V Semester Record Work

⭃

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.

                                                             

Cut the root tips upto 1cm ↓ fix in Carnyos fixative (3:1 – alcohol and acetic acid) ↓ transfer to 70% ethanol and refrigerate ↓ treat with 1N HCL for minutes ↓ add 45% Acetocaramine stain ↓ warm gently over the spirit lamp ↓ transfer the root tips to slide ↓ Add 1 drop of Acetocaramine stain ↓ Cover with cover slip Tap with needle or pencil head ↓ Observe under a microscope

Draw the box on left plain page

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.

 

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.

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

Genotype:                 TT          X    tt

        

 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) B) 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.

1^st 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.

2^nd As  sumption:

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   ⭃
     Crossrsrossrscombinantsmbinants

  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

                                           = 

                                                       = 126 X124   X 100

                                                               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

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        

                            ________3.5____________18.4________________

 →                                     →

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        

                            ________7.2_________________5___________

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

Gametes               R                X         r

  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:

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.

Semester III Record

Note: Some of the diagrams are in either right or left side orientation, While drawing draw in correct orientation

Anomalous secondary Growth in Achyranthes Stem
The young stem has a wavy outline with alternate ridges and furrows.
Epidermis: Made up of single row of tubular cells. The cells are closely arranged with thick outer walls coated with lignin, followed by cutin. Several multicellular hairs are present over the cells of the epidermis.
Cortex: The peripheral hypodermal region is made up of collenchymas below the ridges and chlorenchyma below the furrows.
Endoermis: consists of single row of tangentially elongated parenchyma cells.
Pericycle: Made up of sclerenchyma, parenchymatous cells are also present.
Vascular Bundles: arranged in form of a ring.  They are conjoint, collateral, endarch and open. Medullary rays are present between the vascular bundles.
Apart from the primary vascular bundles, two medullary vascular bundles are present in the pith region. The medullary vascular bundles are conjoint, collateral, endarch and closed. These two bundles lie and grow opposite to each other.
Secondary Growth and Medullary Bundles:
In the pericycle region, extrastelar cambium strips is seen producing secondary vascular bundles. Cambium also produces the conjunctive tissue between the vascular bundles. So phloem of the secondary vascular bundles appears in the form of patches. This phloem is the included phloem.

Anomalous secondary Growth in Boerhaavia
In the primary structure, the stem can be distinguished into epidermis, cortex and stele
Epidermis:
 Single-layered epidermis consists of small, radially elongated cells. Multicellular epidermal hair arise from some cells. A thick cuticle is present on the epidermis.
Cortex:
 It is well-differentiated and consists of few- layered collenchymatous hypodermis followed by chlorenchyma.Collenchyma is 3 to 4 cells deep.
Chlorenchymatous cells are thin-walled, oval, full of chloroplasts and enclose many intercellular spaces. Endodermis is clearly developed and made up of many, tubular, thick-walled cells.
Pericycle:
 Inner to the endodermis is present parenchymatous pericycle but at some places it is represented by isolated patches of sclerenchyma.
Vascular Bundles:
Vascular bundles are present in three rings. In the innermost ring are present two large bundles; in the middle ring the number ranges from 6 to 14 while the outermost ring consists of 15 to 20 vascular bundles.
 Vascular bundles of innermost and middle rings are medullary bundles. All the Vascular bundles are conjoint, collateral, endarch and open.
Two vascular bundles of the innermost ring are large, oval and lie opposite to each other with their xylem facing towards centre and phloem outwards.
Vascular bundles of inner and middle rings may show a little secondary growth. The cambium produces only a little amount of secondary xylem to the inner side and secondary phloem to the outer side. As a result the vascular bundles of these rings become slightly enlarged.
Anomalous Secondary Growth:
In this stem several rings of cambia are seem successively in a centrifugal manner. The abnormal cambial ring produces xylem and conjuctive tissue on the inner side and phloem and parenchyma on the outer side. The resulting tissue gives the appearance of concentric rings of vascular bundles embedded in the conjuctive tissue.

Anomalous secondary Growth in Bignonia Stem
Primary Structure:
In T.S, the young stem exhibits the ridges and furrows in outline.
Epidermis:
 Single-layered epidermis consists of rectangular cells.A thick cuticle is present.  A few multicellular hair are also arising from some cells.
Cortex:
It is well-differentiated into collenchyma and parenchyma. Collenchyma is present below the epidermis in the ridges in young stem but at maturity there develops sclerenchyma.
 Parenchyma is present below the sclerenchyma or collenchyma in the ridges and directly below the epidermis in the grooves. Endodermis is undistinguishable from cortical cells. The cells lack casparian strips.
Pericycle:
 The pericycle shows alternate bands of sclerenchyma and parenchyma.
Conjoint, collateral, endarch and open vascular bundles are arranged around the pith in form of a ring.
Anomalous Secondary Growth
At four places, the cambium produces more amount of secondary phloem on the outside and relatively small amount of secondary xylem on the inner side. As a result four deep furrows of phloem projecting into the secondary xylem are seen

Anomalous Secondary Growth in Dracaena
Primary Structure:
Epidermis: Outer most layer made of single layer of cells with thin cuticle on the outside.
Cortex: Several rows of parechymatous cortex is present.
Stele: Several vascular bundles are scattered irregulary in the ground tissue. The vascular bundles are collateral, and closed.
Anomalous Secondary growth:
During the initiation of secondary growth, the parenchymatous cells of the cortex, external to the primary vascular bundles, become meristematic and forms the cambium in a ring. The cells of cambium divide and produce more cells towards the inner side and few cells towards outside. The cells produced on the inner side develop into vascular bundles and conjuctive tissue.
Each vascular bundle develops from a single cambial initial. The initial divides first by anticlinally to form a row of two or three cells. These cells undergo periclinal division, but the division soon become irregular. This results in the formation of a group of cells.
The peripheral cells develop into xylem elements, whereas the central cells differentiate into phloem elements. This leads to formation of amphivasal vascular bundles.
Formation of cork:
After certain period of secondary growth, the parenchymatous cells below the epidermis give rise to the meristematic initials. The initials divide several times periclinally. The cells thus formed become suberised and form the cork or phellem. This cork, with suberin coated cells and radial or storied arrangement is called as storied cork.

Anomalous Secondary Growth in Beta Vulgaris Root:

Periderm:
1. Outer most layer is consists of large periderm which is made up of cork (phellem), cork cambium(phellogen) and secondary cortex (phelloderm)
2. Endoermis is indistinguishable
3. Rings of vascular bundles are seen
4. Each ring has many bundles arranged very closely
5. Each vascular bundle is collateral
6. Each ring of vascular bundle is separated from the other by bands of storage parenchyma
7. Adjacent vascular bundles are also separated from each other by bands of radial parenchyma

Determination of pollen viability by Evans Blue Method
Aim :
To determine the viability of pollen grains by using Evans Blue stain method.
Principal :
Evans Blue stain penetrates into dead cell resulting in blue staining of cellular contents. The percentage of dead cells in each treatment is determined by scoring them with microscope, manually or by using a hematocytometer.
Chemicals and Requirements:
1% Evans Blue stain (dissolve one gram  Evans Blue stain in hundred ml distilled water and store it in refrigerator), fresh Pollen from any plant,  test tubes, centrifuge tubes
Procedure
Collect fresh pollen grains of any plant in centrifuge tube and add 1% Evans Blue stain solution to it so that it covers the pollen grains. Stir well gently with a needle and incubate for 3 minutes at room temperature. Centrifuge at 1000 RPM for 5 minutes. Discard the stain solution. Add distilled water and centrifuge as above. Repeat the washing twice or until the water is clear.
 Place the pollen grains on a glass slide. Observe them under a microscope to check for the stain and unstained pollens. Count the number of stained and unstained pollens in a single view. Take readings thrice and estimate the average value of pollen grains that have retain their original colour to be reported as viable pollens. Calculate the percentage value of pollen viability. The pollen grains can be analysed with hematocytometer for systematic counting.

% of Viable Pollen grains = 

                          No. of Viable Pollen grains.  X 100
                          Total number of Pollen grains

% of UnViable Pollen grains =

          No. of UnViable Pollen grains.  X 100
          Total number of Pollen grains

Isolation of Cenna Embryo

Aim: Isolation of Cenna embryo from the developing seeds.

Principal: Embryo of dicotyledons bears embryonal axis, two cotyledons, coleorhizza and coleoptiles

Apparatus:

    Petridish, Blade, Slide, Forceps, Water

Procedure:

Take seeds in petridish and add some water and wait for some time. Select the developing seed and keep on the slide and remove the seed coat.

Two green colored cotyledons are seen. Place the seeds on the slide and expand the cotyledons. The embryo with two cotyledons, coleoptiles is seen.

Embryology Slides

Anomocytic Stomata:
(Irregular celled type)
1. There is no difference between the guard cells and the surrounding epidermal cells
2. Four or more cells surround the the guard cells irregularly.
3. Ex – Tridax,Amaranthus, Tagetus.

Anisocytic Stomata
( Unequal- celled type)
1. The guard cells are surrounded by 3 subsidiary cells.
2. Of the three cells, one is distinctly smaller than the other two.
3. Ex –Datura, Brassica, Hibiscus

Paracytic Stomata
( Parallel- celled type)
1. The guard cells are surrounded by two subsidiary cells.
2. The subsidiary cells are arranged parallel to the long axis of the guard cells
3. Ex – Ixora, Hamelia

Diacytic Stomata
( Corss - celled type)
1.  The guard cells are surrounded by two subsidiary cells.
2. The subsidiary cells are arranged at the right angles to the long axis of the guard cells.
3. Ex – Ocimum, Hyptis, Leucas

T.S of Anther
1. It shows sporogenous tissue covered by an anther wall
2. Anther wall consists of epidermis, endothecium, middle layers and tapetum
3. Epidermis is outermost layer, one celled thick
4. Endothecium is present below the epidermis. The cells of this layer are expanded radially to form fibrous thickenings.
5. Beneath endothecium there are thin walled cells arranged in one to five layers.
6. The innermost layer of anther wall is tapetum which encircles the sporogenous tissues.
7. At four corners of the anther, four pollen chambers with sporogenous tissue are seen

Orthotropous
1. The ovule is straight, without any curvature
2. Micropyle, chalaza, funicle and embryo sac lie in a straight line.
3. Ex – Polygonum, Piper

Anatropous
1.  Body of ovule becomes completely inverted to 1800.
2. The micropyle lies close to funicle
3. Micropyle, Embryo sac and chalaza lie on the same line.
4. Ex – Helianthus, Ricinus

Campylotropous
1. The body of the ovule is placed at right angles to the funiculus.
2. The body of ovule bends in such a way that micropyle comes towards funiculus.
3. Micropyle and chalaza do not lie on the same straight line.
4. Ex – Pisum, Mustard

Dicot Embryo T.S
1. The embryo consists of embryonal axis with two large lateral cotyledons.

2. The portion of embryonal axis above the level of cotyledons is called epicotyl and the portion below the level of cotyledons is hypocotyl
3. The epicotyl forms the plumule or embryonic shoot and lower end of hypocotyl forms the radicle.
4. At one end of the embryo, swollen suspensor is seen

Monocot Embryo T.S
1. The embryo consists of shield shaped cotyledon called scutellum and the embryonal axis
2. Plumule is lateral in position and is surrounded by a leaf-sheath called coleoptile.
3. Radicle is surrounded by a root-sheath called coleorhiza

Dicot Root TS
1. It Shows Epidermis, Cortex and Stele
2. Epidermis is outermost layer made of thin walled , rectangular cells. Shows  root hairs
3. Cortex is differentiated into exodermis, General cortex and Endodermis
4. Stele is smaller than cortex composed of pericycle, vascular bundles and medulla or pith
5. Radial vascular bundles are seen
6. Xylem is tetrarch and exarch.

Dicot Stem T S
1. It shows Epidermis, Cortex and Stele
2. Epidermis is outermost layer made of rectangular cells. It shows cuticle and multicellular trichomes
3. Cortex is differentiated into Collenchymatous hypodermis, parenchymatous general cortex and Enodermis
4. The stele shows Pericycle, Vascular bundles, Medulla and medullary rays
5. Wedge or top shaped vascular bundles are arranged in a ring
6. Vascular bundles are conjoint, collateral, open and endarch.

Monocot Stem T S
1. It shows Epidermis, Hypodermis, Ground tissue and Vascular bundles
2. Epidermis is outermost layer composed of rectangular cells.
3. Thick walled sclerenchymatous hyodermis is present below the epidermis
4. Below hypodermis is parenchymatous ground tissue
5. Numerous vascular bundles are scatter in the ground tissue.
6. Vascular bundles are conjoint, collateral, closed and endarch.
7. Xylem vessels are arranged in ‘Y’ Shaped and shows protoxylem lacunae

Palynology

Hibiscus
Family: Malvaceae
Pollen grains monads.
They are isopolar, radially symmetrical, spheroidal in shape.
Exine tecate, sexine thicker than nexine.
Surface beset with robust and prominent spines.

Ocimum
Family: Lamiaceae
Pollen grains are monads
They are radially symmetrical, isopolar.
Apertures - 6 colpate- Hexacolpate.
Colpi broad with blunt ends.
Examine sub - tectate, surface prominently reticulated.

Acacia
Family: Mimosoidae
Pollen grains monads
Radially symmetrical arranged in polyads of 16 - grains.
Each grain spheroidal to sub- spheroidal
Each grain provided with 2 or 3 faint pores often considered as inaperturate.
Exine tecate, sexine thicker than nexine
Surface granular

Grass Pollen grains 
Family: Poaceae
Pollen grains monads
Radially symmetrical, heterpolar, spheroidal.
Distally monoporate; pore with a prominent annulus
Exine tectate, Sexine as thick or thicker than nexine.
Exine pilate (smooth) to faintly granular

Tangential Longitudinal










Section (TLS)
1. In this section, the cells of the axial system are cut longitudinally and those of the ray system at right angles to their orientation.
2. The width and height of the ray are clearly visible
3.
Radial Longitudinal Section (RLS)