Srinivas

Regular practice with AP Inter 1st Year Zoology Study Material Chapter 1 The Living World Questions and Answers helps students stay prepared for examinations.

AP Inter 1st Year Zoology 1st Lesson The Living World Questions and Answers

IV. Very Short Answer Questions

Question 1.
What does ICZN stand for ?
Answer:
International Code of Zoological Nomenclature.

Question 2.
Define taxon. Give two examples of taxa at different hierarchical levels.
Answer:

• A taxon is a level of hierarchy in the system of classifying organisms.
• The two examples of taxa are – The basic level of classification is species, and the highest level of classification is known as kingdom.

Ex: Mango
Kingdom – Plantae
Phylum or Division – Angiospermae
Class – Dicotyledonae
Order – Sapindales
Family – Anacardiaceae
Genus – Mangifera
Species – indica

Ex: Man
Kingdom – Animalia
Phylum or Division – Chordata
Class – Mammalia
Order – primata
Family – Homonidae
Genus – Homo
Spècies – sapiens.

Question 3.
What is biodiversity?
Answer:

• The variety of life of all living organisms, their genetic makeup, and the ecosystems they inhabit is biodiversity.
• The number of species that are known and described ranges between 1.7 to 1.8 million.

Question 4.
What are the basic processes in taxonomy ?
Answer:
Characterisation, Identification, Classification and nomenclature are the processes in taxonomy.

Question 5.
What is nomenclature in taxonomy and which process precedes it ?
Answer:
The process of naming of animals with a distinctive (scientific) names is called nomenclature. The process that precedes nomenclature is characterization, identification, classifiction.

Question 6.
Define the word systematics. What is the title of Linnaeus’s publication ?
Answer:

• The word systematic is derived from the Latin word ‘Systema’ means systematic arrangement of organisms.
• The title of Linnaeus’s publication is Systema Naturae.

Question 7.
What is binomial nomenclature? Give an example.
Answer:
Binomial nomenclature is a system of scientific naming in which the organism has two components. The first name represents the genus, and the second part represents the specific epithet.
Ex: Mango – The scientific name is written as Mangifera indica.

V. Short Answer Questions

Question 1.
Define the following terms.
a) Phylum
b) Class
c) Family
d) Genus
Answer:
(a) Phylum: It includes one or more classes.

• Classes comprising animals like fishes, amphibians, reptiles, birds along with mammals.
• These are based on common features like presence of notochord and a dorsal hollow neural system which are included in phylum chordata.
• In case of plants classes with a few similar characters are assigned to a higher category called Division.

(b) Class: It includes one or more related orders.

• For example the order primata comprising monkeys, gorillas and gibbons is placed in class of mammalia and along with the order Carnivora that includes animals like tigers, cats and dogs.

(c) Family: It includes one or more genera.

• Families are characterized on the basis of both vegetative and reproductive features of plant species.
• For example – Among plants three different genera Solanum, Petunia and Datura are placed in the family Solanaceae.
• Among animals Genus Panthera comprising lions, tigers, leopards is put along with genus Felis (cats) in the family Felidae.

(d) Genus:

• It is a group of related species, resembling one another in certain characters.
• Example: Potato and Brinjal are two different species but both belong to the genus Solanum.
• Lion (Panthera leo), leopard (Panthera pardus) and tiger (P.tigris) with several common features are all species of the genus Panthera.

Question 2.
Illustrate the taxonomic hierarchy with an animal example.
Answer:
The table below is the taxonomic hierarchy with humans as an example of an animal.

Question 3.
What is binomial nomenclature? Write the universal rules of nomenclature.
Answer:

• In binomial nomenclature biologist follow universally accepted principles to provide scientific names to known organisms.
• Each name has two components – the generic name and the specific epithet.
• This naming system given by carolus Linnaeus is being practised by biologists all over the world.
• This naming system using a two word format was found convenient.
• The scientific name of mango is written as Mangifera indicAnswer: In this name Mangifera represents the genus, while indica is species.
• Name of the author appears after the specific epithet i.e., at the end of the biological name and is written in an abbreviated form.
  Ex : Mangifera indica Linn. It indicates that this species was first described by Linnaeus.

The universal rules of nomenclature are –

• Biological names are generally in Latin and written in italics. They are Latinised or derived from Latin irrespective of their origin.
• The first word in a biological name represents the genus while the second component denotes the specific epithet.
• Both the words in a biological name, when handwritten, are separately underlined or printed in italics to indicate their Latin origin.
• The first word denoting the genus starts with a capital letter while the specific epithet starts with a small letter.

Question 4.
Write short notes on Classification. Taxonomy and Systematics.
Answer:
Classification: Classification is the process by which anything is grouped into convenient categories based on some early observable characters.

• For example, we easily recognise groups such as plants or animals or dogs, cats or insects.
• The moment we use any of these terms we associate certain characters with the organism in that group.

Taxonomy:
Based on characteristics, all living organisms can be classified into different taxa. This process of classification is taxonomy.

• Characterization, Identification, Classification and Nomenclature are the processes that are basic to taxonomy.
• External and internal structure along with the structure of cell, development process and ecological information of organisms are essential and form the basis of modern taxonomic studies.
• The science of classifying organisms it involves describing, naming, and grouping living things (plants, animals, and microorganisms) based on their shared characteristics and evolutionary relationships.

Systematics: The systematic arrangement of organisms.

• The word systematic is derived from the Latin word ‘Systema’ which means systematic arrangement of organisms.
• Linnaeus used ‘Systema naturae’ as the title of his publications.
• The scope of systematic was later enlarged to include Identification, Nomenclature and Classification.
• Systematics takes into account evolutionary relationships between organisms.

I. Multiple Choice Questions

Question 1.
What is the first process in taxonomy ?
1. Identification
2. Nomenclature
3. Classification
4. Characterization
Answer:
4. Characterization

Question 2.
The second word in Binomial nomenclature represents ___________
1. genus
2. family
3. species
4. class
Answer:
3. species

Question 3.
Select the correctly written scientific name of the human being.
1. Homo species
2. Homo sapiens
3. Homo sapians
4. Homo genus family
Answer:
2. Homo sapiens

Question 4.
In a taxonomic hierarchy family is placed between
1. class and kingdom
2. order and class
3. genus and order
4. genus and class
Answer:
3. genus and order

Question 5.
The lowest taxonomic category is
1. genus
2. family
3. species
4. taxon
Answer:
3. species

Question 6.
Which of the following is not a taxon in Linnaeus hierarchy ?
1. Class
2. Kingdom
3. Population
4. Family
Answer:
3. Population

Question 7.
Which of the following belongs to the family Muscidae ?
1. Housefly
2. Grasshopper
3. Firefly
4. Cockroach
Answer:
1. Housefly

Question 8.
Nomenclature is governed by the rules of ICZN. Which of the following is contrary to the rules of nomenclature ?
1. Handwritten scientific names should be underlined.
2. Every organism should have a generic name and a specific epithet.
3. Scientific names are in Latin and should be printed in italics.
4. Both genus and species names start with a capital letter.
Answer:
4. Both genus and species names start with a capital letter.

Question 9.
Can you identify the correct sequence of taxonomical categories?
1. Species – Order – Genus – Kingdom
2. Genus – Species – Order – Kingdom
3. Species – Genus – Order – Phylum
4. Genus – Family – Class – Order
Answer:
3. Species – Genus – Order – Phylum

Question 10.
Biological names are printed in
1. Bold letters
2. Capital letters
3. Italics
4. Small letters
Answer:
3. Italics

II. Fill in the Blanks

Question 1.
The book “Systema Naturae” was written by ___________.
Answer:
Linnaeus

Question 2.
The system of providing a scientific name with two components is called ___________.
Answer:
Binomial nomenclatur

Question 3.
The first word in binomial nomenclature represents ___________.
Answer:
Genus

Question 4.
In binomial nomenclature, the word denoting the genus starts with a .
Answer:
Capital letter

Question 5.
The process of classification is called ___________.
Answer:
Taxonomy

Question 6.
Biological names are generally derived from ___________.
Answer:
Latin

Question 7.
The group of individual organisms with fundamental similarities is called ___________.
Answer:
Species

Question 8.
The scientific name of lion is ___________.
Answer:
Panthera leo

III. One Word Answer Questions

Question 1.
Which taxonomic category comprises a group of related species ?
Answer:
Genus

Question 2.
Who proposed binomial nomenclature ?
Answer:
Linnaeus

Question 3.
Who is known as ‘the Darwin of the 20th century’ ?
Answer:
Ernst Mayr

Question 4.
Write the generic name of tiger.
Answer:
Panthera

Question 5.
What is the scientific term used to describe the categories in biological classification?
Answer:
Taxa

Question 6.
What is the highest category in taxonomic hierarchy?
Answer:
Kingdom

Question 7.
Name the Immediate higher category of “family” in the taxonomic hierarchy.
Answer:
Order

Question 8.
Which branch of biology deals with the different kinds of organisms, their diversities, and the relationships among them ?
Answer:
Systematics

Question 9.
Which taxonomic category includes related orders ?
Answer:
Class

Regular practice with AP Inter 1st Year Botany Study Material Chapter 10 Plant Growth and Development Questions and Answers helps students stay prepared for examinations.

AP Inter 1st Year Botany 10th Lesson Plant Growth and Development Questions and Answers

IV. Very Short Answer Questions

Question 1.
Define plasticity? Give an example.
Answer:
Ability of Plants to follow different pathways in response to the environment or phases of life to form different kinds of structures is called plasticity.
Ex: Heterophylly in Cotton, Coriander and Larkspur

Question 2.
What is the disease that formed the basis of the identification of gibberellins in plants? Name the causative fungus of this disease.
Answer:
The “BAKANE” or foolish seedling disease of rice seedlings. It is caused by a fungal pathogen Gibberella fujikuroi.

Question 3.
What is apical dominance? Name the growth hormone that causes it. [March-26]
Answer:

• The growing apical bud inhibits the growth of lateral or axillary buds is called apical dominance.
• It is caused by Auxins.

Question 4.
What is meant by bolting? Which hormone causes bolting?
Answer:

• The sudden elongation of internodes just prior to flowering is called bolting.
• Gibberellins are responsible for bolting.

Question 5.
Define respiratory climactic? Name the PGR associated with it.
Answer:

• The rise in the rate of respiration during ripening of fruits is called respiratory climactic.
• Ethylene enhances the respiration rate during ripening of fruits.

Question 6.
What is ethephon? Write its role in agricultural practices.
Answer:

• The most widely used compound as a source of ethylene is ethephon.
• Ethephon releases ethylene slowly.
• Ethephon hastens fruit ripening in tomatoes and apples and accelerates abscission in flowers and fruits.
• It promotes female flowers in cucumbers, thereby increasing the yield.

Question 7.
Why is abscisic acid also known as stress hormone ?
Answer:

• Abscisic acid (ABA) is a stress hormone.
• ABA stimulates the closure of stomata in the epidermis and increases the tolerance to plants for various kinds of stresses.
• Therefore, It is also called a stress hormone.

Question 8.
Define growth and development.
Answer:

• Growth is an irreversible, permanent increase in size of an organism or its parts or even of an individual cell.
• The changes that an organism goes through during its life cycle is called development.

Question 9.
Name the phases observed in sigmoid growth curve.
Answer:
Lag phase, Log phase and Stationary phase.

Question 10.
What are the very essential conditions required for growth ?
Answer:
Water, oxygen and nutrients.

V. Short Answer Questions

Question 1.
Write a note on agricultural and horticultural applications of auxins.
Answer:

• IBA, NAA and IAA help to initiate rooting in stem cuttings, an application widely used for plant propagation in horticulture.
• Auxins promote flowering in Pineapple.
• Auxins help to prevent fruit and leaf drop at early stages but promote the abscission of older mature leaves and fruits.
• Auxins are responsible for apical dominance and inhibit growth of lateral buds.
• Auxins also induce parthenocarpy in Tomatoes.
• Auxins like 2, 4-D are widely used as herbicides, which kill broad leafed dicot weeds.
• Auxins also control xylem differentiation and help in cell division.

Question 2.
Write the physiological responses of the gibberellins in plants.
Answer:

• Gibberellins have the ability to cause an increase in the length of axis or peduncle (or) pedicels in grapes.
• Gibberellins cause fruits like apples to elongate and improve their shape.
• Gibberellins delay senescence thus the fruits can be left on the tree longer so as to extend the market period.
• GA3 is used to speed up the malting process in the brewing industry.
• Spraying sugarcane crops with Gibberellins increases the length of the stem, thus increasing the yield by as much as 20 tonnes per acre.
• Spraying juvenile conifers with Gibberellins speeds up the maturity period, thus leading to early seed production.
• Gibberellins promote bolting in beet, cabbages and many plants with rosette habit.
• They also promote parthenocarpic fruits in grapes and tomatoes.

Question 3.
Write any four physiological effects of cytokinins in plants.
Answer:

• Cytokinins induce cell division.
• Cytokinins help to produce new leaves, chloroplasts in leaves.
• Cytokinins help in lateral shoot growth and adventitious shoot formation.
• Cytokinins help to overcome apical dominance.
• They promote nutrient mobilization which helps in the delay of leaf senescence.
• Cytokinins help in the opening of stomata by increasing the concentration of K⁺ ions in guard cells.

Question 4.
What are the physiological processes that are regulated by ethylene in plants?
Answer:

• Ethylene influences Horizontal growth of seedlings, swelling of the axis and apical hook formation in dicot seedlings.
• Ethylene promotes senescence and abscission of plant organs, especially of leaves and flowers.
• Ethylene is highly effective in fruit ripening.
• Ethylene enhances the respiration rate during ripening of fruits. This phenomenon is called respiratory climatic.
• Ethylene breaks seed and bud dormancy, initiates germination of peanut seeds and sprouting of potato tubers.
• Ethylene promotes rapid inter node or petiole elongation in deep water rice plants to make the leaves or upper parts of the shoot to remain above water.
• Ethylene promotes root growth and root hair formation, thus helping plants to increase their absorption surface.
• Ethylene is used to initiate flowering and for synchronizing fruit set in pineapples. It also induces flowering in mango.
• It promotes female flowers in cucumbers, thereby increasing the yield.

VI. Long Answer Questions

Question 1.
Define growth, differentiation, development, dedifferentiation, redifferentiation, determinate growth, meristem and growth rate.
Answer:

• Growth : It is an irreversible, permanent increase in size of an organism or its parts or even of an individual cell.
• Differentiation : The process by which cells become specialized in structure and function.
• Development: The changes in an organism undergoes from a single cell to maturity.
• Dedifferentiation: The process where mature cells reverse their state of differentiation and become undifferentiated.
• Redifferentiation : The process where dedifferentiated cells lose their ability to divide and become mature to perform a specific function.
• Determinate growth: Growth that stops after a certain point, meaning the organism or organ stops growing after reaching a certain size.
• Meristem : A region of actively dividing cells in plants, responsible for new growth.
• Growth rate : The increase in growth per unit time.

Question 2.
Describe briefly
a) Arithmetic growth,
b) Geometric growth,
c) Sigmoid growth curve,
d) Absolute and relative growth rates.
Answer:
a) Arithmetic Growth:

• In arithmetic growth, following cell division (mitosis), only one of the daughter cells continues to divide, while the other differentiates and matures.
• Rate of Growth : The growth rate is constant, meaning the increase in size or number is the same over equal time intervals.
  Ex: The elongation of roots at a constant rate is an example of arithmetic growth.
• Graph : A graph of length against time in arithmetic growth results in a straight line (linear curve).

b) Geometric Growth:

• In geometric growth, the daughter cells retain the ability to divide, but the rate of growth slows down due to factors like limited resources.
• Rate of Growth : The initial growth is slow (lag phase), followed by a period of rapid increase (log/exponential phase), and then a slowing phase.
  Ex: The growth of all cells, tissues, and organs generally follows a geometric pattern.
• Graph : A graph of size against time in geometric growth shows a S-shaped curve (sigmoid curve).

c) Sigmoid growth curve :

• If a graph is plotted for geometrical growth, it gives a typical sigmoid or S- curve.
• A Sigmoid curve is a characteristic of living organisms growing in a natural environ¬. ment.
• It consists of three phases, namely Lag phase, log phase, and stationary phase.
• Lag phase: Growth is slow ih the beginning when the cell number is small.
• Log phase: Growth increases rapidly or exponentially.
• Stationary phase : Due to shortage of space and nutrients, growth slows down leading to a stationary phase. This gives an S- shaped curve.

d) Absolute and relative growth rates:

• Quantitative comparisons between the growth of living systems can be of two kinds.
• Absolute growth rate: Measurement and comparison of the total growth per unit time is called the absolute growth rate.
• Relative growth rate: The growth of the given system per unit time expressed on a common basis.
  Ex: Per unit initial parameter is called the relative growth rate.

I. Multiple Choice Questions

Question 1.
The three phases of growth in correct order is
1. Meristematic, maturation, elongation
2. Elongation, meristematic, maturation
3. Meristematic, elongation, maturation
4. Elongation, maturation, meristematic
Answer:
3. Meristematic, elongation, maturation

Question 2.
The phase in which maximum growth can be seen in the sigmoid curve.
1. Log
2. Lag
3. Stationary
4. Lag and Log
Answer:
1. Log

Question 3.
An aquatic plant which shows plasticity
1. Cotton
2. Coriander
3. Buttercup
4. Larkspur
Answer:
3. Buttercup

Question 4.
An example of adenine derivative plant growth regulator is
1. IAA
2. Kinetin
3. ABA
4. Gibberellic acid
Answer:
2. Kinetin

Question 5.
Gibberellic acid is
1. Indole compound
2. Adenine compound
3. Carotenoid derivative
4. Terpene derivative
Answer:
4. Terpene derivative

Question 6.
The foolish seedling disease of rice is caused by
1. Nematode
2. Bacteria
3. Fungus
4. Virus
Answer:
3. Fungus

Question 7.
2, 4-D is used to kill
1. Gymnosperms
2. Dicot weeds
3. Monocot grasses
4. Pteridophytes
Answer:
2. Dicot weeds

Question 8.
Bolting is
1. Yellowing of leaves
2. Internodal elongation prior to flowering
3. Early seed production
4. Re-greening of leaves
Answer:
2. Internodal elongation prior to flowering

Question 9.
Cytokinins help to produce
1. Chloroplast in leaves
2. Stem elongation in sugarcan
3. Synchronized fruit set in pineapple
4. Flowering in pineapple
Answer:
1. Chloroplast in leaves

Question 10.
Ethylene promotes
1. Senescence and abscission of flowers
2. Senescence but not abscission of flowers
3. Abscission of flowers but not senescence
4. Neither senescence nor abscission of flowers
Answer:
1. Senescence and abscission of flowers

II. Fill in the Blanks

Question 1.
Auxin was isolated by F.W.Went from tips of coleoptiles of ___________ seedlings.
Answer:
Oats

Question 2.
The ‘Bakanae’ disease of rice seedlings was caused by a fungal pathogen ___________
Answer:
Gibberella fujikuroi

Question 3.
Skoog and Miller crystallized the cytokinesis promoting active substance that they termed as ___________
Answer:
Kinetin

Question 4.
Inhibitor-b and Dormin were proved to be chemically identical and was named ___________
Answer:
ABA

Question 5.
NAA and 2,4-D are ___________ auxins.
Answer:
Synthetic

Question 6.
2, 4-D is widely used to kill ___________ weeds.
Answer:
Dicot weeds

Question 7.
___________ PGR causes elongation of apple fruits and improves its shape,
Answer:
libberellins

Question 8.
Natural cytokinin zeatin was extracted from ___________ and coconut milk.
Answer:
Corn Kernels

Question 9.
The rise in rate of respiration during ripening of fruits is called as ___________
Answer:
Respiratory climacteric

Question 10.
___________ plant growth regulator is called stress hormone.
Answer:

III. One Word Answer Questions

Question 1.
In which phase of growth is most rapid ?
Answer:
Log phase

Question 2.
Name the hormone that induce rooting in stem cutting.
Answer:
Auxins

Question 3.
Apical dominance is induced by which hormone ?
Answer:
Auxins

Question 4.
Name the auxin that kills dicotyledonous weeds.
Answer:
2,4 D

Question 5.
Which hormone is useful for ripening of fruits ?
Answer:
Ethylene

Question 6.
Name the phytohormone useful for closure of stomata.
Answer:
ABA

Question 7.
Name the fungus responsible for discovery of gibberellins.
Answer:
Gibberella fujikuroi

Question 8.
Which phytohormone induces bolting in rosette plants ?
Answer:
Gibberellins

Question 9.
Name the phytohormone which is isolated from corn kernels and coconut milk.
Answer:
Zeatin

Question 10.
Which phytohormone promotes female flowers in cucumbers ?
Answer:
Ethylene

Regular practice with AP Inter 1st Year Botany Study Material Chapter 9 Respiration in Plants Questions and Answers helps students stay prepared for examinations.

AP Inter 1st Year Botany 9th Lesson Respiration in Plants Questions and Answers

IV. Very Short Answer Questions

Question 1.
Where does anaerobic respiration occur in man and yeast?
Answer:

• In muscles of man during exercise, when oxygen is inadequate for cellular respiration anaerobic respiration occurs.
• In yeasts when oxygen is not available, anaerobic respiration occurs.

Question 2.
What is the common pathway for aerobic and anaerobic respiration? Where does it take place?
Answer:

• The common pathway for aerobic and anaerobic respiration is Glycolysis.
• It occurs in the cytoplasm (cytosol) of the cell.

Question 3.
What is the final acceptor of electrons in aerobic respiration? From which complex does it receive electrons?
Answer:

• The final acceptor of electrons in aerobic respiration is oxygen.
• The oxygen receives electrons from Enzyme complex IV.

Question 4.
Why is the RQ of fats less than that of the carbohydrates?
Answer:

• Fats are poorer in O_2, and the proportion of oxygen to carbon in fats is less when compared to carbohydrates.
• They require more O₂ for complete oxidation. So the number of O₂ used is greater than CO₂ released.
• Thus, the RQ of fats is less than that of carbohydrates.

Question 5.
Name the mobile electron carriers of the respiratory electron transport chain in the inner mitochondrial membrane.
Answer:

• Mobile electron carriers of the respiratory electron transport chain are ubiquinone and cytochrome ‘C’.
• Ubiquinone is present in the inner mitochondrial membrane while cytochrome ‘c’ is attached to the outer surface of the inner mitochondrial membrane.

Question 6.
F₀ – F₁ particles are involved in the synthesis of?
Answer:

• F₀ – F₁ particles are the two major components of ATP synthase or Enzyme complex-V.
• F₀ is an integral membrane protein complex that forms the channel through which protons cross the inner membrane.
• The F₁ head piece is a peripheral membrane protein complex and contains the site for synthesis of ATP from ADP and inorganic phosphate.

Question 7.
What is the end product of glycolysis? Where is it produced?
Answer:

• The end product of Glycolysis is Pyruvic acid.
• It is produced in the cytoplasm.

Question 8.
Write the substrate-level phosphorylation reaction in Krebs cycle.
Answer:

• Succinyl Co.A splits into succinic acid and Co.A in the presence of thiokinase.
• The energy released is utilised to form ATP from ADP and Pi.

Question 9.
What are the end products in alcoholic fermentation?
Answer:
CO₂ and Ethyl alcohol.

Question 10.
How many ATP are produced in aerobic and anaerobic respiration for one glucose.
Answer:

• In Aerobic respiration 38 ATP molecules are produced.
• In Anaerobic respiration only 2 ATP molecules are produced.

V. Short Answer Questions

Question 1.
Distinguish between aerobic and anaerobic respiration.
Answer:

Question 2.
Define RQ? Write the RQ value of fats and carbohydrates.
Answer:

• The ratio of the volume of CO₂ evolved to the volume of O₂ consumed in respiration is called the respiratory quotient.
• The value of respiratory quotient depends on the type of respiratory substrate.
• If carbohydrates are used as respiratory substrates, the volume of CO₂ evolved is equal to the amount of O₂ consumed, so the RQ value is one.
• C₆H₁₂O₆ + 6O₂ + 6 H₂O → 6 CO₂ + 12 H₂O + 686 KCal.
• RQ = 6/6 = 1
• If Fats are used as respiratory substrates, the volume of CO₂ evolved is less than the amount of O₂ consumed, so the RQ value is less than one.
• 2 (C₅₁H₉₈O₆) + 145O₂ → 102 CO₂ + 98 H₂O
• RQ = 102/145 = 0.7
• If proteins are used as respiratory substrate, the RQ value will be around 0.9.

Question 3.
Discuss the respiratory pathway is an amphibolic pathway.
Answer:

• Since respiration involves the breakdown of substrates, the respiratory pathway has been considered as a catabolic pathway.
• But the respiratory pathway is involved in both anabolism and catabolism.
• Fatty acid would be broken down to acetyl Co. A before entering the respiratory pathway when it is used as a substrate.
• When the organism needs to synthesize fatty acids, acetyl Co.A would be withdrawn from the respiratory pathway.
• Thus the respiratory pathway comes into picture both during breakdown and synthesis of fatty acids.
• Similarly during the breakdown and the synthesis of proteins also, respiratory intermediates form the link.
• The breaking-down process within the living organism is called catabolism while synthesis is called anabolism.
• Because the respiratory pathway is involved in both anabolism and catabolism, it is better referred to as an amphibolic pathway.

Question 4.
Explain the process of fermentation in yeast cells.
Answer:
In anaerobic conditions, Pyruvic acid undergoes partial oxidation to form ethyl alcohol called fermentation. It involves two steps.

A) Decarboxylation:
Pyruvic acid undergoes decarboxylation in the presence of Pyruvic decarboxylase to form Acetaldehyde molecules and CO₂, molecules.
2 pyruvic acid → 2 Acetaldehyde + 2 CO₂

B) Reduction:
2 Acetaldehyde undergoes reduction in the presence of Alcoholic dehydrogenase to form Ethyl alcohol. NADPH formed in Glycolysis supplies H+ to this reaction.
2 Acetaldehyde + 2 NADPH + H+ → 2 C₂H₅OH + 2 NAD⁺

VI. Long Answer Questions

Question 1.
Give an account of glycolysis? Where does it occur? What are the end products? Trace the fate of the products in both aerobic and anaerobic respiration.
Answer:

• Glucose undergoes partial oxidation to form 2 molecules of pyruvic acid is called Glycolysis.
• Glycolysis occurs in the “cytoplasm” of the cell and takes place in all living organisms.
• Its end products are 2ATP, 2 NADH + H⁺ and 2 PA.
• The reaction sequence was worked out by Embden, Mayerhoff and Paranas, hence the cycle is known as EMP-pathway.

Various steps of Glycolysis are:

1. Phosphorylation: Glucose is phosphorylated to glucose-6-phosphate in the presence of ATP, catalyzed by the enzyme hexokinase.
2. Isomerization : Glucose-6-phosphate is converted into its isomer fructose-6-phosphate, catalyzed by hexose phosphate isomerase.
3. Phosphorylation: Fructose-6-phosphate is phosphorylated in the presence of ATP to form fructose 1.6 bisphosphate by phosphofructokinase.
4. Cleavage : Fructose 1,6-bisphosphate is split into 2 molecules of triose phosphate namely glyceraldehyde 3-phosphate (G,P or PGAI) and dihydroxyacetone phosphate (DHAP). This interconvertible reaction is catalyzed by Aldolase.
5. Isomerization : DHAP is converted into another G₃P molecule in the presence of isomerase.
6. Dehydrogenation: 3- phosphoglyceraldehyde is oxidized to 1, 3 bisphosphoglyceric acid with the reduction of NAD to NADH + H⁺. This is catalysed by G-3 P dehydrogenase.
7. Dephosphorylation: Phosphoglycerokinase catalyses the formation of 3-phosphoglyceric acid from 1,3-bisphosphoglyceric acid. Two molecules of ATP are produced directly by substrate-level phosphorylation.
8. Intra molecular shift: 3-phosphoglyceric acid is converted into 2 phosphoglyceric acid in the presence of phosphoglyceromutase.
9. Dehydration: 2 PGA molecules lose water molecules in the presence of Enolase to form PEPA molecules.
10. Dephosphorylation: 2 PGA molecules undergo dephosphorylation in the presence of pyruvic kinase to form 2 PA molecules, and 2 ATP molecules are formed.
    • Fate of pyruvic acid depends on the cellular need.
    • There are three major ways in which different cells handle pyruvic acid produced by glycolysis.
    • These are lactic acid fermentation, alcoholic fermentation, and aerobic respiration.
    • Fermentation takes place under anaerobic conditions in many prokaryotes and unicellular eukaryotes.
    • For the complete oxidation of glucose to CO₂ and H₂O organisms adopt Krebs’ cycle which is also called aerobic respiration. This requires O₂ supply.

Question 2.
Explain the reactions of Krebs cycle.
Answer:

• Acetyl CoA is formed from pyruvic acid as a result of oxidative decarboxylation.
• Acetyl CoA acts as a substrate for Krebs cycle, Krebs cycle may also be called Tricarboxylic acid cycle (TCA cycle) or Citric acid cycle or Organic acid cycle.
• It is called Krebs cycle after the scientist Sir Hans Krebs who first elucidated it.
• There are 10 biochemical reactions in Krebs cycle.

1. Condensation :
In this, acetyl Co.A condenses with oxaloacetic acid and water to yield citric acid in the presence of Citrate synthetase and Co.A is released.
Oxaloacetic acid + Acetyl Co.A + H₂O → Citric acid + Co.A

2. Dehydration :
Citric acid loses water molecule to form cis-aconitic acid in the presence of aconitase.
Citric acid → Cis-aconitic acid + H₂O

3. Hydration:
A water molecule is added to cis-aconitic acid to yield isocitric acid in the presence of aconitase.
Cis – aconitic acid + H₂O → isocitric acid

4. Oxidation I :
Isocitric acid undergoes oxidation in the presence of isocitric dehydrogenase to yield Oxalosuccinic acid,
Isocitric Acid + NAD⁺ → Oxalosuccinic acid + NADH + H⁺

5. Decarboxylation :
Oxalosuccinic acid undergoes decarboxylation in the presence of Oxalosuccinic decarboxylase to form a – ketoglutaric acid.
Oxalosuccinic acid → ∝ – ketoglutaric acid + CO₂

6. Oxidative decarboxylation, Oxidation II :
∝ – ketoglutaric acid undergoes oxidation and decarboxylation in the presence of ∝ – ketoglutaric dehydrogenase and condenses with CoA to form succinyl CoA.
∝ – keto Glutaric acid + NAD⁺ + CoA → succinyl CoA + NADH + H⁺ + CO₂

7. Cleavage:
Succinyl CoA splits into succinic acid and CoA in the presence of succinic thiokinase. The energy released is utilized to form ATP from ADP and Pi.
Succinyl CoA + ADP + Pi → Succinic Acid + ATP + CoA

8. Oxidation-III:
Succinic acid undergoes oxidation and forms Fumaric acid in the presence of succinic dehydrogenase.
Succinic Acid + FAD → Fumaric acid + FADH₂

9. Hydration:
A water molecule is added to Fumaric acid in the presence of Fumarase to form Malic acid.
Fumaric acid + H₂O → Malic acid

10. Oxidation IV :
Malic acid undergoes oxidation in the presence of malic dehydrogenase to form oxaloacetic acid.
Malic Acid + NAD⁺ → Oxaloacetate + NADH + H⁺
In the TCA cycle, for every 2 molecules of Acetyl CoA undergoing oxidation, 2 ATP, 6 NADH + H⁺ 2 FADH₂ molecules are formed.

I. Multiple Choice Questions

Question 1.
Glycolysis is also known as ___________ pathway.
1. ETS
2. EMP
3. ENP
4. ELP
Answer:
2. EMP

Question 2.
End product of glycolysis is
1. Pyruvic acid
2. Oxalo acetic acid
3. Citric acid
4. Phosphoenolpyruvic acid
Answer:
1. Pyruvic acid

Question 3.
Fermentation occurs when there is
1. Complete supply of oxygen
2. No supply of oxygen
3. Complete supply of water
4. No supply of water
Answer:
2. No supply of oxygen

Question 4.
In alcoholic fermentation, pyruvate is converted to which among the following.
1. Ethanol, CO₂, NADH
2. CO₂ and methanol
3. CO₂ and Ethanol only
4. CO₂ and Carboxylic acid
Answer:
3. CO₂ and Ethanol only

Question 5.
Which enzyme catalyses the oxidative decarboxylation of pyruvic acid.
1. Pyruvate carboxylase
2. Lactate dehydrogenase
3. Alcohol dehydrogenase
4. Pyruvate dehydrogenase
Answer:
4. Pyruvate dehydrogenase

Question 6.
Where does TCA cycle occurs
1. Cytoplasm
2. Inner membrane of Mitochondria
3. Mitochondrial matrix
4. Stroma of Chloroplast
Answer:
3. Mitochondrial matrix

Question 7.
What is the first formed compound in TCA cycle
1. Acetyl CoA
2. Citric acid
3. Isocitric acid
4. OAA
Answer:
2. Citric acid

Question 8.
Which among the following is synthesized during the conversion of succinyl – CoA to succinic acid inTCA cycle.
1. FADH₂
2. GTP
3. NADH₂
4. NADPH₂
Answer:
2. GTP

Question 9.
How many total NADH + H⁺ are produced from a pyruvic acid after completion of TCA cycle ?
1. 2
2. 3
3. 4
4. 5
Answer:
3. 4

Question 10.
The site of ETC is
1. Outer membrane of mitochondria
2. Cytoplasm
3. Inner membrane of mitochondria
4. Matrix of mitochondria
Answer:
3. Inner membrane of mitochondria

II. Fill in the Blanks

Question 1.
The energy currency of the cell ___________
Answer:
ATP

Question 2.
Glycolysis occurs in ___________ of the cell.
Answer:
Cytoplasm

Question 3.
In muscle cells of animals, when oxygen is inadequate, pyruvic acid is converted to
Answer:
Lactic acid

Question 4.
Pyruvic acid + CoA + NAD + Acetyl CoA + CO₂ + NADH + H⁺. Above reaction is catalyzed by enzyme ___________
Answer:
Pyruvic dehydrogenase

Question 5.
During the conversion of succinyl-CoA to succinic acid in citric acid cycle, a molecule of ___________ is synthesized.
Answer:
GTP

Question 6.
Tricarboxylic acid cycle was discovered by ___________
Answer:
Hans Krebs

Question 7.
ATP synthase is located on ___________ membrane of Mitochondria.
Answer:
Inner

Question 8.
Oxidation of one molecule of FADH₂ gives rise to ___________ molecules of ATP.
Answer:
2

Question 9.
Respiratory pathway is an ___________ pathway as it involves both catabolism and anabolism.
Answer:
Amphibolic

Question 10.
The ratio of the volume of C02 evolved to the volume of 02 consumed in respiration is called ___________
Answer:
Respiratory Quotient

III. One Word Answer Questions

Question 1.
Which enzyme catalyses the phosphorylation of glucose during glycolysis?
Answer:
Hexokinase

Question 2.
Where is the electron transport system located in the mitochondria of a cell?
Answer:
Inner membrane of mitochondria

Question 3.
Name the cofactors required for the activity of pyruvate dehydrogenase.
Answer:
NAD+ and CoA

Question 4.
What is the number of ATP produced when pyruvate is converted to lactate by Fermentation?
Answer:
Zero (0)

Question 5.
In glucose, how many carbon molecules are present? [March-26]
Answer:
6 carbons

Question 6.
What is the end product of anaerobic respiration in yeast?
Answer:
Ethanol + CO₂

Question 7.
Respiratory quotient of carbohydrates is 1. Why?
Answer:
Volume of CO₂ released and volume of O₂ consumed is equal. So the Rq value is one.

Question 8.
Glycolysis occurs in which part of the cell?
Answer:
Cytoplasm

Question 9.
Name the process of conversion of glucose into ethyl alcohol.
Answer:
Fermentation

Question 10.
Which compound serves as a connecting link between glycolysis and Kreb’s cycle?
Answer:
Acetyl CoA

Regular practice with AP Inter 1st Year Botany Study Material Chapter 8 Photosynthesis in Higher Plants Questions and Answers helps students stay prepared for examinations.

AP Inter 1st Year Botany 8th Lesson Photosynthesis in Higher Plants Questions and Answers

IV. Very Short Answer Questions

Question 1.
Name the processes which take place in grana and stroma regions of chloroplast.
Answer:
Light reaction of photosynthesis takes place in the grana of chloroplast. Dark reactions of photosynthesis take place in the stroma of chloroplast.

Question 2.
Where does the photolysis of H₂O occur? What is its significance?
Answer:
Photolysis of H₂O takes place in “Lumen of thylakoids” or in grana thylakoids. Due to this process, electrons, protons and oxygen are released. Oxygen is the basis for all organisms.

Question 3.
Where is the enzyme NADP reductase located? What is released if the proton gradient breaks down?
Answer:
The enzyme NADP reductase is located on the outer side of the thylakoid membrane in the chloroplast. When the proton gradient breaks down, energy is released in the form of NADPH and ATP.

Question 4.
Explain the terms :
(a) PEP Carboxylase
(b) Bundle sheath cells.
Answer:
a) PEP Carboxylase : PEP carboxylase is an enzyme that catalyzes the reaction of phosphoenolpyruvate (PEP) with bicarbonate to form oxaloacetate.

b) Bundle Sheath Cells: Bundle sheath cells are thick-walled cells that surround vascular bundles in C₄ plant leaves and stems.

Question 5.
Mention the components of ATPase enzyme. What is their location? Which part of the enzyme shows conformational change?
Answer:

• The ATPase enzyme consists of two parts: F₀ is embedded in the thylakoid membrane and forms a trans-membrane channel that carries out facilitated diffusion of protons across the membrane.
• F₁ protrudes on the outer surface of the thylakoid membrane that faces the stroma. It provides enough energy to cause a conformational change in the ATPase, involved in the synthesis of ATP.

Question 6.
Distinguish between action spectrum and absorption spectrum.
Answer:

• Action spectrum: A graph showing the rate of photosynthesis at different wavelengths of light.
• Absorption spectrum: A graph showing the absorption of light by the photosynthetic pigments at different wavelengths.

Question 7.
Out of the basic raw materials of photosynthesis, What is reduced? What is oxidized?
Answer:
Of the basic raw materials like CO₂ and H₂O, H₂O is oxidized in light reaction and CO₂ is reduced in the dark phase of photosynthesis.

Question 8.
Define the law of limiting factors proposed by Blackmann.
Answer:
If a process is conditioned as to its rapidly by a number of separate factors, the rate of the process is limited by the factor that is present in a relative minimum value.

Question 9.
What is the primary acceptor of CO₂ in C₃ plants? What is the first stable compound formed in the Calvin cycle?
Answer:

• Primary acceptor of CO₂ in C₃ plants is RUBP.
• The first stable compound formed in Calvin cycle is PGA.

Question 10.
What is the primary acceptor of CO₂ in C₄ plants? What is the first compound formed as a result of primary carboxylation in the C₄ pathway?
Answer:

• Primary acceptor of CO₂ in C₄ plants is PEP (phosphoenol pyruvic acid).
• First stable compound formed in C₄ cycle is OAA (Oxaloacetic Acid).

V. Short Answer Questions

Question 1.
Differentiate between C₃ and C₄ plants.
Answer:

Question 2.
Write short notes on photorespiration.
Answer:
Photorespiration is a metabolic process in plants where oxygen binds to the enzyme RuBisCO instead of carbon dioxide, resulting in the release of carbon dioxide.

• RUBIsCO has a much greater affinity for CO₂ when the CO₂ and O₂ are nearly equal.
• In C₃ plants, Some O₂ binds with RUBIs CO₂ and hence CO₂ fixation is reduced.
• RUBP Binds with Oxygen to form one molecule of Phosphoglycerate and phosphoglycolate (2c). So called C₂ cycle.
• In this proces, release of CO₂ with the utilization of ATP occurs.
• No synthesis of ATP or NADPH.
• It is a wasteful process.
• In C₄ plants, photorespiration does not occur, because they have a mechanism that increases the concentration of CO₂ at the enzyme site.
• In C₄ plants, C₄ acid from the mesophyll is broken down in the bundle sheath cells to release CO₂.
• This results in increasing the intracellular concentration of CO₂.
• In turn this ensures that the RUBIsCO functions as a carboxylase minimising the oxygenase activity.
• So C₄ plants are more effective in productivity and yield and also show tolerance to higher temperatures than C₃ plants.

Question 3.
Describe the mechanism of the C4 pathway.
Answer:

• Kortschak, Hartt and Burr found that 3-PGA is not the initial product of photosynthesis in case of sugarcane, instead it was four carbon compounds like malic and aspartic acids.
• Hatch and Slack confirmed the results of Kortschak, Hartt and Burr while working on sugarcane plants.
• The first stable compound is a four-carbon compound, the pathway is termed as C₄ pathway and plants which undergo this pathway are called C₄ plants.
• This pathway is also called Beta-Carboxylation pathway or Hatch-Slack pathway.

REACTIONS:

In Mesophyll cells:

• CO₂ is accepted by a 3-carbon molecule phosphoenol pyruvic acid, in the form of HCO₃ and is present in the mesophyll cells in the presence of PEP carboxylase and water to form oxaloacetic acid.
• CO₂ phosphoenol pyruvic acid + H₂O → oxaloacetic acid + H₃PO₄^–
• Oxaloacetic acid is reduced to malic acid by using the light generated NADPH + H⁺ in the presence of Malic dehydrogenase.
• Oxaloacetic acid + NADPH + H⁺ → malic acid + NADP⁺

In Bundle sheath cells:

• Malic acid formed in the chloroplast of mesophyll cells is now transported to the chloroplast of bundle sheath cells.
• Malic acid now undergoes oxidative decarboxylation to form Pyruvic acid (3C) in the presence of Malic enzyme. In this NADP+ is reduced to NADPH + H⁺.
• Malic acid + NADP⁺ → pyruvic acid + NADPH + H⁺ + CO₂
• The CO₂ generated is utilized in the Calvin cycle to synthesize sugars
• The Pyruvic acid produced in bundle sheath cells moves to the chloroplast of mesophyll cells and is phosphorylated to Phosphoenol Pyruvic acid in the presence of pyruvate dikinase.
• PEP undergoes phosphorylation, then enters into the cytoplasm of the mesophyll cells.
• Pyruvic acid + 2ATP + Pi → phosphoenol pyruvic acid + 2 AMP + 2 Pi.
• During this pathway two carboxyations and one decarboxylation reaction occurs.
• To synthesize one molecule of Glucose, C₄ plants utilize 30 ATP and 12 NADPH + H⁺.

Question 4.
Draw a neat labelled diagram of chloroplast.
Answeer:

Question 5.
Differentiate between cyclic and non-cyclic photophosphorylation.
Answer:

VI. Long Answer Questions

Question 1.
Describe the process of reactions in the Calvin cycle.
Answer:

• Melvin Calvin and his associates Andrew Benson and lames Basham discovered the CO₂ fixation pathway in Chlorella and hence named Calvin cycle.
• It is also called Photosynthetic carbon reduction (PCR) cycle or reductive pentose phosphate pathway (RPP cycle) or Dark reaction.
• This pathway is studied under three phases namely – A) Carboxylation, B) Reduction and C) Regeneration.

I. Carboxylation Phase:
Fixation of CO₂ into a stable organic intermediate is called carboxylation. It is the most crucial step of the calvin cycle where CO₂ is utilized for carboxylation of RUBP in the presence of RUBIsCO results in the formation of two molecules of 3-PGA.

II. Reduction Phase:
2 molecules of ATP for phosphorylation and two molecules of NADPH for reduction per CO₂ molecule fixed. The fixation of 6 molecules of CO₂ and 6 turns of the cycle are required for the formation of one molecules of Glucose.

III. Regeneration phase:

• To continue the cycle uninterrupted, regeneration of RUBP is crucial. It requires one ATP for phosphorylation to form RUBP.
• Hence for every CO₂ molecule entering the Calvin cycle, 3 molecules of ATP and 2 molecules of NADPH are required.
• To meet this difference in the number of ATP and NADPH used in dark reactions, cyclic phosphorylation takes place.
• To synthesize one Glucose molecule, 18 ATP and 12 NADPH + H⁺ are required.

I. Multiple Choice Questions

Question 1.
Who discovered oxygen?
1. Joseph Priestly
2. T. Engel
3. Jan Ingenhousz
4. Van Neil
Answer:
1. Joseph Priestly

Question 2.
Chloroplast is
1. Single membrane-bound organelle
2. Double membrane-bound organelle
3. Triple membrane-bound organelle
4. Membrane-lacking organelle
Answer:
2. Double membrane-bound organelle

Question 3.
Which is the most abundant plant pigment in the world?
1. Chlorophyll a
2. Chlorophyll b
3. Carotenoids
4. Xanthophylls
Answer:
1. Chlorophyll a

Question 4.
Maximum absorption by chlorophyll a occurs in
1. Blue green region
2. Red green region
3. Blue red region
4. Yellow red region
Answer:
3. Blue-red region

Question 5.
LHC stands for-
1. Late Harvesting Complex
2. Light Harvesting Complex
3. Light Hanging Complex
4. Late Hanging Complex
Answer:
2. Light Harvesting Complex

Question 6.
During photosynthesis the O₂ is released in
1. Lumen of thylakoid
2. Outer side of thylakoid
3. Stroma
4. Cytoplasm
Answer:
1. Lumen of thylakoid

Question 7.
For formation of 1 glucose molecule, how many turns of Calvin cycle is/are needed?
1. 3
2. 1
3. 2
4. 6
Answer:
4. 6

Question 8.
The most crucial step of Calvin cycle is
1. Carbonation
2. Carboxylation
3. Reduction
4. Regeneration
Answer:
2. Carboxylation

Question 9.
The first stable product during CO₂ fixation in C₄ cycle is
1. RBP
2. PEP
3. OAA
4. PGA
Answer:
3. OAA

II. Fill in the blanks

Question 1.
___________ in 1770 performed a series of experiments that revealed the essential role of air in the growth of green plants.
Answer:
Joseph Priestley

Question 2.
Chlorophyll ___________ is the chief pigment associated with photosynthesis.
Answer:
Chlorophyll a

Question 3.
In PS-I, the reaction centre chlorophyll-a has an absorption peak at ___________ nm.
Answer:
700

Question 4.
The splitting of water is associated with the PS-___________
Answer:
PS II

Question 5.
The ___________ enzyme is located on the stroma side of the membrane.
Answer:
NADP reductase

Question 6.
Chemiosmosis is helpful for the synthesis of ___________
Answer:
ATP

Question 7.
The products of light reaction are ATP, NADPH and ___________
Answer:
Oxygen

Question 8.
RuBP carboxylase also has an oxygenation activity, so it would be more correct to call. It ___________
Answer:
RUBISCO

Question 9.
‘Kranz’ anatomy is a characteristic feature of ___________ plants.
Answer:
C₄

Question 10.
___________ is the major limiting factor for photosynthesis.
Answer:
CO₂

III. One Word Answer Questions

Question 1.
What is the primary acceptor of CO₂ in C₃ plants?
Answer:
RUBP

Question 2.
Which is the first formed compound in C₃ plants?
Answer:
PGA

Question 3.
What is the primary acceptor of CO₂ in C₄ plants?
Answer:
PEPA

Question 4.
Name the first formed compound in C₄ plants.
Answer:
OAA

Question 5.
Where does photolysis of water take place in Chloroplast?
Answeer:
Lumen of thylakoid

Question 6.
Who proposed the law of limiting factors in photosynthesis?
Answer:
Blackmann

Question 7.
How many ATP are required for synthesis of one glucose in C3 plants?
Answer:
18 ATP

Question 8.
Where does light reaction take place in chloroplast?
Answer:
Grana or Thylakoids

Question 9.
Where does dark reaction take place in chloroplast?
Answer:
Stroma

Question 10.
Name the first enzyme that is involved in the carboxylation process in C3 Plants.
Answer:
RUBISCO

Regular practice with AP Inter 1st Year Botany Study Material Chapter 6 Biomolecules Questions and Answers helps students stay prepared for examinations.

AP Inter 1st Year Botany 7th Lesson Cell Cycle and Cell Division Questions and Answers

IV. Very Short Answer Questions

Question 1.
Which tissue of animals and plants exhibit meiosis ?
Answer:

• In higher animals, meiosis occurs in gamete mother cells (spermatocytes and oocytes) during gametogenesis.
• In higher plants, meiosis occurs in spore mother cells, microspore mother cells and megaspore mother cells during sporogenesis.

Question 2.
What attributes does a chromatid require to be classified as a chromosome?
Answer:
A chromatid is said to be as a chromosome when it possesses

• Independent existence.
• Its own centromere with one DNA molecule.

Question 3.
Which of the four chromatids of a bivalent at prophase-I of meiosis can involve in crossing over?
Answer:
Non-sister chromatids of homologous chromosomes of a bivalent can involve in crossing over.

Question 4.
Write the sub stages of meiotic prophase-I in which ‘synapsis’ and ‘crossing over’ occurs.
Answer:
Synapsis occurs in Zygotene and crossing over occurs in Pachytene stage.

Question 5.
A cell has 32 chromosomes. It undergoes mitotic division. What will be the chromosome number during metaphase and DNA content (C) during anaphase?
Answer:

• The mitotic cell division occurs in somatic cells of an organism.
• The chromosome number in the daughter cells remains the same as that of the parent cell. So even at metaphase, the chromosome number does not change.
• The DNA content gets doubled at the synthetic phase or interphase and gets divided at anaphase but the chromosome number remains same.

V. Short Answer Questions

Question 1.
In which phase of meiosis the following are formed ? Choose the answers from the hint points given below.
a) Synaptonemal complex
b) Recombination nodules
c) Functioning of the enzyme recombinase
d) Terminalization of chiasmata
e) Interkinesis
f) Formation of dyad of cells
Hints :
i) Pachytene,
ii) After Telophase-I/before meiosis-II,
iii) Zygotene,
iv) Telophase-I/After meiosis-I,
v) Pachytene,
vi) Diakinesis
Answer:
a) Zygotene
b) Pachytene
c) Pachytene
d) Diakinesis
e) After Telophase-I/before Meiosis-II
f) Telophase-1/After Meiosis-I

Question 2.
Mitosis results in producing two daughter cells which are similar to each other. What would be the consequence if each of the following irregularities occurs during mitosis?
a) Nuclear membrane fails to disintegrate
b) Duplication of DNA does not occur.
c) Centromere does not divide
d) Cytokinesis does not occur.
Answer:
a) If a nuclear membrane fails to disintegrate, the replicated chromosomes remain within the same cell. Nucleus does not divide (Endomitosis).

b) If duplication of DNA does not occur there is no further cell division. Every mitotic cell division must be preceded by DNA duplication. Then only separation of sister chromatids is possible in Anaphase.

c) If centromeres do not divide, daughter chromosomes are not formed (polytomy).

d) If cytokinesis does not occur, multinucleate condition arises leading to the formation of syncytium (Ex: liquid endosperm in coconut).

Question 3.
Describe the events in prophase-1 of meiosis.
Answer:
Prophase-I : This is a typically longer and complex phase. It is further subdivided into 5 phases based on the behaviour of chromosomes into

• Leptotene,
• Zygotene,
• Pachytene,
• Diplotene,
• Diakinesis.

i) Leptotene :
During the leptotene stage, chromosomes become gradually visible. This compaction of chromosomes lasts throughout the leptotene stage.

ii) Zygotene:

• Chromosomes start pairing together by a process called synapsis. Such chromosomes are called homologous chromosomes.
• It is followed by the formation of synaptonemal complex.
• Such a complex formed by a pair of synapsed homologous chromosomes is called bivalent or a tetrad of chromatids.

iii) Pachytene:

• Bivalents clearly appear as tetrads. Formation of recombination nodules occur.
• Crossing over is exchange of genetic material between two homologous chromosomes is mediated by recombinase enzymes.
• It leads to recombination of genetic material on the two chromosomes.

iv) Diplotene:

• The dissolution of the synaptonemal complex occurs.
• Separation of bivalents from each other except at the regions of crossing over occurs. Such X- shaped regions are referred to as chiasmata.

v) Diakinesis:

• Terminalisation of chiasmata occurs.
• The chromosomes are fully condensed.
• The meiotic spindle is assembled to prepare the homologous chromosomes for separation.
• During late diakinesis, the nucleolus disappears and nuclear envelope also break down.

Question 4.
Though redundantly described as a resting phase, interphase does not really involve rest. Comment.
Answer:

• Actually, during the interphase, great metabolic activities occur so it is a preparatory phase.
• Division of chromosomes or cytoplasm does not occur.
• During interphase cell enlarges.
• The interphase is divided into three phases: G₁ phase (Gap 1), S-phase (Synthesis), and G₁ phase (Gap 2).

G₁ phase:

• It corresponds to the interval between mitosis and initiation of DNA replication.
• During G₁ phase the cell is metabolically active and continuously grows.

S or synthesis phase:

• It marks the period during which DNA synthesis or replication takes place.
• During this time the amount of DNA per cell doubles.
• If the initial amount of DNA is denoted as 2C, then it increases to 4C.
• However, there is no increase in the chromosome number.

G₂ phase :

• Proteins are synthesized in preparation for mitosis.
• Cell growth continues.
• Hence, interphase is not a resting phase.

VI. Long Answer Questions

Question 1.
Discuss on the statement – telophase is he reverse of prophase.
Answer:
It is true to say that telophase is the reverse of the prophase stage. Based on the following points we can support this fact.

During Prophase:

• Chromatin fibres are shorter and thicker due to coiling and folding which results in thread-like chromosomes.
• Each chromosome consists of two coiled sister chromatids joined by centromeres
• During late prophase, ER, Golgi complex, Nucleolus and nuclear envelope disappear.
• Centrioles migrate towards opposite poles.
• Sharp radiating microtubules appear around each centriole and migrate towards opposite poles.

During Telophase :

• Daughter chromosomes undergo decondensation and uncoiling at each pole.
• The chromatin material gets surrounded by segments of nuclear membrane formed from the elements of ER.
• Cellular components like ER and golgi complex reorganize once again.
• Formation of the new nucleoli can be noticed.
• Astral rays and spindle fibres are gradually disintegrate and are absorbed into the cytoplasm.
• Two daughter nuclei are designed.

Question 2.
Differentiate between the events of mitosis and meiosis.
Answer:

Question 3.
Write a brief note on the following:
a) Synaptonemal complex,
b) Metaphase plate
Answer:
a) Synaptonemal complex:

• Synaptonemal complex is a fibrillar structure that develops during zygotene of prophase-1 of meiosis I.
• It develops between the synapsed homologous chromosomes.
• It helps in stabilization of paired conditions of chromosomes and involves in chiasmata formation and crossing over.
• The synaptonemal complex is attached at both ends through its lateral elements to the inner surface of the nuclear membrane.
• It disappears during diplotene.
• This helps in repulsion activity between homologous chromosomes. .
• As a result the homologues of the bivalents are separated from each other.

b) Metaphase Plate:

• Metaphase plate gets organised during metaphase of mitosis and metaphase of meiosis.
• The spindle fibres organised during metaphase get accumulated at the centre of the cell.
• The plane of alignment of the chromosomes during cell division at metaphase is referred to as the metaphase plate.
• The orientation of the metaphase plate is based on the chromosomal arrangement.
• Metaphase plate is important in equitable and simultaneous distribution of all the chromosomes.

Question 4.
Write briefly the significance of mitosis and meiosis in multicellular organisms.
Answer:
A) Significance of mitosis:

• Equal distribution of chromosomes : Mitosis results in the production of diploid daughter cells with genetic complement usually identical to that of the parent cell.
• The growth in multicellular organisms is due to mitosis only.
• Cell growth results in disturbing the ratio between the nucleus and the cytoplasm.
• It therefore becomes essential for the cell to divide to restore the nucleo-cytoplasmic ratio.
• Repair: The cells of the upper layer of the epidermis, cells of the lining of the gut, and RBC are constantly being replaced.
• Mitotic divisions in the meristematic cells and the apical and the lateral meristems result in continuous growth of plants throughout their lifetime.

B) Significance of meiosis

• Meiosis is responsible for the formation of sex cells or gametes that are responsible for sexual reproduction.
• In sexually reproducing organisms the constant number of chromosomes through generations is maintained by meiosis by producing haploid gametes.
• It is the only means of restoring the chromosome number characteristic of the species.
• In pachytene due to crossing over the hereditary factors from male and female parents get mixed thus providing a new combination of genetic material.
• Variations inherited lead to evolution of species.

I. Multiple Choice Questions

Question 1.
Phase between two successive M-phases
1. Prophase
2. Metaphase
3. Interphase
4. Anaphase
Answer:
3. Interphase

Question 2.
Cell cycle duration of yeast
1. 24 hours
2. 90 minutes
3. 20 minutes
4. 60 minutes
Answer:
2. 90 minutes

Question 3.
During ‘S’ phase of interphase
1. DNA synthesizes, chromosome number remains same
2. DNA synthesizes, chromosome number doubled
3. DNA does not synthesize, chromosome number reduced to half
4. DNA does not synthesize, chromosome number remains same
Answer:
1. DNA synthesizes, chromosome number remains same

Question 4.
A cell in G₁ phase has 16 chromosomes, how many chromosomes does the cell have a G₂ phase.
1. 24
2. 32
3. 8
4. 16
Answer:
4. 16

Question 5.
The cells in the quiescent stage (G₀) exit the cell cycle from the following phase;
1. G₂ phase
2. G₁ phase
3. S phase
4. M phase
Answer:
2. G₁ phase

Question 6.
Identify the correct sequential phases in the karyokinesis.
1. Prophase, metaphase, anaphase, telophase
2. Metaphase, anaphase, telophase, prophase
3. Anaphase, telophase, prophase, metaphase
4. Telophase, prophase, metaphase, anaphase
Answer:
1. Prophase, metaphase, anaphase, telophase

Question 7.
Centromere splitting occurring in
1. Mitotic anaphase and meiotic anaphase – I
2. Mitotic anaphase and meiotic anaphase – II
3. Meiotic anaphase – I and meiotic anaphase- II
4. Mitotic anaphase only
Answer:
2. Mitotic anaphase and meiotic anaphase – II

Question 8.
Compaction of chromosomes start from
1. Zygotene
2. Pachytene
3. Leptotene
4. Diplotene
Answer:
3. Leptotene

Question 9.
Bivalent formation occurs in the following stage.
1. Zygotene
2. Pachytene
3. Leptotene
4. Diplotene
Answer:
1. Zygotene

Question 10.
The X-shaped structures called chiasmata appears in
1. Zygotene
2. Pachytene
3. Leptotene
4. Diplotene
Answer:
4. Diplotene

Question 11.
Terminalisation of chiasmata happens in
1. Zygotene
2. Pachytene
3. Diplotene
4. Diakinesis
Answer:
4. Diakinesis

Question 12.
A bivalent in meiosis-I consists of
1. One chromosome, two chromatids
2. Two chromosomes, two chromatids
3. Two chromosomes, four chromatids
4. Four chromosomes, four chromatids
Answer:
3. Two chromosomes, four chromatids

II. Fill in the Blanks

Question 1.
Most suitable phase of cell division to study the morphology of chromosomes is ____________
Answer:
Metaphase

Question 2.
Prophase – I of meiosis has leptotene, ____________ pachytene, diplotene and diakinesis.
Answer:
Zygotene

Question 3.
The stage between meiosis -I and meiosis – II is ____________
Answer:
Interkinesis

Question 4.
Crossing over happens in the ____________ of prophase – I.
Answer:
Pachytene

Question 5.
In meiosis splitting of centromere in a chromosome happens during ____________
Answer:
Anaphase – II

Question 6.
What is the longest phase in the oocytes of some vertebrates ____________
Answer:
Diplotene

Question 7.
The disc shaped structures at the surface of the centromeres are called ____________
Answer:
Kinetochores

Question 8.
Cell furrow forms during cytokinesis of ____________ cell.
Answer:
Animal

Question 9.
The duration of M-phase in human cells is ____________
Answer:
1 hour

Question 10.
Which division increases the genetic variability in the population ? ____________
Answer:
Meiosis

III. One Word Answer Questions

Question 1.
Name the two phases of cell cycle.
Answer:
M phase and Interphase

Question 2.
What is the approximate duration of human cell cycle?
Answer:
24 Hours

Question 3.
Which of the phases of the cell cycle is of longest duration?
Answer:
Interphase

Question 4.
In which stage of cell cycle does DNA synthesis occur?
Answer:
S-phase

Question 5.
Name the stage of meiosis in which actual reduction in chromosome number occurs.
Answer:
Anaphase-I

Question 6.
What is the DNA content in a cell after completion of ‘S’ phase of interphase, when taken G₁ as in its phase.
Answer:
4C

Question 7.
Centriole duplication occurs in which phase of interphase.
Answer:
S-phase

Question 8.
Is the cell in quiescent stage (G₀) metabolically active or not.
Answer:
Active

Question 9.
What type of cell division is also known as equational division ?
Answer:
Mitosis

Question 10.
How many chromatids does the prophase chromosome has ?
Answer:
Two

Question 11.
An anther has 1200 pollen grains. How many pollen mother cells must have been there to produce them ?
Answer:
300 pollen mother cells

Question 12.
Given that the average duplication time of E.coli is 20 minutes. How much time will two E. coli cells take to become 32 cells ?
Answer:
80 minutes

Question 13.
If a tissue has at a given time 1024 cells. How many cycles of mitosis had the original parental single cell undergone?
Answer:
10 Cycles

Regular practice with AP Inter 1st Year Botany Study Material Chapter 6 Biomolecules Questions and Answers helps students stay prepared for examinations.

AP Inter 1st Year Botany 6th Lesson Biomolecules Questions and Answers

V. Very Short Answer Questions

Question 1.
Give one example for each of amino acids, sugars, nucleotides and fatty acids.
Answer:

• Amino acids : Glycine, Alanine, Serine
• Sugars : Glucose, Ribose, deoxyribose
• Nucleotides : Adenylic acid, Thymidylic acid, Guanylic acid, Uridylic acid, Cytidylic acid
• Fatty acids : Palmitic acid, Arachidonic acid

Question 2.
Explain the zwitterionic form of an amino acid.
Answer:

• The zwitterionic form of an amino acid is when an amino acid has both a positive and a negative charge on the same molecule, but an overall neutral charge.
• This form of amino acid exists under neutral conditions.

Question 3.
Glycine and alanine are different with respect to one substituent on the alpha carbon. What are the other common substituent groups?
Answer:
Common substituent groups on the alpha carbon of Glycine and Alanine are hydrogen, carboxyl group and amino group.

Question 4.
Starch, cellulose, glycogen, and chitin are polysaccharides found among the following.
Choose the one appropriate against each.
a) Cotton fibre ___________
b) Exoskeleton of cockroach ___________
c) Liver ___________
d) Peeled potato ___________
Answer:
a) Cotton fibre : Cellulose
b) Exoskeleton of cockroach : Chitin
c) Liver : Glycogen
d) Peeled potato : Starch

Question 5.
What are primary, secondary metabolites ? Give examples.
Answer:
Primary metabolites:
The metabolites which have identifiable functions and play known roles in normal physiological processes are called primary metabolites.
Ex: Carbohydrates, lipids, proteins, Aminoacids

Secondary metabolites:
The metabolic products that do not have identifiable functions in the host organism are called secondary metabolites. Many of them are useful to human welfare.
Ex: Rubber, drugs, spices, scents, pigments, Alkaloids, Lectins etc.

Question 6.
Distinguish between apoenzyme and cofactor.
Answer:

• Apoenzyme: The protein part of the enzyme is called apoenzyme.
• Cofactor: Non-protein part of a holoenzyme is a co-factor.

Question 7.
How are prosthetic groups different from coenzymes ?
Answer:
Prosthetic Group:
Non-proteinaceous carbon cofactor that is tightly attached to the Apoenzyme is called the prosthetic group.
Ex: Haeme group of peroxidase.

Co-factor:
Non-protein part of the holoenzyme is called cofactor. It may be a metal ion co-factor or an organic cofactor.
Ex: Zn, NAD.

Question 8.
What are competitive enzyme inhibitors? Mention one example.
Answer:
The inhibitor closely resembles the substrate in its molecular structure and inhibits the activity of the enzyme known as ‘competitive inhibitor’.
Ex : Inhibition of succinic dehydrogenase by malonate which closely resembles the substrate succinate in structure.

Question 9.
Why are Oxidoreductases so named?
Answer:
Enzymes which catalyse oxidation and reduction between two substrates S and S’.
Malate + NAD → Oxaloacetate + NADH + H⁺.

V. Short Answer Questions

Question 1.
Schematically represent primary, secondary and tertiary structures of a hypothetical polymer using protein as an example.
Answer:

Primary Structure:

Definition: The primary structure of a protein is the linear sequence of amino acids linked by peptide bonds.

Representation:
Draw a straight line of circles or squares, each representing an amino acid. Label them with their respective one-letter codes (e.g., A for Alanine, R for Arginine, etc.). Connect these with lines to represent peptide bonds.

Secondary Structure:

Definition: The secondary structure refers to the local folding of the polypeptide chain into structures such as alpha helices and beta sheets.

Representation:

For the alpha helix, draw a spiral or coiled structure.
For the beta sheet, draw arrows pointing in the direction of the polypeptide chains. Indicate whether they are parallel or antiparallel by arranging the arrows accordingly.

Tertiary Structure

Definition: The tertiary structure is the overall three-dimensional shape of a single polypeptide chain, formed by the interactions between the side chains of the amino acids.

Representation:

Draw a more complex, folded structure that incorporates both the alpha helices and beta sheets from the secondary structure.
Indicate interactions such as hydrogen bonds, Van der Waals forces, electrostatic interactions, and disulfide bonds with different types of lines or symbols.

Question 2.
Nucleic acid exhibits secondary structure, justify with example.
Answer:

1. One of the secondary structures exhibited by DNA is the Watson-Crick model.
2. According to this model, DNA exists as a double helix.
3. The two polynucleotide strands are antiparallel, i.e., run in opposite directions.
4. The backbone is formed by the sugar-phosphate-sugar chain.
5. N₂ -bases are projected-perpendicular to the back bone, but face inside.
6. Adenine (A) and Guanine (G) of one strand pair with Thymine (T) and Cytosine (C) of other strands, respectively.
7. Two hydrogen bonds are present in between A and T. Three hydrogen bonds are present in between G and C.
8. Each strand looks like a helical staircase. Each step is represented by a pair of N₂ -bases.
9. At each step of ascent, the strand turns 36°
10. Ten steps or ten base pairs are present in one full turn of the helix.
11.  The pitch (coil) would be 34 Å. The distance between two successive base pairs would be 3.4 Å. This form of DNA is called B – DNA.

Question 3.
Explain briefly about polysaccharides.
Answer:

• Polysaccharides are polymeric carbohydrate molecules composed of long chains of monosaccharide units bound together by glycosidic bonds.
• The building blocks of polysaccharide are called monosaccharides.
• Cellulose is a homopolymer as it consists of only one type of monosaccharide called glucose. It is a structural polysaccharide present in the cell walls of plants and other organisms.
• Paper made from plant pulp is cellulose.
• Starch is a homopolymer of glucose and is used as energy storage in plant tissues.
• Glycogen is a branched homopolymer and is used as energy storage in animal cells.
• Inulin is a homopolymer of fructose and is used as energy storage in tuberous roots or stems. Ex: Asteraceae.
• In a polysaccharide chain, the right end is called the reducing end and the left end is called the non-reducing end.
• Complex polysaccharides possess amino-acids and chemically modified sugars (glucosamine, N-acetyl galactosamine etc).
• Exoskeleton of arthropods and the cell wall of fungi have a complex polysaccharide called chitin.

Question 4.
Explain how pH affects enzyme activity with the help of a graphical representation.
Answer:

• Enzymes generally function in a narrow range of pH.
• Each enzyme shows its highest activity at a particular pH, called optimum pH.
• Mostly intracellular enzymes function near neutral pH.
• Different digestive enzymes have different optimum pH.
  Ex: Pepsin at 2.0, trypsin at 8.0.
• Change in pH above or below the optimum value within range reduces the rate of enzyme action.

Question 5.
Explain the mechanism of enzyme action.
Answer:

• The chemical which is converted into a product is called a ‘substrate’.
• Hence enzymes, i.e, proteins with three dimensional structures including an ‘active site’, convert a substrate (S) into a product (P).
• Symbolically, this can be depicted as: S → P

Nature of Enzyme action:

• Each enzyme (E) has a substrate (S) binding site in its molecule so that a highly reactive enzyme – substrate complex (ES) is produced.
• This complex is short – lived and dissociates into its product(s) P and the unchanged enzyme, with an intermediated formation of the enzyme – product complex (EP).
• The formation of the ES complex is essential for catalysis.
• E + S → (ES) (EP) → E + P
• Formation of (ES) complex has been explained with the ‘lock and key’ hypothesis by Emil fischer (1884) and much later with the ‘induced – fit hypothesis’ by Daniel E.Koshland.

The catalytic cycle of an enzyme action :

• First, the substrate binds to the active site of the enzyme,, fitting into the active site.
• The binding of the substrate induces the enzyme to alter its shape, fitting more tightly around the substrate.
• The active site of the enzyme, now in close proximity to the substrate, breaks the chemical bonds of the substrate and the new enzyme – product complex is formed.
• The enzyme releases the products of the reaction and the free enzyme is ready to bind to another molecule of the substrate and runs through the catalytic cycle once again.

Question 6.
Define enzyme inhibition. Write briefly about competitive inhibition, give an example.
Answer:

• The activity of an enzyme is also sensitive to the presence of specific chemicals that bind to the enzyme.
• When the binding of the chemical shuts off enzyme activity, the process is called inhibition and the chemical is called an inhibitor.
• When the inhibitor closely resembles the substrate in its molecular structure and inhibits the activity of the enzyme, it is known as a competitive inhibitor.
• Due to its close structural similarity with the substrate, the inhibitor competes with the substrate for the substrate-binding site of the enzyme.
• Consequently, the substrate cannot bind and as a result, the enzyme action declines.
  Ex : Inhibition of succinic dehydrogenase by malonate which closely resembles the substrate succinate in structure.
• Such competitive inhibitors are often used in the control of bacterial pathogens.

Question 7.
Explain different types of cofactors.
Answer:

• There are a number of non-protein constituents called co-factors which are bound to the apoenzyme to make the holoenzyme catalytically active.
• The protein portion of the enzyme is called the apoenzyme, the non-protein part of the holoenzyme is called co-factor.
• Three kinds of cofactors are :
  • Prosthetic groups
  • Coenzymes
  • Metal ions.

a) Prosthetic groups: Organic compounds that are tightly bound to the apoenzyme.
Ex : Haeme group in peroxidase enzyme.

b) Coenzymes: Organic compounds that are loosely bound to apoenzymes.
Ex : Nicotinamide adenine dinucleotide (NAD) and NADP contain the vitamin niacin, TPP.

c) Metal ions: They form coordination bonds with side chains at the active site and at the same time form one or more coordination bonds with the substrate.
Ex : Zinc is a co-factor for Carboxypeptidase. Copper for Cytochrome oxidase.

VI. Long Answer Questions

Question 1.
What are secondary metabolites? Enlist them indicating their usefulness to man.
Answer:

• Secondary metabolites: Metabolic products that do not have identifiable functions in the host organism are called secondary metabolites.
• Thousands of compounds found in plant, fungal and microbial cells other than primary metabolites are called secondary metabolites’.
  Eg : Alkaloids, flavonoids, rubber, essential oils, antibiotics, coloured pigments, scents, gums, spices, etc.

Some secondary metabolites:

Many secondary metabolites are useful to human welfare’.
Ex: Rubber, drugs, spices, scents and pigments.

1. Rubber :

• Uncured rubber is used for adhesive, insulating and friction tapes.
• Other significant uses of rubber are manufacturing of belts, matting, flooring, medical gloves and much more Used rubber tyres are often recycled to make other items like shoes, bags, coats.

2. Drugs:

In medicine:

• Antidiabetic drug is used to treat diabetes mellitus.
• Antihistamine medicine is used to treat allergies and hypersensitivity reactions and cold.
• Anti-inflammatory drug is intended to reduce inflammation.

In sports :
Anabolic steroids are synthetic substances that stimulate proteins that help in building non-fat muscle mass, helping an athlete become stronger and able to play for longer periods of time.

3. Spices:

1. Cloves : Cloves offer health benefits for reducing intestinal worms, digestive discomfort and can be used topically for toothache.
2. Cardamom : This spice has been shown to reduce cancer development in animal studies and increase cell death of cancer cells in the colon. This herb can be used for its diuretic benefits.
3. Asafoetida It is used as a remedy for asthma and bronchitis. It has antiflatulent and antimicrobial properties.

4. Scents : These are used at various places like retail space with customers, hotel lobby with guests, an office space with clients and employees. Scents smell has a strong influence on the emotions we feel in our daily lives. Fragrances make clothes smell clean, cosmetics pretty’ and households ‘well kept’.

5. Pigments These are used in food colouring, water colour paints, clothing dyes Green tea guards against cardiovascular disease.

Question 2.
What are the processes used to analyse elemental composition, organic constituents and inorganic constituents of living tissue?
Answer:

• Chemical analysis of a living tissue : On elemental analysis of a plant tissue, animal tissue or a microbial paste, a list of elements like carbon, hydrogen, oxygen and several others and their content per unit mass of a living tissue is known.
• The relative abundance of carbon and hydrogen with respect to other elements is higher in any living organism than in earth’s crust.

A comparison of elements present in Non-living and Living matter

• To analyse the organic compounds in a living tissue, any living tissue should be taken and grind it in trichloroacetic acid (Cl₃CCOOH) using a mortar and a pestle.
• The obtained thick slurry should be strained through a cheesecloth or cotton.
• Two fractions can be obtained.
  • Filtrate or acid soluble pool, and
  • Retentate or acid insoluble fraction.
• Thousands of organic compounds can be found in acid soluble pools.
• To analyse a living tissue sample and to identify a particular organic compound, first the compounds should be extracted.
• Then the extract should be subjected to various separation techniques to separate a compound from all other compounds.
• The isolated compound should be purified. All the carbon compounds obtained from living tissue are called ‘biomolecules’.
• Living organisms also contain inorganic elements and compounds in them.
• To analyse inorganic elements and compounds in living organisms a small amount of a living tissue should be weighed (wet weight) and it should be dried.
• As a result of this water evaporated. The remaining material gives dry weight.
• Now, the tissue should be burnt. Due to this, all the carbon compounds are oxidised to gaseous form (CO₂ and water vapour) and are removed.
• The remaining substance is called ‘ash’.
• This ash contains inorganic elements like calcium, magnesium, etc. Inorganic compounds like sulphate phosphate, etc. are also seen in the acid soluble fraction.

A list of representatives inorganic constituents of living tissues

• Therefore, element analysis gives elemental composition of living tissues in the form of hydrogen, oxygen, chlorine, carbon, etc.
• The analysis of compounds gives the analysis of organic and inorganic constituents present in living tissues.

Question 3.
Write an account of the classification of enzymes.
Answer:

• Thousands of enzymes have been discovered, isolated and studied. Most of these enzymes have been classified into different groups based on the type of reactions they catalyse.
• Enzymes are divided into 6 classes each with 4-13 subclasses and named accordingly by a four-digit number.

a) Oxidoreductases/dehydrogenases:

Enzymes which catalyse oxidoreduction between two substrates S and S’.
Ex: S reduced + S’ oxidised → S oxidised + S’ reduced.

b) Transferases

Enzymes catalysing a transfer of a group, G (other than hydrogen) between a pair of substrate S and S’.
Ex: S – G + S’ → S + S’ -G

c) Hydrolases
Enzymes catalysing hydrolysis of ester, ether, peptide, glycosidic, C-C, C-halide or P-N bonds.

d) Lyases
Enzymes that catalyse removal of groups from substrates by mechanisms other than hydrolysis leaving double bonds.

e) Isomerases
Includes all enzymes catalysing inter-conversion of optical, geometric or positional isomers.

f) Ligases
Enzymes catalysing the linking together of 2 compounds.
Ex: Enzymes which catalyse joining of C-O, C-S, C-N, P-0 etc. bonds.

I. Multiple Choice Questions

Question 1.
Chemical analysis of living organisms can be done by the usage of the following chemical.
1. Ethanol
2. Benzene
3. Trichloro acetic acid
4. Acetic acid
Answer:
3. Trichloro acetic acid

Question 2.
Identify the polysaccharide which is a polymer of fructose.
1. Starch
2. Glycogen
3. Cellulose
4. Inulin
Answer:
4. Inulin

Question 3.
Identify the aromatic amino acid.
1. Glutamic acid
2. Tyrosine
3. Lysine
4. Alanine
Answer:
2. Tyrosine

Question 4.
Palmitic acid contains how many carbons ?
1. 16
2. 20
3. 15
4. 19
Answer:
1. 16

Question 5.
Cytidylic acid is a
1. Nitrogen base
2. Nucleotide
3. Nucleoside
4. Nucleic acid
Answer:
2. Nucleotide

Question 6.
Concanavalin A is
1. Drug
2. Lectins
3. Alkaloids
4. Toxins
Answer:
2. Lectins

Question 7.
Identify the secondary metabolite.
1. Amino acid
2. Nucleic acids
3. Carbohydrates
4. Rubber
Answer:
4. Rubber

Question 8.
Which among the following is not a pyrimidine ?
1. Thymine
2. Cytosine
3. Adenine
4. Uracil
Answer:
3. Adenine

Question 9.
The following protein enables glucose uptake into cells.
1. GLUT-4
2. Collagen
3. Antibody
4. Trypsin
Answer:
1. GLUT-4

Question 10.
The following structure is necessary for the many biological activities of proteins.
1. Primary structure
2. Secondary structure
3. Tertiary structure
4. Quaternary structure
Answer:
3. Tertiary structure

Question 11.
Which among the following is not a polymer ?
1. Polysaccharide
2. Protein
3. Nucleic acid
4. Lipid
Answer:
4. Lipid

II. Fill in the Blanks

Question 1.
The pentose sugar in DNA is ___________
Answer:
Deoxyribose sugar

Question 2.
Nucleic acid percentage in the total cellular mass is ___________
Answer:
5 – 7%

Question 3.
Most abundant protein in the biosphere ___________
Answer:
RUBISCO

Question 4.
The molecular weight of macromolecules is greater than ___________ Daltons.
Answer:
1000

Question 5.
The amino acids in protein are linked by ___________ bond.
Answer:
Peptide

Question 6.
Exoskeleton of arthropods is made up of ___________
Answer:
Chitin

Question 7.
The R-group of the serine is ___________
Answer:
CH₂OH

Question 8.
The macromolecule which is found in acid insoluble fraction ___________
Answer:
Lipids

Question 9.
Coenzyme NAD and NADP contain ___________ vitamin.
Answer:
Niacin

Question 10.
Most of the enzymes get damaged/ denatured above ___________ degrees temperature.
Answer:
40

III. One Word Answer Questions

Question 1.
Name the pyrimidine which is absent in the RNA.
Answer:
Thymine

Question 2.
Name the acid which is formed in our skeletal muscle, under anaerobic conditions.
Answer:
Lactic acid

Question 3.
In the presence of carbonic anhydrase how many H₂CO₃ molecules can be formed per sec.
Answer:
6 lakhs

Question 4.
What is the metal ion cofactor for the proteolytic enzyme carboxypeptidase?
Answer:
Zn

Question 5.
Name the non-protein constituent of the enzyme.
Answer:
Cofactor

Question 6.
What is the most abundant protein in the animal world?
Answer:
Collagen

Question 7.
Among the starch and cellulose which one holds iodine and gives blue colour?
Answer:
Starch

Question 8.
Nucleic acids with catalytic power are known as _______ ?
Answer:
Ribozyme

Question 9.
Name the first and last amino acids of a polypeptide chain.
Answer:
N-terminal and C-terminal amino acids.

Question 10.
Enzymes are categorised into how many classes?
Answer:
6

Regular practice with AP Inter 1st Year Botany Study Material Chapter 5 Cell: The Unit of Life Questions and Answers helps students stay prepared for examinations.

AP Inter 1st Year Botany 5th Lesson Cell: The Unit of Life Questions and Answers

IV. Very Short Answer Questions

Question 1.
What is the significance of vacuole in a plant cell ?
Answer:
Vacuole plays an important role in osmoregulation and also contains sap mainly composed of water. Metabolic byproducts, excretions, other waste materials, and pigments like anthocyanins.

Question 2.
What does ‘S’ refer to in 70S & 80S ribosomes ?
Answer:
“S” stands for the sedimentation coefficient’ (expressed in the Svedberg unit). It is indirectly a measure of density and size of ribosbmes.

Question 3.
What is the function of a polysome ?
Answer:
Several ribosomes attach to a single m-RNA and form a chain called polyribosomes (or) polysome. The ribosomes of a polysome translate the genetic message of one m-RNA into proteins.

Question 4.
What is referred to as a satellite in some chromosomes ?
Answer:
A few chromosomes have non- staining secondary constrictions at a constant location. A round terminal part of the chromosome present beyond the secondary constriction is called satellite and such chromosomes are called satellite chromosomes.

Question 5.
What is middle lamella made of? What is its functional significance ?
Answer:
Middle lamella is mainly made of calcium pectate and magnesium pectate. lt holds or glues the different neighbouring cells together.

Question 6.
What is osmosis?

Answer:

V. Short Answer Questions

Question 1.
Briefly describe the cell theory.
Answer:

• In 1838, Mathias Schleiden, a German botanist, examined a large number of plants and observed that all plants are composed of different kinds of cells which form the tissues.
• Theodor Schwann in 1839, a German zoologist, studied different types of animal cells and reported that all cells possess an outer layer called “Plasma membrane”.
• He also concluded that plant cells have a characteristic cell wall that differentiates from animal cells.
• Schwann proposed the hypothesis that the bodies of animals and plants are composed of cells and products of cells.
• On the basis of these points, both Schleiden and Schwann together formulated cell theory. But they failed to explain how new cells are formed.
• Rudolf Virchow (1855) first explained that cells divide and new cells are formed from pre-existing cells (Omnis cellula-e cellula).
• He modified the hypothesis of Schleiden and Schwann to give the cell theory a final shape.
• Cell theory states that, all living Organisms are composed of cells and products of cells.
• Cell is the structural and functional unit of living organisms.
• All cells arise from pre-existing cells.

Question 2.
Give the biochemical composition of plasma membrane. How are lipid molecules arranged in the membrane?
Answer:

• Plasma membrane is chemically composed of lipids, proteins, and carbohydrates.
• So, plasma membrane is lipo-proteinaceous in nature.
• Depending on the location and ease of extraction, membrane proteins can be classified as integral and peripheral.
• Integral proteins are partially or totally buried in the membrane, whereas peripheral proteins lie on the surface of the membrane.
• In the cell membrane lipids are arranged in bilayer.
• The lipids are arranged within the membrane with the polar (hydrophilic) head towards the outer side and the non-polar (hydrophobic) tails towards the inner part, which ensures that the non-polar tail of saturated hydrocarbons is protected from the aqueous environment.
• Widely accepted model of plasma membrane is the “Fluid mosaic model” improved by Singer and Nicolson. According to this, the quasi-fluid nature of lipids, enables lateral movement of proteins within the overall bilayer.

Question 3.
Distinguish between active transport and passive transport?
Answer:

Question 4.
What are the characteristics of a prokaryotic cell?
Answer:

• Most prokaryotes, particularly bacterial cells, have a chemically complex cell envelope.
• Cell envelope consists of glycocalyx, cell wall and cell membrane.
• Glycocalyx may be in loose sheath form called slime layer (or) thick and tough form called capsule.
• There is no well-defined nucleus.
• The genetic material is naked and is not enveloped by nuclear membrane.
• DNA is double stranded, circular, without histones (Naked) called genophore and the area is called nucleoid.
• Prokaryotes like some bacteria possess extra chromosomal DNA called plasmid.
• Prokaryotes lack membrane bound organelles.
• 70S type of Ribosomes are present in the cytoplasm.
• Prokaryotes have invaginations in the cell membrane called ‘Mesosomes’.
• Some bacterial cells are able to move with flagella.
• Flagella has three parts namely basal body, hook and filament which is the longest portion. Flagella play a role in motility.
• Besides flagella, pili and fimbriae are present.
• Pili are elongated tubular structures.
• Fimbriae are small bristle-like fibres and are useful for attachment of bacteria to rocks, in streams and also to the host tissues.
• Division of cells takes place amitotically.

Question 5.
Explain the structure of the cell organelle which contain chlorophyll, with a neat labelled diagram.
Answer:

• The cell organelle which contains chlorophyll pigments is chloroplast.
• These are green coloured plastids found in all green plant parts i.e., mesophyll cells of the leaves.
• Each chloroplast is bounded by a double membrane envelope and the gap present between the two membranes is called peri plastidial space of about 10 nm wide.
• It contains an inner space called stroma containing photosynthetic enzymes, circular, naked d.sDNA, RNA and 70S ribosomes.
• At certain places, flattened membranous sacs are present called thylakoids. They are arranged in stacks like the pile of coins called Grana.
• The large number of lipoprotein membranes present between grana are called stroma lamellae (or) stroma thylakoids.
• The empty space present in each thylakoid is called lumen.
• Chlorophylls and carotenoids which are present in the membranes of thylakoids are responsible for trapping light energy.
• DNA present in a chloroplast helps in self duplication. So they are called semi- autonomous cell organelles.

Functions:

• Chloroplasts are mainly concerned with assimilation of food materials by photosynthetic process.
• Light reaction of photosynthesis takes place in grana and carbon fixation (dark reaction) takes place in the stroma region.

Question 6.
Explain the structure of the cell organelle which is known as ‘Powerhouse of the cell’ with a neat labelled diagram.
Answer:

• ‘Mitochondria’ are commonly called powerhouses of cell.
• Mitochondria occur in all eukaryotic cells.
• Mitochondria is sausage – shaped or cylindrical having a diameter of 0.2 -1.0 Micrometer and a length of 1.0 – 4.1 Micrometer.
• Mitochondria is a double membrane bound cell organelle and the space present between the two membranes is called peri mitochondrial space.
• Outer membrane is continuous while the inner membrane forms a number of infoldings called cristae.
• Many stalked particles present on the.surface of the cristae are called F₀ – F₁ particles.
• Inner space of mitochondria is filled with a fluid matrix which contains circular, naked dsDNA, RNA, 70S ribosomes and respiratory enzymes.
• Krebs cycle of aerobic respiration occurs in matrix and electron transport takes place in cristae.
• Mitochondria divide by fission.

Functions:
Mitochondria are concerned with cellular respiration and are involved in oxidation of food material. They produce energy in the form of ATP and hence are called powerhouses of cell.

Question 7.
Differentiate between rough endoplasmic reticulum (RER) and smooth endoplasmic reticulum (SER).
Answer:

Question 8.
Explain the structure of the nucleus with a neat labelled diagram.
Answer:

• Nucleus as a cell organelle was first described by Robert Brown.
• Nucleus is also a double membrane bound cell organelle and the space present between the two membranes is called perinuclear space.
• Nucleus consists of highly extended and élaborated nucleoprotein fibres called chromatin, nuclear matrix and one (or) more spherical bodies called nucleoli.
• Outer membrane of the nucleolar envelope usually remains continuous with the E.R. and also bears 80S ribosomes on it.
• At a number of places the nuclear envelope is interrupted by minute pores called nuclear pores which connect the nucleoplasm with cytoplasm for the movement of RNA and protein molecules.
• The nuclear matrix (or) the nucleoplasm contains nucleolus and chromatın.
• Chromatin contains DNA and some basic proteins called histones, some non-histone proteins and RNA.
• Chromatin transforms into chromosomes during cell division.
• Nucleoli are spherical structures present in the nucleoplasm.
• The content of the nucleolus is continuous with the rest of the nucleoplasm as it is not a membrane bounded structure.
• Nucleolus is the site for active r-RNA synthesis.
• Larger and more numerous nucleoli are present in the cells which are actively carrying out protein synthesis.

Function :
Nucleus controls and regulates the functions of the cell.

Question 9.
Classify the chromosomes on the basis of centromere location.
Answer:
A chromosome with one centromere is called a monocentric chromosome. Based on the position of centromere, monocentric chromosomes are four types. They are,

a) Metacentric:
Centromere present in the middle point of the chromosome. So both arms of the chromosomes are equal in length. Chromosome appears L-shaped during anaphase of cell division.

b) Sub-metacentric:
Centromere is slightly away from the centre of the chromosome. So, the chromosome has two unequal arms. During anaphase such chromosomes appear as ‘L’ shaped.

c) Acrocentric:
In a chromosome, centromere is situated close to its end forming one extremely short and one very long arm. So chromosomes appear as ‘J’ shaped during anaphase.

d) Telocentric:
Centromere is present at the terminal end of chromosome. So the telocentric chromosome has only one arm. During anaphase it appears ‘I’ shaped.

VI. Long Answer Questions

Question 1.
What structural and functional attributes must a cell have to be called a living cell ?
Answer:

• A living cell must possess a plasma membrane to define its boundaries, contain cytoplasm for essential cellular processes, and have genetic material (DNA) for heredity and cell function.
• Additional attributes include being composed of protoplasm, which may be covered by a cell wall, and possessing at least some sub-cellular organelles like ribosomes.

Structural Attributes:

Plasma Membrane:
This outer covering separates the cell’s internal environment from the outside, allowing selective passage of materials.

Cytoplasm:
The semi-fluid matrix within the cell where various cellular components and chemical reactions occur.

Genetic Material (DNA):
This material, found in chromosomes or a nucleus, carries the cell’s genetic information and is responsible for heredity.

Protoplasm:
The living substance within the cell, bounded by the plasma membrane, and may be covered by a cell wall in some cases.

Sub-cellular Organelles (e.g., ribosomes):
These structures within the cytoplasm carry out specific functions essential for cell life.

Functional Attributes:

Metabolism:
Living cells must be capable of carrying out various metabolic processes, including energy production, nutrient processing, and waste excretion.

Regulation and Homeostasis:
Cells must be able to regulate their internal environment and maintain a stable internal state.

Response to Stimuli:
Living cells must be able to detect and respond to changes in their environment.

Adaptation and Reproduction:
Cells must be able to adapt to their environment through changes in DNA and reproduce to perpetuate life.

Question 2.
Write the functions of the following.
a) Centromere
b) Cell wall
c) Smooth ER
d) Golgi apparatus
Answer:
a) Functions of centromere:

• Centromere (or) primary constriction divides the chromosomes into two equal arms (or) two unequal arms based on its position in the chromosomes.
• Centromere has disc shaped structures on its either side called kinetochores, which are the sites of implantation of spindle fibres during cell division.
• Thus helpful in the formation of daughter chromosomes during cell division.

b) Functions of a cell wall:

• It gives shape to the cell.
• It protects the cell from mechanical damage and infection.
• It helps in cell-to-cell interaction.
• It acts as a barrier to undesirable macro molecules.

c) Functions of smooth E.R :

• Smooth endoplasmic reticulum is the major site for synthesis of lipids.
• In animal cells, lipids – like steroidal hormones are synthesized in smooth endoplasmic reticulum.

d) Functions of golgi apparatus:

• Golgi apparatus is the important site of formation of glycoproteins and glycolipids.
• In plants, golgi apparatus involved in the synthesis of cell wall materials like cellulose, hemi cellulose etc.
• It plays a role in the formation of cell plates during cell division.
• Golgi apparatus is involved in the formation of lysosomes.

Question 3.
Describe the structure of the following with the help of labelled diagrams.
a) Nucleus
b) T.S. of flagellum
Answer:
a) Nucleus:

• Nucleus as a cell organelle was first described by Robert Brown.
• Nucleus is also a double-membrane-bound cell organelle and the space present between the two membranes is called perinuclear space.
• Nucleus consists of highly extended and elaborated nucleoprotein fibres called chromatin, nuclear matrix and one (or) more spherical bodies called nucleoli.
• Outer membrane of the nucleolar envelope usually remains continuous with the E.R. and also bears 80S ribosomes on it.
• At a number of places the nuclear envelope is interrupted by minute pores called nuclear pores which connect the nucleoplasm with cytoplasm
• for the movement of RNA and protein molecules.
• The nuclear matrix (or) the nucleoplasm contains nucleolus and chromatin.
• Chromatin contains DNA and some basic proteins called histones, some non-histone proteins and RNA.
• Chromatin transforms into chromosomes during cell division.
• Nucleoli are spherical structures present in the nucleoplasm.
• The content of the nucleolus ft continuous with the rest of the nucleoplasm as it is not a membrane bounded structure.
• Nucleolus is the site for active r-RNA synthesis.
• Larger and more numerous nucleoli are present in the cells which are actively carrying out protein synthesis.

Function:
Nucleus controls and regulates the functions of the cell.

T.S of flagellum:

• The electron microscopic study of the flagellum shows that they are covered with plasma membrane.
• Their core, called the axoneme, possesses a number of microtubules running parallel to the long axis.
• The axoneme usually has nine doublets of radially arranged peripheral microtubules, and a pair of centrally located microtubules.
• Such an arrangement of axonemal microtubules is referred to as the 9 + 2 array.
• The central tubules are connected by bridges and are also enclosed by a central sheath, which is connected to one of the tubules of each
• peripheral doublets by a radial spoke.
• Thus, there are nine radial spokes.
• The peripheral doublets are also interconnected by linkers.
• Both the cilium and flagellum emerge from centriole-like structures called the basal bodies.

Question 4.
Name two cell- organelles that are double membrane bound. State their functions and draw labelled diagrams of both.
Answer:

Chloroplast:

• The cell organelle which contains chlorophyll pigments is chloro¬plast.
• These are green coloured plastids found in all green plant parts i.e., mesophyll cells of the leaves.
• Each chloroplast is bounded by a double membrane envelope and the gap present between the twO membranes is called peri plastidial space of about 10 nm wide.
• It contains an inner space called stroma containing photosynthetic enzymes, circular, naked dsDNA, RNA and 70S ribo-somes.
• At certain places, flattened membranous sacs are present called thylakoids. They are arranged in stacks like the pile of coins called Grana.
• The large number of lipoprotein membranes present between grana are called stroma lamellae (or) stroma thylakoids.
• The empty space present in each thylakoid is called lumen.
• Chlorophylls and carotenoids which are present in the membranes of thylakoids are responsible for trapping light energy.
• DNA present in a chloroplast helps in self duplication. So they are called semi- autonomous cell organelles.

Functions:

• Chloroplasts are mainly concerned with assimilation of food materials by photosynthetic process.
• Light reaction of photosynthesis takes place in grana and carbon fixation (dark reaction) takes place in the stroma region.

Mitochondria:

• ‘Mitochondria are commonly called powerhouses of cell.
• Mitochondria occur in all eukaryotic cells.
• Mitochondria is sausage – shaped or cylindrical having a diameter of 0.2 -1.0 Micrometer and a length of 1.0 – 4.1 Micrometer.
• Mitochondrion is a double membrane bound cell organelle and the space present between the two membranes is called peri mitochondrial space.
• Outer membrane is continuous while the inner membrane forms a number of infoldings called cristae.
• Many stalked particles present on the surface of the cristae are called F₀ – F₁ particles.
• Inner space of mitochondria is filled with a fluid matrix which contains circular, naked dsDNA, RNA, 70S ribosomes and respiratory enzymes.
• Krebs cycle of aerobic respiration occurs in matrix and electron transport takes place in cristae.
• Mitochondria divide by fission.

Functions:

• Mitochondria are concerned with cellular respiration and are involved in oxidation of food material.
• They produce energy in the form of ATP and hence are called powerhouses of cell.

I. Multiple Choice Questions

Question 1.
Identify the cell organelle that lacks a membrane.
1. Vacuole
2. Nucleus
3. Ribosome
4. Mitochondria
Answer:
3. Ribosome

Question 2.
Identify the cell which does not contain a cell wall.
1. Bacteria
2. Blue-green algae
3. Mycoplasma
4. Plant cell
Answer:
3. Mycoplasma

Question 3.
The protein-lipid ratio in the cell membrane of human erythrocytes.
1. 40 : 52
2. 52 : 40
3. 42 : 50
4. 50 : 42
Answer:
2. 52 : 40

Question 4.
Movement of molecules from lower concentration to higher concentration by utilising energy is
1. Diffusion
2. Osmosis
3. Passive transport
4. Active transport
Answer:
4. Active transport

Question 5.
Middle lamella is made up of
1. Cellulose
2. Calcium pectate
3. Chitin
4. Lignin
Answer:
2. Calcium pectate

Question 6.
Which cell organelle is important site for formation of glycoproteins and glycolipids ?
1. Golgi apparatus
2. Er
3. Ribosomes
4. Peroxisomes
Answer:
1. Golgi apparatus

Question 7.
In a plant cell the vacuole can occupy upto
1. 70% of cell volume
2. 90% of cell volume
3. 80% of cell volume
4. 60% of cell volume
Answer:
2. 90% of cell volume

Question 8.
The protein storing leucoplast is
1. Amyloplast
2. Aleuroplast
3. Elaioplast
4. Chromoplast
Answer:
2. Aleuroplast

Question 9.
Which among the following does not come under the cytoskeleton ?
1. Microtubules
2. Microfilaments
3. Cilia
4. Intermediate filaments
Answer:
3. Cilia

Question 10.
A chromosome has extremely short and very long arm, identify the possible chromosome type.
1. Metacentric
2. Sub-metacentric
3. Acrocentric
4. Telocentric
Answer:
3. Acrocentric

Question 11.
Ribosomes are site for the synthesis of
1. Lipids
2. Proteins
3. Carbohydrates
4. Nucleic acids
Answer:
2. Proteins

Question 12.
Identify the smallest living cell.
1. Cells of all living organisms have a nucleus.
2. Both animal and plant cells have a well-defined cell wall.
3. In prokaryotes, there are no membrane bound organelles.
4. Cells are formed de novo from abiotic materials.
Answer:
2. Both animal and plant cells have a well-defined cell wall.

Question 13.
Identify the cell organelle which is not considered as part of the endomembrane system.
1. ER
2. Golgi complex
3. Peroxisomes
4. Lysosomes
Answer:
3. Peroxisomes

Question 14.
According to Rudolf Virchow new cells generate from
1. Bacterial fermentation
2. De novo synthesis
3. Pre-existing cells
4. Abiotic materials
Answer:
3. Pre-existing cells

Question 15.
Which of the following is correct ?
1. Cells of all living organisms have a nucleus.
2. Both animal and plant cells have a well-defined cell wall.
3. In prokaryotes, there are no membrane bound organelles.
4. Cells are formed de novo from abiotic materials.
Answer:
3. In prokaryotes, there are no membrane bound organelles.

Question 16.
Which of the following is not correct ?
1. Robert Brown discovered the cell.
2. Schleiden and Schwann formulated the cell theory.
3. Virchow explained that cells are formed from pre-existing cells.
4. A unicellular organism carries out its life activities within a single cell.
Answer:
1. Robert Brown discovered the cell.

II. Fill In the Blanks

Question 1.
The smaller subunit of 80S ribosome is ____________
Answer:
40S

Question 2.
Nucleolus is a site for ____________ RNA synthesis.
Answer:
Ribosomal

Question 3.
Chloroplast number in Chlamydomonas cell ____________
Answer:
One

Question 4.
In cell membranes ____________ proteins are partially or totally buried.
Answer:
Integral

Question 5.
The peripheral fibrils of centriole are made up of ____________ protein.
Answer:
Tubulin

Question 6.
The membrane of the vacuole is named as ____________
Answer:
Tonoplast

Question 7.
The contractile vacuole in Amoeba is important for ____________.
Answer:
Osmoregulation/ Excretion

Question 8.
Arrangement of axonemal microtubules in cilia or flagella is ____________ array.
Answer:
9 + 2

Question 9.
Chromatin contains DNA, RNA, Histones and ____________.
Answer:
Non-histone proteins

Question 10.
Small bristles like fibres sprouting out of the bacterial cell are known as ____________.
Answer:
Fimbriae

III. One Word Answer Questions

Question 1.
Mention a single membrane bound organelle which is rich in hydrolytic enzymes.
Answer:
Lysosomes

Question 2.
What is the location of a centromere in a metacentric chromosome?
Answer:
Middle of the chromosome

Question 3.
Which part of the bacterial cell is targeted in gram staining ?
Answer:
Cell wall

Question 4.
Which type of ribosomes are present in chloroplast and mitochondria ?
Answer:
70S

Question 5.
Name the fat-soluble pigments which are present in chromoplast.
Answer:
Carotenoid pigments

Question 6.
Who discovered the ribosomes ?
Answer:
Palade

Question 7.
Name the cell organelle which is called the powerhouse of the cells.
Answer:
Mitochondria

Question 8.
Name the model which was proposed by Singer and Nicolson to explain the cell membrane structure.
Answer:
Fluid Mosaic model

Question 9.
Who discovered the living cell ?
Answer:
Leeuwenhoek

Question 10.
In which form of cellular energy does the mitochondria produce ?
Answer:
ATP

Regular practice with AP Inter 1st Year Botany Study Material Chapter 4 Anatomy of Flowering Plants Questions and Answers helps students stay prepared for examinations.

AP Inter 1st Year Botany 4th Lesson Anatomy of Flowering Plants Questions and Answers

IV. Very Short Answer Questions

Question 1.
A transverse section of a plant shows conjoint, scattered vascular bundles surrounded by sclerenchymatous bundle sheath. What will you identify it as?
Answer:
Monocot stem

Question 2.
Which parts of the leaf surface reduce the rate of transpiration?
Answer:
The waxy cuticle on the leaf surface and the number and position of stomata on the leaf surface both reduce the rate of transpiration.

Question 3.
What are the cells that make the leaves curl in plants during water stress? Give an Example.
Answer:
Bulliform cells.
Ex: Monocot leaves (Grass).

Question 4.
How is the study of plant anatomy useful to us?
Answer:

• Anatomy helps in understanding internal structure or arrangement of various tissues of plants.
• Anatomy helps in understanding structural details of different organs of a plant, anatomical differences found in between different kinds of plants like monocots and dicots, and also helps in knowing adaptations found in plants to diverse environments

Question 5.
Name any two simple tissues found in the ground tissue system.
Answer:
Parenchyma and collenchyma are two simple tissues found in the ground tissue system of plants.

Question 6.
What are the shapes of guard cells in dicot leaf and monocot leaf respectively?
Answer:

• In Dicot leaf- guard cells are Kidney or bean-shaped.
• In Monocot leaves- guard cells are dumbbell-shaped.

Question 7.
What are casparian strips? In which part of the plant anatomy do you find them?
Answer:
Casparian strips are ring-like cell wall thickenings found in the endodermis of the roots of vascular plants.

Question 8.
Write three words to describe the vascular bundle of a dicot stem.
Answer:
Vascular bundles in dicot stem are: conjoint (xylem and phloem are arranged in the same radius), open (a strip of cambium is present in between xylem and phloem) and endarch (protoxylem is present towards the inside and metaxylem is present towards periphery).

V. Short Answer Questions

Question 1.
Write any four differences between monocot leaf and dicot leaf.
Answer:

Question 2.
Cut a transverse section of the young stem of a plant from your garden and observe it under the microscope. How would you ascertain whether it is a monocot stem or a dicot stem? Give reasons.
Answer:
I) The given transverse section of the young stem can be ascertained as a dicot stem based on the following characters.

• The epidermis bear trichomes.
• Cortex is well distinguished into three subzones viz., collenchymatous hypodermis, general cortex, and strachy endodermis.
• Pericycle is present as semilunar patches of sclerenchyma, between endodermis and phloem of vascular bundles.
• A large number of vascular bundles are arranged in a ring around the medulla.
• Vascular bundles are conjoint and open.
• Medulla (or) pith and Medullary rays are present.

II) The given transverse section of the young stem can be ascertained as monocot stem based on the following characters.

• Trichomes are absent in the epidermis.
• Cortex is highly reduced and represented by sclerenchyrhatous hypodermis.
• Endodermis and pericycle are absent.
• A large number of vascular bundles surrounded by a sclerenchymatous bundle sheath are scattered in the ground tissue.
• Vascular bundles are conjoint and closed and phloem parenchyma is absent. In the vascular bundles, water-containing lysigenous cavities are present.
• Peripheral vascular bundles are generally smaller than the centrally located ones.
• Medulla (or) pith and medullary rays are absent.

Question 3.
The stomatal pore is guarded by two kidney-shaped guard cells. Name the epidermal cells surrounding the guard cells. How does a guard cell differ from an epidermal cell? Use a diagram to illustrate your answer.
Answer:

• • The stomatal pore is guarded by two kidney-shaped guard cells.
  • These guard cells are surrounded by the epidermal cells known as subsidiary cells or accessory cells.

Question 4.
Name the three basic tissue systems in the flowering plants. Give the tissue names under each system.
Answer:
The three basic tissue systems of flowering plants are
a) Epidermal tissue system
b) Ground tissue system
c) Vascular tissue system

a) Epidermal tissue system:
It forms the outermost covering of the whole plant body and comprises Epidermis, Stomata, Epidermal appendages like the trichomes and root hairs.

b) Ground tissue system :
All tissue, except epidermis and vascular bundles constitute the ground tissues. It consists of hypodermis, general cortex, endodermis, pericycle, medulla and medullary rays.

c) Vascular tissue system:

• It consists of complex tissues like phloem and xylem, arranged in the form of bundles.
• Vascular Bundles are of Radial (Roots), Collateral (Stem) type.

VI. Long Answer Questions

Question 1.
Describe the internal structure of dorsiventral leaf with the help of labelled diagram.
Answer:
The transverse section of dorsiventral leaf (dicot leaf) shows three parts, namely

I) Epidermis,
II) Mesophyll,
III) Vascular bundle

I) Epidermis:

• A single-layered epidermis is present towards the lower (abaxial) and upper side (adaxial) of the leaf called lower epidermis and upper epidermis respectively.
• Both adaxial and abaxial epidermis of leaves covered by a cuticle.
• Stomata are more towards lower or abaxial epidermis.

II) Mesophyll:
Tissue present between upper and lower epidermis is called Mesophyll. It is of two kinds,

• Palisade parenchyma and
• Spongy parenchyma.

a) Palisade parenchyma:

• It is present below the upper epidermis that is towards the adaxial surface.
• It shows elongated cells which are arranged vertically and parallel to each other.

b) Spongy parenchyma:

• Part of the mesophyll towards the lower epidermis or abaxial surface is called spongy parenchyma.
• Spongy parenchyma is situated below the palisade cells and extends to the lower epidermis.
• It is made up of oval or round-shaped cells with large intercellular spaces.

III) Vascular bundle :

• Vascular bundles are extended in the mesophyll in the form of vein and the midrib.
• The vascular bundles are conjoint, collateral and closed.
• The xylem is present towards the upper side (adaxial) and phloem is present towards lower epidermis (abaxial) in each vascular bundle.
• The vascular bundle is surrounded by a layer of thick-walled cells called bundle sheath. The cells are arranged compactly.

Question 2.
What is a stomatal apparatus? Explain the structure of stomata with a labelled diagram.
Answer:

• Stomata are tiny pores in the epidermis of leaves and other plant parts.
• The stomata consist of minute pores called stoma surrounded by a pair of guard cells.
• Stomata, open and close according to the turgidity of guard cells.
• The cell wall surrounding the pore is tough and flexible.
• Guard cells are bean-shaped and contain chloroplasts.
• They contain chlorophyll and capture light energy.
• The Guard cells are surrounded by subsidiary cells.
• They are the accessory cells to guard cells and are found in the epidermis of plants.
• The stomatal aperture, guard cells and the surrounding subsidiary cells are together called stomatal apparatus.

Question 3.
Describe the T.S. of a dicot stem with the help of neat labelled diagram ?
Answer:
The transverse section of the young primary dicot stem is differentiated into
I) Epidermis
II) Cortex
III) Stele

I) Epidermis:

• It is the outermost protective layer of the stem.
• It is covered by a thin cuticle.
• In young stems epidermis has a few stomata and multicellular hairs called trichomes.

II) Cortex:

• The part below the epidermis is called the cortex.
• It is present between the epidermis and the pericycle of stele.
• It shows three subzones.
  • a) Hypodermis
  • b) General cortex
  • c) Endodermis

a) Hypodermis:

• It is the outermost part of the cortex.
• It is composed of a few layers of collenchymatous cells.
• It provides mechanical strength (tensile) to the young stem.

b) General Cortex:

• It is present between the hypodermis and endodermis.
• It is composed of thin-walled, round parenchyma cells with intercellular spaces.

c) Endodermis:

• The innermost layer of the cortex is called endodermis.
• Due to storage of large quantities of starch in the cells of endodermis, it is also called as “Starch sheath”.

III) Stele:

• The central cylinder concerned with conduction of water and minerals is called stele.
• It is differentiated into
  a) Pericycle
  b) vascular bundles
  c) Medulla (or) Pith.

a) Pericycle:

• It is the outermost layer of the stele.
• It is present in the form of semilunar patches of sclerenchyma in between endodermis and phloem of vascular bundles.

b) Vascular bundles:

• A large number of vascular bundles are arranged in the form of a ring.
• This arrangement is called Eustele.
• Vascular bundles are conjoint, collateral, endarch, and open type.

c) Medulla or Pith:
It is composed of a large number of rounded parenchymatous cells having intercellular spaces.

d) Medullary rays:
In between the vascular bundles there are a few layers of radially placed living parenchyma cells, which constitute medullary rays.

Question 4.
Describe the T.S. of a monocot stem with the help of labelled diagram ?
Answer:
T.S of young monocotyledonous stem shows Epidermis, hypodermis, ground tissue and vascular bundles.

A) Epidermis:

• The outermost layer composed of compactly arranged cells.
• It is covered by a thin layer of cuticle.
• Trichomes are absent.

B) Hypodermis:

Below the epidermis, sclerenchymatous hypodermis is present, which gives mechanical support to the stem.

C) Ground tissue:

• Below the hypodermis, parenchymatous ground tissue is present with intercellular spaces.
• Endodermis and pericycle are absent.

D) Vascular bundles:

• A large number of vascular bundles are scattered in the ground tissue, so-called atactostele.
• Each vascular bundle is Collateral, conjoint, endarch, and closed type.
• Each vascular bundle is surrounded by sclerenchymatous bundle sheath.
• The phloem parenchyma is absent and water-containing cavities are present.
• Medulla and medullary rays are absent.

Question 5.
Describe the internal structure of a dicot root with the help of labelled diagram ?
Answer:
The internal structure of the dicot root shows Epidermis, cortex, and stele.

A) Epidermis:

• The outermost layer is made up of compactly arranged elongated cells called Epiblema.
• Many of the cells produce unicellular root hairs which help in absorption of water from the soil.

B) Cortex:

• It is present in between the epidermis and the stele.
• Below the epidermis, a several-layered parenchymatous general cortex is present.
• The cells are thin walled living.

Endodermis:

• The innermost layer of the cortex is composed of a single layer of barrel-shaped cells.
• The radial and transverse walls are wrapped by ligno-suberised bands called Casparian thickenings.

C) Stele:
It is the central part of the root and shows pericycle, vascular tissues, pith and conjunctive tissue.

i) Pericycle:

• Below the endodermis, a single layer of thin-walled parenchymatous cells without intercellular spaces is present called the pericycle.
• These cells help in producing lateral roots and also become meristematic during secondary growth.

ii) Vascular tissues:

• Xylem and phloem strands are present on different radii so called radial vascular bundles.
• The xylem is in exarch type i.e., protoxylem is towards the periphery and metaxylem is towards the centre.
• The xylem and the phloem are present in two to four patches so it is in a diarch to tetrarch condition.

iii) Conjunctive tissue:

• The parenchymatous tissue present between xylem and the phloem is called conjunctive tissue.
• It becomes meristematic during secondary growth.

D) Pith:
It is small, made up of parenchyma tissue or absent.

I. Multiple Choice Questions

Question 1.
Which of the following is not a component of stomatal apparatus?
1. Guard cell
2. Subsidiary cell
3. Trichome
4. Stomatal aperture
Answer:
3. Trichome

Question 2.
When xylem and phloem are jointly situated along the same radius in a vascular bundle it is called
1. Closed
2. Open
3. Radial
4. Conjoint
Answer:
4. Conjoint

Question 3.
A plant material which shows small pith, two to four xylem patches alternating with phloem.
1. Dicot root
2. Monocot root
3. Dicot stem
4. Monocot stem
Answer:
1. Dicot root

Question 4.
The hypodermis is collenchymatous in
1. Monocot stem
2. Dicot stem
3. Dicot root
4. Both dicot stem and monocot root
Answer:
2. Dicot stem

Question 5.
Phloem parenchyma is absent in
1. Monocot stem
2. Dicot stem
3. Dicot root
4. Both dicot stem and monocot root
Answer:
1. Monocot stem

Question 6.
Casparian strips are made up of water impermeable substance ___________
1. Inulin
2. Suberin
3. Chitin
4. Cutin
Answer:
2. Suberin

Question 7.
In a dicot root conjunctive tissue is generally present between
1. Xylem and phloem
2. Xylem and pericycle
3. Phloem and pericycle
4. Endodermis and pericycle
Answer:
1. Xylem and phloem

Question 8.
Open vascular bundles are seen in
1. Dicot root
2. Monocot root
3. Dicot stem
4. Monocot stem
Answer:
3. Dicot stem

Question 9.
The elongated parenchyma cells arranged vertically and parallel to each other
1. Hypodermis in dicot stem
2. Palisade tissue in dicot leaf
3. Hypodermis in monocot stem
4. Bulliform cells of monocot leaf
Answer:
2. Palisade tissue in dicot leaf

Question 10.
Water-containing cavities in the vascular bundles is a characteristic of
1. Dicot leaf
2. Dorsiventral leaf
3. Dicot stem
4. Monocot stem
Answer:
4. Monocot stem

II. Fill in the Blanks

Question 1.
The special cells that surround the guard cells in the stomatal apparatus are ___________
Answer:
Subsidiary cells

Question 2.
Simple tissues like parenchyma, collenchyma and sclerenchyma are largely present in the ___________ tissue system.
Answer:
Ground

Question 3.
The tissue present between xylem and phloem in a dicot stem is ___________.
Answer:
Cambium

Question 4.
In the anatomy of a dicot or monocot root all the tissues on the inner side of the endodermis such as pericycle, vascular bundles and pith constitute the ___________.
Answer:
Stele

Question 5.
The innermost layer of cortex is ___________
Answer:
Endodermis

Question 6.
The tissue between upper and lower epidermis in a leaf is ___________.
Answer:
Mesophyll

Question 7.
The large colourless cells in the adaxial surface of monocot leaf are ___________.
Answer:
Bulliform cells

Question 8.
The conjoint vascular bundles usually have the phloem located only on the ___________ side of Xylem.
Answer:
Outer

Question 9.
Mechanical strength to the young dicot stem is provided by hypodermis which is made up of ___________ cells.
Answer:
Collenchymatous

Question 10.
The cells of ___________ in dicot stem are rich in starch grains, hence this layer is also referred to as starch sheath.
Answer:
Endodermis

III One Word Answer Questions

Question 1.
To which tissue system does the root hairs and trichomes belong?
Answer:
Epidermal tissue system

Question 2.
What name is given to the vascular bundles of roots where the xylem and phloem are arranged in an alternate manner along different radii?
Answer:
Radial vascular bundles / Alternate

Question 3.
What is the region between epidermis and stele in a dicot stem called?
Answer:
Cortex

Question 4.
Which cells of a grass leaf help in curling to minimise water loss?
Answer:
Bulliform cells

Question 5.
What do you call the loosely arranged tissue situated below the palisade in a dicot leaf?
Answer:
Spongy parenchyma

Question 6.
In which vegetative part (among stem, root and leaf) of a plant body cuticle is absent?
Answer:
Roots

Question 7.
With respect to the presence of cambium, what is the name given to the vascular bundle in a dicot stem?
Answer:
Open vascular bundle

Question 8.
Which is the part in the dicot stem with a large number of rounded, parenchymatous cells with intercellular spaces in the stele?
Answer:
Pith or medulla

Question 9.
What is the shape of sclerenchymatous patches of pericycle in a dicot stem?
Answer:
Semilunar

Question 10.
In a dorsiventral leaf, which epidermis bears more stomata?
Answer:
Lower (Abaxial) epidermis

Regular practice with AP Inter 1st Year Botany Study Material Chapter 3 Morphology of Flowering Plants Questions and Answers helps students stay prepared for examinations.

AP Inter 1st Year Botany 3rd Lesson Morphology of Flowering Plants Questions and Answers

IV. Very Short Answer Questions

Question 1.
Differentiate fibrous roots from adventitious roots.
Answer:

Question 2.
What is meant by pulvinus leaf base ? In members of which angiospermic family do you find them?
Answer:

• Swollen leaf base is called a pulvinus leaf base.
• It is found in plants of Leguminosae- (Fabaceae).

Question 3.
How do dicots differ from monocots with respect to venation?
Answer:

• In Dicots- reticulate venation is seen (net-like arrangement of Veins and Veinlets).
• In monocots- parallel venation is seen (Parallel arrangement of Veins).

Question 4.
How is a pinnately compound leaf different from a palmately compound leaf ? Explain with one example each.
Answer:

Question 5.
Differentiate between Racemose and Cymose inflorescence.
Answer:

Question 6.
Differentiate actinomorphic from zygomorphic flowers.
Answer:

Question 7.
How do the petals in a pea plant are arranged ? What is such a type of arrangement called?
Answer:

• In the flower of the pea plant there are five, unequal petals, the largest standard petal / vexillum, overlaps the two lateral wing petals / alae.
• Wing petals in turn overlap two smallest anterior keel petals/carina.
• This type of corolla is called papilionaceous corolla.
• This type of arrangement of petals is called vexillary or descendingly imbricate aestivation.

Question 8.
What is meant by epipetalous condition? Give an example.
Answer:
Stamens are united with petals which is called epipetalous condition.
Ex: Datura, Solarium

Question 9.
Differentiate between apocarpous and syncarpous ovary.
Answer:

Question 10.
Define placentation. What type of placentation is found in Dianthus?
Answer:

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

Question 11.
What is meant by parthenocarpic fruit?
Answer:

• The fruit that is formed from the ovary without fertilization is called parthenocarpic fruit.
• Such fruits are seedless and commercially useful in juice industries.
  Ex. Banana, Grapes.

Question 12.
What is the type of fruit found in mango ? How does it differ from that of coconut ?
Answer:

• In Mango, the fruit is known as Drupe.
• In Mango, the middle mesocarp is fleshy and edible.
• In coconut, the fruit is drupe, mesocarp is fibrous and not edible.
• In coconut, endosperm is edible.

Question 13.
What is meant by scutellum? In which type of seeds is it present?
Answer:

• Large shield shaped cotyledon is called scutellum.
• It is seen in monocot seeds.

Question 14.
Define with examples endospermic and non-endospermic seeds.
Answer:
Endospermic seeds or albuminous seeds:
The embryo doesn’t consume the entire endosperm. The seeds with endosperm are called endospermic seeds.
Ex :Maize, Wheat, Rice, Castor and Coconut.

Non-endospermic seeds or exalbuminous seeds or non-albuminous seeds:
The embryo consumes the entire endosperm during development. Seeds without endosperm are called non-endospermic seeds.
Ex: Pea, Gram and Beans

Question 15.
Write the floral formula of the Solatium plant.
Answer:
Ebr, Ebrl,

Question 16.
Give the technical description of ovary in Solanum nigrum.
Answer:

• In Solanum nigrum ovary is bicarpellary, syncarpous, bilocular, superior ovary, with many ovules on swollen axile placentation.
• Style terminal, stigma capitate.
• Carpels are arranged obliquely at 45°.

V. Short Answer Questions

Question 1.
Explain different regions of root with neat labelled diagram.
Answer:
A typical young root shows 4 regions.

1. Region of root cap
2. Region of meristematic activity
3. Region of elongation
4. Region of maturation

1. Region of root cap :

• The root is covered at the apex by a thimble-like structure called the root cap.
• It protects the apex of the root as it makes way through the soil.

2. Region of meristematic activity:

• The cells of this region are very small, thin-walled, and with dense protoplasm.
• These cells divide repeatedly by mitosis and produce daughter cells.

3. Region of elongation :

• Cells proximal to the region of the meristematic zone undergo rapid elongation and enlargement.
• This region is responsible for the growth of the root in length.

4. Region of maturation :

• This region is proximal to the region of elongation.
• Cells in the elongation zone gradually differentiate and mature.
• From this region, some of the epidermal cells form very fine and delicate thread-like structures called “root hairs”.
• Root hairs absorb water and minerals from the soil.

Question 2.
Explain different types of phyllotaxy with examples.
Answer:
The pattern of arrangement of leaves on the stem or branches is called phyllotaxy.
Phyllotaxy is usually 3 types based on the number of leaves at each node.
They are
A) Alternate phyllotaxy,
B) Opposite Phyllotaxy, and
C) Whorled Phyllotaxy

A) Alternate phyllotaxy:
A single leaf arises at each node in an alternate manner.
Ex: Hibiscus rosa-sinensis (China rose), Mustard, Sunflower.

B) Opposite phyllotaxy:

• A pair of leaves arise at each node.
• They lie opposite to each other.
  Ex: Calotropis, Guava.

C) Whorled phyllotaxy:
More than two leaves arise at a node and form a whorl.
Ex: Alstonia.

Question 3.
Describe the arrangement of floral members in relation to their insertion on thalamus.
Answer:
Based on the position of calyx, corolla and androecium on thalamus with respect to ovary; flowers are three types.
A) Hypogynous flower,
B) Perigynous flower
C) Epigynous flower.

A) Hypogynous flower:

• Thalamus is convex or cone-shaped.
• Gynoecium occupies the highest position on the thalamus.
• Other floral parts of the flower are situated below the gynoecium.
• Position of the ovary is superior.
  Ex: Mustard, China rose, Brinjal.

B) Perigynous flower :

• Thalamus is saucer or shallow cup-shaped.
• Gynoecium is situated in the centre of the thalamus.
• Other floral parts of the flower are located on the rim of the thalamus almost at the same level.
• Position of the ovary is half inferior / half superior.
  Ex: Plum, Rose, Peach.

C) Epigynous flower:

• Thalamus is deep cup shaped.
• The margins of the thalamus grow upward enclosing the ovary completely.
• Thalamus gets fused. with the ovary wall.
• Other floral parts of the flower arise above the ovary from the margin of the thalamus.
• Position of the ovary is inferior.
  Ex : Guava, Cucumber, Ray and disc florets of sunflower.

Question 4.
The flowers of many angiospermic plants which show sepals and petals, differ with respect to the arrangement of sepals and petals in respective whorls. Explain.
Answer:
The mode of arrangement of sepals/ petals in a floral bud is known as aestivation.
It is of four types.
A) Valvate
B) Twisted
C) Imbricate
D) Vexillary

A) Valvate:
Sepals or petals in a whorl just touch one another at the margin without overlapping.
Ex: Calotropis.

B) Twisted:
One margin of the appendage (sepal/ petal) overlaps that of the next one.
Ex: China rose, Lady’s finger, Cotton.

C) Imbricate:
The margins of sepals/ petals overlap one another but not in any particular direction.
Ex: Cassia, Gulmohur.

D) Vexillary: (Papilionaceous / descendingly imbricate aestivation).

• It is commonly found in members of Fabaceae.
• Corolla has 5 petals.
• Largest standard petal overlaps the two lateral wing petals.
• Wing petals in turn overlap the two smallest anterior keel petals.
  Ex: Pea, Bean.

Question 5.
Describe any four types of placentations found in flowering plants.
Answer:
The arrangement of ovules within the ovary is known as placentation. It is of 5 types.
A) Marginal placentation
B) Parietal placentation
C) Axile placentation
D) Basal placentation
E) Free central placentation

A) Marginal placentation:

• The placenta forms a ridge along the ventral suture of the ovary.
• Ovules are borne on this ridge forming two rows.
  Ex: Pea.

B) Axile placentation:
Placenta is axial and the ovules are attached to it in a multi locular ovary.
Ex: China rose, Tomato, Lemon.

C) Parietal placentation:

• Ovules develop on the inner wall of the ovary or on the peripheral parts.
• Ovary is one chambered (unilocular). But it becomes two chambered due to formation of false septum.
  Ex: Mustard, Argemone.

D) Free central placentation :
Ovules are borne on the central columnar axis from the base without septum
Ex: Dianthns, Primrose.

E) Basal placentation:
Placenta develops at the base of the ovary, single ovule is attached to the placenta.
Ex: Sunflower, Marigold.

Question 6.
Write a brief note on semi-technical description of a typical flowering plant.
Answer:

• The plant is described beginning with its habit, habitat, vegetative characters (root, stem and leaves) and then floral characters (inflorescence, flower and its parts) followed by fruit.
• After describing various parts of a plant, a floral diagram and floral formula are presented.
• The floral formula is represented by some symbols of floral parts.
• In the floral formula.

• Floral formula also indicates the number of free or united (within brackets) members of the corresponding whorl as subscript of the respective symbol.
• It also shows cohesion (union among similar members) and adhesion (union between dissimilar members).
• A floral diagram provides information about the number of parts of a flower, their arrangement and the relation they have with one another.
• The mother axis represents the posterior side of the flower and is indicated as a dot or a circle at the top of the floral diagram.
• Calyx, corolla, androecium and gynoecium are drawn in successive whorls, calyx being the outermost and the gynoecium being in the centre represented by a diagram of T.S. of ovary.
• The bract represents the anterior side of the flower and is indicated at the bottom of the floral diagram.

Question 7.
Give an account of the floral diagram.
Answer:

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

Question 8.
Draw the labelled diagram of the following.
i) Gram seed
ii) V.S. of maize seed
Answer:
i) Gram seed

ii) V.S. of maize seed

VI. Long Answer Questions

Question 1.
What is a flower ? Describe the parts of a typical angiosperm flower.
Answer:
A flower is a reproductive unit of an angiosperm plant that performs sexual reproduction. It is a modified stem with a condensed axis. A typical flower has four different kinds of whorls arranged successively on the swollen end of the pedicel called thalamus or receptacle. The parts of a flower include : Calyx, Corolla, Androecium and Gynoecium. Calyx and corolla are called accessory organs whereas Androecium and Gynoecium are called reproductive organs.

1. Calyx:

• It is the outermost whorl of a flower. It is made up of units called sepals.
• It is generally green in colour, leaf-like and protects the flower in the bud condition.
• The calyx may be gamosepalous(Sepals united) or polysepalous (Sepals free).

2) Corolla:

• It is the whorl present inner to the Calyx.
• It consists of petals. The petals are brightly coloured to attract the insects for pollination.
• It may be gamopetalous (Petals United) or polypetalous (Petals free).

3. Androecium:

• It is the whorl present next to the corolla.
• The androecium mainly consists of stamens which are the male reproductive unit of a flower.
• A stamen is composed of two parts i.e. anther and filament.
• The anther is the bilobed structure and each lobe has two chambers called pollen sacs.
• The stalk of the stamen is called a filament. Inside the anther, pollen grains are formed.
• A sterile stamen is called staminode.
• Stamens of a flower may be united with other floral parts.
• When stamens are attached to petals, they are Epipetalous (Brinjal), when attached to the perianth- they are called Epiphyllous (Lily).
• The stamens in a flower may be free (Polyandrous) or may be united among themselves.
• They may be Monadelphous- the filaments of all the stamens may be united into Diadelphous- into two bundles (Pea)
• Polyadelphous- into more than two bundles (Citrus).
• In Salvia, didynamous stamens are seen meaning two longer and two shorter stamens in a flower.
• In mustard- Tetradynamous stamens are seen, meaning four are longer and two are shorter.

4. Gynoecium:

• The innermost whorl and female reproductive part of a flower is called gynoecium.
• It consists of carpels. A carpel is composed of three parts: stigma, style and ovary.
• Ovary is the enlarged basal part on which lies the style.
• The style connects the ovary to the stigma.
• The stigma is the receptive surface for pollen grains.
• Each ovary bears one or more ovules attached to a flattened cushion like placenta.
• When more than one carpel is present, they may be free (Lotus, Rose) and are called apocarpous or fused they are syncarpous (Mustard, Tomato).
• After fertilization, the ovary develops into fruit and ovules develop into seeds.

Question 2.
Write about the key characteristics of Solanaceae ?
Answer:

• It is commonly called the ‘potato family and includes about 3200 species belonging to 85 genera.
• Members of this family are widely distributed in mesophytic habitats of tropical, subtropical and even temperate zones.

Vegetative Characters:

• Habit: Plants mostly herbs, some shrubs (Cestrum).
• Root system : Tap root system.
• Stem : Aerial, erect, herbaceous or rarely woody, cylindrical, solid or hollow, hairy or glabrous. Underground stem is tuber in Potato.
• Leaves: Alternate phyllotaxy, exstipulate, petiolate, simple or rarely pinnately compound leaf with reticulate venation.
• Inflorescence : Cymose, axillary as in Solanum.
• Flower : Actinomorphic, bisexual, pentamerous, hypogynous.
• Calyx: Sepals five, gamosepalous, persistent, valvate aestivation.
• Corolla : Petals five, gamopetalous, valvate aestivation.
• Androecium : Stamens five, epipetalous

Gynoecium : Bicarpellary, syncarpous, bilocular, superior ovary, placenta is swollen with many ovules on axile placentation, style terminal, stigma capitate. Carpels are arranged obliquely at 45°.

• Fruit: Berry (Capsicum, Solanum, Lycopersicon) or capsule (Datura, Nicotiana).
• Seed : Many, endospermic, dicotyledonous
• Floral Formula : Ebr, Ebrl,

I. Multiple Choice Questions

Question 1.
The type of root system commonly found in dicot plants is
1. Tap root system
2. Fibrous root system
3. Adventitious root system
4. Rhizoids
Answer:
1. Tap root system

Question 2.
A few millimeters above the root cap, what region can be found
1. Elongation
2. Maturation
3. Meristematic activity
4. Root hairs
Answer:
3. Meristematic activity

Question 3.
The stem develops from
1. Hypocotyl
3. Epicotyl
3. Radicle
4. Plumule
Answer:
4. Plumule

Question 4.
Arrangement of flowers on floral axis is called
1. Placentation
2. Phyllotaxy
3. Inflorescence
4. Aestivation
Answer:
3. Inflorescence

Question 5.
In a cymose inflorescence, the main axis
1. Terminates with a flower
2. Has unlimited growth but lateral branches ends with a flower
3. Has limited growth, do not ends with a flower
4. Shows acropetal arrangement of flowers
Answer:
1. Terminates with a flower

Question 6.
Arrangement of ovules within the ovary is known as
1. Aestivation
2. Placentation
3. Cohesion
4. Adhesion
Answer:
2. Placentation

Question 7.
The basal bulged part of the carpel is
1. Filament
2. Ovary
3. Style
4. Stigma
Answer:
2. Ovary

Question 8.
Name the aestivation where the standard petal overlaps the two wing petals, which then overlap the two smaller keel petals.
1. Twisted
2. Valvate
3. Vexillary
4. Alternate
Answer:
3. Vexillary

Question 9.
The main characteristic feature of a drupe.
1. Stony mesocarp
2. Stony endocarp
3. Fleshy seed coat
4. Stony pericarp
Answer:
2. Stony endocarp

Question 10.
After fertilization, ovule develops into
1. Seed
2. Fruit
3. Flower
4. Root
Answer:
1. Seed

Question 11.
Non-endospermic monocot seed
1. Rice
2. Maize
3. Wheat
4. Orchid
Answer:
4. Orchid

Question 12.
The placentation in Solanaceae family members
1. Parietal placentation
2. Basal placentation
3. Marginal placentation
4. Axile placentation
Answer:
4. Axile placentation

Question 13.
Persistent calyx can be seen in
1. Pea
2. Brinjal
3. Lily
4. Brassica
Answer:
2. Brinjal

II. Fill In the Blanks

Question 1.
Radicle develop into ____________ of the plant.
Answer:
Primary Root system

Question 2.
Reticulate venation is a characteristic feature of ____________ Plants.
Answer:
Dicots

Question 3.
The flowers in racemose inflorescence are arranged in ____________ succession.
Answer:
Acropetal

Question 4.
The first and outermost whorl of a flower is ____________.
Answer:
Calyx

Question 5.
Epiphyllous conditions can be seen in ____________.
Answer:
Lily
Question 6.
The function of stigma is to receive ____________.
Answer:
Pollen grains

Question 7.
The sheath enclosing the radicle in a monocot seed is called ____________.
Answer:
Coleorhiza

Question 8.
Seeds with an endosperm are called ____________ seeds.
Answer:
Endospermic / Albuminous

Question 9.
The ovary of Solanum nigrum is bicarpellary, syncarpous and ____________ locular.
Answer:
Bi

Question 10.
In semi-technical description of a flower G stands for ____________ ovary.
Answer:
Superior

III. One Word Answer Questions

Question 1.
What is the role of root hair?
Answer:
Absorption of water

Question 2.
What is the swollen leaf base in leguminous plants called?
Answer:
Pulvinus

Question 3.
What is the arrangement of veins in the leaf called?
Answer:
Venation

Question 4.
What type of phyllotaxy is seen in Alstonia?
Answer:
Cyclic or whorled

Question 5.
What is the flower with both androecium and gynoecium called?
Answer:
Bisexual

Question 6.
In which type of flower does the gynoecium occupy the highest position?
Answer:
Hypogynous

Question 7.
What is the function of sepals in the bud stage?
Answer:
Protection of inner floral leaves

Question 8.
In vexillary aestivation, what is the largest petal called?
Answer:
Standard or vexillum

Question 9.
What is a sterile stamen called?
Answer:
Staminode

Question 10.
What connects the ovary to the stigma?
Answer:
Style

Question 11.
What is the middle fleshy part of the pericarp in mango fruit called?
Answer:
Mesocarp

Question 12.
How many cotyledons are present in the embryo of wheat and maize?
Answer:
One

Question 13.
Name the two layers of the dicot seed coat?
Answer:
Testa and tegmen

Question 14.
What is the large shield-shaped cotyledon in grass seed called?
Answer:
Scutellum

Question 15.
In the floral formula, “Br” stands for?
Answer:
Bracteate

Regular practice with AP Inter 1st Year Botany Study Material Chapter 2 Plant Kingdom Questions and Answers helps students stay prepared for examinations.

AP Inter 1st Year Botany 2nd Lesson Plant Kingdom Questions and Answers

IV. Very Short Answer Questions

Question 1.
Differentiate between antheridium and archegonium.
Answer:

Question 2.
What are the two stages found in the gametophyte of mosses? Mention the structures from which these two stages develop.
Answer:

• The gametophyte of a moss consists of two stages.
  • Juvenile, green, filamentous, protonema
  • The adult leafy gametophore.
• Protonema develops by the germination of spores.
• Gametophores develop from the buds on protonema.

Question 3.
Name the stored food materials found in Phaeophyceae and Rhodophyceae.
Answer:

• In Phaeophyceae – Laminarin or mannitol.
• In Rhodophyceae – Floridian starch

Question 4.
Name the pigments responsible for brown colour of Phaeophyceae and red colour of Rhodophyceae.
Answer:

• The brown colour of Phaeophyceae is due to fucoxanthin. (a xanthophyll)
• Red colour of Rhodophyceae is due to r-phycoerythrin. (a phycobilin)

Question 5.
Name different methods of vegetative reproduction in Bryophytes.
Answer:

• In liverworts, vegetative reproduction takes place by special structures called gemmae.
• They develop in small receptacles called gemma cups.
  Ex. Marchantia
• In mosses, vegetative reproduction is carried out by fragmentation, gemmae and budding.
  Ex. Funaria.

Question 6.
Name the composite structure in which nucellus is protected by envelops in Gymno- sperms.
Answer:
The Ovule

Question 7.
Name the Gymnosperms which contain mycorrhizae and coralloid roots respectively.
Answer:

• Pinus is the gymnosperm which contains mycorrhiza.
• Cycas is the gymnosperm which contains coralloid roots.

Question 8.
Name the four classes of Pteridophyta with one example each.
Answer:

• Psilopsida – Psilotum
• Lycopsida – Selaginella, Lycopodium
• Sphenopsida – Equisetum
• Pteropsida – Pteris,’Dryopteris

Question 9.
What are the first organisms to colonise rocks? Give the generic name of the moss which provides peat.
Answer:

• Mosses along with Lichens are the first organisms to colonise rocks.
• Sphagnum, a moss provides peat that is used as a fuel.

Question 10.
Why are Bryophytes called the Amphibians of Plant Kingdom ?
Answer:
Bryophytes are called amphibians of the plant kingdom because they survive in moist soils and are dependent on water for sexual reproduction.

V. Short Answer Questions

Question 1.
Differentiate between red algae and brown algae.
Answer:

Question 2.
Differentiate between liverworts and mosses.
Answer:

Question 3.
What is meant by homosporous and heterosporous pteridophytes? Give two examples.
Answer:

• Pteridophytes which produce the same kind of spores are known as homosporous pteridophytes.
  Ex: Lycopodium, Pteris
• Pteridophytes,which produce two kinds of spores i.e microspores and megaspores are called heterosporous pteridophytes.
  Ex: Selaginella, Salvinia.
• Small spores are called microspores that develop into male gametophytes and large spores are called megaspores which develop into female gametophytes.

Question 4.
Give a brief account of prothallus.
Answer:

• In pteridophytes, haploid spores liberated from the sporangia of sporophyte develop into inconspicuous, small, multicellular, free-living photosynthetic, thalloid gametophytes called prothallus.
• These gametophytes require cool, damp, shady places to grow.
• The prothallus bears sex organs male and female called antheridia and archegonia.
• The sex organs are multicellular, jacketed and sessile.
• Water is required for the transfer of antherozoids.
• Fusion of male gamete with egg takes place in the archegonium.
• After fertilization, zygote develop into Embryo.

Question 5.
Write a note on the economic importance of algae and bryophytes.
Answer:
i) Economic importance of Algae :

• At least a half of the total carbon dioxide fixation on earth is carried out by algae through photosynthesis.
• Being photosynthetic, they increase the level of dissolved oxygen in their immediate environment.
• They are important as primary producers of energy rich compounds which form the basis of the food cycles of all aquatic animals.
• Many species of Porphyra, Laminaria and Sargassum are used as food.
• Certain marine brown and red algae produce large amounts of hydrocolloids (water holding substances) e.g., algin (brown algae) and carrageenan (red algae) which are used commercially.
• Agar, one of the commercial products obtained from Gelidium and Gracilaria is used to grow microbes and in preparation of ice-creams and jellies.
• Chlorella are unicellular alga used as food supplements even by space travelers.

ii) Economic importance of Bryophytes :

• Some mosses provide food for herbaceous mammals, birds and other animals.
• Species of Sphagnum, (peat moss) provides peat that has long been used as fuel, and because of its capacity to hold water as packing material for trans-shipment of living material.
• Mosses along with lichens are the first organisms to colonise rocks and hence are of great ecological importance.
• They decompose rocks making the substrate suitable for the growth of higher plants, hence play a significant role in plant succession.
• Since mosses form dense mats on the soil, they reduce the impact of falling rain and prevent soil erosion.

Question 6.
Draw labelled diagrams of:
a) Female thallus and male thallus of a liverwort
b) Gametophyte and sporophyte of Funaria.
Answer:
a)

b)

VI. Long Answer Questions

Question 1.
Give the salient features of Pteridophytes.
Answer:

• The pteridophytes include club mosses, horsetails, ferns, etc.
• They are embryophytic, archegoniate, vascular (tracheophyte) cryptogams.
• The pteridophytes are commonly found in cool, damp, shady places.
• The main plant body is a sporophyte which is differentiated into true roots, stems and leaves.
• The roots are adventitious.
• The leaves in pteridophytes are small (microphylls) as in Selaginella or large (macrophylls) as in ferns.
• The sporophytes bear sporangia that are subtended by leaf-like appendages called sporophylls.
• The sporangia produce spores by meiosis in spore mother cells.
• In the majority of the pteridophytes, all the spores are of similar kind; such plants are termed homosporous.
  Ex: Lycopodium, Pteris.
• Genera like Selaginella and Salvinia, which produce two kinds of spores, megaspores and microspores, are described as heterosporous.
• The spores germinate to give rise to inconspicuous, small but multicellular, free-living, mostly photosynthetic, thalloid gametophytes called Prothallus.
• The gametophytes bear male and female sex organs called antheridia and archegonia, respectively.
• The sex organs are multicellular, jacketed and sessile.
• In heterosporous plants, the megaspores and microspores germinate and give rise to female and male gametophytes respectively.
• Water is required for the transfer of antherozoids (the male gametes) released from the antheridia, to the mouth of the archegonium.
• Fusion of motile male gamete with the egg present in the archegonium results in the formation of a zygote. This is called zooidogamous oogamy.
  The development of zygote into young embryo takes place within the female gametophytes.
• This event is a precursor to the seed habit and is considered an important step in evolution.
• The embryo thereafter produces a multicellular, well differentiated sporophyte which is the dominant phase in the pteridophytes.
• Pteridophytes show heteromorphic alternation of generations with a diplo-haplontic life cycle.
• Both the sporophyte and gametophyte are independent, but the sporophyte is dominant.

Question 2.
Describe the important characteristics of Gymnosperms.
Answer:

• The gymnosperms (gymnos: naked, sperma : seeds) are plants in which the ovules are not enclosed by any ovary wall and remain exposed, both before and after fertilization.
• The seeds are naked.
• Gymnosperms include medium-sized trees or tall trees and shrubs.
• The giant redwood tree Sequoia is one of the tallest trees.
• The roots are generally tap roots.
• Roots in Pinus, have a fungal association in the form of mycorrhiza, while in Cycas, small specialized roots called coralloid roots are associated with N₂– fixing cyanobacteria.
• The stem is unbranched (Cycas) or branched (Pinus, Cedrus).
• The leaves may be simple or compound.
• In Cycas, the pinnate leaves persist for a few years.
• In conifers, the needle-like leaves reduce the surface area.
• The gymnosperms are heterosporous; they produce haploid microspores and megaspores.
• The two kinds of spores are produced within sporangia that are borne on sporophylls, which are arranged spirally along an axis to form lax or compact strobili or cones.
• The strobili bearing microsporophylls and microsporangia are called microsporangiate or male strobili.
• The microspores develop into a male gametophyte called a pollen grain.
• The cones bearing megasporophylls with ovules or megasporangia are called microsporangiate or female strobili.
• The male or female cones or strobili may be borne on the same tree (Pinus).
• However, in Cycas male cones and megasporophylls are borne on different trees.
• The megaspore mother cell is differentiated from one of the cells of the nucellus.
• The nucellus is protected by envelopes and the composite structure is called an ovule.
• The ovules are borne on megasporophylls, which may be clustered to form the female cones.
• The megaspore mother cell divides meiotically to form four megaspores.
• One of the megaspores enclosed within the megasporangium develops into a multicellular female gametophyte that bears two or more archegonia.
• The multicellular female gametophyte is also retained within megasporangium.
• The male and the female gametophytes, do not have an independent free-living existence.
• The pollen grains are released from the microsporangium. They are carried by air currents and come in contact with the opening of the ovules borne on megasporophylls.
• The pollen tube carrying the male gametes grows towards archegonia in the ovules and discharges their contents near the mouth of the archegonia.
• Following fertilization, the zygote develops into an embryo, and the ovules into seeds. These seeds are not covered.

I. Multiple Choice Questions

Question 1.
Which of the following have been excluded from plant kingdom (Plantae) in the modern system of classification (Whittaker’s classification) ?
1. Algae and Fungi
2. Fungi and some members of Monera and Protista
3. Members of Monera only
4. Algae and some members of Protista
Answer:
2. Fungi and some members of Monera and Protista

Question 2.
Which of the following pieces of evidence is very crucial to establish evolutionary relationships in Phylogenetic systems of classification ?
1. Morphological evidence
2. Anatomical evidences
3. Fossil evidence
4. Cytological evidences
Answer:
3. Fossil evidence

Question 3.
Pick out the right examples for filamentous and colonial forms of algae respectively.
1. Ulothrix and Spirogyra
2. Ulothrix and Volvox
3. Volvox and Fucus
4. Spirogyra and Fucus
Answer:
2. Ulothrix and Volvox

Question 4.
Which of the following alga is not suitable as human food ?
1. Chlorella
2. Porphyra
3. Chara
4. Laminaria
Answer:
3. Chara

Question 5.
The alga Porphyra belongs to the class
1. Rhodophyceae
2. Phaeophyceae
3. Chlorophyceae
4. Cyanophyceae
Answer:
1. Rhodophyceae

Question 6.
The shape of archegonium in bryophytes is
1. Flask shape
2. Club shape
3. Saucer shape
4. Cup shape
Answer:
1. Flask shape

Question 7.
The sporophyte in mosses develops from
1. Microspore
2. Megaspore
3. Zygote
4. Zygospore
Answer:
3. Zygote

Question 8.
Heterospory in Pteridophytes is commonly seen in species of
1. Selaginella and Salvinia
2. Lycopodium and Equisetum
3. Equisetum and Psilotum
4. Lycopodium and Psilotum
Answer:
1. Selaginella and Salvinia

Question 9.
Which of the following is synonymous to the male gametophyte in Gymnosperms?
1. Megaspore
2. Microspore / pollen grain
3. Megasporangia
4. Microsporophyll
Answer:
2. Microspore / pollen grain

Question 10.
Smallest living angiosperm
1. Wolffia
2. Typha
3. Cuscuta
4. Cedrus
Answer:
1. Wolffia

II. Fill in the Blanks

Question 1.
Natural classification system based on natural affinities was given by __________.
Answer:
George Bentham and J.D. Hooker

Question 2.
Using numbers and codes for characters and utilising computers to process the data is seen in __________ taxonomy.
Answer:
Numerical

Question 3.
The fusion of two gametes dissimilar in size in Eudorina is termed as __________.
Answer:
Anisogamy

Question 4.
At Least half of the total carbon dioxide fixation on earth is carried out by __________ through photosynthesis.
Answer:
Algae

Question 5.
In green algae the members have one or more storage bodies called __________ located in the chloroplasts.
Answer:
Pyrenoids

Question 6.
In phaeophyceae, the plant body is usually attached to the substratum by a __________.
Answer:
Holdfast

Question 7.
The food is stored in the form of __________ in Rhodophyceae (red algae).
Answer:
Floridean starch

Question 8.
__________ moss provides peat and is also used as packing material to hold water.
Answer:
Sphagnum

Question 9.
__________ are green multicellular asexual buds on the thalli of liverworts.
Answer:
Gemmae

Question 10.
In mosses the first stage is the __________ stage, which directly develops from a Spore.
Answer:
Protonema

Question 11.
In Pteridophytes, the main plant body is a __________.
Answer:
Sporophyte

Question 12.
The free living, photosynthetic, thalloid gametophyte of Pteridophytes is called __________.
Answer:
Prothallus

Question 13.
Selaginella belongs to __________ class of Pteridophyta.
Answer:
Lycopsida

Question 14.
Mycorrhizal association with a fungus is seen in the Gymnosperm
Pinus

III. One Word Answer Questions

Question 1.
Which taxonomy is based on cytological information like chromosome number, structure and behaviour?
Answer:
Cytotaxonomy

Question 2.
What do you call the fusion between one large, non-motile (static) female gamete and a smaller motile male gamete in Volvox and Fucus ?
Answer:
Oogamy

Question 3.
What is the commercial product obtained from red algae Gelidium and Gracilaria?.
Answer:
Agar agar

Question 4.
Which are the two chlorophyll pigments found in chloroplasts of Phaeophyceae algae?
Answer:
Chlorophyll a and c

Question 5.
Which is the predominant red pigment present in the body of red alga?
Answer:
r-Phycoerythrin

Question 6.
In which algae the vegetative cells have a cellulose cell wall covered by algin?
Answer:
Phaeophyceae

Question 7.
Give one example for a liverwort with a thalloid plant body.
Answer:
Marchantia

Question 8.
Which stage of the mosses bears sex organs?
Answer:
Gametophyte, leafy stage

Question 9.
Majority of the Pteridophytes produce spores that are similar. What do you call this condition?
Answer:
Homosporous condition or Homospory

Question 10.
Which important step in evolution does the heterospory lead to?
Answer:
Seed-Habit

Question 11.
Name the tallest tree in Gymnosperms.
Answer:
Sequoia

Question 12.
What type of roots in Cycas show symbiosis with Nitrogen fixing cyanobacteria?
Answer:
Coralloid roots

Question 13.
In which group of plants the seeds are enclosed in a fruit?
Answer:
Angiosperms

Regular practice with AP Inter 1st Year Botany Study Material Chapter 1 Biological Classification Questions and Answers helps students stay prepared for examinations.

AP Inter 1st Year Botany 1st Lesson Biological Classification Questions and Answers

IV. Very Short Answer Questions

Question 1.
What is the nature of the cell wall in Diatoms?
Answer:
In Diatoms, the cell wall forms two thin overlapping shells, which fit together like a soapbox. The cell walls are embedded with silica and thus they are indestructible.

Question 2.
What do the terms “Phycobiont” and “Mycobiont” refer to?
Answer:
The algal component in a lichen is called a phycobiont and the fungal component in a lichen is called a mycobiont.

Question 3.
What do the terms “Algal blooms” and “Red tides” signify?
Answer:

• The luxuriant growth of cyanobacterial members in stagnant polluted waters due to excess of nitrogen and phosphorus from fertilizers and sewage are called algal blooms.
• Red dinoflagellates like Gonyaulax undergo rapid multiplication that they make the sea red. Hence it is called “red tides”.

Question 4.
How are “Viroids” different from “Viruses”?
Answer:

Question 5.
State two economically important uses of Heterotrophic bacteria?
Answer:

1. Making curd from milk.
2.  Production of antibiotics.
3. Nitrogen fixation in legume roots.

Question 6.
Some plants are autotrophic, Can you think of some plants that are partially heterotrophic?
Answer:
Dionaea (Venus fly trap), Utricularia (Bladderwort), Cuscuta is a parasite.

Question 7.
Give the main criteria used for classification by Whittaker? [March-26]
Answer:
The main criteria for the classification of whittaker include cell structure, thallus organisation, mode of nutrition, mode of reproduction and phylogenetic relationships.

V. Short Answer Questions

Question 1.
What are the characteristic features of Euglenoids?
Answer:

1. These are freshwater, flagellated organisms, found in stagnant water.
2. They are surrounded by a flexible protein rich layer called pellicle.
3. They have two unequal flagella, a short and long one.
4. They are photosynthetic in the presence of light.
5. They behave as heterotrophs when deprived of light.
6. Reproduction takes place by binary fission.
7. This pigments of Euglenoids are Identical to those present in Higher Plants.
   Ex: Euglena.

Question 2.
Give the salient features and importance of Chrysophytes?
Answer:
Salient features

1. This group includes diatoms and desmids or golden algae.
2. They are found in fresh water as well as in marine environments.
3. They are microscopic, and float passively in water currents.
4. Most of them are photosynthetic.
5. In diatoms, the cell wall contains two thin overlapping shells, which fit together like a soap box.
6. Cell walls are embedded with silica and thus the walls are indestructible.
7. After death diatoms leave large amounts of cell wall deposits in their habitat and form diatomaceous earth.

Importance:

• The diatomaceous soil is used in polishing, filtration of oils and syrups.
• Diatoms are the chief producers of the ocean.

Question 3.
Give a brief account of Dinoflagellates?
Answer:

1. These are marine and photosynthetic forms.
2. They appear yellow, green, blue or red based on the pigments in the cell.
3. The cell wall contains stiff cellulosic plates on the outer surface.
4. These are biflagellated. One lies longitudinally and the other is transversely in a furrow between the wall plates.
5. Very often, red dinoflagellates (Ex: Gonyaulax) undergo such rapid multiplication that they make the sea appear red (red tides).
6. Toxins released by such large numbers may even kill other marine animals such as fishes.

Question 4.
Write the role of Fungi in our daily life.
Answer:
Uses:

1. Yeast are used to make bread and beer.
2. Some fungi like Penicillium are the source of antibiotics.
3. Agaricus are edible mushrooms.

Diseases:

1. Fungi spoils bread. Ex: Rhizopus.
2. White spots appear on the leaves of mustard by Albugo.
3. Puccinia causes rust in wheat.

VI. Long Answer Questions

Question 1.
Give the salient features and comparative account of different classes of fungi studied by You?
Answer:
Kingdom fungi includes heterotrophic organisms.

Salient Features:

• All the fungal members are filamentous except yeast (unicellular).
• Body of the fungi is called mycelium.
• Each thread-like slender structure in the mycelium is called hypha.
• In some fungi (Rhizopus), hyphae are aseptate and multinucleated, and are called “coenocytic hyphae”.
• Cell wall is made up of chitin and polysaccharides.
• Reserve food materials are glycogen and oil.
• They live as “saprophytes”, parasites and symbionts. (Lichens and mycorrhizae).
• Reproduction takes place vegetatively by fragmentation or fission or by budding.
• Asexually by producing spores like conidia or sporangiospores or zoospores.
• Plasmogamy, karyogamy and meiosis are the sequential steps in their sexual reproduction.

Kingdom fungi are divided into 4 classes based on their morphology of mycelium, mode of spore formation and fruiting bodies. They are

1. Phycomycetes
2. Ascomycetes.
3. Basidiomycetes
4. Deuteromycetes

1. Phycomycetes:

• They are found in aquatic habitats and on decaying wood in moist and damp places or as obligate parasites on plants.
• The mycelium is aseptate and coenocytic.
• Asexual reproduction takes place by zoospores or by aplanospores.
• Zygospores are formed by the fusion of two gametes which may be similar or dissimilar or oogamous.
  Ex: Mucor, Rhizopus, Albugo.

2. Ascomycetes (Sac Fungi):

• They are unicellular (Yeast) or multicellular (Penicillium).
• They live as saprophytes or decomposers or parasites or coprophilous (Grow on dung).
• Mycelium is branched and septate
• The asexual reproduction occurs by conidia formed on conidiophores.
• They reproduce sexually by producing Ascospores in Asci.
  Ex: Aspergillus, Claviceps, Neurospora.
• Neurospora are used extensively in biochemical genetic work.

3. Basidiomycetes (Bracket fungi or puffballs)

• They grow in soil, on logs and tree stumps and in living plant bodies as parasites.
• The mycelium is branched and septate.
• Vegetative reproduction occurs by fragmentation.
• Sex organs are absent but plasmogamy occurs by the fusion of two vegetative cells of different strains or genotypes.
• The dikaryotic mycelium produces basidia which in turn produce basidiospores.
  Ex: Agaricus, Ustilago, polyporus.

4. Deuteromycetes (Imperfect fungi):

• Some members live as saprophytes or parasites and most of the members live as decomposers and help in mineral cycling.
• The mycelium is branched and septate.
• They reproduce asexually by conidia.
  Ex: Alternaria, Colletotrichum, Trichoderma.

Question 2.
Describe briefly different groups of Monerans you have studied.
Answer:
Kingdom Monera includes Archaebacteria, Eubacteria, Cyanobacteria and Mycoplasmas.

Archaebacteria:

• These are special Monerans.
• They live in extreme habitats like salty areas, (halophiles), hot springs (thermoacidophiles) and marshy areas (methanogens).
• Their cell wall contains Pseudomurein.
• Their cell membrane contains a branched chain of lipids. So they can even survive in extreme conditions.
• Methanogens are responsible for the production of methane gas from dung.

Eubacteria:

• Bacteria are abundant microorganisms, present everywhere.
• They also live in extreme habitats like hot springs, deserts, snow and deep oceans.
• Bacteria appear in 4 shapes
  • spherical – coccus
  • Rod shaped – Bacillus
  • Comma shaped – Vibrio
  • Spiral shaped – Spirilla.
• Bacteria are surrounded by a cell wall made up of peptidoglycan.
• The infoldings of the plasma membrane are called mesosomes.
• Genetic material is not covered by nuclear membrane.
• Except ribosomes other cell organelles are absent.
• Motile bacteria contain one or more flagella.
• Based on the nutrition bacteria are 2 types,
  • Autotrophs
  • Heterotrophs
• Autotrophs are two types:
  • Photosynthetic autotrophs.
  • Chemosynthetic autotrophs.
• Heterotrophic bacteria are of two types:
  • Photosynthetic heterotrophs
  • Chemosynthetic heterotrophs (Saprophytes or decomposers, Parasites).
• Bacteria reproduce mainly by binary fission. During unfavourable conditions they produce spores – endospores.
• Bacteria reproduce sexually by transferring genetic material from one bacterium to another bacterium.

Cyanobacteria:

• These are also known as blue-green algae.
• Cyanobacteria are unicellular, colonial or filamentous aquatic or terrestrial algae.
• These are photosynthetic autotrophs, due to the presence of chlorophyll.
• They show oxygenic photosynthesis.
• The colonies and trichomes or filaments are generally surrounded by a gelatinous sheath.
• They often form blooms in polluted water bodies.
• Some of these organisms can fix atmospheric nitrogen in specialised cells called heterocysts.
  Ex. Nostoc and Anabaena.

Mycoplasma:

• They do not have a cell wall, and they are pleomorphic.
• They are the smallest living cells and can survive without O₂.
• Many mycoplasmas are pathogenic to animals and plants.

I. Multiple Choice Questions

Question 1.
Two kingdom classification was given by
1. Whittaker
2. Linnaeus
3. Aristotle
4. Theophrastus
Answer:
2. Linnaeus

Question 2.
Bacteria that live in most harsh habitats are
1. Bacteria
2. Cyanobacteria
3. Mycoplasmas
4. Archaebacteria
Answer:
4. Archaebacteria

Question 3.
Nitrogen Fixing Cyanobacterium is
1. Rhizobium
2. Nostoc
3. Chlorella
4. Methanogens
Answer:
2. Nostoc

Question 4.
Smallest living monera cells that lack a cell wall are
1. Cyanobacteria
2. Protozoans
3. Mycoplasma
4. Bacteria
Answer:
3. Mycoplasma

Question 5.
Diatomaceous Earth is Indestructible due to cell walls embedded by
1. Calcium
2. Silica
3. Zinc
4. Phosphorus
Answer:
2. Silica

Question 6.
Protists that form plasmodium are
1. Euglenoids
2. Slime moulds
3. Dinoflagellates
4. Diatoms
Answer:
2. Slime moulds

Question 7.
Which is the correct sequence of the sexual cycle of fungi
1. Mitosis-Meiosis-Fertilization
2. Plasmogamy-Karyogamy-Meiosis
3. Meiosis- Plasmogamy-Karyogamy
4. Karyogamy-Plasmogamy-Meiosis
Answer:
2. Plasmogamy-Karyogamy-Meiosis

Question 8.
The viruses which infect bacteria are known as
1. Zoophages
2. Bacteriophages
3. Cyanophages
4. Zymophages
Answer:
2. Bacteriophages

Question 9.
Fungus that is extensively used in biochemical and genetic work
1. Neurospora
2. Ustilago
3. Colletotrichum
4. Saccharomyces
Answer:
1. Neurospora

Question 10.
One of the following are very good pollution indicator
1. Fungi
2. Mycoplasma
3. Lichens
4. Golden algae
Answer:
3. Lichens

II. Fill in the Blanks

Question 1.
In five kingdom classification Bacteria are included in __________ kingdom.
Answer:
Monera

Question 2.
Methanogens are present in gut of several ruminant animals such as cows and buffaloes and they are responsible for the production of __________.
Answer:
Methane (biogas)

Question 3.
The colonies of cyanobacteria are generally surrounded by __________ sheath.
Answer:
Gelatinous

Question 4.
Bacteria reproduce mainly by __________.
Answer:
Binary fission

Question 5.
Diatoms are the chief __________ in the oceans.
Answer:
Producers

Question 6.
Pellicle is found in __________ organisms.
Answer:
Euglenoid

Question 7.
The network of hyphae is known as __________.
Answer:
Mycelium

Question 8.
Agaricus belongs to __________ fungi.
Answer:
Basidiomycetes

Question 9.
__________ fungi is known as Imperfect fungi.
Answer:
Deuteromycetes

Question 10.
Viroids were discovered by __________.
Answer:
T.O.Diener

III. One Word Answer Questions

Question 1.
Among five kingdom classification, eukaryotes are placed in how many kingdoms ?
Answer:
Four

Question 2.
What is the cell wall composition of fungi?
Answer:
Chitin

Question 3.
Give one example of Nitrogen fixing cyanobacteria?
Answer:
Nostoc or Anabaena

Question 4.
What is the protist responsible for Red tides?
Answer:
Gonyaulax

Question 5.
What are the smallest living cells which can survive without oxygen?
Answer:
Mycoplasma

Question 6.
What is the causative organism for Sleeping sickness disease?
Answer:
Trypanosoma

Question 7.
Which of the protists show both autotrophic and heterotrophic nutrition?
Answer:
Euglenoids (Euglena)

Question 8.
Agaricus belongs to which class of the fungi ?
Answer:
Basidiomycetes

Question 9.
Which organism causes mad cow disease?
Answer:
Prion

Question 10.
Who proposed the name “Contagium vivum fluidum”[Infectious living fluid]?
Answer:
Beijereinck