Foundation Bhawan Bharti Biology

98 Foundation Science: Biology for Class 10 Fig. 9.3 Types of simple permanent tissues (a) Parenchyma cells in transverse section (b) Parenchyma cells in longitudinal section (c) Sclerenchyma cells in TS (d) Sclerenchyma cells in LS Experiment 2.2 Objective A. To examine the prepared slide of striped muscle fibres Observations You will find that the cells are cylindrical. The location of the nucleus is interesting, it is not in the centre of the cylinder. The nucleus may be seen just near the outside of the cell. When you examine the cytoplasm carefully, you will find striations arranged in a parallel fashion. Fig. 9.4 Striated or voluntary muscle fibre (a) Isolated skeletal fibre (b) Fibre details

Practicals 99 Objective B. To examine the prepared slide of nerve cells. Observations When you focus on an individual nerve cell, you will see a buldging part which is called cyton. In the centre of the cyton you will find a nucleus. If the material was specially prepared, you may be able to locate granules around the nucleus. Try to count the number of processes arising from the cyton. The smaller processes arising from the cyton look like roots and are called dendrites. A long, cylindrical process arising from the other side of the cyton is called an axon. It ends in a bulb called axon bulb. Fig. 9.5 Structure of a nerve cell VIVA VOCE 1. Define tissue. A tissue is a group of cells with similar structure organised to do a common function. 2. What is the difference between meristematic and permanent tissue? Meristematic tissue cells are capable of dividing, while permanent tissue cells are not. 3. Name the plant parts where parenchymatous cells are present. Parenchymatous cells are present in the soft parts of root, stem, leaves, flowers and fruits. 4. What are the examples of simple permanent tissue? Parenchyma, collenchyma, and sclerenchyma. 5. Name the plant tissue which is dead at maturity. Sclerenchyma. 6. Name the plant tissue which is mainly responsible for mechanical strength. Sclerenchyma. 7. Why are the sclerenchymatous cells hard? They are hard because their cell wall is thickened with lignin. 8. In which tissue is the cell wall perforated with pits? Sclerenchyma.

100 Foundation Science: Biology for Class 10 3. CHEMICAL ANALYSIS OF FOODSTUFFS AND THEIR ADULTERANTS Experiment 3.1 Objective To test the presence of starch in the given food sample Apparatus and materials required Test tubes, test-tube stand, test-tube holder, spirit lamp, dropper, filter paper, iodine solution, distilled water, and foodstuff (potato, rice, wheat or maize grains) Theory Starch, a complex carbohydrate, is composed of Test tube 15%–20% amylose and 80%–85% amylopectin. It is found in different kinds of cereals such as rice, wheat, Filtrate + Blue-black maize, etc. After reacting with iodine solution starch 2 drops of colour forms a dark, blue-black compound. The appearance of iodine solution blue-black colour is due to the presence of amylose in starch. Procedure 1. Take a few small, freshly cut pieces of potato or a Fig. 9.6 Test for the presence of few grains of rice or wheat or maize in a clean test starch in foodstuff tube. 2. Pour 10 mL distilled water into the test tube. 3. Now, boil the contents of the test tube for about 5 minutes. 4. Allow the test tube to cool. 5. Filter the contents of the test tube through a filter paper. 6. Test the obtained filtrate for the presence of starch by the following method. Observation Experiment Observation Inference Take 2 mL filtrate in a clean The colour of the filtrate Appearance of dark blue-black test tube. Add 2 drops of changes to dark blue-black. colour shows the presence of iodine solution to it by starch. a dropper. Precautions 1. Use test-tube holder for holding the test tubes and keep the mouth of the test tube away from yourself while heating. 2. Use clean test tubes. 3. Do not use too much of iodine solution. Experiment 3.2 Objective To test the presence of the adulterant metanil yellow in dal (pulse) Apparatus and materials required Test tubes, test-tube stand, test-tube holder, conc. HCl, mortar-pestle, filter paper, distilled water and a sample of dal

Practicals 101 Theory Metanil is a cheap dye which is commonly used in colouring non-food items like clothes. Government of India has introduced “Prevention of Food Adulteration Act” to prevent the use of harmful chemicals such as this dye in foodstuffs. Procedure 1. Grind 3–5 g of dal in a mortar-pestle. 2. Take this powdered dal in a clean test tube. 3. Pour 10 mL distilled water into the test tube and shake it well. 4. Filter the contents of the test tube through a filter Fig. 9.7 Test for the presence of metanil paper and use the filtrate to test for metanil yellow in pulse (dal) sample yellow by the following method. Inference Observation Experiment Observation Take 2 mL filtrate in a clean Yellow colour of the The change in the colour test tube. Add 2 drops of mixture changes to pink. from yellow to pink shows conc. HCl to it with the the presence of metanil help of a dropper. yellow. Precautions 1. Always use clean test tubes. 2. Use test-tube holder at the time of adding conc. HCl and keep the mouth of the test tube away from yourself. 3. Do not add excess conc. HCl. 4. Handle the bottle of conc. HCl carefully. VIVA VOCE 1. What are the major components of food? Carbohydrates, proteins and fats are the major components of food. 2. What is starch? Starch is a polymer of glucose and is a complex carbohydrate. 3. What are food adulterants? Cheap and inferior materials mixed with foodstuff are called food adulterants. 4. In which form is carbohydrate stored in plants? Starch 5. Name some good sources of starch. Potato, rice, wheat and maize 6. What harm can be caused by food adulteration? Several diseases in human beings, and even death may be caused by consuming adulterated food. 7. Name some common sources of protein. Pulses, eggs and milk 8. In which form is carbohydrate stored in animals? Glycogen 9. Name the disease caused by eating arhar dal adulterated with khesari dal. Lathyrism 10. Most food items are marked with ISI, FPO and Agmarks. What are their full forms? ISI–Indian Standard Institute, FPO–Food Products Order, and Agmarks—Agricultural Marketing

102 Foundation Science: Biology for Class 10 4. TRANSPORTATION IN PLANTS—ENDOSMOSIS IN RAISINS Experiment Objective To determine the percentage of water absorbed by raisins Apparatus and materials required A bowl, forceps, a common balance, a weight box, blotting paper and a few raisins Theory When raisins are placed in water (a hypotonic solution), they absorb water by a process called osmosis and swell up. Osmosis is the net movement of solvent molecules from a region of their higher concentration to a region of their lower concentration through a semipermeable membrane. The movement of water into the cells of the raisins through their cell membranes continues until the cells become turgid. This process is called endosmosis. If these swollen raisins are now kept in a concentrated salt or sugar solution (a hypertonic solution), water from the turgid raisins will come out of the cells and they will shrink. This process is called exosmosis. The percentage of water absorbed by the raisins due to endosmosis can be measured by knowing the difference of their initial weight and final weight. Procedure 1. Take three raisins and weigh them on the common balance. Let this value be W1 . 2. Keep these raisins in a bowl containing water for 2 hours. 3. Take the raisins out of water and gently dry them with the help of blotting paper. 4. Weigh the soaked swollen raisins again on the common balance. Let this value be W2 . Fig. 9.8 Experiment to show endosmosis Observations 1. Weight of dry raisins = W1 . 2. Weight of swollen raisins = W2 . 3. Weight of the water absorbed by raisins = W2 -W1 . 4. Percentage (%) of water absorbed by raisins = W2 -W1 ´ 100. W1 Result The soaked swollen raisins weigh more than the dry raisins. This is because the raisins absorbed water by the process of endosmosis. Precautions 1. The raisins should be weighed accurately.

Practicals 103 2. The raisins should be immersed completely in water. 3. Before weighing the soaked raisins, these should be dried gently with the help of blotting paper. VIVA VOCE 1. Define osmosis. The net movement of solvent molecules from a region of their higher concentration to a region of their lower concentration through a semipermeable membrane is called osmosis. 2. By which process do water molecules diffuse out from a living cell? Exosmosis 3. Which molecules can move freely across the semipermeable membrane of plant cells? Water molecules 4. How long does endosmosis continue? Endosmosis continues until the concentration of water molecules becomes equal inside and outside the cell. 5. What is endosmosis? The movement of water into the cell across a semipermeable membrane is called endosmosis. 6. Why should we dry the raisins with blotting paper gently after taking them out of water? Any extra water present on the outer surface of the raisins will increase the final weight and give an incorrect result. 7. What is a hypotonic solution? A solution in which the concentration of solute molecules is less than the concentration of solute molecules in a cell sap is called a hypotonic solution. 8. How is osmosis different from diffusion? In osmosis a semipermeable membrane is present between the two solutions, while in diffusion there is no such membrane. 5. DIVERSITY OF PLANTS Experiment Objective To study the characteristics of Spirogyra, Agaricus, Moss, Fern, Pinus (either with male or female cone) and an angiospermic plant. Draw and give two identifying features of the groups they belong to. Apparatus and materials required A slide of Spirogyra, specimen of Agaricus, moss, fern, Pinus with a male female cone, an angiospermic plant, like mustard, hand lens and a compound microscope. Theory Diverse organisms have a wide range of sizes, structures, forms, shapes and distributions on the earth. There are more than 10 million organisms on the earth. About 1.7 million of them (1.2 million animals and 0.5 million plants) have been identified, scientifically named and classified. Spirogyra (green algae) Characteristics 1. Spirogyra is a green alga having a filamentous, unbranched, multicellular and threadlike structure. 2. Each filament has a large number of rectangular (length being more than breadth) cells. 3. Each cell has two parts: the thick, two-layered cell wall (outer wall made up of pectin and the inner wall is cellulosic) and the protoplasm.

104 Foundation Science: Biology for Class 10 4. The filaments of Spirogyra are slimy to touch due to the dissolution of their outer pectin layer. 5. The cytoplasm has a large vacuole at the centre and 1–16 ribbon-shaped, spirally coiled chloroplasts. Each chloroplast has a number of small round bodies called pyrenoids. 6. A large nucleus is suspended in the centre of the cell by a number of cytoplasmic strands. Fig. 9.9 (a) A part of the filament of Spirogyra (b) Detailed structure of a cell Characteristic features of the group 1. The green algae Spirogyra belongs to the group Thallophyta. The members of this group have an undifferentiated body called thallus. 2. No vascular system is found in the members. 3. Algae are autotrophic, i.e., they synthesize food by photosynthesis as they have the chlorophyll pigments. Agaricus (mushroom) Characteristics 1. Agaricus is a common, white, fleshy edible mushroom. 2. It grows in the rainy season on damp logs of wood, trunks of trees and decaying organic matter. 3. It is a saprophytic fungus. 4. The body is umbrella-shaped and is divided into a fleshy stalk, or stipe, and a fleshy pileus, or cap. 5. The pileus is dome-shaped, present at the top of stipe. The under surface of the pileus has many radiating strips called gills. Fig. 9.10 (a) Agaricus (b) Gills in enlarged view

Practicals 105 6. A membranous, ringlike structure called annulus is present on the stalk which covers the young basidiocarp. 7. The function of the basidiocarp is to produce and disperse spores. Characteristic features of the group 1. A mushroom is a fungus that belongs to the group Thallophyta. 2. The modes of nutrition in fungi are saprophytic or parasitic. 3. They do not possess chlorophyll, hence depend either on dead organic matter or on other living organisms for food. Funaria (moss) Characteristics 1. Mosses are commonly found growing in tufts on moist and shady walls, damp soil and on tree trunks. 2. The main plant body is a gametophyte (haploid) which is green, erect (1–3 cm high) and sparsely branched. 3. The plant body is differentiated into rootlike structures called rhizoids, axis, or stem and spirally arranged leaves. 4. The rhizoids are branched and multicellular which fix the plant to the soil and absorb water and minerals. Fig. 9.11 (a) Funaria (moss) plant bearing sporophyte (b) External features of a leaf

106 Foundation Science: Biology for Class 10 5. The plant is monoecious or dioecious, i.e., bear both male and female sex organs on the same plant. 6. The mature plant bears sporophyte which consists of foot, seta and capsule for asexual reproduction. Identifying features of the group 1. Mosses belong to Bryophyta. 2. Proper root and shoot systems are absent. Vascular tissues are absent in this group. 3. Rhizoids present in this group function as roots. Dryopteris (fern) Characteristics 1. Dryopteris is commonly found in shady and moist areas in tropical, subtropical and warm, temperate regions. 2. The plant body is a sporophyte (diploid) which is differentiated into roots, rhizome (underground stem) and leaves. 3. The primary root is short-lived. It is replaced by adventitious roots which grow from the rhizome. 4. The rhizome represents the modified stem. It is a creeping structure and its surface is covered with leaf bases and numerous thin brown hair called ramenta. 5. The leaves are large and bipinnately compound. The entire leaf is called a frond. It has a rigid, scaly petiole elongated to form a rachis bearing two rows of leaflets. Young leaves show circinate vernation (coiled inwards like a spring). 6. The lower (ventral) surface of mature leaves bear spore-producing structures called sori. Such sori-bearing leaves are called sporophyll. 7. Each sorus has many saclike sporangia (spore-bearing structures), which produce spores. Fig. 9.12 (a) External features of Dryopteris (fern) (b) Ventral veiw of a pinna-showing sori

Practicals 107 Identifying features of the group 1. Ferns belong to Pteridophyta. 2. The plant body is differentiated into root and shoot systems. 3. Vascular system is present in the members of this group. Pinus Characteristics 1. Pinus is commonly found on temperate and tropical hills. 2. The adult plant is a tall, evergreen tree with widespread branches giving a typical pyramidal shape. 3. The plant body is differentiated into tap root, stem and leaves. 4. The stem is thick, cylindrical and bears two types of branches—long shoots, or branches of unlimited growth and dwarf shoots, or branches of limited growth. 5. The leaves are of two types (dimorphic)—scale leaves which are nonphotosynthetic, found at the base of dwarf shoot and foliage leaves which are needle-like, green, photosynthetic, and found on the dwarf shoot. 6. The male and female reproductive parts in the form of male and female cones are present on the same plant, i.e., the plant is monoecious. Male cone (staminate strobilus) 1. The male cones are present in clusters (15–140) on long branches. 2. These are small dark brown, compact, oval structures which develop earlier than the female cones. Fig. 9.13 Pinus (a) Long and dwarf shoots with male and female cones (b) A part of stem showing two types of leaves and branches (c) Cluster of male cones (d) Female cone in different stages

108 Foundation Science: Biology for Class 10 3. Each male cone has a centrally located cone axis with many microsporophylls (space-producing structures) arranged spirally on it. 4. Each microsporophyll bears two microsporangia, or pollen sac, on its lower surface. 5. The microsporongia release winged-pollen grains which are carried by the wind to ovules. Female cone (ovulate strobilus) 1. The female cones are born in groups of 2–4 at the tips of long branches. 2. They are maroon-coloured, compact and bigger in size than the male cones. 3. Each female cone consists of a central cone axis covered with spirally arranged megasporophylls. 4. Megasporangia in the form of two naked sessile ovules are present on the dorsal surface of each megasporophyll. 5. The female cones take 3 years to mature. The first-year cones are very small and greenish in colour. The second-year cones are larger and woody with compact sporophylls which get separated during the third year due to elongation of the cone axis. Identifying features of the group 1. Pinus is a Gymnosperm. 2. The plant body has root and shoot systems with vascular tissues, but flowers are absent. 3. The members of this group bear naked seeds on the scales of cones. A dicotyledonous Angiosperm (mustard) Characteristics 1. The plant is an annual herb. Fig. 9.14 Brassica campestris (mustard). (a) A flowering plant (b) A fruit-showing seeds

Practicals 109 2. The plant body consists of the vascular shoot and root systems. 3. The root is a tap root. 4. The shoot system consists of stem, leaves, flowers and fruits. 5. The stem is green, erect, herbaceous, branched, solid and smooth, bearing prominent nodes and internodes. 6. The leaves are sessile, alternate and dorsiventral with lobed margin. 7. The flowers are yellow, tetramerous (4 petals and 4 sepals) and bisexual. All the four whorls of a flower, i.e., calyx, corolla, androecium (male) and gynoecium (female) are present. 8. After fertilization ovary of the flower develops into a fruit and the ovules present inside the ovary develop into seeds. The seeds are used for extracting mustard oil. Identifying features of the group 1. The angiospermic plants have well-developed root system and shoot system. 2. The plants bear flowers, fruits and seeds. 3. These plants are either monocotyledonous (seeds with one cotyledon) or dicotyledonous (seeds with two cotyledons). Monocots have leaves with a scattered arrangement of vascular bundles. Dicots have reticulate venation in leaves and their vascular bundles are arranged in ring. VIVA VOCE 1. What are algae? Algae are chlorophyll-bearing thallophytes which are usually aquatic. 2. What are thallophytes? The plants which are not differentiated into roots, stems and leaves, and are without vascular tissues are called thallophytes. 3. If an alga is devoid of chlorophyll, would you call it a fungus? No, there are certain algae which lack chlorophyll and are parasitic in nature. 4. What is the role of pyrenoids present in the chloroplast? Pyrenoids store starch and proteins of the cell. 5. Why is the Spirogyra commonly known as pond silk? The filaments of Spirogyra are slimy and silklike in texture. 6. What makes the Spirogyra filaments slimy to touch? The outer cell wall of Spirogyra is made of pectin which dissolves in water and forms a slimy mucilaginous envelope around the filament. 7. What is the most characteristic feature of Spirogyra? It has ribbon-shaped, spirally arranged chloroplasts. 8. Why are fungi heterotrophic? Fungi are heterotrophic because they obtain their food either from dead organic matter (saprophytes) or from living organisms. 9. What is the botanical name of edible mushroom? Agaricus campestris. 10. Can you grow mushrooms in your house? Yes, they can be grown in wooden trays containing decaying organic matter. 11. Why are mushrooms called saprophytes? Because they derive food from dead and decaying organic matter. 12. How do fungi differ from algae? Algae are autotrophic and have a cell wall made of cellulose, while fungi are heterotrophic and have cell wall made of chitin.

110 Foundation Science: Biology for Class 10 13. What are bryophytes? Bryophytes are nonvascular, thalloid, autotrophic plants with multicellular sex organs surrounded by a sterile jacket. 14. What is the dominant phase in Funaria? The gametophytic phase. 15. What type of leaves do you find in mosses? Small, sessile and spirally arranged on the stem. 16. What are sporophylls? Sporophylls are vegetative leaves bearing sporangia. 17. What is circinate vernation? Young leaves of Pinus remain coiled like a spring and open slowly from the base upwards as they mature. This is called circinate vernation. 18. How are pteridophytes different from bryophytes? A pteridophyte is differentiated into root, stem, leaves, has vascular tissues (xylem and phloem), and the sporophyte is independent. All these are absent in bryophytes. 19. What are gymnosperms? Gymnosperms are a group of plants with naked ovules (not enclosed in ovary) and naked seeds. 20. How many types of branches are found in Pinus? Two types of branches—long shoots and dwarf shoots. 21. Where are the foliage leaves found in Pinus? On the dwarf shoots. 22. What is a male cone? It is a compact aggregation of several microsporophylls bearing microsporangia. 23. In how many years does the female cone of Pinus mature? In 3 years. 24. What characters of Pinus classify it as a gymnosperm? The ovules in Pinus are exposed. After fertilization they develop into naked seeds. Reproductive organs remain in the form of male and female cones. Due to these characters Pinus is classified as a gymnosperm. 25. What are angiosperms? All flower-bearing plants are called angiosperms. 26. How will you define a flower? A flower may be defined as a modified shoot which is meant essentially for the reproduction of the plant. 27. Which group of angiosperms has reticulate venation in leaves and vascular bundles arranged in a ring? Dicots. 28. What are the characteristic features of monocots? Parallel venation in leaves, scattered vascular bundles in stem and trimerous flowers (floral whorls in groups of 3) are the characteristic features of monocots. 6. ADAPTATION IN ANIMALS Experiment Objective Observe and draw the specimens of earthworm, cockroach, bony fish and bird showing a specific feature and an adaptive feature

Practicals 111 1. Earthworm Examine the body of an earthworm. It is cylindrical. The body is divided into a number of segments. It has no other appendages. The anterior half has a band of skin called clitellum. A fresh specimen will be moist. The shape of the earthworm makes it a suitable burrower of the soil. Mouth Setae Clitellum Female genital aperture Genital Male genital pore papillae Anus Annuli Prostomium Peristomium Setae (a) (b) Fig. 9.15 Pheretima posthuma (earthworm). (a) Ventral view of the entire worm (b) Dorsal view of the anterior end 2. Cockroach Examine the body of a cockroach. You will find that it has 6 legs. Each leg has joints. A cockroach has two compound eyes. At the posterior end it has two anal cerci. The body is dorsoventrally flattened and is divided into 3 parts: head, thorax, and abdomen. The jointed legs make it a runner and its wings enable it to fly for a short distance. Note that the outer wings are leathery, but the inner ones are thin and membranous. Fig. 9.16 Periplaneta americana (cockroach) (a) Dorsal view of male (b) Ventral view of female

112 Foundation Science: Biology for Class 10 3. Bony fish Examine a bony fish. Note the position of the eyes, shape of the mouth, scales on the body, fins on the dorsal and ventral surfaces. Also examine the shape of the tail fin. You can also feel the endoskeleton by grabbing it firmly. The contour of the body is streamlined so that it can swim without much resistance. Fins also help in swimming. The gills are located just behind the head on the lateral sides. The gills are covered by an operculum. Gills enable the fish to obtain oxygen dissolved in water, and expel the carbon dioxide produced in the body. Fig. 9.17 Labeo rohita 4. Bird Examine a pigeon. Observe the position of the beak. See the types of plumage, forelegs and wings. Hindlegs have claws. Try to look for teeth in the jaws. You will find none. There are scales in the hindlegs. The body is light. The body plan of the bird helps it to fly long distances. Hold the bird gently with your hands. You will feel that its body temperature is higher than yours. Fig. 9.18 Columba livia (pigeon) 7. EXTERNAL FEATURES OF PLANT PARTS Experiment Objective To study the external features of root, stem, leaf and flower of monocot and dicot plants

Practicals 113 Apparatus and materials required Glass slides, forceps, hand lens, scissors, dissecting microscope, a complete monocot plant such as onion or paddy or wheat or maize, and a complete dicot plant such as mustard or sunflower or pea Theory The flowering plants, or angiosperms, are differentiated into root, stem, leaves and flowers. They bear seeds enclosed in a fruit. They are divided into monocotyledons and dicotyledons on the basis of the kind of seeds they bear. Monocotyledons bear seeds which have a single cotyledon. The seeds of dicotyledons have two cotyledons. Procedure 1. Take a monocot plant. Separate root, stem, a leaf and a flower of this plant with the help of scissors and place these parts on different slides separately with forceps. 2. Then take a dicot plant and repeat the process. 3. Now observe and compare the external features of root, stem, a leaf and a flower of the monocot and dicot plants using hand lens and subsequently by dissecting microscope. Observation Root Identifying features of root 1. The part of a plant that generally develops from the radicle of embryo is called root. 2. It fixes the plant firmly into the ground and provides rigidity against wind and water. 3. It absorbs water and minerals from soil. 4. It grows towards the centre of gravity, i.e., it is positively geotropic. 5. It possesses unicellular root hairs. 6. It normally grows away from light, i.e., it is negatively phototropic. 7. It does not bear buds, leaves and flowers, and lacks nodes and internodes. 8. The root has four regions from the apex to the base: (i) Root cap (ii) Region of cell division (apical meristem) (iii) Region of elongation (iv) Region of maturation Fig. 9.19 Regions of the root

114 Foundation Science: Biology for Class 10 9. The root cap protects the growing root apex while the main growing region of the root lies just behind the root cap. Monocot Root 1. In monocots, primary root does not persist for a longer period. It is soon replaced by a cluster of long, threadlike roots which originate from the base of the stem. These roots are called fibrous roots. 2. Roots developing from any other part of the plant than radicle are called adventitious roots. 3. Fibrous root is a type of adventitious root. 4. Due to absence of secondary growth in thickness these roots remain slender. Fig. 9.20 Types of root (a) Monocot root (b) Dicot root Dicot Root 1. In most of the dicots, root develops directly from the radicle. 2. It grows longer, thickens and is known as primary root. 3. It persists and becomes stronger to form tap root. 4. It generally produces lateral branches called secondary roots. 5. Branches of the secondary roots are called tertiary roots. 6. Tap root along with its branch system is called tap root system. Stem Identifying features of stem 1. The part of the plant that develops from the plumule of embryo is called stem. 2. It forms the axis and is the ascending part of the plant. 3. It is differentiated into nodes and internodes. 4. It bears leaves and branches at the nodes. The part of stem that lies between two nodes is called internode. 5. It is positively phototropic, i.e., grows towards light and negatively geotropic, i.e., grows away from the gravity. 6. The shoot (stem and its branches) is usually green and photosynthetic. 7. The apex of stem is called shoot tip. It bears apical bud which is responsible for elongation of the plant. Shoot apex lacks cap. 8. Stem bears either unicellular or multicellular hair, or trichomes.

Practicals 115 9. The main function of the stem is to support leaves and branches and hold them in a position to receive maximum light. Thus it forms the main skeleton of the plant. Monocot Stem 1. It is aerial, erect, herbaceous or woody, usually unbranched. 2. It is usually differentiated into solid nodes and hollow internodes. In maize, internodes are also solid. 3. In some members stem is modified into underground organs like rhizome (e.g., ginger), corm (e.g., Colocasia) or bulb (e.g., onion). Fig. 9.21 External features of plants (a) A monocot plant (onion) (b) A dicot plant (mustard) Dicot Stem 1. It is normally long, erect, herbaceous or woody, cylindrical and branched. 2. It has distinct nodes and internodes. Both the nodes and internodes are solid. 3. Sometimes it is creeping and modified into tendril. 4. It is often four-angled (quadrangular) or five-angled (pentangular). 5. In potato, the underground stem is modified into tubers. Leaf Identifying features of leaf 1. It is the lateral appendage of the stem that arises at the node. 2. It bears a bud in its axil. 3. It is attached to the stem with the help of a structure called the leaf base. 4. A stalk called petiole develops from the leaf base which bears a green flattened structure called lamina. 5. Lamina, or leaf blade, has midrib, veins, leaf apex and leaf margin.

116 Foundation Science: Biology for Class 10 6. The leaves are grouped into two categories—simple and compound—on the basis of incision. Simple leaves have a single lamina. When the incision of the lamina goes down to the midrib, the leaf becomes compound having a number of leaf segments called leaflets. 7. The main functions of leaves are synthesis of food (photosynthesis), transpiration, and exchange of gases through its pores called stomata. 8. Sometimes leaves get modified for storage, defence, support, reproduction and trapping insects. Monocot leaf 1. Leaves are arranged isobilaterally, i.e., Fig. 9.22 Venation in leaves (a) Reticulate (b) Parallel both surfaces are similar. 2. The venation (arrangement of veins and veinlets on the lamina) is parallel. In monocots veins run parallel to each other from base to the tip of the lamina. Veinlets connecting the adjacent longitudinal veins are inconspicuous. 3. Leaves are usually long and narrow, running parallel to the stem. 4. Leaves are mostly simple. 5. Leaf sheath (expansion of leaf base into a broad sheath) is usually present. Dicot Leaf 1. Leaves are arranged dorsiventrally, i.e., upper and lower surfaces are distinctly different. 2. In dicot leaves, venation is reticulate, i.e., irregularly distributed to form a network. 3. In most dicots, the leaf base bears two lateral appendages called stipules. 4. Leaves are either simple or pinnately compound. 5. Leaf sheath is usually absent. Flower Identifying features of flower 1. The reproductive part of an angiospermic (higher) plant is flower, which develops from floral buds. 2. The flower is considered to be a modified shoot. 3. The stalk of the flower is called pedicel and the tip of the pedicel continues as an enlarged axis called thalamus or receptacle. 4. All the floral parts are arranged on the thalamus in a definite sequence. 5. A typical flower consists of four sets of floral parts, or whorls: calyx (sepals), corolla (petals), androecium (stamens) and gynoecium (carpels). 6. The first two whorls, i.e., calyx and corolla are not directly involved in reproduction and are called accessory whorls. 7. The inner two whorls, i.e., androecium and Fig. 9.23 Diagram of different parts of a flower gynoecium are directly concerned with sexual reproduction and are called essential whorls.

Practicals 117 8. Sepals form the outermost whorl called calyx. They are usually green and leaflike, and arise at the base of the flower. 9. Petals form the corolla. They are generally brightly coloured and sometimes fragrant to attract insects. 10. The third whorl androecium is the male reproductive part of the flower and consists of stamens. Each stamen consists of a slender filament and an anther at the tip. 11. Gynoecium, or pistil, is the centrally placed fourth whorl which bears the female reproductive organ called carpel. Each pistil consists of a basal swollen ovary, a narrow stalklike style, and stigma at the tip. The ovary contains one or many ovules. Monocot Flower 1. Calyx and corolla are not distinct in monocot flowers. Instead, perianth is present which is composed of tepals. 2. In monocots, flowers appear in clusters. 3. The flower is typically trimerous (each whorl is in multiple of three). 4. Stamens are usually versatile, i.e., filament is attached to the back of anther at a point only. Dicot Flower 1. Dicot flowers usually have distinct floral parts, i.e., calyx, corolla, androecium and gynoecium. 2. Calyx is composed of sepals, and corolla is composed of petals. 3. Flower is mostly pentamerous (each whorl in multiple of five), sometimes tetramerous (each whorl in multiple of four). 4. In dicots, flowers usually appear separately. 5. Stamens are usually basifixed, i.e., filament is attached to the base of the anther. VIVA VOCE 1. Name the part of the plant which develops from the radicle in dicots. Tap root or primary root 2. Name the structure which protects the root tip. Root cap 3. Which part of the embryo forms the root in a plant? Radicle 4. Why is root said to be positively geotropic? Because it grows towards the centre of gravity. 5. What is the primary function of root? It absorbs water and minerals from the soil. 6. Which type of root is found in monocots? Fibrous root 7. Name the root that develops from any unusual part of the plant body. Adventitious root 8. Name the part of the plant which develops from plumule of embryo. Stem 9. Name the part of the stem that lies between two nodes. Internode

118 Foundation Science: Biology for Class 10 10. What is the main function of stem? The main function of stem is to support leaves and branches and hold them in a position to receive maximum light. 11. In which type of stem are internodes usually hollow? Monocot stem 12. What is the difference between simple leaf and compound leaf? Simple leaf has single lamina whereas compound leaf has many leaflets. 13. What is the importance of leaves? The leaves help in photosynthesis, transpiration and exchange of gases. 14. What is the difference between monocot leaves and dicot leaves? Monocot leaves are isobilateral having parallel venation while dicot leaves are dorsiventral having reticulate venation. 15. What is venation? The arrangement of veins and veinlets on the lamina is called venation. 16. What is pedicel? The stalk of the flower is called pedicel. 17. Define flower. Flower may be defined as a modified shoot which is essentially meant for reproduction of the plant. 18. What are angiosperms? All flower-bearing plants are called angiosperms. 19. Which type of flowers are usually found in monocots and dicots? Trimerous flowers in monocots, and pentamerous flowers in dicots 20. Name the reproductive organs of a flower. Stamen and carpel 21. Name the outermost whorl of a flower. Calyx 22. Where are perianth found? In monocot flowers 23. What is perianth? It is the outer part of monocot flower in which calyx and corolla are not always distinguishable. 8. LIFE CYCLE OF MOSQUITO Experiment Objective To study the life cycle of mosquitoes Apparatus and materials required Specimens of different stages of life of mosquito, compound microscope Theory Mosquitoes are small harmful insects, found everywhere, especially in damp, dark places. They breed in stagnant water. Eggs develop into larvae, larvae into pupa, and pupa develop into imago, or an adult. The different stages are clearly distinguishable.

Practicals 119 In India, Culex and Anopheles are mostly found. Observation Eggs Collect eggs from stagnant water bodies such as the nearby drain or ditch, and keep them in a wide-mouthed bottle containing some water. Take the eggs by means of a dropper and put them on the glass slide and examine under the compound microscope. Eggs are laid by the female mosquito in the night in stagnant water. Eggs are slightly brown and extremely small. But on taking a closer look they can be identified as rafts composed of cigar-shaped bodies glued together, or free boat-shaped bodies. Cigar-shaped eggs belong to Culex and boat-shaped ones belong to Anopheles. At a time, Culex lays up to 300 eggs, and Anopheles, 100 eggs. Eggs float on the surface of the water. Examine the sample and note the shape and size of the eggs. Draw a sketch of the shape of the eggs and infer whether they belong to Culex or Anopheles. After 2–3 days, you will notice that these eggs are ruptured, and from each one of them an elongated creature emerges. This is called larva. (a) (b) Fig. 9.24 Eggs of (a) Culex and (b) Anopheles Larvae Larvae are elongated, hairy, segmented creatures that move about and feed upon algae growing in water. They swim (wriggle) in a characteristic jerking manner and hence they are also called wrigglers. The anterior part of the body has somewhat indistinct head equipped with mouth parts, compound eyes, and antennae. The body is divided into ten segments. Each segment has a few bristles, or hairs. The posterior part of the larva bears gills, respiratory siphons, and the comb. Examine the larvae under the compound microscope to observe the movement of the jaws and the body. You can count the number of segments in the body and observe gills and siphons located at the posterior end. Draw a sketch of a larva. Brush Brush Antenna Head Antenna Compound eye Compound eye Thorax Thorax Bristles Abdomen Abdomen Tracheal gills Comb Tracheal gills Respiratory siphon (a) (b) Fig. 9.25 Larvae of (a) Culex and (b) Anopheles

120 Foundation Science: Biology for Class 10 If you see the water surface of the cylinder or the container in which the larvae are kept, you will find larvae hanging by the air-water interface. The posterior part of the larva is always in contact with the meniscus, as respiratory siphons have to draw atmospheric air for respiration. The gills remain submerged in water to draw dissolved oxygen of water. The larva of Anopheles lies parallel to the water surface. The head of the larva of Culex hangs downward at an angle. Larval life lasts for two weeks during which it casts its skin four times. After that the larva is transformed into another form called pupa. Pupae Pupal life is sluggish. There is no feeding, no movement except for occasional tumbling in water. That is why a pupa is called a tumbler. The head of the pupa is large, formed by the fusion of the head and thorax. Therefore, it is called cephalothorax, which bears on its dorsal surface two respiratory trumpets. Legs and wings, which are in the early stages of formation, lie on the ventrolateral surfaces. The pupal life lasts for a week. Fig. 9.26 Pupae of (a) Culex and (b) Anopheles Imago (young adult) At the end of pupal life, the skin (cuticle) of the pupa splits along mid-dorsal line, making it easy for the imago to come out. The wings of the newly emerged imago are dried and spread out, and it flies off to lead an aerial life. Fig. 9.27 (a) Adult Culex (b) Adult Anopheles

Practicals 121 FOR CLASS 10 1. PREPARATION OF TEMPORARY MOUNT Experiment Objective To prepare a temporary mount of a leaf peel to show stomata Apparatus and Materials Required A potted Tradescantia or Bryophyllum plant, forceps, needles, watch glasses, glass slides, a dropper, coverslips, a brush, blotting paper, safranin, glycerine and a compound microscope Theory Stomata are small openings found widely scattered on the epidermis of leaves and young stems. They are mostly found on the lower surface of a dicot leaf and on both the surfaces of a monocot leaf. Stomata regulate the exchange of gases and water vapour between the atmosphere and leaves. Procedure 1. Remove a healthy leaf from the potted plant. 2. Remove a part of the peel from the lower surface of the leaf. You can do this by folding the leaf over and gently pulling the peel apart using forceps. Keep the peel in a watch glass containing water. 3. Put a few drops of safranin stain in a watch glass. 4. After 2–3 minutes take out the peel and place it on a clean glass slide. 5. Put a drop of glycerine over the peel and place a clean coverslip gently over it with the help of a needle. 6. Remove the excess stain and glycerine with the help of blotting paper. 7. Observe the slide under the low-power and high-power magnifications of the compound microscope. Observations 1. The epidermal cells are visible. These are irregular in outline and have no intercellular spaces. 2. Many small pores (stomata) are seen scattered among the epidermal cells. 3. Each pore is guarded by two bean-shaped guard cells, each containing chloroplasts and a nucleus.

122 Foundation Science: Biology for Class 10 Fig. 9.28 (a) Mounting a leaf peel (b) Epidermal layer in the peel taken from a dicot leaf showing open stomata (c) High-power magnification of stomata 4. The inner concave boundary of each guard cell is thick, whereas its outer boundary is thin. 5. The stomata may be open or closed. The guard cells regulate the opening and closing of the stomata. Result Stomata are present in the epidermal cells of the lower surface of the leaf. Precautions 1. Cut the peel to a proper size and avoid folding it. 2. Always place the peel at the centre of the slide and hold the slide at the edges. 3. Do not overstain or understain the peel. 4. Always handle the peel with a brush as a needle may damage the cells. 5. Take care to prevent the peel from drying by using glycerine. 6. Place the coverslip gently, avoiding any air bubbles. 7. Remove excess stain and glycerine with a blotting paper. VIVA VOCE 1. Why do we take an epidermal peel from the lower surface of the leaf? More stomata are present on the lower surface of a dicot leaf than on the upper surface. 2. What are the functions of stomata? Exchange of gases and transpiration 3. Where are stomata located in monocot plants? Stomata are found on both surfaces of the leaves in monocot plants. 4. Which stain is used while preparing a temporary mount of a leaf peel? Safranin 5. Name the cells which surround a stoma. Guard cells 6. How are the opening and closing of stomata regulated? Turgidity of the guard cells regulate the opening and closing of stomata. 7. How do the stomata of dicots and monocots differ? In dicots the guard cells are kidney-shaped, while they are dumb-bell-shaped in monocots. 8. Why are there no stomata in submerged aquatic plants? Submerged aquatic plants remain inside water, and the diffusion of gases occurs through their general body surface.

Practicals 123 9. Why do we use glycerine for mounting a leaf peel? To prevent the cells from drying up 10. How are stomata differentiated from the surrounding epidermal cells? Compared to the epidermal cells, the guard cells are smaller, different in shape and have thicker inner walls. 2. PHOTOSYNTHESIS Experiment Objective To show experimentally that light is essential for photosynthesis Apparatus and Materials Required A healthy potted plant, a Petri dish, a beaker containing water, forceps, a water bath, a piece of wire gauze, a tripod, a burner, a box of matches, alcohol a strip of black paper, iodine solution and clips. Theory Photosynthesis is a biochemical process by which green plants synthesize simple sugar in the presence of sunlight using carbon dioxide from the atmosphere and water from the soil. This simple sugar (glucose) is later converted to starch. 6CO2 + 12H2 O ¾C¾hl¾oro¾phy¾ll ® C6 H12 O6 + 6H2 O+ 6O2 Sunlight The most important factor for photosynthesis is light. The rate of photosynthesis depends on the quantity and quality of light. The chlorophyll molecules in green leaves absorb light, get excited and emit electrons. The emitted electrons are used in the production of adenosine triphosphate (ATP). Finally the solar energy is converted into chemical energy and stored in the glucose produced. The rate of photosynthesis is the maximum in the presence of red and blue lights, while in green light the rate is minimum because green light is reflected by the chlorophyll molecules. Procedure 1. Take the potted plant and keep it in a dark place for 2–3 days so that the leaves get destarched. 2. Cover a part of one of its leaves with the strip of black paper. Make sure that you cover both the sides of the leaf. 3. Now place this plant in sunlight for 3–4 hours. 4. Pluck the selected covered leaf and remove the black paper covering it. 5. Place this leaf in the beaker containing water and boil it for about 10 minutes. 6. Take out the leaf and now boil it in alcohol, using the water bath, for 10 minutes. This removes the chlorophyll. 7. Take out the leaf and wash it under running water. 8. Place this leaf in the Petri dish and put a few drops of iodine solution on it. Now observe the change in colour.

124 Foundation Science: Biology for Class 10 Fig. 9.29 Experiment to show that light is essential for photosynthesis Observations The leaf turns blue-black except in the covered region. As this covered region did not receive light, photosynthesis did not occur. Hence no starch was formed there. The uncovered region received light and starch was formed there due to photosynthesis. Result Light is essential for photosynthesis. Precautions 1. Before starting the experiment, the leaf must be destarched. 2. The leaf must be covered with black paper properly to prevent the entry of light. 3. Boiling the leaf in alcohol should be done in the water bath. VIVA VOCE 1. What happens when light falls on chlorophyll molecules? The chlorophyll molecules absorb the light, get excited and emit electrons. 2. What does destarching mean? When green plants are kept in darkness for 2–3 days, photosynthesis does not occur. Sugar is not synthesized and hence not stored as starch. This is called destarching the plants. 3. What are the factors that affect the rate of photosynthesis? Light, carbon dioxide and temperature affect the rate of photosynthesis. 4. Is the rate of photosynthesis same throughout the day? No, the rate of photosynthesis varies with the amount and intensity of light available at different times during the day. 5. What is the source of the oxygen liberated during photosynthesis? Water 6. Why do we use a water bath for boiling the leaf in alcohol? The vapours of alcohol may catch fire if heated directly, without using a water bath. 7. Why is the leaf boiled in alcohol? To remove all the chlorophyll pigments and for decolouring the leaf 8. Which chemical is used to test the presence of starch? Iodine solution 9. Why does the uncovered portion of the leaf turn blue-black after putting iodine solution on it? The uncovered portion receives light and synthesizes starch by photosynthesis. 10. Why is the rate of photosynthesis reduced considerably in green light? Green light is reflected by the chlorophyll molecules and is not used in photosynthesis.

Practicals 125 3. RESPIRATION Experiment Objective To show experimentally that carbon dioxide is released during respiration Apparatus and Materials Required A conical flask, a beaker, a cork with a hole, a glass tube bent at right angles at two places, a small test tube, KOH solution, thread, coloured water, Vaseline and germinating seeds of gram or pea Theory Respiration is a catabolic process which involves the breakdown of food or complex organic molecules into simpler products, with the release of energy. This process can take place either in the presence of oxygen (aerobic respiration) or in its absence (anaerobic respiration). The overall reaction mechanism of aerobic respiration involves the oxidation of carbohydrate and the subsequent production of CO2 , H2 O and energy. C6H12 O6 + 6O2 ¾¾® 6CO2 + 6H2 O + Energy Procedure 1. Take the conical flask and place some germinating gram or pea seeds in it. 2. Insert the shorter end of the glass tube through the hole in the cork and fix it on the conical flask. 3. Before fixing the cork, hang a test tube containing KOH solution inside the conical flask with the help of a thread. 4. Take coloured water in the beaker and keep the longer end of the glass tube dipped inside it. 5. Make the conical flask airtight by applying vaseline on its rim. 6. Note the initial level of water in the tube. 7. Observe and note the rise in the water level after an hour, without disturbing the apparatus. Fig. 9.30 Experiment to show that CO2 is released during respiration Observations Water level rises up in the bent tube. Result The rise in the level of water indicates that CO2 is produced by germinating seeds during respiration. Actually, the germinating seeds respire and produce CO2 , which is absorbed by KOH solution. This creates a vacuum in the conical flask. The air present in the bent glass tube moves into the conical flask. This pulls the water in the bent tube further up. Precautions 1. Keep the conical flask airtight. 2. Fix the shorter end of the glass tube in such a way that it does not touch the seeds. 3. Use freshly prepared KOH solution.

126 Foundation Science: Biology for Class 10 VIVA VOCE 1. Why do we select germinating seeds for studying respiration? Germinating seeds respire actively. 2. Why is KOH solution used in this experiment? KOH solution absorbs the CO2 released during respiration. 3. Why do we use coloured water in this experiment? To record the rise in the level of water in the glass tube correctly 4. What is the difference between a catabolic process and an anabolic process? In a catabolic process complex molecules are broken down into smaller and simpler molecules, while in an anabolic process complex molecules are formed from simpler molecules. 5. Which cell organelle is associated with aerobic respiration? Mitochondrion 6. What are respiratory substrates? The substances oxidized during respiration are called respiratory substrates, e.g., carbohydrates, fats and proteins. 7. Define fermentation. The incomplete anaerobic breakdown of a respiratory substrate with the production of CO2 and energy is called fermentation. 8. Differentiate between aerobic and anaerobic respiration. Aerobic respiration takes place in the presence of O2 and involves the complete oxidation of food, releasing CO2, H2O and energy. Anaerobic respiration takes place in the absence of O2 and involves the incomplete oxidation of food, releasing ethyl alcohol and CO2. 9. What is the difference between the processes of respiration and photosynthesis? Respiration is a catabolic process, whereas photosynthesis is an anabolic process. 10. Which type of respiration releases more energy? Aerobic respiration 11. Which substance is used to make the conical flask airtight? Vaseline (petroleum jelly) 4. ASEXUAL REPRODUCTION—BINARY FISSION AND BUDDING Experiment Objective To study (a) binary fission in Amoeba and (b) budding in yeast with the help of prepared permanent slides Apparatus and Materials Required Permanent slides of Amoeba showing binary fission and yeast in budding, and a compound microscope Theory Reproduction is one of the basic characteristics of a living organism. An organism reproduces to produce more of its own kind. Reproduction may be either asexual or sexual.

Practicals 127 The type of reproduction that takes place without gametes forming is called asexual reproduction. Sexual reproduction is the type of reproduction in which both the male and female gametes are involved. Asexual reproduction is common in lower plants and some lower animals. It is a process of rapid multiplication in which the new organisms produced are genetically identical to the parent. Sexual reproduction is common in higher plants and most animals. The organisms produced by this method are not genetically identical to the parents. Asexual reproduction may be of various types such as binary fission, multiple fission, budding, fragmentation, sporulation and vegetative propagation. Procedure 1. Observe each permanent slide first under the low-power magnification and then under the high-power magnification of a compound microscope. 2. Draw diagrams and compare their features. Observations Binary fission in Amoeba 1. In this division, two similar individuals are formed from a single parent. 2. A mature Amoeba cell is larger. Its nucleus elongates and gradually divides amitotically into two. 3. The division of the nucleus is followed by the division of the cytoplasm. 4. Thus two new amoebae are formed from a single parent and the parent’s identity is lost. Fig. 9.31 Binary fission in Amoeba Budding in yeast 1. In this type of asexual reproduction, bulblike projections called buds arise from the parent body. 2. Mature yeast cells are larger, and spherical or oval in shape. Fig. 9.32 Budding in yeast

128 Foundation Science: Biology for Class 10 3. One or more bulblike projections (buds) arise from the cell membrane. 4. The nucleus of the parent cell divides and one of the daughter nuclei passes into the bud. 5. The bud is finally separated from the parent body and grows into a new individual. 6. The parent’s identity is maintained in budding. Result The prepared slides show asexual reproduction in which only one individual is involved in the production of new individuals. Precautions 1. Focus the slides properly. 2. Study the slides first under low-power magnification and then under high-power magnification of the compound microscope. 3. Draw diagrams as seen under the microscope. VIVA VOCE 1. Define binary fission. Binary fission is the division of a single parent cell into two similar daughter cells. 2. What is budding? Budding is a type of asexual reproduction in which a new individual develops within the body wall or from the cell membrane of the parent by forming a bulblike outgrowth, called bud. 3. What is the basic difference between binary fission and budding? In binary fission the parent’s identity is lost, while it is maintained in budding. 4. What are the different modes of reproduction? There are two modes of reproduction—asexual and sexual. 5. What happens during sexual reproduction? In sexual reproduction, the male and female partners produce gametes, which later fuse to form new individuals. 6. Name a multicellular organism that reproduces by budding. Hydra 7. Which organelle divides first during binary fission? Nucleus 8. Name the type of nuclear division that occurs during binary fission. Amitosis 9. By which mode of reproduction are new individuals produced rapidly? Asexual reproduction 10. What type of organism is yeast? Yeast is a unicellular fungus. 5. IDENTIFICATION OF DICOT EMBRYO Experiment Objective To identify the different parts of an embryo of a dicot seed (Pea, gram or red kidney bean) Apparatus and Materials Required Permanent slides of dicot embryo showing different stages, and a compound microscope

Practicals 129 Theory During sexual reproduction male gamete fuses with female gamete and zygote is formed. After some rest period zygote divides, redivides and finally develops into an embryo. In the earlier stages of development of embryo there is no difference between monocots and dicots, but their development differs in later stages. The process of development of mature embryo from zygote is called embryogeny. Procedure 1. Observe each permanent slide in sequence from early stages to maturity first under low-power magnification and then under high-power magnification of a compound microscope. 2. Draw diagrams and identify the different parts. 3. Label the different regions of the embryo. Observation 1. Zygote divides transversely forming a basal cell towards the micropyle and a terminal cell towards the chalaza. 2. The basal cell divides by transverse divisions and finally forms 6–10 celled suspensor. 3. The uppermost cell of suspensor swells up to form a vesicular cell while the lowest cell of the suspensor is called hypophysis. Hypophysis forms part of the radicle and root cap. 4. The terminal embryonal cell divides by transverse and vertical divisions and forms 16-celled globular embryo. 5. The globular embryo later becomes heart-shaped due to its differentiation into cotyledons. 6. The embryo and cotyledons become larger and curved. Vesicle Cotyledons Suspensor Stem tip Heart-shaped embryo Root region Suspensor Suspensor cell Inner integument of ovule Embryonal cell Outer integument of ovule Hypophysis Globular embryo Fig. 9.33 Different stages in the development of dicot embryo Result The prepared permanent slides show successive stages of the development of dicot embryo. The young dicot embryo is globular but changes to heart-shaped structure at maturity due to differentiation into cotyledons. Precautions 1. Focus the slides properly.

130 Foundation Science: Biology for Class 10 2. Study the slides first under low-power magnification and then under high-power magnification of the compound microscope. 3. Draw diagrams of the stages as seen under the microscope. VIVA VOCE 1. What is the term used for the process of development of mature embryo from zygote? Embryogeny 2. Where is vesicular cell present? A vesicular cell is present at the top of the suspensor. 3. What is hypophysis? The lowest cell of the suspensor is called hypophysis. 4. What shape is a young dicot embryo? Globular 5. How many cells are present in suspensor? six to ten cells 6. Which part of the suspensor forms some portion of radicle? Hypophysis 7. Why does globular embryo change to heart-shaped structure on maturity? Due to its differentiation into cotyledons 8. Where is embryo located? Embryo is located inside ovule. 9. How many cells are present in young globular embryo of dicots? Sixteen 10. Is there any difference between the development of dicot and monocot embryo? There is no difference in the development of dicot and monocot embryo during earlier stages but their development differs in later stages. 6. STUDY OF HOMOLOGY AND ANALOGY Experiment Objective To study homology and analogy with the help of preserved specimens of organs of animals Apparatus and Materials Required Preserved specimens of different organs of animals, compound microscope Theory In animals, organs that are functionally dissimilar but anatomically, or structurally, similar are called homologous organs. Different modes of life have created the differences, i.e., modified the organs to enable them to survive. Analogous organs are those which are functionally similar but structurally dissimilar. Observation Homologous organs If you externally examine the wings of the flying mammal bat and the forelimb of a man, you will not find any similarity.

Practicals 131 But examining the bones one by one, you will find that each of them has arm bone (humerus), hand bones (radius-ulna), wrist bones (carpals), palm bones (metacarpals), and fingers (phalanges). Of course, in terms of proportions of growth of each constituent bone, there are differences. For example, the fingers of bat are much longer. What this comparative study suggests is that basically the forelimbs of these two creatures are made up of the same parts, that is, they are anatomically similar. These organs need not perform the same function, as you see that bat uses it for flying and man uses it for handling tools. Hence, the forelimb of man and the wing of bat are homologous organs. Similarly, forearms of cat and man are homologous. Fig. 9.34 Homologous organs—forelimbs of some mammals Analogous organs Observe the internal structure of the wings of butterfly, or see its preserved specimen, observe the shape and size. You will find that it is membranous and is made up of thin cuticle. There are veins in the wing but there is no skeleton. Now, take the preserved specimen of a bat and a bird, and examine their wings. You will find skeletal support. What does this type of comparative study indicate? It shows that the basic structures of wings of butterfly, bird and bat are different. In other words, they are anatomically different, although externally they look alike. Wings in these animals are used for flying. Such organs that differ anatomically and in embryonic mode of origin but perform similar function are said to be analogous organs. Fig. 9.35 Analogous organs—wings F

132 Foundation Science: Biology for Class 10 •EXERCISES• FOR CLASS 9 Multiple-Choice Questions 10. If the cheek cells are placed on a dirty slide, what do Pick the correct option. we observe under the microscope? (a) Cells appear coloured 1. What type of cells will you observe in an onion peel (b) Cells show staining examined under a microscope? (c) Cells are not visible clearly (a) Dead cells (d) None of these (b) Guard cells and stomata (c) Typical plant cells 11. The oval dense structure present in the centre of a (d) All of these cheek cell is (a) cytoplasm (b) cell membrane 2. On adding a drop of iodine solution to an onion peel, (c) vacuole (d) nucleus (a) the cells will shrink 12. The region between the nucleus and cell membrane (b) the cells will swell up (c) the cells will turn yellow in a cheek cell is occupied by (d) the food stored as starch in the cells will turn (a) protoplasm (b) cytoplasm blue-black in colour (c) vacuole (d) none of these 3. Why do we cover the onion peel placed on a glass 13. Which of the following is not found in cheek cells? slide with a coverslip? (a) To protect the objective of the microscope (a) Cytoplasm (b) Nucleus (b) To protect the onion peel cells (c) To protect the glass slide (c) Cell membrane (d) Cell wall (d) To focus the specimen 14. Which of the following features makes plant cells autotrophic? (a) Plastids (b) Cell walls 4. Which type of cells are there in the onion peel? (c) Lack of centrioles (d) Mitochondria (a) Guard cells (b) Oval cells 15. Compared to the onion peel cells, cheek cells are more irregular in shape due to (c) Epidermal cells (d) None of these (a) presence of cell membrane (b) lack of large vacuole 5. An onion is a modified (b) stem (c) lack of cell wall (a) root (d) rhizome (d) lack of plastids (c) leaf 6. When you observe the onion peel under the high 16. In the following figure of a cheek cell which part magnification of the microscope after observing it represents the nucleus? under low magnification what differences do you see? I (a) Cell size appears larger II III (b) Fewer cells are seen (c) Cell organelles look magnified (d) All of these 7. Why do we keep the onion peel on a drop of (a) I (b) II water? (c) III (d) None of these (a) To keep the cells living and transparent (b) To prevent displacement of the specimen 17. Which of the following is an example of simple (c) To make the cells larger (d) None of these permanent tissue in plants? (a) Parenchyma (b) Collenchyma 8. What kind of cells are the human cheek cells? (c) Sclerenchyma (d) All of these (a) Dead animal cells (b) Living animal cells 18. Which of the following plant tissues generally have (c) Cells without nucleus (d) All of these oval or spherical and thin-walled cells? 9. Why is the flat end of a sterile toothpick used to (a) Collenchyma (b) Sclerenchyma scrape the inside surface of the cheek? (a) To prevent injury and infection (c) Parenchyma (d) None of these (b) To get more cells (c) To avoid saliva 19. Which of the following simple permanent tissues (d) All of these does not have closely packed cells? (a) Sclerenchyma (b) Parenchyma (c) Collenchyma (d) All of these

Practicals 133 20. Intercellular spaces are usually found between I II III IV (a) parenchyma cells (b) collenchyma cells (a) I (b) II (c) III (d) IV (c) sclerenchyma cells (d) none of these 21. The main functions of parenchyma tissues are 30. A nerve cell is (b) nonstriped (a) storage and assimilation of food (a) striped (d) elongated (b) providing mechanical strength (c) storage of waste products (c) syncytial (d) all of these 31. A muscle cell is 22. When parenchyma cells contain chlorophyll they (a) provided with an end bulb (b) sheathed are called (c) myelinated (d) contractile (a) collenchyma (b) sclerenchyma (c) chlorenchyma (d) none of these 32. A nerve cell differs from a muscle cell in 23. Which of the following tissues has lignified cells? (a) genetic constitution (b) the kinds of proteins in the cytoplasm (a) Collenchyma (b) Parenchyma (c) being noncontractile (d) the features stated in (b) and (c) (c) Sclerenchyma (d) Chlorenchyma 24. Mature sclerenchyma cells are 33. How many dendrites are there in a hexapolar nerve (a) living (b) dead cell? (c) not packed closely (d) thickened with cellulose (a) Five (b) Four (c) Six (d) Seven 34. A solution that has a higher solute concentration 25. The main function of sclerenchyma is to than another solution is (a) synthesize food (b) store food (a) isotonic (b) hypotonic (c) give mechanical support (d) store waste products (c) saturated (d) hypertonic 35. When a cell is placed in a hypertonic solution the net movement of water molecules is 26. Which tissue is found in abundance in (a) into the cell (b) out of the cell fibre-yielding plants like jute and flax? (c) into the vacuole (d) out of stomata (a) Collenchyma (b) Sclerenchyma 36. A plant cell placed in water will (a) swell up and become turgid (c) Parenchyma (d) All of these (b) swell up and burst (c) lose water and become flaccid 27. In the following figure of a nerve cell, which part is (d) shrink and die an axon? I II IV 37. Raisins placed in water swell up due to III (a) plasmolysis (b) adsorption (c) exosmosis (d) endosmosis (a) I (b) II (c) III (d) IV 38. A membrane which allows solvent molecules to pass through it, but not the solute molecules is called 28. Which of the following represents the shape of cells (a) impermeable membrane lining the intestine? (b) semipermeable membrane (c) permeable membrane (d) none of the above 39. Which of the following is a partially permeable membrane in a cell? (a) Cell wall (b) Cytoplasm I II III IV (c) Cell membrane (d) All of these (a) I (b) II (c) III (d) IV 40. Endosmosis in a plant cell takes place when it is immersed in 29. Which of the following types of muscle cells (a) an isotonic solution (b) a hypotonic solution constitutes the heart muscle? (c) a hypertonic solution (d) a saturated solution

134 Foundation Science: Biology for Class 10 41. In osmosis, the net movement of solvent molecules is 50. Which of the following is the characteristic feature of Spirogyra? (a) from a region of their lower concentration to a (a) Thin cell wall region of their higher concentration (b) Spiral chloroplast (c) Star-shaped chloroplast (b) from a region of their higher concentration to a (d) Filamentous structure region of their lower concentration 51. Spirogyra represents (c) always into the cell (a) multicellular organization (d) always out of the cell (b) filamentous algae (c) photosynthetic aquatic organism 42. When a raisin is placed in a concentrated sugar (d) all of these solution, it 52. Spirogyra is commonly known as pond silk due to its (a) filamentous structure (a) swells up (b) shows no change (b) silklike texture (c) spiral chloroplast (c) shrinks (d) dies (d) presence in ponds 43. What is the term used to describe the process in 53. A permanent slide showing green spiral chloroplast like in the following figure, will be of which water from a swollen raisin comes out when the raisin is placed in a concentrated sugar solution? (a) Endosmosis (b) Exosmosis (c) Active transport (d) Reverse osmosis 44. A partially permeable membrane of a cell facilitates the process of (a) diffusion (b) plasmolysis (c) osmosis (d) imbibition 45. A raisin placed in a concentrated salt solution, shrinks because (a) salt enters its cells (a) Agaricus (b) Spirogyra (c) Moss (d) Fern (b) water comes out of its cells to establish an equilibrium (c) the cytoplasm of its cells begins to decompose (d) salt comes out of its cells 46. At the end of the experiment for determining the percentage of water absorbed by raisins, the raisins are wiped just before weighing. This is to ensure that (a) our hands do not get wet (b) the raisins lose water before weighing (c) only water absorbed by the raisins is weighed (d) the weighing scale does not get wet 47. 5 g raisins are soaked in 25 mL of ice-chilled water and 54. Agaricus is commonly called 25 mL of tap water separately at room temperature. What will be the result after one hour? (a) bread mould (b) black mould (a) The raisins in ice-chilled water will absorb (c) mushroom (d) bracket fungi more water than the raisins in tap water. 55. Which of the following commonly grows on (b) The raisins in tap water will absorb more water than the raisins in ice-chilled water. decaying organic matter during the rainy season? (c) The amounts of water absorbed by the raisins (a) Chlamydomonas (b) Mould in both the conditions will be equal. (c) Spirogyra (d) Agaricus (d) No water will be absorbed by the raisins in either condition. 56. Which of the following is a feature of Agaricus? (a) An umbrella-shaped, white, fleshy structure 48. The following photograph shows the process of (b) A mycelial plant body (c) Saprophytic habit (d) All of these 57. In mushrooms, sexual reproduction occurs by the formation of spores on club-shaped structures called (a) mycelia (b) basidia (a) endosmosis (b) exosmosis (c) hyphae (d) sporangia (c) plasmolysis (d) absorption 58. Which of the following is the group of simplest land 49. Spirogyra is an example of (b) brown algae (a) blue-green algae (d) green algae plants? (c) red algae (a) Algae (b) Fungi (c) Bryophytes (d) Pteridophytes

Practicals 135 59. True roots and leaves are absent in 71. What do earthworms eat? (a) mosses (b) ferns (a) Soil with organic matter (b) Leaves (c) Pinus (d) angiosperms (c) Insects (d) Bacteria 60. Which of the following grow close together forming 72. Which characteristic feature of an earthworm makes it different from leeches? a velvety, matlike cover over the substratum? (a) The mode of locomotion (b) The mode of feeding (a) Ferns (b) Mosses (c) Musculature of its body (d) All of the above (c) Lichens (d) Fungi 61. The most developed seedless plants are (a) mosses (b) ferns (c) gymnosperms (d) angiosperms 73. An earthworm lacks (a) haemoglobin 62. Which of the following characters are found in (b) teeth ferns? (c) nephridia (d) pharynx (a) The plant body is differentiated into an aerial 74. Leeches do not have (b) distinct clitellum shoot system and an underground root system. (a) testes (d) suckers (c) gut (b) The leaves are divided into leaflets. 75. The body of an earthworm is (c) The plants have vascular tissues. (d) All of the above (a) thin and moist (b) hard and dry 63. Plants that have seeds, but lack flowers and fruits are (c) yellow (d) black (a) pteridophytes (b) ferns (c) gymnosperms (d) mosses 76. Earthworms lack (a) nerves 64. Which of the following is a conifer? (b) heart (c) kidney (d) haemoglobin (a) Cycas (b) Pea (c) Pinus (d) Mango 77. The largest cell of the body of an earthworm is in its 65. What are the differences between the male and (a) testis (b) ovary female cones of Pinus? (c) intestine (d) coelomic fluid (a) The male cones are fewer than the female cones. 78. Which structure in an earthworm is responsible for (b) The male cones are larger than the female absorption? cones. (a) Typhlosole (b) Gizzard (c) The male cones are smaller and many more than the female cones. (c) Hepatic caeca (d) Pharynx (d) None of these 79. Which of the following is a true worm? (a) Flatworm (b) Glow worm 66. The seeds remain enclosed in a fruit in (c) Lugworm (d) Roundworm (a) mosses (b) ferns 80. Cockroaches are closely related to (c) gymnosperms (d) angiosperms (a) crickets (b) mosquitoes 67. A rice plant is an example of a (c) beetles (d) houseflies (a) dicot (b) monocot 81. Cockroaches live in (a) bright light (c) gymnosperm (d) fern (c) dry places (b) dark places (d) ponds 68. The leaves of a monocot plant have (a) reticulate venation 82. The protective layer of the body of a cockroach is (b) swelling at the base (c) parallel venation made up of (d) all of these (a) keratin (b) tannin (c) chitin (d) cartilage 69. The floral parts are usually pentamerous (5 in number) in 83. The female and male cockroaches are (a) dicots (b) monocots (a) equal in size (b) winged (c) gymnosperms (d) all of these (c) wingless (d) omnivorous 70. Where do earthworms live? 84. Cockroaches are (a) omnivorous (a) In cowdung (b) In the soil (c) carnivorous (b) insectivorous (d) sanguinivorous (c) In the intestine of birds (d) In human excreta

136 Foundation Science: Biology for Class 10 85. A female cockroach lays eggs in a 99. All fish are (a) scaly (a) nest (b) mesh (c) devoid of medulla (b) scaleless (d) active balancers (c) cocoon (d) water bag 86. Which is the mode of feeding of cockroaches? 100. The heart of a fish is (a) Sucking blood (b) Sucking milk (a) ventral (b) three-chambered (c) nonmuscular (d) nonpulsatile (c) Biting and chewing (d) Lapping liquid food 87. Cockroaches are known to be 101. The scales of fish are (a) cursorial (b) active fliers (a) respiratory (b) protective (c) excretory (d) glandular (c) sedentary (d) creepy insect 88. The male and female cockroaches can be 102. The organ which regulates the buoyancy of fish in distinguished by their water is called (a) antennae (b) eyes (a) air sac (b) trachea (c) anal cerci (d) anal styli (c) swim bladder (d) barb 89. The pores through which air enters the body of a 103. Fish lack cockroach are called (a) liver (b) endocrine gland (c) limbs (d) mucous gland (a) spiracles (b) ostia (c) anus (d) cloaca 104. Birds lack 90. During copulation, cockroaches (a) scales (b) urinary bladder (a) take to flight (c) wings (d) air sacs (b) fight (c) come in head to tail contact 105. Birds have beaks for (b) fighting (d) come in tail to tail contact (d) all of these (a) building nests (c) catching food 91. A baby cockroach is called 106. Which of the following statements is correct? (a) caterpillar (b) nymph (a) All birds have special vocal sacs called syrinx. (b) All birds sing equally well. (c) wriggler (d) tumbler (c) All birds display courtship equally well. (d) All birds have solid, heavy bones. 92. The upper lip of cockroach is called (a) maxilla (b) stipes (c) mandible (d) ligula 107. Which of the following statements is correct? 93. The total number of segments comprising head and (a) All birds can see well at night. (b) A swan has to constantly move its legs to float thorax in cockroach is in water. (a) 6 (b) 3 (c) Birds lay eggs in water. (d) Some birds are cold-blooded. (c) 9 (d) 8 94. The most distant part of the leg of a cockroach is 108. Which of the following statements is correct? (a) coxa (b) tibia (a) A sparrow is smaller than a pigeon. (b) Crows and ravens belong to the same species. (c) trochanter (d) tarsus (c) A kingfisher is larger than an ostrich. (d) The beak of a parrot is stronger than that of a kite. 95. The heart of a cockroach is (a) ventrally placed (b) tubular (c) laterally placed (d) four-chambered 96. Which of the following is correct? 109. Flightless birds are (a) All fish have a bony endoskeleton. (b) Some fish have a cartilaginous endoskeleton. (a) very light (b) very heavy (c) Gills in all fish are covered by an operculum. (d) All fish have gills as well as lungs. (c) lighter than game birds (d) totally wingless 110. The greatest variety of birds occurs in (a) Australia (b) South America 97. Which of the following statements is correct? (c) India (d) North America (a) All fish are jawless. (b) All fish are toothless. 111. Which of these organs of a bird is not found in (c) Some fish have additional breathing organs. (d) Fish have haemoglobin in its RBC. human? (a) Pecten (b) Intestine (c) Liver (d) Kidney 98. All fish lack 112. Which of the following is the largest bird? (a) eyes (c) sexual organs (b) swim bladders (a) Penguin (b) Ostrich (d) limbs (c) Peacock (d) Crane

Practicals 137 113. Which gland in a bird helps it arrange feathers? (a) Green gland (b) Preen gland (c) Tear gland (d) Liver 114. Keel in the breastbone of a bird serves to (a) grasp food (b) digest food (c) attach flight muscles (d) perch on the branch 115. In which part of the body of a bird would you find pecten? (a) Eye (b) Nose (c) Leg (d) Ear 116. Which of the following protozoans cannot move (a) I (b) II (c) III (d) IV about? 120. Where is the respiratory organ located in the fish shown below? (IV) Dorsal fin Eye (I) I II III IV (a) I (b) II (c) III (d) IV 117. Which of the following animals possesses a water Anal fin (III) Operculum (II) vascular system? (a) I (b) II (c) III (d) IV 121. Which of the following holds true for root? (a) Positively phototropic (b) Negatively geotropic (c) Positively geotropic and negatively phototropic (d) Positively geotropic and positively phototropic 122. Root differs from stem due to (a) presence of hairs (b) absence of nodes (c) presence of buds (d) thickness (a) I (b) II (c) III (d) IV 123. Absorption of water and minerals is the function of (a) root (b) stem 118. Where is the male genital aperture in the (c) leaf (d) flower earthworm shown below? 124. Region of cell division in roots is located just below (a) root cap (b) region of elongation IV (c) region of maturation (d) root hairs I Setae II III Annuli 125. Fibrous root is a type of (b) Adventitious root (a) tap root (d) tertiary root (c) secondary root (a) I Clitellum Genital (c) III (d) IV 126. The major function of stem is papillae (a) absorption of water (b) conduction of water (b) II (c) to hold branches and leaves (d) photosynthesis 119. Which of the following part is absent in a female 127. Parallel venation is the characteristics of cockroach? (a) dicot leaves (b) monocot leaves

138 Foundation Science: Biology for Class 10 (c) both dicot and monocot leaves (b) divided into ten parts (d) none of these (c) provided with compound eyes, wings and legs 128. All the floral parts are arranged on (d) provided with four equal-sized wings (a) petiole (b) thalamus 136. Eggs of mosquitoes are (a) red (c) stamen (d) petals 129. Calyx and corolla are known as (b) rounded (a) essential whorls (b) accessory whorls (c) shelled (c) secondary whorls (d) reproductive whorls (d) oval- or cigar-shaped 130. Flower is a modified (b) leaf 137. Which is correct? (a) vegetative bud (d) axis (c) shoot (a) Eggs of all types of mosquitoes are of the same shape. 131. Which of the following constitute male (b) Larvae of all types of mosquitoes reproduce in reproductive part of a flower? water. (a) Calyx (b) Corolla (c) Larvae of mosquitoes never cast off skin. (d) Larvae are voracious eaters. (c) Androecium (d) Gynoecium 132. The basal swollen portion of the pistil is called 138. The larva of mosquito (a) breathes air (a) filament (b) ovary (b) respires by gills (c) stigma (d) anther 133. The presence of trimerous flowers is the (c) wriggles in water characteristics of (a) dicots (b) monocots (d) has all the above characteristics (d) unisexual plants (c) bisexual plants 139. The pupa of mosquito is called (a) tumbler (b) grub 134. Which one of the following is true for dicot plants? (c) maggot (d) tadpole (a) Parallel venation 140. Which is incorrect with respect to mosquitoes? (b) Trimerous flowers (a) Different stages of development in mosquito (c) Pentamerous flowers and reticulate venation are distinguishable. (d) None of the above (b) Adult mosquitoes feed on algae. (c) Mosquitoes spread diseases such as malaria 135. The body of an adult mosquito is (a) divided into two parts and dengue. (d) Mosquitoes have respiratory trumpets. FOR CLASS 10 Multiple-Choice Questions 6. The maximum loss of water in transpiration is from Pick the correct option. (a) lenticels (b) cuticle 1. Small openings found widely scattered on the (c) stomata (d) hydathodes epidermis of leaves are called 7. Stomata remain open during the day because the (a) lenticels (b) intercellular spaces guard cells (a) help in exchange of gases (c) stomata (d) none of these (b) have thin walls (c) photosynthesize and produce sugars 2. In dicot leaves, stomata are generally more on the (d) are bean-shaped (a) upper surface (b) lower surface 8. Which of the following are function of the stomata? (a) Absorption (c) petiole (d) veins (b) Translocation (c) Exchange of gases and transpiration 3. In monocot leaves, stomata are present on the (d) All of these (a) lower surface (b) upper surface 9. Presence of more stomata on the lower surface of a (c) lower and upper surfaces dicot leaf helps in (d) petiole (a) enhancement of transpiration (b) reduction of transpiration 4. The stomatal aperture remains surrounded by (c) unequal transpiration from the two surfaces (d) enhancement of photosynthesis (a) cuticle (b) epidermal cells (c) guard cells (d) lenticels 5. Stomata remain open when guard cells are 10. Dumb-bell-shaped guard cells are found in (a) flaccid (b) turgid (a) gymnosperms (b) dicots (c) bean-shaped (d) dumb-bell-shaped (c) monocots (d) xerophytes

Practicals 139 11. Stomatal openings are under the control of 21. The rate of photosynthesis is independent of (a) epidermal cells (a) quality of light (b) duration of light (b) palisade cells (c) intensity of light (d) none of these (c) spongy parenchyma cells 22. The oxygen liberated during photosynthesis is from (d) guard cells (a) carbon dioxide (b) sugar 12. Which side of the wall of a guard cell is thicker? (c) water (d) chlorophyll (a) Lateral (b) Inner 23. Leaves are green because they (a) absorb blue and red light (c) Outer (d) All of these (b) absorb green light (c) do not absorb, but reflect green light 13. In the following diagram of the lower surface of a (d) absorb and reflect green light leaf, what should be labelled as stoma? 24. A balance between CO2 and O2 levels in plants is maintained by (a) transpiration (b) translocation (c) photosynthesis (d) nutrition (a) I (b) II (c) III (d) IV 25. The favoured respiratory substrate is 14. At which wavelength (colour) of light does the (a) glucose (b) sucrose maximum photosynthesis occur in plants? (c) maltose (d) glycogen (a) Red (b) Green 26. Carbon dioxide is released as a product during (c) White (d) Ultraviolet (a) photosynthesis (b) respiration 15. At which wavelength of light does the least (c) transpiration (d) ascent of sap photosynthesis occur in plants? 27. Respiration is (a) an anabolic process (a) Violet (b) Blue (c) Green (d) Red (b) a cyclic pathway (c) a catabolic process (d) an aerobic process 16. The rate of photosynthesis is the highest when a plant is exposed to 28. Anaerobic and aerobic respiration release (a) continuous high light intensity (a) ethyl alcohol (b) water (b) continuous low light intensity (c) energy (d) lactic acid (c) alternating high and low light intensities (d) intermittent light 29. For the complete oxidation of glucose to carbon dioxide and water organisms undergo 17. The light energy absorbed in photosynthesis helps to (a) aerobic respiration (a) activate chlorophyll (b) split water (b) anaerobic respiration (c) fermentation (c) reduce carbon dioxide (d) synthesize glucose (d) all of these 18. The rate of photosynthesis depends upon the (a) quality of light (b) quantity of light 30. Germinating seeds help study the rate of respiration as they (c) quality and quantity of light (a) photosynthesize rapidly (b) absorb CO2 (d) none of these (c) respire actively (d) release O2 19. If a portion of a green leaf of a potted plant is covered with a black paper strip and the potted 31. The experimental set-up shown below to plant is exposed to sunlight for a few hours, what demonstrate that CO2 is released during will happen in the covered portion? respiration will not yield the expected result because (a) Respiration will stop. (b) Respiration will be enhanced. (c) Starch will not be synthesized. (d) Starch will be synthesized. 20. The rate of photosynthesis is reduced considerably (a) germinating seeds are not immersed in water (b) the delivery tube is dipped in water in green light because (a) green light does not activate chlorophyll molecules (b) chlorophyll molecules absorb only blue and red light (c) green light is reflected by the chlorophyll molecules (d) none of the above happens

140 Foundation Science: Biology for Class 10 (c) the flask is not airtight 42. Which of the following best represents multiple (d) there is no KOH solution in the flask fission? 32. Binary fission is the mode of reproduction in (a) algae (b) fungi (c) Amoeba (d) yeast 33. The division of one cell into two new similar I II III IV daughter cells is called (a) I (b) II (c) III (d) IV (a) binary fission (b) multiple fission 43. Embryo is formed from (a) male gamete (c) sporulation (d) budding (b) female gamete 34. Which of the following is the simplest method of (c) zygote (d) vegetative cell asexual reproduction? 44. The first division of zygote is (a) Budding (b) Sporulation (a) transverse (b) vertical (c) Binary fission (d) Multiple fission (c) longitudinal (d) none of the above 35. In binary fission, the genome divides 45. Suspensor is formed by transverse divisions of (a) amitotically (b) mitotically (a) terminal cell (b) basal cell (c) meiotically (d) none of these (c) hypophysis (d) vesicle 36. Binary fission takes place in 46. The lowest cell of suspensor is called (a) unfavourable conditions (b) favourable conditions (a) vesicle (b) radicle (c) hot conditions (d) all conditions (c) hypophysis (d) root cap 47. The length of the suspensor is (a) one-celled (b) two-celled 37. The formation of bulblike outgrowths that become (c) three-celled (d) six to ten-celled detatched from the body of the parent is called 48. The uppermost cell of suspensor swells and forms (a) globular embryo (a) binary fission (b) budding (b) heart-shaped embryo (c) sporulation (d) grafting 38. Yeasts are examples of unicellular (c) vesicular cell (d) radicle (a) algae (b) fungi (c) bacteria (d) prokaryotes 49. Hypophysis forms part of 39. The buds of yeasts are (a) stem (b) radicle and root cap (a) external (c) external and internal (b) internal (c) embryo (d) vesicle (d) none of these 50. The mature dicot embryo appears 40. Which of the following is the correct diagram (a) heart-shaped (b) globular showing binary fission in Amoeba? (c) straight (d) none of these I II III IV 51. What is the basis of having homologous organs? (a) Organisms living in the same habitat have (a) I (b) II (c) III (d) IV homologous organs. (b) Organisms living in different habitat have 41. What does the figure given below show? homologous organs. (c) Organisms lead a sedentary life. (d) Organisms are very agile. Nucleus Daughter Bud 52. Study of homologous organs suggests that nuclei (a) evolution has stopped (b) evolution is very rapid Vacuole (c) there is some kind of attempt to exploit different habitats (a) Amoeba undergiong binary fission. (d) evolution has not taken any advantage of the (b) Yeast undergoing budding. habitats (c) Amoeba undergoing budding. (d) Yeast undergoing binary fission. 53. Comparative anatomy elucidates (a) the path of evolution (b) speed of evolution (c) pattern of evolution (d) both (a) and (c) F

Practicals 141 • ANSWERS • FOR CLASS 9 1. (c) 2. (d) 3. (a) 4. (c) 5. (b) 6. (d) 7. (a) 8. (b) 9. (a) 10. (c) 11. (d) 12. (b) 13. (d) 14. (a) 15. (c) 16. (c) 17. (d) 18. (c) 19. (b) 20. (a) 21. (d) 22. (c) 23. (c) 24. (b) 25. (c) 26. (b) 27. (c) 28. (a) 29. (c) 30. (d) 31. (d) 32. (b) 33. (a) 34. (d) 35. (b) 36. (a) 37. (d) 38. (b) 39. (c) 40. (b) 41. (b) 42. (c) 43. (b) 44. (c) 45. (b) 46. (c) 47. (b) 48. (a) 49. (d) 50. (b) 51. (d) 52. (b) 53. (b) 54. (c) 55. (d) 56. (d) 57. (b) 58. (c) 59. (a) 60. (b) 61. (b) 62. (d) 63. (c) 64. (c) 65. (c) 66. (d) 67. (b) 68. (c) 69. (a) 70. (b) 71. (a) 72. (d) 73. (b) 74. (b) 75. (a) 76. (c) 77. (b) 78. (a) 79. (d) 80. (a) 81. (b) 82. (c) 83. (d) 84. (a) 85. (c) 86. (c) 87. (a) 88. (d) 89. (a) 90. (d) 91. (b) 92. (c) 93. (c) 94. (d) 95. (b) 96. (b) 97. (c) 98. (d) 99. (d) 100. (a) 101. (b) 102. (c) 103. (c) 104. (b) 105. (d) 106. (a) 107. (b) 108. (a) 109. (b) 110. (b) 111. (a) 112. (b) 113. (b) 114. (c) 115. (a) 116. (d) 117. (a) 118. (c) 119. (d) 120. (b) 121. (c) 122. (b) 123. (a) 124. (a) 125. (b) 126. (c) 127. (b) 128. (b) 129. (b) 130. (c) 131. (c) 132. (b) 133. (b) 134. (c) 135. (c) 136. (d) 137. (d) 138. (d) 139. (a) 140. (b) FOR CLASS 10 1. (c) 2. (b) 3. (c ) 4. (c) 5. (b) 6. (c) 7. (c) 8. (c) 9. (b) 10. (c) 11. (d) 12. (b) 13. (c) 14. (a) 15. (c) 16. (d) 17. (a) 18. (c) 19. (c) 20. (c) 21. (d) 22. (c) 23. (c) 24. (c) 25. (a) 26. (b) 27. (c) 28. (b) 29. (a) 30. (c) 31. (d) 32. (c) 33. (a) 34. (c) 35. (a) 36. (b) 37. (b) 38. (b) 39. (a) 40. (d) 41. (b) 42. (d) 43. (c) 44. (a) 45. (b) 46. (c) 47. (d) 48. (c) 49. (b) 50. (a) 51. (b) 52. (c) 53. (d) v

Question Bank 1. Nutrition A. Very-Short-Answer Questions (viii) Projections of cytoplasm surrounding food in Amoeba 1. Name the organelle where photosynthesis takes place. 5. What will be the result if two green plants are kept 2. What role is played by stomata during separately, one in continuous light and the other in photosynthesis? dark, oxygen-free condition? 3. From where does a plant obtain water for 6. How does an autotroph differ from a heterotroph? photosynthesis? 4. What is similar among animals, fungi and some 7. Write the sequence of important events taking place bacteria? during photosynthesis. 5. In which form is food stored in plants and in animals? 8. What would be the consequence if all green plants are eliminated from earth? 6. Why are chlorophyll molecules essential for photosynthesis? 9. If CO2 is not released by the plants during day time, what inference would you draw and why? 7. Give two examples of heterotrophs. 8. Why are heterotrophs called consumers? 10. Do you agree with this statement that all plants release oxygen during day and CO2 during night? 9. What are saprophytes? Explain. 10. Give two examples of saprophytes. 11. How does the movement of food take place inside 11. Name the watery substance released in our mouth the alimentary canal? during eating. 12. Name the substrates on which enzymes lipase, 12. What does saliva contain? amylase, pepsin and trypsin act. 13. Where does peristaltic movement occur? 13. Mention the roles played by gastric glands found in 14. Name the structures through which food reaches the wall of stomach. the stomach from our mouth. 14. Why does small intestine play an important role in 15. Which is the longest part of the alimentary canal? the absorption of digested food? 16. From where is bile secreted? 17. Where is bile stored? 15. What is the role of mouth in digestion of food? 18. Which type of organ is vermiform appendix? 19. Where are villi located? 16. What is alimentary canal in human beings? B. Short-Answer Questions 17. What is mucus? What will be the result if it is not secreted by the gastric glands? 1. Why is nutrition essential for an organism? 18. The small intestine in herbivores is longer than that 2. In which respect are leeches, ticks and Cuscuta in carnivores. Why? similar? 19. What is the importance of emulsification of fats? 3. What will be the effect of the following conditions 20. What role does salivary amylase play? on the rate of photosynthesis? 21. What is the function of pancreas? (i) Dry conditions (ii) Closed stomata 22. What happens to the undigested food? (iii) Rainy days (iv) Sunny days C. Long-Answer Questions (v) Good manuring 4. Give proper terms for the following. 1. Describe different factors which affect (i) Site of photosynthesis photosynthesis. (ii) Organisms dependent on other organisms for their food 2. Describe the mechanism of digestion of fats, proteins and carbohydrates in human beings. (iii) The process which converts solar energy into chemical energy 3. Discuss the role of different glands in human digestive system. (iv) Organisms capable of synthesizing food (v) Cells surrounding stoma 4. Describe different types of heterotrophic nutrition. (vi) The muscular movement taking place in How do they differ from each other and from autotrophic nutrition? oesophagus (vii) An enzyme that breaks down protein in 5. How can you prove the essentiality of CO2 and sunlight for photosynthesis? stomach 142

Question Bank 143 D. Crossword Puzzle (c) 6CO2 + 12H2O ® C6H12O6 + 6O2 + 6H2O 1 (d) 6CO2 + 6H2O ® C6H12O6 + 6CO2 + 6H2O 3 4. Which of the following statements are true about autotrophs? 2 (a) They convert carbon dioxide and water into carbohydrates in presence of light. 4 7 (b) They store carbohydrates in the form of starch. 5 (c) They occupy second trophic level in food chains. (d) They are chlorophyll-bearing organisms. 6 5. Which of the following events takes place during 8 photosynthesis? Down (a) Trapping of light energy by chlorophyll molecules 1. Organelle for photosynthesis 3. Mechanism for food production in green plants (b) Conversion of light energy to chemical energy 5. Chemical nature of bile 7. Bile and pancreatic ducts open here (c) Reduction of CO2 to carbohydrates (d) Release of CO2 in the atmosphere Across 6. In which of the following groups of organisms food 2. Organism making its own food material is broken down outside the body? 4. Muscular and tubular part of alimentary canal 6. A parasitic plant (a) Cuscuta, Amoeba, green plants 8. Middle part of small intestine (b) Amoeba, Paramoecium, Cuscuta (c) Yeast, green plants, mushroom E. Diagrammatic Questions (d) Mushroom, bread mould, yeast 1. Draw a cross section of a leaf and locate the regions 7. In human body, food is finally digested in of chloroplast, stomata and air spaces. (a) large intestine (b) small intestine 2. With the help of diagrams show how food is engulfed by Amoeba. (c) stomach (d) oesophagus 3. Label the following parts after drawing the diagram 8. Which of the following protects the inner lining of of human alimentary canal: mouth, oesophagus, stomach from hydrochloric acid? liver, stomach, small intestine, appendix. (a) Mucus (b) Bile 4. Draw a diagram to show the network of blood capillaries in the villi of small intestine. (c) Pepsin (d) Amylase F. Objective Questions 9. Select the incorrect statement(s). (a) Heterotrophs contain chlorophyll pigments. (b) Heterotrophs can convert solar energy to chemical energy. (c) Heterotrophs are unable to synthesize food. (d) Heterotrophs occupy first trophic level in food chain. 10. During photosynthesis oxygen is evolved from I. Pick the correct option/options. More than one option (a) CO2 (b) water may be correct. (c) glucose (d) chlorophyll 11. The food in autotrophs is reserved in the form of 1. Which of the following enzymes get mixed with (a) protein (b) fatty acid food in our mouth? (c) glycogen (d) starch (a) Trypsin (b) Cellulose 12. Which of the following is the correct sequence representing human alimentary canal? (c) Pepsin (d) Amylase (a) Mouth ® oesophagus ® stomach 2. The mode of nutrition in mushrooms is (a) heterotrophic (b) autotrophic ® small intestine ® large intestine (c) holozoic (c) parasitic (b) Mouth ® small intestine ® stomach 3. The most appropriate equation for photosynthesis ® oesophagus ® large intestine will be (a) 6CO2 + 6H2O+ Sunlight (c) Mouth ® stomach ® oesophagus ® C6H12O6 + O2 + 6H2O ® small intestine ® large intestine (b) 6CO2 + 12H2O+ Sunlight + Chlorophyll (d) Mouth ® stomach ® oesophagus ® C6H12O6 + 6O2 + 6H2O ® large intestine ® small intestine

144 Foundation Science: Biology for Class 10 13. The muscular and tubular part of the alimentary 25. The inner lining of the small intestine has numerous canal is called finger-like projections called (a) pharynx (b) small intestine (a) jejunum (b) duodenum (c) oesophagus (d) stomach (c) ileum (d) villi 14. Which of the following functions is carried out by 26. The undigested food is finally removed from the pancreatic juice? body through (a) Lipase emulsifies fats and trypsin digests proteins. (a) large intestine (b) small intestine (b) Lipase digests carbohydrates and trypsin (c) anus (d) villi digests fats. II. Fill in the blanks. (c) Lipase emulsifies both carbohydrates and fats. 1. Nutrition provides ...... to the body. (d) Trypsin digests both proteins and fats. 15. Pancreatic juice is carried by the pancreatic duct into 2. Since autotrophic plants are able to produce food, they are also called ...... . (a) large intestine (b) liver (c) duodenum (d) stomach 3. ...... are the main sites of photosynthesis. 16. Bile is produced by (b) liver 4. Chlorophyll is mainly found in the ...... . (a) pancreas (d) stomach 5. The opening and closing of stomata is regulated (c) small intestine by ...... . 17. Iodine solution is used to test the presence of (a) proteins (b) fats 6. ...... energy is used in splitting water molecules into hydrogen and oxygen. (c) starch (d) enzymes 18. Which part of alimentary canal receives bile? 7. During photosynthesis light energy is converted into ...... . (a) Oesophagus (b) Pharynx (c) Large intestine (d) Small intestine 8. Heterotrophic organisms obtain food from ...... organisms. 19. Absence of salivary amylase in the saliva will not affect 9. Amoeba exhibits ...... nutrition. (a) breakdown of protein in mouth (b) breakdown of fat in mouth 10. The oral cavity opens into the ...... . (c) breakdown of starch in mouth 11. ...... pairs of salivary glands are present in human mouth. (d) assimilation of vitamins in mouth 20. The process of photosynthesis requires 12. ...... is the muscular and tubular part of the (a) chlorophylls and light only alimentary canal. (b) chlorophylls and CO2 only 13. ...... is the muscular partition between the chest cavity and the abdominal cavity. (c) chlorophylls, CO2 and H2O only (d) chlorophylls, CO2, H2O and light 14. ...... serves as a storehouse of food where partial digestion takes place. 21. Nitrogen, an essential element used in the synthesis of proteins, is obtained by plants in the form of 15. ...... is the first part of small intestine. (a) atmospheric nitrogen (b) nitrates and nitrites 16. Bile is yellowish ...... juice. (c) amino acids (d) peptides 22. Water used in photosynthesis by terrestrial plants is 17. Pancreatic juice contains ...... for the breakdown of fat. taken up from (a) air 18. Herbivores have longer small intestine to digest (b) soil ...... . (c) either from air or from soil (d) none of the above 19. The digested food is taken up by the walls of the ...... . 23. The mode of nutrition in parasitic organisms is 20. The undigested material is removed from the body (a) heterotrophic via ...... . (b) autotrophic (c) both autotrophic and heterotrophic III. Write Yes/No. (d) none of the above 1. Is the general requirement for energy and material 24. Which of the following parts is the site of the complete common in all organisms? digestion of carbohydrates, proteins and fats? 2. Are carbohydrates stored in the form of starch in plants? (a) Oesophagus (b) Stomach 3. Is it true that light energy is not directly absorbed (c) Small intestine (d) Large intestine by chlorophyll molecules?

Question Bank 145 4. Are guard cells devoid of chloroplast? V. Match the columns. B 1. A 5. Are autotrophs dependent upon heterotrophs for (a) Autotroph their food requirements? (i) Chlorophyll (b) Parasite (ii) Heterotrophic 6. Is nitrogen an essential element in the synthesis of (c) Food reserve in proteins? nutrition plants (iii) Leaves 7. Is Cuscuta a parasitic plant? (d) Consumers (iv) Cuscuta (e) Saprophyte 8. Does food mix thoroughly with saliva in our mouth (v) Starch (f) Digestion in food when we chew properly? (vi) Green plants vacuoles 9. Does gall bladder store pepsin? (g) Photosynthesis (h) Stomata 10. Is large intestine the site of complete digestion of proteins? B IV. Mark the statements true (T) or false (F). (vii) Mushroom (a) Bile (viii) Paramoecium (b) Pepsin 1. Carbon and energy requirements of the autotrophic (c) Pancreatic juice organisms are fulfilled by photosynthesis. 2. A (d) Undigested food (e) Inner lining of 2. Oxygen is released during photosynthesis. (i) Leech (ii) Amoeba small intestine 3. CO2 is not essential for photosynthesis. (iii) Liver (f) Parasitic nutrition 4. The alimentary canal and the glands associated with (iv) Salivary amylase (g) Extensive coiling (v) Gastric glands (h) Pseudopodia digestion constitute the human digestive system. (i) Peptic ulcer (vi) Trypsin (j) Mouth 5. Stomach serves as a storehouse of food where (vii) Large intestine complete digestion takes place. (viii) Villi (ix) Small intestine 6. Bile produced by liver is acidic in nature. (x) Stomach 7. Ileum is the last and main part of the small intestine. 8. Gastric glands are present in small intestine. 9. Carnivores like tigers have a shorter small intestine because meat is easier to digest 10. Small intestine receives the secretions of liver and pancreas. F • ANSWERS • A. 1. Chloroplast 6. (d) 7. (b) 8. (a) 9. (a), (b), (d) 10. (b) 2. CO2 enters the leaves and O2 is released into air 11. (d) 12. (a) 13. (c) 14. (a) 15. (c) 16. (b) through stomata. 17. (c) 18. (d) 19. (a), (b), (d) 20. (d) 21. (b) 3. Water is absorbed from soil. 22. (b) 23. (a) 24. (c) 25. (d) 26. (c) 4.They cannot prepare their own food. II. 1. nutrients 2. producers 3. Chloroplasts 5. Starch in plants and glycogen in animals 4. leaves 5. guard cells 6. Light 7. chemical energy 8. other 9. holozoic 6. Chlorophyll absorbs light energy from sunlight. 10. pharynx 11. Three 12. Oesophagus 13. Diaphragm 14. Stomach 15. Duodenum 7. Mushrooms and human beings 16. alkaline 17. lipase 18. cellulose 19. intestine 20. anus 8. They obtain food from other sources. III. 1. Yes 2. Yes 3. No 4. No 5. No 6. Yes 9. They obtain food from dead organisms. 7. Yes 8. Yes 9. No 10. No 10. Bread moulds, yeasts 11. Saliva IV. 1. T 2. T 3. F 4. T 5. F 12. Mucin and salivary amylase 13. Oesophagus 6. F 7. T 8. F 9. T 10. T 14. Pharynx and oesophagus 15. Small intestine V. 1. (i)(g) (ii)(d) (iii)(h) (iv)(b) (v)(c) (vi)(a) 16. Liver 17. Gall bladder 18. Vestigial (vii)(e) (viii)(f) 19. On the wall of the ileum in the small intestine 2. (i)(f) (ii)(h) (iii)(a) (iv)(j) (v)(b) (vi)(c) (vii)(d) (viii)(e) (ix)(g) (x)(i) D. Down: 1. Chloroplast 3. Photosynthesis 5. Alkaline 7. Duodenum Across: 2. Autotroph 4. Oesophagus 6. Cuscuta 8. Jejunum F. I. 1. (d) 2. (a) 3. (b) 4. (a), (b), (d) 5. (a), (b), (c) v

146 Foundation Science: Biology for Class 10 2. Respiration A. Very-Short-Answer Questions D. Crossword Puzzle 1. By which respiratory process is lactic acid formed? 3 2. Name the process through which glucose is broken 1 down to pyruvic acid. 2 3. Where does anaerobic respiration take place? 4 4. Name the organism which converts pyruvate to 5 ethanol. 6 5. Where does the second stage of aerobic respiration take place? Down 6. Name the organs through which gaseous exchange 1. Voice box in human beings takes place in plants. 3. Anaerobic respiration by yeast 5. Respiratory organ of fish 7. Name the respiratory organs found in animals. Across 8. Do the hair present inside our nose play any role? 2. Windpipe 9. Where is pharynx located? 4. Product of glycolysis 5. The opening leading to the voice box 10. Name the structure which prevents food from 6. Common respiratory substrate entering the passage to the lungs. E. Diagrammatic Questions 11. Why are cartilage rings present inside trachea? 1. Draw a flow chart to show different ways of 12. How are bronchi formed? oxidation of glucose. 13. Name the structures formed by the division of 2. Draw an experimental set-up to demonstrate that bronchioles. CO2 is evolved during fermentation. 14. Name the respiratory pigments of human beings. 3. Draw a diagram to show the structure and position of stomata on a leaf. B. Short-Answer Questions 4. Draw a clean diagram of human respiratory system 1. What is the energy currency in the living and label the following structures: organisms? How is it produced? Glottis, larynx, trachea, bronchus, bronchioles and diaphragm 2. Why do animals need more energy as compared to plants? 5. Draw a diagram to show an alveolus and the surrounding blood capillaries. 3. Describe how the guard cells regulate the opening and closing of stomata? F. Objective Questions 4. What will happen if fish are taken out of water? I. Pick the correct option/options. More than one option Explain with reason. may be correct. 5. What will be the consequence if the leaves of a 1. Which part of human cell is involved in conversion healthy potted plant were coated with vaseline? Explain with reason. of pyruvic acid into lactic acid during deficiency of 6. What happens to pyruvate when oxygen is oxygen? available? (a) Golgi body (b) Mitochondria 7. Why do our leg muscles cramp when we run for a long period? (c) Cytoplasm (d) Lysosome 8. What will happen if we inhale carbon monoxide in 2. Which of the following gases turns lime water place of oxygen? milky when we blow air into it from mouth? 9. With regard to energy production, which type of respiration is more efficient? Why is it so? (a) Oxygen (b) Carbon dioxide 10. How does exchange of gases take place in small (c) Nitrogen (d) Carbon monoxide organisms? C. Long-Answer Questions 1. Describe the three pathways of glucose breakdown in living organisms. 2. What is breathing? How does it take place in man? 3. Discuss the exchange of gases in plants. 4. Describe the structure and functions of different organs involved in human respiratory system.

Question Bank 147 3. Which of the following processes is responsible for 10. Which of the following statement(s) is/are not cramps in the muscles of sportsmen? correct? (a) Nonconversion of glucose to pyruvate (a) Bronchus is divided into bronchioles. (b) Conversion of pyruvate to lactic acid due to (b) The opening leading to the larynx is called deficiency of oxygen glottis. (c) Conversion of pyruvate to glucose in presence (c) During inhalation, ribs move inward and of oxygen diaphragm is raised. (d) Conversion of pyruvate to ethanol (d) Oxygen has more affinity with haemoglobin than CO2. 4. Which of the following equations represents 11. Choose the correct statement(s). anaerobic respiration in yeast? (a) Glucose ¾c¾yto¾plas¾m® Pyruvate (a) Alveoli increase the surface area for exchange of gases. (b) Glucose ¾c¾yto¾plas¾m® Ethanol + CO2 (c) Glucose ¾c¾yto¾plas¾m® Pyruvate (b) Oxygen from alveolar air diffuses into blood (d) Glucose ¾m¾ito¾cho¾ndr¾ia® Lactic acid and CO2 from blood diffuses into alveolar air. ¾m¾ito¾cho¾ndr¾ia® Pyruvate ¾c¾yto¾plas¾m® Ethanol + CO2 (c) From the larynx the air goes to pharynx. ¾m¾ito¾cho¾ndr¾ia® Lactic acid ¾c¾yto¾plas¾m® Ethanol + CO2 (d) The trachea is not connected to bronchi. 12. Which of the following is the correct representation of aerobic respiration? (a) Glucose ¾c¾yto¾plas¾m® Ethanol 5. Which of the following statement(s) is/are correct? ¾m¾ito¾cho¾ndr¾ia® CO2 + H2O (a) Fermentation is a form of aerobic respiration. (b) Glucose ¾c¾yto¾plas¾m® Pyruvate (b) Aerobic respiration involves mitochondria. (c) Lactic acid is formed in human muscle cells ¾m¾ito¾cho¾ndr¾ia® CO2 + H2O during anaerobic respiration. (c) Glucose ¾c¾yto¾plas¾m® Pyruvate + Energy (d) Pyruvate can be converted into ethanol in mitochondria. (d) Glucose ¾m¾ito¾cho¾ndr¾ia® CO2 + H2O + Energy ¾m¾ito¾cho¾ndr¾ia® Pyruvate ¾c¾yto¾plas¾m® CO2 + H2O + Energy 6. Which of the following represents the correct 13. Which of the following overall reactions is true sequence of air passage during inhalation? representation of respiration? (a) C6H12O6 + O2 ® 6CO2 + 6H2O (a) Nostrils ® larynx ® pharynx ® alveoli (b) 6C6H12O6 + 6O2 ® 6CO2 + 6H2O+ Energy ® lungs (c) C6H12O6 + O2 ® 6CO2 + 12H2O (d) C6H12O6 + 6O2 ® 6CO2 + 6H2O+ Energy (b) Nostrils ® trachea ® pharynx ® larynx ® lungs 14. Pyruvate is not a (b) 3-carbon molecule (a) 6-carbon molecule (d) 5-carbon molecule (c) Nostrils ® pharynx ® larynx ® trachea ® alveoli (c) 4-carbon molecule (d) Nostrils ® alveoli ® pharynx ® larynx ® lungs 7. The opening and closing of stomatal pore is 15. The amount of energy released after aerobic respiration is regulated by (a) less than that after anaerobic respiration (a) O2 concentration (b) CO2 concentration (b) much more than that after anaerobic respiration (c) temperature (d) turgidity of guard cells (c) almost equal to that after anaerobic respiration 8. Choose the correct statement(s) amongst the (d) about half of that after anaerobic respiration following. 16. Which of the following is not a 3-carbon molecule? (a) For regulating different life processes energy is required. (a) Pyruvate (b) Ethanol (b) Living organisms grow with time. (c) Lactic acid (d) All of the above (c) Living organisms need to repair and maintain 17. As compared to terrestrial organisms, the rate of their structures. breathing in aquatic organisms is (d) Molecules within the cells do not exhibit movement. (a) slightly slower (b) much faster (c) almost the same (d) much slower 9. During respiration exchange of O2 and CO2 takes 18. Which of the following respiratory structures has place in cartilage rings to prevent it from collapsing? (a) pharynx and larynx (b) trachea and lung (a) Nasal cavity (b) Pharynx (c) alveoli of lungs (d) larynx and throat (c) Larynx (d) Trachea