (d) diadelphous (e) Polyadelphous

(a) (b) (c)

(d)

(e)

Fig.1.2 Adhesion of Stamens: Epipetalous/Epiphyllous

Fig.1.3 Attachment of filament to anther: (a) Basifixed (b) Adnate(c) Dorsifixed (d) Versatile

(a) (b) (c) (d)

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Fig.1.5 Position of ovary: (a) Epigynous (b–c) Perigynous (d) Hypogynous

(b) (c) (d)

Fig.1.4 Dehiscence pattern of anther: (a) Porous (b) Longitudinal

(a) (b)

Fig.1.6 Cohesion of carpels: (a) Apocarpous (b–c) Syncarpous

(a) (b) (c)

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EXERCISE 1

Fig.1.7 Placentation: (a) Marginal (b) Axile (c) Parietal (d) Free central (e) Basal

(a) (b) (c)

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Aim: To calculate percentage of pollen germination

Principle: In nature, pollen grains germinate on the compatible stigmas of the carpel. Pollengrains can also be induced to germinate in a synthetic medium. During germination, intine(inner wall) of pollen grain emerges out as pollen tube through one of the germ pores in exine(outer wall).

Requirement: Calcium nitrate, boric acid, sucrose, distilled water, petridish, slides, coverslips,brush, needle, microscope, and mature pollen grains of Tradescantia/balsam/Jasmine/lily/pomegranate/grass/Vinca/China rose/Petunia.

Exercise 2

Procedure(i) Prepare the pollen germination medium by dissolving 10g sucrose,

30mg calcium nitrate and 10mg boric acid in 100ml of distilled water.Alternatively 10% sucrose solution can also be used.

(ii) Take a drop of medium or 10% sucrose solution on a cover slip andsprinkle mature pollen grains on the drop.

(iii) Invert the cover glass on to a slide

(iv) After 10 minutes, observe the slide under microscope.

(v) Count (a) total number of pollen grains seen in the microscope field,and (b) the number of pollen grains that have germinated.

ObservationSeveral pollen grains germinate and put forth pollen tubes. Count the totalnumber of pollen grains and the number of germinated pollen grains in 3-5different microscope fields. Tabulate your observations and calculate thepercentage of pollen germination.

Name of the plant used as source of pollen……………………………Number of pollen grains in a field of microscope = N

Number of germinated pollen grains in a field of microscope = n

Percent pollen germination = ×100nN

or 100

Nn

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EXERCISE 2

DiscussionAlthough pollen grains of many species germinate in this medium, thepercentage of germinations and the time taken for germination varies indifferent species. Draw a germinating pollen grain and label.

Number of Total number of pollen (N) Total number of pollen germinated (n) % pollen germination

observation ×100n

N

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2.

3.

4.

5.

Average

Questions

1. How many pollen tubes emerge from a pollen grain?

2. What does the pollen tube carry?

3. Can you explain as to why some pollen grains fail to germinate?

4. Why do we use sucrose as the medium for pollen germination?

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Aim: To study pollen tube growth on stigma

Principle: Pollen grains germinate and form pollen tubes after they get deposited by the processof pollination on compatible stigma. Pollen tube, made up of cellulose, is an extension of theinner wall of pollen grain (intine). It emerges through one of the germ pore and passes throughtissues of stigma and style to reach the ovule. The growing pollen tube is observed by stainingwith cotton blue.

Requirement: 5–6 excised styles with stigma of Petunia/grass/maize/sunflower/Abelmoschus(Lady's finger), beaker, water, slides, cover slips, cotton blue stain, microscope, brush, needle.

Exercise 3

Fig.3.1 Growth of pollen tubein the style of a carpel

Pollen grains

Pollen tube

Style

Procedure(i) Place the stigmas in boiling water in a beaker

for softening the tissues for 5–10 minutes.

(ii) Stain with cotton blue for 3–5 minutes andwash with water to remove excess stain.

(iii) Mount one stigma in a drop of glycerine on aslide. Place a cover slip on the stigma and gentlypress the cover slip on the material. Observethe slide under a microscope.

(iv) If you fail to observe pollen tubes mountanother stigma.

ObservationLook for long blue-coloured tubular structurestraversing through the tissues of stigma andstyle (Fig. 3.1).

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EXERCISE 3

DiscussionPollen tubes are seen amidst the stylar tissue. Many pollen tubes may beseen. Trace the origin of pollen tubes to the pollen grains present aroundthe surface of the stigma.

Questions

1. Can pollen grains of one plant species germinate on stigma of other species?Give reasons.

2. Do all pollen tubes reach the ovules?

3. Are all the pollen tubes of equal length? If not, why?

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Aim: To study the discrete stages of gametogenesis in mammalian testis and ovary

Principle: In all male and female organisms gamete formation takes place in their gonads, i.e .,testis and ovary respectively. The process of gamete formation, called gametogenesis involvesmeiotic cell division. The gametogenic development in testis is called spermatogenesis and inovary it is oogenesis. They exhibit marked differences and can be examined in transverse section(T.S.) of these organs.

Requirement: Permanent slides of T.S. of testis and ovary, compound microscope,lens-cleaning paper and cleaning fluid

Procedure(i) Clean the slide and microscope’s eye and objective lenses with the

help of lens cleaning paper using any cleaning fluid.(ii) Place the slide on the stage of the microscope and observe first under

lower magnification and then in higher magnification. Observe variousstages of gamete development.

(iii) Record your observations in the notebook and draw labelled diagrams.

ObservationT.S. of testis

Exercise 4

Fig. 4.1 T.S. of mammalian testis

Seminiferous tubule

Spermatozoa

GerminalEpithelium

Spermatogonia

(i) You will observe a large numberof seminiferous tubules underlower magnification. Observe acomplete tubule in highermagnification and view variousstages of gamete developmentfrom periphery towards lumen(Fig. 4.1) and identify thefollowing types of cells namely,Germinal epithelium,Spermatogonial cells, Primaryspermatocytes, Secondaryspermatocytes, Spermatids andSpermatozoa.

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EXERCISE 4

(ii) In T.S. of testis the space between tubules are filled with blood vesselsand a specific cell type called Leydig's cell or Interstitial cells.

T.S. of Ovary

(i) In the section of ovary, there is a massof tissue lined with germinalepithelium. Inside that you willobserve an ovum, which is a cellsurrounded by one to several layersof follicular cells. As the ovummatures, the number of surroundingfollicular cell layer increases (Fig. 4.2).

(ii) In the later stage of folliculardevelopment a cavity called antrumappears.

(iii) The cavity gets further enlarged andthe follicle grows bigger. This is thestage of Graafian follicle ready to release the ovum (ovulation).

(iv) In the next stage, you may notice a Corpus luteum, and/or Corpusalbicans, which differ from each other and also from Graafian follicle intheir features.

DiscussionSpermatogenesis is a continuous process after attainment of puberty, andthat is why gamete development and spermatozoa are observed in a singleseminiferous tubule. In case of ovary, the follicular development stages areobserved.

Questions

1. What would happen if meiosis fails to occur in gametocyte?

2. At which stage of follicular development, is ovum released?

3. Spermatogenesis is a continuous process. Justify the statement.

4. Draw a labelled diagram of T.S. of testis.

5. Draw a labelled diagram of T.S. of ovary.

6. What would happen if sperms are devoid of their tail?

7. What are the consequences of failure of ovulation?

Fig. 4.2 Section of mammalian ovary

Graafian Follicle

Corpus luteum

Corpus albicans

Antrum

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Aim: To study and identify various stages of female gametophyte development in the ovaryof a flower

Principle: In flowering plants, female gametophyte (embryo sac) is a microscopic structuresituated deep inside the ovule. An ovule generally has one female gametophyte. Developmentof female gametophyte begins with megaspore mother cell. Most common type of femalegametophyte is the monosporic, 8-nucleate, 7-celled type.

Requirements: Permanent slides of V.S. of ovary, photographs/chart or models showingstages of female gametophyte development and microscope

Procedure(i) In a V.S. of ovary we generally find several ovules. Carefully observe

each ovule and locate as many stages of female gametophytedevelopment as possible.

(ii) Draw the diagrams as observed under microscope.

Exercise 5

Embryo sac

Fig. 5.1 V.S. of an ovule

Chalaza

Outer integument

Inner integument

Micropyle

Funiculus

Observation(i) Record the features of ovule

like number of integuments,nucellus and micropylarand chalazal poles. Fig 5.1shows the femalegametophyte (embryo sac)as seen in a V.S. of an ovule.Different stages ofdevelopment of femalegametophyte are shown inFig. 5.2.

(ii) Observe the placement ofembryo sac close to themicropylar pole.

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EXERCISE 5

Fig. 5.2 Stages of gametophyte development: (a) megaspore with 2 nucleus(b) 4-nucleate stage (c) 8- nucleate stage (d) 8- nucleate stageshowing 3+2+3 distribution of nuclei (e) mature embryo sac.

(a) (b) (c)

Egg

Synergids

Central cell

Secondary nucleus

Antipodals(d) (e)

(iii) Note the contents of embryo sac, namely, an egg apparatus(2 synergids and egg) at micropylar end, secondary nucleus in thecenter and three antipodal cells at the chalazal end (Fig. 5.2).

Questions

1. Explain the difference between gamete and a gametophyte.

2. Name two differences between synergids and egg.

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Aim: Preparation and study of mitosis in onion root tips

Principle: Somatic growth in plants and animals takes place by the increase in the number ofcells. A cell divides mitotically to form two daughter cells wherein the number of chromosomesremains the same (i.e., unchanged) as in the mother cell. In plants, such divisions rapidly takeplace in meristematic tissues of root and shoot apices, where the stages of mitosis can beeasily observed. In animals, mitotically dividing cells can be easily viewed in the bone marrowtissue of a vertebrate, epithelial cells from gills in fishes and the tail of growing tadpole larvaeof frog.

Requirement: Onion bulbs, wide mouth glass tubes/jar/bottle, glacial acetic acid, ethanol2-4% acetocarmine/acetoorcein stain, N/10 HCl, spirit lamp/hot plate, slide, cover slips,blotting paper, molten wax/nail polish and compound microscope

ProcedureGrowing of root tips

Select a few medium-sized onion bulbs. Carefully remove the dry rootspresent. Grow root tips by placing the bulbs on glass tubes (of about 3–4cm. diameter) filled with water. Care should be taken so that the stem portionof the bulb (basal part) just touches the water. A few drops of water may beadded periodically to compensate evaporation losses. New roots may take3–6 days to grow. Cut 2–3 cm long freshly grown roots and transfer them tofreshly prepared fixative, i.e., aceto-alcohol (1:3:: glacial acetic acid : ethanol).Keep the root tips in the fixative for 24 hours and then transfer them to 70%ethanol (for preservation and use in future). Onion root-tip cells have a cellcycle of approximately 24-hour duration, i.e., they divide once in 24 hours,and this division usually takes place about two hours after sunrise. Therefore,roots grown on water should be cut only at that time to score maximumnumber of dividing cells.

Preparation of slide

Take one or two preserved roots, wash them in water on a clean and grease-free slide. Place one drop of N/10 HCl on the root tip followed by 2–3 dropsof aceto-carmine or aceto-orcein stain on it. Leave the slide for 5–10 minutes

Exercise 6

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EXERCISE 6

on a hot plate (or warm it slightly on spirit lamp). Care should be taken thatthe stain is not dried up. Carefully blot the excess stain using blotting paper.Now cut the comparatively more stained (2–3 mm) tip portion of the rootand retain it on the slide and discard the remaining portion. After(10–20 seconds) put one or two drops of water and blot them carefully usingblotting paper. Again put a drop of water on the root tip and mount a coverslip on it avoiding air bubbles. Place the slide in between the folds of blottingpaper using the fingers in such a way that the cover slip mounted on theslide is properly held. Now slowly tap the cover slip using the blunt end of apencil so that the meristematic tissue of the root tip below the cover slip isproperly squashed and spread as a thin layer of cells. Carefully seal themargins of the cover slip using molten paraffin wax or nail polish. Thispreparation of onion root tips cells is now ready for the study of mitosis.

Study of slide

Place the slide on the stage of a good quality compound microscope. Firstobserve it under the lower magnification (10 X objective) to search for thearea having a few dividing cells. Examine the dividing cells under highermagnification of the microscope to observe the detailed features of mitosis.

ObservationThe stages of mitosis can be broadly categorised into two parts: karyokinesis(division of nucleus) followed by cytokinesis (division of cytoplasm, andultimately of the cell). Those cells, which are not in the phases of cell divisionare considered to be in interphase. You may observe that most of the cellsin a microscope field are in interphase

InterphaseThe cells are mostly rectangular, oval or even circular in shape, with almostcentrally situated densely stained nucleus. The chromatic (coloured) materialof the nucleus is homogeneous and looks granular. The boundary of thenucleus is distinct. One or few nucleoli (sing: nucleolus) can also be observedinside the nucleus (Fig. 6.1a).

Stages of Mitosis(a) Prophase

Intact nuclear outline is seen. The chromatin (seen as a homogeneousmaterial in the nucleus at interphase) appears as a network of fine threads(chromosomes). Nucleoli may or may not be visible (Fig. 6.1b).

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If the cell under observation is in the early stage of prophase then thechromatin fibres (chromosomes) are very thin. However, in the cellsat late prophase, comparatively thicker chromatin fibres would bevisible. Besides this, in the late prophase the nuclear membranemay not be noticed.

(b) Metaphase

The nuclear membrane disappears. Chromosomes are thick and are seenarranged at the equatorial plane of the cell (Fig. 6.1c). Each chromosome at

Fig.6.1 Interphase (a) and stages of mitosis (b - e) – actual microscopicview on left side and its diagrammatic representation on theright hand side

a. Interphase

b. Prophase

c. Metaphase

d. Anaphase

e. Telophase

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EXERCISE 6

this stage has two chromatids joined together at the centromere, which canbe seen by changing the resolution of the microscope. Nucleolus is notobserved during metaphase.

(c) Anaphase

This stage shows the separation of the chromatids of each chromosome. Thechromatids separate due to the splitting of the centromere. Each chromatidnow represents a separate chromosome as it has its own centromere. Thechromosomes are found as if they have moved towards the two poles of thecell. The chromosomes at this stage may look like the shape of alphabets 'V','J' or 'I' depending upon the position of centromere in them. Different anaphasecells show different stages of movement of chromosomes to opposite poles,and they are designated to represent early, mid and late anaphase (Fig. 6.1d).

(d) Telophase

Chromosomes reach the opposite poles, lose their individuality, and looklike a mass of chromatin (Fig. 6.1e). Nuclear membrane appears to form thenuclei of the two future daughter cells.

CytokinesisIn plants, a cell plate is formed in the middle after telophase. The plate canbe seen to extend outwards to ultimately reach the margin of the cell anddivide the cell into two. Such cell plates are characteristic of plant cells(Fig. 6.2). However, in an animal cell, the two sides of the cell show inpushingsor constrictions formed from the peripheral region in the middle of the cell,which grow inward and meet to divide the cell into two daughter cells.

Draw labelled diagrams of all the phases of mitosis.

Fig. 6.2 Cytokinesis

DiscussionMitotic index (MI) is defined as a ratio of the total number of dividing cells (n)and the total number of cells (N) in a particular focus chosen randomly under

the microscope and is calculated as MI = ×100n

N . By randomly selecting

5 to 10 such foci, one can estimate the mitotic index for a given type.

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Questions

1. Suggest names of a few tissues, which are suitable for the study of mitosis.

2. Why is mitosis also known as equational division?

3. What shape would a metacentric and a sub-metacentric chromosome exhibitduring the anaphase stage?

4. How does cytokynesis differ in plant and animal cells?

Features Interphase Karyokinesis Cytokinesis

Prophase Metaphase Anaphase Telophase

1. Cell morphology

2. Nuclear morphology

3. Chromosomes/chromatids

Tabulate your observations in the tabular form given below

The effect of different samples of water (polluted or contaminated) can beassayed on the mitotic-index (an indicative feature of somatic growth rate inthem).

Further, the impact of different types of pollutants on different phases ofmitosis can also be assayed.

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Aim: Study of stages of meiosis using permanent slides

Principle: Meiosis is a type of cell division in which the number of chromosomes is halved(from diploid to haploid) in the daughter cells, i.e., the gametes. The division is completed intwo phases, meiosis I and meiosis II. Meiosis I is a reductional division in which the chromosomesof homologous pairs separate from each other. Meiosis II is equational division resulting in theformation of four daughter cells. Stages of meiosis can be observed in a cytological preparationof the cells of testis tubules or in the pollen mother cells of the anthers of f lower buds.

Requirement: Permanent slides of meiosis and compound microscope

Exercise 7

ProcedurePlace the slide on the stage of the microscope and search for the dividingcells using lower magnification. When dividing cells are located observe themunder higher magnification.

ObservationObserve various stages of meiosis and identify them on the basis of the specificfeatures given in the table 7.1. A significant number of cells will be in theInterphase. These cells have a centrally positioned densely stained nucleus.In case of slide of animal tissue a few mitotically dividing spermatogonialcells may also be seen.

Unlike the prophase of mitosis, it is a comparatively complex phasecharacterised by a number of events. Five sub-phases can beidentified in it.

(a) Leptotene (leptos = slender tene = band or thread)

(i) The nuclear membrane and nucleolus are not distinctlyobservable (Fig. 7.1 a).

(ii) Fine network of thin threads are seen uniformly distributedin the nucleus.These are chromatin threads, which may beobserved as more prominent structures in the later stages.

Meiosis I

1. Prophase I

Table 7.1 Different stages of meiosis and their features

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Fig. 7.1 Sub-phase of Prophase I (a-d) – actual microscopic view on leftside and its diagrammatic representation on the right hand side

(a) Leptotene

(b) Zygotene

(c) Pachytene

(d) Diplotene-Diakinesis

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EXERCISE 7

2. Metaphase I

3. Anaphase I

4. Telophase I

Meiosis II

1. Prophase II

(b) Zygotene (Zygon = paired)

This stage is characterised by the pairing of the homologouschromosomes, which can be seen as paired chromatin threads(bivalents) (Fig. 7.1b).

(c) Pachytene (pachy = thick)

The chromatin threads get condensed and appear shortened andthick. Pairs of homologous chromosomes can be seen. Eachchromosome has two chromatids and thus each bivalent consistsof four chromatids. This configuration is called tetrad (Fig. 7.1c).

(d) Diplotene (diplos = double)

The homologous chromosomes (each made up of two chromatids)show distinct separation from each other except at few regions whereattachments are seen (Fig. 7.1d). These are chiasmata (sing.chiasma) representing the site of exchange of the parts betweentwo homologous chromosomes (i.e. crossing over).

(e) Diakinesis (Dia = opposite; kinesis= separation or movement)

(i) The homologous pair of chromosomes appear more shortened,thick and prominent (Fig. 7.1d).

(ii) Chiasmata can be still observed.

(iii) All the homologous pairs appear scattered in the cell.

Homologous chromosomes are still in pairs, and are arranged alongthe equatorial plane of the cell (Fig. 7.2a). At this stage, the numberof bivalents can be counted. Chiasmata may still be seen in a fewbivalents.

The chromosome pairs appear to have moved towards the twoopposite poles of the cell. At the later stage, the anaphase - I mayshow the assembly of chromosomes at two poles (Fig. 7.2b). Thisresults into the reduction of number of chromosomes to half.

This stage can be identified by the presence of two chromatids ineach chromosome.

The chromosomes present at the two poles appear decondensedand form two distinct nuclei (Fig. 7.2c).

Note: After the telophase I stage there may or may not be cytokinesis.Thereafter the cell enters into the second meiotic division.

(i) Distinct thread- like chromatin fibres or rod- shaped chromosomeare seen.

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Fig.7.3 Phases of Meosis II (a,b) – actual microscopic view on left sideand its diagrammatic representation on the right hand side.

(a) Metaphase II

(b) Anaphase II

(a) Metaphase I(b) Anaphase I

(c) Telophase I

Fig. 7.2 Phases of Meosis I (a-c) – actual microscopic view on left sideand its diagrammatic representation on the right hand side.

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EXERCISE 7

2. Metaphase II

3. Anaphase II

4. Telophase II

Questions

1. What is the significance of meiosis?

2. What is synapsis and crossing over?

3. How can anaphase I and anaphase II be distinguished from each other?

4. Indicate distinguishing feature of metaphase I of meiosis and metaphase of mitosis.

5. How many daughter cells are produced at the end of meiosis?

6. The daughter cells produced at the end of meiosis are genetically different. Explain.

7. What is the significance of synapsis?

This phase is similar to that of mitotic division

(i) The chromosomes having two chromatids attached at thecentromere are observed arranged at the equatorial plane of thecell.

Note: Metaphase II of meiosis can be differentiated from metaphase-Ion the basis of the following features:

(ii) Each chromosome of metaphase II has two chromatid(Fig. 7.3a) whereas in metaphase I these are pairedhomologous chromosomes each having two chromatids thusforming tetrad.

(iii) In the metaphase I of meiosis, a few chiasmata are observed,where as no chiasmata are observed during metaphase II.

The two chromatids of each chromosome after separation appearto lie at the two poles of the cell (Fig. 7.3b).

Note: Anaphase II can also be distinguished from the anaphase I ofmeiotic division on the basis of chromatids: In anaphase I, eachchromosome has two distinct chromatids, but in anaphase II, eachchromosome is represented by one chromatid only.

The separated chromosomes appear decondensed and form nuclei(Fig. 7.3c).

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Aim: To study the blastula stage of embryonic development in mammals, with the help ofpermanent slide, chart, model or photograph

Principle: The zygote undergoes a few cycles of mitotic divisions to form a solid ball of cells calledmorula. The cells continue to divide and at a later stage a cavity is formed within it. This stage isblastula. The internal structural details of blastula can be observed in its transverse section.

Requirement: Permanent slide, chart/model of T.S. of blastula, compound microscope,lens cleaning fluid and paper

Exercise 8

ProcedureObserve the slide under lower magnification of the microscope. In case ofchart/models/photographs, note the feature of blastula in your practicalrecord and draw labelled diagram.

ObservationIn transverse section, the blastula appears as a sphere with a cavity, calledblastocoel within it (Fig. 8.1). Notice an outer layer of blastomeres calledtrophoblasts. A cellular mass, adhered to the trophoblast is present on oneend of the blastula. It is called inner cell mass.

Fig.8.1 Blastula stage of a mammal

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EXERCISE 8

Questions

1. What are the differences between blastula and morula?

2. What are the main structures you observe in T.S. blastula?

3. Match the stages in column I with features in column II

Column I Column II

(a) Trophoblast (i) Dividing cells of the morula(b) Morula (ii) Outer layer of blastula(c) Blastocoel (iii) Solid ball of cells

(iv) Cavity

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Aim: To verify Mendel's Law of Segregation

Principle: When two pure lines with contrasting forms of a particular character (phenotypes) arecrossed to produce the next generation (F1 generation), all the members of the progeny are of onlyone phenotype i.e. of one of the two parents. The phenotype that appears is called dominant, andthe one that does not appear is called recessive. When the F1 plants are selfed, the progeny i.e. theF2 generation is in the ratio of 3 dominant: 1 recessive (¾: ¼ or 75%: 25%). This reappearance ofthe recessive phenotype in F2 generation verifies law of segregation.

Requirement: 64 yellow and 64 green plastic beads, all of exactly same shape and size, (whenbeads are not available, pea seeds may be coloured using paint, these beads represent thegametes of a specific trait), plastic beakers/petri dishes and a napkin/hand towel

ProcedureStudents have to work in pairs to perform the experiment. The followingsteps are to be strictly followed in the sequence mentioned below.

(i) Put 64 yellow beads in one beaker/petridish and 64 green beads inthe other to represent respectively male and female gametes. Let theyellow bead be indicated by ‘Y’ and green bead by ‘y’.

(ii) Take a bead from each container and place them together (it representsfertilisation) on the napkin spread before you on the table. (One studentto take out beads and to put in the hands of the other student who willput them on the table).

(iii) Just like the previous step, continue to pick beads and arrange themin pairs. Thus 64 pairs of beads are obtained representing the 64heterozygous F

1 progeny.

Note that all the F1 individuals are represented by one yellow and onegreen bead.

(iv) Put 32 F1 progeny in one petridish and the remaining 32 in another

petridish (representing the F1 males and females).

(v) Stir the beads of each petridish with a pencil/pen for about 10 timestaking care that no bead falls off.

Exercise 9

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EXERCISE 9

(vi) To obtain the F2 generation, one student would withdraw one beadfrom one beaker labelled male and one from the other beaker labelledfemale keeping his/her eyes closed (to ensure randomness), and put

them together in the stretched palm of the partner, who will put themtogether on the napkin spread over the table. Continue this process till

all the beads are paired. Thus 64 offsprings of F2 are obtained.

(vii) Note the genotype (YY or Yy or yy) of each pair, and their possible

phenotype.

(viii) Have six repeats of the experiment (steps i to vii) with partners changing

their roles. Pool all the data from the six repeats together.

(ix) Calculate the genotypic and phenotypic ratios of your pooled data.

Note that larger the sample size, more accurate is the result.

ObservationRecord the result in the following table:

Generation Repeat No. Total no. of Genotype (s) Phenotype (s)

individuals YY Yy yy

F1 1.

2.

3.

4.

5.

6.

Total

F2 1.

2.

3.

4.

5.

6.

Total

Phenotypic Ratio: in F1…….

in F2…….

Genotypic Ratio: in F1…….

in F2…….

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Questions

1. Do you expect the same results in terms of 3:1 ratio in F2 if you had started with

smaller number of beads (say 10 beads)?

DiscussionThe results are so because each diploid individual contains two copies ofevery gene - one copy on each of the two homologous chromosomes. Thesetwo copies of the gene may be of similar type (YY or yy) or are dissimilar Yy.The former (YY or yy) are called homozygous for that particular character,and the Yy are called heterozygous ones. The pure lines in the above crossare homozygous ones, which contributed only one copy of their gene (as aresult of meiosis) to their F1 progeny to restore its diploid nature with genotypeYy (heterozygous) where only one form (allele) is expressed (dominant) andthe other form (allele) is not expressed (recessive). This is the phenomenonof Dominance.

When the F1 individuals are crossed together to raise the F2 generation,each F

1 individual produces two types of gametes: 50% having dominant

allele, and the remaining 50% having recessive allele. These gametes undergorandom fusion during fertilisation to produce the F2 generation. Accordingto simple probability of mixing of opposite sex gametes (sperms and ova),offsprings of three genotypes are likely to appear as follows: [(half of gametesof Y type + half of remaining gamete y type) X (half gametes of Y type + halfof remaining gamete of y type)] = One-fourth of F2 individuals of YY phenotype+ half of F

2 individual Yy type + one-fourth of F

2 individul of yy type. Among

these proportion of dominant phenotype would be � YY+ � Yy = � yellowand recessive phenotype � yy i.e. green phenotypes in 3:1 or 75%:25%ratio.

This ratio of 3:1 in the F2 suggests that in the F

1 heterozygotes, the

recessive allele does not get destroyed and remains only in the recessive(dormant) state to get an opportunity to express itself when it has separatedfrom the influence of the dominant allele (Y). This is called Law of Segregationof the alleles.© N

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Aim: To verify the Mendel’s Law of Independent Assortment

Principle: In a dihybrid cross, the segregation of one gene pair is independent of the segregationof the other pair. It means that genes of two different traits assort independently to give aprobability ratio equal to segregration probability ratio of one allele pair X segregation probabilityratio of other allele pair, which comes to, (3:1) X (3:1) = 9:3:3:1

Requirement: Plastic beakers; 64 plastic beads each of yellow, green, red and white to represent,yellow and green colour of seed coat and red and white flowers respectively and napkin/handtowel

Exercise 10

ProcedureStudents are to work in pair.

The following steps are to be followed sequentially:

(i) Place 64 beads of each colour in four separate beakers.

(ii) Put the beakers containing the yellow and red beads on your left side,and those containing the green and white beads on your right side.The beakers on your left side represent plants bearing yellow seed andred flower (dominant character YY, RR). Beakers on the right siderepresent plants bearing green seeds and white flowers (recessivecharacter yy, rr). These are the two parental types having contrastingforms of two different characters.

(iii) Stir the beads in each beaker with a pencil/pen. Each bead nowrepresents alleles in the male and female gametes.

(iv) Pick up one yellow, one green, one red and one white bead, and putthem together on the napkin spread on the table.

(v) Continue picking up and putting together of the beads of all coloursas mentioned in the previous step, till all the beads are utilised.

(vi) Note that in all, 64 such 4-bead clusters are obtained representing theF

1 individuals. Ascertain their genotype and phenotype.

(vii) Next step is to cross these F1 individuals to raise the F

2 generation. Let

us suppose half of the 4-bead clusters (32 clusters) represent the maleparents and the remaining half (32 clusters) the female parents. Nowput the 32 red and 32 white beads together in one beaker (numbered-I), and similarly put 32 yellow and 32 green beads together in other

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beaker (numbered-II). These two beakers represent F1 female. Similarlyput remaining 32 red + 32 white beads in beaker numbered-III, and32 yellow and 32 green one in beaker numbered-IV to represent the F

1

male. The arrangement can be presented as below

(viii) Stir the beads in each beaker with a pencil. In order to raise the F2

generation, pick up (with eyes closed) one bead from the beaker-I offemale and one bead from the beaker-III of the male, and put into thepalm of the partner student. Similarly, pick up one bead each from thebeaker-II of female and beaker IV of male to put in the palm of thepartner. This partner would now keep all the four beads together (torepresent the F2 individual). Continue this process till all beads areutilised. At the end, 64 F

2 individuals (each represented by a 4-bead

cluster) are obtained.

(ix) Determine the genotype and phenotype of each of the 64 F2 individualsand write down the number of individuals of different genotypes andphenotypes in the tabular form (given below), remembering that Y(yellow seed colour) is dominant over y (green seed) and R (red flower)is dominant over r (white flower).

(x) Repeat the whole procedure (steps i to ix) six times, and tabulate yourresults.

ObservationTabulate the results as follow:Symbol (-) indicates the presence of corresponding dominant or recessiveallele e.g. Y or y and R or r.

Summarise your results (adding together the data of all the six repeats)

F1 Generation

(a) Total number of individuals: _________________________

(b) Phenotype (s) _________________________

(c) Genotype (s) _________________________

Female F1 Male F1

32 red + 32 white (Beaker I) 32 red+32 white (Beaker III)

32 yellow + 32 green (Beaker II) 32 yellow + 32 green (Beaker IV)

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EXERCISE 10

Generation Total No. Genotype Phenotype

& repeat of offsprings Y-R- Y-rr yyR- yyrr Yellow Yellow Green Green

No. Red White Red white

F1

1.

2.

3.

4.

5.

6.

Total

F2

1.

2.

3.

4.

5.

6.

Total

F2 Generation

(a) Total number of individuals _________________________

(b) Phenotypes _________________________

(c) Number of individuals in each phenotypic class:

Number Phenotype

__________________ _____________________

__________________ _____________________

__________________ _____________________

__________________ _____________________

(d) Phenotypic ratio _____________________

(e) Genotypic ratio _____________________

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Questions

1. Linked traits fail to assort independently. Explain.

2. How is independent assortment of alleles important from the point of view ofvariation?

(f) Number of individuals of each genotypic class:

Number Genotype

__________________ _____________________

__________________ _____________________

__________________ _____________________

__________________ _____________________

__________________ _____________________

(g) Genotypic Ratio ______________________________________

DiscussionThe four phenotypic classes in the F2 generation are in ratio of 9:3:3:1 asexpected from the Law of Independent Assortment. The genotypic ratiowould be (1:2:4:2): (2:1):(2:1):1.

Note

1. In case six repeats of the experimental procedure are not feasible dueto time limitations, either the number of repeats be slashed down tothree or the data from single repeat of six different pair of students maybe pooled together to make the final calculations.

2. This Law of Independent Assortment was later found to be true only fortraits present on two different homologous pair of chromosomes, thatis, the two are not linked together. The linked traits do not assortindependently, rather they are inherited together (linked) except whencrossingover separates them.

3. It is quiet likely that you may not find your data exactly in theexpected ratio, instead almost approximate to it. The statisticalsignificance of this deviation from the exact expected ratio due to

probality can be checked using chi-square (χ2) test, about which

you will study in higher classes.

Aim: Preparation and analysis of Pedigree Charts

Principle: The Mendelian concept of dominance and segregation can also be studied in humansby preparing and then analysing the pedigree charts. The internationally approved symbols forindicating males and females, marriages, various generations (I, II, III), etc., are given below.

Exercise 11

Requirement: Information about characters/traits in a family for more than one generation

ProcedureSelect a family in which any one of the monogenic traits such as tonguerolling, widow's peak, blood groups’, red-green colour blindness, dimple in

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the cheek, hypertrichosis of ear, hitch-hiker's thumb, etc., is found. Ask theperson exhibiting the trait to tell in which of his/her parents, grand parents(both maternal and paternal), their children and grand children the trait inquestion is present. Among surviving individuals the trait may also beexamined. The information made available is the basis for the preparation ofpedigree chart using the appropriate symbols. A careful examination of thepedigree chart would suggest whether the gene for the character is autosome-linked dominant or recessive, X - chromosome linked dominant or recessive,

Y- chromosome linked or not.

Explanation

1. Autosome Linked Dominant traits: These are the traits whoseencoding gene is present on any one of the autosomes, and the wild-type allele is recessive to its mutant allele, i.e., the mutant allele isdominant.

The pedigree-chart can be of the undernoted pattern (Fig. 11.2), wherethe female being interviewed is exhibiting the trait, and is indicated byan arrow-mark in the chart.

The characteristic features of inheritance of such type of traits are:

(a) Transmission of traits occurs from parents of either sex.

(b) Males and females are equally affected.

(c) The pedigree is vertical, i.e., the trait is marked to be present in each ofthe generations.

(d) Multiple generations are characteristically affected.

Brachydactyly, polydactyly, dimple in the cheek are some of the commontraits of this type.

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EXERCISE 11

2. Autosomal Recessive trait: These are the traits whose mutant allele isrecessive to its wild type allele.

The pedigree chart can be more or less of the pattern given below (Fig.11.3), where the lady (marked by the arrow) is showing the trait. The bar

in the example represents the presence of corresponding dominant orrecessive allele for the specific trait.

Suppose the given trait is albinism. Denote its dominant allele as ‘A’that produces pigments, and the recessive allele as ‘a’ that fails to synthesisethe pigment, melanin. The female (our subject in generation III) is thereforeof genotype aa. She must have received each of her ‘a’ allele from both theparents (generation-II), who are therefore themselves normal but are definitelyof genotype Aa, and are carriers of the trait. The allele a must also have beenpresent in her grand parents too, of course in heterozygous condition alsoto make them carriers (generation-I)

Albinism in the subject’s children (generation-IV) suggests her husbandtoo to be of genotype Aa, a carrier. Marriage of her albino daughter to analbino man is bound to produce all her grand-children albino (gen-V).

The following are the salient features of the inheritance of such type of traits.

(a) Occur in equal proportions in multiple male and female siblings, whoseparents are normal but carriers;

(b) The siblings are homozygous for the defective allele, but their parents,though some may appear normal, are obviously heterozygous, i.e.,are merely carriers of the trait.

(c) Consanguinity (marriage between man and woman genetically relatedto each other, such as cousins) occasionally results in the appearanceof such traits.

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3. X-Linked Dominant traits: These are the traits whose encoding geneis present on the X- chromosome, and the mutant allele of which isdominant over its wild-type allele.

Such traits are very rare, and are almost difficult to find in thepopulation. One example is oral-facial-digital syndrome (DucheneMuscular Dystrophy), which results in absence of teeth, cleft (bifid) tongueassociated with mental retardation. The pedigree chart may appear asfollows (Fig. 11.4):

The possible genotypes of the above pedigree can be written as follows(Fig 11.5):

Fig. 11.5 Genotypes of individuals shown in Fig. 11.4

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EXERCISE 11

Here, the dominant mutant allele is denoted by ‘D’, and its recessive wildtype allele is denoted by ‘d’. Remember that human females have twoX-chromosomes (XX), and the males have only one X and one Y chromosome.Males receive their lone X-chromosome from their mother, and theY-chromosomes from their father, whereas females receives one of herX-chromosome from her mother, and the other X from her father.

The characteristics of such inheritance are:

(a) The trait appears in almost all the generations, and the inheritance isvertical.

(b) If the female is affected, then about half of her sons are affected.

(c) If the male is affected then all of his daughters would be affected, butnone of his sons are affected.

(d) In short, the pedigree resembles the pattern of inheritance of autosomaldominants, except that there is no male-to-male transmission.

4. X-linked Recessive traits: These are the traits whose encoding gene ispresent on the X-chromosome and its mutant allele is recessive to itswild-type allele.

Red-green colour blindness and hemophilia, are some of its well knownexamples. The characteristic features of such inheritance are:

(a) Females express the trait only when they are homozygous for themutant allele, whereas the males do so even when they are hemizygousfor it.

The pedigree chart would appear as the following one (Fig. 11.6):

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(b) About half of the sons of the carrier (heterozygous for the trait) femalesare affected. In case of homozygous females showing the trait, fiftypercent of her daughters and all of her sons are likely to be affected.Therefore, the males are most affected in the population.

(c) Affected persons are related to one another through the maternal sideof their family.

(d) Any evidence of male-to-male transmission of the trait rules out theX- linked inheritance.

5. Y-chromosome linked traits: These are the traits whose gene is presenton the Y-chromosome. The females do not have any Y-chromosome,whereas all the males must have a Y-chromosome to be a male, and thisY-chromosome they get from their father. Therefore, any trait linked tothe Y- chromosome must be present only in males, and certainly not inany of the females. This is why these traits are also called male-sex limitedtraits. All the sons of the affected male would express the trait whereasnone of his daughters would do so.

The pattern of the pedigree chart would be as follows (Fig 11.7):

Hypertrichosis of the ear (presence of hairs on pinna) is one most commonexample of such traits.

Questions

1. How will you differentiate between autosome linked dominant and sex chromosomelinked dominant pedigree chart? Explain.

2. Discuss the differences in the patterns of autosome linked recessive and sex-chromosome linked pedigree.

Note: Students may be asked to prepare the pedigree-chart from given data and analyse thepattern of inheritance. The work may be done as a project.

Aim: To perform emasculation, bagging and tagging for controlled pollination

Principle: Conventional plant breeding programmes involve bringing under human controlreproductive processes that lead to seed and fruit formation. For this controlled pollination isdesirable using male and female parent having desired traits. One of the process that can be easilybrought under human control is emasculation. For this the knowledge of flower structure,mechanism of pollination, fertilisation and physiology of f lowering is essential for this. Inemasculation technique the stamens are removed before anthesis to obtain female parent and pollenfrom the desired male parent is transferred on to its stigma.

Requirement: Ornamental plants/ wild plants bearing large bisexual flower, magnifying lens,tweezers, small sharp scissors, brush, alcohol, rubber bands, paper bags, paper clips and tags

Procedure(i) Select a flower in bud condition where antheses has not occurred. Open

the bud carefully and remove the stamens (Fig. 12.1). Mark this asfemale parent plant.

Exercise 12

Fig. 12.1 Showing process of Emasculation

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Questions

1. Why is emasculation performed before anthesis?

2. What are the advantages of using a bag containing minute pores?

Fig. 12.2 Bagging of an emasculated flower

(ii) Cover the emasculated flower with a plasticbag to protect it from undesired pollen(Bagging) (Fig. 12.2). The bag should beheld securely in place with a paper clip/string/thread. Select the size of bag inaccordance with the flower size. Bagsmust be transparent with minute pores.

(iii) Bring into physical contact anthers of adesired male plant containing maturepollen grains with the stigmatic surfaceof emasculated female flower (Fig. 12.3).Use tweezers/brush if necessary to dustthe stigmatic surface with pollen.

(iv) Cover the pollinated flower again with thebag immediately. For identification, labelthe female parent (Tagging). Eachpollinated flower should bear a labelcontaining the name of the seed parent,the letter X (to signify a cross), the nameof the pollen parent, and the date on whichthe cross was effected.

Fig. 12.3 Showing cross pollination on anemasculated flower

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Aim: Staining of nucleic acid by acetocarmine

Principle: Acetocarmine combines with nucleic acid present in the nuclei of cells to form a deepred conjugate.

Requirements: Onion bulb, onion root tips, 2 to 4% acetocarmine/acetoorcein stains, slideand coverslips, brush/needle, pair of fine scissors, filter paper and microscope

Procedure(i) Peel off epidermis from the fleshy leaf of onion and put it on a slide.

Add a few drops of water over it to avoid desication.

(ii) Cut out a small piece (about 0.5 cm size) of the epidermal peel anddiscard the remaining portion.

(iii) Wipe out the water with a filter paper.

(iv) Put 2 drops of acetocarmine on the epidermal peel and heat gently ona spirit lamp.

(v) Apply a coverslip over the peel avoiding air bubbles and wrinkles ofthe material.

(vi) Wipe out the excess stain with help of blotting paper.

(vii) Examine the material under low magnification of a microscope.

ObservationRecord your observations with regard to shape of cell, the number of nucleiand their position in the cell. Draw a diagram based on your preparationand label its parts.

DiscussionNuclei in cells are extremely rich in nucleic acid which exist in a conjugatedform with protein to form nucleoproteinous structures, called chromatinfibres/chromosomes.

Exercise 13

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Questions

1. What are the building blocks of the nucleic acid?

2. What is DNA and how is it different from RNA?

3. Name different nitrogenous bases present in the nucleic acid.

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Aim: To identify common disease-causing organisms and the symptoms of the diseases

Principle: There are quite a large number of organisms that are parasitic/pathogenic to humans.These organisms substantialy damage the human body and cause diseases, which may even be fatalsometimes. These organisms exhibit characteristic features in their external morphology. Symptomsof the diseases caused by them are also specific.

Requirement : Preserved specimens/permanent slides/photographs of Ascaris, Entamoeba,Plasmodium, Ring-worm fungus and compound microscope

ProcedureObserve the preserved specimens/slides/photographs and note down thefeatures in the practical record book. Take care to observe all the minutedetails and draw labelled diagrams of the pathogens.

Exercise 14

Observation

A. EntamoebaObserve the following features of the parasite in the slide or photograph:

(i) It is unicellular.

(ii) Shape of the cell is irregular due topseudopodia.

(iii) A single nucleus is present eccentrically inthe cell.

(iv) *In the nucleus a peripheral ring of granuleof nucleoprotein and central karyosome areobserved. Rest of the space in the nucleuslooks empty (Fig. 14.1).

(v) A few food vacuoles may be seen in thecytoplasm. Contractile vacuoles are absent.

(vi) *Mature quadrinucleated cysts may bepresent.

Fig.14.1 An Entamoeba

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Note: Entamoeba is an intestinal parasite in humans and causes amoebicdysentery. The symptoms of the disease are frequent loose, mucus filledwatery stools, abdominal pain and spasms.

Systematic positionPhylum – Protozoa

Class – Rhizopoda

Type – Entamoeba histolytica

* Distinctive feature of the pathogen

B. Plasmodium vivax(i) It is an intracellular endoparasite seen easily within the RBC of the

infected person.

(ii) It is unicellular.

(iii) The most diagnostic stage of the parasite is "signet ring" stage in theerythrocytes, within which it appears as a roundedbody (Fig. 14.2).

(iv) It has a big vacuole inside, and the cytoplasm is accumulated at oneplace containing the nucleus. Because of the above mentioned features,the parasite appears as a ring.

Search the stage in the blood film slide, find the signet-ring stage, anddraw its labeled diagram.

Note: It is a protozoan parasite causing malaria in humans. When an infectedfemale anopheles mosquito bites a healthy person, it injects the infectivestage, sporozoite, into the peripheral blood vessels. The infective stageundergoes several rounds of multiplication in liver and erythrocytes.

Symptoms: Intermittent high fever with chills followed by profuse sweatingat an interval of alternate days.

Systematic positionPhylum – Protozoa

Class – Sporozoa

Type – Plasmodium vivax

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EXERCISE 14

C. AscarisThe external features of round worm are as follows:

(i) Body long (20 to 40 cm), cylindrical (5 to 6 mmdiameter) with no segmentation (Fig. 14.3).

(ii) Sexes are separate; the females are longer thanthe males.

(iii) Both the ends are pointed; posterior end of maleis ventrally curved.

(iv) Mouth is situated at the anterior end, and issurrounded by three lips, one present mid-dorsally and rest two lips are situatedventrolaterally (for viewing these lips a magnifyinglens is needed).

(v) Single longitudinal lines are present on the dorsal,ventral and on the two lateral sides, all along thelength of the body. Out of these the lateral linesare comparatively more distinct than the otherslines.

(vi) Excretory pore is present on the ventral surfaceslightly behind the anterior end.

(vii) In addition to the ventrally curved posterior tip,the male worm has a pair of penial spicules veryclose to the cloacal opening.

(viii) In case of female specimen a female genitalaperture is present mid-ventrally about one thirddistance from the anterior end.

Systematic positionPhylum – Aschelminthes

Class – Nematoda

Type – Ascaris lumbricoides

Note: Round worm or Ascaris is one of the common parasite found in theintestine of human beings.

Symptoms: (a) Irregular bowel, (b) Occasional vomiting, (c) Anaemia

Fig.14.3 Ascaris (a) Female (b) Male

Female genitalaperture

Mouth

Penial spicule

(a)

(b)

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Trichophyton (Ringworm fungus)It is a fungus that feeds on keratin of the skin of human beings. The featuresas observed under the microscope are:

1. Texture of hyphae is waxy, glabrous to cotton like.

2. Unstained hyphae are white, yellowish brown to reddish brown in colour.

Systematic positionKingdom – Fungi

Class – Deuteromycetes

Type – Trichophyton rubrum

SymptomsRingworm is a contagious fungal infection of the skin. Infected area of skinis itchy, red, raised, scaly patches (with sharply defined edges). It is morered on the periphery than in the center creating a ring like appearance.

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Aim: To study the texture of soil samples

Principle: Texture is one of the most important physical properties of soil. The soil texture is basedupon division of the size of soil particles into three size fractions viz., Sand (2–0.05mm averageparticle diameter), Silt (0.05–0.002mm) and Clay ( less then 0.002mm). If one of these fractionsdominates the properties of a soil, the name of that fraction is included in the name of the texture.A soil which has all of these fractions in nearly equal proportion is called a loam soil.

The four terms—sand, silt, clay andloam— are combined in various ways to name12 different textural classes. The 12 texturalclasses and the percentages of sand, silt andclay fractions that are included in each areshown in textural triangle (Fig. 15.1).

Texture affects several physicochemicalproperties of soil like density, capillary andnon-capillary pore spaces, water holdingcapacity, aeration, temperature and also theroot penetration.

Requirement : Oven/stove dried soilsamples, balance, weights, mechanical sieveset and blotting sheets/old newspapers

ProcedureThree methods are suggested here. Any one of these may be followed.

Method I

(i) Collect about 300–500g of soil from two different locations. Label them as sample A and B.

(ii) Dry the samples in an oven, or stove or in sun to remove the soil moisture (capillary andbound water).

(iii) Select the 3 sieves of different mesh sizes (2mm, 0.05mm and 0.002mm). Arrange them ina collecting chamber as shown in Fig. 15.2.

(iv) Place 200g of the soil in the Ist sieve (sieve of 2mm mesh) and close the lid. To sieve the soil,shake the set manually for 5–10 minutes and collect the three soil fractions.

Exercise 15

Fig.15.1 Soil Textural Triangle© N

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Soil sample Percentage (%) Texture class

Sand Silt Clay

A

B

Note for Teachers: The sieve sets contain a number and an abbreviation BSS/ASTM/ISS on each sieve. In the given table (Table No. 15.1) the corresponding aperture size ofthe sieves is listed. For example, BSS 30 sieve aperture size will be 500 microns.

Fig.15.2 Sieve set

2mm mesh

0.05mm mesh

0.002mm mesh

Collecting chamber

Lid

the precentage lines of silt run parallel to the clay side of the triangle and, (iii)

perentage lines of sand run parallel to the silt silde of the triangle. In reading

the textural triangle, any two particle fractions will locate the textural class at

the point where these two intersect.

(v) Weigh the soil fractions viz-sand, silt and claycollected in the 3 compartments

Wt of soil sample taken – .........g

Wt of sand fraction – .........g

Wt of silt fraction – .........g

Wt of clay fraction – .........g

The weight of three fractions should be equal to thetotal weight of sample taken for analysis.

ObservationsCalculate the percentages of the various soil fractions

and tabulate:

Calculate the percentages of sand, silt and clay

fractions.

Use the textural triangle now. Note that the three

sides of the textural triangle represent 0 to 100% of sand,

silt and clay respectively. Note that (i) the percentage

lines for clay run paralled to the base line of sand, (ii)

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EXERCISE 15

Appendix 1

Table 15.1 Mesh No. and the corresponding Aperture size

Sl. No. BSS Mesh ASTM Mesh ISS Mesh Aperture

1 4 5 480 4.75 mm

2 5 6 340 3.35 mm

3 6 7 280 2.80 mm

4 7 8 240 2.36 mm

5 8 10 200 2.00 mm

6 10 12 170 1.70 mm

7 12 14 140 1.40 mm

8 14 16 120 1.18 mm

9 16 18 100 1.00 mm

10 18 20 85 850 micron

11 22 25 70 710 micron

12 25 30 60 600 micron

13 30 35 50 500 micron

14 36 40 40 425 micron

15 44 45 35 355 micron

16 52 50 30 300 micron

17 60 60 25 250 micron

18 72 70 20 212 micron

19 85 80 18 180 micron

20 100 100 15 150 micron

21 120 120 12 125 micron

22 150 140 10 106 micron

23 170 170 9 90 micron

24 200 200 8 75 micron

25 240 230 6 63 micron

26 300 270 5 53 micron

27 350 325 4 45 micron

28 400 380 3 38 micron

29 25 micron

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Method II

Texture by Feel

The texture of the soil sample can also be estimated by feeling it in the dry,moist and wet states. Sand is coarse and gritty, silt feels smooth like flourand clay is sticky and plastic. The smallest soil particles that one can see arecoarse silt. Feel the known texture samples first, then feel the unknown onesand decide their textures.

Procedure(i) Feel the dry soil first. Does it crumble easily or is it hard to break?

Hard soil samples contain a moderate amount of clay.

(ii) Take in your palm a lump of soil sample about the size of a one-rupee

coin and wet it to the consistency of modeling clay. Try to press it intoa ribbon between the thumb and forefinger. An alternate test is to form

a wire by rolling the wet soil until it is about 1/8" in diameter.

(iii) If a long wire or ribbon can be formed readily the soil is plastic andprobably contains over 40% of clay. Its texture must therefore be clay/

silty clay/or sandy clay. If a ribbon or wire can be formed easily but

also breaks easily, the soil sample is probably a clay loam/silty clay

loam/or sandy clay loam. A heavy loam/silt loam/or sandy loamsample may form ribbon or a wire if the moisture content is just right

but these will be still weaker than the ribbons and wires formed by the

clay loam samples.

(iv) Next determine whether sand or silt is dominant. If there is a gritty feel

without the smooth floury touch of silt, choose a texture-name that

includes the word ‘sandy’. If the smooth floury feel predominates andthere is not much gritty feel, choose one of the ‘silty’ texture names.

Use the name without a prefix if neither smoothness nor grittiness

predominates (simply clay or sand or silt). Often this can best be

determined by adding more water until the soil is in a wet state.

If the soil is very sandy, you must choose between sandy loam, loamy

sand and sand. In the moist state, sandy loam samples will have some

tendency to stick together but loam sand and sand samples will not do so.Use the wet state to determine whether a sample is sand or loamy sand.

After handling wet sand, your hands will be moist but clean loamy sand will

make the hands slightly soiled.

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EXERCISE 15

Method III

Requirements: Soil samples, balance, weights, glass rod, standard sieves of2mm and 0.5mm mesh size, blotting sheets/old newspapers, evaporatingdish and water

Procedure(i) Collect 200-300g of soil samples from different sites, and dry them as

suggested previously to remove the moisture.(ii) Sieve the sample through a 2mm sieve to remove stones, pebbles,

roots etc.(iii) Take 100-150 g of the sieved soil sample and further sieve it through a

0.05 mm sieve to separate the sand fraction (collected in the sieve) fromsilt and clay (collected on a blotting sheet). Weigh the amount of sandfraction and silt + clay fraction.

(iv) Take a large evaporating dish (a shallow clay plate, glass trough or ashallow iron plate) and record its weight.

(v) Add the clay and silt fraction to the dish and note the weight.(vi) Add water to the dish leaving half an inch space empty at the top and

stir the liquid thoroughly with a glass rod taking care that the contentsdo not spill out. Allow it to stand for several hours. Decant off the cloudysupernatant liquid (clay fraction). Repeat the process three to four timesuntil the decanted liquid is quite clear.

(vii) Dry the silt left in the evaporating dish to dryness. Cool the dish andweigh it.

ObservationRecord your observation in the following table:

Subtract the weight of silt fraction from the weight of silt + clay fraction. Thedifference will be the weight of clay decanted.

Calculate the % of sand, silt and clay fraction of the soil and express the texture.

A B

Weight of the soil sample taken

Weight of sand fraction

Weight of silt & clay fraction

Weight of silt fraction

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Questions

1. Which type of soil is better for root-penetration and better aeration?

2. Among sandy and clay soil which one has higher water holding capacity? Explain.

3. If the clay content is high, will it affect soil fertility? Explain.

4. Which type soil has poor nutrient status and high leaching?

5. What kind of plants grow in smooth texture soil? Name two plants that grow inheavy-textured soil.

DiscussionCorrelate the texture with the plants growing in the area from which the soilsample has been collected. Discuss how the texture of soil can affect the rootpenetration, tillage, soil aeration, moisture content, water holding capacityand other aspects related to plant growth. In sandy soil the non-capillarypore spaces will be more and the capillary pore spaces will be less. Thecondition will be reverse in case of clay soil. The pore space in turn determineswater holding capacity, percolation rate, aeration, root penetration and soilflora and fauna. Clay particles are anionic colloides and adsorb mineralnutrients and minimise their leaching.

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Aim: To determine the water-holding capacity of soils

Principle: Water holding capacity of the soil is the amount of water retained in the capillaryspaces of the soil after the percolation of gravitational water into the deeper layers. Water holdingcapacity depends upon the capillary pore spaces in the soil. Sandy soil has very low water holdingcapacity, whereas clayey soils have very high water holding capacity.

Requirement: Soil samples from different sites (garden, road side, bank of river, paddy fieldetc.), Gooch crucible (china clay crucible with perforated bottom), filter-paper, pestle andmortar, petridish, beaker, glass rod, balance and blotting paper

Exercise 16

Procedure(i) Dig a small pit about 10cm x 10cm x 10cm, Scoop 100–300 g of soil

from the pit and collect it in a small polythene bag.

(ii) Remove the pebbles and large lumps from the soil sample.

(iii) Pass the soil through a coarse sieve to remove small lumps and deaddecaying leaves and twigs.

(iv) Spread the soil into a thin layer on a sheet of blotting paper or oldnewspaper and sun dry it for 2–3 hours or dry it in a pan kept onstove. Alternatively dry the soil sample in oven at 1080C for 1 hour.

(v) With the help of pestle and mortar grind the sample into fine powder.

(vi) Put a small disc of blotting paper at the base of the Gooch crucible.Weigh the crucible along with the blotting paper and note its weight.

(vii) Transfer the soil sample into the crucible. Tap the rim of the cruciblegently several times with the help of glass rod so that soil is compactlyfilled and forms a uniform layer at the top. Add more soil if necessary.

(viii) Weigh the crucible along with soil sample and note its weight.

(ix) Fill the petridish with water and place two small glass rods in it parallelto and at a small distance from each other.

(x) Place the crucible on the two glass rods in such a manner that itsbottom is in contact with water.

(xi) Leave the set up undisturbed till water appears at the upper surface ofthe soil. Wait till entire soil surface is wet.

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(xii) Remove the crucible and allow all the gravitational water to flow outfrom the bottom. When no more water percolates, wipe the bottom drywith the blotting paper.

(xiii) Weigh the crucible and note its weight.

ObservationRecord your observation in the following table.Calculate the % water holding capacity of the soil as follows.

Weight of crucible + blotting paper: A g

Weight of crucible + blotting paper+ soil sample before experiment: B g

Weight of dry soil: B - A= Cg

Weight of crucible + blotting paper+ wet soil sample after experiment: D g

Weight of wet soil after the experiment: D - A= Eg

Mass of water absorbed by soil: E - C= Ng

% Water holding capacity:

Sample No. Wt. of Crucible Wt. of Crucible Wt. of soil Wt. of crucible Wt. of wet Amount of % water+ blotting + blotting paper sample + blotting paper soil (D-A) water holdingpaper (A) + soil sample (B) (B-A) = (C) + wet soil (D) = E absorbed capacity

(E-C) = N

A Garden soil

B Road side soil

C……...

D……...

Tabulate your results as shown below

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EXERCISE 16

Questions

1. What are heavy soil and light soil?

2. Give examples of a plant seen in heavy soil and light soil.

3. How does pore space determine the % water holding capacity of soil?

4. Why is clay soil often referred to as physiologically dry soil?

5. Which type of soil is suitable for cultivation of crop plants?

6. How can water-holding capacity of soil be improved?

7. Dead decomposed organic matter is usually added in the fields before thecultivation of crops. Apart from providing the mineral nutrients, what additionalrole does organic matter play in the cultivation of crop plants?

DiscussionCompare % water holding capacity of soil collected from different habitatconditions. The variation in water holding capacity is due to varyingproportion of sand, silt and clay in the soil of different habitats. Soil withvery high proportion of sand have very low water holding capacity due tolarge pore spaces between the particles which enables the water to percolatefreely into deeper layers leaving upper layers practically dry. In clay soil,due to very small size of the pore spaces (fine capillaries) the water is retainedin the capillary spaces as capillary water. In these soil the water does notpercolate freely. Soil with more or less equal proportion of sand, silt and clay(loam soil) combines the properties of sand and clay and therefore hasoptimum water holding capacity and optimum soil-air for root growth.

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Aim: To study the ecological adaptations in plants living in xeric and hydric conditions

Principle: Successful adjustment of plants and animals under prevailing environmental conditionsis known as adaptation. For terrestrial plants, the habitats vary from extremely dry conditions as indeserts to extremely wet conditions as in marsh lands. For aquatic plants the habitats may vary fromdeep water bodies like oceans and lakes to shallow ponds and pools. The plants are adapted todiurnal, seasonal or annual fluctuations of the habitat conditions. For land plants the main limitingfactor is the availability of soil water whereas, for aquatic plants the main limiting factors are thefluctuations in water level, availability of gases like CO2 and O2 and the light intensity. Adaptationof land plants are primarily for conservation of available soil water, avoidance of bright sunlightand intense heat and for aquatic plants, adaptation are for conservation of gases and efficientutilization of available sunlight.On the basis of availability of water, plants are classified as:(a) Xerophytes : These are plants growing in extreme dry conditions throughout the year. Forexample, plants growing in deserts (psammophytes), on rock (lithophytes) or alpine plantsgrowing above 14000 feet altitude.(b) Mesophytes: These are plants growing in soils with optimum soil water conditions prevailingfor major part of the year.(c) Hydrophytes: These are aquatic plants growing in fresh to marine water.

The morphological, anatomical and physiological attributes of terrestrial plants are differentfrom the aquatic plants.

Requirement: Plant specimens from xeric and hydric habitat conditions. The specimens fromxeric condition may include a few cacti, succulents (Euphorbia, Bryophyllum, Kelancho) cycasleaves, pine needles, twigs of Acacia, Nerium, Parkinsonia, Casuarina etc. The aquatic plants:Salvinia, Eichornia, Pistia, Hydrilla, Vallisnaria, Utricularia, Lymnophila; some reeds like Typha,Phragmites, amphibious plants like Marsilea and halophyte like Rhizophora. Beakers, glassjars,microscope, slide, coverslips and rajor blades

Exercise 17

ProcedurePrepare temporary stained transverse sections of leaf, stem and root of thespecimens. Study the morphological and anatomical features of the plants

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EXERCISE 17

collected and look for the following adaptations. Write the name of the plantin which a particular adaptation is observed.

ObservationsRecord your observation in the given tables:

Xerophytes

Adaptations

1. Conservation of Water

2. Storage of Water

3. Prevention of loss ofwater by transpiration

4. Prevention of excessiveheat

5. Efficient mechanism ofwater absorption

Modifications(Morphological/Anatomical)

a. Leaves few or absent or representedby spines only

b. Petiole modified into leaf likestructure

c. Stem reduced, branching sparsed. In some cases stem flattened, leaf

like, green, photosynthetic innature

Thick, fleshy and succulentleaves as well as stem

a. Intercellular spaces reducedb. Spongy parenchyma/ palisade

parenchyma presentc. Stomata on lower surface,

sunken in stomatal pitsd. Leaves needle likee. Thick cuticle on leaf surface

a. Leaves covered with dense hairs;b. Leaf surfaces shiny or glaborousc. Leaf blade remains rolled during

the day

a. Long and profusely branched rootsb. Dense root hairsc. Well developed xylem

Examples (from thespecimen collected)

------

--------

---------

-------------

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Questions

1. Give three adaptive features of water hyacinth suitable to aquatic life.

2. What are the features present in plants of xeric habitat for the prevention ofloss of water?

3. What is the importance of succulent leaves and stem for a xerophytic plant?

4. Why is air stored between tissues in aquatic plants?

Hydrophytes

Adaptations

1. Buoyancy and resistanceto currents of water

2. Transpiration

3. Absorption of water

4. Gaseous circulation andstorage of air

5. Mechanical tissues

Modifications(Morphological/Anatomical)

a. Leaves long and cylindricalb. Petioles flexible to withstand

currents of water and to carry theleaf blade on the surface of water

c. Petioles are modified into airpockets

d. Leaf blade pale green in colour,finely dissected

e. Leaf blade waxy with thin cuticle

a. Stomata absent

b. Stomata present on upper surfaceof leaves

a. Poorly developed rootsb. Root hairs absentc. Roots with air pocket to help in

buoyancy

Parenchymatous tissue of stem, roots,petioles and leaves modified intoaeranchyma in the form of air channelsin

a. Rootb. Stemc. Petioled. Leaf

a. Poorly developed xylemb. Poorly developed sclerenchymac. Sclerides present

Examples

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Aim: To study the adaptations in animals living in xeric and hydric conditions

Principle: The aquatic ecosystem exhibits a different pattern of abiotic factors as compared to thosein terrestrial ecosystem. The temperature of the water, penetration of sun light, the physicochemicalcharacteristics of water body affect the growth and survival of the biotic community. In order to over-come the cumulatory effects of these factors, certain morphological and anatomical features, as wellas physiological processes develop in the organisms. These modifications in animals are called adaptivefeatures. We will study adaptations in selected animals living in aquatic and xeric condition.

Requirement: Animal specimen/models of xeric (rat, camel, squirrel) and hydric (fish, frog,prawn, etc.) conditions

Exercise 18

ProcedureObserve the animals provided and note down their adaptive features in theobservation table with example.

ObservationsHydric adaptations

Features

Body colour

Body contour

Locomotory

Adaptations Example

(a) On dorsal surface(b) On ventral surface

(a) Streamlined(b) Disappearance of neck constriction(c) Tail enlargement(d) Position of external nostrils(e) Loss of external ears(f) Position of eyes(g) Presence of eye protecting membrane

(a) Fins or fin-like expansions of the body wall(b) Loss of limbs(c) Webbed feet

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Integument

Mouth

Respiratory organs

Presence of dermal/epidermal derivatives(a) Scales(b) Hairs(c) Mucous glands(d) Oil glands

(a) Position(b) Presence of teeth

a. Upper jawb. Lower jaw

(a) Gills/lungs(b) Cutaneous

Xeric adaptations

Features

Moisture getting

Moisture Conservations

Body colour

Body contour

Skin

Limbs

Scrotum

Adaptations

(a) Preference for juices as food(b) Hygroscopic skin

(a) Storage of water in body(b) Avoidance of evaporation (non-perspiring)

(a) Protective mimicry(b) Predating mimicry

(a) Position ofa. Nostrils directly upwardb. Reduction to pin-head size

(b) Position of eyesa. Covering of eyesb. Size

(a) Hard(b) Spiny(c) Poison glands

(a) Speed(b) Long slender(c) Padded feet

Present or Absent

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EXERCISE 18

Questions

1. Name the features that helps a frog for aquatic life.

2. What are the adaptations present in xeric animals for conservation of water?

DiscussionYou may have noticed many features in the body of aquatic animalswhich support their life. As the different aquatic bodies vary to agreat extent, there are many other adaptive features you may notice.For example the aquatic organism in ponds, lakes, river and sea.

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Aim: To determine the pH of different water and soil samples

Principle: The pH value of a water/soil sample can be determined by (i) indicator dye method,(ii) electrometric method using a pH meter and (iii) colorimetric method. For routine purposesthe indicator dye method using universal pH indicator solution (containing a wide range of pHindicator dyes) or paper strips containing the pH indicators are preferred though it is not asaccurate as the electrometric method.

Requirement: Soil or water samples A, B and C collected from different sites (for examplesoils from road side, garden, humus rich sites; water samples from borewell, handpump, pond,sewage), balance, weights, filter paper, distilled water, measuring cylinder (50 mL), droppers,cavity tile, funnel, beakers (100 mL), funnel stand, universal pH indicator solution andpH indicator paper (narrow range and broad range)

Exercise 19

Procedure(i) Weigh 10 g of the soil sample A. Add 50 mL of distilled water to soil

sample to make a soil solution.

(ii) Filter the soil solution through a filter paper and collect the filtrate in abeaker. Label it as soil solution -A.

(iii) Take a clean dry porcelain cavity tile. Place 5 drops of soil solution Ain three cavities of the tile as shown in Fig. 19.1.

Soil - A Soil - B Soil - C

Universal pH indicatorsolution

pH indicator paper(Broad range)

pH indicator paper(Narrow range)

Fig. 19.1 Porcelain cavity tile

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EXERCISE 19

(iv) To the 5 drops of soil solution present in one cavity add 5 drops of

universal pH indicator solution. Note the colour developed and compareit with the colour chart given on the universal pH indicator

solution bottle.

(v) To the soil solution present in the second cavity, dip a small strip of broad

range pH indicator paper (pH 2-11). Note the colour and compare withthe colour chart given on the broad range indicator paper and get a rough

estimate of pH of the sample solution.

(vi) Choose a suitable narrow range pH indicator paper (for e.g. If the pH of

soil is determined by you as 8.0, choose a narrow range 7.0 to 9.0) anddip a small strip of it in the soil solution present in the third cavity. Note

the colour developed and determine the pH to the nearest possible valuewith the help of the colour chart.

Repeat the same steps for determining the pH of sample B and C. Follow thesame procedure for water samples collected from different sites.

ObservationRecord your observations in the given table.

DiscussionBased on the pH values obtained, categorise the samples into acidic, basic,neutral type.

Record the plant species present in the site from which the samples are collected.

Note for teachers: The colour developed should be noted against direct sunlight. Also, sometimes the soil solution colour may interfere with the readings.Thus one has to be careful while making the observations.

pH value as determined by Soil Samples

A B C

Universal indicator solution

Broad range indicator paper

Narrow range indicator paper

Table: Measurement of pH of soil samples A, B and C

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Questions

1. What will be the pH of chalk (calcareous) soil?

2. pH measurement with indicator paper is not very accurate. Comment.

3. Water logged soils are acidic. Comment.

4. Why are soil around mineral mining areas acidic?

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Aim: To study turbidity of water samples

Principle: Various characters that control the quality of water are taste, smell, colour, amount ofdissolved nutrients, dissolved O2 and CO2, pH and different types of plants and animals and theirdensity. Turbidity of the water body determines the depth upto which light can penetrate and thusaffects the distribution and photosynthesis of phytoplanktons and macrophytes. More turbid thewater body less is the thickness of its photic zone.In polluted water bodies turbidity is due to:1. Effluents: A water body which receives domestic sewage, run off from adjacent agricultural

fields and liquid wastes from nearby small and large industries remains turbid.2. Planktons: A water body may be turbid due to very high density of phytoplanktons and

zooplanktons, especially when the water body is rich in nutrients.

I. Secchi’s Disc method

Requirement: Secchi’s Disc, rope of moderate thickness, meter rod, black and whitepaints; paint br ush. Prepare a Secchi's disc by taking an iron disc of about 6 inches diameter, towhich a weight is attached in the centre on one side and an iron hook on the other side. Tie a plasticrope of sufficient length to the hook. Divide the upper surface of the disc into 4 equal segments andpaint two of these white and the other two segments black in such a way that black and whitesegments alternate with each other (Fig. 20.1).

Exercise 20

Procedure(i) Visit a nearby pond.

(ii) Reach to the center of the pond in a small boat.

(iii) Slowly immerse the Secchi disc into water vertically holdingthe rope tightly in the hand till the black and white segmentsof the disc just begin to disappear. On reaching to aparticular depth, the disc becomes completely invisible.Mark the length of the rope when the disc just disappears(say A cm).

Fig. 20.1 A Secchi’s Disc

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(iv) Slowly pull up the disc and find out the length of the rope where theblack and white segments of the disc just reappear (say B cm).

(v) Find out the mean length (X) of the rope by the following method.

(vi) Repeat the process at different sites of the pond.

Water body Depth at which Disc Depth at which Depth of Photic

disappears (A cm) disc reappears (B cm) zone

Pond Site 1

Site 2

Site 3

-

-

Tabulate the results in the given table

ObservationsThe value X represents the depth of the photic zone upto which sunlight

penetrates in the water body and photosynthesis takes place.

DiscussionGreater the value of 'X' less turbid is the water. In crystal clear deep lakes, thevalue of 'X' will be very high indicating, thereby, that the water body does nothave large quantities of flocculating silt or organic matter residues. This maybe due to no discharge of effluents or domestic sewage into the water body.The high clarity of water is also an indication of very less density of phyto andzooplanktons. These water bodies are called as non-productive or oligotrophic,while highly turbid water bodies are eutrophic in nature.

PrecautionsStudents are advised to perform this experiment under the strict supervisionof teacher to prevent incidents due to drowning.

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EXERCISE 20

II. Measurement of turbidity using measuring cylinder

Requirements: Water samples from different sources, three measuringcylinders (500mL) of the same height.

Procedure(i) Collect about 2 liters each water samples from different sources.

(ii) Transfer 500ml of water sample in the measuring cylinders of samevolume and height.

(iii) Mark the three cylinders A, B and C and leave them undisturbedovernight.

ObservationsObserve the amount of sediment settled at the bottom of each cylinder andalso note whether the water above the sediment is still turbid.

DiscussionDo all the samples show same amount of sediments?

Which sample shows maximum sedimentation and correlate it with thesource of the sample?

Find out whether in all the cylinders, water above the sediment is clear orturbid. Explain with reasons.

Draw conclusions on the basis of the observations.

Water sample Thickness of sediment Clarity of water—turbid/

semiturbid/clear

'A'

'B'

'C'

Record your observations in the following table:

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Questions

1. Is turbid water fit for drinking? Explain.

2. Why is the penetration of sunlight in any water body important?

3. Green plants are seen only in photic zone. Comment.

4. It is a common practice to use alum for clearing turbid waters. Explain.

5. Turbidity of water body varies with season. Comment.

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Aim: To analyse living organisms in water samples

Principle: The productivity and the trophic status of a water body is determined by assessing thenumber and type of organisms (micro as well as macro) present in the water body. Water body withvery high density of phytoplankton per unit area is a productive water body. Such water bodies areusually turbid and have high amounts of nutrients and dissolved oxygen. These water bodies supportfairly large number of organisms of different trophic levels. This is in contrast to non-productivewater bodies, which have very low density of organisms per unit area, fairly transparent waters withlow mineral concentration and dissolved oxygen and also fewer trophic levels. The status of healthof a water body can be determined by analysing water samples for the number and type of organismspresent in it at a given time. Such assays also help us to find out whether a water body is polluted assome of the organisms are strong indicators of water pollution.

Requirement: Water samples from different water bodies (lake, pond, river etc.), beakers,a few vials or small test tubes, slides and cover slips, watch glasses, dropper, compoundmicroscope and 5% FAA (Formalin Aceto Alcohol 5:5:90:Formalin: Acetic acid: Ethanol) aspreservative

Exercise 21

Procedure(i) Collect about a liter of water sample from nearby water body (pond lake,

reservoir, river etc).

(ii) Add about 5 ml of FAA to fix and preserve the living organisms present ineach sample at the place of collection.

(iii) In the laboratory, transfer the water sample into a measuring cylinder ofone litre capacity. Label each water sample to indicate the site from whichthe water sample has been collected.

(iv) Leave the water samples undisturbed for 48-72 hours.

(v) Decant off the clear water, leaving concentrated sediment at the bottom.

(vi) Transfer the sediment into a vial or a small test tube. Cork and label eachvial for future use.

(vii) With the help of a dropper, transfer a few drops of sediment liquid from avial into a watch glass. Dilute the sediment with water if the sediment ishighly concentrated.

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Questions

1. Why do you find few organisms in polluted water? Explain.

2. Why is FAA (Formaline Aceto Alcohol) added after collecting the water sample?

3. Name at least one phytoplankton and zooplankton commonly found in pollutedwater.

(viii) With the help of a dropper transfer a drop of water from the watch glasson the center of a slide and mount it. Blot the excess water using blottingpaper.

(ix) Prepare a few more slides of each water sample in the same way.

(x) Observe each slide, first under lower magnification and then under highermagnification.

Observations1. Record the different types of organisms present.

2. Count the number of organisms under each field of microscope.

3. Some of the commonly found organisms of water bodies are given inAnnexure 2.

DiscussionPrepare a list of organisms observed in each water sample and make anassessment of type and density of different organisms in each water sample.Polluted waters may contain very few types of organisms but in very highdensity. The non-polluted waters will have large variety of organisms inlow density.

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EXERCISE 21

Cosmarium

Desmickum

Spirogyra

Stigeoelomium

Draparnaldiopsis

Moubeotia

Nitella

Dinobryon Euglena

Annexure 2

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Gymnodinum

Ceratium

Peridinium

Calothrix

Anabaena

Cylindespermum

Rivularia

Scytonema Fischerella

Gleotrichia

Cyanomonas

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EXERCISE 21

VolvoxChlamydomonas

Golenkinia

Coelastrum

Closterium

Pedrastrum

Hydrodictyon Chlorella

AnkistrodesmusStaurastrumScenedesmus

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Microcystis

Synechococcus

Nostoc

Spirulina

Lyngbya

PhormidiumSynechocystis

Oscillatoria

Gloeocapsa

M e l o s i r a

Chae toce ros

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EXERCISE 21

R h i z o s e l i n a

B i d d u l p h i a

B a c i l l a r i a

A m p h o r a

N i t z s c h i a

Centronella

Synedra

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Pleurosigma

Lauderia

SkeletonemaCoscinodiseus

Somphonema

Navicula

Fragilaria

CocconeisAsterionella

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Aim: To determine the amount of Suspended Particulate Matter (SPM) in air at different sitesin a city

Principle: Environmental pollution is the unfavorable alteration of our surroundings wholly orlargely as a by-product of man's action through direct or indirect effects of changes in energypatterns, radiation levels, chemical and physical constitutions of environment and abundance oforganisms. Substances that cause pollution to the environment are called pollutants. They are theresidues of things that man makes, uses and throws away. These residues pollute soil, water and air.The atmosphere in highly populated area is very rich in dust, smoke and SPM all due to vehicularexhausts and industrial emission.

Requirement: A few freshly cut broad leaves, Vaseline, laboratory balance, weights, brush,paper clips and twine thread

Exercise 22

ProcedureThis experiment is an outdoor activity and may be

conducted by assigning 2–3 students into a group.

(i) Collect a few locally available broad leaves from a

nearby tree plant (Canna, Peepal, etc.).

(ii) Wash the leaves gently in running water to remove

any dust settled on their surfaces.

(iii) Blot dry the surface area of the leaves. To calculatethe area of the leaf, trace the outline of the leaf on

graph paper (Fig 22.1). Within the traced areacalculate the total number of full squares, 1/2, 1/3

and 2/3 squares and individual small squares. Add

all the squares to get the total leaf area. Multiplytheir value with two to obtain total area of both the

surfaces.

(iv) Take 8–10 feet long twine thread and tie five leaves

leaving a foot distance in between. Apply an

extremely thin layer of vaseline on both surfaces ofeach leaf. Make a bundle of these leaves and pack

Fig. 22.1 Calculating the area of aleaf on a graph paper

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them in polythene bags. Ensure that the outer surface of polythenebag does not have any vaseline sticking on it.

(v) Make three such bundles of smeared leaves, each bundle containing 5leaves.

(vi) Mark bundles as A, B and C and carefully weigh each bundle of leavesalong with the polythene bags.

(vii) Select three spots (X, Y and Z) near by your school. Spots selectedshould be in a manner that spot 'X' has very heavy vehicular traffic,the spot Y has moderate traffic and spot 'Z' has little or no vehiculartraffic. At spot 'X' expose each leaf of bundle 'A' by stretching theattached thread and tie the two ends to two poles or branches of treespreferably at 10 feet height above ground. Keep leaves exposed forabout two hours.

(viii) After exposure at spot 'X', collect the leaves and carefully re-bundleexposed leaves and place them along with the string in the polythenecover 'A'.

Site Leaf bundle Weight of leaves (g) Weight of Total leaf areasample suspended (cm2) of five

particle (W2-W1) leaves

Before Afterexposure (W1) exposure (W2)

X 'A'

Y 'B'

Z 'C'

Record your findings in the following table:

(ix) Repeat the same process at spot 'Y' and 'Z' exposing leaves of 'B' and'C' bundles respectively.

(x) At the end of the experiment, return back to the laboratory. Reweigheach bundle of exposed leaves along with their respective polythenecover.

- Calculate the amount of suspended particles deposited in mg cm2 ofleaf at each spot.

- Compare the results of three different spots and interpret.

Since the weight of suspended particles will be in milligrams or even lessit is advised to use a very sensitive laboratory balance.

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Aim: To study plant population density by quadrat method

Principle: Density represents the numerical strength of a certain plant species in the communityper unit area. The number of individuals of the species in any unit area is its density. The unit areamay be as small as 5 square cm to as large as 10 square metre depending on the size and nature ofthe plant community under study. For herbaceous vegetation a metre square quadrat is normallyused. Density which gives an idea of degree of competition is calculated as follows.

Total number of individual(s) of the species in all the sampling unit (S)Density=

Total number of sampling units studied (Q)

The value thus obtained is then expressed as number of individuals per unit area. When themeasured unit area is divided by the number of individuals the average area occupied by eachindividual is obtained.

Requirement: Cotton/nylon thread (five meters), 4 nails and a hammer

Exercise 23

Procedure(i) In the selected site of study, make a 1 m X 1 m quadrat with the help of

nails and thread. Hammer the nails firmly and make sure that thevegetation is not damaged while laying the quadrat.

(ii) List the names of the plant species seen in the quadrat (if the name isnot known mark these as species A or B etc., and the same species ifseen in other quadrats assign the same alphabet).

(iii) Count the number of individuals of each species present in the quadratand record the data as shown in the table.

(iv) Similarly make nine more quadrats randomly in the site of study andrecord the names and number of individuals of each species.

ObservationsRecord the total number of species seen in the ten quadrats. This will givean idea about the composition of the vegetation.

There will be difference in the species composition in the quadrats made inshady areas, exposed areas with bright sunlight, dry or wet areas etc.

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DiscussionPlants growing together exhibit mutual relationships among themselves andalso with the environment. Such a group of plants in an area represent acommunity. The number of individuals of a species varies from place toplace, making it necessary to take many random sample areas for reliableresults. Density values are significant because they show relative importanceof each species. With increasing density the competition stress increasesand the same is reflected in poor growth and lower reproductive capacity ofthe species. Data on population density are often very essential in measuringthe effects of reseeding, burning, spraying and successional changes.

Discuss the vegetation composition of the area (herbs/shrubs) andcomment on the dominant component species.

Questions

1. What factors influence the population density?

2. What is the significance of quadrat method?

3. What conclusion can be drawn if density of a plant species is low?

Plant Quadrats employed in study & no. of Total No. of Total no. Density (D)Species individuals in each quadrat individuals (S) of Quadrats

studied (Q)

I II III IV V VI VII VIII IX X

A 2 5 7 10 3 27 10 27/10 = 2.7

Z 1 2 4 8 3 2 20 10 20/10 = 2.0

Table 23.1: Density studies of the given vegetation

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Aim: To study plant population frequency by quadrat method

Principle: Frequency is concerned with the degree of uniformity of the occurrence of individualsof a species within a plant community. It is measured by noting the presence of a species in randomsample areas (quadrats) which are distributed as widely as possible throughout the area of study.Frequency is the number of sampling units (as %) in which a particular species (A) occurs. Thefrequency of each species (sps. A or sps. B or sps. X etc) is expressed in percentage and is calculatedas follows.% Frequency or Number of sampling units (quadrats) in which the species occurs

= Frequency Index Total number of sampling units (quadrats) employed for the study

Requirements: Cotton/nylon thread of 5 metres, 4 nails and a hammer

Exercise 24

Procedure(i) In the selected site of study, make a 1 m X 1 m quadrat with the help of

nails and thread. Hammer the nails firmly and make sure that thevegetation is not damaged while laying the quadrat.

(ii) List the names of the plant species seen in the quadrat (if the name is notknown mark these as species A or B etc. and if the same species is seen inother quadrats assign the same alphabet)

(iii) Similarly lay nine more quadrats randomly in the site of study and recordthe names of individuals of each species.

(iv) Calculate the percentage frequency of occurrence using the formula given.

ObservationsRecord the total number of species seen in the ten quadrats. This will give anidea about the composition of the vegetation.

There will be difference in the species composition in the quadrats made inshady areas, exposed areas with bright sunlight, dry or wet areas etc.

Observe that the frequency of occurrence is not the same for all species.

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DiscussionVariation in distribution of a species is caused by factors like soil conditions,quantity and dispersal of gemmules, vegetative propagation, grazing, predation,diseases and other biotic activities. Also frequency values differ in differentcommunities. They are influenced by micro-habitat conditions, topography,soil and many other environmental characteristics. Thus unless frequency isnot correlated with other characters such as density, frequency alone does notgive correct idea of the distribution of a species.

Frequency determinations by means of sample areas are often needed inorder to check general impressions about the relative values of species. Manyspecies having low cover or population density also rate low in frequency, butsome may have high frequency because of their uniform distribution. Usuallyif the cover and population density are high, the frequency will be high. Theplants with high frequency are wide in distribution.

Questions

1. If frequency of a plant is high, what will be your interpretation?

2. Can many micro-habitat in an area affect frequency of a species? Comment.

Plant Number of quadrats employed No. of quadrats PercentageSpecies in the study (Q) in which the of frequency

species is present (N) F=N/Q X 100

I II III IV V VI VII VIII IX X

A √ √ √ √ √ 5 5/10 100 = 50%

B √ 1 1/10 100 = 10%

C √ √ √ √ 4 4/10 100 = 40%

Table 24.1: Frequency studies for the given vegetation

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Aim: Study of homologous and analogous organs in plants and animals

Principle: In plants and animals there are several organs or parts thereof, apparently alike intheir function and appearance, but markedly different from each other in their origin andanatomical structure. These organs are called analogous organs, and the seeming similarityamong them is the result of convergence, that is, adaptation to similar habitat and identicalecological niche.

On the other hand, there are organs or parts thereof, which apparently are quite dissimilarto each other in appearance and perform different functions, but have the same origin andanatomy. The differences in their function and also in their appearances are the result ofdivergence, due to adaptive radiation to different habit, habitat and ecological niche. Theseorgans are called homologous organs.

Requirement: Plant specimens showing tendrils, thorns, etc., as given in the text or any otherlocally available plants, a plant with normal stem, potato and onion bulb, prickly pear, specimensof phylloclade, cladode, wings of bird, cockroach and bat, and cervical, thoracic and lumbarvertebrae of a mammal/lizard

Exercise 25

Observations

1. Homologous Organs in Plants

(i) Tendrils of passion flower and thorns of pomegranate

Tendrils of passionfruit and thorns ofpomegranate arestructurally andfunctionally differentbut they have similarorigin i.e. they arisefrom axillary bud(Fig. 25.1a & b).

Fig. 25.1(a) Tendrils of passion fruit (b) Thorns of pomegranate

(a) (b)

Tendril

Thorn

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(ii) Tendrils of Vitis and thorns ofCarissa

Tendrils of Vitis and thorns ofCarissa originate from the terminalbud, but they are functionallydifferent (Fig. 25.2 a & b).

(iii) Tendrils of baloon vine(Cardiospermum) and bulbils ofAgave.

Both are modifications of floral bud,but they perform different functions.Tendrils help in climbing but bulbilsare meant for reproduction(Fig. 25.3 a & b).

(iv) Scale leaves of onion and spinesof prickly pear (Opuntia)

Both the scale leaves and spines aremodifications of leaves but arestructurally and functionallydifferent. Scale leaves of onion arethick and fleshy and store food. Onthe other hand spines of cactus aredefensive organs (Fig. 25.4 a & b).

2. Analogous Organs in Plants

(i) Stem tendrils and leaf tendrils

All tendrils are analogous with oneanother, being structurally andfunctionally similar, irrespective oftheir origin.

Example: Tendrils of pea andtendrils of Vitis. Tendrils of pea aremodification of leaf and in Vitis it isthe modification of terminal bud(Fig. 25.5 a & b).

Fig. 25.2 (a) Tendrils of Vitis (b) Thorns of Carissa

(a) (b)

Tendril Thorns

Fig. 25.3 (a) Tendrils of baloon vine (b) Bulbils of Agave

(a) (b)

Tendril

Fig. 25.5 (a) Tendrils of pea (b) Tendrils of Vitis(a) (b)

Tendril

Tendril

Fig. 25.4 (a) Scale leaves of onion (b) Spines of cactus

Spines

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EXERCISE 25

(ii) Thorns and spines

Thorns and spines are analogous structures beingdefensive in function. Thorns are modifications ofaxillary or terminal buds, and spines aremodifications of leaves.

e.g: Thorns of pomegranate and spines ofprickly pear.

(iii) Modified underground stems and modified roots

Modified stems (rhizome, corm, tuber) areanalogous to modified roots (carrot, radish) as theyperform similar function of storage of food but theirorigin is different. Rhizome of ginger, potato tuber,Colocasia are stems and beetroot, radish etc. areroots. (Fig. 25.6 a & b)

(iv) Phylloclade, cladode and leaves

They perform the same function i.e. theyphotosynthesise but phylloclade and cladode aremodifications of stem. Phylloclade of Opuntia,Parkinsonia, Asparagus and leaves of any localplant like mango are analogous organs.(Fig. 25.7 a & b)

3. Homologous Organs in Animals

(i) Wings of birds, and forelimb of mammals/reptiles/frog: All have the same bony elements (humerus radio-ulna, carpals, metacarpals and phalanges), butperform different (flying in birds, for holding or walkingetc. in other) functions. (Fig. 25.8 a & b)

4. Analogous Organs in Animals

(i) Wings of dragonfly/cockroach/butterfly and ofbirds. (Fig. 25.9 a & b)

(ii) Mandible of cockroach and mandible (lower jaw) ofa vertebrate. (Fig. 25.10 c & d)

Fig. 25.7 (a) Phylloclade (b) Cladode ofruscus

(a) (b)

Questions

1. Suggest examples of homologous and analogous organs other than what aregiven in the manual.

2. Why are stem and leaf tendrils considered as analogous organs?

Fig. 25.6 (a) Modified root of carrot(b) Rhizome of ginger

(a) (b)

Note: Students and teachers are suggested to discuss more examples.

Spine

Fig. 25.8 Fore limb of (a) human (b) bat

(a) (b)

Fig. 25.9 Wing of (a) dragonfly (b) bird

(a) (b)

Fig. 25.10 Mandible of (a) cockroach(b) rabbit

(a) (b)

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Investigations are more open-ended than practical exercises involving a search to understandthe unknown and begins with a question or a hypothesis. You are not instructed exactly whatto do, but are given only general guidance. These give you more opportunity to plan yourwork. For example, you might investigate what traits you and your classmates inherit fromyour parents and forefathers (both maternal and paternal).

Projects are even more open-ended than investigations. These are practical investigationscarried out by an individual or a group of students. Projects are largely your own initiative. Italso requires evaluation of your findings, redefining ideas and designing further investigations.This may lead to evidence that enables answering the question posed at the outset. Some ofthese projects would take about few hours to complete. Other may take few weeks. Some arelaboratory based, others involve fieldwork. Many could be carried out at home.

Investigatory projects are part of obligatory assignment involving purely experimentalprocedures so that you report on, duplicate, or adapt something that someone else has alreadydiscovered. It may involve some other form of investigation also. For example, you may undertaketo investigate the richness and patterns of biodiversity (flora and fauna) in your school campusand prepare a mural of it or to investigate the effects of physical fitness on your pulse rate.

Investigatory Project Work

Choosing an Investigatory ProjectYou may be guided by your teacher for your choice of topic. The more originalor new the project is, the better it would be. But it must be realistic in terms ofthe time available and at a level attained in the higher secondary biology.

You must review the available literature to find out what type of work hasbeen done. This will help you to reject some of the alternatives, and possiblycause you to modify others. It may also be the source of new ideas.

By doing these investigatory projects you will gain experience of researchbesides providing opportunity for learning skills such as photography,electronics, etc.

Identifying the Objectives of the ProjectHaving identified a possible project, you should be able to identify and list thetentative objectives you hope to attain by completing that investigation. For example,

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INVESTIGATORY PROJECT WORK

¨ Suppose your project involves studying the biodiversity of birds in yourdistrict/state, examine the data in the light of some questions (say, how dothe birds in Rajasthan differ from those in Assam or Bihar?) yourinvestigation might attempt to answer.

¨ Suppose your project involves investigating leaf mosaics revealing thecomplexity of the growth correlations that lead to efficient light interception,suggest also the factors that might affect this type of study.

Keep the aim of your project simple. Investigate only one factor at a timeand never allow yourself to be side-tracked. Remember that time may be tooshort for follow-up and any fascinating secondary aspects that you may comeacross.

Designing ProjectsHaving established the objectives of your chosen project, you must have anexperimental design. This will allow you to collect the data you need in ascientific way to test the hypothesis. For example, if your project involvesinvestigating the hypothesis that stale milk contains more bacteria than freshmilk, devise the procedure you would adopt to carry out your investigation.

Planning InvestigationsHaving decided your topic for scientific investigation, you should give carefulthought to the plan of your investigation in some detail. These may include

¨ What hypothesis can you make?

¨ How can you ensure that the experimental tests and measurement you

carry out are accurate and reliable?

¨ What controls do you need?

• How many variables are you investigating? Correctly identify key variablesas independent and dependent.

¨ Are your variables discrete or continuous?

¨ Identify appropriate control variable for fair test.

¨ How many repeat observations or samples will you require?

¨ What instruments/equipment or techniques will you use to obtain relevantinformation? Identify suitable materials and equipment to be used.

¨ If your investigation requires the use of a questionnaire, design andstandardise before implementation.

¨ Is your intended procedure safe and ethically permitted, i.e., taking care of

the distress or suffering of living organisms and damage to the environment?

¨ How will you collect your data?

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¨ How do you plan to analyse your results? Would you employ statistical orother methods? Are scale range, interval, number of values chosen areadequate and reasonable ?

Executing the ProjectFollowing planning, a brief description of the expected procedures has to beapproved in advance by the teacher. Having decided what controls you needto use, list the components of your experiment and decide what quantities ofsubstances to use, how to set the experiment. You should also decide whattype of readings or measurements you are going to make, how often and howmany. Note the source of error, if any, that you come across.

¨ Handle instruments and equipments appropriately to give accuracy.

¨ Repeat measurement.

¨ Keep proper controls and the variables constant.

Reporting/Writing of ProjectA format, such as given below, can be followed.

(i) Title of the investigatory project: Write the title of the project, forexample, ‘Inheritance pattern of eye colour’.

(ii) Objectives: Express as clearly as possible the effect of one variable thatthe experiment is designed to investigate.

(iii) Materials needed: This might be just a list, or a diagram if a particularpiece of apparatus was used.

(iv) Method: Describe the procedure stepwise including the precautionstaken, if any.

(v) Result: A suitable chart or table for recording and organising yourreadings or measurements should be made out before you start theexperiment.

(vi) Analysis and interpretation: Observation data are factual, and maynot be as expected by you.

(vii) Discussion: Discuss briefly the implication of your results and suggestextensions of any kind that can be undertaken.

(viii) Conclusion: In view of the results obtained and related work done onthe topic of the project, write conclusion briefly.

(ix) References: Any work related to the project which you have come acrossthrough books/articles or any other source should be written as reference,for example: Michael Michalco (2001), Cracking Creativity, Berkeley, TenSpeed Press.

This write up is meant to train the students in scientific methods. In otherwords, it accentuates the spirit of enquiry and investigation in young minds.

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The operational aspects of doing a project include choosing a hypothesis orproblem to be investigated, collecting data in a designed manner, analysingthe data in a scientific way, drawing conclusions which are justified anddiscussing the results in the light of known knowledge and bringing out itsimportance. Finally it includes the scientific way of communicating the findings.

While your discovery during the investigatory project may not merit a NobelPrize it may help you discover something, a fact or relationship that wasunknown to you and that was not recorded in any book available to you.Scientists refer to this as an independent discovery. Your investigation willcertainly give a sample of the thrill of discovery.

Following are pages on procedural guideline about a few suggestiveinvestigatory project work.

1. Investigating the pH of a water sampleBackground information

Monitoring the physico-chemical properties of water is of vital importance.Normal maximum permissible limit of pH for our life and health is 6.5–8.5.

Abnormal levels of pH and their consequences are given below:pH 3 to 5 is too acidic for most organisms to survive, when the pH of water fallsbelow 4.5 most of the fishes die, leaving only a small number of acid-tolerantinsects such as water boatman and whirligig. These insects (beetles) can surviveand multiply even at pH 3.5. Similarly, pH>8.5 is too basic for most organismsto survive.

Materials needed

¨ Universal indicator test paper (broad range, narrow range PH 2–11)

¨ Water sample

2. Investigating the biochemical (also called biological oxygendemand [BOD]) of a water sample as pollution indicator.

Background information

A dissolved oxygen (DO) test measures the current status of oxygen in a waterbody. This is a useful starting point.

However, DO content can vary considerably from day-to-day as affected bymany factors like temperature, wind velocity, eutrophication, pollution, etc.The unpolluted water is characteristically rich in DO and low in BOD.

Higher the BOD, lower would be DO. Conversely, the polluted water hashigh values of BOD. Water for drinking should have a BOD less than 1. TypicalBOD value for raw sewage run from 200–400 mg of oxygen/litre. The maximum

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Concentration Indicator organism Site 1 Site 2

Organisms in order oftendency to disappearas degree of pollutionincreases

Red sledge worm (Tubifex worm)

Larvae of midge (Chironomus)

Blood Worm

Leech (Hirudinea)

Water louse, water skaters(Asellus)

Fresh water shrimp (Gammarus)

Water boatman

Diving beetle (Dytiscus)

Caddisfly larva (Ochrotricha)

Damselfly larva (Lestes)

Stonyfly nymph (Isoperla)

Mayfly nymph (Stenonema)

Snail (Lymnaea)

Clams (Corbicula)

Fungi

Bacteria (anaerobic)

Utricularia

Chara

Water fern (Salvinia)

Water velvet (Azolla)

Water meal (Wolffia)

Lesser duckweed (Lemna minor)

Greater duckweed (Spirodela)

Diatom

A

R

X

X

X

X

X

X

X

X

X

X

X

none

none

X

X

X

X

X

X

X

X

X

X

A

A

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

3. Population density of plants(i) Identify any 5 weeds from your locality.

(ii) Collect information about them from various sources. Focus on theireconomic importance especially their medicinal importance (collectsamples for herbarium preparation).

permissible limit of BOD followed by Central Pollution Control Board (India)for national water quality monitoring purposes is less than or equal to 3mg/L.

For example, in a study the prevalence of some organisms were done attwo different sites in a water body. The result can be tabulated on the basis offollowing facts:

A = Abundant; R = Rare; X = Very few

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(iii) Study their distribution in different localities by quadrat method.

(iv) Register their data and draw comparisons of their distribution byhistograms.

(v) Try to analyse the differences in distribution and density.

(vi) Correlate their presence with their habitat/adaptability.

4. To make an inventory of local tree, shrub and herbInformations can be listed in following categories:

(i) Avenue trees

(ii) Wind Breakers

(iii) Road dividers

(iv) Sound barriers

(v) Medicinal and other uses

5. Agrochemicals and their effectsThe project may be carried out in a survey mode with a questionnaire preparedwith the help of the teacher to cover the following aspects.

(i) List of pesticides used, amount used/hectare or acre, periodicity of spray,name of the crop plant grown, recommended dose and the dose employed,known effects on pest, whether the chemical pesticide is biodegradableor not, alternate ecofriendly biocides.

(ii) List of fertilisers used, cost incurred/acre/year, recommended dosage/time of use and the dosage used, known effects on fertility of soil, anydecrease in crop productivity, use of ecofriendly biofertilisers (VAM fungi,leaf mold, green manure, dung, etc.).

6. Ecological role of some animals observed in a local areaRecord the various plant species growing in the area under study—trees,shrubs, annual/perennial herbs, etc.

Note the season when flower/seed is formed.

Note the various types of insects, birds, reptiles, amphibians, mammals, etc.,and record their role as a/an

(i) Herbivore

(ii) Pollinator

(iii) Agent in seed dispersal

(iv) Prey

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(v) Predator

(vi) Vectors for transmission of diseases

(vii) Any other

07. Study the effect of a local industry on environment(i) Select an industry of your choice.

(ii) Note the source of energy used, product formed, raw materials used(locally available or imported) mode of transport used to move the finalproduct.

(iii) Possible types of pollutants released by the industry (air/water/soil).

(iv) Measures undertaken by the management to comply with the standardset by Central Pollution Control Board (CPCB), PCBs, etc.

(v) Awareness about ISO 2000.

(vi) Impact assessment carried out or not.

08. Study of the effect of chemicals and pollutants on the MitoticIndex of the mitotically dividing onion root tip cells

This study may include

(i) Growing of onion root tip cells in the solution of pollutant/chemicaland also in normal water as control.

(ii) Preparation and observation of slide for the study of mitotic index bothin experimental and control set-ups.

(iii) Analysis of the effect of pollutant/chemical by comparing the data ofmitotic index between experimental and control variable.

09. Study of the genetic markers in the human populationIn this investigation a few selected inherited traits can be investigated in thefamily members of a small population in the locality. The compilation andanalysis of data will provide an about prevalence of trait in the saidpopulation.

10. Inventory of weeds in aquatic bodies/agricultural fields

11. Inventory of birds in your locality, their ecological role asscavangers, pollinators, etc.

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12. Impact of local industry on the environment and the remedialmeasures taken by the industry

The guidelines and a brief outline of a few projects have been given with apurpose to design and perform such investigation. Students and teachers canthink and design investigatory projects based on almost all concepts aboutwhich experimental protocol have been given in the manual. However, a smalllist of suggestive projects are also given below. Please note that these are onlysuggestive and it is expected that students and teachers will take up manymore types of investigatory projects depending on the specificity of their area,need and problems.

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Aim: To study the effect of pH on seed germination

Principle: pH is one of the most important factors that controls the composition of flora andfauna in different terrestrial and aquatic ecosystems. pH of soils is essentially controlled by theamount and type of various minerals and also by the quality and quantity of humus (dead,decaying organic matter) present in it.

Seed germination is controlled by pH of the germinating medium. Seeds of different speciesprefer a specific range of pH for maximum germination. pH not only controls the germinationof seeds but also growth and development, reproduction and various other metabolic activities,of the plant.

Objectives: After completing the project, the students will be able to1. Plan out an experiment and understand the use of appropriate chemicals, apparatus and

equipment, and learn the preparation of solutions.2. Understand research methodology.3. Generate, analyse and interpret the data and draw conclusions.4. Conceive and choose other different themes related to pH and plant growth.

Materials required:1. 125 seeds each of sunflower, mustard, green moong, alfa alfa, fenugreek and barley (selection

of seeds of different species may be made as per their availability)2. Phosphate buffers3. Distilled water4. Petridishes of 7.5 cm diameter (15 pairs)5. Blotting papers cut into circular discs to the size of petridishes

Procedure(i) Prepare a range of pH buffers using Na

2HPO

4 and KH

2PO

4.

(ii) Wash the seeds with water and blot them dry.

(iii) Select an appropriate place in the laboratory where there is sufficientlight. Arrange petridishes in three horizontal lines, with 5 dishes in eachline. Arrange petridishes horizontally in three rows A, B and C with sevendishes in each row.

Investigatory Project 1

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INVESTIGATORY PROJECT 1

No and % of seed germination Total % of seedgermination

pH 1 2 3 4 5 6 7 A B C

4

5

6

7

8

(iv) Place one blotting paper disc in each petridish.

(v) Wet the blotting papers with small quantities of buffer solution of 4.0pH for petridish No. 1, 5.0 pH for petridish No. 2 and so on till all theblotting papers in petridishes No. 1–5 in row 'A' are wet with appropriatebuffer solution.

(vi) Repeat the process for rows B and C also.

(vii) Spread selected seeds in each of the five petridishes in such a way thateach petridish contains 25 seeds in row 'A'.

(viii) Repeat the process for rows B and C using two other types of seeds.

(ix) Cover the petridishes and record your observation after every 24 hoursfor 7 days. Tabulate your results as given.

ObservationObserve the emergence of radicle as an indicator of germination and record inthe table. Calculate the percentage of germination every day.

Use the data of the table for graphic presentation.

Inferences and conclusionThe inferences and conclusion may be drawn on the basis of observations and the points givenbelow

• Find out at what pH range seeds of different species had maximum % of germination.

• Find out at what pH seeds failed to germinate or showed minimum % of germination.• Is there any general pattern of seed germination with regard to pH ranges?

• What are the common features exhibited by the various types of seeds under varied pHranges? For example, did all the type of seeds show maximum germination in acidic range oralkaline range or did pH preference varied between acidic and alkaline ranges?

• Did you observe any relationship among time period, seed germination and pH range?

Table: Percentage of ................ seed germination

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Aim: Quantitative analysis of phytoplankton in a water body

Principle: The species composition and the density of the phytoplanktons determine theproductivity status of a water body. Phytoplanktons are the principal producers in a waterbody. Based on the density of phytoplanktons, the water bodies are classified into non-productive or oligotrophic and highly productive or eutrophic. Oligotrophic water bodiessupport only a few species, whereas eutrophic water bodies support large number of species.Further, the species composition of the phytoplanktons indicates the status of health of thewater body. Through phytoplankton assays, limnologists make an estimate of degree of pollutionin the water body. High density of cyanophycean algae, diatoms, volvocales, etc., are theindicators of high degree of pollution. It should be noted that density and the species compositionof phytoplanktons exhibit diurnal, seasonal and annual fluctuations. It therefore becomesimportant to monitor water bodies at regular intervals for drawing specific conclusions relatedto their ecology.

It is in this context, the procedure for phytoplankton analysis on qualitative (speciescomposition) and quantitative estimation (density/unit area) is suggested for students whowant to enter into fascinating realms of aquatic ecology.

Objectives: After completing the project, the students will be able to1. Plan out an experiment.2. Identify and quantify phytoplanktonic forms present in an aquatic ecosystem.3. Interpret the data and draw conclusions.4. Recognise the indicator species of pollution.

Requirements: Plankton net with plankton bucket, graduated plastic bucket (15 L), slides,cover slips, compound microscope, watch glasses, dropper, and 5% F.A.A (FormaldehydeAcetic acid, Alcohol)

Investigatory Project 2

Procedure(i) Plankton net resembles the butterfly net in several aspects. Plankton

net, however, is prepared from bolting silk cloth which is readily availableat shops dealing in scientific equipments and chemicals. Procure aboutone metre of bolting silk cloth of 40 mesh size and stich out of it a 40 cm

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INVESTIGATORY PROJECT 2

long cone with a diameter of about 20 cm at the mouth and a diameterof 3–4 cm at the other end (both the sides open). Fasten to the mouth ofthe cone a circular iron ring (with handle) of about 20 cm diameter withthe help of twine thread. Fasten a small steel bucket (plankton bucket)or glass specimen tube of 50 ml capacity at the lower end.

(ii) Visit a nearby pond, pool or river bank carrying along with you theplankton net fitted with the plankton bucket, graduated plastic bucketand 5–10 ml of F.A.A.

(iii) Since this is a group activity, ask your friend to hold the plankton netfirmly a few centimetres above the water surface.

(iv) Immerse and fill the plastic bucket with water completely upto 15–litre –graduated mark and filter the water through the plankton net. Repeatthe process several times (say10 times).

(v) At the end calculate the quantity of water in litres (X) filtered bymultiplying the amount of water in one bucket and number of bucketsof water filtered.

(vii) During this process of filtering, the planktons are collected in planktonbucket. Only the water free of planktons escapes through the mesh ofthe net.

Splash a few buckets of water against the net from outside taking care thatno water enters into the cone from the mouth. This will wash all the planktonssticking against the inside wall of the net into the plankton bucket.

Detach the plankton bucket from the net and add a few drops (1–2 ml) of5% F.A.A. to the plankton concentrate. Transfer the concentrate collected intoa suitable specimen tube and cork it. Note the volume of the concentrate (Y).

In the laboratory1. With the help of 1 ml pipette, draw 1 ml of concentrate and transfer it

dropwise into the watch glass. Count the total number of drops that make1 ml of concentrate (A).

2. Transfer one drop of plankton concentrate from the watch glass on a cleanslide. Cover it with square shaped cover slip. (For convenience divide thearea of the cover slip into parts with the help of lines drawn by Indian ink).

ObservationObserve the slide under microscope and count the number of total organisms(B) by moving the slide from one corner of the cover slip to another horizontallyas well as vertically till the entire sample under the cover slip is completed.With the help of following calculations find out the total number of differentorganisms per litre of water.

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Unit cells/L = 1000×(ABY)

Unit/L =x

Where A = number of drops in 1ml concentrate

B = number of organisms counted in 1 drop of concentrate

X = total amount of water filtered

Y = total volume of concentrate after filtration

Note: Another alternate method is the use of haemocytometer to calculate the

density of organisms under the guidance of teacher.

Inferences and conclusionFind out the density and composition of organisms in different water samples(polluted/non-polluted).

Note the common organisms in both the water samples and those specificto each sample.

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Notes

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Notes

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Documents

Laboratory Manual Biology not to be republished · PDF fileInvestigatory Project Work Project 1 : To study the effect of pH on seed germination Project 2 : ... Biology, like any other