Follow the instructions given
Write the Record in physics or chemistry Record Book
Draw the diagrams on the left plain page and corresponding
notes on the ruled page
If diagrams are not clear see the similar diagram on the net
or draw from a good text book
Study of Mitosis in plant cells by Root Tip Squash Technique.
Aim: To observe the mitotic cell division and different stages of mitosis in Onion root tips.
Materials Required : Watch glass, holders, Slide, Cover slips, Needles, root tips, 70% ethanol, acetic acid, 1% acetocarmine, distilled water,1N HCL, Dropper and microscope.
Method of preparing Acetocaramine:
Acetocaramine is used for staining the root tips. 1% Acetocaramine is prepared for staining.
Preparation:
Take 55ml of distilled water in a beaker or conical flask. Add 45ml of glacial acetic acid to prepare 45% of acetic acid. Heat the solution until it reaches its boiling point. To that add 1 gm of Orcein powder, after heating the solution, cool it overnight. Filter the solution and use for staining.
Procedure:
Take a medium sized onion bulb (Allium cepa). Trim off the old roots at its base and place the onion bulb on a bottle of water with its cut base dipping in the water. Notice the emergence of a number of new roots after a few days.
As the onion roots grow to a length of 2-3 cms or more the terminal tips (about 5mm) are harvested between 9:00 am to 12 noon. And immediately fixed in Carnyos fixative of 3:1 – alcohol 70% and acetic acid 30% solution. This fixes the different stages of mitosis permanently.
After 15 minutes the roots are transferred to 70% ethanol and refrigerated.
The root tips are washed with water and are treated with 1N HCL in a watch glass for 5 minutes to soften the tissue. The root tips are removed from the HCL and placed in 45% Acetocaramine stain in a watch glass.
Warm the slide gently over the spirit lamp for about one minute. It is kept aside for 5 minutes so that stain is taken up by the cells. The tips of the root appear in dark red colour.
The stained material is placed on a slide and the dark red root tips are cut using a blade and remaining part of the root is discarded. Add one drop of acetocaramine stain. The root tips are covered with cover slip and topped by using a needle or pencil, so that, the material is squashed properly and evenly.
Observe the slide under a compound microscope in 10x objective. Scan and narrow down to a region containing diving cells and switch to 40x for a better view.
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Observation Mitosis occurs in somatic or vegetative cells hence it is also called as somatic cell division. Mitotic division results in the formation of two identical daughter cells. Hence it is also called as equatorial division. Mitosis includes karyokinesis (Nuclear division) and cytokinesis (cytoplasmic divison) Karyokinesis consists of 4 phases namely – i. Prophase, ii. Metaphase, iii. Anaphase, iv. Telophase. | |||||
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Prophase:
1. 1. The chromosomes are long, slender and spread extensively in the nucleoplasm.
2. 2.They are gradually condensing to short, thick, stout and rod shaped.
3. 3. Each chromosome has two arms called the chromatids which are united at the centromere.
4. 4. Nuclear membrane and nucleolus are dissolving slowly.
5. 5. The chromosomes appear to be randomly scattered in the cytoplasm.
Metaphase:
1. 1. Bipolar spindle is organised between two poles of the cell.
2. 2. Three types of spindle fibres are seen – continuous fibres, chromosomal fibres and interzonal fibres
3. 3. Chromosomal fibres are attached to the centromere of the each chromosomes and bring them to the middle of the cell forming the equatorial plane.
4. 4.The centromeres of all the chromosomes lie on equatorial plane and their arms are floating freely.
Anaphase:
1. Due to contraction of spindle fibres, the centromere of each chromosome divides and the two chromatids are separated, thus daughter chromosomes are seen
2. The spindle fibres are pulling the daughter chromosomes to the opposite poles
3. During the movement of the chromosomes to the poles, the centromeres lie ahead followed by arms.
4. The chromosomes appear in shape of V, L, or I
Telophase:
1. The daughter chromosomes are seen at the respective opposite poles.
2. They are decondensing to form the chromatin material
3. The nuclear member and nucleolus re-appears
Meiosis
Smear Preparation of Onion Flower Buds to study Meiosis
Aim: To observe the meiotic cell division and different stages of meiosis in Onion flower buds
Materials Required: Flower buds of Onion of appropriate size, Carnoy’s fluid, Acetocarmine stain, Glass slides, cover slips
Procedure:
Select appropriate flower buds of different size from the buds of Onion. Fix them in Carnoy’s fluid for 12 – 24 hours. Take a preserved flower bud and place it on a glass slide. Separate the anthers and discard the other parts of the bud. Add 1 or 2 drops of acetocarmine stain and squash the anthers.
Leave the material in the stain for 5 minutes. Place a cover slip over them and tap it gently with a needle or pencil. Warm slightly over the flame of a spirit lamp. Put a piece of blotting paper on the cover slip and apply uniform pressure with your thumb.
Observe the slide under the microscope for different meiotic stages.
Meiosis occurs in two stages called 1) Meiosis – I and 2) Meiosis – II
Meiosis I is heterotypic division. During this stage, the diploid parent nucleus divides into two daughter nuclei each having haploid set of chromosomes.
Meiosis II is homeotypic divison. In this the two haploid nuclei divide mitotically and form four haploid nuclei. Thus, from a single parent cell containing diploid (2n) chromosomes, four haploid (n) daughter cells are formed.
Draw the respective stage from the diagram
given below on the left plain page and notes on
the ruled page.
given below on the left plain page and notes on
the ruled page.
Meiosis I or First Meiotic Division
1. Prophase I
A) Leptotene:
i. Nucleus is enlarged in size
ii. Chromosomes are long and slender and show bead like structures called ‘chromomeres’.
iii. The chromosomes are arranged parallel to each other and well separated
B) Zygotene:
i. The homologous chromosomes are in pairs and are called as bivalents. The process of pairing is known as synapasis
ii. Synapsis may be proterminal, procentric or Random
iii. Nucleolus is enlarged
C) Pachytene:
i. Each chromosome is having two chromatids. Thus, in each bivalent, four chroomatids are seen. These are called pachytene tetrads
ii. X shaped structures called the chiasmata are seen between non-sister chromatids
iii.
D) Diplotene:
i. Homologous chromosomes are moving away from each other
ii. But they are attached at chiasma, which appear in ‘X’shaped structures.
iii. Chromosomes are condensed and thick
E) Diakinesis:
i. Chiasmata begin to move towards the chromosome ends, this is called terminalisation.
ii. The bivalents are very thick and short and are present in the periphery of nucleus.
iii. The nucleolus and nuclear membrane disappear.
iv. Chromosomes are released into the cytoplasm.
2. Metaphase I
i. Bipolar spindle apparatus present
ii. Spindle fibres are attached to the centromere of the each chromosomes and bring them to the middle of the cell forming the equatorial plane.
iii. Centromeres are directed towards the opposite poles and their arms towards the equator.
iv. The homologous chromosomes are fused by the chiasmata at the telomeric ends.
3. Anaphase I
i. Due to contraction of spindle fibres each homologous chromosome moves towards the respective poles.
ii. Two genomes (chromosome sets) are separated and carried to their respective poles
4. Telophase I
i. The chromosomes arrives at the poles.
ii. The chromosomes start elongating by lessening their coils
iii. Two daughter nuclei are organised each possessing a haploid (n) number of chromosomes
Genetics Problems
Write the Cross and Punnet square for each problem on left side plain page.
1. 1. When a tall plant is selfed it produced 64 plants having tall and dwarf phenotype. How many are tall and how many are dwarf?
Phenotype: Tall plants X Dwarf plants
Gametes T X t
Tt
Selfing of F1 parents:
F1 x F1
Tt x Tt
Phenotypic ratio : 3:1 (Tall Plants 3, Dwarf plants 1)
Genotypic ratio : TT : Tt: tt
1 : 2 : 1
Number of plants = 64
Number of dwarf plants = 64/4 = 16
Number of tall plants = 16 X 3 = 48
Result : Out of 64 plants, 48 are tall plants and 16 are dwarf plants.
2) What will be the result of selfing F1 generation in a cross when round and yellow seeded pea plants (YYRR) are crossed with green and wrinkled (yyrr) seeded pea plants.
Phenotype: Yellow Round X Green Wrinkled
Genotype : YYRR X yyrr ------------- Parents
Gametes : YR X yr
F1 Generation YyRr
Yellow Round
Self cross of F1 Generation : F1 X F1
Result : -
Phenotypic ratio = 9:3:3:1
Yellow Round : Yellow Wrinkled : Yellow Round : Green Wrinkled
9 : 3 : 3 : 1
Genotypic ratio = 1:2:2:4:1:2:1:2:1
YYRR:YYRr:YyRR:YyRr:YYrr:Yyrr:yyRR:yyRr:yyrr
1 : 2: 2 : 4 : 1: 2 : 1: 2: 1
1 – YYRR : - Homozygous Yellow and Round
2 – YYRr :- Homozygous Yellow and Heterozygous Round
Write the Remaining Genotypes
3) When round and yellow seeded pea plants (YYRR) are crossed with green and wrinkled (yyrr) seeded plants F1 are yellow and round seeded plants (YyRr). What will be the result when this F1 is crossed with round and yellow seeded plants?
Phenotype : Round and Yellow X Wrinkled and Green
Genotype : RRYY X rryy --------Parent
Gametes RY X ry
F1 Generation RrYy
Crossing of F1 Generation with Round and Yellow seeded plants
RrYy X RRYY
Result:
Phenotypic ratio : all are round and yellow seeded plants
Genotypic ratio : RRYY : RRYy : RrYY : RrYy
1 : 1 : 1 : 1
Write the names of Genotypes
4) In Garden peas tall plant habit ‘T’ is dominant over dwarf ‘t’. Green pods ‘G’ over yellow ‘g’. Bring out a cross between Tall Yellow with Dwarf Green and obtain F1 and F2. Give percentage of Tall Green homozygous among F2. Give the F2 genotypic ratio.
Phenotype : Tall plants and Yellow pods X Dwarf plants with Green pods
Genotype : TTgg X ttGG
Gametes Tg X tG
F1 Generation TtGg
Self Cross of F1 Generation TtGg X TtGg
Result :
F2 Phenotypic Ratio : 9:3:3:1
Tall Green : Tall Yellow : Dwarf Green : Dwarf Yellow
9 : 3 : 3: : 1
F2 Genotypic Ratio: TTGG: TTGg: TtGG : TtGg : TTgg: Ttgg : ttGG : ttGg : ttgg
1 2 2 4 1 2 1 2 1
Write the names of Genotypes
Percentage of Tall plants with Green Pods and F2 generation is
1/16 X 100 = 6.25%
5) In Snapdragon Red Flower ‘R’ is incompletely dominant over white ‘r’, the heterozygous being pink. The normal broad leaves ‘B’ are incompletely dominant over narrow leaves ‘b’. The heterozygous being intermediate leaf breadth. Find out the phenotype of the following crosses:
A. Red flowered Broad leaved plant crossed with White flowered Narrow leaved plant. What will be F1 and F2.
B. Test Cross
C. Back Cross
A) Phenotype : Red Flower Broad Leaves X White Flowered Narrow leaves
Genotype : RRBB X rrbb
Gametes RB X rb
F1 Generation RrBb ------ Pink Flowered and Intermediate leaves
Self Cross of F1 Generation
RrBb X RrBb
Phenotypic Ratio: 1:2:2:4:1:2:1:2:1
Write the names of Phenotypes
B) F1 X Recessive Parent
RrBb X rrbb
Gametes - RB, Rb, rB, rb X rb
| RB | Rb | rB | rb |
rb | RrBb | Rrbb | RrBb | rrbb |
Phenotypic and Genotypic Ration - 1:1:1:1
C) F1 X Dominant Parent
RrBb X RRBB
Gametes - RB, Rb, rB, rb X rb
| RB | Rb | rB | rb |
RB | RRBB | RRBb | RrBB | RrBb |
Write the names of Genotypes for all the three crosses
6) In a pea plant the allele ‘T’ for tallness is dominant over the allele ‘t’ for dwarfness and the allele ‘R’ for round seeds is dominant over allele ‘r’ for wrinkled seeds.
Give the phenotypes of the progeny of the following crosses:
i. TtRr X ttrr ii. TTRR X ttrr iii. TtRr X TtRr
I) TtRr X ttrr
Phenotype : Tall Plants and Round seeds X Dwarf Plants and Wrinkled seeds
Genotype : TtRr X ttrr
Phenotypic Ratio : 1:1:1:1
Tall Round : Tall Wrinkled : Dwarf Round : Dwarf Wrinkled
Genotpic Ratio : TtRR : Ttrr : ttRr : ttrr
1 1 1 1
II) TTRR X ttrr
Phenotype : Tall Plants and Round seeds X Dwarf Plants and Wrinkled seeds
Genotype : TTRR X ttrr
Gametes TR X tr
F1 Generation TtRr
(Tall Plants and Round Seeds)
III) TtRr X TtRr
Phenotypic Ratio: 9:3:3:1
Write the names of Phenotypes
Genotypic Ratio : 1:2:2:4:1:2:1:2:1
Write the names of Genotypes
7) In a plant a cross between Red flowered plant and White flowered plant yields plant of both the colours in equal proportion but a cross between two white flowered plants yields only white flowered plants. What could be the genotypes of the parents and which phenotype is recessive?
If a cross between Red flowered plant and White flowered plant yields plant of both the colours in equal proportion means it should be a test cross.
Phenotype : Red flowered plant X White flowered plant
Genotype : Rr X rr
Gametes R r X r
Rr rr
Red flowered White Flowered
And if a cross between two white flowered plants yields only white flowered plant, then it means that the white flowered phenotype is recessive.
Phenotype : White flowered plant X White flowered plant
Genotype : rr X rr
Gametes : r r
F1Generation : rr
White flowered
White flowered phenotype is recessive.
8. In a pea plant with round seeds is crossed with a dwarf plant having wrinkled seeds. The progeny is obtained in the ratio of
1. One Tall plant with round seeds
2. One Tall plant with wrinkled seeds
3. One Dwarf plant with round seeds
4. One Dwarf plant with wrinkled seeds
Derive the parental genotype.
1st Assumption:
Parents: TTRR X ttrr
Gametes TR X tr
TtRr
The progeny obtained does not match with that of the progeny given in the problem. So, the homozygous parent is not the actual parent.
2nd Assumption:
Parents: TtRr X ttrr
Gametes: TR Tr tR tr X tr
TrRr Ttrr ttRr ttrr
tall plant round seeds tall plant with wrinkled seeds dwarf plant with round seeds dwarf plant with wrinkled seeds
The parental genotype is heterozygous tall plant with round seeds and homozygous dwarf plant with wrinkled seeds.
=
9) A fully heterozygous grey bodied (B+) normal winged (Vg+) female F1 of fruit fly was crossed with black bodied (b), vestigial (Vg) male gave the following results:
Grey Normal - 126
Grey Vestigial - 24
Black Normal - 26
Black Vestigial - 124
a) Does this indicate linkage?
b) If so what is the percentage of crossing over?
c) Draw the cross showing the arrangement of the genetic markers on the chromosome.
B+ - Grey body Vg+ - Normal Winged
B - Black Body Vg – Vestigial wings
Parents B+ B Vg+ Vg X BB Vg Vg
Grey Vestigial B+ B Vg Vg - 24
⭃Cross over or parental types
Black Normal - BB Vg+ Vg - 26
Grey Normal B+ B Vg+ Vg - 126
Black Vestigial c Vg Vg - 124 ⭃
Crossover or recombinants
Crossover or recombinants
As B+ and Vg+ are present on the same chromosome they show linkage. As they enter the gametes together some fruit flies show parental characters, hence they are called parental types or Non-crossovers.
% of Non-Cross over = Number of Parental Recombinants X 100
Total number of progeny
300
= 250 X 100
300
The percentage of Non-Crossovers = 83.33%
% of Crossing over = Number of Recombinants X 100
Total number of progeny
= 24 X26 X 100
300
= 50 X 100
300
The percentage of Crossover = 16.6% centimorgans
Draw the Map distance on the left plain page.
10) The recessive gene ‘sh’ produces shrunken corn kernels and its dominant allele ‘sh+’ produces full plumpy kernels. The recessive gene ‘c’ produces colourless endosperm and its dominant allel ‘c+’produces coloured endosperm. A pure plumpy kernels and coloured endosperm is crossed with shrunken kernels and colourless endosperm. The F1 is crossed with recessive parent and produced the following progeny:
Shrunken Coloured -149
Shrunken Colourless - 4035
Plumpy Colourless - 152
Plumpy Coloured - 4032
a) Does this indicate linkage?
b) What is the crossing over percentage?
c) Construct the genetic map.
Parent - Sh+ C+ X Sh C
Sh+ C+ Sh C
Gametes Sh+ C+ Sh C
F1 Generation Sh+ C+
Sh C
Yes it indicates Linakge.
Sh+ C+ X Sh C
Sh C Sh C
% of Non-Cross over = Number of Parental Recombinants X 100
Total number of progeny
= 4032 X4035 X 100
8368
= 8067 X 100
8368
The percentage of Non-Crossovrs = 96.4%
% of Crossing over = Number of Recombinants X 100
Total number of progeny
= 149 X152 X 100
8368
= 301 X 100
8368
= 3.6%
The percentage of Crossover = 3.6% centimorgans
________7.2_________________5___________
Determination of Stomatal Frequency by Using Epidermal Peelings
60
Rf of Carotenes = 3.8cm
11) In corn a dominant gene ‘C’ produces coloured aleurone, its recessive allele produces colourless aleurone. Another dominant gene ‘Sh’ produces full, plumpy kernels, its recessive allele ‘sh’produces shrunken kernels, due to collapsing of endosperm. A third dominant allele ‘Wx’ produces normal starchy endosperm and its recessive allele ‘wx’ produces waxy starch.
A homozygous plant from a seed with colourless, plumpy and waxy endosperm is crossed to a homozygous plant from a seed with coloured, shrunken and starchy endosperm.
The F1 is test crossed to a colourless, shrunken, waxy strain. The progeny seed exhibit the following phenotypes
1. Colourless, shrunken, starchy - 113
2. Coloured, plumpy, waxy - 116
3. Coloured, shrunken, waxy - 601
4. Colourless, plumpy, starchy - 626
5. Colourless, plumpy, waxy - 2708
6. Coloured, shrunken, starchy - 2538
7. Colourless, shrunken waxy - 2
8. Coloured, plumpy, starchy - 4
a. Construct a genetic map of this region on Chromosome.
b. Calculate the coefficient of coincidence.
Dominant Recessive
i. Coloured aleurone – C Coloureless aleurone – c
ii. Full Endosper - Sh Shrunken endosperm - sh
Starchy Endosperm - Wx Waxy Endosperm - wx
→Phenotypes - Coloured, Shrunken, Starchy Endosperm X Coloureless, Plumpy, Waxy Endosperm
Genotype - C sh Wx X c Sh wx
C sh Wx c Sh wx
Gametes C sh Wx X c Sh wx
F1 Generation: C sh Wx
c Sh wx
(Coloured, Plumpy, Starch Endosperm)
Test Cross : C sh Wx X c sh wx
c Sh wx c sh wx
(Coloured, Plumpy, Starchy Endosperm) (Coloureless, shrunken, waxy endosperm)
Female gametes | Male gametes c sh wx | Progeny | |
Non-Cross Over | C sh Wx c Sh wx | C shWx/ c sh wx ( Coloured, shrunken, starchy) c Sh wx / c sh wx (Colourless, plumpy, waxy) | 2538 5246 2708 |
Single Cross over (C – Sh) | C Sh wx c sh Wx | C Sh wx/ c sh wx ( Coloured, Plumpy, Waxy) c sh Wx / c sh wx (Colourless, shrunken, starchy ) | 116 229 113 |
Single Cross over ( Sh - Wx) | C sh wx c Sh Wx | C sh wx / c sh wx (Coloured, Shrunken, Waxy) c Sh Wx / c sh wx (Colourless, Plumpy, Starchy) | 601 1227 626 |
Double Crossover (C – Sh) (C – Wx) | C Sh Wx c sh wx | C Sh Wx/ c sh wx (Coloured, Plumpy, Starchy) c sh wx / c sh wx (colourless, shrunken, waxy) | 4 6 2 |
Total | 6708 |
% of Non-Cross over = Number of Parental types X 100
Total number of progeny
= 2538 + 2708 X 100
6708
= 5246 X 100
6708
= 78.20%
Frequency (%) of crossing over between C and Sh
229 + 6 X 100
6708
= 3.5 %
Frequency (%) of crossing over between Sh and Wx
1227 + 6 X 100
6708
= 18.4 %
Frequency (%) of crossing over between C and Wx
229+ 1227 X 100
6708
= 21.7 %
The recombination value of C and Wx (21.7) is close to the recombination value of (C-Sh) + (Sh-Wx) = 3.4 + 18.4 = 21.9%.
This shows that gene ‘Sh’is present between the genes ‘C’and ‘Wx’.
On this basis the linkage map of genes C, Sh, Wx can be –
C 3.5 Sh 18.4 Wx
12. A kidney or bean shaped eye is produced by a recessive gene ‘k’ on the third chromosome of Drosophilla. Orange eye colour called ‘cardinal’ is produced by the recessive gene ‘cd’ on the same chromosome. Between those two loci is a third locus with a recessive allele ‘e’ producing ebony body colour. Homozygous kidney, cardinal females are mated to a homozygous ebony males. The tri-hybrid F1 females are then test crossed to produce the F2. Among 4000 F2 progeny are of the following:
1761 - Kidney, cardinal 97 - Kidney
1773 - Ebony 89 - Ebony, cardinal
128 - Kidney, ebony 6 - Kidney, ebony, cardinal
138 - Cardinal 8 - Wild type
a) Determine the linkage relationship in the parents and F1 tri-hybrid.
b) Estimate the map distance.
Dominant Recessive
i. Coloured aleurone – C Coloureless aleurone – c
ii. Full Endosper - Sh Shrunken endosperm - sh
Starchy Endosperm - Wx Waxy Endosperm - wx
Phenotypes - Kideny, Cardinal X Ebony
Genotype - K e+ cd X K+ e cd+
K e+ cd K+ e cd+
Gametes K e+ cd X K+ e cd+
F1 Generation: K e+ cd
K+ e cd+
(Wild Type)
Test Cross : K e+ cd X K e cd
K+ e cd+ K e cd
(Wild type) (Kidney, Cardinal, ebony)
Female gametes | Male gametes c sh wx | Progeny | |
Non-Cross Over | K e+ cd K+ e cd+ | K e+ cd / k e cd ( Kidney, Cardinal) K+ e cd+/ k e cd (Ebony) | 1761 3534 1773 |
Single Cross over | K e cd+ K e+ cd | K e cd+/ k e cd ( Kideny, ebony) K+ e+ cd / k e cd (Cardinal) | 128 266 138 |
Single Cross over | k e+ cd+ K+ e cd | k e+ cd+ / k e cd (Kideny) K+ e cd / k e cd (Ebony, k e cd Cardinal) | 97 186 89 |
Double Crossover (C – Sh) (C – Wx) | k, e, cd K+ e+ cd+ | k, e, cd/ k e cd (Kidney, ebony, cardinal) K+ e+ cd+/ k e cd (wild type) | 6 14 8 |
Total | 4000 |
The linkage relationships in the trihybrid F1 can also be determined directly from the F2 by for the most frequent F2 phenotypes are kidney cardinal (1761) and ebony (1773) indicating that kidney and cardinal were on one chromosome in the F1 and ebony on other
% of Non-Cross over = Number of Parental types X 100
Total number of progeny
= 3534 X 100
4000
= 88.6%
Frequency (%) of crossing over between k and e
226 + 14 X 100
4000
= 7.02 %
Frequency (%) of crossing over between e and cd
186 + 14 X 100
4000
= 5%
Frequency (%) of crossing over between k and cd
226+ 186 X 100
4000
= 11.3 %
The recombination value of k and cd (11.3) is close to the recombination value of (k-e) + (e-cd) = 7.2 + 5 = 11.2%.
This shows that gene ‘e’is present between the genes ‘k’and ‘cd’.
On this basis the linkage map of genes k, e, cd can be –
k 7.02 e 5 cd
13. In a four-o-clock plant, red coloured flowers ‘R’ is incompletely dominant over white coloured flowers ‘r’, the heterozygous plant being pink flowered. If a red flowered four-o-clock plant is crossed with a white flowered one, what will be the flower colour of the F1, of the F2 of the offspring of a cross of the F1 with its
Phenotype Red Flower X White Flower
Genotype RR X rr
F1 Generation Rr
Self Cross of F1 Rr X Rr
Phenotypic Ratio : Red : Pink : White
1 : 2 : 1
Genotypic Ratio: RR : Rr : rr
1 : 2 : 1
Test Cross : F1 X Recessive parent
Rr X rr
Phenotypic Ratio : Pink : White
Genotypic Ratio Rr : rr
1 : 1
Back Cross : F1 X Dominant parent
Rr X RR
Phenotypic ratio : Red : Pink
Genotypic ratio : 1 : 1
Phenotypic ratio : Rr : rr
1 : 1
Spotters: (Draw the Diagrams)
Mitochondria
1. It shows double membrane envelope.
2. The outer membrane is smooth
3. The inner membrane show invaginations called Cristae
4. The Central space is filled with fluid matrix
Chloroplast
1. It is spherical or oval in shape
2. It shows double membrane separated by an empty space called the periplastidial space
3. The inner space is filled with a colourless, homogenous matrix called the stroma
4. A large number of lipoprotein membrane extend from one end to the other and are arrange parallel to each other. These are called thylakoids.
5. At some places the thylakoids are arranged one above the other and appear as stack of coins. These are called grana.
Endoplasmic Reticulum:
1. It shows an network of microtubules extending from the outer nuclear membrane to the plasmamembrane.
2. It shows cisternae, vesicles and tubules
3. Cisternae are long, flattened, unbranched tubules, which are arranged in parallel arrays
4. Vesicles are large, rounded or spherical in shaped
5. Tubules are small, smooth walled branching structures of various sizes and shapes.
Golgi Complex
1. It shows a group of membrane bound structures
2. It shows cisteranae, vacuoles and tubules
3. Cisternae comprises 3-7 flattened sacs
4. Vacuoles are large and spherical
5. Tubules are single unit membrane and filled with amorphous substances
Nucleus:
1. It shows a spherical ball like structure bound by a double membrane envelope called the nuclear membrane
2. Inside the nuclear membrane is homogenous, semi-solid substance called the nucleoplasm.
3. Deeply stained net work like substances are present in the nucleoplasm called the chromatin reticulum
4. One or two spherical, deeply stained bodies called the nucleoli are found inside the nucleus.
Ribosomes
1. It shows ribonucleoprotein granules
2. They are composed of two unequal sub-units, namely larger and smaller
3. Some of they appear as free floating masses in the cytoplasm and some are attached to the surface of endoplasmic membranes.
Lampbrush Chromosomes
1. It shows exceptionally large sized chromosomes
2. It shows an main axis and the lateral loops
3. The four chromatids of the meiotic bivalent are held together by chiasmata and are seen as a long axis
4. The chromonema of these chromatids gives out fine loops at the lateral side giving the appearance of lampbrush
5. Each loop occurs in pairs.
Polytene Chromosomes
1. It shows a multistranded chromosomes
2. Several chromatids lie parallel to each other and are held together at the centromere
3. Each strand shows swellings called the puffs or Balbiani rings.
4. Each chromatids shows bands and interbands similar to a bar code.
Physiology
Determination of Osmotic Potential of Vacuolar Sap by Plasmolysis Method
Aim:
To determine osmotic potential in given tissue by incipient plasmolysis.
Principal:
When a living tissue is placed in hypertonic solution the cells exhibit plasmolysis i.e., the water diffuses into concentrated solution present in the external media. this poricess will continued until an equilibrium is attained between the solution and the cell present in the solution.
Requirements:
Petriplates, Measuring cylinders, beakers, slides, sharp blade, forceps, one molar sucrose solution, Rheo discolor leaves, watch glass, microscope.
Preparation of different Molar concentrations:
35:25 gm of Sucrose is weighed in simple balance and dissolved in 75ml of distilled water. Make its volume to 100ml by addding distilled water. Thus stock solution is prepared. Form this stock solution prepare a series of concentrations of sucrose solutions.
0.1 M Concentration: From the stock solution take 1ml of sucrose solution and add 9ml of distilled water.
O 0.2 M Concentration: 2ml of sucrose solution + 8 ml of distilled water.
0.4 M Concentration: 4ml of sucrose solution + 6 ml of distilled water.
Take each concentration of sucrose solution in respective pre-labelled petridishes and cover them with another set of suitable petridishes.
Procedure
Then peel the lower epidermis of Rheo discolor leaf with the help of fine forceps and keep them in petridishes containing water.
Epidermal peelings are cut into 0.5 X 0.5 cm with the help of a blade and are transferred into each petridishesh labelled earlier i.e., 0.1M, 0.2 M and 0.4 M pre-labelled petridishes and incubate for 30-45 minutes.
After incubation prepare a temporary slide by placing the epidermal peels on the slides. Mount the peels with the same molar solution from which it was taken. Observe the epidermal peels under the microscope for the plasmolysis.
Observarion:
In the beginning epidermal peeling of the controlled petriplate (with water) is taken on to the glass slide and observed under microscope.
All the cells under microscope filed are observed under turgid condition.
Subsequently the peels from 0.1M, 0.2 M and 0.4 M Sucrose solutions are observed under the microscopic filed.
Percentage of Plasmolysis:
Number of cells plasmolysed is counted and recorded in the table. Later the Osmotic potential is calculated by the formula:
% of Plasmolysed Cells = No. of Plasmolysed cells
_____________________ X 100
Total No. of Cells
Osmotic potential = ψs = 22.4 X M X T
___________________
T X Room Temperature
M= molarity of Sucrose solution where 50% cells are plasmolysed.
T = Atmospheric pressure = 273 mm of Hg.
gg
Write on Left side Plane Paper
Normal Incipient Plasmolysed cell
Cell Plasmolysis
Osmotic potential = ψs = 22.4 X 0.5 X 273
___________________
273 + 24
3057.6
_____________ = 10.29
297
Determination of Rate of Transpiration using Cobalt Chloride Method
Aim:
Determining the rate of Transpiration using the cobalt chloride method.
Principle:
The phenomenon of water difussion in form of vapour through aerial parts of plant body is called transpiration. Plants loose water through transpiration through the epidermal layers which consists of stomatal complexes. On the leaves the number of stomata present on the lower epidermis is more than the number of stomata present on upper epidermis.
Requirements:
Potted plants, Petirdishes, Dessicator, Whattman paper, Cobalt chloride solution, Forceps, Scissors, Simple balance, Stopwatch, Clips, Slides.
Procedure:
Whatmann paper (filter paper) is taken and cut into 1X1 cm strips.
Preparation of 5% Cobalt Chloride Solution:
5 gms of Cocl2 is dissolved in 50 ml of distilled water and volume is made to 100 ml by adding distilled water. the filter paper strips are dipped in the Cocl2 solution for two minutes and are transferred to petridish with the help of forceps and theses strips are allowed to dry in hot air oven. the dried filter strips are stored in desicator.
The strips become blue in colour when dried.
Now the filter strips are placed on both surfaces of the leaf of the plotted plants and strips are pressed closure to surface of the leaf by placing two glass slides. To hold the slides tightly, clip them at the sides.
Observation:
After a definite time the filter strips are observed. Note the time taken for change in colour of the paper strips from blue to pink on both surfaces of the leaf.
Results:
The time taken for colour change is less on the lower surface than on the upper surface of the leaf. This is because of stomatal frequency and increased rate of transpiration on the lower surface.
Write on Left side Plane Paper
( Write different timings)
Determination of Stomatal Frequency by Using Epidermal Peelings
Aim:
To calculate stomatal frequency in the given epidermal peelings.
Principle:
S Stomatal frequency is the number of stomata per unit area of leaf surface. Stomatal frequency varies in different plants. It depends on the transpiration rate. Stomatal frequency increases when transpiration rate is high and it decreases when transpiration rate is low.
Requirements:
Slides, Coverslip, Needle, Brush, Forceps, Compound microscope.
Chemicals Required:
Saffranin stain.
Procedure:
Preapare an epidermal peel from the dorsal and ventral surfaces of leaf. keep the strips on the slide and mount a drop of saffranin. Then cover the peel with coverslip. Observe the slide under a microscope first under 10x and then 40x.
Number of stomata on both the surfaces are counted and tabulated. thus, frequency can be calculated using the formula:
Stomatal frequency = No. of Stomata
___________ __ X 10,000
Microscope field
Write on Left side Plane Paper
60
= _____ X 10,000
61,600
= 9.740
40
= _____ X 10,000
61,600
= 6.493
Separation of Chloroplast Pigment by Paper Chromatography
Aim:
To Separate the chloroplast pigments by paper chromatography method.
Requirements:
Hibiscus leaves, Whatmann No.1filter paper (chromatography paper), mortar and pestle, beakers, capillary tuve, chromatographic chamber, pencil, distilled water, rubber cork.
Chemicals:
Acetone, Petroleum ether, Calcium Carbonate, Measuring
cylinder, muslin cloth.
Procedure:
Preparation of Paper:
Cut the chromatography paper i.e.,whatmann no.1 paper
into sheet of 15-20cm length and 2cm width. A line is drawn on the paper 2cm
above the bottom of the paper using the pencil.
Preparation of Extract:
5 gm healthy, Hibiscus leaves are taken and grind them
in motor and pestle after adding a pinch of CaCo3 and 10 ml of 80% acetone.
The extract is filtered using muslin cloth into a 50ml beaker and the
supernatant is collected.
Preparation of Solvent Mixture:
Solvent mixture is prepared by adding 90 ml of
petroleum ether to 10 ml of acetone i.e.,9:1 ratio respectively.
Loading Paper:
Put 5-10 drops of the leaf extract with a capillary tube
on the pencil line in the centre of the paper and allow the pigment to dry by
gently blowing air. Add the next drop on the top of the previous one and repeat
the process. By repeating this for 5-10 minutes results in building up of dry
pigments on one spot.
Pour sufficient amount of solvent mixture filling
about 2cm from the bottom of the chromatography chamber. Keep it as such for 30
min in order to bring the chamber to a saturated condition with solvent vapours.
Place the spotted paper vertically by hanging it from
the rod with the help of clips such that bottom of paper touches the solvent
and spot is just above the solvent level.
Place the lid tightly and allow the solvent movement
for about 30 min. Allow the solvent to reach upto 3/4th part of the
paper.
Remove the chromatograph paper and allow it to dry and
examine the paper when fresh.
Observatoin:
Each pigment component moves upward at its own rate and independent of another type of pigment component.
Result:
The sequence of pigment s from top to bottom is
Carotenoids - Orange Yellow (Max Rf Value)
Xanthophylls - Yellow
Chlorophyll a - Blue Green
Chlorophyll b - Yellow green (Min Rf Value)
Due to least solubility of chlorphyll - b it tranvesls minimum distance on paper or chromatogram. Due to greater solubility of Carotenes it travels maximum on paper.
Rf (Resolution factor) = distance travelled by the pigment or solute
____________________________________
distance travelled by solvent from origin
Write on Left side plane page
Rf of Carotenes = 3.8cm
____ = 0.38
9.9 cm
Rf of Xanthophylls = 3.2 cm
____ = 0.323
9.9 cm
Rf of Chlorophyll a = 2.0 cm
____ = 0.20
9.9 cm
Rf of Chlorophyll b = 0.9 cm
____ = 0.0909
9.9 cm
Estimation of protein content by Biuret Method
Aim:
To estimate the amount of protein present in a given unknown sample.
Principle:
The peptide bonds of protein react with cupric ion under alkaline condition to yeild a purple or violet coloured complex which shows an absorption maximum at 540 nm.
Requirements:
Protein Reference Standard:
Caesin is used as a reference standard protein (10 mg/ml). Weigh accurately 1 gm of caesin and transfer it to a clean and dry 100 ml volumetric flask. Suspend the protein in 50 ml distilled water and dissolve the protein by adding few drops (10ml) of 10N sodium hydroxide solution and make upto the volume to 100 ml.
Biuret Reagent:
Dissolve 1.5 gm of Cupric sulphate (CuSo4) in 250 ml distilled water. separately weigh and dissolve 6 gm of Sodium Potassium Tartarate in 250 ml distilled water. And then mix the Curpic Sulphate and Sodium Potassium Tartarate solutions in a litre beaker. To this solution add 300 ml of 10% Sodium hydroxide solution with constant stirring.
Make up the volume to 1 litre by adding distilled water using a volumetric flask. Store the reagent in a plastic container.
10% NaOH = 10 gm NaOH in 100 ml water.
Test Solution:
Take 1 gm germinating seeds of green gram or Bengal gram ( Remove seed coat). Grind in mortar and pestle. While grinding add 5 ml of 10% Trichloro acetic acid. centrifuge the extract for 10 min at 3000 rpm.
Transfer the supernatant in to a small beaker and add 5 ml of 10% NaOH to the precipitate. Again centrifuge for 5 min at 300 rpm. Again transfer the supernatant into 50 ml beaker/flask and make up the volume to 50 ml with distilled water.
Take 0.4 ml of this solution and estimate the protein concentration.
Procedure:
Take 0.1, 0.2, 0.4, 0.6, 0.8, 1.0, 1.2 ml of standard protein solution separately in 6 test tubes and take one test tube with 0ml of standard protein solution as blank.
To each test tube add distilled water to make the volume to 2ml.
Take 0.4 ml of the sample solution in separate test tube and make up the volume to 2ml with distilled water then add 5ml of Biuret reagent to each of the test tubes and mix the contents.
After 30 min of incubation at room temperature measure the violet colour against the reagent blank at 540 nm in Calrimeter and record the absorbance.
Construct a callibration curve on a graph paper by plotting protein concentration on X-axis and and OD on Y-axis.
Die Back Disease of Citrus:
1. It is caused due to deficiency of Copper
2. It shows large leaves with gummy tissue and longitudinal breaks
3. Fruits are brown, glossy and with splits
4. Affected shoots loose their leaves and lateral shoots produce bunchy appearance.
Blossom End Rot:
1. It is caused due to deficiency of Calcium.
2. brown, leathery rot developing on or near the blossom end of the tomato.
3. The depressed region remain surrounded by dark green tissue and flesh is
orange coloured.
Whiptail Disease of Cauliflower:
1. It is caused due to deficiency of Molybdenum.
2. Chlorosis of leaf is seen and whole leaf may trun white.
3. Leaf blades donot develop properly.
Grey Spec of Oats:
1. It is casued due to deficiency of Manganese.
2. Young leaves have grey/brown necrotic tissue with orange margins.
3. The plants are week, stunted and floppy and pale green or yellow coloured.
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