This document contains laboratory protocols for experiments in cell biology and genetics. It includes procedures for observing mitosis in onion root tip cells using a squash technique, studying meiosis in onion flower bud cells using a permanent slide, examining Barr bodies in human buccal smear cells, and isolating chloroplasts from spinach leaves and mitochondria from yeast cells through differential centrifugation. The document provides detailed methodologies, materials required, and expected observations for each experiment to analyze key cellular processes like the cell cycle, meiosis, and intracellular organelle isolation.

1. CELL BIOLOGY
AND GENETICS
BSc III SEM Biotechnology lab manual

2. LAB MANUAL CELL BIOLOGY & GENETICS ,III SEM BIOTECHNOLOGY,SARDAR HUSSAIN
.ASST PROF,GSC CTA
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CONTENT
SL
NO
TOPIC PAGE
NO.
1 Cell division: Mitosis in onion root tip
2 Cell division: Meiosis in onion flower bud.
3 Isolation of chloroplast.
4 Isolation of Mitochondria.
5 Study of Buccal Smear –Barr body.
6 Use of micrometer and calibration
7 Measurementofonionepidermalcellsandyeastcellsby
micrometry.
8 Genetics problems
9 Pedigreechartofsomecommoncharacterslike-bloodgroup,color
blindness.

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Expt 1: study of MITOSIS in Onion root tip.
Aim: To prepare a temporary squash of Onion root tip to observe different stages of mitosis.
Principle: Mitosisis anequational divisionoccurringin allthesomaticcellof plants andanimals.The
resulting daughter cells carry the same numberof chromosomes as in the mother cell. It helps in growth
and repair of the body. In plants, the mitotic division rapidly occur in meristematic tissues of root and
shootapices.Inanimals,themitoticdivisionscaneasilyviewed inthebonemarrowtissueof vertebrates,
epithelial cell from gills of fishes, tail of growing frog tadpole, etc.
Requirements : Onion Bulb, Tray filled with sand or glasses tubes with water, Glacial acetic acid, Ethyl
alcohol, 2% Acetocarmine stain or Acetoorcein stain, 1 N HCl, 45% Acetic acid, Spirit lamp, Watch glass, Glass
slides, Cover slips, Blotting sheets, Molten wax or nail polish & Compound microscope.
Procedure:
I. Growing root tips:
 Take a medium sized onion bulb and carefully remove the old roots using a sharp blade.
 Place the onion bulbona suitable sized glass bottlefilled with water. Care should be takenso that
basal part of the bulb is always in contact with the water. Keep it there for about 5 days for new
roots to grow.
 Cut a few root tips [about 3 cm long] and transfer to a vial containing freshly prepared fixative
solution[1partglacialaceticacidand3partsofethylalcohol]forabout24hours.Thisprocessis
called fixation.
 Preserve the onion root tips in 70% alcohol.
II. Preparation of slide:
 Take onion root tips into a watch glasses.
 Add 1 N HCl and gently warm on a spirit lamp.
 Transfer the treated onion root tips on to a glass slide and stain with few drops of stain
Acetocarmine or Acetoorcein for about 30 minutes.
 Remove excess of stain by using blotting sheets.
 Add 45% acetic acid and cover the material with cover slip.
 Place the blotting paper on the top of cover slip tightly.
 Gently squash the material by tapping with blunt end of pencil to spread the material into the thin
film.

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 Seal the margins of cover slip using molten paraffin wax or nail polish.
Observation: Observe and record the different stages of mitosis under the microscope.
1. Interphase:Anon-dividingrestingphasewherethenucleusismetabolicallyactive.Inthisphasethe
cell prepare itself for division.
2. Cell division : It includes two major events – Karyokinesis [division of nucleus] and Cytokinesis
[division of cytoplasm]
Mitosis is divided into 4 phases – Prophase, Metaphase, Anaphase and Telophase.
i. Prophase :
 Chromatin becomes condensed thread like structures called chromosomes.
 Nuclear membrane and Nucleolus starts disintegrating.
ii. Metaphase :
 Nucleolus and nuclear membrane disappears.
 Chromosomes are thick and arranged on the equatorial plane of the cell.
 Each chromosome is made up of two chromatids held together at the centromere.
iii. Anaphase :
 Two sisterchromatids of eachchromosomesseparatedue to thesplittingof centromere
and move towards the opposite poles.
 Daughterchromosomes[separatedchromatids]duringtheirmovementappearV,J,L&I
shaped depending upon the position of the centromere.
iv. Telophase :
 Daughter chromosomes reaches the opposite poles.
 They uncoil to form chromatin Fibres.
 Nuclearmembraneandnucleolireappear.Thetwodaughternucleiappearatopposite
poles.
Cytokinesis:
 Aftertelophaseacleavagefurrowappearsaroundtheequatorialplatewhichgraduallydeepens
and divides the cytoplasm into two daughter cells.
 It occur by phragmoplast and cell plate formation in the plant cells.

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Simple protocol
For mitosis take 2-3 root tip in a watch glass.
Add 1-2ml of 1N HCL
Remove from it and put into another watch glass
containing acetocarmine solution for 1min.
Then take on slide add one drop acetocarmine on it.
Put cover slip.
Put small piece of filter paper gently on cover slip
hold it by one hand and tap on it with needle head.
Take out filter paper
Then add stain if necessary by cover slip side wall.
Then heat on spirit lamp gently.
Observed under high power.

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Images adapted from:

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Expt2:studyofMEIOSISinonionflowerbud.
Aim: To study the stages of meiosis.
Principle: Meiosis is a reduction cell division in which chromosome number is reduced to half (2n -> n). It is
also called germ cell division since it occurs in the reproductive cells of gonads. Meiosis includes two
successive divisions – meiosis – I (reduction division) and meiosis-Il (equational division). During interphase
between Meiosis – I and II there is no replication of chromosomes. Meiosis is generally studied by
observing the cytological preparation of the cells of the testis tubules of grasshopper or anthers of an
Onion flower buds.
Requirements: Permanent slides of meiosis and a compound microscope.
Procedure:
 Placethepermanent slideonthestageof themicroscopeand searchforthedividingcells under low
power objective.
 When dividing cells are located observe them under high power objective.
Observation: Record the observations and draw labeled diagrams.
DIFFERENT STAGES OF MEIOSIS AND THEIR FEATURES
1. Meiosis - I:
Sequential phases of Meiosis-I seen during karyokinesis are: Prophase – I, Metaphase -I, Anaphase-I and
Telophase-I.
1. Prophase-I: ThelongestphaseofmeiosisandisfurtherdividedintofivestagesnamelyLeptotene,
Zygotene, Pachytene, Diplotene and Diakinesis.
a. Leptotene (leptose - slender; tene-thread):
 First stage ofprophase-I.
 Chromosomesappearthinthread-likestructureswithbeadlikethickeningsallalongthelength
called chromomeres.
 Thetelomericendsofchromosomesconvergetowardsonesideofthenuclearmembrane
forming a loop like a bouquet.
 Nuclear membrane and nucleolus are intact.
 The centrioles start moving to opposite poles to form asters.
b. Zygotene (Zygon-paired):
 Second stage ofprophase-I.
 Homologouschromosomesundergopairing(orsynapsis)andeachpairedunitiscalledabivalent.
 Chromosomes undergo condensation.
 Nucleoli tend todisappear.
 Asters move apart towards the poles.

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c. Pachytene (Pachy- Thick):
 Third stage ofprophase-I.
 Each bivalent chromosome shows two sister chromatids and hence is called tetrad stage.
 Chromosomes of homologous pair are twisted spirally around each other.
 Exchangeof smallsegmentsorcrossingoverbetweennon-sisterchromatidsofbivalentsoccurs.
d. Diplotene (Diplos -double):
 Fourth stage ofprophase-I.
 Separation or terminalisation of bivalents begins.
 Bivalents are tied up at points of crossing over called chiasmata.
 Nucleolus disappears.
e. Diakinesis (Dia - opposite; Kinesis - separation):
 Last stage ofprophase-l.
 Separations of bivalents continue and are still held together at the region of chiasmata.
 Condensation of bivalent begins once again.
 Nuclear membranedisappears.
2. Metaphase I:
 Bivalents are arranged at the equatorial plate of the spindle.
 Bivalents are attached at the centromeres to the spindles.
 Arms of thechromatidsarepointed inwards.
 Centromeresarepointedtowardsthepoles.
3. Anaphase I:
 The two chromosomes in each bivalent segregate and move to the opposite poles.
 There is no splitting of centromere.
 Segregationofhomologouschromosomesandtheirmigrationtowardsoppositepolescause
reduction in chromosome number from diploidy to haploidy.
4. Telophase I:
 Segregated chromosomes reach opposite poles.
 Spindle fibersdisintegrate.
 Nuclear membrane and nucleoli are formed around the chromosomes at each pole.
 Chromosomes decondense to become a chromatin network.
Cytokinesis: A cleavage furrow appears around the equatorial plate which gradually
Deepens and divides the cytoplasm into two daughter cells.
2. Meiosis - II:
The sequential phases seen during Karyokinesis of meiosis - II are: Prophase - II,

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Metaphase – II, Anaphase – II and Telophase - II.
1. Prophase -II:
 Chromosomes are found in highly condensed state.
 Nuclear membrane breakdown and nucleoli disappear.
 Centrioles migrate to opposite poles.
 Spindle fibersappear.
2. Metaphase -II:
 Chromosomes are arranged on an equatorial or metaphase plate.
 Chromosomes are highlycondensed.
 Centrioles occupy oppositepoles.
 Chromosomes are held at their centromeres by spindle fibers produced by centrioles.
3. Anaphase - II:
 Centromere divide and chromatids move to opposite poles.
 Separated chromatids are now called as chromosomes.
4. Telophase - II:
 Daughter chromosomes reach opposite poles.
 Spindle Fibresdisintegrate.
 Nuclear membrane and nucleoli are formed around the chromosomes at each pole.
 Chromosomes decondense to become chromatin network.
Cytokinesis: A cleavage furrow appears around the equatorial plate which gradually deepens and
Divides the cytoplasm into daughter cells. Thus from two meiotic divisions 4 haploid daughter
Cells are formed.

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Expt 3: BUCCAL SMEAR- BARR BODIES.
Aim: To prepare a smear of buccal mucosal cells of human to study barr bodies.
Principle: Normal females have two egg chromosome one in active and other one in inactive. The inactive X-
Chromosome determine the Sex but does not contain functional genes. This inactive X-Chromosomes takes up
the stain and look like a barr bodies the bear represent the in active x- chromosomes.
Material Required:-
 Spatula
 fitter paper
 slides
 stain
 microscope
Reagent required:-
 Giemsa stain
 Methyl alcohol
 Distilled water.
Procedure:-
 Epithelial cells of buccal mucosa was gently scraped and spread over a clean slide.
 The smear is fixed by adding few drops of Geimsa stain to it and kept for 5-10 minutes.
 This slide is then washed with tap water and observed under microscope for barr bodies.
Result:-
 The buccal smear prepared and observed for barr bodies.

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Experiment no 04: Isolation of
Chloroplast Aim: To isolate chloroplast
from green plants Principal:-
Chloroplast are specialized organelles found in all higher plant cell these organelles contain chlorophyll
hence provided the green color they have a double outer membrane within the stroma are other membrane
structure the thylakoids granum where photosynthesis takes places chloroplast can be isolated in the laboratory
by using centrifugation technique.
Materials Required:
1. Spinach leaves 30 grams.
2. Clean sharp sand
3. 100 ml 0.5 M sucrose (17% w/v)
4. Cheese cloth 12X2
5. Ice bath
Equipment: Ice bath , Graduated cylinder, Mortar and pestle or blender, Funnel, Test tubes, Racks, 15ml
centrifuge tube,Weighing balance, Glass rod, Scalpel.
Procedure:
1. Grind 8gm of deveined spinach with ½ tea spoon sand in mortar.
2. Measure out 16ml of ice cold, 0.5 M sucrose solution and grind to a smooth pulp.
3. Homogenate through 8 layers of clean cheese cloth in a glass funnel into an 16X150mm test tube.
4. Centrifuge the filterate and spinut 50G for 10 minutes.
5. Decant the top 10ml into a clean cold centrifuge tube, discard sediment.
6. Now centrifuge the supernatant at 1000G for 10 minutes to precipitate chloroplast. Carefully decant the
supernatant but save the pellet.
7. Resuspend the pellet into an ice cold 0.5M sucrose with a ice cold glass rod. Keep on eyes all the time
examine suspended organalles under microscope.
Note : Keep all equipment and materials ice cold.
Result: The chloroplast were observed under 45X and 60X

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Experiment no 05: Isolation of Mitochondria
Aim: To isolate Mitochondria from Yeast cells.
Principle:
Materials Required:
1. Yeast culture.
2. Refrigerated Centrifuge.
3. 15ml centrifuge tubes.
4. Sodium Chloride (0.9%).
5. Micropipette.
6. Ice cold lysis buffer.
7. Shaker.
8. Mitochondria storage buffer.
9. Refrigerator.
10. 15ml micro centrifuge tube.
Procedure
1. Aseptically transfer the overnight yeast culture into two 15ml centrifugation tubes.
2. Centrifuge it at 500g for 10 minutes at 4o
C.
3. Carefully remove the supernatant without disturbing the pellet.
4. Carefully rinse the pellet in 1ml sodium chloride (0.9%) using a micropipette.
5. The sodium chloride from the centrifugation tube is discarded using a micropipette.
6. Resuspend the pellet in 1ml of ice cold lysis buffer and mix well using a micropipette.
7. Incubate it at 4o
C on a shaker for 10 minutes.
8. Centrifuge it at 1000g for 10 minutes at 4o
C and carefully remove the supernatant.
9. Resuspend the cell pellet in 1.5 ml ice cold disruption buffer and complete cell disruption by using the blunt end
of a needle.
10. Centrifuge the lysate at 1000g for 10 minutes at 4o
C.
11. Transfer the supernatant to a fresh 15mL tube and also mix the supernatant obtained from the step 7.
12. Centrifuge it at 6000g for 10 minutes at 4o
C and discard the supernatant.
13. Wash the pellet with mitochondria storage buffer.
14. Centrifuge it at 6000g for 20 minutes at 4o
C.
15. Resuspend the pellet in mitochondria storage buffer and store at -20o
C.
Note:
1. Before starting the experiment sterile the laminar air flow chamber using spirit.
2. All the tubes and bottles used in the experiment should labeled properly.
3. Always disinfect your work area when you are finished.

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Experiment no 06: Genetic Problems
1. When a tall (TT) pea plant is crossed with a dwarf (tt) pea plant. What will be the phenotype of F1?
When F1 is selfed, what will be the phenotype and genotype of F2?
2. What result would you expect if cross a purple flowered (Pp) pea plant with a white flowered (PP)
pea plant? Show both the genotype and phenotype of F1 generation.
3. In cats gene for short hair is dominant over gene for long hair. A short haired tom cat is mated with
anaugorafemale.Shebears8kittens,6 shorthairedand2longhaired. Compare withtheexpected
ratio.
4. In cattle polled cattle (hornless) is dominant over horned cattle, if a polled cattle is crossed with
horned one. What are the different types of offspring’s expected?
5. Ifalbinomanmarriesanormalpigmented womenwho hadalbinomothershowthetypesof children
that this couple may have and the proportion of each.
6. InDrosophila,vestigialwingsandebonybodycolorsareduetotwoseparaterecessivegenes.The
dominant alleles are normal wings and grey body color. What type of offspring would you expect
fromacrossbetweenvestigialwingsebonyfemaleandhomozygousnormalmale.IfnormalF1are
allowed to breed among themselves what type of offspring would be expected in F2? Show the
phenotype and genotype of both the generation.
7. In tomato, Yellow fruit and dwarf vine are due to recessive alleles of genes which produce the more
common red fruit and tall vine, if pollen from a pure line dwarf plant bearing red fruit is placed on
the pistil of a pure line tall plant bearing yellow fruit. What type of plant and fruit would be
expectedinF1?Ifthesearecrossedamongthemselves.Whatwouldbetheexpectedprogeny?
8. In a certain plant, blue (B) flower color is dominant to white (b). You have a blue flowered plant and
a white floweredplant.
a) What is the genotype and phenotype of each plant?
b) Whenthe plantsare crossed theplants and allthe offspringare blue.What are the
genotypes of the original plants?
c) What types of genomes will the blue-flowered plants procedure from the above cross; if two
of these blue-flowered plants are crossed to produce a large number of offspring’s?What
genotypes and phenotypes will be produced and in what proportional?
9. In pepper plants, green (G) fruit color is dominant to red (g) and round(R) fruit shape is dominant
to square(r) fruit-shape. These two genes are located on different chromosomes.
10. a) What gamete types will be produced by a heterozygous green round plant?
11. b)If two suchheterozygousplantsarecrossed,what genotypes andphenotypeswillbeseen inthe
offspring and in what proportions?
12. IsitpossibletocrosstwoAgoutirabbitsandproducebothchinchillaandHimalayanprogeny.11.
Twowhitefloweredvarietiesofpeaplantswhencrossed,produced112progenyplants,62With
purple flowers and so with white flowers.
a) What type of interaction is involved?
b) Give a phenotype ratio approximated by the F2,
c) Give the genotype of the parents.

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12. Coat colourin micedepends upon the action of at two genes. A dominant gene I inhibit the expression
of another non-allelic colour gene B and while colour is produced, the gene is expressed onlywhen the
recessiveconditionexists attheinhabitor locusii, thusiiBBproducesblack coatcolourand iibbgenotype
produces brown colour, if dihybrid(heterozygous) white mice are mated together, determine:
i. The ratio of the three phenotypes (white, black and brown) in F1 progeny.
ii. The chance of selecting a phenotype homozygous at both the loci.
a) From among whiteprogeny
b) From among the black progeny
13. InTessepigs,whentwosandyvaritiesarecrossed,progenywillshowredcolour.Whenredcoloured
pigs are selfed. In F2 red, sandyand white coloured pigs are obtained in theratio of 9:6:1. Explain the
concept.

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Experiment no 07: Human Karyotype
The human chromosome number was determined as 46 by I.H. Tjio and A. Levan in 1956. Earlier workers
had thought that the number was 48, that the number is indeed 46 has been confirmed since then by a
large number of workers.
The normal diploid complement of chromosomes in humans consists of 46 chromosomes in humans
consists of 46 chromosomes 44 autosomes and 2 sex chromosomes. In females, the sex chromosomes are
both X chromosomes. In males one sex chromosome is X and the other is Y.
EacheggproducedbythefemalecontainsoneX-chromosome,butthespermproducedbythemalecan
containeitheranXoronYchromosomes,theunionofanegg,whichalwaysbearsanXchromosomeswith
aspermalsobearinganXchromosomesproducesafemaleoffspringcarryingXX-chromosome,theunion
of aneggwithaspermthatbearsaYchromosomeproducesamaleoffspringcarryingXYchromosomes.
The human Y chromosomes is approximately one –third as long as the X and apart from its role in
determiningmalenessitappearstobegeneticallyinactive.MostgenesontheXhavenocounterparton
the Y.
Karyotype is the depiction of the chromosomes of an organism normally from a mitotic cell in metaphase
underthemicroscopeusingspecializedstainingtechniquesofvariouslevelsofsophistication.Itispossible
to depict and describe chromosomes quite accurately for a number of organisms.
Recenttechniquesevenallowstainingofeachchromosomepairwithdifferentfluoresceinstaininhuman
cells.Chromosomesizeisonecriteriontoconstructkaryotypes;theshortarmofhumanchromosomesis
named “P” for ‘petite’. The long arm was called “g” for ‘grande’.
However the long arm is now called by convention “q”. In addition to size the position of the centromere
is important for chromosome description.
Humanchromosomescanbeidentifiedbymorphologicallydifferencesandbythebandingpatternthey
demonstrate. A band is defined as part of a chromosome that is distinguishable from its adjacent
segments by appearing lighter or darker as a result of the new staining methods.
Ahumankaryotypeisthearrangementofsimilarchromosomesontosevenautosomalgroups:A,B,C,D,
E, F and G and sex chromosomes based on size and centromere location.
The criteria bywhich the chromosomes areassigned to the various groups are listed inthefollowing table.

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Group
Chromosome
number
Descriptions
A 1 to 3
Thelongestmetacentric,distinguishedfromeachother
by centromerelocation
B 1 to 3
Thelongestsubmetacentricsseparatedonthebanding
pattern
C 4 and 5
Medium-sizedmetacentric,identifiedonlyaccordingto
the banding
D 6 and 12 Medium –sized acrocentric with satellites
E 13 to 15 Short metacentric 16 and submetacentrics 17 and 18
F 19 and 20 Short metacentric
G 21 and 22 Short acrocentric with satellite
XX: About the size and shape of C group of chromosomes.
Y: Similar in size and shape to the G group chromosomes but do not have satellites.

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Experiment no 08: Inherited disorders
Autosomal Genetic disorders:
1. Down’s syndrome:
DownsyndromewasfirstdescribedbyLongdonDown,a Britishphysicianandhencecalledbythatname.
Individuals with Down syndrome have short stature, loose joints especially in the anklets, broad and short
skull, wide nostrils, large tongue, stubby hands and low mental abilities. The average life expectancy of
Down syndrome babies is low (16-20 years).
Note the presence of an extra chromosome at 21(TRisomy). In meiosis, the pairs of chromosomes should
separate and move to different spots in dividing cell, this event is called disjunction. However occasionally
one pair goes to one spot. This means that in the resulting cells, one will have 24 chromosomes and the
other will have 22 CHromosomes that is called mom-disjunction.
If a sperm or egg with an abnormal number of chromosomes merges with a normal mate, the resulting
fertilized egg will have an abnormal number of chromosomes/. In Down’s syndrome of all cases are caused
by this event. One cell has two 21st chromosomes instead of one, so the resulting fertilized egg has three
21st chromosomes hence the scientific name, trisomy. Recent research has shown that in these cases,
approximately 90% of the abnormal cells are the eggs.
2. Cri-du-chat-syndrome:
Cri-du-chat(cat’scry)syndromeiswellknowngeneticdisordercausedduetodeletion,thiswasdescribed
by Lejeune and Colleagues in 1963. It is called so because the child’s suffering from this have a
characteristic, high pitched cry very similar to a cat.
Thechildrenbornwiththisdisorderhasspecificphysiologicalproblemswhichresultintheirdevelopment
being delayed bothphysicallyand intellectually, theymayalso have healthproblemsbecauseparts of
physiology have not developed correctly.
The syndrome is associated with the malformation of larynx, abnormal physical features and the IQ of the
person will usually be in the range of 20-40, it is also associated with malformations of heart, brain,
kidneys, eyes and skeleton the frequency of this disorder is about 1 in 50000 live births. The deletion occurs
on a segment of short arm of chromosome 5 generally this occurs during gametogenesis. It is called a
deletion syndrome because part of the short arm is missing or deleted. That missing piece must contain a
certain region of the short-arm for cri-du-chat syndrome to result cri-du-chat syndrome is also known as 5P
minute syndrome and at cry syndrome [A chromosome has a narrow point called a centromere separating
the two segments or arms which are called the short and the long arms. The short arm is named p for the
French word ‘petite’ which means small, and the long arm is names q because it follows p in alphabet].
There are more children being diagnosed now the genetic testing is carried out more frequently and is

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moreaccurate. In addition since records of their nature arenot kept inmost countries, theactualincidence
is not known.
Major identifying characters
1. Monotone, weak, cat like cry.
2. Small head (smallhead-microcephaly)
3. High palate
4. Laryngomalacia (poor muscle tone in laryngeal area)
5. Round face
6. Small receding chin(microghathia)
7. Widely spacedeyes
8. Low nasal bridge (saddle nose)
9. Folds of skin over the upper eyelid (epicanthic folds)
10. Distinctive palmarcreases.

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