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Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 7 Cell Cycle

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Samacheer Kalvi 11th Bio Botany Cell Cycle Text Book Back Questions and Answers

I. Multiple Choice Questions
Choose the correct answer
Question 1.
The correct sequence in cell cycle is …………… .
(a) S-M-G1-G2
(b) S-G1-G2-M
(c) G1-S-G2-M
(d) M-G-G2-S
Answer:
(c) G1-S-G2-M

Question 2.
If mitotic division is restricted in G1 phase of the cell cycle then the condition is known as …………… .
(a) S Phase
(b) G2 Phase
(c) M Phase
(d) G₀ Phase
Answer:
(d) G₀ Phase

Question 3.
Anaphase promoting complex APC is a protein degradation machinery necessary for proper mitosis of animal cells. If APC is defective in human cell, which of the following is expected to occur?
(a) Chromosomes will be fragmented
(b) Chromosomes will not condense
(c) Chromosomes will not segregate
(d) Recombination of chromosomes will occur
Answer:
(b) Chromosomes will not condense

Question 4.
In S phase of the cell cycle …………… .
(a) Amount of DNA doubles in each cell
(b) Amount of DNA remains same in each cell
(c) Chromosome number is increased
(d) Amount of DNA is reduced to half in each cell
Answer:
(a) Amount of DNA doubles in each cell

Question 5.
Centromere is required for …………… .
(a) Transcription
(b) Crossing over
(c) Cytoplasmic cleavage
(d) Movement of chromosome towards pole
Answer:
(d) Movement of chromosome towards pole

Question 6.
Synapsis occur between …………… .
(a) mRNA and ribosomes
(b) Spindle fibres and centromeres
(c) Two homologous chromosomes
(d) A male and a female gamete
Answer:
(c) Two homologous chromosomes

Question 7.
In meiosis crossing over is initiated at …………… .
(a) Diplotene
(b) Pachytene
(c) Leptotene
(d) Zygotene
Answer:
(b) Pachytene

Question 8.
Colchicine prevents the mitosis of the cells at which of the following stage …………… .
(a) Anaphase
(b) Metaphase
(c) Prophase
(d) Interphase
Answer:
(b) Metaphase

Question 9.
The paring of homologous chromosomes on meiosis is known as …………… .
(a) Bivalent
(b) Synapsis
(c) Disjunction
(d) Synergids
Answer:
(b) Synapsis

Question 10.
Anastral mitosis is the characteristic feature of …………… .
(a) Lower animals
(b) Higher animals
(c) Higher plants
(d) All living organisms
Answer:
(c) Higher plants

Question 11.
Write any three significance of mitosis.
Answer:
Exact copy of the parent cell is produced by mitosis (genetically identical).

1. Genetic stability – daughter cells are genetically identical to parent cells.
2. Repair of tissues – damaged cells must be replaced by identical new cells by mitosis.
3. Regeneration – Arms of star fish.

Question 12.
Differentiate between Mitosis and Meiosis.
Answer:
Difference Between Mitosis and Meiosis:

Difference Between Mitosis and Meiosis
Mitosis	Meiosis
1. One division	1. Two divisions
2. Number of chromosomes remains the same	2. Number of chromosomes is halved
3. Homologous chromosomes line up separately on the metaphase plate	3. Homologous chromosomes line up in pairs at the metaphase plate
4. Homologous chromosome do not pair up	4. Homologous chromosome pairup to form bivalent
5. Chiasmata do not form and crossing over never occurs	5. Chiasmata form and crossingover occurs
6. Daughter cells are genetically identical	6. Daughter cells are genetically different from the parent cells
7. Two daughter cells are formed	7. Four daughter cells are formed

Question 13.
Given an account of G₀ phase.
Answer:
Some cells exit G₁ and enters a quiescent stage called G₀, where the cells remain metabolically active without proliferation. Cells can exist for long periods in G₀ phase. In G₀ cells cease growth with reduced rate of RNA and protein synthesis. The G₀ phase is not permanent. Mature neuron and skeletal muscle cell remain permanently in G₀ .Many cells in animals remains in G₀ unless called onto proliferate by appropriate growth factors or other extracellular signals. G₀ cells are not dormant.

Question 14.
Differentiate Cytokinesis in plant cells and animal cells.
Answer:
1. Cytokinesis in Plant Cells:
Division of the cytoplasm often starts during telophase. In plants, cytokinesis cell plate grows from centre towards lateral walls centrifugal manner of cell plate formation. Phragmoplast contains microtubules, actin filaments and vesicles from golgi apparatus and ER. The golgi vesicles contains carbohydrates such as pectin, hemicellulose which move along the microtubule of the pharagmoplast to the equator fuse, forming a new plasma membrane and 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.

2. Cytokinesis in Animal Cells:
It is a contractile process. The contractile mechanism contained in contractile ring located inside the plasma membrane. The ring consists of a bundle of microfilaments assembled from actin and myosin. This fibril helps for the generation of a contractile force. This force draws the contractile ring inward forming a cleavage furrow in the cell surface dividing the cell into two.

Question 15.
Write about Pachytene and Diplotene of Prophase I.
Answer:
1. Pachytene: At this stage bivalent chromosomes are clearly visible as tetrads. Bivalent of meiosis I consists 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 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.

2. 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. Chiasmata are chromatin structures at sites where recombination has been taken place. They are specialised 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. 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 lampbrush chromosome.

Samacheer Kalvi 11th Bio Botany Cell Cycle Additional Questions and Answers

I. Multiple Choice Questions
Choose the correct answer:
Question 1.
Most of the neurons in the brain are in …………… stage.
(a) G₁
(b) S
(c) G₂
(d) G₀
Answer:
(d) G₀

Question 2.
Short, constricted region in the chromosome is …………… .
(a) Kinetochore
(b) Centromere
(c) Satellite
(d) Telomere
Answer:
(b) Centromere

Question 3.
Robert Brown discovered the nucleus in the cells of …………… roots.
(a) Mirabilas
(b) Orchid
(c) Moringa
(d) Oryza
Answer:
(b) Orchid

Question 4.
Scientist who described chromosomes for the first time is …………… .
(a) Robert Brown
(b) Anton van Leeuwenhoek
(c) Boveri
(d) Anton Schneider
Answer:
(d) Anton Schneider

Question 5.
Number of chromosomes in onion cell is …………… .
(a) 8
(b) 16
(c) 32
(d) 64
Answer:
(a) 16

Question 6.
Longest part of the cell cycle is …………… .
(a) Prophase
(b) G₁ Phase
(c) Interphase
(d) Sphase
Answer:
(c) Interphase

Question 7.
Eukaryotic cells divides every …………… .
(a) 12
(b) 24
(c) 1
(d) 6
Answer:
(b) 24

Question 8.
Cell cycle was discovered by …………… .
(a) Singer & Nicolson
(b) Prevost & Dumans
(c) Schleider & Schwann
(d) Boveri
Answer:
(b) Prevost & Dumans

Question 9.
G₀ stage is called as …………… stage.
(a) Quiescent
(b) Metabolically active
(c) Synthesis of DNA
(d) Replication
Answer:
(a) Quiescent

Question 10.
…………… protein acts as major check point in phase.
(a) Porins
(b) Kinases
(c) Cyclins
(d) Ligases
Answer:
(c) Cyclins

Question 11.
Replication of DNA occurs at …………… phase.
(a) G₀
(b) G₁
(c) S
(d) G₂
Answer:
(c) S

Question 12.
Condensation of interphase chromosomes into mitotic forms is done by …………… proteins.
(a) MPF
(b) APF
(c) AMF
(d) MAF
Answer:
(a) MPF

Question 13.
Which of the following is also called as direct division?
(a) Amitosis
(b) Meiosis
(c) Mitosis
(d) Reduction division
Answer:
(a) Amitosis

Question 14.
Cells of mammalian cartilage undergoes …………… .
(a) Amitosis
(b) Meiosis
(c) Mitosis
(d) Equational division
Answer:
(a) Amitosis

Question 15.
Yeast cells undergo …………… .
(a) Open mitosis
(b) Closed mitosis
(c) Amitosis
(d) Meiosis
Answer:
(b) Closed mitosis

Question 16.
…………… is the longest phase in mitosis.
(a) Anaphase
(b) Telophase
(c) Prophase
(d) Interphase
Answer:
(c) Prophase

Question 17.
The DNA protein complex present in the centromere is …………… .
(a) Cyclin
(b) Kinesis
(c) MPF
(d) Kinetochore
Answer:
(d) Kinetochore

Question 18.
…………… protein induces the break down of cohesion proteins leading to chromatid separation during mitosis.
(a) APC
(b) MPF
(c) Cyclin
(d) Kinetochore
Answer:
(a) APC

Question 19.
Regeneration of arms of star fish is due to …………… .
(a) Meiosis
(b) Amitosis
(c) Mitosis
(d) Budding
Answer:
(c) Mitosis

Question 20
…………… is called as reduction division.
(a) Meiosis
(b) Mitosis
(c) Amitosis
(d) Budding
Answer:
(a) Meiosis

Question 21.
Bivalents occur at …………… stage.
(a) Zygotene
(b) Leptotene
(c) Pachytene
(d) Diplotene
Answer:
(a) Zygotene

Question 22.
Recombination of chromosomes occur at …………… .
(a) Zygotene
(b) Leptotene
(c) Pachytene
(d) Diplotene
Answer:
(c) Pachytene

Question 23.
Terminalisation of chiasmata occurs at …………… .
(a) Zygotene
(b) Leptotene
(c) Diakinesis
(d) Pachytene
Answer:
(c) Diakinesis

Question 24.
Number of daughter cells formed at the end of Meiosis I is …………… .
(a) 2
(b) 4
(c) 1
(d) 0
Answer:
(a) 2

Question 25.
…………… division leads to genetic variability.
(a) Mitotic
(b) Amitotic
(c) Meiotic
(d) Equational
Answer:
(c) Meiotic

Question 26.
Crossing over occurs at …………… stage.
(a) Leptotene
(b) Zygotene
(c) Pachytene
(d) Diplotene
Answer:
(c) Pachytene

Question 27.
Which of the following is not a mitogen?
(a) Giberellin
(b) Ethylene
(c) Kinetin
(d) Colchicine
Answer:
(d) Colchicine

Question 28.
In plants mitosis occurs at …………… cells.
(a) Sclerenchyma
(b) Meristem
(c) Xylem
(d) Parenchyma
Answer:
(b) Meristem

Question 29.
Which of the following alone is formed in the division of plant cells?
(a) Aster
(b) Centrioles
(c) Spindle
(d) Microtubules
Answer:
(c) Spindle

Question 30.
Amphiastral type cell division is seen in …………… cells.
(a) Fungal
(b) Algal
(c) Plant cells
(d) Animal
Answer:
(d) Animal

II. Very Short Answer Type Questions (2 Marks)

Question 1.
Name the two types of nuclear division.
Answer:
The two types of nuclear division:

1. Mitosis and
2. Meiosis.

Question 2.
Define Cell Cycle.
Answer:
A series of events leading to the formation of new cell is known as cell cycle.

Question 3.
Who discovered the Cell Cycle?
Answer:
Prevost & Dumans in 1824.

Question 4.
Draw a tabular column showing the duration of various phase in the cell cycle of human cell.
Answer:
A tabular column showing the duration of various phase in the cell cycle of human cell:

Cell cycle of a proliferating human cell
Phase	Time Duration (in hrs)
1. G2	1. 11
2. S	2. 8
3. G2	3. 4
4. M	4. 1

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

Question 6.
Which is the longest phase of cell cycle? What happens during that phase?
Answer:
Interphase is the longest phase. Cells are metabolically active and involved in protein synthesis and growth.

Question 7.
Name the phases which comprises the Interphase.
Answer:
The phases which comprises the Interphase:

1. G₁ Phase
2. S Phase and
3. G₂ Phase.

Question 8.
Name the proteins involved in the activation of genes & their proteins to perform cell division.
Answer:
Kinases & Cyclins.

Question 9.
What do you mean by G₀ stage?
Answer:
G₀ stage is called as quiescent stage, where the cells remain metabolically active without proliferation.

Question 10.
What is the role of MPF in Cell cycle?
Answer:
Maturation Promoting Factor (MPF) brings about condensation of interphase chromosomes into the mitotic form.

Question 11.
Distinguish between Karyokinesis & Cytokinesis.
Answer:
Between Karyokinesis & Cytokinesis:

• Karyokinesis: Karyokinesis refers to the nuclear division.
• Cytokinesis: Cytokinesis refers to the cytoplasmic division.

Question 12.
Point out any two cell – types which remain G₀ phase.
Answer:
Mature neurons and Skeletal muscle cells.

Question 13.
Why amitosis is called as incipient cell division?
Answer:
Amitosis is also called incipient cell division. Since there is no spindle formation and chromatin material does not condense.

Question 14.
List out the disadvantages of Amitosis.
Answer:
The disadvantages of Amitosis:

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

Question 15.
Mitosis also called as equational division – Justify.
Answer:
At the end of mitosis the number of chromosomes in the parent and the daughter (Progeny) cells remain the same so it is also called as equational division.

Question 16.
Enumerate the stages of mitosis.
Answer:
Mitosis is divided into four stages prophase, metaphase, anaphase and telophase.

Question 17.
Define an aster.
Answer:
In 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 18.
What is metaphase plate?
Answer:
The alignment of chromosome into compact group at the equator of the cell is known as metaphase plate.

Question 19.
What is Kinetochore?
Answer:
Kinetochore is a DNA – Protein complex present in the centromere DNA, where the microtubules are attached. It is a trilaminar disc like plate.

Question 20.
How will you calculate the length of the S period.
Answer:
Length of the S period = Fraction of cells in DNA replication × generation time.

Question 21.
Which type of cell division occurs in reproductive cells? What will be the result?
Answer:
Meiosis takes place in the reproductive organs. It results in the formation of gametes with half the normal chromosome number.

Question 22.
Define Synapsis.
Answer:
In Zygotene, pairing of homologous chromosomes takes place and it is known as synapsis.

Question 23.
What do you understand by independent assortment?
Answer:
The random distribution of homologous chromosomes in a cell in Metaphase I is called independent assortment.

Question 24.
Define Mitogen. Give an example.
Answer:
The factors which promote cell cycle proliferation is called mitogen.
Example: gibberellin. These increase mitotic rate.

Question 25.
What are mitotic poisons.
Answer:
Certain chemical components act as inhibitors of the mitotic cell division and they are called mitotic poisons.

Question 26.
Distinguish between Anastral & Amphiastral.
Answer:
Between Anastral & Amphiastral:
Anastral:

1. This is present only in plant cells.
2. No asters or centrioles are formed only spindle fibres are formed during cell division.

Amphiastral:

1. This is found in animal cells.
2. Aster and centrioles are formed at each pole of the spindle during cell division.

Question 27.
Draw a simple diagram to show the Amitosis.
Answer:
The Amitosis:

III. Short Answer Type Questions (3 Marks)

Question 1.
What is the role of nucleus in the cell?
Answer:
The role of nucleus in the cell:

• Control activities of the cell.
• Genetic information copied from cell to cell while the cell divides.
• Hereditary characters are passed onto new individuals when gametic cells fuse together in sexual reproduction.

Question 2.
What are restriction points? Mention its role in Cell cycle.
Answer:
The checkpoint called the restriction point at the end of G₁ it determines a cells 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.

Question 3.
Point out the reasons responsible for the arresting of the cell in G₁ phase?
Answer:
Cells are arrested in G₁ due to:

• Nutrient deprivation
• Lack of growth factors or density dependant inhibition
• Undergo metabolic changes and enter into G₀ state.

Question 4.
Write a note on G₀ phase.
Answer:
Some cells exit G₁ and enters a quiescent stage called G₀, where the cells remain metabolically active without proliferation. Cells can exist for long periods in G₀ phase. In G₀ cells cease growth with reduced rate of RN A and protein synthesis. The G₀ phase is not permanent. Mature neuron and skeletal muscle cell remain permanently in G₀. Many cells in animals remains in G₀ unless called onto proliferate by appropriate growth factors or other extracellular signals. G₀ cells are not dormant.

Question 5.
List out the events taking place in S – Phase.
Answer:
S Phase – Synthesis phase – cells with intermediate amounts of DNA Growth of the cell continues as replication of DNA occur, protein molecules called histones are synthesised and attach to the DNA. The centrioles duplicate in the cytoplasm. DNA content increases from 2C to 4C.

Question 6.
Distinguish between Karyokinesis & Cytokinesis.
Answer:
Karyokinesis:

1. Involves division of nucleus.
2. Nucleus develops a constriction at the center and becomes dumbellshaped.
3. Constriction deepens and divides the nucleus into two.

Cytokinesis:

1. Involves division of cytoplasm.
2. Plasma membrane develops a constriction along nuclear constriction.
3. It deepens centripetally and finally divides the cell into two cells.

Question 7.
Explain the differences between closed and open mitosis.
Answer:
Between closed and open mitosis:

1. In closed mitosis, the nuclear envelope remains intact and chromosomes migrate to opposite poles of a spindle within the nucleus. Example: Yeast and slime molds.
2. In open mitosis, the nuclear envelope breaks down and then reforms around the 2 sets of separated chromosome. Example: Most plants and animals cells.

Question 8.
What happens to plant cells at the end of Telophase in Mitosis?
Answer:
In plants, phragmoplast are formed between the daughter cells. Cell plate is formed between the two daughter cells, reconstruction of cell wall takes place. Finally the cells are separated by the distribution of organelles, macromolecules into two newly formed daughter cells.

Question 9.
Bring out the significance of Meiosis.
Answer:
The significance of Meiosis:

• Meiosis maintains a definite constant number of chromosomes in organisms.
• Crossing over takes place and exchange of genetic material leads to variations among species. These variations are the raw materials to evolution. Meiosis leads to genetic variability by partitioning different combinations of genes into gametes through independent assortment.
• Adaptation of organisms to various environmental stress.

Question 10.
Differentiate between the mitosis of Plant Cell & Animal Cell.
Answer:
Plants:

1. Centrioles are absent
2. Asters are not formed
3. Cell division involves formation of a cell plate
4. Occurs mainly at meristem.

Animals:

1. Centrioles are present
2. Asters are formed
3. Cell division involves furrowing and cleavage of cytoplasm
4. Occurs in tissues throughout the body.

Question 11.
Explain briefly about Endomitosis.
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 do not separate to form chromosomes, but remain closely associated with each other. 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 over thousands of intimately associated, or synapsed, chromatids. Example: Polytene chromosome.

Question 12.
How G₀ cells help in Closing Technology?
Answer:
Since the DNA of cells in G₀, do not replicate. The researcher are 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 synchronised with cytoplasm of the recipient egg, which developed into the clone Dolly.

IV. Long Answer Type Questions (5 Marks)

Question 1.
Draw and label the various stages of Prophase I.
Answer:
Label the various stages of Prophase I:

Question 2.
Explain in detail about the various stages of Prophase I.
Answer:
The various stages of Prophase I:
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 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 synaptonemal complex. The complex formed by the homologous chromosomes are called as bivalent (tetrads).

4. Pachytene – At this stage bivalent chromosomes are clearly visible as tetrads. Bivalent of meiosis I consists 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 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.

6. Chiasmata: Chiasmata are chromatin structures at sites where recombination has been taken place. They are specialised 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. 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 lampbrush chromosome.

7. Diakinesis – Terminalisation of chiasmata. Spindle fibres assemble. Nuclear envelope breaks down. Homologous chromosomes become short and condensed. Nucleolus disappears.

Question 3.
Describe the process of Cytokinesis in Plant cell & Animal Cell.
Answer:
1. Cytokinesis in Plant Cell: Division of the cytoplasm often starts during telophase. In plants, cytokinesis cell plate grows from centre towards lateral walls – centrifugal manner of cell plate formation. Phragmoplast contains microtubules, actin filaments and vesicles from golgi apparatus and ER. The golgi vesicles contains carbohydrates such as pectin, hemicellulose which move along the microtubule of the pharagmoplast to the equator fuse, forming a new plasma membrane and the materials which are placed there 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.

2. Cytokinesis in Animal Cells:
It is a contractile process. The contractile mechanism contained in contractile ring located inside the plasma membrane. The ring consists of a bundle of microfilaments assembled from actin and myosin. This fibril helps for the generation of a contractile force. This force draws the contractile ring inward forming a cleavage furrow in the cell surface dividing the cell into two.

Question 4.
What are the significances of Mitosis.
Answer:
Exact copy of the parent cell is produced by mitosis (genetically identical).

1. Genetic stability – Daughter cells are genetically identical to parent cells.
2. Growth – As multicellular organisms grow, the number of cells making up their tissue increases. The new cells must be identical to the existing ones.
3. Repair of tissues – Damaged cells must be replaced by identical new cells by mitosis.
4. Asexual reproduction – Asexual reproduction results in offspring that are identical to the parent. Example Yeast and Amoeba.
5. In flowering plants, structure such as bulbs, corms, tubers, rhizomes and runners are produced by mitotic division. When they separate from the parent, they form a new individual. The production of large numbers of offsprings in a short period of time, is possible only by mitosis. In genetic engineering and biotechnology, tissues are grown by mitosis (i.e. in tissue culture).
6. Regeneration – Arms of star fish

Question 5.
Explain the various phases in Cell Cycle.
Answer:
The different phases of cell cycle are as follows:
1. Interphase: Longest part of the cell cycle, but it is of extremely variable length. At first glance the nucleus appears to be resting but this is not the case at all. The chromosomes previously visible as thread like structure, have dispersed. Now they are actively involved in protein synthesis, at least for most of the interphase. C – Value is the amount in picograms of DNA contained within a haploid nucleus.

2. G₁ Phase: The first gap phase – 2C amount of DNA in cells of G₁ The cells become metabolically active and grows 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 it determines a cells 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:

3. Nutrient deprivation: Lack of growth factors or density dependant inhibition. Undergo metabolic changes and enter into G₀ state. Biochemicals inside cells activates the 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.

4. G₀ Phase: Some cells exit G₁ and enters a quiescent stage called G₀, where the cells remain metabolically active without proliferation. Cells can exist for long periods in G₀ phase. In G₀ cells cease growth with reduced rate of RNA and protein synthesis. The G₀ phase is not permanent. Mature neuron and skeletal muscle cell remain permanently in G₀. Many cells in animals remains in G₀ unless called on to proliferate by appropriate growth factors or other extracellular signals. G₀ cells are not dormant.

5. S phase – Synthesis phase – cells with intermediate amounts of DNA. Growth of the cell continues as replication of DNA occur, protein molecules called histones are synthesised and attached to the DNA. The centrioles duplicate in the cytoplasm. DNA content increases from 2C to 4C.

6. G₂ – The second Gap phase – 4C amount of DNA in cells of G₂ and mitosis. Cell growth continues by protein and cell organelle synthesis, mitochondria and chloroplasts divide. DNA content remains as 4C. Tubulin is synthesised and microtubules are formed. Microtubles organise to form spindle fibre. The spindle begins to form and nuclear division follows.

One of the proteins synthesized only in the G₂ period is known as Maturation Promoting Factor (MPF). It brings about condensation of interphase chromosomes into the mitotic form. DNA damage checkpoints operates in G₁ S and G₂ phases of the cell cycle.

Question 6.
List out the important features of Chromosomes.
Answer:
The four important features of the chromosome are:
1. The shape of the chromosome is specific: The long, thin, lengthy structured chromosome contains a short, constricted region called centromere. A centromere may occur any where along the chromosome, but it is always in the same position on any given chromosome. The number of chromosomes per species is fixed: For example the mouse has 40 chromosomes, the onion has 16 and humans have 46.

2. Chromosomes occur in pairs: The chromosomes of a cell occur in pairs, called homologous pairs. One of each pair come originally from each parent. Example, human has 46 chromosomes, 23 coming originally from each parent in the process of sexual reproduction. Chromosomes are copied: Between nuclear divisions, whilst the chromosomes are uncoiled and cannot be seen, each chromosome is copied. The two identical structures formed are called chromatids.

V. Higher Order Thinking Skills (HOTs)

Question 1.
Given that the average duplication time of E.coli is 20 minutes. How much time will two E.coli cells takes to become 32 cells?
Answer:

One cells takes 80 minutes to form 16 cells. If 2 cells undergoes division simultaneously, it take 160 minutes (2 hours 40 minutes) to form 32 cells.

Question 2.
Complete the cell cycle by filling the gaps with respective phases.
Answer:
X= S phase or Synthesis phase
Y= M phase or Mitosis phase
Z= G₀ phase

Question 3.
Telophase is reverse of prophase – Comment
Answer:
Events in Prophase:

1. Nuclear membrane disappears
2. Nucleolus disappear
3. Spindle fibre begins to form
4. Chromosomes threads condeme to form chromosomes

Events in Telophase:

1. Nuclear membrane reappears
2. Nucleolus reappears
3. Spindle fibre disappears
4. Chromosomes decondeme to form chromosomes

Question 4.
Name the pathological condition when uncontrolled cell division occurs.
Answer:
Uncontrolled cell division & abnormal growth of cells leads to the pathological condition called tumor or cancer.

Question 5.
Microspores are produced in the multiples of four, why?
Answer:
Microspores are haploid spores produced from diploid microspores mother cells. Each microspores mother cell (2n) undergoes meiosis producing four Microspores (n). Because a complete meiotic division yields 4 cells. Thus microspores are produced in multiples of four.

Question 6.
Between Prokaryotes & Eukaryotes, which cell has a shorter cell division time.
Answer:
Prokaryotes like bacteria undergo simple form of cell division called binary fission which will get completed with in a hour, whereas Eukaryotic cell division (mitosis) takes nearly 24 hours to get completed. Hence Prokaryotes have shorter cell division time.

Question 7.
Though Prokaryotic cell division differs from Eukaryotic cell division, both show certain common aspects during cell division. Explain.
Answer:
Whether a cell is Prokaryote or Eukaryote, while undergoing division, the following events must occur in common.

1. Replication of DNA.
2. Cytokinesis at the end of cell division.

Question 8.
An anther has 1204 pollen grains. How many Pollen mother cells must have been there to produce them?Explain.
Answer:
301 – Pollen mother cells: 301 Pollen mother cells undergo meiosis producing 1204 pollen grains. Because at the end of meiosis, each pollen mother cells produces 4 pollen grains.

Question 9.
A cell has 32 chromosomes. It undergoes mitosis. What will be the chromosome number during metaphase?
Answer:
During S phase of interphase, the genetic material of the cell is duplicated. So during metaphase, the chromosome number(chromatid number) will be doubled thus 64 chromosomes (chromatids) will be present.

Question 10.
Why sibilings show disimilarities?
Answer:
Though born to same parents, siblings show dissimilarities and variation due to the crossing over and recombination of chromosomes during meiosis.

Question 11.
Ramu’s met with an accident while riding cycle and got wounded in his leg. After few days, the wound was healed and the skin becomes normal. How?
Answer:
Ramu’s wound was healed because of the mitotic division. As a result of mitosis, new cells are produced and damaged tissues were repaired resulting the damaged skin to become normal.

Question 12.
A flower of tomato plant following the process of sexual reproduction produces 240 viable seeds. What is the minimum number of microspore mother cells involved in this process?
Answer:
60 microspore mother cells are involved in providing 240 pollen grains. Because each microspore mother cell undergoes meiosis producing four pollen grains (i.e. 60 × 4 = 240). Each pollen grain produces two male gametes of which one undergoes true fertilization of ovule producing seeds. Other male gamete participate in double fertilization.

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Samacheer Kalvi 11th Bio Zoology Digestion and Absorption Text Book Back Questions and Answers

I. Multiple Choice Questions
Question 1.
Choose the incorrect sentence from the following:
(a) Bile juice emulsifies the fat.
(b) Chyme is a digestive acidic food in stomach.
(c) Pancreatic juice converts lipid into fatty acid and glycerol.
(d) Enterokinase stimulates the secretion of pancreatic juice.
Answer:wer:
(d) Enterokinase stimulates the secretion of pancreatic juice.

Question 2.
What is chyme ?
(a) The process of conversion of fat into small droplets.
(b) The process of conversion of micelles substances of glycerol into fatty droplet.
(c) The process of preparation of incompletely digested acidic food through gastric juice.
(d) The process of preparation of completely digested liquid food in midgut.
Answer:
(c) The process of preparation of incompletely digested acidic food through gastric juice.

Question 3.
Which of the following hormones stimulate the production of pancreatic juice and bicarbonate?
(a) Angiotensin and epinephrine
(b) Gastrin and insulin
(c) Cholecysokinin and secretin
(d) Insulin and glucagon
Answer:
(c) Cholecysokinin and secretin.

Question 4.
The sphincter of Oddi guards ……………
(a) Hepatopancreatic duct
(b) Common bile duct
(c) Pancreatic duct
(d) Cystic duct
Answer:
(a) Hepatopancreatic duct

Question 5.
In small intestine, active absorption occurs in case of ……………
(a) Glucose
(b) Amino acids
(c) Na
(d) All the above
Answer:
(d) All the above

Question 6.
Which one is incorrectly matched?
(a) Pepsin – stomach
(b) Renin – liver
(c) Trypsin – intestine
(d) Ptyalin – mouth
Answer:
(b) Renin – liver

Question 7.
Absorption of glycerol, fatty acids and monoglycerides takes place by …………….
(a) Lymph vessels within villi
(b) Walls of stomach
(c) Colon
(d) Capillaries within villi
Answer:
(a) Lymph vessels within villi

Question 8.
First step in digestion of fat is ……………..
(a) Emulsification
(b) Enzyme action
(c) Absorption by lacteals
(d) Storage in adipose tissue
Answer:
(a) Emulsification

Question 9.
Enterokinase takes part in the conversion of ……………..
(a) Pepsinogen into pepsin
(b) Trypsinogen into trypsin
(c) Protein into polypetide
(d) Caseinogen into casein
Answer:
(b) Trypsinogen into trypsin

Question 10.
Which of the following combinations are not matched?

Column I	Column II
(a) Bilirubin and biliverdin	(i) intestinal juice
(b) Hydrolysis of starch	(ii) Amylases
(c) Digestion of fat	(iii) Lipases
(d) Salivary gland	(iv)Parotid

Answer:
(a) Bilirubin and biliverdin – (i) intestinal juice

Question 11.
Match column I with column II and choose the correct option

Column – I	Column – II
(P) Small intestine	(i) Largest factory
(Q)  Pancreas	(ii) Absorption of Water
(R)  Liver	(iii) Carrying electrolytic solution
(S)  Colon	(iv) Digestion and absorption

(a) (P – iv) (Q – iii) (R – i) (S – ii)
(b) (P – iii) (Q – ii) (R – i) (S – iv)
(c)(P – iv) (Q – iii) (R – i) (S – ii)
(d) (P – ii) (Q – iv) (R – iii) (S – i)
Answer:
(a) (P-iv ) ( Q – iii ) ( R- i ) ( S – ii)

Question 12.
Match column I with column II and choose the correct option

Column I	Column II
(p) Small intestine	(i) 23 cm
(q) Large intestine	(ii) 4 meter
(r) Oesophagus	(iii) 12.5 cm
(s) Pharynx	(iv)1.5 meter

(a) (P – iv) (Q – ii) (R – i) (S – iii)
(b) (P – ii) (Q – iv) (R – i) (S – iii)
(c) (P – i) (Q – iii) (R – ii) (S – iv)
(d) (P – iii) (Q – i) (R – ii)(S – iv)
Answer:
(b) (P – ii ) ( Q – iv ) ( R – i ) (S – iii)

Question 13.
Match column I with column II and choose the correct option Column – I Column – II

Column I	Column II
(p) Lipase	(i) Starch
(q) Pepsin	(ii) Cassein
(r) Renin	(iii) Protein
(s) Ptyalin	(iv) Lipid

(a) (P – iv) (Q – ii)(R – i)(S – iii)
(b) (P – iii) (Q – iv) (R – ii) (S – i)
(c) (P – iv) (Q – iii) (R – ii) (S – i)
(d) (P – iii) (Q – ii) (R – iv) (S – i)
Answer:
(c) (P – iv ) ( Q – iii ) ( R – ii) ( S – i)

Question 14.
Which of the following is not the function of liver?
(a) Production of insulin
(b) Detoxification
(c) Storage of glycogen
(d) Production of bile
Answer:
(a) Production of insulin

Question 15.
Assertion : (A) Large intestine also shows the presence of villi like small intestine.
Reason : (B) Absorption of water takes place in large intestine.
(a) Both A and B are true and B is the correct explanation of A
(b) Both A and B are true but B is not the correct explanation of A
(c) A is true but B is false
(d) A is false but B is true
Answer:
(d) A is false but B is true

Question 16.
Which of the following is not true regarding intestinal villi?
(a) They possess microvilli.
(b) They increase the surface area.
(c) They are supplied with capillaries and the lacteal vessels.
(d) They only participate in digestion of fats.
Answer:
(d) They only participate ¡n digestion of fats.

II. Short Answer Questions

Question 17.
Why are villi present in the intestine and not in the stomach?
Answer:
In small intestine digestion gets completed and the absorption of digested food materials like glucose, amino acids, fatty acids and glycerol takes place. The food materials are to be retained in the intestine by increasing the surface area. Hence villi are present in the intestine. Stomach is the temporary storing organ of food. In the stomach, HCl, pepsin, renin and lipase are secreted. These are concerned with digestion. Hence villi are not present in the stomach.

Question 18.
Bile juice contains no digestive enzymes, yet It is important for digestion. Why?
Answer:
Liver does not secrete digestive enzymes. It contains bile pigments hilirubin and hiliverdin which are the break down products of haemoglobin of dead RBCs. bile salts, cholesterol and phospholipids. Bile helps in emulsification of fats. Bile salts reduce the surface tension of fat droplets and break them into small globules. Bile also activates lipase to digest lipids.

Question 19.
List the chemical changes that starch molecule undergoes from the time it reaches the small intestine.
Answer:
In the small intestine, starch digestion gets completed. The pancreatic juice contains pancreatic amylases which acts on polysaccharide and convert into disaccharides (maltose). These cretions of the Brunner’s gland along with the secretions of the intestinal glands constitute the intestinal juice or succus entericus. It contains maltose, lactose and sucrose. These convert maltose, lactose and sucrose into monosaccharides, glucose and fructose.

In the small intestine, complex carbohydrates are converted into simple ghicose. fructose and galactose. These are absorbed by active transport.

Question 20.
How do proteins differ from fats in their energy value and their role in the body?
Answer:
The calorific value and physiological fiel value of one gram of protein are 5.65 Kcal and 4 Kcal respectively. Fat has a calorific value of 9.45 Kcal and the physiological fuel value of 9 Kcal per gram. Proteins are the source of amino acids required for growth and repair of body cells. They are stored in the body only to a certain extent. They replace the worn out protoplasm. They are important for the production of many enzymes. hormones and plasma. The catabolism of amino acids releases toxic nitrogenous wastes which are removed by the
kidneys.

Question 21.
Digestive secretions are secreted only when needed. Discuss.
Answer:
Digestive glands are exocrine glands which secrete biological catalysts called enzymes. These enzyme convert the complex. insoluble foodmaterials like carbohydrates, proteins and lipids into simplex, soluble food materials like glucose and fructose, amino acids and fatty acids and glycerol. These digestive ccretions act only when food materials are available in the alimental)’ canal.

Question 22.
Label the given diagram.
Answer:
A – Right and left hepatic duct of liver.
B – Common bile duct.
C – Pancreatic duct (duct of Wirsung)
D – Sphincter of oddi
E – Gall bladder.

In-Text Questions Solved

Question 1.
Though the bile juice of liver has no digestive enzyme but is very essential for proper digestion of food, especially of the fats. Discuss the following?
(a) What is composition of bile?
(b) How it helps in digestion of fats and other nutrients of food?
(c) How it helps in absorption of fats?
Answer:
The bile contains bile pigments (bilirubin and biliverdin) as the break down products of hemoglobin of dead RBCs, bile salts, cholesterol and phospholipids but has no enzymes. Bile helps in emulsification of fats. Bile salts reduce the surface tension of fat droplets and break them into small globules. Bile also activates lipases to digest lipids.

Question 2.
What would happen if HCl is not secreted in the stomach?
Answer:
The gastric juice contains HCl and proenzymes. The proenzyme pepsinogen, on exposure to HCl gets converted into the active enzyme pepsin which converts proteins into proteoses and peptones (peptides). The HCl provides an acidic medium (pH – 1.8) which is optimum for pepsin, kills bacteria and other harmful organisms and avoids putrifaction. So, if HCl is not secreted in stomatch, digestion of protein and destruction of harmful micro organisms will be affected.

Question 3.
What features of the small intestine enables it to absorb digested food efficiently?
Answer:
Absorption is a process by which the end product of digestion passes through the intestinal mucosa into the blood and lymph. The villi in the lumen of ileum are the absorbing units, consisting of a lacteal duct in the middle surrounded by fine network of blood capillaries. The process of absorption involves active, passive and facilitated transport.

Question 4.
What happens to the protein molecules in food, from the time it is swallowed, to the time its products are built up in the cytoplasm of a muscle cell?
Answer:
Proteins and partially digested proteins in the chyme on reaching the intestine are acted upon by the proteolytic enzymes of pancreatic juice. Trypsin hydrolyses proteins into polypeptides and peptones, while chymotrypsin hydrolyses peptide bonds associated with specific amino acids. Proteins are source of amino acids required for growth and repair of body cells. They are stored in the body only to a certain extent; large quantities are excreted as nitrogenous waste.

Textbook Activities Solved

Test for Starch :
Add a few drops of iodine to the given warm food sample. If any starch is present in the given food sample it will change the colour of the iodine from brown to blue black.

2. Test for protein :
Mix the given food sample with 3 mL of water in a test tube. Shake the mixture, and then add a few drops of Biuret solution. If protein is present, the colour of the solution will change to purple.

3. Test for glucose :
Mix the given food sample with 3 mL of water in a test tubes. Shake the mixture, and then add a few drops of Benedict’s solution. Keep the test tube in a water bath and heat carefully. If glucose is present, the colour of the solution will change from blue to green to brick red depending upon the amount of glucose.

Entrance Examination Questions Solved
Choose the correct Answer
Question 1.
How pepsin is differing from trypsin? [DPMT 1993]
(a) It digests protein in acidic medium
(b) It digests protein in alkaline medium
(c) It digests carbohydrate in acidic medium
(d) It digests carbohydrate in alkaline medium
Answer:
(a) It digests protein in acidic medium

Question 2.
Human intestine is large because ………… [DPMT1996]
(a) Bacteria in the food moves slowly
(b) Substances of food digest slowly
(c) It provide more space for the absorption of digested food
(d) It provide more space for the storage of food
Answer:
(c) It provide more space for the absorption of digested food

Question 3.
How the epidermal cells in the stomach of vertebrate animal is protected against HCl? [NCERT1981]
(a) HCl is dilute
(b) Epidermal cells defense the function of HCl
(c) HCl is neutralized in stomach
(d) Epidermal cells covered with secretion of mucus
Answer:
(d) Epidermal cells covered with secretion of mucus

Question 4.
By what the major part of mammalian teeth is made up ?
(a) Root
(b) Pulp
(c) Dentin
(d) Enamel
Answer:
(c) Dentin

Question 5.
Enterokinase takes part in the conversion of what? [BHU 2000]
(a) Pepsinogen into pepsin
(b) Trypsinogen into trypsin
(c) Protein into polypetide
(d) Casernogen into casein
Answer:
(b) Trypsinogen into trypsin .

Question 6.
Secretin stimulates production of ………… [MPPMT 2002]
(a) Saliva
(b) Gastric juice
(c) Bile
(d) Pancreatic juice
Answer:
(d) Pancreatic juice

Question 7.
Pepsin acts in ………… [HPPMT 2001]
(a) Basic medium
(b) Acidic medium
(c) Neutral medium
(d) All type of medium
Answer:
(b) Acidic medium

Question 8.
Enzyme trypsin is secreted by ………… [AFMC 20031
(a) Duodenum
(b) Liver
(c) Pancreas
(d) Stomach
Answer:
(c) Pancreas

Question 9.
The number of teeth that grow twice in the human life is ………… [AFMC 2002, 2004]
(a) 4
(b) 12
(c) 20
(d) 28
Answer:
(c) 20

Question 10.
The number of teeth that grow once in the human life is ………… [DPMT BHU 1986]
(a) 4
(b) 12
(c) 20
(d) 28
Answer:
(d) 28

Question 11.
Cholesterol is synthesised in ………… [M.RPMT 2000]
(a) Bninner’s glands
(b) Liver
(c) Spleen
(d) Pancreas
Answer:
(b) Liver

Question 12.
Largest gland in human body is ………… [JK, CMEE 2003]
(a) Liver
(b) Pancreas
(c) Pituitary
(d) Thyroid
Answer:
(a) Liver

Question 13.
Muscular contraction of alimentary canal are ………… [CMC 2003]
(a) Circulation
(b) Deglutition
(c) Churning
(d) Peristalsis
Answer:
(d) Peristalsis

Question 14.
Fatty acids and glycerol are first absorbed by ………… [BV2000J]
(a) Lymph vessels
(b) Villi
(c) Blood capillaries
(d) Hepatic portal vein
Answer:
(a) Lymph vessels

Question 15.
Trypsin changes ………… [MPPMT 1995]
(a) Proteins into peptones
(b) Fats into fatty acids
(c) Starch and glycogen into maltose
(d) Maltose into its components
Answer:
(a) Proteins into peptones

Question 16.
Secretin hormone is produced by ………… [MPPMT 1995]
(a) Stomach
(b) Liver
(c) Intestine
(d) Pancreas
Answer:
(c) Intestine

Question 17.
Narrow distal part of stomach is ………… [MPPMT 1995]
(a) Cardiac
(b) Pharynx
(c) Duodenum
(d) Pylorus
Answer:
(d) Pylorus

Question 18.
pH suitable for ptyalin actions is ………… [AFMC 1996]
(a) 6 – 8
(b) 7 – 8
(c) 3 – 2
(d) 9 – 3
Answer:
(c) 3 – 2

Question 19.
What will happen if bile duct gets choked? [DPMT 1991]
(a) Feces become dry
(b) Acidic chyme will not be neutralised
(c) There will be little digestion in intestine
(d) Little absorption of fat will occur
Answer:
(b) Acidic chyme will not be neutralised

Question 20.
Digestion of both starch and protein is carried out by ………… [AFMC 19961
(a) Gastric juice
(b) Gastric lipase
(c) Pancreatic juice
(d) Ptyalin
Answer:
(c) Pancreatic juice

Question 21.
What is common among amylase, renin and trypsin? [CPMT 2000]
(a) All proteins
(b) Proteolytic enzymes
(c) Produced in stomach
(d) Act at pH lower than 7
Answer:
(a) All proteins

Question 22.
Enterokinase is ………… [BHU l997]
(a) Pancreatic hormone
(b) Intestine hormone
(c) Pancreatic enzyme
(d) Component of Intestinal juice
Answer:
(d) Component of Intestinal juice

Question 23.
Which enzyme initiates protein digestion? [MPPMT l997
(a) Pepsin
(b) Trypsin
(c) Aminopeptidase
(d) Carboxypeptidase
Answer:
(a) Pepsin

Question 24.
Enzyme which does not directly act upon food substrate is …………
(a) Trypsin
(b) Lipase
(c) Enterokinase
(d) Arnylopsin
Answer:
(c) Enterokinase

Question 25.
Pepsin is secreted by …………… [CPMT 1997]
(a) Peptic cells
(b) Zymogen cells of stomach
(c) Zymogcn cells of duodenum
(d) Pancreas
Answer:
(a) Peptic cells

Question 26.
Pepsinogen is activated by …………..
(a) Chymotrypsin
(b) Trypsin
(c) HCl
(d) Pepsin
Answer:
(c) HCl

Question 27.
Contraction of gall bladder is induced by …………
(a) Gastrin
(b) Cholecystokinin
(c) Secretin
(d) Enterogastrone
Answer:
(b) Cholecystokinin

Question 28.
Hormone that stimulates stomach to secrete gastric juice is …………
(a) Run
(b) Enterokinase
(c) Enterogastron
(d) Gastrin
Answer:
(d) Gastrin

Question 29.
Water is largely absorbed in ………… [CPMT 1999]
(a) Stomach
(b) Oesophagus
(c) Small intestine
(d) Colon
Answer:
(d) Colon

Question 30.
HCl is secreted by ………… [DPMT 2002]
(a) Zymogen cells
(b) Kupifer’s cells
(c) Oxyntic cells
(d) Mucous cells
Answer:
(c) Oxyntic cells

Question 31.
Jaundice is a disease of ………… [APMEE 1999]
(a) Kidney
(b) Liver
(c) Pancreas
(d) Duodenum
Answer:
(b) Liver

Question 32.
Which is different? [BHU 1999]
(a) Gastrin
(b) Secretin
(c) Ptyalin
(d) Glucagon
Answer:
(C) Ptyalin

Question 33.
Gastrin is ………… [BHU 1999]
(a) Hormone
(b) Enzyme
(c) Nutrient
(d) Digestive secretion
Answer:
(a) Hormone

Question 34.
Saliva contains enzyme ………… [CPAÍT 2003]
(a) Enterokinase
(b) Ptyalin/ Amylase
(c) Chymotrypsin
(d) Lipase
Answer:
(b) Ptyalin) Amylase

Question 35.
In human being cellulose is digested by …………
(a) Enzyme
(b) Symbiotic bacteria
(c) Symbiotic protozoAnswer:
(d) None of the above
Answer:
(b) Symbiotic bacteria

Question 36.
Enzyme lactase occurs in ………… [MPPMT 2000]
(a) Saliva
(b) Pancreatic juice
(c) Intestinal juice
(d) Stomach
Answer:
(c) Intestinal juice

Question 37.
Protein I enzyme is absent in ………… [MPPMT 2000]
(a) Saliva
(b) bile
(c) Pancreatic juice
(d) Intestinal juice
Answer:were:
(b) Bile

Question 38.
Dental formula shows ………… [MPPMT 2000]
(a) Structure of teeth
(b) Monophyodont or diphyodont condition
(c) Number and type of teeth in both jaws
(d) Number and type of teeth in one half of both jaws
Answer:
(c) Number and type of teeth ¡n both jaws

Question 39.
pH of gastric juice / stomach is …………
(a) 1.5-3.0
(b) 5.0-6.8
(c) 7.0 -9.0
(d) 6.0 -8.0
Answer:
(a) 1.5 -3.0

Question 40.
In case of taking food rich in lime juice, the action of ptyalin on starch is ………… [AuMS 2000]
(a) Enhanced
(b) Reduced
(c) Unaffected
(d) Stopped
Answer:
(b) Reduced

Question 41.
Bile salts take part in ………… [AMU 2000]
(a) Digestion of carbohydrates
(b) Brokedowri of proteins
(c) Emulsification of fat
(d) Absorption of glycerol
Answer:
(c) Emulsification of fat

Question 42.
Digestive juice contains catalytic agents called ………… [PMT 000]
(a) Vitamins
(b) Hormones
(c) Enzymes
(d) Nitrates
Answer:
(c) Enzymes

Question 43.
Which is not the function of liver? [DPMT 2001]
(a) Production of insulin
(b) Detoxification
(c) Storage of glycogen
(d) Production of bile
Answer:
(a) Production of insulin

Question 44.
Fat absorbed from gut is trAnswer:ported in blood …………
(a) Micelles
(b) Liposomes
(c) Chemomicrons
(d) Chlymicrons
Answer:
(a) Micelles

Question 45.
In small intestine, active absorption occurs in case of ………… [AMU 2001]
(a) Glucose
(b) Amino acids
(c) Na+
(d) All the above
Answer:
(d) All the above

Question 46.
Which one is not matched? [Har. PMT 2002]
(a) Pepsin – stomach
(b) Renin – liver
(e) Trypsin – intestine
(d) Ptyalin – mouth
Answer:
(b) Renin – liver

Question 47.
What is cholecystokinin?
(a) Bile pigment
(b) Gastro-intestinal hormone
(c) Enzyme
(d) Lipid
Answer:
(b) Gastro-intestinal hormone

Question 48.
Secretion of gastric juice is controlled by ………… [CPMT2002J
(a) Enterogesterone
(b) Cholecystokinin
(c) Gastrin
(d) Pepsin
Answer:
(c) Gastrin

Question 49.
Which one is wisdom teeth? [CPMT 2002]
(a) Third molar, four in number
(b) Third molar, two in number
(c) Second molar, four in number
(d) Second molar, two in number
Answer:
(a) Third molar, four in number

Question 50.
in humans, digestion is ………… [BHU 2002]
(a) Intercellular
(b) Intracellular
(c) Extracellular
(d) Both A and B
Answer:
(b) intracellular

Question 51.
Gall bladder takes part in ………… [RPMT 2002]
(a) Secretion of bile
(b) Storage of bile
(c) Formation of bile salts
(d) Formation of enzymes
Answer:
(b) Intracellular .

Question 51.
Gall bladder takes part in ………… [RPMT 2002]
(a) Secretion of bile
(b) Storage of bile
(c) Formation of bile salts
(d) Formation of enzymes
Answer:
(b) Storage of bile

Question 52.
Renin acts on milk protein and changes ………… [JIPMER 2002]
(a) Caesinogen into caesin
(b) Caesin into paracaesin
(c) Caesinogen into paracaesin
(d) Paracaesin into Caesinogen
Answer:
(a) Caesinogen into caesin

Question 53.
Glucose is stored in liver as ………… [AFMC 2003]
(a) Starch
(b) Glycogen
(c) Cellulose
(d) Sucrose
Answer:
(b) Glycogen

Question 54.
Absorption of glycerol, fatty acids and monoglycerides takes place by …………
(a) Lymph vessels within villi
(b) Walls of stomach
(c) Colon
(d) Capillaries within villi
Answer:
(a) Lymph vessels within villi

Question 55.
Which ones are bile salts?
(a) Haemoglobin and biliverdin
(b) Bilirubin and biliverdin
(c) Bilirubin and Haemoglobin
(d) Sodium glycolate and taurocholate
Answer:
(d) Sodium glycolate and taurocholate

Question 56.
Ptyalin is inactivated by a component of gastric juice called ……………. [Har PMT 2003]
(a) Pepsin
(b) HCl
(c) Renin
(d) Mucus
Answer:
(b) HCl

Question 57.
Epithelial cells involved in absorption of digested food have on their free surface ……………. [A IEEE 2003]
(a) Zymogen granules
(b) Pinocytic vesicles
(c) Phagocytic vesicles
(d) Microvilli
Answer:
(d) Microvilhi

Question 58.
First step in digestion of fat is …………… [BIIU 2003]
(a) Emulsification
(b) Enzyme action
(c) Absorption by lacteals
(d) Storage in adipose tissue
Answer:
(a) Emulsification

Question 59.
DNA-ase and RNA-ase are enzymes produced by …………… [BHU 2003]
(a) Salivary glands
(b) Pancreas
(c) Stomach
(d) Intestine
Answer:
(b) Pancreas

Question 60.
Carboxypeptidase is secreted by …………….
(a) Pancreas
(b) Stomach
(c) Salivary glands
(d) Intestine
Answer:
(a) Pancreas

Question 61.
Secretin and Cholecystokinin are digestive hormone. They are secreted in …………
(a) Pyloric stomach
(b) Duodenum
(c) Ileum
(d) Oesophagus
Answer:
(b) Duodenum

Question 62.
Crown of teeth is covered by ………… [AFMC 2005]
(a) Dentin
(b) Enamel
(c) (a) and (b) both
(d) None of these
Answer:

Question 63.
Both the crown and root of a teeth is covered by a layer of bony hard substance called ………… [J&K CET 2005]
(a) Enamel
(b) Dentin
(c) Bony socket
(d) Cementum
Answer:
(d) Cementum

Question 64.
Lysozymes are found in ………… [MPPMT 2004]
(a) Saliva
(b) Tears
(c) (a) and (b) both
(d) Mitochondria
Answer:
(c) (a) and (b) both

Question 65.
Which of the following is not present in pancreatic juice? [HPPMT 2005]
(a) Trypsinogen
(b) Chymotrypsin
(c) Parasitic
(d) lipase
Answer:
(c) Parasitic

Question 66.
Which of the following statement is not correct? [NEET 2015]
(a) Brunner’s glands are present in the submucosa of stomach and secrete pepsinogen
(b) Goblet cells are present in the mucosa of intestine and secrete mucus.
(c) Oxyntic cells are present in the mucosa of stomach and secrete HCI.
(d) Acini are present in the pancreas and secrete carboxypeptidse.
Answer:
(a) Brunner’s glands are present in the submucosa of stomach and secrete pepsinogen

Question 67.
Which hormones stimulate the production of pancreatic juice and bicarnates? [NEET20I6]
(a) Cholecystokinin and secretin
(b) Insulin and glucagon
(c) Angiotensin and epinephrine
(d) Gastnn and Insulin
Answer:
(a) Cholecystokinin and secretin

Question 68.
In the stomach, gastric acid is secreted by the ………… [AIPMT / NEET 2016]
(a) gastrin secreting cells
(b) parietal cells
(c) peptic cells
(d) acidic cells
Answer:
(b) parietal cells

Question 69.
The enzymes that is not present is succus entericus is ……………… [RE-A IPMTNEE T 2015]
(a) lipase
(b) maltase
(c) nucleases
(d) nucleosîdase
Answer:
(c) nucleases

Question 70.
Which of the following are not polymerase? [NEET 2017]
(a) Proteins
(b) Polysaccharides
(c) Lipids
(d) Nucleic acids
Answer:
(c) Lipids3

Question 71.
A baby aged two years is admitted to play school and passes through a dental check-up. The dentist observed that the boy had twenty teeth. Which teeth were absent? [NEET 2017]
(a) Canines
(b) Pre-Molars
(c) Molars
(d) Incisors
Answer:
(b) Pre-Molars

Question 72.
Which cells of crypts of Leiberkuhn’ secrete antibacterial lysozyme? [NEET 2017]
(a) Paneth cells
(b) Zymase cells
(c) Kupifer cells
(d) Argentaffin cells
Answer:
(a) Paneth cells

Question 73.
The hepatic portal veins drains blood to liver from ………… [NEET 2017]
(a) Stomach
(b) Kidneys
(c) Intestine
(d) Heart
Answer:
(c) Intestine

Question 74.
Which of the following options best represent the enzyme composition of pancreatic juice? [NEET2O17]
(a) Amylase, pepsin, trypsinogen, maltase
(b) Peptidase, Amylase, pepsin, renin
(c) Lipase, amylase, trypsinogen, procarboxypeptidase
(d) Amylase, peptidase, trypsinogen, renin.
Answer:
(c) Lipase, amylase, trypsinogen, procarboxypeptidase

Question 75.
Good vision depends on adequate intake of carotene rich food. Select the best option from the following statements. [NEET 2017]
(a) Vitamin A derivatives are formed from carotene.
(b) The photo pigments are embedded in the membrane discs of the inner segments.
(c) Retinal is a derivative of vitamin A
(d) Retinal is light absorbing part of all the visual photopigments.
OPTION
(a) a, c and d
(b) a and e.
(c) b, e and d
(d) a and b
Answer:
(a) a, c and d

Samacheer Kalvi 11th Bio Zoology Digestion and Absorption Additional Questions & Answers

I. Multiple Choice Questions
Choose the correct Answer
Question 1.
Which of the following is the last phase of the process of digestion?
(a) ingestion
(b) assimilation
(c) eqestion
(d) digestion
Answer:
(c) eqestion

Question 2.
Which of the following teeth are the cutting teeth?
(a) incisors
(b) canines
(c) premolars
(d) molars
Answer:
(a) incisors

Question 3.
Plague formed on the teeth are mineral salts of ……………..
(a) sodium
(b) magnesium and manganese
(c) potassium
(d) calcium and magnesium
Answer:
(d) calcium and magnesium

Question 4.
Which muscle regulates the opening of oesophagus into the stomach?
(a) pyloric sphincter
(b) cardiac sphincter
(c) anal sphincter
(d) epiglottis
Answer:
(a) pyloric sphincter

Question 5.
The ‘V’ shaped part of the small intestine is …………
(a) ileum
(b) jejunum
(c) duodenum
(d) colon
Answer:
(c) duodenum

Question 6.
…………….. is the longest part of the small intestine
(a) Ileum
(b) Jejunum
(c) Duodenum
(d) Rectum
Answer:
(a) Ileum

Question 7.
Villi are concerned with
(a) secretion of enzymes
(b) secretion of mucus
(c) digestion of food
(d) absorption of digested food
Answer:
(d) absorption of digested food.

Question 8.
Peyer’s patches produce
(a) monocytes
(b) lymphocytes
(c) basophils
(d) neutrophils
Answer:
(b) lymphocytes

Question 9.
Which of the following are non-functional in human beings?
(a) Small intestine
(b) Duodenum
(c) lleum
(d) Vermiform appendix
Answer:
(d) Vermiform Appendix

Question 10.
…………… is formed of loose connective tissue containing nerves, blood, lymph vessels and sympathetic nerve fibres that control the secretions of intestinal juice.
(a) Serosa
(b) Sub-mucosa
(c) Mucosa
(d) Muscularis
(b) Sub-mucosa

Question 11.
………… secrete HCl in the stomach.
(a) Peptic cells
(b) Goblet cells
(c) Oxyntic cells
(d) Zymogen cells
Answer:
(c) Oxyntic cells

Question 12.
…………. stores bile secreted by the liver.
(a) Pancreatic duct
(b) Duodenum
(c) Hepatic duct
(d) Gall bladder
Answer:
(d) Gall bladder

Question 13.
Detoxification is one of the functions of –
(a) small intestine
(b) liver
(c) pancreas
(d) stomach
Answer:
(b) liver

Question 14.
Which is the antibacterial agent present in the saliva?
(a) Ptyalin
(b) Mucus
(c) Lysozyme
(d) Bicarbonates
Answer:
(c) Lysozyme

Question 15.
Salivary amylase hydrolyses …………
(a) proteins
(b) carbohydrate
(c) fats
(d) vitamins
Answer:
(b) carbohydrate

Question 16.
Pepsin is a enzyme ………….
(a) amylolytic
(b) lipolytic
(c) proteolytic
(d) proenzyme
Answer:
(c) proteolytic

Question 17.
Which enzyme acts on milk protein in infants in the presence of calcium ions?
(a) Pepsin
(b) Chymotrypsin
(c) Lipase
(d) Renin
Answer:
(d) Renin

Question 18.
Which is the active enzyme?
(a) Pepsinogen
(b) Trypsin
(c) Trypsinogen
(d) Chymotrypsinogen
Answer:
(b) Trypsin

Question 19.
What are the break down products of haemoglobin?
(a) Bile salts
(b) Bilirubin and biliverdin
(c) Phospholipids
(d) Cholesterol
Answer:
(d) Cholesterol

Question 20.
Which of the following does not contain any enzyme?
(a) Gastric juice
(b) Bile
(c) Pancreatic juice
(d) Succus entericus
Answer:
(b) Bile

Question 21.
Identify the odd one ……….
(a) maltose
(b) lactose
(c) sucrose
(d) sucrase
Answer:
(d) sucrase

Question 22.
At the end of digestion, fats are converted into ………..
(a) amino acids
(b) glycerol and fatty acids
(c) glucose
(d) galactose
Answer:
(b) glycerol and fatty acids

Question 23.
Protein coated fat globules are called …………
(a) micelles
(b) villi
(c) microvilli
(d) chylomicrons
Answer:
(d) chylomicrons

Question 24.
The calorific value of carbohydrates is ………….
(a) 9.45
(b) 4.1
(c) 3.5
(d) 6.5
Answer:
(b) 4.1

Question 25.
Retaining feces in the rectum due to poor intake of fibre in the diet is called ………..
(a) vomiting
(b) indigestion
(c) constipation
(d) diarrhoea
Answer:
(c) constipation

Question 26.
Which one of the following occurs due to reverse peristalsis?
(a) Vomiting
(b) Diarrhoea
(c) Constipation
(d) Indigestion
Answer:
(a) Vomiting

Question 27.
The failure of the liver to break down haemoglobin is called as ……………….
(a) gall stone
(b) jaundice
(c) appendicitis
(d) hernia
Answer:
(c) appendicitis

Question 28.
Degeneration and destruction of liver cells resulting in abnormal blood vessel and bile duct leading to the formation of fibrosis is known as ………………
(a) gall stones
(b) jaundice
(c) hernia
(d) liver cirrhosis
Answer:
(d) liver cirrhosis

Question 29.
What causes obstruction in the cystic duct, hepatic duct and hepato-pancreatic duct?
(a) Gall stones
(b) Jaundice
(c) Hernia
(d) Liver cirrhosis
Answer:
(a) Gall stones

Question 30.
Appendicitis is the inflammation of the ……………..
(a) caecum
(b) liver
(c) appendix
(d) rectum
Answer:
(c) appendix

Question 31.
Diarrhoea happens due to ……………….
(a) taking in of more volume of water
(b) defective liver
(c) inflammation of the appendix
(d) inability of colon to absorb fluid from the feces
Answer:
(d) inability of colon to absorb fluid from the feces

Question 32.
The erosion of the mucosa in the stomach or duodenum is referred to as ……………….
(a) stomach ache
(b) peptic ulcer
(c) duodenal cancer
(d) diarrhoea
Answer:
(b) peptic ulcer

Question 33.
The storage of excess of body fat in adipose tissue is ……………
(a) peptic ulcer
(b) kwashiorkor
(c) jaundice
(d) obesity
Answer:
(d) obesity

II. Fill in the Blanks

Question 1.
Breakdown of macromolecules of food into micromolecules is ………….
Answer:
Digestion.

Question 2.
The tearing teeth are called as ……………….
Answer:
Canines

Question 3.
The deposition of calcium and magnesium salts on the teeth forms a hard layer called ……………..
Answer:
Tartar/calculus/plaque.

Question 4.
The oral cavity leads into a short common passage for food and air called
Answer:
Pharynx.

Question 5.
Two masses of lymphoid tissues called are located at the sides of the pharynx.
Answer:
Tonsils.

Question 6.
The opening of the stomach into the duodenum is guarded by the ……………
Answer:
Pyloric sphincter.

Question 7.
The inner wall of stomach has many folds called which unfolds to accommodate a large meal.
Answer:
Gastric rugae.

Question 8.
The wall of the duodenum has glands which secrete mucus and enzymes.
Answer:
Brunner’s.

Question 9.
The ileal mucosa has numerous vascular projections called …………..
Answer:
Villi.

Question 10
are the mucus secreting goblet cells and lymphoid tissue which produce lymphocytes.
Answer:
Peyer‘s patches.

Question 11.
The wall of the small intestine bears crypts between the base of the villi called …………….
Answer:
Crypts of Leiberkuhn.

Question 12.
Caecum and vermiform appendix are the important site for digestion in herbivores.
Answer:
Cellulose.

Question 13.
The dilations of colon are called ……………..
Answer:
Haustra

Question 14.
The enlargement of anal column causes …………..
Answer:
Piles / haemorrhoids

Question 15
………. is the outermost layer of the wall of the alimentary canal.
Answer:
Serosa.

Question 16.
The enzyme secreting cells of gastric glands are called cells.
Answer:
Peptic or zymogen.

Question 17.
The cells secrete HCl in the stomach.
Answer:
Parietal or oxyntic.

Question 18.
The is the largest gland in our body.
Answer:
Liver.

Question 19.
Each lobe of liver is covered by a thin connective tissue sheath called the ……………
Answer:
Glisson’s capsule.

Question 20.
The detoxifies toxic substances.
Answer:
Liver.

Question 21.
The opening of the hepato-pancreatic duct into the duodenum is guarded by a sphincter called the sphincter of
Answer:
Oddi.

Question 22
gland is both exocrine and endocrine.
Answer:
Pancreas.

Question 23.
The successive waves of muscular contraction by which bolus passes down through the oesophagus is called
Answer:
Peristalsis.

Question 24
activates pepsinogen.
Answer:
HCl.

Question 25.
Renin converts caseinogen into casein is the presence of ions.
Answer:
Calcium.

Question 26
activates trypsinogen.
Answer:
Enterokinase.

Question 27.
Trypsin hydrolyses proteins into polypeptides and
Answer:
Peptones.

Question 28.
mucosa into the blood and lymph.
Answer:
Absorption.

Question 30.
The chylomicrons are transported into the within the intestinal villi.
Answer:
Lacteals.

Question 31.
Large intestine absorbs more amount of vitamins, some minerals and certain
drugs.
Answer:
Water.

Question 32.
The eqestion of feces through the anal opening is called ……………..
Answer:
Defecation

Question 33
are the major source of cellular fuel which provides energy.
Answer:
Carbohydrates.

Question 34
are the best reserved food stored in our body.
Answer:
Lipids.

Question 35.
Marasmus is the deficiency disease.
Answer:
Protein.

Question 36
is the digestive disorder in which the food is not properly digested leading to a feeling of fullness of stomach.
Answer:
Indigestion.

Question 37
is caused due to hepatitis viral infection.
Answer:
Jaundice.

Question 38.
The crystallized cholesterol in the bile forms …………….
Answer:
Gall stones.

Question 39
…………. is a structural abnormality in which superior part of the stomach protrudes slightly
Answer:
Hiatus hernia.

Question 40.
Treatment for diarrhoea is known as therapy.
Answer:
Oral hydration.

Question 41
refers to an eroded area of the tissue lining in the stomach or duodenum.
Answer:
Peptic ulcer.

Question 42.
Ulcer is mostly due to infection caused by the bacterium
Answer
Helicobacter pylori.

Question 43
…………… is caused due to the storage of excess of body fat in adipose tissue.
Answer:
Obesity.

III. Answer the following Questions

Question 1.
What are the uses of food?
Answer:
The food we eat provides energy and organic substances for growth and the replacement of worn out and damaged tissues. It regulates and coordinates the various activities that take place in the body.

Question 2.
What are the components of food?
Answer:
The components of food are carbohydrates, proteins, lipids, vitamins, minerals, fibre and water.

Question 3.
Why do we need a digestive system?
Answer:
The food that we eat are macromolecules, and inabsorbable. These are to be broken down . into smaller micro-molecules in absorbable forms. This is done by digestive system.

Question 4.
Why do plants not require a digestive system?
Answer:
Plants are autotrophs. They prepare their own food using CO₂, H₂O in the presence of sunlight trapped by the chlorophyll pigment present in the leaves. There is no need of digestive system for plants as they use the starch as such.

Question 5.
What is the function of the digestive system?
Answer:
The function of the digestive system is to bring the nutrients, water and electrolytes from the external environment into every cell in the body through the circulatory system.

Question 6.
Define:
(a) Ingestion : The taking in of food is called ingestion.
(b) Digestion : The break down of the macromolecules into micromolecules by the action of digestive enzymes is called digestion.
(c) Absorption : The taking in of digested food materials into the blood stream is called absorption.
(d) Assimilation : The conversion of absorbed food materials into components of cells is called assimilation.
(e) Egestion : The elimination of the undigested substances from the body is called egestion.

Question 7.
What is thecodont dentition?
Answer:
Every tooth in human beings is embedded in a socket in the jaw bone. This type of attachment is called the codont dentition.

Question 8.
What is diphyodont dentition?
Answer:
Human beings and many mammals form two sets of teeth during their life time, a set of 20 temporary milk teeth which gets replaced by a set of 32 permanent teeth. This type of dentition is called diphyodont dentition.

Question 9.
What is heterodont dentition?
Answer:
Human beings have four different types of teeth namely incisors, canines, premolars and molars. This is known as heterodont dentition.

Question 10.
What is known as the dental formula of human beings?
Answer:
The arrangement of teeth in each half of the upper and lower jaw in the order of I, C, P and M can be represented by the dental formula. The dental formula of man is 2123 / 2123.

Question 11.
What is gingivitis?
Answer:
Mineral salts like calcium and magnesium are deposited on the teeth and form a hard layer of ‘tartar’ or calculus called plaque. If the plaque formed on teeth is not removed regularly, it would spread down the tooth into the narrow gap between the gums and enamel and causes inflammation, called gingivitis, which leads to redness and bleeding of the gums and to bad smell.

Question 12.
What is enamel?
Answer:
The hard chewing surface of the teeth is made of enamel and helps in mastication of food.

Question 13.
What is the function of tongue?
Answer:
Tongue helps in intake of food, chew and mix food with saliva, to swallow food and also to speak. The upper surface of the tongue has small projections called papillae with taste buds.

Question 14.
What is the function of the cardiac sphincter?
Answer:
A cardiac sphincter or gastro esophageal sphincter regulates the opening of oesophagus into the stomach.

Question 15.
What is gastro oesophagus reflex disorder?
Answer:
If the cardiac sphincter does not contract properly during the churning action of the stomach the gastric juice with acid may flow back into the oesophagus and cause heart bum, resulting in GERD (Gastro Oesophagus Reflex Disorder).

Question 16.
What is the function of the pyloric sphincter?
Answer:
The opening of the stomach into the duodenum is guarded by the pyloric sphincter. It periodically allows partially digested food to enter the duodenum and also prevents regurgitation of food.

Question 17.
What are gastric rugae?
Answer:
The inner wall of stomach has many folds called gastric rugae which unfolds to accommodate a large meal.

Question 18.
Write a short note on the small intestine.
Answer:
The small intestine assists in the final digestion and absorption of food. It is the longest part of the alimentary canal and has three regions, a ‘U’ shaped duodenum (25 cm long), a long coiled middle portion jejunum (2.4m long) and a highly coiled ileum (3.5 m long).

The wall of the duodenum has Brunner’s glands which secrete mucus and enzymes. Ileum is the longest part of the small intestine and opens into the caecum of the large intestine.

The ileal mucosa has numerous vascular projections called villi which are involved in the process of absorption and the cells lining the villi produce numerous microscopic projections called microvilli giving a brush border appearance that increase the surface area enormously.

Along with Villi, the ileal mucosa also contain mucus secreting goblet cells and lymphoid tissue known as Peyer’s patches which produce lymphocytes. The wall of the small intestine bears crypts between the base of villi called crypts of Leiberkuhn.

Question 19.
Explain the structure of the large intestine.
Answer:
The large intestine consists of caecum, colon and rectum. The caecum is a small blind pouch like structure that opens into the colon and it possesses a narrow finger like tubular projection called vermiform appendix.

Both caecum and vermiform appendix are large in herbivorous animal and act as an important site for cellulose digestion with the help of symbiotic bacteria. The colon is divided into four regions – an ascending, a transverse, a descending part and a sigmoid colon.

The colon is lined by dilations called haustra (singular – haustrum) (Figure). The “S” shaped sigmoid colon (pelvic colon) opens into the rectum. Rectum is concerned with temporary storage of feces. The rectum open out through the anus. The anus is guarded by two anal sphincter muscles. The anal mucosa is folded into several vertical folds and contains arteries and veins called anal columns. Anal column may get enlarged and causes piles or haemorrhoids.

Question 20.
Explain the internal structure of the gut.
Answer:
The wall of the alimentary canal from oesophagus to rectum consists of four layers (Figure) namely serosa, muscularis, sub-mucosa and mucosa. The serosa (visceral peritoneum) is the outermost layer and is made up of thin squamous epithelium with some connective tissues. Muscularis is made of smooth circular and longitudinal muscle fibres with a network of nerve cells and parasympathetic nerve fibres which controls peristalsis.

The submucosal layer is formed of loose connective tissue containing nerves, blood, lymph vessels and the sympathetic nerve fibres that control the secretions of intestinal juice. The innermost layer lining the lumen of the alimentary canal is the mucosa which secretes mucous.

Question 21.
Explain the salivary glands.
Answer:
There are three pairs of salivary glands in the mouth. They are the largest parotids gland in the cheeks, the sub-maxillary sub-mandibular in the lower jaw and the sublingual beneath the tongue. These glands have ducts such as Stenson’s duct, Wharton’s duct and Bartholin’s duct or duct of Rivinis respectively (Figure). The salivary juice secreted by the salivary glands reaches the mouth through these ducts. The daily secretion of saliva from salivary glands ranges from 1000 to 1500 mL.

Question 22.
Write a short note on gastric glands.
Answer:
The wall of the stomach is lined by gastric glands. Chief cells or peptic cells or zymogen cells in the gastric glands secrete gastric enzymes and Goblet cells secrete mucus. The Parietal or oxyntic cells secrete HCl and an intrinsic factor responsible for the absorption of Vitamin B₁₂ called Castle’s intrinsic factor.

Question 23.
Explain the structure of liver.
Answer:
The liver, the largest gland in our body, is situated in the upper right side of the abdominal cavity, just below the diaphragm. The liver consists of two major left and right lobes; and two minor lobes. These lobes are connected with diaphragm. Each lobe has many hepatic lobules (functional unit of liver) and is covered by a thin connective tissue sheath called the Glisson’s capsule.

Liver cells (hepatic cells) secrete bile which is stored and concentrated in a thin muscular sac called the gall bladder. The duct of gall bladder (cystic duct) along with the hepatic duct from the liver forms the common bile duct.

The bile duct passes downwards and joins with the main pancreatic duct to form a common duct called hepato-pancreatic duct. The opening of the hepato-pancreatic duct into the duodenum is guarded by a sphincter called the sphincter of Oddi (Figure). Liver has high power of regeneration and liver cells are replaced by new ones every 3-4 weeks.

Question 24.
Write the functions of liver.
Answer:
Apart from bile secretion, the liver also performs several functions.

• Destroys aging and defective blood cells.
• Stores glucose in the form of glycogen or disperses glucose into the blood stream with the help of
• pancreatic hormones.
• Stores fat soluble vitamins and iron.
• Detoxifies toxic substances.
• Involves in the synthesis of non-essential amino acids and urea.

Question 25.
Write on the secretions of the pancreas.
Answer:
The second largest gland in the digestive system is the Pancreas, which is a yellow coloured, compound elongated organ consisting of exocrine and endocrine cells. It is situated between the limbs of the ‘U’ shaped duodenum.

The exocrine portion secretes pancreatic juice containing enzymes such as pancreatic amylase, trypsin and pancreatic lipase and the endocrine part called Islets of Langerhans secretes hormones such as insulin and glucagon. The pancreatic duct directly opens into the duodenum.

Question 26.
Explain the digestion in the buccal cavity.
Answer:
The smell, sight and taste as well as the mechanical stimulation of food in the mouth, triggers a reflex action which results in the secretion of saliva. The mechanical digestion starts in the mouth by grinding and chewing of food. It is called mastication.

The saliva contain water, electrolytes (Na⁺, K⁺, CL^– and HCO₃), salivary amylase (ptyalin), antibacterial agent lysozyme and a lubricating agent mucus (a glycoprotein). The mucus in saliva prepares the food for swallowing by moistening, softening, lubricating and adhering the masticated food into a bolus.

About 30 percent of polysaccharide, starch is hydrolyzed by the salivary amylase enzyme into disaccharides (maltose). The bolus is then passed into the pharynx and then into the oesophagus by swallowing or deglutition. The bolus further passes down through the oesophagus to the stomach by successive waves of muscular contraction called peristalsis. The gastro oesphageal sphincter controls the passage of food into the stomach.

Question 27.
Explain the digestion in the stomach.
Answer:
Food remains in the stomach for 4 to 5 hours, the rhythmic peristaltic movement chums and mixes the food with gastric juice and make it into a creamy liquid called chyme. The gastric secretion is partly controlled by autonomic reflexes.

The secretion of gastric juice begins when the food is in the mouth. The gastric juice contains HCl and proenzymes. The proenzyme pepsinogen, on exposure to HCl gets converted into the active enzyme pepsin which converts proteins into proteoses and peptones (peptides).

The HCl provides an acidic medium (pH 1.8) which is optimum for pepsin, kills bacteria and other harmful organisms and avoids putrifaction. The mucus and bicarbonates present in the gastric juice play an important role in lubrication and protection of the mucosal epithelium from the eroding nature of the highly acidic HCl (Figure). Another proteolytic enzyme found in gastric juice of infants is renin which helps in the digestion of milk protein, caseinogen to casein in the presence of calcium ions. This enzyme secretion gradually reduces with aging.

Question 28.
Explain the digestion in the small intestine.
Answer:
The bile, pancreatic juice and intestinal juice are the secretions released into the small intestine. Movements generated by the muscularis layer of the small intestine helps in the thorough mixing of the food with various secretions in the intestine and thereby facilitate digestion.

The pancreatic juice contains enzymes such as trypsinogen, chymotrypsinogen, carboxypeptidases, pancreatic amalyses, pancreatic lipases and nucleases. Trypsinogen is activated by an enzyme, enterokinase, secreted by the intestinal mucosa into active trypsin, which in turn activates the enzyme chymotrypsinogen in the pancreatic juice.

The bile contains bile pigments (bilirubin and biliverdin) as the break down products of hemoglobin of dead RBCs, bile salts, cholesterol and phospholipids but has no enzymes. Bile helps in emulsification of fats. Bile salts reduce the surface tension of fat droplets and break them into small globules. Bile also activates lipases to digest lipids.

Proteins and partially digested proteins in the chyme on reaching the intestine are acted upon by the proteolytic enzymes of pancreatic juice. Trypsin hydrolyses proteins into polypeptides and peptones, while chymotrypsin hydrolyses peptide bonds associated with specific amino acids. The pancreatic amylase converts glycogen and starch into maltose.

Lipase acts on emulsified fat (triglycerides) and hydrolyses them into free fatty acid and monoglycerides. Monoglycerides are further hydrolysed to fatty acid and glycerol. Nucleases in the pancreatic juice break the nucleic acid into nucleotides and nucleosides.

The secretions of the Brunner’s gland along with the secretions of the intestinal glands constitute the intestinal juice or succus entericus. The enzymes in the intestinal juice such as maltase, lactase, sucrase (invertase), dipeptidases, lipases, nucleosidases act on the breakdown products of bile and pancreatic digestion.

The mucus along with the bicarbonate ions from the pancreas provides an alkaline medium (pH 7.8) for the enzymatic action. As a result of digestion, all macromolecules of food are converted into their corresponding monomeric units.

Question 29.
Explain the process of absorption of the digested food.
Answer:
Absorption is a process by which the end product of digestion passes through the intestinal mucosa into the blood and lymph. The villi in the lumen of ileum are the absorbing units, consisting of a lacteal duct in the middle surrounded by fine network of blood capillaries. The process of absorption involves active, passive and facilitated transport.

Small amounts of glucose, amino acids and electrolytes like chloride ions are generally absorbed by simple diffusion. The passage of these substances into the blood depends upon concentration gradients. However, some of the substances like fructose are absorbed with the help of the carrier ions like Na⁺.

This mechanism is called facilitated transport. Nutrients like amino acids, glucose and electrolytes like Na⁺ are absorbed into the blood against the concentration gradient by active transport. The insoluble substances like fatty acids, glycerol and fat soluble vitamins are first incorporated into small, spherical water soluble droplets called micelles and are absorbed into the intestinal mucosa where they are re-synthesized into protein coated fat globules called chylomicrons which are then transported into the lacteals within the intestinal villi and eventually empty into lymphatic duct.

The lymphatic ducts ultimately release the absorbed substances into the blood stream. While the fatty acids are absorbed by the lymph duct, other materials are absorbed either actively or passively by the capillaries of the villi (Figure). Water soluble vitamins are absorbed by simple diffusion or active transport. Transport of water depends upon the osmotic gradient. Absorption of substances in the alimentary canal takes place in mouth, stomach, small intestine and large intestine.

However maximum absorption takes place in the small intestine. Absorption of simple sugars, alcohol and medicines takes place in the stomach. Certain drugs are absorbed by blood capillaries in the lower side of the tongue and mucosa of mouth. Large intestine is also involved in absorption of more amounts of water, vitamins, some minerals and certain drugs.

Question 30.
Write a note on assimilation.
Answer:
Absorbed substances are transported through blood and lymph to the liver through the hepatic portal system. From the liver, nutrients are transported to all other regions of the body for utilization. All the body tissues utilize the absorbed substance for their activities and incorporate into their protoplasm, this process is called assimilation.

Question 31.
Write a paragraph on egestion.
Answer:
The digestive waste and unabsorbed substances in the ileum enter into the large intestine and it mostly contains fibre called roughage. The roughage is utilized by symbiotic bacteria in the large intestine for the production of substances like vitamin K and other metabolites. All these substances are absorbed in the colon along with water.

The waste is then solidified into fecal matter in the rectum. The fecal matter initiates a neural reflex causing an urge or desire for its removal. The egestion of feces through the anal opening is called defecation. It is a voluntary process and is carried out by a peristaltic movement.

Question 32.
Write on carbohydrates and lipids.
Answer:
Carbohydrates are sugar and starch. These are the major source of cellular fuel which provides energy. The caloric value of carbohydrate is 4.1 calories per gram and its physiological fuel value is 4 Kcal per gram. Lipids are fats and derivatives of fats, are also the best reserved food stored in our body which is used for production of energy. Fat has a caloric value of 9.45 Kcal and a physiological fuel value of 9 Kcal per gram.

Question 33.
Write a paragraph on proteins.
Answer:
Proteins are source of amino acids required for growth and repair of body cells. They are stored in the body only to a certain extent; large quantities are excreted as nitrogenous waste. The caloric value and physiological fuel value of one gram of protein are 5.65 Kcal and 4 Kcal respectively. According to ICMR (Indian Council of Medical Research) and WHO (World Health Organization), the daily requirement of protein for an average Indian is 1 gm per 1 kg body weight.

Question 34.
Explain the protein deficiency diseases.
Answer:
Growing children require more amount of protein for their growth and development. Protein deficient diet during early stage of children may lead to protein energy malnutrition such as Marasmus and Kwashiorkor. Symptoms are dry skin, pot-belly, oedema in the legs and face, stunted growth, changes in hair colour, weakness and irritability.

Marasmus is an acute form of protein malnutrition. This condition is due to a diet with inadequate carbohydrate and protein. Such children are suffer from diarrhoea, body becomes lean and weak (emaciated) with reduced fat and muscle tissue with thin and folded skin.

Question 35.
What is indigestion?
Answer:
It is a digestive disorder in which the food is not properly digested leading to a feeling of fullness of stomach. It may be due to inadequate enzyme secretion, anxiety, food poisoning, over eating and spicy food.

Question 36.
What is constipation?
Answer:
In this condition, the feces are retained within the rectum because of irregular bowel movement due to poor intake of fibre in the diet and lack of physical activities.

Question 37.
What is vomiting?
Answer:
It is reverse peristalsis. Harmful substances and contaminated food from stomach are ejected through the mouth. This action is controlled by the vomit center located in the medulla oblongata. A feeling of nausea precedes vomiting.

Question 38.
Write a short note on jaundice.
Answer:
It is the condition in which liver is affected and the defective liver fails to break down haemoglobin and to remove bile pigments from the blood. Deposition of these pigments changes the colour of eye and skin yellow. Sometimes, jaundice is caused due to hepatitis viral infections.

Question 39.
Write a short note on liver cirrhosis.
Answer:
Chronic disease of liver results in degeneration and destruction of liver cells resulting in abnormal blood vessel and bile duct leading to the formation of fibrosis. It is also called deserted liver or scarred liver. It is caused due to infection, consumption of poison, malnutrition and alcoholism.

Question 40.
Write a short note on gall stones.
Answer:
Any alteration in the composition of the bile can cause the formation of stones in the gall bladder. The stones are mostly formed of crystallized cholesterol in the bile. The gall stone causes obstruction in the cystic duct, hepatic duct and also hepato-pancreatic duct causing pain, jaundice and pancreatitis.

Question 41.
What is appendicitis?
Answer:
It is the inflammation of the vermiform appendix, leading to severe abdominal pain. The treatment involves the removal of appendix by surgery. If treatment is delayed the appendix may rupture and results in infection of the abdomen, called peritonitis.

Question 42.
What is hiatus hernia or diaphragmatic hernia?
Answer:
It is a structural abnormality in which superior part of the stomach protrudes slightly above the diaphragm. The exact cause of hiatus hernias is not known. In some people, injury or other damage may weaken muscle tissue, by applying too much pressure (repeatedly) on the muscles around the stomach while coughing, vomiting, and straining during bowel movement and lifting heavy object.

Heart bum is also common in those with a hiatus hernia. In this condition, stomach contents travel back into the oesophagus or even into oral cavity and causes pain in the center of the chest due to the eroding nature of acidity.

Question 43.
Write a short note on diarrhoea.
Answer:
It is the most common gastrointestinal disorder worldwide. It is sometimes caused by bacteria or viral infections through food or water. When the colon is infected, the lining of the intestine is damaged by the pathogens, thereby the colon is unable to absorb fluid.

The abnormal frequency of bowel movement and increased liquidity of the fecal discharge is known as diarrhoea. Unless the condition is treated, dehydration can occur. Treatment is known as oral hydration therapy. This involves drinking plenty of fluids – sipping small amounts of water at a time to rehydrate the body.

Question 44.
Write a paragraph on peptic ulcer.
Answer:
It refers to an eroded area of the tissue lining (mucosa) in the stomach or duodenum. Duodenal ulcer occurs in people in the age group of 25 – 45 years. Gastric ulcer is more common in persons above the age of 50 years. Ulcer is mostly due to infections caused by the bacterium Helicobacter pylori. It may also be caused due to uncontrolled usage of aspirin or certain anti-inflammatory drugs. Ulcer may also be caused due to smoking, alcohol, caffeine and psychological stress.

Question 45.
What is obesity?
Answer:
It is caused due to the storage of excess of body fat in adipose tissue. It may induce hypertension, atherosclerotic heart disease and diabetes. Obesity may be genetic or due to excess intake of food, endocrine and metabolic disorders.

Question 46.
Write a note on BMI.
Answer:
Degree of obesity is assessed by body mass index (BMI). A normal BMI range for adult is 19 – 25; above 25 is considered as obese. BMI is calculated as body weight in Kg, divided by the square of body height in meters. For example, a 50 Kg person with a height of 160 cms would have a BMI of 19.5.
That is BMI = 50 / 1.62= 19.5

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Samacheer Kalvi 11th Bio Botany Taxonomy and Systematic Botany Text Book Back Questions and Answers

I. Multiple Choice Questions
Choose the correct answer
Question 1.
Specimen derived from non – original collection serves as the nomenclatural type, when original specimen is missing. It is known as …………… .
(a) Holotype
(b) Neotype
(c) Isotype
(d) Paratype
Answer:
(b) Neotype

Question 2.
Phylogenetic classification is the most favoured classification because it reflects …………… .
(a) Comparative Anatomy
(b) Number of flowers produced
(c) Comparative cytology
(d) Evolutionary relationships
Answer:
(d) Evolutionary relationships

Question 3.
The taxonomy which involves the similarities and dissimilarities among the immune system of different taxa is termed as …………… .
(a) Chemotaxonomy
(b) Molecular systematics
(c) Serotaxonomy
(d) Numerical taxonomy
Answer:
(c) Serotaxonomy

Question 4.
Which of the following is a flowering plant with nodules containing filamentous nitrogen – fixing micro – organisms?
(a) Crotalaria juncea
(b) Cycas revoluta
(c) Cicer arietinum
(d) Casuarina equisetifolia
Answer:
(a) Crotalaria juncea

Question 5.
Flowers are zygomorphic in …………… .
(a) Ceropegia
(b) Thevetia
(c) Datura
(d) Solarium
Answer:
(a) Ceropegia

Question 6.
What is the role of national gardens in conserving biodiversity – discuss.
Answer:
Botanical Gardens plays the following important roles.

1. Gardens with aesthetic value which 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. Garden is used for self-instruction or demonstration purposes.

Question 7.
Where will you place the plants which contain two cotyledons with cup shaped thalamus?
Answer:
The plants which contain two cotyledons with cup shaped thalamus:

• Class: Dicotyledonae
• Subclass: Polypetalae
• Series: Calyciflorae

Question 8.
How does molecular markers work to unlock the evolutionary history of organisms?
Answer:
Molecular taxonomy using molecular markers like RAPD’s etc, helps in establishing the relationship between the members of different taxonomic groups at DNA level. Thus it helps to unlock the evolutionary history of  organisms.

Question 9.
Give the floral characters of Clitoria ternatea.
Answer:
Flower: Bracteate, bracteolate, bracteoles usually large, pedicellate, heterochlamydeous, complete, bisexual, pentamerous, zygomorphic and hypogynous.

Question 10.
How will you distinguish Solanaceae members from Liliaceae members?
Answer:

Characters	Solanaceae members	Liliaceae members
Root	Branched taproot	Adventitious fibrous root
Stem	Herbaceous	bulbous / rhizomatous
Leaf	Reticulate venation	Parallel venation
Inflorescence	Solitary and axillary cyme	Simple or branched raceme
Flowers	pentamerous	Trimerous
Androecium	Stamens 5, epipetalous	Stamens 6, epiphyllous
Gynoecium	Bicarpellary	Tricarpellary

Textbook Activities Solved

Question 1.
Write common name and scientific name of 10 different plants around your home.
Answer:

Common name of plants	Scientific name of plants
1. Mango	1. Mangifera indica
2. Banana	2. Musa paradisiaca
3. Shoe flower	3. Hibiscus rosa – sinensis
4. Neem	4. Azadirachta indica
5. Coconut	5. Cocus nucifera
6. Rice	6. Oryza sativa
7. Onion	7. Allium cepa
8. Tomato	8. Solanum lycopersicum
9. Carrot	9. Dacus carota
10. Brinjal	10. Solanum melongena

Question 2.
Can you identify this?

(a) Name the family
(b) Write the binomial
(c) List the economic uses
Answer:
(a) Liliaceae family
(b) Aloe vera
(c) List the economic uses:

• The gel – like part of Aloe vera contain proteole enzymes which repairs and rejunuveates the skin.
• It acts as a great conditioner for hair.
• Aloe juice acts as coolant and soothes gastric problems like ulcer.

Entrance Examination Questions Solved

Choose the correct answer:
Question 1.
Leaves become modified into spines in …………… . [AIPMT – 2015]
(a) Silk Cotton
(b) Opuntia
(c) Pea
(d) Onion
Answer:
(b) Opuntia

Question 2.
Keel is the characteristic feature of flower of …………… . [AIPMT – 2015]
(a) Tomato
(b) Tulip
(c) Indigofera
(d) Aloe
Answer:
(c) Indigofera

Question 3.
Perigynous flowers are found in …………… .[AIPMT – 2015]
(a) Rose
(b) Guava
(c) Cucumber
(d) China rose
Answer:
(a) Rose

Question 4.
Which one of the following statements is correct? [AIPMT – 2014]
(a) The seed in grasses is not endospermic
(b) Mango is a parthenocarpic fruit
(c) A proteinaceous aleurone layer is present in maize grain
(d) A sterile pistil is called a staminode
Answer:
(c) A proteinaceous aleurone layer is present in maize grain

Question 5.
An example of edible underground stem is …………… . [AIPMT – 2014]
(a) Carrot
(b) Groundnut
(c) Sweet potato
(d) Potato
Answer:
(d) Potato

Question 6.
Placenta and pericarp are both edible portions in …………… . [AIPMT – 2014]
(a) Apple
(b) Banana
(c) Tomato
(d) Potato
Answer:
(c) Tomato

Question 7.
When the margins of sepals or petals overlap one another without any particular direction, the condition is termed as …………… . [AIPMT – 2014]
(a) Vexillary
(b) Imbricate
(c) Twisted
(d) Valvate
Answer:
(b) Imbricate

Question 8.
An aggregate fruit is one which develops from …………… . [AIPMT – 2014]
(a) Multicarpellary syncarpous gynoecium
(b) Multicarpellary apocarpous gynoecium
(c) Complete inflorescence
(d) Multicarpellary superior ovary
Answer:
(b) Multicarpellary apocarpous gynoecium

Question 9.
Non – albuminous seed is produced in …………… .
(a) Maize
(b) Castor
(c) Wheat
(d) Pea
Answer:
(d) Pea

Question 10.
Seed coat is not thin, membranous in …………… .
(a) Coconut
(b) Groundnut
(c) Gram
(d) Maize
Answer:
(a) Coconut

Question 11.
In china rose the flower are …………… .
(a) Actinomorphic,. Epigynous with valvate aestivation
(b) Zygomorphic, hypogynous with imbricate aestivation
(c) Zygomorphic, epigynous with twisted aestivation
(d) Actinomorphic, hypogynous with twisted aestivation
Answer:
(d) Actinomorphic, hypogynous with twisted aestivation

Question 12.
Placentation in tomato and lemon is …………… .
(a) Marginal
(b) Axile
(c) Parietal
(d) Free central
Answer:
(b) Axile

Question 13.
Vexillary aestivation is characteristic of the family …………… .
(a) Solanaceae
(b) Brassicaceae
(c) Fabaceae
(d) Asteraceae
Answer:
(c) Fabaceae

Question 14.
Phyllode is present in …………… . [AIPMT Prelims – 2012]
(a) Australian Acacia
(b) Opuntia
(c) Asparagus
(d) Euphorbia
Answer:
(c) Asparagus

Question 15.
How many plants in the list given below have composite fruits that develop from an inflorescence? Walnut, poppy, radish,pineapple, apple, tomato …………… . [AIPMT Prelims – 2012]
(a) Two
(b) Three
(c) Four
(d) Five
Answer:
(a) Two

Question 16.
Cymose inflorescence is present in …………… . [AIPMT Prelims – 2012]
(a) Trifolium
(b) Brassica
(c) Solanum
(d) Sesbania
Answer:
(b) Brassica

Question 17.
Which one of the following organism is correctly matched with its three characteristics? [AIPMT Mains – 2012]
(a) Pea : C₃ pathway, Endospermic seed,Vexillary aestivation
(b) Tomato : Twisted aestivation, Axile placentation, Berry
(c) Onion : Bulb, Imbricate aestivation, Axile placentation
(d) Maize : C₃ pathway, Closed vascular bundles, scutellum
Answer:
(c) Onion : Bulb, Imbricate aestivation, Axile placentation

Question 18.
How many plants in the list given below have marginal placentation? Mustard, Gram, Tulip, Asparagus, Arhar,sun hemp, Chilli, Colchicine, Onion,Moong, Pea, Tobacco, Lupin …………… . [AIPMT Mains – 2012]
(a) Four
(b) Five
(c) Six
(d) Three
Answer:
(c) Six

Question 19.
The Eyes of the potato tuber are …………… . [AIPMT Prelims – 2011]
(a) Axillary buds
(b) Root buds
(c) Flower buds
(d) Shoot buds
Answer:
(a) Axillary buds

Question 20.
Which one of the following statements is correct? [AIPMT Prelims – 2011]
(a) Flower of tulip is a modified shoot
(b) In tomato, fruit is a capsule
(c) Seeds of orchids have oil – rich endosperm
(d) Placentation in primrose is basal
Answer:
(d) Placentation in primrose is basal

Question 21.
A drup develops in …………… . [AIPMT Prelims-2011]
(a) Tomato
(b) Mango
(c) Wheat
(d) Pea
Answer:
(b) Mango

Samacheer Kalvi 11th Bio Botany Taxonomy and Systematic Botany Additional Questions and Answers

I. Multiple Choice Questions
Choose the correct answer:
Question 1.
Who is called as the father of Botany?
(a) Linnaeus
(b) Theophrastus
(c) Darwin
(d) Thales
Answer:
(b) Theophrastus

Question 2.
…………… is the lowest level of classification.
(a) Kingdom
(b) Class
(c) Order
(d) Species
Answer:
(d) Species

Question 3.
Who proposed the concept of “idos” or species?
(a) Theophrastus
(b) Stebbians
(c) Darwin
(d) Plato
Answer:
(d) Plato

Question 4.
Which type of species develop by the process of evolution?
(a) Taxonomic species
(b) Morphological species
(c) Biological species
(d) Phylogenetic species
Answer:
(c) Biological species

Question 5.
Who first proposed the early elementary rule of naming plants?
(a) A.P.de Candolle
(b) Linnaeus
(c) Alphonse de Candolle
(d) Simpson
Answer:
(b) Linnaeus

Question 6.
Isolation species can also be called as …………… Species.
(a) Biological
(b) Taxonomical
(c) Phylogenetic
(d) Morphological
Answer:
(a) Biological

Question 7.
18^th International Botanical congress was held at …………… .
(a) Sydney
(b) Leningard
(c) Melbourne
(d) London
Answer:
(c) Melbourne

Question 8.
In 2017, …………… International Botanical congress was held at Shenzhen.
(a) 17^th
(b) 18^th
(c) 19^th
(d) 20^th
Answer:
(c) 19^th

Question 9.
The vernacular name of Albizia amara in South Tamilnadu is …………… .
(a) Thurinji
(b) Kurinji
(c) Nithyakalyani
(d) Usilai
Answer:
(d) Usilai

Question 10.
…………… is a descriptive phrase of a plant.
(a) Vernacular name
(b) Binomial
(c) Polynomial
(d) Botanical name
Answer:
(c) Polynomial

Question 11.
…………… introduced the concept of Binomial nomenclature.
(a) Linnaeus
(b) Gaspard Bauhin
(c) Darwin
(d) Wallace
Answer:
(b) Gaspard Bauhin

Question 12.
Duplicate specimen of holotype is …………… .
(a) Lectotype
(b) Isotype
(c) Neotype
(d) Syntype
Answer:
(b) Isotype

Question 13.
…………… are the tools for identifying unfamiliar plants.
(a) Flora
(b) Keys
(c) Monograph
(d) Catalogues
Answer:
(b) Keys

Question 14.
…………… is a complete global account of a taxon of any rank.
(a) Flora
(b) Keys
(c) Monograph
(d) Catalogues
Answer:
(c) Monograph

Question 15.
The first botanical garden was established and maintained by …………… .
(a) Linnaeus
(b) Babylonians
(c) Theophrastus
(d) Stebbins
Answer:
(c) Theophrastus

Question 16.
First modern botanical garden was established by …………… .
(a) Theophrastus
(b) Linnaeus
(c) Luca Ghini
(d) Stebbins
Answer:
(c) Luca Ghini

Question 17.
…………… is the largest botanical garden in world.
(a) Royal Botanical garden
(b) Madras Presidency College
(c) Indian Botanical Garden
(d) New York Botanical garden
Answer:
(a) Royal Botanical garden

Question 18.
Who is called as father of taxonomy?
(a) Engler & Prantl
(b) Linnaeus
(c) Theophrastus
(d) Darwin
Answer:
(b) Linnaeus

Question 19.
Number of stamen(s) in monandria is …………… .
(a) 4
(b) 3
(c) 2
(d) 1
Answer:
(d) 1

Question 20.
Sexual system of classification is also called as …………… .
(a) Natural system
(b) Artificial system
(c) Phylogenetic
(d) APG system
Answer:
(b) Artificial system

Question 21.
Number of series under Polypetalae.
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(c) 3

Question 22.
Which series includes epigynous flowers with inferior ovary?
(a) Heteromerae
(b) Disaflorea
(c) Inferae
(d) Thalanifloreae
Answer:
(c) Inferae

Question 23.
Undistinguished sepal and petal is called as …………… .
(a) Petaloid
(b) Staminode
(c) Perianth
(d) Sepaloid
Answer:
(c) Perianth

Question 24.
Which is not a family of gymnospermae?
(a) Gnetaceae
(b) Equistae
(c) Coniferae
(d) Cycadaceae
Answer:
(b) Equistae

Question 25.
Which is not a monocotyledon character?
(a) One cotyledon
(b) Parallel venation
(c) Pentamerous
d) Fibrous root
Answer:
(c) Pentamerous

Question 26.
Number of divisions in Engler and Prantl classification?
(a) 10
(b) 11
(c) 12
(d) 13
Answer:
(d) 13

Question 27.
“The evolution and classification of flowering plants” – book was written by …………… .
(a) Engler & Prantl
(b) Bentham & Hooker
(c) Cronquist
(d) Theophrasthus
Answer:
(c) Cronquist

Question 28.
Which of the following is not a clade of APG – classification?
(a) Early angiosperm
(b) Early gymnosperm
(c) Monocots
(d) Eudicots
Answer:
(b) Early gymnosperm

Question 29.
The term biosystematics was introduced by …………… .
(a) Bauhin
(b) Camp & Gilly
(c) Cronquist
(d) Smith
Answer:
(b) Camp & Gilly

Question 30.
Taxonomy based on chromosomal number & characteristics is called …………… .
(a) Serotaxonomy
(b) Cytotaxonomy
(c) Chemotaxonomy
(d) Molecular taxonomy
Answer:
(b) Cytotaxonomy

Question 31.
Classification based on protein content is called …………… .
(a) Serotaxonomy
(b) Cytotaxonomy
(c) Chemotaxonomy
(d) Molecular taxonomy
Answer:
(a) Serotaxonomy

Question 32.
DNA bar coding was introduced by …………… .
(a) Stebbins
(b) Hebert
(c) Camp & Gilly
(d) Darwin
Answer:
(b) Hebert

Question 33.
Biosystematics is also called as …………… .
(a) α – taxonomy
(b) S – taxonomy
(c) Ω – taxonomy
(d) β – taxonomy
Answer:
(c) Ω – taxonomy

Question 34.
Outcome of cladistics analysis is …………… .
(a) Monogram
(b) Monograph
(c) Cladogram
(d) Cladograph
Answer:
(c) Cladogram

Question 35.
Taxa comprising all the descendents of a common ancestor.
(a) Monophyletic group
(b) Diphyletic group
(c) Paraphyletic group
(d) Polyphyletic group
Answer:
(a) Monophyletic group

Question 36.
Papilionaceous corolla is seen in …………… family.
(a) Apocyanaceae
(b) Fabaceae
(c) Solanaceae
(d) Liliaceae
Answer:
(b) Fabaceae

Question 37.
Stipitate ovary is seen in …………… .
(a) Solanaceae
(b) Liliaceae
(c) Fabaceae
(d) Apocyanaceae
Answer:
(a) Solanaceae

Question 38.
Number of genus in fabaceae is …………… .
(a) about 131
(b) about 741
(c) about 751
(d) about 761
Answer:
(b) About 741

Question 39.
The characteristic fruit of fabaceae is …………… .
(a) Regima
(b) Legume
(c) Hespiridium
(d) Berry
Answer:
(b) Legume

Question 40.
In Arachis hypogea, the fruit development is …………… .
(a) Syncarpic
(b) Apocarpic
(c) Geocarpic
(d) Photocarpic
Answer:
(c) Geocarpic

Question 41.
Which of the following plants root is a Immunomodulator?
(a) Glycirrhiza glabra
(b) Dalbergia latifolia
(c) Mucum pruriens
(d) Crotolaria jurcea
Answer:
(a) Glycirrhiza glabra

Question 42.
Indigo dye is obtained from plant.
(a) Lupin
(b) Avuri
(c) Sesban
(d) Agathi
Answer:
(b) Avuri

Question 43.
Which plant is commonly called as the “Flame of the forest”?
(a) Clitoritematea
(b) Butea frondosa
(c) Lupinus hirsutus
(d) Butea monosperma
Answer:
(b) Butea frondosa

Question 44.
Which year is declared as the “International year of the pulses”.
(a) 2017
(b) 2018
(c) 2015
(d) 2016
Answer:
(d) 2016

Question 45.
Which is called as “Night Shade family”?
(a) Solanaceae
(b) Fabaceae
(c) Apocyanaceae
(d) Liliaceae
Answer:
(a) Solanaceae

Question 46.
Rhiphidium inflorescence is seen in …………… .
(a) Solanum nigrum
(b) Solanum tuberosum
(c) Datura
(d) Withania somnifera
Answer:
(a) Solatium nigrum

Question 47.
The fruit of Datura metal is …………… .
(a) Spinescent Capsule
(b) Regma
(c) Legume
(d) Capsule
Answer:
(a) Spinescent capsule

Question 48.
Plicate inflorescence is seen in …………… .
(a) Solanum nigrum
(b) Datura metal
(c) Petunia hybrida
(d) Solanum tuberosum
Answer:
(b) Datura metal

Question 49.
…………… drug is used to treat asthma & whooping cough.
(a) Atropine
(b) Stramonium
(c) Anabasine
(d) Normicotine
Answer:
(b) Stramonium

Question 50.
Inflorescence in Aloe is …………… .
(a) Compound Spadix
(b) Spike
(c) Paricle
(d) Solitary
Answer:
(b) Spike

Question 51.
Carpels are obliquely placed in …………… .
(a) Fabaceae
(b) Solanaceae
(c) Liliaceae
(d) Apocyanaeae
Answer:
(b) Solanaceae

Question 52.
Septal glands are present in the gynoeciums of …………… .
(a) Solanaceae
(b) Liliaceae
(c) Fabaceae
(d) Apocyanaeae
Answer:
(b) Liliaceae

Question 53.
…………… is an alkaloid that induces polyploidy.
(a) Nictonine
(b) Stramonium
(c) Atropine
(d) Colchicine
Answer:
(d) Colchicine

Question 54.
The leaves of …………… is used in hemorrhoidal salves & shampoos.
(a) Aloe barbadense
(b) Aloevera
(c) Allium sativum
(d) Allium cepa
Answer:
(b) Aloevera

Question 55.
Botanical survey of India has …………… regional centres in India.
(a) 10
(b) 11
(c) 12
(d) 13
Answer:
(b) 11

Question 56.
Synstamenous condition is seen in …………… .
(a) Haemodorum
(b) Ruscus
(c) Paris quadrfolia
(d) Maianthenum
Answer:
(b) Ruscus

Question 57.
Scapigerous inflorescence is seen in …………… .
(a) Allium sativum
(b) Allium cepa
(c) Aloevera
(d) Maenodorum
Answer:
(b) Allium cepa

Question 58.
Number of stamens in Schizanthus is …………… .
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(a) 2

Question 59.
Extra axillary scorpiod cyme is called …………… .
(a) Spike
(b) Monochasical cyme
(c) Helicoid cyme
(d) Rhiphidium
Answer:
(d) Rhiphidium

II. Very Short Answer Type Questions (2 Marks)

Question 1.
Define Taxonomy.
Answer:
Taxonomy is “the science dealing with the study of classification including the bases, principles, rules and procedures”.

Question 2.
List out the various rank or taxa of taxonomic hierarchy.
Answer:
Kingdom, Phylum, Class, Order, Family, Genus, Species

Question 3.
Which is the lowest taxon in classification? Define.
Answer:
Species is the lowest taxon in classification. It is defined as the group of individuals which are closely resembling each other and interbreed among themselves producing fertile offspring.

Question 4.
Define Nomenclature.
Answer:
Assigning name for a plant is known as Nomenclature.

Question 5.
What are vernacular names? Give an example.
Answer:
Vernacular names are known as common names. Example: Albizia amara L. is called as Usilai in South Tamil Nadu and Thurinji in North Tamil Nadu.

Question 6.
What is Author citation?
Answer:
Author citation refers to valid name of the taxa accompanied by the author’s name who published the name validly. Example: Solanum nigrum L.

Question 7.
Define e – Flora.
Answer:
Electronic Floras (e – floras) is the digitized form of a flora published online.
Example: e – Flora China. This provides the information and also functions as an identification tool.

Question 8.
When a neotype specimen is selected?
Answer:
Neotype Specimen is derived from non – original collection selected as the type, when original specimen is missing or destroyed.

Question 9.
What do you mean by taxonomical aids?
Answer:
Tools, techniques, procedures and stored information that are useful in identification and classification of organisms are called taxonomical aids.

Question 10.
Differentiate Regional Flora from continental flora.
Answer:
Regional Flora from continental flora.

1. Regional Flora: Flora covering a large geographical area or a botanical region Ex: flora of Madras Presidency.
2. Continental Flora: Flora covering the entire continent. Ex: flora of Europaea.

Question 11.
What is Herbarium specimen?
Answer:
Herbarium Specimen is defined as a pressed and dried plant sample that is permanently glued or strapped to a sheet of paper along with a documentation label.

Question 12.
Name the major classes of Bentham & Hooker Classification.
Answer:
The major classes of Bentham & Hooker Classification:

• Class 1 – Dicotyledonae
• Class 2 – Gymnospermae
• Class 3 – Monocotyledonae

Question 13.
How Cronquist classified the angiosperms?
Answer:
Cronquist classified the angiosperms into two main classes Magnoliopsida and Liliopsida.

Question 14.
Cronquist classification is a failure. Justify.
Answer:
Cronquist classification system is not very useful for identification and cannot be adopted in herbaria due to its high phylogenetic nature.

Question 15.
Which is the most recent classification of flowering plants? How many versions it had been published so far?
Answer:
Angiosperm phylogeny group classification (APG) is the recent classification of flowering plants. APG I, APG II, APG III, APG IV are the four versions.

Question 16.
Name any 4 sub classes of Liliopsida.
Answer:
4 sub classes of Liliopsid:

1. Alismatidae
2. Arecidae
3. Commelinidae
4. Zingiberidae

Question 17.
Why the classification undergoes changes very often?
Answer:
Classification reflects the state of our knowledge at a given point of time. It will continue to change as we acquire new information.

Question 18.
Point out the aims of chemotaxonomy.
Answer:
The aims of chemotaxonomy:

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

Question 19.
Define Biosystematics.
Answer:
Biosystematics is an “Experimental, ecological and cytotaxonomy” through which life forms are studied and their relationships are defined.

Question 20.
Name few molecular markers used in molecular taxonomy.
Answer:
Allozymes, mitochondrial DNA, micro satellites, RFLP (Restriction Fragment Length Polymorphism), RAPD (Random amplified polymorphic DNA), AFLPs (Amplified Fragment Length Polymorphism), Single nucleotide Polymorphism – SNP, microchips or arrays.

Question 21.
List out the significance of DNA bar coding?
Answer:
The significance of DNA bar coding:

1. DNA bar coding greatly helps in identification and classification of organism.
2. It aids in mapping the extent of biodiversity.

Question 22.
State the demerits of RAPD analysis.
Answer:
RAPD analysis has the major disadvantage that results are difficult to replicate and in the homology of similar bands in different taxa may be nuclear.

Question 23.
Define Cladistics.
Answer:
The method of classifying organisms into monophyletic group of a common ancestor based on shared apomorphic characters is called Cladistics.

Question 24.
Define Cladogram.
Answer:
The outcome of a cladistic analysis is a cladogram, a tree – shaped diagram that represent the best hypothesis of phylogenetic relationships.

Question 25.
The genetic sequence used to identify a plant is known as “DNA tags” or “DNA barcodes”.
Answer:
The system of naming the organism with two names, generic name and specific (species)
name is known as binomial system of nomenclature, e.g. Pavo cistatus – Indian pea fowl.

Question 26.
Compare the Gynoecium of Pisum sativum and Datura metal.
Answer:
Gynoecium of Pisum sativum:

1. Mono Carpellary
2. Unilocular
3. Ovules on marginal placentation
4. Feathery stigma

Gynoecium of Datura metal:

1. Bicarpellary
2. Tetralocular
3. Ovules on axile placentation
4. Bilobed stigma

Question 27.
Write the floral formula of Pisum sativum.
Answer:

Question 28.
Name binomial name of any two oil plants of Fabaceae.
Answer:
Two oil plants of Fabaceae:

1. Arachis hypogea (Ground nut) and
2. Pongamia pinnata (Pungam).

Question 29.
Explain the classical taxonomical tools.
Answer:
Extra axillary scorpiod cyme is called rhiphidium.

Question 30.
Name the type of fruit seen in Capsicum and Datura.
Answer:
The type of fruit seen in Capsicum and Datura:

1. Capsicum – Berry
2. Datura – Capsule

Question 31.
What is atropine?
Answer:
Atropine is a powerful alkaloid obtained from Atropa belladonna root is used in belladonna plasters, tinctures etc, for relieving pain and also for dilating pupils of eyes for eye – testing.

Question 32.
What is Stramonium?
Answer:
Stramonium is a drug obtained from the leaves and roots of Datura stramonium and used to treat asthma and whooping cough.

Question 33.
Which stimulated Engler and Prantl to prepare phylogenic classification?
Answer:
The publication of the Origin of Species (1859) by Charles Darwin has given stimulus for the emergence of phylogenetic system of classification.

III. Short Answer Type Questions (3 Marks)

Question 1.
How dichotomous key helps in identification of plants?
Answer:
Dichotomous key consists of a sequence of two contrasting statements. A pair of contrasting statements is known as couplet. Each statement is known as lead. The plant is correctly identified with keys by narrowing down the characters found in plant.

Question 2.
Differentiate between Taxonomy & Systematics.
Answer:
Taxonomy:

1. Discipline of classifying organisms into taxa
2. Governs the practices of naming, describing, identifying and specimen preservation.
3. Classification + Nomenclature = Taxonomy

Systematics:

1. Broad field of biology that studies the diversification of species
2. Governs the evolutionary history and phylogenetic relationship in addition to taxonomy
3. Taxonomy + Phylogeny = Systematics

Question 3.
Write a note on Binomial nomenclature.
Answer:
binomial nomenclature, the first one is called genus name and second one is specific epithet. Example: Mangifera indica, Mangifera is a genus name and indica is specific epithet.

Question 4.
Enumerate the steps involved in herbarium preparation.
Answer:
Preparation of herbarium specimen includes the following steps.

1. Plant Collection
2. Documentation of field site data
3. Preparation of plant specimen
4. Mounting herbarium specimen
5. Herbarium labels
6. Protection of herbarium sheets against mold and insects

Question 5.
Why do we need the classification of organisms?
Answer:
The classification of organisms:

1. Understanding the classification of organisms can gives an insight into other fields and has significant practical value.
2. Classification helps us to know about different taxa, their phylogenetic relationship and exact position.
3. It helps to train the students of plant sciences with regard to the diversity of organisms and their relationship with other biological branches.

Question 6.
Classification is a essential part of biology – Justify.
Answer:
Classification is essential to biology because there is a vast diversity of organisms to sort out and compare. Unless they are organized into manageable categories it will be difficult for identification.

Question 7.
Linnaeus classification is also called sexual system of classification. Why?
Answer:
Linnaeus classification is mostly based on sexual characters like number, union, length and distribution of stamens and also on carpel characters. Hence it is called sexual system of classification.

Question 8.
Write a note on Monochlamydeae.
Answer:
Plants with incomplete flowers either apetalous or with undifferentiated calyx and corolla are placed under Monochlamydeae. The sepals and petals are not distinguished and they are called perianth. Sometimes both the whorls are absent. Monochlamydeae includes 8 series and 36 families.

Question 9.
In chemotaxonomy, how the chemicals are categorised?
Answer:
The chemical characters can be divided into three main categories:

1. Easily visible characters like starch grains, silica etc.
2. Characters detected by chemical tests like phenolics, oil, fats, waxes etc.
3. Proteins

Question 10.
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 11.
What is Molecular taxonomy?
Answer:
Molecular Taxonomy is the branch of phylogeny that analyses hereditary molecular differences, mainly in DNA sequences, to gain information and to establish genetic relationship between the members of different taxonomic categories.

Question 12.
Point out the uses of molecular taxonomy.
Answer:
The uses of molecular taxonomy:

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

Question 13.
How RFLP helps in taxonomical studies?
Answer:
RFLP (Restriction Fragment Length Polymorphism): RFLP’s is a molecular method of genetic analysis that allows 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 14.
Define DNA barcoding.
Answer:
DNA barcoding is a taxonomic method that uses a very short genetic sequence from a standard part of a genome. The genetic sequence used to identify a plant is known as “DNA tags” or “DNA barcodes”. Paul Hebert in 2003 proposed “DNA barcoding” and he is considered as ‘Father of barcoding’.

Question 15.
In which organelle of plant cell does the barcode genes are located? Name the genes.
Answer:
Chloroplast, the genes are matK and rbcL.

Question 16.
Differentiate between Monophyletic group and paraphyletic group.
Answer:
1. Monophyletic Group: Taxa comprising all the descendants of a common ancestor

2. Paraphyletic Group: Taxon that includes an ancestor but not all of the descendants of that ancestor.

Question 17.
Why do we need Cladistics?
Answer:
Cladistics:

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

Question 18.
Write a note on the petals of papilionaceous Corolla.
Answer:
The outer most petal is large called standard petal or vexillum. Lateral 2 petals are lanceolate and curved. They are called wing petals or alae. Anterior two petals are partly fused and are called keel petals or carina which encloses the stamens and pistil.

Question 19.
Draw the floral diagram of Pisum sativum.
Answer:

Question 20.
Write systematic position of Solanaceae based on APG classification.
Answer:

Kingdom	Plantae
1. Clade	1. Angiosperms
2. Clade	2. Eudicot
3. Clade	3. Asterids
4. Clade	4. Solanales
5. Family	5. Solanaceae

Question 21.
Mention the diagnostic features of Liliaceae member.
Answer:
The diagnostic features of Liliaceae member:

1. Perennial herbs often with bulbous stem / rhizomes
2. Radical leaves
3. Perianth showy
4. Stamens six
5. Ovary superior

Question 22.
Write systematic position of liliaceae based of Bentham and Hooker Classification?
Answer:
Systematic Position:

Kingdom	Plantae
1. Class	1. Monocotyledons
2. Series	2. Coronarieae
3. Order	3. Liliales
4. Family	4. Liliaceae

IV. Long Answer Type Questions (5 Marks)

Question 1.
List out the principles of ICN.
Answer:
International Code of Nomenclature is based on the following six principles.

1. Botanical nomenclature is independent of zoological and bacteriological nomenclature.
2. Application of names of taxonomic group is determined by means of nomenclatural types.
3. Nomenclature of a taxonomic group is based on priority of publication.
4. Each taxonomic group with a particular circumscription, position and rank can bear only one correct name, the earliest that is in accordance with the rules except in specified cases.
5. Scientific names of taxonomic groups are treated as Latin regardless of their derivation.
6. The rules of nomenclature are retroactive unless expressly limited.

Question 2.
Explain the role of Botanical garden in taxonomy.
Answer:
Botanical gardens play the following important roles.

1. Gardens with aesthetic value which 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. Garden is used for self-instruction or demonstration purposes.
4. It can integrate information of diverse fields like Anatomy, Embryology, Phytochemistry, Cytology, Physiology and Ecology.
5. Act as a conservation centre for diversity, rare and endangered species.
6. It offers annual list of available species and a free exchange of seeds.
7. Botanical garden gives information about method of propagation, sale of plant material to the general public.

Question 3.
Enumerate the uses of Herbarium.
Answer:
The uses of Herbarium:

1. Herbarium provides resource material for systematic research and studies.
2. It is a place for 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 opportunity for documenting biodiversity and studies related to the field of ecology and conservation biology.

Question 4.
Point out the characters of Early angiosperm according to APG Classification.
Answer:
The characters of Early angiosperm according to APG Classification:

1. Seeds always with two cotyledons.
2. Presence of ethereal oils.
3. Leaves are always simple net – veined
4. Each floral whorls with many parts
5. Perianth usually spirally arranged or parts in threes
6. Stamens with broad filaments
7. Anthers tetrasporangiate
8. Pollen monosulcate
9. Nectaries are rare
10. Carpels usually free and
11. Embryo very small

Question 5.
Draw a flow chart depicting the Bentham and Hooker Classification.
Answer:
Bentham and Hooker system of classification

Question 6.
Draw an outline of Engler & Prantl Classification.
Answer:

Question 7.
Define biosystematics & list out the aim of biosystematics.
Answer:
1. Biosystematics: Biosystematics is an “Experimental, ecological and cytotaxonomy” through which life forms are studied and their relationships are defined.
2. Aims of Biosystematics: The aims of biosystematics are as follows:

• To delimit the naturally occurring biotic community of plant species.
• To establish the evolution of a group of taxa by understanding the evolutionary and phylogenetic trends.
• To involve any type of data gathering based on modem concepts and not only on morphology and anatomy.
• To recognize the various groups as separate biosystematics categories such as ecotypes, ecospecies, cenospecies and comparium.

Question 8.
Distinguish between classical taxonomy & modern taxonomy.
Answer:
Classical Taxonomy:

1. It is called old systematics or Alpha (α) – taxonomy or Taxonomy
2. It is pre – Darwinean
3. Species is considered as basic unit and is static
4. Classification is mainly based on morphological characters
5. This system is based on the observation of a few samples / individuals

Modern Taxonomy:

1. It is called Neosystematics or Biosystematics or Omega (Ω) taxonomy
2. It is post – Darwinean
3. Species is considered as dynamic entity and ever changing
4. Classification is based on morphological, reproductive characters and phylogenetic (evolutionary) relationship of the organism
5. This system is based on the observation of large number of samples / individuals

Question 9.
List out the significance of Molecular Taxonomy.
Answer:
The significance of Molecular Taxonomy:

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 10.
Explain Clitoria ternatea in botanical terms. Draw floral diagram.
Answer:

1. Habit: Twining climber
2. Root: Branched tap root system having nodules.
3. Stem: Aerial, weak stem and a twiner.
4. Leaf: Imparipinnately compound, alternate, stipulate showing reticulate venation. Leaflets are stipellate.
   Petiolate and stipels are pulvinated.
5. Inflorescence: Solitary and axillary
6. Flower: Bracteate, bracteolate, bracteoles usually large, pedicellate, heterochlamydeous, complete, bisexual, pentamerous, zygomorphic and hypogynous.
7. Calyx: Sepals 5, synsepalous, green showing valvate aestivation. Odd repel is anterior in position.
8. Corolla: Petals 5, white or blue apopetalous, irregular papilionaceous corolla showing, descendingly imbricate aestivation.
9. Androecium: Stamens 10, diadelphous (9) + 1 nine stamens fused to form a bundle and the tenth stamen is free. Anthers are dithecous, basifixed, introse and dechising by longitudinal slits.
10. Gynoecium: Monocarpellary, uni – locular, with many ovules on marginal placentation, ovary superior, style simple and incurved with feathery stigma.
11. Fruit: Legume
12. Seed: Non – endospermous, reniform.
13. Floral Formula:
    

Question 11.
Explain Datura metal in botanical terms. Draw floral diagram.
Answer:

1. Habit: Large, erect and stout herb.
2. Root: Branched tap root system.
3. Stem: Stem is hollow, green and herbaceous with strong odour.
4. Leaf: Simple, alternate, petiolate, entire or deeply lobed, glabrous exstipulate showing unicostate reticulate venation.
5. Inflorescence: Solitary and axillary cyme.
6. Flower: Flowers are large, greenish white, bracteate, ebracteolate, pedicellate, complete, heterochlamydeous, pentamerous, regular, actinomorphic, bisexual and hypogynous.
7. Calyx: Sepals 5, green synsepalous showing valvate aestivation. Calyx is mostly persistant, odd sepal is posterior in position.
8. Corolla: Petals 5, greenish white, sympetalous, plicate (folded like a fan) showing twisted aestivation, funnel shaped with wide mouth and 10 lobed.
9. Androecium: Stamens 5, free from one another, epipetalous, altemipetalous and are inserted in the middle of the corolla tube. Anthers are basifixed, dithecous, with long filament, introse and longitudinally dehiscent.
10. Gynoecium: Ovary bicarpellary, syncarpous superior ovary, basically bilocular but tetralocular due to the formation of false septum. Carpels are obliquely placed and ovules on swollen axile placentation. Style simple long and filiform, stigma two lobed.
11. Fruit: Spinescent capsule opening by four special valves with persistent calyx.
12. Seed: Endospermous
13. Floral Formula:
    

Question 12.
Explain Allium cepa in botanical terms. Draw floral diagram.
Answer:
Botanical description of Allium cepa:

1. Habit: Perennial herb with bulb.
2. Root: Fibrous adventitious root system
3. Stem: Underground bulb
4. Leaf: A cluster of radical leaves emerges from the underground bulb, cylindrical and fleshy having sheathy leaf bases with parallel venation.
5. Inflorescence: Scapigerous i.e. 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.
6. Flower: Small, white, bracteate, bracteolate, pedicellate, complete, trimerous, actinomorphic and hypogynous. Flowers are protandrous.
7. Perianth: Tepals 6, white, arranged in two whorls of three each, syntepalous showing valvate aestivation.
8. Androecium: Stamens 6, arranged in two whorls of three each, epitepalous, apostamenous / free and opposite to tepals. Anthers dithecous. basifixed, introse and dehiscing longitudinally.
9. Gynoecium: Tricarpellary and syncarpous. Ovary superior, trilocular with two ovules in each locule on axile placentation. Style simple, slender with simple stigma.
10. Fruit: A loculicidal capsule.
11. Seed: Endospermous
12. Floral Formula:
    
    

Question 13.
List out the economic importance of plants & their uses of Fabaceae.
Answer:

Question 14.
List out the economic importance of plants & their uses of Solanaceae.
Answer:

Question 15.
List out the economic importance of plants & their uses of Liliaceae.
Answer:
Economic Importance of the Family Liliaceae:

V. Higher Order Thinking Skills (HOTs)

Question 1.
Neem is highly valued tree in Indian medicine. It is called by the name veppu in Malayalam, Arishta in Sanskrit, Vembu in Tamil, Nimbo in Portuguese. Suggest a solution for this varied naming problem considering as a taxonomist.
Answer:
As a taxonomist, this can be solved by using Binomial nomenclature. According to ICBN, every plant is given a scientific name which can be used in common all throughout the world. Thus neem is named as Azadirachta indica.

Question 2.
According to Binomial nomenclature, Human beings are named as Homosapiens. Following this, write the binomials for Brinjal and Rosewood.
Answer:
The binomials for Brinjal and Rosewood:

1. Brinjal – Solatium melongena
2. Rosewood – Dalbergia latifolia

Question 3.
Officially, every state in the Republic of India has its own flower, fruit etc. If Andhra Pradesh has Lotus as its state flower, what is the state flower of Tamil Nadu? Mention its family.
Answer:
State flower of Tamilnadu is Gloriosa superba belonging to Liliaceae family.

Question 4.
Peanut is a geocarpic fruit – Comment on the statement.
Answer:
In peanut (Arachis hypogea), after fertilisation, the stipe of ovary grows down into the soil, later develops & matures into fruit. Such a underground developed fruit is called geocarpic fruit.

Question 5.
You are given an entire plantlet of Clitoria ternatea. Give possible reasons to say that it is a dicot plantlet.
Answer:
Clitoria ternatea belongs to Dicots because:

1. The root is a tap root
2. Leaves show reticulate venation

Question 6.
Give possible terms explaining the gynoecium of a flower whose C.S. of ovary is given below.
Answer:

Tricarpellary, Trilocular, Syncarpous, 2 ovules in 1. Each ovule has axile placentation.

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Tamilnadu Samacheer Kalvi 11th Bio Botany Solutions Chapter 4 Reproductive Morphology

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Samacheer Kalvi 11th Bio Botany Reproductive Morphology Text Book Back Questions and Answers

Choose the correct answer
Question 1.
Vexillary aestivation is characteristic of the family …………… .
(a) Fabaceae
(b) Asteraceae
(c) Solanaceae
(d) Brassicaceae
Answer:
(a) Fabaceae

Question 2.
Gynoecium with united carples is termed as …………… .
(a) apocarpous
(b) multicarpellary
(c) syncarpous
(d) none of the above
Answer:
(c) syncarpous

Question 3.
Aggregate fruit develops from …………… .
(a) multicarpellary, apocarpous ovary
(b) multicarpellary, syncarpous ovary
(c) multicarpellary ovary
(d) whole inflorescence
Answer:
(a) multicarpellary, apocarpous ovary

Question 4.
In an inflorescence where flowers are borne laterally in an acropetal succession the position of the youngest floral bud shall be …………… .
(a) proximal
(b) distal
(c) intercalary
(d) anywhere
Answer:
(b) distal

Question 5.
A true fruit is the one where …………… .
(a) only ovary of the flower develops into fruit
(b) ovary and calyx of the flower develops into fruit
(c) ovary, calyx and thalamus of the flower develops into fruit
(d) all floral whorls of the flower develops into fruit
Answer:
(a) only ovary of the flower develops into fruit

Question 6.
Find out the floral formula for a bisexual flower with bract, regular, pentamerous, distinct calyx and corolla, superior ovary without bracteole?
Answer:

Question 7.
Give the technical terms for the following:
(a) A sterile stamen
(b) Stamens are united in one bunch
(c) Stamens are attached to the petals
Answer:
(a) A sterile stamen – Staminode
(b) Stamens are united in one bunch – Monadelphous
(c) Stamens are attached to the petals – Epipetalous (petalostemonous)

Question 8.
Explain the different types of placentation with example.
Answer:
The different types of placentation with example:

1. Marginal: It is with the placentae along the margin of a unicarpellate ovary. Example: Fabaceae.
2. Axile: The placentae arises from the column in a compound ovary with septa. Example: Hibiscus, tomato and lemon.
3. Superficial: Ovules arise from the surface of the septa. Example: Nymphaeceae.
4. Parietal: It is the placentae on the ovary walls or upon intruding partitions of a unilocular, compound ovary. Example: Mustard, argemone and cucumber.
5. Free – central: It is with the placentae along the column in a compound ovary without septa. Example: Caryophyllaceae, Dianthus and primrose.
6. Basal: It is the placenta at the base of the ovary. Example: Sunflower (Asteraceae) Marigold.

Question 9.
Differentiate between aggregate fruit with multiple fruit.
Answer:
1. Aggregate fruit:
Aggregate fruits develop from a single flower having an apocarpous pistil. Each of the free carpel is developed into a simple fruitlet. A collection of simple fruitlets makes an aggregate fruit. An individual ovary develops into a drupe, achene, follicle or berry. An aggregate of these fruits borne by a single flower is known as an etaerio. Example: Magnolia, Raspberry, Annona and Polyalthia.

2. Multiple or Composite fruit: A multiple or composite fruit develops from the whole inflorescence along with its peduncle on which they are borne.

• Sorosis: A fleshy multiple fruit which develops 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. Example: Pineapple, Jack fruit and Mulberry.
• Syconus: A multiple fruit which develops from hypanthodium inflorescence. The receptacle develops further and converts into fleshy fruit which encloses a number of true fruit or achenes which develops from female flower of hypanthodium inflorescence. Example: Ficus.

Question 10.
Explain the different types of fleshy fruit with suitable example?
Answer:
The fleshy fruits are derived from single pistil, where the pericarp is fleshy, succulent and differentiated into epicarp, mesocarp and endocarp. It is subdivided into the following:

1. Berry: Fruit develops from bicarpellary or multicarpellary, syncarpous ovary. Here the epicarp is thin, the mesocarp and endocarp remain undifferentiated. They form a pulp in which the seeds are embedded. Example: Tomato, date palm, grapes and brinjal.
2. Drupe: Fruit develops from monocarpellary, superior ovary. It is usually one seeded. Pericarp is differentiated into outer skinny epicarp, fleshy and pulpy mesocarp and hard and stony endocarp around the seed. Example: Mango and coconut.
3. Pepo: Fruit develops from tricarpellary inferior ovary. Pericarp terns leathery or woody which encloses, fleshy mesocarp and smooth endocarp. Example: Cucumber, watermelon, bottle gourd and pumpkin.
4. Hesperidium: Fruit develops from multicarpellary, multilocular, syncarpous, superior ovary. The fruit wall is differentiated into leathery epicarp with oil glands, a middle fibrous mesocarp. The endocarp forms distinct chambers, containing juicy hairs. Example: Orange and lemon.
5. Pome: It develops from multicarpellary, syncarpous, inferior ovary. The receptacle also develops along with the ovary and becomes fleshy, enclosing the true fruit. In pome the epicarp is thin skin like and endocarp is cartilagenous. Example: Apple and pear.
6. Balausta: A fleshy indehiscent fruit developing from multicarpellary, multilocular inferior ovary whose pericarp is tough and leathery. Seeds are attached irregularly with testa being the edible portion. Example: Pomegranate.

Textbook Activity Solved

Prepare a diet chart to provide 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.
Diet Chart for an Adolescent:

Samacheer Kalvi 11th Bio Botany Reproductive Morphology Additional Questions and Answers

I. Multiple Choice Questions
Choose the correct answer:
Question 1.
Placentation in tomato and lemon is …………… .
(a) parietal
(b) marginal
(c) free – central
(d) axile
Answer:
(d) axile

Question 2.
The coconut water and the edible part of coconut are equivalent to …………… .
(a) endosperm
(b) endocarp
(c) mesocarp
(d) embryo
Answer:
(a) endosperm

Question 3.
Geocarpic fruits are seen in …………… .
(a) carrot
(b) groundnut
(c) radish
(d) turnip
Answer:
(b) groundnut

Question 4.
Keel is characteristic petal of the flowers of …………… .
(a) Gulmohar
(b) Cassia
(c) Calotropis
(d) Bean
Answer:
(d) Bean

Question 5.
When the calyx is coloured and showy, it is called …………… .
(a) petaloid
(b) sepaloid
(c) bract
(d) spathe
Answer:
(a) petaloid

Question 6.
Bracts are modified leaves which bear flowers in their axils. Identify the plant which has a large showy brightly coloured bract …………… .
(a) Jasmine
(b) Euphorbia
(c) Hibiscus
(d) Bougainvillea
Answer:
(d) Bougainvillea

Question 7.
A flower which can be divided into equal vertical halves, by more than one plane of division is …………… .
(a) zygomorphic
(b) cyclic
(c) actinomorphic
(d) heteromorphic
Answer:
(c) actinomorphic

Question 8.
In Theobroma cocoa, the inflorescence arise from …………… .
(a) terminal shoot
(b) axillary part
(c) trunk of plant
(d) leaf node
Answer:
(c) trunk of plant

Question 9.
The type of inflorescence seen in Caesalpinia is …………… .
(a) corymb
(b) compound corymb
(c) capitulum
(d) umbel
Answer:
(a) corymb

Question 10.
Head is the characteristic of …………… family.
(a) Fabaceae
(b) Malvaceae
(c) Asteraceae
(d) Solanaceae
Answer:
(c) Asteraceae

Question 11.
Thyrsus is a type of …………… inflorescence.
(a) raceme
(b) cyme
(c) mixed
(d) special
Answer:
(c) mixed

Question 12.
Number of whorls in a complete flower is …………… .
(a) one
(b) two
(c) three
(d) four
Answer:
(d) four

Question 13.
Monoclinous flower will have …………… .
(a) androecium
(b) gynoecium
(c) both androecium & gynoecium
(d) none
Answer:
(c) both androecium & gynoecium

Question 14.
If unisexual and bisexual flowers are seen in same plant then the plant is said to be …………… .
(a) polyphyllous
(b) polygamous
(c) hermaphroditic
(d) dioecious
Answer:
(b) polygamous

Question 15.
…………… is a raceme of cymes.
(a) Verticil
(b) Cyathium
(c) Umbel
(d) Thyrsus
Answer:
(d) Thyrsus

Question 16.
Unit of perianth is …………… .
(a) petal
(b) sepal
(c) tepal
(d) stamen
Answer:
(c) tepal

Question 17.
Number of floral parts per whorl is called …………… .
(a) curosity
(b) atrocity
(c) merosity
(d) porosity
Answer:
(c) merosity

Question 18.
What is the green cap – like part of brinjal fruit?
(a) Corolla
(b) Perianth
(c) Calyx
(d) Pistil
Answer:
(c) Calyx

Question 19.
Butterfly shaped corolla is seen in …………… type.
(a) rosaceous
(b) caryophyllaceous
(c) cruciform
(d) papilionaceous
Answer:
(d) papilionaceous

Question 20.
Inflorescence seen in Daucas carota is …………… .
(a) umbel
(b) corymb
(c) compound umbel
(d) compound corymb
Answer:
(c) compound umbel

Question 21.
Arrangement of sepals and petals in flower bud is called …………… .
(a) adhesion
(b) aestivation
(c) placentation
(d) cohesion
Answer:
(b) aestivation

Question 22.
Which is not a part of pistil?
(a) Style
(b) Stigma
(c) Connective tissue
(d) carpel
Answer:
(c) Connective tissue

Question 23.
The type of calyx in brinjal is …………… .
(a) caducous
(b) deciduous
(c) persistent
(d) fugacious
Answer:
(c) persistent

Question 24.
Sterile stamen is called …………… .
(a) pistillode
(b) sessile
(c) staminode
(d) apostamen
Answer:
(c) staminode

Question 25.
Other name for gynoecium is …………… .
(a) carpel
(b) pistil
(c) style
(d) overy
Answer:
(b) pistil

Question 26.
Cavity found inside the ovary is called …………… .
(a) lobule
(b) locule
(c) lacuna
(d) labium
Answer:
(b) locule

Question 27.
Which part of saffron flower is used as flavouring agent?
(a) Carpel
(b) Anther
(c) Style
(d) Stigma
Answer:
(d) Stigma

Question 28.
If the ovary is inferior, then the flower is …………… .
(a) hypogynous
(b) epigynous
(c) perigynous
(d) epihypogynous
Answer:
(b) epigynous

Question 29.
Fabaceae members show …………… placentation.
(a) basal
(b) parietal
(c) superficial
(d) marginal
Answer:
(d) marginal

Question 30.
The side of the flower facing mother axis is called as …………… side.
(a) anterior
(b) posterior
(c) lateral
(d) dorsi – ventral
Answer:
(b) posterior

Question 31.
Which of the following option represents calyx?
(a) C
(b) Ca
(c) K
(d) Ka
Answer:
(c) K

Question 32.
…………… are the products of pollination & fertilization.
(a) Seeds
(b) Ovules
(c) Fruits
(d) Vegetables
Answer:
(c) Fruits

Question 33.
Study of fruits is called as …………… .
(a) honology
(b) pomology
(c) horticulture
(d) apology
Answer:
(b) pomology

Question 34.
Fruit wall can also be called as …………… .
(a) endocarp
(b) epicarp
(c) pericarp
(d) mericorp
Answer:
(c) pericarp

Question 35.
Which part of the apple fruit does we eat?
(a) Perianth
(b) Involucre
(c) Thalamus
(d) Bracteole
Answer:
(c) Thalamus

Question 36.
The false septum seen in siliqua fruits is …………… .
(a) frenulum
(b) micropyle
(c) raphae
(d) replum
Answer:
(d) replum

Question 37.
The type of fruit in Ricinus in …………… .
(a) lomentum
(b) cremocarp
(c) regma
(d) nut
Answer:
(c) regma

Question 38.
Jack fruit is an example for …………… .
(a) syconus
(b) siliqua
(c) sorosis
(d) nut
Answer:
(c) sorosis

Question 39.
Which of the following is not a schizocarpic fruit?
(a) Cremocarp
(b) Regma
(c) Samara
(d) Carcerulus
Answer:
(c) Samara

Question 40.
After fertilization …………… modifies into seed.
(a) ovary
(b) ovule
(c) carpel
(d) stigma
Answer:
(b) ovule

Question 41.
In groundnuts, which part nourishes the embryo?
(a) Endosperm
(b) Albumin
(c) Cotyledons
(d) Carpel
Answer:
(c) Cotyledons

Question 42.
…………… are the means for perpetuation of species.
(a) Fruits
(b) Seeds
(c) Corolla
(d) Flowers
Answer:
(b) Seeds

Question 43.
…………… is the second whorl of the flower.
(a) Calyx
(b) Corolla
(c) Gynoecium
(d) Perianth
Answer:
(b) Corolla

Question 44.
…………… is a ripened ovule.
(a) Carpel
(b) Pistil
(c) Seed
(d) Fruit
Answer:
(c) Seed

Question 45.
Imperfect flowers will have …………… essential whorl(s).
(a) only 1
(b) 2
(c) none
(d) 4
Answer:
(a) only 1

II. Very Short Answer Type Questions (2 Marks)

Question 1.
How will you define inflorescence?
Answer:
An inflorescence is a group of flowers arising from a branched or unbranched axis with a definite pattern.

Question 2.
Where does the inflorescence axis arise in cauliflorous type of inflorescence?
Answer:
In cauliflorous type, inflorescence developed directly from a woody trunk. Example: Theobroma cocoa.

Question 3.
Name any two mixed inflorescences.
Answer:
Two mixed inflorescences:

1. Thyrsus and
2. Verticillaster.

Question 4.
When a flower is said to be complete?
Answer:
A flower is said to be complete when it has all the four whorls (calyx, corolla, androecium & gynoecium).

Question 5.
What is a sessile flower?
Answer:
A flower without a pedicel or stalk is said to be sessile flower.

Question 6.
Define merosity.
Answer:
Number of floral parts per whorl is called merosity.

Question 7.
Write the units of (a) Perianth and (b) Calyx.
Answer:
The units of (a) Perianth and (b) Calyx:

1. (a) Perianth – tepals and
2. (b) Calyx – sepals

Question 8.
Name the three types of petals in papilionoceous corolla.
Answer:
The three types of petals in papilionoceous corolla:

1. Vexillum (standard)
2. wings (alae)
3. petals (carina)

Question 9.
What is a staminode? Give example.
Answer:
Sterile stamen is called staminode. e.g. Cassia

Question 10.
Define Pollinium.
Answer:
When the pollen grains are fused together as a single main, it is said to be pollinium.

Question 11.
List out the parts of a pistil.
Answer:
Ovary, style and stigma.

Question 12.
Define aestivation.
Answer:
Arrangement of sepals and petals in a floral bud.

Question 13.
What is mother axis?
Answer:
The branch that bears the flower is called mother axis.

Question 14.
What do you understand by the term “Pomology”?
Answer:
The branch of horticulture that deals with the study of fruits and their cultivation is called pomology.

Question 15.
How the seeds are classified based on endosperm?
Answer:
(a) Albuminous seed or Endospermous seed.
(b) Ex – Albuminous seed or non – Endospermous seed.

Question 16.
What is Spathe?
Answer:
In spadix, entire inflorescence is covered by a brightly coloured or hard bract called a spathe.

Question 17.
Differentiate Apocarpous and Syncarpous ovary.
Answer:
Apocarpous and Syncarpous ovary:

1. Apocarpous: A pistil contains two or more distinct carpels. Example: Annona
2. Syncarpous: A pistil contains two or more carpels which are connate. Example: Citrus and Tomato

Question 19.
Give examples for following fruit types: (a) Berry (b) Hesperidium
Answer:
(a) Berry: Tomato
(b) Hesperidium: Orange

III. Short Answer Type Questions (3 Marks)

Question 1.
Distinguish between Monoecious & Dioecious.
Answer:
Between Monoecious & Dioecious:

1. Monoecious: Both male and female flowers are present in the same plant, e.g., Coconut
2. Dioecious:  Male and female flowers are present on separate plants, e.g., Papaya

Question 2.
Explain Bilateral symmetry.
Answer:
In bilateral symmetry the flower can be divided into equal halves in only one plane. Zygomorphic flower can efficiently transfer pollen grains to visiting pollinators. Example: Pisum.

Question 3.
Differentiate Apopetalous from Sympetalous.
Answer:
Apopetalous from Sympetalous:

1. Apopetalous (or) Polypetalous: Petals are distinct, e.g., Hibiscus.
2. Sympetalous (or) Gamopetalous: Petals are fused, e.g., Datura.

Question 4.
Define Placentation & mention their types.
Answer:
The mode of distribution of placenta inside the ovary is called placentation. Types of placentation: Marginal, Axile, Superficial, Parietal, Free – central and Basal.

Question 5.
Write the floral formula for the Hibiscus rosasinensis.
Answer:

Question 6.
What are Parthenocarpic fruit?
Answer:
Development of fruits without fertilization are called Parthenocarpic fruit. They are seedless fruits. Example: Banana.

Question 7.
From which type of flowers does the aggregate fruit develops?
Answer:
Aggregate fruits develop from a single flower having an apocarpous pistil. Each of the free carpel is develops into a simple fruitlet. A collection of simple fruitlets makes an aggregate fruit.

Question 8.
Classify seeds based on their cotyledons.
Answer:
Based on the number of cotyledons present, two types of seeds are recognized.

1. Dicotyledonous seed: Seed with two cotyledons.
2. Monocotyledonous seed: Seed with one cotyledon.

Question 9.
List out any 3 significances of seed.
Answer:
3 significances of seed:

1. The seed encloses and protects the embryo for next generation.
2. Seeds of various plants are used as food, both for animals and human.
3. Seeds are the products of sexual reproduction so they provide genetic variations and recombination in a plant.

Question 10.
What is the importance of inflorescence.
Answer:
Function of inflorescence is to display the flowers for effective pollination and facilitate seed dispersal. The grouping of flowers in one place gives a better attraction to the visiting pollinators and maximize the energy of the plant.

Question 11.
Draw the line diagram for the following inflorescence.
Answer:
(a) Simple Dichasium:

(b) Compound Dichasium:

IV. Long Answer Type Questions (5 Marks)

Question 1.
Explain the various types of Schizocarpic fruit.
Answer:
This fruit type of intermediate between dehiscent and indehiscent fruit. The fruit instead of dehiscing rather splits into number of segments, each containing one or more seeds. They are of following types:

1. Cremocarp: Fruit develops from bicarpellary, syncarpous, inferior ovary and splitting into two one seeded segments known as mericarps. e.g., Coriander and Carrot.
2. Carcerulus: Fruit develops from bicarpellary, syncarpous, superior ovary and splitting into four one seeded segments known as nutlets, e.g., Leucas, Ocimum and Abutilon.
3. Lomentum: The fruit is derived from monocarpellary, unilocular ovary. A leguminous fruit, constricted between the seeds to form a number of one seeded compartments that separate at maturity, e.g., Desmodium, Arachis and Mimosa.
4. Regma: They develop from tricarpellary, syncarpous, superior, trilocular ovary and splits into one – seeded cocci which remain attached to carpophore, e.g., Ricinus and Geranium.

Question 2.
Explain the different types of flowers based on the position of ovary.
Answer:
Based on the position of ovary, a flower can be classified as:

1. Hypogynous: The term is used for sepals, petals and stamens attached at the base of a superior ovary, e.g. Malvaceae.
2. Epihypogynous: The term is used for sepals, petals and stamens attached at the middle of the ovary (half – inferior), e.g. Fabaceae and Rosaceae.
3. Epigynous: The term is used for sepals, petals and stamens attached at the tip of an inferior ovary, e.g. Cucumber, apple and Asteraceae.
4. Perigynous: The term is used for a hypanthium attached at the base of a superior ovary.
5. Epiperigynous: The term is used for hypanthium attached at the apex of an inferior ovary.

Question 3.
Classify the anthers based on their mode of attachment.
Answer:
The anthers based on their mode of attachment:

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

Question 4.
Define aestivation. Explain its types with example.
Answer:

Question 5.
Distinguish between racemose and cymose inflorescence.
Answer:
Racemose inflorescence

1. Main axis of unlimited growth
2. Flowers arranged in an acropetal succession
3. Opening of flowers is centripetal
4. Usually the oldest flower at the base of the inflorescence axis.

Cymose inflorescence:

1. Main axis of limited growth
2. Flowers arranged in a basipetal succession
3. Opening of flowers is centrifugal
4. Usually the oldest flower at the top of the inflorescence axis.

Question 6.
Write in detail about head inflorescence.
Answer:
Head: A head is a characteristic inflorescence of Asteraceae and is also found in some members of Rubiaceae. Example: Neolamarkia cadamba and Mitragyna parvifolia; and in some members of Fabaceae – Mimosoideae, example: Acacia nilotica, Albizia lebbeck, Mimosa pudica (sensitive plant). Torus contains two types of florets:

1. Disc floret or tubular floret.
2. Ray floret or ligulate floret.

Heads are classified into two types:

1. Homogamous head: This type of inflorescence exhibits single kind of florets. Inflorescence has disc florets alone. Example: Vernonia, Ageratum or Ray florets alone. example: Launaea, Sonchus.
2. Heterogamous head: The inflorescence possesses both types of florets. Example: Helianthus, Tridax.

Disc florets at the centre of the head are tubular and bisexual whereas the ray florets found at the margin of the head which are ligulate pistilate (unisexual).

Question 7.
List out the significance of fruits.
Answer:
The significance of fruits:

1. Edible part of the fruit is a source of food, energy for animals.
2. They are source of many chemicals like sugar, pectin, organic acids, vitamins and minerals.
3. The fruit protects the seeds from unfavourable climatic conditions and animals.
4. Both fleshy and dry fruits help in the dispersal of seeds to distant places.
5. In certain cases, fruit may provide nutrition to the developing seedling.
6. Fruits provide source of medicine to humans.

Question 8.
Draw a flow chart depicting the classification of fruits.
Answer:

Question 9.
Explain in detail about any two special inflorescence.
Answer:
The inflorescences do not show any of the development pattern types are classified under special type of inflorescence.
1. Cyathium: Cyathium inflorescence consists of small unisexual flowers enclosed by a common involucre which mimics a single flower. Male flowers are organized in a scorpioid manner. Female flower is solitary and centrally located on a long pedicel. Male flower is represented only by stamens and female flower is represented only by pistil. Cyathium may be actinomorphic (Example: Euphorbia) or zygomorphic (Example: Pedilanthus) Nectar is present in involucre.

2. Hypanthodium: Receptacle is a hollow, globose structure consisting unisexual flowers present on the inner wall of the receptacle. Receptacle is closed except a small opening called 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 middle below. Example: Ficus sp. (Banyan and Pipal).

V. Higher Order Thinking Skills (HOTs)

Question 1.
Brinjal fruit has persistent calyx. Have you ever noticed the same in any other fruits? Name them.
Answer:
Tomato, Lady’s finger, guava and chilli also have persistent calyx.

Question 2.
Whether parthenocarpic fruits develop endosperm? Why?
Answer:
No, parthenocarpic fruits are developed without fertilization, but endosperm will form only after fertilization. So parthenocarpic fruits do not have endosperm.

Question 3.
Ovary develops into fruit after fertilization. While eating an Apple which part do you eat? Explain.
Answer:
Apple belongs to false fruit. In false fruits, apart from the ovary, non – carpellary parts also develop into fruit. In apple, the thalamus develops into fleshy edible part.

Question 4.
Cremocarp and Carcerulus both are schizocarpic fruits yet they differ. How?
Answer:
Cremocarp:

1. Fruit develops from syncarpous inferior ovary.
2. Ripened fruit split into two, one – seeded segments called mericarps.

Carcerulus:

1. Fruit develops from syncarpous superior ovary.
2. Ripened fruit split into four, one – seeded segments called nutlets.

Question 5.
Mango and coconut are ‘drupe’ type of fruits. In Mango, the edible part is fleshy mesocarp. What does the milk of tender coconut represent?
Answer:
Endosperm is the liquid (milk) potable part of tender coconut, which is rich in nutrients and is formed as a result of triple fusion.

Question 6.
Pick out correct ratio of the male flower to female flower in cyathium inflorescence and explain it?
(a) one : one
(b) one : many
(c) many : many and
(d) many : one.
Answer:
(a) One : one – Cyathium has single female flower represented by pistil and male flower represented by stamen.

Question 7.
Pollen differs from pollinium. How?
Answer:
Pollen are microspores which produces male gametes, whereas pollinium refers to the single mass of fused pollen grains.

Question 8.
Sunflower is not a flower – Justify your answer.
Answer:
Sunflower is actually an inflorescence not a single flower. The inflorescence of sunflower is capitulum composed of disc florets and ray florets.

Question 9.
Flower is a modified shoot for reproduction – Give possible evidence.
Answer:
Modified shoot for reproduction:

• Floral leaves (sepals and petals) are modified leaves.
• Floral and vegetative buds both emerge either in terminal or axillary position.
• Foliage leaves and floral leaves have identical arrangement on stem.

Question 10.
Is tomato a fruit or vegetable? Explain.
Answer:
Yes, tomato is a fruit. Because it develops from the ripened ovary and bears seeds, whereas vegetable refers to all other plant parts like root, stem and leaves.

Question 11.
What is caruncle? Where it is seen? How it helps the plant?
Answer:
Caruncle is the fleshy outgrowth at the base of seed. Usually caruncle helps in the seed dispersal, particularly by ants (Myrmecophily).

Question 12.
Both the prefixes (Uni – and Mono -) have the same meaning i.e. one in number. Does it mean that unisexual and monoecious are the same?
Answer:
That unisexual and monoecious:

1. Unisexual refers to the sex of flower (i.e. whether it has anther or carpel).
2. Monoecious refers to the plant bearing both the sexes in their flowers.

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Samacheer Kalvi 11th Bio Zoology Organ and Organ Systems in Animals Text Book Back Questions and Answers

Question 1.
The clitellum is a distinct part in the body of earthworm Lampito mauritii, it is found in …………….
(a) Segments 13 – 14
(b) Segments 14 – 17
(c) Segments 12 – 13
(d) Segments 14 – 16
Answer:
(b) Segments 14 -17

Question 2.
Sexually, earthworms are ……………..
(a) Sexes are separate
(b) Hermaphroditic but not self-fertilizing
(c) Hermaphroditic and self-fertilizing
(d) Parthenogenic
Answer:
(b) Hermaphroditic but not self-fertilizing

Question 3.
To sustain themselves, earthworms must guide their way through the soil using their powerful muscles. They gather nutrients by ingesting organic matter and soil, absorbing what they need into their bodies. Say whether the statement is true or false: The two ends of the earthworm can equally ingest soil.
(a) True
(b) False
Answer:
(b) False

Question 4.
The head region of Cockroach pairs of and shaped eyes occur.
(a) One pair, sessile compound and kidney shaped
(b) Two pairs, stalked compound and round shaped
(c) Many pairs, sessile simple and kidney shaped
(d) Many pairs, stalked compound and kidney shaped
Answer:
(a) One pair, sessile compound and kidney shaped

Question 5.
The location and numbers of malpighian tubules in Periplaneta ……………
(a) At the junction of midgut and hindgut, about 150.
(b) At the junction of foregut and midgut, about 150.
(c) Surrounding gizzard, eight.
(d) At the junction of colon and rectum, eight.
Answer:
(a) At the junction of midgut and hindgut, about 150.

Question 6.
The type of vision in Cockroach is ………………
(a) Three dimensional
(b) Two dimensional
(c) Mosaic
(d) Cockroach do not have vision
Answer:
(c) Mosaic

Question 7.
How many abdominal segments are present in male and female cockroaches?
(a) 10, 10
(b) 9, 10
(c) 8, 10
(d) 9, 9
Answer:
(a) 10,10

Question 8.
Which of the following have an open circulatory system?
(a) Frog
(b) Earthworm
(c) Pigeon
(d) Cockroach
Answer:
(d) Cockroach

Question 9.
Buccopharyngeal respiration in frog …………….
(a) is increased when nostrils are closed
(b) stops when there is pulmonary respiration
(c) is increased when it is catching fly
(d) stops when mouth is opened.
Answer:
(b) Stops when there is pulmonary respiration

Question 10.
Kidney of frog is ………………
(a) Archinephros
(b) Pronephros
(c) Mesonephros
(d) Metanephros
Answer:
(c) Mesanephros

Question 11.
Presence of gills in the tadpole of frog indicates that ………….
(a) fishes were amphibious in the past
(b) fishes evolved from frog-like ancestors
(c) frogs will have gills in future
(d) frogs evolved from gilled ancestor
Answer:
(d) frogs evolved from gilled ancestor

Question 12.
Choose the wrong statement among the following …………….
(a) In earthworm, a pair of male genital pore is present.
(b) Setae help in locomotion of earthworms.
(c) Muscular layer in the body wall of earthworm is made up of circular muscles and longitudinal muscles.
(d) Typhlosole is part of the intestine of earthworm.
Answer:
(a) In earthworm, a pair of male genital pore is present.

Question 13.
Which of the following are the sense organs of Cockroach?
(a) Antennae, compound eyes, maxillary palps, anal cerci
(b) Antennae, compound eye, maxillary palps and tegmina
(c) Antennae, ommatidia, maxillary palps, stemumv and anal style
(d) Antennae, eyes, maxillary palps, tarsus of walking legs and coxa
Answer:
(c) Antennae, ommatidia, maxillary palps, sternumv and anal style

Question n 14.
What characteristics are used to identify the earthworms?
Answer:
In gardens, earthworms can be traced by their fecal deposits known as worm castings on the soil surface. The earthworms can be identified using the following characteristics:

• Long and cylindrical narrow body.
• Bilateral symmetry
• It is light brown in colour with purple tinge at the anterior end.
• The division of body into many segments or metameres.
• The dorsal surface of the body is marked by a dark mid dorsal line.
• In mature worms, segments 14-17 may be found swollen with a glandular thickening of the skin called the clitellum.

Question 15.
What are earthworm casts?
Answer:
The undigested particles of food along with earth passed out through the anus of the earthworm are called worm casting.

Question 16.
How do earthworms breathe?
Answer:
In earthworms, respiration takes place through the body wall by the moist skin diffusion, oxygen diffuses through the skin into the blood while carbon dioxide from the blood diffuses out.

Question 17.
Why do you call cockroach a pest?
Answer:
Cockroaches destroy food and contaminates it with their offensive odour. They are carriers of a number of bacterial diseases. The cockroach allergen can cause asthma to sensitive people.

Question 18.
Comment on the functions of alary muscles?
Answer:
Alary muscles are the triangular muscles that are responsible for blood circulation in the cockroach. Each segment has one pair and a pumped anteriorly to sinuses again.

Question 19.
Name the visual units of the compound eyes of cockroach.
Answer:
Ommatidia

Question 20.
How does the male frog attracts the female for mating?
Answer:
Male frog has a pair of vocal sacs and a nuptial pad on the ventral side of the first digit of each forelimb. Vocal sacs assist in amplifying the croaking sound of frog. It makes a characteristic sound and attracts the female.

Question 21.
Write the types of respiration seen in frog.
Answer:
Frog respires on land and in the water by two different methods. In water, skin acts as aquatic respiratory organ (cutaneous respiration). Dissolved oxygen in the water gets exchanged through the skin by diffusion. On land, the buccal cavity, skin and lungs act as the respiratory organs. In buccal respiration on land, the mouth remains permanently closed while the nostrils remain open.

The floor of the buccal cavity is alternately raised and lowered, so air is drawn into and expelled out of the buccal cavity repeatedly through the open nostrils. Respiration by lungs is called pulmonary respiration. The lungs are a pair of elongated, pink coloured sac-like structures present in the upper part of the trunk region (thorax). Air enters through the nostrils into the buccal cavity and then to the lungs. During aestivation and hibernation gaseous exchange takes place through skin.

Question 22.
Differentiate between peristomium and prostoinium in earthworm.
Answer:
Peristomium:
The first segment of the body of earthworm is called peristomium.

Prostomium:
A small flap overhanging the mouth is called prostomium or upper lip.

Question 23.
Give the location of clitellum and spermathecal openings in Lampito mauritii.
Answer:
In mature earthworms, 14 – 17th segments are swollen with a glandular thickening of the skin called the clitellum. Permathecal openings are three pairs of small ventrolateral apertures lying intersegmentally between the grooves of the segment 6 / 7,  7 / 8 and 8 / 9.

Question 24.
Differentiate between tergum and a sternum
Answer:
Tergum:
Tergum is the covering each segment of cockroach on the dorsal side.

Sternum:
The sternum is the covering of each segment of cockroach on the ventral side.

Question 25.
Head of cockroach is called hypognathous. Why?
Answer:
The mouth parts of cockroach are directed downwards. The head is small, triangular lies at right angle to the longitudinal body axis. Hence it is called hypognathous.

Question 26.
What are the components of blood in frog?
Answer:
The blood consists of plasma [60%] and blood cells [40%], red blood cells, white blood cells, and platelets. RBCs are loaded with red pigment, nucleated and oval in shape. Leucocytes are nucleated, and circular in shape.

Question 27.
Draw a neat labeled diagram of the digestive system of frog.
Answer:
   

Question 28.
Explain the reproductive system of frog.
Answer:
The male frog has a pair of testes which are attached to the kidney and the dorsal body wall by folds of peritonium called mesorchium. Vasa efferentia arise from each testis. They enter the kidneys on both side and open into the bladder canal. Finally, it communicates with the urinogenital duct that comes out of kidneys and opens into the cloaca (Fig. 1).

Female reproductive system (Fig. 2) consists of paired ovaries, attached to the kidneys, and dorsal body wall by folds of peritoneum called mesovarium. There is a pair of coiled oviducts lying on the sides of the kidney. Each oviduct opens into the body-cavity at the anterior end by a funnel like opening called ostia. Unlike the male frog, the female frog has separate genital ducts distinct from ureters. Posteriorly the oviducts dilated to form ovisacs before they open into cloaca. Ovisacs store the eggs temporarily before they are sent out through the cloaca.

In-Text Questions Solved

Question 1.
How do the earthworms sense activity in their habitat without eyes, ears or a nose?
Answer:
The earthworm’s receptors are stimulated by a group of slender columnar cells connected with nerves. The Photoreceptors (sense of light) are found on the dorsal surface of the body. Gustatory (sense of taste) and olfactory receptors (sense of smell) are found in the buccal cavity. Tactile receptors (sense of touch), chemoreceptors (detect chemical changes) and thermoreceptors (changes in temperature) are present in the prostomium and the body wall.

Question 2.
Respiratory system of cockroach is formed of spiracles and tracheal interconnections. Why is it said to be more efficient than that of earthworm? Why inspiration of cockroach is said to be a passive process while it is an active process in man?
Answer:
The respiratory system of cockroach is well developed compared with other terrestrial insects. Branched tubes known as trachea open through 10 pairs of small holes called spiracles or stigmata

Question 3.
Arthropod eyes are called compound eyes because they are made up of repeating units, the ommatidia, each of which functions as a separate visual receptor. What is the difference between compound eyes and simple eyes? Why is mosaic vision with less resolution seen in cockroaches?
Answer:
f an eye is made of many eyes, it is a compound eye. It has many visual receptors unlike a simple eye which has only one visual receptor. The photoreceptors of the cockroach consists of a pair of compound eyes at the dorsal surface of the head. Each eye is formed of about 2000 simple eyes called the ommatidia (singular: ommatidium), through which the cockroach can receive several images of an object. This kind of vision is known as mosaic vision with more sensitivity but less resolution.

Question 4.
Why three chambered heart of frog is not as efficient has the four chambered heart of birds and mammals?
Answer:
A 4-chambered heart can pump blood more powerfully and efficiently. This helps in better oxygenation of the blood, better circulation and better purification of the blood.

Entrance Examination Questions Solved

Question 1.
The body cells in cockroach discharge their nitrogenous waste in the haemolymph mainly in the form of ……….. (NEET 2015)
(a) Calcium carbonate
(b) Ammonia
(c) Potassium urate
(d) Urea
Answer:
(c) Potassium urate

Question 2.
Frog’s heart when taken out of the body continues to beat for sometime. Select the best option from the following statements. (NEET 2017)
(a) Frog is a poikilotherm.
(b) Frog does not have any coronary circulation.
(c) Heart is “myogenic” in nature.
(d) Heart is autoexcitable Options
Answer:
(d) Heart is autoexcitable Options

Samacheer Kalvi 11th Bio Zoology Organ and Organ Systems in Animals Additional Questions & Answers

I. Multiple Choice Questions
Choose the correct answer

Question 1.
Which of the following is found in upper layers of the soil?
(a) Perionyx exavatus
(b) Octochaetona thurstoni
(c) Lampito mauritii
(d) Eudrilus eugeniae
Answer:
(c) Lampito mauritii

Question 2.
The body setae of earthworms are concerned with
(a) sensory function
(b) protection
(c) excretion
(d) locomotion
Answer:
(d) locomotion

Question 3.
The female genital aperture lies on the ventral side in the segment.
(a) 18^th
(b) 10^th
(c) 14^th
(d) 8^th
Answer:
(c) 14^th

Question 4.
The pair of male genital apertures are situated later ventrally in the segment.
(a) 18^th
(b) 10^th
(c) 14^th
(d) 9^th
Answer:
(a) 18^th

Question 5.
In earthworm, what is present in the 6th segment?
(a) oesophagus
(b) intestine
(c) gizzard
(d) anus
Answer:
(c) gizzard

Question 6.
Which of the following is not found in the prostomium?
(a) Tactile receptors
(b) chemoreceptors
(c) Thermoreceptors
(d) Olfactory receptors
Answer:
(d) Olfactory receptors

Question 7.
In protandrous condition, what is correct?
(a) ova mature earlier than sperms
(b) sperms mature earlier than ova
(c) both sperms and ova mature at the same time
(d) both sperms and ova do not mature at all
Answer:
(b) sperms mature earlier than ova –

Question 8.
The mouth parts of cockroach are of type.
(a) eating and chewing
(b) chewing and sicking
(c) sucking and chewing
(d) biting and chewing
Answer:
(d) biting and chewing

Question 9.
Which of the following is right?
(a) Anal styles are present only in male cockroach.
(b) Anal styles are present in both male and female cockroaches
(c) Anal cerei are present in male cockroach only
(d) Anal cerei are present in female cockroach only
Answer:
(a) Anal styles are present only in male cockroach.

Question 10.
Which is responsible for circulation of blood in cockroach?
(a) spiracular muscles
(b) alary muscles
(c) haemocytes
(d) ostia
Answer:
(b) alary muscles

Question 11.
Cockroach excretes as the nitrogenous waste from the body.
(a) urea
(b) ammonia
(c) uric acid
(d) creatinine
Answer:
(c) uric acid

Question 12.
In cockroach, maiphigian tubules are concerned with
(a) digestion
(b) respiration
(c) excretion
(d) secretion of fluids
Answer:
(c) excretion

Question 13.
The secretion of collaterial glands in cockroach is to produce
(a) ova
(b) sperms
(c) pigments
(d) oothoea
Answer:
(d) oothoea

Question 14.
Which of the following is not a feature of frog?
(a) presence of webbed deet
(b) absence of teeth
(c) smooth and moist skin
(d) slender body
Answer:
(b) absence of teeth

Question 15.
When the frog is ¡n water respiration occurs.
(a) pulmonary
(b) gill
(c) buccal
(d) cutaneous
Answer:
(d) cutaneous

Question 16.
In frog, aortic arch divides into three aortic arches namely …………….
(a) carotid, systemic and sinus venosus
(b) carotid, systemic and pulmo cutaneous
(c) carotid, dorsal aorta and pulmo cutaneous
(d) carotic, truneous arteriosus and pulmo cutaneous
Answer:
(b) carotid, systemic and pulmo cutaneous

Question 17.
Frogs excrete and hence they are called ……………
(a) urea, urecotelic
(b) uric acid. urecotelic
(c) ammonia, ammonotalic
(d) urea, urotelic
Answer:
(c) ammonia, ammonotalic

Question 18.
Which of the following change happens during metamorphosis of tadpole?
(a) gills disappear and lungs develop
(b) lungs disappear and gills develop
(c) gills remains as such
(d) lungs disappear
Answer:
(a) gills disappear and lungs develop

II. Fill In the blanks

Question 1.
………… are the earthworms found in deeper layers of the soil.
Answer:
Endogeics

Question 2.
The segments of the earthworm are ………..
Answer:
metameres

Question 3.
The first segment of earthworm is called …………..
Answer:
prostomium

Question 4.
The last segment of the earthworm is called ……………
Answer:
pygidium

Question 5.
The swollen segments 14 to 17 in mature earthworm is called the …………..
Answer:
clitellum

Question 6.
The body setac are absent in the , last segments and allatum ……………
Answer:
first

Question 7.
The are present from the O segment onwards in the earthworm …………..
Answer:
dorsal

Question 8.
Nephrogenous eliminate wastes …………….
Answer:
metabolic/nitrogenous

Question 9.
found in the coelomic fluid of earthworm plays an important role in regeneration, immunity and wound healing.
Answer:
Coelomocytes

Question 10.
The dorsal wall of the intestine of earthworm is tblded into the cavity as the …………..
Answer:
typhiosole

Question 11.
In earthworm, the vessel has no valves and non-contractile …………….
Answer:
Ventral

Question 12.
In earthworms excretion is effected by …………….
Answer:
nephridia

Question 13
nephridia are present in the 5th – 9th segments.
Answer:
Pharyngeal of tufted

Question 14.
Special cells on the coelomic wall of the intestine called are excretory in function in earthworms.
Answer:
chioragogen cells

Question 15.
In earthworms, the male genital aperture contains two pairs of for copulation.
Answer:
penial setae

Question 16.
The secretion of the prostate gland serves to cement the spermatozoa into bundles known as
Answer:
spermatophores

Question 17.
are present in the 7th, 8th and 9th segment of earthworm.
Answer:
Spermatheca

Question 18.
The process of producing compost using earthworm is called ……………
Answer:
vermi composting

Question 19.
The first pair of wings of cockroach is called ………..
Answer:
alytra of fegmina

Question 20.
In both male and female cockroach, genital apertures are surrounded by scierites called …………….
Answer:
gonapophysis

Question 21.
helps in grinding of food particles in cockroach.
Answer:
Gizzard

Question 22.
In between the foregut and midgut of cockroach are present.
Answer:
Aepatic caecae

Question 23.
In between the midgut and the hindgut of cockroach are present which help in …………….
Answer:
maiphigian tubules excretion

Question 24.
The respiratory organs of cockroach are ……………..
Answer:
trachea

Question 25.
The trachea of cockroach open through 10 pairs of small holes called ……………
Answer:
spiracles/stigmata

Question 26.
The opening and closing of spiracles are regulated by ……………..
Answer:
sphincter spiracular muscles

Question 27.
The blood of cockroach is called as ……………
Answer:
Haemolymph.

Question 28.
The blood from the sinuses enters the heart of cockroach through the …………..
Answer:
ostia

Question 29.
in cockroach, there is a accessory at the base of each antenna which also pumps blood.
Answer:
pulsatile vesicle

Question 30.
The receptors are located in the antenna, maxillary paips and cerci.
Answer:
thigmo

Question 31.
are found on the first four tarsal segments on the legs of cockroach.
Answer:
Thermoreceptors

Question 32.
The receptor chordotonal is found on the which respond lo air or earth borne vibrations.
Answer:
Anal cerci

Question 33.
The eye of cockroach is made up of simple’eyes called the ……………..
Answer:
ommatidia

Question 34.
The collaterial glands present behind the ovaries of female reproductive system cockroach produce ………….
Answer:
ootheca

Question 35.
The young cockroach are called …………….
Answer:
nymphs

Question 36.
The temperature of the frog varies with the varying environmental temperature and hence it is called …………..
Answer:
poikulothermic

III. Answer the following Questions

Question 1.
Classify earthworms based on their ecological strategies.
Answer:

• Earthworms are classified as epigeics, anecics and endogeics based on their ecological strategies.
• Epigeics are the surface dwellers e.g., Perionyx excavaus and Eudrilus eugeniae.
• Anecics are found in the upper layers of the soil e.g., Lampiro mauritii, Lumbricus terrestris.
• Endogeics are found in deeper layers of the soil e.g., Octochaetona thursoni.

Question 2.
Explain the morphology of earthworm, Lampito mauritii.
Answer:

• Lampilo mauritii has a long and cylindrical narrow body which is laterally symmetrical.
• it is light brown in colour with purplish length at the anterior end.
• The body is divided into 165 – 190 segments.
• The dorsal surface of the body is marked by a dark mid dorsal line along the longitudinal axis of the body.
• The ventral surface has genital opening.
• The mouth is found in the centre of the first segment called peristomium. Overhanging
• the mouth is a small flag called upper lip or prostomium.
• The last segment called pygidiurn has the anus.
• The segments 14-17 are swollen with a glandular thickening of the skin called the clitellum.
• The body of earthworm is divided into pre clitellar region (1 – 13), clitellar region (14 – 17) and post clitellar region (after 17).
• Chitinous body selae are present in all segments except the first, last and clitellar segments. They are locomoting in function.
• The dorsal pores are present from the 10th segment onwards. The coelomic fluid
• communicates to the exterior through these pores and keeps the body surface moist and free from harmful microorganism.
• Between the segment 6/7, 7/8 and 8/9, supernatural openings are present.
• The female genital aperture lies on the ventral side in the 14th segment and a pair of male genital apertures are situated determinately in the 1 8th segment.
• Nephridiopores are numerous and found throughout the body except a free anterior segments.

Question 3.
Explain the internal structure of the earthworm.
Answer:
The body wall of the earthworm is very moist, thin, soft, skinny, elastic and consists of the cuticle, epidermis, muscles and coelomic epithelium. The epidermis consists of supporting cells, gland cells, basal cells and sensory cells.

Question 4.
Write a note on coelom of earthworm.
Answer;
A spacious body cavity called the coelom is seen between the alimentary canal and the body wall. The coelom contains the coelomic fluid and serves as a hydrostatic skeleton, in which the coelomocytes are known to play a major role in regeneration. immunity and wound healing. The coelomic fluid of the earthworm is milky and alkaline, which consists of granulocytes or cicocytes. amoebocytes, mucocytes and leucocytes.

Question 5.
Explain the digestive system of earthworm.
Answer:
The digestive system of the earthworm consists of the alimentary canal and the digestive glands. The alimentary canal runs as a straight tube throughout the length of the body from the mouth to anus. The mouth opens into the buccal cavity which occupies the 1 and 2 segments.

The buccal cavity leads into a thick muscular pharynx,which occupies the 3^th and 4^th segments and is surrounded by the pharyngeal glands. A small narrow tube, oesophagus lies in the 5^th segment and continues into a muscular gizzard in the 6th segment. The gizzard helps in the grinding of soit particles and decaying leaves.

Intestine starts from the 7^th segment and continues till the last segment. The dorsal wall of the intestine is folded into the cavity as the typhlosole. This fold contains blood vessels and increases the absorptive area of the intestine. The inner epithehum consists of columnar cells and glandular cells. The alimentary canal opens to the exterior through the anus. The ingested organic rich soil passes through the digestive tract where digestive enzymes breakdown complex food into smaller absorbable units.

The simpler molecules are absorbed through the intestinal membrane and are utilized. The undigested particles along with earth are passed out through the anus, as worm castings or vermicasts. The pharyngeal or salivary gland cells and the glandular cells of the intestine are supposed to be the digestive glands which secrete digestive enzymes for digestion of food.

Question 6.
What is typhiosole?
Answer:

• The dorsal wall of the intestine of earthworm is folded into the cavity as the typhiosole.
• This fold contains blood vessels and increases the absorptive area of the intestine.

Question 7.
Explain the nervous system of earthworm.
Answer:
The bibbed mass of nervous tissue called supra-pharyngeal ganglia lies on the dorsal wall of the pharynx in the 3 segment. It is referred as the brain. In the 4th segment, the sub-pharyngeal ganglia is found. The brain and the sub-pharyngeal are connected by a pair of circum-pharyngeal connectives. The double ventral nerve cord runs backward from the sub-pharyngeal ganglion. The brain and other nerves in the ring integrates sensory inputs and command muscular responses of the body.

Question 8.
Explain the receptors in earthworm.
Answer:
The receptors are simulated by a group of slender columnar cells connected with nerves. The photoreceptor are found on the dorsal surface of the body. Gustatory and olfactory receptors are found in the buccal cavity. Tactile receptors, chernoreceptors and thermoreceptors are present in the prostomium and the body wall.

Question 9.
Explain the excretory system of the earthworm
Answer:
Excretion is the process of elimination of metabolic waste products from the body. In earthworm, excretion is effected by segmentally arranged, minute coiled, paired tubules called nephridia.
There are three types of nephridia:
1. pharyngeal or tufted nephridia – present as paired tufts in the 5^th segments.

2. Microncphridia or Integurnentary nephridia attached to the lining of the body wall from the 14th segment to the last which open on the body surface.

3. Megancphridia or septal nephridia – present as pair on both sides of intersegmental septa of the 1 9th segment to the last and open into intestine (Figure). The mcganephridium has an internal funnel like opening called the nephrostome. which is fully ciliated. The nephrostorne is in the preceding segment and the rest of the tube is in the succeeding segment. This tube consists of three distinct divisions, the ciliated, the glandular and the muscular region.

The waste material collected through the ciliated funnel is pushed into the muscular part of nephridium by the ciliated region. The glandular part extracts the waste from the blood and finally the wastes exit out through the nephridiopore. Bcsides nephridia. special cells on the coelomic wall of the intestine, called chloragogen cells arc present. They extract the nitrogenous waste from the blood of the intestinal wall. into the body cavity to he sent out through the ncphridia.

Question 10.
Explain the male reproductive system of the earthworm.
Answer:
In the male reproductive system, two pairs of testes are present in the 10^th and 11^th segments. The testes give rise to the germ cells or spermatogonia, which develops into spermatozoa in the two pairs of seminal vesicles. Two pairs of seminal ftrnnels called ciliary rosettes are situated in the same segments as the testes. The ciliated funnels of the same side are connected to a long tube called vas dcfcrcns.

The vasa deferentia run upto the 18^th segment where they open to the exterior through the male genital aperture. The male genital aperture contains Iwo pairs of penial setae for copulation. A pair of prostate glands lies in the 18 – 9th segments. The secretion of the prostate gland serves to cement the spermatozoa into bundles known as spermatophores.

Question 11.
Explain the female reproductive system of earthworm.
Answer:
The female reproductive system consists of a pair of ovaries lying in the 3’ segment. Each ovary has finger like projections which contain ova in linear series. Ovarian funnels are present beneath the ovaries which continue into the oviducts.

They join together and open on the ventral side as a single median female genital pore in the 14^th segment. Spermathecae or seminal receptacles are three pairs lying in segments 7^th, 8^th and 9^th opening to the exterior on the ventral side between 6th & 7^th 7^th & 8^th and 8^th & 9^th segments. They receive spermatozoa from the partner and store during copulation.

Question 12.
In earthworm, self fertilization does not take place though it has both male and female reproductive system. Why?
Answer:
The male and female sex organs mature at different times Sperms develops earlier than the production of ova (protandrous). Hence, self fertilization does not take place in earthworm.

Question 13.
Write a short note on the life cycle of earthworni.
Answer:
Lampito mauritii begins its life cycle, from the fertilized eggs. The eggs are held in a protective cocoon. These cocoons have an incubation period of about 14 – 18 days after which they hatch to release juveniles (Figure). The juveniles undergo changes into non clitellate forms in phase I after about 15 days.

which then develops a clitellum, called the clitellate at the end of the growth phase II taking 15 – 17 days to complete. During the reproductive stage, earthworms copulate, and later shed their cocoons in the soil after about 10 days. The life cycle of Lampito mauritii takes about 60 days to complete.

Question 14.
What are earthworm knowii as friends of farmers?
Answer:
They make burrows in the soil and make it porous which helps in respiration and penetration of developing plant roots. They help in recycling of dead and decayed plant material by feeding on them. Hence they arc called as friends of farmers.

Question 15.
Write a short note on vermitech and

1. Vermiculture
2.  Vermicomposting
3. Vermiwash
4.  Wormery

Answer:
1. Vermiculture :
Artificial rearing or cultivation of earthworms involves new technology for the betterment of human beings. This process is known as vermiculture.

2. Vermicomposting :
The process of producing compost using earthworms is called vermicomposting.

3. Vermiwash :
Vermiwash is a liquid manure or plant tonic obtained from earthworm It is used as a foliar spray and helps to induce plant growth. It is a collection of excretory products and mucus secretion of earthworms along with micronutrients from the soil organic molecules.

4. Wormery or wormbin :
Earthworm can be used for recycling of waste food, leaf, litter and biomass to prepare a good fertilizer in a container known as wormery or wormbin. It makes superior compost than conventional composting methods. Vermiculture, vericomposting, vermiwash and wormery are inter-related and independent process, collectively referred as vermitech.

Question 16.
Explain the morphology of cockroach.
Answer:
The body of cockroach is compressed dorso – ventrally, bilaterally symmetrical and segmented. The body is divisible into head, thorax and abdomen. The entire body is covered by a hard, brown, chitinous exoskeleton. In each segment.

exoskeleton has hardened plates called scierites which are joined by a delicate and elastic articular membrane or arthrodial membrane. The sclerites on the dorsal side are called tergites, and those on the ventral side are called stemites and those on the lateral sides are called pleurites.

Head :
Head is small, triangular lies at right angle to the longitudinal axis. The mouth parts are directed downwards so it is hypognathous. The head has a pair of large, sessile and reniform compound eyes, a pair of antennae and appendages around the mouth part.

Antennae have sensory organs. The cockroach has a biting and chewing type (mandibulate or orthopterus type) of mouth parts. It includes a labrum, a pair of mandibles, a pair of maxillae, a labium and a hypopharynx or lingua.

Thorax :
The thorax consists of three segments, prothorax, mesothorax and metathorax. The head is connected with thorax by a short neck or cervicum. Each thoracic segment has three pairs of walking legs. Each leg consists of five segments, coxa, trochanter, femur, tibia and tarsus.

The tarsus has five movable joints or podomeres or tarsomeres. A pair of forewings arises from the mesothorax called elytra or tegmina. It protects the hind wings when at rest. The second pair membranous wings arises from the metathorax and are used in flight.

Abdomen :
The abdomen has 10^th segments. Each segment has a dorsal tergurn, ventral sternum and between them a narrow membranous pleuron. In female the 7^th, segment is boat shaped and together with the gth and 9th sterna forms a brood pouch or genital pouch. Its anterior parts contain female gonopore, spermathecal pores, collaterial glands and the posterior parts constitutes the oothecal chamber in which the cocoons are formed.

in males, the genital pouch lies at the hind end of the abdomen bound dorsally by 9^th and 10^th terga and ventrally by the 9^th sternum. It contains dorsal anus and ventral male genital pore. In both sexes, genital apertures are surrounded by scierites called goriapophysis.

Male has a pair of short and slender anal styles in the 9th stenium which are absent in feniale. In both sexes, the 10th segment has a pair of jointed filanientous structures called anal cerci which is a sense organ that is receptive to vibrations in the air and land. The 7th sternum of male has a pair of large and oval apical lobes orgynovalvular plates which form a keel like structure which distinguishes the male from female.

Question 17.
Explain the digestive system of cockroach.
Answer:
The digestive system of cockroach consists of the alimentary canal and digestive glands. The alimentary canal is present in the body cavity and is divided into three regions: foregut, midgut and hindgut (Figure). The foregut includes pre-oral cavity, mouth, pharynx and oesophagus. This in turn opens into a sac like structure called the crop which is used for storing food.

The crop is followed by the gizzard or proventriculus which has an outer layer of thick circular muscles and thick inner cuticle forming six highly chitinous plates called ‘tecth’. gizzard helps in the grinding of the food particles. The midgut isa short and narrow tube behind the gizzard and is glandular in nature. At the junctional region of the gizzard are eight fingers like tubular blind processes called the hapatic caecae or enteric caecae.

The hindgut is marked by the presence of 100 – 150 yellow coloured thin filamentous maiphigian tubules which arc helpful in removal of the excretory products from the haemolymph. The hindgut is broader than the midgut and is differentiated into ileum. colon, and rectum.

The rectum opens out through the anus. Digestive glands of cockroach consist of the salivary glands. the glandular cells and hcpatic caccac. A pair of salivary glands is found on either side of the crop in the thorax. Thc.glandular cells of the midgut and hepatic or gastric caeeae produce digestive juices.

Question 18.
Explain the circulatory system of cockroach.
Answer:
Periplaneta has an open type of circulatory system (Figure). Blood vessels are poorly developed and opens into the haemocoel in which the blood or haemolymph flows freely. Visceral organs located in the haemocoel are bathed in blood. The haemolynìph is colourless and consists of plasma and haemocytes which are ‘phagocytic’ in nature. Heart is an elongated tube with muscular wall lying mid dorsally beneath the thorax.

The heart consists of 13 chambers with ostia on either side. The blood from the sinuses enters the heart through the ostia and is pumped anteriorly to sinuses again. The triangular muscles that are responsible for blood circulation in the cockroach are called alary muscles (13 pairs). One pair of these muscles is found in each segment on either side of the heart. In cockroach, there is an accessory pulsatile vesicle at the base of each antenna which also pumps blood.

Question 19.
Explain the nervous system of cockroach.
Answer:
The nervous system of cockroach consists of a nerve ring and a ganglionated double ventral nerve cord. suh-oesophageal ganglion. circum – oesophageaÌ connectives and double ventral nerve cord (see figure). The nerve ring is present around the ocsophagus in the head capsule and is formed by the supra-oesophagial ganglion called the ‘brain’. The brain is mainly a sensory and an endocrine centre and lies above the oesophagus. Sub – oesophageal ganglion is the motor centre that controls the movements of the mouth parts, legs and wings.

it lies below the oesophagus and formed by the fusion of the paired gangalia of mandibular, maxillary and labial segments of the head. A pair of circum – oesophageal connectives is present around the oesophagus. connecting the supra oesophageal ganglia with the sub-ocsophageal ganglion.

The double ventral nerve cord is solid. ganglionated and arises from the sub-oesophageal ganglion and extends up to the 7th abdominal segment. Three thoracic ganglia are one in each thoracic segment and six abdominal ganglia in the abdomen.

Question 20.
Write about the sense organs of cockroach.
Answer:
In cockroach, the sense organs are antennae, compound eyes, labrum, maxillary alps, labial paips and anal cerci. The receptor for touch (thigmo receptors) is located in the antenna, maxillary palps and cerci. The receptor for smell (olfactory receptors) is found on the antennae. The receptor for taste (gustatory receptors) is found on the paips of maxilla and labium. Thermoreceptors are found on the first four tarsal segments on the legs.

The receptor chordotonal is found on the anal cerci which respond to air or earth borne vibrations. The photoreceptors of the cockroach Consists of a pair of compound eyes at the dorsal surface of the head. Each eye is formed of about 2000 simple eyes called the ommatidia (singular: ommatidium), through which the cockroach can receive several images of an object. This kind of vision is known as mosaic vision with more sensitivity
but less resolution.

Question 21.
Differentiate male and female cockroaches.
Answer:

Question 22.
Explain excretion in cockroach.
Answer:
The Malpighian tubules are the main excretory organs of cockroach which help in eliminating the nitrogenous wastes from the body in the form of uric acid. Cockroach excretes uric acid, so it is uricotelic. In addition, fat body, nephrocytes. cuticle, and urecose glands are also excretory in function.

The malpighian tubules are thin, long, tilamentous, yellow coloured structures attached at the junction of midgut and hindgut. These are about 100-150 in number and are present in 6-9 bundles. Each tubule is lined by glandular and ciliated cells and the waste is excreted out through the hindgut.

The glandular cells of the malpighian tubules absorb water. salts, and nitrogenous wastes from the haernolymph and transfer them into the lumen of the tubules. The cells of the tubules reabsorb water and certain inorganic salts. By the contraction of the tubules nitrogenous waste is pushed into the ileum, where more water is reabsorbed. It moves into the rectum and almost solid uric acid is excreted along with the fecal matter.

Question 23.
Explain the male reproductive system of cockroach.
Answer:
Cockroach is dioecious or unisexual. They have well developed reproductive organs. The male reproductive system consists of a pair of testes, vasa deferentia, an ejaculatory duct, utricular gland, phallic gland and the external genitalia. A pair of three lobed testes lies on the lateral side of the 4th and 6th abdominal segments. From each testis anses a thin vas deferens.

which opens into the ejaculatory duct through the seminal vesicles. The ejaculatory duct is an elongated duct which opens out by the male gonopore lying ventral to the anus.

A utricular or mushroom shaped gland is a large accessory reproductive gland, which opens into the anterior part of the ejaculatory duct. The seminal vesicles are present on the ventral surface of the ejaculatory duct, These sacs store the sperms in the form of bundles called spermatophores.

The duct of phallic or conglobate gland also opens near the gonopore, whose function is uncertain. Surrounding the male genital opening are few chitinous and asymmetrical structures called phallomeres or gonapophyses which help in copulation.

Question 24.
ExplaIn the female reproductive system in cockroach.
Answer:
The female reproductive system of cockroach consists of a pair of ovaries, vagina, genital pouch, collaterial glands. spermathecae and the external genitalia. A pair of ovaries lies laterally in the 2^nd and 6^th abdominal segment. Each ovary is formed of a group of eight ovarian tubules or ovarioles, containing a chain of developing ova. The lateral oviducts of each ovary unite into a broad median common oviduct known as vagina, which opens into the genital chamber.

The vertical opening of the vagina is the female genital pore. A pair of spermathccae is present in the 6^th segment, which opens by a median aperture in the dorsal wall of the genital pouch. During copulation, the ova descend to the genital chamber, where they are fertilized by the sperms. A pair of white and branched collaterial glands present behind the ovaries forms a hard egg case called Ootheca around the eggs. Genital pouch is formed by the 7, 8^th and 9^th abdominal sterna.

The genital pouch has two chambers, a genital chamber into which the vagina opens and an oothecal chamber where oothecae are formed. Three pairs of plate like chitinous structures called gonapophyses are present around the female genital aperture. These gonapophyses guide the ova into the ootheca as ovipositors.

Question 25.
Write a short note on ootheca.
Answer:
A pair of white and branched collaterial glands present behind the ovaries forms a hard egg case called oothcca around the eggs. Ootheca is a dark reddish to blackish brown capsule about 12 mm long which contains nearly 16 fertilized eggs and dropped or glued to a suitable surface, usually in crack or crevice of high relative humidity near a food source. On an average, each female cockroach produces nearly 15 – 40 oothecae in its life span of about one to two years.

Question 26.
Explain the morphological features of frog.
Answer:
The body of a frog is streamlined to help in swimming. It is dorso-ventrally flattened and is divisible into head and trunk. Body is covered by a smooth, slimy skin loosely attached to the body wall. The skin is dark green on the dorsal side and pale ventrally.

The head is almost triangular in shape and has an apex which forms the snout. The mouth is at the anterior end and can open widely. External nostrils are present on the dorsal surface of the snout, one on each side of the median line. Eyes are large and project above the general surface of the body.

They lie behind the external nostrils and are protected by a thin movable lower eyelid, thick immovable upper eyelid and a third transparent eyelid called nictitating membrane. This membrane protects the eye when the frog is tinder water.

A pair of tympanic membranes forms the ear drum behind the eyes on either side. Frogs have no external ears, neck and tail are absent. Trunk bears a pair of fore limbs and a pair of hind limbs. At the posterior end of the dorsal side, between the hind limbs is the cloacal aperature.

This is the common opening for the digestive, excretory and reproductive systems. Fore limbs are short, stumpy, and helps to bear the weight of the body. They are also helpful for the landing of the frog aller leaping. Each forelimb consists of an upper arm.

fore arm and a hand. Hand bears four digits. Hind limbs are large, long and consist of thigh. shank and foot. Foot bears five long webbed toes and one small spot called the sixth toe. These are adaptations for leaping and swimming. When the animal is at rest, the hind limbs are kept folded in the form of letter ‘Z’ Sexual dimorphism is exhibited clearly during the breeding season.

The male frog has a pair of vocal sacs and a copulatory or nuptial pad on the ventral side of the first digit of each forelimb. ‘cal sacs assist in amplifying the croaking sound of frog. Vocal sacs and nuptial pads are absent in the female frogs.

Question 27.
Differentiate between a frog and toad.
Answer:

Question 28.
Explain the digestive system of frog.
Answer:
The alimentary canal consists of the buccal cavity, pharynx, oesophagus, duodenum, ileum and the rectum which leads to the cloaca and opens outside by the cloacal aperture. The wide mouth opens into the buccal cavity. On the floor of the buccal cavity lies a large muscular sticky tongue.

The tongue is attached in front and free behind. The free edge is forked. When the frog sights an insect it flicks out its tongue and the insect gets glued to the sticky tongue. The tongue is immediately withdrawn and the mouth closes. A row of small and pointed maxillary teeth is found on the inner region of the upper jaw.

In addition vomerine teeth are also present as two groups, one on each side of the internal nostrils. The lower jaw is devoid of teeth. The mouth opens into the buccal cavity that leads to the oesophagus through the pharynx. Oesophagus is a short tube that opens into the stomach and continues as the intestine, rectum and finally opens outside by the cloaca. Liver secretes bile which is stored in the gall bladder. Pancreas, a digestive gland produces pancreatic juice containing digestive enzymes.

Food is captured by the bilobed tongue. Digestion of food takes place by the action of hydrochloric acid and gastric juices secreted from the walls of the stomach. Partially digested food called chyme is passed from the stomach to the first part of the intestine, the duodenum. The duodenum receives bile from the gall bladder and pancreatic juices from the pancreas through a common bile duct.

Bile emulsifies fat and pancreatic juices digest carbohydrates, proteins and lipids. Final digestion takes place in the intestine. Digested food is absorbed by the numerous finger-like folds in the inner wall of intestine called villi and microvilli. The undigested solid waste moves into the rectum and passes out through the cloaca.

Question 29.
Explain the circulatory system of frog.
Answer:
Blood vascular system consists of a heart with three chambers, blood vessels and blood. Heart is covered by a double- walled membrane called pericardium. There are two thin walled anterior chambers called auricles (Atria) and a single thick walled posterior chamber called ventricle. Sinus venosus is a large, thin walled, triangular chamber, which is present on the dorsal side of the heart. Truncus arteriosus is a thick walled and cylindrical structure which is obliquely placed on the ventral surface of the heart.

It arises from the ventricle and divides into right and left aortic trunk, which is further divided into three aortic arches namely carotid, systemic and pulmo-cutaneous. The Carotid trunk supplies blood to the anterior region of the body. The systemic trunk of each side is joined posteriorly to form the dorsal aorta. They supply blood to the posterior part of the body. Pulmo-cutaneous trunk supplies blood to the lungs and skin.

Sinus venosus receives the deoxygenated blood from the body parts by two anterior precaval veins and one post caval vein. It delivers the blood to the right auricle; at the same time left auricle receives oxygenated blood through the pulmonary vein. Renal portal and hepatic portal systems are seen in frog.

Question 30.
Explain the nervous system of frog.
Answer:
The Nervous system is divided into the Central Nervous System [CNS], the Peripheral Nervous System [PNS], and the Autonomous Nervous System [ANS], Peripheral Nervous System consists of 10 pairs of cranial nerves and 10 pairs of spinal nerves. Autonomic Nervous System is divided into sympathetic and parasympathetic nervous system. They control involuntary functions of visceral organs.

Question 31.
Explain the structure of brain of frog ?
Answer:
Brain is situated in the cranial cavity and covered by two meninges called piamater and duramater. The brain is divided into forebrain, midbrain and hindbrain. Fore brain (Prosencephalon) is the anterior most and largest part consisting of a pair of olfactory lobes and cerebral hemisphere (as Telen-cephalon) and a diencephalon. Anterior part of the olfactory lobes is narrow and free but is fused posteriorly. The olfactory lobes contain a small cavity called olfactory ventricle.

The mid brain (Mesencephalon) includes two large, oval optic lobes and has cavities called optic ventricles. The hind brain (Rhombencephalon) consists of the cerebellum and medulla oblongata. Cerebellum is a narrow, thin transverse band followed by medulla oblongata. The medulla oblongata passes out through the foramen magnum and continues as spinal cord, which is enclosed in the vertebral column.

Question 32.
Explain the excretory system of Frog.
Answer:
Elimination of nitrogenous waste and salt and water balance are performed by a well developed excretory system. It consists of a pair of kidneys, ureters, urinary bladder and cloaca. Kidneys are dark red, long, flat organs situated on either sides of the vertebral column in the body cavity.

Kidneys are Mesonephric. Several nephrons are found in each kidney. They separate nitrogenous waste from the blood and excrete urea, so frogs are called ureotelic organisms. A pair of ureters emerges from the kidneys and opens into the cloaca. A thin walled unpaired urinary bladder is present ventral to the rectum and opens into the cloaca.

Question 33.
Write the short note on the development in frog.
Answer:
Within few days of fertilization, the eggs hatch into tadpoles. A newly hatched tadpole lives off the yolk stored in its body. It gradually grows larger and develops three pairs of gills. The tadpole grows and metamorphosis into an air-breathing carnivorous adult frog (Figure). Legs grow from the body, and the tail and gills disappear. The mouth broadens, developing teeth and jaws, and the lungs become functional.

Question 34.
Write a note on the economic importance of frog.
Answer:
Economic importance of Frog:
Frog is an important animal in the food chain; it helps to maintain our ecosystem. So ‘frogs should be protected’.
Frog are beneficial to man, since they feed on insects and helps in reducing insect pest population. Frogs are used in traditional medicine for controlling blood pressure and for its anti aging properties. In USA. Japan, China and North East of India, frogs are consumed as delicious food as they have high nutritive value.

Question 35.
Frog respires through gills, lungs, skin and buccal cavity’. Justify.
Answer:
Frog is amphibious. It respires through lungs (pulmonary respiration) when it is on land. Buccal respiration is there in frog when it is on land. When it is in water, it respires through skin (cutaneous respiration). The larva of frog, tadpole respires through gills. During development, gills disappear and lungs develop and the tadpole metamorphoses into an adult frog. Hence, frog respires through gills, lungs, buccal cavity and skin in its life cycle.

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Tamilnadu Samacheer Kalvi 11th Bio Zoology Solutions Chapter 3 Tissue Level of Organisation

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Samacheer Kalvi 11th Bio Zoology Tissue Level of Organisation Text Book Back Questions and Answers

Choose the correct answer:
Question 1.
The main function of the cuboidal epithelium is –
(a) Protection
(b) Secretion
(c) Absorption
(d) Both (b) and (c)
Answer:
(d) Both (b) and (c)

Question 2.
The ciliated epithelium lines the –
(a) Skin
(b) Digestive tract
(c) Gall bladder
(d) Trachea
Answer:
(d) Trachea

Question 3.
What type of fibres are found in connective tissue matrix?
(a) Collagen
(b) Areolar
(c) Cartilage
(d) Tubular
Answer:
(a) Collagen

Question 4.
Prevention of substances from leaking across the tissue is provided by –
(a) Tight junction
(b) Adhering junction
(c) Gap junction
(d) Elastic junction
Answer:
(a) Tight junction

Question 5.
Non-shivering thermogenesis in neonates produces heat through –
(a) White fat
(b) Brown fat
(c) Yellow fat
(d) Colourless fat
Answer:
(b) Brown fat

Question 6.
Some epithelia are pseudostratified. What does this mean?
Answer:
Pseudostratified epithelial cells are columnar, but unequal in size. Although the epithelium is single layered yet it appears to be multilayered due to the fact that nuclei lie at different levels in different cells.

Question 7.
Differentiate white adipose tissue from brown adipose tissue.
Answer:
Adipose tissue:

• Adipose tissue is the group of fat cells.
• It stores fats.
• It releases energy while fasting.

Brown Adipose Tissue:

• Adipose tissue which contains abundant mitochondria are called brown adipose tissue.
• It is used to warm the blood stream to warm the body.
• It produces heat by non-shivering thermogenesis.

Question 8.
Why blood is considered as a typical connective tissue?
Answer:
Blood is considered as a typical connective tissue because it is the fluid connective tissue containing plasma, RBCs, WBCs and platelets. It functions as the transport medium for the cardiovascular system carrying nutrients, nitrogenous wastes and respiratory gases throughout the body.

Question 9.
Differentiate between elastic fibres and elastic connective tissue.
Answer:
Elastic fibres:

• It contains elastin and other proteins and glycoproteins.
• It attaches muscles and bones and one bone to another bone.
• It withstands tensile stress when pulling force is applied in one direction or in many directions.

Elastic connective tissue:

• It contains high proportion of elastic fibres.
• It is found in the walls of large arteries, ligaments associated with vertebral column and within the walls of the bronchial tubes.
• It allows recoil of tissues after stretching.

Question 10.
Name any four important functions of epithelial tissue and provide at least one example of a tissue that exemplifies each function.
Answer:
1. Secretion and absorption :
Cuboidal epithelium in kidney tubules, ducts. Columnar epithelium found in the digestive tract.

2. Filtration :
Squamous epithelium found in the glomerulus of kidney.

3. Ciliated epithelium :
Found in the bronchi, uterine tubes propels the materials due to ciliary actions.

Question 11.
Write the classification of connective tissue and their functions.
Answer:
There are four main classes of connective tissues.

• Connective tissue proper
• Cartilage
• Bones
• Blood

The major functions of connective tissues are binding and support, protection, insulation and transportation of substances.

Question 12.
What is an epithelium? Enumerate the characteristic features of different epithelia.
Answer:
Epithelial tissue is sheet of cells that covers the body surface or lines the body cavity.

• Simple epithelium is single layered.
• Squamous epithelium is made of flattened cells with irregular boundaries.
• Columnar epithelium is made of column like cells with round to oval nuclei at the base.
• Ciliated epithelium has ciilia at the free end.
• Compound epithelium is made of multi-layered cells.

In-Text Questions Solved

Question 1.
Stratified epithelia are “built” for protection or to resist abrasion. What are the simple epithelia better at?
Answer:
The simple epithelia are better at absorption, secretion of mucus, enzymes and other substances.

Question 2.
What type of connective tissue is damaged when one get cut on his index finger accidently?
Answer:
The Areolar connective tissue is damaged when finger gets cut.

Question 3.
The stored lipids are in the form of adipose tissue. Are they coloured? Why?
Answer:
The white adipose tissue is called white fat. The adipose that has abundant mitochondria is called Brown fat.

Question 4.
You are looking at a slide of a tissue through the compound microscope and you see striped branching cells that connect with one another. What type of muscle are you viewing?
Answer:
I am viewing the skeletal muscle.

Question 5.
A player has sustained a severe injury during football practice and was told that he has a torn knee cartilage. Can he expect a quick uneventful recovery? Explain your response.
Answer:
The knee cartilage is an important connective tissue. Since the knee moves during locomotion, a quick, uneventful recovery cannot take place. Complete rest to the knee joint is necessary.

Question 6.
An overweight high school student, is overheard telling her friend that she is going to research how she can transform some of her white fat to brown fat. What is her rationale here (assuming it is possible)?
Answer:
The white fat stores nutrients while the brown fat warms the body. The student feels that she may bring down her weight by converting brown fat to white fat.

Textbook Activities Solved

Question 1.
Students are asked to identify the unlabelled slides of tissues and to classify them. Similar exercise can also be accomplished by projecting unlabelled histological images on a screen. They can identify the slides of different tissues through microscope.
Answer:
Do it yourself.

Question 2.
The preparation of smear of stratified squamous epithelia from the inner lining of cheek allows the students to make their own slides using biological stain. They will have the experience of examining their cheek cells.
Answer:
Do it yourself.

Entrance Examination Questions Solved

Question 1.
Transitional epithelium occurs in ………… (MHTCET 2008)
(a) Blood vessels
(b) Trachea
(c) Kidney
(d) Ureter/urinary bladder
Answer:
(d) Ureter/urinary bladder

Question 2.
The study of tissues is known as …………….. (MPPMT 2010)
(a) Physiology
(b) Ecology
(c) Histology
(d) Anatomy
Answer:
(c) Histology

Question 3.
Find out the wrong match :
(a) Eosinophils Allergic response
(b) Basophils Secrete histamine and serotonin
(c) Monocytes Secrete heparin
(d) Lymphocytes Immune response
Answer:
(c) Monocytes Secrete heparin

Question 4.
The outer covering of cartilage is called (WB 2010)
(a) Peritoneum
(b) Periosteum
(c) Endosteum
(d) Perichondrium
Answer:
(d) Perichondrium

Question 5.
Skin is …………… (CPMT2010)
(a) Cubiodal epithelium
(b) Stratified epithelium
(c) Columnar epithelium
(d) Pseudostratified epithelium
Answer:
(b) Stratified epithelium

Question 6.
Match the animals listed in column – I to blood listed in column – II. (KCET 2010)

Column -I	Column – II
(i) Plasma and cells are colourless	(P) Man
(ii) Plasma colourless and nucleated RBC	(Q) Earth worm
(ii) Plasma colourless and enucleated RBC	(R) Cockroach
(iv) Plasma red and nucleated colourless RBC	(S) Frog
(v) Plasma and RBS have haemoglobin	–

(a) (P – iii), (Q – iv), (R – i), (S – ii)
(b) (P – iv), (Q – v), (R – iii), (S – ii)
(c) (P – i), (Q – iv), (R – ii), (S – iii)
(d) (P – v), (Q – iii), (R – i), (S – iv)
Answer:
(a) (P – iii), (Q – iv), (R – i), (S – ii)

Question 7.
Matrix of bone and cartilage can be distinguished by the presence of –
(a) Lacunae
(b) Chromatophares
(c) Haversian canals
(d) Adipose cells
Answer:
(c) Haversian canals

Question 8.
Which type of tissue forms glands? (MPPMT- 2010)
(a) Epithelial
(b) Muscular
(c) Nervous
(d) Connective
Answer:
(a) Epithelial

Question 9.
Which of the following blood cells help in blood coagulation?
(a) RBCs
(b) Lymphocytes
(c) Thrombocytes
(d) Basophils
Answer:
(c) Thrombocytes

Question 10.
Fibroblasts macrophages and mast cells are present in –
(a) Cartilage tissue
(b) Areolar tissue
(c) Adipose tissue
(d) Glandular epithelium
Answer:
(b) Areolar tissue

Question 11.
Which type of epithelium is involved in a function to move particles or mucus in specific direction? (HPPMT 2010)
(a) Squamous epithelium
(b) Cuboidal epithelium
(c) Columnar epithelium
(d) Ciliated epithelium
Answer:
(d) Ciliated epithelium

Question 12.
Which of these is not found in connective tissue? (MPPMT2010)
(a) Collagen fibres
(b) Basement membrane
(c) Hyaluronic acid
(d) Fluid
Answer:
(b) Basement membrane

Question 13.
Multi-lobed nucleus and granular cytoplasm are characteristics of which of the WBCs?
(a) Neutrophils
(b) Monocytes
(c) Lymphocytes
(d) Eosinophils
Answer:
(a) Neutrophils

Question 14.
Which one of the following plasma proteins is involved in the coagulation of blood? (2011)
(a) globulin
(b) Fibrinogen
(c) albumin
(d) Serum amylase
Answer:
(b) Fibrinogen

Question 15.
Which of the following is not a connective tissue? (CPMT – 2010)
(a) Blood
(b) bone
(c) Lymph
(d) Nerve
Answer:
(d) Nerve

Question 16.
The ciliated columnar epithelial cells in humans are known to occur in –
(a) Bile duct and oesophagus
(b) Fallopian tubes and urethra
(c) Eustachian tube and stomach lining
(d) Bronchioles and fallopian tubes
Answer:
(d) Bronchioles and fallopian tubes

Samacheer Kalvi 11th Bio Zoology Tissue Level of Organisation Additional Questions & Answers

 Multiple Choice Questions
Choose the correct answer
Question 1.
Groups of cells that are similar in structure and perform a common function are called –
(a) tissues
(b) organs
(c) cells
(d) organ systems
Answer:
(a) tissues

Question 2.
Which of the following have flattened cells?
(a) cuboidal epithelium
(b) columnar epithelium
(c) squamous epithelium
(d) ciliated epithelium
Answer:
(c) squamous epithelium

Question 3.
Microvilli and Goblet cells are the modifications of –
(a) cuboidal epithelium
(b) columnar epithelium
(c) squamous epithelium
(d) ciliated epithelium
Answer:
(b) Columnar epithelium

Question 4.
Which of the following is not exocrine gland?
(a) Sweat glands
(b) Sebaceous glands
(c) Mammary glands
(d) Thyroid gland
Answer:
(d) Thyroid gland

Question 5.
Pancreas is the example of glands –
(a) Merocrine
(b) Holocrine
(c) Apocrine
(d) Epithelial
Answer:
(a) Merocrine

Question 6.
Which is the site of production of blood cells?
(a) Cartilage
(b) Blood
(c) PLasma
(d) Bone marrow
Answer:
(d) Bone marrow

Question 7.
Biceps and Triceps are the examples of –
(a) Smooth muscle
(b) Cardiac muscle
(c) Striped muscle
(d) Involuntary muscle
Answer:
(c) Striped muscle

Question 8.
The walls of internal organs are made up of –
(a) Smooth muscle
(b) involuntary muscle
(c) Skeletal muscle
(d) Cardiac muscle
Answer:
(a) Smooth muscle

Question 9.
Bone cells are called as –
(a) Neurons
(b) Epithelial cells
(c) Osteoblasts
(d) Chondrocytes
Answer:
(c) Osteoblasts

Question 10.
Cartilage is the –
(a) Loose connective tissue
(b) Dense connective tissue
(c) Areolar connective tissue
(d) Specialized connective tissue
Answer:
(d) Specialized connective tissue

Question 11.
Salivary gland is –
(a) Unicellular, glandular cells
(b) Multicellular, glandular cells
(c) Unicellular, sensory cells
(d) Multicellular, sensory cells
Answer:
(c) Unicellular, sensory cells

Question 12.
lines gall bladder.
(a) ciliated epithelium
(b) columnar epithelium
(c) non – ciliated epithelium
(d) pseudo – stratified epithel lurn
Answer:
(c) non – ciliated epithelium

Question 13.
Dry epidermis of the skin is formed as –
(a) keratinized stratified squamous epithelium
(b) non – keratinized stratified squamous epithelium
(c) stratified cuboidal epithelium
(d) stratified columnar epithelium
Answer:
(a) keratinized stratified squamous epithelium

Question 14.
The walls of the Bronchial tubes have –
(a) Dense irregular connective tissues
(b) Reticular connective tissue
(c) elastic connective tissue
(d) Adipose tissue
Answer:
(c) elastic connective tissue

Question 15.
Bones have –
(a) Osteocytes
(b) Fibroblasts
(c) Adipocytes
(d) Myofibrils
Answer:
(c) Adipocytes

II. Answer the following Questions

Question 1.
Define tissues.
Answer:
Group of cells that are similar in structure and perform a common or related functions are called tissues.

Question 2.
What is the study of tissues called?
Answer:
Histology.

Question 3.
Differentiate Simple epithelium and compound epithelium.
Answer:
Simple epithelium:

• It consists of a simple layer.
• It helps in protection, absorption, filtration, excretion, secretion and sensory reception.

Compound epithelium:

• It is multilayered.
• It provides protection against chemical and mechanical stresses.

Question 4.
Explain the types of simple epithelium.
Answer:
Simple epithelium is a simple layered sheet of cells that covers the body surface or lines the body cavity.
Types:
1. Squamous epithelium:
It is made of flattened cells with irregular boundaries. It is found in glomeruli, air sacs of lungs, lining of heart, blood vessels.

2. Cuboidal epithelium:
It is made of cube like cells. It is found in kidney tubules, ducts and glands. It is important for secretion and absorption.

3. Columnar epithelium :
It is made of column like cells. It lines the digestive tract. It is important for secretion and absorption.

4. Ciliated epithelium :
It has cilia at the free end. It is found in bronchi, uterine tubes. It is helpful in propelling materials.

5. Glandular epithelium :
Cuboidal or columnar epithelium specialized for secretion is called glandular epithelium. E.g., goblet cells and salivary gland.

Question 5.
Distinguish between exocrine glands and endocrine glands.
Answer:
Exocrine glands:

• These glands release their products through ducts.
• These secrete mucous, saliva, ear wax, oil, milk, digestive enzymes etc. e.g., salivary glands

Endocrine glands:

• These are ductless gland and their secretions are released directly into the blood.
• These secrete hormones, e.g., Pituitary gland

Question 6.
Classify multicellular exocrine glands based on their structure.
Answer:

Question 7.
Classify exocrine glands based on mode of secretion.
Answer:

Question 8.
Explain compound epithelium.
Answer:

• Compound epithelium is made up of multilayered cells.
• These protect organs against chemical and mechanical stresses.
• These cover the dry surface of the skin, moist surface of the buccal cavity, pharynx, inner lining of ducts of salivary glands and pancreatic ducts.

Question 9.
Classify compound epithelium.
Answer:

Question 10.
Write a short note on specialized junctions of epithelia.
Answer:
All cells of epithelia are held together with little intercellular material forming specialized junctions. These provide structural and functional links between the cells. Three types of cell junctions, tight, adhering and gap junctions are found in animal tissues.

Tight junctions help to stop substances from leaking across the tissue. Adhering junctions cement the neighbouring cells together. Gap junctions facilitate the transfer of ions, small and big molecules between the adjoining cells by connecting the cytoplasm of these cells.

Question 11.
Write a short note on connective tissue.
Answer:
Connective tissue develops from the mesoderm. Proper, cartilage, bones and blood are the four main classes of connective tissues. Binding, support, protection, insulation and transportation of substances are the major functions of connective tissue.

Question 12.
What are the types of proper connective tissues?
Answer:
Loose connective tissue and dense connective tissues.

Question 13.
Write a short note on loose connective tissues.
Answer:
In this tissue, the cells and fibres are loosely arranged in semifluid ground substances, e.g., fibroblasts, macrophages, fat cells and mast cells. Areolar connective tissue present beneath the skin acts as a support framework for epithelium. It acts as a reservoir of water and salts for the surrounding body tissues. Hence, these are called tissue fluid.

Adipose tissue is similar to areolar tissue in structure and function. It is located beneath the skin, surrounding the kidneys, eyeball, heart etc. Adipocytes store fat. It is called white fat. The adipose tissue which contains a lot of mitochondria is called brown fat or brown adipose tissue. Reticular connective tissue is filled with fibroblasts called reticular cells. These cells store fats and the excess nutrients.

Question 14.
Distinguish between tendons and ligaments.
Answer:
Tendons:
Tendons attach skeletal muscles to bones

Ligaments:
Ligaments attach one bone to another.

Question 15.
Explain specialised connective tissues.
Answer:
Cartilage :
The intercellular material of cartilage is solid and pliable and resists compression. Cells of cartilage (chondrocytes) are enclosed is small cavities within the matrix secreted by them. Cartilage is present in the tip of nose, outer ear joints, ear pinna, between adjacent bones of the vertebral column, limbs and hands on adults.

Bones :
Bones have a hard and non-pliable ground substance rich in calcium salts and collagen fibres. Bones support and protect softer tissues and organs. Osteoblasts are present in the spaces called lacunae.

Blood :
It is the fluid connective tissue. It contains RBCs, WBCs and platelets. It functions as a transport medium for nutrients, wastes and respiratory gases.

Question 16.
Explain the types of muscle.
Answer:
Each muscle is made of long, cylindrical fibres. They are composed of fine fibrils called myofibrils. Muscle fibres contract and relax. Skeletal muscle is attached to skeletal bones. It is striped or striated. It is the voluntary muscle. The smooth muscle fibres are fusiform and do not have striations. It is an involuntary muscle. Cardiac muscle tissue is present in the heart. It is striated and branched and involuntary.

Question 17.
Write a note on neural tissue.
Answer:

• Neurons are units of the neural system. The neuroglial cells protect and support the neurons.
• Neurons transmit sensations as electric impulses.

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Samacheer Kalvi 11th Chemistry Environmental Chemistry Textual Evaluation Solved

Samacheer Kalvi 11th Chemistry Environmental Chemistry Multiple Choice Questions

Question 1.
Th gaseous envelope around the earth is known as atmosphere. The region lying between an altitude of 11-50 km is –
(a) Troposphere
(b) Mesophere
(c) Thermosphere
(d) Stratosphere
Answer:
(d) Stratosphere

Question 2.
Which of the following is natural and human disturbance in ecology?
(a) Forest fire
(b) Floods
(c) Acid rain
(d) Greenhouse effect
Answer:
(a) Forest fire

Question 3.
Bhopal Gas Tragedy is a case of –
(a) thermal pollution
(b) air pollution
(c) nuclear pollution
(d) land pollution
Answer:
(b) air pollution

Question 4.
Haemoglobin of the blood forms carboxyhaemoglobin with –
(a) Carbon dioxide
(b) Carbon tetrachioride
(c) Carbon monoxide
(d) Carbonic acid
Answer:
(c) Carbon monoxide

Question 5.
Which sequence for greenhouse gases is based on GWP?
(a) CFC > N₂O > CO₂ > CH₄
(b) CFC > CO₂ > N₂O > CH₄
(c) CFC > N₂O > CH₄ > CO₂
(d) CFC > CH₄ > N₂O > CO₂
Answer:
(c) CFC > N₂O > CH₄ > CO₂

Question 6.
Photo chemical smog formed in congested metropolitan cities mainly consists of –
(a) Ozone, SO₂ and hydrocarbons
(b) Ozone, PAN and NO₂
(c) PAN, smoke and SO₂
(d) Hydrocarbons, SO₂ and CO₂
Answer:
(b) Ozone, PAN and NO₂

Question 7.
The pH of normal rain water is –
(a) 6.5
(b) 7.5
(c) 5.6
(d) 4.6
Answer:
(c) 5.6

Question 8.
Ozone depletion will cause –
(a) forest fires
(b) eutrophication
(c) bio magnification
(d) global warming
Answer:
(d) global warming

Question 9.
Identify the wrong statement in the following.
(a) The clean water would have a BOD value of more than 5 ppm
(b) Greenhouse effect is also called as Global warming
(c) Minute solid particles in air is known as particulate pollutants
(d) Biosphere is the protective blanket of gases surrounding the earth
Answer:
(a) The clean water would have a BOD value of more than 5 ppm

Question 10.
Living in the atmosphere of CO is dangerous because it –
(a) Combines with O₂ present inside to form CO₂
(b) Reduces organic matter of tissues
(c) Combines with haemoglobin and makes it incapable to absorb oxygen
(d) Dries up the blood
Answer:
(c) Combines with haemoglobin and makes it incapable to absorb oxygen

Question 11.
Release of oxides of nitrogen and hydrocarbons into the atmosphere by motor vehicles is prevented by using –
(a) grit hamber
(b) scrubbers
(c) trickling filters
(d) catalytic convertors
Answer:
(c) trickling filters

Question 12.
Biochemical oxygen Demand value less than 5 ppm indicates a water sample to be
(a) highly polluted
(b) poor in dissolved oxygen
(c) rich in dissolved oxygen
(d) low COD
Answer:
(c) rich in dissolved oxygen

Question 13.
Match the list I and list II and select the correct answer using the code given below in the list:

Answer:

Question 14.
Match the list I and list II and select the correct answer using the code given below in the list

Answer:

Question 15.
Assertion (A) : if BOD level of water in a reservoir is more than 5 ppm it is highly polluted.
Reason(R) : High biological oxygen demand means high activity of bacteria in water
(a) (i)
(b) (ii)
(c) (iii)
(d) (iv)
Answer:
(d) (iv)

Question 16.
Assertion (A) : Excessive use of chlorinated pesticide causes soil and water pollution.
Reason (R) : Such pesticides arc non-biodegradable.
(a) (i)
(b) (ii)
(c) (iii)
(d) (iv)
Answer:
(a) (i)

Question 17.
Assertion (A) : Oxygen plays a key role in the troposphere.
Reason (R) : Troposphere is not responsible for all biological activities
(a) (i)
(b) (ii)
(c) (iii)
(d) (iv)
Answer:
(d) (iv)

Samacheer Kalvi 11th Chemistry Environmental Chemistry Short Answer Questions

Question 18.
Dissolved oxygen in water is responsible for aquatic life. What processes are responsible for the reduction in dissolved oxygen in water’?
Answer:

• Organic matter such as leaves, grass, trash can pollute water. Microorganisms present in water can decompose these organic matter and consume dissolved oxygen in water.
• Eutrophication is a process by which water bodies receive excess nutrients that stimulates excessive plant growth. This enhanced plant growth in water bodies is called algal bloom.
• The growth of algae in extreme abundance covers the water surface and reduces the oxygen concentration in water. Thus, bloom-infeded water inhibits the growth of other. living organisms in the water body.
• This process in which the nutrient rich water support a dense plant population, kills animal life by depriving it of oxygen and results in loss of biodiversity is known as eutrophication.

Question 19.
What would happen, if the greenhouse gases were totally missing in the earth’s atmosphere?
Answer:

• The primary greenhouse gases in Earth’s atmosphere are water vapour, carbon dioxide, methane, nitrous oxide and ozone.
• Naturally occurring greenhouse gases allow solar radiations to reach the earth’s surface, while trapping radiations from the earth on its way back out to space.
• There would he no life on Earth without the warmth provided by this natural greenhouse gases.
• In the absence of greenhouse gases. the average temperature of the earth will decrease drastically. As a result. life on Earth would be impossible.

Question 20.
Define smog.
Answer:

• Smog is a combination of smoke and fog which form droplets that remains suspended in the air.
• Smog is a chemical mixture of gases that forms a brownish yellow haze. It mainly consists of ground level ozone, oxides of nitrogen, volatile organic compounds, SO₂, acidic aerosols and some other gases.

Question 21.
Which is considered to be earth’s protective umbrella? Why?
Answer:

• At high altitudes in the atmosphere consists of a layer of ozone (O₂) which acts as an umbrella for harmful UV radiations. Ozone is considered to be earth’s protective umbrella.
• It protects us from harmful effects of UV-radiations of the sun such as skin cancer.
• Ozone layer prevent the UV radiations to reach the earth surface. So it acts as an umbrella for the Earth.

Question 22.
What are bio-degradable and non-biodegradable pollutants?
Answer:

• The pollutants which can be easily decomposed by the natural biological processes are called biodegradable pollutants. For example plant wastes, animal wastes.
• The pollutants which cannot be decomposed by the natural biological processes are called non-biodegradable pollutants. For example, metal wastes such as Hg and Pb, D.D.T. plastics, nuclear vastcs.

Question 23.
From where does ozone come in the photochemical smog?
Answer:

• Photochcrnical smog is formed by the combination of smoke, dust and fog with air pollutants in the presence of sunlight.
• Chemically it is oxidising in nature because of high concentration of oxidising agents such as NO₂ and O₂. So it is also called oxidising smog.
• Photochemical smog is formed by following reactions:
• N₂ + O₂ → 2NO
  2NO + O₂ → 2NO₂
  
  (O) + O₂ → O₃
  O₃ + NO → NO₂ + O₂
  
• NO and O₃ arc strong oxidizing agents and they can react with unburnt hydrocarbons in polluted air to form formaldehyde, acrolcin and PAN.

Question 24.
A person was using water supplied by corporation. Due to shortage of water he started using underground water. He felt laxative effect. What could be the cause?
Answer:
Drinking water containing moderate level of sulphatcs is harmless. But excessive concentration (>500 ppm) of suiphates in drinking water causes laxative effect.

Question 25.
What is green chemistry?
Answer:

• Green chemistry is a chemical philosophy encouraging the design of products and processes that reduces or eliminates the use and generation of hazardous substances.
• Efforts to control environmental pollution resulted in development of science for the synthesis of chemicals favorable to environment.
• Green chemistry means science of environmentally favorable chemical synthesis.

Question 26.
Explain how does greenhouse effect cause global warming.
Answer:

• The earth’s atmosphere allows most of the visible light from the sun to pass through and reach the earth’s surface. As earth’s surface is heated by sunlight, it radiates a part of this energy back towards the space as longer IR wavelengths.
• Some of the heat is trapped by CH₂, CO₂. CFCs and water vapour present in the atmosphere. They absorb IR radiations and block a large portion of earth’s emitted radiations.
• The radiations thus absorbed is partly remitted to the earth’s surface. Therefore the earth’s surface gets heated up by a phenomenon called greenhouse effect.
• Thus greenhouse effect is defined as the heating up of the earth surface due to trapping of infrared radiations reflected by earth’s surface by CO₂ layer in the atmosphere. The heating up of the earth through the greenhouse effect is called global warming.

Question 27.
Mention the standards prescribed by BIS for quality of drinking water.
Answer:
Standard characteristics prescribed for deciding the quality of drinking water by BIS are as follows:

Question 28.
How does classical smog differ from photochemical smog?
Answer:
Classical smog:

• Classical smog is caused by coal-smoke and fog.
• It occurs in cold humid climate.
• The chemical composition is the mixture of SO₂, SO₃ gases and humidity.
• Chemically it is reducing in nature because of high concentration of SO, and so it is also called reducing smog.
• It is primarily responsible for acid rain.
• It also causes bronchial irritation.

Photochemical smog:

• Photochemical smog is cause by photochemical oxidants.
• It occurs in warm, dry and sunny climate.
• The chemical composition is the mixture of NO₂ and O₃ gases.
• Chemically it is oxidising in nature because of high concentration of oxidising agents such as NO₂ and O₃
• and so it is also called oxidising smog.
• It causes irritation to eyes, skin and lungs and increase the chances of asthma.
• It causes corrosion of metals, stones

Question 29.
What are particulate pollutants? Explain any three.
Answer:
1. Particulate pollutants are small solid particles, and liquid droplets suspended in air.
Examples:
dust, pollen, smoke, soot and liquid aerosols.

2. Types of Particulates:
Particulates in the atmosphere may be of two types:

• viable particulate and
• non-viable particulate.

3.The viable particulates are small size living organisms such as bacteria, fhngi, moulds and algae which are dispersed in air.

4. The non-viable particulates are small solid particles and liquid droplets suspended in air. There are four types of non-viable particulates in the atmosphere. They are
(a) Smoke
(b) Dust
(c) Mist
(d) Fumes

5. Smoke:
Smoke particulate consists of solid particles formed by combustion of organic matter. For example, cigarette smoke, oil smoke, smokes from burning of fossil fuels, garbage and dry leaves.

6. Dust:
It is composed of fine solid particles produced during crushing and grinding of solid materials. For example, sand from sand blasting, saw dust from wood works and fly ash from power generating units.

7. Mist:
They are formed by particles of sprayed liquids and condensation of vapours in air. For example, sulphuric acid mist, herbicides and insecticides sprays can form mists.

8. Fumes:
They are obtained by condensation of vapours released during sublimation, distillation, boiling and calcination and by several other chemical reactions.
For example:
organic solvents, metals and metallic oxides.

Question 30.
Even though the use of pesticides increases the crop production, they adversely affect the living organisms. Explain the function and the adverse effects of the pesticides.
Answer:
1. Pesticides are the chemicals that are used to kill or stop the growth of unwanted organims. But these pesticides can affect the health of human beings. Pesticides are classified as
(a) insecticides,
(b) Fungicides and
(c) Herbicides.

(a) Insecticides:
Insecticides like DDT, BHC, Aidrin can stay in soil for a long period of time and are absorbed by soil. They contaminate root crops like carrot, radish.

(b) Fungicides:
Organomercury compounds dissociate in soil to produce mercury which is highly toxic.

(c) Herbicides:
They are used to control unwanted plants and are also known as weed killers. Eg, Sodium chlorate, sodium nitrate. They are toxic to mammals.

Question 31.
Ethane bums completely in air to give CO₂. while in a limited supply of air gives CO. The same gases are found in automobile exhaust. Both CO and CO₂ are atmospheric pollutants

1. What is the danger associated with these gases?
2. How do the pollutants affect the human body?

Answer:
Danger associated with CO and CO₂ & health hazards to human body
(a) Carbon monoxide binds with haemoglohin and form carboxyhaemoglobin which impairs normal oxygen transport by blood and hence the oxygen carrying capacity of blood is reduced. This oxygen deficiency results in headache, tension, dizziness, loss of consciousness, blurring of eyesight and cardiac arrest.

(b) Increase in CO₂ level in the atmosphere is responsible for global warming. It causes headache and nausea.

Question 32.
On the basis of chemical reactions involved, explain how do CFC’s cause depletion of ozone layer in stratosphere?
Answer:
(i) The chioro-Iluoro derivatives of methane and ethane are named Freons (CFC’s). They slowly pass from troposphere to stratosphere. They stay for a very longer period of about 50-100 years. In the presence of UV radiations, CFC’s break up into chienne free radicals.

Cl• + O₃ → { ClO }^{ \bullet } + O₂
ClO• + O → { Cl }^{ \bullet } + O₂

(iii) Chlorine radical is regenrated in the course of the reaction. Due to this continuous attack of Cl free radicals, thinning of ozone layer takes place which leads to the formation of ozone hole.

(iv) Li is estimated that for every reactive chlorine atom generated in the stratosphere 1,00,000 molecules of ozone are depleted.

Question 33.
How is acid rain formed? Explain its effect.
Answer:
1. Rain water has a pli of 5.6 due to the dissolution of CO., into it. Oxides of sulphur and nitrogen in the atmosphere may be absorbed by droplets of water that make up the clouds and get chemically converted into sulphuric acid and nitric acid. Due to this the pH of rain water drops below the level of 5.6. Hence it is called acid rain.

2. Acid rain is a by-product of sulphur and Nitrogen oxides in the atmosphere. Burning of fossil fuels in power stations, furnaces and petrol, diesel in motor engines produce SO₂ and NO₂ gases. They are converted into H₂SO₄ and HNO₃ by the reaction with oxygen and water.

3. 2SO₂ + O₃ + 2H₂O → 2H₂SO₄
4NO₂ + O₂+ 2H₂O → 4HNO₃

Harmful effects of acid rain:
1. Acid rain causes damage to buildings made us of marbles. This attack on marble is termed as stone leprosy.
CaCO₃ + H₂SO₂ CaSO₄ + H₂O + CO₂↑

2. Acid rain affects plant and animal life in aquatic ecosystem.

3. It is harmful For agriculture, as it dissolves in the earth and removes the nutrients needed for the growth of plants.

4. It corrodes water pipes resulting in the leaching of heavy metals such as iron, lead and copper into drinking water which have toxic effects.

5. it causes respiratory ailment in humans and animals.

Question 34.
Differentiate the following:

1. BOD and COD
2. Viable and non-viable particulate pollutants

Answer:
Biochemical oxygen demand (BOD):

• The total amount of oxygen (in milligrams) consumed by microorganisms in decomposing the waste in one litre ut water at 20°C for a period of 5 days is called biochemical oxygen demand (BOD).
• its value is expressed in ppm.
• DOD is used as a measure of degree ofwater pollution.
• BOD is only a measurement of consumed oxygen by microorganims to decompose the organic matter.
  Clean water would have BOD value less than 5 ppm.

Chemical ox gen demand (COD):

• Chemical oxygen demand is defined as the amount of oxygen required by the organic matter in a sample of water for its oxidation by a strong oxidising agent like K₇Cr₂O₇ in acidic medium for a period of 2 hours.
• Its value is expressed in mg / litre.
• COD is a measure of amount of organic compounds in a water sample.
• COD refers to the requirement of dissolved oxygen for both the oxidation of organic and inorganic constituents.
• Clean water would have COD value greater than 250 mg/litre.

2. Viable and non-viable particulate pollutants
Viable pollutants:

• The viable particulates are small size living organisms such as bacteria, fungi. moulds, algae which are dispersed in air.
• They are all organic particulates.
• They contain living organisms.
•  Eg. ftingi, bacteria, algae, moulds. Viable particles are the particles with at least one microorganism affecting the sterility of the product.

Non-viable pollutants:

• The non-viable particulates are small solid particles and liquid droplets suspended in air.
• They are all inorganic particulates.
• They contain non-living organisms.
• Eg. Smoke, dust, mist, fumes.
• Non-viable particles are the particles without microorganisms but act as transporting agent for viable particles.

Question 35.
Explain how oxygen deficiency is caused by carbon monoxide in our blood? Give its effect.
Answer:

1. Carbon monoxide binds with haemoglobin and form carboxy the haemoglobin which impairs normal oxygen transport by blood and hence the oxygen carrying capacity of blood is reduced.
2. This oxygen deficiency results in headache, dizziness, tension, loss of consciousness, blurring of eyesight and cardiac arrest.

Question 36.
What are the various methods you suggest lo protect our environment from pollution?
Answer:
Methods to control environmental pollution:

• Waste management Environmental pollution can be controlled by proper disposal of wastes.
• Recycling A large amount of disposed waste materials can he reused by recycling the waste, thus it reduces the landfill. –
• By substitution of less toxic solvents for highly toxic ones are used in industrial processes.
• By growing more trees.
• By using fuels with lower sulphur content.
• By control measures in vehicle emissions which are adequate.

Samacheer Kalvi 11th Chemistry Environmental Chemistry Additional Questions Solved

I. Choose the correct answer

Question 1.
Which one of the following is bio-degradable pollutant?
(a) DDT
(b) Plastics
(c) Mercury
(d) Wood
Answer:
(d) Wood

Question 2.
Which one of the following gases is not present in troposphere?
(a) N₂O₂
(b) CO₂
(c) N₂
(d) water vapours
Answer:
(c) N₂

Question 3.
Which of the following pair of’ oxides is responsible for acid rain?
(a) SO₃ + NO₂
(b) CO₂ + CO
(c) N₂O + CH₄
(d) O₂ + H₂
Answer:
(a) SO₃ + NO₂

Question 4.
Which one of the following is produced as a result of incomplete combustion of coal’?
(a) CO₂
(b) CO
(c) SO₂
(d) SO₃
Answer:
(b) CO

Question 5.
Which of the following is not a greenhouse gas’?
(a) CO
(b) O₃
(c) CH₄
(d) Water vapours
Answer:
(a) CO

Question 6.
Photochemical smog occurs in warm, dry and sunny climate. One of the following is not among-st the components of photochemical smog. Identify it.
(a) NO₂
(b) O₃
(c) SO₂
(d) Unsaturated hydrocarbons
Answer:
(c) SO₂

II. Match the following
1.

Answer:

Answer:

Answer:

Answer:

Answer:

Answer:

III. Fill in the blanks

Question 1.
An example for biodegradable pollutant is ………
Answer:
plant waste.

Question 2.
An example for non-biodegradable pollutant ………
Answer:
metal waste.

Question 3
……… is the lowest layer of atmosphere.
Answer:
Troposphere.

Question 4.
Ozone is present in ……… layer of the atmosphere.
Answer:
Stratosphere.

Question 5
……… is called blue planet.
Answer:
Earth.

Question 6.
About 80% of the mass of the atmosphere is present in ………
Answer:
Troposphere.

Question 7.
……… will reduce the oxygen carrying capacity of blood.
Answer:
CO

Question 8
……… gas is used in the process of photosynthesis.
Answer:
CO₂

Question 9.
……… gas potentially damage plant leaves and retard photosynthesis.
Answer:
NO₂

Question 10.
……… is formed by incomplete combustion of coal.
Answer:
CO₂

Question 11.
……… binds with haemoglobin and reduce the oxygen carrying capacity of blood.
Answer:
CO

Question 12.
……… is carcinogenic and causes irritation in eyes and mucous membrane.
Answer:
PAH

Question 13.
The earths surface get heated up by a phenomenon called ………
Answer:
Greenhouse effect

 

Question 14.
The main constituent of layer responsible for global warming is ………
Answer:
CO₂

Question 15.
Earth’s average surface temperature would be only about ………
Answer:
-18°C

Question 16.
The pH of rain water is normally ………
Answer:
5.6

Question 17.
……… pair of compounds is present in acid-rain water.
Answer:
HNO₃ + H₂SO₄

Question 18.
……… is an example of viable particulate.
Answer:
Bacteria.

Question 19.
……… is an example of non-viable particulate matter.
Answer:
Cement dust.

Question 20.
Particulate pollutants will result in the health hazard named as ………
Answer:
pneimoconiosis.

Question 21.
Coal miners may suffer from ………
Answer:
Black lung disease.

Question 22.
Textile workers may suifer from ………
Answer:
White lung disease.

Question 23.
………….. aftect childrcns brain. interfers with maturation of RBC’s and even causes cancer.
Answer:
Lead.

Question 24.
……… can be used to reduce particulate pollutant.
Answer:
electrostatic precipitator

Question 25.
……… is the combination of smoke and fog.
Answer:
Smog.

Question 26.
Classical smog called London smog contains ………
Answer:
Coal-smoke and fog.

Question 27.
Classical smog is otherwise called ………
Answer:
reducing smog.

Question 28.
Photochemical smog is otherwise called ………
Answer:
Los angeles smog.

Question 29.
The three main components of photochemical sinog are ………
Answer:
Hydrogen sulphide. dust and PAN.

Question 30.
……… plantation can metabolise nitrogen oxide and control photochemical smog.
Answer:
Pinus tree.

Question 31.
……… acts as an umbrella for the earth and prevent harmful UV radiations.
Answer:
ozone.

Question 32.
……… pair of compounds is found to be highly responsible for depletion of ozone layer.
Answer:
Nitric oxide + CFC

Question 33.
Freons are ………
Answer:
Chiorofluoroalkanes.

Question 34.
Ozone layer is depicted by the reactive ………
Answer:
Chlorine atom.

Question 35.
Examples for water borne diseases are ………
Answer:
Dysentery and cholera.

Question 36.
The standard pH of drinking water is ………
Answer:
6.5 to 8.5

Question 37.
The essential elements for soil are ………
Answer:
N,P,K

Question 38
……… is a better alternative for carnicogentic benzene.
Answer:
Xylene.

Question 39.
The alternate solvent used instead of tetrachioroethYlefle in dry cleaning is ………
Answer:
Liquefied CO₂

Question 40.
……… is used for bleaching clothes in laundry.
Answer:
H₂O₂

Question 41.
……… is used to bleach paper.
Answer:
H₂O₂

Question 42.
……… is the most safest pesticide.
Answer:
Neem based pesticide.

Question 43.
……… acid is most abundant ¡n acid rain.
Answer:
H₂SO₄

Question 44.
……… causes less pollution.
Answer:
CO₂

Question 45.
Besides CO₂ the other greenhouse gas is ………
Answer:
CH₄

Question 46.
BOD is a measure of ………
Answer:
Organic pollutant in water.

Question 47.
The pollutant released in Bhopal gas tragedy was ………
Answer:
Methyl isocyanate.

Question 48.
The greatest affinity for haemoglobin in shown by ………
Answer:
CO

Question 49.
Eutrophication causes reduction in ………
Answer:
dissolved oxygen.

IV Choose the odd one out

Question 1.
(a) Plant waste
(b) DDT
(c) Plastic
(d) Nuclear waster
Answer:
(a) Plant waste. It is biodegradable pollutant whereas others are non-biodegradable pollutants.

Question 2.
(a) Plant waste
(b) Animal wastes
(c) Paper
(d) Nuclear waste
Answer:
(d) Nuclear waste. It is non-biodegradable pollutant whereas others are biodegradable wastes.

Question 3.
(a) N₂O₂
(b) CO₂
(c) H₂O (Vap)
(d) O₃
Answer:
(d) O₃ It is present in stratosphere whereas others are present in troposphere.

Question 4.
(a) O₂⁺
(b) O⁺
(c) N₂
(d) NO⁺
Answer:
(c) N₂. ills present in mesosphere whereas others are present in thermosphere.

Question 5.
(a) N₂
(b) O₂
(c) O₃
(d) N₂O₂
Answer:
(d) N₂O₂ It is present in troposphere whereas others are present in stratosphere.

Question 6.
(a) CH₄
(b) CO
(c) CO₂
(d) CFC
Answer:
(b) CO. It is a poisonous gas whereas others are responsible for green house effect.

V. Choose the correct pair

Question 1.
(a) CCF : Green house effect
(b) CO : Carcinogenic
(c) PAH : Acid rain
(d) NO : Lung injury
Answer:
(a) CCF : Green house effect

Question 2.
(a) NO₂ : Green house effect
(b) CFC : asthma and lung injury
(c) PAH : carcinogenic
(d) CO₂ : green house effect
Answer:
(a) NO₂ : Green house effect

Question 3.
(a) Classical smog : NO₂ and O₂
(b) London smog : SO₂, SO₂ and humidity
(c) Photochemical smog : CO₂ and CO
(d) Los Angel smog : NO₂ and NO₃
Answer:
(b) London smog : SO₂ , SO₂ and humidity

VI. Choose the incorrect pair

Question 1.
(a) Photochemical smog : NO₂ and O₂
(b) Classical smog : SO₂, SO₃ and humidity
(c) Smog : Smoke and fog
(d) Non viable particulate : Algae, Fungi
Answer:
(d) Non viable particulate : Algae, Fungi

Question 2.
(a) Viable particulate : bacteria, fungi
(b) Non viable particulate : smoke, dust
(c) Acid rain : HCl + HNO₂
(d) Photochemical smog : NO₂ + O₃
Answer:
(c) Acid rain : HCl + HNO₂

Question 3.
(a) Lead : Damage to kidney, liver
(b) Sulphate : Laxative effect
(c) Nitrage : Blue baby syndrome
(d) TDS : Damage to bone and teeth
Answer:
(d) TDS : Damage to bone and teeth

Question 4.
(a) Insecticides : DDT. BCHs
(b) Herbicides : Organo mercury compounds
(c) Herbicides : Sodium chlorate, sodium arsenite
(d) Industrial waste : Mercury. copper
Answer:
(b) Herbicides : Organo mercuiy compounds

VII. Assertion & Reason
Assertion (A) : Depletion of ozone layer causes skin cancer.
Reason (R) : Depletion olozone layer will allow more UV rays to reach the earth surface and cause skin cancer.
(a) Both (A) and (R) are correct and (R)is the correct explanation of (A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of (R).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Question 2.
Assertion (A) : UV radiations damages the fish productivity.
Reason (R) : UV radiations affect the growth of phytoplankions as a result food chain in ocean is disturbed.
(a) Both (A) and (R) are correct hut (R) is not the correct explanation of (A).
(b) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(b) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Question 3.
Assertion (A) : The pH ol acid rain is less than 5.6.
Reason (R) : CO, present in the atmosphere dissolves in rain water and forms carbonic acid,
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(b) Both (A) and (R) are correct hut (R) is not correct explanation of (A).
(c) (A) is correct bitt (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Samacheer Kalvi 11th Chemistry Environmental Chemistry 2 Mark Questions and Answers

Question 1.
What is meant by environmental pollution?
Answer:
Any undesirable change in our environment that have harmful effects on plants. animals and human beings is called environmental pollution.

Question 2.
Write a note on constituents of earth’s atmosphere.
Answer:
Earth’s atmosphere is a layer of gases retained by earth’s gravity. it contains roughly 78% Nitrogen, 21% Oxygen. 0.93% Argon, 0.04% Carbon dioxide. traces of other gases and little amount of water vapour. This mixture is commonly known as air.

Question 3.
What is called troposphere? How many layers are present in it? Give their names.
Answer:

1. The lowest layer of the atmosphere is called the troposphere and il extends from 0 10 km from the earth surface.
2. About 80% of the mass of the atmosphere is in this layer. 3 layers are present in it.
3. They are –
   • Hydrosphere
   • Lithosphere
   • Biosphere

Question 4.
Write about hydrosphere (or) Why Earth is called as Blue planer?
Answer:

• Hydrosphere include all types of water sources like oceans, seas, rivers, lakes. streams, underground water, polar ice-caps, clouds etc.
• It covers about 75% of the earth’s surface. Hence earth is called as Blue planet.

Question 5.
What is lithosphere?
Answer:
Lithosphere includes soil, rocks and mountains which are solid components of Earth.

Question 6.
What is biosphere?
Answer:
Biosphere includes the lithosphere. hydrosphere and atmosphere integrating the living organisms present in the lithosphere, hydrosphere and atmosphere.

Question 7.
What is air pollution?
Answer:
Air pollution is defined as any undesirable change in air which adversely affects living organisms. Air pollution is limited to troposphere and stratosphere.

Question 8.
What are the types of air pollutants? Give examples.
Answer:
1. Air pollutants exists in two major forms namely:
(a) Gaseous air pollutants
(b) Particulates

2. Gaseous air pollutants:
Oxides of sulphur, oxides of nitrogen, oxides of carbon and hydrocarbons are the gaseous air pollutants.

3. Particulate pollutants:
Particulate pollutants are small solid particles and liquid droplets suspended in air.
Exarnple:
dust, pollen, smoke, soot, aerosols.

Question 9.
Define greenhouse effect.
Answer:
Greenhouse effect is defined as the heating up of the earth’s surface due to trapping of infrared radiations reflected by earth’s surface by the CO₂ layer in the atmosphere.

Question 10.
What is Global warming?
Answer:
The heating up of the earth through the greenhouse effect is called Global warming.

Question 11.
When rain water is named as acid rain?
Answer:
Rain water normally has a pH of 5.6 due to dissolution of atmospheric CO₂ particles into it. Oxides of sulphur and nitrogen in the atmosphere are absorbed by droplets of water that make up clouds and get chemically converted into sulphuric acid and nitric acid. As a result the pH of rainwater drops below 5.6 and hence it is called acid rain.

Question 12.
What is stone leprosy? How is it formed?
Answer:

• The attack on the marble of buildings by acid rain is called stone leprosy.
• Acid rain causes extensive damage to buildings made up of marble CaCO + H₂SO₄ CaSO₄ + H₂O + CO₂↑

Question 13.
What are fumes? Give one example.
Answer:
Fumes are one of the non-viable particulate pollutants air. They are obtained by condensation of vapours released during sublimation, distillation, boiling and calcination. For example, organic solvents, metals and metallic oxides form fume particles.

Question 14.
What are the techniques to reduce particulate pollutants?
Answer:
The particulates from air can be removed by using electrostatic precipitators, gravity settling chambers, wet scrubbers or by cyclone collectors. These techniques are base on washing away or settling of the particulate matter.

Question 15.
How will you control photochemical smog?
Answer:
The formation of photochemical smog can be suppressed by preventing the release of nitrogen oxide and hydrocarbons into the atmosphere from the motor vehicles by using catalytic convertors in engines. Plantation of certain trees like Pinus, Pyrus Querus vitus and Juniparus can metabolise nitrogen oxide.

Question 16.
What is meant by water pollution’?
Answer:
Water pollution is defined as the addition of foreign substances or factors like heat which degrades the quality of water so that it becomes an health hazard or unfit for use.

Question 17.
What are the sources of water pollution? Give examples.
Answer:

• The water pollutants originate form both natural as well as human activities. The source of water pollution are classified as point and non-point sources.
• Easily identified source of water pollution is called a point source. For example, Municipal and industrial discharge pipes.
• Non-point source cannot be identified easily. For example, agricultural runoffs, mining wastes, acid rain, storm water drainage and construction sediments.

Question 18.
What is BOD’?
Answer:
The total amount of oxygen (in milligrams) consumed by microorganisms in decomposing the waste in one litre of water at 20°C for a period of 5 days is cal Led biochemical oxygen demand (BOD) and its value is expressed in ppm.

Question 19.
What is COD?
Answer:
Chemical oxygen demand (COD) is defined as the amount of oxygen required by the organic matter in a sample of water for its oxidation by a strong oxidising agent like K₂Cr₂O₇ in acidic medium for a period of 2 hours.

Question 20.
What are total dissolved solids (TDS)?
Answer:

• Most of the salts are soluble in water. It include cations like calcium, magnesium, sodium, potassium, iron and anions like carbonate, bicarbonate, chloride, sulphate, phosphate and nitrate.
• Use of drinking water having TDS (total dissolved solids) concentration higher than 500 ppm causes possibilities of irritation in stomach and intestine.

Question 21.
What are the constituents of soil?
Answer:
Soil is a thin layer of organic and inorganic material that covers the earth’s rocky surface. Soil constitutes the upper crust of the earth, which supports land, plants and animals.

Question 22.
Define – soil pollution.
Answer:
Soil pollution is defined as the build up of persistent toxic compounds, radioactive materials, chemical salts and disease causing agents in soil which have harmful effects on plant growth and animal health. Soil pollution affects the structure and fertility of soil, ground water quality and food chain in biological ecosystem.

Question 23.
Explain how industrial waste affects the soil.
Answer:

• Industrial activities have been the biggest contributor to soil pollution especially the mining and manufacturing activities.
• Industrial wastes include cyanides. chromates, acids, alkalis and metals like mercury, copper, zinc, cadmium and lead.
• These industrial wastes in the soil surface lies for a long time and makes it unsuitable for use.

Question 24.
What is green chemistry?
Answer:

• Efforts to control environment pollution have resulted in the development of chemicals favorable for environment and this branch of science is called green chemistry.
• It is a chemical philosophy encouraging the design of products and processes that reduces or eliminates the use and generation of hazardous substances.

Question 25.
Write a note about dry cleaning of clothes.
Answer:
Solvents like tetrachioroethylene is used in dry cleaning of clothes, pollute the ground water and are carcinogenic. In place of tetrachloroethylene liquefied CO₂ with suitable detergents an alternate solvent used. Liquefied CO₂ is not that much harmful for the groundwater. Nowadays H₂O₂ is used for bleaching clothes in laundry that give better results and utilises less water.

Question 26.
Carbon monoxide gas is more dangerous than carbon dioxide gas. Why?
Answer:
Carbon monoxide gas combines with haemoglobin to form a very stable compound known as carboxy haemoglobin. When its concentration in blood reaches 3-4%, the oxygen carrying capacity of the blood is greatly reduced. This results into headache, nervousness and sometimes death of the person. On the other hand CO₂ does not combine with haemoglobin and hence it is less harm ful than CO.

Question 27.
Which gases are responsible for greenhouse effect? List some of them.
Answer:
CO, is mainly responsible for greenhouse effect. Other greenhouse gases are methane, nitrous oxide, water vapours, CFCs and ozone.

Question 28.
A large number of fishes are suddenly found floating dead on a lake. There is no evidence of toxic dumping but you find an abundance of phytoplankton. Suggest a reason for the fish killing.
Answer:
Excessive phytoplankton (organic pollutants such as leaves, grass trash etc.) growth which is present in water is biodegradable. Bacteria decomposes this organic matter in water. During this process when large number of bacteria decomposes the organic matter, they consume the dissolved oxygen in water. When the level of dissolved oxygen falls below 6 ppm then the fishes cannot survive and they die.

Question 29.
How carbon monoxide acts as a poison for human beings’?
Answer:
Carbon monoxide is a poisonous gas because it combines with haemoglobin of RBC to form carboxyhaernoglobin as:
CO + Haemoglobin ⇌ Carboxyhaemoglobin
It inhibits the transport of oxygen to different pans of the body. Thus the body becomes oxygen-starved.

Question 30.
Give three examples in which the principles green chemistry has been applied.
Answer:

• In dry-cleaning, use of liquefied CO₂ in place of non-environment friendly tetrachioroethene (Cl₂C = CCl₂).
• Use of H₂O₂ in bleaching in place of chlorine.
• In the manufacture of chemicals like ethanal by using environment-friendly chemicals and conditions. ,

Samacheer Kalvi 11th Chemistry Environmental Chemistry 3 Marks Questions and Answers

Question 1.
Write about Bhopal gas tragedy.
Answer:

• On 3rd Decembers. 1984 at Bhopal city in India, by the early morning, an explosion at Union carbide pesticide plant released a cloud of toxic gas (Methyl isocyanate) CH₃NCO into the air.
• Since the gas is twice as heavy as air, it did not drift away but formed a blanket over the surrounding area.
• It attacked people’s lungs and affect their breathing staying there or in the nearby areas. Thousands of peopic died and lives of many were ruined. The lungs, brain, eyes, muscles as well as gastrointestinal, neurological and immune system of those people who survived were severely affected.

Question 2.
Explain how the oxides of sulphur pollute the atmospheric air. Give its haririful eflects.
Answer:

• Sulphur dioxide and sulphur trioxide are produced by burning sulphur containing fossil fuels and by roasting of suiphide ores.
• SO₂ is a poisonous gas for both animals and plants. SO₂ causes eye irritation, coughing and respiratory diseases like asthma, bronchitis.
• SO₂ is oxidised to more harmful SO₃ gas in the presence of particulate matter present in the polluted air:
  
  SO₂ combines with atmospheric water vapour to form H₂SO₄ which comes down along with rain in the form of acid rain:
• Acid rain causes stone leprosy. affect aquatic ecosystem, corrode water pipes and causes respiratory ailment in humans and animals.

Question 3.
How oxides of nitrogen are harmful?
Answer:
(i) Oxides of nitrogen are produced during high temperature combustion processes, by oxidation of nitrogen in air and it is formed the combustion of fuels such as coal, diesel and petrol.

(iii) The oxides of nitrogen are converted into nitric acid which comes down in the form of acid rain. They also form reddish brown haze in heay traffic.
(iv) Nitrogen dioxide potentially damage plant leaves and retard photosynthesis.
(v) NO₂ is a respiratory irritant and it can cause asthma and lung injury.

Question 4.
Explain how hydrocarbons pollute the atmospheric air.
Answer:

• The compounds composed of carbon and Hydrogen only are called hydrocarbons. They are mainly produced naturally and also by incomplete combustion of automobile fuels.
• They are potential cancer causing (carcinogenic) agents.
• For example polynuclear aromatic hydrocarbons (PAH) are carcinogenic, they cause irritation in eyes and mucous membranes.

Question 5.
Write flotes about great London smog.
Answer:

• The great smog of London (1952) was an instance of severe air pollution that affected the London from 5th December 9^th of December, 1952 and then dispersed quickly when the whether changed.
• It causes major disruption by reducing the visibility and even penetrating indoor areas.
• Government medical reports estimated that 4000 people had died as a direct result of smog and 100.000 were made ill by the smog effect on their respiratory tract.

Question 6.
What are the effects of classical smog?
Answer:

• Smog is primarily responsible for acid rain.
• Smog results in poor visibility and it affects air and road transport.
• It also causes bronchial irritation.

Question 7.
Explain how oxides of nitrogen are introduced directly into the stratosphere?
Answer:

• Nitrogen oxides are introduced directly into the stratosphere by the supersonic jet aircraft engines in the form of exhaust gases.
• These oxides arc released by combustion of fossil fuels and nitrogen fertilisers. Inert nitrous oxide in the stratosphere is photochemically converted into more reactive nitric oxide.
• Oxides of nitrogen catalyses the decomposition of ozone and are themselves regenerated. Ozone gets depleted as follows:
  
  Thus NO is regenerated in the chain reaction.

Question 8.
How chemical wastes pollute water?
Answer:

• A whole variety of chemicals from industries, such as metals and solvents are poisonous to fish and other aquatic life.
• Toxic pesticides can accumulate in fish and shell fish and poison the people who eat them.
• Detergents and oil float spoils the water bodies.
• Acids from mine drainage and salts from various sources can also contaminate water sources.

Question 9.
Explain the presence of fluoride in water and its hazardous effects?
Answer:

• Fluoride ion deficiency in drinking water causes tooth decay.
• The fluoride ions make the enamel on teeth much harder by converting hydroxyapatite
  [3(Ca₃(PO₄)₂. Ca(OFl)₂]. the enamel on the surface of the teeth into much harder fluorapatite [3(Ca₃(PO₃)₂. CaF₂]
• Fluoride ion concentration above 2 ppm causes brown mottling of teeth. Excess fluoride causes damage to bones and teeth.

Question 10.
Explain the harmful effects of

1. lead
2. Nitrate in drinking water.

Answer:

1. Drinking water containing lead contamination above 50 ppm can cause damage to liver, kidneys and reproductive system.
2. Use of drinking water having concentration of nitrates higher than 45 ppm may cause (blue baby syndrome) methemoglobinemia disease in chiidrens.

Question 11.
Explain how styrene is produced by traditional and greener routes?
Answer:
1. Traditional route:
This method involve two steps-Carcinogenic benzene reacts with ethylene to form ethyl benzene. After that ethyl benzene undergoes dehydrogenation using Fe₂O₃ / Al₂O₃ to give styrene.

2. Greener route:
To avoid carcinogenic benzene. greener routes to start with cheaper and
environment friendly xylene.

Question 12.
How acetaldehyde is commercially prepared by green chemistry?
Answer:
Acetaldehyde is commercially prepared by one step oxidation of ethene in the presence of ionic catalyst in aoiieniis medium with 9% yield.

Question 13.
What do you mean by ozone hole? What are its consequences?
Answer:
Depletion of ozone layer creates some sort of holes in the blanket of ozone which surrounds us in the atmosphere and this is known as ozone hole.

• With the depletion of the ozone layer, UV radiations filters into the troposphere which leads to ageing of skin. cataract, sunburn, skin cancer etc.
• By killing many of the phytoplanktons, it can damage the fish productivity.
• Evaporation rate increases through the surface and stomata of leaves which can decrease the moisture content of the soil.

Question 14.
What is photo chemical smog? What arc its effects? How can it be controlled?
Answer:
It is a kind of smog formed in warm, dry and sunny climate. It is formed when sunlight is absorbed by SO₂ oxides of nitrogen and hydrocarbons. It act as an oxidising agent.

Effects of photo chemical smog:

• It produces irritation in eyes and also in respiratory system.
• It can damage many materials such as metals, stones, building materials etc.
• NO₂ present in photochemical smog gives it brown colour which reduces the visibility.
• it is harmful to fabrics, crops and ornamental plants.

Control of photochemical smog:

• By using catalytic converters in automobilës.
• By spraying certain compounds into the atmosphere which generate free radicals that can easily combine with the free radicals that initiates the reaction forming toxic compounds of photochernical smog.
• Certain plants such as Pinus, Juniparus, Pyrus could be also helpful in this matter.

Question 15.
(a) Define eutrophication and pneumoconiosis.
(b) Write differences between photochemical smog and classical smog.
Answer:
(a) Eutrophication:
When the growth of algae increases in the surface of water, dissolved oxygen in water is greatly reduced. This phenomenon is known as eutrophication. Due to this growth of fishes gets inhibited.

(b) Pueumoconiosis:
It is a disease which irritates lungs. It causes scarring or fibrosis of the lungs.

Pholochemical smog:

• It is formed as a result of photochernical decomposition of nitrogen dioxide and chemical reactions involving hydrocarbons.
• It takes place during dry warm season in presence of sunlight.
• It is oxidising in nature.

Classical smog:

• It’s formed due to condensation of SO₂ vapours on particles of carbon in cold climate.
• it is generally formed during winter when there is severe cold.
• It is reducing in nature.

Samacheer Kalvi 11th Chemistry Environmental Chemistry 5 Marks Questions and Answers

Question 1.
Explain about the different layers of Earth’s atmosphere.
Answer:
Earth’s atmosphere can he divided into different layers with characteristic altitude and temperature.

Question 2.
What arc the health effects of particulate pollutants?
Answer:
1. Dust, mist, fumes etc. are air borne particles which are dangerous for human health. Particulate pollutants bigger than 5 microns are likely to settle in the nasal passage whereas particles of about 10 microns enter the lungs easily and causes scaring or fibrosis of lung lining.

They irritate the lungs and causes cancer and asthma. This disease is called pneumoconiosis. Coal miners may suffer from black lung disease. Textile workers may suffer from white lung disease.

2. Lead particulates affect children’s brain. interferes with maturation of RBC’s and even causes cancer.

3. Particulates in the atmosphere reduces the visibility by scattering and absorption of sun light. It is dangerous for aircraft and motor vehicles.

4. Particulates provide nuclei for cloud formation and increases fog and rain.

5. Particulates deposit on plant leaves and hinder the intake of CO₂ from the air and affect photosynthesis.

Question 3.
What arc the effects of photochernical smog?
Answer:

• The three main components of photochemical smog are nitrogen oxide, ozone and oxidized hydrocarbons like HCHO, CH₂ = CH – CRO and PAN.
• Photo chemical smog causes irritation to eyes. skin and lungs. It also increases the chance of asthma.
• high concentration of ozone and NO can cause nose and throat irritation, chest pain, difficulty in breathing.
• PAN is toxic to plants, attack younger leaves and cause bronzing and glazing of their surfaces.
• It causes corrosion of metals, stones, building materials and painted surfaces.

Question 4.
Explain about the environmental impacts of ozone depletion.
Answer:
1. The formation and destruction of ozone is a regular natural process, which never disturbs the equilibrium level of ozone in the stratosphere. Any change in the equilibrium level of ozone in the atmosphere will adversely affect the Life in biosphere in the following ways.
2. Depletion of ozone layer will allow more UV rays to reach the earth surface and would cause skin cancer and also decreases the immunity level in human beings.
3. UV radiations atlecis plant proteins which lead to harniful mutation in plant cells.
4. UV radiations affect the growth of phytoplankton and as a result ocean food chain is disturbed and it even damages the fish productivity.

Question 5.
Explain the list of major water pollutants and their sources.
Answer:

• Microorganisms – Domestic sewage, domestic waste water, dung heap
• Organic wastes – Domestic sewage, animal excreta, food processing factory wastc. detergents and decayed animals and plants
• Plant nutrients – Chemical fertilisers
• Heavy metals – Heavy metal producing factories
• Sediments – Soil erosion by agriculture and strip-mining
• Pesticides – Chemicals used for killing insects, fungi and weeds
• Radioactive substances – Mining of uranium containing minerals
• Heat – Water used for cooling in industries

Question 6.
Describe about the causes of water pollution.
Answer:
Causes of water pollution –
1. Microbiological pollutants:
(a) Disease causing microorganisms like bacteria, viruses and protozoa are most senous water pollutants. They come from domestic sewage and animal excreta.
(b) Fish and shellfish can become contaminated from them and people who eat them will also become ill.
(c) Dysentery and cholera are water borne diseases.
(d) Human excrcta contain bacteria such as Escherichia coll and Streptococcus farcical- -is which causes gastrointestinal diseases.

2. Organic wastes:
Organic matter such as leaves, grass, trash can also pollute water. Water pollution is cause by excessive phytoplankton growth within water.

3. Chemical wastes:
A whole variety of chemicals from industries such as metals and solvents are poisonous to fish and other aquatic life. Detergents and oils float spoils the water bodies. Acids from mine drainage and salts form various sources can also contaminate water sources.

Question 7.
What are the harmful effects of chemical water pollutants’?
Answer:

1. Cadmium and mercury can cause kidney damage.
2. Lead poisoning can lead to severe damage of kidneys. liver and brain. It also affects the central nervous system.
3. Polychiorinated biphenyl causes skin diseases and are carcinogenic in nature.

Question 8.
Explain about green chemistry in day-to-day life.
Answer:
1. Dry cleaning of clothes:
Solvents like tetrachloroethylene used in dry cleaning of clothes, pollute the ground water and are carcinogenic. In place of tetrachloro ethylene, liquefied CO₂ with suitable detergent is an alternate solvent used. Liquefied CO₂ is not harmful to the ground water. Nowadays H₂O₂ is used for bleaching clothes in laundry, gives better result and utilises less water.

2. Bleaching of paper:
Conventional method of bleaching was done with chlorine. Nowadays H₂O₂ can be used for bleaching paper in the presence of catalyst.

3. Synthesis of chemicals:
Acetaldehyde is commercially prepared by one step oxidation of ethene in the presence of ionic catalyst in aqueous medium with 90% yield.

4. Instead of petrol, methanol is used as a fuel in automobiles.
5. Neem based pesticides have been synthesis-ed, which are more safer than the chlorinated hydrocarbons.

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Tamilnadu Samacheer Kalvi 11th Chemistry Solutions Chapter 14 Haloalkanes and Haloarenes

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Samacheer Kalvi 11th Chemistry Haloalkanes and Haloarenes Textual Evaluation Solved

Samacheer Kalvi 11th Chemistry Haloalkanes and Haloarenes Multiple Choice Questions

Question 1.
The IUPAC name of  is ………….
(a) 2-Bromopent – 3 – ene
(b) 4-Bromopent – 2 – ene
(c) 2-Bromopent – 4 – ene
(d) 4-Bromopent – 1 – ene
Answer:
(b) 4 – Bromopent – 2 – ene

Question 2.
Of the following compounds. which has the highest boiling point’?
(a) n-Butyl chloride
(b) Isobutyl chloride
(c) t-Butyl chloride
(d) n-propyl chloride
Answer:
(a) n-Butyl chloride

Question 3.
Arrange the following compounds in increasing order of their density.
(A) CCl₄
(B) CHCl₃
(C) CH₂Cl₂
(D) CH₃Cl
(a) D<C<B<A
(b) C<B<A<D
(c) A<B<C<D
(d) C<A<B<D
Answer:
(a) D<C<B<A

Question 4.
With respect to the position of – Cl in the compound CH₃ – CH = CH – CH₂ – Cl, it is classified as ………..
(a) Vinyl
(b) Allyl
(c) Secondary
(d) Aralkyl
Answer:
(b) Allyl

Question 5.
What should be the correct IUPAC name of diethyl chioromethane?
(a) 3-Chioropentane
(b) 1-Chloropentane
(c) 1-Chloro- 1, 1, diethylmethanc
(d) 1-Chloro- 1 -ethylpropane
Answer:
(a) 3-Chioropentane

Question 6.
C-X bond is strongest in …………
(a) Chioromeihane
(b) lodomethane
(c) Bromomethane
(d) Fluoromethanc
Answer:
(d) Fluoromethane

Question 7.
In the reaction  X + N₂, X is …………

Answer:

Question 8.
Which of the following compounds will give racemic mixture on nucleophilic substitution by OH^– ion?

(a) (i)
(b) (ii) and (iii)
(c) (iii)
(d) (i) and (ii)
Answer:
(c) (iii)

Question 9.
The treatment of ethyl formate with excess of RMgX gives –

Answer:
(c) R—CHO

Question 10.
Benzene reacts with Cl₂ in the presence of FeCl₂ and in absence of sunlight to form –
(a) Chiorobenzene
(b) Benzyl chloride
(c) Benzal chloride
(d) Benzene hexachloride
Answer:
(a) Chiorobenzene

Question 11.
The name of C₂F₄Cl₂ is –
(a) Freon – 112
(b) Freon – 113
(c) Freon – 114
(d) Freon – 115
Answer:
(c) Freon – 114

Question 12.
Which of the following reagent is helpful to differentiate ethylene dichloride and ethylidene chloride?
(a) Zn / methanol
(b) KOH / ethanol
(c) Aqueous KOH
(d) ZnCl₂ / Cone. HCl
Answer:
(e) Aqueous KOH

Question 13.
Match the compounds given in Column I with suitable items given in Column II.
Answer:

Code:
(a) A→2,B→4,C→1,D→3
(b) A→3,B→2,C→4,D→1
(c) A→1,B→2,C→3,D→4
(d) A→3,B→1,C→4,D→2
Answer:
(d) A→3,B→1,C→4,D→2

Question 14.
Assertion : in monohaloarenes, electrophilic substitution occurs at oriho and para positions.
Reason : Halogen atom is a ring deactivator.
(a) If both assertion and reason arc true and reason is the correct explanation of assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.
(c) If assertion is true but reason is false.
(d) If both assertion and reason are false.
Answer:
(b) If both assertion and reason are true but reason is not the correct explanation of assertion.

Question 15.
Consider the reaction, CH₃CH₂CH₂Br + NaCN → CH₃CH₂CH₂CN + NaBr This reaction will be the fastest in …………
(a) ethanol
(b) methanol
(c) DMF (N, N’ – dimethyl forrnamide)
(d) water
Answer:
(c) DMF (N, N’ – dimethyl fonuarnide)

Question 16.
Freon-12 is manufactured from tetrachioromethane by …………
(a) Wurtz reaction
(b) Swarts reaction
(c) DMF (N, N’ – dimethyl tòrmamide)
(d) water
Answer:
(c) DMF (N, N’ – dimethyl formarnide)

Question 16.
Frcon-12 is manufactured from tetrachioromethane by …………
(a) Wurtz reaction
(b) Swans reaction
(c) Haloform reaction
(d) Gattennann reaction
Answer:
(b) Swarts reaction

Question 17.
The most easily hydrolysed molecule under S_N¹ condition is …………
(a) allyl chloride
(b) ethyl chloride
(c) isopropyl chloride
(d) benzyl chloride
Answer:
(a) benzyl chloride

Question 18.
The carbocation formed in S_N¹ reaction of alkyl halide in the slow step is …………
(a) sp³ hybridised
(b) sp² hybridised
(c) sp hybridised
(d) none of these
Answer:
(b) sp² hybridised

Question 19.
The major products obtained when chiorobenzene is nitrated with HNO₃ and cone. H₂SO₄
(a) 1-chloro-4-nitrobenzenc
(b) 1-chloro-2-n itrobenzene
(c) 1-chloro-3-n itroberizene
(d) 1-chloro- 1 -nitrobenzene
Answer:
(a) 1 -chloro-4-nitrobenzene

Question 20.
Which one of the following is most reactive towards nucleophilic substitution reaction?

Answer:

Question 21.
Ethylidene chloride on treatment with aqueous KOH gives …………
(a) acetaldehyde
(b) ehtylene glycol
(c) formaldehyde
(d) glyoxal
Answer:
(a) acetaldehyde

Question 22.
The raw material for Rasching process is …………
(a) chiorobenzene
(b) phenol
(c) benzene
(d) anisole
Answer:
(c) benzene

Question 23.
Chloroform reacts with nitric acid to produce …………
(a) nitro-toluene
(b) nitro-glycerine
(c) chloropicrin
(d) chioropicric acid
Answer:
(c) chioropicrin

Question 24
Acetone X, X is …………
(a) 2-propanol
(b) 2-methyl-2-propanol
(c) 1 -propanol
(d) acetonol
Answer:
(b) 2-methyl-2-propanol

Question 25.
Silver propionate when refluxed with Bromine in carbon tetrachioride gives …………
(a) propionic acid
(b) chioroethane
(c) bromoethane
(d) chioropropane
Answer:
(c) bromoethane

Samacheer Kalvi 11th Chemistry Haloalkanes and Haloarenes Short Answer Questions

Question 26.
Classify the following compounds in the form of alkyl. allylic, vinyl, benzylic halides:
(a) CH₃-CH = CH-Cl
(b) C₆H₅CH₂I

(d) CH₂ = CH – Cl
Answer:
(a) CH₃-CH = CH-Cl – Allylic halide
(b) C₆H₅CH₂ – Benzylic halide

(d) CH₂ = CHCl – Vinyl halide

Question 27.
Why chlorination of methane is not possible in dark?
Answer:

• Chlorination of methane is a free radical substitution reaction.
• Before chlorine reacts with methane, the Cl-Cl single bond must break to form free radicals and this can only be done in the presence of ultraviolet light.
• In dark, chlorine free radicals formation is not possible and so chlorination of methane is not possible in dark.
• The ultraviolet light is a source of energy and it being used to break of Cl-Cl and
  produce Cl free radical Free radical’s which can attack methane. in dark this is not possible.

Question 28.
How will you prepare n-propyl iodide from n-propyl bromide?
Answer:
Finkeistein reaction:
n-propyl bromide on heating with a concentrated solution of sodium iodide in dry acetone gives n-propyl iodide. This S_N² reaction is called Finkelstein reaction.

Question 29.
Which alkyl halide from the following pair is –
1. chiral
2. undergoes faster S_N² reaction?

Answer:
1.

Here the C*is chiral carbon atom and it is surrounded by four different groups. Br
2.  i.e., 1-chiorohutane undergoes faster S_N² reaction. CH₃-CH₂-CH₂Cl is a primary alkyl halide and so it undergoes faster S_N² reaction.

Question 30.
How does chiorobenzene react with sodium in the presence of ether? What is the name of the reaction?
Answer:
Chiorobenzene when heated with sodium in ether solution will forrn biphenyl as the product. This reaction is called fitting reaction.

Question 31.
Give reasons for polarity of C-X bond in haloalkancs.
Answer:

• Carbon halogen bond is a polar bond as halogens are more electronegative than carbon.
• The carbon atom exhibits a partial positive change (δ⁺) and halogen atom acquires a partial negative change. (δ^–)
• 

Question 32.
Why is it necessary to avoid even traces of moisture during the use of Grignard reagent?
Answer:
Grignard reagents are highly reactive substances. They react with any source of proton to form hydrocarbons. Even water is sufficiently acidic to convert it into the corresponding hydrocarbon. So it is necessary to avoid even traces of moisture with the Grignard reagent as they arc highly reactive.

Question 33.
What happens when acetyl chloride is treated with excess of CH₃ MgI?
Answer:
When acetyl chloride is treated with excess of Grignard reagent, the product formed is tertiary butyl alcohol.

Question 34.
Arrange the following alkyl halideïniiicreasing order of bond enthalpy of RX: CH₃Br, CH₃F, CH₃Cl, CH₃I
Answer:
increasing order of bond enthalpy of RX is …………
CH₃ I bond enthalpy 234 kJ mol^-1
CH₃ Br bond enthalpy 293 kJ mol^-1
CH₃ Cl bond enthalpy 351 kJ mol^-1
CH₃ F bond enthalpy 452 IcI mol^-1
CH₃ F > CH₃ Cl> CH₃Br < CH₃ I

Question 35.
What happens when chloroform reacts with oxygen in the presence of sunlight?
Answer:
Chloroform undergoes oxidation in the presence of sunlight and air to form phosgene (carboxyl chloride)

Since phosgene is very poisonous, therefore its presence makes chloroform unfit for use as anaesthetic.

Question 36.
Write down the possible isomers of C₅H₁₁Br and give their IUPAC and common names.
Answer:
C₅H₁₁Br : 8 isomers are possible.

Question 37.
Mention any three methods of preparation of haloalkanes from alcohols.
Answer:
Haloalkanes are prepared from alcohols by treating with –

1. HCl
2. PCl₅
3. SO₂Cl₂

Answer:
1. Reaction of alcohol with hydrogen halide:

2. Reaction of alcohol with phosphorous halide:

3. Reaction of alcohol with thionyl chloride:

Question 38.
Compare S_N¹ and S_N² reaction mechanisms.
Answer:
S_N¹ Mechanism:
1. It is a unimolecular nucleophilic substitution reaction of first order.
2. It takes place in two steps.
3. It leads to racemisation.
4. It mostly take place in tertiary alkyl halides.
5. The rate of the reaction depends only on the concentration one of substrate and so it is a first order reaction.
Step – I

Step – II

S_N² Mechanism:
1. It is a bimolecular nucleophilic substitution reaction of second order.
2. It takes place in one step.
3. It leads to invertion of configuration.
4. It mostly take place in primary alkyl halides.
5. The rate of the reaction depends on the concentration of both the substrate as well as the nucleophile and so it is a second order reaction.

Question 39.
Compare S_N¹ and S_N² reaction mechanisms.

Answer:

Question 40.
Discuss the aromatic nucicophilic substitution reactions of chiorobenzene.
Answer:
Aromatic nucleophilic substitution reactions:
Dow’s process:

Question 41.
Account for the following –
1. t-butyl chloride reacts with aqueous KOH by S_N¹ mechanism while n-butyl chloride reacts with S_N² mechanism.
2. p-dichlorobenzene has higher melting point than those of o-and m-dichlorobenzene.
Answer:
1. In t-butyl chloride, there is niore steric hindrance and it involves formation of a stable tertiary carbocation. Therefore it reacts with KOH by S_N¹ mechanism rather than S_N² mechanism because S_N¹ mechanism is faourable in case of steric crowding and is directly proportional to partial positive charge on carbon atom. In n-butyl chloride, there is least steric hindrance and involves formation of less stable primary carbocation. Thus it takes place in one step and is favoured by S_N² mechanism.

2. Melting point of p – dichiorobenzene is higher than that of ortho and meta-dichiorobenzene. This is due to the fact that is has a symmetrical structure and therefore, its molecules can easily pack closely in the crystal lattice. p-dichlorobcnzene being more symmetrical fits closely in the crystal lattice and has stronger intermolecular attraction than o & m isomers. So p-isomer has high melting point than the corresponding o & m-isomers.

Question 42.
In an experiment ethyliodide in ether is allowed to stand over magnesium pieces. Magnesium dissolves and product is formed
(a) Name the product and write the equation for the reaction.
(b) Why all the reagents used in the reaction should be dry? Explain.
(c) How is acetone prepared from the product obtained in the experiment?
Answer:
(a) When ethyl iodide in ether is allowed to stand over magnesium pieces. the product formed will be Ethyl magnesium iodide.

(b) Gngnard reagent are highly reactive compounds and react with any source of proton to give hydrocarbons. Even when water is there, it is sufficiently acidic to convert it into the corresponding hydrocarbon. So it is necessary to avoid even traces of moisture with the grignard reagent as they are highly reactive. That is why the all reagents used in the reaction should be dry.

Methyl magnesium iodide reacts with acetyl chloride to form acetone.

Question 43.
Write a chemical reaction useful to prepare the following:
1. Freon- 12 from carbon tetrachioride.
2. Carbon tetrachioride from carbon disulphide.
Answer:
1. Freon-12 from carbon tetrachioride:
Freon- 12 is prepared by the action of hydrogen fluoride on carbon tetrachioride in the presence of catalytic amount of antimony pentachloride. This is called “swartz reaction”.

2. Carbon tetrachioride from carbon disuiphide:
Carbon disuiphide reacts with chlorine gas in the presence ofanhydrous AlCl₃ as catalyst to give carbon tetrachloridc.

Question 44.
What are Freons? Discuss their uses and environmental effects.
Answer:
Freons are the chiorofluoro derivatives of methane and ethane.
Freon is represented as Freon – cba
Where, c = number of carbon atoms, b = number of hydrogen atoms, a = total number of fluorine atoms.
CF₂Cl₂ = 0
c = 1 – 1 = 0
H = 0 +1 = 1
F = 2
So Freon – 12 is CF₂Cl₂

Uses of Freons:

• Freons are used as refrigerants in refrigerators and air conditioners.
• it is used as a propellant for aerosols and foams.
• it is used as propellant for foams to spray out deodorants, shaving creams and insecticides.

Environmental effects of Freons:

1. Freon gas is a very powerful greenhouse gas which means that it traps the heat normally radiated from the earth out into the space. This causes the earth’s temperature to increase, resulting in rising sea levels, droughts. stronger storms, flash floods and a host of other very unpleasant effect.

2. As freon moves throughout the air, its chemical ingredients causes depletion of ozone layer. Depletion of ozone increases the amount of ultraviolet radiations that reaches the earths surface, resulting in serious risk to human health. High levels of ozone, in turn, causes resoiratory problems and can also kill olants.

Question 45.
Predict the products when bromoethane is treated with the fòllowing:
1. KNO₂
2. AgNO₂
Answer:
1. When bromoethane is treated with KNO₂ the product formed is Ethyl nitrite.

2. When bromoethane is treated with AgNO₂ nitroethane will be formed as the product.

Question 46.
Explain the mechanism of S_N¹ reaction by highlighting the stereochemistry behind it.
Answer:
1. S_N¹ stands for unimolecular nucleophilic substitution reaction of first order reaction.
2. The rate of the following S_N¹ reaction depends upon the concentration of alkyl halide
(RX) and it is independent of the concentration of the nucicophile (OH^– ) used.
3. R-Cl + OH^– R – OH + Cl^–
4. This S_N¹ reaction follows first order kinetics and it occurs in two steps:
5. S_N¹ reaction mechanism takes place in tertiary butyl bromide with aqueous KOH a follows:

6. the 2 steps of the reaction are:
Step 1:
Formation of carbocation:
The polar C-Br bond breaks first forming a carbocation and bromide ion. This step is slow and hence it is the rate determining step.

The carbocation has two equivalent lobes of the vaccant 2p orbital, So it can react equally fast form either face.

Step 2:
The nucleophile immediately reacts with the carbocation. This step is fast and hence does not affect the rate of the reaction.

The nucleophile OH can attack carbocation from both the sides.

Question 47.
Write short notes on the the following:
1. Raschig process
2. Dows Process
3. Darzens process
Answer:
1. Raschig process:
Chiorobenzene is commercially prepared by passing a mixture of benzene vapour in air and HUt over heated cupric chloride.

2. Dow’s process:
Chiorobenzene is boiled with Sodium hydroxide to get Phenol. This reaction is called Dow’s process.

3. Darzens process:
Ethanol reacts with SOCl, in the presence of pyridine to form chloroethane. This reaction is called Darzens process.

Question 48.
Starting from CH₃MgI. How will you prepare the following’?

1. Acetic acid
2. Acetone
3. Ethyl acetate
4. Isopropyl alcohol
5. Methyl cyanide

Answer:
1. Acetic acid from grignard reagent:

2. Acetone from Gngnard reagent:

3. Ethyl acetate from Grignard reagent:

4. Isopropyl alcohol from Grignard reagent:

5. Methyl cyanide form Grignard reagent:

Question 49.
Complete the following reactions:

Answer:

Question 50.
Explain the preparation of the following compounds:

1. DDT
2. Chloroform
3. Biphenyl
4. Chioropicrin
5. Freon-12

Answer:
1. DUT

2. Chloroform:

3. Biphenyl:
Fittig reaction:

4. Chioropicrin.

5. Freon – 12.

Question 51.
An organic compound (A) with molecular formula C₂H₅Cl reacts with KOH gives compounds (B) and with alcoholic KOH gives compound (C). Identify (A), (B) and (C)
Answer:

1. An organic compound (A) of molecular formula C₂H₅Cl is identified as Chioroethane with molecular formula CH₂-CH₂Cl from the formula.

2. Chioroethane reacts with aqueous KOH to give ethanol, i.e., Aqueous C₂H₅OH as (B) by substitution reaction.

3. ChIooethane reacts with alcoholic KOH to give ethene C₂H₄ as (C) by elimination reaction.


Question 52.
Simplest alkene (A) reacts with HCl to form compound (B).Compound (B) reacts with ammonia to form compound (C) of molecular formula C₂H₂N. Compound (C) undergoes carbylarnine test. Identify (A), (B), and (C).
Answer:

1. The simplest alkene (A) is CH₂ = CH₂, ethene.

2. Ethene reacts with HCl to give Chioroethane CH₂-CH₂Cl as (B) by addition reaction.

3. Chioroethane reacts with ammonia to give Ethylamine CH₃-CH₂ NH₂ as (C). It is a primary amine and Carbylamine test is the characteristic test for 1° amine.

Question 53.
A hydrocarbon C₃H₆ (A) reacts with HBr to form compound (B). Compound (B) reacts with aqueous potassium hydroxide to give (C) of molecular formula C₃H₈O. What are (A) (B) and (C). Explain the reactions.
Answer:

1. The hydrocarbon with molecular formula C₃H₆ (A) is identified as propene,
CH₃-CH = CH₂

2. Propene reacts with HBr to form brornopropane CH₃—CH₂—CH₂Br as (B).

3. 1 -bromopropanc react with aqueous potassium hydroxide to give 1 -propanol
CH₃ – CH₂ CH₂OH as (C).

4. 2-bromo propane reacts with aqueous KOH to give 2-propanol as (C)

Question 54.
Two isomers (A) and (B) have the same molecular formula C₂H₄Cl₂. Compound (A) reacts with aqueous KOK, gives compound (C) of molecular formula C₂H₄O. Compound (B) reacts with aqueous KOH, gives compound (D) of molecular formula C₂H₆O₂. Identif’ (A), (B), (C) and (D).
Answer:

1. The compounds (A) and (B) with the molecular formula C₂H₄Cl₂ are and respectively with IUPAC names: 1, 1-dichloroethane (A) and 1.

2. 1, 1-dichloroethane reacts with aqueous KOH tp give CH₃CHO Acetaldehyde as (C).

3. 1 ,2-dichloroethane reacts with aqueous KOH to give ethylene glycol  as (D).

[Evaluate Yourself ]

Question 1.
Write the IUPAC name of the following:

Answer:

Question 2.
Write the structure of the following compounds:

1. 1-Bromo-4-cthylcyclo hexane
2. 1 4-Dichlorobut-2-ene
3. 2 Chloro-3-methyl pentane

Answer:
1. 1-Bromo-4 ethylcyclo hexane

2. 1. 4-Dichioro but-2-ene

3. 2-Çhloro-3-methyl pentane

Question 3.
Write all possible chain isomers with molecular formula C₅H₁₁Cl
Answer:
C₅H₁₁Cl:
8 isomers arc possible. These are –

Question 4.
neo-pentyl bromide undergoes nucleophilic substitution reactions very slowly. Justify.
Answer:
1.  neo-pentyl bromide undergoes nucleophilic substitution reactions very slowly due to steric hindrance of many alkyl groups. When bromide is attached to neo-pentyl carbon atom, the heterolytic cleavage of C-Br takes place very slowly and substitution is also a very slow reaction.

2. Due to bulky neo-pentyl group, it becomes difficult for a nucleophile to attack from the back side of carbon having C-Br bond.

3. Splitting of C-Br gives a primary carbocation which is very less stable. So neo-pentyl bromide undergoes nucleophilic substitution reactions very slowly.

Question 5.
Why Grignard reagent should be prepared in anhydrous condition’?
Answer:
Grignard reagent are highly reactive species and react with any source of proton to give hydrocarbons. Even when water is there, it is sufficiently acidic to convert it into the corresponding hydrocarbon. So it is necessary to avoid even traces of moisture while preparing grignard reagent as they are highly reactive and that is why gngnard reagent should be prepared in anhydrous conditions.

Question 6.
Haloalkanes undergo nucleophilic substitution reaction whereas haloarenes undergo electrophilic substitution reaction. Comment.
Answer:
Haloalkanes undergo nucleophilic substitution reactions due to high clectronegativity of the halogen atom. The C-X bond in haloalkanes is slightly polar, thereby the C atom acquires a slight positive charge in C-X bond. [lence C atom is a good target for attack for nucleophiles. Therefore C-X atom of haloalkanes is replaced by a nucleophile easily.

On the other hand in haloarenes, the halogen atom releases electron to the benzene nucleus relatively electron rich with respect to the halogen atom. As a result the electrophilic attack occurs at ortho and para positions. Hence haloarenes undergo electrophilic substitution reactions.

Question 7.
Chloroform is kept with a little ethyl alcohol in a dark coloured bottle. Why?
Answer:
1. Chlorofòrm is slowly oxidised by air in the presence of light to an extremely poisonous gas, carboxyl chloride (phosgene). it is therefore stored in closed dark coloured bottles completely filled so that air is kept out.

2. With the use of 1% ethanol we can stabilise chloroform, because ethanol can convert the poisonous COCl₂ gas into non poisonous diethyl carbonate.
COCl₂ + 2C₂H₅OH CO(OC₂H₅)₂ + 2HCl

Question 8.
What is the IUPAC name of the insecticide DDT? Why is their use banned in most of the countries?
Answer:
1. The IUPAC name of the insecticide DDT isp, p-dichloro-diphenyl trichioroethane.
2. Even DDT is an effective insecticide. Now-adays it is banned because of its long term toxic effects.
3. DDT is very persistent in the environment and it has a high affinity for fatty tissues. As a result, DDT gets accumulated in animal tissue fat, in particular that of birds of prey with subsequent thinning of their eggs shells and impacting their rate of reproduction. That is why DDT is banned in most of the countries.

Samacheer Kalvi 11th Chemistry Haloalkanes and Haloarenes Additional Questions Solved

I. Choose The Correct Answer.

Question 1.
Which of the following is an example for polyhalo compounds?
(a) Vnyl iodide
(b) Chiorobenzene
(c) Allyl chloride
(d) Chloroform
Answer:
(d) Chloroform

Question 2.
Which of the following is a secondary haloalkane?
(a) Bromoethane
(b) 2-Chioropropane
(c) 2-Jodo-2-methylpropane
(d) 1-Chloropropane
Answer:
(b) 2-Chioropropane

Question 3.
How many isomers are possible for the formula C₄H₉Cl?
(a) 3
(b) 2
(c) 4
(d) 5
Answer:
(c) 4

Question 4.
How many isomers are possible for the formula C₅H₁₁Br?
(a) 11
(b) 8
(c) 4
(d) 5
Answer:
(b) 8

Question 5.
Which of the following is called Lucas reagent?
(a) Conc. H₂SO₄ + Anhydrous CuSO₄
(b) Conc.HCl + Anhydrous ZnCl₂
(c) Dil.HCl + AlCl₃
(d) Conc.HCl + ConcHNO₂
Answer:
(b) Conc.HCl + Anhydrous ZnCl₂

Question 6.
Which of the following mechanism is followed in the halogenation of alkanes in the presence of U-V light?
(a) Nucleophilic substitution
(b) Electrophilic addition
(c) Free radical substitution
(d) Elimination reaction
Answer:
(c) Free radical substitution

Question 7.
The reactivity of alcohols with haloacid is –
(a) 3° > 2° > 1°
(b) 1° > 2° > 3°
(c) 2° > 3° > 1°
(d) 3° > l° > 2°
Answer:
(a) 3° > 2°> 1°

Question 8.
Which of the following reagent is not used to convert alcohol to haloalkane?
(a) H-X
(b) PX₅
(c) CCl₄
(d) SOCl₂
Answer:
(c) CCl₄

Question 9.
What is the name of the reaction in which bromoethane is converted to iodoethane by reacting with NaI in acetone?
(a) Hunsdicker reaction
(b) Dow’s process
(c) Finkelstein reaction
(d) Swarts reaction
Answer:
(c) Finkeistein reaction

Question 10.
Identify the correct order of boiling point of haloalkanes?

Answer:

Question 11.
Which of the following pair functional groups represents ambident nucleophiles?
(a)-SH-&-OH
(b)-CN-&-NO₂
(c)-Br-&-Cl
(d)-O-&-CHO
Answer:
(b)-CN & NO₂

Question 12.
Which one following mechanism will be followed when Tartiary butyl chloride is treated with alcoholic KOH?
(a) S_N¹ mechanism
(b) E₁ mechanism
(c) S_N² mechanism
(d) E₂ mechanism
Answer:
(b) E₁ mechanism

Question 13.
Which one of the following is used for producing pesticìdes?
(a) CHI₃
(b) CHCl₃
(c) CCl₃ NO₂
(d) CCl₄
Answer:
(b) CHCl₃

Question 14.
Which one of the following react with gringnard reagent followed by hydrolysis will yield primary alcohol?
(a) CH₃ CHO
(b) HCHO
(c) CH₃ COCH₃
(d) CO₂
Answer:
(b) HCHO

Question 15.
Which one of the following reacts with CH₃MgI followed by hydrolysis and gives isopropyl alcohol?
(a) CH₃ COCH₃
(b) CH₃ CHO
(c) HCHO
(d) CNCl
Answer:
(b) CH₃ CHO

Question 16.
Which one of the following reacts with CH₃ Mg I followed by hydrolysis to yield tert. butyl alchol ?
(a) CH₃CHO
(b) HCHO
(c) CH₃COOC₂H₅
(d) CH₃COCH₃
Answer:
(d) CH₃COCH₃

Question 17.
Which one of the following reacts with CH₃ Mg I followed by acid hydrolysis to yield acetic acid ?
(a) CNCl
(b) CH₃COOC₂H₅
(c) HCOOC₂H₅
(d) CO₂
Answer:
(d) CO₂

Question 18.
Which one of the following reagent react with methyl magnesium iodide followed by acid hydrolysis to give ethyl acetate?
(a) Chiorodimethyl ether
(b) Ethyl chloroformate
(c) Ethyl formate
(d) Acetaldehyde
Answer:
(b) Ethyl chioroformate

Question 19.
Which one of the following is used as fibre-swelling agent in textile processing?
(a) Chiorohenzene
(b) Chloroform
(c) Chlorai
(d) Chloroethane
Answer:
(a) Chiorobenzene

Question 20.
Which one of the fillowing is a gemdihalide?
(a) CH₃CHCl₂

(c) CH₃-CH₂Cl
(d) C₆H₄Cl₂
Answer:
(a) CH₃CHCl₂

Question 21.
Which of the following reagent is used to distinguish gem-dihalides and vicinal dihalides?
(a) Alcoholic KOH
(b) Aqueous KOH
(c) FeCl₃ / Cl₂
(d) Ethanol
Answer:
(b) Aqueous KOH

Question 22.
Which one of the following is used in the conversion of ethyliden dichioride to Acetylene?
(a) Zn + Methanol
(b) KOH + Ethanol
(c) Aqueous NaOH
(d) Alcoholic KOH
Answer:
(b) KOH + Ethanol

Question 23.
Which one of the following is used as a metal cleaning solvent?
(a) Isopropylidene chloride
(b) Methylene chloride
(c) Chloroform
(d) lodoform
Answer:
(b) Methylene chloride

Question 24.
Which one of the following is used as an insecticide and as a soil sterilising agent?
(a) Chloroform
(b) Chlorai
(c) Chioropicrin
(d) Tetrachioromethane
Answer;
(c) Chloropicrin

Question 25.
Which one of the following is used to test primary amincs?
(a) Schiff’s test
(b) Carbylarnine test
(c) Dye test
(d) Silver mirror test
Answer:
(b) Carbyianiine test

Question 26.
Which one of the following is used as propellant for aerosols and foams?
(a) Freons
(b) Methylidene chloride
(c) Chlorai
(d) Chloroform
Answer:
(a) Freons

Question 27.

the product X is –

Question 28.
Which one of the following will undergo S_N¹ reaction faster?

Answer:

Question 29.
Which one of the following compounds does not undergo nucleophilic substitution reactions at all?
(a) Ethyl bromide
(b) Vinyl chloride
(c) Benzyl chloride
(d) isopropyl chloride
Answer:
(b) Vinyl chloride

II. Match the following.

Question 1.

Answer:

Question 2.

Answer:

Question 3.

Answer:

Question 4.

Answer:

Question 5.

Answer:
“17” />

Question 6.

Answer:

III. Fill in the blanks.

Question 1.
is used in the treatment of typhoid ……….
Answer:
chloramphenicol

Question 2.
………. is used in the treatment of malaria
Answer:
Chioroquine

Question 3.
………. is used as an anesthetic ……….
Answer:
Halothane

Question 4.
………. is used for cleaning electronic equipments ……….
Answer:
Trichioroethylene

Question 5.
The IUPAC name of CH₂=CH-CH₂Cl is ……….
Answer:
3-Chioro-prop- 1-ene

Question 6.
The structure of Vinyl iodide is ……….
Answer:
CH₂=CHI

Question 7.
2-iodo-2-methylpropane belongs to type ……….
Answer:
3°haloalkanes

Question 8.
The structure of 2-Iodo-2-methylpropanc is ………
Answer:

Question 9.

the IUPAC name of this compound is ……..
Answer:
1-Bromo-2, 2-dimethyipropane

Question 10.
The IUPAC name of CH₂=CHCl is ………
Answer:
Chioroethene

Question 11.
The IUPAC name of is ………
Answer:
2-Bromo-3-chloro-2, 4-dimethylpentane.

Question 12.
The IUPAC name of is ………
Answer:
3-Chloro-2-methyl-prop- 1 -ene

Question 13.
The reactivity of haloacids (HCl , HBr, HI) with alcohol is in the order ………
Answer:
HI > HBr > HCl

Question 14.
The decreasing order of bond length among alkyl halides (CH₂I , CH₂ Br , CH₃F , CH₃Cl ) is in the order ………
Answer:
CH₃F < CH₃Cl < CH₃Br < CH₃I

Question 15.
The bond strength of C-X for the C-Cl , C-Br, C-I , C-F decreases in the order is ………
Answer:
C-F < C-Cl<C-Br < C-I

Question 16.
The catalyst used in Darzen halogenation of alcohol is ………
Answer:
Pyridine

Question 17.
In Finkeistein reaction, the mechanism followed is
Answer:
S_N²

Question 18.
Silver salt of fatty acid is converted to bromo alkanc by ………
Answer:
Hunsdicker reaction

Question 19.
In Swarts reaction, chioroalkane is converted to ………
Answer:
Fluoroalkane.

Question 20.
The conversion of bromoalkane to fluroalkane by heating with AgF is called ………
Answer:
Swartz reaction

Question 21.
The decreasing order of boiling point of haloalkanes CH₃ Br , CH₃Cl , CH₃F CH₃I is ………
Answer:
CH₃I > CH₃Br > CH₃CI > CH₃F

Question 22.
The correct increasing order of boiling point of haloalkanes CH₃Cl , CHCl₃ , CH₂Cl₂ , CCl₄ is ………
Answer:
CCl₄ > CHCl₃ > CH₂Cl₂ > CH₃Cl

Question 23.
Haloalkanes reacts with aqueous solution of KOH to form ………
Answer:
Alcohol

Question 24.
Haloalkane reacts with alcoholic solution of KOH to form ………
Answer:
Alkene

Question 25.
Ethyl bromide reacts with alcoholic AgCN to form ………
Answer:
CH₃CH₂NC

Question 26.
Ethyl bromide reacts with alcoholic KNO₂ to form ………
Answer:
Ethyl nitrite

Question 27.
The reaction in which sodium alkoxide react with haloalkane to form ether in called ………
Answer:
Williamson’s synthesis

Question 28.
Primary alkyl halide react with aqueous NaOH follows ………
Answer:
S_N²

Question 29.
Tertiary butyl bromide reacts with aqueous KOH follows ………
Answer:
S_N² mechanism

Question 30.
Ethyl bromide reacts with alcoholic KOH following mechanism. ………
Answer:
E₁

Question 31.
The product formed when tertiary butyl chloride is treated with alcoholic KOH is ………
Answer:
Iso-butylene

Question 32.
When 2-bromobutane react with alcoholic KOH, the products formed are ………
Answer;
1-butene & 2-butene

Question 33.
The product formed when iodoethane is treated with HI in the presence of red phosphorous is ………
Answer:
CH₃-CH₃

Question 34.
……… is used as an antiseptic
Answer:
Iodoform

Question 35.
……… is used for extinguishing the fire caused by oil or petrol under the commercial name pyrene
Answer:
Tetrachioromethane

Question 36.
……… Ethyl formate reacts with methyl magnesium iodide followed by acid hydrolysis to yield
Answer:
Acetaldehyde

Question 37.
……… Ethyl-methyl ether ethene is obtained by the action of methyl magnesium iodide with
Answer:
Chioro-aimethyl-ether

Question 38.
The hybridised state of carbon in haloarenes is ………
Answer:
sp²

Question 39.
The catalyst used in the preparation of chiorobenzene from benzene is ………
Answer:
FeCl₃

Question 40.
In the Gattermann reaction of preparation of chiorobenzene from benzene, the catalyst used is ………
Answer:
Cu / HCl

Question 41.
The conversion of benzene diazoniurn chloride to chiorobenzene in the presence of Cu₂Cl₂ + HCl is named as ………
Answer:
Sandmeyer reaction

Question 42.
Fluorobenzene is prepared from benzene diazonium chloride by ………
Answer:
BaIz-Scheimann reaction

Question 43.
Conversion of benzene to chiorobenzene in the presence of CuCl₂ / HCl is named as ………..
Answer:
Gattermann reaction

Question 44.
The conversion of chiorobenzene to phenol by the action of NaOH is called ………..
Answer:
Dow’s process

Question 45.
In Wurtz fittig reaction, chiorohenzene is converted to by reacting it with ethyl chloride.
Answer:
Ethylbenzene

Question 46.
The product obtained in fittig reaction of chiorobenzene is ………..
Answer:
Biphenyl

Question 47.
The reagent used in the conversion of Chiorobenzene to Benzene is ………..
Answer:
Ni-Al/NaOH

Question 48.
The catalyst used in the preparation of Phenyl magnesium chloride from chiorobenzene is ………..
Answer:
THF

Question 49.
Iso-propylidene chloride is an example of ………..
Answer:
vicinal dihalide

Question 50.
The IUPAC name of is ………..
Answer:
1, 2-dibromo-2-methylpropane

Question 51.
The reagent used in the conversion of ethylene dichioride is ………..
Answer:
Zn + CH₃OH

Question 52.
Chloroform is converted to methylene-chioride by the action of ………..
Answer:
Zn + HCl and H₂ / Ni

Question 53.
The reagents used in the preparation of chloroform are ………..
Answer:
Ethanol + Bleaching powder

Question 54.
The formula of Chioropicrin is ………..
Answer:
CCl₃NO₂

Question 55.
The product formed when methylamine react with chloroform and alkali is ………..
Answer:
CH₃NC

Question 56.
The product formed when CCI4 reacts with hot water vapours is
Answer:
COCl₂

Question 57.
The formula of Freon -11 is ………..
Answer:
CFCl₃

Question 58.
The formula of Freon – 12 is ………..
Answer:
CF₂Cl₂

Question 59.
The catalyst used in the preparation of CCl₂F₂ from CCl₄ and HF is ………..
Answer:
SbCl₅

Question 60.
The reagents used to prepare DDT are ………..
Answer:
Chlorai and Chiorobenzene

Question 61.
……….. is used to kill various insects like housefly and mosquitoes.
Answer:
(c) DDT

Question 62.
The name of CFCl₃ is
Answer:
Freon-11

Question 63.
The treatment of acetone with excess of RMgX gives:
Answer:

Question 64.
The most easily hydrolysed molecule under S_N² reaction is………..
Answer:
Ten. butyl chloride

Question 65.
. The product ‘X’ is ………..
Answer:
Ethanol

Question 66.
The product X is ………..
Answer:
CH₄

Question 67.
On heatrng CHCl₃ with aqueous NaOH solution, the product formed is
Answer:
HCOONa

Question 68.
Depletion of ozone layer is caused by
Answer:
Freons

Question 69.
Chloropicrin is used as
Answer:
soil sterilizing agent

Question 70.
lodoform can be used as
Answer:
Antiseptic.

Question 71.
in oil fire extinguisher, the compound used pyrene is chemically
Answer:
CCl₄

Question 72.
Reaction of ethyl chloride with sodium metal leads to the formation of
Answer:
n-Butane

Question 73.
When chloroform is treated with primary amine and KOH, we get ………..
Answer:
Offensive odours

IV. Choose the odd one out.

Question 1.
(a) CH₃Br
(b) CH₃-CH₂Br
(c) (CH₃)₃C-Br
(d) CH₃-CH₂-CH₂Br
Answer:
(c) (CH₃)₃C-Br. It is a haloalkane whereas others are 1° haloalkanes.

Question 2.
(a) Finkeistein reaction
(b) Wurtz reaction
(c) Swarts reaction
(d) Friedel crafts alkylation
Answer:
(d) Friedel crafts alkylation. It is used to prepare aromatic hydrocarbon whereas others are used to prepare alkanes.

Question 3.
(a) PCl₅
(b) SOCl₂
(c) HCl
(d) HF
Answer:
(cl) HF It is not used to prepare directly fluoro alkane whereas others are used to prepare directly chioro alkanes.

Question 4.
(a) Aerosol spray propellant
(b) Metal cleaning agent
(c) Anaesthetic agent
(d) Solvent in paint remover
Answer:
(c) Anaesthetic agent. It is not the use of methylene chloride whereas others are uses of methylene chloride.

V. Choose the correct pair.

Question 1.
(a) Chiorobenzene + chloral : DDT
(b) Chloroform + HNO₃ : Phosgene
(c) Chloroform + Zn / HCl : Methyl isocyanide
(d) Methane + 4Cl₂ : Carbon tetra chloride
Answer:
(c) Chlorobenzene + chloral : DDT

Question 2.
(a) Chloroform : Analgesic
(b) Freon : Propellant
(c) Chioropicrin : Antiseptic
(d) DDT : Soil sterilizing agent
Answer:
(b) Freon : Propellant

Question 3.
(a) Freon : Refrigerant
(b) DDT : Antiseptic
(c) Methylene : Soil sterilizing agent
(d) Iodotorni : Anaesthetic
Answer:
(a) Freon : Refrigerant

Question 4.
(a) HCOH + CH₃MgI : Secondary alcohol
(b) CH₃CHO + CH₃MgI : Tertiary alcohol
(c) CH₃COCH₃ + CH₃MgI : Primary alcohol
(d) CO₂ + CH₃MgI : Acetic acid
Answer:
(d) CO₂ + CH₃MgI : Acetic acid

Question 5.
(a) (C₃H)₃C-Cl + alcoholic KOH : S_N¹ reaction
(b) (CH₃)₃C-Cl + alcoholic KOH : E₁ reaction
(c) (CH₃)₃C-Cl + aqueous KOH : S_N² reaction
(cl) CH₃-Cl + aqueous KOH : S_N¹ reaction
Answer:
(b) (CH₃)₃C-Cl + alcoholic KOH : E₁ reaction

VI. Choose the incorrect pair.

Question 1.

Answer:

Question 2.
(a) CH₃I > CH₃Br > CH₃Cl > CH₃F : decreasing order of boiling point
(b) CCl₄ > CHCl₃> CH₂Cl₂ > CH₃Cl : increasing order of boiling point
(c) CH₃-CH₂-CH₂Cl <CH₃ CH₂ Cl <CH₃Cl : increasing order of boiling point

Answer:
(c) CH₃-CH₂-CH₂Cl <CH₃ CH₂Cl <CH₃Cl : increasing order of boiling point

Question 3.
(a) (CH₃)₃ C-Br + aqueous KOH : S_N¹ reaction
(b) (CH₃)₃ C-Br + alcoholic KOH: E₁ reaction
(c) CH₃Br – aqueous KOH : E₂ reaction
(d) CH₃Br + aqueous KOH : S_N² reaction
Answer:
(c) CH3Br + aqueous KOH : E₂ reaction

Question 4.
(a) 1-chioro propane + Alcoholic KOH : Propene
(b) Tert.butyl bromide + Alcoholic KOH : Isobutylenc
(c) CH₃-CH₂I + HI + Red p : Ethanc
(d) CH₃CHO + CH₃ Mg I : Ten. Butyl alcohol
Answer:
(d) CH₃CHO + CH₃ Mg I : Ten. Butyl alcohol

Question 5.
(a) HCHO + CH₃MgI : Primary alcohol
(b) CH₃CHO + CH₃MgI : Secondary alcohol
(c) CH₃COCH₃ + CH₃MgI : Aromatic alcohol
(d)CO₂ + CH₃MgI : Acetic acid
Answer:
(c) CH₃COCH₃ + CH₃MgI : Aromatic alcohol

VII. Assertion & Reason.

Question 1.
Assertion (A): The C-I in CH₃X is weak.
Reason (R): Larger the size, greater is the bond length and weaker is the bond formed.
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(b) Both (A) and (R) are correct but (R) is the correct explanation of (A).
(c) (A) is correct but (R) is wrong.
(a) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Question 2.
Assertion (A): Haloalkanes have higher boiling point and melting point than the parent alkanes having the same number of carbon.
Reason (R): The intermolecular forces of attraction and dipole-dipole interactions are
stronger in haloalkanes.
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of (A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Question 3.
Assertion (A) : Among isomenc halides, the boiling point decreases with increase in branching in alkyl group.
Reason (R) : With the increase in branching, the molecule attains spherical shape with less
surface area and less forces of interaction.
(a) Both (A) and (R) are correct and (R) is not the correct explanation of (A).
(b) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Question 4.
Assertion (A): The melting point of para halóbenzene is higher than that of ortho and meta isomers.
Reason (R): The higher melting point of p-isomer is due to its symmetry which leads to more close packing of its molecules in the crystal and subsequently p-isomer have strong intermolecular attractive forces.
(a) both (A) and (R) are correct and (R) is the correct explanation of(A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of(A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) both (A) and (R) are correct and (R) is the correct explanation of(A).

Question 5.
Assertion (A) : Haloarenes are insoluble in water.
Reason (R) : Haloarenes are able to form hydrogen bonds with water.
(a) Both (A) and (R) are correct but (R) is the correct explanation of (A).
(b) both (A) and (R) are correct and (R) is not the correct explanation of (A).
(e) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct
Answer:
(c) (A) is correct but (R) is wrong.

Question 6.
Assertion (A) : Haloarenes do not undergo nucleophilic substitution reactions readily.
Reason (R) : The C-X bond in aryl halides is short and stronger and also the aromatic ring is a center of high electron density.
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(h) Both (A) and (R) are correct but (R) is not the correct explanation of (A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Question 7.
Assertion (A): Chloroform vapours can be used as an anaesthetic.
Reason (R): Chloroform vapours depresses the central nervous system and cause unconsciousness.
(a) Both (A) and (R) are correct but (R) is not the correct explanation of (A).
(b) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(c) (A) is correct but (R) is wrong.
(d) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Question 8.
Assertion (A): Nowadays chloroform is not used as an anaesthetic.
Reason (R): Chloroform undergoes oxidation in the presence of light and air to form highly poisonous phosgene.
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).
(b) Both (A) and (R) are correct but (R) is not the correct explanation of (A).
(c) (A) is correct but (R) is wrong.
(cl) (A) is wrong but (R) is correct.
Answer:
(a) Both (A) and (R) are correct and (R) is the correct explanation of (A).

Question 9.
Assertion (A): DDT is banned now-a-days.
Reason (R): DDT has a long term toxic effect.
(a) Both assertion and reason are true and reason is the correct explanation of assertion.
(b) Both assertion and reason are true but reason is not the correct explanation of assertion.
(c) Assertion is true but reason is false.
(d) Assertion is false but reason is true.
Answer:
(a) Both assertion and reason are true and reason is the correct explanation of assertion.

VIII. Choose the correct statement.

Question 1.
(a) Halo alkanes have higher boiling point than the parent alkane with same number of carbons because of strong inter molecular forces of attraction.
(b) The boiling point of halo alkanes decreases with the increase of halogen atoms.
(c) The boiling point of mono halo alkanes decreases with the increase in the number of carbon atoms.
(d) Halo alkanes are soluble in water.
Answer:
(a) Halo alkanes have higher boiling point than the parent alkane with same number of carbons because of strong inter molecular forces of attraction.

Question 2.
(a) Halo alkanes are soluble in water.
(b) The boiling point of halo alkanes increase with the increase in the number of halogen atoms.
(c) The melting point of mono halo alkane decrease with the increase in the number of carbon atoms.
(d) The density of alkyl halides are lesser than those of hydrocarbons of comparable molecular weight.
Answer:
(b) The boiling point of halo alkanes increase with the increase in the number of halogen atoms.

Question 3.
(a) Williamson’s synthesis of ether is an example of nucleophilic substitution reaction.
(b) Reaction of methyl bromide with aqueous potassium hydroxide is an example ofelimination reaction.
(c) Reaction of Tertiary butyl bromide with alcoholic KOH is an example of SN2 reaction.
(d) Reaction of Tertiary butyl bromide with alcoholic KOH is an example of E, reaction.
Answer:
(a) Williamson’s synthesis of ether is an example of nucleophilic substitution reaction.

Samacheer Kalvi 11th Chemistry Haloalkanes and Haloarenes 2 Marks Questions and Answers

Question 1.
Write the IUPAC names of –

1. CH₂=CHCI
2. CH₂=CH-CH₂Br.

Answer:

1. CH₂=CHCl : Chloroethene
2. CH₂=CH-CH₂Br : 3-Bromo-prop- 1-ene

Question 2.
Write the structural formula of the following compounds:

1. 2-Chloro-2-Methylpropane
2. 1 -Bromo-2, 2-Dimethylpropane

Answer:
1. 2-Chloro-2-Methylpropane

2. 1 -Brorno-2, 2-dimethyipropane

Question 3.
How many isomers are possible for the formula C₃H₇F? Give their structures and names.
Answer:
C₃H₇F – 2 isomers are possible.
1. CH₃-CH₂-CH₂F – 1- fiuoropropene

Question 4.
Write the isomeric structures and names for the formula C₂H₄Cl₂.
Answer:

 Ethylene dichioride (or) 1, 2-dichioroethane.

Ethylidene chloride (or) 2-dichioroethane.

Question 5.
Draw the structures of –

1. 1-brorno-2, 3-dichiorobutane
2. 2-bromo-3-chloro-2, 4-dimethyl pentane

Answer:
1. 1 -bromo-2, 3-dichiorobutane

2. 2-bromo-3-chloro-2, 4-dimethy 1 pentane

Question 6.
What happens when HI reacts with ten. butyl alcohol?
Answer:

Question 7.
Explain the action of
(i) PCl₅
(ii) PCl₃ with ethanol.
Answer:

Question 8.
How does HBr react with propene?
Answer:
Propene react with HBr and follows Markovnikov’s rule.

Question 9.
How methane reacts with Cl₂ in the presence of light?
Answer:

Question 10.
Explain-Finkelstein reaction.
Answer:
Chioro (or) bromoalkane on heating with sodium iodide in dry acetone gives jodo alkane. This reaction is called Finkeistein reaction.
Answer:

Question 11.
Explain Swans reaction.
Answer:
Chioro (or) bromoalkanes on heating with AgF give fluoroalkanes. This reaction is called Swans reaction.

Question 12.
What happens when silver propionate reacts with Br, in CCl₄?
Answer:
Silver salt of fatty acids (CH₃CH₂COOAg), e.g., silver propionate treated with Br₂ / CCl₄ gives bromoalkane. This reaction is called Hunsdicker reaction.

Question 13.
Haloalkanes have higher boiling point and melting point than the parent alkane. Justify this statement.
Answer:
Haloalkanes have higher boiling point than the parent alkane having the same number of carbon atoms because the intermolecular forces of attraction and dipole-dipole interactions are comparatively stronger in haloalkanes.

Question 14.
CCl₄ > CHCl₃ > CH₂Cl₂ > CH₃Cl is the decreasing order of boiling point in haloalkanes. Give reason.
Answer:
The boiling point of chioro, bromo and iodoalkanes increases with increase in the number of halogen atoms. So the correct decreasing order of boiling point of haloalkanes is:
CCl₄> CHCl₃ > CH₂Cl₂> CH₃Cl.

Question 15.
Arrange the following in increasing order of boiling point. Give reason.
(CH₃CH₂CH₂Cl, CH₃CH₂Cl, CH₃Cl)
Answer:
The correct order of boiling point of haloalkanes is:
CH₃CH₂CH₂Cl > CH₃CH₂Cl > CH₃Cl.
The boiling point of monoalkanes increase with the increase in the number of carbon atoms.

Question 16.
Why haloalkanes arc insoluble in water but soluble in organic solvents?
Answer:
Haloalkanes are polar covalent compounds, soluble in organic solvents, but insoluble in water because they are unable to form hydrogen bond with water molecules.

Question 17.
What happens when bromoethane is treated with moist silver oxide?
Answer:
When bromoethane is treated with moist silver oxide, ethanol will be formed as product:

Question 18.
Complete the following reactions:

Answer:

Question 19.
Explain the action of sodium hydrogen suiphide with bromoethane?
Answer:

Question 20.
Explain Williamson’s synthesis.
Answer:
Williamson’s synthesis:
Haloalkanes when boiled with sodium alkoxide gives the corresponding ether.

Question 21.
Explain the action of alcoholic potash with bromoethane.
Answer:
Elimination reaction takes place when alcoholic potash reacts with bromoethane:

Question 22.
State-Saytzeff’s rule.
Answer:
Some haloalkanes when treated with alcoholic KOH yield a mixture of olefins with different amounts. It is explained by Saytzeff’s rule which states that in a dehydrohalogenation reaction, the preferreed product is that alkene which has more number ofalkyl group attached to the doubly bonded carbon atom.

Question 23.
How will you convert 1-chioropropanc to propene?
Answer:

Question 24.
What is Grignard reagent? How is it prepared from ethyl bromide?
Answer:
When a solution of haloalkane in either is treated with magnesium, we will get alkyl magnesium halide known as Grignard reagent, ethyl magnesium bromide is prepared from ethyl bromide as:

Question 25.
How will you prepare ethyl lithium’?
Answer:
When bromoethane is treated with an active metal like lithium in the presence of dry ether, then ethyl lithium will be formed.

Question 26.
What is tetraethyl Lead? How is it prepared?
Answer:

Question 27.
Convert bromoethane to ethane.
Answer:

Question 28.
How is iodoethane converted to ethane?
Answer:

Question 29.
Starting from CH₃MgI, how will you prepare acetaldehyde?
Answer:

Question 30.
How will you get acetone from methyl manesium iodide?
Answer:

Question 31.
Explain the action of Ethyl chioroformate with Methyl magnesium iodide.
Answer:
Ethyl chioroformate reacts with Grignard reagent to form esters as follows:

Question 32.
Write the IUPAC names of:

Answer:
1 – bromo-4-fluoro-2-iodobenzene
1 – bromo-2-fluoro-4-iodobenzene

Question 33.
How is benzene directly converted to chiorobenzene?
Answer:

Question 34.
Explain Sandmeyer’s reaction.
Answer:
Sandmeyer’s reaction:
The reaction in which benzene-diazonium chloride is converted to chiorobenzene on heating it with Cu₂Cl₂/HCl is known as sand Meyer’s reaction.

Question 35.
Explain Gattcrrnann reaction.
Answer:
Gattermann reaction:
The reaction in which benzene diazonium chloride react with copper in the presence of HCl, chlorobenzene is formed, called Gatterrnann reaction.

Question 36.
How will you prepare iodobenzene?
Answer:

Question 37.
Explain-Balz-Schiernann reaction.
Answer:
Fluorobenzene is prepared by treating benzene diazonium chloride with fluoro boric acid. This reaction produces diazonium fluoroborate which on heating produce fluorobenzene. this reaction is called Balz-Schiernann reaction.

Question 38.
p-dichlorobenzcnc has higher melting point than ortho and meta dichloro benzene. Why?
Answer:
The melting point of p-dichloro benzene is higher than the melting point of the corresponding ortho and mcta isomers. The higher melting which leads to more close packing of its molcules in the crystal lattice and consequently strong intermolecular attractive forces which requires more energy for melting.
p-dichlorobenzene > o-dichlorobenzene > m-dichlorobenzene

Question 39.
Explain Wurtz-fìtting reaction.
Answer:
Wurtz-fitting reaction:
Chlorobenzene and chioroethane are heated with sodium in ether solution to form ethylbenzene. This reaction is called Wurtz-fìtting reaction.

Question 40.
How will you get benzene from chlorobenzene?
Answer:
Chiorobenzene is reduced with Ni-Al Alloy in the presence of NaOH to give benzene.

Question 41.
Explain about the preparation of phenyl magnesium chloride.
Answer:

Question 42.
How will you prepare ethylidcne dichioride from acetylene?
Answer:

Question 43.
What is gern-dihalide? Give one example and explain its preparation.
Answer:
1. If two halogen atoms are attached to one carbon atom of alkyl halide, it is named as gem dihalide. e.g., CH₃-CHCl₂ (ethylidene dichloride.)

2. Preparationof CH₃-CHCl₂:

Question 44.
Explain the action of zinc and HCl on chloroform.
Answer:

Question 45.
How does nickel react with chloroform?
Answer:

Question 46.
Convert methane to methylene chloride.
Answer:

Question 47.
Explain Carbylamine reaction. (or) Give the characteristic test for primary amine.
Answer:
Chlorofonn reacts with aliphatic or aromatic primary amines and alcoholic caustic potash to give foul smelling alkyl isocyanide (carbylamine).

Question 48.
How will you prepare carbon tetrachioride?
Answer:
The reaction of methane with excess of chlorine in the presence of sunlight give carbon tetrachioride as major product.

Question 49.
How will you convert carbon tetrachioride to chloroform?
Answer:

Question 50.
What are freons? How are they named? Give two examples.
Answer:
(i) The chioro-fluoro derivatives of methane and ethane are called freons.
(ii) Freon is represented as freon-cba
c = number of C atoms –
b = number of H atoms + 1
a = total number ofF atoms

Question 51.
What are ambident nucleophiles’? Explain with an example.
Answer:
Nucleophiles which can attack through two different sites are called ambident nucicophiles. For example, cyanide group is a resonance hybrid of two contributing structures and therefore it can act as a nucleophile in two different ways:

It can attack through carbon to form cyanides and through nitrogen to form isocyanides or carbylamines.

Question 52.
Which compound in each of the following pairs will react faster in S_N² reaction with OH ion?
1. CH₃Br or CH₃I
2. (CH₃)₃CCl or CH₃Cl
Answer:
1. Since I^– ion a better leaving group than Br^– ion, therefore CH₃I will react faster than CH₃Br in S_N² reaction with OH^– ion.
2. On steric grounds 1° alkyl halides are more reactive than tert alkyl halide in S_N² reactions. Therefore, CH₃Cl will react at a faster rate than (CH₃)₃CCl in a S_N² reaction with OH^– ion.

Question 53.
The treatment ofalkyl chlorides with aqueous KOH solution leads to the formation of alcohols but in the presence of alcoholic KOH solution, alkenes are the major product. Explain.
Answer:
In aqueous solution, KOH is almost completely ioniscd to give OH^– ions which being a strong nucleophile brings about a substitution reaction of alkyl halides to form alcohols. In aqueous solution, OH^– ions are highly hydrated. This solvation reduces the basic character of OH^– ions which therefore, abstract fails to abstract a hydrogen atom from the n-carbon of the alkyl chloride to form an alkene. In contrast, an alcoholic solution of KOH contains alkoxide (RO^–) ions which being a much stronger base than OH^– ions preferentially eliminates a molecule of HCl from an alkyl chloride to form alkenes.

Question 54.
Give one example of each of the following reactions:
1. Wurtz Reaction
2. Wurtz – Fittig reaction
Answer:
1. Wurtz Reaction:
It involves conversion of alkyl halides into alkane.
Example:

2. Wurtz-Fittig reaction:
It involves the reaction of an aryl halide and alkyl halide to form the corresponding hydrocarbon.

Question 55.
How will you distinguish between the following pair of compounds:
1. Chloroform and carbon tetrachioride,
2. Benzyl alcohol and chiorobenzene.
Answer:
1. On heating chloroform and carbon tetrachloride with aniline with ethanoic acid and potassium hydroxide separately, chloroform forms a pungent smelling isocyanide compound but carbon tetrachloride does not form this compound.

2. On adding sodium hydroxide and silver nitrate to both the compounds, benzyl chloride forms a white precipitate but chiorobenzene does not form any white precipitate.

Samacheer Kalvi 11th Chemistry Haloalkanes and Haloarenes 3 Marks Questions And Answers

Question 1.
Give one example with structure and name for each of the following compounds.
(a) Primary haloalkane
(b) Secondary haloalkane
(c) Tertiary haloalkane
Answer:
(a) Primary haloalkane

(b) Secondary haloalkane

(c) Tertiary haloalkane

Question 2.
Write the possible isomers for the formula C₄H₉Cl with structures and names.
Answer:
C₄H₉Cl (4 isomers):

Question 3.

Answer:
For a compound with molecular formula C₄H₉Cl – 3 isomers (1°, 2°, 3°) arc possible. Among the isomeric alkyl halides, the boiling point decreases with the increase in branching in the alkyl chain. This is because with increase in branching, the molecule attains spherical shape with less surface area. As a result, the intermolecular forces become weak, resulting in lower boiling point. Therefore the boiling point decreases in the order:

Hence, (CH₃)₃ C-Cl has the lowest boiling point.

Question 4.
Explain ammonolysis of haloalkanes. (or) How excess of haloalkanc react with alcoholic ammonia?
Answer:
With excess of haloalkanes, ammonia react to give primary, secondary, tertiary amines along with quarternary ammonium salt:

Question 5.
Explain how bromoethane reacts with –
1. alcoholic KCN
2. alcoholic AgCN
Answer:
1. Bromoethanc reacts with alcoholic KCN to form ethyl cyanide.

2. Bromoethane reacts with alcoholic AgCN to form alkyl isocyanide.

Question 6.
Explain the hydrolysis of 2-bromobutane with aqueous KOB.
Answer:

• 2-bromobutane is optically active and it undergoes S_N¹ reaction with aqueous KOH.
• The product obtained will be an optically inactive racemic mixture.
• As nucleophilic reagent OH^– ion can attack the carbocation from both sides to form equal proportions of dextro and levo rotatory optically active isomers, it results in the formation of optically inactive racemic mixture.

Question 7.
Explain the action of alcoholic KOH with 2-bromobutanc.
Answer:

According to Saytzcff’s rule, when 2-bromobutanc reacts with alcoholic KOK, yields a mixture of olefins in different amounts.

(ii) In a dehydrohalogenation reaction, the preferred product is that alkene which has more number of alkyl groups attached to the doubly bonded carbon alkene.

Question 8.
Mention the uses of chloroform.
Answer:

• Chloroform is used as a solvent in pharmaceutical industry.
• It is used for producing pesticides and drugs.
• It is used as an anaesthetic.

Question 9.
What are the uses of carbon tetrachioride?
Answer:

• Carbon tetrachioride in used as a dry cleaning agent.
• It is used as a solvent for oils, fats and waxes.
• As the vapours of CCl₄ is non-combustible, it is used under the name pyrene for extinguishing the fire caused by oil (or) petrol.

Question 10.
What are ogranometallic compounds? Give one example. Explain the nature of carbon-metal bond.
Answer:

• Organometallic compounds are organic compounds in which there ¡s a direct carbon-metal bond.
• Example :  CH₃MgI. methyl magnesium iodide.
• The carbon-magnesium bond in Grignard reagent is covalent but highly polar. The carbon atom is more electronegative than magnesium. Hence, the carbon atom has partial negative charge and magnesium atom has a partial positive charge.
  

Question 11.
How would you prepare acetic acid from methyl magnesium iodide?
Answer:
Solid carbon dioxide reacts with grignard reagent to form addition product which on hydrolysis yields acetic acid.

Question 12.
Explain the nature of C-X bond in haloarenes and its resonance structure.
Answer:

1. In haloarenes, the carbon atom is sp² hybridised. The sp² hybridised orbitais arc shorter and holds the electron pair bond more.
2. Halogen atom contains p-orbital with lone pair of electrons which interacts with π – orbitaIs of berizene ring to form extended conjugated system of π – orbitals.
3. The delocalisation of these electrons give double bond character to C-X bond. The resonance structure of halobenzene are given as:
   

Question 13.
Compare the bond length C-X in haloarenes and C-X in haloalkanes.
Answer:
1. Due to this double bond character of C-X bond in haloarenes, the C-X bond length is shorter length and stronger than in haloalkanes.
2. Example:

Question 14.
What are the uses of chlorobenzene?
Answer:

• Chiorobenzene is used in the manufacture of pesticides like DDT.
• It is used as high boiling solvent in organic synthesis.
• It is used as a fibre – swelling agent in textile processing.

Question 15.
What are polyhalogen compounds? Give its types with example.
Answer:

1. Carbon compounds containing more than one halogen atom are called polyhalogen compounds.
2. They are classified as
   • gem dihalides
   • Vicinal dihalides.

(a) Gem dihalide:
In this compound, two halogen atoms are attached to one carbon atom.
e.g, CH₃CHCl₂ – ethylidene chloride.

(b) Vicinal dihalide:
In this compound, two halogen atoms are attached to two adjacent carbon atoms.
1,2-dichloroethane

Question 16.
Give two examples for –
1.  gem dihalide
2. vicinal dihalide.
Answer:
1. Gem dihalides:

2. Vicinal dihalides:

Question 17.
Explain two methods of preparation of ethylene diehioride.
Answer:
1. Addition of Cl₂ to Ethylene:

2. Action of PCl₅ on Ethylene glycol:

Question 18.
How would you distinguish gern-dihalides and iinal dihalides?
Answer:
(i) Gem-dihalides on hydrolysis with aqueous KOH gives an aldehyde or a ketone whereas vicinal dihalides on hydrolysis with aqueous KOH give glycols.

The above reaction can be used to distinguish between gem-dihalides and vicinal dihalides.

Question 19.
Explain the action of metallic zinc with –
(i) Ethylidene dichioride
(ii) Ethylene dichloridc.
Answer:

Question 20.
What happens when alcoholic KOH is treated with
(i) Ethylidene dichioride
(ii) Ethylene dich I onde?
Answer:
(i) Gem-dihalides and vicinal dihalides both on treatment with alcoholic KOH give alkynes.

Question 21.
What are the uses of methylene chloride?
Answer:
Methylene chloride is used as:

• Aerosol spray propellant.
• Solvent in paint remover.
• Process solvent in the manufacture of drugs.
• A metal cleaning agent.

Question 22.
Explain the laboratory preparation of chloroform.
Answer:
Haloform reaction:
Chloroform is prepared in the laboratory by the reaction between ethyl alcohol with bleaching powder followed by the distillation of the final product chloroform. Bleaching powder acts as a source of chlorine and calcium hydroxide. The reaction take place in 3 steps.
Step 1:
Oxidation-

Step 2:
Chlorination-

Step 3:
Hydrolysis-

Question 23.
What is chloropicrin? How is it obtained? Mention its uses.
Answer:
1. Chioropicrin is CCl₃NO₂.

2. Chloroform reacts with nitric acid to form chioropicrin (trichioro-nitromethane):

3. Chioropicrin is used as an insecticide and soil sterilising agent.

Question 24.
What are the uses of freons?
Answer:

• Freons are used as refrigerants in refrigerators and air conditioners.
• It is used as a propellant for foams and aerosols.
• It is used as a propellant for foams to spray out deodorants, shaving creams and insecticides.

Question 25.
What is DDT? How is it prepared’?
Answer:
(i) DDT is p, p’-dichloro-diphenyi-trichloroethane
(ii) DDT can be prepared by heating a mixture of chloro benzene with chlorai in the presence of conc. H₂SO₄

Question 26.
Mention the uses of DDT.
Answer:

• DDT is used lo control certain insects which carries diseases like malaria and yellow fever.
• It is used in farms to control some agricultural pests.
• It is used in building construction as pest control agent.
• It is used to kill certain kind of insects like housefly and mosquitoes due to its high and specific toxicity.

Question 27.
Write the equations for the preparation of I-iodobutane from:
(i) 1 -butanol
(ii) 1 -chiorobutane
(iii) but- 1-ene
Answer:

Question 28.
Explain why:

1. the dipole moment of chiorobenzene is lower than that of cyclohexyl chloride?
2. alkyl halides though polar, are immiscible with water?
3. Gringard reagents should be prepared under anhydrous conditions?

Answer:

1. Chlorobenzene is stabilised by resonance and there is negative charge on ‘Cl’ in 3 out of 5 resonating structures, therefore it has a lower dipole moment than cyclohexyl chloride in which there is no such negative charge.
2. Alkyl halides cannot form H-bond with water and cannot break H-bonds between water molecules, therefore they are insoluble in water.
3. Grignard reagents react with H₂O to form alkanes, therefore they are prepared under anhydrous conditions.

Question 29.
Explain as to why haloarenes arc much less reactive than haloalkanes towards nucleophilic substitution reactions?
Answer:
Haloarenes are much less reactive than haloalkanes towards nucleophilic substitution reactions due to the following reasons:

1. Resonance effect:
In haloarenes the electron pair on the halogen atom is in conjugation with the π – electrons of the ring and the following resonating structures are possible. C-Cl bond acquires a partial double bond character due to resonance. As a result, the bond cleavage in haloarenes is difficult than in case of haloalkanes and therefore they are less reactive towards nucleophilic substitution reactions.

2. The C-Cl bond length in haloalkanes is 177 pm while in haloarenes it is 169 pm. Since it is difficult to break shorter bond than a longer bond. Therefore, haloarenes are less reactive than haloalkanes towards nucleophilic substitution reactions.

Question 30.
Do the following conversions:
(i) Methyl bromide to acetone
(ii) Benzyl alcohol to 2-phenylacetic acid
Answer:

Question 31.
Give reasons for the following:

1. Ethyl iodide undergoes S_N² reactions faster than ethyl bromide.
2. (±) 2-Butanol is optically inactive.
3. C-X bond length in halobenzene is smaller than C-X bond length in CH₃-X.

Answer:

1. Because in ethyl iodide, iodide being the best leaving group among all the halide ions. Rate of S_N² reaction ability of leaving group.
2. (±) 2-hutanol is a racemic mixture which is optically inactive due to the external compensation.
3. Due to resonance in halobenzene, it has a smaller bond length value as compared to CH₃-X.

Question 32.
Write the structure of diphenyl. How is it prepared from chlorobcnzene?

Answer:

Samacheer Kalvi 11th Chemistry Haloalkanes and Haloarenes 5 Marks Question And Answers

Question 1.
Explain the classification of organic compounds with example.
Answer:

Question 2.
Explain S_N² mechanism with suitable examples.
Answer:

• S_N² reaction means bimolecular nucleophilic substitution reaction of second order.
• The rate of S_N² reaction depends upon the concentration of both alkyl halides and the nucleophile.
• This reaction involves the formation of a transition state in which both the reactant molecules are partially bonded to each other. The attack of nucleophile occurs from the back side.
• The carbon at which substitution occurs has inverted configuration during the course of reaction just as an umbrella has the tendency to invert in a wind-storm. This inversion of configuration is called walden inversion.
  

Question 3.
S_N² reaction of an optically active haloalkane is accompanied by inversion of configuration at the asymmetric centre. Prove it.
Answer:
1. S_N² reaction of an optically active haloalkane (e.g., 2-bromo-octane) is accompanied by inversion of configuration at the asymmetric center.

2. When 2-bromooctane is heated with sodium hydroxide, 2-octanol is formed with inversion of configuration. (-) 2-bromo-octane on heating with sodium hydroxide gives (+) 2-octanol is formed in which -OH group occupies a position opposite to what bromine had occupied in the optically active haloalkane.

Question 4.
Explain E₂ reaction mechanism with a suitable example.
Answer:

• E₂ reaction stands bimolecular elimination reaction of second order. The rate of E₂ reaction depends on the concentration of alkyl halide and the base.
• Primary alkyl halide undergoes this reaction in the presence of alcoholic KOH.
• It is a one step process in which the abstraction of the proton from the 3 carbon atom and expulsion of halide from the x carbon atom occur simultaneously.
• The mechanism that follows is shown below:
  

Question 5.
Describe E₁ reaction mechanism with a suitable example.
Answer:
1. E₁ stands for unimolecular elimination reaction of first order.
2. Tertiary alkyl halides undergoes elimination reaction in the presence of alcoholic KOH.
3. It takes place in two steps.
4.  Step 1.
Heterolytic fission to yield a carbocation:

5. Step 2.
Elimination of a proton from the fl-carbon to produce an alkene:

Question 6.
Starting from methyl magnesium iodide how would you prepare –

1. Ethanol
2. 2-propanol
3. Tert-butyl alcohol.

Answer:
1. Ethanol:
Formaldehyde reacts with CH₃MgI to give an addition product which on hydrolysis yields ethanol.

2. 2-Propanol :
Acetaldehyde react with CH₃MgI to give an addition product which on hydrolysis yields 2-propanol.

Question 7.
Starting from methyl magnesium iodide, how would you prepare –

1. Ethyl methyl ether
2. methyl cyanide
3. methane.

Answer:
1. Ethyl methyl ether:
Lower halogenated ether reacts with grignard reagent to form higher ether.

2. Methyl cyanide:
Grignard reagent reacts 4rith cyanogen chloride to form alkyl cyanide.

3.  Methane:
Grignard reagent reacts with waler to give methane as product.

Question 8.
Describe electrophilic substitution reaction of chiorohenzene with equations.
Answer:

Question 9.
An organic compound Ⓐ of molecular formula C₃H₆ react with HBr in the presence of peroxide to give C₃H₇Br as Ⓑ Ⓑ on reaction with aqueous KOH gives © with molecular formula C₃H₈O. Identify Ⓐ Ⓑ and ©
Answer:
1. An organic compound Ⓐ of molecular formula C₃H₆ is CH₃-CH=CH₂, Propene

2. Propene Ⓐ reacts ith HRr in the presence of peroxide to give 1-bromopropane

3. 1-Bromopropane on reaction with aqueous KOH undergoes hydrolysis to give  Propan- 1-ol. CH₃-CH₂-CH₂OH as ©

Question 10.
An organic compound Ⓐ of molecular formula C₂H₆O reacts with thionyl chloride in the presence of pyridine gives Ⓑ C₂H₅Cl. Ⓑ on reaction with alcoholic KOH gives Ⓒ, C₂H₄. Ⓒ on treatmeni with Cl₂ gives C₂H₄Cl₂ as Ⓓ. Identify Ⓐ Ⓑ Ⓒ Ⓓ and explain the reaction.
Answer:
1. An organic compound Ⓐ of molecular formula C₂H₆O is Ethanol CH₃-CH₂OH.

2. Ethanol reacts with thionyl chloride in the presence of pyridine to give CH₃-CH₂Cl Ⓑ as product.

3. Ethyl chloride on treatment with alcoholic KOH, undergoes dehydrohalogenation to give C₂H₄, ethylene Ⓒ as the product.

4. Ethylene on reaction with Cl₂ yield ethylene dichioride as Ⓓ as the product.

Question 11.
The simplest aromatic hydrocarbon C₆H₆ reacts Ⓑ on treatment with sodium hydroxide will (C₆H₅OH), Phenol, Ⓒ as the product. Also Cl₂ to give Ⓐ which on reaction with sodium hydroxide gives Ⓑ.Ⓑ of molecular formula C₆H₆O. Ⓑ on treatment with ammonia will give C₆H₇N as D. Identify Ⓐ, Ⓑ,Ⓒ, and explain the reactions involved.
Answer:
1. The aromatic hydrocarbon Ⓑ is Benzene, C₆H₆.

2. Benzene reacts with Cl, to give chiorobenzene (C₆H₅Cl), as Ⓑ as the product.

3. Chiorobenzene Ⓑ reacts with sodium hydroxide to give (C₆H₅OH) phenol, Ⓒ as the product.

4. Chiorobenzene reacts with ammonia to give Aniline, C₆H₅NH₂ Ⓓ as the product.
Question 12.
An organic compound Ⓐ of molecular formula C₂H₂ reacts with HCl to give C₂H₄Cl₂ as Ⓑ. Ⓑ on reaction with aqueous KOH will give C₂H₄O as Ⓒ Identify Ⓐ, Ⓑ,Ⓒ and explain the reactions involved.
Answer:
1. An organic compound Ⓐ of molecular formula C₂H₂ is CH ≡ CH (Acetylene)

2. Acetylene Ⓐ reacts with HCl to give ethylene dichioride Ⓑ as the product.

3. Ethylidene dichioride on reaction with aqueous KOH will give acetaldehyde, © as the product.

Question 13.
Two isomers of formula C₄H₉Br are Ⓐ and B. Ⓐ on reaction with alcoholic KOH gives of molecular formula C₈H₈ by E₁ reaction. Ⓑ on reaction with alcoholic KOH gives Ⓓ and Ⓔ as products by Saytzefì’s rule. Idcnti’ A, B, C, D, E.
Answer:
1. C₄H₉Br: Two isomers may be there: Tert butyl bromide and

2. Ⓐ on reaction with alcoholic KOH gives iso butylene © as the product.

3. 2-bromobutane Ⓑ on reaction with alcoholic KOH follows Saytzeff s rule to give a mixture of olefins in different amounts.

Question 14.
A simple aromatic hydrocarbon Ⓐ reacts with Cl₂ to give Ⓑ of molecular formula C₆H₅Cl. Ⓑ on reaction with ethyl chloride along with sodium metal gives © of formula C₈H₁₀. © alone reacts with Na metal in the presence of ether to give Ⓓ C₁₂H₁₀. Identify Ⓐ Ⓑ Ⓒ & Ⓓ
Answer:
1. The simple aromatic hydrocarbon is C₆H₆. Benzene (A)

2. Benzene reacts with Cl₂ to give chiorobenzene.

3. Chiorobenzene reacts with ethyl chloride to form ethyl benzene, © as the product

4. Chiorobenzene on reaction Na metal gives Biphenyl compound © as the product.

Question 15.
An organic compound Ⓐ of molecular formula CH₂O reacts with methyl magnesium iodide followed by acid hydrolysis to give Ⓑ of moLecular formula C₂H₆O. Ⓑ on reaction with PCl₅ gives Ⓑ. © on reaction with alcoholic KOK gives Ⓓ an alkene as the product. Identif’ Ⓐ Ⓑ Ⓒ Ⓓ and explain the reactions involved.
Answer:
1. Ⓐ of molecular formula CH₂O is identified as HCHO, formaldehyde.

2. Formaldehyde reacts with CH₃MgI followed by hydrolysis to give ethanol, CH₃-CH₂OH B as the product.

3. Ethanol Ⓑ reacts with PCl₅ to give C₂H₅Cl, Ethyl chloride © as the product.

4. CH₃-CH₂Cl © on reaction with alcoholic KOH undergoes dehydrohalogenation to give ethylene CH₂=CH₃ Ⓓ as the product.

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