TS Inter 1st Year Botany Question Paper May 2022

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TS Inter 1st Year Botany Question Paper May 2022

Time: 3 Hours
Maximum Marks: 60

Section – A (10 × 2 = 20)

Note:

• Answer ANY TEN questions.
• Each Question carries TWO marks.
• ALL are very short answer type questions.

Question 1.
What does ICBN stand for ?
Answer:
ICBN stands for International Code for Botanical Nomenclature.

Question 2.
How are Viroids different from Viruses ?
Answer:

Viroids	Viruses
Infectious agents which contain only Nucleic acid (RNA) are called viroids.	Infectious agents which contain both Nucleic acid and protein coat are called viruses.
e.g : Potato spindle tuber virus.	e.g : TMV

Question 3.
Name two diseases caused by Mycoplasmas.
Answer:
a) Witches broom disease in Plants.
b) Pleuropneumonia in Cattle.
c) Mycoplasmal Urethritis in Human beings.

Question 4.
Who discovered the cell and what was the book written by him?
Answer:
‘Robert Hooke’ discovered the cell. ‘Micrographia’ was the book written by him.

Question 5.
Which group of plants is called amphibians of plant kingdom ? Name the branch of Botany which deals with them.
Answer:
Bryophytes are called as amphibians of the plant kingdom because these plants though live in soil but they need water for sexual reproduction. The sperm of bryophyte (antherozoids) are flagellate and need water to swim to the eggs.

Question 6.
Differentiate actinomorphic from zygomorphic flower.
Answer:

Actinomorphic flower	Zygomorphic flower
“A flower that can be cut into two equal halves in any vertical plane”. Ex: Datura	“A flower that can be cut into two equal halves in one vertical plane”. Ex: Bean

Question 7.
What is the morphology of cup like structure in Cyathium? In which family it is found ?
Answer:
In Cyathium, cup like structure is formed from Involucre of Bracts. It is seen in “Euphorbiaceae” family.

Question 8.
What is meant by epipetalous condition ? Give an example.
Answer:
When the statements are attached to the petals, which is called epipetalous condition. Ex : Datura.

Question 9.
Write the floral formula of solanum plant.
Answer:
Br or Ebr, Ebrl, O, K(5) C(5)A5 G(2).

Question 10.
What is Natural system of plant classification? Name the scientists who followed it.
Answer:

1. All the important, mostly morphological characters were taken into consideration and plants were classified accordingly in natural system of classification.
2. Bentham and Hooker(1862-1893) proposed natural system of classification in their book “Genera Plantarum”.

Question 11.
What is ‘Omega Taxonomy’ ?
Answer:
Taxonomy based on Information from other branches, i.e., Embryology, cytology, palynology, phytochemistry, serology etc, along with morphological characters is called omega taxonomy.

Question 12.
An anther has 1200 pollen grains. How many pollen mother cells must have been there to produce them ?
Answer:
300 pollen mother cells.

Question 13.
Mention a single membrane bound organelle which is rich in hydrolytic enzymes.
Answer:
Lysosome.

Question 14.
Hydrophytes show reduced xylem. Why ?
Answer:
Lysosome.

Question 15.
Define population and community.
Answer:
A group of individuals belongs to the same species occuring in an area is called population. An assemblage of all the populations belonging to different species in an area is called community.

Section – B (6 × 4 = 24)

Note :

• Answer ANY SIX questions.
• Each Question carries FOUR marks.
• ALL are of short answer type questions.

Question 16.
Differentiate between red algae and brown algae.
Answer:

Red algae	Brown algae
1. They belongs to the class Rhodbphyceae.	1. They belongs to the class Phaeophyceae.
2. Most of them are marine and some are fresh water forms.	2. They live in fresh water, brackish and salt water.
3. The Thallus is multicellular.	3. The Thallus range from simple branched filamentous form to profusely branched forms.
4. The major pigments are chlorophyll a, d and r- phyco erythrin.	4. The major pigments are chlorophyll a, c, carotexnoids and xanthophylls (fucoxanthin).
5. Flagella are absent.	5. Flagella are 2, unequal lateral.
6. Cellwall is made up of cellu-lose, pectin and poly sulphate esters.	6. Cellwall is made up of cellulose and algin.
7. Food materials are stored in the form of floridean starch.	7. Food materials are stored in the form of mannitol and laminarin.
8. Asexual reproduction is by non – motile spores.	8. Asexual reproduction is by Biflagellate zoospores.
9. Sexual reproduction occurs by non – motile gametes. Ex : Polysiphonia, Porphyra	9. Sexual reproduction occurs by motile gametes. Ex : Ectocarpus, Laminaria, Sargassum.

Question 17.
Give the salient features and importance of Chrysophytes.
Answer:

1. Chrysophytes includes diatoms and desmids.
2. They are found in fresh water as well as marine environment.
3. They are microscopic and float passively in water currents.
4. Most of them are photosynthetic.
5. In diatoms, the cell walls form two thin overlapping shells.
6. Epitheca over hypotheca which fit together as in a soapbox.
7. The walls are embedded with silica and thus the walls are indestructible.
8. The cell wall deposits form diatomaceous earth/kieselguhr.

Importance : Diatomaceous earth is used in polishing, Alteration of oils and syrups.

Question 18.
Give a brief account of Dinoflagellates.
Answer:
a) Dinoflagellates are mostly marine and photosynthetic. They appear yellow, green, brown, blue or red depending on the pigment in their cells.
b) The cell wall has stiff cellulose plates on the outer surface.
c) They have two flagella and produce spinning movements. So these protists are also called “Whirling whips”.
d) The nucleus has condensed chromosomes which are without histones, called mesokaryon.
e) Some dinoflagellates like Noctiluca show bioluminescence.
f) Red dinoflagellates like Gonyaulax undergo rapid multiplication and make the sea appear red (Red tides in Mediterranean sea).
g) Toxins produced by them may kill fishes.

Question 19.
Explain binomial nomenclature.
Answer:
Naming the plants with two words is called Binomial nomen – clature. It was first inroduced by Gaspard Bauhin (1623) but the credit goes to Linnaeus (1753).

In Binomial nomenclature, each and every plant has a name with two words, they should be in Latin. The first word in Genus and the second word is Species. The generic name will be in noun form and always starts with capital letter. The specific name will be in adjective form and starts with small letter. The Botanical name should be printed in Italics and should be underlined. For ex : Solanum tuberosum – Potato plant in which Solanum is the generic name and tuberosum is the species name.

Question 20.
Describe the arrangement of floral members in relation to their insertion on thalamus.
Answer:
i) Epigynous flower : In this fiower, thalamus is cup shaped or flask shaped in which ovary is completely inserted. All other floral parts are present on the upper parts of thalamus. The ovary is inferior and other floral parts are superior. This condition is called epigyny. Example : Cucurbita.

ii) Hypogynous flower : In this flower, thalamus is slightly swollen and forms a cushion like disc, The gynoecium is situated at the top of thalamus and other floral parts arise below the level of ovary. So that ovary is superior and other parts are inferior. This condition is called hypogyny. Example : Citrus, brassica.

iii) Perigynous flower : In this flower, thalamus is disc shaped or cup shaped or flask shaped. Ovary is present in the centre of thalamus while other floral parts occur on the periphery and at the same level. So that ovary is half superior while other parts are inferior, this condition is called perigyny.

Question 21.
Describe any four types of placentations found in flowering plants.
Answer:
The various types of placentations are marginal placentation, axile placentation, parietal placentation, basal placentation, and free central placentation. It is described as how the ovules are arranged inside the ovary. In flowering plants, placentation can take a variety of forms.

Types of Placentation :

Marginal placentation :

• Ovules are born in two rows on the placenta-formed ridge that crosses the ventral suture of the ovary.
• Example: Pea

Axile placentation:

• Ovules in a multilocular ovary are connected to the axial placenta.
• Lemon is a good illustration.

Parietal placentation :

• Ovules develop on the inner or outer wall of the ovary.
• Initially single-chambered, it develops a fake septum to become two-chambered.
• An illustration is a mustard.

Basal placentation :

• The placenta forms at the base of the ovary and are connected to one ovule.
• A good example is a marigold.

Free central placentation:

• Ovules are born on the central axis because there are no septa.
• One illustration is primrose.

Placenta :
The placenta serves as the link between the mother and the foetus. The placenta makes it easier for the foetus to receive oxygen and nutrients. Additionally, it is in charge of removing waste products and carbon dioxide from the foetus. The placenta is a disk-shaped organ that can grow up to 22 cm long. The placenta contains a lot of blood vessels as well.

Question 22.
Describe any two special types of inflorescences.
Answer:
An inflorescence may either be of racemose or cymose types. Racemose or Indefinite or Indeterminate Type of Inflorescence: ADVERTISEMENTS: The arrangement in which the youngest flower is present near the apex and older towards the base, i.e., in acropetal succession.

Question 23.
“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:
Mode of the arrangement of sepals or petals in the floral bud, with respect to the members of the same whorls, is called aestivation. The main types of aestivation are valvate, imbricate, twisted and vexillary.

Valvate : When sepals or petals just touch one another at the margin, without overlapping (eg. Calotropis).

Twisted: One margin of sepals or petals overlap with that of next sepal or petal and so on (eg. China rose, ladies finger, cotton).
Imbricate : Margins of sepals or petals overlap one another but not in any particular direction (eg. gulmohur, Casia).
Vexillary/papilionaceous: Among five petals, the largest petal (standard) overlaps the two lateral petals (wings) which in turn overlap two smallest anterior petals (Keel petals). This is observed in pea and bean flowers.

Question 24.
Give an account of floral diagram.
Answer:

• Floral diagram provides information about the number of parts of a flower and their arrangement and the relation they have with one another.
• The mother axis represent the posterior side of the flower and is indicated as a dot (or) a circle at the top of the floral diagram.
• Calyx, corolla, androecium and gynoecium are drawn in successive whorls.
• The bract represent the anterior side of the flower and is indicated at the bottom of the floral diagram.

Question 25.
Describe the essential organs of Solanaceae.
Answer:
Androecium and Gynoecium are the essential organs.
Androecium : Stamens 5, epipetalous, alternating with petals anthers are dithecous, basifixed, Introrse and dehisce longitudinally ‘Datura) or porous (Solanum).

Gynoecium : Bicarpellary, syncarpous, bilocular, superior ovary with many ovules on swollen axile placentation. Style is terminal and stigma is capitate. Carpels are arranged obliquly at 45°. (Posterior carpel to the right and anterior carpel to the left).

Question 26.
Describe the cell organelle which contains chlorophyll pig-ments.
Answer:
Androecium and Gynoecium are essential organs.

Androecium : 5 stamens, epipetalous, alternate to petals. The anthers are large, dithecous, basifixed, Introse and dehiscence. Longi – tudinally (Datura) or porous (Solanum).

Gynoecium : Bicarpellary, syncarpous, superior ovary with numerous ovules arranged on swollen axile placentation. Ovary is oblique in position due to tilting of posterior carpel to the right and anterior carpel to the left at an angle of 45°. The style is terminal and the stigma is capitate.

Question 27.
Though redundantly described as a resting phase, interphase does not really involve rest. Comment.
Answer:
The interphase, also called phase of non apparent division though called the resting phase, is the time during which the cell is preparing for division by undergoing both cell growth and DNA Replication. The
interphase is divided into three further phases. They are

A) G₁ phase : It is the interval between mitosis and Initiation of DNA replication. In this cell is metabolically active and continuously grows.

B) ‘S’ phase : In this, DNA replication occurs. The amount of DNA per cell doubles. However three is no increase in the number of chromosomes.

C) G₂ phase: Proteins are synthesized in preparation for mitosis while ceil growth continues. So Interphase is not a resting phase.

Question 28.
Enumerate the morphological adaptations of hydrophytes.
Answer:
Morphological adaptations:
A. Roots : Plenty of water is available to hydrophytes. So . roots are of secondary importance, less significant and hence poorly developed.
a) Roots are absent (Wolffia and Ceratophyllum) or poorly developed (Hydrilla) or submerged leaves act as roots as in Salvinia.
b) Root caps are absent or replaced by root pockets (Pistia). Amphibious plants that grow in mud possess distinct root caps.
c) If roots are present they are fibrous, adventitious, reduced in length, unbranched or poorly branched. Balancing roots are present in Pistia and Fichhomia.

B. Stems :
a) Stem is long, slender and flexible in submerged forms like Hydrilla arid Potamogeton. Stem in Eichhomia (free – floating form) is thick, short and spongy.
b) Stem modified into rhizome is Nymphaea and Nelumbo.

C. Leaves :
In submerged forms, the leaves are like thick, long, ribbon shaped (Vallisneria) or long and linear (Potamogeton) or finely dissected (Ceratophyllum). In Nymphaea, Nelumbo and Victoria, the floating leaves are large, flat and coated with wax on their upper surfaces.

Question 29.
Write a brief account on classification of xerophytes.

Section – C (2 × 8 = 16)

Note :

• Answer ANY TWO questions.
• Each Question carries EIGHT marks.
• ALL are of long answer type questions.

Question 30.
Describe the process of Fertilization in angiosperms.
Answer:
The fusion of male and female gametes is called fertilization.
The process of fertilization in angiosperms is described under the following five steps.

A) Entry of the pollen tube into the ovule : The pollen tube enters into the ovule in 3 ways.

1. Pologamy : Pollen tube enters into the ovule through microphyle. e.g. : Ottelia, Hibiscus.
2. Chalazogamy : Pollen tube enters into the ovule through chalaza. e.g. : Casiarina.
3. Mesogamy : Pollen tube enters into the ovule through integu-ments or funiculus, e.g. : Cucurbita.

B) Entry of pollen tube into the embryo sac : The pollen tube enters into the embryo sac only through the microphylar region either by destroying one of the synergids or in between egg cell and synergid. The entry of pollen tube is directed by filiform apparatus.

C) Dicharge of male gametes or sperms: After entry of pollen tube into the embryosac, the male gametes are liberated by one of the following ways.
a) Tip of the pollen tube may burst out
b) Degeneration of the tip of the pollen tube.
c) Formation of an apical pore at the tip of the pollen tube. The pollen tube finally releases the intact male gametes and vegetative nucleus.

D) Gametic Fusion : One of sperm nucleus [first sperm] fuses with egg cell and forms a dipliod zygote (2n). This fusion is called syngamy or fertilization. It was first discovered by Strasburger (1884).

E) Triple fusion and Double fertilization : The second sperm nucleus fuses with the secondary nucleus of the embryosac and forms Primary endosperm nucleus [PEN], This fusion is called as Triple fusion.

It was first observed by Nawaschin in Lilium and Fertillaria. In angiosperms the two male gametes release into the embryo sac take part in two fertilizations. The first sperm nucleus combines with egg cell to form zygote and the second sperm nucleus combines with secondary nucleus to form primary endosperm nucleus. With the occurrence of two fertili- zations, this phenomenon is, i.e., Double (fertilization) includes syngamy as well as triple fusion. It results in the formation of fertile endospermic seeds.

Question 31.
With a neat, labelled diagram, describe the parts of a mature angiosperm embryo sac. Mention the role of synergids.
Answer:
Mature angiosperm embryosac shows three parts.
They are :
1) Egg apparatus
2) Antipodals
3) Central cell.

1. Egg Apparatus : Three cells present towards the micropyle of the embryosac together called egg apparatus. Of which, the central, largest one is called egg and two lateral cells are called synergids. Synergids show finger like projections towards the micropyle called filliform apparatus.
2. Antipodals : Three cells present towards the chalazal end of the ovule are called antipodals. They are also referred to as vegetative ’ells of the embryosac and disintegrates before or after fertilisation.
3. Central cell : It is the largest cell of the embryosac. It is formed by the fusion of two polar nuclei. It is also called secondary nucleus. It shows central vacuole and 2 haploid polamuclei.

Role of synergies :

1. The filliform apparatus of the synergids absorbs food materials from the Nu cells and supplies to embryosac.
2. It also secretes some chemicals which direct the growth of the pollen tube towards embryosac.

Question 32.
Justify the statement, “Mitochondria are power houses of the cell”.
Answer:
Structure of mitochondria
Mitochondria is a double membrane-bound organelle in which the inner layer has folds called cristae helping in increasing the surface area. This membrane contains oxysomes, which take part in ATP synthesis, through the process of oxidative phosphorylation.

Mitochondria as power houses
Mitochondria are the primary sites of aerobic respiration. The synthesised ATP is not utilised by the cell instantly but stored in the mitochondria. When required, the ATP is hydrolysed releasing energy which can be used for various other metabolic activities. As the mitochondria are the sites of energy production and storage, they are called powerhouses of the cell.

Question 33.
Describe the internal structure of a dicot root.
Answer:
A thin transerverse section of dicot root shows three parts namely
(i) Epidermis
(ii) Cortex and
(iii) stele.

(i) Epidermis – It is the outer most layers made of thin walled cells. Some cells protrude in the form of unicelluler root hairs. So called Epiblema. it protects the inner parts. Root hairs help in absorption of water from the soil.

(ii) Cortex – It consists of several layers of thin walled parenchyma cells with inter cellular spaces. The Innermost layer of cortex is called Endodermis. It comprises a single layer of barrel shaped cells without intercellular spaces. The tangential as well as the radial walls of Endoderm cells show suberin thickenings called casparian strips. Some cells opposite to protoxylem lack these strips. Some cells opposite to protoxylem lack these strips called passeage cells. They help in the movement of water and dissolved salts from cortex into xylem.

(iii) Stele : – It is the central part, consists of 4 layers.

(d) Pericycle : It is single layered, made of thin walled parenchyma cells, present next to endodermis. It produces lateral roots and become vascular cambium during secondary growth.

(e) Vascular Bundle : Xylem and phloem constitutes vascular Bundle. They are arranged on different radius (in alternate manner) so called Radial vascular Bundle, Xylem is exarch, where protoxylem is towards periphery and metaxylem is towards the center. Xylem is diarch to tetrach condition. Xylem helps in conduction of water and minerals and phloem helps in conduction of food materials.

(f) Medulla : It is absent or small, made of parenchyma cells. When present, it helps in the storage of food and water.