⭃
Follow the instructions given
Draw the box on left plain
page
Anaphase:
Telophase:
Genetics Problems
Write the Cross and Punnet square for each problem on left side plain page.
Genotype: TT X tt
Gametes T X t
Tt
Gametes : YR X yr
Crossing of F1 Generation with Round and Yellow seeded plants :
:
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.
|
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.
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
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
|
|
|
|
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
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.
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
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
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
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
________3.5____________18.4________________
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
→ →
________7.2_________________5___________
Gametes R X r
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:
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.
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