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NSEB 201 9-2035. The following characters are found in many trees thatbelong to temperate forests.31. Two ecological pyramids are represented in thediagrams A and B:ABChoose the correct statement/s from the following:i.ii.i.Pollen shed occurs at the beginning of growingseason before the leaves develop.ii.Pollen shed is also timed to avoid high humidityand rain.Identify the type of pollination.A is based on biomass and B is based onenergy at every level.In B, the producers are very small in size andproduce enough food for first order consumersand the turnover of producers is much morerapid than that of herbivore.(a) Entomophily(b) Anemophily(c) Ornithophily(d) ChiropterophilyAnswer (b)36. A son with Klinefelter syndrome is born to a motherwho is phenotypically normal. The father has X linkedskin defect (Anhidrotic ectodermal dysplacia). Butthe son has patches of normal as well de defectiveskin. This can be explained as:iii. In A, the size of the producer is huge andsupports large number of herbivores.iv. In B, the producers have longer life span and inA the producers have shorter life span.i.Non- disjunction of X chromosome took placeduring oogenesis and the son inherited two Xchromosomes.Answer (c)ii.32. What is the probability that , in an organism with adiploid number 20, a sperm will be formed whichcontains all 10 chromosomes that come from themother?Non- disjunction of X and Y chromosomes tookplace during spermatogenesis.iii. Mosaic phenotype caused by random inactivationof X chromosome resulted in different patcheson skin.(a) i and iv(b) Only ii(c) ii and iii(d) Only iv 1 (a) 2 20 1 (b) 2 20 1 (d) 4 1 (c) 4 Answer (b)10iv. X linked gene might have crossed over to Y andthe son inherited the skin disorder.10(a) i and ii(b) ii and iii(c) i and iv(d) ii and ivAnswer (d)37. An aphid is fed on a herbaceous plant and its styletis removed by anesthetizing the insect. The fluid inthe stylet is analysed for its chemical content. Whichof the following will be the correct observation/s?33. The nuclei in the tender coconut water and the hardwhite pulp of coconut are respectively(a) Triploid, Diploid(b) Diploid, Diploid(c) Triploid, Triploid(d) Triploid, Haploidi.The main component will be starch if it is apotato plant. Sugars like sucrose and fructosealso will be found.ii.The main component will be fructose when theplant bears sweet fruits.Answer (c)34. The following are the T.S. of different types of ovaries.The types of placentation in I, II, III and IV arerespectively-iii. The contents will be minerals from xylem as wellas sucrose from phloem.IIIIII(a) Axile, free central, axile, basaliv. The contents will be mostly sucrose.IV(a) i, ii and iv(b) Marginal, free central, axile, basal(b) iii only(c) Marginal, free central, axile, free central(c) iv only(d) Basal, axile, free central, axile(d) i, ii and iiiAnswer (b)Answer (c)6

NSEB 201 9-2038. The floral characters that cannot be identified byfloral diagram and floral formula are respectively:42. Enzyme A has higher km value than enzyme B,although both can achieve the same Vmax. Henceit can be concluded that,(a) Position of ovary and monadelphous stamens(a) Enzyme A requires higher substrateconcentration and has lower affinity to substratethan enzyme B(b) Epipetalous stamens and position of ovary(c) Position of ovary and aestivation in calyx andcorolla(b) Enzyme A requires lower substrate concentrationand has lower affinity to substrate than enzymeB(d) Gamopetalous condition and number of loculesin ovaryAnswer (c)(c) Enzyme A requires higher substrateconcentration and has higher affinity to substratethan enzyme B.39. If for convenience, the biochemical pathway ofphotosynthesis is represented briefly by the followingequation;(d) Enzyme A requires lower substrate concentrationand has higher affinity to substrate than enzymeBLightCO2 2H2 A CH2 O H2 O 2AThen, A can represent:i.ii.Answer (a)Oxygen utilised by land plants and in blue-greenalgae.43. Average molecular weight of amino acid isconsidered to be 110 Da.Oxygen utilized by phototrophic bacteria andsulphur by cyanobacteria.A homodimeric membrane protein is found to have amolecular weight of 44,000 Da. How many aminoacids are present in each monomer of the protein?iii. Oxygen utilised by angiosperm and sulphur inphototrophic bacteria.iv. Oxygen utilized by all eukaryotes and sulphur byall prokaryotes.(a) 400(b) 300(c) 200(d) 100(a) i and ii(b) ii and ivAnswer (c)(c) i and iii(d) i, iii and iv44. The graphs show the data on sex determination ofthe progeny which is dependent on temperature.40. Which of the following sets of tissues represents theground tissue of plants?Case I(a) Epidermis, sclerenchyma fibres, xylem vessels,phloem sieve tube members50(b) Parenchyma of cortex of stem, mesophyll cellsof leaf, collenchyma of young stem, sclereids inthe pulp of guava0100Case IIPercent femalePercent female10050FTMTTPTemperature 0100Case IIIPercent femaleAnswer (c)50MTTPFTTemperature 0FT T P MT TP FTTemperature MT Male favouring Temperature; FT Femalefavouring Temperature; Tp Temperature obtainingratios at equilibrium(c) Parenchyma of pith of stem, epidermis of leaf,epidermis of young stem, root hair(d) Collenchyma of hypodermis of young stem, corkcells of the bark, parenchyma of pith, cortex ofyoung rootA few statements regarding the data are made.i.Case I: At a mid range temperature, 3:1 is apredicted the male: female ratio.ii.Case II: The number of males will be muchhigher at lower temperatures.Answer (b)41. The function of contractile vacuole is to pump outexcess water from the cell. In Paramecium, theactivity of contractile vacuole was found to increasewhen transferred from one medium to another. Henceit can be concluded that the transfer was from:iii. Case III: The number of females: males willalways be higher at temperature extremes.(a) Isotonic to hypotonic solutioniv. Case I and II are likely to face ratio imbalanceat mid ranges of temperatures.(b) Hypotonic to isotonic solutionThe correct statement/s is/are:(c) Hypotonic to hypertonic solution(a) ii only(b) i and ii(d) Isotonic to hypertonic solution(c) ii and iii(d) iii and ivAnswer (a)Answer (c)7

NSEB 201 9-2048. Predict the phenotype of a promoter mutant (lacP)for the lac operon.45. The given table showing the recombinationfrequencies between different gene loci on the samechromosome. Recombination frequencies are directlyrelated to the distance between the two genes.Higher the recombination frequency, greater thedistance between the two loci. However, even if theactual distance exceeds 50 units, the recombinationfrequency does not exceed 50%. Select the mostprobable arrangement of genes based on the databelow:(a) The lac genes would be expressed efficientlyonly in the absence of lactose(b) The lac genes would be expressed efficientlyonly in the presence of lactose(c) The lac genes would be expressed continuously.(d) The lac genes would never be expressedefficientlyGene Recombinationpairs frequencyab50ac7adbcbdcdAnswer (d)49. A tall plant with red flowers is crossed with a dwarfplant that produces white flowers. In F1 all plants aretall with pink flowers. The F1 plant is crossed withdwarf parent that bears white flower. Four types ofprogenies were produced in a ratio of 102 : 98 :103 : 99. The progenies expressed:22505015(a) All the five characters of parents and F1(a) d-c-a-b / b-a-c-d(b) b-d-a-c / c-a-d-b(b) All the characters except pink colour of theflower(c) c-d-a-b / b-a-d-c(d) d-a-c-b / b-c-a-d(c) All characters except white colour of the flowerAnswer (a)(d) All characters except red46. The compartmentalization of the cytoplasm by themembranes of the endoplasmic reticulum (ER)results in:Answer (d)50. Thyroid gland produces hormones which control therate of metabolism in animals. Which of the followingwould occur if the thyroid of cattle is fed to a mandeficient in thyroid secretion?(a) Increasing the surface area available forbiochemical synthesis(b) Providing a structural frameworki.It will speed up his metabolism(c) Facilitating cell mobilityii.It will lower the rate of his metabolism(d) Maintaining cell fluidity and cell dynamicsiii. It will regulate the hormone secretionAnswer (a)iv. It will not have any effect on hormone secretion47. Cross pollination will take place when:Which of the following are correct option?i.ii.The flowers are Cleistogamous (flowers neveropen).(a) i and iv(b) ii and iii(c) i and iii(d) ii and ivAnswer (c)The flowers show Herkogamy (physical barrierbetween anther and style).51. Removal of which of the components from the givenfood chain will not result in complete collapse of thefood chain?iii. The flowers show Dichogamy (stamens andcarpels of bisexual flowers mature at differenttimes).Producers Primary consumers Secondaryconsumers Decomposersiv. The plants are Dioecious (plants havingunisexual flower).(a) Producers and primary consumersChoose the correct option:(b) Primary consumers and secondary consumers(a) ii only(b) i and iii(c) Secondary consumers and decomposers(c) ii and iii(d) ii, iii and iv(d) Producers and decomposersAnswer (b)Answer (d)8

NSEB 201 9-2052. Suppose a leaf containing chlorophyll ‘a’ molecule isirradiated at its absorption maxima i.e. 450 nm and662 nm. The fluorescence emission of this leaf wouldbe at 668 and 723 nm. If now the leaf is irradiatedwith either 400 nm or 550 nm wavelength of lightthen, the fluorescence emission of this leaf wouldprobably be at:(a) 668 and 723 nm(b) 610 and 700(c) 610 nm only(d) 610 and 668 nm56. Which of the following diploids produce -galactosidase, in the absence of lactose?(a) p lacZ– lacl / p lacZ lacl–(b) p– lacZ– lacl– / p– lacZ lacl–(c) p lacZ lacl– / p lacZ lacl–(d) p oc lacZ– lacl / p o lacZ lacl Answer (c)57. How many meiotic divisions will be required for theformation of 80 zygotes in an angiospermic plant?Answer (c)53. Patterns of diffusion for two molecules A and B fora living cell are shown in the graph.Rate of solute movement (V)Vmax(a) 40(b) 100A(c) 80Protein-mediatedtransport(facilitated diffusion)(d) 160Answer (b)BSimple diffusion58. A food chain in a terrestrial ecosystem is shownSun Grass Rabbit SnakeThe food chain is incomplete due to:Solute concentrationA and B most likely could be respectively :(a) Absence of tertiary consumer(a) Na and Glucose(b) O2 and CO2(b) Absence of decomposers(c) Glucose and O2(d) O2 and glycerol(c) Absence of quaternary consumerAnswer (c)(d) Absence of parasitic component54. During transmission of impulses across the nervemembrane; a simple impulse dies out just before thesynapse, whereas several impulses reaching thesynapse within a short period “Fire” the impulse intothe next neuron. The reason for simple neuron to dieout maybe that the:Answer (b)59. Eutrophication is considered bad for a water bodybecause it leads to:(a) Increase in demand for carbon dioxide(b) Increase in demand for oxygen(a) Synapse gets fatigued by continuous work(c) Increase in demand for nitrogen(b) Impulse is unable to produce the adequatequantity of neurotransmitters(d) Change in pH of the water body(c) Speed at which impulse travels is lessAnswer (b)(d) Dendrites of nerve fibres take time to acceptsignal for nerve impulse60. Cells of E. coli are placed in a solution with 12%NaCl. Which effect would be visible after 24 h ofincubation?Answer (b)55. Double fertilization is not found in(a) Plasmolysis(a) Cucumber(b) Plasmoptysis(b) Rice(c) Osmotic lysis(c) Pinus(d) Swelling of cells(d) MangoAnswer (c)Answer (a)9

NSEB 201 9-2061. Enzyme ‘x’ is a polypeptide in nature. When addedto solvent ‘s’ it acquires following conformationX SSSSSSSState BS(b) Coral reefs, deserts and fertile agricultural areaS(c) Estuaries, oceans and grasslandS(d) Oceans, swamp waters and coral reefsSSSState AS(a) Oceans, deserts and pondsSAnswer (d)S64. Generalized profile of a soil in which a plant isgrowing is shown. The region/s rich humus will be:State CWhich of the following is correct?P(a) Enzyme will be most active in state C(b) The solvent acts as a denaturant for the proteinmoleculeQ(c) Further addition of solvent will lead toprecipitation of the proteinR(d) Further addition of solvent will lead to breakingof polypeptide bonds of the proteinSAnswer (b)62. Which of the following correctly represents simplifiedmodel of energy and mineral movement in anecosystem?Producers(a) SunNutrientpool(a) P only(b) P and QDecomposers(c) R onlyProducers(b) SunHerbivores(d) S onlyHerbivoresAnswer (b)NutrientpoolDecomposersProducers(c) eGiven below are four DNA sequence. Only one strandof the double stranded DNA has been represented.Which one of the following can possibly code for theabove tri-peptide?DecomposersProducers(d) Sun65. The following represents a tri-peptide (3 amino acids)stretch of a protein sequence:Herbivores(a) 5' AAA GTA CGC 3'NutrientpoolDecomposers(b) 5' TTT CAT GCG 3'(c) 5' GCG TAC TTT 3'Answer (a)(d) 5' CGC AUG AAA 3'63. Various communities can be classified based ontheir metabolic characteristics such as productivityand respiration. Communities P, Q and R in thegraph respectively represent:Answer (b)66. In a diploid organism the total DNA content of asperm was found to be 'C'. What will be the DNAcontent of its cell that is at Metaphase I of meiosis?RCommunityproduction(gm/m2/day)(a) C(b) 0.5CPQ(c) 2C(d) 4CCommunity respiration2(gm/m /day)Answer (d)10

NSEB 201 9-2067. In a plant, the color of a flower is determined by theconversion of a white pigment into a red pigment thatis controlled by the product of gene ‘B’. Product ofthe gene ‘A’ is responsible for bringing the whitepigment into the cell for conversion. The process isschematically represented in the figure.69. Bacteriophages are viruses that infect bacterial cells.In a given experiment bacteriophages were grown inthe presence of radioisotopes 14C and 32P. Thesebacteriophages were used to infect bacterial cells.Following infection, radioisotopes present in thebacterial cells were analyzed. The radioactivity in thebacterial cell will be observed due to the presenceof:Alleles 'a' and 'b' are non-functional mutant alleles ofgenes 'A' and 'B', respectively. Two parental plantswith white flowers are crossed. F1 progeny have redflowers only. When the F1 progeny is self-pollinated,the F2 progeny has plants that have either red orwhite flowers.(a) Only32P(c) Both32P(b) Onlyand14C14C(d) Either32Por14CAnswer (c)White pigment70. A water strider can walk on the surface of waterwithout even getting its claws wet. The insect cando it due to which property of water?Gene AWhite pigment(a) Specific gravity(b) Surface tension(c) Specific heat(d) Anomalous behaviourAnswer (b)71. The spider silk has a predominant component called‘spiderwin’, with five times the strength of steel,weight for weight. The elasticity of the web strandsis due to the presence of :Gene BRed pigmentConsidering that the two genes are on twoindependent chromosomes, what is the expectedratio of the two phenotypes in the F2 progeny?(a) Beta sheets(b) Alpha helices(c) Disordered loops(d) Sugar residues(a) 3 Red : 1 whiteAnswer (b)(b) 9 Red : 7 white72. The hepatocytes of an elephant, in comparison to thehepatocytes of a mouse are :(c) 1 Red : 1 white(d) 15 Red : 1 white(a) Twice as bigAnswer (b)(b) Five times bigger(c) Twenty times bigger (d) Of the same size68. The following pedigree represents the inheritance ofa rare disorder caused due to an autosomalrecessive allele. Filled square indicates affectedmale.Answer (d)73. Plant scientists are worried that C4 crops such ascorn and sugarcane may suffer stiffer competitionfrom C3 weeds since there is a global(a) Increase in temperature(b) Increase in CO2 content of atmosphere(c) Decrease in rainfall(d) Increase in genome contamination of C4 cropsAnswer (b)74. During menstrual cycle there are two surges inestrogen concentration of blood. The first and majorsurge is just prior to the ovulation phase and thenext one is in :What is the probability that the daughter in the thirdgeneration carries the allele responsible for thedisorder?(a) 1/2(b) 2/3(a) Menstruation phase (b) Early follicular phase(c) 3/4(d) 1/4(c) Mid-luteal phaseAnswer (c)Answer (c)11(d) Late-luteal phase

NSEB 201 9-2078. Acid precipitation refers to rain, snow or fog with apH lower or more acidic than pH 5.6. It resultsprimarily by the presence of which of the followingcomponents in the atmosphere?75. Match the following examples with the evolutionaryphenomena, namely, convergent evolution (p),divergent evolution (q) and adaptive radiation (r).i.ii.Sugar gliders of Australia and European flyingsquirrel.(a) CO and CO2Squirrel species on opposite rims of GrandCanyon.(b) Sulphur and nitrogen oxides(c) Lend and phosphorous oxidesiii. Sharks and dolphins.(d) Ozone and hydrocarbonsiv. Darwin’s finches.(a) i-p, ii-q, iii-p, iv-r(b) i-r, ii-r, iii-p, iv-rAnswer (b)(c) i-r, ii-q, iii-p, iv-q(d) i-q, ii-r, iii-p, iv-p79. Peroxisomes are often noticed in proximity ofmitochondria. This is due to the fact that theproducts can be transported to mitochondria. Whichof the following functions is most relevant to thisexplanation?Answer (a)76. Most of the drugs are eliminated by nephronsthrough:(a) Filtration at loop of Henle(a) Peroxisomes use oxygen to break fatty acidsdown into smaller molecules that are then usedas fuel for cellular respiration.(b) Tubular reabsorption at proximal convolutedtubules(c) Tubular secretion at distal convoluted tubules(b) Peroxisomes oxidise alcohol to detoxify it inliver.(d) Tubular secretion at collecting ductAnswer (c)(c) Peroxisomes transfer hydrogen from toxins tooxygen rendering them harmless.77. Consumption of salty food results in increased thirstand a cascade of events. Select and arrange thesequence of events:(d) Peroxisomes produce H2O2 and also convert itto water.(i) Increased reabsorption of water.Answer (a)(ii) High Na in blood.80. Homology suggests a common ancestry, whileanalogy suggests:(iii) Increased release of aldosterone.(iv) Increased ADH in blood.(a) Monophyletic origin(v) Passing out more concentrated urine.(b) Character displacementChoose the correct sequence.(a) (ii), (iv), (i), (v)(b) (i), (iii), (iv), (ii), (v)(c) (iii), (i), (iv), (v), (ii)(d) (ii), (iii), (vi), (i)(c) Polyphyletic origin(d) Adaptation to common environmentAnswer (d)Answer (a) 12

i. Starfish ii. Hydra iii. Spider iv. Planaria v. Crab (a) i, iii and iv (b) ii, iii and v (c) iii and v (d) i and v Answer (c) 26. Cabbage, cauliflower, broccoli, kohlrabi, kale, brussels sprouts have all sprung from the wild mustard plant through: (a) Variations and natural selection (b) Induced mutations and their propagation (c) Induced .