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Target Publications Pvt. Ltd. Biology Board Answer paper: october 2014 Biology SECTION – I [BOTANY] Q.1. Select and write the most appropriate answer from the given alternatives for each sub-question: i. (B) HbS [1] ii. (C) complementary DNA [1] iii. (A) high proteins [1] iv. (D) Streptomyces erythreus [1] v. (C) Sugarcane [1] vi. (B) 1 [1] vii. (D) CFCs [1] Q.2. (A) i. Answer in ‘one’ sentence each: Vinegar is obtained from Acetobacter aceti. [1] ii. The first CO2 acceptor in C3 pathway is Ribulose −1, 5 − diphosphate (RUDP). [1] iii. Fermentation is the process of anaerobic breakdown (oxidation) of an extracellular organic substrate by the action of enzymes secreted by micro-organisms (which respire anaerobically) resulting in the formation of some important products. [1] The deposition of pesticides in the fatty tissues of fishes and aquatic birds which feed on them is called Bioaccumulation. [1] v. Methanogenic bacteria (Methanobacillus; Methanococcus) convert organic acids into methane. [1] vi. Carbon dioxide (CO2) gas is mainly responsible for global warming. [1] iv. (B) Dividing generative cell Pollen tube Tube nucleus Germinating angiospermic pollen grain (C) i. (Proportionate diagram) [½] (Three labels) [1½] Answer the following (Any TWO): a. Genetic modification of organisms has unpredictable effects when such organisms are introduced into the ecosystem. b. Cross pollination between GM plants and wild varieties leads to contamination of gene pools of wild varieties. c. Consumption of GM foods may lead to allergies. d. GM microbes may escape from the laboratory and prove to be hazardous. e. Manipulation of living organisms requires regulation. 1 [½] [½] [½] [½] Target Publications Pvt. Ltd. ii. Board Answer Paper: October 2014 Gibberellins are a group of growth hormones produced by fungi and higher plants. Uses: a. Promote growth of plant by stem elongation. b. Induce parthenocarpy in apple, pear, etc. c. Breaking dormancy d. Induce flowering (Any two uses) [½ × 2] iii. a. b. c. Structurally, the ATP consists of nitrogenous purine base called Adenine, a pentose sugar called Ribose (C5H10O5) and 3 Phosphate molecules. Adenine and ribose sugar together form Adenosine. Adenosine with 1 phosphate, 2 phosphates and 3 phosphates are called AMP, ADP and ATP respectively. Between the 2nd and 3rd molecules of phosphate, a high energy phosphate bond is present. For the formation of these bonds, energy is given by either light (Photophosphorylation) or oxidative process of respiration (oxidative phosphorylation). [1] [1] [½] [½] [½] High energy bond Adenine P P P [½] Triphosphate Ribose sugar Adenosine iv. Q.3. (A) i. Structure of ATP Conservation and management of forests can be done by following ways: a. By effective control of wild fire using suitable measures. b. By regulating grazing of animals in forests lands. c. By protecting forests from pests and pathogens. d. Economic use of timber and fuel wood to avoid its wastage. e. Forest conservation through law, adopting programmes, reforestation and afforestation. f. Encouraging agro-forestry ,educating people about hazards of deforestation,etc. (Any four points) [½ × 4] Attempt the following (Any TWO): Enzyme β-galactosidase Permease Transacetylase ii. Function β -galactosidase → Glucose + Galactose Lactose ⎯⎯⎯⎯⎯ Entry of lactose in the cell Transfers acetyl group from Acetyl CoA to β-galactosidase PCR is used in: a. DNA cloning. b. gene amplification. c. DNA-based phylogeny or functional analysis of gene. d. Diagnosis of hereditary disease. e. DNA fingerprinting. f. Diagnosis of infectious diseases and cancer. 2 [2] [1] [1] [1] [½] [½] [½] [½] [½] [½] Target Publications Pvt. Ltd. iii. a. b. c. d. Biology In selective breeding method, plant breeders search seed or germplasm banks for existing varieties of crops which are naturally high in nutrients. Then, they crossbreed these high-nutrient varieties with high-yielding varieties of crops, to provide a seeds with high yields and increased nutritional value. Crops with sufficient amounts of nutrients must be bred to have a measurable positive impact on human health. Such crops must be developed with the involvement of nutritionists and should have extra nutrients, as storage, processing and cooking of the food affects their available nutrient levels. Examples: 1. Hybrid maize with almost double the quantity of amino acids like lysine and tryptophan. 2. Wheat variety, Atlas-66 with high protein content. (B) [½] [½] [½] [½] [½] [½] Upper Epidermis Mesophyll cell Bundle sheath cell Vascular Bundle Stoma Lower Epidermis Kranz anatomy (Proportionate diagram) (Any four labels) Q.4. Vegetative reproduction is a kind of asexual reproduction which occurs with the help of vegetative organs like root, stem, leaf or bud. A. Root Tuber: e.g. Sweet Potato (Ipomoea batatas) i. It is a modification of root for vegetative reproduction. ii. These roots develop from the nodes of stem. iii. They become tuberous and fleshy for storage of food. In addition to the storage of food, these root also develop adventitious buds on their surface which sprout under favourable condition to produce ‘leafy shoot’ and adventitious roots. iv. Under suitable environmental conditions, these leafy shoots separate and develop into new plants. [1] [½] [½] [1] Creeping stem Scar of detached root Tuberous root [1] [2] Adventitious roots Adventitious roots Root tuber of sweet potato 3 Target Publications Pvt. Ltd. B. Board Answer Paper: October 2014 Stem tuber : e.g. Solanum tuberosum i. Stem tuber is a modified stem for vegetative reproduction. ii. In case of potato, some lateral branches are produced from under ground part of stem which grow down in the soil. iii. The tip of these branches store food and hence are swollen. iv. A stem tuber has many notches on its surface called ‘eye’. v. The eyes that are seen on the surface of the tuber represent nodes. vi. Each ‘eye’ is actually a node and is made up of one or more axillary buds substended by a leaf scar. vii. [½] [½] Under favourable conditions, axillary buds develop into new plants. Apical shoot Stem tuber [1] Adventitious root Scar at distal end of tuber Seed potato Cluster of buds Leaf scar Adventitious root “Eye” Stem tuber of potato C. Vegetative propagation by runner: e.g. Cynodon i. Runner is a slender, prostrate, sub aerial branch with short or long internode and creeps horizontally on the soil. ii. Runner develops from the axillary bud in the axil of the lowest leaf. iii. After creeping some distance, away from the parent plant, it produces shoot from upper side and roots from lower side of the nodes. iv. Many runners are produced by the parent plant which may get detached from parent plant to develop into new plants. [½] [½] Leaf Node Aerial branch Node Scale leaf Runner Adventitious roots Runner in Cynodon 4 [1] Target Publications Pvt. Ltd. Biology OR Let the gene for tall habit of pea plant be represented by ‘T’ and dwarf habit be represented by ‘t’. Let the gene for round-seed be represented by ‘R’ and that of wrinkled seed be represented by gene ‘r’. Then, the genotypes of the parents would be: i. Homozygous tall, round-seeded − TTRR ii. Homozygous dwarf, wrinkled seeded – ttrr Phenotype of parents Genotype Tall Round TTRR Gametes TR TR F1 generation tr tr Dwarf Wrinkled ttrr × tr TR TR TtRr Tall round TtRr Tall round TtRr Tall round TtRr Tall round tr [1] All Tall round Phenotype of F1 generation Genotype of F1 generation TtRr Selfing of F1 TtRr Gametes [1] TR Tr × tR TR F2 generation TR Tr tR tr TTRR Tall Round TTRr Tall Round TtRR Tall Round TtRr Tall Round tr TtRr TR Tr Tr tR tR tr tr TTRr Tall Round TtRR Tall Round TtRr Tall Round TTrr Tall Wrinkled TtRr Tall Round TtRr Tall Round Ttrr Tall Wrinkled ttRr Dwarf Round ttrr Dwarf Wrinkled Ttrr Tall Wrinkled ttRR Dwarf Round ttRr Dwarf Round [2] F2 progeny obtained Tall Round; Tall Wrinkled; Dwarf Round; Dwarf Wrinkled Phenotypic ratio → 9 : 3 : 3 : 1 Genotypic ratio→ 1 : 2 : 2 : 4 : 1 : 2 : 1 : 2 : 1 TTRR TTRr TtRR TtRr ttRR ttRr TTrr Ttrr ttrr The above example of a dihybrid cross between homozygous tall, round-seeded pea plant and dwarf, wrinkled-seeded pea plant explains the ‘Law of Independent Assortment’. The law of independent assortment states that “when two parents differing from each other in two or more pairs of contrasting characters are crossed, then the inheritance of one pair of character is independent of the other pair of character.” 5 [1] [1] [1] Target Publications Pvt. Ltd. Board Answer Paper: October 2014 SECTION – II [ZOOLOGY] Q.5. Select and write the most appropriate answer from the given alternatives for each sub-question: i. (B) Ethological isolaion [1] ii. (C) Mice [1] iii. (A) Caterpillar [1] iv. (D) T.S of thyroid gland [1] v. (B) Spongilla [1] vi. (A) Insulin [1] vii. (C) mercury [1] Q.6. (A) i. Answer in only ‘One’ sentence each of the following questions: Copper ions released from Copper−T suppress the sperm motility and hence the fertilizing capacity of sperms. [1] ii. Methyl isocyanate (MIC) gas caused Bhopal gas tragedy in 1984. [1] iii. a. b. c. iv. v. vi. The Sinu-auricular (SA) node is the pacemaker of the heart. It generates wave of contraction which spread along the wall of atria and bring about their contraction. It has highest degree of rhythmicity and is the first to originate the cardiac impulses and determine the rate of heart beat. [½] [½] The modes of transmission of amoebiasis are: a. Faeco − oral route. b. Through contaminated food and water. [½] [½] Vaccine is an antigenic preparation used to stimulate the production of antibodies and induce immunity against several diseases. [1] The process of evolution which results in transformation of original species to many different varieties is called adaptive radiation. [1] (B) Tunica externa Elastic membrane Tunica media Lumen Tunica interna (endothelium) T.S. of an Artery 6 (Proportionate diagram) [½] (Any three labels) [1½] Target Publications Pvt. Ltd. (C) i. Biology Attempt any TWO of the following questions: Cromagnon Man (Homo sapiens fossilis): a. It is the closest ancestor of modern man. b. It lived in France and Spain and made paintings inside caves and ornaments from ivory. c. It was an omnivore having the aesthetic sense. d. The Cro-magnons were about 180 cm in height with large skull, rounded forehead, distinct chin, narrow nose, eyebrow ridges thin and broad face. e. The cranial capacity was about 1600 c.c. f. They lived in caves with families. g. They were expert in hunting and painting. They made tools and weapons such as spearheads, bows and arrows. h. They made ornaments from ivory and decorated their bodies. i. They used hides of animals to protect their body and buried their dead according to their customs. (Any four points) [½ × 4] ii. Parents Female × 2n = 32 (Diploid) Meiosis n = 16 (haploid) n = 16 (haploid) [2] Male (n = 16) (haploid) Mitosis n = 16 (haploid) fertilization [2] Without 2n = 32 Fertilization (diploid) Female (n=16) Male (Drone) Haploid number of chromosomes in drone of honeybee iii. iv. Goals of Human Genome Project: a. Determine the sequence and number of all the base pairs (three billion) in the human genome – gene mapping. b. Identify all the genes (25000) present in human DNA. c. Determine the function of all the genes and identify the various genes that cause genetic disorders. d. Store the information in data base. e. Improve tools for data analysis. f. Find out possibilities of transfer of technology developed during HGP to industry. g. Address ethical, legal and social issues (ELSI) that may arise from the project. h. To have greater understanding of process of human evolution. i. To understand more about genetic structure functions, gene mutation, expression and methods to control them. (Any four points) [½ × 4] a. b. c. d. Major products of apiculture are honey and beeswax. Honey has high nutritive value. It contains levulose, dextrose, maltose, enzymes, minerals, vitamins and water. Honey also has medicinal importance. Honey is used as an antiseptic and sedative.It is used in gastrointestinal disorders. Honey bees are pollinating agents. They help to increase crop yield by pollination. 7 [2] Target Publications Pvt. Ltd. e. f. g. Q.7. (A) i. Board Answer Paper: October 2014 Beeswax is used in cosmetics, paints and ointments. Honey is used in manufacturing of cakes and as a flavouring agent. Venom from sting is used in the treatment of rheumatoid arthritis and snake bite. (Any four points) [½ × 4] Attempt any TWO of the following: Structure of X and Y chromosomes a. X chromosomes is longer than Y chromosomes. b. X chromosome contains large amount of euchromatin and small amount of heterochromatin, whereas the Y chromosome contains small amount of euchromatin and large amount of heterochromatin. c. Non-homologous region of X-chromosome contains more genes than that of nonhomologous region of Y chromosome. d. Non-homologous region of X-chromosome contains X-linked genes while nonhomologous region of Y chromosome contains Y-linked genes. [2] [½] [½] [½] [½] Non-homologous parts [1] ii. a. b. c. d. e. f. iii. a. b. c. 8 Homologous parts Sex chromosomes (X and Y) HIV virus is the causative organism for AIDS. AIDS begins with HIV infection. People infected with HIV may have no symptoms for 10 years or longer, but they can still transmit the infection to other during this symptom-free period. If the infection is not detected and treated, the immune system gradually weakens and AIDS develops. Acute HIV infection progresses over time (usually a few weeks to months) to asymptomatic HIV infection (no symptoms) and then to early symptomatic HIV infection with CD4 T-cell count below 200 cells/mm3). Almost all people infected with HIV, if not treated, will develop AIDS. A small group of patients develop AIDS very slowly, or never at all. These patients are called nonprogressors, and many seem to have a genetic difference that prevents the virus from damaging their immune system. The symptoms of AIDS are primarily the result of infections that do not normally develop in individuals with healthy immune systems. Theses are called opportunistic infections. People with AIDS have had their immune system damaged by HIV and are very susceptible to these opportunistic infections. Common symptoms are:Chills, Fever, Sweat (particularly at night), Swollen lymph glands, weakness and weight loss. The radioactive waste is normally disposed by land filling. For land filling, deep pits are dug in desert areas or in the sea bottom to dispose the radioactive waste. These pits are capped by concrete to avoid contamination. [½] [½] [½] [½] [½] [½] [1] [1] [1] Target Publications Pvt. Ltd. Biology (B) Dorsal root White Matter Dorsal root ganglion Spinal nerve Dorsal Horn Ramus Dorsalis Central canal Ramus Ventralis (Lateralis) Ventral Horn Ramus Communicans Grey Matter Ventral root Sympathetic ganglion Formation of Typical Spinal Nerve (Proportionate diagram) [1] (Any four labels) [2] Q.8. Excretory system of man consists of: A. Kidneys (a pair) B. Ureters (a pair) C. Urinary Bladder D. Urethra Right Kidney Left Kidney Ureter [1] Urinary bladder Urethra Excretory system of man A. Kidneys: Kidneys are a pair of dark red, bean shaped organs about 10 cm long, 5 cm wide and 4 cm thick. They are attached to the dorsal body wall at the level of 12th thoracic to 3rd lumbar vertebra in the abdominal cavity. They are retroperitoneal as they have peritoneal covering only on the anterior surface. They are situated on either side of the vertebral column. The right kidney is slightly lower than left kidney due to presence of liver. Left kidney is slightly back than right kidney due to the position of stomach in front of it. On the concave surface there is a notch called hilus or hilum. Through the hilum, blood vessels, nerves, ureter, lymphatic vessels enter or exit. Renal arteries enter in both the kidneys while renal veins leave the kidneys. Each kidney is covered by three layers of tissues such as outer renal fascia, middle adipose tissue (perirenal fat) and inner renal capsule. Functions: i. Removal of nitrogenous waste ii. Maintenance of water–salt balance [Homeostasis] iii. Removal of excess of foreign substances like drugs and pigments. iv. Osmoregulation. 9 [½] [½] [½] [½] Target Publications Pvt. Ltd. B. C. D. Board Answer Paper: October 2014 Ureters: From the hilum of each kidney, a thin, narrow and muscular duct emerges out which is called as ureter. It measures about 40cm in length The ureters open into urinary bladder. Functions: They carry urine from kidneys upto the urinary bladder by peristalsis. Urinary bladder: It is single, large and pear shaped structure. It is located in pelvic cavity just behind the pubis bone. It is thick walled and highly muscular. It has a layer of detrusor muscles. These muscles help in the expansion during the time of storage of urine. Functions: It stores urine temporarily (500 ml to 1 litre of urine) and expels it at intervals through urethra. Urethra: It starts from the lower part of the urinary bladder and opens to the exterior. Its opening is guarded by sphincter muscle called urethral sphincter. Through urethral sphincter micturition (urination) takes place. In males, urethra is long (20 cm) and has to pass through the penis. In females, it is short (4 cm) and opens in vestibule. Acute kidney injury (AKI) is characterized by: i. Oliguria (decreased urine production; < 400 ml per day in adults) ii. Anuria (absence of urine), iii. Accumulation of urea and other metabolic waste products, iv. Body water and body fluid disturbances, v. Electrolyte derangement. If all of the above signs are seen in urine analysis, then the person may be suffering from Acute kidney injury (AKI). [½] [½] [½] [½] [½] [½] [1] OR Structure of human sperm: i. The human sperm is a haploid, elongated and motile male gamete. ii. It is microscopic, measuring about 0.060 mm in length. iii. The body of sperm is divided into four parts, viz., head, neck, middle piece and tail. a. Head: It is flat and oval consisting of large haploid nucleus and an acrosome formed from Golgi complex. The acrosome secretes an enzyme hyalourinidase which helps in penetration of egg during fertilization. The anterior half of the nucleus is covered by a fibrous sheath. b. Neck: It is a narrow constricted region consisting of two centrioles (proximal and distal) and cytoplasm. The proximal centriole plays an important role in first cleavage of the zygote and distal centriole forms the axial filament present in the tail of the sperm. c. Middle piece: Middle piece has many mitochondria spirally coiled around axial filament. The mitochondria provide energy for the sperm during its movement, hence it is called the powerhouse of the sperm. 10 [½] [½] [½] Target Publications Pvt. Ltd. d. Biology Tail: The tail is long, slender and tapering. It is made up of cytoplasm. The axial filament passes through the tail. [½] Acrosome Head Neck Middle piece Nucleus Proximal centriole Distal centriole Spiral mitochondria [1] Ring centriole Axial filament Tail piece or flagellum Structure of Sperm Structure of Ovum: Nucleus Polar body Corona radiata Perivitelline space [1] Vitelline membrane Zona pellucida Structure of ovum(unfertilized) i. The ovum discharged by the ovary during ovulation is actually a secondary oocyte. ii. It is round, non-motile, haploid female gamete. iii. It is the largest cell of the body and measures about 0.1 mm or 100 µ in diameter. iv. The human egg is almost free of yolk, hence called microlecithal. v. It shows abundant cytoplasm called ooplasm. vi. Ooplasm has a large eccentric nucleus called germinal vesicle. vii. Ovum is surrounded by plasma membrane called vitelline membrane. [½] [½] viii. The ovum lack centrioles and therefore does not undergo further divisions. 11 Target Publications Pvt. Ltd. Board Answer Paper: October 2014 ix. The ovum contains two poles. The pole which shows presence of polar body is called animal pole while the opposite side is termed vegetal pole. x. The ovum shows two covering layers viz. an outer, thicker, cellular corona radiata and an inner, thinner, non-cellular zona pellucida. xi. The zona pellucida is secreted by ovum itself while corona radiata is formed of radially elongated follicular cells which are glued together by hyaluronic acid. If a couple is unable to conceive, the modern techiques available to overcome this problem are: i. Zygote intra-fallopian transfer (ZIFT) : Zygote or blastocyst upto 8- blastomeres is transferred into the fallopian tube, the embryo moves to the uterus where it becomes implanted and continues further development. ii. Gamete intra−fallopian transfer (GIFT) : In GIFT, ovum collected from a donor is transferred into the fallopian tube of another female who can provide suitable environment for fertilization and development. 12 [½] [½] [½] [½]