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Students get through the TN Board 11th Bio Botany Important Questions Chapter 7 Cell Cycle which is useful for their exam preparation.

TN State Board 11th Bio Botany Important Questions Chapter 7 Cell Cycle

Answer the following short answers.

Question 1.
Who coined the word “Cell” and “protoplasm”?
Answer:
Robert Hooke, Jan Evangelista purkyne (J.E.Purkinje)

Question 2.
Mentions any two roles of the nucleus.
Answer:

1. Control activities of the cell.
2. Genetic information copied from cell to cell while the cell divides.

Question 3.
Define the term haploid.
Answer:
In meiosis, the daughter cells contain half the number of chromosomes of the parent cell and are known as a haploid state (n).

Question 4.
Define the term Cytokinesis.
Answer:
Whichever division takes place, it is normally followed by division of the cytoplasm to form separate cells, called cytokinesis.

Question 5.
What is C – Value?
Answer:
C-Value is the amount in picograms of DNA contained within a haploid nucleus.

Question 6.
What do you know about clone dolly?
Answer:
Since the DNA of cells in G₀, does not replicate. The researcher is able to fuse the donor cells from a sheep’s mammary glands into G₀ state by culturing in the nutrient-free state. The G₀ donor nucleus synchronized with the cytoplasm of the recipient egg, which developed into the clone, Dolly.

Question 7.
Define maturation promoting factor (MPF).
Answer:
One of the proteins synthesized only in the G₂ period is known as Maturation Promoting’Factor (MPF). It brings about the condensation of interphase chromosomes into the mitotic form.

Question 8. Define Amitosis.
Answer:
Amitosis is also called direct or incipient cell division. Here there is no spindle formation and chromatin material does not condense.

Question 9.
Mention two drawbacks of amitosis.
Answer:

1. Causes unequal distribution of chromosomes.
2. Can lead to abnormalities in metabolism and reproduction.

Question 10.
Explained closed mitosis.
Answer:
In closed mitosis, the nuclear envelope remains intact and chromosomes migrate to opposite poles of a spindle within the nucleus.

Question 11.
How does an aster form?
Answer:
In an animal cell, the centrioles extend a radial array of microtubules towards the plasma membrane when they reach the poles of the cell. This arrangement of microtubules is called an aster. Plant cells do not form asters.

Question 12.
What do you mean by anaphase-promoting complex cyclosome (APC/C)
Answer:
A ubiquitine ligase is activated called the anaphase-promoting complex cyclosome (APC/C) leads to degradation of the key regulatory proteins at the transition of metaphase to anaphase. APC is a cluster of proteins that induces the breaking down of cohesion proteins which leads to the separation of chromatids during mitosis.

Question 13.
List out various stages of prophase I of Meiotic cell division
Answer:
Leptotene, zygotene, pachytene, diplotene and diakinesis.

Question 14.
Define Chiasmata.
Answer:
The homologous chromosomes remain attached at one or more points where crossing over has taken place. These points of attachment where ‘X’ shaped structures occur at the sites of crossing over is called Chiasmata.

Question 15.
What is meant by metaphase plate?
Answer:
Bivalent (pairs of homologous chromosomes) aligned at the equator of the cell known as the metaphase plate.

Question 16.
Define interkinesis.
Answer:
The stage between the two meiotic divisions is called interkinesis which is short-lived.

Question 17.
Mention any two of the significance of meiosis.
Answer:
This maintains a definite constant number of chromosomes in organisms. Crossing over takes place and the exchange of genetic material leads to variations among species. These variations are the raw materials for evolution. Meiosis leads to genetic variability by partitioning different combinations of genes into gametes through independent assortment.

Question 18.
Explain the term ‘Mitogen’.
Answer:
The factors which promote cell cycle proliferation are called mitogen. Plant mitogens include gibberellin, ethylene, Indole acetic acid, kinetin. These increase the mitotic rate.

Question 19.
Define Anastral condition in cell division.
Answer:
This is present only in plant cells! No asters or centrioles are formed only spindle fibers are formed during cell division.

Question 20.
Explain Amphiastral condition.
Answer:
Aster and centrioles are formed at each pole of the spindle during cell division. This is found in animal cells.

Answer In brief.

Question 1.
Enumerate any three important features of the chromosome.
Answer:

1. The shape of the chromosome is specific: The long, thin, lengthy structured chromosome contains a short, constricted region called a centromere. A centromere may occur anywhere along the chromosome, but it is always in the same position on any given chromosome.
2. The number of chromosomes per species is fixed: for example, the mouse has 40 chromosomes, the onion has 16 and humans have 46.
3. Chromosomes occur in pairs: The chromosomes of a cell occur in impairs, called homologous pairs. One of each pair come originally from each parent, eg. human has 46 chromosomes,
4. coming originally from each parent in the process of sexual reproduction.

Question 2.
Tabulate the time duration in hours of the different phases of the cell cycle.
Answer:

Question 3.
Explain G₁ phase of cell cycle.
Answer:
The first gap phase-2 C amount of DNA in cells of G₁. The cells become metabolically active and grow by producing proteins, lipids, carbohydrates, and cell organelles including mitochondria and endoplasmic reticulum. Many checkpoints control the cell cycle. The checkpoint called the restriction point at the end of G₁ determines a cell’s fate whether it will continue in the cell cycle and divide or enter a stage called G₀ as a quiescent stage and probably as specified cell or die. Cells are arrested in G₁ due to:

1. Nutrient deprivation.
2. Lack of growth factors or density-dependent inhibition.
3. Undergo metabolic changes and enter into G₀ state.

Biochemicals inside cells activate cell division. The proteins called kinases and cyclins activate, genes and their proteins to perform cell division. Cyclins act as major checkpoint which operates in G₁ to determine whether or not a cell divides.

Question 4.
Distinguish between Karyokinesis and cytokinesis.
Answer:

Question 5.
Describe the prophase of Mitotic cell division.
Answer:

1. Prophase is the longest phase in mitosis. Chromosomes become visible as long thin thread-like structure, condenses to form compact mitotic chromosomes. In-plant cells initiation of spindle fibers takes place, the nucleolus disappears. The nuclear envelope breaks down. Golgi apparatus and endoplasmic reticulum are not seen.
2. In the animal cell, the centrioles extend a radial array of microtubules towards the plasma membrane when they reach the poles of the cell. This arrangement of microtubules is called an aster. Plant cells do not form asters.

Question 6.
How Do you calculate the length of the S period?
Answer:
A Culture of animal cells in which the cell cycles were asynchronous was incubated with 3H-Thymidine for 10 minutes. Autoradiography showed that 50% of the cells were labeled. If the cell cycle time (generation time) was 16 hrs how long was the S period?
Length of the S period Fraction of cells in DNA replication x generation time
Length of the S period == 0.5 x 16 hours = 8 hours

Question 7.
Explain the process of cytokinesis in plant cells.
Answer:
Division of the cytoplasm often starts during telophase. In plants, cytokinesis cell plate grows from center towards lateral walls – centrifugal manner of cell plate formation. Phragmoplast contains microtubules, actin filaments, and vesicles from the Golgi apparatus and ER. The Golgi Vesicles contain carbohydrates such as pectin, hemicellulose which move along the microtubule of the phragmoplast to the equator fuse, forming a new plasma membrane arid the materials which are placed their becomes new cell wall. The first stage of cell wall construction is a line dividing the newly forming cells called a cell plate. The Cell plate eventually stretches right across the cell forming the middle lamella. Cellulose builds up on each side of the middle lamella to form the cell walls of two new plant cells.

Question 8.
Explain the sequences of Anaphase I Telophase I in Meiotic cell division.
Answer:
Anaphase I: Homologous chromosomes are separated from each other. Shortening of spindle fibers takes place. Each homologous chromosome with its two chromatids and undivided centromere move towards the opposite poles of the cells. The actual reduction in the number of chromosomes takes place at this stage. Homologous chromosomes which move to the opposite poles are either 1 paternal or maternal in origin. Sister chromatids remain attached with their centromeres.
Telophase I: Haploid set of chromosomes are present at each pole. The formation of two daughter cells, each with the haploid number of chromosomes. Nuclei are reassembled. Nuclear envelope forms around the chromosome and the chromosomes become uncoiled. Nucleolus reappears.

Question 9.
Give the differences between mitosis in plant and animal cells.
Answer:

Question 10.
What is Endomitosis? Explain with example.
Answer:
The replication of chromosomes in the absence of nuclear division and cytoplasmic division resulting in numerous copies within each cell is called endomitosis. Chromonema does not separate to form chromosomes but remains closely associated with each other. The nuclear membrane does not rupture. So no spindle formation. It occurs notably in the salivary glands of Drosophila and other flies. Cells in these tissues contain giant chromosomes (polyteny), each consisting of thousands of intimately associated, or synapsed, chromatids, eg. Polytene chromosome.

Answer In detail.

Question 1.
Draw and label the events of the cell cycle.
Answer:

Question 2.
Draw the schematic diagram of anaphase-promoting complex cyclosome and explain briefly.
Answer:
A ubiquitine ligase is activated called the anaphase-promoting complex cyclosome (APC/C) leads to degradation of the key regulatory proteins at the transition of metaphase to anaphase. APC is a cluster of proteins that induces the breaking down of cohesion proteins which leads to the separation of chromatids during mitosis.

Question 3.
Explain the events that take place at metaphase and anaphase stages of somatic cell division – with diagram.
Answer:
Metaphase: Chromosomes (two sister chromatids) are attached to the spindle fibers by kinetochore of the centromere. The spindle fibers is made up of tubulin. The alignment of the chromosome into the compact group at the equator of the cell is known as the metaphase plate. This is the stage where chromosome morphology can be easily studied.
The kinetochore is a DNA-Protein complex present in the centromere DNA where the microtubules are attached. It is a trilaminar disc-like plate.
The spindle assembly checkpoint decides the cell to enter anaphase.
For diagram refer to Figure 7.2 (Metaphase only).
Anaphase: Each chromosome split simultaneously and two daughter chromatids begin to migrate towards two opposite poles of a cell.
Each centromere splits longitudinally into two, freeing the two sister chromatids from each other. Shortening of spindle fiber and longitudinal splitting of centromere creates a pull that divides the chromosome into two halves.
Each half receives two chromatids (that is sister chromatids are separated). When the sister chromatids separate the actual partitioning of the replicated genome is complete.
For diagram refer to Figure 7.2 (Anaphase only).

Question 4.
Explain the different stages of prophase I of meiotic cell division.
Answer:

1. Prophase I: Prophase I is of longer duration and it is divided into 5 substages – Leptotene, Zygotene, Pachytene, Diplotene, and Diakinesis.
2. Leptotene: Chromosomes are visible under a light microscope. Condensation of chromosomes takes place. Paired sister chromatids begin to condense.
3. Zygotene: Pairing of homologous chromosomes takes place and it is known as synapsis. Chromosome synapsis is made by the formation of the synaptonemal complex. The complex formed by the homologous chromosomes is called bivalent (tetrads).
4. Pachytene: At this stage, bivalent chromosomes are clearly visible as tetrads. Bivalent of meiosis I consist of 4 chromatids and 2 centromeres. Synapsis is completed and, recombination nodules appear at a site where crossing over takes place between non-sister chromatids of the homologous chromosome. Recombination of homologous chromosomes is completed by the end of the stage but the chromosomes are linked at the sites of crossing over. This is mediated by the enzyme recombinase.
5. Diplotene: Synaptonemal complex disassembled and dissolves, the homologous chromosomes remain attached at one or more points where crossing over has taken place.
   These points of attachment where ‘X’ shaped structures occur at the sites of crossing over is called Chiasmata.
6. Chiasmata are chromatin structures at sites where recombination has been taken place. They are specialized chromosomal structures that hold the homologous chromosomes together.
   Sister chromatids remain closely associated whereas the homologous chromosomes tend to separate from each other but are held together by chiasmata. This substage may last for days or years depending on the sex and organism.
7. The chromosomes are very actively transcribed in females as the egg stores up materials for use during embryonic development. In animals, the chromosomes have prominent loops called the lampbrush chromosome.
8. Diakinesis: Terminalisation of chiasmata. Spindle fibers assemble. The nuclear envelope breaks down. Homologous chromosomes become short and condensed. Nucleolus disappears.

Question 5.
Enumerate the differences between mitosis and meiosis.
Answer:

Question 6.
Draw and label the various stages of mitosis cell division.
Answer:

Choose the correct answer.

1. The word cell coined by:
(a) Robert brown
(b) Leeuwenhoek
(c) Anton
(d) Robert Hooke
Answer:
(d) Robert Hooke

2. Which of the following is incorrect about the role of the nucleus?
(a) Control activities of the cell
(b) Genetic information copied from cell to cell
(c) Gametic cells fused together in sexual reproduction
(d) Characters passed on to new individuals
Answer:
(c) Gametic cells fused together in sexual reproduction

3. Match the following with phase and time duration:

(a) (i)-(c), (ii)-(a), (iii)-(d), (iv)-(b)
(b) (i)-(c), (ii)-(b), (iii)-(d), (iv)-(a)
(c) (i)-(a), (ii)-(c), (iii)-(d), (iv)-(b)
(d) (i)-(b), (ii)-(d), (iii)-(a), (iv)-(c)
Answer:
(a) (i)-(c), (ii)-(a), (iii)-(d), (iv)-(b)

4. Name the types of nuclear divisions?
(a) Mitosis
(b) Mitosis and Meiosis
(c) A Mitosis
(d) Meiosis
Answer:
(b) Mitosis and Meiosis

5. The checkpoint is otherwise called:
(a) Important point
(b) collecting point
(c) restriction point
(d) Controlling point
Answer:
(c) restriction point

6. In S phase DNA count increases from 2C to:
(a) 3C
(b) 4C
(c) 5C
(d) 6C
Answer:
(b) 4C

7. Which of the following, Microtubules are formed?
(a) G₀
(b) G₁
(c) S
(d) G₂
Answer:
(d) G₂

8. Which animal regenerates the parts of its body?
(a) starfish
(b) amoeba
(c) shark
(d) jellyfish
Answer:
(a) starfish

9. Pick the incorrect statement of the significance of meiosis:
(a) maintain a definite constant number of chromosome
(b) Adaption of an organism to various stress
(c) crossing over takes place
(d) The chromosome number increases in the Organism.
Answer:
(d) The chromosome number increases in the Organism.

10. Which one of the following is not a mitotic poison?
(a) Cyanide
(b) A zide
(c) 2, 4, dinitrophenol
(d) Polyamines
Answer:
(d) Polyamines

11. Insulin and steroid hormones are examples of:
(a) Inhibiting factors
(b) Growth factors
(c) Limiting factors
(d) Synthetic factors
Answer:
(b) Growth factors

12. Identify the correct statement for equational division:
(a) The number of chromosomes in the parent and daughter cells remains the same
(b) Chromosome in the parent and daughter are different
(c) Double the number of chromosomes
(d) It depends upon the division
Answer:
(a) The number of chromosomes in the parent and daughter cells remains the same

13. The number of chromosomes in humans are:
(a) 36
(b) 40
(c) 46
(d) 16
Answer:
(c) 46

14. The chromosomes of a cell occur in pairs called:
(a) haploid pair
(b) homologous pair
(c) diploid pair
(d) Tetraploid
Answer:
(b) homologous pair

15. The amount in picograms of DNA contained within a haploid nucleus is called:
(a) B-value
(b) X – value
(c) Y-value
(d) C – value
Answer:
(d) C – value

16. In meiosis, the daughter cells contain half the number of chromosomes of the parent cell and is known as:
(a) Triploid
(b) Haploid
(c) Tetraploid
(d) Diploid
Answer:
(b) Haploid

17. The major checkpoint which operates in G1 to determine whether or not a cell divides is:
(a) Kinase
(b) protease
(c) cyclins
(d) pepsin
Answer:
(c) cyclins

18. A cell division in which no spindle formation and non condensation of chromatin material is called:
(a) Mitosis
(b) Amitosis
(c) Meiosis
(d) None of the above
Answer:
(b) Amitosis

19. Karyokinesis involves in:
(a) division of cytoplasm
(b) division of cells
(c) division of nucleus
(d) none of the above
Answer:
(c) division of nucleus

20. The sequence of stages of mitotic cell division is as follows:
(a) Metaphase, Anaphase, Prophase and Telophase
(b) Telophase, Anaphase, Prophase and Metaphase
(c) Prophase, Anaphase, Telophase and metaphase
(d) Prophase, Metaphase, Anaphase and Telophase
Answer:
(d) Prophase, Metaphase, Anaphase and Telophase

21. The longest phase in mitosis is:
(a) Metaphase
(b) Telophase
(c) Prophase
(d) Anaphase
Answer:
(c) Prophase

22. The cells, which do not form asters during cell division are:
(a) Animal cells
(b) Plant cells
(c) Virus
(d) None of the above
Answer:
(b) Plant cells

23. Mitosis cell division occurs during:
(a) Cogenesis
(b) Gametogenesis
(c) Somatic growth
(d) Spermatogenesis
Answer:
(c) Somatic growth

24. The Golgi vesicles contain carbohydrates such as:
(a) Pectin
(b) Actin
(c) Glucose
(d) Fructose
Answer:
(a) Pectin

25. A sexual reproduction is prominent is:
(a) Starfish
(b) Dolphin
(c) Yeast
(d) Delonix regia
Answer:
(c) Yeast

26. Synapsis takes place in meiosis during:
(a) Pachytene
(b) Metaphase
(c) Diplotene
(d) Zygotene
Answer:
(d) Zygotene

27. Chiasmata is the point where:
(a) Grossing over takes place
(b) Nuclear division takes place
(c) Cytokinesis occur
(d) Nucleolus disappears
Answer:
(a) Grossing over takes place

28. The random distribution of homologous chromosomes in a cell in metaphase I of meiotic cell division is called:
(a) Segregation
(b) Independent assortment
(c) Linkage
(d) None of the above
Answer:
(b) Independent assortment

29. The stage between two meiotic divisions is called:
(a) Karyokinesis
(b) cytokinesis
(c) Interphase
(d) Interkinesis
Answer:
(d) Interkinesis

30. The factors which promote cell cycle proliferation is called:
(a) Mitotic poison
(b) action
(c) mitogen
(d) recombinase
Answer:
(c) mitogen

Students get through the TN Board 11th Bio Botany Important Questions Chapter 6 Cell: The Unit of Life which is useful for their exam preparation.

TN State Board 11th Bio Botany Important Questions Chapter 6 Cell: The Unit of Life

Answer the following short answers.

Question 1.
What is primary magnification?
Answer:
The first magnification of the microscope is done by the objective lens which is called primary magnification.

Question 2.
Name the two kinds of an electron microscope.
Answer:
There are two kinds of electron microscopes namely:

1. Transmission Electron Microscope (TEM).
2. Scanning Electron Microscope (SEM).

Question 3.
What is the main use of TEM?
Answer:
This is the most commonly used electron microscope that provides two-dimensional images. The components of the microscope are as follows: (i) Electron Generating System, (ii) Electron Condenser, (Hi) Specimen Objective, (iv) Tube Lens, (v) Projector. It is used for studying the detailed structure of viruses, mycoplasma, cellular organelles, etc.

Question 4.
List out the name of microscopes in which the light of illumination is visible light.
Answer:
Light Microscope, Darkfield Microscopy, Phase contrast Microscope.

Question 5.
What are the constituents of protoplasm?
Answer:
It is primarily made of water contents and various other solutes of biological importance such as glucose,fatty acids, amino acids, minerals,vitamins, hormones and enzyme.

Question 6.
Define cohesiveness of protoplasm.
Answer:
Particles or molecules of protoplasm are adhered with each other by forces, such as Vander Waal’s bonds, that hold long chains of molecules together.

Question 7.
Nome the three types of cells.
Answer:
The three types of cells are Prokaryotes, Mesokaryotes, and Eukaryotes.

Question 8.
Define mesokoryotes.
Answer:
These organisms share some of the characters of both prokaryotes and eukaryotes. In other words, these are organisms intermediate between pro and eukaryotes.

Question 9.
Explain primary cell wall.
Answer:
It is the first layer of inner to middle lamellae, primarily consisting of a loose network of cellulose microfibrils in a gel matrix. It is thin, elastic, and extensible.

Question 10.
How does a ‘glycocalyx’ form?
Answer:
The Carbohydrate molecules of the cell membrane are short-chain polysaccharides. These are either bound with ‘glycoproteins’ or ‘glycolipids’ and form a glycocalyx.

Question 11.
Define Endocytosis and exocytosis.
Answer:
Cell surface membranes are able to transport individual molecules and ions. There are processes in which a cell can transport a large number of solids and liquids into the cell (endocytosis) or out of the cell (exocytosis).

Question 12.
What do you understand by the term phagocytosis?
Answer:
The particle is engulfed by a membrane, which folds around it and forms a vesicle. The enzymes digest the material and products are absorbed by cytoplasm.

Question 13.
Define cristae of mitochondria.
Answer:
The inner membrane is convoluted (infoldings), called the crista (plural: cristae). Cristae contain most of the enzymes for the electron transport system.

Question 14.
Name any three odors plastids.
Answer:
The three odors plastids are Chloroplast, Phaeoplast, Rhodoplast.

Question 15.
What is the main function of polysome?
Answer:
The function of polysomes in the formation of several copies of a particular polypeptide during protein synthesis.

Question 16.
What are Glyoxysomes?
Answer:
Glyoxysome is a single membrane-bound organelle. It is a sub-cellular organelle and contains enzymes of glyoxylate pathway, fi-oxidation of fatty acid occurs in glyoxysomes of germinating seeds, eg. Castor seeds.

Question 17.
What is the main function of .plant vacuoles?
Answer:
The major function of plant vacuole is to maintain water pressure known as turgor pressure, which maintains the plant structure. Vacuoles organize themselves into a storage/sequestration compartment.

Question 18.
What are the reserve materials present in prokaryotes?
Answer:
In prokaryotes, reserve materials such as phosphate granules, cyanophycean granules, glycogen granules, poly-hydroxy butyrate granules, sulfur granules, carboxysomes, and gas vacuoles are present.

Question 19.
Explain briefly the Holocentric chromosomes.
Answer:
Holocentric chromosomes have centromere activity distributed along the whole surface of the chromosome during mitosis. The holocentric condition can be seen in Caenorhabditis elegans (transparent nematode) Mid many insects.

Question 20.
What do you know about microphotographs?
Answer:
Images of structures observed through microscopes can be further magnified, projected, and saved by attaching a camera to the microscope by a microscope coupler or; eyepiece adaptor. Picture taken using an inbuilt camera in a microscope is called microphotography or microphotograph.

Answer In brief.

Question 1.
Describe briefly the Darkfield microscope.
Answer:

1. The darkfield microscope was discovered by Z. Sigmondy (1905). Here the field will be dark but the object will be glistening so the appearance will be bright.
2. A special effect in an ordinary microscope is brought about by means of a special component called Patch Stop Carrier’.
3. It is fixed in the metal ring of the condenser component
4. Patch top is a small glass device that has a dark patch at the center of the disc leaving a small area along the margin through which the light passes.
5. The light passing through the margin will travel oblique like a hollow cone and strikes the object in the periphery, therefore the specimen appears glistening in a dark background.

Question 2.
Explain the principle and function of a scanning electron microscope.
Answer:

1. This is used to obtain the three-dimensional image and has a lower resolving power than TEM.
   In this, electrons are focused by means of lenses into a very fine point.
2. The interaction of electrons with the specimen results in the release of different forms of radiation (such as Auger electrons, secondary electrons, backscattered electrons) from the surface of the specimen.
3. These radiations are then captured by an appropriate detector, amplified, and then imaged on a fluorescent screen.
4. The magnification is. 2,00,000 times and resolution is 5-20 nm.

Question 3.
Enumerate the functions of the cell wall.
Answer:

1. Offers definite shape and rigidity to the cell.
2. Serves as a barrier for several molecules to enter the cells.
3. Provides protection to the internal protoplasm against mechanical injury.
4. Prevents the bursting of cells by maintaining the osmotic pressure.
5. Plays a major role by acting as a mechanism of defense for the cells.

Question 4.
What do you know about signal transduction in cells? Explain briefly.
Answer:

1. The process by which the cell receives information from outside and responds is called signal transduction.
2. Plants, fungi, and animal cells use nitric oxide as one of the many signaling molecules.
3. The cell membrane is the site of chemical interactions of signal transduction.
4. Receptors receive the information from the first messenger and transmit the message through series of membrane proteins.
5. It activates a second messenger which stimulates the cell to carry out specific functions.

Question 5.
List out the functions of Golgi bodies.
Answer:

1. Glycoproteins and glycolipids are produced.
2. Transporting and storing lipids.
3. Formation of lysosomes.
4. Production of digestive enzymes.
5. Cell plate and cell wall formation
6. Secretion of Carbohydrates for the formation of plant cell walls and insect cuticles.

Question 6.
How do the grana in chloroplast form? Mention their structure and function.
Answer:

1. Grana (singular: Granum) are formed when many of these thylakoids are stacked together
   like a pile of coins.
2. Light is absorbed and converted into chemical energy in the granum, which is used in the stroma to prepare carbohydrates. Thylakoids contain chlorophyll pigments.
3. The chloroplast contains osmophilic granules, 70s ribosomes, DNA (circular and non-histone), and RNA.
4. This chloroplast genome encodes approximately 30 proteins involved in photosynthesis including the components of photosystem I & II, cytochrome bf complex, and ATP synthase.
5. One of the subunits of Rubisco is encoded by chloroplast DNA.
6. It is the major protein component of chloroplast stroma, the single most abundant protein on earth.
7. The thylakoid contains small, rounded photosynthetic units called quantosomes.
8. It is a semi-autonomous organelle and divides by fission.

Question 7.
Explain the types of chromosomes based on the position of the centromere.
Answer:

1. Based on the position of the centromere, chromosomes are called telocentric (terminal centromere), Acrocentric (terminal centromere capped by telomere), Sub metacentric
2. The eukaryotic chromosomes may be tod-shaped (telocentric and acrocentric), L-shaped (sub-metacentric), and V-shaped (metacentric).
   

Question 8.
Describe the structure of flagellum in bacteria.
Answer:
The gram-positive bacteria contain only two basal rings. S-ring is attached to the inside of peptidoglycan and M-ring is attached to the cell membrane. In Gram-negative bacteria, two pairs of rings proximal and distal ring are connected by a central rod. They are L- Lipopolysacchride ring P- Peptidoglycan ring, S-Super membrane ring, and M-membrane ring. The outer pair of L and P rings are attached to the cell wall and the inner pair of S and M rings attached to the cell membrane.

Question 9.
Explain the structure and function of Cilia.
Answer:

Cilia (plural) are short cellular, numerous microtubule bound projections of the plasma membrane.
Cilium (singular) is a membrane-bound structure made up of basal body, rootlets, basal plate, and shaft. The shaft or axoneme consists of nine pairs of microtubule doublets, arranged in a circle along the die periphery with two central tubules, (9 + 2) arrangement of microtubules is present. Microtubules are made up of tubulin. The motor protein dynein connects the outer microtubule pair and links them to the central pair. Nexin links the peripheral doublets of microtubules.

Answer In detail.

Question 1.
Enumerate the physical properties of protoplasm.
Answer:
Physical Properties of Protoplasm: The protoplasm exists either in a semisolid (jelly-like) state called ‘gel’ due to suspended particles and various chemical bonds or maybe a liquid state called ” ‘sol’.
The colloidal protoplasm which is in gel form can change into sol form by solation and the sol can change into a gel by gelation. These gel-sol conditions of the colloidal system are the prime basis for the mechanical behavior of cytoplasm.

1. Protoplasm is translucent, odorless, and polyphasic fluid.
2. It is a crystal colloid solution which is a mixture of chemical substances forming crystalloid
   i.e. true solution (sugars, salts, acids, bases) and others forming colloidal solution (Proteins and lipids).
3. It is the most important property of the protoplasm by which it exhibits three main phenomena namely Brownian movement, amoeboid movement, and cytoplasmic streaming or cyclosis.
   The viscosity of protoplasm is 2-20 centipoises. The Refractive index of the protoplasm is 1.4.
4. The pH of the protoplasmic around 6.8, contains 90% water (10% in dormant seeds)
5. Approximately 34 elements are present in protoplasm but only 13 elements are main or universal elements i.e. C, H, O, N, Cl, Ca, P, Na, K, S, Mg, I, and Fe. Carbon, Hydrogen, Oxygen, and Nitrogen form 96% of protoplasm.
6. Protoplasm is neither a good nor a bad conductor of electricity. It forms a delimiting the membrane on coming in contact with water and solidifies when heated.

Cohesiveness: Particles or molecules of protoplasm are adhered with each other by forces, such as Vander Waal’s bonds, that hold long chains of molecules together. This property varies with the strength of these forces.
Contractility: The contractility of protoplasm is important for the absorption and removal of water especially stomatal operations.
Surface tension: The proteins and lipids of the protoplasm have less surface tension, hence they are found at the surface forming the membrane. On the other hand, the chemical substances (NaCl) have high surface tension, so they occur in deeper parts of the cell protoplasm.

Question 2.
Describe the structure and function of mitochondria.
Answer:

1. It was first observed by A. Kolliker (1880). Altmann (1894) named it as Bioplasts. Later Benda (1897, 1898), named as mitochondria. They are ovoid, rounded, rod shape and pleomorphic structures.
2. Mitochondrion consists of a double membrane, the outer and inner membrane.
3. The outer membrane is smooth, highly permeable to small molecules and it contains proteins called Porins, which form channels that allow free diffusion of molecules smaller than about 1000 daltons and the inner membrane divides the mitochondrion into two compartments, the outer chamber between two membranes and the inner chamber filled with matrix.
4. The inner membrane is convoluted (infoldings), called the crista (plural: cristae). Cristae contain most of the enzymes for the electron transport system.
5. The inner chamber of the mitochondrion is filled with proteinaceous material called mitochondrial matrix.
6. The inner membrane consists of stalked particles called elementary particles or Fernandez Moran particles, F1 particles, or Oxysomes.
7. Each particle consists of a base, stem, and round head. In the head, ATP synthase is present for oxidative phosphorylation.
8. The inner membrane is impermeable to most ions, small molecules and maintains the proton gradient that drives oxidative phosphorylation.
   
9. Mitochondria contain 73% of proteins, 25-30% of lipids, 5-7 % of RNA, DNA (in traces), and enzymes (about 60 types).
10. Mitochondria are called the Powerhouse of a cell, as they produce energy-rich ATP.
11. All the enzymes of Kreb’s cycle are found in the matrix except succinate dehydrogenase. Mitochondria consist of circular DNA and 70S ribosome.
12. They multiply by fission and replicates by strand displacement model. Because of the presence of DNA, it is a semi-autonomous organelle.
13. The unique characteristic of mitochondria is that they are inherited from female parents only. Mitochondrial DNA comparisons are used to trace human origins.
14. Mitochondrial DNA is used to track and date recent evolutionary times because it mutates 5 to 10 times faster than DNA in the nucleus.

Question 3.
Explain the structure of Ribosomes. Mention the types of ribosomes.
Answer:
Ribosomes were first observed by George Palade (1953) as dense particles or granules in the electron microscope. Electron microscopic observation reveals that ribosomes are composed of two rounded subunits, united together to form a complete unit. Mg²⁺ is required for the structural cohesion of ribosomes. Biogenesis of ribosomes is de nova formation, auto replication, and nucleolar origin. Each ribosome is made up of one small and one large subunit. Ribosomes are the sites of protein synthesis in the cell. The ribosome is not a membrane-bound organelle.

Question 4.
Give details of the special types of chromosomes in plant and animal cells.
Answer:

1. Special types of chromosomes are found only in certain special tissues. These chromosomes are larger in size and are called giant Chromosomes in certain plants and they are found in the suspensors of the embryo.
2. The polytene chromosome and lampbrush chromosome occur in animals and are also called giant chromosomes.
3. Polytene chromosomes observed in the salivary glands of Drosophila (fruit fly) by C.G. Balbiani in 1881. In larvae of many flies, midges (Diptera), and some insects the interphase chromosomes duplicate and reduplicate without nuclear division.
4. A single chromosome that is present in multiple copies forms a structure called a polytene chromosome which can be seen in the light microscope.
5. They are genetically active. There are distinct alternating dark bands and light inter-bands. About 95%of DNA are present in bands and 5% in inter-bands. The polytene chromosome has extremely large puff called Balbiani rings which is seen in chironortious larvae. It is also known as chromosomal puff. Puffing of bands are the sites of intense RNA synthesis,
6. As this chromosome occurs in the salivary gland it is known as salivary gland chromosomes, Polyteny is achieved by repeated replication of chromosomal DNA several times without nuclear division and the daughter chromatids aligned side by side and do not separate (called endomitosis).
7. Gene expression, transcription of genes, and RNA synthesis occur in the bands along the polytene chromosomes. Maternal and paternal homologs remain associated side by side is called somatic pairing.
8. Lampbrush chromosomes occur at the diplotene stage of the first meiotic prophase in oocytes of an animal Salamandar and in the giant nucleus of the unicellular alga Acetabularia.
9. It was first observed by Flemming in 1882. The highly condensed chromosome forms the chromosomal axis, from which lateral loops of DNA extend as a result of intense RNA synthesis.

Question 5.
Describe the structure of eukaryotic flagella and explain the movement of the flagellum.
Answer:

1. Eukaryotic Flagella are enclosed by a unit membrane and it arises from a basal body Flagella is composed of outer nine pairs of microtubules with two microtubules in its center (9+2 arrangement).
2. Flagella are microtubule projections of the plasma membrane. The flagellum is longer than cilium (as long as 200pm). The structure of the flagellum has an axoneme made up of microtubules and protein tubulin.
   
3. Movement: Outer microtubule doublet is associated with axonemal dynein which generates force for movement. The movement is ATP-driven. The interaction between tubulin and dynein is the mechanism for the contraction of cilia and flagella. Dynein molecules use energy from ATP to shift the adjacent microtubules. This movement bends the cilium or flagellum.

Question 6.
Describe the structure and functions of Lysosomes.
Answer:
Lysosomes were discovered by Christian de Duve (1953), these are known as suicidal bags. They are spherical bodies enclosed by a single unit membrane. They are found in eukaryotic cells. Lysosomes are small vacuoles formed when small pieces of the Golgi body are pinched off from its tubules.
They contain a variety of hydrolytic enzymes, that can digest material within the cell. The membrane around the lysosome prevents these enzymes from digesting the cell itself.

Functions:

Intracellular digestion: They digest carbohydrates, proteins, and lipids present in the cytoplasm.
Autophagy: During the adverse condition, they digest their own cell organelles like mitochondria and endoplasmic reticulum
Autolysis: Lysosome causes self-destruction of cells on the insight of disease they destroy the cells.
Ageing Lysosomes have autolytic enzymes that disrupt intracellular molecules.
Phagocytosis: Large cells or contents are engulfed and digested by macrophages, thus forming a phagosome in the cytoplasm. These phago some fuse with lysosome for further digestion.
Exocytosis: Lysosomes release their enzymes outside the cell to digest other cells.

Choose the correct answer.

1. The word cell was first used by:
(a) Robert brown
(b) Robert Hooke
(c) Zemike
(d) Robert schwann
Answer:
(b) Robert Hooke

2. Idea of cell theory was first proposed by:
(a) Matthias Schleiden
(b) Theodor schwann
(c) H.J. Dutrochet
(d) Rudolf Virchow
Answer:
(c) H.J. Dutrochet

3. Phase the contrast microscope was invented by:
(a) Zemike
(b) Robert brown
(c) Sigmondy
(d) Robert Hooke
Answer:
(a) Zemike

4. Source of illumination for image formation in dark field microscope is:
(a) Electron
(b) ultraviolet light
(c) X- rays
(d) Visible light
Answer:
(d) Visible light

5. Who coined the term “protoplasm”?
(a) Corti
(b) Van mohl
(c) Purkinje
(d) Fisher
Answer:
(c) Purkinje

6. Indicate the wrong statement:
(a) All organisms are made up of cells
(b) All metabolic reactions take place inside the cell
(c) The structure and function of the cell is controlled by DNA
(d) Group of cells with different structures are called tissue
Answer:
(d) Group of cells with different structures are called tissue

7. Approximately the number of elements present in protoplasm is:
(a) 28
(b) 34
(c) 38
(d) 24
Answer:
(b) 34

8. These organisms with the primitive nucleus are called:
(a) Mesokaryotes
(b) prokaryotes
(c) Eukaryotes
(d) None of the above
Answer:
(b) prokaryotes

9. Cell wall is not present in:
(a) Bacteria
(b) Fungi
(c) Animal cell
(d) plant cell
Answer:
(c) Animal cell

10. The carbohydrate molecules of the cell membrane are:
(a) Long-chain polysaccharides
(b) Short-chain polysaccharides
(c) Long-chain glycoproteins
(d) Short-chain glycolipids.
Answer:
(b) Short-chain polysaccharides

11. One of the many signaling molecules used by plants, fungi, and animal cell is:
(a) sodium chloride
(b) cupric oxide
(c) Nitric oxide
(d) None of the above
Answer:
(c) Nitric oxide

12. Cytoplasm helps the movement of cellular materials around die cell through a process called:
(a) Cytoplasmic streaming
(b) Brownian movement
(c) Active movement
(d) none of the above
Answer:
(a) Cytoplasmic streaming

13. The name endoplasmic reticulum was given by:
(a) Camillo
(b) K.R.Porter
(c) Nickolson
(d) S.B.Roberts
Answer:
(b) K.R.Porter

14. The functions of the Golgi body includes:
(a) helping cell division
(b) initialing protein synthesis
(c) Transporting and storing lipids
(d) None of the above
Answer:
(c) Transporting and storing lipids

15. Mitochondria are called the powerhouse of the cell as they:
(a) Synthesis lipid
(b) Involve in protein synthesis
(c) initiate oxidation metabolism
(d) Produce energy-rich ATP
Answer:
(d) Produce energy-rich ATP

16. Mitochondrial DNA mutates …………. times faster than DNA in the nucleus.
(a) 15 to 20
(b) 30 to 40
(c) 5 to 10
(d) 10 to 15
Answer:
(c) 5 to 10

17. Plastids were classified into various types according to their structure pigments and function:
(a) Robert Hooke
(b) A.F.tJ. Schimper
(c) A. Kolliker
(d) Altmann
Answer:
(b) A.F.tJ. Schimper

18. B – oxidation of fully acids occurs in glyoxysomes of germinating seeds of:
(a) Paddy
(b) Brinjal
(c) Ladies finger
(d) Caster
Answer:
(d) Caster

19. Centriole consists of nine triplet peripheral fibrils made up of
(a) tubulin
(b) fibrils
(c) annuli
(d) none of the above.
Answer:
(a) tubulin

20. Match the following:

(a) (i)-(d), (ii)-(b), (iii)-(a), (iv)-(c)
(b) (i)-(c), (ii)-(d), (iii)-(a), (iv)-(b)
(c) (i)-(b), (ii)-(a), (iii)-(d), (iv)-(c)
(d) (i)-(b), (ii)-(c), (iii)-(a), (iv)-(d)
Answer:
(b) (i)-(c), (ii)-(d), (iii)-(a), (iv)-(b)

21. During cell division, chromation is condensed into an organised form called:
(a) nucleolus
(b) Euchromation
(c) Nuclear pores
(d) Chromosomes
Answer:
(d) Chromosomes

22. Chromatin is made up of
(a) DNA, Polysaccharides, and RNA
(b) DNA, lipids, and RNA
(c) DNA, protein, and RNA
(d) RNA, Glucose, and lipid
Answer:
(c) DNA, protein, and RNA

23. Chromosomes having terminal centromere are called:
(a) Acrocentric
(b) Telocentric
(c) Metacentric
(d) Submetacentric
Answer:
(b) Telocentric

24. Polytene chromosome in the salivary glands of drosophila was first observed by:
(a) Flemming
(b) C.G. Balbiani
(c) Harry Beevers
(d) A. Kolliker
Answer:
(b) C.G. Balbiani

25. Lampbrush chromosome occur at the diplotene stage of first meiotic prophase in oocytes of:
(a) Frog
(b) Moth
(c) Sherk
(d) Salamandra
Answer:
(d) Salamandra

26. The main function of Bacterial flagellum is:
(a) Locomotion
(b) Protection
(c) Feeding
(d) none of the above
Answer:
(a) Locomotion

27. The molecules involved in the mechanism for the contraction of cilia and flagella are:
(a) Aetin and peptin
(b) Flagellin
(c) Tubulin and dynein
(d) none of the above
Answer:
(c) Tubulin and dynein

28. The technique of staining the cells and tissue is called:
(a) microphotography
(b) histochemistry
(c) anatomy
(d) geochemistry
Answer:
(b) histochemistry

29. The stain used for staining mitochondria of the cell is:
(a) Sudan black
(b) Eosin
(c) cotton blue
(d) Janus green
Answer:
(d) Janus green

30. Cilia are short cellular, numerous microtubules bound projections of:
(a) mitochondrial membrane
(b) Cell wall
(c) Plasma membrane
(d) Nuclear membrane
Answer:
(c) Plasma membrane

Students get through the TN Board 11th Bio Botany Important Questions Chapter 5 Taxonomy and Systematic Botany which is useful for their exam preparation.

TN State Board 11th Bio Botany Important Questions Chapter 5 Taxonomy and Systematic Botany

Answer the following short answers.

Question 1.
Define taxonomy.
Answer:
Davis and Heywood (1963) defined taxonomy as “the science dealing with the study of classification including the bases, principles, rules, and procedures”.

Question 2.
Describe a genus with an example.
Answer:
Genus consists of multiple species which have similar characters but differ from toe species of another genus, eg. Helimthmt Tridax.

Question 3.
What are the types of species?
Answer:
There are different types of species and they are as follows:

1. Process of evolution – Biological Species
2. Product of evolution – Morphological Species and Phylogenetic Species

Question 4.
Differentiate between Anamorph and Telomorph.
Answer:

Question 5.
Write Common name and Scientific name of any two plants.
Answer:

Question 6.
What is holotype nomenclature?
Answer:
A specimen or illustration originally cited by the author in the protologue. It is a definitive reference source for identity. Citation of holotype and submission of it is one of the criteria for valid publication of a botanical name.

Question 7.
Define Flora.
Answer:
Flora is the document of all plant species in a given geographic area. Flora consists of a total number of plant species in an area and gives information about the flowering season, fruiting season, and distribution for the given geographic area.

Question 8.
Mention any two international botanical gardens.
Answer:

1. New York Botanic garden, USA.
2. Royal Botanic Garden, Kew – England.

Question 9.
Define Chemotaxonomy.
Answer:
Chemotaxonomy is the scientific approach to the classification of plants on the basis of their biochemical constituents.

Question 10.
What are the aims, of chemotaxonomy?
Answer:

1. To develop taxonomic characters which may improve the existing system of plant classification.
2. To improve present-day knowledge of phytogeny of plants.

Question 11.
Define Serotaxonomy.
Answer:
The classification of very similar plants by means of differences in the proteins they contain, to solve taxonomic problems is called serotaxonomy.

Question 12.
What is meant by DNA barcodes?
Answer:
The genetic sequence used to identify a plant is known as “DNA tags” or “DNA barcodes”.

Question 13.
What do you know about cladistics?
Answer:
The method of classifying organisms into monophyletic group of a common ancestor based on shared apomorphic characters is called cladistics (from Greek, klados- branch).

Question 14.
Explain monadelphous stamens.
Answer:
In Aeschynomene Aspera, the stamens are fused to form two bundles each containing five stamens (5)+5. Stamens are monadelphous.

Question 15.
Mention any two oil plants with their scientific name.
Answer:

1. Arachis hypogea (Groundnut).
2. Pongamia pinnata (Pungam).

Question 16.
Mention any two medicinal plants with their useful parts of the plant.
Answer:

Question 17.
What is meant by scapigerous Inflorescence?
Answer:
The Inflorescence axis (peduncle) arising from the ground bearing a cluster of flowers at its apex. Pedicels are of equal length, arising from the apex of the peduncle which brings all flowers at the same level.

Question 18.
List any two economic uses of plants under the family Liliaceae.
Answer:

Question 19.
List any two ornamental plants under the family Fabaceae.
Answer:
Butea frondosa (Flame of the forest), Clitoria ternatea, Lathyrus odoratus (Sweet pea), and Lupinus hirsutus (Lupin).

Question 20.
What is meant by root module?
Answer:
Taproot system, roots are nodulated, have tubercles containing nitrogen-fixing bacteria (Rhizobium leguminosarum)

Answer In brief.

Question 1.
Explain the differences between taxonomy and systematics.
Answer:

Question 2.
What is Botanical nomenclature? Explain the international code of botanical nomenclature.
Answer:
Assigning a name for a plant is known as Nomenclature. This is based on the rules and recommendations of the International Code of Botanical Nomenclature. ICB deals with the names of existing (living)and extinct (fossil) organisms. The elementary rule of the naming of the plant was first proposed by Linnaeus in 1737and 1751 in his Philosophia Botanica. In 1813 a detailed set of rules regarding plant nomenclature was given by A.P. de Candolle in his famous ‘ work “Theorie elementaire de la botanique”. Then the present ICBN was evolved by following the same rules of Linnaeus, A.P. de Candolle, and his son Alphonse de Candolle.

Question 3.
Mention any three roles of Botanical gardens.
Answer:

1. Gardens with an aesthetic value attract a large number of visitors. For example, the Great Banyan Tree (Ficus benghalensis) in the Indian Botanical Garden at Kolkata.
2. Gardens have a wide range of species and supply taxonomic material for botanical research.
3. It can integrate information from diverse fields like Anatomy, Embryology, Phytochemistry, Cytology, Physiology, and Ecology.

Question 4.
List out the uses of the herbarium.
Answer:

1. Herbarium provides resource material for systematic research and studies.
2. It is a place for the orderly arrangement of voucher specimens.
3. Voucher specimen serves as a reference for comparing doubtful newly collected fresh specimens.
4. Voucher specimens play a role in studies like floristic diversity, environmental assessment, ecological mechanisms, and survey of unexplored areas.
5. Herbarium provides an opportunity for documenting biodiversity and Studies related to the field Of ecology and conservation biology.

Question 5.
Explain Bentham and the hooker system of Glassification.
Answer:

1. A widely followed natural system of classification Considered the best was proposed by two English botanists George Bentham (1800 – 1884) and Joseph Dalton Hooker (1817-1911).
2. The classification was published in a three-volume work as “Genera Plantation”
   (18624 883) describing 202 families and 7569 genera and 97,205 Species.
3. In this system, the seeded plants were classified into 3 major classes such as Dicotyledonae, Gymnospermae, and Monocotyledonae.

Question 6.
What do you know about karyotaxonomy?
Answer:

1. Chromosomes are the carriers of genetic information. Increased knowledge about the chromosomes has been used for extensive biosystematic studies and resolving many taxonomic problems.
2. Utilization of the characters and phenomena of cytology for the explanation of the taxonomic problems is known as cytotaxonomy or karyotaxonomy.
3. The characters of chromosomes such as number, size, morphology, and behavior during meiosis have proved to be of taxonomic value.

Question 7.
What is RFLP? Explain briefly.
Answer:
RFLPs are a molecular method of genetic analysis that allows the identification of taxa based on unique patterns of restriction sites in specific regions of DNA. It refers to differences between taxa in restriction sites and therefore the lengths of fragments of DNA following cleavage with restriction enzymes.

Question 8.
What is the significance of molecular taxonomy?
Answer:

1. It helps to identify a very large number of species of plants and animals by the use of conserved molecular sequences.
2. Using DNA data evolutionary patterns of biodiversity are now investigated.
3. DNA taxonomy plays a vital role in phytogeography, which ultimately helps in genome mapping and biodiversity conservation.
4. DNA- based molecular markers used for designing DNA-based molecular probes, have also been developed under the branch of molecular systematics.

Question 9.
Write briefly about the diagnostic features of Solanaceae family.
Answer:

1. Leaves alternate, exstipulate
2. Flowers actinomorphic, pentamerous
3. Calyx often persistence / accrescent
4. Stamens 5, epipetalous, poricidal in dehiscence
5. Carpels 2, ovary superior, 2 chambered, obliquely placed, falsely four-chambered placenta swollen, ovule numerous,
6. Fruits berry or capsule, vascular bundles with both outer and inner phloem (Bicollateral vascular bundle)

Question 10.
Write down the floral formula of the following plants.
Answer:

1. Datura metal,
2. Solanum nigrum,
3. Allium cepa.
   

Answer In detail.

Question 1.
What is the concept of Species? Explain briefly the various concepts.
Answer:
Species are the fundamental unit of taxonomic classification. The Greek philosopher Plato proposed the concept of “eidos” or species and believed that all objects are shadows of the “eidos”. According to Stebbins (1977) species is the basic unit of the evolutionary process. A species is a group of individual organisms which have the following characters.

1. A population of organisms that closely resemble each other more than the other population.
2. They descend from a common ancestor.
3. In sexually reproducing organisms, they interbreed freely in nature, producing fertile offspring.
4. In asexually reproducing organisms, they are identified by their morphological resemblance.
5. In the case of fossil organisms, they are identified by morphological and anatomical resemblance. Species concepts can be classified into two general groups. The concept emphasizing the process of evolution that maintains the species as a unit and that can result in evolutionary divergence and speciation. Another concept emphasizes the product of evolution in defining a species.

Question 2.
Describe the method of preparation of herbarium specimen.
Answer:

1. Plant Collection: Plant specimen with flower or fruit is collected.
2. Documentation of field site data: Certain data are to be recorded at the time of plant collection. It includes date, time, country, state, city, specific locality information, latitude, longitude, elevation, and landmark information. These data will be typed onto a herbarium label.
3. Preparation of plant specimen: Plant specimen collected from the field is pressed immediately with the help of a portable field plant press. Plant specimen is transferred to a standard plant press (12” x 18”) which between two outer 12” * 18” frames and secured by two straps.
4. Mounting herbarium specimen: The standard size of herbarium sheet is used for mounting the specimen (29cm x 41cm). specimens are affixed to herbarium sheet with standard white glue or solution of Methylcellulose.
5. Herbarium label: Herbarium label size is generally 4-5” wide and 2-3” tall. A typical label contains all information like habit, habitat, vegetation type, landmark information, latitude, longitude, image document, collection number, date of collection, and name of the collector.
6. Protection of herbarium sheets against mold and insects: Application of 2% Mercuric chloride, Naphthalene, DDT, carbon disulfide. Fumigation using formaldehyde. Presently deep freezing(-20°C) method is followed throughout the world.

Question 3.
Explain the basis of molecular taxonomy with its uses.
Answer:

1. Molecular Taxonomy is the branch of phytogeny that analyses hereditary molecular differences, mainly in DNA sequences, to gain information and to establish genetic relationships between the members of different taxonomic categories.
2. The advent of DNA cloning and sequencing methods has contributed immensely to the development of molecular taxonomy and population genetics over the years.
3. These modern methods have revolutionized the field of molecular taxonomy and population genetics with improved analytical power and precision.
4. The results of a molecular phylogenetic analysis are expressed in the form of a tree called a phylogenetic tree.
5. Different molecular markers like allozymes, mitochondrial DNA, microsatellites, RFLP (Restriction Fragment Length Polymorphism), RAPD (Random amplified polymorphic DNA), AFLPs (Amplified Fragment Length Polymorphism), single nucleotide polymorphism-SNP, microchips, or arrays are used in the analysis.

Uses-of molecular taxonomy:

1. Molecular taxonomy helps in establishing the relationship of different plant groups at the DNA level.
2. It unlocks the treasure chest of information on the evolutionary history of organisms.

Question 4.
Write briefly about Cladistic analysis.
Answer:

1. Cladistics is one of the primary methods of constructing phylogenies or evolutionary histories. Cladistics uses shared, derived characters to group organisms into clades.
2. These clades have at least one shared derived character found in their most recent common ancestor that is not found in other groups hence they are considered more closely related to each other.
3. These shared characters can be morphological such as, leaf, flower, Suit, seed, and so on; behavioral, like the opening of flowers nocturnal/diurnal; molecular like DNA or protein sequence, and more.
4. Cladistic accepts only monophyletic groups. Paraphyletic and polyphyletic taxa are occasionally considered when such taxa conveniently treated as one group for practical purposes.
   eg. dicots, Sterculiaceae. Polyphyletic groups are rejected by cladistic.

(a) Monophyletic group: Taxa comprising all the descendants of a common ancestor.

(b) Paraphyietic group: Taxon that includes an ancestor but not all of the descendants of that ancestor.

(c) Polyphyletic group: Taxa that includes members from two different lineages.

Need for Cladistics:

1. Cladistics is now the most commonly used and accepted method for creating a phylogenetic system of classifications.
2. Cladistics produces a hypothesis about the relationship of organisms to predict the morphological characteristics of organisms.
3. Cladistics helps to elucidate the mechanism of evolution.

Question 5.
Explain the general characters of the family Solanaceae.
Answer:

1. Distribution: Family Solanaceae includes about 88 genera and about 2650 species, of these Solanum, is the largest genus of the family with about 1500 species. Plants are worldwide in distribution but more abundant in South America.
2. Habit: Mostly annual herbs, shrubs, small trees (Solanum violaceum) lianas with prickles (Solanum trilobatum) present in some taxa, many with stellate trichomes; rarely vines (Lycium, Sinensis)
3. Root: Branched tap root system.
4. Stem: Herbaceous or woody; erect or twining, or creeping; sometimes modified into tubers (Solanum tuberosum) often with collateral vascular bundles.
5. Leaves: Alternate, simple, rarely pinnately compound (Solanum tuberosum and (Lycopersicon esculentum) exstipulate, opposite or sub-opposite in the upper part, unicostate reticulate venation.
6. Inflorescence; Generally axillary or terminal cymose (Solanum) or solitary flowers (Datura stramonium). Extra axillary scorpioid cyme called rhiphidium (Solanum nigrum) solitary and axillary (Datura and Nicotiana) umbellate cyme (Withania somnifera).
7. Flowers: Bracteate (Petunia), or ebracteate (Withania) pedicellate, bisexual, heterochlamydeous, actinomorphic, or weakly zygomorphic due to oblique position of ovary pentamerous, hypogynous. ,
8. Calyx: Sepals 5, synsepalous, valvate, persistent rarely the sepals are 4 or 6. Often enlarging to envelop the fruit (Physalis, Withania).
9. Corolla: Petals 5, sympetalous, rotate, tubular (Solanum) or bell-shaped (Atropa) or infundibuliform YPemwi’aj usually alternate with sepals; rarely bilipped and zygomorphic (Schizanthus) usually valvate, sometimes convolute (Datum).
10. Androecium: Stamens 5, epipetalous, filaments usually Unequal in length, stamens only 2 in Schizanthus, 4 and didynamous in (Salpiglossis) Anthers dithecous, dehisce longitudinally or poricidal. …
11. Gynoecium: Bicarpeliary, syncarpous obliquely placed, ovary superior, bilocular but looks tetralocular due to the formation of false septa, numerous ovules in each locule on axile placentation.
12. Fruit: A capsule or berry. In Lycopersicon esculentum, Capsicum, the fruit is a berry and in species of Datura and Petunia, the fruit is a capsule.
13. Seed: Endospermous.

Question 6.
Give the floral characters of Pisum sativum.
Answer:

Activity

Text Book Page No. 128

Write common names and scientific names of 10 different plants around your home.
Answer:

Text Book Page No,136

Prepare a herbarium of 5 common weed plants found inside your school campus /nearby garden/wasteland.
Answer:
Common weed plants: i) Tridax procumbence, (ii) Vinca rosea, (Hi) Lamium amplexicaula, . (iv) Poa annua, (v) Stellaria media.

Text Book Page No.160

Can you identify this?

(a) Name the family.
Answer:
Asparagales

(b) Write the binomial.
Answer:
Aloe vera.

(c) List the economic uses.
Answer:
(i) Cosmetic and medicine.
(ii) Soothing and moisturing agent.
(iii) Protection for human from sunburn.
(iv) Treatment of wounds and bums.
(v) Aloe vera gel is used commercially as an ingredient in yogurts.
(vi) As dietary supplement.
(vii) Skin treatment in Ayurvedic medicine.

Choose the correct answer.

1.18th International Botanical congress was held in 2011 at:
(a) London, U.K
(b) Melbourne, Australia
(c) Newyork, U.S.A
(d) Sydney, Australia
Answer:
(b) Melbourne, Australia

2. The lowest of classification is:
(a) Genus
(b) Kingdom
(c) species
(d) Family
Answer:
(c) species

3. Flora is the term used for:
(a) the document of all plant species.
(b) the document of single species in a given geographic area.
(c) the document of only endemic species of plants in a given area.
(d) none of the above.
Answer:
(a) the document of all plant species.

4. The first botanical garden was established by Theophrastus at:
(a) London
(b) Sydney
(c) Athens
(d) Singapore
Answer:
(c) Athens

5. Which is the largest Botanical garden in the world.
(a) Indian Botanical Garden, Kolkata, India
(b) Botanical Garden at Athens.
(c) National Botanical Garden, Lucknow, India
(d) Royal Botanical Garden Kew, England
Answer:
(d) Royal Botanical Garden Kew, England

6. Who was called as Father of Taxonomy?
(a) E.K. Janaki Ammal
(b) Carolus Linnaeus
(d) Theophrastus
(c) Heywood
Answer:
(b) Carolus Linnaeus

7. Plants with one stamen are grouped under:
(a) Tetrandria
(b) Diandria
(c) Monandria
(d) Pentandria
Answer:
(c) Monandria

8. The family coniferae is included under the class:
(a) Dicotyledonae
(b) Monocotyledonae
(c) Gymnospermae
(d) None of the above
Answer:
(c) Gymnospermae

9. Chemotaxomony is mainly based on:
(a) The chemical characters of the plant
(b) the morphological characters of the plant
(c) The phylogenetic characters of the plant
(b) none of the above
Answer:
(a) The chemical characters of the plant

10. The classification based on the characters of chromosome, such as number, size, morphology and behaviour during meiosis is known as:
(a) serotaxonomy
(b) Chemotaxonomy
(c) karyotaxonomy
(d) none of the above
Answer:
(c) karyotaxonomy

11. RAPD( Random amplified polymorphic DNA) is a method:
(a) to identify the morphomatic character of a plant
(b) to identify specific regions of DNA
(c) to identify genetic sequence of a plant
(d) to identify genetic markers using a randomly synthesised primer.
Answer:
(d) to identify genetic markers using a randomly synthesised primer.

12. Genetic sequence used to identify a plant is known as:
(a) DNA tags
(b) Polymorphic DNA
(c) Fragment of DNA
(d) Genome
Answer:
(a) DNA tags

13. Metformin used for treatment of diabetes is exacted from:
(a) Hibiscus rosasinensis
(b) Galega officinalis
(c) Arachis hypogea
(d) none of the above
Answer:
(b) Galega officinalis

14. Earlier classification emphasized on:
(a) Reproductive characters
(b) Vegetative characters
(c) both
(d) Anatomical characters
Answer:
(c) both

15. Naming a plant is called:
(a) systematic botany
(b) Taxonomy
(c) nomenclature
(d) Cytology
Answer:
(c) nomenclature

16. Bentham and Hooker’s classification is:
(a) Phylogenetic system of classification
(b) artificial system of classification
(c) natural system of classification
(d) sexual system of classification
Answer:
(c) natural system of classification

17. Carolus Linnaeus proposed classification based on:
(a) Artificial system
(b) natural system
(c) phylogenetic system
(d) modem system
Answer:
(a) Artificial system

18. The largest group of plant kingdom is
(a) Cryptogams
(b) Angiosperms
(c) Gymnosperms
(d) phanerogams
Answer:
(b) Angiosperms

19. Number of volumes in genera plantorum
(a) two
(b) three
(c) four
(d) five
Answer:
(b) three

20. The family gnetaceae is included under
(a) monochlamydae
(b) monotyledons
(c) dicotyledons
(d) Gymnosperms
Answer:
(d) Gymnosperms

21. Bentoam and Hooker are associated with
(a) Indian botanical garden
(b) Royal botanical garden
(c) American botanical garden
(d) French botanical garden
Answer:
(b) Royal botanical garden

22. The division of angiosperme is achieved by:
(a) floral characters
(b) anatomy
(c) physiology
(d) ecology
Answer:
(a) floral characters

23. Classification of plants into different groups is known as:
(a) Morphology
(b) Physiology
(c) Pathology
(d) Systematic botany
Answer:
(d) Systematic botany

24. Syngenesious anthers and epipetalous stamens are found in:
(a) Liliaceae
(b) malvaceae
(c) Solanaceae
(d) cruciferae.
Answer:
(d) cruciferae.

25. Synandrous condition is common in the family:
(a) Lumbelliferae
(b) Rosaceae
(c) malvaceae
(d) Cucurbitaceae
Answer:
(d) Cucurbitaceae

26. Polyadelphous condition is found in:
(a) Leguninaceae
(b) rutaceae
(c) compositae
(d) Liliaceae
Answer:
(b) rutaceae

27. Indefinite stamens are characteristic of the family:
(a) malvaceae
(b) graniceae
(c) labiatac
(d) cruciferae
Answer:
(a) malvaceae

28. Classical taxonomy is also termed as:
(a) B taxonomy
(b) systematics
(c) description and taxonomy
(d) experimental taxonomy
Answer:
(c) description and taxonomy

29. Match the following:

(a) (i)-(c), (ii)-(d), (iii)-(a), (iv)-(b)
(b) (i)-(d), (ii)-(c), (iii)-(a), (iv)-(b)
(c) (i)-(b), (ii)-(a), (iii)-(d), (iv)-(c)
(d) (i)-(b), (ii)-(c), (iii)-(a), (iv)-(d)
Answer:
(a) (i)-(c), (ii)-(d), (iii)-(a), (iv)-(b)

30. FAO declared the year for pulses as:
(a) 2017
(b) 2015
(c) 2018
(d) 2016
Answer:
(d) 2016

Students get through the TN Board 11th Bio Botany Important Questions Chapter 4 Reproductive Morphology which is useful for their exam preparation.

TN State Board 11th Bio Botany Important Questions Chapter 4 Reproductive Morphology

Answer the following short answers.

Question 1.
Define floriculture.
Answer:
Floriculture is a branch of Horticulture. It deals with the cultivation of flowers and ornamental crops.

Question 2.
What is axillary inflorescences?
Answer:
Inflorescence present in the axile of the nearest vegetative leaf.

Question 3.
What is meant by spadix inflorescences?
Answer:
An inflorescence with a fleshy or thickened central axis that possesses many unisexual sessile flowers in acropetal succession.

Question 4.
Define Capitulum.
Answer:
The capitulum is a determinate or indeterminate, group of sessile or subsessile flowers arising on a receptacle, often subten&d by an involucre.

Question 5.
What do you know about Helicoid?
Answer:
Axis develops on only one side and forms a coil structure at least at the earlier development stage. Example: Hamelia, potato.

Question 6.
Define Cymule.
Answer:
A small, simple dichasium is called a cymule

Question 7.
Explain briefly about coenanthium.
Answer:
Circular disc-like fleshy open receptacle that bears pistillate flowers at the center and staminate flowers at the periphery. Example: Dorstenia

Question 8.
Define bisexual flower.
Answer:
When a flower contains both androecium and gynoecium is called a perfect flower.

Question 9.
What is meant by polygamous plants?
Answer:
The condition in which bisexual and unisexual (staminate/pistillate) flowers occur in the same plaint is called polygamous.

Question 10.
Explain Actinomorphic flower.
Answer:
The flower shows two mirror images when cut in any plane or radius through the center.
Normally there are more than two planes of symmetry.

Question 11.
What do you know about merosity?
Answer:
A number of floral parts per whorl are called merosity. Perianth merosity is the number of perianth parts per whorl.

Question 12.
Describe briefly deciduous calyx.
Answer:
Calyx that falls after the opening of flower (anthesis) eg. Nelumbo.

Question 13.
What type of calyx is present in Ocimum?
Answer:
Two lipped calyces is present in Ocimum.

Question 14.
Define infundibuliform corolla.
Answer:
Petals fused to form funnel-shaped corolla. Tube gradually widens into limbs, eg. Datura, Ipomoea.

Question 15.
Mention three parts of the stamen.
Answer:

1. Filament,
2. Anther,
3. Connective.

Question 16.
Define the term “Connation”.
Answer:
Connation refers to the fusion of stamens among themselves.

Question 17.
Explain briefly about the Apocarpous ovary.
Answer:
A pistil contains two or more distinct carpels, eg. Annona

Question 18.
Describe Gynobasic style with an example.
Answer:
Arising from the base of the ovary, eg. Lamiaceae (Ocimum), characteristic of Boraginaceae.

Question 19.
What is meant by pomology?
Answer:
The branch of horticulture that deals with the study of fruits and their cultivation are called pomology.

Question 20.
Write down two important functions of seed.
Answer:

1. The seed encloses and protects the embryo for the next generation.
2. It contains food for the development of the embryo.

Answer In brief.

Question 1.
Distinguish between racemose and cymose inflorescence.
Answer:

Question 2.
Explain the term “Hypanthodium” with a suitable example.
Answer:
A receptacle is a hollow, globose structure consisting of unisexual flowers present on the inner wall of the receptacle. The receptacle is closed except for a small opening called the ostiole which is covered by a series of bracts. Male flowers are present nearer to the ostiole, female and neutral flowers are found in a mixed manner from the middle below, eg. Ficus sp. (Banyan and Pipal).

Question 3.
Describe briefly the three types of sympetalous zygomorphic corolla.
Answer:

1. Bilabiate: Corolla with two lips. eg. Ocimum, Leucas, Adhatoda. Tubular corolla with a single strap-shaped limb. eg. Ray floret of Helianthus.
2. Personate: Corolla made up of two lips with the upper arched and the lower protruding into the corolla throat, eg. Antirrhinum, Linaria.
3. Ligulate: Tubular corolla with a single strap-shaped limb, eg. Ray floret of Helianthus.

Question 4.
What is meant by anther dehiscence? Explain the different kinds of anther dehiscence.
Answer:
It refers to the opening of anther to disperse pollen grains.

1. Longitudinal: Anther dehisces along, a suture parallel to the long axis of each anther lobe, eg. Datura, china rose, cotton.
2. Transverse: Anther dehisces at right angles to the long axis of anther lobe. eg. Malvaceae.
3. Poricidal: Anther dehisces through pores at one end of the thecae, eg. Ericaceae, Solanum, potato, brinjal, Cassia.
4. Valvular: Anther dehisces through a pore covered by a flap of tissue, eg. Lauraceae, Cinnamomum.

Question 5.
Explain the different types of stigma with suitable examples.
Answer:
A stigma is a structure present at the tip of a pistil which receives the pollen grains,

1. Discoid: A disk-shaped stigma is called discoid.
2. Capitate: Stigma appearing like ahead, eg. Alchemilla.
3. Globose: Stigma is spherical in shape is called globose.
4. Plumose stigma: Stigma feathery which is unbranched or branched as in Asteraceae, Poaceae.

Question 6.
Describe the salient features of the floral diagram.
Answer:
A floral formula is a simple way to explain the salient features of a flower. The floral diagram is a representation of the cross-section of the flower, It represents floral whorls arranged as viewed from above. The floral diagram shows the number and arrangement of bract, bracteoles, and floral parts, fusion, overlapping, and placentation.

Question 7.
Distinguish between true fruit and false fruit.
Answer:

Question 8.
Mention any three functions of fruit.
Answer:

1. The edible part of the fruit is a source of food, energy for animals.
2. They are a source of many chemicals like sugar, pectin, organic acids, vitamins, and minerals.
3. The fruit protects the seeds from unfavorable climatic conditions and animals.

Question 9.
Write briefly about the types of multiple fruits for example.
Answer:
A Multiple or composite fruit develops from the whole inflorescence along with its peduncle on which they are borne.
Sorosis: A fleshy multiple fruits that develop from a spike or spadix. The flowers fused together by their succulent perianth and at the same time, the axis bearing them become fleshy or juicy and the whole inflorescence forms a compact mass. eg. Pineapple, Jack fruit, Mulberry.
Syconus: A multiple fruits that develops from hypanthodium inflorescence. The receptacle develops further and converts into fleshy fruit which encloses a number of true fruit or achenes which develop from female flower of hypanthodium inflorescence, eg. Ficus.

Question 10.
What are the types of seed, based on the presence and absence of the endosperm? Explain with a suitable example.
Answer:
Based on the presence or absence of the endosperm the seed is of two types.

1. Albuminous or Endospermous seed: The cotyledons are thin, membranous, and mature seeds have endosperm persistent and nourish the seedling during its early development, eg. Castor, sunflower, maize.
2. Ex-albuminous or non-endospermous seed: Food is utilized by the developing embryo and so the mature seeds are without endosperm; In such seeds, cotyledons store food and become thick and fleshy, eg. Pea, Groundnut.

Answer In detail.

Question 1.
Write an essay on cymose inflorescence.
Answer:
The central axis Stops growing and ends in a flower, further growth is by means of axillary buds. Old flowers present at the apex and young flowers at the base.

1. Simple cyme (solitary): Determinate inflorescence consists of a single flower. It may be terminal or axillary, eg. terminal in Trillium grandiflorum and axillary in Hibiscus.
2. Monochasial Cyme (uniparous): The main axis ends with a flower. From two lateral bracts, only one branch grows further. It may be helicoid (bostryx) or Scorpioid (cincinnus).
   (a) Helicoid: Axis develops on only one side and forms a coil structure at least at the earlier development stage, eg. Hamelia, potato.
   (b) Scorpioid: Axis develops on alternate sides and often becomes a coil structure, eg. Heliotropium.
3. Simple dichasium (Biparous): A central axis ends in a terminal flower; further growth is produced by two lateral buds. Each cymose unit consists of three flowers of which the central one is an old one. This is true cyme. eg. Jasminum.
4. Compound dichasium It has many flowers. A terminal old flower develops lateral simple dichasial cymes on both sides. Each compound dichasium consists of seven flowers, eg. Clerodendron. A small, simple dichasium is called a cymule.
5. Polychasial Cyme (multiparous): The central axis ends with a flower. The lateral axes branches repeatedly, eg. Nerium.
6. Sympodial Cyme: In monochasial cyme, successive axes at first develop in a zigzag manner, and later it develops into a straight pseudo axis. eg. Solanum americanum.

Question 2.
Explain the parts of a flower with a diagram.
Answer:
In a plant, which part would you like the most? Of course, it is a flower, because of its colour and fragrance. The flower is a significant diagnostic feature of angiosperms. It is a modified condensed reproductive shoot. The growth of the flower shoot is determined. There are two whorls, accessory and essential. Accessory whorl consists of calyx and corolla and the essential whorl comprises androecium and gynoecium. The flower is said to be complete when it contains all four whorls. An Incomplete flower is devoid of one or more whorls.

1. Pistil The female reproductive organ of a flower is Gynoecium or pistil. Each member is carpel.
2. Petal Innermost accessory whorl of the flower is corolla. Each member is called a petal.
3. Sepal: Outermost whorl of the flower is calyx. Each member is called sepal.
4. Perianth (perigonium) Undifferentiated calyx and corolla. Individual members are called tepal.
5. Bract Subtending leaf or leaf-like the structure of any flower is called Bract.
6. Stamen Male organ of a flower is the androecium. Each member is the stamen.
7. Thalamus (torus or receptacle) The part of the flower on which other floral parts are attached.
8. Bracteole: A smaller bract present on the side of the pedicel is called bracteole or bractlet. A whorl of bracteoles at the base of the calyx is called epicalyx.
9. Pedicel Stalk of the flower. The flower is pedicellate or sessile depending upon presence or absence. The flowers with a short, rudimentary pedicel are called subsessile flowers.

Question 3.
What is Aestivation? Explain different types of aestivation.
Answer:

1. Valvate Margins of sepals or petals do not overlap but just touch each other, eg. Calyx in members of Malvaceae, Calotropis, Annona.
2. Twisted or convolute or contorted One margin of each petal or sepal overlapping on the other petal, eg. Petals of china rose.
3. Imbricate: Sepals and petals irregularly overlap on each other; one member of the whorl is exterior, one interior, and the rest of the three having one margin exterior and the other interior, eg. Cassia, Delonix. There are three types: (i) Ascendingly imbricate, (ii) Quincuncial, (iii) Vexillary.
4. Quincuncial It is a type of imbricate aestivation in which two petals are external and two internal and one petal with one margin internal and the other margin external, eg. Guava, calyx of Ipomoea, Catharanthus.
5. Vexillary Large posterior petals both margins overlap lateral petals. Lateral petals other margin overlaps anterior petals, eg. Pea, bean.
   

Question 4.
Describe the difference between anthers in plants.
Answer:
Anther types:

1. Monothecal: One lobe with two microsporangia. They are kidney-shaped in a cross-section, eg. Malvaceae.
   Some other types: Hfcplostemonous stamens are uniseriate and equal in number to the petals and opposite the sepals (antisepalous).
   Obhaplostemonous: Stamens are uniseriate, a number equal to petals and opposite the petals (antipetalous)
   Diplostetnonous: Stamens are biseriate, outer antisepalous, inner antipetalous. eg. Murraya.
   Obdiplostemonous Stamens are biseriate, outer antipetalous, inner antisepalous. eg. Caryophyllaceae.
   Poiystemonou Numerous stamens are normally many more than the number of petals.
2. Dithecal: It is a typical type, having two lobes with four microsporangia. They are butterfly-shaped in cross-section, eg. Solanaceae.

Anther attachment:

1. Basifixed (Innate) Base of the anther is attached to the tip of filament, eg. Brassica, Datura.
2. Dorsifixed Apex of filament is attached to the dorsal side of the anther, eg. Citrus, Hibiscus.
3. Versatile: Filament is attached to the anther at the midpoint, eg. Grasses.
4. Adnate: Filament is continued from the base to the apex of anther, eg. Verbena, Ranunculus, Nelumbo.

Anther dehiscence: It refers to the opening of anther to disperse pollen grains.

1. Longitudinal: Anther dehisces along a suture parallel to long axis of each anther lobe, eg. Datura, china rose, cotton.
2. Transverse: Anther dehisces at right angles to the long axis of anther lobe. eg. Malvaceae.
3. Poricidal Anther dehisces through pores at one end of the thecae, eg. Ericaceae, Solarium, potato, brinjal, Cassia.
4. Valvular: Anther dehisces through a pore covered by a flap of tissue, eg. Lauraceae, Cinnamomum.

Anther dehiscing direction: It shows the position of anther opening relative to the anther of the flower.

1. Introrse: Anther dehisces towards the center of the flower, eg. Dianthus.
2. Extrorse: Anther dehisces towards the periphery of the flower, eg. Argemone.

Question 5.
Explain the floral diagram and floral formula of Hibiscus Rosa Sinensis.
Answer:

Question 6.
Write an essay on fleshy fruits and their kind.
Answer:
The fruits are derived from single pistils where the pericarp is fleshy, succulent, and differentiated
into epicarp, mesocarp, and endocarp. It is subdivided into the following.

1. Ben Fruit develops from bi carpellary or multi carpellary, syncarpous ovary. Here the epicarp is thin, the mesocarp and endocarp remain undifferentiated. They form a pulp in which the seeds are embedded, eg. Tomato, Date Palm, Grapes, Brinjal.
2. Drupe: Fruit develops from the monocarpellary, superior ovary. It is usually one-seeded. The pericarp is differentiated into outer skinny epicarp, fleshy and pulpy mesocarp, and hard and stony endocarp around the seed. eg. Mango, Coconut.
3. Pepo: Fruit develops from tri carpellary, inferior ovary. Pericarp terns leathery or woody which encloses, fleshy mesocarp and smooth endocarp. eg. Cucumber, Watermelon, Bottle gourd, Pumpkin.
4. Hesperidiur Fruit develops from the multi carpellary, multilocular, syncarpous, superior ovary. The fruit wall is differentiated into a leathery epicarp with oil glands, a middle fibrous mesocarp. The endocarp forms distinct chambers, containing juicy hairs, eg. Orange, Lemon. ,
5. Pome: it develops from multi carpellary, syncarpous, inferior ovary. The receptacle also develops along with the die ovary and becomes fleshy, enclosing the true fruit. In pome the epicarp is thin skin-like and the endocarp is cartilaginous. eg. Apple, Pear.
6. Balausta: A fleshy indehiscent fruit developing from multi carpellary, multilocular inferior ovary whose pericarp is tough and leathery. Seeds are attached irregularly with testa being the edible portion, eg. Pomegranate.

Activity

Text Book Page No. 121

Prepare a diet chart to provide a balanced diet to an adolescent (a school going child) which includes food items (fruits, vegetable, and seeds) which are non – expensive and are commonly available.
Answer:
Diet chart for school going child

Choose the correct answer.

1. Axillary inflorescence is present in:
(a) Nerium oleander
(b) Theobroma cocoa
(c) Hibiscus rosa sinensis
(d) Couropita guianensis
Answer:
(c) Hibiscus rosa sinensis

2. An unbranched indeterminate inflorescence with sessile flowers is categorized as:
(a) spikelet
(b) spike
(c) simple raceme
(d) none of the above
Answer:
(b) spike

3. Umbeltypeofinflorescenceisseenin:
(a) Allium cepa
(b) Caesal pinia
(c) Cauliflower
(d) Opuntia
Answer:
(a) Allium cepa

4. Cyathium inflorescence consists of:
(a) Small bisexual flowers
(b) small unisexual flower
(c) small anthers
(d) none of the above
Answer:
(b) small unisexual flower

5. Which of the following is a monoecious plant?
(a) Musa
(b) Coconut
(c) Magnifera
(d) Papaya
Answer:
(b) Coconut

6. A piant with both male flowers and bisexual flower is termed as:
(a) androdioecious
(b) gynodioecious
(c) andromonoecious
(d) trioecious
Answer:
(c) andromonoecious

7. Flowers that lack any plane of symmetry and cannot be divided into equal halves in any plane are seen in:
(a) bean
(b) Datura
(c) Cassia
(d) Canna indica
Answer:
(d) Canna indica

8. A flower which is composed of distinct outer calyx and the inner corolla is termed as:
(a) Homochlamydous
(b) dichlamydeous
(c) achlamydeous
(d) none of the above
Answer:
(b) dichlamydeous

9. Match the following:

(a) (i)-(d), (ii)-(b), (iii)-(a), (iv)-(c)
(b) (i)-(b), (ii)-(c), (iii)-(d), (iv)-(a)
(c) (i)-(b), (ii)-(a),(iii)-(d), (iv)-(c)
(d) (i)-(b), (ii)-(d), (iii)-(a), (iv)-(c)
Answer:
(d) (i)-(b), (ii)-(d), (iii)-(a), (iv)-(c)

10. Synsepalous condition is present in the flowers of:
(a) Annona
(b) Papaya
(c) Datura
(d) palmyra
Answer:
(c) Datura

11. Crciform type of corolla is present in:
(a) radish
(b) dianthus
(c) tea
(d) rose
Answer:
(a) radish

12. Flowerpetals fused to form a bell-shaped corolla are termed as:
(a) tubular
(b) Rotate
(c) Infundibuliform
(d) campanulate
Answer:
(d) campanulate

13. Corolla with two lips is present in:
(a) adhatoda
(b) linaria
(c) helianthus
(d) allium
Answer:
(a) adhatoda

14. Aestivation is the term used for:
(a) arrangement of flowers in the inflorescence
(b) arrangement of leaves in the stem.
(c) arrangement of sepals and petals in the flower bud
(d) none of the above
Answer:
(c) arrangement of sepals and petals in the flower bud

15. Match the following:

(a) (i)-(d), (ii)-(b), (iii)-(a), (iv)-(c)
(b) (i)-(b), (ii)-(c), (iii)-(d), (iv)-(a)
(c) (i)-(b), (ii)-(a), (iii)-(d), (iv)-(c)
(d) (i)-(b),(ii)-(c), (iii)-(a), (iv)-(d)
Answer:
(b) (i)-(b), (ii)-(c), (iii)-(d), (iv)-(a)

16. Didynamous condition in which two stamens with long filaments and two with short filaments and two with short filaments, is seen in:
(a) Ipomoea
(b) Ocimum
(c) Mimosa
(d) Datura
Answer:
(a) Ipomoea

17. Innate condition of anther attachment is named for the condition:
(a) the base of anther is attached to the tip of the filament
(b) the apex of the filament is attached to the dorsal side of the anther.
(c) the filament is continued from the base to the apex of anther
(d) none of the above
Answer:
(a) the base of anther is attached to the tip of the filament

18. Poricidal type of anther dehiscence is present in:
(a) Cinnamomum
(b) Nelumbo
(c) Brinjal
(d) Datura
Answer:
(c) Brinjal

19. A condition in which the gynoecium has four carpels is termed as:
(a) bicarpellary
(b) multicarpellary
(c) unicarpellary
(d) tetracarpellary
Answer:
(d) tetracarpellary

20. Match the following.

(a) (i)-(d), (ii)-(c), (Hi)-(a), (iv)-(b)
(b) (i)-(b), (ii)-(c), (iii)-(d), (iv)-(a)
(c) (i)-(b), (ii)-(a), (iii)-(d), (iv)-(c)
(d) (i)-(b), (ii)-(c), (iii)-(a), (iv)-(d)
Answer:
(a) (i)-(d), (ii)-(c), (Hi)-(a), (iv)-(b)

21. In Malvaceae family the sepals, petals, and stamens are attached at the base of a superior ovary. This condition is known as:
(a) Perigynous
(b) Epigynous
(c) Epiperigynous
(d) Hypogynous
Answer:
(d) Hypogynous

22. Axile placentation is present in:
(a) Hibiscus
(b) mustard
(c) marigold
(d) cucumber
Answer:
(a) Hibiscus

23. Drupe fruit develops from:
(a) tricatpellary inferior ovary
(b) monocarpellary, superior ovary
(c) Bicarpellery ovary
(d) bicarpellary, syncarpous ovary
Answer:
(b) monocarpellary, superior ovary

24. Multiple or composite fruit develops from:
(a) the whole flower
(b) Single flower
(c) the whole inflorescence
(d) none of the above
Answer:
(c) the whole inflorescence

25. Match the following.

(a) (i)-(d), (ii)-(b), (iii)-(a), (iv)-(c)
(b) (i)-(d), (ii)-(c), (iii)-(a), (iv)-(b)
(c) (i)-(b), (ii)-(a), (iii)-(d), (iv)-(c)
(d) (i)-(b), (ii)-(c), (iii)-(a), (iv)-(d)
Answer:
(c) (i)-(b), (ii)-(a), (iii)-(d), (iv)-(c)

26. In an inflorescence where flowers are borne laterally in an accropeted succession, the position of the youngest floral bud shall be:
(a) proximal
(b) Distal
(c) Intercalary
(d) Anywhere
Answer:
(a) proximal

27. Edible part of jack fruit is:
(a) whole fruit
(b) mesocarp
(c) perianth and seeds
(d) perianth and rachis
Answer:
(c) perianth and seeds

28. Non-endospermous seeds are present in:
(a) Groundnut
(b) Maize
(c) sunflower
(d) castor
Answer:
(a) Groundnut

29. Indicate the correct statement.
(a) Edible part of the fruit is poisonous to animals
(b) Edible part of the fruit is a source of food energy for animals.
(c) Edible part of the fruit is the exclusive source of medicine.
(d) Edible part of the fruit is the only source of fodder.
Answer:
(b) Edible part of the fruit is a source of food energy for animals.

30. In angiosperms embryo seed represents:
(a) female gametophyte
(b) male gametophyte
(c) sporophyte
(d) none of the above
Answer:
(a) female gametophyte

Students get through the TN Board 11th Bio Botany Important Questions Chapter 3 Vegetative Morphology which is useful for their exam preparation.

TN State Board 11th Bio Botany Important Questions Chapter 3 Vegetative Morphology

Answer the following short answers.

Question 1.
Define reproductive morphology.
Answer:
It includes Flower/inflorescence, Fruit, and Seed.

Question 2.
Write down any two primary functions of the stem.
Answer:

1. Provides support and bears leaves, flowers, and fruits.
2. It transports water and mineral nutrients to the other parts from the root.

Question 3.
Mention three important components of vegetative morphology.
Answer:

1. Habit,
2. Habitat and
3. Life span.

Question 4.
What are herbs? Give two examples.
Answer:
Herbs are soft-stemmed plants with less wood or no wood. Phyllanthus amarus, Cleome viscosa.

Question 5.
What is the two plant habitat?
Answer:
Depending upon where plants grow habitats may be classified into major categories:

1. Terrestrial and
2. Aquatic.

Question 6.
Define perennial plants give two examples.
Answer:
A plant that grows for many years that flowers and set fruits for several seasons during the life span.

Question 7.
What is meant by Angiosperms?
Answer:
Flowering plants are called “Angiosperms” or Magnoliophyta.

Question 8.
Give two primary functions of the root system.
Answer:

1. Absorb water and minerals from the soil.
2. Help to anchor the plant firmly in the soil.

Question 9.
Explain fusiform root with one example.
Answer:
These roots are swollen in the middle and tapering towards both ends. eg. Raphanus sativus.

Question 10
What is meant by thorn climbers? Give one example.
Answer:
Climbing or reclining on the support with the help of thorns as in Bougainvillea and Carissa.

Question 11.
Explain Phylloclade.
Answer:
Phylloclade is a characteristic adaptation of xerophytes where the leaves often fall off early and modified into spines or scales to reduce transpiration.

Question 12.
Define Bulb.
Answer:
It is a condensed conical or convex stem surrounded by fleshy scale leaves.

Question 13.
What is meant by Cladode?
Answer:
Cladode is a flattened or cylindrical stem similar to Phylloclade but with one or two internodes only.

Question 14.
Define Venation.
Answer:
The arrangement of veins and veinlets on the leaf blade or lamina is called venation. Internally, the vein contains vascular tissues.

Question 15.
What is meant by Phyllotaxy?
Answer:
The mode of arrangement of leaves on the stem is known as phyllotaxy (Gk. Phyllon = leaf; taxis = arrangement).

Question 16.
Give two examples for parallel venation.
Answer:

1. Canna,
2. Bamboo,

Question 17.
Explain opposite phyllotaxy.
Answer:
In this type, each node possesses two leaves opposite to each other.

Question 18.
What is meant by phyllode?
Answer:
Phyllodes are flat, green-colored leaf-like modifications of petioles or rachis. The leaflets or lamina of the leaf are highly reduced or caducous.

Question 19.
Define Heterophylly.
Answer:
The occurrence of two different kinds of leaves in the same plant is called heterophylly.

Question 20.
Give two examples of evergreen plants.
Answer:
Mimiisops, Calophyllum.

Answer In brief.

Question 1.
What is plant morphology? Explain the types of morphology.
Answer:
The study of various external features of the organism is known as morphology. Plant morphology is also known as external morphology deals with the study of the shape, size, and structure of plants and their parts (roots, stems, leaves, flowers, fruits, and seeds). The study of morphology is important in taxonomy. Morphological features are important in determining the productivity of crops. Morphological characters indicate the specific habitats of living as well as the fossil plants and help to correlate the distribution in space and time of fossil plants. Morphological features are also significant for phylogeny.
Plant Morphology can be studied under two broad categories:

1. Vegetative morphology – It includes the shoot system and root system.
2. Reproductive morphology – It includes Flower/inflorescence, Fruit, and Seed.

Question 2.
Explain the adventitious root system with an example.
Answer:

1. Root developing from any part of the plant other than the radicle is called adventitious root. It may develop from the base of the stem or nodes or internodes. eg. Monstera deliciosa, Ficus benghalensis, Piper nigrum.
2. In most of monocots, the primary root of the seedling is short-lived and lateral roots arise from various regions of the plant body. These are a bunch of thread-like roots equal in size which are collectively called fibrous root systems generally found in grasses, eg. Oryza sativa, Eleusine coracana, Pennisetum americanum.

Question 3.
What are the storage roots? Explain each type with a suitable example.
Answer:
Tuberous root: These roots are swollen without any definite shape. Tuberous roots are produced singly and not in clusters, eg. Ipomoea batatas.
Fasciculated root: These roots are in the cluster from the base of the stem. eg. Dahlia, Asparagus, Ru&ltia.
Nodulose root: In this type of root swelling occurs only near the tips. eg. Mtmnfct (arrow root) Curcuma amada (mango ginger), Curcuma loHga (turmeric).
Monilifonn or Beaded root: These roots swell at frequent intervals giving them a beaded appearance, eg. Vitis, Portulaca, Momordica Indian spinach.
Annulatedroot: These roots have a series of ring-like swelling on their surface at regular intervals. eg. Ipecac (Psychotria)

Question 4.
List out the characteristic features of this stem.
Answer:

1. The stem is usually the aerial portion of the plant
2. It is positively phototropic and negatively geotropic.
3. It has nodes and internodes.
4. The stem bears vegetative bud for vegetative growth of the plant, and floral buds for reproduction, and ends in a terminal bud.
5. The young stem is green and thus carries out photosynthesis.
6. During reproductive growth, the stem bears flowers and fruits.
7. Branches arise exogenously.
8. Some stems bear multicellular hairs of different kinds.

Question 5.
What is the secondary function of the stem?
Answer:

1. Food storage – eg. Solanum tuberosum, Colocasia, and Zingiber officinale.
2. Perennation / reproduction – eg. Zingiber officinale, Curcuma longa.
3. Water storage – eg. Opuntia.
4. Buoyancy – eg. Neptunia.
5. Photosynthesis – eg. Opuntia, Ruscus, Casuarina, Euphorbia, Caralluma.
6. Protection – eg. Citrus, Duranta, Bougainvillea, Acacia, Fluggea, Carissa.
7. Support – eg. Passiflora,Bougainvillea, Vitis, Cissus quadrangularis.

Question 6.
What are Bulbils? Explain different types with suitable examples.
Answer:

1. Bulbils are modified and enlarged buds, meant for propagation.
2. When bulbils detach from the parent plant and fall on the ground, they germinate into new plants and serve as a means of vegetative propagation.
3. In Agave and Allium, proliferum floral buds get modified into bulbils.
4. In Lilium bulbiferum and Dioscorea bulbifera, the bulbils develop in the axil of leaves.
5. In Oxalis, they develop just above the swollen root.

Question 7.
What is meant by Rhizome? Give at least three examples.
Answer:
This is an underground stem that grows horizontally with several lateral growing tips. Rhizome possess conspicuous nodes and internodes covered by scale leaves, eg. Zingiber officinale, Canna, Curcuma longa, Maranta arundinacea, Nymphaea, Nelumbo.

Question 8.
List out the primary function of the leaf.
Answer:
Primary functions:

1. Photosynthesis,
2. Transpiration,
3. Gaseous exchange,
4. Protection of buds,
5. Conduction of water and dissolved solutes.

Question 9.
Describe Palmately reticulate venation. Mention its types for example.
Answer:
Palmately reticulate venation (multicostate): In this type of venation there are two or more principal veins arising from a single point and they proceed outward or upwards. The two types of palmate reticulate venation are

1. Divergent type: When all principal veins originate from the base and diverge from one another towards the margin of the leaf as in Cucurbita, Luffa, Carica papaya, etc.
2. Convergent: When the veins converge to the apex of the leaf, as in Indian plum (Zizyphus),
   bay leaf (Cinnamomum)
   
   (a) Pinnately reticulate, (b) Palmately reticulate (Divergent), (c) Palmately reticulate (Convergent)

Question 10.
Give an account of storage leaves with suitable examples.
Answer:
Some plants of saline and xerophytic habitats and members of the family Crassulaceae commonly have fleshy or swollen leaves. These succulent leaves store water, mucilage or food material. Such storage leaves resist desiccation, eg. Aloe, Agave, Bryophyllum, Kalanchoe, Sedum, Sueada, Brassica oleracea (cabbage-variety capitata).

Answer In detail.

Question 1.
Describe the regions of the root with a suitable diagram.
Answer:

The root tip is covered by dome-shaped parenchymatous cells called root caps. It protects the meristematic cells in the apex. In Pandanus multiple root cap is present. In Pistia instead of a root cap, a root pocket is present. A few millimeters above the root cap the following three distinct zones have been classified based on their meristematic activity.

1. Meristematic Zone,
2. Zone of Elongation,
3. Zone of Maturation.

Question 2.
What are the vital functions of root? Explain each function with a suitable example.
Answer:
Epiphytic or velamen root: Some epiphytic orchids develop a special kind of aerial roots which hang freely in the air. These roots develop a spongy tissue called velamen which helps in the absorption of moisture from the surrounding air. eg. Vanda, Dendrobiutn, Aerides.
Foliar root: Roots are produced from the veins or lamina of the leaf for the formation of the new plant, eg. Bryophyllum, Begonia, Zamioculcas.
Sucking or Haustorial roots: These roots are found in parasitic plants. Parasites develop adventitious roots from the stem which penetrate into the tissue of the host plant and suck nutrients, eg. Ciiscuta (dodder), Cassytha, Orobanche (broomrape), Viscum (mistletoe), Dendrophthoe.
Photosynthetic or assimilatory roots: Roots of some climbing or epiphytic plants develop chlorophyll and turn green which help in photosynthesis, eg. Tinospora, Trapa natans (water chestnut), Taeniophyllum.

Question 3.
Explain different types of stem.
Answer:
The majority of angiosperm possess upright, vertically growing erect stem. They are (i) Excurrent, (ii) Decurrent, (iii) Caudex, (iv) Culm

1. Excurrent: The main axis shows continuous growth and the lateral branches gradually becoming shorter towards the apex which gives a conical appearance to the trees, eg. Polyalthia longifolia, Casuarina.
2. Decurrent: The growth of the lateral branch is more vigorous than that of the main axis. The tree has a rounded or spreading appearance, eg. Mangifera indica, Azadirachta indica, Tamarindus indicus, Aegle marmelos.
3. Caudex: It’s an unbranched, stout, cylindrical stem, marked with scars of fallen leaves, eg. Cocus Nucifera, Borassusflabellifermis, Areca catechu.
4. Culm: Erect stems with distinct nodes and usually hollow internodes clasped by leaf sheaths.
   eg. Majority of grasses including Bamboo.

Question 4.
Describe the various sub-aerial stem modifications with suitable examples.
Answer:
Subaerial stem found in plants with the weak stem in which branches lie horizontally on the ground. These are meant for vegetative propagation. They may be subaerial or partially subterranean.
Runner: This is a slender, prostrate branch creeping on die ground and rooting at the nodes, eg. Centella (Indian pennywort), Oxalis (wood sorrel), lawn grass (Cynodon dactylon).
Stolon: This is also a slender, lateral branch originating from the base of the stem. But it first grows obliquely above the ground, produces a loop and bends down towards the ground. When touches the ground it produces roots and becomes an independent plantlet. eg. Mentha piperita (peppermint), Fragaria indica (wild strawberry).
Sucker: Sucker develops from an underground stem and grows obliquely upwards and gives rise to a separate plantlet or new plant, eg. Chrysanthemum, Musa, Bambusa.
Offset: Offset is similar to runner but found in aquatic plants especially in rosette leaved forms.
A short thick lateral branch arises from the lower axil and grows horizontally leafless for a short distance, then it produces a bunch of rosette leaves and roots at nodes, eg. Eiehhornia (water hyacinth), Pistia (water lettuce).

Question 5.
Explain the parts of the leaf with a suitable diagram.
Answer:
Three main parts of a typical leaf are:
(i) Leaf base (Hypopodium), (ii) Petiole (Mesopodium), (iii) Lamina (Epipodium).

1. The part of the leaf attached to the node of the stem is called the leaf base. Usually, it protects growing buds at its axil. In legumes, the leaf base becomes broad, thick, and swollen which is known as a pulvinus. eg. Clitoria, Lablab, Cassia, Erythrina, Butea, Peltophorum.
   In many monocot families such as Arecaceae, Musaceae, Zingiberaceae, and Poaceae the leaf base extends, into a sheath and clasps part or whole of the internode. Such leaf base also leaves permanent scars on the stem when they fall. eg. Arecaceae.
   
2. Petiole (stipe or mesopodium): Tt is the bridge between leaf and stem. Petiole or leaf stalk is a cylindrical or subcylindrical or flattened structure of a leaf that joins the lamina with the stem. A leaf with petiole is said to be petiolate. eg. Ficus, Hibiscus, Mangifera, Psidium. Leaves that do not possess petiole is said to be sessile, eg. Calotropis, Gloriosa.
3. The expanded flat green portion of the leaf is the blade or lamina. It is the seat of photosynthesis, gaseous exchange, transpiration, and most of the metabolic reactions of the plant. The lamina is traversed by the midrib from which arise numerous lateral veins and thin veinlets. The lamina shows great variations in its shape, margin, surface, texture, color, venation, and incision.
4. In most of the dicotyledonous plants, the leaf base bears one or two lateral appendages called the stipules. Leaves with stipules are called stipulate. The leaves without stipules are called exstipulate or estipulate. The stipules are commonly found in dicotyledons. In some grasses (Monocots) an additional outgrowth is present between leaf base and lamina. It is called Ligule. Sometimes, small stipule-like outgrowths are found at the base of leaflets of a compound leaf. They are called stipels. The main function of the stipule is to protect the leaf in the bud condition.

Question 6.
What are the types of pinnately compound leaves? Explain each type with, suitable example.
Answer:
A pinnately compound leaf is defined as one in which the rachis, bears laterally a number of leaflets, arranged alternately or in an opposite manner, as in tamarind, Cassia.

1. Unipinnate: The rachis is simple and unbranched which bears leaflets directly on its sides in alternate or opposite manner, eg. Rose, Neem. Unipinnate leaves are of two types.
   When the leaflets are even in number, the leaf is said to be paripinnate. eg. Tamarind, Abrus, Sesbania, Saraca, Cassia.
   When the leaflets are odd in number, 1he leaf is said to be imparipinnate. eg. Rose, Neem \ (Azadirachta), Chinese box (Murraya).
2. Bipinnate: The primary rachis produces secondary rachis which bears the leaflets. The secondary rachii are known as pinnae. Number of pinnae varies depending on the species, eg. Delonix, Mimosa, Acacia nilotica, Caesalpinia.
3. Tripinnate: When the rachis branches thrice the leaf is called tripinnate. (i.e) the secondary rachii produce the tertiary rachii which bear the leaflets, eg. Moringa, Oroxylum.
4. Decompound: When the rachis of the leaf is branched several times it is called decompound, eg. Daucus carota, Coriandrum sativum, Foeniculum vulgare.

Choose the correct answer.

1. Vegetative morphology of plant includes:
(a) Shoot system, root system, and inflorescence
(b) the root system, flower, and seed
(c) shoot system and root system
(d) flower, fruit, and seed.
Answer:
(c) shoot system and root system

2. Phyllanthus amarus belongs to the group:
(a) Slmibs
(b) Herbs
(c) Climbers
(d) Trees
Answer:
(b) Herbs

3. The root system of the plant is generally:
(a) positively geotropic and negatively phototrophic in nature
(b) negatively geotropic and positively phototrophic in nature
(c) Positively geotropic and negatively phototrophic in nature
(d) negatively geotropic and negatively phototrophic in nature
Answer:
(c) Positively geotropic and negatively phototrophic in nature

4. Otym Sativa has:
(a) Tab toot system
(b) fibrous root system
(c) Adventitious and tap root system
(d) taproot with the secondary root system
Answer:
(b) fibrous root system

5. Match the following:

(a) (i)-(d), (ii)-(b), (iii)-(a), (iv)-(c)
(b) (i)-(d), (ii)-(c), (iii)-(a), (iv)-(b)
(c) (i)-(b), (ii)-(a), (iii)-(d), (iv)-(c)
(d) (i)-(b), (ii)-(c), (iii)-(a), (iv)-(d)
Answer:
(b) (i)-(d), (ii)-(c), (iii)-(a), (iv)-(b)

6. Foliar root is present in:
(a) randa
(b) Bryophyllum
(c) Delonix regia
(d) piper betel
Answer:
(b) Bryophyllum

7. Match the following:

(a) (i)-(d), (ii)-(b), (iii)-(a), (iv)-(c)
(b) (i)-(d), (ii)-(c), (iii)-(b), (iv)-(a)
(c) (i)-(b), (ii)-(a), (iii)-(d), (iv)-(c)
(d) (i)-(c), (ii)-(a), (iii)-(d), (iv)-(b)
Answer:
(d) (i)-(c), (ii)-(a), (iii)-(d), (iv)-(b)

8. cladode is present in:
(a) Bambusa
(b) Musa
(c) Asparagus
(d) Citrus
Answer:
(c) Asparagus

9. Rhizome is the modification of:
(a) Stem
(b) Root
(c) Undergroimd stem
(d) Undergroimd bulb
Answer:
(c) Undergroimd stem

10. Petiole is present in:
(a) Calotropis
(b) Hibiscus
(c) Gloriosa
(d) None of the above
Answer:
(b) Hibiscus

11. Ficus religiosa has:
(a) Pinnately parallel venation
(b) Palmately reticulate venation
(c) Multicostate venation
(d) Pinnately reticulate venation
Answer:
(d) Pinnately reticulate venation

12. Pinnately compound leaf is present in:
(a) Cassia
(b) Cucurbita
(c) begonia
(d) acalypha
Answer:
(a) Cassia

13. The part of the root which is most active in water absorption is called:
(a) root cap
(b) maturation zone
(c) meristematic zone
(d) zone of elongation
Answer:
(b) maturation zone

14. Venation is a term used to describe the pattern of arrangement of:
(a) floral organs
(b) veins and veinlets in a lamina
(c) flower in inflorescence
(d) all of them
Answer:
(b) veins and veinlets in a lamina

15. Epiphytic roots are found in;
(a) Indian rubber
(b) Orchid
(c) Tinospora
(d) Cuscuta
Answer:
(b) Orchid

16. Potatoes are borne on:
(a) Primary roots
(b) lateral roots
(cl Adventitious roots
(d) axil of scaly leaves
Answer:
(d) axil of scaly leaves

17. Winged petiole is found in:
(a) citrus
(b) radish
(c) acacia
(d) peepal
Answer:
(a) citrus

18. Fibrous root in ficus benghalensis develop from:
(a) stem
(b) node
(c) intemode
(d) none of the above
Answer:
(a) stem

19. Foliar roots are present in:
(a) vanda
(b) Bombax
(c) Bryophyllum
(d) Ficus pumila
Answer:
(c) Bryophyllum

20. Which one of the following is not a characteristic of the root.
(a) presence of root cap
(b) presence of chlorophyll
(c) absence of buds
(d) presence of unicellular hair
Answer:
(b) presence of chlorophyll

21. ………… are the vegetative organs of the flowering plant.
(a) Leaves, stem, fruits
(b) Roots, stem, flowers
(d) Roots, stem, leaves.
(c) Roots, leaves, flowers
Answer:
(d) Roots, stem, leaves.

22. Which is not a stem modification?
(a) Rhizome of ginger
(b) Corn of colocasia
(c) Pitcher of nepenthes
(d) Tuber of potato
Answer:
(c) Pitcher of nepenthes

23. In One of the following stem performs the function of storage and propagation,
(a) Wheat
(b) Ginger
(c) Radish
(d) Paddy
Answer:
(b) Ginger

24. An underground specialised sheet with a reduced disc-like stem covered by fleshy leaves is:
(a) Rhizome
(b) Rhizosphere
(c) Bulb
(d) Bulbil
Answer:
(c) Bulb

25. A phyllode i

s modified:
(a) leaf
(b) Stem
(c) Root
(d) Branch
Answer:
(a) leaf

26. A fibrous root system is better adapted than a tap root system for:
(a) Storage food
(b) Anchorage plant to soil
(c) absorption of water and organic food
(d) Transport of water and organic food
Answer:
(b) Anchorage plant to soil

27. Arrangement of leaves on a stem is called:
(a) Venation
(b) Vernation
(c) Phyllotaxy
(d) Axis
Answer:
(c) Phyllotaxy

28. The pitcher in nepenthes is a modification of:
(a) Stem
(b) root
(c) branch
(d) leaf
Answer:
(d) leaf

29. Leaf spines are present in:
(a) bombax
(b) asparagus
(c) mango
(d) citrus
Answer:
(b) asparagus

30. Heterophylly is found in:
(a) Limnophila heterophylla
(b) Calophyllum
(c) Erythrina
(d) Cabbage
Answer:
(a) Limnophila heterophylla

Students get through the TN Board 11th Bio Botany Important Questions Chapter 2 Plant Kingdom which is useful for their exam preparation.

TN State Board 11th Bio Botany Important Questions Chapter 2 Plant Kingdom

Answer the following short answers.

Question 1.
Define photosynthesis.
Answer:
Plants are unique living entities as they are endowed with the power to harvest the light energy from the sun and to convert it to chemical energy in the form of food through the astounding reaction, photosynthesis.

Question 2.
What is an alternation of generation?
Answer:
Alternation of generation is common in all plants. Alternation of the haploid gametophytic phase (n) with the diploid sporophytic phase (2n) during the life cycle is called alternation of generation.

Question 3.
What do you know about Halophyte algae?
Answer:
Dunaliella salina grows in salt pans (Halophytic algae), which can tolerate high concentration of salt content.

Question 4.
Mention any two vegetative reproduction in algae.
Answer:
Vegetative reproduction includes fission in unicellular forms the cell divides mitotically to produce two daughter cells, eg. Chlamydomonas.
Fragmentation: fragments of parent thallus grow into new individual, eg. Uloihrix.

Question 5.
Describe the thallus of red algae.
Answer:
The thallus is multicellular, macroscopic and diverse in form.

Question 6.
What is Garpogohium?
Answer:
The female sex organ of red algae is called carpogonium.

Question 7.
Mention any two characteristic features of bryophytes.
Answer:

1. The plant body of bryophyte is gametophyte and is not differentiated into root, stem and leaf-like structure.
2. Most of them are primitive land dwellers. Some of them are aquatic (Riella, Ricciocarpus).

Question 8.
Mention any three uses of Pteridophyte.
Answer:
Cut flower arrangements, Food, Biofertilizer.

Question 9.
Define the term Solenostele.
Answer:
The stele is perforated at a place or places corresponding to the origin of the leaf trace.

Question 10.
What is meant by Amber?
Answer:
Amber is a plant secretion that is an efficient preservative that doesn’t get degraded and hence can preserve remains of extinct life forms. The amber is produced by Pinites succinifera, a Gymnosperm.

Question 11.
Why do you call some plant fossil plants?
Answer:
The term ’form genera’ is used to name the fossil plants because the whole plant is not recovered as fossils instead organs or parts of the extinct plants are obtained in fragments.

Question 12.
What are the three classes of Gymnosperms?
Answer:

1. Cycadopsida,
2. Coniferopsida,
3. Gnetopsida

Question 13.
Define the term Gymnosperm.
Answer:
Gymnosperms (Gr. Gymnos = naked; Sperma = seed) are naked seed-producing plants. They were dominant in the Jurassic and cretaceous periods of the Mesozoic era.

Question 14.
What are the non-vascular cryptogams?
Answer:
Vascular tissue like xylem and phloem are completely absent, hence called ‘Non-vascular cryptogams’.

Question 15.
Mention any two fossil gymnosperms.
Answer:
Medullosa and Lepido-carpon.

Question 16.
List out morphological features of the dicot plant.
Answer:

1. Reticulate venation is present in the leaves.
2. Presence of two cotyledons in the seed.
3. Primary root radicle persists as the taproot.
4. Flowers tetramerous or Pentamerous.
5. Tricolpate (3 furrow) pollen is present.

Question 17.
What do you know about pyrenoids?
Answer:
Storage bodies called pyrenoids are present in the chloroplast and store starch.

Question 18.
What are Epiphytic algae?
Answer:
A few algae grow on the surface of aquatic plants and are called epiphytic algae (Coleochaete, and Rhodymenia).

Question 19.
What are Bryophytes?
Answer:
Bryophytes are the simplest land inhabiting cryptogams and are restricted to moist, shady habitats.
They lack vascular tissue and hence called ‘Non- vascular cryptogams’.

Question 20.
Mention the five subdivisions of Pteridophytes.
Answer:

1. Psilophytopsida,
2. Psilotopsida,
3. Lycopsida,
4. Sphenopsida,
5. Pteropsida.

Answer In brief.

Question 1.
List the life cycle patterns in plants.
Answer:

Question 2.
Name the 11 classes of Algae by which F.E. Fritsch classified.
Answer:
F.E. Fritsch proposed a classification for algae-based on pigmentation, types of flagella, reserve food materials, thallus structure and reproduction. He published his classification in the book “The structure and reproduction of Algae” (1935). He classified algae into 11 classes namely Chlorophyceae, Xanthophyceae, Chrysophyceae, Bacillariophyceae, Cryptophyceae, Dinophyceae, Chloromonodineae, Euglenophyceae, Phaeophyceae, Rhodophyceae, Cyanophyceae.

Question 3.
Describe the gametophyte phase of non-vascular cryptogams.
Answer:

1. The gametophyte is a conspicuous, long-lived phase of the life cycle.
2. Thalloid forms are present in liverworts and Hornworts.
3. In Mosses leaf-like, stem-like structures are present.
4. In Liverworts thallus grows prostrate on the ground and is attached to the substratum by means of rhizoids.
5. Two types of rhizoids are present namely smooth-walled and pegged. Multicellular scales are also present.
6. In Moss, the plant body is erect with a central axis bearing leaf-like expansions. Multicellular rhizoids are present.

Question 4.
Describe the sporophyte of non-vascular cryptogam.
Answer:

1. The embryo divides and gives rise to the sporophyte.
2. The sporophyte is dependent on the gametophyte.
3. It is differentiated into three recognizable parts namely foot, seta and capsule.
4. The foot is the basal portion and is embedded in the gametophyte through which water and nutrients are supplied for the sporophyte.
5. The diploid spore mother cells found in the capsule region undergo meiotic division and give rise to haploid spores. Bryophytes are homosporous.
6. In some sporophytes, elaters are present and help in the dispersal of spores, eg. Marchantia.
7. The spores germinate to produce gametophyte.

Question 5.
Write briefly about the economic importance of non-vascular cryptogams.
Answer:

1. A large amount of dead thallus of Sphagnum gets accumulated mid compressed, hardened to form peat. In northern Europe, peat is used as fuel in a commercial scale (Netherlands).
2. Apart from this Nitrates, brown dye and tanning materials are derived from peat.
3. Sphagnum and peat are also used in horticulture as packing material because of their water holding capacity.
4. Marchantia polymorpha is used to cure pulmonary tuberculosis.
5. Sphagnum, Bryum and Polytrichum are used as food.
6. Bryophytes play a major role in soil formation through succession and help in soil conservation.

Question 6.
List out any four, characteristic features of pteridophytes.
Answer:

1. The plant body is a sporophyte (2n) and it is the dominant phase.
2. It is differentiated into root, stem and leaves. Roots are adventitious.
3. The stem shows monopodial or dichotomous branching.
4. Leaves may be microphyllous or megaphyllous.

Question 7.
Distinguish between Gymnosperms and Angiosperms.
Answer:

Question 8.
List out any three Economic importance of algae.
Answer:

1. Agar Agar – cell wall material used for media preparation in the microbiology lab.
2. Packing canned food, cosmetic, textile paper industry.
3. Carrageenan – Preparation of toothpaste, paint, blood coagulant
4. Alginate – Ice cream, paints, flameproof fabrics.

Question 9.
Describe the Vegetable reproduction of non-vascular cryptogams.
Answer:

1. Vegetative reproduction takes place by die formation of adventitious buds (Riccia fluitans) tubers develop in Anthoceros.
2. In some forms small detachable branches or brood bodies are formed, they help in vegetative reproduction as in Bryopteris fruticulosa.
3. In Marchantia propagative organs called gemmae are formed and help in reproduction.

Question 10.
List the salient features of Angiosperms.
Answer:

1. Vascular tissue (Xylem and Phloem) is well developed.
2. Flowers are produced instead of a cone.
3. The embryosac (Ovule) remains enclosed in the ovary.
4. Pollen tube helps in fertilization, so water is not essential for fertilization.
5. Double fertilization is present. The endosperm is triploid.
6. Angiosperms are broadly classified into two classes namely Dicotyledons and Monocotyledons.

Answer In detail.

Question 1.
Explain the asexual and sexual reproduction in algae with a suitable example.
Answer:
Asexual reproduction takes place by the production of zoospores (Ulothrix, Oedogonium),

1. Aplanospore – Thin walled non motile spores, eg. Vaucheria.
2. Autospores – Spores that look similar to the parent cell. eg. Chlorella.
3. Hypnospore – Thick walled aplanospore. eg. Chlamydomonas nivalis. .
4. Tetraspores – Diploid thallus of Polysiphonia produce haploid spores after meiosis).

Sexual reproduction in algae are of three types:

1. Isogamy (Fusion of morphologically and Physiologically similar gametes eg. Ulothrix.
2. Anisogamy (Fusion of either morphologically or physiologically dissimilar gametes eg. Pandorina.
3. Oogamy (Fusion of both morphologically and physiologically dissimilar gametes, eg. Sargassum. The life cycle shows a distinct alternation of generation.

Question 2.
Give an account of Phaeophyceae.
Answer:
The members of this class are called ‘Brown algae’.

1. The majority of the forms are found in marine habitats.
2. Pleurocladia is a freshwater form.
3. The thallus is filamentous (Ectocarpus) frond-like (Dictyota) or maybe giant kelps (Laminaria and Macrocystis).
4. The thallus is differentiated into leaf-like photosynthetic part Called fronds, a stalk-like structure called stipe and a holdfast which attach thallus to the substratum.
5. The Pigments include Chlorophyll a, c, Carotenoids and Xanthophylls.
6. A golden-brown pigment called fucOxanthin is present and it gives shades of colour from olive green to brown to the algal members of this group.
7. Mannitol and Laminarin are the reserve food materials. Motile reproductive structures are present.
8. Two laterally inserted unequal flagella are present. Among these one is whiplash and another is tinsel.
9. Although sexual reproduction ranges from isogamy to Oogamy, Most of the forms show the Oogamous type.
10. Alternation of generation is present (isomorphic, heteromorphic or diplontic). eg. Sargassum, Laminaria, Fucus and Dictyota.

Question 3.
Explain the method of sexual reproduction in non-vascular cryptogams.
Answer:

1. Sexual reproduction is oogamous. Antheridia and Archegonia are produced in a protective covering and are multicellular.
2. The antheridia produce biflagellate antherozoids which swim in a thin film of water and reach the archegonium and fuse with the egg to form a diploid zygote.
3. Water is essential for fertilization.
4. The zygote is the first cell of the sporophyte generation. It undergoes mitotic division to form the multicellular undifferentiated embryo. The embryogeny is exoscopic (the first division of the zygote is transverse and the apex of the embryo develops from the outer cell). The embryo divides and gives rise to the sporophyte.
5. The sporophyte is dependent on the gametophyte.
6. It is differentiated into three recognizable parts namely foot, seta and capsule. Foot is the basal portion and is embedded in the gametophyte through which water and nutrients are supplied for the sporophyte.
7. The diploid spore mother cells found in the capsule region undergo meiotic division and give rise to haploid spores. Bryophytes are homosporous.
8. In some sporophytes elaters are present and help in the dispersal of spores eg. Marchantia.
9. The spores germinate to produce gametophyte.

Question 4.
Give an account of types of steles.
Answer:
The term stele refers to the central cylinder of vascular tissues consisting of xylem, phloem; pericycle and sometimes medullary rays with pith. There are two types of steles (i) Protostele, (ii) Siphonostele.
Protostele In protostele xylem surrounds phloem. The type includes Haplostele, Actinostele, Plectostele, and mixed protostele.

1. Haplostele Xylem surrounded by phloem is known as haplostele. eg. Selaginella.
2. Actinostele Star-shaped xylem core is surrounded by phloem is known as actinostele.
   eg. Lycopodium serratum.
3. Plectostele Xylem plates alternate with phloem plates, eg. Lycopodium clavatum.
4. Mixed protostele Xylem groups uniformly scattered in the phloem, eg. Lycopodium cernuum.

Siphonostele: In siphonostele, xylem is surrounded by phloem with pith at the centre. It includes Ectophloic siphonostele, Amphiphloic siphonostele, Solenostele, Eustele, Atactostele and Polycyclic stele.

1. Ectophloic siphonostele: The phloem is restricted only on the external side of the xylem. Pith is in centre, eg. Osmunda.
2. Amphiphloic siphonostele: The phloem is present on both the sides of xylem. The pith is in the centre, eg. Marsilea.
3. Solenostele: The stele is perforated at a place or places corresponding the origin of the leaf trace.
   (a) Ectophloic solenostele: Pith is in the centre and the xylem is surrounded by phloem, eg. Osmunda.
   (b) Amphiphloic solenostele: Pith is in the centre and the phloem is present on both sides of the xylem. eg. Adiantum pedatum
   (c) Dictyostele: The stele is separated into several vascular strands and each one is called meristele. eg. Adiantum capillus-veneris.
4. Eustele: The stele is split into distinct collateral vascular bundles around the pith, eg. Dicot stem.
5. Atactostele: The stele is split into distinct collateral vascular bundles and are scattered in the ground tissue, eg. the Monocot stem.
6. Polycyclicstele: The vascular tissues are present in the form of two or more concentric cylinders, eg. Pteridium.

Question 5.
Describe the general characteristics features of gymnosperms.
Answer:

1. Most of the gymnosperms are evergreen woody trees or shrubs. Some are lianas (Gnetum)
2. The plant body is sporophyte and is differentiated into root, stem and leaves.
3. A well-developed tap root system is present. Coralloid roots of Cycas have a symbiotic association with blue-green algae. In Pinus the roots have mycorrhizae.
4. The stem is aerial, erect and branched or unbranched (Cycas) with leaf scars.
5. In conifers two types of branches namely branches of limited growth (Dwarf shoot) and Branches of unlimited growth (Long shoot) are present.
6. Leaves are dimorphic, foliage and scale leaves are present. Foliage leaves are green, photosynthetic and borne on branches of limited growth. They show xerophytic features.
7. The xylem consists of tracheids but in Gnetum and Ephedra Vessels are present.
8. Secondary growth is present. The wood may be Manoxylic (Porous, soft, more parenchyma with wide medullary ray – Cycas) or Pycnoxylic (compact with narrow medullary ray-Pinus).
9. They are Heterosporous. The plant may be monoecious (Pinus) or dioecious (Cycas).
10. Microsporangia and Megasporangia are produced on Microsporophyll and Megasporophyll respectively.
11. Male and female cones are produced.
12. Anemophilous pollination is present.
13. Fertilization is siphon gamous and pollen tube helps in the transfer of male nuclei.

Question 6.
List out the economic importance of gymnosperms.
Answer:
Economic importance of gymnosperms

Choose the correct answer.

1. Who proposed plant kingdom?
(a) Eichler
(b) G.M.Smith
(c) Chamberlain
(d) Aristotle
Answer:
(a) Eichler

2. Volvox, §pirogyra are the examples for:
(a) Diplontic
(b) Haplontic
(c) Epiphytic
(d) Haplodiplotic
Answer:
(b) Haplontic

3. Father of Indian phycology is:
(a) Went
(b) F.E. Fritsch
(c) Ainsworth
(d) M.O.P. Iyengar
Answer:
(d) M.O.P. Iyengar

4. Identify the incorrect habitat:
(a) Marine – Sargassum
(b) Freshwater – Oedogonium
(c) Soil – Ulothrix
(d) Salt pass – Dunaliella
Answer:
(c) Soil – Ulothrix

5. The study of Algae is called:
(a) Phycology
(b) Mycology
(c) Bryology
(d) Virology
Answer:
(a) Phycology

6. Which is the proteinaceous body found is
(a) Pyrenoids
(b) Siliceous walls
(c) Carrageenan
(d) Alginate
Answer:
(a) Pyrenoids

7. Oedogonium belongs to the class:
(a) Cryptophyceae
(b) Rhodophyceae
(c) Phaeophyceae
(d) Chlorophyceae
Answer:
(d) Chlorophyceae

8. Identify the incorrect pair.
(a) Green algae – Chlorophyceae
(b) Red algae – Rhodophyceae
(c) Brown algae -Phaeophyceae
(d) Blue algae – Dinophyceae
Answer:
(d) Blue algae – Dinophyceae

9. Match the following:

(a) (i)-(d), (ii)-(b), (iii)-(a), (iv)-(c)
(b) (i)-(d), (ii)-(c), (iii)-(b), (iv)-(a)
(c) (i)-(b), (ti)-(a), (iii)-(d), (iv)-(c)
(d) (i)-(b), (ii)-(c), (iii)-(a), (iv)-(d)
Answer:
(b) (i)-(d), (ii)-(c), (iii)-(b), (iv)-(a)

10. Amylum stars, root bulbils and amorphous bulbils are the vegetative reproductive organs of:
(a) Chora
(b) Oedogonium
(c) Selaginella
(d) Pinus
Answer:
(a) Chora

11. In which sporophyte elaters are present and help in dispersal:
(a) Riccia
(b) Anthoceros
(c) Bryopteris
(d) Marchantia
Answer:
(d) Marchantia

12. Who classified bryophytes in to 3 classes.
(a) Proskauer
(b) Kashyap
(c) G.M.Smith
(d) Chamberlain
Answer:
(a) Proskauer

13. Sphagnum is used in:
(a) Agriculture
(b) Horticulture
(c) Sericulture
(d) Monoculture
Answer:
(b) Horticulture

14. Pulmonary tuberculosis is caused by:
(a) Funaria hygrometrica
(b) Sphagnum
(c) Marchantia polynmorpha
(d) Riccia
Answer:
(c) Marchantia polynmorpha

15. Selaginalla is the example for:
(a) Plectostele
(b) Haplostele
(c) Mixed protostele
(d) Actinostele
Answer:
(b) Haplostele

16. ‘Walking fem’ is the common name for:
(a) Pinus
(b) Cycas
(c) Adiantum
(d) Selaginella
Answer:
(c) Adiantum

17. Identify the incorrect statement.
(a) presence of cambium in gymnosperm as in monocotyledons
(b) presence integument around the ovule
(c) both plant group produce seeds.
(d) pollentube helps in the transfer of male nucleus in both.
Answer:
(a) presence of cambium in gymnosperm as in monocotyledons

18. Match the following:

(a) (i)-(d), (ii)-(b), (iii)-(a), (iv)-(c)
(b) (i)-(d), (ii)-(c), (iii)-(a), (iv)-(b)
(c) (i)-(a), (ii)-(d), (iii)-(b), (iv)-(c)
(d) (i)-(a), (ii)-(c), (iii)-(b), (iv)-(d)
Answer:
(b) (i)-(d), (ti)-(c), (iii)-(a), (iv)-(b)

19. Coralloid roots are found in:
(a) Araucaria
(b) Ephedra
(c) Cycas
(d) Pinus
Answer:
(c) Cycas

20. Foliage leaves are otherwise called:
(a) Axillary leaves
(b) Scale leaves
(c) Sessile leaves
(d) Assimilatory leaves
Answer:
(d) Assimilatory leaves

21. Algae having oil as reserve food belongs to:
(a) Xanthophyceae
(b) Rhodophyceae
(c) Chlorophyceae
(d) Phaeophyceae
Answer:
(a) Xanthophyceae

22. Antheridia and Archegonia are sex organs of:
(a) Moss
(b) Mucor
(c) Spirogyra
(d) Puccinia
Answer:
(a) Moss

23. Archegonicphore is found in:
(a) Funaria
(b) Marchantia
(c) Chara
(d) Adiantum
Answer:
(b) Marchantia

24. Which one of the following in Spyrogyra is different based on its nucleus?
(a) Zygospore
(b) Azygospore
(c) Aplanspore
(d) Akinete
Answer:
(a) Zygospore

25. Nostoc fixes dinitrogen in symbiotic association with the following:
(a) Alnus
(b) Gunnera
(c) Anthocerus
(d) Casurina
The correct combination is:
(a) I & II (b) II & III (c) I & III (d) I & IV
Answer:
(b) II & III

26. Select the wrong statement:
(a) Isogametes are similar in structure, function and behaviour
(b) Anisogametes differ either in structure, function and behaviour
(c) In oomycetes , female gamete is smaller and motile, while male gamete is larger and nonmotile.
(d) Cjilamydomonas exhibits both isogamy and anisogamy and focus shows oogamy.
Answer:
(c) In oomycetes , female gamete is smaller and motile, while male gamete is larger and nonmotile.

27. The ladder like structure found in spirogyra is due to:
(a) Asexual reproduction
(b) Lateral conjugation
(c) Direct conjugation
(d) Sealiriform conjugation
Answer:
(d) Sealiriform conjugation

28. Transgenic plants are the ones:
(a) grown in artificial medium, after hybridization in the field.
(b) produced by a somatic embryo in artificial medium.
(c) generated by introducing foreign DNA into a cell and regenerating a plant from that cell.
(d) produced after protoplast fusion in artificial medium.
Answer:
(c) generated by introducing foreign DNA into a cell and regenerating a plant from that cell.

29. Which of the following is a living fossil?
(a) Spirogyra
(b) Moss
(c) Cycas
(d) Saccharomyces
Answer:
(c) Cycas

30. Which of the following species propagates through leaf-tip?
(a) Funaria
(b) Walking fern
(c) Moss
(d) Marchantia
Answer:
(b) Walking fern

Students get through the TN Board 11th Bio Botany Important Questions Chapter 1 Living World which is useful for their exam preparation.

TN State Board 11th Bio Botany Important Questions Chapter 1 Living World

Answer the following short answers.

Question 1.
Define Biosphere.
Answer:
Earth has formed some 4.6 billion years ago. It is a life-supporting planet with landforms like mountains, plateaus, glaciers, etc., Life on earth exists within a complex structure called the biosphere.

Question 2.
What is the role of the DNA molecule?
Answer:
DNA is essential for the regulation of life and is made up of carbon, hydrogen, oxygen, nitrogen, and phosphorus, and thus nonliving and living things exist together to make our planet unique.

Question 3.
What is consciousness?
Answer:
AH organisms are capable of sensing their environment and respond to various physical, chemical, and biological stimuli. Animals sense their surroundings by sense organs. This is called consciousness.

Question 4.
Explain Prions.
Answer:
Prions were discovered by Stanley B.Prusiner in the year 1982 and are proteinaceous infectious particles. They are the causative agents for about a dozen fatal degenerative disorders of the central nervous system of humans and other animals. For example Creutzfeldt – Jakob Disease (CJD), Bovine Spongiform Encephalopathy (BSE) – commonly known as mad cow disease and scrapie disease of sheep.

Question 5.
What is Plasmid?
Answer:
Plasmids are extrachromosomal double-stranded, circular, self-replicating, autonomous elements. They contain genes for fertility, antibiotic-resistant and heavy metals. It also helps in the production of bacteriocins and toxins which are not found in the bacterial chromosome.

Question 6.
Mention the need for classification?
Answer:
Classification is essential to achieve the following needs:

1. To relate things based on common characteristic features.
2. To define organisms based on the salient features.
3. Helps in knowing the relationship amongst different groups of organisms.
4. It helps in understanding the evolutionary relationship between organisms.

Question 7.
What is Pili?
Answer:
Pili or fimbriae are hair-like appendages found on the surface of the cell wall of gram-negative bacteria (eg. Enterobacterium). The pili are 0.2 to 20 pm long with a diameter of about 0.025pm. In addition to normal pili, special types of pili that help in conjugation called sex pili are also found.

Question 8.
Define “Pruteen”.
Answer:
“Pruteen” is a single cell protein derived from Methylophilus and Methylotropus.

Question 9.
What are Mycobiont and Phycobiont?
Answer:
The symbiotic association between algae and fungi is called lichens. The algal partner |is called Phycobiont or Photobiont and the fungal partner is called Mycobiont. Algae provide nutrition for fungal partners in turn fungi provide protection and also help to fix the thallus to the substratum through rhizines. Asexual reproduction takes place through fragmentation, Soredia, and Isidia. Phycobionst reproduces by akinetes, hormogonia, aplanospore, etc., Mycobionts undergo sexual reproduction and produce ascocarps.

Question 10.
Define Homeostasis.
Answer:
Property of self-regulation and tendency to maintain a steady-state within an external environment which is liable to change is called Homeostasis. It is essential for the living organism to maintain internal conditions to survive in the environment.

Question 11.
Mention any four living characteristics of viruses.
Answer:

1. Presence of nucleic acid and protein.
2. Capable of mutation
3. Ability to multiply within living cells.
4. Able to infect and cause diseases in living beings.

Question 12.
Explain the term Viroid.
Answer:
Viroid is a circular molecule of ssRNA without a capsid and was discovered by T.O.Diener in the year 1971. The RNA of viroid has a low molecular weight. Viroids cause citrus exocytic and potato spindle tuber disease in plants.

Question 13.
What is meant by red tide?
Answer:
Red tide is caused by the toxic bloom of Dinoflagellates like Gymnodinium breve and Gonyaulax tamarensis. A major red tide incident in the west coast of Florida in the year 1982 killed hundreds and thousands of fishes.

Question 14.
What are Polysomes?
Answer:
The ribosomes are the site of protein synthesis. The number of ribosomes per cell varies from 10,000 to 15,000. The ribosomes are 70S type and consist of two subunits (50S and 30S). The ribosomes are held together by mRNA and form polyribosomes or polysomes.

Question 15.
Write briefly on purple sulfur bacteria.
Answer:
For bacteria belonging to this group, the hydrogen donor is Thiosulphate and Bacteriochlorophyll. Chlorophyll containing chlorosomes are present, eg. Chromatium.

Question 16.
Match the following:

Answer:
(i)-(c);
(ii)-(a);
(iii)-(d);
(iv)-(b)

Question 17.
What is meant by aflatoxin?
Answer:
Aspergillus, Rhizopus, Mucor, and Penicillium ate involved in the spoilage of food materials. Aspergillus flavus infest dried foods and produce a carcinogenic toxins called aflatoxin.

Question 18.
Explain the term mycorrhizae.
Answer:
The Symbiotic association between fungal mycelium and roots of plants is called mycorrhizae. In this relationship, fungi absorb nutrition from the root, and in turn, the hyphal network of mycorrhizae forming fungi helps the plant to absorb water and mineral nutrients from the soil.

Question 19.
Name any two fungal diseases of plants and their causative organism.
Answer:

1. A blast of Paddy: Magnaporthe grisea.
2. Red rot of sugarcane: Colletotrichum falcatum.

Question 20.
Explain Oomycetes with an example.
Answer:
Coenocytic mycelium is present. The cell wall is made up of Glucan and Cellulose. Zoospore with one whiplash and one tinsel flagellum is present. Sexual reproduction is Oogamous. eg. Albugo.

Answer In brief.

Question 1.
Differentiate extrinsic and intrinsic.
Answer:

Question 2.
List out the classification of viruses.
Answer:

Question 3.
Name any three plant, animal, and human diseases caused by viruses.
Answer:
Plant diseases: (i) Tobacco mosaic, (ii) Cauliflower mosaic, (iii) Sugarcane mosaic.
Animal diseases: (i) Foot and mouth disease of cattle, (ii) Rabies of dog, (iii) Encephalomyelitis of horse
Human diseases: (i) Common cold, (ii) Hepatitis B, (iii) Cancer.

Question 4.
Distinguish catabolism and anabolism.
Answer:

Question 5.
List out any three methods of sexual reproduction, in fungi.
Answer:

1. Sexual reproduction is present but sex organs are absent. Somatogamy or spermatisation results in plasmogamy.
2. Karyogamy is delayed and the dikaryotic phase is prolonged.
3. Karyogamy takes place in basidium and it is immediately followed by meiotic division.

Question 6.
Describe briefly the Viral genome.
Answer:

1. Each virus possesses only one type of nucleic acid either DNA or RNA. The nucleic acid may be in a linear or circular form.
2. Generally nucleic acid is present as a single unit but in wound tumor virus and in influenza virus it is found in segments.
3. The viruses possessing DNA are called ‘Deoxyviruses’ whereas those possessing RNA are called ‘Riboviruses’. The majority of animal and bacterial viruses are DNA viruses (HIV is the animal virus that possesses RNA).
4. Plant viruses generally contain RNA (Cauliflower Mosaic virus possesses DNA). The nucleic acids may be single-stranded or double-stranded. On the basis of the nature of nucleic acid, viruses are classified into four categories. They are Viruses with ssDNA (Parvoviruses), dsDNA (Bacteriophages), ssRNA (TMV), and dsRNA(wound tumour virus).

Question 7.
Explain the structure of T₄ bacteriophage.
Answer:

1. The T₄ phage is tadpole-shaped and consists of head, collar, tail, base plate, and fibers.
2. The head is hexagonal which consists of about 2000 identical protein subunits.
3. The long helical tail consists of an inner tubular core that is connected to the head by a collar.
4. There is a base plate attached to the end of the tail. The base plate contains six spikes and tail fibers. These fibers are used to attach the phage to the cell wall of the bacterial host during replication.
5. A dsDNA molecule of about 50 pm is tightly packed inside the head. The DNA is about 1000 times longer than the phage itself.

Question 8.
Draw and label the ultrastructure of a bacterial cell.
Answer:

Question 9.
What is meant by Ammonification? Explain the bacteria involved in the process.
Answer:

Question 10.
Explain Actinomycetes with an example.
Answer:
Actinomycetes are also called ‘Ray fungi’ due to their mycelia-like growth. They are anaerobic or facultative anaerobic microorganisms and are gram-positive. They do not produce an aerial mycelium. Their DNA contains high guanine and cytosine content (eg. Streptomyces).

Answer In detail.

Question 1.
Draw the structure of different types of viruses and explain.
Answer:
Viruses are ultramicroscopic particles. They are smaller than bacteria and their diameter range from 20 to 300 nm (1 nm = 10^-9 meters). Bacteriophage measures about 10 – 100 nm in size. The size of TMV is 300 x 20 nm.
Generally, viruses are of three types based on shape and symmetry.

1. Cuboid symmetry – eg. Adenovirus, Herpes virus.
2. Helical symmetry – eg. Influenza virus, TMV.
3. Complex or Atypical – eg. Bacteriophage, Vaccinia virus.

Question 2.
Explain the types of Respiration in Bacteria.
Answer:
Two types of respiration are found in Bacteria. They are (i) Aerobic respiration, (ii) Anaerobic respiration.

1. Aerobic respiration These bacteria require oxygen as a terminal acceptor and will not grow under anaerobic conditions (i.e. in the absence of O2) eg. Streptococcus.
   Obligate aerobes: Some Micrococcus species are obligate aerobes (i.e. they must have oxygen to survive),
2. Anaerobic respiration: These bacteria do not use oxygen for growth and metabolism but obtain their energy from fermentation reactions, eg. Clostridium.

Facultative anaerobes: There are bacteria that can grow either using oxygen as a terminal electron acceptor or anaerobically using fermentation reaction to obtain energy. When a facultative anaerobe such as E. coli is present at a site of infection like an abdominal abscess, it can rapidly consume all available 02 and change to anaerobic metabolism producing an anaerobic environment and thus allow the anaerobic bacteria that are present to grow and cause disease, eg. Escherichia coli and Salmonella.
Capnophilic Bacteria: Bacteria that require CO₂ for their growth are called as capnophilic bacteria, eg. Campylobacter.

Question 3.
Describe the various steps in Gram’s staining procedure.
Answer:

Question 4.
Give a concise account of asexual reproduction in bacteria.
Answer:
Bacteria reproduce asexually by Binary fission, conidia, and endospore formation. Among these Binary fission is the most common one.
Binary fission: Under favorable conditions, the cell divides into two daughter cells. The nuclear material divides first and it is followed by the formation of a simple median constriction which! finally results in the separation of two cells.
Endospores: During unfavorable conditions bacteria produce endospores. Endospores are produced in Bacillus megaterium, Bacillus sphaericus, and Clostridium tetani. Endospores are thick-walled resting spores. During the favorable conditions, they germinate and form bacteria.

Question 5.
Enumerate the general characteristic features of fungi.
Answer:

1. The majority of fungi are made up of thin, filamentous branched structures called hyphae. A number of hyphae get interwoven to form mycelium. The cell wall of fungi is made up of a polysaccharide called chitin (polymer of N-acetyl glucosamine).
2. The fungal mycelium is categorized into two types based on the presence or absence of septa (figure). In lower fungi, the hypha is aseptate, multinucleate and is known as coenocytic mycelium (eg. Albugo). In higher fungi, a septum is present between the cells of the hyphae. eg. Fusarium:
   
3. The mycelium is organized into loosely or compactly interwoven fungal tissues called plectenchyma. It is further divided into two types prosenchyma and pseudoparenchyma. In the former type, the hyphae are arranged loosely but parallel to one another In the latter hyphae are compactly arranged and lose their identity.
4. In holocarpic forms, the entire thallus is converted into reproductive structure whereas in Eucarpic some regions of the thallus are involved in the reproduction other regions remain vegetative. Fungi .reproduce both by asexual and sexual methods. The asexual phase is called Anamorph and the sexual phase is called Teleomorph. Fungi having both phases are called Holomorph.
5. In general sexual reproduction in fungi includes three steps (a) Fusion of two protoplasts (plasmogamy), (b) Fusion of nuclei (karyogamy), and (c) Production of haploid spores through meiosis.

Question 6.
Write an essay on the Beneficial activities of fungi.
Answer:
Fungi provide delicious and nutritious food called mushrooms. They recycle the minerals by decomposing the litter thus adding fertility to the soil. The dairy industry is based on a single-celled fungus called yeast. They deteriorate the timber. Fungi cause food poisoning due to the production of toxins. The Beneficial and harmful activities of fungi are discussed below:
Beneficial activities:

1. Food: Mushrooms likeLentinus edodes, Agaricusbisporus, Volvariella volvaceae are consumed ‘ for their high nutritive value. Yeasts provide vitamin B and Eremothecium ashbyii is a rich source of Vitamin B₁₂.
2. Medicine: Fungi produce antibiotics that arrest the growth or destroy the bacteria. Some of the antibiotics produced by fungi include Penicillin (Penicillium notatum), Cephalosporins,
   (Acremonium chrysogenum), Griseofulvin (Penicillium griseofulvum). Ergot alkaloids (Ergotamine) produced by Claviceps purpurea are used as vasoconstrictors.
3. Industries: Production of organic acid: For the commercial production of organic acids fungi are employed in the Industries. Some of the organic acids and fungi which help in the production of organic acids are Citric acid and Gluconic-acid – Aspergillus niger, Itaconic acid – Aspergillus terreus, Kojic acid – Aspergilhis oryzae.
4. Bakery and Brewery: Yeast (Saccharomyces cerevisiae) is used for the fermentation of sugars to yield alcohol. Bakeries utilize yeast for the production of Bakery products like Bread, buns, rolls. etc. Penicillium roquefortii and Penicillium camemberti are employed in cheese production.
5. Production of enzymes: Aspergillus oryzae, Aspergillus niger is employed in the production of enzymes like Amylase, Protease, Lactase etc., ’Rennet’ which helps in the coagulation of milk in cheese manufacturing is derived from Mucor spp.
6. Agriculture: Mycorrhiza forming fungi like Rhizoctonia, Phallus, Scleroderma helps in the absorption of water and minerals. Fungi like Beauveria bassiana, Metarhizium anisopliae are used as Biopesticides to eradicate the pests of crops. Gibberellin, produced by a fungus Gibberella fujikuroi induce plant growth and is used as growth promoter.

Activity

Text Book Page No. 25

Collect some root nodules of leguminous crops. Draw diagram. Wash, it in tap water and prepare d smear by squeezing the content into a clean slide. Follow the Gram staining method and identify the bacteria.
Answer:
Azotobacter, Clostridium, rhizobium are observed.

Text Book Page No. 37

Get a button mushroom. Draw a diagram of the fruit body. Take a thin longitudinal section passing through the gill and observe the section under a microscope. Record your observations.
Answer:


Keep a slice of bread in a clean plastic tray or plate. Wet the surface with little water. Leave the setup for 3 or 4 days. Observe the moldy growth on the surface of the bread. Using .a needle removes some mycelium and place it on a slide and stain the mycelium using lactophenol blue. Observe the mycelium and sporangium under the microscope and Record your observation and identify the fungi and its group based on characteristic features:
The bread mold fungi such as Mucor, Rhizopus can be observed.

I. Choose the correct answer.

1. A sexual reproduction in living organism occurs by the production of:
(a) conidia formation
(b) budding
(c) binnary fission
(d) all the above
Answer:
(d) all the above

2. Animal sense their surroundings by sense organ is called:
(a) Metabolism
(b) Irritability
(c) Consciounsness
(d) Anabolism
Answer:
(c) Consciounsness

3. Who obtained the crystalline protein sediment from infected tobacco juice?
(a) Edward Jennev
(b) W.M. Stanley
(c) Ivanowsky
(d) F.W. Twort
Answer:
(b) W.M. Stanley

4. Which of the following is the diameter of bacteriophage?
(a) 10 – 100 nm
(b) 20 – 100 nm
(c) 10 – 200 nm
(d) 20 – 30 nm
Answer:
(a) 10 – 100 nm

5. How many classes are there in viruses?
(a) 5
(b) 3
(c) 1
(d) 6
Answer:
(c) 1

6. The shape of TMV is:
(a) round shape
(b) cuboid shape
(c) cylindrical shape
(d) rod shape
Answer:
(d) rod shape

7. Identify the incorrect statement of living characters of viruses.
(a) Absence of metabolism
(b) Presence of nucleic acid
(c) Multiply within the cells
(d) Capable of mutation
Answer:
(a) Absence of metabolism

8. Cyanophages means:
(a) Viruses infecting red algae
(b) Viruses infecting brown algae
(c) Viruses infecting blue-green algae
(d) Viruses infecting green algae
Answer:
(c) Viruses infecting blue-green algae

9. Which one of the following is a bacterial disease?
(a) Common cold
(b) Cancer
(c) Rabies
(d) Tetanus
Answer:
(d) Tetanus

10. Who proposed the five kingdom classification?
(a) R.H. Whittaker
(b) Linnaeus
(c) Haeckel
(d) Aristotle
Answer:
(a) R.H. Whittaker

11. Actinomycetes, cyanobacteria are examples of:
(a) Protostar
(b) Fungi
(c) Monera
(d) Animalia
Answer:
(c) Monera

12. What are the three domains of life according to Carl Woese and co-workers?
(a) Prokaiyota, Eukaryota, Archaea
(b) Archaea, Bacteria, Protozoa
(c) Archaea, Fungi, Bacteria
(d) Bacteria, Archaea, Eukarya
Answer:
(d) Bacteria, Archaea, Eukarya

13. Who coined the word Bacterium?
(a) Christian Gram
(b) C.G.Ehrenberg
(c) Griffith
(d) Lederberg
Answer:
(b) C.G.Ehrenberg

14. Eschrichia coli and Salmonella are examples of:
(a) Aerobes
(b) Obligate aerobes
(c) Facultative anaerobes
(d) Anaerobes
Answer:
(c) Facultative anaerobes

15. Which among the bacteria breakdown hydrocarbons?
(a) Pseudomonas putida
(b) Lacto bacillus
(c) Bifido bacterium
(d) Nitrosomonas
Answer:
(a) Pseudomonas putida

16. Match the following disease with their pathogen:

(a) (i)-(d); (ii)-(c); (iii)-(b); (iv)-(a)
(b) (i)-(c); (ii)-(a); (iii)-(b); (iv)-(d)
(c) (i)-(a)-, (ii)-(d); (iii)-(b); (iv)-(c)
(d) (i)-(c); (ii)-(d); (iii)-(b); (iv)-(a)
Answer:
(d) (i)-(c); (ii)-(d); (iii)-(b); (iv)-(a)

17. PHB means:
(a) Poly Hexo Butyrate
(b) Poly Hydrolxl Butyle
(c) Poly -13 Hydroxl Butyrate
(d) Poly Hedfal Butyrate
Answer:
(c) Poly -13 Hydroxl Butyrate

18. Which of the statement is incorrect in blue-green algae?
(a) The thallus is unicellular
(b) The reserve food material is cyanophycean starch
(c) Absence of mucilage around the thallus
(d) Sexual reproduction is absent
Answer:
(c) Absence of mucilage around the thallus

19. Nostoc and Anabaena are examples of:
(a) Biofertilizer
(b) Biological fuel
(c) Biotic factor
(d) Pro-Biotic
Answer:
(a) Biofertilizer

20. Father of the Indian mycology is:
(a) P.A.Michali
(b) C.H.Blackley
(c) Sir Edwin John Butler
(d) A.F. Blakeshec
Answer:
(c) Sir Edwin John Butler

21. In Fungi reproduction sexual phase is called:
(a) Anamorph
(b) Holomorph
(c) Allelomorph
(d) Teleomorph
Answer:
(d) Teleomorph

22. Odd one out (Type of ascocarp):
(a) Deistothecium
(b) Perithecium
(c) Apothecium
(d) Basidium
Answer:
(d) Basidium

23. Which one of the following is a rich source of vitamin B12?
(a) Agaricus bisporus
(b) Aspergillus oryzae
(c) Eremothecium ashbyii
(d) Clavicepspurpurea
Answer:
(c) Eremothecium ashbyii

24. Pick out the incorrect pair:
(a) Blast of paddy – Magnaporthe grisea
(b) Red rot of sugarcane – Albugo Candida
(c) Rust of wheat – Pucciniagraminis
(d) peach of leaf curl – Taphrina deformans
Answer:
(b) Red rot of sugarcane – Albugo Candida

25. The symbiotic association between fungal mycelium and roots of higher plants is called:
(a) Lichen
(b) Symbiotic
(c) Monotropa
(d) Mycorrhizac
Answer:
(d) Mycorrhizac

26. Each virus posses only one type of nucleic acid:
(a) Either DNA or RNA
(b) DNA only
(c) Neither DNA nor RNA
(d) RNA only
Answer:
(a) Either DNA or RNA

27. Tobacco mosaic virus is a:
(a) rhomboid-shaped helical virus
(b) rod-shaped helical virus
(c) rectangular-shaped virus
(d) triangular shaped helical virus
Answer:
(b) rod-shaped helical virus

28. Indicate the correct statement:
(a) Virion is a circular molecule of ss RNA without a capsid
(b) Virion is a phage that injects linear DNA into the host cell
(c) Virion is an intact infective virus particle, which is non – replicating outside a host cell
(d) Virion is discovered by J.W Randles
Answer:
(c) Virion is an intact infective virus particle, which is non – replicating outside a host cell

29. Transfer of DNA from one bacterium to another is called:
(a) Conjugation
(b) Transformation
(c) Transduction
(d) Binary fission
Answer:
(b) Transformation

30. Fusion of two somative cells of the hypae is called:
(a) Anisogamy
(b) Somatogamy
(c) Oogamy
(d) Isogamy
Answer:
(b) Somatogamy

Students get through the TN Board 12th Chemistry Important Questions Chapter 1 Metallurgy which is useful for their exam preparation.

TN State Board 12th Chemistry Important Questions Chapter 1 Metallurgy

Question 1.
Name a few metals that occur natively.
Answer:
Silver, gold, platinum, copper occur in nature in significant amounts.

Question 2.
Distinguish between ore and minerals.
Answer:
Both minerals and ores are naturally occurring compounds of the metal. It is possible to extract metal in large quantities and in an easy manner from ore but it is not so in a mineral. All ores are minerals but all minerals are not ores.

Question 3.
What do you understand by the term “concentration of the ore”?
Answer:
The ore particles are associated with earthy impurities like sand, clay etc. These impurities have to be removed before it is subjected to extraction. The removal of impurities from the ore by specific methods is known as the concentration of the ore.

Question 4.
Write the name and formula of the common ores of aluminium, iron and copper.
Answer:

Question 5.
The names of a few ores are given. Write the composition of the ore. (i) Calamine, (ii) Zincite, (Hi) Galena, (iv) Cerrusite, (v) Cassiterite, (vi) Silver glance, (vii) Chloroargyrite.
Answer:
(i) Calamine: ore of Zn: ZnCO₃
(ii) Zincite : ore of Zn: ZnO
(Hi) Galena: ore of Pb: PbS
(iv) Cerrusite: ore of Pb: PbCO₃
(v) Cassiterite: ore of Sn: SnO₂
(vi) Silver glance: ore of Ag: Ag₂S
(vii) Chlorargyrite: ore of Ag: AgCl

Question 6.
Briefly give the principle involved in gravity separation of concentrating the ore.
Answer:
This method uses the difference in specific gravity between the ore particles and impurities (gangue). The crushed ore is treated with a rapidly flowing current of water. Lighter gangue particles are washed away, leaving behind heavier ore particles.

Question 7.
Why is the froth floatation process selected for the concentration of sulphide ores?
Answer:
The sulphide ore particles are preferentially wetted by oil, becomes lighter and thus rise to the surface along with the froth while the gangue particles are preferentially wetted by water, becomes heavier and settle down at the bottom of the tank. In this way, sulphide ore particles are separated and hence concentrated.

Question 8.
Name the chemicals used as frothing agents in the froth floatation process.
Answer:
Pine oil and eucalyptus oil are used as a frothing agent in the froth floatation process.

Question 9.
What is the role of depressant in the froth floatation process?
Answer:
The depressants help in the separation of two sulphide ores. For example, in the case of an ore containing PbS (galena) and ZnS (Zinc blende), the depressant is sodium cyanide (NaCN). It prevents ZnS from coming to the froth but allows PbS to come with froth in the froth floatation process ZnS forms a soluble complex with NaCN.

Question 10.
Explain the term leaching.
Answer:
It is a chemical separation of the ore and impurities. It involves the treatment of the ore with a suitable reagent to make it soluble while impurities remain insoluble. The insoluble impurities are separated by filtration.

Question 11.
Explain how, gold ore is concentrated by leaching, (or) How is leaching used to concentrate native ores of gold?
Answer:
The crushed ore of Au is leached with an aerated dilute solution of NaCN. Gold is converted into a soluble cyanide complex. The gangue, aluminosilicate remains insoluble.

Question 12.
What is cementation? Explain with an equation.
Answer:
Gold can be recovered by reacting the deoxygenated leached solution with zinc. In this process, the gold is reduced to its elemental state (zero oxidation state) and the process is called cementation.

Question 13.
What is the role of ammonia in leaching ores containing nickel or copper or cobalt?
Answer:
When a crushed ore containing nickel, copper and cobalt is treated with aqueous ammonia under suitable pressure, ammonia selectively leaches these metals by forming their soluble complexes viz. [Ni(NH₃)₆]²⁺, [Cu(NH₃)₄]²⁺, and [CO(NH₃)₅H₂O]³⁺ respectively from the ore leaving behind the gangue, iron (III) oxides/hydroxides and aluminosilicate.

Question 14.
Explain how bauxite ore is leached in the extraction of aluminium.
Answer:
The bauxite ore is treated with aqueous sodium hydroxide and heated to a temperature of around 470 – 520 K, at 35 atm. The aluminium dissolves forming a soluble complex, sodium meta aluminate leaving behind impurities.
Al₂O₃ (S) + 2NaOH (aq) + 3H₂O (l) → 2Na [Al(OH)₄] aq.
The hot solution is separated from its impurities and CO₂ gas is passed. Pure alumina gas precipitated
2Na [Al(OH)₄] (aq) + CO₂ (g) → Al₂O₃. H₂0 + 2 NaHCO₃ (aq)
The precipitate is washed and dried.

Question 15.
How is sulphide ores leached?
Answer:
The sulphide ores (ZnS, PbS) are treated with hot, aqueous sulphuric acid. The insoluble sulphide is converted to soluble sulphate and elemental sulphur which gets precipitated.
2ZnS (s) + 2H₂SO₄ (aq) + O₂ (g) → 2ZnSO₄ (aq) + 2S (s) + H₂O

Question 16.
Explain the principle involved in the magnetic separation of the ore with an example.
Answer:

This method is based on the difference in magnetic properties of the ore and the impurities. For example, tinstone, an ore of tin can be separated from its magnetic impurities (wolframite). The crushed ore is poured onto an electromagnetic separator consisting of a belt moving over two rollers of which one is magnetic. The magnetic part of the ore is attracted towards the magnet and falls as a heap close to the magnetic region while the non-magnetic part falls away.

Question 17.
Give chemical equations involved in the process of roasting.
Answer:
Roasting converts sulphide ores to their oxides.

It also removes volatile impurities like moisture and non-metals as their oxides.

Question 18.
Explain the term ‘Roasting’ in metallurgy.
Answer:
Roasting is a process in which sulphide ores are converted to oxide ores by heating the ore in excess of oxygen in a reverberatory furnace below the melting point of the metal.
eg: Galena when roasted gets converted to lead monoxide.
2PbS + 3O₂ → 2PbO + SO₂
During roasting volatile and non-metallic impurities are removed. The ore is rendered porous so that further steps in the extraction process becomes easy.

Question 19.
What is calcination? Explain with an example.
Answer:
Calcination is the process in which the concentrated ore is heated strongly in the absence of air. During this process, the water of crystallisation present in hydrated oxides escapes as moisture; This method is carried out in a limited supply of air. For example, calcination of the carbonate ore, carbon dioxide example is expelled.

Question 20.
Distinguish between calcination and roasting.
Answer:

1. Both processes are used to convert ores into metallic oxides. During roasting sulphide ores are converted to oxides and during calcination carbonate ores are converted to oxides.
2. Roasting is carried out by heating the ore with excess oxygen below the melting point of the ore whereas calcination is done by heating the ore in a limited supply of oxygen.
3. During roasting volatile impurities are removed and non-metallic impurities escape as their oxides. During calcination, the water of crystallisation escapes.
4. Both processes render the ore porous which makes further steps of the extraction process easiest.

Question 21.
Write chemical equations involved in the calcination process?
Answer:
Calcination is the process that involves the conversion of carbonate ores to oxides.

The water of crystallisation of hydrated salts escapes as water vapour.

Question 22.
Give a brief account of the reduction of metal oxides to metals.
Answer:

1. Metal oxides can be reduced to metals by using a suitable reducing agent. The choice of the reducing agent depends on the nature of the metal. The reducing agents used commonly are carbon, carbon monoxide, hydrogen and reactive metals like sodium.
2. Carbon is used as a reducing agent in the reduction of oxide ores such as ZnO, Mn₃O₄ and Cr₂O₃. The ore and coke are heated strongly in a blast furnace.
   
3. Carbon monoxide is used to reduce ferric oxide in the extraction of iron.
   Fe₂O₃ (s) + 3CO (g) → 2Fe (s) + 3CO₂ (g) ↑
4. Gaseous hydrogen is used to reduce oxides of less electropositive metals (Fe, Pb, Cu) than hydrogen.
   Ag₂O + H₂ → 2Ag + H₂O
   Fe₃O₄ + 4H₂ → 3Fe + 4H₂O
5. Aluminium is used to reduce metal oxides like Cr₂O₃.
   
   Active metals such as sodium, potassium and calcium are also used as reducing agents.
   B₂O₃ + 6Na → 2B + 3Na₂O
   Rb₂O₃ + 3Mg → 2Rb + 3MgO

Question 23.
What do you understand by the terms (i) Flux, (ii) Slag? Explain with examples.
Answer:

1. Flux is a chemical substance that is added to the concentrated ore, to remove impurities present. The impurities combine with the flux and form a fusible mass slag.
2. Acidic impurities are removed by using a basic flux and basic impurities are removed by adding acidic fluxes.
3. Silicon dioxide is an acidic flux and limestone (CaO) is a basic flux.
4. CaO(s), the basic flux removes the gange silicon dioxide (SiO₂), as calcium silicate (slag) in the extraction of iron.
   

Slag: Slag is a compound formed when flux and gangue can combine with each other. The slag is a feasible mass, which removes, the impurities in the ore. In the extraction of iron, FeO is added as a flux to remove the gangue (SiO₂) as ferrous silicate (slag).

Question 24.
Give chemical equations involved in the extraction of copper from copper pyrites.
Answer:

1. Smelting of the concentrated ore:
   Cu₂S and FeS formed in equation (1) is called mattle.
2. The remaining Cu₂S is further oxidised to cuprous oxide, which is subsequently converted to metallic copper.
   Cu₂S + 3O₂ → 2Cu₂S + 2SO₂
   2Cu₂O + Cu₂S → 6Cu + SO₂

Question 25.
Complete and balance the following equations.
Answer:

Question 26.
Briefly outline the thermodynamic principle involved in metallurgy.
Answer:
For any process, Gibb’s free energy change ΔG should be negative, i.e.,
ΔG = ΔH – TΔS should be negative where ΔH and ΔS are enthalpy and entropy changes respectively. The extraction of the metals from their oxides is carried out by using different reducing agents. A suitable reducing agent is chosen based on thermodynamics.
Let us consider the reduction of metallic oxide by carbon.

Thermodynamically, the reduction of metal oxide (equation -1), with a given reducing agent (equation -2) can occur if the free energy change for the coupled reaction is negative. Hence the reducing agent in chosen in such a way to produce a large negative value for the coupled reactions.

Question 27.
What is the Ellingham diagram? Explain.
Answer:
Ellingham’s diagram is used to obtain the ΔG° values at various temperatures for the reduction of metal oxides by treating reduction as an equilibrium process.
The diagram is made by plotting temperature in the x-axis and the standard free energy of formation (ΔG°) of the metal oxide in the ‘y ’ axis. The resultant plot is a straight line with ΔS as slope and ΔH as ‘y’ intercept. The graphical representation of the variation of standard free energy of formation of various metal oxides with temperature is known as the Ellingham diagram.

Question 28.
Briefly explain the salient features of the Ellingham diagram.
Answer:
(i) Ellingham diagram consists of plots that represent the variation of standard free energy of formation of metal oxides versus temperature.
(ii) These plots help is in predicting the feasibility of thermal reduction of ore. Consider the formation of metal oxide.
2xM (s) + O₂ (g) → 2MxO (s)
In this reaction, there is a decrease in the value of ΔS° as M.rO is a solid and 02 is a gas. i.e., ΔS° is negative. Thus, if the temperature is increased, TΔS° becomes more negative. As in the equation ΔG° = ΔH° – TΔS°
TΔS° is substracted, therefore ΔG° becomes less negative i.e., ΔG° is likely to increase with temperature and this trend is confirmed from the curves of metal oxides.
The following observations are made from the curves.
(a) The slope of the curves of formation of metal oxides is positive because ΔG° becomes less negative and increases with temperature.
(b) Each curve is a straight line except when some change takes place in phase. The temperature at which such a change occur is indicated by an increase in slope on the positive side.
(c) In the case of CO, ΔG° decreases as ΔS° increases. This is indicated by the downward trend.
(d) Any metal oxide with lower ΔG° is more stable than a metal oxide with higher ΔG°. This implies that the metal oxide placed higher in the diagram can be reduced by the metal lower in the diagram.

Question 29.
Mention the limitations of the Ellingham diagram.
Answer:

1. Ellingham diagram is constructed based only on thermodynamic considerations. It gives information about the thermodynamic feasibility of a reaction. It does not tell anything about the rate of the reaction. Moreover, it does not give any idea about the possibility of other reactions that might be taking place.
2. The interpretation of ΔG is based on the assumption that the reactants are in equilibrium with the product which is not always true.

Question 30.
Based on the Ellingham diagram explain.
Answer:

1. Aluminium can be used as a reducing agent for the reduction of chromic oxide.
   In the Ellignhem diagram, the formation of chromium oxide lies above that of the aluminium, meaning that Al₂O₃ is more stable than Cr₂O₃. Hence aluminium can be used as a reducing agent for the reduction of chromic oxide.
2. Aluminium cannot be use4 to reduce MgO or CaO.
   The formation of MgO or CaO lies below that, that of aluminium. Hence, it cannot be used to reduce oxides of Mg or Ca.

Question 31.
Why is the reduction of a metal oxide easier if the metal is in the liquid state at the temperature of reduction?
[OR]
Account for the fact that the reduction of a metal oxide is easier if the metal is formed in the liquid state at the temperature of reduction.
Answer:
Entropy is higher when the metal is in the liquid state than when it is in the solid-state. Therefore, the value of entropy change (ΔS°) of the reduction process is more positive, when the metal is formed in the liquid state. The value of TΔS° increases and that of ΔH° remains constant, the value of ΔG° becomes more negative and hence reduction becomes easier.

Question 32.
The choice of a reducing agent in a particular reduction depends on the thermodynamic factor. How far do you agree with this statement?
Answer:
Support your opinion with two examples.
From the Ellingham diagram, it is evident that any metal oxide with a lower AG° value is more stable than the metal oxide with a higher AG° value. This implies that the metal oxide placed lower in the diagram cannot be reduced by a metal involved in the formation of oxide placed higher in the diagram. However, the reverse can readily take place.
Thus, Al₂O₃ cannot be reduced by Cr however Cr₂O₃ can be reduced by Al.
At temperature 1733K, AG° formation of Al₂O₃ and Cr₂O₃ is given below.

The reduction of Al₂O₃ by Cr may be obtained by subtracting equation (i) and (ii).

ΔG° = 400 kJ mol-1 This reaction is thermodynamically not feasible since ΔG° has a positive value. But the reverse reaction is possible as it has a negative ΔG° value. Similarly, both aluminium and zinc can reduce FeO to Fe, but Fe cannot reduce Al₃O₂ Or ZnO.

Question 33.
Based on the Ellingham diagram given below predict a suitable reducing agent for the reduction of Cu₂O.
Answer:

It is seen in the diagram that the Cu₂0 curve lies almost at the top while the lines sharing the formation of CO₂ from C, and the formation of CO from carbon lie much below it. So it is easy to reduce Cu₂O to the metal by heating it with coke at temperatures after 500 – 600 K.
Cu₂O + C → 2Cu + CO

Question 34.
Show by using the Ellingham diagram given below to predict the possible temperature at which ZnO is reduced to Zn by carbon.
Answer:

Ellingham diagram reveals that the lines involving the formation of ZnO and carbon monoxide cross each other at about 1270 K. There is a sudden increase in the value of ΔG° for the formation of ZnO above 1180°C. This is due to the fact that zinc begins to boil at this temperature. Above, 1270 K, ΔG° for the reaction.

is considered negative and thus ZnO is reduced by coke.

Question 35.
Out of carbon and carbon monoxide which is a better reducing agent for ZnO?
Answer:
From Ellingham’s diagram, it is clear that the free energy of formation of CO from C is lower at a temperature above 1120 K while that of CO₂ from carbon is lower above 1323K than that of free energy of formation of ZnO. However, the free energy of the formation of CO₂ from CO is always higher than that of ZnO. Hence, C is a better reducing agent of ZnO.

Question 36.
Explain why is the extraction of copper from pyrites difficult than that from its oxide ore through reduction.
Answer:
The standard free energy formation (ΔG°) for Cu₂S is more negative than those of CS₂ and H₂S. Therefore neither carbon nor hydrogen can reduce Cu₂S to copper. But the standard free energy of formation of Cu₂0 (ΔG°) is less negative than that of CO and hence carbon can readily reduce Cu₂O to Cu.
Alternatively,
ΔG° / T line for CO has a negative slope and there is no compound CS analogous to CO with a steep negative ΔG°/T line. Thus, carbon is a good reducing agent for oxides and not for sulphides. Therefore sulphide ores are normally roasted in air to form oxides before reducing than with carbon.

Question 37.
Given the following ores indicate which of them Can be concentrated by magnetic separation method, (i) Bauxite, (ii) Kaolinite, (iii) Haematite, (iv) siderite, (v) iron pyrites, (vi) copper pyrites, (vii) galena, (viii) cassiterite, (ix) horn silver, (x) zincite, (xi) magnetite.
Answer:
Ores that are magnetic in nature can be separated from non-magnetic gangue particles by magnetic separation ores of iron such as Haematite (Fe₂O₃), Magnetite (Fe₃O₂), Siderite (FeCO₃) and Iron pyrites (FeS₂) being magnetic can be separated by magnetic separation.

Question 38.
Explain the principles involved in each of the following processes of refining metals. (i) Distillation, (ii) Liquation, (iii) Electrolyte refining, (iv) zone refining, (v) vapour phase method.
Answer:

1. Distillation: This method is used for the purification of volatile metals like zinc, mercury, cadmium etc. The impure metal is heated in a retort and the vapours are condensed in separate receivers. The pure metal distils leaving behind the non-volatile having higher boiling point) impurities in the retort.
2. Liquation: This method is used for the purification of metals in which the melting points of metals are lower than these of the impurities. The crude metal is heated in an inert atmosphere on a sloping hearth of a reverberatory furnace when the metal flows down into the receiver leaving behind infusible impurities on the hearth. Metals like tin and lead one purified by this method.
3. Electrolyte refining:
   Electrolytic refining is a process of purification of an impure metal by electrolysis. The impure metal is used as an anode and a pure metal is used as a cathode. The electrolyte is a suitable aqueous solution of the salt of the metal. During electrolysis, the metal dissolves at the anode, migrates to the cathode and gets deposited as a pure metal, eg: Electrolytic refining of silver:
   Cathode: Pure silver Anode: Impure silver rods Electrolyte: Acidified aqueous solution of silver nitrate.
   When a current is passed through the electrodes the following reactions will take place.
   Reaction at anode 2Ag (s) → Ag⁺ (aq) + e^–
   Reaction at cathode Ag⁺ (aq) + e^– → Ag(s)
   During electrolysis, at the anode, the silver atoms lose electrons and enter the solution. The positively charged silver cations migrate towards the cathode and get discharged by gaining electrons and deposited on the cathode. Other metals such as copper, zinc etc., can also be refined by this process in a similar manner.
4. Zone refining: This method is based on the principles of fractional crystallisation. When an impure metal is melted and allowed to solidify, the impurities will prefer to be in the molten region, i.e., impurities are more soluble in the melt than in the solid-state metal. In this process, the impure metal is taken in the form of a rod. One end of the rod is heated using a mobile induction heater which results in the melting of the metal on that portion of the rod. When the heater is slowly moved to the other end the pure metal crystallises while the impurities will move on to the adjacent molten zone formed due to the movement of the heater. As the heater moves further away, the molten zone containing impurities also moves along with it. The process is repeated several times by moving the heater in the same direction again and again to achieve the desired purity level. This process is carried out in an inert gas atmosphere to prevent the oxidation of metals. Elements such as germanium (Ge), silicon (Si) and gallium (Ga) that are used as semiconductor are refined using this process.
   (v) Vapour phase method:
   The basic requirement is that the metal, when treated with a suitable reagent form a volatile compound which as decomposition gives the pure metal. eg: Nickel forms a volatile nickel carbonyl on heating with carbon monoxide, which on decomposition gives pure nickel.

Question 39.
Explain how zinc is purified by electrolytic refining.
Answer:
The impure zinc is made the anode while the cathode consists of pure sheets of zinc. The electrolyte is an acidified (H₂SO₄) solution of zinc sulphate (ZnSO₄). On passing electric current pure zinc deposits on the cathode.

Question 40.
State the role of silver in the metallurgy of copper.
Answer:
During roasting, copper pyrites are converted to a mixture of FeO and Cu₂O.

To remove FeO (basic impurity) an acidic flux, SiO₂ is added during smelting. FeO combines with SiO₂ to form ferrous silicate (FeSiO₃) as slag and floats over the molten matter and hence can be easily removed.

Question 41.
What is the role of graphite in the electrometallurgy of aluminium?
Answer:
In the electrometallurgy of aluminium, a fused mixture of alumina, cryolite and fluorspar (CaF₂) is electrolysed using graphite as anode and graphite lined iron as a cathode. During electrolysis, Al is liberated at the cathode while CO and CO₂ one liberated at the anode.
At Cathode: Al⁺³ (melt) → Al (l)
At anode: C + O^-2 (melt) → CO + 2e^–
C + 2O^-2 (melt) → CO₂ + 4e^–
Instead of graphite, some metal is used as an anode, then O₂ liberated will not only oxidise the metal of the electrode but would also convert the same Al liberated at the cathode back to Al₂O₃.
Since graphite is much cheaper than any metal, it is used as a cathode. Thus the role of graphite in the electrometallurgy of Al is to prevent the liberation of 02 at the anode. Which may otherwise convert Al to Al₂O₃.

Question 42.
Mention the application of the following metals, (i) Al, (ii) Cu, (iii) Fe, (iv) Au.
Answer:
Al: Aluminium is the most abundant metal and is a good conductor of electricity and heat. It also resists corrosion. The following are some of its applications.

1. Many heat exchangers/sinks and our day to day cooking vessels are made of aluminium.
2. It is used as wraps (aluminium foils) and is used in packing materials for food items.
3. Aluminium is not very strong, However, its alloys with copper, manganese, magnesium and silicon are lightweight and strong and they are used in the design of aeroplanes and other forms of transport.
4. As aluminium shows high resistance to corrosion, it is used in the design of chemical reactors, medical equipment, refrigeration units and gas pipelines.
5. Aluminium is a good electrical conductor and cheap, hence used in electrical overhead electric cables with a steel core for strength.

Cu:

1. Copper is used for making coins and ornaments along with gold and other metals.
2. Copper and its alloys are used for making wires, water pipes and other electrical parts.

Fe:

1. Iron and its alloys are used everywhere including bridges, electricity pylons, bicycle chains, cutting tools and rifle barrels.
2. Cast iron is used to make pipes, valves and pumps stoves etc…
3. Magnets can be made of iron and its alloys and compounds.
4. An important alloy of iron is stainless steel, and it is resistant to corrosion. It is used in architecture, bearings, cutlery, surgical instruments and jewellery. Nickel steel is used for making cables, automobiles and aeroplane parts. Chrome steels are used for manufacturing cutting tools and crushing machines.

Au:

1. Gold is used for coinage and has been used as the standard for monetary systems in some countries.
2. It is used extensively in jewellery in its alloy form with copper. It is also used in electroplating to cover other metals with a thin layer of gold which are used in watches, artificial limb joints, cheap jewellery, dental fillings and electrical connectors.
3. Gold nanoparticles are also used for increasing the efficiency of solar cells and also used as catalysts.

TN 12th Chemistry Important Questions State Board English Medium 2021-2022.

TN State Board 12th Chemistry Important Questions and Answers

• Chapter 1 Metallurgy Important Questions
• Chapter 2 p-Block Elements – I Important Questions
• Chapter 3 p-Block Elements – II Important Questions
• Chapter 4 Transition and Inner Transition Elements Important Questions
• Chapter 5 Coordination Chemistry Important Questions
• Chapter 6 Solid State Important Questions
• Chapter 7 Chemical Kinetics Important Questions
• Chapter 8 Ionic Equilibrium Important Questions
• Chapter 9 Electro Chemistry Important Questions
• Chapter 10 Surface Chemistry Important Questions
• Chapter 11 Hydroxy Compounds and Ethers Important Questions
• Chapter 12 Carbonyl Compounds and Carboxylic Acids Important Questions
• Chapter 13 Organic Nitrogen Compounds Important Questions
• Chapter 14 Biomolecules Important Questions
• Chapter 15 Chemistry in Everyday Life Important Questions

Students get through the TN Board 12th Bio Zoology Important Questions Chapter 12 Environmental Issues which is useful for their exam preparation.

TN State Board 12th Bio Zoology Important Questions Chapter 12 Environmental Issues

Very short answer questions

Question 1.
Define Pollution.
Answer:
Pollution is defined as any undesirable change in the physical, chemical, and biological characteristics of an environment.

Question 2.
What are pollutants?
Answer:
Pollutants are the agents which cause pollution.

Question 3.
What are the different basic groups of pollutants?
Answer:
The pollutants are classified into two major groups namely Non-degradable and degradable pollutants.

Question 4.
Name the main sources of carbon monoxide pollution.
Answer:
The main sources of carbon monoxide pollution include Automobile exhausts, fumes from factories and the burning of firewood.

Question 5.
Define SMOG.
Answer:
SMOG is defined as a type of air pollution caused by tiny particles in the air.

Question 6.
Mention any two cost-effective treatments of air pollution?
Answer:
The cost-effective treatment of air pollution includes:

1. Growing indoor plants and
2. Using high-performance biofilters.

Question 7.
Explain Air Quality Index (AQI).
Answer:
Air Quality Index is the number used by Government agencies to communicate to the public how polluted the air is at a given time.

Question 8.
Name any two examples of the non-point sources of pollutants.
Answer:

1. Agriculture chemical run-off.
2. Dumping of plastics in water bodies.

Question 9.
Define Noise pollution.
Answer:
Noise pollution is defined as the sound that is unwanted and undesirable or can disrupt one’s quality of life.

Question 10.
Name the chemicals present in mosquito repellents.
Answer:
The chemicals present in mosquito repellents are:

1. DEET (n-n-diethylenetatoluamide)
2. Alletrin

Question 11.
Define Biomagnification.
Answer:
Biomagnification is the process in which a non-degradable substance, that enters die – food chain does not get metabolized and instead gets transferred up the tropic levels of the food chain with the magnified level of that substance.

Question 12.
What is ‘4R’?
Answer:
‘4R’ means Refuse, Reduce, Reuse and Recycle. These are the mantra for remedy II for the plastic waste problem.

Short answer questions

Question 1.
What are plastics? Explain their sources.
Answer:
Plastics are low molecular weight organic polymers, that are not degradable in the natural environment.
The main sources of this plastic include:

1. The packaging material used in supermarkets.
2. Packing material in retail outlets.
3. Manufacturing industries.
4. Households, hotels, hospitals, and restaurants, and transport companies.
5. In municipal solid waste, plastic waste constitutes a major part.

Question 2.
Recycling and disposal of e-waste involve significant risk – Justify.
Answer:
The recycling and disposal of e-waste may involve risks to the health of workers and communities in developed countries.

1. Great care must be taken to avoid unsafe exposure in recycling operations.
2. The leading of materials like heavy metals from landfills and incinerator ashes may cause health problems to humans and animals.

Question 3.
How does water pollution affect on the ecosystem?
Answer:
Water pollution generally affects the aquatic ecosystems in the following ways:

1. It affects existing riches and habitats.
2. It affects the survival of organisms.
3. In certain cases, soil fertility is affected, leading to the state of uninhabitable.
4. It results in Eutrophication.

Question 4.
What are three sources of water pollution?
Answer:
The three sources of water pollution are:

1. Municipal waste
2. Industrial waste
3. Agricultural waste

Question 5.
Water pollution disrupts the nature of food chains as well as food webs – Justify.
Answer:

1. Pollutants such as lead and cadmium are taken up by primary consumers, where they can be lethal or get accumulated.
2. When these animals are consumed by secondary consumers, their pollutants are get enhanced by a process called biomagnification.
3. The food chain may get disrupted at any trophic level due to the magnification and accumulation of these chemicals.

Question 6.
Explain the Government Act and regulations to prevent water pollution.
Answer:

1. Water Act, 1974, section 17- 40 – Prohibits the pollution of a stream or well by disposal of polluting matter.
2. The Central / State pollution control Boards – They have the power to advise the central / state governments on various matters concerned with the prevention and. control of pollution of water.
3. The Ministry of Environment, forest and climate change (MoEFCC) is the nodal agency of the central government for planning, promotion, coordination, and overseeing the implementation of India’s environmental and forestry policies and programs.

Question 7.
How do you control ozone layer depletion?
Answer:

1. Phase down or ban the use of CFCs (CFC-free refrigerants).
2. The usage of chemicals such as halons and halocarbons ought to be minimized.
3. Creating awareness about ozone-depleting agents.

Long answer questions

Question 1.
Give an account of medical waste.
Answer:
Medical waste refers to any waste that contains infectious material generated by hospitals, medical research centers, laboratories, pharmaceutical companies, and veterinary clinics.
Medical’wastes constitute fluid like blood, urine, body parts, and other contaminants. They also include used syringes and needles, culture dishes, glasswares, bandages gloves, scalpels, swabs, and tissues.
Hie social and legal responsibilities of – people working in health care centers are the safe disposal of medical waste.
The disposal methods include incineration, chemical disinfection, autoclaving, encapsulation microwave irradiation. Landfill and burying inside premises as per norms are the final disposal of these wastes.

Question 2.
Describe the prevention methods of water pollution.
Answer:
The followings are the prevention strategies of water pollution:

1. Regulate or control pollutant discharge at the point of generation.
2. Treatment of wastewater by scientific methods before discharge into municipal treatment sources.
3. Setting up of Sewage Treatment Plants (STP) and Effluent Treatment Plants (ETP) where ever needed.
4. Regulate or restrict the use of chemical fertilizers and pesticides in agriculture. Create public awareness and involve people in the action plan.

Question 3.
Enumerate the effects of noise pollution.
Answer:
A link between noise and the health of humans and other organisms was established by USEPA (the United States Environmental Protection Agency). The noise pollution effects are:

1. It causes increased stress and tension nervousness, irritability anxiety depression, and panic attacks.
2. It also causes peptic ulcers, severe headaches, memory loss.
3. Noise pollution from offshore activities and poor activities affects marine animals.
4. Noise from firecrackers frightens animals and birds.
5. In humans, heart disease, high blood pressure, stress-related illness, sleep disruption, hearing loss, and productivity loss are some of the problems related to noise pollution.

Question 4.
Explain the causes and effects of ozone layer depletion.
Answer:
Causes:

1. Anthropogenic activities mainly cause ozone layer depletion.
2. Chlorofluorocarbons (CFCs) is the main cause of ozone layer depletion by releasing excess chlorine and bromine. Other ozone-depleting substances are hydrobromic fluorocarbons and methyl bromide.

Effects:

1. It allows UY rays to penetrate into the atmosphere, resulting in premature aging of skin and wrinkling of the skin.
2. It may cause suppression of the immune system.
3. It causes skin cancer (melanoma) and chronic effects leading to eye damage.
4. The free radicals and reactive oxygen and photons can damage DNA itself.

Choose the correct answers:

1. Match the following:

(a) (p)-(iv); (q)-(iii); (r)-(ii); (s)-(i)
(b) (p)-(iv); (q)-(iii); (r)-(i); (s)-(ii)
(c) (p)-(ii); (q)-(i); (r)-(iv); (s)-(iii)
(d) (p)-(iii); (q)-(iv); (r)-(ii); (s)-(i)
Answer:
(b) (p)-(iv); (q)-(iii); (r)-(i); (s)-(ii)

2. Match the following:

(a) (p)-(iv); (q)-(iii); (r)-(ii); (s)-(i)
(b) (p)-(ii); (q)-0); (r)-(iv); (s)-(iii)
(c) (p)-(iii); (q)-(iv); (r)-(ii);(s)-(i)
(d) (p)-(iv); (q)-(iii); (r)-(i); (s)-(ii)
Answer:
(c) (p)-(iii); (q)-(iv); (r)-(ii); (s)-(i)

3. Match the following:

(a) (p)-(ii); (q)-(i); (r)-(iv); (s)-(iii)
(b) (p)-(iv); (q)-(iii); (r)-(ii); (s)-(i)
(c) (p)-(iii); (q)-(iv); (r)-(ii); (s)-(i)
(d) (p)-(iv); (q)-(iii); (r)-(i); (s)-(ii)
Answer:
(a) (p)-(ii); (q)-(i); (r)-(iv); (s)-(iii)

4. Match the following:

(a) (p)-(iv); (q)-(i); (r)-(ii); (s)-(iii)
(b) (p)-(iv); (q)-(iii); (r)-(i); (s)-(ii)
(c) (p)-(iv); (q)-(iii); (r)-(ii); (s)-(i)
(d) (p)-(ii); (q)-(i); (r)-(iv); (s)-(iii)
Answer:
(b) (p)-(iv); (q)-(iii); (r)-(i); (s)-(ii)

5. Medical waste is classified under:
(a) Solid waste
(b) Organic waste
(c) Hazardous waste
(d) Chemical waste
Answer:
(c) Hazardous waste

6. Which of the following is an anaerobic ‘ process?
(a) Vermicomposting
(b) Biogas generation
(c) Segregation
(d) Nitrification
Answer:
(b) Biogas generation

7. How much garbage is collected per day from Chennai city?
(a) 5400 MT
(b) 4800 MT
(c) 2400 MT
(d) 6800 MT
Answer:
(a) 5400 MT

8. The method not used in the physical methods of waste treatment.
(a) Flotation
(b) Chlorination
(c) Sedimentation
(d) Filteration
Answer:
(b) Chlorination

9. Choose the odd man out:
(a) Nammalvar
(b) Sultan Md. Ismail
(c) Thiagarajan
(d) Jadav Payeng
Answer:
(c) Thiagarajan

10. Identify the odd one:
(a) Food waste
(b) Paper waste
(c) Plastic waste
(d) Organic waste
Answer:
(c) Plastic waste

11. Choose the odd one:
(a) Dilute and disperse
(b) Segregation and treatment
(c) Delay and decay
(d) Concentrate and confine
Answer:
(a) Dilute and disperse

12. Find out the odd one:
(a) Bandages
(b) Sealpels
(c) Glasswares
(d) Swabs and tissues
Answer:
(b) Sealpels

13. Which of the following is a correct pair?

Answer:
(c)

14. Choose the incorrect pair:

Answer:
(d)

15. Find out the correct pair:

Answer:
(a)

16. Indicate the incorrect pair:

Answer:
(c)

17. Assertion: Organic farming creates an eco-friendly pollution-free environment.
Reason: In organic farming, only natural and organic agro-inputs are used.
(a) Both Assertion and Reason are correct, Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are correct, Reason is not the correct explanation of Assertion.
(c) Assertion is correct, Reason is not correct.
(d) Assertion is not correct, Reason is correct.
(e) Both Assertion and Reason are not correct.
Answer:
(a) Both Assertion and Reason are correct, Reason is the correct explanation of Assertion.

18. Assertion: DDT concentration is more in fish-eating birds than the actual concentration of the aquatic system.
Reason: This is due to a phenomenon called Biomagnification.
(a) Both Assertion and Reason are correct, Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are correct, Reason is not the correct explanation of Assertion.
(c) Assertion is correct, Reason is not correct.
(d) Assertion is not correct, Reason is correct.
(e) Both Assertion and Reason are not correct.
Answer:
(a) Both Assertion and Reason are correct, Reason is the correct explanation of Assertion.

19. Assertion: The permissible limit of noise in commercial areas is 65 decibels during the day and 45 decibels during the night.
Reason: The threshold pain of noise pollution is 120 decibels.
(a) Both Assertion and Reason are correct, Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are correct, Reason is not the correct explanation of Assertion.
(c) Assertion is correct, Reason is not correct.
(d) Assertion is not correct, Reason is correct.
(e) Both Assertion and Reason are not correct.
Answer:
(d) Assertion is not correct, Reason is correct.

20. Assertion: Acid rain is caused by Sulphur dioxide and nitric oxide.
Reason: During rain, the precipitation contains acidic components such as sulphuric acid or nitric acid.
(a) Both Assertion and Reason are correct, Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are correct, Reason is not the correct explanation of Assertion.
(c) Assertion is correct, Reason is not correct.
(d) Assertion is not correct, Reason correct.
(e) Both Assertion and Reason are not correct.
Answer:
(b) Both Assertion and Reason are correct, Reason is not the correct explanation of Assertion.

21. Which of the following statement is correct?
(a) Thinning of the stratospheric ozone layer is caused by all GHGs.
(b) Thinning of the stratospheric ozone layer is caused by CO₂ emission.
(c) Thinning of the stratospheric ozone layer is caused by Cf Cs.
(d) None of the above.
Answer:
(a) Thinning of the stratospheric ozone layer is caused by all GHGs.

22. Choose the incorrect statement:
(a) Trees are the best remedy for urban particulate matter.
(b) Forests act as sink for CO₂.
(c) Forests act as lungs of planet.
(d) All the above statements are not correct.
Answer:
(d) All the above statements are not correct.

23. Indicate the correct statement:
(a) Noise pollution can cause severe stomach ache.
(b) The threshold pain of noise pollution is about 120 decibels.
(c) Marine animals are not affected by noise pollution.
(d) None of the above.
Answer:
(b) The threshold pain of noise pollution is about 120 decibels.

24. Choose the incorrect statement:
(a) Limit generation is the most important consideration in managing radioactive wastes.
(b) Dilution and dispersion are adopted for wastes having high radioactivity.
(c) Much of the radioactivity in nuclear reactors and accelerator is very short-lived.
(d) Concentrate and confine process is another method of radioactive material disposal.
Answer:
(b) Dilution and dispersion are adopted for wastes having high radioactivity.