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

TN State Board 11th Bio Zoology Important Questions Chapter 2 Kingdom Animalia

Answer the following.

Question 1.
Define levels of Organization.
Answer:
All members of Kingdom Animalia are metazoans (multicellular animals) and exhibit different patterns of cellular organization. The cells of the metazoans are not capable of independent existence and exhibit division of labour. Among the metazoans, cells may be functionally isolated, or similar kinds of cells may be grouped together to form tissues, organ, and organ systems.

Question 2.
Name the layers of cells found in sponges.
Answer:
In sponges, the outer layer is formed of pinacocytes (plate-like cells that maintain the size and structure of the sponge) and the inner layer is formed of choanocytes. These are flagellated collar cells that create and maintain water flow through the sponge thus facilitating respiratory and digestive functions.

Question 3.
What is the tissue level of Organization?
Answer:
In some animals, cells that perform similar functions are aggregated to form tissues. The cells of a tissue integrate in a highly coordinated fashion to perform a common function, due to the presence of nerve cells and sensory cells. This tissue level of organization is exhibited in diploblastic animals like cnidarians. The formation of tissues is the first step towards the evolution of body plan in animals (Hydra – Coelenterata).

Question 4.
What is the organ level of the Organization from which level of animals show this type of Organization?
Answer:
Different kinds of tissues aggregate to form an organ to perform a specific function. Organ, level of organization is a further advancement over the tissue level of organization and appears for the first time in the Phylum Platyhelminthes and seen in other higher phyla.

Question 5.
Make out the difference between the incomplete and complete digestive systems.
Answer:
The digestive system of Platyhelminthes has only a single opening to the exterior which serves as both mouth and anus and hence called an incomplete digestive system. From Aschelminthes to Chordates, all animals have a complete digestive system with two openings, the mouth, and the anus.

Question 6.
Name the embryonic layers of animals and on the basis of the origin and development.
Answer:
During embryonic development, the tissues and organs of animals originate from two or three embryonic germ layers. On the basis of origin and development, animals are classified into two categories: Diploblastic and Triploblastic.

Question 7.
What are diploblastic animals?
Answer:
Animals in which the cells are arranged in two embryonic layers the external ectoderm and internal endoderm are called diploblastic animals. In these animals, the ectoderm gives rise to the epidermis (the outer layer of the body wall) and the endoderm gives rise to the gastrodermis (tissue lining the gut cavity). An undifferentiated layer present between the ectoderm and endoderm is the mesoglea (Corals, Jellyfish, Sea anemone).

Question 8.
Define triploblastic animals for example.
Answer:
Animals in which the developing embryo has three germinal layers are called triploblastic, animals and consists of outer ectoderm (skin, hair, neuron, nail, teeth, etc), inner endoderm (gut, lung, liver), and middle mesoderm (muscle, bone, heart). Most of the triploblastic animals show organ system level of organization (Flatworms to Chordates).

Question 9.
What is meant by symmetry?
Answer:
Symmetry is the body arrangement in which parts that lie on the opposite side of an axis are identical. An animal’s body plan results from the animal’s pattern of development.

Question 10.
Define asymmetry for example.
Answer:
The simplest body plan is seen in sponges. They do not display symmetry and are asymmetrical. Such animals lack a definite body plan or are irregular shaped and any plane passing through the center of the body does not divide them into two equal halves (Sponges).

Question 11.
Write about the radial symmetry with a suitable diagram.
Answer:
Symmetrical animals have paired body parts that are arranged on either side of a plane passing through the central axis. When any plane passing through the central axis of the body divides an organism into two identical parts, it is called radial symmetry. Such radially symmetrical animals have a top and bottom side but no dorsal (back) and ventral (abdomen) side, no right and left side. They have a body plan in which the body parts are organized in a circle around an axis. It is the principal symmetry in diploblastic animals. Cnidarians such as sea anemone and corals are radially symmetrical. However, triploblastic animals like echinoderms eg. Starfish have five planes of symmetry and show Pentamerous radial symmetry:

Question 12.
What is bilateral and biradial symmetry? Give examples.
Answer:
Animals that possess two pairs of symmetrical sides are said to be biradially symmetrical. Biradial symmetry is a combination of radial and bilateral symmetry as seen in ctenophores. There are only two planes of symmetry, one through the longitudinal and sagittal axis and the other through the longitudinal and transverse axis. eg. Comb jellyfish – Pleurobrachia.
Animals that have two similar halves on either side of the central plane show bilateral symmetry. It is an advantageous type of symmetry in triploblastic animals, which helps in seeking food, locating mates and escaping from predators more efficiently. Animals that . have dorsal and ventral sides, anterior and posterior ends, right and left sides are bilaterally symmetrical.

Question 13.
Which animals are called acoelomates? Give example.
Answer:
Animals that do not possess a body cavity are called acoelomates. Since there is nobody cavity in these animals their body is solid without a perivisceral cavity, this restricts the free movement of internal organs (eg, Flatworms).

Question 14.
What is called coelomates?
Answer:
Eucoelom or true coelom is a fluid-filled cavity that develops within the mesoderm and is lined by mesodermal epithelium called the peritoneum. Such animals with a true body cavity are called coelomates or coelomates.

Question 15.
What is notochord? How the animals are classified based on. this presence or absence of notochord?
Answer:
The notochord is a mesodermally derived rod-like structure formed on the dorsal side during embryonic development in some animals. Based on the presence or absence of notochord, animals are classified as chordates (Cephalochordates, Urochordates, Pisces to Mammalia) and non chordates (Porifera to Hemichordata).

Question 16.
How the kingdom Animalia is classified broadly into sub-kingdoms?
Answer:
The animal kingdom is divided into two sub-kingdoms, the Parazoa and Eumetazoa based on their organization.

1. Parazoa: These include the multicellular sponges and their cells are loosely aggregated and do not form tissues or organs.
2. Eumetazoa: These include multicellular animals with well-defined tissues, which are organized as organs and organ systems. Eumetazoans include two taxonomic levels called grades. They include Radiata and Bilateria.

Question 17.
Write about the ‘Division level’ Classification of Bilateria.
Answer:
The eumetazoans other than Radiata, show organ level of organization and are bilaterally symmetrical and triploblastic. The grade Bilateria includes two taxonomic levels called Division.
Division: 1. Protostomia (Proto: first; stomium: mouth): Protostomia includes the eumetazoans in which the embryonic blastopore develops into the mouth. This division includes three subdivisions namely acoelomata, pseudocoelomata and schizocoelomata.
Division: 2. Deuterostomia (deuteron: secondary; stomium: mouth):, Eumetazoans in which anus is formed from or near the blastopore and the mouth is formed away from the blastopore.
It includes only one subdivision Enterocoelomata. They have a true coelom called enterocoel, formed from the archenteron.

Question 18.
Write about the special features of the phylum Porifera.
Answer:

1. These pore-bearing animals are commonly called sponges. They are aquatic, mostly marine, asymmetrical and a few species live in freshwaters.
2. They are primitive, multicellular, sessile animals with a cellular level of organization in which the cells are loosely arranged. They are either radially symmetrical or asymmetrical animals.
3. They possess a water transport system or canal system where water enters through minute pores called Ostia lining the body wall through which the water enters into a central cavity (spongocoel) and goes out through the osculum.
4. This water transport system is helpful in food gathering, circulation, respiration, and removal of waste.
5. Choanocytes or collar cells are special flagellated cells lining the spongocoel and the canals.
6. Nutrition is holozoic and intracellular.
7. All sponges are hermaphrodites.
8. They also reproduce asexually by fragmentation or gemmule formation and sexually by the formation of gametes.
9. Development is indirect with different types of larval stages such as parenchyma and amphiblastula.
10. eg. Sycon (Scypha), Spongilla (freshwater sponge).

Question 19.
Write about the canal system found in Porifera.
Answer:
They possess a water transport system or canal system where water enters through minute pores called Ostia lining the body wall through which the water enters into a central cavity (spongocoel) and goes out through the osculum.
This water transport system is helpful in food gathering, circulation, respiration, and removal of waste.

Question 20.
What is the function of cnidoblasts in phylum Cnidaria?
Answer:
The name Cnidaria is derived from cnidocytes or cnidoblasts with stinging cells or nematocyst on tentacles. Cnidoblasts are used for anchorage, defense, and capturing prey.

Question 21.
Differentiate polyp and medusa of coelenterates or cnidarians.
Answer:
Cnidarians exhibit two basic body forms, polyp, and medusa. The polyp forms are sessile and cylindrical (eg. Hydra, Adamsia), whereas the medusa is umbrella-shaped and free swimming.
The polyp represents the asexual generation and medusa represents the sexual generation. Polyps produce medusa asexually and medusa forms polyps sexually.

Question 22.
Write short notes about Ctenophora.
Answer:

1. Ctenophora are exclusively marine, radially symmetrical, diploblastic animals with tissue level of organization.
2. Though they are diploblastic, their mesoglea is different from that of cnidaria. It contains amoebocytes and smooth muscle cells.
3. They have eight external rows of ciliated comb plates (comb jellies) which help in locomotion, hence commonly called comb jellies or sea walnuts.
4. Bioluminescence (the ability of a living organism to emit light) is well marked in ctenophores.
5. They possess special cells called lasso cells or colloblasts which help in food capture.
6. Digestion is both extracellular and intracellular.
7. Sexes are not separate (monoecious). They reproduce only by sexual means.
8. (Fertilization is external and development is indirect and includes a larval stage called cydippid larva, eg. Pleurobrachia and Ctenoplana.

Question 23.
What is the type of digestive system found among flatworms?
Answer:
Some of the parasitic flatworms absorb nutrients directly from the host through their body surface. However, flatworms like liver fluke have an incomplete digestive system which means it has an only a single opening to the exterior which serves as both mouth and anus.

Question 24.
List out the characters of the phylum Annelida.
Answer:

1. Annelids are the first segmented animals to evolve.
2. They are aquatic or terrestrial, free-living but some are parasitic.
3. They are triploblastic, bilaterally symmetrical, schizocoelomates, and exhibit organ system level of body organization.
4. The coelom with coelomic fluid creates a hydrostatic skeleton and aids in locomotion.
5. Their elongated body is metamerically segmented and the body surface is divided into segments or metameres.
6. Internally the segments are divided from one another by partitions called septa. This phenomenon is known as metamerism.
7. Aquatic annelids like Nereis have lateral appendages called parapodia, which help in swimming.
8. The circulatory system is of closed type and the respiratory pigments are hemoglobin and chlorocruorin.
9. The nervous system consists of paired ganglion connected by the lateral nerves to the double ventral nerve cord.
10. They reproduce sexually. Development is direct or indirect and includes a trochophore larva, eg. Lampito mauritii (Earthworm).

Question 25.
Write an account on the common characters of Arthropoda.
Answer:

1. This is the largest phylum of the Kingdom Animalia and includes the largest class called Insecta.
2. They are bilaterally symmetrical, segmented, triploblastic and schizocoelomate animals with organ system grade of body organization.
3. They have jointed appendages that are used for locomotion feeding and are sensory in function,
4. The body is covered by cjiitinous exoskeleton for protection and to prevent water loss, It is shed off periodically by a process called molting or ecdysis.
5. The body consists of a head, thorax, and abdomen with a body cavity called haemocoel.
6. Respiratory organs are gills, book gills, book lungs or trachea.
7. The circulatory system is of open type.
8. Sensory organs like antennae, eyes (compound and simple), statocysts (organs of balance/ equilibrium) are present.
9. Excretion takes place through malpighian tubules, green glands, coxal glands, etc.
10. They are mostly dioecious and oviparous; fertilization is usually internal.
11. Life history includes many larval stages followed by metamorphosis. eg. Limulus.

Question 26.
Define metamerism in Annelids.
Answer:
The body of the annelids are metamerically segmented and the body surface is divided into segment or metameres. Internally the segments are divided from one another by partitions called septa. This phenomenon is known as metamerism.

Question 27.
Define moulting or ecdysis.
Answer:
Body is covered by chitinous exoskeleton for protection and to prevent water loss, It is shed off periodically by a process called moulting or ecdysis.

Question 28.
Write few examples for the phylum, Arthopoda.
Answer:

1. Limulus (King crab, a living fossil)
2. Palamnaeus (Scorpion)
3. Eupagarus (Hermit crab)
4. Apis (Honey bee)
5. Musca (House fly)
6. Vectors – Anopheles, Culex, Aedes (mosquitoes)
7. Economically important insects – Apis- (Honey bee), Bombyx (Silk worm)
8. Laccifer (Lac insects)
9. Living fossils Limulus-(King crab)
10. Gregarious pest – Locusta (Locust)

Question 29.
Write the salient features of the phylum mollusca.
Answer:

1. This is the second-largest animal phylum.
2. Molluscs are terrestrial or aquatic (marine or fresh water) and exhibit organ system level of body organisation.
3. They are bilaterally symmetrical (except univalves), triploblastic and coelomate animals.
4. Body is covered by a calcareous shell and is unsegmented with a distinct head, muscular foot and a visceral hump or visceral mass.
5. A soft layer of skin forms a mantle over the visceral hump. The space between the visceral mass and mantle (pallium) is called the mantle cavity.
6. A number of feather like gills (ctenidia) are present, which are respiratory in function.
7. The digestive system is complete and mouth contains a rasping organ called radula with transverse rows of chitinous teeth for feeding (radula is absent in bivalves.
8. The sense organs are tentacles, eyes and ospharidium (to test the purity of water and present in bivalves and gastropods).
9. Excretory organs are nephridia.
10. Open type of circulatory system is seen except for cephalopods such as squids, cuttle fishes and octopuses. Blood contains haemocyanin, a copper containing respiratory pigment.
11. Development is indirect with a veliger larva.
12. eg. Pila (Apple snail).

Question 30.
Write the functions of the following: (i) Ctenidia, (ii) Ospharidiam.
Answer:

1. Ctenidia: Mandle cavity has a number of feather like gills (ctenidia) are present, which are respiratory in function.
2. Ospharidiam: The anterior head regim of molluscs has ospharidium to test the purity of water and presence in bivalves and gasropods.

Question 31.
Why ore certain marine animals termed as echinoderms?
Answer:
All Echinoderms are marine animals. These animals have a mesodermal endoskeleton of calcareous ossicles and hence the name Echinodermata.

Question 32.
What is the most distinctive feature of echinoderms and write its importance in them?
Answer:
The most distinctive feature of echinoderms is the presence of the water vascular system or ambulacral system with tube feet or podia, which helps in locomotion, capture and transport of food and respiration.

Question 33.
What are the fundamental distinct features of all chordates?
Answer:
All chordates possess three fundamental distinct features at some stage of their life cycle, they are:

1. Presence of elongated rod like notochord below the nerve cord and above the alimentary canal. It serves as a primitive internal skeleton. It may persist throughout life in lancelets and lampreys. In adult vertebrates, it may be partially or completely replaced by backbone or vertebral column.
2. A dorsal hollow or tubular fluid filled nerve cord lies above the notochord and below the dorsal body wall. It serve’s to integrate and co-ordinate the body functions. In higher chordates, the anterior end of the nerve cord gets enlarged to form the brain and the posterior part becomes the spinal cord, protected inside the vertebral column.
3. Presence of pharyngeal gill slits or clefts in all chordates at some stage of their life cycle.
   It is a series of gill slits or clefts that perforates the walls of pharynx and appears during the development of every chordate. In aquatic forms, pharyngeal gill slits are vascular, lamellar and form the gills for respiration. In terrestrial chordates, traces of non-functional gill clefts appear during embryonic developmental stages and disappear later. Besides the above said features, chordates are bilaterally symmetrical, triploblastic, coelomates with organ system level of organisation; they possess post anal tail, closed circulatory system with a ventral myogenic heart except in Amphioxus.

Question 34.
Give a brief account on the characters of Tunicates.
Answer:
They are exclusively marine and are commonly called sea squirts. Mostly sessile, some pelagic or free swimming, exist as solitary and colonial forms. Body is unsegmented and covered by a test or tunic. Adult forms%re sac like. Coelom is absent, but has an atrial cavity surrounding the pharynx. Notochord is present only in the tail region of the larval stage, hence named urochordata. Alimentary canal is complete and circulatory system is of open type. The heart is ventral and tubular.

Respiration is through gill slits and clefts. Dorsal tubular nerv e cord is present only in the larval stage and a single dorsal ganglion is present in the adults. Mostly hennaphrodites, development indirect and includes a free swimming tadpole larva with chordate characters. Retrogressive metamorphosis is seen. eg. Ascidia.

Question 35.
Give an account of the General features of subphylum caphalochorelata.
Answer:
Cephalochordates are marine forms, found in shallow waters, leading a burrowing mode of life. They are small fish like coelomate forms with chordate characters such us notochord, dorsal tubular nerve cord and pharyngeal gill slits throughout their life. Closed type of circulatory system is seen without heart. Excretion is by protonephridia. Sexes are separate, Fertilization is external, eg. Branchiostoma.

Question 36.
How are the vertebrates are further divided by divisions? Write the differences between them.
Answer:

1. Subphylum Vertebrata is divided into two divisions, Agnatha and Gnathostomata.
2. Agnatha includes jawless fish-like aquatic vertebrates without paired appendages. Notochord persists in the adult.
3. Agnatha includes one important class – Cyclostomata.
4. Gnathostomata includes jawed vertebrates with paired appendages. Notochord is replaced partly or wholly by the vertebral column.
5. Gnathostomata includes jawed fishes (Pisces) and Tetrapoda ( amphibia, reptilia, aves and mammals).The superclass Pisces includes all fishes which are essentially aquatic forms with paired fins for swimming and gills for respiration. Pisces includes cartilaginous fishes (Chondrichthyes) and bony fishes (Osteicthyes).

Question 37.
What is the special character seen in cyclostomes during spawning?
Answer:
Cyclostomes are marine but migrate to fresh waters for spawning (anadromous migration). After spawning within a few days they die. The larvae (ammocoete) after metamorphosis returns to the ocean.

Question 38.
Give some examples for bony fishes coming under Osteichthyes.
Answer:
Exocoetus, Labeo, Catla, Echeneis, Pterophyllam.

Question 39.
Write the general characters of amphibians with some examples.
Answer:

1. Amphibians are the first vertebrates and tetrapods to live both in aquatic as well as terrestrial habitats.
2. They are poikilothermic.
3. Their body is divisible into the head and trunk and most of them have two pairs of limbs; tail may or may not be present.
4. Their skin is smooth or rough, moist, pigmented and glandular.
5. Eyes have eyelids and the tympanum represents the ear.
6. Respiration is by gills, lungs and through the skin.
7. The heart is three-chambered.
8. Kidneys are mesonephric. Sexes are separate and fertilization is external.
9. They are oviparous and development is indirect. They show hibernation and aestivation.
10. eg. Bufo (Toad), Rana (Frog).

Question 40.
What are the general features of Reptilians?
Answer:

1. They are mostly terrestrial animals and their body is covered by dry, and comified skin with epidermal scales or scutes.
2. Reptiles have three chambered heart but four chambered in crocodiles.
3. All are cold blooded amniotes.
4. Most reptiles lay cleidoic eggs with extraembryonic membranes like amnion, allantois, chorion and yolk sac.
5. Excretion by metanephric kidneys and are uricotelic.
6. They are monoecious. Internal fertilization takes place and all are oviparous.
7. eg. Calotes (Garden Lizard), Draco (Flying Lizard), Crocodilus (Crocodile).

Question 41.
Describe the special characteristic features of Aves.
Answer:

1. Aves are commonly known as birds. The characteristic feature of Aves is the presence of feathers and the ability to fly except for flightless birds, eg. Ostrich, Kiwi.
2. The forelimbs are modified into wings, and the hind limbs are adapted for walking, running, swimming and perching.
3. The skin is dry and devoid of glands except the oil gland or preen gland at the base of the tail.
4. The exoskeleton consists of epidermal feathers, scales, claws on legs and the homy, covering on the beak.
5. The endoskeleton is fully ossified (bony) and the long bones are hollow with air cavities (pneumatic bones).
6. The pectoral muscles of flight (pectoralis major and pectoralis minor) are well developed.
7. Respiration is by compact, elastic, spongy lungs that are continuous with air sacs to supplement respiration.
8. The heart is four chambered.
9. The urinary bladder is absent.
10. Sexes are separate with well marked sexual dimorphism.
11. All birds are oviparous. Eggs are megalecithal and cleidoic. Fertilization is internal.
12. eg. Corvus (Crow), Columba (Pigeon), Pavo (Peacock).

Question 42.
Write the general characters of the class mammalia.
Answer:

1. Their body is covered by hair, a unique feature of mammals. Some of them are adapted to fly or live in water.
2. The presence of mammary glands is the most unique feature of mammals.
3. They have two pairs of limbs adapted for walking, running, climbing, burrowing, swimming and flying.
4. Their skin is glandular in nature, consisting of sweat glands, scent glands and sebaceous glands.
5. Exoskeleton includes homy epidermal horns, spines, scales, claws, nails, hooves and bony dermal plates.
6. Teeth are thecodont, heterodont and diphyodont.
7. External ears or pinnae are present.
8. The heart is four-chambered and possesses a left systematic arch. Mature RBCs are circular, biconcave and non nucleated.
9. Mammals have a large brain when compared to other animals.
10. They show greatest intelligence among all animals. Their kidneys are metanephric and are ureotelic.
11. All are homeothermic, sexes are separate and fertilization is internal.
12. eg. Platypus, Kangaroo, Monkey, Elephant.

Question 43.
Name two different types of larval stages in Porifera.
Answer:
Two different types of larval stages of porifera are parenchymula and amphiblastula.

Choose the correct answer.

1. Biradial symmetry is seen in:
(h) Star fish
(b) Comb jelly fish
(c) Sea anemone
(d) Sponge
Answer:
(b) Comb jelly fish

2. The special flagellated cells lining the spongocoel is:
(a) Choanocytes
(b) Cridocytes
(c) Nematocyst
(d) Lasso cells
Answer:
(a) Choanocytes

3. The minute pores lining the body wall of Porifera are called:
(a) Osculum
(b) Podia
(c) Ostia
(d) Gills
Answer:
(c) Ostia

4. The central body cavity of poriferans are:
(a) Gastrocoel
(b) Coelom
(c) Haemocoel
(d) Spongocoel
Answer:
(d) Spongocoel

5. The free-swimming ciliated larval form of criclaria is:
(a) Planula larva
(b) Parenchymula larva
(c) Amphiblastula larva
(d) Veliger larva
Answer:
(a) Planula larva

6. In case of flatworms the specialized excretory cells are named as:
(a) Nematocysts
(b) Flame cells
(c) Nephridia
(d) Malphigian tubules
Answer:
(b) Flame cells

7. Nereis have lateral appendages called:
(a) Parapodia
(b) Body setae
(c) Foot
(d) Tube feet
Answer:
(a) Parapodia

8. The special cells of ctenophora helps in food capture is:
(a) Cnidoblasts
(b) Choanocytes
(c) Flamecells
(d) Colloblasts
Answer:
(d) Colloblasts

9. ………… are the organs of balance in Arthropods.
(a) Nematocysts
(b) Statocysts
(c) Choanocytes
(d) Cochlea
Answer:
(b) Statocysts

10. ……….. is the largest phylum of the kingdom Animalia.
(a) Annelida
(b) Arthropoda
(c) Aschelminthes
(d) Echinodermata
Answer:
(b) Arthropoda

11. The second largest animal phylum is:
(a) Ctenophora
(b) Arthropoda
(c) Mollusca
(d) Coelenterata
Answer:
(c) Mollusca

12. The anterior head region of molluscs has got this organ which helps to test the purity of water:
(a) Ostia
(b) Ospharidiam
(c) Ossicles
(d) Gills
Answer:
(b) Ospharidiam

13. The larva of Nereis is:
(a) Planula
(b) Tomaria larva
(c) Trocophore larva
(d) Miracidium
Answer:
(c) Trocophore larva

14. Presence of water vascular system is the most distinctive feature of the Phylum:
(a) Mollusca
(b) Sponges
(c) Echinodermata
(d) Arthropoda
Answer:
(c) Echinodermata

15. The mantle cavity of the molluscs has got number of feather like gills, which are respiratory and excretory in function are:
(a) Book lungs
(b) Trachea
(c) Ambulacral system
(d) Ctenidia
Answer:
(d) Ctenidia

16. The phylum Hemichordata are mostly tubiculous and commonly called:
(a) Flat worms
(b) Round worms
(c) Tongue worms
(d) Parasitic worms
Answer:
(c) Tongue worms

17. The free swimming larva of Hemichordata are called:
(a) Tomaria larvae
(b) Planula larvae
(c) Trochophore larvae
(d) Cercaria larvae
Answer:
(a) Tomaria larvae

18. The tunicates are normally called:
(a) Sea squirts
(b) Sea anemone
(c) Sea-walnuts
(d) Sea urchin
Answer:
(a) Sea squirts

19. The chondrichthyes has got this type of gills helps for respiration:
(a) Filamentous gills
(b) Lamelliform gills
(c) Filiform gills
(d) Ambnlacral system
Answer:
(b) Lamelliform gills

20. The excretory organ of Chondrichthyes are:
(a) Pronephric kidneys
(b) Opisthonephric kidneys
(c) Mesonephric kidneys
(d) Metanephric kidneys
Answer:
(b) Opisthonephric kidneys.

21. The excretory organ of Osteichthyes are:
(a) Mesonephnc kidneys
(b) Opisthonephric kidneys
(c) Holonephric kidneys
(d) Metanephric kidneys
Answer:
(a) Mesonephnc kidneys

22. The eggs of Aves are of type.
(a) Microlecithal
(b) Mesolecithal
(c) Megalecithal
(d) Homolecithal
Answer:
(c) Megalecithal

23. The mature RBCs are non – nucleated in:
(a) Molluscs
(b) Mammals
(c) Fishes
(d) Birds
Answer:
(b) Mammals

24. Match:

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

25. Match:

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

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

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

Answer the following.

Question 1.
Define Ecosystem?
Answer:
The ecosystem is a community of living organisms (plants and animals), a non-living environment (including minerals, climate, soil, water; sunlight), and their interrelationships.

Question 2.
Define biological diversity?
Answer:
The presence of a large number of species in a particular ecosystem is called ‘biological diversity or in short ‘biodiversity’.

Question 3.
What is called classification?
Answer:
Classification is a process by which things are grouped in convenient categories, based on easily observable characters. The scientific term used for these categories is taxa.

Question 4.
What are the key characters of living organisms?
Answer:
The key characters of living organisms are cellular organization, nutrition, respiration, metabolism, growth, response to stimuli, movement, reproduction, excretion, adaptation, and homeostasis.

Question 5.
Define taxonomy?
Answer:
Based on their characteristics, all living organisms can be classified into different taxa. This science of classification is called taxonomy.

Question 6.
What are the basic principles of classification?
Answer:

1. To identify and differentiate closely related species:
2. To know the variation among the species.
3. To understand the evolution of the species.
4. To create a phylogenetic tree among the different groups
5. To conveniently study the living organisms

Question 7.
What are the main criteria for systematics?
Answer:
The main criteria of systematics are identifying, describing, naming, arranging, preserving, and documenting the organisms. Apart from the above-said features, the evolutionary history of the species and the environmental adaptations and interrelationship between species are also being investigated in systematics.

Question 8.
How did Aristotle classify the animals?
Answer:
Based on the presence or absence of red blood cells. Aristotle classified the animals into two as Enaima with blood and those without blood as Anaima.

Question 9.
What is cladistics?
Answer:
Biologists initiated studies on the evolutionary and genetic relationships among organisms, which led to the emerge of phylogenetic classification or cladistics. It is an evolutionary classification based on how a common ancestry was shared. Cladistic classification summarizes the genetic differences between all species in the ‘phylogenetic tree’.

Question 10.
Define cladogram?
Answer:
Ernst Haeckal introduced the method of representing evolutionary relationships with the help of a tree diagram known as cladogram.

Question 11.
What are the points to be followed to draw a cladogram?
Answer:
Phylogenetic system of classification takes into account ancestral characters (traits of basic body design which would be in the entire group) and derived characters (traits whose structure and functions differ from those of ancestral characters). One or more derived characters which appeared during evolution resulted in the formation of new subspecies. In a cladogram, each evolutionary step produces a branching and all the members of the branch would possess the derived character which will not be seen in organisms below the particular branch point. Arranging organisms on the basis of their similar or derived characters which differ from the ancestral characters produced a phylogenetic tree or cladogram.

Question 12.
What is the Five Kingdom classification? Who proposed this?
Answer:
Five kingdom classification defined by R.H. Whittaker as Monera, Protista, Fungi, Plantae, and Animalia based on the cell structure, mode of nutrition, mode of reproduction, and phylogenetic relationships.

Question 13.
Which aspect paved the way for three domain classification?
Answer:
Classification has come a long way and now takes into account even molecular level DNA and RNA identification. The advancement in molecular techniques and biochemical assays has led to a new classification – The “Three Domain” classification.

Question 14.
Who proposed three-domain classification? What are the features of this classification?
Answer:
Three domain classification was proposed by Carl Woese. This system emphasizes the separation of prokaryotes into two domains. Bacteria and Archaea, and all the eukaryotes are placed into the domain Eukarya. Archaea appears to have more in common with die Eukarya than the Bacteria. Archaea differ from bacteria in cell wall composition and differs from bacteria and eukaryotes in membrane composition and rRNA types.

Question 15.
What are the three domains of living things? Write it with examples.
Answer:
Three domains are Archaea, Bacteria, Eukarya.

Question 16.
Why domain Archaea are called extremophiles?
Answer:
The domain Archaea includes single-celled organisms, the prokaryotes which have the ability to grow in extreme conditions like volcano vents, hot springs, and polar ice caps, hence are also called extremophiles. They are capable of synthesizing their food without sunlight and oxygen by utilizing hydrogen sulfide and other chemicals from volcanic vents.

Question 17.
Define the following terms – (i) Halophiles, (ii) Methanogens, (iii) Thermoacidophiles.
Answer:

1. Halophiles: Organisms that live in salty environments are called Halophiles.
2. Methanogens: Some of the extremophiles that produce methane are called methanogens.
3. Thermoacidophiles: Thermoacidophiles thrive in acidic environments and at high temperatures.

Question 18.
Name the two broad classifications of Bacteria?
Answer:
Bacteria are of two types. They are beneficial probiotic bacteria and harmful pathogenic bacteria.

Question 19.
What is the role of Cyanobacteria dulling eat ly geologic periods?
Answer:
Cyanobacteria are photosynthetic blue-green algae during geological periods which produce oxygen. These had played a key role in the changes of atmospheric oxygen levels from anaerobic to aerobic during the early geologic periods.

Question 20.
Classify the organisms as per the seven kingdom system of classification?
Answer:
In 1987, Cavalier-Smith revised the six kingdom system to the Seven Kingdom system. According to that, the classification is divided into two Super Kingdoms (Prokaryota and Eukaryota) and seven kingdoms, two Prokaryotic Kingdoms (Eubacteria and Archaebacteria) and five Eukaryotic Kingdoms (Protozoa, Chromista, Fungi, Plantae, and Animalia).

Question 21.
Define species?
Answer:
Species are the basic unit of classification in the taxonomic hierarchical system. It is a group of animals having similar morphological features (traits) and is reproductively isolated to produce fertile offspring.

Question 22.
Differentiate monotypic and polytypic genus?
Answer:
Genus: It is a group of closely related species which have evolved from a common ancestor. In some genus, there are only one species which is called a monotypic genus.
eg. Red panda is the only species in the genus Ailurus: Ailurus fulgens. If there are more than one species in the genus it is known as polytypic genus, eg. ‘Cats’ come under the genus Felis, which has a number of closely related species.

Question 23.
Define family?
Answer:
It is a taxonomic category that includes a group of related genera with less similarity as compared to genus and species. For example, the family Felidae includes the genus Felis (cats) and the genus, Panthera (lions, tigers, leopards).

Question 24.
Define class?
Answer:
This category includes one or more related orders with some common characters, eg. order Primata comprising monkeys, apes, and man is placed in the Class Mammalia, along with the order Carnivora which includes dogs and cats.

Question 25.
Write the resultant animal when there is a cross between (i) a Male donkey with a female horse, (ii) a male lion with a female tiger, (iii) a male tiger with a female lion.
Answer:

1. Male donkey with female horse results in Mule. (Sterile)
2. Male lion with female tiger results in Liger.
3. Male tiger with female lion results in Tigon.

Question 26.
What is Binomial nomenclature? What is its significance?
Answer:
Biologists follow universally accepted principles to provide scientific names to known organisms. Each name has two components, a generic name, and a specific epithet. This system of naming the organism is called Binomial Nomenclature.
Classification and grouping were done to facilitate a deeper understanding of the unique characteristics of each organism and its interrelationship among closely related species.
It plays a vital role in the arrangement of known species based on their similarities and dissimilarities.

Question 27.
Write down the binomial names for the following (i) National Bird of India, (ii) National Animal of India, (iii) Tamil Nadu State Bird.
Answer:

1. National Bird of India – Pavo cristatus (Indian Peafowl)
2. National Animal of India Panthera tigris (tiger)
3. Tamil Nadu State Bird – Chalcophaps indica (Emerald dove)

Question 28.
When does the system of trinomial nomenclature is followed?
Answer:
When members of any species have large variations then a trinomial system is used. On the basis of dissimilarities, this species gets classified into subspecies.

Question 29.
What are the basic rules followed for naming the animals scientifically? (or) Describe the rules of Nomenclature.
Answer:

1. The scientific name should be italicized in printed form and if handwritten, it should be underlined separately.
2. The generic name’s (Genus) first alphabet should be in uppercase.
3. The specific name (species) should be in lowercase.
4. The scientific names of any two organisms are not similar.
5. The name dr abbreviated name of the scientist who first publishes the scientific name may be written after the species name along with the year of publication, eg. Lion-Felis Leo Linn., 1758 or Felis Leo L., 1758.
6. If the species name is framed after any person’s name the name of the species shall end with ‘i’, ‘ii’ or ‘ae’. eg. A new species of a ground-dwelling lizard (Cyrtodactylus) has been discovered and named after scientist Varad Giri, Cyrtodactylus varadgirii.

Question 30.
What are the tools used for the study of taxonomy?
Answer:
Tools and taxonomical aids may be different for the study of plants and animals. Herbarium and Botanical garden may be used as tools for the study of plant taxonomy. In the case of animal studies, the classical tools are Museum, Taxonomical Keys, and Zoological and Marine parks.

Question 31.
What are the components of taxonomical tools?
Answer:
The important components of the taxonomical tools are field visits, survey, identification, classification, preservation, and documentation.

Question 32.
Differentiate museum from the zoological park?
Answer:
Museum: Biological museums have a collection of preserved plants and animals for study and ready reference. Specimens of both extinct and living organisms can be studied.
Zoological parks: These are places where wild animals are kept in protected environments under human care which enables us to study their food habits and behavior.

Question 33.
What are the recent molecular tools used for taxonomical studies?
Answer:
DNA hybridization (measures the degree of genetic similarity between pools of DNA sequences), DNA fingerprinting (to identify an individual from a sample of DNA by looking at unique patterns in their DNA), Restriction Fragment Length Polymorphisms (RFLP) analysis (difference in homologous DNA sequences that can be detected by the presence of fragments of different lengths after digestion of the DNA samples), and Polymerase Chain Reaction (PCR) sequencing (to amplify a specific gene or portion of a gene,) are used as taxonomical tools.

Question 34.
Give the abbreviations for the following: ALIS, DAISY, ABIS, SPIDA, Draw wing.
Answer:

Question 35.
Write about some taxonomical tools.
Answer:

1. Neo taxonomical tools: This is based on Electron Microscopy images to study the molecular structures of cell organelles.
2. Ethology of taxonomical tools: Based on the behavior of the organisms it can be classified. For example sound of birds, bioluminescence, etc.
3. e-Taxonomic resources: INOTAXA is an electronic resource for digital images and descriptions of the species which was developed by the Natural History Museum, London. INOTAXA means Integrated Open TAXonomic Access.

Question 36.
What is Tautonymy?
Answer:
The practice of naming the animals in which the generic name and species name are the same is called Tautonymy. eg. Naja naja (The Indian Cobra).

Question 37.
What kind of adaptations extremophiles have to live in extreme conditions.
Answer:

1. Extremophiles are capable of synthesizing their food without sunlight and oxygen by utilizing hydrogen sulfide and other chemicals from volcanic vents.
2. Methanogens produced methane.
3. Halophiles live in salty environments and thermoacidophiles thrive in acidic environments and high temperatures.

Question 38.
Why T. aquaticus is used in PCR?
Answer:
T. aquaticus: Thermus aquaticus.
PCR: Polymerase chain reaction.
The DNA polymerase found in Thermus aquaticus remains stable even at very high temperatures.
Because of this stability, it can be used in the process known as PCR.

Question 39.
If you discover a new species how will you name them nominally?
Answer:
Before naming a new species there are certain pre-requisites that need to be followed. Initially, a holotype must be designated. A holotype is nothing but a single specimen that serves as the identifier for the entire new Species. The holotype must possess the key features that differentiate the new species from the existing ones i.e., it must be unique. The details of the origin of the holotype, the environment in which it was collected and its paratypes must be provided.
Secondly, a description of the species must be provided. The definition must be in terms of behavioral, anatomical, and genetic marks of the new species.
Thirdly, the species must be allocated a new name. Species are always identified by both a generic name and a species name. The generic name of the species is the genus to which it belongs to and then the new species name is added. There are several rules and conventions involved in creating and managing the names of the new species.
Finally, the details of the newly found species must be published in an internationally accessible format and archived in multiple locations.

Question 40.
Why pig-nosed frog in India is called Bhupathy’s purple frog?
Answer:
Scientists have discovered a new and unusual species of frog in the Western Ghats mountain. range in India.
The frog has shiny, purple skin. The scientists have called the new species as Bhupathy’s purple frog, in honor of their colleague Dr. Subramaniam Bhupathy, a respected herpetologist who lost his life in the Western Ghats in 2014.

Question 41.
Where did they discover a freshwater jellyfish in Tamilnadu?
Answer:
Freshwater jellyfish are extremely rare. Ishaan Abraham Pichamuthu, the nine-year-old boy, discovered a new species of freshwater jellyfish thriving in the depths of Kodaikanal lake.

Question 42.
What are the newly discovered species in India?
Answer:

1. 55 vertebrates and 258 invertebrates are the new animal species discovered in India.
2. Insects 97 species. Fishes 27 species, Amphibians 12 species, Platyhelminthes 10 species, Crustacea 9 species, Reptiles 6 species have been discovered and described by the scientist.
3. Moths and butterflies 61 species, Beetles 38 species also identified.
4. Most of these new species were from biological hotspots of the country, – the Himalayas, the Western Ghats, and the Andaman and Nicobar Islands.

Question 43.
How to save endangered species.
Answer:

1. Preventing listed species from being killed or harmed.
2. Protecting habitat essential to the survival of endangered species.
3. Creating plans to restore a healthy population of endangered animals.
4. Never purchase products made from threatened or endangered species.
5. Don’t use toxic herbicides or pesticides.
6. Support zoos and other wildlife parks.
7. Defending and strengthening the endangered species act.
8. Advocating for increased funding for conservation programs that benefits endangered species.
9. Slow down while driving in forest areas and watch out for wild animals cross the road.
10. Preserving water bodies in habitats of endangered species to ensure then sustaining.

Question 44.
What are the different taxonomical tools? Explain in detail.
Answer:
Tools and taxonomical aids may be different for the study of plants and animals. Herbarium and Botanical garden may be used as tools for the study of plant taxonomy. In the case of animal studies, the classical tools are Museum, Taxonomical Keys, and Zoological and Marine parks.
The important components of the taxonomical tools are field visits, survey, identification, classification, preservation, and documentation. Many tools are being used for taxonomical studies, amongst them, some of the important tools are discussed below:
The classical taxonomical tools:
Taxonomical Keys: Keys are based on a comparative analysis of the similarities and dissimilarities of organisms. There are separate keys for different taxonomic categories.
Museum: Biological museums have a collection of preserved plants and animals for study and ready reference. Specimens of both extinct and living organisms can be studied.
Zoological parks: These are places where wild animals are kept in protected environments under human care which enables us to study their food habits and behavior.
Marine parks: Marine organisms are maintained in a protected environment. Printed taxonomical tools consist of identification cards, descriptions, field guides, and manuals.
Molecular taxonomical tools:
Technological advancement has helped to evolve molecular taxonomical tools from classical tools to molecular tools. The accuracy and authenticity is more significant in the molecular tools. The following methods are being used for taxonomical Classification.
Molecular techniques and approaches such as DNA barcoding (short genetic marker in an organism’s DNA to identify it as belonging to a particular species), DNA hybridization (measures the degree of genetic similarity between pools of DNA sequences), DNA fingerprinting (to identify an individual from a sample of DNA by looking at unique patterns in their DNA), Restriction Fragment Length Polymorphisms (RFLP) analysis (difference in homologous DNA sequences that can be detected by the presence of fragments of different lengths after digestion, of the DNA samples), and Polymerase Chain Reaction (PCR) sequencing (to amplify a specific gene or portion of a gene,) are used as taxonomical tools.
Automated species identification tools:
It consists of Cyber tools, eg. DAISY, ALIS, ABIS, SPIDA, Draw wing, etc.

Neo taxonomical tools: This is based on Electron Microscopy images to study the molecular structures of cell organelles.
Ethology of taxonomical tools: Based on the behavior of the organisms it can be classified. For example sound of birds, bioluminescence, etc.
e-Taxonomic resources: INOTAXA is an electronic resource for digital images and descriptions of the species which was developed by the Natural History Museum, London. INOTAXA means Integrated Open TAXonomic Access.

Choose the correct answer.

1. The term bio-diversity was first introduced by:
(a) A. G. Tansley
(b) Walter Rosen
(c) Aristotle
(d) AP de Candle
Answer:
(b) Walter Rosen

2 is called as the father of taxonomy:
(a) Carolus Linnaeus
(b) Aristotle
(c) Theophrastus
(d) John Ray
Answer:
(b) Aristotle

3. The word taxonomy was coined by:
(a) AP de Candolle
(b) Ernst Haeckel
(c) Carl Woese
(d) Smith
Answer:
(a) AP de Candolle

4. The father f modern taxonomy is:
(a) John Ray
(b) Huxley
(c) Darwin
(d) Carolus Linnaeus
Answer:
(d) Carolus Linnaeus

5. According to Aristotle, the animals with red blood cellš are called:
(a) Anaima
(b) chromista
(c) Enaima
(d) Protozoa
Answer:
(c) Enaima

6. Five kingdom classification was proposed by:
(a) Walter Rosen
(b) R.H. Whittaker
(c) AG Tansley
(d) Bauhin
Answer:
(b) R.H. Whittaker

7. Three domain classification was proposed by:
(a) Carl Woese
(b) Huxley
(c) Varad Gin
(d) Aristotle
Answer:
(a) Carl Woese

8. Prokaryotes that have the ability to grow in extreme conditions are called:
(a) Probiotic
(b) Pathogenic
(c) Extremophiles
(d) Archaea
Answer:
(c) Extremophiles

9. The cell wall of bacteria contains:
(a) Glycogen
(b) Peptidoglycans
(c) Polypeptides
(d) Histones
Answer:
(b) Peptidoglycans

10. Seven Kingdom system was proposed by:
(a) Cavalier-Smith
(b) Adams
(c) AG Tansley
(d) Walter Rosen
Answer:
(a) Cavalier-Smith

11. Mating between the male lion and female tiger results in:
(a) Mule
(b) Tigon
(c) Liger
(d) Feus
Answer:
(c) Liger

12. The scientffic name of human is:
(a) Homo erectus
(b) Homo sapiens
(c) Emberi
(d) Umano
Answer:
(b) Homo sapiens

13. The binomial name of Indian National Bird is:
(a) Pavo cristatus
(b) Homo sapiens
(c) Ailurusfulgens
(d) Fells maigarita
Answer:
(a) Pavo cristatus

14. The binomial name of National animal of India is:
(a) Panthera tigris
(b) Corvus splendens
(c) Felis domestica
(d) Liger
Answer:
(a) Panthera tigris

15. ‘Origin of species’ the book that explains the evolutionary connections of species by the process of natural selection, written by:
(a) Linnaeus
(b) Aristotle
(c) Darwin
(d) Lamark
Answer:
(c) Darwin

16. India has about …………. number of threatened species of animals as per the data given IUCN.
(a) 172
(b) 72
(c) 127
(d) 721
Answer:
(a) 172

Students get through the TN Board 11th Bio Botany Important Questions Chapter 15 Plant Growth and Development which is useful for their exam preparation.

TN State Board 11th Bio Botany Important Questions Chapter 15 Plant Growth and Development

Very short answer questions

Question 1.
Give the definition for Growth.
Answer:
Growth is defined as an irreversible permanent increase in size, shape, number, volume, and dry weight.

Question 2.
Explain the term ‘Monocarpic perennial’.
Answer:
Monocarpic perennials produce flowers only once during their lifetime but the plants survive for many years. Example: Bamboo.

Question 3.
Define the grand period of growth and mention its phases.
Answer:
The total period from the initial to the final stage of growth is called the grand period of growth. They are:

1. Lag phase,
2. Log phase,
3. Decelerating phase and
4. Maturation phase.

Question 4.
Plot a graph depicting the constant linear growth.
Answer:

Question 5.
Compare absolute growth rate with relative growth rate.
Answer:
Comparison between Absolute and Relative Growth Rates

Question 6.
Define Differentiation.
Answer:
The process of maturation of meristematic cells to specific types of cells performing specific functions is called differentiation.

Question 7.
What is Re-differentiation?
Answer:
Differentiated cells, after multiplication again lose the ability to divide and mature to perform specific functions. This is called redifferentiation.

Question 8.
What are Plant Growth Regulators (PGR’s)?
Answer:
Plant Growth Regulators are defined as organic substances which are synthesized in minute quantities in one part of the plant body and transported to another part where they influence specific physiological processes.

Question 9.
Mention any four Phytohormones.
Answer:
Auxins, gibberellins, cytokinin, ethylene, abscisic acid.

Question 10.
Define the term Bioassay.
Answer:
Bioassay means-testing of substances for their activity in causing a growth response in a living plant or its part.

Question 11.
Explain Anti-auxins.
Answer:
Anti-auxin compounds when applied to the plant inhibit the effect of auxin. Example: 2, 4, 5-Tri Iodine Benzoic Acid (TIBA) and Napthylpthalamine.

Question 12.
Apical dominance – Explain.
Answer:
Suppression of growth in the lateral bud by apical bud due to auxin produced by apical bud is termed as apical dominance.

Question 13.
Mention the precursors of (a) Auxin, (b) Gibberellin, (c) Cytokinin and (d) Abscisic acid.
Answer:

Question 14.
How Gibberellins are transported from their site of production?
Answer:
The transport of gibberellins in plants is non-polar. Gibberellins are translocated through the phloem and also occur in the xylem due to lateral movement between vascular bundles.

Question 15.
What do you mean by the term – Basipetal transport and Acropetal transport?
Answer:
Basipetal means transport through the phloem from shoot to root and acropetal means transport through the xylem from root to shoot.

Question 16.
Define Bolting.
Answer:
Rosette plants (genetic dwarfism) plants exhibit excessive intermodal growth when they are treated with gibberellins. This sudden elongation of a stem followed by flowering is called bolting.

Question 17.
Which plants are affected by Bakanae’s disease? Name the causative organism.
Answer:
Bakanae’s disease affects the rice plants. It is caused by a fungus called Gibberella fujikuroi.

Question 18.
Biennials usually flower at the second year of growth. Is it possible to make them flower in the first year itself? How?
Answer:
Many biennials usually flower during the second year of their growth. For flowering to take place, these plants should be exposed to the cold season. Such plants could be made to flower without exposure to the cold season in the first year itself when they are treated with gibberellins.

Question 19.
Where cytokinin is synthesized in plants?
Answer:
Cytokinin is formed in root apex, shoot apex, buds, and young fruits.

Question 20.
Give an account on neem Coty Ledon assay.
Answer:
Bioassay (Neem Cotyledon Assay): Neem cotyledons are measured and placed in cytokinin solution as well as in ordinary water. Enlargement of cotyledons is an indication of cytokinin activity.

Question 21.
State Richmond Lang effect.
Answer:
Application of cytokinin delays the process of aging by nutrient mobilization. It is known as the Richmond Lang effect.

Question 22.
Write a note on Bioassay of Ethylene.
Answer:
Ethylene can be measured by gas chromatography. This technique helps in the detection of the exact amount of ethylene from different plant tissues like lemon and orange.

Question 23.
What are climacteric fruits? Give example.
Answer:
In most of the plants, there is a sharp rise in respiration rate near the end of the development of fruit, called climacteric rise. Such fruits are called climacteric fruits. Example: Tomato, Apples, Banana, Mango.

Question 24.
ABA is called the stress hormone – Justify.
Answer:
It inhibits the shoot growth and promotes the growth of the root system. This character protects the plants from water stress. Hence, ABA is called the stress hormone.

Question 25.
Define photoperiodism. Name the person who coined this term.
Answer:
The physiological change on flowering due to relative length of light and darkness (photoperiod) is called Photoperiodism. The term photoperiodism was coined by Gamer and Allard.

Question 26.
Define the term photo neutrals.
Answer:
There are a number of plants that can flower in all possible photoperiods. They are also called photo neutrals or indeterminate plants. Example: Potato, Rhododendron, Tomato, and Cotton.

Question 27.
Point out the importance of photoperiodism.
Answer:
Importance of photoperiodism

1. The knowledge of photoperiodism plays an important role in hybridization experiments.
2. Photoperiodism is an excellent example of physiological pre-conditioning that is using an external factor to induce physiological changes in the plant.

Question 28.
Vernalisation – Define.
Answer:
Many species of biennials and perennials are induced to flower by low-temperature exposure ‘ (0°C to 5°C). This process is called Vernalization.

Question 29.
How vernalisation is carried out?
Answer:
The seeds are first soaked in water and allowed to germinate at 10°C to 12°C. Then seeds are transferred to low temperature (3°C to 5°C) from few days to 30 days. Germinated seeds after this treatment are allowed to dry and then sown. The plants will show quick flowering when compared to untreated control plants.

Question 30.
Define Seed germination and state its types.
Answer:
The activation and growth of an embryo from seed into seedling during favorable conditions is called seed germination. There are two methods of seed germination. Epigeal and hypogeal.

Question 31.
What are photoelastic seeds?
Answer:
There are many seeds that respond to light for germination and these seeds said to be photoelastic.

Question 32.
How can we overcome dormancy in photoelastic seeds?
Answer:
The dormancy of photoelastic seeds can be broken by exposing them to red light.

Question 33.
Senescence-Define.
Answer:
Old age is called senescence in plants. Senescence refers to all collective, progressive and deteriorative processes which ultimately lead to complete loss of organization and function.

Question 34.
What field do phytogerontology deal with?
Answer:
The branch of botany that deals with aging, abscission, and senescence is called Phytogerontology.

Question 35.
Mention the types of senescence.
Answer:
Leopold (1961) has recognized four types of senescence:

1. Overall senescence
2. Top senescence
3. Deciduous senescence
4. Progressive senescence

Question 36.
Where the abscission zone is formed?
Answer:
Leaf abscission takes place at the base of the petiole which is marked internally by a distinct zone of few layers of thin-walled cells arranged transversely. This zone is called the abscission zone or abscission layer. An abscission layer is greenish-grey in color and is formed by rows of cells of 2 to 15 cells thick.

Question 37.
Expand PCD and define it.
Answer:
Death of the plant or its parts consequent to senescence is called Programmed Cell Death (PCD).

Short answer questions

Question 1.
Draw a graph of the sigmoid curve.
Answer:

Question 2.
Correlate the terms light & etiolation in the plant.
Answer:
Light has its own contribution to the growth of the plant. Light is important for growth and photosynthesis. Light stimulates healthy growth. The absence of light may lead to yellowish in color. This is called etiolation.

Question 3.
List out the internal factors that affect growth.
Answer:
a. Genes are intracellular factors for growth.
b. Phytohormones are intracellular factors for growth. Example: auxin, gibberellin, cytokinin.
c. C/N ratio.

Question 4.
Give an account on Dedifferentiation.
Answer:
The living differentiated cells which had lost capacity to divide, regain the capacity to divide under certain conditions. Hence, dedifferentiation is the regaining of the ability of cell division by the differentiated cells. Example: Interfascicular cambium and Vascular cambium.

Question 5.
Explain synergistic & Antagonistic effects of plant hormones.
Answer:
Synergistic and Antagonistic effects

1. Synergistic effects: The effect of one or more substances in such a way that both promote each other’s activity. Example: Activity of auxin and gibberellins or cytokinins.
2. Antagonistic effects: The effect of two substances in such a way that they have opposite effects on the same process. One accelerates and the other inhibits. Example: ABA and gibberellins during seed or bud dormancy. ABA induces dormancy and gibberellins break it.

Question 6.
Name any three synthetic auxins.
Answer:
1. 2,4-Dichloro Phenoxy Acetic Acid (2,4-D)
2. 2,4,5-Trichloro Phenoxy Acetic Acid (2,4,5-T)
3. Naphthalene Acetic Acid (NAA).

Question 7.
Describe the procedure of the Avena curvature test.
Answer:
When the Avena seedlings have attained a height of 15 to 30 mm, about 1mm of the coleoptile tip is removed. This apical part is the source of natural auxin. The tip is now placed on agar blocks for few hours. During this period, the auxin diffuses out of these tips into the agar. The auxin containing agar block is now placed on one side of the decapitated stump of Avena coleoptile. The auxin from the agar blocks diffuses down through coleoptile along the side to which the auxin agar block is placed. An agar block without auxin is placed on another decapitated coleoptile. Within an hour, the coleoptiles with auxin agar block bend on the opposite side where the agar block is placed. This curvature can be measured.

Question 8.
How auxin is useful in tissue culture technique?
Answer:
Auxin is responsible for the initiation and promotion of cell division in the cambium, which is responsible for secondary growth and tumor. This property of induction of cell division has been exploited for tissue culture techniques and for the formation of callus.

Question 9.
Mention the role of ethylene in the agriculture field.
Answer:

1. Ethylene normally reduces flowering in plants except in Pineapple and Mango.
2. It increases the number of female flowers and decreases the number of male flowers.
3. Ethylene spray in the cucumber crops produces female flowers and increases the yield.

Question 10.
Give an answer for the following with regard to abscisic acid.
(a) Chemical structure (b) Precursor (c) Bioassay
Answer:
(a) Chemical structure – carotenoid structure
(b) Precursor – Mevalonic acid pathway or Xanthophyll
(c) Bioassay – Rice coleoptile.

Question 11.
How ABA involves stomatal closure.
Answer:
ABA helps in reducing the transpiration rate by closing stomata. It inhibits K+ uptake by guard cells and promotes the leakage of malic acid. It results in the closure of stomata.

Question 12.
Enumerate the practical application of vernalization.
Answer:

1. Vernalization shortens the vegetative period and induces the plant to flower earlier.
2. It increases the cold resistance of the plants.
3. It increases the resistance of plants to fungal disease.
4. Plant breeding can be accelerated.

Question 13.
Write a note on Epigeal & hypogeal germination.
Answer:

1. Epigeal Germination: During epigeal germination, cotyledons are pushed out of the soil. This happens due to the elongation of the hypocotyl. Example: Castor and Bean.
2. Hypogeal Germination: During hypogeal germination cotyledons remain below the soil due to rapid elongation of epicotyls. Example: Maize.

Question 14.
Discuss seed viability.
Answer:
Viability: Usually seeds remain viable or living only for a particular period. Viability of seeds ranges from a few days (Example: Oxalis) to more than a hundred years. Maximum viability (1000 years) has been recorded in lotus seeds. Seeds germinate only within the period of viability.

Question 15.
Comment on seed dormancy and its reason.
Answer:
The condition of a seed when it fails to germinate even in suitable environmental conditions is called seed dormancy. There are two main reasons for the development of dormancy: Imposed dormancy and innate dormancy. Imposed dormancy is due to low moisture and low temperature. Innate dormancy is related to the properties of the seed itself.

Question 16.
Which is the final stage of senescence? Define it.
Answer:
The final stage of senescence is abscission. Abscission is a physiological process of shedding of organs like leaves, flowers, fruits, and seeds from the parent plant body.

Question 17.
Mention the role of hormones in Abscission.
Answer:
All naturally occurring hormones influence the process of abscission. Auxins and cytokinins retard abscission, while abscisic acid (ABA) and ethylene induce it.

Question 18.
Why Abscission has to take place?
Answer:
Significance of abscission

1. Abscission separates dead parts of the plant, like old leaves and ripe fruits.
2. It helps in the dispersal of fruits and continuing the life cycle of the plant.
3. Abscission of leaves in deciduous plants helps in water conservation during summer.
4. In lower plants, shedding of vegetative parts like gemmae or plantlets helps in vegetative reproduction.

Long answer questions

Question 1.
Enumerate the characteristics of growth.
Answer:

1. Growth increases in protoplasm at the cellular level.
2. Stem and roots are indeterminate in growth due to continuous cell division and are called an open form of growth.
3. The primary growth of the plant is due to the activity of apical meristem where new cells are added to the root and shoot apex causing linear growth of the plant body.
4. The secondary vascular cambium and cork cambium add new cells to cause an increase in girth.
5. Leaves, flowers, and fruits are limited in growth or of determinate or closed-form growth.

Question 2.
Explain the three phases of growth.
Answer:
Phases of growth
There are three phases of growth, 1. Formative phase, 2. Elongation phase and 3. Maturation phase

1. Formative phase: Growth in this phase occurs in meristematic cells of the shoot and root tips. These cells are small in size, have dense protoplasm, large nucleus, and small vacuoles. Cells divide continuously by mitotic cell division. Some cells retain the capability of cell division while other cells enter the next phase of growth.
2. Elongation Phase: Newly formed daughter cells are pushed out of the meristematic zone
   and increase the volume. It requires auxin and food supply, deposition of new cell wall materials (intussusception), the addition of protoplasm, and development of central vacuole take place.
3. Maturation Phase: During this stage cells attain mature form and size. Thickening and differentiation take place. After differentiation, the cells do not grow further.

Question 3.
Give a detailed account of the geometric growth rate.
Answer:
This growth occurs in many higher plants and plant organs and is measured in size or weight. In-plant growth, geometric cell division results if all cells of an organism or tissue are active mitotically. Example: Round three in the given figure 15.5, produces 8 cells as 23 5 8 and after round 20 there are 220 5 1,048,576 cells. The large, plant, or animal parts are produced this way. In fact, it is common in animals but rare in plants except when they are young and small.

An exponential growth curve can be expressed as,
W₁ = W₀e^rt
W₁ = Final size (weight, height and number)
W₀ = Initial size at the beginning of the period
r = Growth rate
t = Time of growth
e = Base of the natural logarithms
Here r is the relative growth rate and also a measure of the ability of the plant to produce new plant material, referred to as efficiency index. Hence, the final size of W₁ depends on the initial size W₀.

Question 4.
List out the characteristics of phytohormones.
Answer:
Characteristics of phytohormones

1. Usually produced in tips of roots, stems, and leaves.
2. The transfer of hormones from one place to another takes part through conductive systems.
3. They are required in trace quantities.
4. All hormones are organic in nature.
5. There are no specialized cells or organs for their secretion.
6. They are capable of influencing physiological activities leading to promotion, inhibition, and modification of growth.

Question 5.
List out the physiological effects of Auxin.
Answer:
Physiological Effects

1. They promote cell elongation in stem and coleoptile.
2. At higher concentrations, auxins inhibit the elongation of roots but induce more lateral roots. Promotes growth of root-only at extremely low concentrations.
3. Suppression of growth in the lateral bud by apical bud due to auxin produced by apical bud is termed as apical dominance.
4. Auxin prevents abscission.
5. It is responsible for the initiation and promotion of cell division in the cambium, which is responsible for secondary growth and tumor. This property of induction of cell division has been exploited for tissue culture techniques and for the formation of callus.
6. Auxin stimulates respiration.
7. Auxin induces vascular differentiation.

Question 6.
Mention the role of Auxin in Agri-field.
Answer:

1. It is used to eradicate weeds. Example: 2,4-D and 2,4,5-T.
2. Synthetic auxins are used in the formation of seedless fruits (Parthenocarpic fruit).
3. It is used to break the dormancy in seeds.
4. Induce flowering in Pineapple by NAA & 2,4-D.
5. Increase the number of female flowers and fruits in cucurbits.

Question 7.
Mention the role of Gibberellin in agriculture.
Answer:

1. The formation of seedless fruits without fertilization is induced by gibberellins. Example: Seedless tomato, apple, and cucumber.
2. It promotes the formation of male flowers in Cucurbitaceae. It helps in crop improvement.
3. Uniform bolting and increased uniform seed production.
4. Improves the number and size of fruits in grapes. It increases yield.
5. Promotes elongation of inter-node in sugarcane without decreasing sugar content.
6. Promotion of flowering in long-day plants even under short-day conditions.
7. It stimulates seed germination.

Question 8.
Explain the physiological effect of Ethylene.
Answer:

• Ethylene stimulates respiration and ripening in fruits.
• It stimulates radial growth in stem and root and inhibits linear growth.
• It breaks the dormancy of buds, seeds, and storage organiser.
• It stimulates the formation of the abscission zone in leaves, flowers, and fruits. This makes the leaves shed prematurely.
• Inhibition of stem elongation (shortening the internode).
• In low concentrations, ethylene helps in root initiation.
• Growth of lateral roots and root hairs. This increases the absorption surface of the plant roots.
• The growth of fruits is stimulated by ethylene in some plants. It is more marked in climacteric fruits.
• Ethylene causes epinasty.

Question 9.
List out the physiological effects of Abscisic acid.
Answer:

• It helps in reducing the transpiration rate by closing stomata. It inhibits K+ uptake by guard cells and promotes the leakage of malic acid. It results in the closure of stomata.
• It spoils chlorophylls, proteins, and nucleic acids of leaves making them yellow.
• Inhibition of cell division and cell elongation.
• ABA is a powerful growth inhibitor. It causes 50% inhibition of growth in Oat coleoptile.
• It induces bud and seed dormancy.
• It promotes the abscission of leaves, flowers, and fruits by forming abscission layers.
• ABA plays an important role in plants during water stress and during drought conditions. It results in loss of turgor and closure of stomata.
• It has anti-auxin and anti-gibberellin properties.
• Abscisic acid promotes senescence in leaves by causing loss of chlorophyll pigment decreasing the rate of photosynthesis and changing the rate of proteins and nucleic acid synthesis.

Question 10.
Classify and explain the plants based on photoperiodism.
Answer:

1. Long day plants: The plants that require long critical day length for flowering are called long-day plants or short night plants. Example: Pea, Barley, and Oats.
2. Short long-day plants: These are long-day plants but should be exposed to short-day lengths during the early period of growth for flowering. Example: Wheat and Rye.
3. Short-day plants: The plants that require a short critical day length for flowering are called short-day plants or long night plants. Example: Tobacco, Cocklebur, Soybean, Rice, and Chrysanthemum.
4. Long short day plants: These are actually short-day plants but they have to be exposed to long days during their early periods of growth for flowering. Example: Some species of Bryophyllum and Night jasmine.
5. Intermediate day plants: These require a photoperiod between long days and short days for flowering. Example: Sugarcane and Coleus.
6. Day-neutral plants: There are a number of plants that can flower in all possible photoperiods. They are also called photo neutrals or indeterminate plants. Example: Potato, Rhododendron, Tomato, and cotton.

Question 11.
Describe the concept of Phytochrome.
Answer:
Phytochrome is a bluish biliprotein pigment responsible for the perception of light in a photo physiological process. Butler et al, (1959) named this pigment and it exists in two interconvertible forms: (i) red light-absorbing pigment which is designated as P_r and (ii) far-red light-absorbing pigment which is designated as P_fr. The P_r form absorbs red light in 660 nm and changes to P_fr. The P_fr form absorbs far-red light in 730 nm and changes to P_r. The P form is biologically inactive and it is stable whereas P_fr form is biologically active and it is very unstable. In short-day plants, P_r promotes flowering and P_fr inhibits the flowering whereas in long-day plants flowering is promoted by P_fr and inhibited by P_r form. P_fr is always associated with a hydrophobic area of membrane systems while P_r is found in the diffused state in the cytoplasm. The interconversion of the two forms of phytochrome is mainly involved in flower induction and also additionally plays a role in seed germination and changes in membrane conformation.

Question 12.
Explain the theories involved in the mechanism of vernalization.
Answer:
Mechanism of Vernalization:
Two main theories to explain the mechanism of vernalization is:
i. Hypothesis of phasic development,
ii. The hypothesis of hormonal involvement.

1. The hypothesis of phasic development: According to Lysenko, the development of an annual seed plant consists of two phases. The first phase is a thermostat, which is a vegetative phase requiring low temperature and suitable moisture. The next phase is the photo stage which requires high temperature for the synthesis of florigen (flowering hormone).
2. The hypothesis of hormonal involvement: According to Purvis (1961), the formation of a substance A from its precursor, is converted into B after chilling. The substance B is unstable. At a suitable temperature, B is converted into stable compound D called Vemalin. Vemalin is converted to F (Florigen). Florigen induces flower formation. At high-temperature B is converted to C and devemalization occurs.

Question 13.
Discuss the external factors that affect seed germination.
Answer:
External Factors:
a. Water: It activates the enzymes which digest the complex reserve foods of the seed. If the water content of the seed goes below a critical level, seeds fail to germinate.
b. Temperature: Seeds fails to germinate at very low and high temperature. The optimum temperature is 25°C to 35°C for most tropic species.
c. Oxygen: It is necessary for germination. Since aerobic respiration is a physiological requirement for germination most will germinate well in air containing 20% oxygen.
d. Light: There are many seeds that respond to light for germination and these seeds said to be photoelastic.
e. Soil Conditions: Germination of seed in its natural habit is influenced by soil conditions such as holding capacity, mineral composition, and aeration of the soil.

Question 14.
Illustrate the various methods of breaking seed dormancy.
Answer:
Methods of breaking dormancy:
The dormancy of seeds can be broken by different methods. These are:

1. Scarification: Mechanical and chemical, treatments like cutting or chipping of hard tough
   seed coat and use of organic solvents to remove waxy or fatty compounds are called as Scarification. •
2. Impaction: In some seeds, water and oxygen are unable to penetrate micropyle due to blockage by cork cells. These seeds are shaken vigorously to remove the plug which is called Impaction.
3. Stratification: Seeds of rosaceous plants (Apple, Plum, Peach, and Cherry) will not germinate until they have been exposed to well-aerated, moist conditions under low temperatures (0°C to 10°C) for weeks to months. Such treatment is called Stratification.
4. Alternating temperatures: Germination of some seeds is strongly promoted by alternating
   daily temperatures. An alternation of low and high temperature improves the germination of seeds. .
5. Light: The dormancy of photoblastic seeds can be broken by exposing them to red light.

Question 15.
Give a detailed account of different types of senescence.
Answer:

1. Overall senescence: This kind of senescence occurs in annual plants when the entire plant gets affected and dies. Example: Wheat and Soybean. It also occurs in few perennials also. Example: Agave and Bamboo.
2. Top senescence: It occurs in aerial parts of plants. It is common in perennials, underground and root system remains viable. Example: Banana and Gladiolus.
3. Deciduous senescence: It is common in deciduous plants and occurs only in the leaves of plants, the bulk of the stem and root system remains alive. Example: Elm and Maple.
4. Progressive senescence: This kind of senescence is gradual. First, it occurs in old leaves followed by new leaves then stems, and finally root system. It is common in annuals.

Question 16.
Explain the physiology of senescence.
Answer:
Physiology of Senescence

• Cells undergo changes in structure.
• Vacuole of the cell acts as a lysosome and secretes hydrolytic enzymes.
• The starch content is decreased in the cells.
• Photosynthesis is reduced due to loss of chlorophyll accompanied by synthesis and accumulation of anthocyanin pigments, therefore the leaf becomes red.
• There is a marked decrease in protein content in the senescing organ.
• RNA content of the leaf particularly rRNA level is decreased in the cells due to increased activity of the enzyme RNAase.
• DNA molecules in senescencing leaves degenerate by the increased activity of enzyme DNAase.

Question 17.
What are the physiological changes occurring in plants during Abscission?
Answer:
Morphological and Anatomical changes during abscission:
Leaf abscission takes place at the base of the petiole which is marked internally by a distinct zone of few layers of thin-walled cells arranged transversely. This zone is called the abscission zone or abscission layer. An abscission layer is greenish-grey in color and is formed by rows of cells of 2 to 15 cells thick. The cells of the abscission layer separate due to the dissolution of the middle lamella and primary wall of cells by the activity of enzymes pectinase and cellulase resulting in the loosening of cells. Tyloses are also formed blocking the conducting vessels. Degrading of chlorophyll occurs leading to the change in the color of leaves, leaf detachment from the plant, and leaf fall. After abscission, the outer layer of cells becomes suberized by the development of periderm.

Higher Order Thinking Skills (HOTs)

Question 1.
A farmer grows cucumber plants in his field. He wants to increase the number of female flowers in them. Which plant growth hormone can be applied to achieve this.
Answer:
Auxin can be used to increase the number of female flowers.

Question 2.
In the figure of the sigmoid and curve given below, label the segments A, B, and C.

Answer:
A = lag phase, B = log phase and C = Steady state phase.

Question 3.
In most plants, the terminal bud suppresses the development of lateral buds. What is this phenomenon called? Name the phytohormone that promotes this phenomenon.
Answer:
Suppression in the development of lateral buds due to terminal bud is called apical dominance. Auxins are the phytohormones that can promote this phenomenon.

Question 4.
The physiological effects of ethylene are both positive and negative. Comment.
Answer:

Question 5.
Light plays an important role in the life of all organisms. Name any two physiological processes in plants that are affected by light.
Answer:
In plants, light plays a vital role in photosynthesis and growth.

Question 6.
Classify the following plants into long-day plants, short-day plants, and day-neutral plants, Wheat, sunflower, maize, tobacco, oats, chrysanthemum.
Answer:
Long Day plants – Wheat, Oats.
Short Day plants – Tobacco, chrysanthemum.
Day Neutral plants – Sunflower, maize.

Choose the correct answer.

1. Identify the monocarpic perennial plant from the following list.
(a) Paddy
(b) Bean
(c) Bamboo
(d) Coconut
Ans.
(c) Bamboo

2. Growth rate is maximum in ……….. phase.
(a) Lag
(b) Log
(c) Decelerating
(d) Maturation
Ans.
(b) Log

3. Proper plant growth occurs at a temperature between …………
(a) 25°C to 28°C
(b) 28°C to 45°C
(c) 25°C to 35°C
(d) 28°C to 30°C
Ans. (d) 28°C to 30°C

4. The process by which differentiated cell regain the ability of cell division ………….
(a) Differentiation
(b) Dedifferentiation
(c) Redifferentiation
(d) Totipotency
Ans.
(b) Dedifferentiation

5. The term Auxin was first used by ………….
(a) Kogl Smith
(b) Darwin
(c) F.W. Went
(d) Kurosawa
Ans.
(c) F.W. Went

6. Auxin was first isolated from ……….
(a) Com grain oil
(b) Human blood
(c) Human urine
(d) Rice bran oil
Ans.
(c) Human urine

7. Which is NOT a natural auxin?
(a) IAA
(b) PAA
(c) IPA
(d) NAA
Ans.
(d) NAA

8. The amino which is a precursor of IAA is ……….
(a) Methionine
(b) Valine
(c) Isoleuine
(d) Tryptophan
Ans.
(d) Tryptophan

9. Identify the wrong statement regarding the physiological effects of Auxin.
(i) Auxin prevents abscission
(ii) Auxin inhibits respiration
(iii) Auxin promotes cell elongation
(iv) Auxin breaks seed dormancy
(a) (i) only
(b) (ii) only
(c) (iii) only
(d) (i), (ii) and (iv)
Ans.
(b) (ii) only

10. How many number of giberellins were discovered so far?
(a) 50
(6) 70
(c) 100
(d) 80
Ans.
(c) 100

11. Foolish seedling disease affects ………….
(a) Maize
(b) Rice
(c) Sorghum
(d) Wheat
Ans.
(b) Rice

12. Bakanae’s disease was first noticed by …………
(a) F.W. Went
(b) Kurosawa
(c) Cocken
(d) Denny
Ans.
(b) Kurosawa

13. Apical dominance is due to the effect of ………….
(a) Auxin
(b) Gibberellin
(c) Ethylene
(d) Cytokinin
Ans.
(a) Auxin

14. Match the following:
(a) a- iii, b – iv, c – ii, d – i
(b) a – i, b- iii, c – iv, d – ii
(c) a – iii, b – ii, c – iv, d – i
(d) a – ii, b – iii, c – iv, d – i
Ans.
(a) a- iii, b – iv, c – ii, d – i

15. The most widely occuring cytokinin in plants is ………..
(a) Iso propyl adenine
(b) Iso pentenyl adenine
(c) Indole propionic acid
(d) Iso propionic adenine
Ans.
(b) Iso pentenyl adenine

16. Which of the following does not act as a precursor of ethylene?
(a) Fumaric acid
(b) Malic acid
(c) Linolenic acid
(d) Methionine
Ans.
(b) Malic acid

17. Identify the non-climacteric fruit.
(a) Tomato
(b) Grapes
(c) Apples
(d) Mango
Ans.
(b) Grapes

18. ……….. is a stress hormone.
(a) Ethylene
(b) Cytokinin
(c) Auxin
(d) Abscissic acid
Ans.
(d) Abscissic acid

19. Which of the following plant hormone functions against auxin?
(a) Gibberellin
(b) Cytokinin
(c) Ethylene
(d) Abscissic acid
Ans.
(d) Abscissic acid

20. Closure of stomato can be induced by ………..
(a) Abscissic acid
(b) Ethylene
(c) NAA
(d) Cytokinin
Ans.
(a) Abscissic acid

21. Maryland mammoth requires …………. hours of light.
(a) 8
(b) 10
(c) 12
(d) 15.05
Ans.
(c) 12

22. Identify the day neutral plants.
(a) Pea
(b) Wheat
(c) Tomato
(d) Soyabean
Ans.
(c) Tomato

23. The Pr form absorbs red light in ………… nm.
(a) 730
(b) 620
(c) 660
(d) 635
Ans.
(c) 660

24. The term vernalisation was first used by …………
(a) Chailakyan
(b) Gamer & Allard
(c) Lysenko
(d) F.W. Went
Ans.
(c) Lysenko

25. …………. seeds show a maximum viability of 1000 years.
(a) Orange
(b) Oxalis
(c) Lotus
(d) Coconut
Ans.
(c) Lotus

26. Shaking the seeds vigorously to remove the plug in the microphyle is the method of …………..
(a) Stratification
(b) Impaction
(c) Scarification
(d) Emaciation
Ans.
(b) Impaction

27. After abicission, outer layer of cells becomes ………….. by periderm.
(a) Lignified
(b) Pectinised
(c) Suberized
(d) Stratified
Ans.
(c) Suberized

Students get through the TN Board 11th Bio Botany Important Questions Chapter 14 Respiration which is useful for their exam preparation.

TN State Board 11th Bio Botany Important Questions Chapter 14 Respiration

Very short answer questions

Question 1.
Define the term Respiration.
Answer:
During the night, plants take up oxygen and release carbon dioxide and as a result, carbon dioxide will be abundant around the tree. This process of CO₂ evolution is called respiration.

Question 2.
What do you mean by Respiratory substrate? Give example.
Answer:
Respiration is a biological process in which oxidation of various food substances like carbohydrates, proteins, and fats take place and as a result of this, energy is produced where O₂ is taken in and CO₂ is liberated. The organic substances which are oxidised during respiration are called respiratory substrates.

Question 3.
Write the overall equation of respiration.
Answer:
C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + Energy
(686 K cal or 2868 KJ)

Question 4.
How Blackman classified respiration? Name them.
Answer:
Depending upon the nature of respiratory substrate, Blackman divided respiration into,

1. Floating respiration and
2. Protoplasmic respiration.

Question 5.
Compensation point – Define.
Answer:
At dawn and dusk, the intensity of light is low. The point at which CO₂ released in respiration is exactly compensated by CO₂ fixed in photosynthesis that means no net gaseous exchange takes place, it is called the compensation point. At this moment, the amount of oxygen released from photosynthesis is equal to the amount of oxygen utilized in respiration. The two common factors associated with the compensation point are CO₂ and light (Figure 14.2). Based on this there are two types of compensation points. They are CO₂ compensation point and light compensation point. C₃ plants have compensation points ranging from 40-60 ppm (parts per million) CO₂ while those of C₄ plants range from 1-5 ppm CO₂.

Question 6.
Name the types of compensation points.
Answer:
(a) CO₂ compensation point, (b) Light compensation point.

Question 7.
ATP is the universal energy currency of the cell – Justify.
Answer:
Respiration is responsible for the generation of ATP. The discovery of ATP was made by Karl Lohman (1929). ATP is a nucleotide consisting of a base-adenine, a pentose sugar-ribose, and three phosphate groups. Out of three phosphate groups, the last two are attached by high-energy-rich bonds (Figure 14.3). On hydrolysis, it releases energy (7.3 Kcal or 30.6 KJ/ATP) and it is found in all living cells and hence it is called the universal energy currency of the cell. ATP is an instant source of energy within the cell. The energy contained in ATP is used in Lipman (1941).

Question 8.
Who discovered ATP? Name any two other high-energy compounds in the cell similar to ATP?
Answer:
ATP was discovered by Karl Lohman in 1929. GTP (Guanosine Triphosphate), UTP (Uridine Triphosphate) are some other energy-rich compounds in the cell.

Question 9.
Expand the term (a) ATP, (b) ADP.
Answer:
AMP – Adenosine Monophosphate, ADP – Adenosine Diphosphate.

Question 10.
Compare Aerobic respiration with anaerobic respiration.
Answer:

Question 11.
Mention the stages of aerobic respiration.
Answer:

1. Glycolysis-conversion of glucose into pyruvic acid in the cytoplasm of the cell.
2. Link reaction-conversion of pyruvic acid into acetyl coenzyme-A in the mitochondrial matrix.
3. Krebs cycle-conversion of acetyl coenzyme A into carbon dioxide and water in the mitochondrial matrix.

Question 12.
Where does glycolysis occur? Mention the scientists who described it?
Answer:
Glycolysis occurs in the cytoplasm of the cell. Glycolysis was described by Gustav Embden, Otto Meyerhoff, and J. Pamas.

Question 13.
Mention the two phases of Glycolysis.
Answer:
(a) Preparatory phase or hexose phase, (b) Pay off phase or triose phase.

Question 14.
Define Glycolysis.
Answer:
Glycolysis: (Gr: Glykos 5 Glucose, Lysis 5 Splitting) Glycolysis is a linear series of reactions in which 6-carbon glucose is split into two molecules of 3-carbon pyruvic acid. The enzymes which are required for glycolysis are present in the cytoplasm (Figure 14.6). The reactions of glycolysis were worked out in yeast cells by three scientists Gustav Embden (German), Otto Meyerhoff (German), and J Parnas (Polish) and so it is also called as EMP pathway.

Question 15.
What is Substrate/Trans-phosphorylation?
Answer:
The direct transfer of phosphate moiety from substrate molecule to ADP and is converted into ATP is called substrate phosphorylation or direct phosphorylation or trans-phosphorylation.

Question 16.
Define the term Enolation.
Answer:
In glycolysis, at step nine, 2-phosphoglycerate is dehydrated into phosphoenol pyruvate by the enzyme enolase. As a result, an enol group is formed within the molecule. This process is called enolation.

Question 17.
List the products of Glycolysis.
Answer:
(a) Pyruvic acid (2 molecules), (b) 2 ATP molecules and (c) 2 NADH, molecules.

Question 18.
Who discovered the TCA cycle? Where does it occur?
Answer:
The tricarboxylic Acid (TCA) cycle was discovered by Sir Hans Adolf Krebs in 1937. TCA cycle occurs in the mitochondrial matrix.

Question 19.
Draw the structure of oxysomes.
Answer:

Question 20.
Why TCA cycle is called so?
Answer:
TCA cycle starts with the condensation of acetyl CoA with oxaloacetate in the presence of water to yield citrate or citric acid. Therefore, it is also known as the citric acid Cycle (CAC) or Tri Carboxylic Acid (TCA) cycle.

Question 21.
Krebs cycle is an amphibolic pathway – Comment.
Answer:
The Krebs cycle is primarily a catabolic pathway, but it provides precursors for various biosynthetic pathways thereby an anabolic pathway too. Hence, it is called the amphibolic pathway. It serves as a pathway for the oxidation of carbohydrates, fats, and proteins.

Question 22.
Mention the four multi-protein complexes in ETC.
Answer:
(a) Complex – I – NADH dehydrogenase
(b) Complex – II – Succinic dehydrogenase
(c) Complex – III – Cytochrome be, complex
(d) Complex – IV – Cytochrome c oxidase

Question 23.
Write a brief note on Ubiquinone.
Answer:
Ubiquinone and cytochrome bc₁, the complex is structurally and functionally similar to plastoquinone and cytochrome b₆, f complex respectively in the photosynthetic electron transport chain.

Question 24.
Mitochondria are called powerhouses of the cell. Why?
Answer:
Complete oxidation of a glucose molecule in aerobic respiration results in the net gain of 36 ATP molecules in plants. Since a huge amount of energy is generated in mitochondria in the form of ATP molecules they are called the powerhouse of the cell. In the case of aerobic prokaryotes due to lack of mitochondria, each molecule of glucose produces 38 ATP molecules.

Question 25.
Name any two ETC inhibitors.
Answer:
2, 4 DNP – cyanide.

Question 26.
Define Respiratory Quotient.
Answer:
The ratio of the volume of carbon dioxide given out and the volume of oxygen taken in during respiration is called Respiratory Quotient or Respiratory ratio. RQ value depends upon respiratory substrates and their oxidation.

Question 27.
Calculate the respiratory quotient for the following equation.
C₆H₁₂O₆ → 2CO₂2 + 2C₂H₅OH + Energy
Answer:
Volume of O₂ used = O. Volume of CO₂ released = 2

Question 28.
What is the significance of the respiratory quotient?
Answer:
Significance of RQ

1. RQ value indicates which type of respiration occurs in living cells, either aerobic or anaerobic.
2. It also helps to know which type of respiratory substrate is involved.

Question 29.
Define fermentation and mention its types.
Answer:
Some organisms can respire in the absence of oxygen. This process is called fermentation or anaerobic respiration. There are three types of fermentation:

1. Alcoholic fermentation
2. Lactic acid fermentation
3. Mixed acid fermentation

Question 30.
Give an account on lactic acid fermentation.
Answer:
Some bacteria (Bacillus), fungi and muscles of vertebrates produce lactic acid from pyruvic acid:

Question 31.
Observe the given diagram given below and answer the questions.

(a) Name the apparatus set up.
(b) Mention the purpose of the apparatus.
Answer:
(a) The apparatus set up is Kuhne’s fermentation tube.
(b) Kuhne’s fermentation tube is used to demonstrate alcoholic fermentation.

Question 32.
Name the scientists who discovered HMP shunt. Also, expand the term HMP shunt.
Answer:
HMP shunt refers to Hexose Monophosphate shunt which was discovered by Warburg, Dickens, and Lipmann.

Short answer questions

Question 1.
Compare floating respiration with protoplasmic respiration.
Answer:

Question 2.
Draw the structure of an ATP molecule.
Answer:

Question 3.
Describe the structure of ATP.
Answer:
Respiration is responsible for the generation of ATP. The discovery of ATP was made by Karl Lohman (1929). ATP is a nucleotide consisting of a base-adenine, a pentose sugar-ribose, and three phosphate groups.
Out of three phosphate groups, the last two are attached by high-energy-rich bonds (Figure 14.3). On hydrolysis, it releases energy (7.3 Kcal or 30.6 KJ/ATP) and it is found in all living cells and hence it is called the universal energy currency of the cell. ATP is an instant source of energy within the cell. The energy contained in ATP is used in the synthesis of carbohydrates, proteins, and lipids. The energy transformation concept was established by Lipman (1941).

Question 4.
How Redox reaction occurs in NADH₂ and FADH₂.
Answer:
NAD⁺ + 2e⁺ + 2H⁺ → NADH + H⁺
FAD + 2e^– + 2H⁺ → FADH₂
When NAD⁺ (Nicotinamide Adenine Dinucleotide-oxidised form) and FAD (Flavin Adenine Dinucleotide) pick up electrons and one or two hydrogen ions (protons), they get reduced to NADH + H⁺ and FADH₂ respectively. When they drop electrons and hydrogen off they go back to their original form. The reaction in which NAD⁺ and FAD gain (reduction) or lose (oxidation) electrons is called redox reaction (Oxidation-reduction reaction). These reactions are important in cellular respiration.

Question 5.
Explain simply the stages of respiration.
Answer:
Stages of Respiration:

1. Glycolysis-conversion of glucose into pyruvic acid in the cytoplasm of the cell.
2. Link reaction-conversion of pyruvic acid into acetyl coenzyme-A in the mitochondrial matrix.
3. Krebs cycle-conversion of acetyl coenzyme A into carbon dioxide and water in the mitochondrial matrix.
4. Electron transport chain and oxidative phosphorylation remove hydrogen atoms from the products of glycolysis, link reaction, and Krebs cycle release water molecule with energy in the form of ATP in the mitochondrial inner membrane.

Question 6.
Write the overall equation for glycolysis.
Answer:

Question 7.
Give an account of pyruvic oxidation.
Answer:
Two molecules of pyruvate formed by glycolysis in the cytosol enter into the mitochondrial matrix. In aerobic respiration, this pyruvate with coenzyme A is oxidatively decarboxylated into acetyl CoA by pyruvate dehydrogenase complex. This reaction is irreversible and produces two molecules of NADH + H⁺ and 2CO₂. It is also called transition reaction or Link reaction. The reaction of pyruvate oxidation is

Question 8.
Draw and label the structure of mitochondria.
Answer:

Question 9.
How fats and protein enter the Krebs cycle?
Answer:
When fats are a respiratory substrate they are first broken down into glycerol and fatty acid. Glycerol is converted into DHAP and acetyl CoA. This acetyl CoA enters into the Krebs cycle. When proteins are the respiratory substrate they are degraded into amino acids by proteases. The amino acids after deamination enter into the Krebs cycle through pyruvic acid or acetyl CoA and it depends upon the structure.

Question 10.
Mention the types of electron transport chain inhibitors and their action.
Answer:
Electron transport chain inhibitors:

1. 2,4 DNP (Dinitrophenol) – It prevents the synthesis of ATP from ADP, as it directs electrons from Co Q to Q₂
2. Cyanide – It prevents the flow of electrons from Cytochrome a₃ to O₂
3. Rotenone – It prevents the flow of electrons from NADH + H⁺ / FADH₂ to Co Q
4. Oligomycin – It inhibits oxidative phosphorylation.

Question 11.
Give a brief account of Alcoholic fermentation.
Answer:
The cells of roots in waterlogged soil respire by alcoholic fermentation because of lack of oxygen by converting pyruvic acid into ethyl alcohol and CO₂. Many species of yeast (Saccharomyces) also respire anaerobically.
This process takes place in two steps:

Question 12.
List out the industrial uses of alcoholic fermentation.
Answer:

1. In bakeries, it is used for preparing bread, cakes, biscuits.
2. In beverage industries for preparing wine and alcoholic drinks.
3. In producing vinegar and in tanning, curing of leather.
4. Ethanol is used to make gasohol (a fuel that is used for cars in Brazil).

Question 13.
Enumerate the characteristics of anaerobic respiration.
Answer:

1. Anaerobic respiration is less efficient than aerobic respiration.
2. a Limited number of ATP molecules is generated per glucose molecule.
3. It is characterized by the production of CO₂ and is used for Carbon fixation in photosynthesis.

Question 14.
Give a comparative study on glycolysis and fermentation.
Answer:
Comparison between glycolysis and fermentation:

Question 15.
Point out the importance of the pentose phosphate pathway.
Answer:

1. HMP shunt is associated with the generation of two important products, NADPH and pentose sugars, which play a vital role in anabolic reactions.
2. Coenzyme NADPH generated is used for reductive biosynthesis and counter damaging the effects of oxygen free radicals
3. Ribose-5-phosphate and its derivatives are used in the synthesis of DNA, RNA, ATP, NAD⁺, FAD, and Coenzyme A.
4. Erythrose is used for the synthesis of anthocyanin, lignin, and other aromatic compounds.

Long answer questions

Question 1.
Explain the phases of glycolysis.
Answer:
1. Preparatory phase: Glucose enters the glycolysis from sucrose which is the end product of photosynthesis. Glucose is phosphorylated into glucose-6- phosphate by the enzyme hexokinase and subsequent reactions are carried out by different enzymes. At the end of this phase fructose-1, 6 – bisphosphate is cleaved into glyceraldehyde-3- phosphate and dihydroxyacetone phosphate by the enzyme aldolase. These two are isomers. Dihydroxyacetone phosphate is isomerized into glyceraldehyde-3- phosphate by the enzyme triosephosphate isomerase, now’ two molecules of glyceraldehyde 3 phosphate enter into pay off phase. During the preparatory phase, two ATP molecules are consumed in step-1 and step-3.
2. Pay-off phase: Two molecules of glyceraldehyde-3- phosphate oxidatively phosphorylated into two molecules of 1,3 – bisphosphate glycerate. During this reaction, 2NAD⁺ is reduced to 2NADH + H⁺ by glyceraldehyde- 3- phosphate dehydrogenase at step 6. Further reactions are carried out by different enzymes and at the end, two molecules of pyruvate are produced. In this phase, 2ATPs are produced at step 7 and 2 ATPs at step 10. The direct transfer of phosphate moiety from substrate molecule to ADP and is converted into ATP is called substrate phosphorylation or direct phosphorylation or transphosphorylation. During the reaction at step 9, 2phospho glycerate dehydrated into Phospho enol pyruvate a water molecule is removed by the enzyme enolase. As a result, the enol group is formed within the molecule. This process is called Enolation.

Question 2.
Draw a Flow chart depicting the steps of glycolysis.
Answer:

Question 3.
Point out the significance of Kreb’s Cycle.
Answer:
Significance of Krebs cycle:

1. TCA cycle is to provide energy in the form of ATP for metabolism in plants.
2. It provides carbon skeleton or raw material for various anabolic processes.
3. Many intermediates of the TCA cycle are further metabolized to produce amino acids, proteins, and nucleic acids.
4. Succinyl CoA is the raw material for the formation of chlorophylls, cytochrome, phytochrome, and other pyrrole substances.
5. α-ketoglutarate and oxaloacetate undergo reductive animation and produce amino acids.
6. It acts as a metabolic sink which plays a central role in intermediary metabolism.

Question 4.
Draw a Flow chart of Kreb’s cycle.
Answer:

Question 5.
Compare alcoholic fermentation with lactic acid fermentation.
Answer:
Comparison of alcoholic fermentation and lactic acid fermentation

Question 6.
Demonstrate alcoholic fermentation using Kuhne’s apparatus.
Answer:
Demonstration of alcoholic fermentation:
Take a Kuhne’s fermentation tube which consists of an upright glass tube with a side bulb. Pour 10% sugar solution mixed with baker’s yeast into the fermentation tube the side tube is filled plug the mouth with a lid. After some time, the glucose solution will be fermented. The solution will give out an alcoholic smell and the level of solution in the glass column will fall due to the accumulation of CO₂ gas. It is due to the presence of the zymase enzyme in yeast which converts the glucose solution into alcohol and CO₂. Now introduce a pellet of KOH into the tube, the KOH will absorb CO₂, and the level of solution will rise in the upright tube.

Question 7.
What are the factors that affect respiration?
Answer:
Factors Affecting Respiration Answer: Internal Factors

1. The amount of protoplasm and its state of activity influence the rate of respiration.
2. The concentration of respiratory substrate is proportional to the rate of respiration.

External Factors

1. Wounding of plant organs stimulates the rate of respiration in that region.
2. Some chemical substance acts as inhibitors. Example: Cyanides.
3. Rate of respiration decreases with decreasing amount of water. Proper hydration is essential for respiration.
4. Light is an indirect factor affecting the rate of respiration.
5. The optimum temperature for respiration is 30°C. At low temperatures and very high temperatures rate of respiration decreases.
6. When a sufficient amount of O₂ is available the rate of aerobic respiration will be optimum and anaerobic respiration is completely stopped. This is called the Extinction point.
7. A high concentration of CO₂ reduces the rate of respiration.
8. A plant or tissue transferred from water to salt solution will increase the rate of respiration. It is called salt respiration.

Higher Order Thinking Skills (HOTs)

Question 1.
Glycolysis and Kreb’s cycle both are energy-yielding pathways in aerobic respiration. How these two pathways differ among themselves?
Answer:

Question 2.
Anaerobic respiration is usually noticed in lower organisms. A human being is an advanced organism whether anaerobic respiration occurs in a man? If so when and where does it takes place.
Answer:
Yes, anaerobic respiration occurs man at certain cases. Anaerobic respiration (lactic acid fermentation) occurs in the muscle cells of man during heavy exercise, continuous workouts running, etc.

Question 3.
Complete the formula of the respiratory quotient by naming A and B. Also, mention what type of substrate have a respiratory question of 1 and > 1?
Respiratory Quotient = \(\frac{A}{B}\)
Answer:

Respiratory Quotient will be equal to unity if the substrate is a carbohydrate.
Respiratory Quotient will be more than unity if the substrate is an organic acid.

Question 4.
The Flow Chart given below depicts the preparatory phase of the glycolysis pathway. Complete it by filling the missing steps A, B, C and also indicate whether ATP is being used up or released at Step D.

Answer:
A = Glucose – 6 – Phosphate.
B = Fructose -1,6- Bisphosphate.
C = Dihydroxy Acetone Phosphate.
In step D, ATP is utilized to phosphorylate Fructose – 6 – phosphate to Fructose —1,6 — Bisphosphate.

Question 5.
The energy yield in terms of ATP is higher in aerobic respiration than during anaerobic respiration – Discuss.
Answer:
In aerobic respiration, oxidation of one molecule of glucose generates 36 ATP molecules. On the other hand, during anaerobic respiration, the glucose is incompletely oxidized producing only 2 ATP molecules. Thus it is clear that aerobic respiration is more ATP beneficial than anaerobic respectively.

Question 6.
Answer the following questions in concern with the TCA cycle.
(a) Where does it take place?
(b) When and by whom it was discovered?
(c) Name the first formed product of the cycle.
(d) How many ATP molecules are generated per cycle?
(e) TCA cycle is an amphibolic pathway. Say yes or no. Why?
Answer:
(a) TCA cycle takes place in mitochondrion.
(b) TCA cycle was discovered by Sir Hans Adolf Krebs in 1937.
(c) The first formed product of the TCA cycle is citric acid or citrate.
(d) Only one ATP molecule is generated per TCA cycle.
(e) Yes. TCA is an amphibolic pathway because it involves both anaerobic and catabolic reactions.

Question 7.
Mention the fate of pyruvic acid in a cell, under the given circumstances.

Answer:
A = Acetyl Coenzyme A
B = Ethyl alcohol.
C = Lactic acid.

Choose the correct answer.

1. Which of the following is NOT a respiratory substrate in floating respiration?
(a) Carbohydrate
(b) Fat
(c) Organic acids
(d) Protein
Answer:
(d) Protein

2. The term respiration was coined by
(a) Pepys
(b) Calvin-Benson
(c) Meyerhoff
(d) Karl Lohman
Answer:
(a) Pepys

3. Amount of energy released when an ATP molecule is hydrolysed is …………
(a) 2.7 kcal
(b) 7.3 kcal
(c) 2.8 kcal
(d) 6.5 kcal
Answer:
(b) 7.3 kcal

4. ………… is called as the universal energy currency of the cell.
(a) ATP
(b) GTP
(c) UTP
(d) AMP
Answer:
(a) ATP

5. Which step is irrelevant with respect to aerobic respiration?
(a) Glycolysis
(b) Pyruvate oxidate
(c) Fermentation
(d) TCA cycle
Answer:
(c) Fermentation

6. Anaerobic respiration occurs only in …………
(a) Mitochondria
(b) Golgi bodies
(c) Nucleus
(d) Cytoplasm
Answer:
(d) Cytoplasm

7. In glycolysis, Glucose is phosphorylated to glucose-phosphate by the enzyme ………….
(a) Aldolase
(b) Phosphofructo isomerase
(c) Hexokinase
(d) Exolase
Answer:
(c) Hexokinase

8. Glucose is a ………….. carbon compound.
(a) Six
(b) Five
(c) Three
(d) Four
Answer:
(a) Six

9. Which of the following step is common in both aerobic and anaerobic respiration?
(a) Pyruvate oxidation
(b) Glycolysis
(c) ETC
(d) TCA cycle
Answer:
(6) Glycolysis

10. Net gain of ATP’s at the end of glycolysis is ………….
(a) 2
(b) 4
(c) 6
(d) 0
Answer:
(a) 2

11. Which statement is NOT – correct in concern with glycolysis?
(i) Preparatory phase is also called the hexose phase.
(ii) Pay off phase is also called the hexose phase.
(iii) Two ATP’s are consumed in the preparatory phase.
(iv) Glycolysis is also called EMP pathway.
(a) Only (i)
(b) Only (ii)
(c) (iii) and (iv)
(d) Only (iii)
Answer:
(b) Only (ii)

12. Pyruvic oxidation occurs in ……….
(a) Cytoplasm
(b) Mitochondrial matrix
(c) Inner membrane of mitochondria
(d) Both cytoplasm and mitochondrial membrane
Answer:
(b) Mitochondrial matrix

13. Kreb’s cycle is ……… in nature.
(a) Anabolic
(b) Catabolic
(c) Amphoteric
(d) Amphibolic
Answer:
(d) Amphibolic

14. F1 particles are also referred as ……….
(a) Polysomes
(b) Oxysomes
(c) Mesosomes
(d) Liposomes
Answer:
(b) Oxysomes

15. On oxidation in mitochondrian one molecule of NADH₂ yield …………. ATP’s.
(a) 2
(b) 4
(c) 3
(d) 1
Answer:
(c) 3

16. Which cell organelles are referred as “powerhouses”?
(a) Golgi bodies
(b) Endoplasmic reticulum
(c) Nucleus
(d) Mitochondria
Answer:
(d) Mitochondria

17. Identify the electron transport chain inhibitors prevent e^– flow from cytochrome a₃ to O₂.
(a) 2, 4 DNP
(b) Cyanide
(c) Oligomycin
(d) Rotenone
Answer:
(b) Cyanide

18. Pick out the correct statements.

(a) Only (i)
(b) Only (b)
(c) Both (i) and (iii)
(d) Both (b) and (iv)
Answer:
(a) Only (i)

19. Respiratory quotient of glucose in presence of oxygen is …………
(a) Unity
(b) Infinity
(c) Less than unity
(d) Zero
Answer:
(a) Unity

20. Number of CO₂ molecules generated in Kreton cycle is
(a) 2
(b) 4
(c) 6
(d) 8
Answer:
(b) 4

21. Calculate the respiratory quotient from the following equation
C₄H₆O₅ + 3O₂ → 4CO₂ + 3H₂O + Energy
(a) Unity
(b) More than unity
(c) Less than unity
(d) Zero
Answer:
(b) More than unity

22. Identify the wrong statement regarding fermentation.
(i) Fermentation can also be called anaerobic respiration.
(ii) In anaerobic respiration, O₂ is not evolved.
(iii) Sargassum undergoes anaerobic respiration.
(iv) In anaerobic respiration, CO₂ is evolved.
(a) Only (iv)
(b) Only (ii)
(c) Only (iii)
(d) Both (ii) & (iv)
Answer:
(c) Only (iii)

23. ………. is an alternate way for glucose break down.
(a) Glycolysis
(b) Fermentation
(c) Respiration
(d) HMP shunt
Answer:
(d) HMP shunt

24. …….. is used for the synthesis of anthocyanin.
(a) Ribulose
(b) Erythrose
(c) Sedoheptulose
(d) Xylulose
Answer:
(b) Erythrose

25. Mention the optimum temperature for respiration?
(a) 25°C
(b) 30°C
(c) 26°C
(d) 28°C
Answer:
(b) 30°C

Students get through the TN Board 11th Bio Botany Important Questions Chapter 13 Photosynthesis which is useful for their exam preparation.

TN State Board 11th Bio Botany Important Questions Chapter 13 Photosynthesis

Very short answer questions

Question 1.
Who was Stephen Hales?
Answer:
Stephen Hales is the father of plant physiology. He was the person to propound that plants obtain nourishment from air & light.

Question 2.
Give the overall equation of photosynthesis.
Answer:

Question 3.
What happens to water & carbon dioxide during photosynthesis?
Answer:
Water is oxidised to oxygen. CO₂ is reduced to carbohydrates.

Question 4.
Define anaerobic photosynthesis.
Answer:
In some bacteria, oxygen is not evolved and is called non-oxygenic and anaerobic photosynthesis. Examples: Green sulphur, Purple sulphur and green filamentous bacteria.

Question 5.
What is Bioluminescence?
Answer:
Bioluminescence is the production and emission of light by a living organism. Bioluminescence is rare in true plants.

Question 6.
What are Quantasomes?
Answer:
In chloroplast, Inner surface of lamellar membrane consists of a small spherical structure called as Quantasomes.

Question 7.
Define the term photosynthetic pigment.
Answer:
A photosynthetic pigment is a pigment that is present in chloroplasts or photosynthetic bacteria ‘ which captures the light energy necessary for photosynthesis.

Question 8.
Apart from chlorophyll, other pigments are called accessory pigments. Why?
Answer:
Chlorophyll ‘a’ is the primary pigment that acts as a reaction centre and all other pigments act as accessory pigments and trap solar energy and then transfer it to chlorophyll ‘a’.

Question 9.
What is a Phytol tail? Mention its role.
Answer:
a. Phytol tail is the lipophilic tail of chlorophyll molecule,
b. It helps in anchoring the chlorophyll to lamellae.

Question 10.
Mention the minerals used in the biosynthesis of Chlorophyll a.
Answer:
Mg, Fe, Cu, Zn, Mn, K and Nitrogen.

Question 11.
How chlorophyll b differs from chlorophyll a?
Answer:
Chlorophyll ‘b’ differs from Chlorophyll ‘a’ in having CHO (aldehyde) group instead of CH₃ (Methyl) group at the 3rd C atom in II Pyrrole ring.

Question 12.
Carotenoids are shield pigments – Comment.
Answer:
Carotenoids are yellow to orange pigments, mostly tetraterpens and these pigments absorb light strongly in the blue to the violet region of the visible spectrum. These pigments protect chlorophyll from photooxidative damage. Hence, they are called shield pigments.

Question 13.
Give an account of Xanthophylls.
Answer:
Yellow (C₄₀H₅₆O₂) pigments are like carotenes but contain oxygen. Lutein is responsible for the yellow colour change of leaves during the autumn season. Examples: Lutein, Violaxanthin and Fucoxanthin.

Question 14.
Name the two forms of phycobilins and also give an example.
Answer:
Phycobilins exist in two forms. They are:
a. Phycocyanin found in Cyanobacteria.
b. Phycoerythrin found in red algae.

Question 15.
Define Quantum.
Answer:
Light as a particle is called a photon. Each photon contains an amount of energy known as quantum.

Question 16.
How will you define Quantasomes?
Answer:
Quantasomes are the morphological expression of physiological photosynthetic units, located on the inner membrane of thylakoid lamellae.

Question 17.
Mention the events occurring in the photo-oxidation phase of light reaction.
Answer:
Photo-oxidation Phase:

• Absorption of light energy.
• Transfer of energy from accessory pigments to the reaction centre.
• Activation of Chlorophyll ‘a’ molecule.

Question 18.
Mention the events of the Photochemical phase of light reaction.
Answer:
Photo Chemical Phase:

• Photolysis of water and oxygen evolution
• Electron transport and synthesis of assimilatory power.

Question 19.
Define Photophosphorylation.
Answer:
Phosphorylation takes place with the help of light generated electron and hence it is known as
photophosphorylation.

Question 20.
Compare fluorescence with Phosphorescence.
Answer:

Question 21.
Draw the diagram representing the oxygen-evolving complex.
Answer:

Question 22.
Compare oxidative phosphorylation with substrate-level phosphorylation.
Answer:
Phosphorylation taking place during respiration is called oxidative phosphorylation and ATP produced by the breakdown of the substrate is known as substrate-level phosphorylation.

Question 23.
State Chemiosmotic theory.
Answer:
The chemiosmotic theory was proposed by P. Mitchell (1966). According to this theory, electrons are transported along the membrane through PS I and PS II and connected by the Cytochrome b6-f complex.

Question 24.
What are the assimilatory powers produced during the light reaction?
Answer:
ATP and NADPH + H⁺

Question 25.
Why PCR cycle is called as C₃ cycle?
Answer:
The first product of the PCR pathway is a 3- carbon compound (Phospho Glyceric Acid) and so it is also called as C₃ Cycle. It takes place in the stroma of the chloroplast.

Question 26.
Name the three stages of dark reaction.
Answer:

1. Carboxylation (fixation),
2. Reduction (Glycolytic Reversal) and
3. Regeneration.

Question 27.
Give an account on RUBISCO.
Answer:
RUBISCO – RUBP Carboxylase Oxygenase enzyme, is the most abundant protein found on earth. It constitutes 16 % of the chloroplast protein. It acts as carboxylase in the presence of CO₂ and oxygenase in the absence of CO₂.

Question 28.
C₄ plants are of ecological benefit – Comment.
Answer:
C₄ plants account for about 30% of terrestrial carbon fixation. Increasing the proportion of C₄ plants on earth could assist the biosequestration of CO₂ and represent an important climate change avoidance strategy.

Question 29.
Compare the features of dimorphic chloroplasts.
Answer:
The characteristic feature of C₄ plants is the presence of dimorphic chloroplast: Bundle sheath chloroplast: Larger chloroplast, thylakoids not arranged in granum and rich in starch. Mesophyll Chloroplast: Smaller chloroplast, thylakoids arranged in granum and less starch.

Question 30.
Define Photorespiration.
Answer:
Photorespiration is the excess respiration taking place in photosynthetic cells due to the absence of CO₂ and an increase of O₂.

Question 31.
What is CO₂ compensation plant?
Answer:
When the rate of photosynthesis equals the rate of respiration, there is no exchange of oxygen and carbon dioxide and this is called a carbon dioxide compensation point.

Question 32.
State Blackman’s law of limiting factor.
Answer:
“At any given point of time, the lowest factor among essentials will limit the rate of photosynthesis”.

Question 33.
List out the external and internal factors that affect photosynthesis.
Answer:

• External factors: Light, carbon dioxide, temperature, water, mineral and pollutants.
• Internal factors: Pigments, protoplasmic factor, accumulation of carbohydrates, anatomy of leaf and hormones.

Question 34.
Enumerate the anatomical features of the leaf that affects photosynthesis.
Answer:
The thickness of cuticle and epidermis, distribution of stomata, presence or absence of Kranz anatomy and relative proportion of photosynthetic cells affect photosynthesis.

Question 35.
Name the photosynthetic apparatus of bacteria.
Answer:
Bacteria have a special type of photosynthetic apparatus called chlorosomes and chromatophores.

Short answer questions

Question 1.
Give an account of the biosynthesis of chlorophyll.
Answer:
Chlorophyll is synthesized from intermediates of respiration and photosynthesis. Succinic acid an intermediate of the Krebs cycle is activated by the addition of coenzyme A and it reacts with a simple amino acid glycine and the reaction goes on to produce chlorophyll ‘a’.
Biosynthesis of chlorophyll ‘a’ requires Mg, Fe, Cu, Zn, Mn, K and nitrogen. The absence of any one of these minerals leads to chlorosis.

Question 2.
List out the types of radiations in the Electromagnetic spectrum.
Answer:
The electromagnetic spectrum consists of 8 types of radiations such as cosmic rays, gamma rays, X rays, U-V rays, Visible light spectrum, infrared rays, electric rays and radio rays.

Question 3.
Write a brief note on Emerson’s Red Drop.
Answer:
Emerson conducted an experiment in Chlorella using only one wavelength of light (monochromatic light) at a time and he measured quantum yield. He plotted a graph of the quantum yield in terms of O, evolution at various wavelengths of light. His focus was to determine at which wavelength the photochemical yield of oxygen was maximum. He found – that in the wavelength of 600 to 680 the yield was constant but suddenly dropped in the region above 680 nm (red region). The fall in the photosynthetic yield beyond the red region of the spectrum is referred to as Red drop or Emerson’s first effect.

Question 4.
Explain the term Emerson’s Enhancement Effect.
Answer:
Emerson found that the monochromatic light of longer wavelength (far red light), when supplemented with a shorter wavelength of light (red light), enhanced photosynthetic yield and recovered red drop. This enhancement of photosynthetic yield is referred to as Emerson’s Enhancement Effect.

Question 5.
Give the conclusions of Hills Reaction.
Answer:
Conclusions of Hill’s Reaction:

1. During photosynthesis, oxygen is evolved from water.
2. Electrons for the reduction of CO₂ are obtained from water.
3. A reduced substance produced, later helps to reduce CO₂.
   2H₂O + 2A → 2 AH₂ + O₂

Question 6.
Define Dark Reaction.
Answer:
Fixation and reduction of CO₂ into carbohydrates with the help of assimilatory power produced during light reaction. This reaction does not require light and is not directly light-driven. Hence, it is called as Dark reaction or Calvin-Benson cycle.

Question 7.
Illustrate S’ state mechanism.
Answer:
S’ State Mechanism consists of a series of 5 states called as S₀, S₁, S₂, S₃ and S₄. Each state acquires a positive charge by a photon (hv) and after the S₄ state, it acquires 4 positive charges, four electrons and the evolution of oxygen. Two molecules of water go back to the S₀. At the end of photolysis 4H⁺, 4e^– and O₂ are evolved from water.

Question 8.
Kranz Anatomy – Define.
Answer:
Kranz Anatomy is a German term meaning a halo or wreath. In C₄ plants vascular bundles are surrounded by a layer of bundle sheath. The bundle sheath is surrounded by a ring of mesophyll cells.

Question 9.
Point out the significances of C₄ cycle.
Answer:
Significance of C₄ cycle

1. Plants having C₄ cycle are mainly of tropical and sub-tropical regions and are able to survive in an environment with low CO₂ concentration.
2. C₄ plants are partially adapted to drought conditions.
3. Oxygen has no inhibitory effect on C₄ cycle since PEP carboxylase is insensitive to O₂.
4. Due to the absence of photorespiration, CO₂ Compensation Point for C₄ is lower than that of C₃ plants.

Question 10.
Point out the significances of photorespiration.
Answer:
Significance of photorespiration

1. Glycine and Serine synthesised during this process are precursors of many biomolecules like chlorophyll, proteins, nucleotides.
2. It consumes excess NADH + H⁺ generated.
3. Glycolate protects cells from Photooxidation.

Question 11.
Mention the role of light intensity in photosynthesis.
Answer:
The intensity of light plays a direct role in the rate of photosynthesis. Under low intensity the photosynthetic rate is low and at higher intensity photosynthetic rate is higher. It also depends on the nature of plants. Heliophytes (Bean Plant) require higher intensity than Sciophytes (Oxalis).

Question 12.
Classify photosynthetic bacteria and give example.
Answer:
Photosynthetic bacteria are classified into three groups:

1. Green sulphur bacteria. Example: Chlorobacterium and Chlorobium.
2. Purple sulphur bacteria. Example: Thiospirillum and Chromatium.
3. Purple non-sulphur bacteria. Example: Rhodopseudomonas and Rhodospirillum.

Long answer questions

Question 1.
Mention the significance of photosynthesis.
Answer:
Significance of Photosynthesis:

1. Photosynthetic organisms provide food for all living organisms on earth either directly or indirectly.
2. It is the only natural process that liberates oxygen in the atmosphere and balances the oxygen level.
3. Photosynthesis balances the oxygen and carbon cycle in nature.
4. Fuels such as coal, petroleum and other fossil fuels are from preserved photosynthetic plants.
5. Photosynthetic organisms are the primary producers on which all consumers depend for energy.
6. Plants provide fodder, fibre, firewood, timber, useful medicinal products and these sources come by the act of photosynthesis.

Question 2.
Describe the structure of chloroplasts.
Answer:
Chloroplasts are the main site of photosynthesis and both the energy-yielding process (Light reaction) and fixation of carbon dioxide (Dark reaction) that takes place in the chloroplast. It is a double-wall membrane-bounded organelle, discoid or lens-shaped, 4 -10 pm in diameter and 1-33 pm in thickness. The membrane is a unit membrane and the space between them is 100 to 200 A. A colloidal and proteinaceous matrix called stroma is present inside.
A sac-like membranous system called thylakoid or lamellae is present in the stroma and they are arranged one above the other forming a stack of a coin-like structure called granum (plural grana). Each chloroplast contains 40 to 80 grana and each granum consists of 5 to 30 thylakoids. Thylakoids found in granum are called grana lamellae and in the stroma are called stroma lamellae. Thylakoid disc size is 0.25 to 0.8 micron in diameter. A thinner lamella called the Fret membrane connects grana. Pigment system I is located on the outer thylakoid membrane facing stroma and Pigment system II is located on the inner membrane facing the lumen of the thylakoid. Grana lamellae have both PS I and PS II whereas stroma lamellae have only PS I. Chloroplast contains 30-35% Proteins, 20-30% phospholipids, 5-10% chlorophyll, 4-5% Carotenoids, 70S ribosomes, circular DNA and starch grains. The inner surface of the lamellar membrane consists of a small spherical structure called as Quantasomes. The presence of 70S ribosome and DNA gives them the status of semi-autonomy and proves the endosymbiotic hypothesis which says chloroplast evolved from bacteria. Thylakoid contains pigment systems that produce ATP and NADPH + H⁺ using solar energy. The stroma contains an enzyme which reduces carbon dioxide into carbohydrates.

Question 3.
Tabulate the different types of photosynthetic pigments.
Answer:

Question 4.
Give an account of Chlorophyll.
Answer:
Chlorophyll ‘ a’ is the primary pigment that acts as a reaction centre and all other pigments act as accessory pigments and trap solar energy and then transfer it to chlorophyll ‘a’. Chlorophyll molecules have a tadpole-like structure. It consists of the Mg-Porphyrin head (Hydrophilic Head) and (Lipophilic tail) Phytol tail. The Porphyrin head consists of four pyrrole rings linked together by C-H bridges. Each pyrrole ring comprises four carbons and one nitrogen atom. Porphyrin ring has several side groups which alter the properties of the pigment. Different side groups are indicative of various types of chlorophyll. The Phytol tail made up of 20 carbon alcohol is attached to carbon 7 of the Pyrrole ring IV. It has a long propionic acid ester bond. A long lipophilic tail helps in anchoring chlorophyll to the lamellae.

Question 5.
Explain the steps involved in paper chromatography.
Answer:
Separation of Chloroplast pigments by paper Chromatography method:
Step 1. Extract chlorophyll pigment from the leaves using 80% Acetone.
Step 2. Allow concentrating by evaporation.
Step 3. Apply few drops on one end above 2 cm from the edge of a chromatographic paper.
Step 4. A solvent with a mixture of Petroleum ether and acetone in the ratio of 9:1 is prepared and poured into the development chamber.
Step 5. Place the strip above the solvent by placing one end of the strip touching the solvent. Observation: After one hour of observing the chromatographic paper. You can find the pigments being separated into four distinct spots.

Question 6.
Enumerate the properties of light.
Answer:
Properties of Light

1. Light is a transverse electromagnetic wave.
2. It consists of oscillating electric and magnetic fields that are perpendicular to each other and perpendicular to the direction of propagation of the light.
3. Light moves at a speed of 3 x 108 ms^-1.
4. Wavelength is the distance between successive crests of the wave.
5. Light as a particle is called a photon. Each photon contains an amount of energy known as quantum.
6. The energy of a photon depends on the frequency of the light.

Question 7.
Differentiate between Photosystem I and Photosystem II.
Answer:

Question 8.
Explain the various complexes in the Electron transport chain.
Answer:
The electron transport chain in each photosystem involves four complexes:

• Core Complex (CC): CC I in PS I the reaction centre is P700, CC II in PS II the reaction centre is P680
• Light-Harvesting Complex or Antenna complex (LHC):
• Two types: LHC I in PS I and LHC II in PS II.
• Cytochrome b₆ f complex: It is the non-pigmented protein complex connecting PS I and PS II. Plastoquinone (PQ) and Plastocyanin (PC) are intermediate complexes acting as mobile or shuttle electron carriers of the Electron Transport Chain. PQ acts as a shuttle between PS II and Cytochrome b₆– f complex and PC connects
• Cytochrome b₆-f and PS I complex.
• AT Pave complex or Coupling factor: It is found on the surface of the thylakoid membrane. This complex is made up of CF₁ and CF₀ factors. This complex utilizes energy from ETC and converts ADP and inorganic phosphate (Pi) into ATP.

Question 9.
Explain cyclic photophosphorylation.
Answer:
Cyclic photophosphorylation refers to the electrons ejected from the pigment system I (Photosystem I) and again cycled back to the PS I. When the photons activate P700 reaction centre photosystem II is activated. Electrons are raised to a high energy level. The primary electron acceptor is Ferredoxin Reducing Substance (FRS) which transfers electrons to Ferredoxin (Fd), Plastoquinone (PQ), cytochrome b₆-f complex, Plastocyanin (PC) and finally back to chlorophyll P700 (PS I). During this movement of electrons, Adenosine Di Phosphate (ADP) is phosphorylated, by the addition of inorganic phosphate and generates Adenosine Tri Phosphate (ATP). Cyclic electron transport produces only ATP and there is no NADPFI + H⁺ formation. At each step of electron, transport, the electron loses potential energy and is used by the transport chain to pump H⁺ ions across the thylakoid membrane. The proton gradient triggers ATP formation in the ATP synthase enzyme situated on the thylakoid membrane. Photosystem I need the light of a longer wavelength (> P700 nm). It operates under low light intensity, less CO₂ and under anaerobic conditions which makes it considered as earlier in evolution.

Question 10.
Explain the stages occurring during non-cyclic electron transport.
Answer:
In oxygenic species, non-cyclic electron transport takes place in three stages.

1. Electron transport from water to P680: Splitting of water molecule produce electrons, protons and oxygen. Electrons lost by the PS II (P680) are replaced by electrons from the splitting of a water molecule.
2. Electron transport from P680 to P700: Electron flow starts from P680 through a series of electron carrier molecules like pheophytin, plastoquinone (PQ), cytochrome b6 – f complex, plastocyanin (PC) and finally reaches P700 (PS I).
3. Electron transport from P700 to NADP⁺: PS I(P700) is excited now and the electrons pass to a high energy level. When electron travels downhill through ferredoxin, NADP+ is reduced to NADPH + H⁺.

Question 11.
Point out the Bio-energetics of light reaction.
Answer:
Bio energetics of light reaction:

• To release one electron from the pigment system requires two quanta of light.
• One quantum is used for the transport of electron from water to PS I.
• The second quantum is used for the transport of electron from PS I to NADP
• Two electrons are required to generate one NADPH + H⁺.
• During Non-Cyclic electron transport two NADPH + H⁺ are produced and it requires 4 electrons.
• Transportation of 4 electrons requires 8 quanta of light.

Question 12.
Differentiate between Cyclic & Non-cyclic photophosphorylation.
Answer:
Differences between Cyclic Photophosphorylation and Non-Cyclic Photophosphorylation

Question 13.
Draw a Flow chart depicting Calvins cycle.
Answer:

Question 14.
Explain the three phases of the Dark reaction.
Answer:
Phase 1- Carboxylation (Fixation)
The acceptor molecule Ribulose 1,5 Bisphosphate (RUBP) a 5 carbon compound with the help of RUBP carboxylase oxygenase (RUBISCO) enzyme accepts one molecule of carbon dioxide to form an unstable 6 carbon compound. This 6C compound is broken down into two molecules of 3-carbon compound phospho glyceric acid (PGA).

Phospho glyceric acid is phosphorylated by ATP and produces 1,3 bis phospho glyceric acid by PGA kinase. 1,3 bis phospho glyceric acid is reduced to glyceraldehyde 3 Phosphate (G-3-P) by using the reducing power NADPH + H⁺. Glyceraldehyde 3 phosphate is converted into its isomeric form dihydroxy acetone phosphate (DHAP).

Phase 3 – Regeneration: Regeneration of RUBP involves the formation of several intermediate compounds of 6-carbon, 5-carbon, 4-carbon and 7- carbon skeleton. Fixation of one carbon dioxide requires 3 ATPs and 2 NADPH + H⁺, and the fixation of 6 CO₂ requires 18 ATPs and 12 NADPH + H⁺ during C₃ cycle. One 6 carbon compound is the net gain to form hexose sugar.

Question 15.
Explain the phases of C₄ pathway.
Answer:
C₄ pathway is completed in two phases, first phase takes place in the stroma of mesophyll cells, where the CO₂ acceptor molecule is a 3-Carbon compound, phosphoenol pyruvate (PEP) to form 4-carbon Oxalo acetic acid (OAA). The first product is a 4-carbon and so it is named as C₄ cycle. Oxalo acetic acid is a dicarboxylic acid and hence this cycle is also known as a dicarboxylic acid pathway. Carbon dioxide fixation takes place in two places one in mesophyll and another in bundle sheath cell (dicarboxylation pathway). It is the adaptation of tropical and subtropical plants growing in warm and dry conditions. Fixation of C02 with minimal loss is due to the absence of photorespiration. C₄ plants require 5 ATP and 2 NADPH + H⁺ to fix one molecule of CO₂.

Oxaloacetic acid (OAA) is converted into malic acid or aspartic acid and is transported to the bundle sheath cells through plasmodesmata.
Stage: II Bundle Sheath Cells: Malic acid undergoes decarboxylation and produces a 3 carbon compound Pyruvic acid and CO₂. The released CO₂ combines with RUBP and follows the Calvin cycle and finally, sugar is released to the phloem. Pyruvic acid is transported to the mesophyll cells.

Question 16.
Differentiate between C₃ & C₄ plants.
Answer:
Differences between C₃ and C₄ plants

Question 17.
Explain in detail about Crassulacean Acid Metabolism.
Answer:
Crassulacean Acid Metabolism or CAM cycle:
It is one of the carbon pathways identified in succulent plants growing in semi-arid or xerophytic condition. This was first observed in Crassulaceae family plants like Bryophyllum, Sedum, Kalanchoe and is the reason behind the name of this cycle. It is also noticed in plants from other families. Examples: Agave, Opuntia, Pineapple and Orchids. The stomata are closed during the day and are open during the night (Scotoactive). This reverse stomatal rhythm helps to conserve water loss through transpiration and will stop the fixation of CO₂ during the daytime. At night time CAM plants fix CO₂ with the help of Phospho Enol Pyruvic acid (PEP) and produce oxalo acetic acid (OAA). Subsequently, OAA is converted into malic acid like C₄ cycle and gets accumulated in vacuole increasing the acidity. During the daytime stomata are closed and malic acid is decarboxylated into pyruvic acid resulting in the decrease of acidity. CO₂ thus formed enters into Calvin Cycle and produces carbohydrates.
Significance of CAM Cycle

1. It is advantageous for succulent plants to obtain CO₂ from malic acid when stomata are, closed.
2. During the daytime, stomata are closed and CO₂ is not taken but continue their photosynthesis.
3. Stomata are closed during the daytime and help the plants to avoid transpiration and water loss.

Question 18.
State the differences between photorespiration and dark respiration.
Answer:
Differences between Photorespiration and Dark Respiration

Question 19.
Explain the Internal factors that affect photosynthesis.
Answer:
Internal Factors

1. Photosynthetic Pigments: It is an essential factor and even a small quantity is enough to carry out photosynthesis.
2. Protoplasmic factor: Hydrated protoplasm is essential for photosynthesis. It also includes enzymes responsible for Photosynthesis.
3. Accumulation of Carbohydrates: Photosynthetic end products like carbohydrates are accumulated in cells and if translocation of carbohydrates is slow then this will affect the rate, of photosynthesis.
4. Anatomy of leaf: Thickness of cuticle and epidermis, distribution of stomata, presence or absence of Kranz anatomy and relative proportion of photosynthetic cells affect photosynthesis.
5. Hormones: Hormones like gibberellins and cytokinin increase the rate of photosynthesis.

Question 20.
Compare photosynthesis in plants & bacteria.
Answer:
Differences between Photosynthesis in Plants and Photosynthesis in Bacteria

Higher Order Thinking Skills (HOTs)

Question 1.
The picture given below is an organelle of a plant cell. Identify the picture and answer the questions.

(a) Name the organelle.
(b) Mention the role of the organelle in the cell.
(c) How do you call the stack of coin-like structures present on it?
(d) Whether it shows semi-autonomy? If yes how? If no, why?
Answer:
(a) Chloroplast.
(b) Chloroplast is the site of photosynthesis.
(c) Grana made of thylakoids.
(d) Yes, the presence of DNA, 70S ribosomes gives them the status of semi-autonomy.

Question 2.
Succulents are known to keep their stomata closed during the day to check transpiration.
How do they meet their photosynthetic CO₂ requirements?
Answer:
Succulents undergo a special carbon pathway called Crassulacean Acid Metabolism (CAM pathway). At night time, CAM plants for CCE with the help of Phosphoenol Pyruvic Acid (PEP) and produce Oxalo Acetic Acid (OAA). Subsequently, OAA is converted into malic acid like C₄ cycle and produces CO₂. The CO₂ thus formed enters the Calvins cycle and produces carbohydrates.

Question 3.
An increase in temperature decreases photosynthetic rale – Justify.
Answer:
In general, the optimum temperature for photosynthesis is 25°C to 35°C. An increase in temperature will lead to the closure of stomata thereby inhibiting the gaseous exchange and also inactivate the enzymes responsible for photosynthesis.

Question 4.
“Photosynthesis is a redox reaction”. Comment.
Answer:
Photosynthesis is an oxidation and reduction process where, water is oxidised to release oxygen and CO₂ is reduced to form carbohydrates.

Question 5.
Carrots, Capsicum, Tomatoes etc are organe/red coloured fruits.
(a) Name the pigment responsible for the colour.
(b) State whether it is a photosynthetic pigment.
(c) Mention its role in plants.
Answer:
(a) Carotenoids are the yellow to orange pigments responsible for fruit colours.
(b) Yes, it is a photosynthetic pigment.
(c) They absorb the light energy and transfer it to the chlorophyll. They also protect chloro¬phyll from photo-oxidative damage. Hence also called shield pigments.

Question 6.
Paddy is a C₃ plant. C₃ plants utilise ATP and NADPH₂ molecules to generate oxygen molecules. How many ATP and NADPH₂ molecules would a C₃ plant consume to generate 12 molecules of oxygen.
Answer:
C₃ plants utilise 3 ATP’s and 2 NADPH₂ molecules to evolve one oxygen molecule. To generate 12 molecules of oxygen, 36 ATP’s and 24 NADPH₂ molecules are utilised.

Question 7.
Give the meaning for the following terminologies:
(a) Decarboxylation (b) Phosphorylation (c) Photolysis
Answer:
(a) Decarboxylation : Removal of C02 from a molecule.
(b) Phosphorylation: Synthesis of ATP by the addition of inorganic phosphate to ADP.
(c) Photolysis: Splitting of water molecules using light energy.

Question 8.

The above equation represents the first step of a cyclic pathway occurring in plant cells.
(а) Name the pathway.
(б) Where does this pathway occur inside the cell?
(c) How many carbon molecules are seen in RUBP & PGA.
(d) State the role of RUBISCO in this step.
(e) Under which condition does this pathway proceed.
Answer:
(a) Photorespiration or C₂ cycle.
(b) C₂ cycle takes place in chloroplast, peroxisome and mitochondrion.
(c) RUBP is a 5C compound and PGA is a 3C compound
(d) RUBISCO plays the role of oxygenase enzyme.
(e) C₂ cycle occurs in photosynthetic cells due to the absence of CO₂ and increased O₂.

Choose the correct answer.

1. Photosynthetic organisms used only percent of solar light.
(a) 0.2
(b) 0.6
(c) 0.1
(d) 0.8
Answer:
(a) 0.2

2. ………… is famously called as father of plant physiology.
(a) Joseph Priestley
(b) Lavoisier
(c) Stephen Hales
(d) Van Helmont
Answer:
(c) Stephen Hales

3. In Green sulphur bacteria, is the hydrogen donor.
(a) H₂O
(b) H₂S
(c) H₂O₂
(d) HCl
Answer:
(b) H₂S

4. Ruben & Kamen used radioactive oxygen to prove the evolution of oxygen from water.
(a) 180
(b) 160
(c) 140
(d) 120
Answer:
(a) 180

5. In photosynthesis ………….. is reduced into carbohydrates.
(a) H₂O
(b) Chlorophyll
(c) CO₂
(d) O₂
Answer:
(c) CO₂

6. Which of the following performs anaerobic photosynthesis?
(a) Green sulphur bacteria
(b) Cyanobacteria
(c) Purple sulphur bacteria
(d) Green filamentous bacteria
Answer:
(b) Cyanobacteria

7. The colloidal proteinaceous matrix of chloroplast is
(a) Thylakoid
(b) Stroma
(c) Grana
(d) Lamellae
Answer:
(b) Stroma

8. Each granum has thylakoids.
(a) 8-30
(b) 40 – 80
(c) 5-30
(d) 40 – 70
Answer:
(c) 5-30

9. Which of the following is a primary pigment?
(a) Chlorophyll a
(b) Carotene
(c) Xanthophyll
(d) Phycoerythrin
Answer:
(a) Chlorophyll a

10. The chlorophyll pigment found in xanthophycean algae is
(a) Chlorophyll b
(b) Chlorophyll c
(c) Chlorophyll d
(d) Chlorophyll e
Answer:
(d) Chlorophyll e

11. The nature of phytol tail in chlorophyll is
(a) Hydrophilic
(b) Lipophilic
(c) Hydrophobic
(d) Lipophobic
Answer:
(b) Lipophilic

12. The porphyrin head of chlorophyll has ………. pyrrole rings.
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

13. Which of the following mineral is NOT required for the biosynthesis of chlorophyll a?
(a) Mn
(b) Mg
(c) Mo
(d) Cu
Answer:
(c) Mo

14. In chlorophyll C, which of the following component is absent?
(a) Porphyrin head
(b) Phytol tail
(c) Pyrrole ring
(d) Methyl group
Answer:
(b) Phytol tail

15. …………. pigments are also called as shield pigments.
(a) Carotenoids
(b) Chlorophyll b
(c) Phycobilins
(d) Phycoerythrin
Answer:
(a) Carotenoids

16. ………….. is responsible for the yellow colour change of leaves during autumn.
(a) Cutin
(b) Lutein
(c) Phycobilin
(d) Carotein
Answer:
(b) Lutein

17. The colour of the light is determined by its …………..
(a) Intensity
(b) Refractive power
(c) Wavelength
(d) Reflection
Answer:
(c) Wavelength

18. Visible spectrum ranges between ………….
(a) 390 – 763 nm
(b) 370 – 700 nm
(c) 450 – 700 nm
(d) 357 – 736 nm
Answer:
(a) 390 – 763 nm

19. Electromagnetic spectrum consists of …………. types of radiation.
(a) 2
(b) 4
(c) 6
(d) 8
Answer:
(d) 8

20. Light as a particle is called …………
(a) Neutron
(b) Quantum
(c) Photon
(d) Quantasome
Answer:
(c) Photon

21. Usually …………… chlorophyll molecules are considered as physiological units of photosynthesis.
(a) 300 – 700
(b) 200-300
(c) 240-750
(d) 200-280
Answer:
(b) 200-300

22. Who coined the term Quantasomes?
(a) Steinmann
(b) Park & Biggins
(c) Emerson & Arnold
(d) Von Mayer
Answer:
(b) Park & Biggins

23. In Emerson’s first effect, the photosynthetic yield was dropped in the region above ………..
(a) 720 nm
(b) 620 nm
(c) 680 nm
(d) 600 nm
Answer:
(c) 680 nm

24. In photosynthetic reactions ………… is considered as assimilatory power.
(a) NADPH
(b) FADPH
(c) ATP
(d) GTP
Answer:
(c) ATP

25. Antenna molecules refers to …………
(a) Light-harvesting complex
(b) Central core complex
(c) PS II
(d) Oxygen evolving complex
Answer:
(a) Light-harvesting complex

26: Chlorophyll and carotenoid ratio in PS II is …………
(a) 3 to 7:1
(b) 20 to 30:1
(c) 7 to 30:1
(d) 10 to 30:1
Answer:
(a) 3 to 7:1

27. Phosphorylation taking place during respiration is called …………
(a) Substrate level phosphorylation
(b) Oxidative phosphorylation
(c) Reductive phosphorylation
(d) Photophosphorylation
Answer:
(b) Oxidative phosphorylation

28. Which of the following statement is NOT true regarding cyclic photophosphorylation?
(a) The primary electron acceptor is FRS
(b) It produces only ATP molecules
(c) It produces only NADPH + H⁺ molecules.
(d) Electrons ejected from PSI again cycled back to PSI
Answer:
(c) It produces only NADPH + H⁺ molecules

29. Chemiosmotic theory was proposed by …………
(a) Mitchell
(b) Hatch & Slack
(c) Calvin
(d) Priestley
Answer:
(a) Mitchell

30. How many ATP molecules are utilized by C₃ plants to evolve one oxygen molecule.
(a) 3
(b) 4
(c) 8
(d) 12
Answer:
(a) 3

31. The first formed product of C₃ cycle is …………
(a) Succinic acid
(b) Phosphoglyceric acid
(c) Oxalo Acetic acid
(d) Malic acid
Answer:
(6) Phosphoglyceric acid

32. RUBP is a ……….. carbon compound.
(a) Three
(b) Four
(c) Five
(d) Seven
Answer:
(c) Five

33. Number of dicot species performing C₄ pathway is …………
(a) 200
(b) 300
(c) 800
(d) 1000
Answer:
(b) 300

34. The CO₂ acceptor molecule in C₃ plants is …………
(a) PEP
(b) PGA
(c) OAA
(d) RUBP
Answer:
(d) RUBP

35. C₂ cycle refers to
(a) CAM cycle
(b) PCO cycle
(c) PCR cycle
(d) DCA cycle
Answer:
(b) PCO cycle

36. Identify the mismatched pair:
(i) Phosphorylation reaction – phosphorous
(ii) Photolysis of water – Manganese & Chlorine
(iii) Plastocyanin formation – Copper and Zinc
(iv) Chlorophyll formation – Magnesium, Iron, Nitrogen
(a) (i) only
(b) (iii) only
(c) Both (i) & (iii)
(d) All the above
Answer:
(b) (iii) only

37. Photosynthetically Active Radiation (PAR) is between
(a) 700 – 760 nm
(b) 400-700 nm
(c) 500-600 nm
(d) 350 – 760 nm
Answer:
(b) 400 – 700 nm

38. In atmosphere the percentage of CO₂ is …………
(a) 0.3%
(b) 0.7%
(c) 0.1%
(d) 0.6%
Answer:
(a) 0.3%

39. The inhibitory effect of oxygen in photosynthesis was first discovered by ……………..
(a) Warburg
(b) Van Helmont
(c) Dutrochet
(d) Desaussure
Answer:
(a) Warburg

40. In general, the optimum temperature for photosynthesis is …………
(a) 26°C to 39°C
(b) 25°C to 40°C
(c) 55°C
(d) 25°C to 35°C
Answer:
(d) 25°C to 35°C

Students get through the TN Board 11th Bio Botany Important Questions Chapter 12 Mineral Nutrition which is useful for their exam preparation.

TN State Board 11th Bio Botany Important Questions Chapter 12 Mineral Nutrition

Very short answer questions

Question 1.
Differentiate between Micronutrient and Macronutrient.
Answer:

Question 2.
State the “Law of minimum” proposed by Liebig.
Answer:
The “law of minimum” states that the productivity of soil depends on the number of essential elements present in minimum quantity.

Question 3.
Mention any four unclassified minerals.
Answer:
Sodium, Cobalt, Silicon and Selenium.

Question 4.
Potassium and Osmotic potential – Comment.
Answer:
Potassium (K) plays a key role in maintaining the osmotic potential of the cell. The absorption of water, movement of stomata and turgidity are due to osmotic potential.

Question 5.
In which form do the following minerals are absorbed by plants.
(a) Nitrogen
(b) Magnesium
(c) Boron
(d) Phosphorous
Answer:

Question 6.
Name any two calcium deficiency disease in plants.
Answer:
(a) Blackheart of celery,
(b) Hooked leaf tip in beet, Musa and tomato.

Question 7.
Nitrogen is a Macronutrient – Justify.
Answer:
Nitrogen (N) is required by the plants in the greatest amount. It is an essential component of proteins, nucleic acids, amino acids, vitamins, hormones, alkaloids, chlorophyll and cytochrome.

Question 8.
What are Siderophores?
Answer: Siderophores (iron carriers) are Iron chelating agents produced by bacteria. They are used to chelate ferric iron (Fe³⁺) from the environment and host.

Question 9.
Given below are the plant diseases, name the deficient mineral responsible for the disease.
(a) Khaira disease
(b) Internal cork of apple
(c) Die back of Citrus
(d) Sand drown of tobacco
Answer:

Question 10.
Define Calmodulin.
Answer:
Calmodulin is a Ca²⁺ modulating protein in eukaryotic cells. It is a heat-stable protein involved in fine metabolic regulations.

Question 11.
When does an essential mineral is considered “toxic”?
Answer:
An increase of mineral nutrients more than the normal concentration causes toxicity. A concentration, at which 10 % of the dry weight of tissue is reduced, is considered toxic.

Question 12.
Give a brief note on Aluminium toxicity.
Answer:
Aluminium toxicity causes precipitation of nucleic acid, inhibition of ATPase, inhibition of cell division and binding of the plasma membrane with Calmodulin.

Question 13.
Define nitrogen fixation. Mention its types.
Answer:
The process of converting atmospheric nitrogen (N₂) into ammonia is termed as nitrogen fixation. Nitrogen fixation can occur by two methods: 1. Biological; 2. Non-Biological.

Question 14.
How does nitrogen fixation occur non-biologically?
Answer:
Non – Biological nitrogen fixation:

• Nitrogen fixation by a chemical process in the industry.
• Natural electrical discharge during lightning fixes atmospheric nitrogen.

Question 15.
Name the types of Autotrophic nutrition.
Answer:
(a) Photosynthetic autotrophs and (b) Chemosynthetic autotrophs.

Question 16.
What are obligate parasites? Give example.
Answer:
Obligate (Total) parasites completely depend on the host for survival. Example: Cuscuta.

Question 17.
What is a Lichen?
Answer:
Lichens is a mutual association of Algae and Fungi. Algae prepare food and fungi absorbs water and provides thallus structure.

Short answer questions

Question 1.
List out the criteria for being essential minerals.
Answer:
Arnon and Stout (1939) gave criteria required for essential minerals:

1. Elements are necessary for growth and development.
2. They should have a direct role in the metabolism of the plant.
3. It cannot be replaced by other elements.
4. Deficiency makes the plants impossible to complete their vegetative and reproductive phase.

Question 2.
Mention the role of Sulphur in plants.
Answer:
Sulphur (S): Essential component of amino acids like cystine, cysteine and methionine, a constituent of coenzyme A, Vitamins like biotin and thiamine, a constituent of proteins and ferredoxin. Plants utilise sulphur as sulphate (SO₄^–) ions.
Deficiency symptoms: Chlorosis, anthocyanin formation, stunted growth, rolling of leaf tip and reduced nodulation in legumes.

Question 3.
Give a short note on EDTA.
Answer:
EDTA – Ethylene Diamine Tetra Acetic acid is a chelating agent.
Plants growing in alkaline soil though supplied with all nutrients including iron show iron deficiency. To rectify this, iron is made into a soluble complex by adding EDTA (Fe-EDTA Complex) which can be easily absorbed by plants.

Question 4.
Write a brief note on Manganese toxicity.
Answer:
Increased Concentration of Manganese will prevent the uptake of Fe and Mg, prevent translocation of Ca to the shoot apex and cause their deficiency. The symptoms of manganese toxicity are the appearance of brown spots surrounded by chlorotic veins.

Question 5.
Write a note on Hydroponics and Aeroponics.
Answer:
Hydroponics or Soilless culture: Von Sachs developed a method of growing plants in the nutrient solution. The commonly used nutrient solutions are Knop solution (1865) and Arnon and Hoagland Solution (1940). Later the term Hydroponics was coined by Goerick (1940) and he also introduced commercial techniques for hydroponics. In hydroponics roots are immersed in the solution containing nutrients and air is supplied with help of a tube.
Aeroponics: This technique was developed by Soifer Hillel and David Durger. It is a system where roots are suspended in air and nutrients are sprayed over the roots by a motor-driven rotor.

Question 6.
Mention the three possible ways by which ammonia is converted into amino acids during nitrogen metabolism.
Answer:
Ammonia is converted into amino acids by the following processes:
(a) Reductive amination, (b) Trans-amination and (c) Catalytic amination.

Question 7.
Write the equation for reductive amination.
Answer:

Question 8.
Transamination – Write a note.
Answer:
Transfer of amino group (NH₃⁺) from glutamic acid glutamate to keto group of a keto acid. Glutamic acid is the main amino acid from which other amino acids are synthesised by transamination. Transamination requires the enzyme transaminase and coenzyme pyridoxal phosphate (derivative of vitamin B6 – pyridoxine).

Question 9.
Give an account on GOGAT – pathway.
Answer:
Glutamate amino acid combines with ammonia to form the amide glutamine.

Glutamine reacts with α ketoglutaric acid to form two molecules of glutamate.

Question 10.
Write a note on Saprophytes.
Answer:
Saprophytes derive nutrients from dead and decaying matter. Bacteria and fungus are the main saprophytic organisms. Some angiosperms also follow the saprophytic mode of nutrition. Example: Neottia. Roots of Neottia (Bird’s Nest Orchid) associate with mycorrhizae and absorb nutrients as a saprophyte. Monotropa (Indian Pipe) grow on humus-rich soil found in thick forests. It absorbs nutrient through mycorrhizal association.

Question 11.
What is Mycorrhizae?
Answer:
Mycorrhizae: Fungi associated with roots of higher plants including Gymnosperms. Example: Pinus.

Long answer questions

Question 1.
Write in detail about functions, mode of absorption and deficiency symptoms of any five macronutrients.
Answer:

Question 2.
Give in detail about the vitality of Boron and Zinc in plant nourishment.
Answer:
Zinc (Zn): Essential for the synthesis of Indole acetic acid (Auxin) activator of carboxylases, alcohol dehydrogenase, lactic dehydrogenase, glutamic acid dehydrogenase, carboxypeptidases and tryptophan synthetase. It is absorbed as Zn²⁺ ions.
Deficiency: Little leaf and mottle leaf due to deficiency of auxin, Inter veinal chlorosis, stunted growth, necrosis and Khaira disease of rice.
Boron (B): Translocation of carbohydrates, uptake and utilisation of Ca⁺⁺, pollen germination, nitrogen metabolism, fat metabolism, cell elongation and differentiation. It is absorbed as borate BO^3- ions.
Deficiency: Death of root and shoot tips, premature fall of flowers and fruits, the brown heart of beet root, the internal cork of apple and fruit cracks.

Question 3.
Explain in detail the symbiotic nitrogen fixation with nodulation.
Answer:
i. Nitrogen fixation with nodulation:
Rhizobium bacterium is found in leguminous plants and fixes atmospheric nitrogen. This kind of symbiotic association is beneficial for both the bacterium and the plant. Root nodules are formed due to bacterial infection. Rhizobium enters into the host cell and proliferates, it remains separated from the host cytoplasm by a membrane.
Stages of Root nodule formation:

1. Legume plants secret phenolics which attracts Rhizobium.
2. Rhizobium reaches the rhizosphere and enters into the root hair, infects the root hair and leads to curling of root hairs.
3. The infection thread grows inwards and separates the infected tissue from normal tissue.
   
4. A membrane bound bacterium is formed inside the nodule and is called a bacteroid.
5. Cytokinin from bacteria and auxin from host plant promotes cell division and leads to nodule formation.

Question 4.
Describe the various stages of the nitrogen cycle.
Answer:
The nitrogen cycle consists of the following stages:
1. Fixation of atmospheric nitrogen: Di-nitrogen molecule from the atmosphere progressively gets reduced by the addition of a pair of hydrogen atoms. A triple bond between two nitrogen atoms (N = N) is cleaved to produce ammonia. The nitrogen fixation process requires a Nitrogenase enzyme complex, Minerals (Mo, Fe and S), anaerobic condition, ATP, electron and glucose 6 phosphates as H⁺ donor. The nitrogenase enzyme is active only in anaerobic condition. To create this anaerobic condition a pigment known as leghaemoglobin is synthesized in the nodules which act as an oxygen scavenger and removes the oxygen. Nitrogen-fixing bacteria in root nodules appear pinkish due to the presence of this leghaemoglobin pigment.

Overall equation:
N₂ + 8e^– + 8H⁺ + 16 ATP → 2NH₃⁺ + H₂ + 16 ADP + 16 Pi
2. Nitrification: Ammonia (NH₃⁺) is converted into Nitrite (NO₂^–) by Nitrosomonas
bacterium. Nitrite is then converted into Nitrate (NO₃^–) by Nitrobacter bacterium. Plants are more adapted to absorb nitrate (NO₃^–) than ammonium ions from the soil.

3. Nitrate Assimilation: The process by which nitrate is reduced to ammonia is called nitrate assimilation and occurs during the nitrogen cycle.

4. Ammonification: The decomposition of organic nitrogen (proteins and amino acids) from dead plants and animals into ammonia is called ammonification. Organism involved in this process are Bacillus ramosus and Bacillus Vulgaris.
5. Denitrification: Nitrates in the soil are converted back into atmospheric nitrogen by a process called denitrification. Bacteria involved in this process are Pseudomonas, Thiobacillus and Bacillus subtilis.

Question 5.
Explain the types of parasitic mode of nutrition in angiosperms.
Answer:
Parasitic mode of nutrition in angiosperms:
Organisms deriving their nutrient from another organism (host) and causing the disease to the host are called parasites.
a. Obligate or Total parasite – Completely depends on the host for their survival and produces haustoria.

1. Total stem parasite: The leafless stem twine around the host and produce haustoria. Example: Cuscuta (Dodder), a rootless plant growing on Zizyphus, Citrus and so on.
2. Total root parasite: They do not have stem axis and grow in the roots of host plants to produce haustoria. Example: Rafflesia, Orobanche and Balanophora.

b. Partial parasite – Plants of this group contain chlorophyll and synthesize carbohydrates. Water and mineral requirements are dependent on a host plant.

1. Partial Stem Parasite: Example: Loranthus and Viscum (Mistletoe) Loranthus grows on fig and mango trees and absorb water and minerals from the xylem.
2. Partial root parasite: Example: Santalum album (Sandalwood tree) in its juvenile stage produces haustoria which grow on roots of many plants.

Higher Order Thinking Skills (HOTs)

Question 1.
In metro cities like Chennai, there is a lack of space for a garden. Suggest a technique by which plants can be grown without the need for soil and land space.
Answer:
The technique which can be used is hydroponics in which nutrient solution is used for growing plants.

Question 2.
Carnivorous plants like Nepenthes and venus fly trap have nutritional adaptation. Which nutrient do they especially obtain and from where?
Answer:
Plants like nepenthes and venus fly trap etc. are found grouping in nitrogen-deficient soil. To compensate, their body parts are modified to trap insects from which they obtain sufficient nitrogen for their survival. Hence, they are carnivorous in nature.

Question 3.
Excess of manganese in the soil leads to deficiency of Mg, Fe and Ca. Justify.
Answer:
Increased concentration of Manganese will prevent the uptake of Fe and Mg and prevent translocation of Ca to the shoot apex and cause their deficiency. This condition is referred to as manganese toxicity.

Question 4.
Name the crucial enzyme found in the root nodules of leguminous plants for nitrogen fixation. Also, name the pigment which is highly essential for the activation of the enzyme.
Answer:
Root nodules of legume plants contained nitrogenase enzyme for nitrogen fixation.
Leghaemoglobin is the pink coloured pigment that creates an anaerobic condition for the activation of the nitrogenase complex.

Question 5.
Plants require nutrients. If we supply these in excess will they be beneficial to plants? If yes, how? If no, why?
Answer:
No, an excess supply of nutrients is not advised for the healthier growth of plants. Due to the fact that increased concentration of nutrients may lead to the toxicity of particular minerals which prevents the uptake of other minerals and also affect various other metabolic activities of the cells, leading to the poor growth of the plant.

Question 6.
X is a primitive, eukaryotic chlorophyllous organism and Y is a eukaryotic, achlorophyllous organism. Both X and Y live mutually together benefitting each other.
(a) Name this mutual association.
(b) What X and Y are?
(c) Explain their relationship?
Answer:
(a) Lichens.
(b) X = Algae, Y = fungi
(c) In lichens, Algae prepare food and fungi absorbs water and provides thallus structure.

Choose the correct answer.

1. Identify the micronutrient.
(a) Potassium
(b) Phosphorous
(c) Manganese
(d) Magnesium
Answer:
(c) Manganese

2. Which mineral is essential for cell wall formation in Equisetuml?
(a) Boron
(b) Silicon
(c) Cellulose
(d) Carbon
Answer:
(b) Silicon

3. ‘Law of Minimum’ was proposed by ……………
(a) Van Helmont
(b) Von Sachs
(c) Wood word
(d) Liebig
Answer:
(d) Liebig

4. Identify the wrong statement(s).
(i) Molybdenum is a micronutrient.
(ii) Carbon, Hydrogen, Nitrogen are skeletal elements.
(iii) Manganese is the activator for RUBP carboxylase.
(iv) Potassium maintains osmotic potential of the cell.
(a) (i) and (iv)
(b) (ii) and (iii)
(c) Only (ii)
(d) Only (iv)
Answer:
(b) (ii) and (iii)

5. Siderophores are ……….. chelators.
(a) B
(b) Fe³⁺
(c) Ca²⁺
(d) Cl^–
Answer:
(b) Fe³⁺

6. Match the following:
(i) Boron                            1. Nitrogen metabolism
(ii) Molybdenum               2. Formation of Porphyrin
(iii) Zinc                             3. Translocation of sugars
(iv) Iron                             4. Biosynthesis of auxin
(a) (i) – 3, (ii) – 1, (iii) – 4, (iv) – 2
(b) (i) – 2, (ii) – 3, (iii) – 4, (iv) – 1
(c) (i) – 3, (ii) – 4, (iii) – 2, (iv) – 1
(d) (i) – 4, (ii) – 1, (iii) – 2, (iv) – 3
Answer:
(a) (i) – 3, (ii) – 1, (iii) – 4, (iv) – 2

7. Exanthema in Citrus is due to ………… deficiency.
(a) Mo
(b) Cu
(c) B
(d) Zn
Answer:
(b) Cu

8. Mottle leaf disease is a ………… deficiency disease.
(a) Gibberellins
(b) Cytokinin
(c) Auxin
(d) Ethylene
Answer:
(c) Auxin

9. Pollen germination requires ………… mineral.
(a) Copper
(b) Molybdenum
(c) Chlorine
(d) Boron
Answer:
(d) Boron

10. ………… toxicity leads to precipitation of nucleic acids.
(a) Manganese
(b) Iron
(c) Aluminium
(d) Boron
Answer:
(c) Aluminium

11. The term soilless culture refers to …………
(a) Aeroponics
(b) Aquaculture
(c) Hydroponics
(d) Drip irrigation
Answer:
(c) Hydroponics

12. The term hydroponics was coined by …………..
(a) Amon
(b) David Durger
(c) Liebig
(d) Goerick
Answer:
(d) Goerick

13. Which of the following is a free-living bacterium?
(a) Rhizobium
(b) Clostridium
(c) Escherichia
(d) Cyanobacteria
Answer:
(b) Clostridium

14. Legume plants secrete………… which attracts Rhizobium.
(a) Toluenes
(b) Phenolics
(c) Octanes
(d) Xylenes
Answer:
(b) Phenolics

15. In N₂, how many bonds are there between two nitrogen atoms?
(a) Two
(b) Four
(c) Three
(d) One
Answer:
(c) Three

16. Leghaemoglobin pigments removes ……….. to activate nitrogenase enzyme.
(a) CO
(b) CO₂
(c) O
(d) N
Answer:
(c) O

17. Which bacterium is NOT involved in Denitrification?
(a) Pseudomonas
(b) Thiobacillus
(c) Bacillus subtilis
(d) Nitrobacter
Answer:
(d) Nitrobacter

18. On reaction with a-ketoglutaric acid, ammonia yields ………….
(a) Glutamate
(b) Malate
(c) Glutamine
(d) Oxoglutarate
Answer:
(a) Glutamate

19. Pyridoxal phosphate is a derivative of vitamin ……………
(a) A
(b) B₁₂
(c) B₆
(d) D
Answer:
(c) B₆

20. Identify the correct pair:
(i) Cuscuta – Stem parasite
(ii) Viscum – Total parasite
(Hi) Dionaea – Saprophyte
(iv) Lichen – Fungi and bacteria
(a) (i) only
(b) (ii) only
(c) Both (i) and (ii)
(d) Both (iii) and (iv)
Answer:
(a) (i) only

21. Assertion: Loranthus is a partial parasite.
Reason: Partial parasite depends on the host for water & minerals only.
(a) A is right, R is wrong
(b) Both A and R are correct
(c) A is right R, explains A
(d) A is the wrong R, explains A
Answer:
(c) A is right R, explain A

22. Pitcher plant is the common name for …………
(a) Drosera
(b) Nepenthes
(c) Utricularia
(d) Dionaea
Answer:
(b) Nepenthes

23. Insectivorous plants are usually seen in ………… deficient soil.
(a) Sulphur
(b) Nitrogen
(c) Carbon
(d) Potassium
Answer:
(b) Nitrogen

24. ………… is a pioneer species in xeric succession.
(a) Lichens
(b) Mycorrhizae
(c) Cyanobacteria
(d) Pteridophyte
Answer:
(a) Lichens

Students get through the TN Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants which is useful for their exam preparation.

TN State Board 11th Bio Botany Important Questions Chapter 11 Transport in Plants

Very short answer questions

Question 1.
Define transport in plants. Mention the tissues involved in transportation.
Answer:
Transport is the process of moving water, minerals, and food to all parts of the plant body. Conducting tissues such as xylem and phloem play an important role in transport.

Question 2.
What is the need for transport in plants?
Answer:
Water absorbed from roots must travel up to leaves by xylem for food preparation by photosynthesis. Likewise, food prepared from leaves has to travel to all parts of the plant including roots for growth and other processes.

Question 3.
Compare Active transport with Passive transport.
Answer:

Question 4.
Define the term diffusion.
Answer:
The net movement of molecules from a region of their higher concentration to a region of their lower concentration along a concentration gradient until an equilibrium is attained.

Question 5.
How the diffusing molecules will move?
Answer:
In diffusion, the movement of molecules is continuous and random in order in all directions.

Question 6.
Name the transport proteins of the plasma membrane involved in facilitated diffusion.
Answer:
Channel protein and carrier protein.

Question 7.
Give a brief account of Porin.
Answer:
Porin is a large transporter protein found in the outer membrane of plastids, mitochondria, and bacteria which facilitates smaller molecules to pass through the membrane.

Question 8.
State the role of Channel Protein.
Answer:
Channel protein forms a channel or tunnel in the cell membrane for the easy passage of molecules to enter the cell. The channels are either open or remain closed.

Question 9.
Apart from water, what are the substrates that are transported through aquaporins?
Answer:
Glycerol, Urea, CO₂, NH₃, Metalloids, and Reactive Oxygen Species (ROS).

Question 10.
How does a carrier protein function?
Answer:
Carrier protein acts as a vehicle to carry molecules from outside of the membrane to inside the cell and vice versa. Due to the association with molecules to be transported, the structure of carrier protein gets modified until the dissociation of the molecules.

Question 11.
On which basis, the carrier proteins are classified? Mention its types.
Answer:
There are three types of carrier proteins on the basis of handling of molecules and direction of transport. They are,

1. Uniport,
2. Symport and
3. Antiport.

Question 12.
Mention the drawbacks of diffusion.
Answer:
In diffusion, there is a lack of control over the transport of selective molecules. There is a possibility of harmful substances entering the cell by a concentration gradient.

Question 13.
Co-transport and counter transport differ from each other. Justify.
Answer:
In co-transport, two molecules are transported together in the same direction whereas, in counter¬transport two molecules are transported in opposite directions to each other.

Question 14.
State any two vital roles of water in plants.
Answer:

1. Water maintains the internal temperature of the plant.
2. It maintains the turgidity of the cell.

Question 15.
List out a few imbibing.
Answer:
Gum, starch, proteins, cellulose, agar, and gelatin.

Question 16.
Define Imbibition.
Answer:
The phenomenon in which a substance uptake the water and swell up is called Imbibition. The substance is referred to as Imbibant. Example: Swelling of water-soaked seeds.

Question 17.
List out the substances that are transported by facilitated diffusion.
Answer:
Sugars, amino acids, nucleotides, ions, and cell metabolites.

Question 18.
How imbibition, is important for plants?
Answer:
Significance of imbibition:

1. During the germination of seeds, imbibition increases the volume of the seed enormously and leads to the bursting of the seed coat.
2. It helps in the absorption of water by roots at the initial level.

Question 19.
Define Water potential.
Answer:
Water potential is the potential energy of water in a system compared to pure water when both temperature and pressure are kept the same. Water potential is denoted by the symbol ¥ (psi).

Question 20.
Define Osmotic Pressure.
Answer:
When a solution and its solvent (pure water) are separated by a semipermeable membrane, pressure is developed in the solution, due to the presence of dissolved solutes. This is called osmotic pressure (OP).

Question 21.
Mention the symbolic representation of water potential and osmotic pressure.
Answer:
Water potential – ψ (psi) and Osmotic pressure – π (pi).

Question 22.
Mention the alternate terminologies and symbolic representation of solute potential and Matric potential.
Answer:

Question 23.
Expand and Define TP.
Answer:
TP stands for Turgor pressure. When a plant cell is placed in pure water (hypotonic solution) the diffusion of water into the cell takes place by endosmosis. It creates a positive hydrostatic pressure on the rigid cell wall by the cell membrane. Henceforth the pressure exerted by the cell membrane towards the cell wall is Turgor Pressure (TP).

Question 24.
Mention the pressures that act upon the cell to make it a full turgid.
Answer:
Turgor pressure and wall pressure make the cell fully turgid. TP + WP = Turgid.

Question 25.
Define Diffusion Pressure Deficit (DPD).
Answer:
The difference between the diffusion pressure of the solution and its solvent at a particular temperature and atmospheric pressure is called Diffusion Pressure Deficit (DPD).

Question 26.
Why DPD is also called suction pressure?
Answer:
Increased DPD favors endosmosis or it sucks the water from the hypotonic solution, hence it is called suction pressure.

Question 27.
Define Osmosis.
Answer:
Osmosis represents the movement of water or solvent molecules through a selectively permeable membrane from the place of its higher concentration (high water potential) to the place of its’ lower concentration (low water potential).

Question 28.
State the significance of plasmolysis.
Answer:
Plasmolysis is exhibited only by living cells and so it is used to test whether the cell is living or dead.

Question 29.
What happens if a plant cell is treated with a hypertonic solution?
Answer:
When a plant cell is kept in a hypertonic solution, water leaves the cell due to exosmosis. As a result of water loss, protoplasm shrinks and the cell membrane is pulled away from the cell wall and finally, the cell becomes flaccid. This process is named plasmolysis.

Question 30.
Give an example for plasmolysis and also mention the types of plasmolysis.
Answer:
Wilting of plants noticed under the condition of water scarcity is an indication of plasmolysis. Three types of plasmolysis occur in plants: /) Incipient plasmolysis, ii) Evident plasmolysis, and iii) Final plasmolysis.

Question 31.
Draw a simplified diagram depicting Reverse Osmosis.
Answer:

Question 32.
If a cell in the cortex with DPD of 5 atm is surrounded by hypodermal cells with DPD of 2 atm, what will be the direction of movement of water?
Answer:
Water will move from low DPD to high DPD (hypodermis 2 atm to cortex 5 atm).

Question 33.
Arrange in the correct sequence in concern with the pathway of water, in roots, (cortex, root hair, xylem, epidermal cell, endodermis, and pericycle).
Answer:
Root hairs → epidermal cell → cortex → endodermis → pericycle → xylem.

Question 34.
Mention the possible routes of water movement across root cells.
Answer:
There are three possible routes of water.
They are i) Apoplast, ii) Symplast and iii) Transmembrane route.

Question 35.
What are the objections to the osmotic active absorption theory?
Answer:
Objections to osmotic theory: (i) The cell sap concentration in the xylem is not always high. (ii) Root pressure is not universal in all plants, especially in trees.

Question 36.
Name any two respiratory inhibitors.
Answer:
Potassium cyanide (KCN) and chloroform.

Question 37.
Define Ascent of sap.
Answer:
The water within the xylem along with dissolved minerals from roots is called sap and its upward transport is called ascent of sap.

Question 38.
State Relay Pump theory.
Answer:
Relay pump theory of Godlewski (1884)
Periodic changes in osmotic pressure of living cells of the xylem parenchyma and medullary ray act as a pump for the movement of water.

Question 39.
What do you mean by the term ‘Embolism”?
Answer:
Gas bubbles expanding and displacing water within the xylem element is called cavitation or embolism. .

Question 40.
Which is the most widely accepted theory to prove the Ascent of sap? Who proposed it.
Answer:
Cohesion-tension theory or Transpiration pull theory proposed by Dixon and Jolly (1894- 1924).

Question 41.
Define Transpiration.
Answer:
The loss of excess of water in the form of vapor from various aerial parts of the plant is called transpiration.

Question 42.
State any two theories regarding the mechanism of stomatal movement.
Answer:
Starch-sugar interconversion theory and Active potassium transport ion concept.

Question 43.
Draw and label the structure of stomata.
Answer:

Question 44.
What is an anti transpirant?
Answer:
The term antitranspirant is used to designate any material applied to plants for the purpose of retarding transpiration. An ideal antitranspirant checks the transpiration process without disturbing the process of gaseous exchange.

Question 45.
List out the uses of antitranspirants.
Answer:

1. Antitranspirants reduce the enormous loss of water by transpiration in crop plants.
2. Useful for seedling transplantations in nurseries.

Question 46.
What are hydathodes?
Answer:
Hydathodes are the stomata-like pores present in plants that grow in moist and shady places.

Question 47.
Define the term “Translocation of organic solutes”.
Answer:
The phenomenon of food transportation from the site of synthesis to the site of utilization is known as the translocation of organic solutes. The term solute denotes food material that moves in a solution.

Question 48.
What do you mean by Phloem loading?
Answer:
The movement of photosynthates (products of photosynthesis) from mesophyll cells to phloem sieve elements of mature leaves is known as phloem loading.

Question 49.
List the merits of the Munch-Mass Flow hypothesis.
Answer:

1. When a woody or herbaceous plant is girdled, the sap contains high sugar-containing exudates from the cut end.
2. A positive concentration gradient disappears when plants are defoliated.

Question 50.
Define – Flux and its types.
Answer:
The movement of ions into and out of cells or tissues is termed transport or flux. Entry of the ion into the cell is called influx and exit is called efflux.

Short answer questions

Question 1.
Draw a Flow Chart illustrating various types of cell-to-cell transport.
Answer:

Question 2.
Enumerate the significance of diffusion.
Answer:
Significance of diffusion in Plants

1. Gaseous exchange of O₂ and CO₂ between the atmosphere and stomata of leaves takes place by the process of diffusion. O₂ is absorbed during respiration and CO₂ is absorbed during photosynthesis.
2. In transpiration, water vapour from intercellular spaces diffuses into the atmosphere through stomata by the process of diffusion.
3. The transport of ions in mineral salts during passive absorption also takes place by this process.

Question 3.
How semipermeable and selectively permeable membranes differ from each other?
Answer:

Question 4.
Give an account on Aquaporin.
Answer:
Aquaporin is a water channel protein embedded in the plasma membrane. It regulates the massive amount of water transport across the membrane. Plants contain a variety of aquaporins. Over 30 types of aquaporins are known from maize. Currently, they are also recognized to transport substrates like glycerol, urea, CO₂, NH₃, metalloids, and Reactive Oxygen Species (ROS) in addition to water. They increase the permeability of the membrane to water. They confer drought and salt stress tolerance.

Question 5.
Give an account of Matric’s potential.
Answer:
Matric potential represents the attraction between water and the hydrating colloid or gel-like organic molecules in the cell wall which is collectively termed as matric potential. Matric potential is also known as imbibition pressure. The matric potential is maximum (most negative value) in dry material. Example: The swelling of soaked seeds in water.

Question 6.
How DPD differs in various conditions of a cell?
Answer:

• DPD in normal cell: DPD = OP – TP.
• DPD in the fully turgid cell: Osmotic pressure is always equal to turgor pressure in a fully turgid cell.
• OP = TP or OP-TP =0. Hence DPD of fully turgid cell is zero.
• DPD in the flaccid cell: If the cell is in flaccid condition there is no turgor pressure or TP=0.
  Hence DPD = OP.

Question 7.
Compare Hypertonic, Hypotonic, and Isotonic solutions.
Answer:

Question 8.
Give an account of Endosmosis and Exosmosis.
Answer:

1. Endosmosis: Endosmosis is defined as the osmotic entry of solvent into a cell or a system when it is placed in pure water or hypotonic solution. For example, dry raisins placed in the water, it swells up due to turgidity.
2. Exosmosis: Exosmosis is defined as the osmotic withdrawal of water from a cell or system when it is placed in a hypertonic solution. Exosmosis in a plant cell leads to plasmolysis.

Question 9.
Give an account of Deplasmolysis.
Answer:
The effect of plasmolysis can be reversed, by transferring them back into the water or hypotonic solution. Due to endosmosis, the cell becomes turgid again. It regains its original shape and size. This phenomenon of the revival of the plasmolyzed cell is called deplasmolysis. Example: Immersion of dry raisin in water.

Question 10.
Give an account of Reverse Osmosis and its uses.
Answer:
Reverse Osmosis follows the same principles of osmosis, but in the reverse direction. In this process movement of water is reversed by applying pressure to force the water against a concentration gradient of the solution. In regular osmosis, the water molecules move from the higher concentration (pure water = hypotonic) to the lower concentration (saltwater = hypertonic). But in reverse osmosis, the water molecules move from the lower concentration (saltwater = hypertonic) to higher concentration (pure water = hypotonic) through a selectively permeable membrane.
Uses: Reverse osmosis is used for the purification of drinking water and desalination of seawater.

Question 11.
Differentiate between the types of Plasmolysis.
Answer:
Difference between plasmolysis types.

Question 12.
Write in brief about Non-Osmotic active absorption.
Answer:
Bennet-Clark (1936), Thimann (1951), and Kramer (1959) observed absorption of water even if the concentration of cell sap in the root hair is lower than that of the soil water. Such a movement requires an expenditure of energy released by respiration (ATP). Thus, there is a link between water absorption and respiration. It is evident from the fact that when respiratory inhibitors like KCN, Chloroform are applied there is a decrease in the rate of respiration and also the rate of absorption of water.

Question 13.
Explain the pulsation theory of J.C. Bose.
Answer:
Bose invented an instrument called the Crescograph, which consists of an electric probe connected to a galvanometer. When a probe is inserted into the inner cortex of the stem, the galvanometer showed high electrical activity. Bose believed a rhythmic pulsating movement of the inner cortex like a pump (similar to the beating of the heart) is responsible for the ascent of sap. He concluded that cells associated with xylem exhibit pumping action and pumps the sap laterally into xylem cells.

Question 14.
List out the setbacks of Root Pressure Theory.
Answer:
The following objections have been raised against root pressure theory:

1. Root pressure is totally absent in gymnosperms, which includes some of the tallest plants.
2. There is no relationship between the ascent of sap and root pressure. For example, in summer, the rate of the ascent of sap is more due to transpiration in spite of the fact that root pressure is very low. On the other hand, in winter when the rate of ascent of sap is low, a.high root pressure is found.
3. The ascent of sap continues even in the absence of roots.
4. The magnitude of root pressure is about 2atm, which can raise the water level up to few feet only, whereas the tallest trees are more than 100m high.

Question 15.
Describe the structure of stomata.
Answer:
The epidermis of leaves and green stems possess many small pores called stomata. The length and breadth of stomata are about 10-40 p and 3 – 10 p respectively. Mature leaves contain between 50 and 500 stomata per mm2. Stomata are made up of two guard cells, special semilunar or kidney-shaped living epidermal cells in the epidermis. Guard cells are attached to surrounding epidermal cells known as subsidiary cells or accessory cells. The guard cells are joined together at each end but they are free to separate to form a pore between them. The inner wall of the guard cell is thicker than the outer wall. The stoma opens to the interior into a cavity called the sub-stomatal cavity which remains connected with the intercellular spaces.

Question 16.
Explain the theory of photosynthesis in guard cells and state its demerits.
Answer:
Theory of Photosynthesis in guard cells:
Von Mohl (1856) observed that stomata open in light and close in the night. According to him, chloroplasts present in the guard cells photosynthesize in the presence of light resulting ‘ in the production of carbohydrate (Sugar) which increases osmotic pressure in guard cells.
It leads to the entry of water from other cells and the stomatal aperture opens. The above process vice versa at the night lead to the closure of stomata.

Demerits

1. The chloroplast of guard cells is poorly developed and incapable of performing photosynthesis.
2. The guard cells already possess much amount of stored sugars.

Question 17.
Draw a diagrammatic representation of the steward scheme of stomatal movement.
Answer:

Question 18.
Why wilting occurs? Explain its types.
Answer:
Excessive loss of water through transpiration leads to wilting. In general, there are three types of wilting as follows,
a. Incipient wilting: The water content of plant cells decreases but the symptoms are not visible.
b. Temporary wilting: On hot summer days, the freshness of herbaceous plants reduces turgor pressure in the daytime and regains it at night.
c. Permanent wilting: The absorption of water virtually ceases because the plant cell does not get water from any source and the plant cell passes into a state of permanent wilting.

Question 19.
Give an account of types of structural modifications of leaves for reducing transpirational – loss.
Answer:
Leaf structure: Some anatomical features of leaves like sunken stomata, the presence of hairs, cuticle, the presence of hydrophilic substances like gum, mucilage help to reduce the rate of transpiration. In xerophytes the structural modifications are remarkable. To avoid transpiration, as in Opuntia the stem is flattened to look like leaves called Phylloclade. Cladode or cladophyll in Asparagus is a modified stem capable of limited growth looking like leaves. In some plants, the petioles are flattened and widened, to become phyllodes example Acacia melanoxylon.

Question 20.
Comment on various chemicals inducing stomatal closure.
Answer:
Carbon-di-oxide induces stomatal closure and acts as a natural antitranspirant. Further, the advantage of using CO₂ as an antitranspirant is its inhibition of photorespiration. Phenyl Mercuric Acetate (PMA), when applied as a foliar spray to plants, induces partial stomatal closure for two weeks or more without any toxic effect. The use of abscisic acid highly induces the closing of stomata. Dodecenyl succinic acid also affects on stomatal closure.

Question 21.
Classify translocation based on direction.
Answer:
Phloem translocates the products of photosynthesis from leaves to the area of growth and storage, in the following directions,
Downward direction: From leaves to stem and roots.
Upward direction: From leaves to developing buds, flowers, fruits for consumption and storage. Germination of seeds is also a good example of upward translocation.
Radial direction: From cells of the pith to cortex and epidermis, the food materials are radially translocated.

Question 22.
What do you understand by the term source and sink in plant physiology?
Answer:
The source is defined as any organ in plants which are capable of exporting food materials to the areas of metabolism or to the areas of storage. Examples: Mature leaves and germinating seeds.

A sink is defined as any organ in plants that receives food from a source. Example: Roots, tubers, developing fruits, and immature leaves.

Question 23.
Define Phloem unloading with its steps.
Answer:
From sieve elements, sucrose is translocated into sink organs such as roots, tubers, flowers, and fruits, and this process is termed phloem unloading. It consists of three steps:

1. sieve element unloading: Sucrose leave from sieve elements.
2. Short distance transport: Movement of sucrose to sink cells.
3. Storage and metabolism: The final step when sugars are stored or metabolized in sink cells.

Question 24.
Enumerate the objections of the Munchi Mass Flow hypothesis.
Answer:

1. Munchi Mass Flow hypothesis explains the unidirectional movement of solute only. However, bidirectional movement of solute is commonly observed in plants.
2. The osmotic pressure of mesophyll cells and that of root hair do not confirm the requirements.
3. This theory gives a passive role to sieve tube and protoplasm, while some workers demonstrated the involvement of ATP.

Question 25.
Donnan equilibrium. Add a note.
Answer:
Within the cell, some of the ions never diffuse out through the membrane. They are trapped within the cell and are called fixed ions. But they must be balanced by the ions of opposite charge. Assuming that a concentration of fixed anions is present inside the membrane, more cations would be absorbed in addition to the normal exchange to maintain the equilibrium. Therefore, the cation concentration would be greater in the internal than in the external solution. This electrical balance or equilibrium controlled by electrical as well as diffusion phenomenon is known as the Donnan equilibrium.

Long answer questions

Question 1.
Write in detail about types of carrier protein.
Answer:
There are three types of carrier proteins classified on the basis of handling of molecules and the direction of transport. They are i. Uniport, ii. Symport, iii. Antiport.

1. Uniport: In this molecule of a single type move across a membrane independent of other molecules in one direction.
2. Symport or co-transport: The term symport is used to denote an integral membrane protein that simultaneously transports two types of molecules across the membrane in the same direction.
3. Antiport or Counter Transport: An antiport is an integral membrane transport protein that simultaneously transports two different molecules, in opposite directions, across the membrane.

Question 2.
Write a note on Solute potential and Pressure Potential.
Answer:
1. Solute Potential (ψ_S): Solute potential, otherwise known as osmotic potential denotes the effect of dissolved solute on water potential. In pure water, the addition of solute reduces its free energy and lowers the water potential value from zero to negative. Thus the value of solute potential is always negative. In a solution at standard atmospheric pressure, water potential is always equal to solute potential (ψ_W = ψ_S).
2. Pressure Potential (ψ_P): Pressure potential is a mechanical force working against the effect of solute potential. Increased pressure potential will increase water potential and water enters cells and cells become turgid. This positive hydrostatic pressure within the cell is called Turgor pressure. Likewise, withdrawal of water from the cell decreases the water potential and the cell becomes flaccid.

Question 3.
With the help of a diagram explain the possible route of water across root cells.
Answer:
There are three possible routes of water. They are i) Apoplast, ii) Symplast and iii) Transmembrane route.

1. Apoplast: The apoplast (Greek: apo = away; plast = cell) consists of everything external to the plasma membrane of the living cell. The apoplast includes cell walls, extracellular spaces, and the interior of dead cells such as vessel elements and tracheids. In the apoplast pathway, water moves exclusively through the cell wall or the non-living part of the plant without crossing any membrane. The apoplast is a continuous system.
   
2. Symplast: The symplast (Greek: sym = within; plast = cell) consists of the entire mass of cytosol of all the living cells in a plant, as well as the plasmodesmata, the cytoplasmic channel that interconnects them. In the symplastic route, water has to cross the plasma membrane to enter the cytoplasm of the outer root cell; then it will move within adjoining cytoplasm through plasmodesmata around the vacuoles without the necessity to cross more membrane, till it reaches xylem.
3. Transmembrane route: In the transmembrane pathway water sequentially enters a cell on one side and exits from the cell on the other side. In this pathway, water crosses at least two membranes for each cell. Transport across the tonoplast is also involved.

Mechanism of Water Absorption Kramer (1949) recognized two distinct mechanisms which independently operate in the absorption of water in plants. They are, i) active absorption ii) passive absorption.

Question 4.
Point out the differences between Active absorption and Passive absorption.
Answer:

Question 5.
Explain about Cohesion – Tension Theory.
Answer:
Cohesion-tension theory was originally proposed by Dixon and Jolly (1894) and again put forward by Dixon (1914, 1924). This theory is based on the following features:

1. Strong cohesive force or tensile strength of water: Water molecules have the strong mutual force of attraction called cohesive force due to which they cannot be easily separated from one another. Further, the attraction between a water molecule and the wall of the xylem element is called adhesion. These cohesive and adhesive force works together to form an unbroken continuous water column in the xylem. The magnitude of the cohesive force is much high (350 atm) and is more than enough to ascent sap in the tallest trees.
2. Continuity of the water column in the plant: An important factor that can break the water column is the introduction of air bubbles in the xylem. Gas bubbles expanding and displacing water within the xylem element is called cavitation or embolism. However, the overall continuity of the water column remains undisturbed since water diffuses into the adjacent xylem elements for continuing ascent of sap.
3. Transpiration pull or Tension in the unbroken water column: The unbroken water column from leaf to root is just like a rope. If the rope is pulled from the top, the entire rope will move upward. In plants, such a pull is generated by the process of transpiration which is known as transpiration pull. Water vapour evaporates from mesophyll cells to the intercellular spaces near stomata as a result of active transpiration. The water vapours are then transpired through the stomatal pores. Loss of water from mesophyll cells causes a decrease in water potential. So, water moves as a pull from cell to cell along the water potential gradient. This tension, generated at the top (leaf) of the unbroken water column, is transmitted downwards from the petiole, stem and finally reaches the roots. The cohesion theory is the most accepted among plant physiologists today.

Question 6.
Describe the types of transpiration in plants.
Answer:
Types of Transpiration: Transpiration is of the following three types:

1. Stomatal transpiration: Stomata are microscopic structures present in high numbers on the lower epidermis of leaves. This is the most dominant form of transpiration and being responsible for most of the water loss (90 – 95%) in plants.
2. Lenticular transpiration: In stems of woody plants and trees, the epidermis is replaced by periderm because of secondary growth. In order to provide gaseous exchange between the living cells and the outer atmosphere, some pores which look like lens-shaped raised spots are present on the surface of the stem called Lenticels. The loss of water from lenticels is very insignificant as it amounts to only 0.1 % of the total.
3. Cuticular transpiration: The cuticle is a waxy or resinous layer of cutin, a fatty substance covering the epidermis of leaves and other plant parts. Loss of water through the cuticle is relatively small and it is only about 5 to 10 % of the total transpiration. The thickness of cuticle increases in xerophytes and transpiration is very much reduced or totally absent.

Question 7.
Give an account of Starch – Sugar Interconversion Theory.
Answer:

1. According to Lloyd (1908), the turgidity of guard cells depends on the interconversion, of starch and sugar. It was supported by Loftfield (1921) as he found guard cells containing sugar during the daytime when they are open and starch during the night when they are closed.
2. Sayre (1920) observed that the opening and closing of stomata depend upon the change in pH of guard cells. According to him stomata open at high pH during the daytime and become closed at low pH at night. The utilization of CO₂ by photosynthesis during the light period causes an increase in pH resulting in the conversion of starch to sugar. Sugar increase in cell favors endosmosis and increases the turgor pressure which leads to opening of stomata. Likewise, the accumulation of CO₂ in cells during the night decreases the pH level resulting in the conversion of sugar to starch. Starch decreases the turgor pressure of the guard cell and stomata close.
3. The discovery of enzyme phosphorylase in guard cells by Hanes (1940) greatly supports the starch-sugar interconversion theory. The enzyme phosphorylase hydrolyses starch into sugar and high pH followed by endosmosis and the opening of stomata during light. The vice versa takes place during the night.
   
4. Steward (1964) proposed a slightly modified scheme of starch-sugar interconversion theory. According to him, Glucose-1-phosphate is osmotically inactive. Removal of phosphate from Glucose-1-phosphate converts to Glucose which is osmotically active and increases the concentration of guard cells leading to the opening of stomata.

Objections to Starch-sugar interconversion theory

1. In monocots, the guard cell does not have starch.
2. There is no evidence to show the presence of sugar at a time when starch disappears and stomata open.
3. It fails to explain the drastic change in pH from 5 to 7 by change of CO₂

Question 8.
Describe the K⁺ Transport theory on transpiration.
Answer:
The theory of K⁺ transport theory was proposed by Levit (1974) and elaborated by Raschke (1975). According to this theory, the following steps are involved in the stomatal opening:

In light

1. In the guard cell, starch is converted into organic acid (malic acid).
2. .Malic acid in the guard cell dissociates to malate anion and proton (H⁺).
3. Protons are transported through the membrane into nearby subsidiary cells with the
   exchange of K⁺ (Potassium ions) from subsidiary cells to guard cells. This process involves an electrical gradient and is called ion exchange.
4. This ion exchange is an active process and consumes ATP for energy.
5. Increased K⁺ ions in the guard cell are balanced by CL ions. An increase in solute concentration decreases the water potential in the guard cell.
6. Guard cell becomes hypertonic and favours the entry of water from surrounding cells.
7. Increased turgor pressure due to the entry of water opens the stomatal pore.

In Dark

1. In the dark photosynthesis stops and respiration continues with the accumulation of CO₂ in the sub-stomatal cavity.
2. Accumulation of CO₂ in the cell lowers the pH level.
3. Low pH and a shortage of water in the guard cell activate the stress hormone Abscisic acid (ABA).
4. ABA stops further entry of K⁺ ions and also induces K⁺ ions to leak out to subsidiary cells from guard cells.
5. Loss of water from the guard cell reduces turgor pressure and causes the closure of stomata.

Question 9.
Draw the structure of hydathode and explain Guttation.
Answer:

During high humidity in the atmosphere, the rate of transpiration is much reduced. When plants absorb water in such a condition root pressure is developed due to excess water within the plant. Thus excess water exudates as a liquid from the edges of the leaves and are called guttation. Example: Grasses, tomato, potato, brinjal, and Alocasia. Guttation occurs through stomata-like pores called hydathodes generally present in plants that grow in moist and shady places. Pores are present over a mass of loosely arranged cells with large intercellular spaces called epithem. This mass of tissue lies near vein endings (xylem and Phloem). The liquid coming out of the hydathode is not pure water but a solution containing a number of dissolved substances.

Question 10.
Explain Ganong’s Potometer experiment and its purpose.
Answer:

Ganongs potometer is used to measure the rate of transpiration indirectly. In this, the amount of water absorbed is measured, and assumed that this amount is equal to the amount of water transpired.
The apparatus consists of a horizontal graduated tube that is bent in opposite directions at the ends. One bent end is wide and the other is narrow. A reservoir is fixed to the horizontal tube near the wider end.
The reservoir has a stopcock to regulate water flow. The apparatus is filled with water from the reservoir. A twig or a small plant is fixed to the wider arm through a split cock. The other bent end of the horizontal tube is dipped into a beaker containing colored water. An air bubble is introduced into the graduated tube at the narrow end keep this apparatus in bright sunlight and observe. As transpiration takes place, the air bubble will move towards the twig. The loss is compensated by water absorption through the xylem portion of the twig. Thus, the rate of water absorption is equal to the rate of transpiration.

Question 11.
Describe the ringing experiment with a diagram.
Answer:
The experiment involves the removal of all the tissue outside to vascular cambium (bark, cortex, and phloem) in woody stems except the xylem. Xylem is the only remaining tissue in the girdled area which connects the upper and lower part of the plant. This setup is placed in a beaker of water. After some time, it is observed that swelling on the upper part of the ring appears as a result of the accumulation of food material. If the experiment continues within days, the roots die first. It is because the supply of food material to the root is cut down by the removal of phloem. The roots cannot synthesize their food and so they die first. As the roots gradually die the upper part (stem), which depends on root for the ascent of sap, will ultimately die.
Ringing experiment.

Question 12.
Write in detail about Passive Absorption of minerals salts.
Answer:
1. Ion-Exchange: Ions of external soil solution were exchanged with same charged (anion for anion or cation for cation) ions of the root cells. There are two theories explaining this process of ion exchange namely: i. Contact exchange and ii. Carbonic acid exchange.

1. Contact Exchange Theory: According to this theory, the ions adsorbed on the surface of root cells and clay particles (or clay micelles) are not held tightly but oscillate within a small volume of space called oscillation volume. Due to the small space, both ions overlap each other’s oscillation volume and exchange takes place.
2. Carbonic Acid Exchange Theory: According to this theory, soil solution plays an important role by acting as a medium for ion exchange. The CO₂ released during respiration of root cells combines with water to form carbonic acid (H₂CO₃). Carbonic acid dissociates into H⁺ and HCO₃^– in the soil solution. These H⁺ ions exchange with cations adsorbed on clay particles and the cations from micelles get released into soil solution and gets adsorbed on root cells.
   
   

Question 13.
Describe Lundegardh’s Cytochrome Pump Theory.
Answer:
Lundegardh’s Cytochrome Pump Theory:
Lundegardh and Burstrom (1933) observed a correlation between respiration and anion absorption. When a plant is transferred from water to a salt solution the rate of respiration increases which is called anion respiration or salt respiration. Based on this observation Lundegardh (1950 and 1954) proposed cytochrome pump theory which is based on the following assumptions:

1. The mechanism of anion and cation absorption is different.
2. Anions are absorbed through the cytochrome chain by an active process, cations are absorbed passively.
3. An oxygen gradient is responsible for oxidation at the outer surface of the membrane and reduction at the inner surface.

According to this theory, the enzyme dehydrogenase on the inner surface is responsible for the formation of protons (H⁺) and electrons (e^–). As electrons pass outward through the electron transport chain there is a corresponding inward passage of anions. Anions are picked up by oxidized cytochrome oxidase and are transferred to other members of the chain as they transfer the electron to the next component.
The theory assumes that cations (C⁺) move passively along the electrical gradient created by the accumulation of anions (A^–) at the inner surface of the membrane.
Main defects of the above theory are:

1. Cations also induce respiration.
2. Fails to explain the selective uptake of ions.
3. It explains absorption of anions only.
   

Question 14.
Explain Protein-Lecithin Theory.
Answer:
Bennet-Clark’s Protein-Lecithin Theory:
In 1956, Bennet-Clark proposed that the carrier could be a protein associated with phosphatide called lecithin. The carrier is amphoteric (the ability to act either as an acid or a base) and hence both cations and anions combine with it to form the Lecithinion complex in the membrane. Inside the membrane, the Lecithin-ion complex is broken down into phosphatidic acid and choline along with the liberation of ions. Lecithin again gets regenerated from phosphatidic acid and choline in the presence of the enzyme choline acetylase and choline esterase. ATP is required for the regeneration of lecithin.

Higher Order Thinking Skills (HOTs)

Question 1.
Why during rainy seasons, the wooden doors and windows are difficult to close and open? Give the phenomenon behind this and also define the phenomenon.
Answer:
The swelling of wooden windows, doors due to high humidity during the rainy season is due, to imbibition. Imbibition refers to the uptake of water and swelling of substances that are not water-soluble.

Question 2.
While making dry fishes at home high salt concentration is applied. Why? Name the phenomenon.
Answer:
The high salt concentration is applied to extract the excess water from the fishes thereby inhibiting the microbial growth and increase the shelf life. This phenomenon is called exosmosis.

Question 3.
Observe the histogram and answer the following question.

(a) Which type of transpiration does ‘A’ and ‘C’ represent?
(b) Name ‘B’ and also define the plant part. Which is responsible for ‘B’ type of transpiration?
Answer:
(a) A – Stomatal transpiration, C – Cuticular transpiration and B – Lenticular transpiration. Lenticels are the lend-shaped raised spots present on the surface of the stem.

Question 4.
Plants are highly adaptable to the environment where they survive Opuntia which lives in xeric condition shows phylloclade adaptation. Which part of Opuntia is modified as phylloclade? Why does it modify?
Answer:
In Opuntia, the stem is flattered to look like leaves called phylloclade. It is modified to avoid transpirational loss of water.

Question 5.
Nowadays, water scarcity is becoming a prime problem. To compensate for the need, various strategies are being carried, out by the Governments at national and international levels. One such effective technology is the Desalination of seawater. Which principle is followed in their technology. Define it.
Answer:
Desalination of seawater is being effectively earned out by Reverse Osmosis. In Reverse osmosis, the water molecules are forced by applying pressure from lower concentration to higher concentration through a selectively permeable membrane.

Question 6.
Observe the diagram, it is a plant cell undergoing plasmolysis.
(a) Which stage of plasmolysis does the cell represent?

(b) Whether it is reversible?
(c) What happens if this occurs in the leaf cells?
Answer:
(a) Evident plasmolysis.
(b) Yes, evident plasmolysis is reversible.
(c) Under evident plasmolysis, wilting of leaves appears.

Choose the correct answer.

1. …………. is a downhill process using physical forces.
(a) Short distance transport
(b) Translocation
(c) Active transport
(d) Passive transport
Answer:
(d) Passive transport

2. The smell of the room spray can be felt everywhere inside a closed room. This is because of ………….
(a) Osmosis
(b) Passive transport
(c) Diffusion
(d) Imbibition
Answer:
(c) Diffusion

3. A mixture of ………… and potassium permanganate is used for fumigation.
(a) Acetaldehyde
(b) Calcium oxide
(c) Formalin
(d) Vinegar
Answer:
(c) Formalin

4. ROS stands for …………
(a) Reduction Oxygen Species
(b) Reactive Oxygen Syndrome
(c) Reducive Oxygen Species
(d) Reactive Oxygen Species
Answer:
(d) Reactive Oxygen Species

5. Peter Agre discovered aquaporin in ………….
(a) RBC
(b) WBC
(c) Platelets
(d) Plasma membrane
Answer:
(a) RBC

6. Protoplasm is made of ……….. of water.
(a) 80-90%
(b) 85-90%
(c) 60-80%
(d) 75-85%
Answer:
(c) 60-80%

7. ………… does not act as an imbibant.
(a) Protein
(b) Starch
(c) Stone
(d) Gum
Answer:
(c) Stone

8. Which physiological process can be observed in a germinating seed?
(a) Diffusion
(b) Osmosis
(c) Imbibition
(d) Facilitated diffusion
Answer:
(c) Imbibition

9. Water potential is measured in ………….
(a) Watt
(b) Joule
(c) Calories
(d) Pascal
Answer:
(d) Pascal

10. Water potential can be determined by solute potential and ………….
(a) Matric potential
(b) Pressure potential
(c) Osmotic potential
(d) Osmotic potential
Answer:
(b) Pressure potential

11. Osmotic pressure is represented by the Greek letter ………….
(a) α
(b) π
(c) ψ
(d) θ
Answer:
(b) π

12. If osmotic pressure has a positive value, the osmotic potential has ………… value.
(a) Positive
(b) Negative
(c) Neutral
(d) Zero
Answer:
(b) Negative

13. Which combination of pressures makes a cell full turgid?
(a) TP + WP
(b) OP -TP
(c) SP + OP
(d) WP + SP
Answer:
(a) TP + WP

14. In a hypertonic solution, which of the following substance concentration will be high?
(a) Solute
(b) Solvent
(c) Both a & b
(d) None
Answer:
(a) Solute

15. Dry raisins kept in the water begin to swell. It is a perfect example for ………….
(a) Osmosis
(b) Plasmolysis
(c) Endosmosis
(d) Final plasmolysis
Answer:
(c) Endosmosis

16. Among the following physiological processes, which one occurs only in the living cells?
(a) Diffusion
(b) Osmosis
(c) Exosmosis
(d) Plasmolysis
Answer:
(d) Plasmolysis

17. To revive a plasmolyzed cell, it should be treated with ………. solution.
(a) Isotonic
(b) Hypertonic
(c) Hypotonic
(d) Neutral
Answer:
(c) Hypotonic

18. Principle behind the desalination of sea water is …………
(a) Endosmosis
(b) Diffusion
(c) Reverse osmosis
(d) Deplasmolysis
Answer:
(c) Reverse osmosis

19. The final destination of water entering the root hair is ………….
(a) Endodermis
(b) Pericycle
(c) Xylem
(d) Cortex
Answer:
(c) Xylem

20. The nature of cell sap is ………….
(a) Hypertonic
(b) Hypotonic
(c) Isotonic
(d) Apoplast
Answer:
(a) Hypertonic

21. Relay pump theory was put forth by ………….
(a) Atkins and Preistley
(b) Strasburger and Overton
(c) Godlewski
(d) J.C. Bose
Answer:
(c) Godlewski

22. Who is called as father of plant physiology?
(a) J.C. Bose
(b) Stephen Hales
(c) Dixon
(d) Unger
Answer:
(b) Stephen Hales

23. An instrument deviced by J.C. Bose for proving the pulsating movement of cortex is ……….
(a) Seismograph
(b) Galvanometer
(c) Crescograph
(d) Radiograph
Answer:
(c) Crescograph

24. In embolism of xylem, water is displaced by …………
(a) Callose
(b) Photosynthates
(c) Gas bubbles
(d) Porin proteins
Answer:
(c) Gas bubbles

25. Transpiration is a kind of ………. occur through plant body.
(a) Condensation
(b) Sublimation
(c) Evaporation
(d) Percipitation
Answer:
(c) Evaporation

26. Rate of water movement through xylem is …………
(a) 65 cm/second
(b) 75 cm/min
(c) 12 cm/min
(d) 82 cm/sec
Answer:
(b) 75 cm/min

27. ………… is a fatty substance covering the epidermis of leaves.
(a) Mucin
(b) Cutin
(c) Porin
(d) Suberin
Answer:
(b) Cutin

28. Approximately, a corn plant transpires ……….. litres of H₂O per day.
(a) 2
(b) 8
(c) 16
(d) 450
Answer:
(a) 2

29. Starch – sugar interconversion theory was proposed by ………..
(a) Loft field
(b) Lloyd
(c) Von Mohl
(d) Hanes
Answer:
(a) Lloyd

30. ABA stands for ……..
(a) Abscisic acid
(b) Ascorbic acid
(c) Acetyl Butyric Acid
(d) Acetic acid
Answer:
(a) Abscisic acid

31. Pick out the natural antitranspirant.
(a) O₂
(b) SO₂
(c) CO₂
(d) CO
Answer:
(c) CO₂

32. Guttation occurs through …………
(a) Stoma
(b) Epithem
(c) Hydathodes
(d) Epidermis
Answer:
(c) Hydathodes

33. CO₂ inhibits …………
(a) Transpiration
(b) Photorespiration
(c) Both a and b
(d) Respiration
Answer:
(c) Both a and b

34. Match the following:

(a) 1 – (iii), 2 – (i), 3 – (iv), 4 – (ii)
(b) 1 – (i), 2 – (ii), 3 – (i), 4 – (iii)
(c) 1 – (ii), 2 – (i), 3 – (iv), 4 – (iii)
(d) 1 – (i), 2 – (iii), 3 – (ii), 4 – (iv)
Answer:
(a) 1 – (iii), 2 – (i), 3 – (iv), 4 – (ii)

35. …………. is used to measure the rate of transpiration.
(a) Ganongs respiroscope
(b) Ganongs potometer
(c) Arc auxanometer
(d) None of these
Answer:
(b) Ganongs potometer

36. Electro-Osmotic theory was propounded by ……….
(a) Mason and Masked
(b) Fenson and Spanner
(c) Curtis
(d) Levit
Answer:
(b) Fenson and Spanner

37. Translocation of photosynthetic products is a ……… movement.
(a) unidirectional
(b) upward
(c) downward
(d) multi-directional
Answer:
(d) multidirectional

Students get through the TN Board 11th Bio Botany Important Questions Chapter 10 Secondary Growth which is useful for their exam preparation.

TN State Board 11th Bio Botany Important Questions Chapter 10 Secondary Growth

Very short answer questions

Question 1.
Which meristem is responsible for secondary growth in dicots? Mention its types.
Answer:
Secondary growth in dicots is the responsibility of lateral meristems. They are vascular cambium and cork cambium.

Question 2.
Define vascular cambial ring.
Answer:
The interfascicular cambium joins with the intrafascicular cambium on both sides to form a continuous ring. It is called a vascular cambial ring.

Question 3.
Name the two types of initials developed by vascular cambium.
Answer:
Fusiform initials and ray initials.

Question 4.
What happens to the primary xylem and phloem during secondary growth?
Answer:
Due to the continued formation of secondary xylem and phloem through vascular cambial activity, both the primary xylem and phloem get gradually crushed.

Question 5.
Differentiate between porous and non-porous wood.
Answer:

Question 6.
State the types of the lateral meristem, their role in woody plants.
Answer:

Question 7.
Springwood – comment.
Answer:
In the spring season, cambium is very active and produces a large number of xylary elements having vessels/tracheids with a wide lumen. The wood formed during this season is called springwood or earlywood.

Question 8.
Define the term Dendroclimatoiogy.
Answer:
Dendroclimatology is a branch of dendrochronology concerned with constructing records of past climates and climatic events by analysis of tree growth characteristics, especially growth rings.

Question 9.
Which study does dendrochronology deal with?
Answer:
The determination of the age of a tree by counting the annual rings is called dendrochronology.

Question 10.
What do you mean by the term “Autumn Wood”?
Answer:
In winter, the cambium is less active and forms fewer xylary elements that have narrow vessels/tracheids and this wood is called autumn wood or latewood.

Question 11.
All growth rings are not annual. Why?
Answer:
Sometimes annual rings are called growth rings but all the growth rings are not annual. In some trees, more than one growth ring is formed within a year due to climatic changes.

Question 12.
State the location of tyloses.
Answer:
In many dicot plants, the lumen of the xylem vessels is blocked by many balloon-like ingrowths from the neighboring parenchymatous cells. These balloon-like structures are called tyloses.

Question 13.
List out the substances that are accumulated in tyloses.
Answer:
In fully developed tyloses, starchy crystals, resins, gyms, oils, tannins, or colored substances are found.

Question 14.
What are tylosoids?
Answer:
In angiosperms, the sieve tubes are blocked by tylose-like ingrowths from the neighbouring parenchymatous cells. Example: Bombax. These are called tylosoids.

Question 15.
Why the heartwood is unable to conduct water?
Answer:
The heartwood stops conducting, water. As vessels of the heartwood are blocked by tyloses, water is not conducted through them. Due to the presence of tyloses and their contents, the heartwood becomes colored, dead, and the hardest part of the wood.

Question 16.
Give the proper botanical term for heartwood and sapwood?
Answer:
Heartwood – Duramen and Sapwood – alburnum.

Question 17.
Write a brief note on Haematoxylin.
Answer:
Hematoxylin is a dye, obtained from the heartwood of Haematoxylum campechianum used to stain plant materials for observation under a microscope, especially the nucleus of the cell.

Question 18.
From where does the Canada balsam is obtained? Mention its use.
Answer:
Canada balsam is obtained from the resin ducts of a gymnospermic plant called Abies balsamea. It is used as a mounting medium for microscopic slide preparation.

Question 19.
List out the names of the plants and the types of bast fibers produced from that plant.
Answer:

Question 20.
What is periderm? Mention its components.
Answer:
The periderm is a protective tissue of secondary origin that replaces the epidermis and often the primary cortex. The periderm consists of phellem, phellogen, and phelloderm.

Question 21.
What is Phelloderm?
Answer:
Phelloderm (Secondary cortex) is a tissue resembling cortical living parenchyma produced centripetally (inward) from the phellogen as a part of the periderm of stems and roots in seed plants.

Question 22.
What are Lenticels? State its uses.
Answer:
Lenticel is raised opening or pore on the epidermis or bark of stems and roots. It is formed during secondary growth in stems.
Lenticel is helpful in the exchange of gases and transpiration called lenticular transpiration.

Question 23.
Define filling tissue.
Answer:
When phellogen is more active in the region of lenticels, a mass of loosely arranged thin-walled parenchyma cells is formed. It is called complementary tissue or filling tissue.

Question 24.
Enumerate any two commercial barks.
Answer:
(a) Cinchona bark used for antimalarial drug production.
(b) Cinnamon bark used as a spice.

Short answer questions

Question 1.
What do you understand by the term longitudinal growth?
Answer:
The plant organs originating from the apical meristems pass through a period of expansion in length and width. The roots and stems grow in length with the help of apical meristems. This is called primary growth or longitudinal growth.

Question 2.
Compare Intra fascicular & Inter fascicular cambium.
Answer:

Question 3.
Comment on Storied cambium.
Answer:
If the fusiform initials are arranged in horizontal tiers, with the end of the cells of one tier appearing at approximately the same level, as seen in the tangential longitudinal section (TLS), it is called storied (stratified) cambium.

Question 4.
How pseudo-annual rings are formed?
Answer:
Additional growth rings are developed within a year due to adverse natural calamities like drought, frost, defoliation, flood, mechanical injury, and biotic factors during the middle of a growing season, which results in the formation of more than one annual ring. Such rings are called pseudo- or false-annual rings.

Question 5.
List out the differences between springwood and autumn wood.
Answer:

Question 6.
Classify and explain the types of angiospermic woods based on the diameter of xylem vessels.
Answer:
On the basis of the diameter of xylem vessels, two main types of angiosperm woods are recognized.

1. Diffuse porous woods: Diffuse porous woods are woods in which the vessels or pores are rather uniform in size and distribution throughout an annual ring. Example: Acer.
2. Ring porous woods: The pores of the earlywood are distinctly larger than those of the latewood. Thus rings of wide and narrow vessels occur. Example: Quercus.

Question 7.
Differentiate between diffuse-porous wood and ring-porous wood.
Answer:

Question 8.
Give an account on Phellogen.
Answer:
Phellogen (Cork Cambium):
It is a secondary lateral meristem. It comprises homogenous meristematic cells, unlike vascular cambium. It arises from the epidermis, cortex, phloem, or pericycle (extrastelar in origin). Its cells divide periclinally and produce radially arranged files of cells. The cells towards the outer side differentiate into phellem (cork) and those towards the inside as phellogen (secondary cortex).

Question 9.
Define the term Rhytidome.
Answer:
Rhytidome is a technical term used for the outer dead bark which consists of periderm and isolated cortical or phloem tissues formed during successive secondary growth. Example: Quercus.

Question 10.
Compare ring barks with scale barks.
Answer:
If the phellogen forms a complete cylinder around the stem, it gives rise to ring barks. Example: Quercus. When the bark is formed in overlapping scale-like layers, it is known as scale bark. Example: Guava.

Question 11.
Enumerate the beneficial aspects of bark.
Answer:
Bark protects the plant from parasitic fungi and insects, prevents water loss by evaporation, and guards against variations of external temperature. It is insect repellent, decay proof, fireproof and is used in obtaining drugs or spices. The phloem cells of the bark are involved in the conduction of food while secondary cortical cells involved in storage.

Question 12.
(i) Draw and label the structure of lenticel.
Answer:

(ii) Draw the C.S. of Wood showing the annual ring.
Answer:

Long answer questions

Question 1.
Differentiate between secondary growth in dicot stem and root.
Answer:

Question 2.
Depict the tissue lineage during secondary growth in the dicot stem and root.
Answer:

Question 3.
List out the differences between Vascular cambium and cork cambium.
Answer:

Question 4.
Differentiate between Phellem and Phelloderm.
Answer:

Question 5.
Differentiate between Heartwood and Sapwood.
Answer:

Question 6.
Why the study of Growth rings is significant?
Answer:
Importance of Studying Growth Rings:

• Age of wood can be calculated.
• The quality of timber can be ascertained.
• Radio-Carbon dating can be verified.
• Past climate and archaeological dating can be made.
• Provides evidence in forensic investigation.

Higher Order Thinking Skills (HOTs)

Question 1.
What do heartwood and sap wood stand for?
Answer:
Heartwood refers to the dark central part of the wood.
Sapwood refers to the pale peripheral part of the wood.

Question 2.
Arrange the following in the sequence you would find them in a plant starting from ‘ periphery-Phellem, Phellogen, phelloderm.
Answer:
Phellem (cork) → Phellogen (cork cambium) → Phelloderm (secondary cortex).

Question 3.
A transverse section of the trunk of a tree shows concentric rings known as growth rings.
How these rings are formed? Mention its significance.
Answer:
Growth rings denote the combination of earlywood and latewood formed during the spring season and winter season respectively and the rings are more evident due to the high density of latewood. Growth rings (annual rings) in a cut stem can be used to calculate and determine the age of a tree.

Question 4.
If one debarks a tree, what parts of the plant are being removed?
Answer:
The term bark refers to all the tissues outside the vascular cambium i.e., periderm, cortex, primary phloem, and secondary phloem. If a person debarks a tree all these layers of the plant are removed.

Question 5.
Given down is the list of commercial products obtained from plants. Mention the parts of the plant which are used to obtain them.
(a) Rubber (b) Cinnamon (c) Turpentine
Answer:

Choose the correct answer.

1. Longitudinal growth refers to ……….. .
(a) Primary growth
(b) Secondary growth
(c) Tertiary growth
(d) Abnormal growth
Answer:
(a) Primary growth

2. Secondary growth is absent in ………… .
(a) Gymnosperms
(b) Angiosperms
(c) Dicots
(d) Monocots
Answer:
(d) Monocots

3. Interfascicular cambium originates from ………… .
(a) Procambium
(b) Vascular cambium
(c) Pith
(d) Medullary rays
Answer:
(d) Medullary rays

4. ……….. is also called as wood.
(a) Primary xylem
(b) Primary Phloem
(c) Secondary xylem
(d) Pith
Answer:
(c) Secondary xylem

5. Soft wood lacks ………… .
(a) Tracheids
(b) Pith
(c) Vessels
(d) Medullary rays
Answer:
(c) Vessels

6. ……….. produces the primary plant body.
(a) Procambium
(b) Apical meristem
(c) Lateral meristem
(d) Vascular cambium
Answer:
(b) Apical meristem

7. Dendrochronology deals with determining the ………… .
(a) Growth of trees
(b) Age of trees
(c) Branching pattern of trees
(d) Growth and climatic influence on trees
Answer:
(b) Age of trees

8. Diffuse porous wood is seen in …………. .
(a) Quercus
(b) Maple
(c) Pinus
(d) Acer
Answer:
(d) Acer

9. The balloon-like swellings blocking the lumen of xylem vessels are called ………….. .
(a) Xyloses
(b) Callose
(c) Tyloses
(d) Tylosoids
Answer:
(c) Tyloses

10. Sapwood can also be termed as ……….. .
(a) Duramen
(b) Porous wood
(c) Phellogen
(d) Alburnum
Answer:
(d) Alburnum

11. Hematoxylin especialls stains ………… .
(a) Mitochondria
(b) Nucleus
(c) Golgi bodies
(d) Ribosomes
Answer:
(b) Nucleus

12. Sun hemp is obtained from ………… .
(a) Crotolaria iuncea
(b) Corchorus capsularis
(c) Linum ustitaissimum
(d) Cannabis sativa
Answer:
(a) Crotolaria iuncea

13. The non-living cork cells are with ………… walls.
(a) Lignified
(b) Suberized
(c) Cutinised
(d) Cellulose rich
Answer:
(b) Suberized

14. ………….. gives rise to the secondary cortex.
(a) Phelloderm
(b) Endoderm
(c) Phellogen
(d) Periderm
Answer:
(a) Phelloderm

15. Scale barks can be observed in …………. .
(a) Guava
(b) Quercus
(c) Coconut
(d) Neem
Answer:
(a) Guava

16. Which of the following is termed as Cork?
(a) Phellem
(b) Periderm
(c) Phellogen
(d) Phelloderm
Answer:
(a) Phellem

17. Polyderm is found in ………..
(a) Stem
(b) Roots
(c) Underground stem
(d) Both (a) and (c)
Answer:
(d) Both (a) and (c)

18. …………… are formed only during secondary growth.
(a) Stomata
(b) Endodermis
(c) Pith
(d) Lenticel
Answer:
(d) Lenticel

19. Turpentine is obtained from ………. species.
(a) Cinchona
(b) Acacia
(c) Pinus
(d) Cinnamomum
Answer:
(c) Pinus

20. Rubber is a ………..
(a) Latex
(b) Resin
(c) Alkaloid
(d) Drug
Answer:
(a) Latex

Students get through the TN Board 11th Bio Botany Important Questions Chapter 9 Tissue and Tissue System which is useful for their exam preparation.

TN State Board 11th Bio Botany Important Questions Chapter 9 Tissue and Tissue System

Very short answer questions

Question 1.
How primary meristem differs from secondary meristem?
Answer:

Question 2.
What are simple tissues? Mention its types.
Answer:
Simple tissues are composed of one type of cells only. The cells are structurally and functionally similar. It is of three types.

1. Parenchyma,
2. Collenchyma and
3. Sclerenchyma

Question 3.
Define Idioblasts.
Answer:
Parenchyma cells that store resin, tannins, crystals of calcium carbonate, calcium oxalate are called idioblasts.

Question 4.
Mention a few places in plants where collenchyma cells can be observed?
Answer:
Hypodermis of dicot stem, petiole, pedicle.

Question 5.
Classify collenchyma based on cell wall pectinisation.
Answer:
Angular collenchyma, lacunar collenchyma, lamellar collenchyma, Annular collenchyma.

Question 6.
What are Bone cells? Give an example.
Answer:
Rod-shaped sclereids with dilated ends are called bone cells or Osteosclereids.
Example: Seed coat of Pisum.

Question 7.
Fibres are supporting tissues. Justify.
Answer:
Fibres provide mechanical strength and support and protect the plants from strong winds. Hence fibres are called supporting tissues.

Question 8.
Define Wood fibres.
Answer:
Wood Fibres: These fibres are associated with the secondary xylem tissue. They are also called xylary fibres. These fibres are derived from the vascular cambium.

Question 9.
What are Bast fibres?
Answer:
Bast fibres or Extra Xylary Fibres: These fibres are present in the phloem. Natural Bast fibres are strong and cellulosic. Fibres obtaining from the phloem or outer bark of jute, kenaf, flax and hemp plants. The so-called pericyclic fibres are actually phloem fibres.

Question 10.
Write a note on surface fibres.
Answer:
Surface fibres are produced from the surface of the plant organiser Cotton and silk cotton are examples. They occur in the test of seeds.

Question 11.
Define Complex tissue & mention its types.
Answer:
A complex tissue is a tissue with several types of cells but all of them function together as a single unit. It is of two types – xylem and phloem.

Question 12.
What do you understand by Centrarch Xylem?
Answer:
Protoxylem is located in the centre surrounded by the metaxylem is called Centrarch. In this type, only one vascular strand is developed. Example: Selaginella sp.

Question 13.
State the Xylem elements.
Answer:
Xylem consists of Four Types of Cells:

1. Tracheids
2. Vessels or Trachea
3. Xylem Parenchyma and
4. Xylem Fibres.

Question 14.
What is the relation between Haberlandt & complex tissue?
Answer:
Haberlandt in 1914 used the term Leptome for phloem and hadrome for xylem.

Question 15.
When do you call the perforation plate in vessels as simple?
Answer:
In vessels, due to the dissolution of the entire cell wall, a single pore is formed at the perforation plate. Such a perforation is called a simple perforation plate. Example: Mangifera.

Question 16.
Draw the annular and reticulate types of secondary wall thickening in tracheids.
Answer:

Question 17.
Where fibre-tracheids can be seen?
Answer:
Between fibres and normal tracheids, there are many transitional forms that are neither typical fibres nor typical tracheids. The transitional types are designated as fibre-tracheids.

Question 18.
What are axial parenchyma & Ray parenchyma?
Answer:
Parenchyma arranged longitudinally along the long axis is called axial parenchyma. Ray parenchyma is arranged in radial rows.

Question 19.
Mention the components of Phloem.
Answer:
Phloem consists of Four Types of Cells:

1. Sieve elements,
2. Companion cells
3. Phloem parenchyma and
4. Phloem fibres

Question 20.
Companion cells are absent in which groups of plants?
Answer:
Gymnosperms and angiospenns.

Question 21.
Write a brief note on Bast fibres.
Answer:
The fibres of sclerenchyma associated with phloem are called phloem fibres or bast fibres. They are narrow, vertically elongated cells with very thick walls and a small lumen. Among the four phloem elements, phloem fibres are the only dead tissue. These are the strengthening as well as supporting cells.

Question 22.
Define the term Syncyte and give examples.
Answer:
Syncyte: Cell which is formed by fusion of cell is called Syncyte.
Example: Vessels (Dead syncyte), sieve tube (living syncyte).

Question 23.
Mention the cell wall chemicals of parenchyma.
Answer:
Cellulose & Pectin.

Question 24.
Name the three types of tissue system as proponed by Sachs.
Answer:
1. Epidermal tissue system (derived from protoderm), 2. Ground tissue system (derived from ground meristem), 3. Vascular tissue system (derived from procambium).

Question 25.
Mention any two plants having multi seriate epidermis.
Answer:
Ficus & Nerium.

Question 26.
What are passage cells? State its function.
Answer:
The endodermal cells, which are opposite to the protoxylem elements, are thin-walled without Casparian strips. These cells are called passage cells. Their function is to transport water and dissolved salts from the cortex to the protoxylem.

Question 27.
Write a brief note on pericycle.
Answer:
The pericycle is a single or few-layered parenchymatous found inner to the endodermis. It is the outermost layer of the stele. Rarely thick-walled sclerenchymatous. In angiosperms, the pericycle gives rise to lateral roots.

Question 28.
List out the storage products seen in the medulla.
Answer:
Tannins, Phenols, calcium oxalate crystals, fatty substances, starch.

Question 29.
Draw and label the open vascular bundle.
Answer:

Question 30.
The vascular bundle of the monocot stem is said to be closed. Why?
Answer:
The vascular bundle of the monocot stem is said to be closed since there is no cambium present between the xylem & phloem.

Question 31.
Define Stele.
Answer:
Stele: All the tissues present inside the endodermis comprise the stele. It includes the pericycle and vascular system.

Question 32.
Where starch sheath is seen? Why it is called so?
Answer:
In the stem, the innermost layer of the cortex is called the endodermis. Since starch grains are abundant in these cells, it is also called a starch sheath. The starch sheath is homologous to the endodermis of the root.

Question 33.
Define Eustele.
Answer:
In the dicot stem, vascular bundles are arranged in a ring around the pith. This type of stele is called eustele.

Question 34.
What is a Bundle cap?
Answer:
In the stem of the sunflower (Helianthus), a few layers of sclerenchyma cell occur in patches outside the phloem in each vascular bundle. This patch of sclerenchyma cell is called Bundle cap or Hardbast.

Question 35.
Mention the nature of the vascular bundle of the dicot stem.
Answer:
In the dicot stem, the vascular bundle is conjoint, collateral, open & endarch.

Question 36.
What are medullary rays?
Answer:
The pith extends between the vascular bundles. These extensions of the pith between the vascular bundles are called primary pith rays or primary medullary rays.

Question 37.
What are protoxylem lacunae?
Answer:
In the mature vascular bundle of the monocot stem, the lowest protoxylem disintegrates and forms a cavity called a protoxylem lacuna.

Question 38.
Define mesophyll tissue & mention its types.
Answer:
The ground tissue system that lies between the epidermal layers of leaf is known as mesophyll tissue. Often it is differentiated into palisade parenchyma on the adaxial (upper) side and spongy parenchyma on the abaxial (lower) side.

Question 39.
Mention the functions of palisade & spongy parenchyma.
Answer:
Palisade parenchyma performs photosynthesis spongy parenchyma facilitates gaseous exchange.

Question 40.
Where the respiratory cavity is located?
Answer:
The air space that is found next to the stomata is called the respiratory cavity or substomatal cavity.

Question 41.
What is border parenchyma?
Answer:
In the dicot leaf, vascular bundles are surrounded by a compact layer of parenchymatous cells called bundle sheath or border parenchyma.

Question 42.
What are silica cells?
Answer:
Some of the epidermal cells of the grass are filled with silica. They are called silica cells.

Question 43.
Define Kranz Sheath.
Answer:
In C4 grasses, the bundle sheath cells are living and involve in C4 photosynthesis. This sheath is called Kranz sheath.

Question 44.
What do you mean by Hydathode?
Answer:
A hydathode is a type of epidermal pore, commonly found in higher plants. Structurally, hydathodes are modified stomata, usually located at leaf tips or margins, especially at the teeth.

Question 45.
Guttation – Explain.
Answer:
Hydathodes discharge liquid water with various dissolved substances from the interior of the leaf to its surface. This process is called guttation. Example many grasses.

Question 46.
Define Halophiles.
Answer:
Plants that grow in the salty environment are called Halophiles.

Short answer questions

Question 1.
Who proposed the Tunica-Corpus theory? Add a note on it.
Answer:
Tunica-Corpus theory is proposed by A. Schmidt (1924). Two zones of tissues are found in the apical meristem.

1. The tunica: It is the peripheral zone of the shoot apex, that forms the epidermis.
2. The corpus: It is the inner zone of the shoot apex, that forms the cortex and stele of the shoot.

Question 2.
Histogen theory of Stem. Explain.
Answer:
Histogen theory is proposed by Hanstein (1868) and supported by Strassburger. The shoot apex comprises three distinct zones.

1. Dermatogen: It is a outermost layer. It gives rise to the epidermis.
2. Periblem: It is a middle layer. It gives rise to the cortex.
3. Plerome: It is the innermost layer. It gives rise to stele.

Question 3.
Explain briefly Korper Kappe theory.
Answer:
Korper Kappe theory is proposed by Schuepp. There are two zones in the root apex:

1. The Korper zone forms the body.
2. Kappe zone forms the cap. This theory is equivalent to the tunica corpus theory of shoot apex. The two divisions are distinguished by the type of T (also called Y divisions). Korper is characterised by inverted T divisions and kappe by straight T divisions.

Question 4.
Write a note on the Quiescent centre concept.
Answer:
The quiescent centre concept was proposed by Clowes (1961) to explain root apical meristem activity. This centre is located between the root cap and differentiating cells of the roots. The apparently inactive region of cells in the root promeristem is called the quiescent centre. It is the site of hormone synthesis and also the ultimate source of all meristematic cells of the meristem.

Question 5.
Enumerate the functions of parenchyma.
Answer:
Parenchyma may store various types of materials like water, air, ergastic substances. The turgid parenchyma cells help in giving rigidity to the plant body. Partial conduction of water is also maintained through parenchymatous cells.

Question 6.
How buoyancy is maintained in aquatic plants?
Answer:
In aquatic plants like Nymphae, the parenchyma cells contain air in their intercellular spaces which provides aeration and buoyancy.

Question 7.
Compare angular collenchyma with lacunar collenchyma.
Answer:
Types of Collenchyma

1. Angular collenchyma: It is the most common type of collenchyma with the irregular arrangement and thickening at the angles where cells meet. Example: Hypodermis of Datura and Nicotiana.
2. Lacunar collenchyma: The collenchyma cells are irregularly arranged. The cell wall is thickening on the walls bordering intercellular spaces. Example: Hypodermis of Ipomoea.

Question 8.
Draw and label (a) Branchysclereids, (b) Osteosclereids.
Answer:

Question 9.
Differentiate between Exarch and Endarch condition.
Answer:

Question 10.
Compare primary phloem with secondary phloem.
Answer:

Question 11.
Point out the angiospermic families that do not possess xylem vessels.
Answer:
Winteraceae, Tetracentraceae and Trochodendraceae are the vesselless angiospermic families.

Question 12.
Write a short note on Phloem parenchyma.
Answer:
The parenchyma cells associated with the phloem are called phloem parenchyma. These are living cells. They store starch and fats. They also contain resins and tannins in some plants. The primary phloem consists of axial parenchyma and the secondary phloem consists of both axial and ray parenchyma. They are present in Pteridophytes, Gymnosperms and Dicots.

Question 13.
Classify meristem based on function.
Answer:
Based on function, meristems are classified into 3 types.

1. Peridenn giving rise to epidermis.
2. Procambium giving rise to primary vascular tissues.
3. Ground meristem giving rise to cortex a pitch.

Question 14.
Give an account of the piliferous layer.
Answer:
The outer layer of the root is known as the piliferous layer or epiblema. It is made up of a single layer of parenchyma cells which are arranged compactly without intercellular spaces. It is devoid of epidermal pores and cuticle. Root hair is always single-celled, it absorbs water and mineral salts from the soil. Another important function of the piliferous layer is protection.

Question 15.
What are bulliform cells? How it helps the plants?
Answer:
In leaves, some cells of the upper epidermis (Example: Grasses) are larger and thin-walled. They are called bulliform cells or motor cells. These cells are helpful for the rolling and unrolling of the leaf according to the weather change.

Question 16.
Draw the stoma with a dumb-bell shaped guard cell.
Answer:

Question 17.
Define Trichoblasts.
Answer:
The Piliferous layer of the root has two types of epidermal cells, long cells and short cells.’ The short cells are called trichoblasts. Trichoblasts are elongate into root hairs.

Question 18.
What are Prickles? Mention its benefit to plants.
Answer:

(a) Prickles: Prickles, are one type of epidermal emergences with no vascular supply. They are stiff and sharp in appearance. (Example: Rose), (b) Prickles also provide protection against animals and they also check excessive transpiration.

Question 19.
Distinguish between Extrasteiar & Intrastelar ground tissue.
Answer:

Question 20.
Define Pith.
Answer:
The central part of the ground tissue is known as the pith or medulla. Generally, this is made up of thin-walled parenchyma cells with intercellular spaces. The cells in the pith generally store starch, fatty substances, tannins, phenols, calcium oxalate crystals, etc.

Question 21.
Draw & label bicollateral vascular bundle.
Answer:

Question 22.
What are Casparian strips?
Answer:
The radial and the inner tangential walls of endodermal cells are thickened with suberin and lignin. This thickening was first noted by Robert Caspary in 1965. So these thickenings are called Casparian strips.

Question 23.
Differentiate between Protoxylem & Metaxylem.
Answer:

Question 24.
Draw & label the ground plan of T.S. of Dicot root.
Answer:

Question 25.
What is the role of Casparian strips in roots?
Answer:
The main function of Casparian strips in the endodermal cells is to prevent the re-entry of water into the cortex once water entered the xylem tissue.

Question 26.
State the importance of cambium.
Answer:
Cambium consists of brick-shaped and thin-walled meristematic cells. It is one to four layers in thickness. These cells are capable of forming new cells during secondary growth.

Question 27.
What are dorsiventral leaves? Give an example.
Answer:
In dorsiventral leaves, the mesophyll is differentiated into palisade and spongy parenchyma, the former occurring on the upper side and the latter on the lower side. Example: Sunflower.

Question 28.
What are isobilateral leaves? Give an example.
Answer:
In the isobilateral leaf, palisade is present on both sides of the leaf and in between them, spongy parenchyma is present. Example: Nerium.

Question 29.
Compare the characters of Palisade parenchyma & spongy parenchyma.
Answer:

Long answer questions

Question 1.
Enumerate the characters of meristematic tissue.
Answer:
The characters of meristematic tissues:

• The meristematic cells are isodiametric and may be, oval, spherical or polygonal in shape.
• They have generally dense cytoplasm with a prominent nucleus.
• Generally, the vacuoles in them are either small or absent.
• Their cell wall is thin, elastic and essentially made up of cellulose.
• These are the most actively dividing cells.
• Meristematic cells are self-perpetuating.

Question 2.
Classify meristem based on position with a simplified diagram.
Answer:

1. Apical meristem: Present in apices of root and shoot. It is responsible for the increase in the length of the plant, it is called primary growth.
2. Intercalary meristem: Occurs between the mature tissues. It is responsible for the elongation of internodes.
3. Lateral meristem: Occurs along the longitudinal axis of stem and root. It is responsible for secondary tissues and the thickening of the stem and root. Example: vascular cambium and cork cambium.
   Different types of meristems

Question 3.
Elaborate on the different types of economically important fibres.
Answer:
Economically fibres may be grouped as follows:

1. Textile Fibres: Fibres utilized for the manufacture of fabrics, netting and cordage etc.
   a. Surface Fibres: Example: Cotton.
   b. Soft Fibres: Example: Flax, Jute and Ramie
   c. Hard fibres: Example: Sisal, Coconut, Pineapple, Abaca etc.
2. Brush fibre: Fibres utilized for the manufacture of brushes and brooms.
3. Rough weaving fibres: Fibres utilized in making baskets, chairs, mats etc.
4. Papermaking fibres: Wood fibres utilized for paper making.
5. Filling fibres: Fibres used for stuffing cushions, mattresses, pillows, furniture etc. Example: Bombax and Silk cotton.

Question 4.
List out the differences between Meristematic tissue & permanent tissue.
Answer:

Question 5.
Sieve cells & Sieve tubes both are phloem elements. Yet they differ. How?
Answer:

Question 6.
How tracheids differ from fibres?
Answer:

Question 7.
Point out the functions of the Epidermal tissue system.
Answer:
Functions of Epidermal Tissue System:

1. This system in the shoot checks excessive loss of water due to the presence of a cuticle.
2. The epidermis protects the underlying tissues.
3. Stomata is involved in transpiration and gaseous exchange.
4. Trichomes are also helpful in the dispersal of seeds and fruits and provide protection against animals.
5. Prickles also provide protection against animals and they also check excessive transpiration.
6. In some rose plants, they also help in climbing.
7. Glandular hairs repel herbivorous animals.

Question 8.
Draw and label the transverse section of the monocot stem.
Answer:

Question 9.
Draw a labelled diagram of T.S. of Dicot leaf.
Answer:

Question 10.
Differentiate between stoma & hydathode.
Answer:

Higher Order Thinking Skills (HOTs)

Question 1.
Protoxylem is the first formed Xylem. If the protoxylem is surrounded by phloem what kind of arrangement of xylem would you call it? Give an example.
Answer:
If the protoxylem is surrounded by phloem the vascular bundle is said to concentric and amphicribral. Example: Ferns.

Question 2.
Observe the diagram given below and answer the questions.
(a) Mention the parts A, B, C
(b) Name the part of the plant where we can see this structure more in number.

Answer:
(a) A – Stoma, B – Guard cell and C – Subsidiary cell.
(b) Stomata are seen more in number on the lower epidermis of leaves.

Question 3.
In a Biology practical class, your subject teacher has placed a glass slide showing the transverse section of the dicot stem. State any two possible reasons to call the slide is a T.S of dicot stem.
Answer:
In the dicot stem, the stele is Eustele (he. vascular bundles are arranged in the form of a ring). Presence of endarch xylem and central pith.

Question 4.
While eating pear fruit it is usually seen that some stone-like structures get entangled in the teeth, what are these stone-like structures called?
Answer:
The stone-like structures that we feel while eating pear fruit are the branchy sclereids or stone cells which are isodiametric sclereids with the hand cell wall.

Question 5.
A certain tissue in a green plant somehow get blocked and the leaves wilted. What was the tissue that got blocked?
Answer:
The tissue that got blocked may be the xylem. It is through the xylem the water and other minerals are transported from root to leaves and other parts. So if the xylem is blocked, the water and mineral supply is stopped, leading to wilting of leaves that eventually may lead to the death of the plant.

Question 6.
The cross-section of a plant material shown the following features on viewing under the microscope.
(a) Radically arranged vascular bundles.
(b) Four xylem arms with protoxylem facing outer side. To which organ should it be
arranged.
Answer:
Dicot roots show radically arranged vascular hurdles with tetrarch and exarch xylem.

Question 7.
Write the precise function of:
(a) Aerenchyma (b) Collenchyma (c) Sieve tube
Answer:
(a) Aerenchyma – Provides buoyancy in aquatic plants.
(b) Collenchyma – Mechanical strength & support.
(c) Sieve tube – Food conduction

Choose the correct answer.

1. The term meristem is coined by
(a) Schmidt
(b) Clowes
(c) Nageli
(d) Schleiden
Answer:
(c) Nageli

2. The study of tissues is called as
(a) Anatomy
(b) Cytology
(c) Histology
(d) Embryology
Answer:
(c) Histology

3. Primary growth is the responsibility of meristem.
(a) Apical
(b) Primary
(c) Intercalary
(d) Lateral
Answer:
(a) Apical

4. Identify the mismatch pair:
(i) Nageli – Xylem
(ii) Leptome – Phloem
(iii) Tunica Corpus theory – Hanstein
(iv) Tracheids – Sanio
(a) (i) – only
(b) (iii) – only
(c) Both (i) and (iii)
(d) None
Answer:
(b) (iii) – only

5. Who proposed Korper kappe theory?
(a) Schmidt
(b) Hanstein
(c) Nageli
(d) Schuepp
Answer:
(d) Schuepp

6. Parenchyma storing calcium carbonate crystals are called …..
(a) Leucoplasts
(b) Elaioplasts
(c) Idioblasts
(d) Chromoplasts
Answer:
(c) Idioblasts

7. ………. is seen in all organs of plant.
(a) Chlorenchyma
(b) Sclerenchyma
(c) Parenchyma
(d) Collenchyma
Answer:
(c) Parenchyma

8. ………… is absent in roots.
(a) Chlorenchyma
(b) Sclerenchyma
(c) Parenchyma
(d) Collenchyma
Answer:
(d) Collenchyma

9. ………….. sclereids are seen in the seed coat of pisum.
(a) Macro
(b) Osteo
(c) Tricho
(d) Branchy
Answer:
(a) Macro

10. Fibres forming mesocarp of coconut is called as
(a) Surface fibres
(b) Soft fibres
(c) Mesocarp fibres
(d) Septate fibres
Answer: (c) Mesocarp fibres

11. ………… is called as Leptome.
(a) Parenchyma
(b) Phloem
(c) Xylem
(d) Fibres
Answer:
(6) Phloem

12. Multiple perforation plate is seen in …………. .
(a) Lepidodendron
(b) Limnophyton
(c) Mangifera
(d) Liriodendron
Answer: (d) Liriodendron

13. ………… species of gymnosperms have vessels.
(a) Cycas
(b) Selaginella
(c) Gnetum
(d) Conifer
Answer:
(c) Gnetum

14. ………. Only living cells among Xylem elements are
(a) Vessels
(b) Tracheids
(c) Xylem Parenchyma
(d) Xylem fibres
Answer: (c) Xylem Parenchyma

15. In sieve elements, mature sieve plates are blocked by ………… .
(a) Suberin
(b) Lignin
(c) Callose
(d) Pectin
Answer: (c) Callose

16. Campanion cells are present only in …………. .
(a) Pteridophytes
(b) Angiosperms
(c) Gymnosperms
(d) Dicots
Answer:
(b) Angiosperms

17. Rolling & unrolling of leaves due to whether change are controlled by ………… .
(a) Sensory cells
(b) Motor cells
(c) Subsidory cells
(d) Trichomes
Answer:
(b) Motor cells

18. Sunken stoma is seen in ………. .
(a) Cycas
(b) Nerium
(c) Both (a) and (b)
(d) None
Answer:
(c) Both (a) and (b)

19. Root hairs originated from ………… .
(a) Trichomes
(b) Epidermis
(c) Pericycle
(d) Trichoblasts
Answer:
(d) Trichoblasts

20. Which is NOT a part of intrastelar ground tissue?
(a) Pericycle
(b) Medullary ray
(c) Pith
(d) Cortex
Answer:
(d) Cortex

21. Usually ………… form the hypodermis.
(a) Chlorenchyma
(b) Sclerenchyma
(c) Collenchyma
(d) Parenchyma
Answer:
(c) Collenchyma

22. In angiosperms, ……….. gives rise to lateral roots.
(a) Pith
(b) Endodermis
(c) Pericycle
(d) Trichoblasts
Answer:
(c) Pericycle

23. In cucurbitaceae, the vascular bundles are
(a) Bicollateral
(b) Collateral closed
(c) Concentric
(d) Radial
Answer:
(a) Bicollateral

24. Major function of epiblema is
(a) Transport
(b) Support
(c) Protection
(d) Conduction of food
Answer:
(c) Protection

25. Innermost layer of cortex is
(a) Epiblema
(b) Endodermis
(c) Pericycle
(d) Medullary rays
Answer:
(b) Endodermis

26. In T.S. of bean root, the xylem is
(a) Polyarch
(b) Hexarch
(c) Tetrarch
(d) Wedge shaped
Answer:
(c) Tetrarch

27. Phellogen arises from ……….., in dicot roots.
(a) Endodermis
(b) Piliferous layer
(c) Pericycle
(d) Periderm
Answer:
(c) Pericycle

28. Eustele is seen in …………. stem.
(a) Maize
(b) Grass
(c) Paddy
(d) Sunflower
Answer:
(d) Sunflower

29. In monocot stem, the xylem vessels are in the form of ………… .
(a) X
(b) Y
(c) W
(d) +
Answer:
(b) Y

30. Hardbast is composed of …………. .
(a) Parenchyma
(b) Collenchyma
(c) Sclerenchyma
(d) Prosenchyma
Answer:
(c) Sclerenchyma

31. Silica cells seen in epidermis contain
(a) Magnesium
(b) Calcium
(c) Silica
(d) Sand
Answer:
(c) Silica

32. Guttation occurs through
(a) Stomata
(b) Lenticles
(c) Cuticle
(d) Hydathodes
Answer:
(d) Hydathodes

33. Kranz sheath in granes perform photosynthesis.
(a) C₃
(b) C₄
(c) C₂
(d) C₁₂
Answer:
(b) C₁₂

34. Halophiles survive in environment.
(a) Dry
(b) Aquatic
(c) Saline
(d) Cold
Answer:
(c) Saline

35. Who coined the term Hadrome?
(a) Hanstein
(b) Nageli
(c) Hofmeister
(d) Haberlandt
Answer:
(d) Haberlandt

36. Match the following:

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

Students get through the TN Board 11th Bio Botany Important Questions Chapter 8 Biomolecules which is useful for their exam preparation.

TN State Board 11th Bio Botany Important Questions Chapter 8 Biomolecules

Answer the following short answers.

Question 1.
What is meant by the cellular pool?
Answer:
The cell components are made of a collection of molecules called a cellular pool, which consists ’ of both inorganic and organic compounds.

Question 2.
What are Micronutrients? Give any two examples.
Answer:
Micronutrients, which are required in trace amounts, eg. Cobalt, zinc, boron, copper, molybdenum and manganese) and are essential for enzyme action.

Question 3.
Mention any four properties of water.
Answer:

1. Adhesion and cohesion property.
2. High latent heat of vaporisation.
3. High melting and boiling point.
4. Universal solvent.

Question 4.
Define primary metabolites.
Answer:
Primary metabolites are those that are required for the basic metabolic processes like photosynthesis, respiration, protein and lipid metabolism of living organisms.

Question 5.
Match the following:

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

Question 6.
Define monosaccharide.
Answer:
Monosaccharides are relatively small molecules constituting a single sugar unit. Glucose has a chemical formula of CgH^Og. It is a six-carbon molecule and hence is called hexose.

Question 7.
Define glycosidic bond.
Answer:
The bond formed between the glucose and fructose molecule by the removal of water is called a glycosidic bond. This is another example of a strong covalent bond.

Question 8.
What is chitin made up of?
Answer:
Chitin is a homo polysaccharide with amino acids added to form mucopolysaccharide. The ’ basic unit is a nitrogen-containing glucose derivative known as N-acetyl glucosamine. It forms the exoskeleton of insects and other arthropods. It is also present in the cell walls of fungi.

Question 9.
How does Herbivores digest cellulose?
Answer:
Herbivores can digest them With the help of bacteria present in the gut which produces en2yme cellulase. This is an example of mutualism.

Question 10.
What are steroids? Explain with an example.
Answer:
These are complex compounds commonly found in the cell membrane and animal hormones, eg. Cholesterol which reinforces the structure of the cell membrane in animal cells and in an unusual group of cell wall deficient bacteria —Mycoplasma.

Question 11.
Define the term amphoteric.
Answer:
(NH₂), an acidic carboxylic group (COOH) and a hydrogen atom (H) and side-chain or variable R group* The amino acid is both an acid and a base is called amphoteric.

Question 12.
What do you know about the primary structure of a protein?
Answer:
The primary structure is the linear arrangement of amino acids in a polypeptide chain.

Question 13.
What is meant by an ionic bond?
Answer:
It is formed between any charged groups that are not joined together by a peptide bond. It is stronger than a hydrogen bond and can be broken by changes in pH and temperature.

Question 14.
Explain briefly anabolic reactions.
Answer:
Anabolic (building up of organic molecules). Synthesis of proteins from amino acids and synthesis of polysaccharides from simple sugars are examples of anabolic reactions.

Question 15.
What are the lock and key mechanism?
Answer:
The substrate binds to the specially formed pocket in the enzyme – the active site, this is called the lock and key mechanism of enzyme action. As the enzyme and substrate form an ES complex, the substrate is raised in energy to a transition state and then breaks down into products plus an unchanged enzyme.

Question 16.
What are competitive inhibitors? Give an example.
Answer:
Molecules that resemble the shape of the substrate and may compete to occupy the active site of the enzyme are known as competitive inhibitors, eg. the enzyme that catalyses the reaction between carbon dioxide and the CO₂ acceptor molecule in photosynthesis, known as ribulose biphosphate carboxylase oxygenase (RUBISCO) is competitively inhibited by oxygen/carbon-di-oxide in the chloroplast.

Question 17.
Define the terms nucleotide and nucleoside?
Answer:
DNA and RNA are polymers of monomers called nucleotides, each of Which is composed of a nitrogen base, a pentose sugar and a phosphate. A purine or a pyrimidine and ribose or deoxyribose sugar are called nucleoside.

Question 18.
Mention any two sulphur-containing amino acids.
Answer:
Methionine and cysteine.

Question 19.
What is meant by Plectonemic coiling of DNA?
Answer:
Plectonemic coiling – the two strands of the DNA are wrapped around each other in a helix, making it impossible to simply move them apart without breaking the entire structure.

Question 20.
Give a short note on RNA.
Answer:
tRNA (transfer RNA): Translates the code from mRNA and transfers amino acids to the ribosome to build proteins. It is highly folded into an elaborate 3D structure and comprises about 15% of total RNA. It is also called soluble RNA.

Answer In brief.

Question 1.
Explain Polysaccharide with an example.
Answer:

1. These are made of hundreds of monosaccharide units. Polysaccharides also called “Glycans”.
2. Long-chain of branched or unbranched monosaccharides is held together by glycosidic bonds.
3. Polysaccharide is an example of a giant molecule, a macromolecule and consists of only one type of monomer.
4. Polysaccharides are insoluble in water and are sweet. Cellulose is an example built from repeated units of glucose monomer.
5. Depending on the function, polysaccharides are of two types: Storage Polysaccharide and structural Polysaccharide.

Question 2.
Describe the test for reducing sugars.
Answer:
Aldoses and ketoses are reducing sugars. This means that, when heated with an alkaline solution of copper (II) sulphate (a blue solution called benedict’s solution), the aldehyde or ketone group reduces Cu²⁺ ions to Cu⁺ ions forming a brick-red precipitate of copper (I) oxide. In the process, the aldehyde or ketone group is oxidised to a carboxyl group (- COOH). This reaction is used as a test for reducing sugar and is known as Benedict’s test. The results of benedict’s test depend on the concentration of the sugar. If there is no reducing sugar it remains blue.

1. Sucrose is not a reducing sugar.
2. The greater the concentration of reducing sugar, the more is the precipitate formed and greater is the colour change.

Question 3.
Define triglycerides. Explain with examples.
Answer:
Triglycerides are composed of a single molecule of glycerol bound to 3 fatty acids. These include fats and oils. Fatty acids are long-chain hydrocarbons with a carboxyl group at one end which binds to one of the hydroxyl groups of glycerol, thus forming an ester bond. Fatty acids are structural unit of lipids and are carboxylic acid of long-chain hydrocarbons. The hydrocarbon can vary in length from 4-24 carbons and the fat may be saturated or unsaturated. In saturated fatty acids, the hydrocarbon chain is single-bonded.
eg. Palmitic acid, Stearic acid and unsaturated fatty acids.
eg. Oleic acid, linoleie acid) the hydrocarbon chain is double bonded (one/two/three). In general solid fats are saturated and oils are unsaturated, in which most are globules.

Question 4.
Enumerate the properties of enzymes.
Answer:
Properties of Enzyme:

1. All are globular proteins.
2. They act as catalysts and effective even in small quantity.
3. They remain unchanged at the end of the reaction.
4. They are highly specific.
5. They have an active site where the reaction takes place.
6. Enzymes lower the activation energy of the reaction they catalyse.

Question 5.
What are Allosteric enzymes? Explain with a suitable example.
Answer:
They modify enzyme activity by causing a reversible change w the structure of the enzyme active site. This in turn affects the ability of the substrate to bind to the enzyme. Such compounds are called Allosteric inhibitors, eg. The enzyme hexokinase which catalysis glucose to glucose-6 phosphate in glycolysis is inhibited by glucose 6 phosphate. This is an example of a feedback allosteric inhibitor.

Question 6.
Distinguish between nucleoside and nucleotide with two examples.
Answer:

Question 7.
Explain ribosomal RNA. Add a note on its function.
Answer:
RNA (ribosomal RNA): Single-stranded, metabolically stable, make up the two subunits of ribosomes. It constitutes 80% of the total RNA. It is a polymer with varied length from 120-3000 nucleotides and gives ribosomes their shape. Genes for rRNA are highly conserved and employed for phylogenetic studies.

Question 8.
Explain the process of negative feedback inhibition with a schematic diagram.
Answer:
Negative Feedback Inhibition: When the end product of a metabolic pathway begins to accumulate, it may act as an allosteric inhibitor of the enzyme controlling the first step of the pathway. Thus the product starts to switch off its own production as it builds up. The process is- self – regulatory. As the product is used up, its production is switched on once again. This is called end-product inhibition.

Question 9.
Explain the Michaelis-Menton Constant (km) with graphical representation.
Answer:
The rate of reaction is directly proportional to the enzyme concentration.

When the initial rate of reaction of an enzyme is measured over a range of substrate concentrations (with a fixed amount of enzyme) and the results plotted on a graph. With increasing substrate concentration, the velocity increases – rapidly at lower substrate concentration.

Question 10.
What is the three types of chemical bond in protein structure? Explain them with an example.
Answer:
A hydrogen bond is formed between some hydrogen atoms of oxygen and nitrogen in a polypeptide chain. The hydrogen atoms have a small positive charge and oxygen and nitrogen have a small negative charge. Opposite charges attract to form hydrogen bonds. Though these bonds are weak, a large number of them maintains the molecule in 3D shape.
Ionic Bond: It is formed between any charged groups that are not joined together by a peptide bond. It is stronger than a hydrogen bond and can be broken by changes in pH and temperature.
Disulfide Bond; Some amino acids like cysteine and methionine have sulphur. These form a disulphide bridge between sulphur atoms and amino acids.
Hydrophobic Bond: This bond helps some protein to maintain structure. When globular proteins are in solution, their hydrophobic groups point inwards away from water.

Answer In detail.

Question 1.
Describe the structure of a protein with a neat diagram.
Answer:

1. Protein is synthesised on the ribosome as a linear sequence of amino acids which are held together by peptide bonds. After synthesis, the protein attains conformational change into a specific 3D form for proper functioning. According to the mode of folding, four levels of protein’ organisation have been recognised namely primary, secondary, tertiary and quaternary.
2. The primary structure is the linear arrangement of amino acids in a polypeptide chain.
3. Secondary structure arises when various functional groups are exposed on the outer surface of the molecular interaction by forming hydrogen bonds. This causes the amino acid chain to twist into a coiled configuration called a-helix or to fold into flat β-pleated sheets.
4. Tertiary protein structure arises when the secondary level proteins fold into a globular structure called domains.
5. Quaternary protein structure may be assumed by some complex proteins in which more than one polypeptide forms a large multi-unit protein. The individual polypeptide chains of the protein are called subunits and the active protein itself is called a multimer.
6. eg. Enzymes serve as a catalyst for chemical reactions in the cell and are non-specific. Antibodies are complex glycoproteins with specific regions of attachment for various organisms.

Question 2.
What are the types of cofactors? Explain each of them.
Answer:

1. Many enzymes require non-protein components called cofactors for their efficient activity. Cofactors may vary from simple inorganic ions to complex organic molecules. They are of three types: inorganic ions, prosthetic groups and coenzymes.
2. Holoenzyme: Active enzyme with its non-protein component.
3. Apoenzyme: The inactive enzyme without its non-protein component.
4. Inorganic ions help to increase the rate of the reaction catalysed by enzymes. Example: Salivary amylase activity is increased in the presence of chloride ions.
5. Prosthetic groups are organic molecules that assist in the catalytic function of an enzyme. Flavin adenine dinucleotide (FAD) contains riboflavin (Vit B2), the function of which is to accept hydrogen. ‘Haem’ is an iron-containing prosthetic group with an iron atom at its centre.
6. Coenzymes are organic compounds that act as cofactors but do not remain attached to the enzyme. The essential chemical components of many coenzymes are vitamins, eg. NAD, NADP, Coenzyme A, ATP.
   

Question 3.
Explain the structure of DNA with Watson and Crick modal.
Answer:

1. Watson and Crick shared the Nobel Prize in 1962 for their discovery, along with Maurice Wilkins, who had produced the crystallographic data supporting the model, Rosalind Franklin (1920-1958) had earlier produced the first clear crystallographic evidence of a helical structure. James Watson and Francis Crick of Cavendish laboratory in Cambridge built a scale model of the double-helical structure of DNA which is the most prevalent form of DNA, the B-DNA. This is the secondary structure of DNA.
2. As proposed by James Watson and Francis Crick, DNA consists of a right-handed double helix with 2 helical polynucleotide chains that are coiled around a common axis to form a right-handed B form of DNA. The coils are held together by hydrogen bonds which occur between complementary pairs of nitrogenous bases. The sugar is called 2-deoxyribose because there is no hydroxyl at position 2′. Adenine and thiamine base pairs have two hydrogen bonds while guanine and cytosine base pairs have three hydrogen bonds.
   Chargaff’s Rule:
   • A = T; G = C
   • A+G = T + C
   • A : T = G : C =1
3. As published by Erwin Chargaff in 1949, a purine pairs with pyrimidine and vice versa. Adenine (A) always pairs with Thymine (T) by a double bond and Guanine (G) always pairs with Cytosine (C) by a triple bond.

Question 4.
Explain any two factors affecting the rate of enzyme reaction, with the help of graphical representation.
Answer:
Enzyme Reactions: Enzymes are sensitive to environmental condition. It could be affected by temperature, pH, substrate concentration and enzyme concentration. The rate of enzyme reaction is measured by the amount of substrate changed or amount of product formed, during a period of time.
Temperature: Heating increases molecular motion. Thus the molecules of the substrate and enzyme move more quickly resulting in a greater probability of occurrence of the reaction. The temperature that promotes maximum activity is referred to as optimum temperature.

pH: The optimum pH is that at which the maximum rate of reaction occurs. Thus the pH change leads to an alteration of enzyme shape, including the active site. If extremes of pH are encountered by an enzyme, then it will be denatured.

Substrate Concentration: For a given enzyme concentration, the rate of an enzyme reaction increases with increasing substrate concentration.
Enzyme Concentration: The rate of reaction is directly proportional to the enzyme concentration:

Question 5.
Describe the structure and functions of various other polysaccharides.
Answer:

Choose the correct answer.

1. Which is the most abundant component in living organisms.
(a) Minerals
(b) Macromolecules
(c) Water
(d) Protein
Answer:
(c) Water

2. In a water molecule, the hydrogen and oxygen atom stick together by:
(a) Monovalent bond
(b) Covalent bond
(c) Hydrogen bond
(d) None of the above
Answer:
(b) Covalent bond

3. Morphine is the first alkaloid to be found from a plant called:
(a) Vinca rosea
(b) Sweet pea
(c) Delonix regia
(d) Papaver somniferum
Answer:
(d) Papaver somniferum

4. Indicate a macromolecule:
(a) Amino acid
(b) Protein
(c) Nucleotide
(d) Glucose
Answer:
(b) Protein

5. The number of sugar units present in oligosaccharides:
(a) 14 to 15
(b) 6 to 8
(c) 2 to 10
(d) 11 to 12
Answer:
(c) 2 to 10

6. Sucrose is a
(a) Polysaccharide
(b) Disaccharide
(c) Monosaccharide
(d) Triglyceride
Answer:
(b) Disaccharide

7. A test for the presence of starch by adding a solution of iodine gives:
(a) Greenish blue colour
(b) Reddish green colour
(c) blue-black colour
(d) Violet-pink colour
Answer:
(c) blue-black colour

8. Glycogen is not seen in the organs of the human body:
(a) Muscle fibre
(b) Liver
(c) Brain
(d) Kidney
Answer:
(c) Brain

9. Chitin is composed of
(a) Mucopolysaccharides
(b) Oligopolysaceharides
(c) Glycoprotein
(d) Dipolysaccharides
Answer:
(a) Mucopolysaccharides

10. Match the following.

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

11. Lipids do not include:
(a) Steroid
(b) Waxes
(c) Enzymes
(d) phospholipids
Answer:
(c) Enzymes

12. A molecule of glycerol bound to have:
(a) 5 fatty acids
(b) 6 fatty acids
(c) 4 fatty acids
(d) 3 fatty acids
Answer:
(d) 3 fatty acids

13. Indicate saturated fatty acids:
(a) Palmitic acid
(b) Oleic acid
(c) Linoleic acid
(d) None of the above
Answer:
(a) Palmitic acid

14. Phospholipids serve as a major structural component of:
(a) Feathers
(b) Cell membrane
(c) Leaves
(d) Skin
Answer:
(b) Cell membrane

15. Cholesterol is an example of:
(a) Membrane lipids
(b) Triglycerides
(c) Steroids
(d) Adipose tissue
Answer:
(c) Steroids

16. The term ‘protein’ was coined by:
(a) Watson
(b) Gerardus Johannes Mulder
(c) Erwin Chargaff
(d) Maurice Wilkins
Answer:
(b) Gerardus Johannes Mulder

17. First protein insulin was sequenced by;
(a) Fred Sanger
(b) Robert Brown
(c) Robert Hooke
(d) Christian Anfinsen
Answer:
(a) Fred Sanger

18. Protein is synthesized in:
(a) Mitochondria
(b) Golgi body
(c) Lysosome
(d) Ribosome
Answer:
(d) Ribosome

19. A linear arrangement of amino acids is a polypeptide chain is seen in:
(a) Secondary structure of the protein
(b) Primary structure of the protein
(c) Tertiary structure of the protein
(d) Quaternary protein structure
Answer:
(b) Primary structure of the protein

20. Protein denaturation is due to:
(a) Exposure to pressure
(b) Exposure to light
(c) Exposure to heat
(d) None of the above
Answer:
(c) Exposure to heat

21. In a polypeptide chain hydrogen bone is formed between some hydrogen atoms of:
(a) Oxygen and methane
(b) Ethylene and nitrogen
(c) Nitrogen and methane
(d) Oxygen and nitrogen
Answer:
(d) Oxygen and nitrogen

22. Disulfide bond is seen between some amino acids like:
(a) Glycine and alanine
(b) Serine and proline
(c) Cysteine and methionine
(d) Aspartate and glutamate
Answer:
(c) Cysteine and methionine

23. Synthesis of polysaccharides from simple sugars is termed as:
(a) Catabolic reaction
(b) Anabolic reaction
(c) Hydrolytic reaction
(d) Oxidative reaction
Answer:
(b) Anabolic reaction

24. Indicate the correct statement:
(a) The rate of reaction is indirectly proportional to the enzyme concentration.
(b) The rate of reaction is directly proportional to the enzyme concentration.
(c) The rate of reaction is indirectly proportional to an increase in temperature
(d) None of the above.
Answer:
(c) The rate of reaction is indirectly proportional to an increase in temperature

25. The increased concentration of malonate inhibits the reaction of the enzyme, succinic dehydrogenase. This type of inhibitors is termed as:
(a) Competitive inhibitors
(b) Non-competitive inhibitors
(c) Irreversible inhibitors
(d) None of the above
Answer:
(a) Competitive inhibitors

26. NADP serves as:
(a) Apoenzyme
(b) Holoenzyme
(c) Coenzyme
(d) None of the above
Answer:
(c) Coenzyme

27. Formation of new chemical bonds using ATP as a source of energy is the mode of action of the enzymes.
(a) Hydrolase
(b) Isomerase
(c) Lyase
(d) Ligase
Answer:
(d) Ligase

28. DNA and RNA are polymers of monomers called:
(a) Nucleoside
(b) Nucleotide
(c) Pyrimidine
(d) Dinucleotide
Answer:
(b) Nucleotide

29. Which of the RNA constitutes 80% of the total RNA:
(a) mRNA
(b) tRNA
(c) rRNA
(d) None of the above
Answer:
(c) rRNA

30. Who got a noble prize for the finding of the helical structure of DNA?
(a) Rosalind Franklin and Erwin Chargaff
(b) Maurice Wilkins and Rosalind Franklin
(c) James Watson and Francis Crick
(d) Robert Hooke and Robert Brown
Answer:
(c) James Watson and Francis Crick