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TS Inter 1st Year Botany Study Material Chapter 8 Taxonomy of Angiosperms

Sandhya

January 10, 2026

TS Inter 1st Year Botany Study Material Chapter 8 Taxonomy of Angiosperms

Very Short Answer Type Questions

Question 1.

What is Omega Taxonomy?

Answer:

Taxonomy based on information from other branches such as Embryology, Cytology, Palynology, Phytochemistry etc., in addition to morphological characters is called Omega Taxonomy

Question 2.

What is Natural system of plant classification? Name the scientists who followed it. [May 14]?

Answer:

1.The system in which plants are grouped on the basis of their natural relationships taking all possible morphological characters into consideration is known as Natural system of classification.
2.de Jussieu, de Candolle and Bentham and Hooker followed it.

Question 3.

Explain the scope and significance of “Numerical Taxonomy”?

Answer:

1.Numerical Taxonomy uses mathematical methods to evaluate observable differences and similarities between taxonomic groups. It is now easily carried out using the computers is based on all observable characters.
2.In this process no. and codes are assigned to all the characters and the data are then processed. Each character is given equal importance and at the same time hundreds of characters can be considered.

Question 4.

What is geocarpy? Name the plant which exhibits this phenomenon. [May 17, Mar. 17 – A.P ; Mar. 15 – T.S]

Answer:

Development of fruit inside the soil is known as geocarpy. Eg: Arachis hypogea (groundnut).

Question 5.

Name the type of pollination mechanism found in members of Fabaceae. [Mar. 14]

Answer:

1.Piston mechanism for cross pollination.
2.It is entomophilous, occurs with the help of insects.

Question 6.

Write the floral formula of solanum plant. [Mar 2020]?

Answer:

Question 7.

Give the technical description of ovary in solanum nigrum?

Answer:

1.Bicarpellary, syncarpous, bilocular, superior ovary.
2.Placentata swollen with many ovules on axile placentation. Style Carpels are arranged obliquely at 45°.

Question 8.

Give the technical description of anthers of Allium cepa?

Answer:

Anthers are dithecous, basifixed, introse and dehiscence is longitudinal.

Short Answer Type Questions

Question 1.

Write a brief note on semi technical description of a typical flowering plant. [Mar. – 2019]?

Answer:

The plant is described beginning with its habit, habitat, vegetative characters, floral characters followed by fruit. After describing various parts of a plant, a floral diagram and a floral formula are presented.

Floral formula indicates the number of free or united members of corresponding whorl as subscript of the respective symbols. It also shows cohesion and adhesion.
Floral diagram provides information about the no. of parts of a flower, their arrangement and their relation with one another.

Question 2.

Describe the non-essential floral parts of plants belonging to Fabaceae?

Answer:

The non-essential floral parts are calyx and corolla.
Calyx :
Sepals five in number, Gamosepalous (sepals united), valvate aestivation. The odd sepal is anterior in position.
Corolla :
Petals five in number, Polypetalous (Petals are free), descending imbricate aestivation. Corolla is Papilonaceous. The posterior petal is the largest and is called Vexillum or Standard Petal The two lateral petals are called Wings or Alae. The two boat shaped petals beneath the wings on the anterior side are called Keel or Carina. The keel petals are fused and enclose the essential organs.

Question 3.

Give an account of floral diagram?

Answer:

Floral diagram :
It provides information about the number of parts of a flower, their arrangement and the relation they have with one another. The mother axis represents the posterior side of the flower and is indicated as a dot or a circle at the top of the floral diagram. Successive whorls represent calyx, corolla, androecium and gynoecium. Gynoecium at the centre represents T.S. of ovary. The bract represents the anterior side of the ovary

Question 4.

Describe the essential floral parts of plants belonging to Liliaceae?

Answer:

The essential floral parts are Androecium and Gynoecium.
Androecium :
Stamens are six, arranged in two whorls of three each. They are free or epiphyllous, anthers are dithecous, basifixed, introrse and dehiscence is longitudinal.
In Allium, the stamens are inserted at the base of the tepals which are also fused at the base.
Gynoecium :
Overy is tricarpellary, syncarpous, superior, trilocular with numerous ovules on axile placentation. The style is terminal and stigma is trifid and capitate. The ovary has septal nectaries, one on each septum.

Question 5.

Write a brief account on the class of Dicotyledanae of Bentham and Hooker s classification?

Answer:

The class Dicotyledonae are characterised by tap root, reticulate venation, tetramerous or pentamerous flowers and two cotyledons in a seed. On the basis of the number of whorls in the Perianth and the condition of petals, the dicotyledons are divided into three sub-classes namely Polypetalae, Gamopetalae and Monochlamydae. Polypetalae was divided into three series namely 1) Thalamiflorae, 2) Disciflorae and, 3) Calyciflorae, Gamopetalae was divided into three series viz. 1. Inferae 2. Heteromerae and 3. Bicarpellatae. Monochlamydae was divided into eight series.

Question 6.

Explain Floral formula?

Answer:

The floral formula is represented by some symbols of floral parts.
In the floral formula :
a. Br stands for bracteate; Ebr for ebracteate (Bracts absent)
b. Brl stands for bracteolate; Ebrl for ebracteolate (Bracteoles absent)
c. ⊕ stands for actinomorphic; % for zygomorphic

e. K stands for calyx; C for corolla.
f. A for androecium and G for gynoecium.
g. G stands for superior ovary and G¯¯¯¯ for inferior ovary and G – for half inferior or half superior.
Floral formula also indicates the number of free or united (within brackets) members of the corresponding whorl as subscript of the respective symbol.
It also shows cohesion (union among similar members) and adhesion (union between dissimilar members).

Question 7

Give economic importance of plants belonging to Fabaceae. [May ’17, May 14 ; Mar. ’13]

Answer

Pulses :
Pulses like red gram, black gram, green gram, bengal gram are rich source of proteins.
Oil :
Groundnut oil from groundnut (Arachis hypogaea) seeds and soyabean oil from seeds of soyabean (Glycine max) are used in cooking.
Vegetables :
Pods of bean (Dolichos), soyabean (Glycine max) are used as vegetables.
Fodder :
Many plants are used as fodder (Crotalaria & Phaseolus).
Fibres :
from Crotalaria (sun-hemp) is used in making ropes.
Medicine :
Seeds of Trigonella (Menthi) are used as condiment and medicine. The leaves are used as vegetables.
Commercially & Products :
Commercially important products obtained are viz., blue dye (Indigofera tinctoria), yellow dye (Butea monosperma)
Medicine :
Medicine is obtained from Derris indica.
Green manure :
Sesbania and Tephrosia are used as green manure.
Wood :
Wood from (Pterocarpus santalinus) Red sanders is used for making musical instrument.

Long Answer Type Questions

Question 1.

Describe the characteristics of plants belonging to Fabaceae?

Answer:

Vegetative Characters:
Habit :
Many of them are herbs. Some are shrubs, trees, weak stemmed twinners or tendril climbers.
Root System :
Tap root system having root nodules. These nodules contain symbiotic nitrogen fixing bacteria – Rhizobium.
Stem :
Aerial, prostrate or erect, herbaceous or woody climbers.
Leaves :
Cauline, alternate, stipulate, base pulvinate, petiolate, simple or pinnately compound leaf, venation reticulate.
Floral Characters:
Inflorescence :
Axillary or terminal, raceme.
Flower :
Bracteate, bracteolate or ebracteolate, pedicillate, zygomorphic, complete, bisexual, pentamerous, perigynous cup-shaped thalamus.
Calyx :
Sepals five, gamosepalous, imbricate aestivation, odd sepal anterior.
Corolla :
Petals five, polypetalous, papilionaceous consisting of a large posterior petal (standard or vexillum), two laterals (wings or Alae) two anterior fused petals (keel or Carina) enclosing essential organs, vexillary / descendingly imbricate aestivation.
Androecium :
Stamens 10, usually diadelphous [(9) +1] as in Pisum or monadelphous as in Crotalaria, Arachis, anthers dithecous, introse.
Gynoecium :
Monocarpel I ary, unilocular half superior ovary with many ovules on marginal placentation; Style single, long, terminal; stigma simple.
Pollination :
Protandry in flowers prevent self pollination. But Lathyrus, Pisum are self pollinated. Pollination is entomophily and occurs by Piston mechanism.
Fruit :
Generally legume or pod. In Arachis the pods are indehiscent and geocarpic.

Question 2.

Write about the key characteristics of Solanaceae?

Answer

1.Mostly plants are herbs.
2.Presence of hairs on the plant
3.Bicollateral vascular bundles in the stem
4.Adnation of petioles and pedicels with the internode
5.Pentamerous, actinomrophic and hypogynous flowers
6.Presence of Persistant calyx
7.Epipetalous stamens
8.Bicarpellary, syncarpous, superior ovary with obliquely placed carpels
9.Presence of false septum in the ovary
10.Swollen axile placentation
11.Fruit is a berry or capsuie
12.Presence of curved embryos

Question 3.

Give an account of the family Liliaceae?

Answer:

Class- Monocotyledonae
Series- Coronariae
Family- Liliaceae
I. Vegetative characters:

Habitat and habit :
Mesophytes (Allium, Lilium) and also xerophytes (Asparagus, Ruscus, Aloe)are found in this family. Mostly perennial herbs, shrubs or trees are found in some species of Dracaena, Yucca and Aloe. Few are climbers (Gloriosa, Smilax).
Root system :
Adventitious root system. In Asparagus, fasciculated tuberous roots are present,
Stem :
Mostly perennial underground stems. It may be a bulb (Allium, Lilium, Scilla), rhizome (Gloriosa), or a corm (Colchicum). Gloriosa and Smilax are tendrillar climbers. Cladophylls are present in Ruscus and Asparagus.
Leaf :
The leaves are radical (Allium, Lilium) or cauline (Smilax, Gloriosa). Phyllotaxy is usually alternate (Gloriosa) or whorled (Trillium)Simple, petiolate, stipulate exstipulate, parallel venation (exceptionally reticulate in Smilax). Leaves are succulent in Aloe, Yucca and reduced to scales in Asparagus, Ruscus. Stipules in Smilax and leaf apex in Gloriosa are modified into tendrils. Epiphyllous buds present at leaf apex in Scilla.

II. Floral Characters:

Inflorescence :
Simple raceme (Asparagus) or panicle (Aloe) or umbel (Allium). Solitary, terminal (Lilium) or Solitary, axillary (Gloriosa).
Flower :
Bracteate, ebracteolate, pedicellate, actinomorphic, complete, bisexual,
homochlamydeous, trimerous, and hypogynous.
A) Perianth :
Tepals 6 in two whorls, polyphyllous (Lilium) or gamophyllous (Aloe), petaloid. Odd tepal of outer whorl is anterior, Valvate aestivation.
B) Androecium :
Stamens 6 in two whorls of 3 each, free or epiphyllous. Anthers are dithecous, introrse, basifixed and dehisce longitudinally.
C) Gynoecium :
Tricarpellary and syncarpous. Ovary is superior and trilocular with several ovules on axile placentation. Style is terminal and stigma is trifid, capitate. The ovary has septal nectaries.

Floral formula :

Pollination :
Flowers are protandrous (Allium) or protogynous (Colchicum) or Herkogamy (Gloriosa) to prevent self pollination. Cross pollination is by entomophily. Yucca has a symbiotic type of pollination by a specific moth, Pronuba yuccasella.
Fruit :
Berry or capsule. Seed is monocotyledonous.

III. Economic Importance :

Edible :
Allium cepa (bulb), Asparagus (roots), cloves of Allium sativum as spice.
Medicinal plants :
Allium cepa [bulb – bactericidal), Allium sativum (cloves – gastric and heart.)
Colchicine :
Colchicum, autumnale [mutagen obtained from corm]
Fibre yielding plants :
Leaves of Yucca and Dracaena.
Ornamental plants :
Gloriosa, Lilium, Asparagus, Dracaena.

Question 4.

Write the characteristics of plants that are necessary for classification. Describe them in brief?

Answer:

Depending upon flowering plants are divided into Non-flowering plants (cryptogamae) and Flowering plants (phaenerogamae)
Sub kingdom :
Cryptogamae (cryptogams): Cryptogams are flowerless and seedless spore plants. They never bear flowers, fruits and seeds. They reproduce asexually by spores and sexually by gametes.
Basing upon the plant body cryptogams are divided into three divisions.
(a) Thallophyta (b) Bryophyta (c) Pteridophyta.
Division Thallophyta :
These are simplest and most primitive plants. All of them have thallus, which is not differentiated into root, stem and leaves.
Division Thallophyta is divided into Algae & Fungi.
Sub division Algae :
Algae are green, photoautotrophic and usually aquatic thallophytes.
Sub division Fungi :
Fungi are non-green (achlorophyllous) heterotrophic . thallophytes.
Division Bryophyta :
Bryophytes are green, autotrophic, embryophytic, nonvascular cryptogams. They are first land plants. They are the amphibians of plant kingdom.
Division Pteridophyta :
Pteridophytes are green, autotrophic, embryophytic, vascular plants.
Sub kingdom Phanerogamae (Phanerogams) :
These are called flowering plants. They bear flowers or cones for reproduction. These phanerogams have one division Spermatophyta.
Division Spermatophyta :
There spermatophytes are seed plants without or with fruits. It is divide into two 1. Gymnospermae 2. Angiospermae.
Gymnospermae (Gymnosperms) :
These are seed plants without ovary and fruit. Seeds are naked without fruit wall.
Angiospermae (Angiosperms) :
These are seed plants with ovaries and fruits. The seeds are enclosed in the fruit wall (pericarp). These are fruit bearing flowering plants.
Depending upon the number of cotyledon in the seed angiosperms are divided into Dicotyledonae and Monocotyledonae.
Class :
Dicotyledonae (Dicot plants) :
Fruit bear seeds with two cotyledons. Tap root system, leaf with reticulate venation, tetra on pentamerous flowers are present.
Class :
Monocotyledonae (Monocot plants):
Fruit bear seeds with one cotyledons; Fibrous root system; leaf with parallel venation; trimerous flowers are present.

Question 5.

Describe a typical flowering plant in the taxonomic perspective?

Answer:

I. Vegetative characters :

Habit –
1.Herb / Shrub / Tree
2.annual/biennial / perennial
Habitat- Hydrophyte/Mesophyte/Xerophyte
Root – Tap root system / Adventitious root system, special character like root modification
Stem –
1.Aerial/ underground / sub – aerial
2.Erect / twiner procumbent / prostrate
3.Herbaceous / woody
4.Special characters and modification
Leaf –
1.Cauline / ramal / radical
2.Phyllotaxy – Alternate / opposite / circular.
3.Simple / compound leaf
4.Stipulate / exstipulate
5.Leaf base – adnate / pulvinate
6.Petiolate, sessile
7.Reticulate / parallel venation
8.Any special characters or modification.
II. Floral characters:
Inflorescence :
1.Axillary/terminal
2.Racemose / cymose / special type
3.Inflorescence type
Flower in general
1.Bracteate /Ebracteate
2.Pedicillate / sessile
3.Bracteolate / ebracteolate
4.Actinomorphic / zygomorphic
5.Complete / incomplete
6.Bisexual / unisexual
7.Pentamerous / tetramerous / trimerous
8.Hypogynous / perigynous / epigynous
9.Dichlamydeous /monochlamydeous
10.Heterochlamydeous / homochlamydeous
11.Cyclic / spiral
Perianth:
a) Calyx :
1.No. of sepals
2.United (poly) free (gamo)
3.Aestivation – valvate / twisted / intricate
4.Persistent / deciduous
5.Odd sepal – anterior or posterior
b) Corolla :
1.No. of petals
2.United (poly) / free (gamo)
3.Shape : tubular / ligulate / papilionaceous/ funnel shape.
4.Aestivation – valvate / twisted / imbricate
c) Androecium :
1.No. of stamens – definite / indefinite
2.Arrangement – one whorl / two whorls
3.Free / united (monoadelphous / diadelphous / polyadelphous / syngenicious)
4.Length – didynamous / tetradynamous
5.Epiphyllous/ epipetalpus
6.Extrose / introse
7.Fertile / sterile
8.Anthers : dithecous / monothecous
9.Basifixed / dorsifixed versatile.
10.Dehiscence – longitudinal / transverse / porous
Gynoecium :
1.Mono / bi/ tri / tetra / penta / multicarpellary
2.Carpels – United /free
3.Mono / bi/ tri / tetra / penta / multilocular
4.Superior / half superior / inferior ovary
5.Placentation – Axile / central / basal/ marginal
6.Style – terminal / basal / lateral
7.Stigma – capitate / dumbel shaped / bifid
Fruit :
1.Simple / compound / Aggregate
2.Fruit type

Question 6.

Give an account of Bentham and Hooker’s classification of plants?

Answer:

Bentham and Hooker classification is a natural system of classification. It was published in 3 volumes of Genera Plantarum.
They divided flowering plants into 3 classes – Dicotyledonae, Gymnospermae and Monocotyledonae.
Class :
Dicotyledonae :

Tap root system 2. Reticulate venation. 3. Flowers tetramerous or pentamerous 4. Seed has two cotyledons Cl. Dicotyledons is divided into 3 sub classes – polypetalae, Gamopetalae and Monochlamydae.

Subclass : Polypetalae :
Perianth in two whorls.
Petals are free.
Sub cl: Polypetalae is divided into 3 series based on the nature of thalamus.
A. Series Thalamiflorae :
Thalamus is elongated, conical or convex.
It has 5 cohorts.
B. Series Disciflorae :
Thalamus is disc shaped.
It has 4 cohorts.
C. Series Calyciflorae :
Thalamus is cup shaped.
It has 5 cohorts.
Family Fabaeceae belongs to order Rosales.

Subclass : Gamopetalae :
Perianth in two whorls
Petals are united.
Stamens are epipetalous.
S. Class :
Gamopetalae is divided into 3 series based on the nature of ovary and merosity of flower
A. Series Inferae :
Ovary is inferior.
It includes 3 cohorts.
B. Series Heteromerae :
Ovary is superior.
Carpels are more than two.
It has 3 cohorts.
C. Series Bicarpellatae :
Ovary is superior.
No. of carpels are two.
It has 4 cohorts.
Family Solanaceae belongs to order Polemoniales.
Subclass :
Monochlamydeae :
Perianth is not divisible into calyx and corolla. It has 8 series. No cohorts. Families are directly under series.
Class: Gymnospermae :
Seeds are naked. It is divided into 3 families – Cycadaceae, Coniferae & Gnetaceae.
Class : Monocotyledonae :
Adventitious root system, 2. Parallel venation, 3. Trimerous flowers and 4. Seed has one cotyledon. It has 7 series. No cohorts. Families are placed directly under the series.
Thus flowering plants are grouped into 202 natural orders now called as families. Of these 165 natural orders belong to Dicotyledonae. 3 to Gymnosperms and 34 belong to Monocotyledonae.

Question 7.

What is taxonomy? Give a brief account of different types of plant classification?

Answer:

Classification of plants into groups based on their similarities and their relationships is called Taxonomy. It deals with characterization, identification, nomenclature and classification.
On the basis of criteria taken into account, classification systems are 3 types. They are

Artificial system of classification :
It is based on one or few comparable characters like morphology or natural habits.
Eg: 1: Classification of plants on the basis of form into herbs, shrubs, trees etc., was done by Theophrastus.
Eg: 2 : Sexual system of Linnaeus.
Natural system of classification :
Plants are grouped on the basis of their natural relationship taking into consideration all possible morphological characteristics.
Eg : Classification of de Candolle
Bentham and Hooker system
Phylogenetic system of classification :
These systems are proposed after the publication of “Origin of species” and the announcement of the “theory of evolution” by Charles Darwin. Hence they are also called post-Darwinism classifications. They reflect the genetic and evolutionary relationships among the taxa and show them in the form of a phylogenetic tree. Ex : Classification of Engler and Prantl in their “Die Naturlichen Plazenfamilien”. J. Hutchinson in his book “Families of Flowering Plants”. Latest phylogenetic classification is APG (Angiospermic Phylogenetic Group) system.
Other Types:
Numerical Taxonomy :
Uses mathematical methods to evaluate observable differences and similarities between taxonomic groups. Numerical taxonomy which is now easily carried out using computers is based on all observable characteristics.
Cytotaxonomy :
A branch of taxonomy that uses cytological characters like chromosome number, and structure in solving taxonomic problems.
Chemotaxonomy :
A branch of taxonomy that uses phytochemical data to solve the problems of taxonomy.

TS Inter 1st Year Maths 1A Solutions Chapter 3 Matrices Ex 3(a)

Junaid

January 9, 2026

TS Inter 1st Year Maths 1A Solutions Chapter 3 Matrices Ex 3(a)

TS Inter 1st Year Maths 1A Matrices Solutions Exercise 3(a)

Question 1.
Write the following as a single matrix.
(i) [2 1 3] + [0 0 0]
Answer:
[2 1 3] + [0 0 0] = [2 + 0 1 + 0 3 + 0]
= [2 1 3]
(ii) $\left[\begin{array}{r} 0 \\ 1 \\ -1 \end{array}\right]+\left[\begin{array}{r} -1 \\ 1 \\ 0 \end{array}\right]$
Answer:
$\left[\begin{array}{r} 0 \\ 1 \\ -1 \end{array}\right]+\left[\begin{array}{r} -1 \\ 1 \\ 0 \end{array}\right]$ = $\left[\begin{array}{r} 0-1 \\ 1+1 \\ -1+0 \end{array}\right]$ = $\left[\begin{array}{r} -1 \\ 2 \\ -1 \end{array}\right]$
(iii) $\left[\begin{array}{ccc} 3 & 9 & 0 \\ 1 & 8 & -2 \end{array}\right]+\left[\begin{array}{ccc} 4 & 0 & 2 \\ 7 & 1 & 4 \end{array}\right]$

Answer:

(iv) $\left[\begin{array}{rr} -1 & 2 \\ 1 & -2 \\ 3 & -1 \end{array}\right]+\left[\begin{array}{rr} 0 & 1 \\ -1 & 0 \\ -2 & 1 \end{array}\right]$
Answer:

Question 2.
If A = $\left[\begin{array}{cc} -1 & 3 \\ 4 & 2 \end{array}\right]$ , B = $\left[\begin{array}{cc} 2 & 1 \\ 3 & -5 \end{array}\right]$ , X = $\left[\begin{array}{ll} \mathbf{x}_1 & \mathbf{x}_2 \\ \mathbf{x}_3 & \mathbf{x}_4 \end{array}\right]$ and A + B = X then find the values of x₁, x₂, x₃ and x₄.

Answer:
A + B = X

⇒ x₁ = 1, x₂ = 4, x₃ = 7, x₄ = – 3.

Question 3.
If A = $\left[\begin{array}{ccc} -1 & -2 & 3 \\ 1 & 2 & 4 \\ 2 & -1 & 3 \end{array}\right]$ B = $\left[\begin{array}{ccc} 1 & -2 & 5 \\ 0 & -2 & 2 \\ 1 & 2 & -3 \end{array}\right]$ and C = $\left[\begin{array}{ccc} -2 & 1 & 2 \\ 1 & 1 & 2 \\ 2 & 0 & 1 \end{array}\right]$ then find A + B + C.

Answer:
A + B + C =

Question 4.
If A = $\left[\begin{array}{ccc} 3 & 2 & -1 \\ 2 & -2 & 0 \\ 1 & 3 & 1 \end{array}\right]$ , B = $\left[\begin{array}{ccc} -3 & -1 & 0 \\ 2 & 1 & 3 \\ 4 & -1 & 2 \end{array}\right]$ and X = A + B then find X.

Answer:
X = A + B

Question 5.
If $\left[\begin{array}{cc} x-3 & 2 y-8 \\ z+2 & 6 \end{array}\right]$ = $\left[\begin{array}{cc} 5 & 2 \\ -2 & a-4 \end{array}\right]$ then find the values of x, y, z and a. [May 2006, Mar. 14]

Answer:
Given $\left[\begin{array}{cc} x-3 & 2 y-8 \\ z+2 & 6 \end{array}\right]$ = $\left[\begin{array}{cc} 5 & 2 \\ -2 & a-4 \end{array}\right]$
We have x – 3 = 5, 2y – 8 = 2, z + 2 = – 2, a – 4 = 6
⇒ x = 8, y = 5, z = – 4, a = 10

II.
Question 1.
If $\left[\begin{array}{ccc} x-1 & 2 & 5-y \\ 0 & z-1 & 7 \\ 1 & 0 & a-5 \end{array}\right]$ = $\left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 4 & 7 \\ 1 & 0 & 0 \end{array}\right]$ then find the values x, y, z and a.
Answer:
Given $\left[\begin{array}{ccc} x-1 & 2 & 5-y \\ 0 & z-1 & 7 \\ 1 & 0 & a-5 \end{array}\right]$ = $\left[\begin{array}{lll} 1 & 2 & 3 \\ 0 & 4 & 7 \\ 1 & 0 & 0 \end{array}\right]$
we have x – 1 = 1, 5 – y = 3, z – 1 = 4,
a – 5 = 0
⇒ x = 2, y = 2, z = 5, a = 5

Question 2.
Find the trace of $\left[\begin{array}{ccc} 1 & 3 & -5 \\ 2 & -1 & 5 \\ 1 & 0 & 1 \end{array}\right]$

Answer:
Trace of the given matrix
= 1 – 1 + 1 = sum of the diagonal elements
= 1

Question 3.
If A = $\left[\begin{array}{ccc} 0 & 1 & 2 \\ 2 & 3 & 4 \\ 4 & 5 & -6 \end{array}\right]$ and B = $\left[\begin{array}{ccc} -1 & 2 & 3 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array}\right]$ find A – B and 4A – 5B.

Answer:

Question 4.
If A = $\left[\begin{array}{lll} 1 & 2 & 3 \\ 3 & 2 & 1 \end{array}\right]$ and B = $\left[\begin{array}{lll} 3 & 2 & 1 \\ 1 & 2 & 3 \end{array}\right]$ find 3B – 2A.

Answer:

TS Inter 1st Year Maths 1A Solutions Chapter 3 Matrices Ex 3(b)

Junaid

January 9, 2026

TS Inter 1st Year Maths 1A Solutions Chapter 3 Matrices Ex 3(b)

TS Inter 1st Year Maths 1A Matrices Solutions Exercise 3(b)

I.
Question 1.
Find the following products wherever possible.
i) [-1 4 2] $\left[\begin{array}{l} 5 \\ 1 \\ 3 \end{array}\right]$

Answer:
It is a product of 1 × 3 and 3 × 1 matrices and the resulting is an 1 × 1 matrix.
[-1 4 2] $\left[\begin{array}{l} 5 \\ 1 \\ 3 \end{array}\right]$
= [-1 × 5 + 4 × 1 + 2 × 3]_{1 × 1} = [5]_{1 × 1}

ii) $\left[\begin{array}{lll} 2 & 1 & 4 \\ 6 & -2 & 3 \end{array}\right]\left[\begin{array}{l} 1 \\ 2 \\ 1 \end{array}\right]$
Answer:
It is a product of 2 × 3 and 3 × 1 matrices and the resulting is an 2 × 1 matrix.

iii) $\left[\begin{array}{cc} 3 & -2 \\ 1 & 6 \end{array}\right]\left[\begin{array}{cc} 4 & -1 \\ 2 & 5 \end{array}\right]$
Answer:
Product of 2 × 2 and 2 × 2 matrices and the resulting is an 2 × 2 matrix.

iv) $\left[\begin{array}{lll} 2 & 2 & 1 \\ 1 & 0 & 2 \\ 2 & 1 & 2 \end{array}\right]\left[\begin{array}{ccc} -2 & -3 & 4 \\ 2 & 2 & -3 \\ 1 & 2 & -2 \end{array}\right]$
Answer:
Product of 3 × 3 and 3 × 3 matrices and the resulting is an 3 × 3 matrix.

v) $\left[\begin{array}{ccc} 3 & 4 & 9 \\ 0 & -1 & 5 \\ 2 & 6 & 12 \end{array}\right]\left[\begin{array}{lll} 13 & -2 & 0 \\ 0 & 4 & 1 \end{array}\right]$
Answer:
Product of A_{3 × 3} and B_{2 × 3} matrices. Matrix multiplication is not confirmable since the number of columns of A ≠ number of rows of B.

vi) $\left[\begin{array}{c} 1 \\ -2 \\ 1 \end{array}\right]\left[\begin{array}{lll} 2 & 1 & 4 \\ 6 & -2 & 3 \end{array}\right]$
Answer:
Product of A_{3 × 1} and B_{2 × 3} matrices. Matrix multiplication is not confirmable since the number of coloumn of A ≠ number of rows of B.

vii) $\left[\begin{array}{rr} 1 & -1 \\ -1 & 1 \end{array}\right]\left[\begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right]$
Answer:
Product of 2 × 2 and 2 × 2 matrices and the resulting is an 2 × 2 matrix.

viii) $\left[\begin{array}{ccc} 0 & c & -b \\ -c & 0 & a \\ b & -a & 0 \end{array}\right]\left[\begin{array}{ccc} a^2 & a b & a c \\ a b & b^2 & b c \\ a c & b c & c^2 \end{array}\right]$
Answer:
Product of 3 × 3 and 3 × 3 matrices and the resulting matrix is an 3 × 3 matrix.

Question 2.
If A = $\left[\begin{array}{rrr} 1 & -2 & 3 \\ -4 & 2 & 5 \end{array}\right]$ and B = $\left[\begin{array}{ll} 2 & 3 \\ 4 & 5 \\ 2 & 1 \end{array}\right]$ do AB and BA exist ? If they exist find them. Do A and B commute with respect to
multiplication.

Answer:
Given A = $\left[\begin{array}{rrr} 1 & -2 & 3 \\ -4 & 2 & 5 \end{array}\right]$ and B = $\left[\begin{array}{ll} 2 & 3 \\ 4 & 5 \\ 2 & 1 \end{array}\right]$
We have the product of A_2×3 and B_3×2 matrices and resulting AB is a product matrix of order 2 × 2. Similarly the product of B_3×2 and A_2×3 matrices results a product matrix BA of order 3 × 3.

Since AB ≠ BA, we have A and B are not com¬mutative with respect to multiplication of matrices.

Question 3.
Find A² where A = $\left[\begin{array}{cc} 4 & 2 \\ -1 & 1 \end{array}\right]$

Answer:

Question 4.
If A = $\left[\begin{array}{ll} \mathrm{i} & 0 \\ 0 & \mathrm{i} \end{array}\right]$ find A².

Answer:

Question 5.
If A = $\left[\begin{array}{cc} \mathrm{i} & 0 \\ 0 & -\mathrm{i} \end{array}\right]$ ; B = $\left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right]$ and C = $\left[\begin{array}{ll} 0 & \mathrm{i} \\ \mathrm{i} & 0 \end{array}\right]$ and I is the unit matrix of order 2, then show that
i) A² = B² = C² = – I

Answer:

ii) AB = – BA = – C (March 2008)
Answer:

Question 6.
If A = $\left[\begin{array}{ll} 2 & 1 \\ 1 & 3 \end{array}\right]$ and B = $\left[\begin{array}{lll} 3 & 2 & 0 \\ 1 & 0 & 4 \end{array}\right]$ find AB. Find BA if it exists.
Answer:
Given A = $\left[\begin{array}{ll} 2 & 1 \\ 1 & 3 \end{array}\right]$ and B = $\left[\begin{array}{lll} 3 & 2 & 0 \\ 1 & 0 & 4 \end{array}\right]$ are matrices of 2 × 2 and 2 × 3. The resulting matrix AB is of the form 2 × 3.

Question 7.
If A = $\left[\begin{array}{cc} 2 & 4 \\ -1 & K \end{array}\right]$ and A² = 0 then find the value of K. (May 2011, Mar. ’14, ’05)

Answer:

II.
Question 1.
If A = $\left[\begin{array}{lll} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array}\right]$ there find A⁴.

Answer:

Question 2.
If A = $\left[\begin{array}{ccc} 1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{array}\right]$ then Find A³.

Answer:

Question 3.
If A = $\left[\begin{array}{rrr} 1 & -2 & 1 \\ 0 & 1 & -1 \\ 3 & -1 & 1 \end{array}\right]$ then find A³ – 3A² – A – 3I, where I is a unit matrix of order 3. (March 2011)

Answer:

Question 4.
If I = $\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$ and E = $\left[\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right]$ show that (aI + bE) = a³I + 3a²bE, where I is a unit matrix of order 2. (Mar. 2015-A.P)(May ’05)|

Answer:

III.
Question 1.
If A = _diag[a₁ a₂ a₃] then for any Integer n ≥ 1 show that Aⁿ = _diag $\left[\mathrm{a}_1^{\mathrm{n}}, \mathrm{a}_2^{\mathrm{n}}, \mathrm{a}_3^{\mathrm{n}}\right]$

Answer:

We prove this result by using mathematical induction suppose n = 1 then
A’ = $\left[\begin{array}{ccc} \mathrm{a}_1 & 0 & 0 \\ 0 & \mathrm{a}_2 & 0 \\ 0 & 0 & \mathrm{a}_3 \end{array}\right]$ = A
The result is true for n = 1.
Suppose the result for n = k then

The result is true for n = k + 1.
So by the principle of mathematical induc-tion the statement is true ∀ n ∈ N.

Question 2.
If θ – Φ = $\frac{\pi}{2}$ , then show that $\left[\begin{array}{cc} \cos ^2 \theta & \cos \theta \sin \theta \\ \cos \theta \sin \theta & \sin ^2 \theta \end{array}\right]\left[\begin{array}{cc} \cos ^2 \phi & \cos \phi \sin \phi \\ \cos \phi \sin \phi & \sin ^2 \phi \end{array}\right]$ = 0

Answer:

Question 3.
If A = $\left[\begin{array}{rr} 3 & -4 \\ 1 & -1 \end{array}\right]$ , then show that Aⁿ = $\left[\begin{array}{cc} 1+2 n & -4 n \\ n & 1-2 n \end{array}\right]$ for any integer n ≥ 1, by using mathematical induction.

Answer:
We shall prove the result by mathematical induction.

∴ The given result is true for n = k + 1
∴ By Mathematical induction the given result is true for all positive integral values of n.

Question 4.
Given examples of two square matrices A and B of the same order for which AB = 0 but BA ≠ 0.

Answer:

Question 5.
A trust fund has to invest Rs. 30,000 in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication determine how to divide Rs. 30,000 among the two types of bonds, if the trust fund must obtain an annual total interest of (a) Rs. 1800 and (b) Rs. 2,000.

Answer:
Let the first bond be ‘x’, then the second bond will be 30,000 – x.
Rate of interest are 5% and 7% means 0.05 and 0.07.
a) [x 30,000 – x] $\left[\begin{array}{l} 0.05 \\ 0.07 \end{array}\right]$ = [1800]
⇒ 0.05x + 0.07 (30,000 – x) = 1800
⇒ – 0.02x + 0.07 (30,000) = 1800
⇒ – 0.02x + 2100 = 1800
⇒ – 0.02x = – 300
⇒ x = $\frac{300}{0.02}$ = 300 × $\frac{100}{2}$ = 15, 000
First bond = 15, 000
Second bond = 30,000 – x
= 30,000 – 15,000 = 15,000

(b) [x 30,000 – x] $\left[\begin{array}{l} 0.05 \\ 0.07 \end{array}\right]$ = [2000]
⇒ 0.05x + 0.07 (30,000 – x) = 2000
⇒ – 0.02x + 0.07 (30,000) = 2000
⇒ – 0.02x + 2100 = 2000
⇒ x = $\frac{100}{0.02}$ = 5, 000
∴ Second bond = 30,000 – x
= 30,000 – 5,000
= 25,000

TS Inter 1st Year Maths 1A Matrices Solutions Exercise 3(d)

Junaid

January 8, 2026

TS Inter 1st Year Maths 1A Matrices Solutions Exercise 3(d)

I.
Question 1.
Find the determinants of the following matrices.

(i) $\left[\begin{array}{cc} 2 & 1 \\ 1 & -5 \end{array}\right]$

Answer:
Let A = $\left[\begin{array}{cc} 2 & 1 \\ 1 & -5 \end{array}\right]$ then determinant A
= det A = |A| = 2(-5) – 1(1)
= -10 – 1
= -11

(ii) $\left[\begin{array}{cc} 4 & 5 \\ -6 & 2 \end{array}\right]$

Answer:
Let A = $\left[\begin{array}{cc} 4 & 5 \\ -6 & 2 \end{array}\right]$ then
det A = 4(2) – 5(-6)
= 8 + 30 = 38

(iii) $\left[\begin{array}{cc} \mathrm{i} & 0 \\ 0 & -\mathrm{i} \end{array}\right]$
Answer:
Let A = $\left[\begin{array}{cc} \mathrm{i} & 0 \\ 0 & -\mathrm{i} \end{array}\right]$ then
det A = i(-1) – 0 = -i² = 1 (∵ i² = -1)

(iv) $\left[\begin{array}{lll} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right]$
Answer:

(v) $\left|\begin{array}{rrr} 1 & 4 & 2 \\ 2 & -1 & 4 \\ -3 & 7 & 6 \end{array}\right|$
Answer:
Let A = $\left[\begin{array}{rrr} 1 & 4 & 2 \\ 2 & -1 & 4 \\ -3 & 7 & 6 \end{array}\right]$
Then det A = 1 $\left|\begin{array}{rr} -1 & 4 \\ 7 & 6 \end{array}\right|$ – 4 $\left|\begin{array}{ll} -2 & 4 \\ -3 & 6 \end{array}\right|$ + 2 $\left|\begin{array}{rr} 2 & -1 \\ -3 & 7 \end{array}\right|$
= 1(-6 – 28) – 4(12 + 12)+ 2(14 – 3)
= 1 (- 34) – 4(24) + 2(11)
= -34 – 96 + 22
= -108

(vi) $\left[\begin{array}{rrr} 2 & -1 & 4 \\ 4 & -3 & 1 \\ 1 & 2 & 1 \end{array}\right]$
Answer:
Let A = $\left[\begin{array}{rrr} 2 & -1 & 4 \\ 4 & -3 & 1 \\ 1 & 2 & 1 \end{array}\right]$
Then det A = 2 $\left|\begin{array}{rr} -3 & 1 \\ 2 & 1 \end{array}\right|$ + 1 $\left|\begin{array}{ll} 4 & 1 \\ 1 & 1 \end{array}\right|$ + 4 $\left|\begin{array}{rr} 4 & -3 \\ 1 & 2 \end{array}\right|$
= 2(- 3 – 2)+ 1(4 – 1) + 4(8 + 3)
= 2(-5) + 3 + 4(11)
= – 10 + 3 + 44
= 37

(vii) $\left[\begin{array}{rrr} 1 & 2 & -3 \\ 4 & -1 & 7 \\ 2 & 4 & -6 \end{array}\right]$
Answer:
Let A = $\left[\begin{array}{rrr} 1 & 2 & -3 \\ 4 & -1 & 7 \\ 2 & 4 & -6 \end{array}\right]$
Then det A = 1 $\left|\begin{array}{rr} -1 & 7 \\ 4 & -6 \end{array}\right|$ – 2 $\left|\begin{array}{rr} 4 & 7 \\ 2 & -6 \end{array}\right|$ – 3 $\left|\begin{array}{rr} 4 & -1 \\ 2 & 4 \end{array}\right|$
= 1(6 – 28) – 2(- 24 – 14) – 3(16 + 2)
= -22 + 76 – 54 = 0
[Note : Since R₁ and R₂ are proportional, we have det A = 0.]

(viii) $\left[\begin{array}{lll} a & h & g \\ h & b & f \\ g & f & c \end{array}\right]$
Answer:
Let A = $\left[\begin{array}{lll} a & h & g \\ h & b & f \\ g & f & c \end{array}\right]$
Then det A = a $\left|\begin{array}{ll} b & f \\ f & c \end{array}\right|$ – h $\left|\begin{array}{ll} \mathrm{h} & \mathrm{f} \\ \mathrm{g} & \mathrm{c} \end{array}\right|$ – g $\left|\begin{array}{ll} h & b \\ g & f \end{array}\right|$
= a(bc – f²) – h(ch – fg) + g(fh – bg)
= abc – af² – ch² + fgh + fgh – bg²
= abc + 2fgh – af² – bg² – ch²

(x) $\left[\begin{array}{ccc} 1^2 & 2^2 & 3^2 \\ 2^2 & 3^2 & 4^2 \\ 3^2 & 4^2 & 5^2 \end{array}\right]$
Answer:
Let A = $\left[\begin{array}{ccc} 1^2 & 2^2 & 3^2 \\ 2^2 & 3^2 & 4^2 \\ 3^2 & 4^2 & 5^2 \end{array}\right]=\left[\begin{array}{ccc} 1 & 4 & 9 \\ 4 & 9 & 16 \\ 9 & 16 & 25 \end{array}\right]$
Then det A = 1(225 – 256) – 4(100 – 144) + 9(64 – 81)
= -31 + 176 – 153 = -8

Question 2.
If A = $\left[\begin{array}{rrr} 1 & 0 & 0 \\ 2 & 3 & 4 \\ 5 & -6 & x \end{array}\right]$ and det A = 45 then find x.

Answer:
det A = 45
⇒ $\left|\begin{array}{rrr} 1 & 0 & 0 \\ 2 & 3 & 4 \\ 5 & -6 & x \end{array}\right|$ = 45
⇒ 1(3x + 24) = 45
⇒ 3x = 21
⇒ x = 7

II.
Question 1.
Show that $\left|\begin{array}{lll} \mathrm{b c} & \mathrm{b}+\mathrm{c} & 1 \\ \mathrm{c a} & \mathrm{c}+\mathrm{a} & 1 \\ \mathrm{a b} & \mathrm{a}+\mathrm{b} & 1 \end{array}\right|$ = (a – b)(b – c)(c – a).

Answer:
Operating R₂ – R₁, R₃ – R₁, on the given determinant
LHS = $\left|\begin{array}{ccc} b c & b+c & 1 \\ c(a-b) & a-b & 0 \\ b(a-c) & a-c & 0 \end{array}\right|$
= (a – b)(a – c) $\left|\begin{array}{ccc} b c & b+c & 1 \\ c & 1 & 0 \\ b & 1 & 0 \end{array}\right|$
= (a – b)(a – c)(1)(c – b)
= (a – b)(b – c)(c – a) (exponding on 3rd column)
= RHS

Question 2.
Show that $\left|\begin{array}{ccc} b+c & c+a & a+b \\ a+b & b+c & c+a \\ a & b & c \end{array}\right|$ = a² + b² + c² – 3abc (Mar. 2008; May 2007)

Answer:

= (a + b + c) [(c – b) (c – a) – (a – b) (b – a)]
= (a + b + c) [c² – bc – ac + ab + a² – 2ab + b²]
= (a + b + c) [a² + b² + c² – ab – bc – ca]
= a² + b² + c² – 3abc

Question 3.
Show that $\left|\begin{array}{ccc} \mathrm{y}+\mathrm{z} & \mathrm{x} & \mathrm{x} \\ \mathrm{y} & \mathrm{z}+\mathrm{x} & \mathrm{y} \\ \mathrm{z} & \mathrm{z} & \mathrm{x}+\mathrm{y} \end{array}\right|$ = 4xyz.
Answer:
R₁ – (R₂ + R₃) on the given determinant gives

= 2[z(xy) – y(-xz)]
= 2[2xyz] = 4xyz = RHS

Question 4.
If $\left|\begin{array}{ccc} a & a^2 & 1+a^3 \\ b & b^2 & 1+b^3 \\ c & c^2 & 1+c^3 \end{array}\right|$ = 0 and $\left|\begin{array}{ccc} \mathrm{a} & \mathrm{a}^2 & 1 \\ \mathrm{b} & \mathrm{b}^2 & 1 \\ \mathrm{c} & \mathrm{c}^2 & 1 \end{array}\right|$ ≠ 0, then show that abc = -1. (Mar. ’14)

Answer:

⇒ abc + 1 = 0
⇒ abc = -1

Question 5.
Without expanding the determinant, prove that
(i) $\left|\begin{array}{lll} \mathrm{a} & \mathrm{a}^2 & \mathrm{b c} \\ \mathrm{b} & \mathrm{b}^2 & \mathrm{c a} \\ \mathrm{c} & \mathrm{c}^2 & \mathrm{a b} \end{array}\right|=\left|\begin{array}{ccc} 1 & \mathrm{a}^2 & \mathrm{a}^3 \\ 1 & \mathrm{b}^2 & \mathrm{b}^3 \\ 1 & \mathrm{c}^2 & \mathrm{c}^3 \end{array}\right|$

Answer:

(ii) $\left|\begin{array}{ccc} \mathrm{a x} & \mathrm{b y} & \mathrm{c z} \\ \mathrm{x}^2 & \mathrm{y}^2 & \mathrm{z}^2 \\ 1 & 1 & 1 \end{array}\right|=\left|\begin{array}{ccc} \mathrm{a} & \mathrm{b} & \mathrm{c} \\ \mathrm{x} & \mathrm{y} & \mathrm{z} \\ \mathrm{y z} & \mathrm{z x} & \mathrm{x y} \end{array}\right|$
Answer:

(iii) $\left|\begin{array}{lll} 1 & b c & b+c \\ 1 & c a & c+a \\ 1 & a b & a+b \end{array}\right|=\left|\begin{array}{lll} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{array}\right|$ (Board Model Paper)
Answer:

(∵ R₂ – R₁; R₃ – R₁)
= (b – a) (c² – a²) – (c – a) (b² – a²)
= (b – a) (c – a) (c + a) – (c – a) (b – a) (b + a)
= (b – a) (c – a) (c + a – b – a)
= (b – a) (c – a) (c – b)
= (a – b) (b – c) (c – a)
LHS = RHS

Question 6.
If Δ₁ = $\left|\begin{array}{ccc} \mathrm{a}_1^2+\mathrm{b}_1+c_1 & \mathrm{a}_1 \mathrm{a}_2+\mathrm{b}_2+c_2 & \mathrm{a}_1 \mathrm{a}_3+\mathrm{b}_3+c_3 \\ \mathrm{b}_1 \mathrm{b}_2+c_1 & \mathrm{b}_2^2+\mathrm{c}_2 & \mathrm{b}_2 \mathrm{b}_3+\mathrm{c}_3 \\ \mathrm{c}_3 c_1 & \mathrm{c}_3 \mathrm{c}_2 & \mathrm{c}_3^2 \end{array}\right|$ and Δ₂ = $\left|\begin{array}{lll} \mathrm{a}_1 & \mathrm{b}_1 & \mathrm{c}_1 \\ \mathrm{a}_2 & \mathrm{b}_2 & \mathrm{c}_2 \\ \mathrm{a}_3 & \mathrm{b}_3 & \mathrm{c}_3 \end{array}\right|$ , then find the value of $\frac{\Delta_1}{\Delta_2}$ .

Answer:

Question 7.
If Δ₁ = $\left|\begin{array}{ccc} 1 & \cos \alpha & \cos \beta \\ \cos \alpha & 1 & \cos \gamma \\ \cos \beta & \cos \gamma & 1 \end{array}\right|$ and Δ₂ = $\left|\begin{array}{ccc} 0 & \cos \alpha & \cos \beta \\ \cos \alpha & 0 & \cos \gamma \\ \cos \beta & \cos \gamma & 0 \end{array}\right|$ and Δ₁ = Δ₂ then show that cos²α + cos²β + cos²γ = 1.

Answer:
Given $\left|\begin{array}{ccc} 1 & \cos \alpha & \cos \beta \\ \cos \alpha & 1 & \cos \gamma \\ \cos \beta & \cos \gamma & 1 \end{array}\right|$
= (1 – cos²γ) – cos α (cos α – cos β cos γ) + cos β (cos α cos γ – cos β)
= 1 – cos²γ – cos²α + cos β cos α cos γ + cos α cos β cos γ – cos²β
= 1 – (cos²α + cos²β + cos²γ) + 2 cos α cos β cos γ

Δ₂ = $\left|\begin{array}{ccc} 0 & \cos \alpha & \cos \beta \\ \cos \alpha & 0 & \cos \gamma \\ \cos \beta & \cos \gamma & 0 \end{array}\right|$
= – cos α (0 – cos γ cos β) + cos β (cos α cos γ)
= cos α cos β cos γ + cos α cos β cos γ
= 2cos α cos β cos γ
Also given Δ₁ = Δ₂
⇒ 1 – (cos²α + cos²β + cos²γ) + 2 cos α cos β cos γ
= 2 cos α cos β cos γ
⇒ 1 – (cos²α + cos²β + cos²γ) = 0
∴ cos²α + cos²β + cos²γ = 1

III.
Question 1.
Show that
$\left|\begin{array}{ccc} a+b+2 c & a & b \\ c & b+c+2 a & b \\ c & a & c+a+2 b \end{array}\right|$ = 2(a + b + c)³

Answer:

Question 2.
Show that $\left|\begin{array}{lll} a & b & c \\ b & c & a \\ c & a & b \end{array}\right|^2$ = $\left|\begin{array}{ccc} 2 b c-a^2 & c^2 & b^2 \\ c^2 & 2 a c-b^2 & a^2 \\ b^2 & a^2 & 2 a b-c^2 \end{array}\right|$ = (a³ + b³ + c³ – 3abc)². (May 2014, Mar. 01′)

Answer:
Let Δ = $\left|\begin{array}{lll} a & b & c \\ b & c & a \\ c & a & b \end{array}\right|$ = a(bc – a²) – b(b² – ac) + c(ab – c²)
= abc – a³ – b³ + abc + abc – c³
= – (a³ + b³ + c³ – 3abc)
⇒ Δ² = (a³ + b³ + c³ – 3abc)² …………..(1)

From (1) and (2) the result is proved.

Question 3.
Show that $\left|\begin{array}{ccc} a^2+2 a & 2 a+1 & 1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{array}\right|$ = (a – 1)³. (March 2007)

Answer:
Apply operations R₁ – R₂ and R₂ – R₃ we get

Question 4.
Show that $\left|\begin{array}{ccc} \mathrm{a} & \mathrm{b} & \mathrm{c} \\ \mathrm{a}^2 & \mathrm{b}^2 & \mathrm{c}^2 \\ \mathrm{a}^3 & \mathrm{b}^3 & \mathrm{c}^3 \end{array}\right|$ = abc(a – b)(b – c)(c – a)

Answer:
LHS = abc $\left|\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{array}\right|$
= abc $\left|\begin{array}{ccc} 0 & 0 & 1 \\ a-b & b-c & c \\ a^2-b^2 & b^2-c^2 & c^2 \end{array}\right|$ (Use operations C₁ – C₂, C₂ – C₃)
= abc [(a – b) (b² – c²) – (b – c) (a² – b²)]
= abc [(a – b) (b – c) (b + c) – (b – c) (a – b) (a + b)]
= abc (a – b) (b – c) [b + c – a – b]
= abc (a – b) (b – c) (c – a)

Question 5.
Show that $\left|\begin{array}{ccc} -2 a & a+b & c+a \\ a+b & -2 b & b+c \\ c+a & c+b & -2 c \end{array}\right|$ = 4(a + b)(b + c)(c + a)

Answer:

= 0 (∵ R₁ & R₃ are similar)
∴ (a + b) is a factor of Δ.
Similarly putting b + c = 0 and c + a = 0 we shall find that b + c and c + a are also factors of Δ.
∵ Δ is a 3rd degree expression in a, b, c.

Let Δ = k (a + b) (b + c) (c + a)
Where k ≠ 0 is a scalar.
Put a = 1, b = 1, c = 1 then
= k(1 + 1) (1 + 1) (1 + 1)
= 8k -2(4 – 4) – 2(-4 – 4) + 2(4 + 4)
= 8k
⇒ -16 + 16 = 8k ⇒ k = 4
Δ = 4(a + b) (b + c) (c + a)
Here $\left|\begin{array}{ccc} -2 a & a+b & c+a \\ a+b & -2 b & b+c \\ c+a & c+b & -2 c \end{array}\right|$ = 4(a + b)(b + c)(c + a)

Question 6.
Show that $\left|\begin{array}{lll} \mathrm{a}-\mathrm{b} & \mathrm{b}-\mathrm{c} & \mathrm{c}-\mathrm{a} \\ \mathrm{b}-\mathrm{c} & \mathrm{c}-\mathrm{a} & \mathrm{a}-\mathrm{b} \\ \mathrm{c}-\mathrm{a} & \mathrm{a}-\mathrm{b} & \mathrm{b}-\mathrm{c} \end{array}\right|$ = 0
Answer:
R₁ + (R₂ + R₃) given
$\left|\begin{array}{ccc} 0 & 0 & 0 \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{array}\right|$
= 0 (∵ If one row or column elements of a square matrix are zeroes then the value of the determinant of that matrix is equal to zero)
= RHS.

Question 7.
Show that $\left|\begin{array}{lll} 1 & a & a^2-b c \\ 1 & b & b^2-c a \\ 1 & c & c^2-a b \end{array}\right|$ = 0

Answer:
Make operations R₂ – R₁, R₃ – R₁ then the given determinant.

Question 8.
Show that $\left|\begin{array}{lll} x & a & a \\ a & x & a \\ a & a & x \end{array}\right|$ = (x + 2a)(x – a)².

Answer:

TS Inter 1st Year Maths 1A Matrices Solutions Exercise 3(c)

Junaid

January 8, 2026

TS Inter 1st Year Maths 1A Matrices Solutions Exercise 3(c)

Question 1.
If A = $\left[\begin{array}{rrr} 2 & 0 & 1 \\ -1 & 1 & 5 \end{array}\right]$ and B = $\left[\begin{array}{rrr} -1 & 1 & 0 \\ 0 & 1 & -2 \end{array}\right]$ then find (AB’)’

Answer:
We have (AB)’ = B’A’
and (AB’)’ = (B’)’ A’ = BA’ (∵ (B )’ = B)

Question 2.
If A = $\left[\begin{array}{rr} -2 & 1 \\ 5 & 0 \\ -1 & 4 \end{array}\right]$ and B = $\left[\begin{array}{rrr} -2 & 3 & 1 \\ 4 & 0 & 2 \end{array}\right]$ then find 2A + B’ and 3B’ – A.

Answer:

Question 3.
If A = $\left[\begin{array}{cc} 2 & -4 \\ -5 & 3 \end{array}\right]$ then find A + A’ and A. A’ (May 2007) (Board Model Paper)

Answer:

Question 4.
If A = $\left[\begin{array}{ccc} -1 & 2 & 3 \\ 2 & 5 & 6 \\ 3 & x & 7 \end{array}\right]$ is a symmetric matrix then find x.

Answer:
A matrix ‘A’ is said to be symmetric if A’ = A

Question 5.
If A = $\left[\begin{array}{ccc} 0 & 2 & 1 \\ -2 & 0 & -2 \\ -1 & x & 0 \end{array}\right]$ is a skew symmetric matrix, find x. (May 2014, 11)

Answer:
A matrix A is said to be skew symmetric if A’ = – A
$\left[\begin{array}{ccc} 0 & -2 & -1 \\ 2 & 0 & \mathrm{x} \\ 1 & -2 & 0 \end{array}\right]=\left[\begin{array}{ccc} 0 & -2 & -1 \\ 2 & 0 & 2 \\ 1 & -\mathrm{x} & 0 \end{array}\right]$
from equality of matrix x = 2

Question 6.
Is $\left[\begin{array}{ccc} 0 & 1 & 4 \\ -1 & 0 & 7 \\ -4 & -7 & 0 \end{array}\right]$ a symmetric or skew symmetric?

Answer:
Let A = $\left[\begin{array}{ccc} 0 & 1 & 4 \\ -1 & 0 & 7 \\ -4 & -7 & 0 \end{array}\right]$ then A is symmetric if A’ = A and skew symmetric if A’ = – A
i.e., A’ = $\left[\begin{array}{ccc} 0 & -1 & -4 \\ 1 & 0 & -7 \\ 4 & 7 & 0 \end{array}\right]=\left[\begin{array}{ccc} 0 & 1 & 4 \\ -1 & 0 & 7 \\ -4 & -7 & 0 \end{array}\right]$ = -A
∴ The matrix A is a skew symmetric matrix.

II.
Question 1.
If A = $\left[\begin{array}{cc} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right]$ , show that A . A’ = A’ . A = I₂. (March 2007)

Answer:

Question 2.
If A = $\left[\begin{array}{ccc} 1 & 5 & 3 \\ 2 & 4 & 0 \\ 3 & -1 & -5 \end{array}\right]$ and B = $\left[\begin{array}{ccc} 2 & -1 & 0 \\ 0 & -2 & 5 \\ 1 & 2 & 0 \end{array}\right]$ , then find 3A – 4B’.

Answer:

Question 3.
If A = $\left[\begin{array}{rr} 7 & -2 \\ -1 & 2 \\ 5 & 3 \end{array}\right]$ and B = $\left[\begin{array}{rr} -2 & -1 \\ 4 & 2 \\ -1 & 0 \end{array}\right]$ then find AB’ and BA’.

Answer:

Question 4.
For any square matrix A; show that A A’ is symmetric. (March 2015-A.P)

Answer:
By definition a matrix is said to be symmetric if A’ = A.
∴(A A’)’ = (A’)’ A’ = A A’
[(∵ (AB)’ = B’A’ and (A’)’ = A]
Hence AA’ is a symmetric matrix.

TS Inter 1st Year Botany Study Material Chapter 5 Morphology of Flowering Plants

Sandhya

January 3, 2026

TS Inter 1st Year Botany Study Material Chapter 5 Morphology of Flowering Plants

Very Short Answer Type Questions

Question 1.

Differentiate fibrous roots from adventitious roots. [Mar. – 2020]?

Answer:

1) Adventitious roots :
The roots that arise from plant parts other than radicle.
Ex: Climbing roots, Velamen roots, Respiratory roots etc.

2) Fibrous roots :
Large number of roots which originate from the base of the stem after the loss of short lived primary root. Ex: Monocot plants.

Question 2.

Define modification. Mention how root is modified in banyan tree and mangrove plants ?

Answer:

1) Modification :
A permanent morphological change in a plant organ to perform a special function depending upon environment.

2) In banyan tree, roots are modified as prop roots (pillar roots) to give additional support for branches.

While in mangrove plants (Rhizophora and Avicennia) pneumatophores or respiratory roots come out of the ground and grow vertically upwards to get oxygen for respiration.

Question 3.

What type of specialized roots are found in epiphytic plants ? What is their function ?

Answer:

1. Velamen roots are found in epiphytic plants like Vanda.
2. They absorb moisture from the atmosphere.

Question 4.

How does the sucker of Chrysanthemum differ from the stolon of jasmine?

Answer:

Sucker of Chrysanthemum is the lateral branch arises from the basal and underground portion of stem. It grows obliquely upward giving rise to leafy shoot.
Stolon of jasmine is an obliquely downward growing slender lateral branch that arises from the base of the main axis and produces adventitious roots on touching the ground.

Question 5.

What is meant by pulvinus leaf base? In members of which Angiospermic family do you find them? [Mar. ’14 – A.P. ; Mar. ’14]?

Answer:

The swollen leaf base is called pulvinus leaf base.
It is found in some members of leguminaceae family.

Question 6.

Define venation. How do dicots differ from monocots with respect to venation? [Mar. ’15 – A.P.]

Answer:

The arrangement of veins and the veinlets in the lamina of leaf is termed as venation.
Leaves of dicots have reticulate venation, whereas, leaves of monocots possess parallel venation.

Question 7.

How is a pinnately compound leaf is different from a palmately compound leaf? Explain with one example, each.?

Answer:

Pinnately compound leaf: It consists of a number of leaflets on a common axis called rachis. Ex: Neem.
Palmately compound leaf: If consists of leaflets attached at a common point, i.e., at the tip of the petiole. Ex: Bombax ceiba (silk cotton).

Question 8.

Which organ is modified to trap insects in insectivorous plants? Give two examples. [Mar. 2019, ’13]?

Answer:

Leaves are modified to trap insects in insectivorous plants.
Example : Nepenthes (Pitcher plant), Dionea (Venus fly-trap).

Question 9.

Differentiate between Racemose and Cymose inflorescences. [Mar. ’15 – T.S.]?

Answer:

Racemose inflorescence: The main axis (Peduncle) continues to grow and produce flowers in an acropetal succession.
Cymose inflorescence: The main axis ends in a flower due to limited growth and flowers are borne in basipetal succession.

Question 10.

What is the morphology of cup like structure in Cyathium? In which family it is found? [Mar. – 2018, Mar. ’15 – A.P.]

Answer:

In Cyathium, involucre of bracts form cup like structure.
It is found in family Euphorbiaceae.

Question 11.

What type of inflorescence is found in fig trees ? Why does the insect B/astophaga visits the inflorescence of fig tree?

Answer:

Hypanthodium is found in fig trees.
The insect Blastophaga visits for pollination and lays its eggs in the gall flowers.

Question 12.

Differentiate actinomorphic from zygomorphic flower. [May ’14]?

Answer:

1. Actinomorphic flower :
A flower that can be divided into two equal radial halves in any radial plane passing through the centre Ex: Hibiscus.’

2. Zygomorphic flower :
A flower that can be divided into two similar halves only in one particular vertical plane Ex: Bean.

Question 13.

How do the petals in pea plant are arranged? What is such type of arrangement called?

Answer:

In pea plant there are five petals. The largest (standard) petal overlaps the two lateral (wings) petals that inturn overlap the two smallest anterior petals (keel).
This arrangement is called vexillary or papilionaceous.

Question 14.

What is meant by Epipetalous condition ? Give an example. [May ’17, Mar. ’17 – A.P.]?

Answer:

Epipetalous condition : A condition in which stamens attached to the petals.
Ex: Brinjal, Datura.

Question 15.

Differentiate between apocarpous and syncarpous ovary. [Mar. – 2018]?

Answer:

1) Apocarpous ovary :
More than one carpel is present in gynoecium and they are free. Eg : Lotus and Rose

2) Syncarpous condition :
Carpels are fused. Eg: Mustard and Tomato.

Question 16.

Define placentation. What type of placentation is found in Dianthus? [Mar. – 2020, Mar. 15 – T.S.]

Answer:

The arrangement of ovules within the ovary is known as placentation.
In Dianthus, free central placentation is present.

Question 17.

What is meant by parthenocarpic fruit? How is it useful?

Answer:

A fruit formed without fertilization of the ovary is called parthenocarpic fruit.
They are without seeds. Ex: Banana.

Question 18.

What is the type of fruit found in mango? How does it differ from that of coconut?

Answer:

The type of fruit found in mango is drupe. In which the pericarp is well differentiated into an outer thin epicarp, a middle fleshy edible mesocarp and an inner stony hard endocarp.
In coconut, also fruit is a drupe in which the mesocarp is fibrous.

Question 19.

Why certain fruits are called false fruits ? Name two examples of plants having false fruits.?

Answer:

Certain fruits that develop from floral parts other than the ovary called false fruits.
Ex: Apple (Thalamus) Cashew (Pedicel) Strawberry (Thalamus.)

Question 20.

Name any two plants having single seeded dry fruits.?

Answer:

Dry indehiscent fruits are single seeded.
Coryza (caryopsis) and Tridax (Cypsela).

Question 21.

Define schizocarpic dry fruits. Give an example?

Answer:

The dry fruits which split into one Seeded bits called mericarps are known as Schizocarpic dry fruits.
Ex: Acacia, Castor.

Question 22.

Define mericarp. In which plant you find it?

Answer:

One seeded bits formed after splitting of Schizocarpic dry fruits are called mericarp.
Ex: Acacia, Castor.

Question 23.

What are aggregate fruits? Give two examples.?

Answer:

Bunch of fruitlets developed from multicarpellary, apocarpous ovary are called aggregate fruits.
Ex: Custard apple (Annona squamosa), Strawberry.

Question 24.

Name a plant that has single fruit developing from the entire inflorescence. What is such a fruit called?

Answer:

Single fruit that develops from an entire inflorescence is called composite fruit. Ex: Pineapple, Jack fruit.

Short Answer Type Questions

Question 1.

Explain different regions of root with neat labeled diagram.?

Answer:

Root has four regions. They are
a) Root cap
b) Region of rheristematic activity
c) Region of elongation
d) Region of maturation.

a) Root cap :
The tip of the root is covered by a thimble-like structure called the root cap. It gives protection to the root tip as it penetrates into the soil.

b) Region of meristematic activity :
Above the root cap, region of meristematic activity is present. It has meristematic cells. These cells are small, thin walled with dense protoplasm. They divide repeatedly.

c) Region of elongation :
Above the region of meristematic activity, region of elongation is present. The new cells formed grows and elongates. It is responsible for growth of length.

d) Region of maturation :
Behind the region of elongation region of maturation is present. Depending upon the function young cells differentiates into permanent cells. As it matures, it is called region of maturation.

From this region, some of the epidermal cells forms unicellular root hairs. The main function of root hairs is absorption of water from the soil. They are short lived.

Question 2.

Justify the statement: “Underground parts of plants are not always roots”?

Answer:

Normally roots are underground whereas stems are aerial. But in some plants stem grow below the soil. They are called underground stems. They are rhizome corm, stem tuber and bulb.

These underground parts of the plants can be recognised as stem due to presence of nodes, internodes, scale leaves, axillary buds and terminal buds. They can be even identified by their anatomical structures.

For example, stem tubers (Potato), these are underground branches which store food at the tip and becomes tuberous. The tuber is covered by brown coloured layer. It bears many ‘eye’ like structures. These eyes represent the nodes. Each eye has leaf scar and axillary bud. Scar represents the position of scale leaf. Eye help in vegetative propagation.

Question 3.

Explain with examples different types of phyllotaxy?

Answer:

The mode of arrangement of leaves on the stem and branches is called phyllotaxy. It is three types. They are
a) Alternate phyllotaxy :
In this type only one leaf arises at each node in alternate manner.
Eg : Hibiscus, rosa-sinensis (china rose), mustard, sunflower.

b) Opposite phyllotaxy :
In this type, a pair of leaves arise at each node and lie opposite to each other. Eg : Calotropis, Guava.

c) Whorl phyllotaxy :
In this type, more than two leaves arise at a node and form a whorl. Eg : Nerium, Alstonia.

Question 4.

How do leaf modifications help plants?

Answer:

The main function of leaf is photosynthesis and transpiration. In some plants leaves change in their structure to perform new function other than photosynthesis. This is called leaf modification.

1. Tendrils :
In weak stemmed plant, the entire leaf or any part of the leaf is modified into tendrils. They provide mechanical support and help in climbing. Eg: Pea.

2. Spine :
In some plants, leaves are modified into sharp pointed spines. They help in reducing the rate of transpiration in xerophytic plants and also for defence.
Eg: Cacti.

3. Storage leaves :
The fleshy leaves of onion and garlic store food materials.

4. Phyllode :
In some plants such as Australian acacia, the leaves are pinnately compound in which the leaflets are small and short lived. The petioles of these plants expand into green structure performing photosynthesis. These are called phyllode.

5. Insectivorous leaves :
In plants growing in nitrogen deficient soils, leaves are modified to trap insects for their nitrogen requirement.
Eg : Nepenthes (Pitcher plant)
Dionea (Venus fly trap)

6. Vegetative propagation :
In some plants leaves produce buds called epiphyllous buds. They help in vegetative propagation. Eg : Bryophyllum.

Question 5.

Describe any two special types of inflorescences?

Answer:

(Note : Write any two of the following ��
Verticellaster, Cyathium and Hypanthodium are special types of inflorescence.

Verticellaster:

It is a special type of inflorescence found in the family Lamiaceae (Labiatae).
In this type, flowers arise in the axil of leaves arranged opposite to each other at every node.
In the axil of leaf, the flowers are developed initially in dichasial cyme and later in monochasial scorpoid cyme.
Flowers are crowded round the node like a false whorl (verticel). Hence it is called ‘Verticellaster’.
Eg: Leucas and Leonotis.

Cyathium :

This is a single flower like special inflorescence found in family Euphorbiaceae.
The inflorescence is covered by a deep cup like involucre of bracts.
At the centre of this cup there is a single female flower represented by tricarpellary syncarpous ovary.
Surrounding this female flower many male flowers are arranged in monochasial cyme.
Male flowers are represented by single stalked stamen. Male and female flowers are achlamydeous arranged in centrifugal manner. Eg: Euphorbia, Poinsettia.

Hypanthodium :

It is fruit like inflorescence.
In this peduncle is modified into a deep cup like fleshy structure with an apical opening.
The male flowers located near the opening and the female flowers are at the bottom while in between them the sterile female flowers called gall flowers are present.
Pollination in these plants takes place by an insect called Blastophaga which lay its eggs in the gall flowers. After fertilisation the whole inflorescence becomes into a fig fruit.

Question 6.

Describe the arrangement of floral members in relation to their insertion on thalamus.?

Answer:

Depending upon the arrangement of floral members in relation to their insertion on thalamus, flowers are divided into three types. They are
1) Hypogynous :
Thalamus is conical. The gynoecium occupies the highest position. The remaining floral members like calyx, corolla and androecium are at the base of the gynoecium. In this the ovary is called “Superior”.
Ex : Hibiscus, Datura, Mustard, Brinjal etc.

2) Perigynous :
Thalamus is concave or saucer shaped. Gynoecium is situated at the centre. The remaining floral members like calyx, corolla and androecium are arranged along the margin, almost at the same level. In this, the ovary is said to be half inferior or half superior. Ex: Tephrosia, plum, rose, peach etc.

3) Epigynous flower :
In this, thalamus is deep cup like structure, inside it gynoecium is present. The walls of the thalamus and ovary are fused. The remaining floral members are arranged along the margins of the thalamus, i.e. above the level of ovary. So, the ovary is called inferior. Ex : Tridax, guava, cucumber, ray floret of sunflower.

Question 7.

“The flowers of many angiospermic plants which show sepals and petals, differ with respect to the arrangement of sepals and petals in respective whorls’. Explain?

Answer:

The arrangement of sepals and petals in floral bud is known as aestivation. It is of different types.

1) Valvate aestivation :
When sepals or petals in a whorl just touch one another at the margin without overlapping is called valvate aestivation. Eg : Calotropis.

2) Twisted aestivation :
When sepals or petals margin in a whorl overlap one another it is said to be twisted aestivation. Eg: Corolla of hibiscus, cotton, lady’s finger etc.

3) Imbricate aestivation :
If the margins of sepals or petals overlap one another but not in any particular direction is called imbricate aestivation. Eg : Cassia, gulmohur.

4) Vexillary or Papilionaceous aestivation :
In this, there are five petals. The largest petal towards posterior side is called Vexillum or Standard Petal. It overlaps the two lateral petals called Alae or wing petal. These overlap the two smallest petals called keel petals towards anterior side, which are boat shaped. Ex : Bean, Pea.

Question 8.

Describe any four types of placentations found in flowering plants.?

Answer:

The arrangement of ovules within the ovary is known as placentation. They are Marginal placentation: Placenta forms a ridge along the ventral suture of the ovary. Ovules are borne on this ridge forming two rows. Eg: Pea

Axial placentation :
When the placenta is axial and ovules are attached to it in multilocular ovary, it is called axile placentation. Eg: China rose, rose, tomato and lemon.

Parietal placentation :
Ovules born on the inner wall of the ovary or on a parietal part, it is called parietal placentation. Ovary is one chambered but it becomes two chambered due to the formation of the false septum. Eg: Mustard and Argemone. Free central placentation : When the ovules are borne on the central axis without septa, it is known as free central placentation. Eg : Dianthus, primrose.

Basal placentation :
Single ovule is attached to placenta at the base of the ovary. It is called Basal placentation. Eg : Sunflower, marigold.

Question 9.

Describe in brief fleshy fruits by you studied.?

Answer:

In fruits where pericarp becomes fleshy at the time of ripening are called fleshy fruits. Pericarp can be divided into three layers namely outer epicarp, middle mesocarp and inner endocarp. Basing upon the nature of pericarp, fleshy fruits are divided into five types. They are

1) Drupe :
It is one seeded fleshy fruit developed from monocarpellary, superior ovary. The fruit is characterised by stony endocarp. So it is known as Drupe.

In Mango, the outer epicarp is thin, middle mesocarp is fleshy and edible. The inner endocarp is hard stony.
In coconut, the outer epicarp is thin, middle mesocarp is fibrous and inner endocarp is hard stony. The edible part is the endosperm of seed (Copra).

2) Berry :
It is a fleshy fruit having one or more seeds. In this, epicarp is thin. Mesocarp and endocarp are fused to form pulp. Seeds are hard. These fruits develop from bi to multicarpellary syncarpous gynoecium: Eg : Guava, grapes, tomato.

3) Pome :
It is a fleshy fruit developed from inferior ovary of bi or multicarpellary syncarpous gynoecium. It is surrounded by fleshy thalamus. The endocarp is cartilagenous. Eg: Apple.

4) Pepo :
It is developed from tricarpellary syncarpous unilocular inferior ovary. The epicarp is like a rind., the mesocarp is fleshy and the endocarp is smooth. Eg : Cucumber.

5) Hesperidium :
It is developed from multicarpellary syncarpous, multilocular and superior ovary. In this epicarp is leathery with many volatile oil glands. Mesocarp is papery and endocarp has many chambers filled with juicy hairs. Eg : Citrus.

Question 10.

Describe with examples the various dry fruits studied by you?

Answer:

When the fruit wall or pericarp is dry or non-fleshy they are called dry fruits. They are of three types (i) Dry dehiscent (ii) Dry indehiscent (iii) Schizocarpic.
i) Dry dehiscent :
The dry fruit which break open and liberate the seeds are called dry dehiscent fruits. They are of different types.

a) Legume :
The fruits which break dorsiventrally into two halves liberating the seeds are called legumes. It is a characteristic fruit of family fabaceae. Eg : Pea, bean etc.

b) Capsule :
It is a dry fruit which liberates seeds in different ways at maturity. Eg : Cotton, Datura.

ii) Dry indehiscent fruits :
These dry fruits are normally one seeded and never dehisce even at maturity. The seeds are liberated only after the disintegration of the pericarp. They are of following types.

1) Caryopsis :
In this the pericarp and seed coat fuse together. It is a characteristic, . fruit of family poaceae. Ex: Grass, Rice.

2) Nut :
This single seeded dry fruit has a stony pericarp. The pericarp and seed coat remains free. Eg : Cashew.

3) Cypsela :
The single seeded fruit characterized by persistent pappus like calyx. Eg: Tridax.

iii) Schizocarpic fruits :
The fruit which split into one-seeded bits called mericarps are called schizocarpic fruits. Eg: Acacia, Castor.

Long Answer Type Questions

Question 1.

Define root. Mention the types of root systems. Explain how root is modified [Mar. 17 A.P & T.S ; Mar. 15 – A.P & T.S ; Mar. 13]?

Answer:

The part of the plant body present below the soil is called root. It is developed from radicle. There are two types of root sytems.

Tap root system
Fibrous root system

In some plants root is modified to perform new function suitable for the environment. It is called root modification. They are
1) Storage roots :
The roots which store food materials are called storage roots or tuberous roots. In biennial plants, the tap root is modified into storage roots. Depending upon the shape, they are
a) Spindle shape (fusiform) Eg: Radish
b) Cone shape (Conical) Eg: Carrot
c) Top shape (Napiform) Eg: Beetroot
Adventitious roots of sweet potato and fibrous roots in Asparagus store food materials.

2) Prop roots or Pillar roots :
In plants like banyan trees, branches are large and heavy.
From the branches roots arise, they hang in air for sometime and later fixes into the soil.
They are called prop roots or pillars roots.
They act like pillars and gives support.

3) Stilt roots :
In plants like maize and sugarcane roofs arise from the lower nodes of the stem. They are called stilt roots. They give support to plant.

4) Pneumatophores or Respiratory roots :
The Mangrove plants which grow in swampy areas suffer from lack of oxygen as the soils are water lodged. In these plants root comes out of soil and grow vertically upwards. These roots are called respiratory roots or pneumatophores as they take oxygen from air for respiration.

5) Photosynthetic roots :
In Taeniophyllum (epiphyte) the stem and leaves are absent. The roots become green and perform photosynethsis. Such roots are called photosynthetic roots.

6) Velamen roots or Epiphytic roots :
The plants which grow on the branches of big trees for sunlight are
called epiphytes. They have roots which hang freely
in the air. They absorb moisture from the atmosphere. These roots are called velamen roots
or epiphytic roots. Eg : Vanda.

7) Nodular roots :
In members of Fabaceae, the roots show small nodule like structures. Hence it is called nodular roots. In the nodule, Rhizobium bacteria is present. It fixes atmospheric nitrogen into soil. Plant and Rhizobium show symbiotic association.

8) Parasitic roots or Kaustoria :
The plants which depend upon the other plants completely or incompletely for their food and water are called Parasites. They produce parastic roots. They are 1) Complete parasites 2) Partial parasites.
a) Complete parasitic piamts :
These are leafless. So the haustorial root enters into both xylem and phloem to obtain both water and food from the host plant. Ex : Cuscuta, Rafflesia.

b) Partial parasitic plants :
These plants bear leaves. So they can prepare food by photosynthesis. The haustorial roots penetrate only into the xylem tissue of the host to absorb water.

Question 2.

Explain how stem is modified variously to perform different functions. [Mar. 2020, 2019, 14, May 2017 ’14]?

Answer:

When a permanent change occurs in the structure of stems to perform new functions suitable for the environment, it is called ‘stem modification’.

It is of three types. They are
I. Aerial stem modification
II. Sub aerial stem modification
III. Underground stem modification

I. Aerial stem modification :
Modification of aerial stems, vegetative buds and reproductive buds of a plant is called aerial stem modification.
a) Tendril :
Wiry delicate organ useful for climbing are called tendril. Axillary bud modified into tendril in gourds (cucumber, pumpkin, watermelon) or terminal bud in grapevines.

b) Thorn :
A woody pointed structure meant for protection are called thorns. Axillary bud modified into thorn. Eg : Bougainvillea. Terminal bud modified into thorn. Eg : Carissa.

c) Hook :
It is a woody, curved structure which helps in climbing. Ex : Artabotry.

d) Phylloclade :
Leaf like stem performing photosynthesis are called phylloclade. In order to reduce transpiration the leaves are modified into scales, spines etc.
Ex : In Opuntia, fleshy green flattened stem
In Euphorbia, fleshy green cylindrical stem In Casuarina green needle like stem.
Cladode or Cladophyll is a phylloclade of limited growth. Ex : Asparagus.

e) Bulbils :
Buds which show vegetative propagation are called bulbils. Vegetative buds in Dioscorea, Floral buds in Agave.

II. Sub aerial stem modification :
In some weak stemmed plants the stem remains partly aerial and partly underground. These are specialised for vegetative propagation. There are four types.

a) Runner:
Underground stems in some grasses and strawberry and sub-aerial stems in oxalis spread to new nitches and form new plants when older parts die.

b) Stolon :
In some plants slender branches arises grow obliquely downwards, produce roots at the point of contact with soil. These branches are called stolons. Ex : Nerium, Jasmine etc.

c) Suckers :
In plants like banana, pine apple, chrysanthemum part of the stem is in the soil. Underground branches grow obliquely upwards giving rise to leafy shoots. These branches are called suckers.

d) Offset :
In aquatic plants like Pistia and Eichhornia, a lateral branch of one internode length is called offset. At each node it bears a rosette of leaves and balancing roots at the base.

III. Underground stem modification :
Functions of the underground stems are storage of food materials, perennation through unfavourable seasons and vegetative reproduction. So these are called multipurpose stem modification.

a) Rhizome :
It is an underground stem which grows horizontally. It is branched. Ex: Zingiber (ginger).

b) Corm :
It is an underground stem which grows vertically. It is unbranched contractile roots present. Ex : Amorphaphallus (Zaminkand) and colocasia.

c) Stem tuber :
The swollen tip of an underground branch is called stem tuber. It bears many eyes. These eyes represent the nodes. Ex ; Potato.

d) Bulb :
It is a small reduced underground stem. Food is stored in the leaf bases. Ex: Onion.

Question 3.

Explain different types of racemose inflorescences.?

Answer:

Types of racemose inflorescence.
1) Simple Raceme :
Peduncle is unbranched, grows indefinitely on its numerous pedicillate, bracteate flowers are arranged in accropetal manner.
Ex: Crotalaria.

2) Compound Raceme :
Peduncle is branched. Each branch resembles a simple raceme. Ex: Mangifera. It is also called Panicle.

3) Simple Corymb :
Peduncle is unbranched and grows indefinitely. On it numerous pedicillate, bracteate flowers are arranged in accropetal manner. The lower flowers have long pedicels and upper flowers have shorter pedicels. Thus all the flowers are brought more or less to the same height. Eg: Cassia.

4) Compound Corymb :
Peduncle is branched and each branch is produced into a simple corymb. Eg: Cauliflower.

5) Simple Umbel :
The peduncle is condensed and unbranched. Many bracteate and pedicellate flowers arise at the tip. At the base of flowers, all the bracts form a whorl called ‘involucre’. Ex: Onion.

6) Compound Umbel :
Peduncle is branched. Each branch produces a simple umbel at its tip. Ex : Carrot.

7) Simple Spike :
Peduncle is unbranched. On it bracteate, sessile flowers are arranged, accropetally. Ex : Achyranthes.

8) Compound Spike :
Peduncle is branched. Each branch is similar to simple spike. Ex: Grass (Poaceae family).

9) Simple Spadix :
Peduncle is unbranched. On it sessile, unisexual and neuter flowers are arranged in acropetal succession. It is protected by modified bract called ‘Spathe’. Ex: Colocasia.

10) Compound Spadix :
Peduncle is branched. Each branch is similar to simple spadix. Ex: Musa, Cocos.

11) Head inflorescence :
Unisexual and bisexual sessile flowers are arranged centripetally on a condensed peduncle. Such an arrangement of flowers is called Head inflorescence. Ex: Tridax, Sunflower (Asteraceae family).

Intext Question Answers

Question 1.

In which plant, the underground stem grows horizontally in soil and helps in perennation?

Answer:

Zingiber (ginger)
Curcuma (turmeric)

Question 2.

Needle like phylloclades are found in which plant?

Answer:

Casuarina

Question 3.

Why do plants like Nepenthes trap insects?

Answer:

For their nitrogen requirement

Question 4.

What is the characteristic inflorescence found in members of Asteraceae?

Answer:

Head inflorescence

Question 5.

Can you name a plant that has least number of flowers in its inflorescence?

Answer:

Single flower in Hibiscus and Datura

Question 6.

Which family shows naked flowers?

Answer:

Euphorbiaceae family

Question 7.

In which flowers of the fig trees does the insect Blastophaga lay its eggs?

Answer:

Gall flowers (Sterile female flowers)

Question 8.

What type of symmetry is shown by the flowers of Canna?

Answer:

Asymmetry (irregular)

Question 9.

On which side of the flower do the flowers of pea have the keel petals?

Answer:

Anterior side

Question 10.

What is the ratio of overlapping margins of petals to overlapped ones in imbricate aestivation?

Answer:

1 : 1

Question 11.

How many ovules are found attached in basal placentation?

Answer:

One ovule

Question 12.

Which part of the flower in cashew plant forms the false fruit?

Answer:

Pedicel

Question 13.

Which plant has hard, stony endocarp and fleshy edible mesocarp?

Answer:

Mango

Question 14.

What is the morphology of ’spathe’ in Spadix inflorescence?

Answer:

Bract

Question 15.

What is the type of fruit known as if it develops from apocarpous ovary of a single flower?

Answer:

Aggregate fruit

TS Inter 1st Year Botany Study Material Chapter 4 Plant Kingdom

Sandhya

January 3, 2026

TS Inter 1st Year Botany Study Material Chapter 4 Plant Kingdom

Very Short Answer Type Questions

Question 1.

What is the basis of classification of Algae?

Answer:

Pigmentation and the type of stored food are the basis for classification of algae.
Based on above criteria algae are divided into 3 main classes-chlorophyceae Phaeophyceae and Rhodophyceae.

Question 2.

When and where does reduction division take place in the life cycle of a liverwort, a moss, a fern, a gymnosperm and an angiosperm?

Answer:

Liverwort – Sporophyte (capsule) – spore mother cell.
Moss – Sporophyte (capsule) – spore mother cells.
Gymnosperm – Sporophyte (Micro and mega sporangia) – Micro and megaspore
Angiosperm – Sporophyte (Anthers and ovule) – Spore mother cells

Question 3.

Differentiate between syngamy and triple fusion.?

Answer:

1) Syngamy :
One of the two male gametes released from pollen tube fuses with the egg cell to form a diploid zygote. This is also called true or real fertilisation.

2) Triple fusion :
One of the two male gametes released from pollen tube fuses with the diploid secondary nucleus to produce the triploid Primary Endosperm Nucleus (PEN).

Question 4.

Differentiate between antheridium and archegonium.?

Answer:

The antheridium is male sex organ whereas archegonium is female sex organ.
Antheridium produces many antherozoids (sperms), while archegonium produces an egg cell.

Question 5.

What are the two stages found in the gametophyte of mosses? Mention the structures from which these two stages develop?

Answer:

Protonema and Gametophore.
Protonema develops directly from the spore and adult gametophore develops from protenema.

Question 6.

Name the stored food materials found in Phaeophyceae and Rhodophyceae. [May ’14]?

Answer:

The stored food materials found in Phaeophyceae are laminarin or mannitol.
The stored food material found in Rhodophyceae is floridian starch.

Question 7.

Name the pigments responsible for brown colour of Phaeophyceae and red colour of Rhodophyceae.?

Answer:

The brown colour of Phaeophyceae depends upon the amount of xanthophyll pigment, fucoxanthin present in it.
The red colour of Rhodophyceae is due to red pigment r – phycoerythrin.

Question 8.

Name different methods of vegetative reproduction in Bryophytes. [Mar. 15 – A.P.]?

Answer:

Fragmentation, gemmae and budding.

Question 9.

Name the integumented megasporangium found in Gymnosperms. How many female gametophytes are generally formed inside the megasporangium?

Answer:

Ovule
One female gametophyte with 2 or more archegonia is formed inside the megasporangium.

Question 10.

Name the Gymnosperms which contain mycorrhiza and corolloid roots respectively.?

Answer:

Pinus contains mycorrhizal roots.
Cycas, contains corolloid robts.

Question 11.

Mention the ploidy of any four of the following.?

a) Protonemal cell of a moss
b) Primary endosperm nucleus in a dicot
c) Leaf cell of a moss
d) Prothallus of a fern
e) Gemma cell in Marchantia
f) Meristem cell of monocot
g) Ovum of a liverwort
h) Zygote of a fern.

Answer:

a) Haploid
b) Triploid
c) Haploid
d) Haploid
e) Haploid
f) Diploid
g) Haploid
h) Diploid

Question 12.

Name the four classes of Pteridophyta with one example each.?

Answer:

Cl : Psilopsida Ex : Psilotum
Cl: Lycopsida Ex : Selaginello, Lycopodium
Cl : Sphenopsida Ex : Equisetum
Cl : Pteropsida Ex : Dryopteris, Pteris, Adiantum

Question 13.

What are the first organisms to colonise rocks? Give the generic name of the moss which provides peat.?

Answer:

Mosses and lichens are the first organisms to colonise rocks
Sphagnum, (a moss) provides peat.

Question 14.

Mention the fern characters found in Cycas.?

Answer:

Fern characters found in Cycas are

Young leaves exhibit circinate vernation.
Presence of ramenta.
Male gametes are multiciliated.
Archegonia are present in the female gametophyte.

Question 15.

Why are Bryophytes called the amphibians of the plant kingdom?

Answer:

Bryophytes are called the amphibians of the plant kingdom because these plants live in moist soil and are dependent on water for sexual reproduction.

Question 16.

Name an algae which show?
a) Haplo-diplontic and
b) Diplontic types of life cycles.

Answer:

a) Ectocarpus and Kelps (Laminaria) – haplo-diplontic life cycle.
b) Fucus – diplontic life cycle.

Question 17.

Give examples for unicellular, colonial and filamentous algae.?

Answer:

Unicellular algae Ex : Chlamydomonas.
Colonial algae Ex : Volvox
Filamentous algae Ex : Spirogyra, Ulothrix.

Short Answer Type Questions

Question 1.

Differentiate between red algae and brown algae. [Mar. ’14]?

Answer:

Red algae	Brown algae
a) Members of Rhodophyceae are commonly called red algae.	a) Members of Phaeophyceae are commonly called brown algae.
b) They possess chlorophyll a, d, and phycoerythrin.	b) They posses chlorophyll a, c, carotenoids, and xanthophyll.
c) Red colour is due to Phycoerythrin pigment.	c) Brown colour is due to xanthophyll pigment.
d) Reserve food material is in the form of Floridean starch.	d) Reserve food material is in the form of mannitol (or) laminaria.

Question 2.

Differentiate between liverworts and mosses.?

Answer:

Liverworts	Mosses
1) They have a thallus-like dorsoventrally flattened body.	1) These are differentiated into stem-like and leaf-like structures.
2) Rhizoids unicellular.	2) Rhizoids multicelluar and branched.
3) Sporangium is differentiated into foot, seta and capsule. In some cases foot and seta may be absent.	3) Sporangium is differentiated into foot, seta and capsule.
4) Sporangium does not synthesise its food.	4) The sporangium synthesise its own food.
5) Elaters in the capsule help in spore dispersal.	5) Peristomial teeth help in spore dispersal.
6) Columella lacking	6) Columella is found.
7) Protonema and gametophore are absent.	7) Gametophyte has two stages. They are 1) Protonema 2) Gametophore.

Question 3.

What is meant by homosporous and heterosporous pteridophytes? Give two examples.?

Answer:

The plants which produce only one kind of spores are called homosporous Ex: Psilotum, Lycopodium. The plants which produce two kinds of spores, macro or megaspores and microspores are called heterosporous Ex: Selaginella, Salvinia.

Question 4.

What is heterospory? Briefly comment on its significance. Give two examples.?

Answer:

Producing two types of spores is called heterospory 1) Microspores 2) Megaspores.

Significance :
In heterosporous plants, the megaspores and microspores germinate and give rise to female and male gametophyte respectively. Male gametes are transferred to the egg of female archegonium. The female gametophytes retain on the parent sporophyte for variable period. The development of zygotes into young embryos takes place within the female gametophytes. This event is a precursor to the seed habit. It is considered as an important step in evolution.
Example : Selaginella and salvinia.

Question 5.

Write a note on economic importance of Algae and Bryophytes. [March 2019]?

Answer:

Economic importance of Algae :

About 50% of carbon fixation is done by algae by photosynthesis. Thus by photosynthesis 02 is released into environment.
Algae are primary producers for all aquatic animals in food cycle.
Many species of Porphyra, Laminaria and Sargassum are among the 70 species of marine algae used as food.
Brown algae and Red algae produce large amounts of hydrocolloids (water holding substances) & algin (brown algae) and carrageen (red algae) which are used commercially.
Agar obtained from Gelidium and Gracilaria is used to grow microbes and preparations of ice-creams and jellies.
Iodine is extracted from kelps like Laminaria.
Chlorella and Spirullina are unicellular algae used as food supplements even by space travellers.

Economic importance of bryophytes (mosses) :

Mosses provide food for herbaceous mammals, birds and other animals.
Sphagum provides peat used as fuel, because of its water holding capacity, they are used as packing material for trans-shipment of living material.
Mosses along with lichens are the first organisms to colonise rocks and hence they have great ecological importance.
They decompose rock making it suitable for the growth of higher plants, hence, they play important role in plant succession.
Mosses form dense mats on the soil, they reduce the impact of falling rain and prevent soil erosion.

Question 6.

How would you distinguish Monocots from Dicots?

Answer:

Monocots	Dicots
1) Monocot seeds have one cotyledon.	1) Dicot seeds have two cotyledons.
2) Adventitious root system is present.	2) Tap root system is present.
3) Leaves show parallel venation	3) Leaves show reticulate venation.
4) Sheathing leaf base is present.	4) Sheathing leaf base is absent.
5) Leaves are isobilateral.	5) Leaves are dorsiventral.
6) Secondary growth is absent.	6) Secondary growth takes place.

Question 7.

Give a brief account of prothallus. [Mar. – 2020]?

Answer:

Haploid spores give rise to gametophytic prothallus in pteridophytes.
Prothallus is small, thin, green and autotrophic thallus like structure.
Prothallus grows in cool, damp, shady places as it requires water for fertilization.
Prothallus bears sex organs. The male sex organs are called antheridia and the female sex organs are called archegonia.
These sex organs are multicellular, jacketed and sessile.
Antheridia produce male gametes called antherozoids. Archegonia has egg cell which is a female gamete.
Antherozoids require water to reach egg.
Zygote develops into embryo within the female gamete. This event is a precursor to the seed habit. It in considered as an important step in evolution.

Question 8.

Draw labelled diagrams of
a) Female thallus and male thallus of a liverwort
b) Gametophyte and sporophyte of Funaria.

Answer:

Long Answer Type Questions

Question 1.

Name three groups of plants that bear archegonia. Briefly describe the life cycle of any one of them.?

Answer:

Bryophytes, pteridophytes and gymnosperms bear archegonia.

Life cycle of a Moss plant:

The gametophyte is a dominant phase.
Moss plant is a haploid gametophore.
It produces gametangia on separate branches of the same plant and hence monoecious.
The club shaped antheridia or antheridial branches produce biflagellate male gametes called “antherozoids”.
The flask shaped archegonia on archegonial branches produce eggs (female gamete) in their venter.
Antherozoids liberated swim in water to reach egg in the venter.
Union of one male gamete (antherozoid) and egg unite results in diploid zygote. This is called fertilization.
Zyeote is the first cell for sporophytic generation. Zygote develops into embryo. Embryo is retained within the archegonium.
It develops into semi parasitic sporophyte consisting of foot, seta and capsule.
The spore mother cells in the spore sac of the capsule undergo meiosis and forms a number of haploid spores of one kind. Hence Funaria is homosporous.
Spore is the first cell for gametophytic generation under favourable conditions, the spore liberated germinates to give rise to filamentous protonema.
The buds arising from the aerial branches of protonema develop into independent gametophores.

Question 2.

Describe the important characteristics of Gymnosperms.?

Answer:

The important characters of Gymnosperms are :

All gymnosperms are perennial, growing as woody trees or bushy shrubs.
Vascular tissues are arranged into vascular bundles.
Flowers are absent, however, microsporophylls and megasporophyll usually aggregate to form distinct cones or strobili called male cones and female cones respectively.
Gymnosperms are heterosporous: Gymnosperms produce two types of spores- Microspores and megaspores. Microspores are produced in Microsporangium. They are called pollen grain. Megaspores are produced in Megasporangia. The megasporangia are integumented and are called ovules.
Ovules are borne on megasporophyll and have three layered integument with an opening called micropyle.
Pollination is indirect.
Fertilization is affected by pollen tube produced by the male gametophyte. It is called siphonogamy.
Endosperm is formed before fertilization. That means female gametophyte is considered as endosperm. It is haploid.
Seeds are naked i.e., without any seed coat.

Question 3.

Give the salient features of pteridophytes?

Answer:

The main plant body of pteridophyte is sporophyte. It is differentiated into true roots, stem and leaves.
Vascular tissues are present. So pteridophytes are commonly called vascular cryptogams.
Roots are adventitious.
Stem is underground rhizome.
The leaves of pteridophytes are small (microphyllous) as in selaginella or large fronds (macrophyllous) as in ferns.
Ferns show circinate vernation and the petioles are covered with brown multicellular hairs called ramenta.
The stele may be protostele or siphonostele or solenostel6.
One of the important characters of pteridophyte is that the sprophyte has become the dominant part of the life cycle while the gametophyte is reduced.
Gametophyte is small and inconspicuous and it is produced from haploid spores known as prothallus.
The asexual generation or the sporophyte may be homosporous (all spores are similar) or heterosporous (two different types of spores) i.e., microspores or megaspores.
Prothallus (gametophytes) are monoecicus or dioecious.
Sex organs are Antheridia and Archegonia.
Male gametes are Antherozoids formed from Antheridia. Antherozoids are uninucleate, spirally coiled, biflagellate or multiflagellate structures.
Union of male gamete and female gamete results in diploid zygote.
Zygote develops into Embryo stage.
True fruit and seeds are not formed at any stage.

Question 4.

Give an account of plant life cycles and alternation of generations?

Answer:

Plant life cycles :
1) Haplontic type of life cycle :

The dominant phase is gametophyte. It is photosynthetic free living.
Sporophytic generation is represented by zygote.
Zygote is a resting stage not a free living sporophyte.
Thus the life cycle having only free living gametophyte without free living sporophyte is called Haplontic type of life cycle. Ex :Algae like chlamydomonas, Volvox, Spirogyra etc.

2) Diplontic type of life cycle :

The dominant phase is sporophyte. It is photosynthetic independent plant.
Gametophytic generation is represented by gametes.
Thus the life cycle having only independent sporophyte without gametophyte is called Diplontic type of life cycle.
The life cycle having only independent sporophyte with few celled or many celled staged gametophyte is called diplo-haplontic type of life cycle.
Ex: Pteridophytes and seed-bearing plants.

3) Haplo-diplontic type of life cycle :

The dominant phase is gametophytic generation. It is independent.
Sporophytic generation is photosynthetic dependent on gametophytic generation.
The life cycle having both dominant gametophytic generation and dependent sporophytic generation is called Haplo-diplontic type.

Alternation of generations:

In the life cycle of an organism two phases are present. They are Gametophytic phase and Sporophytic phase.
Haploid spore is the first cell for gametophytic generation.
Haploid spore divides mitotically and forms haploid gametophyte.
Gametophyte shows sexual reproduction. It forms male and female gametes.
The fusion of male and female gametes results in diploid zygote.
Diploid zygote is the first cell for sporophytic generation.
Diploid zygote divides mitotically and forms diploid sporophyte.
Diploid sporophyte shows asexual reproduction. It undergoes meiosis and forms haploid spores.

Thus during the life cycle of plants gamete producing haploid gametophyte alternates with spore producing diploid sporophyte. This is known as alternation of generation.

Question 5.

Both Gymnosperms and Angiosperms bear seeds, then why are they classified separately?

Answer:

Even though both Gymnosperms and Angiosperms bear seeds, they are classified separately because of nature of seeds.

In Gymnopserms:

The plant bears ovules which are not covered by any ovary wall. They remain exposed.
Pollination is direct.
Seeds formed are not covered by seed-coat. They are naked seeded plants.
Female gametophyte is considered as Endosperm as it is nutritive in function.
Endosperm is formed before fertilisation.
It is haploid.
Poly embryonic condition is present.

In Angiosperms:

The plant bears ovules which are present inside the ovary. They are not exposed.
Pollination is indirect.
Seeds formed are covered by seed – coat. They are closed seeded plants.
Endosperm is formed after double fertilisation and triple fusion.
Endosperm is triploid.
Single embryo is present.

Intext Question Answers

Question 1.

How far does SelagineJIa, one of the few living members of Lycopodiales (Pteridophytes) fall short of seed habit.?

Answer:

Selaginella is heterosporous. It produces two kinds of spores. They are macrospores and microspores. Macrospores germinate and give rise to female gametophyte whereas Microspores give rise to male gametophyte. Fusion of male gamete with the egg present in the archegonium results in the formation of Zygote. The development of zygotes into young embryos takes place with the female gametophyte. This is the precusor to the seed habit.

Question 2.

Each plant or group of plants has some phylogenetic significance in the relation of evolution. Cycas, one of few living members of Gymnosperms is called as the “relic of past”. Can you establish a phylogenetic relationship of Cycas with any other group of plants that justifies the above statement?

Answer:

Cycas, one of Gymnosperms shows close resemblance with fern (pteridophyte) on one hand and angiosperms on the other hand. Thus occupying a position intermediate between the two.

Certain primitive characters in cycas are similar to ferns. They are

Stem when young is underground and subterranean.
Leaf bases are persistent on the stem.
Young leaves show circinate vernation.
The sporophylls are leaf-like.
Ramenta are present.
Xylem consists of tracheids only and there are no vessels.
Phloem lacks companion cells.
Microsporangia occur in sori on the abaxial side of microsporophyll.
Archegonia are still retained in the female gametophyte.
Sperms are multiciliate.

Question 3.

The male and female reproductive organs of several pteridophytes and Gymnosperms are comparable to floral structures of angiosperms. Make an attempt to compare the various reproductive parts of Pteridophytes and Gymnosperms with reproductive structures of Angiosperms.?

Answer:

Reproductive structures of Pteridophytes are strobili or cone.	Reproductive structures of Gymnosperms are strobili or cone.	Reproductive structures of Angiosperms are flowers.
Cone is not differentiated as male cone and female cone.	Male cone and female cone are present.	Flowers may be unisexual as male flowers and female flowers. (or) Flowers may be bisexual.
Mostly homosporous Sporophylls bear sporangia which produce spores. Some are heterosporous Microsporophyll bearing Microsporangia produce Microspores. Macrosporophyll-bearing Macrosporangia produce Macrospores.	The male cone bearing Microsporophyll and Microsporangia are called microsporangiate or male strobili. (This is similar to male flower) They produce Microspores.The female cone bearing Megasporophyll and Megasporophyll with ovule or integumented mega-sporangia are called megasporangiate or female strobili (This is similar to female flower)	The male sex organs are called stamen or Microsprophyll.Anther represents microsporangium.Pollengrains represent microspores.Female sex organ carpels represent Megasporophyll. Carpel is with ovule or integumented Megasporangia.

Question 4.

The plant body in higher plants is well differentiated and well developed. Roots are organs used for the purpose of absorption. What are the equivalent of roots in the less developed lower plants?

Answer:

Rhizoids. They are unicellular or multicellular hair-like structures that penetrate the moist soil and absorb the water for the plants.

TS Inter Practical Hall Tickets 2026: Download TG Intermediate Practical Admit Card at manabadi.co.in

Ramu

January 2, 2026

TS Inter Practical Hall Tickets 2026: Download TG Intermediate Practical Admit Card at manabadi.co.in

The Telangana State Board of Intermediate Education (TSBIE), also known as TGBIE, is set to conduct the Intermediate Practical Examinations 2026 for both 1st Year and 2nd Year students. As the exams approach, students are eagerly waiting for the release of the TS Inter Practical Hall Tickets 2026. According to the latest updates, the board will make the hall tickets available online, and students can conveniently download them from the popular education portal manabadi.co.in. or Parents will receive a WhatsApp message containing a download link.

In this article, we provide complete details about TS Inter Practical Admit Card 2026, including the release date, download process, exam schedule, important instructions, and frequently asked questions.

TG Inter Practical Hall tickets 2026 Out

TS Inter Practical Hall Tickets 2026 – Overview

The hall ticket is a mandatory document for appearing in the Intermediate Practical Examinations. Without it, students will not be allowed to enter the examination hall. The Telangana Board has taken several digital initiatives in recent years to simplify exam-related processes, and downloading hall tickets online is one such step.

The Telangana State Board of Intermediate Education (TGBIE), Hyderabad has announced an important digital initiative for the smooth conduct of the Intermediate Public Examinations – 2026. In a first-of-its-kind move, the Board will send TS Inter Practical Hall Tickets 2026 download links directly to parents’ registered WhatsApp numbers, ensuring transparency, accuracy, and early verification of student details.

Students appearing for the Intermediate First Year and Second Year Practical Examinations 2026 can also access their hall tickets online through trusted education portals such as manabadi.co.in. This new system aims to minimize errors, improve communication, and strengthen coordination between parents, students, colleges, and district authorities.

TS Inter Practical Hall Tickets 2026 – Latest Official Update

As per the official announcement by TGBIE, preview hall tickets for Intermediate students will be made available before the commencement of practical examinations. Parents will receive a WhatsApp message containing a download link, enabling them to preview and verify the hall ticket details of their children.

This facility applies to:

Intermediate 1st Year Students (General & Vocational)
Intermediate 2nd Year Students (General & Vocational)

The preview system allows sufficient time for correction of errors, thereby avoiding last-minute issues during examinations.

TS Inter Practical Examination Schedule 2026

The main practical examinations for both General and Vocational streams will take place in February 2026. Additionally, specific dates have been allocated for English practicals and internal assessments, such as Ethics and Human Values. Refer to the detailed schedule below.

Exam Type	Year / Class	Date(s)	Time / Session
English Practical	1st Year	January 21, 2026 (Wednesday)	Check the TS Inter 1st Year Hall Ticket
English Practical	2nd Year	January 22, 2026 (Thursday)	Check the TS Inter 2nd Year Hall Ticket
Ethics and Human Values	Both Years	January 23, 2026 (Friday)	10:00 AM to 1:00 PM
Environmental Education	Both Years	January 24, 2026 (Saturday)	10:00 AM to 1:00 PM
General Practicals (Lab)	Both Years	February 2, 2026 to February 21, 2026	9:00 AM to 12:00 Noon / 2:00 PM to 5:00 PM

Steps to Download TS Inter Practical Hall Tickets 2026 at Manabadi

Visit the official website manabadi.co.in
Click on the link “TS Inter Practical Hall Tickets 2026”
Select 1st Year or 2nd Year
Enter the required login details
Click on Download Hall Ticket
Verify all details carefully
Save and take a printout for examination use

Details Printed on TS Inter Practical Hall Ticket 2026

Parents and students must thoroughly verify the following information on the hall ticket preview:

Student Name
Hall Ticket Number
Photograph & Signature
College Name and Code
Subjects & Practical Timetable
Medium of Instruction
Group / Stream
Examination Center Details

Frequently Asked Questions (FAQs)

When will TS Inter Practical Hall Tickets 2026 be released?
They will be released before the practical exams, with preview links sent to parents via WhatsApp.
Is the hall ticket mandatory for practical exams?
Yes, students must carry the hall ticket to the examination center.
Can hall tickets be downloaded without WhatsApp?
Yes, students can download hall tickets directly from manabadi.co.in.
Who should be contacted for corrections?
Students should contact their College Principal or DIEO/Nodal Officer.

TS Inter 1st Year Botany Study Material Chapter 3 Science of Plants – Botany

Sandhya

January 2, 2026

TS Inter 1st Year Botany Study Material Chapter 3 Science of Plants – Botany

Very Short Answer Type Questions

Question 1.

Explain how the term Botany has emerged. [Mar. – 2009]?

Answer:

The term Botany had its origin in the Greek language ‘Bous’ refers to cattle and ‘Bouskein’to cattle feed.’
In course of time, Bouskein was transformed into Botane and later into Botany.

Question 2.

Name the books written by Parasara and mention the important aspects discussed in those books. [Mar. ’20. ’17]?

Answer:

Krishi Parasaram and Vrikshayurveda were the books written by Parasara. (1300 B.C.)
Krishi Parasaram the oldest book dealt with agriculture and weeds;, while Vrikshayurveda is about different types of forests, external and internal characters of plants including medicinal plants.

Question 3.

Who is popularly known as “Father of Botany”? What was the book written by him?

Answer:

Theophrastus is popularly known as Father of Botany.
Historia plantarum was the book written by him.

Question 4.

Who are Herbalists? What are the books written by them?

Answer:

Herbalists are botanists of Renaissance period of 16th and 17th centuries who identified and described medicinal plants living in natural surroundings.
The books written by them are called Herbais.

Question 5.

What was the contribution of Carolus Von Linnaeus’for the development of plant taxonomy?

Answer:

Carolus Von Linnaeus, the Sweedish Botanist popularised the Binomial Nomenclature System.
He also proposed the sexual system of classification.

Question 6.

Why is Mendel considered as the Father of Genetics?

Answer:

Mendel (1866) proposed the laws of inheritance based on his hybridization experiments on pea plant.
He marked the beginning of Genetics. Hence he is popular as the Father of Genetics.

Question 7.

Who discovered the cell and what was the book written by him? [Mar. ’14]

Answer:

Robert Hooke (1665) discovered the cell.
Micrographia was the book written by him.

Question 8.

What is Palaeobotany? What is its use? [May ’17, Mar. ’15 – T.S. : Mar. ’13]

Answer:

Palaeobotany is the study of fossil plants.
It helps in understanding the course of evolution in plants.

Question 9.

Name the branches of Botany which deal with the chlorophyllous autotrophic thallophytes and non-chlorophyllous heterotrophic thallophytes?

Answer:

The study of chlorophyllous autotrophic thallophytes (Algae) is Phycology.
The study of non-chlorophyllous heterotrophic thallophytes (Fungi) is Mycology.

Question 10.

What are the groups of plants that live as symbionts in lichens ? Name the study of lichens.?

Answer:

Algal members (Phycobionts) and fungal members (Mycobionts) live as symbionts in lichens.
The study of lichens is called Lichenology.

Question 11.

Which group of plants is called vascular cryptogams? Name the branch of Botany which deals with them. [Mar. – 2018]?

Answer:

Pteridophytes are called vascular cryptogams.
The branch which deals with pteridophytes is called Pteridology.

Question 12.

Which group of plants is called amphibians of plant kingdom? Name the branch of Botany which deals with them.?

Answer:

Bryophytes are called amphibians of plant kingdom.
The branch which deals with bryophytes is called Bryology.

Short Answer Type Questions

Question 1.

Explain in brief the scope of Botany in relation to agriculture, horticulture and medicine?

Answer:

Agriculture, horticulture and medicine have recorded great progress through experiments in hybridization and genetic engineering.
New techniques of plant breeding are useful to develop hybrid varieties in crop plants like rice, wheat, maize, sugarcane etc.
The role of minerals in plant nutrition and the importance of hormones in plant growth helped in the development of agriculture.
Antibiotics like penicillin are obtained from fungi.
There are many plants like Arnica, Cinchona, Neem, Datura, Digitalis, Rauwolfia, Withania, Ocimum, Belladona, Aloe etc., which have medicinal values.
Using genetic engineering technique, cloned DNA s are produced which prepare hormoneslike insulin, interferon and vaccines.

Question 2.

Explain the scope of Botany taking plant physiology as example?

Answer:

The efforts made in plant physiology have helped the development of agriculture.
It provided the knowledge about role of minerals in plant nutrition and importance of hormones in plant growth.
Auxins at low concentration can form roots, so it is applied in agriculture and horticulture.
Gibberelins induce seed germination.
Cytokinins are used to enhance the shelf life period of leafy vegetables like spinach, lettuce etc.
Abscisic acid is used for delaying the sprouting of potato tubers under storage.
Ethylene accelerates the ripening of fruits like apple, banana, watermelons etc.

Question 3.

What are the different branches of Botany that deal with morphology of plants? Give their salient features?

Answer:

Morphology deals with the study and description of different organs of a plant. It is a fundamental requisite for classification of plants. It can be divided into two parts.
a) External Morphology :
It is the study and description of external characters of plant organs like root, stem, leaf, flower, fruit and seeds etc.

b) Internal Morphology :
It is the study of internal structure of different plant organs. It has two branches.

1) Histology :
It is the study of different tissues present in the plant body.

ii) Anatomy :
It deals with the study of gross internal details of plant organs like root, stem, leaf, flower etc.

Long Answer Type Questions

Question 1.

Give a comprehensive account on the scope of Botany in different fields giving an example for each.?

Answer:

Man has been using plants for various purposes like food, clothes and shelter.
The global population is increasing rapidly. So to meet the demands of food and other challenges man depends on plants.
The problem of increasing population can be solved by increasing the crop production through “Green Revolution”.
Biotechnology is based upon the principles of molecular genetics, microbiology and biochemistry.
Biotechnology is applied for the production of medicine, chemicals, food, biofertilizers, biopesticides, disease resistant and pest resistant crops.
Agriculture, forestry, horticulture, floriculture have recorded great progress through experiments in hybridisation and genetic engineering.
New techniques of plant breeding are useful to develop hybrid varieties in crop plants like rice, wheat, maize, sugarcane etc.
Plant Taxonomy helps to study the diversity of plant kingdom by dividing plants into groups.
Plant pathology helped in prevention and eradication of several plant diseases.
Plant physiology helped the development of agriculture by providing knowledge about the role of minerals in plant nutrition and importance of hormones in plant growth.
Algae like spirulina and chlorella are good source of single celled proteins and vitamins.
Fungi like penicillin are good source for antibiotic.
Plants having medicinal value are Arnica, Cinchona, Neem, Datura, Rauwolfia, Withania, Ocimum, Belladona etc.
Plant fossils produce fuels like coal, coke, gasoline, petrol etc.
Recently bio-diesel is produced from jatropa and other petro plants belonging to the family Euphorbiaceae. Experiments in tissue and organ culture have made it possible to produce large number of plants within a short duration of time.
Industries like cloth mills, paper mills, sugar mills could be developed due to Botany.
Commercially important products like timber, fibres, beverages like coffee and tea, condiments, rubber, gums, resins, dyes, and essential and aromatic oils are obtained from plants.
Green plants reduce pollution, control greenhouse effect.
The hazardous effect of frequent use of chemical fertilizers have been reduced by using biofertilizers like Azolla, Nostoc, Anabaena, Rhizobium etc.
The Algae like Chlorella is used as food for astronauts in space research programmes.
Sand-binding plants help to check soil erosion and also control floods.
Several seaweeds are used in the extraction of iodine, agar-agar, etc.

TS Inter 1st Year Botany Study Material Chapter 2 Biological Classification

Sandhya

January 2, 2026

TS Inter 1st Year Botany Study Material Chapter 2 Biological Classification

Very Short Answer Type Questions

Question 1.

What is the nature of cell walls in diatoms?

Answer:

In Diatoms, the cell walls form two thin overlapping shells, epitheca over hypotheca which fit together as in a soap box.
The walls are embedded with silica and are indestructible. The cell walls left behind by diatoms in their habitat and accumulate over billions of years as diatomaceous earth or kieselguhr.

Question 2.

How are Viroids different from Viruses?

Answer:

Viruses	Viroids
1) It is a nucleoprotein particle.	1) It is a free RNA particle.
2) Nucleic acid can be DNA or RNA.	2) Viroid is formed only by RNA.
3) Viruses infect all types of living organisms.	3) Viroids infect only plants.

Question 3.

What do the terms phycobiont and mycobiont signify? [Mar. ’17, A.P. : Mar. ’13]?

Answer:

1) Phycobiont :
The group of Algae that live as symbionts in lichens.

2) Mycobiont :
The group of fungi that live as symbionts in lichens.

Question 4.

What do the terms ‘algal bloom’ and ‘red tides’ signify?

Answer:

1. Algal bloom :
Excessive growth of algae mostly cyanophyceae members due to the enrichment of excessive nutrients in a water body. t

2. Red tides :
Sea appears red due to the rapid multiplication of a dinoflagellate, Gonyaulax Red tides in Meditarrenian sea.

Question 5.

State two economically important uses of heterotrophic bacteria?

Answer:

They help in making curd from milk.
They are helpful in nitrogen fixation in roots of leguminous plants.

Question 6.

What is the principle underlying the use of cyanobacteria in agricultural fields for crop improvement? [Mar. 2019, ’15 A.P]?

Answer:

Cyanobacteria Eg : Nostoc, Anabaena, can fix atmospheric nitrogen in specialised cells called heterocysts.
They improve soil fertility by adding organic matter. ,

Question 7.

Plants are autotrophic. Name some plants which are partially heterotrophic?

Answer:

Insectivorous plants are partially heterotrophic.
Eg : Bladderwort and Venus fly trap.
Parasitic plant, cuscuta is also partially heterotrophic.

Question 8.

Who proposed five kingdom classification ? How many kingdoms of this classification contain eukaryotes?

Answer:

R.H. Whittaker (1969) proposed Five Kingdom Classification.
Four kingdoms namely Protista, Fungi, Plantae and Animalia, consists of eukaryotes, while kingdom Monera consists of Prokaryotes.

Question 9.

Give the main criteria used for classification by Whittaker. [Mar. – 2020, 2018 • Mar. 15 – T.S.]?

Answer:

The main criteria for five kingdom classification of Whittaker are cell structure, thallus organisation, mode of nutrition, reproduction and phylogenetic relationships.

Question 10.

Name two diseases caused by Mycoplasmas. [May ’14]?

Answer:

Witches broom in plants.
Pleuropheumonia in cattle.
Mycoplasmal urethritis in humans.

Question 11.

What are slime moulds? Explain what is meant by plasmodium with reference to slime moulds.?

Answer:

1. Slime moulds are saprophytic protists.

2. Plasmodium :
An aggregation formed by a slime mould under suitable conditions, which may grow and spread over several feet.

Short Answer Type Questions

Question 1.

What are the characteristic features of Euglenoids?

Answer:

Euglenoids:

These are unicellular, flagellate, fresh water organisms found in stagnant water.
Cell wall is absent.
The body is covered by thin flexible pellicle.
They bear two flagella, usually one long and one short. They swim actively by flagella.
The anterior part of the cell bears an invagination consisting of cytostome (cell mouth), cytopharynx (gullet) and reservoir.
A photosynthetic stigma or eye spot is present in the reservoir.
Chloroplast is present. The pigments in it are identical to those present in higher plants. Performs photosynthesis.
In the absence of sunlight they behave like heterotrophs depending on smaller organisms for food.,
Reproduction is by longitudinal binary fission Palmella stage is found in Euglena.

Question 2.

What are the advantages and disadvantages of two kingdom classification?

Answer:

a) Two kingdom classification with Plantae and Animalia was developed during cinnaeues tissue, that included all plants and animals respectively.

Advantages	Disadvantages
1) Organisms were easily classified into plants and animals and was easy to understand.	1) But a large number of organisms did not fall into either of the two categories. This system did not distinguish between the eukaryotes and prokeryotes; unicellular and multicellular organisms and photo-synthetic and non photosynthetic organisms.
2) All cell wall containing organisms were included in plantae kingdom. So Bacteria, Algae, Fungi, Bryophytes, Pteridophytes, gymno- sperms and angiosperms were placed under plants.	2) This placed together groups which widely differed in other character- sties prokeryotic bacteria and Blue green algae were brought together and placed with other groups which are eukaryotic. It also grouped together the unicellular (eg.: Chlamydomonas) and multicellular, (eg : spirogyra) ones. This systems did not differentiate between the heterographic group, fungi and the autotrophic green plants. Fungi consists of chitin in their cell wall, while green plants have cellulosic cell walls.

Question 3.

Give the salient features and importance of Chrysophytes. [Mar. – 2018, Mar. ’15 – A.P. : Mar. ’13]?

Answer:

This group includes diatoms and desmids (golden algae).
They are green, microscopic, float in water currents.
In diatoms the cell walls form two thin overlapping shells, epitheca and . hypotheca which fit together as in a soap box.
The cell walls are embedded with silica which are indestructible. They pile at the bottom of water reservoir to form diatomaceous earth.
Diatoms are divided into two types based oh symmetry.
i) CentraIe diatoms are radially symmetrical.
ii) Pennales are bilaterally symmetrical.
Asexual reproduction is by binary fission and sexual reproduction is by the formation of gametes.

Question 4.

Give a brief account of Dinoflagellates. [Mar. 2019, ’17 – A.P, Mar. ’15 – T.S]?

Answer:

Dinoflagellates are marine. They appear as yellow, green, brown, blue or red depending upon the pigments present in the cells.
Cell wall is made up of cellulose plates.
Two flagella are present. One lies longitudinally and the other lies transversely in the furrow between the wall plates.
Flagella produce spinning movements. So these are called whirling whips.
Nucleus is called Mesokaryon as chromosomes are condensed without histones.
Example : Nostoc shows bioluminescence.
Gonyaulax make the sea appear red.

Question 5.

Write the role of fungi in our daily life. [Mar. ’14]?

Answer:

Mushroom and toadstools are edible fungus.
Unicellular fungi like yeast are used to make beer and bread.
Fungi like Rhizopus commonly grow on stale bread, pickles, jams, cheese, on moist food stuff and spoils them. They are called moulds.
Fungi causes diseases in plants and animals.
Eg : Wheat rust is caused by puccinia.
Late blight of potato by phytopthora.
Orange rot, Red rot in sugarcane are caused by fungus.
White spots on mustard leaves are due to parasitic fungus. (Albugo)
Some fungi are the source of antibiotics and peninllium.

Long Answer type Questions

Question 1.

Give the salient features and comparative account of different classes of Fungi studied by you.?

Answer:

Question 2.

Describe briefly different groups of Monerans you have studied?

Answer:

Kingdom Monera includes all prokarytes like Archebacteria, Eubacteria, Mycoplasma and Actinomycetes.

Archebacteria :

These are different type of bacteria as they have a different cell wall structure. They can survive in extreme conditions like salty areas (halophiles), hot springs (thermacidophiles) and marshy areas (methanogens).
The cell wall does not contain peptidoglycan as in bacteria but contain pseudomurein.
The cell membrane contains branched lipid which is responsible for their survival in extreme conditions.
Methanogens live in the guts of several ruminant animals like cowand buffaloes and help in their digestion.
They help in the production of,biogas such as methane from the dungs of the animals.

Eubacteria:

They occur everywhere even in extreme habitats.
They live as parasites, and symbionts also.
Basing on the shape bacteria are grouped under four categories. They are 1) Spherical coccus 2) Rod shaped Bacillus 3) Comma, shaped Vibrium 4) Spiral shape Spirillum
In Bacteria, cell wall consists of peptidoglycan also called murein or ‘ mucopeptide.
Infolding of cell membrane called mesosomes responsible for respiration.
Cell organelles are absent except ribosome.
As it is prokaryotic, the genetic material DNA is naked without nuclear membrane.
It shows autotrophic and heterotrophic nutrition.
Excessive growth of cyanobacteria due to nutrients present in sewage causes Algal blooms.
Rapid growth of red dinoflagellate like Gonyaulax make sea appear red or red tides in Mediterianian sea.
Chemo autotrophic bacteria oxidise various inorganic substances.
Chemo heterotrophs are saprophytes which grow on dead organic matter and parasite which causes diseases.
Asexual reproduction is mainly by binary fission or by spores during unfavourable condition. Sexual reproduction is done by transfer of genetic material from one bacteria to other.

Mycoplasma :

Mycoplasma are the smallest living cells and can survive without oxygen.
They do not have any cell wall.
Mostly they are pathogenic in plants and animals. They cause witches broom in plants, pleuropneumonia in cattle and mycoplasmal urethritis in humans.

Actinomycetes :

These are branched filamentous bacteria.
Cell wall contains mycolic acid.
Most of them are saprophytic and decomposers. Mycobacterium and Corynebacteriurn are parasites.
Antibiotics are produced from the genus Streptomyces.

Question 3.

Enumerate the salient features of different groups of protista.?

Answer:

Kingdom Protista includes unicellular, aquatic, eukaryotes. It includes Chrysophytes, Dinoflagellates, Euglenoids, Slime moulds and Protozoans.
1. Chrysophytes:

It includes diatoms and desmids (golden algae).
They are green, microscopic, float in water currents.
In Diatoms the cell walls form two thin overlapping shells, epitheca and hypotheca which fit together as soap box.
The cell walls are embedded with silica which are indestructible. They pile at the bottom of water reservoir to form diatomaceous earth.
Diatoms are divided into two types based on symmetry.
i) Centrale diatoms are radially symmetrical.
ii) Pennales are bilaterally symmetrical.
Asexual reproduction is by binary fission and sexual reproduction is by the formation of gametes.

2. Dinoflagellates:

Dinoflagellates are marine. They appear as yellow, green, brown, blue or red depending upon the pigments present in the cells.
Cell wall is made up of cellulose plates.
Two flagella are present. One lies longitudinally and the other lies transversely in the furrow between the wall plates.
Flagella produce spinning movements. So these are called whirling whips.
Nucleus is called Mesokaryon as chromosomes are condensed without histones.
Example : Nostoc shows bioluminescence.
Gonyaulax make the sea appear red.

3. Euglenoids :

These are unicellular, flagellate, fresh water organisms found in stagnant water.
Cell wall is absent.
The body is covered by thin flexible pellicle.
They bear two flagella, usually one long and one short. They swim actively by flagella.
The anterior part of the cell bears an invagination consisting of Cytostome (cell mouth), Cytopharynx (gullet) and reservoir.
A photosynthetic stigma on eye spot is present inthe reservoir.
Chloroplast is present. The pigments in it are identical to those present in higher plants. Performs photosynthesis.
In the absence of sunlight they behave like heterotrophs depending on smaller organisms for food.
Reproduction is by longitudinal binary fission Palmella stage is found in Euglena.

4. Slime moulds:

They show saprophytic nutrition.
Slime moulds are multinucleated protoplasm surrounded by plasma membrane.
They are aquatic; move along with decaying twigs.
Under favourable Conditions they aggregate to form plasmodium. They may spread upto several feet.
Under unfavourable conditions, plasmodium differentiates and forms fruiting bodies which bear spores at their tips. These spores are wind dispersed and can survive for many years.

5. Protozoans:
All protozoans are heterotrophs and live as parasites. The four major groups of protozoans are given below.
i) Amoeboid protozoans:

These organisms live in fresh water, sea water or moist soil.
They have locomotory organ called pseudopodia or false feet.
Marine forms have silica shells on their surface. Example: Amoeba.

ii) Flagellated protozoans:

These members are either free-living or parasitic.
They have flagella.
They cause diseases like sleeping sickness. Example : Trypanosoma.

iii) Ciliated protozoans:

These are aquatic and actively moving organisms because of cilia.
They have a cavity that opens to outside of the cell surface.
Example : Paramecium.

iv) Sporozoans :
It includes diverse organisms that have an infectious sporelike stage in their life cycle. Example : Plasmodium.

InText Question Answers

Question 1.

State two economically important uses of?
a) Heterotrophic bacteria.
b) Archaebacteria.

Answer:

a) Use of heterotrophic bacteria :

They help in making curd from milk.
They convert dead plants and animals into simpler substances and make them available to plants.

b) Use of archaebacteria :
1) They live in the guts of several ruminant animals such as cow and buffaloes and help in their digestion.

Question 2.

Give a comparative account of the classes of Kingdom Fungi on the basis of the following i) Mode of nutrition ii) Mode of reproduction?

Answer:

Question 3.

Give a brief account of viruses with respect to their structure and nature of genetic material. Also name four common viral diseases?

Answer:

Viruses are acellular, ultramicroscopic, nucleoprotein particles.
Viruses are obligate parasites. They are inert outside the host cell.
Viruses contain nucleic acid and protein.
The protein part forms a coat called capsid. It is made up of small sub units called capsomeres.
The nucleic acid which is genetic material may be DNA or RNA.
No virus contains both DNA and RNA.
Tobacco Mosaic Virus (TMV) and Human Immuno Virus (HIV) are examples for virus having RNA.
Bacteriophages contain DNA as genetic material.
Four common Viral Diseases : 1) AIDS, 2) Influenza 3) Small pox 4) Mumps.

Question 4.

Organise a discussion in your class on the topic. Are viruses living or non-living?

Answer:

The main points to be discussed in class :
a) Viruses can be regarded as living organisms because

They are formed by macromolecules which occur in living beings.
Presence of genetic material
Ability to multiply or reproduce
Occurrence of mutations
Infectivity and host specificity
Occurrence of antigenic property
Viruses are killed by autoclaving and ultraviolet rays.
Viruses are responsible for infectious diseases like common cold, influenza, chicken pox, mumps etc.

b) Viruses can be regarded as non-living organisms because

Protoplasm absents
Ability to get crystallized & TMV
Inability to live independently
High specific gravity which is found only in non-living objects
Absence of respiration
Absence of storing energy system
Absence of growth and division.

Question 5.

Suppose you accidentally find an old preserved permanent slide without a label and in your effort to identify it, you place the slide under microscope and observe the following features?
a) unicellular body
b) well defined nucleus
c) biflagellate condition – one flagellum lying longitudinally and the other transversely.
What would you identify it as ? Can you name the kingdom it belongs to?

Answer:

It is identified as Dinoflagellates.
It belongs to kingdom : Protista.

Question 6.

Polluted water bodies have usually high abundance of plants like IMostoc and Oscillatoria. Give reasons.?

Answer:

Polluted water bodies have excessive growth of plants like Nostoc and Oscillatonia because of the excessive nutrients present in it. It results in algal booms.

Question 7.

Cyanobacteria and heterotrophic bacteria have been clubbed together in Eubacteria of kingdom Monera as per the five kingdom classification, even though the two are vastly different from each other. Is this grouping of the two types of taxa in the same kingdom justified? If so why?

Answer:

Yes. because both are unicellular prokaryotic organisms.

Question 8.

What observable features in Trypanosoma would make you classify under the kingdom Protista?

Answer:

They are aquatic, single-celled eukaryotes.

Question 9.

At a stage of their life cycle, ascomycetous fungi produce fruiting bodies like cleistothecium, perithecium or apothecium. How are these three types of fruiting bodies differ from each other?

Answer:

The fruiting bodies are produced in Fungi.
Ascomycetes are called Ascocarp.
The globose ascocarp without opening is called cleistothecium.
The flask-shaped ascocarp with an apical opening is called perithecium.
The cup or saucer-shaped ascocarp is called apothecium.

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TS Inter 1st Year Botany Study Material Chapter 8 Taxonomy of Angiosperms

Very Short Answer Type Questions

Question 1.

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Question 2.

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Question 3.

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Question 4.

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Question 5.

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Question 6.

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Question 7.

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Question 8.

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Short Answer Type Questions

Question 1.

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Question 2.

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Question 3.

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Question 4.

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Question 5.

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Question 7

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Long Answer Type Questions

Question 1.

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Question 2.

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Question 3.

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Question 4.

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Question 5.

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Question 6.

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Question 7.

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TS Inter 1st Year Maths 1A Solutions Chapter 3 Matrices Ex 3(a)

TS Inter 1st Year Maths 1A Matrices Solutions Exercise 3(a)

Question 4. If A = ⎡⎣⎢3212−23−101⎤⎦⎥\left[\begin{array}{ccc} 3 & 2 & -1 \\ 2 & -2 & 0 \\ 1 & 3 & 1 \end{array}\right], B = ⎡⎣⎢−324−11−1032⎤⎦⎥\left[\begin{array}{ccc} -3 & -1 & 0 \\ 2 & 1 & 3 \\ 4 & -1 & 2 \end{array}\right] and X = A + B then find X.

Question 5. If [x−3z+22y−86]\left[\begin{array}{cc} x-3 & 2 y-8 \\ z+2 & 6 \end{array}\right] = [5−22a−4]\left[\begin{array}{cc} 5 & 2 \\ -2 & a-4 \end{array}\right] then find the values of x, y, z and a. [May 2006, Mar. 14]

Question 2. Find the trace of ⎡⎣⎢1213−10−551⎤⎦⎥\left[\begin{array}{ccc} 1 & 3 & -5 \\ 2 & -1 & 5 \\ 1 & 0 & 1 \end{array}\right]

Question 3. If A = ⎡⎣⎢02413524−6⎤⎦⎥\left[\begin{array}{ccc} 0 & 1 & 2 \\ 2 & 3 & 4 \\ 4 & 5 & -6 \end{array}\right] and B = ⎡⎣⎢−10021030−1⎤⎦⎥\left[\begin{array}{ccc} -1 & 2 & 3 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array}\right] find A – B and 4A – 5B.

Question 4. If A = [132231]\left[\begin{array}{lll} 1 & 2 & 3 \\ 3 & 2 & 1 \end{array}\right] and B = [312213]\left[\begin{array}{lll} 3 & 2 & 1 \\ 1 & 2 & 3 \end{array}\right] find 3B – 2A.

TS Inter 1st Year Maths 1A Solutions Chapter 3 Matrices Ex 3(b)

TS Inter 1st Year Maths 1A Matrices Solutions Exercise 3(b)

I. Question 1. Find the following products wherever possible. i) [-1 4 2]⎡⎣⎢513⎤⎦⎥\left[\begin{array}{l} 5 \\ 1 \\ 3 \end{array}\right]

Question 2. If A = [1−4−2235]\left[\begin{array}{rrr} 1 & -2 & 3 \\ -4 & 2 & 5 \end{array}\right] and B = ⎡⎣⎢242351⎤⎦⎥\left[\begin{array}{ll} 2 & 3 \\ 4 & 5 \\ 2 & 1 \end{array}\right] do AB and BA exist ? If they exist find them. Do A and B commute with respect to multiplication.

Question 3. Find A2 where A = [4−121]\left[\begin{array}{cc} 4 & 2 \\ -1 & 1 \end{array}\right]

Question 4. If A = [i00i]\left[\begin{array}{ll} \mathrm{i} & 0 \\ 0 & \mathrm{i} \end{array}\right] find A2.

Question 7. If A = [2−14K]\left[\begin{array}{cc} 2 & 4 \\ -1 & K \end{array}\right] and A2 = 0 then find the value of K. (May 2011, Mar. ’14, ’05)

II. Question 1. If A = ⎡⎣⎢300030003⎤⎦⎥\left[\begin{array}{lll} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array}\right] there find A4.

Question 2. If A = ⎡⎣⎢15−212−136−3⎤⎦⎥\left[\begin{array}{ccc} 1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{array}\right] then Find A3.

Question 3. If A = ⎡⎣⎢103−21−11−11⎤⎦⎥\left[\begin{array}{rrr} 1 & -2 & 1 \\ 0 & 1 & -1 \\ 3 & -1 & 1 \end{array}\right] then find A3 – 3A2 – A – 3I, where I is a unit matrix of order 3. (March 2011)

Question 4. If I = [1001]\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right] and E = [0010]\left[\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right] show that (aI + bE) = a3I + 3a2bE, where I is a unit matrix of order 2. (Mar. 2015-A.P)(May ’05)|

III. Question 1. If A = diag[a1 a2 a3] then for any Integer n ≥ 1 show that An = diag[an1,an2,an3]\left[\mathrm{a}_1^{\mathrm{n}}, \mathrm{a}_2^{\mathrm{n}}, \mathrm{a}_3^{\mathrm{n}}\right]

Question 3. If A = [31−4−1]\left[\begin{array}{rr} 3 & -4 \\ 1 & -1 \end{array}\right], then show that An = [1+2nn−4n1−2n]\left[\begin{array}{cc} 1+2 n & -4 n \\ n & 1-2 n \end{array}\right] for any integer n ≥ 1, by using mathematical induction.

Question 4. Given examples of two square matrices A and B of the same order for which AB = 0 but BA ≠ 0.

TS Inter 1st Year Maths 1A Matrices Solutions Exercise 3(d)

I. Question 1. Find the determinants of the following matrices.

Question 2. If A = ⎡⎣⎢12503−604x⎤⎦⎥\left[\begin{array}{rrr} 1 & 0 & 0 \\ 2 & 3 & 4 \\ 5 & -6 & x \end{array}\right] and det A = 45 then find x.

II. Question 1. Show that ∣∣∣∣bccaabb+cc+aa+b111∣∣∣∣\left|\begin{array}{lll} \mathrm{b c} & \mathrm{b}+\mathrm{c} & 1 \\ \mathrm{c a} & \mathrm{c}+\mathrm{a} & 1 \\ \mathrm{a b} & \mathrm{a}+\mathrm{b} & 1 \end{array}\right| = (a – b)(b – c)(c – a).

Question 2. Show that ∣∣∣∣b+ca+bac+ab+cba+bc+ac∣∣∣∣\left|\begin{array}{ccc} b+c & c+a & a+b \\ a+b & b+c & c+a \\ a & b & c \end{array}\right| = a2 + b2 + c2 – 3abc (Mar. 2008; May 2007)

III. Question 1. Show that ∣∣∣∣a+b+2cccab+c+2aabbc+a+2b∣∣∣∣\left|\begin{array}{ccc} a+b+2 c & a & b \\ c & b+c+2 a & b \\ c & a & c+a+2 b \end{array}\right| = 2(a + b + c)3

Question 3. Show that ∣∣∣∣a2+2a2a+132a+1a+23111∣∣∣∣\left|\begin{array}{ccc} a^2+2 a & 2 a+1 & 1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{array}\right| = (a – 1)3. (March 2007)

Question 4. Show that ∣∣∣∣∣aa2a3bb2b3cc2c3∣∣∣∣∣\left|\begin{array}{ccc} \mathrm{a} & \mathrm{b} & \mathrm{c} \\ \mathrm{a}^2 & \mathrm{b}^2 & \mathrm{c}^2 \\ \mathrm{a}^3 & \mathrm{b}^3 & \mathrm{c}^3 \end{array}\right| = abc(a – b)(b – c)(c – a)

Question 5. Show that ∣∣∣∣−2aa+bc+aa+b−2bc+bc+ab+c−2c∣∣∣∣\left|\begin{array}{ccc} -2 a & a+b & c+a \\ a+b & -2 b & b+c \\ c+a & c+b & -2 c \end{array}\right| = 4(a + b)(b + c)(c + a)

Question 7. Show that ∣∣∣∣∣111abca2−bcb2−cac2−ab∣∣∣∣∣\left|\begin{array}{lll} 1 & a & a^2-b c \\ 1 & b & b^2-c a \\ 1 & c & c^2-a b \end{array}\right| = 0

Question 4.
If A = $\left[\begin{array}{ccc} 3 & 2 & -1 \\ 2 & -2 & 0 \\ 1 & 3 & 1 \end{array}\right]$ , B = $\left[\begin{array}{ccc} -3 & -1 & 0 \\ 2 & 1 & 3 \\ 4 & -1 & 2 \end{array}\right]$ and X = A + B then find X.

Question 5.
If $\left[\begin{array}{cc} x-3 & 2 y-8 \\ z+2 & 6 \end{array}\right]$ = $\left[\begin{array}{cc} 5 & 2 \\ -2 & a-4 \end{array}\right]$ then find the values of x, y, z and a. [May 2006, Mar. 14]

Question 2.
Find the trace of $\left[\begin{array}{ccc} 1 & 3 & -5 \\ 2 & -1 & 5 \\ 1 & 0 & 1 \end{array}\right]$

Question 3.
If A = $\left[\begin{array}{ccc} 0 & 1 & 2 \\ 2 & 3 & 4 \\ 4 & 5 & -6 \end{array}\right]$ and B = $\left[\begin{array}{ccc} -1 & 2 & 3 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array}\right]$ find A – B and 4A – 5B.

Question 4.
If A = $\left[\begin{array}{lll} 1 & 2 & 3 \\ 3 & 2 & 1 \end{array}\right]$ and B = $\left[\begin{array}{lll} 3 & 2 & 1 \\ 1 & 2 & 3 \end{array}\right]$ find 3B – 2A.

I.
Question 1.
Find the following products wherever possible.
i) [-1 4 2] $\left[\begin{array}{l} 5 \\ 1 \\ 3 \end{array}\right]$

Question 2.
If A = $\left[\begin{array}{rrr} 1 & -2 & 3 \\ -4 & 2 & 5 \end{array}\right]$ and B = $\left[\begin{array}{ll} 2 & 3 \\ 4 & 5 \\ 2 & 1 \end{array}\right]$ do AB and BA exist ? If they exist find them. Do A and B commute with respect to
multiplication.

Question 3.
Find A² where A = $\left[\begin{array}{cc} 4 & 2 \\ -1 & 1 \end{array}\right]$

Question 4.
If A = $\left[\begin{array}{ll} \mathrm{i} & 0 \\ 0 & \mathrm{i} \end{array}\right]$ find A².

Question 7.
If A = $\left[\begin{array}{cc} 2 & 4 \\ -1 & K \end{array}\right]$ and A² = 0 then find the value of K. (May 2011, Mar. ’14, ’05)

II.
Question 1.
If A = $\left[\begin{array}{lll} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{array}\right]$ there find A⁴.

Question 2.
If A = $\left[\begin{array}{ccc} 1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{array}\right]$ then Find A³.

Question 3.
If A = $\left[\begin{array}{rrr} 1 & -2 & 1 \\ 0 & 1 & -1 \\ 3 & -1 & 1 \end{array}\right]$ then find A³ – 3A² – A – 3I, where I is a unit matrix of order 3. (March 2011)

Question 4.
If I = $\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]$ and E = $\left[\begin{array}{ll} 0 & 1 \\ 0 & 0 \end{array}\right]$ show that (aI + bE) = a³I + 3a²bE, where I is a unit matrix of order 2. (Mar. 2015-A.P)(May ’05)|

III.
Question 1.
If A = _diag[a₁ a₂ a₃] then for any Integer n ≥ 1 show that Aⁿ = _diag $\left[\mathrm{a}_1^{\mathrm{n}}, \mathrm{a}_2^{\mathrm{n}}, \mathrm{a}_3^{\mathrm{n}}\right]$

Question 3.
If A = $\left[\begin{array}{rr} 3 & -4 \\ 1 & -1 \end{array}\right]$ , then show that Aⁿ = $\left[\begin{array}{cc} 1+2 n & -4 n \\ n & 1-2 n \end{array}\right]$ for any integer n ≥ 1, by using mathematical induction.

Question 4.
Given examples of two square matrices A and B of the same order for which AB = 0 but BA ≠ 0.

I.
Question 1.
Find the determinants of the following matrices.

Question 2.
If A = $\left[\begin{array}{rrr} 1 & 0 & 0 \\ 2 & 3 & 4 \\ 5 & -6 & x \end{array}\right]$ and det A = 45 then find x.

II.
Question 1.
Show that $\left|\begin{array}{lll} \mathrm{b c} & \mathrm{b}+\mathrm{c} & 1 \\ \mathrm{c a} & \mathrm{c}+\mathrm{a} & 1 \\ \mathrm{a b} & \mathrm{a}+\mathrm{b} & 1 \end{array}\right|$ = (a – b)(b – c)(c – a).

Question 2.
Show that $\left|\begin{array}{ccc} b+c & c+a & a+b \\ a+b & b+c & c+a \\ a & b & c \end{array}\right|$ = a² + b² + c² – 3abc (Mar. 2008; May 2007)

III.
Question 1.
Show that
$\left|\begin{array}{ccc} a+b+2 c & a & b \\ c & b+c+2 a & b \\ c & a & c+a+2 b \end{array}\right|$ = 2(a + b + c)³

Question 3.
Show that $\left|\begin{array}{ccc} a^2+2 a & 2 a+1 & 1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{array}\right|$ = (a – 1)³. (March 2007)

Question 4.
Show that $\left|\begin{array}{ccc} \mathrm{a} & \mathrm{b} & \mathrm{c} \\ \mathrm{a}^2 & \mathrm{b}^2 & \mathrm{c}^2 \\ \mathrm{a}^3 & \mathrm{b}^3 & \mathrm{c}^3 \end{array}\right|$ = abc(a – b)(b – c)(c – a)

Question 5.
Show that $\left|\begin{array}{ccc} -2 a & a+b & c+a \\ a+b & -2 b & b+c \\ c+a & c+b & -2 c \end{array}\right|$ = 4(a + b)(b + c)(c + a)

Question 7.
Show that $\left|\begin{array}{lll} 1 & a & a^2-b c \\ 1 & b & b^2-c a \\ 1 & c & c^2-a b \end{array}\right|$ = 0

Question 1.
If A = $\left[\begin{array}{rrr} 2 & 0 & 1 \\ -1 & 1 & 5 \end{array}\right]$ and B = $\left[\begin{array}{rrr} -1 & 1 & 0 \\ 0 & 1 & -2 \end{array}\right]$ then find (AB’)’

Question 2.
If A = $\left[\begin{array}{rr} -2 & 1 \\ 5 & 0 \\ -1 & 4 \end{array}\right]$ and B = $\left[\begin{array}{rrr} -2 & 3 & 1 \\ 4 & 0 & 2 \end{array}\right]$ then find 2A + B’ and 3B’ – A.

Question 3.
If A = $\left[\begin{array}{cc} 2 & -4 \\ -5 & 3 \end{array}\right]$ then find A + A’ and A. A’ (May 2007) (Board Model Paper)

Question 4.
If A = $\left[\begin{array}{ccc} -1 & 2 & 3 \\ 2 & 5 & 6 \\ 3 & x & 7 \end{array}\right]$ is a symmetric matrix then find x.

Question 5.
If A = $\left[\begin{array}{ccc} 0 & 2 & 1 \\ -2 & 0 & -2 \\ -1 & x & 0 \end{array}\right]$ is a skew symmetric matrix, find x. (May 2014, 11)

Question 6.
Is $\left[\begin{array}{ccc} 0 & 1 & 4 \\ -1 & 0 & 7 \\ -4 & -7 & 0 \end{array}\right]$ a symmetric or skew symmetric?

Question 2.
If A = $\left[\begin{array}{ccc} 1 & 5 & 3 \\ 2 & 4 & 0 \\ 3 & -1 & -5 \end{array}\right]$ and B = $\left[\begin{array}{ccc} 2 & -1 & 0 \\ 0 & -2 & 5 \\ 1 & 2 & 0 \end{array}\right]$ , then find 3A – 4B’.

Question 4.
For any square matrix A; show that A A’ is symmetric. (March 2015-A.P)