Q55Multiple choice
What is erythropoiesis? Which hormone stimulates it? Ans. Formation of RBC is known as erythropoiesis and the hormone enythropoietin stimulates the process. MODEL ANSWERS TO DESCRIPTIVE QUESTIONS 141 ANSWERS TO SA TYPE QUESTIONS 1. Differentiate between:
- a.Open Circulatory System and Closed Circulatory System Ans. Open Circulatory System Closed Circulatory System The blood is pumped out of the heart The blood is circulated within a into sinuses and the cells and tissues network of vessels. are directly bathed in it.
- b.Oviparous and Viviparous. Ans. Oviparous Viviparous Animals which lay eggs are called Animals which give birth to their oviparous young ones are called viviparous
- c.Direct Development and Indirect Development Ans. Direct Development Indirect Development Animals which do not have a larval Animals which have a larval stage, stage in their development are said which do not resemble the adult in to exhibit direct development. their development are said to exhibit indirect development. 2. Fill up the blank spaces appropriately Phylum/Class Excretory Organ Circulatory Organ Respiratory Organ Arthropoda A B Lungs/ Gills/ Tracheal System C Nephridia Closed Skin/Parapodia D Metanephridia Open E Amphibia F Closed Lung Ans. A = Malpighian Tubule/ coxal glands/ D= Mollusca anternary glands/ green glands B = Open E= Feather like gills C = Annelida F= Kidney 3. Give two examples of roots that develop from different parts of the angiospermic plant other than the radicle. Ans. The root that arise from parts of plant other than radicle are called adventitious roots. Pneumatophores – for respiration Stilt roots – for support Prop roots – for support. 4. While eating peach or pear it is usually seen that some stone like structures get entangled in the teeth, what are these stone like structures called? Ans. The structures that get entangled in the teeth while eating fruits like peach and pear are actually the stone cells or brachysclereids which are unbranched, short and isodiametric type of sclereids. These stone cells usually occur in groups and provide grit or stone like hardness that get entangled in the spaces between teeth. 5. Palm is a monocotyledonous plant, yet it increases in girth. Why and how? Ans . A palm tree is a monocotyledonous plant and like all other monocot the stems do not have primary cambium in the vascular bundles. However, with age the tree grows in diameter, though slowly, as a result of growth of the ground tissue. A secondary cambium may be formed in the hypodermal region of the stem. The later forms the conjunctive tissue and patches of meristematic cell. The activity of the meristematic cells result in the formation of secondary vascular bundles. 6. Give the location of hepatic caeca in a cockroach? What is their function? Ans. A ring of 6-8 blind tubules called hepatic caeca are present at the junction of foregut and midgut. Hepatic ceaca secrete digestive juice. 7. Frogs are beneficial for mankind, justify the statement. Ans. Frogs are beneficial for mankind as they can eat some crop pests and protect crop. Frogs maintain ecological balance as they are an important component of food chain and food web in the ecosystem. They are edible in some countries. MODEL ANSWERS TO DESCRIPTIVE QUESTIONS 143 8. Common name of some animals are given in Column A, write their scientific name in Column B. A B a. Tiger Panthera tigris b. Peacock Pavo cristatus c. Housefly Musca domestica 9. When a freshly collected Spirogyra filament is kept in 10% potassium nitrate solution, it is observed that the protoplasm shrinks in size: a. What is this phenomenon called? b. What will happen if the filament is replaced in distilled water? Ans. a. This phenomenon is called plasmolysis. (The shrinkage of protoplast from the cell wall under the influence of a strong solution/ hypertonic solution is called plasmolysis. b. If filaments are replaced in water the protoplast starts swelling. It comes in contact with cell wall and cell regains its original size. The swelling up of plasmolyzed protoplast under the influence of a weak solution or water is called deplasmolysis. 10. How is sulphur important for plants? Name the amino acids in which it is present. Ans. Sulphur, besides being present in some amino acids essential for protein synthesis, is also a constituent of several coenzymes, vitamins and ferrodoxin which are involved in some biochemical pathway. 11. How are organisms like Pseudomonas and Thiobacillus of significance in nitrogen cycle? Ans. Pseudomonas and Thiobacillus carry out denitrification process wherein the nitrate present in the soil is reduced to nitrogen thus contributing to the atmospheric nitrogen. 12. Carefully observe the following figure a. Name the technique shown in the figure and the scientist who demonstrated this technique for the first time? Ans. Hydroponics, Julius von Sachs. b. Name atleast three plants for which this technique can be employed for their commercial production. Ans. Tomato, seedless cucumber, lettuce. c. What is the significance of aerating tube and feeding funnel in this set-up? Ans. Aerating tube ensures adequate aeration of the root for optimum growth of the plant. The funnel is used to release water and nutrients into the container with nutrient solution. This solution needs to be replaced every day or two for maximum growth. 13. Which is the most crucial enzyme found in root nodules for N2 fixation? Does it require a special pink coloured pigment for its functioning? Elaborate? Ans. Nitrogenase. Yes, it does require the presence of a pink coloured pigment in the nodule called leg-haemoglobin for its functioning. This pigment helps in scavenging oxygen as nitrogenase functions under anaerobic condition. 14. Succulents are known to keep their stomata closed during the day to check transpiration. How do they meet their photosynthetic CO2 requirements? Ans. Succulent (water storing) plants such as cacti, euphorbias fix CO2 into organic compound using PEP carboxylase at night, when the stomata are open. PEP + CO2 OAA OAA Malic acid The organic compound (malic acid) accumulates throughout the night and is decarboxylated during the day to produce CO2. 15. Chlorophyll ‘a’ is the primary pigment for the light reaction. What are accessory pigments? What is their role in photosynthesis? Ans. Accessory pigments are those pigments, which assist in photosynthesis by capturing energy from light of different wavelengths. e.g., chlorophyll- b, Xanthophylls and carotenoids. Role in Photosynthesis: a. They absorb wavelength of light not absorbed by chlorophyll ‘a’ and transfer the energy to chlorophyll. b. They also protect chlorophyll a from photo-oxidation. MODEL ANSWERS TO DESCRIPTIVE QUESTIONS 145 16. Nicotiana tabacum, a Short Day Plant, when exposed to more than critical period of light fails to flower. Explain. Ans. a. Some plants require a periodic exposure to alternate light and dark for its flowering response. This phenomenon is termed photoperiodism. b. The requirement of light exposure is critical. The SDP plants, when exposed to light period in excess of critical period fail to flower, c. Those plants which require exposure to light period at critical or more than critical period for its flowering response are called long-day-plant.
- d.Nicotiana tabacum, fails to flower if exposed to more than critical period of light because it is a SDP. 17. What are the structural characteristics of a. Meristematic cells near root tip b. The cells in the elongation zone of the root Ans. a. The meristematic cells near root tip are characterised by: - rich protoplasm, - large conspicuous nucleus - thin and cellulosic cell wall-primary in nature - fewer vacuoles - greater number of mitochondria - numerous (abundant) plasmodesmata b. The cells in the elongation zone of a root are characterized by - increased vacuolation - enlarged size/dimension - deposition of new cellulosic cell walls. 18. A gardener finds some broad-leaved dicot weeds growing in his lawns. What can be done to get rid of the weeds efficiently? Ans. The dicotyledonous plant grow by their apical shoot meristems while grasses (which make lawns) possess intercalary meristem. Certain auxins, such as synthetic 2, 4-Dichlorophenoxyacetic acid (2,4-D) when applied in excess can damage the shoot apical meristems but they do not cause any damage to the intercalary meristems. Thus, when 2, 4-D is sprayed on lawns, only the dicots get killed and the lawns become free of weeds. 19. What is pancreas? Mention the major secretions of pancreas that are helpful in digestion. Ans. Pancreas is a gland having exocrine and endocrine portions involved in secreting digestive enzymes as well as hormones. Major secretions of pancreas involved in digestion are inactive enzymes listed below: a. Trypsinogen b. Chymotrypsinogen c. Procarboxypeptidases d. Amylases e. Lipases f. Nucleases 20. Name the part of the alimentary canal where major absorption of digested food takes place. What are the absorbed forms of different kinds of food materials? Ans. Small intestine is the part of alimentary canal where digested food materials are mainly absorbed. Amino acids (proteins), monosachharides like glucose, fructose galactose, etc. (carbohydrate) and fatty acids and glycerol (fats) are different absorbable forms of food materials. 21. State the different modes of CO2 transport in blood. Ans. Nearly 20-25% of CO2 by RBCs Nearly 70% of CO2 as bicarbonates Nearly 7% of CO2 as dissolved state in plasma. 22. Compared to O2, diffusion rate of CO2 through the diffusion membrane per unit difference in partial pressure is much higher. Explain. Ans. Solubility is an important factor deciding diffusion rate. As the solubility of CO2 is 20-25 times higher than O2, diffusion of CO2 through the diffusion membrane per unit difference in partial pressure is much higher. 23. Differentiate between Blood and Lymph. Ans. Blood is a connective tissue consisting of a fluid matrix, plasma and formed elements (RBCs, WBCs & Platelets). Blood flows in blood vascular system comprising heart, artcries and veins. L ymph is a colourless fluid containing specialized lymphocytes (imparting immunity to the body), but devoid of RBCs. Lymph flows in the lymphatic system and it absorbs fats. MODEL ANSWERS TO DESCRIPTIVE QUESTIONS 147 24. Briefly describe the following: a. Atherosclerosis b. Thrombocytes Ans. Atherosclerosis: Sometimes deposition of calcium, fat, cholesterol and fibrous tissues occurs in the blood vessel (e.g., coronary artery) supplying blood to the heart muscles. This condition makes the lumen of arteries narrower affecting blood supply to heart; which leads to Coronary Artery Disease (CAD) also referred to as atherosclerosis. Thrombocytes: Blood platelets are cell fragments produced from megakaryocytes (they are special cells in the bone marrow) and also referred as thrombocytes. Normal blood contains 1,50,000 - 3,50,000 platelets mm3. Thrombocytes release a variety of substances like thrombokinase, most of which are involved in blood clothig. A significant drop in the count of blood platelets can lead to Clotting disorders which will lead to excessive blood loss from the body. 25. a. What is the major site where RBCs are formed. Ans. Bone marrow b. Which part of the heart is responsible for initiating and maintaining the rhythmic activity of the heart. Ans. Sino-Atrial Node (S.A. Node) c. Name the reptile which has four chambered heart. Ans. Crocodile 26. What is the role played by Renin-Angiotensin in the regulation of kidney function? Ans. Renin is released from JGA on activation due to fall in the glomerula blood pressure/flow. Renin converts angiotensinogen in blood to angiotensin-I and further to angiotensin-II. Angiotensin-II being a powerful vasoconstrictor, increase the glomerular blood pressure and thereby GFR. Angiotensin-II also activates the adrenal cortex to release aldosterone. Aldosterone causes reabsorption of Na+ and water from the distal parts of the tubule. This also leads to an increase in blood pressure and thereby GFR. This is generally known as the Renin - Angiotensin mechanism. 27. The following parts in our body have specific name in Osteology, write against each. Ans. a. Knee cap - patella b. Collar bone - clanicle c. Skull - cranium 28. Write a few lines about Gout. Ans. When metabolic waste-uric acid crystals are accumulated in bones, then it results into inflammation of bone and joints thereby causing pain. This disorder of skeletal system is called gout. 29. What are the point for articulation of pelvic and pectoral girdles? Ans. The components of pelvic girdle are ilium, ischium and pubis. It articulates with femur through acetabulum. The component of pectoral girdle are scapula and clavicle. It is the glenoid cavity of pectoral girdle in which head of humerus articulates. 30. What is the role-played by lutenizing hormones in males and females respectively? Ans. LH stimulates the synthesis and secretion of androgens called male hormones. In females, LH is essential for ovulation. In females, LH induces ovulation of fully mature follicles (graafian follicles) and maintains the corpus luteum formed from the remnants of the graafian follicles after ovulation. 31. What is the role of second messenger in hormone action? Ans. Hormones which do not enter the target cell, interact with specific receptors located on the surface of the target cell membranes and generates second messengers (e.g., CAMP) on the inner side of plasma membrane. The second messenger, in turn, carries out all the hormonal functions. 32. Old people have weak immune system. What could be the reason? Ans. Thymus is degenerated in old individual resulting in a decreased production of thymosins. As a result the immune responses of old persons become weak. ANSWERS TO LA TYPE QUESTIONS 1. With the help of a schematic diagram describe the haplo-diptontic life cycle pattern in plants. Ans. In a sexually reproducing plant there is an alternation of generation between a haploid and a diploid phase of plant bodies. The haploid plant body is termed gametophyte while the diploid plant body is called sporophyte. The gametophyte produces gametes by mitosis while the haploid spores are produced by sporophyte following meiosis (reduction MODEL ANSWERS TO DESCRIPTIVE QUESTIONS 149 division). Two gamete fuse together to produce a zygote which develops into the diploid sporophyte. In a haplodiplontic life cycle pattern, such as in bryophyta or pteridophyta both the phases of life are multicellular. However, in bryophytes, the gametophytes are small, photosynthetic, independent and represent dominant phase. The partly or totally dependent, sporophyte is physically attached to the gametophyte. The (n) spores dispersed by sporophyte germinate into individual gametophytic plants. However, in pteridopyhtes the 2n (diploid) phase is dominant, well organized, independent while the n phase though also freeliving and independent is short lived and photosynthetic. In both of these groups of plants the mobile male gametes, antherozoid produced by sex organ antheridium, travel to archegonium (bearing an egg cell) via the medium of water. Egg cell is non motile hence the reproduction is oogamous. Sporophyte (2n) Zygote (2n) Syn A Meiosis Gametogenesis Haplo-diplontic B Spores (n) Gametophyte (c) (n) Diagram 2. Lichen is usually cited as an example of ‘symbiosis’ in plants where an algal and a fungal species live together for their mutual benefit. What will happen if algal and fungal partners are separated from each other? a. Both will survive and grow normally and independent from each other. b. Both will die c. Algal component will survive while the fungal component will die. d. Fungal component will survive while algal partner will die. Based on your answer how do you justify this association as symbiosis. Ans. Lichen is usually cited as an example of symbiosis in biology where in a fungal and an algal species live together for mutual benefit. The algal component synthesizes the food through photosynthesis which is utilized by the fungal species for its survival. The fungal component in return provides shelter and waste products that are consumed by algal species. Experiments though have shown that algal component can grow independently when separated from fungal species. But same is not true with the fungal component which dies when separated from algal component. This association is, therefore, a typical case of master- slave relationship where fungus (master) has trapped the algal components (slave) for its own survival while giving nothing in return to it . Some authors consider this association as controlled parasitism or helotism due to the fact that sometimes the fungus sends its haustoria into the algal cells to derive nourishment. 3. Distinguish between families fabaceae, solanaceae, liliaceae on the basis of gynoecium characteristics. Also write economic importance of any one of the above family. Ans. a. Gynoecium i. Fabaceae - Monocarpellary, ovary unilocular, marginal placentation ii. Solanaceae - Bicarpellary syncarpous, carpels placed obliquely, bilocular, axile placentation iii. Liliaceae - Tricarpellary, syncarpous, ovary superior, axile placentation b. Economic importance of fabacae. i. Source of pulses (gram, arhar) ii. Edible oil (soyabean, groundnut) iii. dye (indigofera) iv. fibres (sunhemp) v. fodder (Sesbania, Trifolium) vi. ornamental (lupin) vii. Medicine (mulaithi) 4. The arrangement of ovules within the ovary is known as placentation. What does the term placenta refer to? Name the different types of placentation seen in plants. Draw any three types of placentation in flowers as seen in T.S. MODEL ANSWERS TO DESCRIPTIVE QUESTIONS 151 Ans. The ovules are female reproductive structures and borne in the ovary of the flower. The number, structure, their position in the ovary varies in different plants. They also differ in mode of attachment with the ovary wall. At the point of attachment there is a cellular ridge or cushion of cells called placenta. The mode of attachment of ovule to the placenta is known as placentation which is of the following types: (a) Parietal (b) Marginal (c) Axile (d) Free central (e) Basal. (b) (c) (d) Axile Placentation Parietal placentation Free central placentation 5. Explain the digestive system of cockroach with the help of a labelled sketch. Ans. The digestive system consists of alimentary canal and digestive glands. The alimentary canal of cockroach is divided into foregut, midgut and hindgut. The mouth opens into a short tubular pharynx, leading to a narrow tubular passage, the oesophagus, which opens into a sac like crop used for storing food. The crop is followed by a gizzard or proventriculus. Gizzard consists of six - chitinous plates called teeth which helps in grinding food. The entire foregut is lined by cuticle. A ring of six - eight blind tubular structures called hepatic or gastric caecae is present at the junction of foregut and midgut which secrete digestive juice. At the junction of midgut and hindgut is present another ring of yellow colored thin filamentous malpighian tubules which help in removal of excretory products from haemolymph. The hindgut is broader than midgut and is differentiated into ileum, colon and rectum. The rectum opens out through anus. 6. It is observed that deficiency of a particular element showed its symptoms initially in older leaves and then in younger leaves. a. Does it indicate that the element is actively mobilized or relatively immobile? Name two elements which are highly mobile and two which are relatively immobile. Ans. It is actively mobilized. Highly mobile- nitrogen, magnesium Relatively immobile- calcium, boron b. How is the aspect of mobility of elements important to horticulture and agriculture? Ans. Symptoms of deficiency of mobile elements are more pronounced in older leaves and symptoms of deficiency of relatively immobile element appear first in younger leaves. This information can be utilised by horticulturist and agriculturist to get a broad idea of the deficiency elements in plants. 7. Explain the mechanism of breathing with neat labelled sketches Ans. Breathing involves two stages. a. Inspiration: Inspiration is initated by the contraction of diaphragm, which increases the volume of thoracic chamber in the anterioposterior axis. The contraction of external inter-costal muscles lifts up the ribs and the sternum causing an increase in the volume of thoracic chamber in the dorso-ventral axis also. Such Pharynx Salivary gland Salivary reservoir Oesophagus Crop Gizzard Hepatic caeca Mesentron or midgut Malpighian tubules Rectum Ileum Colon an increase in thoracic volume leads to a similar increase in pulmonary volume resulting in decreased intra-pulmonary pressure to less than atmospheric pressure. This causes the movement of external air into the lungs. i.e., inspiration. b. Expiration: The inter-costal muscles return the diaphragm and sternum to their normal positions with relaxation of the diaphragm. This reduces the thoracic volume and thereby the pulmonary MODEL ANSWERS TO DESCRIPTIVE QUESTIONS 153 volume. As a result an increase in intra-pulmonary pressure to slightly above the atmospheric pressure causes the expulsion of air from the lungs i.e., expiration. 8. Explain different types of blood groups and donor compatibility by making a table. Ans. ABO blood grouping is based on the presence or absence of two surface antigens on the RBCs namely A and B. Similarly, the plasma of different individuals contain two natural antibodies anti-A and anti-B. Blood group ‘A’ carries antigen-A and antibodies-B. The donor’s group for blood group A are A and O. Blood group B carries antigen-B and antibodies-A. The donor’s group for blood group B are B and O. Blood group AB carries antigens A and B but no corresponding antibodies so, the compatible donor’s group for blood group AB are A, B, AB and O hence, blood group ‘AB’ is also called as “universal acceptor”. Blood group ‘O’ carries no antigens but carries antibodies both A and B hence its compatible donor’s group is only ‘O’ but it is a compatible donor group for all the blood groups. A, B, AB and O hence, blood group ‘O’ is called as ‘universal donor’. Blood groups and donor compatibility Blood Group Antigen on RBCs Antibodies in Plasma Donor’s Compatibility A A Anti-B A. O B B Anti-A B. O AB A.B Nil AB, A. B. O O Nil Anti - A, B O 9. Discuss the role of Ca2+ ions in muscle contraction. Draw neat sketches to illustrate your answer. Ans. Muscle contraction is initiated by a neural signal, which after reaching neuromuscular junction or motor end plate releases a neurotransmitter, as a result an action potential in the sarcolemma is generated. Action potential spreads through muscle fibre and causes the release of calcium ions into the sarcoplasm. Increase in Ca2+ level leads to the binding of calcium with a subunit of troponin on actin filaments and thereby removes the masking of active sites for myosin. Utilising the energy from ATP hydrolysis, the myosin head now binds to the exposed active site on actin to form a cross-bridge. This pulls the attached actin filaments towards the centre of ‘A’ band. The ‘Z’ line attached to these actins are also pulled inwards thereby causing shortening of the sarcomere, i.e., contraction. A new ATP binds to myosin head and the cross-bridge is broken. The ATP is again hydrolysed by the myosin head and the cycle of cross- bridge formation and breakage is repeated causing further sliding. The process continues till the Ca++ ions are pumped back to the sarcoplasmic cisternae resulting in masking of actin filaments and breakage of all cross bridges. This cause the return of ‘Z’ lines along with filaments back to their original position, i.e., relaxation. 11. Differentiate between pectoral and pelvic girdle. Ans. Pectoral and pelvic girdle help in the articulation of upper and lower limbs respectively. Each girdle is made of two equal halves. Each half of a pectoral girdle consists of clavicle and scapula. Scapula is a large triangular flat bone. There is glenoid cavity at the joint of scapula, clavicle and acromian process, which articulates with the head of humerus to form the shoulder joint. Each half of pelvic girdle is formed by three bones-ilium, ischium and pubis. At the point of their fusion; there is a cavity called acetabulum to which the head of femur articulates. MODEL ANSWERS TO DESCRIPTIVE QUESTIONS 155 12. On an educational trip to Uttaranchal, Ketki and her friends observe that people in many localities have swollen necks. Please help Ketki and her friends to find out the solution to the following questions: a. Which disease are these people suffering from? Ans. People with swollen necks are suffering from goitre. b. How is it caused? Ans. It is caused due to the deficiency of Iodine in our diet. Iodine is essential for the normal rate of hormone synthesis in the thyroid. c. Which hormones could be decreased by this condition? Ans. Tetraiodothyronine or thyroxine (T4) and triiodothyronine (T3) d. What effect does this condition have on pregnancy? Ans. Hypothyroidism during pregnancy causes defective development and maturation of the growing baby leading to - i) stunted growth, ii) mental retardation, iii) low I Q, iv) abnormal skin and v) deaf-mutism.