R SOLUTION OF NEET AIPMT - 2017 (HELD ON 7th MAY SUNDAY 2017) FEEL THE POWER OF OUR KNOWLEDGE & EXPERIENCE Our Top class IITian & Doctor faculty team promises to give you an authentic answer key which will be fastest in the whole country. Set - A (All Code - A/P/W, B/Q/X, C/R/Y, D/S/Z) BIOLOGY 91. Double fertilization is exhibited by 94. What is the criterion for DNA fragments (1) Gymnosperms movement on agarose gel during gel electrophoresis? (2) Algae (1) The larger the fragment size, the farther it (3) Fungi moves (4) Angiosperms (2) The smaller the fragment size, the farther it Answer (4) moves SOLUTION :- (3) Positively charged fragments move to farther Double fertilization is a characteristic feature end exhibited by angiosperms. It involves syngamy (4) Negatively charged fragments do not move and triple fusion. Answer (2) 92. Which of the following are found in extreme saline conditions? SOLUTION :- (1) Archaebacteria During gel electrophoresis, DNA fragments separate (resolve) according to their size through (2) Eubacteria sieving effect provided by agarose gel. (3) Cyanobacteria 95. Attractants and rewards are required for (4) Mycobacteria (1) Anemophily (2) Entomophily Answer (1) (3) Hydrophily (4) Cleistogamy SOLUTION :- Archaebacteria are able to survive in harsh Answer (2) conditions because of branched lipid chain in SOLUTION :- cell membrane which reduces fluidity of cell Insect pollinated plants provide rewards as edible membrane. Halophiles (Archaebacteria) are pollen grain and nectar as usual rewards. While exclusively found in saline habitats. some plants also provide safe place for 93. Select the mismatch : deposition of eggs. (1) Frankia – Alnus 96. Which of the following is made up of dead cells? (2) Rhodospirillum – Mycorrhiza (1) Xylem parenchyma (3) Anabaena – Nitrogen fixer (2) Collenchyma (4) Rhizobium – Alfalfa (3) Phellem Answer (2) (4) Phloem SOLUTION :- Answer (3) Rhodospirillum is anaerobic, free living nitrogen SOLUTION :- fixer. Mycorrhiza is a symbiotic relationship Cork cambium undergoes periclinal division and between fungi and roots of higher plants. cuts off thick walled suberised dead cork cells (Phellem) towards outside and it cuts off thin walled living cells i.e., phelloderm on inner side. H.O : JABALPUR : 1525, Wright Town, Ph. (0761) 4218116, 2400022, 8349992509, RAMPUR : Rampur Chowk, Above PNB RANJHI : Opp. Jain Mandir, Main Road, Ranjhi follow us on : momentumacademy www.momentumacademy.com momentumacademy 97. Which cells of ‘Crypts of Lieberkuhn’ secrete DNA as genetic material. antibacterial lysozyme? 101. Which among the following are the smallest (1) Argentaffin cells (2) Paneth cells living cells, known without a definite cell wall, pathogenic to plants as well as animals and can (3) Zymogen cells (4) Kupffer cells survive without oxygen? Answer (2) (1) Bacillus (2) Pseudomonas SOLUTION :- (3) Mycoplasma (4) Nostoc -Kupffer-cells are phagocytic cells of liver. – Zymogen cells are enzyme producing cells. Answer (3) – Paneth cell secretes lysozyme which acts as SOLUTION :- anti-bacterial agent. Mycoplasmas are smallest, wall-less prokaryotes, pleomorphic in nature. These are – Argentaffin cells are hormone producing cells. pathogenic on both plants and animals. 98. Adult human RBCs are enucleate. Which of the 102. Which of the following options gives the correct following statement(s) is/are most appropriate sequence of events during mitosis? explanation for this feature? (1) Condensation nuclear membrane (a) They do not need to reproduce disassembly crossing over segregation (b) They are somatic cells telophase (c) They do not metabolize (2) Condensation nuclear membrane (d) All their internal space is available for oxygen disassembly arrangement at equator  transport centromere division segregation  (1) Only(d) (2) Only (a) telophase (3) (a), (c) and (d) (4) (b) and (c) (3) Condensation crossing over nuclear Answer (1) membrane disassembly segregation  SOLUTION :- telophase In Human RBCs, nucleus degenerates during (4) Condensation arrangement at equator maturation which provide more space for oxygen centromere division segregation  carrying pigment (Haemoglobin). It lacks most telophase of the cell organelles including mitochondria so respires anaerobically. Answer (2) 99. The hepatic portal vein drains blood to liver from SOLUTION :- (1) Heart (2) Stomach The correct sequence of events during mitosis (3) Kidneys (4) Intestine would be as follows Answer (4) (i) Condensation of DNA so that chromosomes become visible occurs during early to mid- SOLUTION :- prophase. In hepatic portal system, hepatic portal vein carries maximum amount of nutrients from (ii) Nuclear membrane disassembly begins at intestine to liver. late prophase or transition to metaphase. 100. The final proof for DNA as the genetic material (iii) Arrangement of chromosomes at equator came from the experiments of occurs during metaphase, called congression. (1) Griffith (iv) Centromere division or splitting occurs during (2) Hershey and Chase anaphase forming daughter chromosomes. (3) Avery, Mcleod and McCarty (v) Segregation also occurs during anaphase as daughter chromosomes separate and move to (4) Hargobind Khorana opposite poles. Answer (2) (vi) Telophase leads to formation of two daughter SOLUTION :- nuclei. Hershey and Chase gave unequivocal proof which ended the debate between protein and H.O : JABALPUR : 1525, Wright Town, Ph. (0761) 4218116, 2400022, 8349992509, RAMPUR : Rampur Chowk, Above PNB RANJHI : Opp. Jain Mandir, Main Road, Ranjhi follow us on : momentumacademy www.momentumacademy.com momentumacademy 103. Which one of the following statements is correct, SOLUTION :- with reference to enzymes? Biosphere reserve is protected area with (1) Apoenzyme = Holoenzyme + Coenzyme multipurpose activities. It has three zones (a) Core zone – without any human interference (2) Holoenzyme = Apoenzyme + Coenzyme (b) Buffer zone – with limited human activity (3) Coenzyme = Apoenzyme + Holoenzyme (c) Transition zone – human settlement, grazing (4) Holoenzyme = Coenzyme + Cofactor cultivation etc., are allowed. Answer (2) 107. A dioecious flowering plant prevents both: SOLUTION :- (1) Autogamy and xenogamy Holoenzyme is conjugated enzyme in which (2) Autogamy and geitonogamy protein part is apoenzyme while non-protein is (3) Geitonogamy and xenogamy cofactor. Coenzymes are also organic (4) Cleistogamy and xenogamy compounds but their association with Answer (2) apoenzyme is only transient and serve as SOLUTION :- cofactors. When unisexual male and female flowers are 104. During DNA replication, Okazaki fragments are present on different plants the condition is called used to elongate dioecious and it prevents both autogamy and (1) The leading strand towards replication fork geitonogamy. 108. A temporary endocrine gland in the human body (2) The lagging strand towards replication fork is (3) The leading strand away from replication fork (1) Pineal gland (2) Corpus cardiacum (4) The lagging strand away from the replication (3) Corpus luteum (4) Corpus allatum fork Answer (3) Answer (4) SOLUTION :- SOLUTION :- Corpus luteum is the temporary endocrine Two DNA polymerase molecules work structure formed in the ovary after ovulation. It simultaneous at the DNA fork, one on the leading is responsible for the release of the hormones strand and the other on the lagging strand. like progesterone, oestrogen etc. Each Okazaki fragment is synthesized by DNA 109. Match the following sexually transmitted polymerase at lagging strand in 3'5' diseases (Column - I) with their causative agent direction. (Column - II) and select the correct option. New Okazaki fragments appear as the Column - I Column- II replication fork opens further. (a) Gonorrhea (i) HIV As the first Okazaki fragment appears away from (b) Syphilis (ii) Neisseria the replication fork, the direction of elongation would be away from replication fork 3'5' (c) Genital Warts (iii) Treponema direction. (d) AIDS (iv) Human Papilloma 105. Which of the following are not polymeric? virus (1) Nucleic acids (2) Proteins Options : (3) Polysaccharides (4) Lipids (a) (b) (c) (d) Answer (4) (1) (ii) (iii) (iv) (i) SOLUTION :- (2) (iii) (iv) (i) (ii) – Nucleic acids are polymers of nucleotides – Proteins are polymers of amino acids (3) (iv) (ii) (iii) (i) – Polysaccharides are polymers of (4) (iv) (iii) (ii) (i) monosaccharides Answer (1) – Lipids are the esters of fatty acids and alcohol SOLUTION :- 106. The region of Biosphere Reserve which is legally protected and where no human activity is allowed Gonorrhoea – Neisseria (Bacteria) Syphilis – is known as Treponema (Bacteria) (1) Core zone (2) Buffer zone Genital Warts – Human papilloma virus (Virus) (3) Transition zone (4) Restoration zone AIDS – HIV (Virus) Answer (1) H.O : JABALPUR : 1525, Wright Town, Ph. (0761) 4218116, 2400022, 8349992509, RAMPUR : Rampur Chowk, Above PNB RANJHI : Opp. Jain Mandir, Main Road, Ranjhi follow us on : momentumacademy www.momentumacademy.com momentumacademy 110. Transplantation of tissues/organs fails often due 115. The process of separation and purification of to non-acceptance by the patient’s body. Which expressed protein before marketing is called type of immune-response is responsible for such (1) Upstream processing rejections? (2) Downstream processing (1) Autoimmune response (3) Bioprocessing (2) Cell-mediated immune response (4) Postproduction processing Answer (2) (3) Hormonal immune response SOLUTION :- (4) Physiological immune response Biosynthetic stage for synthesis of product in Answer (2) recombinant DNA technology is called SOLUTION :- upstreaming process while after completion of Non-acceptance or rejection of graft or biosynthetic stage, the product has to be transplanted tissues/organs is due to cell subjected through a series of processes which mediated immune response. include separation and purification are collectively referred to as downstream 111. Spliceosomes are not found in cells of processing. (1) Plants (2) Fungi 116. Mycorrhizae are the example of (3) Animals (4) Bacteria (1) Fungistasis (2) Amensalism Answer (4) (3) Antibiosis (4) Mutualism Answer (4) SOLUTION :- SOLUTION :- Spliceosomes are used in removal of introns Mycorrhizae is a symbiotic association of fungi during post-transcriptional processing of hnRNA with roots of higher plants. in eukaryotes only as split genes are absent as 117. Viroids differ from viruses in having : prokaryotes. (1) DNA molecules with protein coat 112. An example of colonial alga is (2) DNA molecules without protein coat (1) Chlorella (2) Volvox (3) RNA molecules with protein coat (3) Ulothrix (4) Spirogyra (4) RNA molecules without protein coat Answer (2) Answer (4) SOLUTION :- SOLUTION :- Viroids are sub-viral agents as infectious RNA Volvox is motile colonial fresh water alga with particles, without protein coat. definite number of vegetative cells. 118. Root hairs develop from the region of 113. Which of the following represents order of (1) Maturation ‘Horse’? (2) Elongation (1) Equidae (2) Perissodactyla (3) Root cap (3) Caballus (4) Ferus (4) Meristematic activity Answer (2) Answer (1) SOLUTION :- SOLUTION :- Horse belongs to order perissodactyla of class In roots, the root hairs arise from zone of mammalia. Perissodactyla includes odd-toed maturation. This zone is differentiated zone thus mammals. bearing root hairs. 119. Coconut fruit is a 114. Which of the following cell organelles is (1) Drupe (2) Berry responsible for extracting energy from carbohydrates to form ATP? (3) Nut (4) Capsule Answer (1) (1) Lysosome (2) Ribosome SOLUTION :- (3) Chloroplast (4) Mitochondrion Coconut fruit is a drupe. A drupe develops from Answer (4) monocarpellary superior ovary and are one SOLUTION :- seeded. Mitochondria are the site of aerobic oxidation of carbohydrates to generate ATP. H.O : JABALPUR : 1525, Wright Town, Ph. (0761) 4218116, 2400022, 8349992509, RAMPUR : Rampur Chowk, Above PNB RANJHI : Opp. Jain Mandir, Main Road, Ranjhi follow us on : momentumacademy www.momentumacademy.com momentumacademy 120. Plants which produce characterstic SOLUTION :- pneumatophores and show vivipary belong to The association of H1 protein indicates the (1) Mesophytes (2) Halophytes complete formation of nucleosome. Therefore (3) Psammophytes (4) Hydrophytes the DNA is in condensed form. Answer (2) 125. DNA fragments are SOLUTION :- (1) Positively charged Halophytes growing in saline soils show (2) Negatively charged (i) Vivipary which is in-situ seed germination (3) Neutral (ii) Pneumatophores for gaseous exchange (4) Either positively or negatively charged 121. Which one of the following is related to Ex-situ depending on their size conservation of threatened animals and plants? Answer (2) (1) Wildlife Safari parks SOLUTION :- (2) Biodiversity hot spots DNA fragments are negatively charged because (3) Amazon rainforest of phosphate group. (4) Himalayan region 126. Capacitation occurs in Answer (1) (1) Rete testis SOLUTION :- (2) Epididymis Ex-situ conservation is offsite strategy for conservation of animals and plants in zoological (3) Vas deferens park and botanical gardens respectively. (4) Female Reproductive tract 122. Select the mismatch : Answer (4) (1) Pinus - Dioecious SOLUTION :- (2) Cycas - Dioecuous Capacitation is increase in fertilising capacity of (3) Salvinia - Heterosporous sperms which occurs in female reproductive tract. (4) Equisetum - Homosporous 127. Which ecosystem has the maximum biomass? Answer (1) (1) Forest ecosystem SOLUTION :- (2) Grassland ecosystem Pinus is monoecious plant having both male and female cones on same plant. (3) Pond ecosystem 123. Which of the following facilitates opening of (4) Lake ecosystem stomatal aperture? Answer (1) (1) Contraction of outer wall of guard cells SOLUTION :- (2) Decrease in turgidity of guard cells High productive ecosystem are (3) Radial orientation of cellulose microfibrils in – Tropical rain forest the cell wall of guard cells – Coral reef (4) Longitudinal orientation of cellulose – Estuaries microfibrils in the cell wall of guard cells – Sugarcane fields Answer (3) 128. A disease caused by an autosomal primary non- SOLUTION :- disjunction is Cellulose microfibrils are oriented radially rather (1) Down’s syndrome than longitudinally which makes easy for the stoma to open. (2) Klinefelter’s syndrome 124. The association of histone H1 with a nucleosome (3) Turner’s syndrome indicates: (4) Sickle cell anemia (1) Transcription is occurring Answer (1) (2) DNA replication is occurring SOLUTION :- (3) The DNA is condensed into a Chromatin Down’s syndrome is caused by non-disjunction Fibre of 21st chromosome. (4) The DNA double helix is exposed Answer (3) H.O : JABALPUR : 1525, Wright Town, Ph. (0761) 4218116, 2400022, 8349992509, RAMPUR : Rampur Chowk, Above PNB RANJHI : Opp. Jain Mandir, Main Road, Ranjhi follow us on : momentumacademy www.momentumacademy.com momentumacademy 129. Life cycle of Ectocarpus and Fucus respectively (1) Tropical Savannah are (2) Tropical Rain Forest (1) Haplontic, Diplontic (3) Grassland (2) Diplontic, Haplodiplontic (4) Temperate Forest (3) Haplodiplontic, Diplontic Answer (2) (4) Haplodiplontic, Haplontic SOLUTION :- Answer (3) The tropical rain forest have five vertical strata SOLUTION :- on the basis of height of plants. i.e., ground Ectocarpus has haplodiplontic life cycle and vegetation, shrubs, short canopy trees, tall Fucus has diplontic life cycle. canopy trees and tall emergent trees. 130. If there are 999 bases in an RNA that codes for 134. The genotypes of a Husband and Wife are IAIB a protein with 333 amino acids, and the base at and IAi. Among the blood types of their children position 901 is deleted such that the length of how many different genotypes and phenotypes the RNA becomes 998 bases, how many codons are possible will be altered? (1) 3 genotypes ; 3 phenotypes (1) 1 (2) 11 (2) 3 genotypes ; 4 phenotypes (3) 33 (4) 333 (3) 4 genotypes ; 3 phenotypes Answer (3) (4) 4 genotypes ; 4 phenotypes SOLUTION :- Answer (3) If deletion occurs at 901st position the remaining SOLUTION :- 98 bases specifying for 33 codons of amino acids Husband (X) Wife will be altered. 131. The pivot joint between atlas and axis is a type (IAIB) (IAi) of Possibilities in children (1) Fibrous joint (2) Cartilaginous joint IAIB, IAi, IBi, IAIA (3) Synovial joint (4) Saddle joint Number of genotypes = 4 Answer (3) Number of phenotypes = 3 SOLUTION :- Synovial joints are freely movable joint which IAIA and IAi  A allow considerable movements. Pivot joint is a type of synovial joint which provide rotational IAIB= AB movement as in between atlas and axis IBi  B vertebrae of vertebral column. 132. A gene whose expression helps to identify 135. Zygotic meiosis is characteristic of transformed cell is known as (1) Marchantia (2) Fucus (1) Selectable marker (3) Funaria (4) Chlamydomonas (2) Vector Answer (4) (3) Plasmid SOLUTION :- (4) Structural gene Chlamydomonas has haplontic life cycle hence Answer (1) showing zygotic meiosis or initial meiosis. SOLUTION :- 136. Which of the following is correctly matched for the product produced by them? In recombinant DNA technology, selectable markers helps in identifying and eliminating non- (1) Acetobacter aceti : Antibiotics transformants and selectively permitting the (2) Methanobacterium : Lactic acid growth of the transformants. (3) Penicillium notatum : Acetic acid 133. Presence of plants arranged into well defined (4) Saccharomyces cerevisiae : Ethanol vertical layers depending on their height can be Answer (4) seen best in : H.O : JABALPUR : 1525, Wright Town, Ph. (0761) 4218116, 2400022, 8349992509, RAMPUR : Rampur Chowk, Above PNB RANJHI : Opp. Jain Mandir, Main Road, Ranjhi follow us on : momentumacademy www.momentumacademy.com momentumacademy SOLUTION :- 140. Which of the following RNAs should be most Saccharomyces cerevisiae is commonly called abundant in animal cell? Brewer’s yeast. It causes fermentation of (1) r-RNA (2) t-RNA carbohydrates producing ethanol. (3) m-RNA (4) mi-RNA 137. Frog’s heart when taken out of the body Answer (1) continues to beat for some time Select the best option from the following statements SOLUTION :- (a) Frog is a poikilotherm rRNA is most abundant in animal cell. It constitutes 80% of total RNA of the cell. (b) Frog does not have any coronary circulation 141. Which among these is the correct combination (c) Heart is “myogenic” in nature of aquatic mammals? (d) Heart is autoexcitable (1) Seals, Dolphins, Sharks Options : (2) Dolphins, Seals, Trygon (1) Only (c) (2) Only (d) (3) Whales, Dolphins, Seals (3) (a) & (b) (4) (c) & (d) (4) Trygon, Whales, Seals Answer (4) Answer (3) SOLUTION :- SOLUTION :- Frog or the vertebrates have myogenic heart Sharks and Trygon (sting ray) are the members having self contractile system or are of chondrichthyes (cartilaginous fish) while autoexcitable; because of this condition, it will whale, Dolphin and Seals are aquatic mammals keep on working outside the body for some time. belong to class mammalia. 138. Which statement is wrong for Krebs’ cycle? 142. With reference to factors affecting the rate of (1) There are three points in the cycle where photosynthesis, which of the following NAD+ is reduced to NADH + H+ statements is not correct? (2) There is one point in the cycle where FAD+ (1) Light saturation for CO2 fixation occurs at is reduced to FADH 10% of full sunlight (3) During conversion of succinyl CoA to succinic (2) Increasing atmospheric CO2 concentration acid, a molecule of GTP is synthesised upto 0.05% can enhance CO2 fixation rate (4) The cycle starts with condensation of acetyl (3) C3 plants responds to higher temperatures group (acetyl CoA) with pyruvic acid to yield citric with enhanced photosynthesis while C4 plants acid have much lower temperature optimum Answer (4) (4) Tomato is a greenhouse crop which can be SOLUTION :- grown in CO2 - enriched atmosphere for higher yield Krebs cycle starts with condensation of acetyl CoA (2C) with oxaloacetic acid (4C) to form citric Answer (3) acid (6C). SOLUTION :- 139. In case of poriferans the spongocoel is lined with In C3 plants photosynthesis is decreased at flagellated cells called : higher temperature due to increased (1) Ostia photorespiration. C4 plants have higher temperature optimum because of the presence (2) Oscula of pyruvate phosphate dikinase enzyme, which (3) Choanocytes is sensitive to low temperature. (4) Mesenchymal cells 143. Asymptote in a logistic growth curve is obtained Answer (3) when SOLUTION :- (1) The value of ‘r’ approaches zero Choanocytes (collar cells) form lining of (2) K = N spongocoel in poriferans (sponges). Flagella in (3) K > N collar cells provide circulation to water in water (4) K < N canal system. Answer (2) H.O : JABALPUR : 1525, Wright Town, Ph. (0761) 4218116, 2400022, 8349992509, RAMPUR : Rampur Chowk, Above PNB RANJHI : Opp. Jain Mandir, Main Road, Ranjhi follow us on : momentumacademy www.momentumacademy.com momentumacademy SOLUTION :- 145. The DNA fragments separated on an agarose gel can be visualised after staining with A population growing in a habitat with limited resources shows logistic growth curve. For (1) Bromophenol blue logistic growth (2) Acetocarmine dN K N   N  (3) Aniline blue rN rN 1     (4) Ethidium bromide dt  K   K  Answer (4) dN SOLUTION :-  where Rate of change in population size, dt Ethidium bromide is used to stain the DNA fragments and will appear as orange coloured r  intrinsic rate of natural increase bands under UV light. N  Population size, 146. Functional megaspore in an angiosperm develops into K N    Environmental resistance. (1) Ovule (2) Endosperm  K  (3) Embryo sac (4) Embryo If K = N then Answer (3) SOLUTION :- K N  0   , and Megaspore is the first cell of female  K  gametophytic generation in angiosperm. It the population reaches asymptote. undergoes three successive generations of free nuclear mitosis to form 8-nucleated and 7-celled 144. Out of ‘X’ pairs of ribs in humans only ‘Y’ pairs embryo sac. are true ribs. Select the option that correctly represents values of X and Y and provides their 147. Among the following characters, which one was explanation : not considered by Mendel in his experiments on pea? (1) X = 12, Y = 7 True ribs are attached (1) Stem – Tall or Dwarf dorsally to vertebral (2) Trichomes – Glandular or non-glandular column and ventrally to the sternum (3) Seed – Green or Yellow (2) X = 12, Y = 5 True ribs are attached (4) Pod – Inflated or Constricted dorsally to vertebral Answer (2) column and sternum SOLUTION :- on the two ends During his experiments Mendel studied seven (3) X = 24, Y = 7 True ribs are dorsally characters. Nature of trichomes i.e., glandular or non-glandular was not considered by Mendel. attached to vertebral 148. Lungs are made up of air-filled sacs the alveoli. column but are free on They do not collapse even after forceful ventral side expiration, because of : (4) X = 24, Y = 12 True ribs are dorsally (1) Residual Volume attached to vertebral (2) Inspiratory Reserve Volume column but are free on (3) Tidal Volume ventral side (4) Expiratory Reserve Volume Answer (1) Answer (1) SOLUTION :- SOLUTION :- In human, 12 pairs of ribs are present in which 7 Volume of air present in lungs after forceful pairs of ribs (1st to 7th pairs) are attached expiration as residual volume which prevents the dorsally to vertebral column and ventrally to the collapsing of alveoli even after forceful expiration. sternum. H.O : JABALPUR : 1525, Wright Town, Ph. (0761) 4218116, 2400022, 8349992509, RAMPUR : Rampur Chowk, Above PNB RANJHI : Opp. Jain Mandir, Main Road, Ranjhi follow us on : momentumacademy www.momentumacademy.com momentumacademy 149. GnRH, a hypothalamic hormone, needed in derivative of vitamin A, is the light-absorbing part reproduction, acts on of all visual photopigments. (1) Anterior pituitary gland and stimulates 153. Which one of the following statements is not valid secretion of LH and oxytocin for aerosols? (2) Anterior pituitary gland and stimulates (1) They are harmful to human health secretion of LH and FSH (2) They alter rainfall and monsoon patterns (3) Posterior pituitary gland and stimulates (3) They cause increased agricultural productivity secretion of oxytocin and FSH (4) They have negative impact on agricultural (4) Posterior pituitary gland and stimulates land secretion of LH and relaxin Answer (3) Answer (2) SOLUTION :- SOLUTION :- Aerosols can cause various problems to Hypothalamus secretes GnRH which stimulates agriculture through its direct or indirect effects anterior pituitary gland for the secretion of on plants. However continually increasing air gonadotropins (FSH and LH). pollution may represent a persistent and largely 150. In Bougainvillea thorns are the modifications of irreversible threat to agriculture in the future. (1) Stipules (2) Adventitious root 154. A decrease in blood pressure/volume will not cause the release of (3) Stem (4) Leaf (1) Renin Answer (3) (2) Atrial Natriuretic Factor SOLUTION :- (3) Aldosterone Thorns are hard, pointed straight structures for protection. These are modified stem (4) ADH 151. Which one from those given below is the period Answer (2) for Mendel’s hybridization experiments? SOLUTION :- (1) 1856 - 1863 (2) 1840 - 1850 A decrease in blood pressure / volume stimulates (3) 1857 - 1869 (4) 1870 - 1877 the release of renin, aldosterone, and ADH while increase in blood pressure / volume stimulates Answer (1) the release of Atrial Natriuretic Factor (ANF) SOLUTION :- which cause vasodilation and also inhibits RAAS Mendel conducted hybridization experiments on (Renin Angiotensin Aldosterone System) Pea plant for 7 years between 1856 to 1863 and mechanism that decreases the blood volume/ his data was published in 1865 (according to pressure. NCERT). 155. Homozygous purelines in cattle can be obtained 152. Good vision depends on adequate intake of by carotene rich food Select the best option from (1) mating of related individuals of same breed the following statements (2) mating of unrelated individuals of same breed (a) Vitamin A derivatives are formed from (3) mating of individuals of different breed carotene (4) mating of individuals of different species (b) The photopigments are embedded in the membrane discs of the inner segment Answer (1) (c) Retinal is a derivative of vitamin A SOLUTION :- (d) Retinal is a light absorbing part of all the visual Inbreeding results in increase in the photopigments homozygosity. Therefore, mating of the related individuals of same breed will increase (1) (a) & (b) (2) (a), (c) & (d) homozygosity. (3) (a) & (c) (d) (b), (c) & (d) 156. The vascular cambium normally gives rise to Answer (2) (1) Phelloderm (2) Primary phloem SOLUTION :- (3) Secondary xylem (4) Periderm Carotene is the source of retinal which is involved Answer (3) in formation of rhodopsin of rod cells. Retinal, a H.O : JABALPUR : 1525, Wright Town, Ph. (0761) 4218116, 2400022, 8349992509, RAMPUR : Rampur Chowk, Above PNB RANJHI : Opp. Jain Mandir, Main Road, Ranjhi follow us on : momentumacademy www.momentumacademy.com momentumacademy SOLUTION :- (1) Stabilizing selection as it stabilizes this During secondary growth, vascular cambium character in the population gives rise to secondary xylem and secondary (2) Directional as it pushes the mean of the phloem. Phelloderm is formed by cork cambium. character in one direction 157. Which of the following statements is correct? (3) Disruptive as it splits the population into two (1) The ascending limb of loop of Henle is one yielding higher output and the other lower impermeable to water output (2) The descending limb of loop of Henle is (4) Stabilizing followed by disruptive as it impermeable to water stabilizes the population to produce higher (3) The ascending limb of loop of Henle is yielding cows permeable to water Answer (2) (4) The descending limb of loop of Henle is permeable to electrolytes SOLUTION :- Answer (1) Artificial selection to obtain cow yielding higher SOLUTION :- milk output will shift the peak to one direction, hence, will be an example of Directional Descending limb of loop of Henle is permeable selection. In stabilizing selection, the organisms to water but impermeable to electrolytes while with the mean value of the trait are selected. In ascending limb is impermeable to water but disruptive selection, both extremes get selected. permeable to electrolytes. 158. Fruit and leaf drop at early stages can be 162. Select the correct route for the passage of prevented by the application of sperms in male frogs : (1) Cytokinins (2) Ethylene (1) Testes Bidder’s canal Kidney  (3) Auxins (4) Gibberellic acid Vasa efferentia Urinogenital duct Cloaca Answer (3) (2) Testes Vasa efferentia Kidney  SOLUTION :- Seminal Vesicle Urinogenital duct  Auxins prevent premature leaf and fruit fall. NAA Cloaca prevents fruit drop in tomato; 2,4-D prevents fruit drop in Citrus. (3) Testes Vasa efferentia Bidder’s canal 159. A baby boy aged two years is admitted to play Ureter Cloaca school and passes through a dental check-up. (4) Testes Vasa efferentia Kidney The dentist observed that the boy had twenty Bidder’s canal Urinogenital duct  teeth. Which teeth were absent? Cloaca (1) Incisors (2) Canines Answer (4) (3) Pre-molars (4) Molars SOLUTION :- Answer (3) SOLUTION :- In male frog the sperms will move from Total number of teeth in human child = 20. Testes Vasa efferentia Kidney  Premolars are absent in primary dentition. Bidder’s canal Urinogenital duct Cloaca. 160. An important characteristic that Hemichordates 163. Which of the following options best represents share with Chordates is the enzyme composition of pancreatic juice? (1) Absence of notochord (1) Amylase, peptidase, trypsinogen, rennin (2) Ventral tubular nerve cord (2) Amylase, pepsin, trypsinogen, maltase (3) Pharynx with gill slits (4) Pharynx without gill slits (3) Peptidase, amylase, pepsin, rennin Answer (3) (4) Lipase, amylase, trypsinogen, procarboxypeptidase SOLUTION :- Answer (4) Pharyngeal gill slits are present in hemichordates as well as in chordates. Notochord is present in SOLUTION :- chordates only. Ventral tubular nerve cord is Rennin and Pepsin enzymes are present in the characteristic feature of non-chordates. gastric juice. Maltase is present in the intestinal 161. Artificial selection to obtain cows yielding higher juice. milk output represents H.O : JABALPUR : 1525, Wright Town, Ph. (0761) 4218116, 2400022, 8349992509, RAMPUR : Rampur Chowk, Above PNB RANJHI : Opp. Jain Mandir, Main Road, Ranjhi follow us on : momentumacademy www.momentumacademy.com momentumacademy