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Test Booklet CodeDATE : 13/09/2020F1KANHARegd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005. Ph.: 011-47623456Time : 3 hrs.Answers & SolutionsMax. Marks : 720forNEET (UG) - 2020Important Instructions :1.The test is of 3 hours duration and Test Booklet contains 180 questions. Each question carries4 marks. For each correct response, the candidate will get 4 marks. For each incorrect response, onemark will be deducted from the total scores. The maximum marks are 720.2.Use Blue / Black Ball point Pen only for writing particulars on this page/marking responses.3.Rough work is to be done on the space provided for this purpose in the Test Booklet only.4.On completion of the test, the candidate must handover the Answer Sheet to the Invigilator before leavingthe Room / Hall. The candidates are allowed to take away this Test Booklet with them.5.The CODE for this Booklet is F1.6.The candidates should ensure that the Answer Sheet is not folded. Do not make any stray marks on theAnswer Sheet. Do not write your Roll No. anywhere else except in the specified space in the TestBooklet/Answer Sheet.7.Each candidate must show on demand his/her Admission Card to the Invigilator.8.No candidate, without special permission of the Superintendent or Invigilator, would leave his/her seat.9.Use of Electronic/Manual Calculator is prohibited.10.The candidates are governed by all Rules and Regulations of the examination with regard to their conductin the Examination Hall. All cases of unfair means will be dealt with as per Rules and Regulations of thisexamination.11.No part of the Test Booklet and Answer Sheet shall be detached under any circumstances.12.The candidates will write the Correct Test Booklet Code as given in the Test Booklet / Answer Sheet inthe Attendance Sheet.

NEET (UG)-2020 (Code-F1)1.4.In light reaction, plastoquinone facilitates thetransfer of electrons from(1) Active immunity is quick and gives fullresponse.(1) PS-I to NADP (2) PS-I to ATP synthase(2) Foetus receives some antibodies frommother, it is an example for passiveimmunity.(3) PS-II to Cytb6f complex(4) Cytb6f complex to PS-I(3) When exposed to antigen (living or dead)antibodies are produced in the host’sbody. It is called “Active immunity”.Answer (3)S o l . After excitement, e– is passed from PS-II (P680)to primary electron acceptor (Pheophytin).From primary e– acceptor, e– is passed toplastoquinone. Plastoquinone (PQ) in turntransfer its e– to Cyt b6f complex. Thereforeplastoquinone facilitates the transfer ofelectrons from PS-II to Cyt b6f complex.2.Identify the wrong statement with referenceto immunity.(4) When ready-made antibodies are directlygiven, it is called “Passive immunity”.Answer (1)S o l . The correct option is (1) because activeimmunity is slow and takes time to give its fulleffective response in comparison to passiveimmunity where pre-formed antibodies areadministered.The sequence that controls the copy numberof the linked DNA in the vector, is termed(1) Palindromic sequence5.(2) Recognition siteExperimental verification of the chromosomaltheory of inheritance was done by(3) Selectable marker(1) Boveri(2) Morgan(4) Ori site(3) Mendel(4) SuttonAnswer (4)Answer (2)S o l . The correct option is (4) because Orisequence is responsible for controlling thecopy number of the linked DNA in the vector.Ori i.e. origin of replication is responsible forinitiation of replication.S o l . Experimental verification of the chromosomaltheory of inheritance was done by Morgan.3.Note:Sutton and Boveri proposed chromosomaltheory of inheritance but it wasexperimentally verified by T.H. Morgan.The specific palindromic sequence which isrecognized by EcoRI is6.(1) 5 - CTTAAG - 3 Match the following concerning essentialelements and their functions in plants3 - GAATTC - 5 (a) Iron(i) Photolysis of water(2) 5 - GGATCC - 3 (b) Zinc(ii) Pollen germination(c) Boron(iii) Required for3 - CCTAGG - 5 chlorophyll(3) 5 - GAATTC - 3 biosynthesis3 - CTTAAG - 5 (d) Manganese(4) 5 - GGAACC - 3 (iv) IAA biosynthesisSelect the correct option3 - CCTTGG - 5 (a)(b)(c)(d)(1) (iii)(iv)(ii)(i)(2) (iv)(i)(ii)(iii)(3) (ii)(i)(iv)(iii)5 - GAATTC - 3 (4) (iv)(iii)(ii)(i)3 - CTTAAG - 5 Answer (1)Answer (3)S o l . The correct option is (3) because the specificpalindromic sequence which is recognised byEco RI is2

NEET (UG)-2020 (Code-F1)S o l . (a) Iron– Essentialfortheformation of chlorophyll(b) Zinc– Needed for synthesis ofauxin7.(c) Boron– Have a role in pollengrain germination(d) Manganese– Is involved in the splittingof water to liberate O 2during photosynthesis10.11.(2) Low pCO2 in alveoli favours the formationof oxyhaemoglobinAnswer (4)(3) Binding of oxygen with haemoglobin ismainly related to partial pressure of O2S o l . The separated DNA fragments can bevisualised only after staining the DNA withEthidium bromide followed by exposure to UVradiation.(4) Partial pressure of CO2 can interfere withO2 binding with haemoglobinAnswer (1)Name the enzyme that facilitates opening ofDNA helix during transcription.S o l . The correct option is (1) because higher H concentration favours the dissociation ofoxygen from oxyhaemoglobin in tissues.(1) DNA polymerase (2) RNA polymerase(4) DNA helicaseIn the alveoli, high pO2, low pCO2, lesser H concentration and lower temperature favourformation of oxyhaemoglobin.Answer (2)S o l . RNA polymerase facilitates opening of DNAhelix during transcription.12.(4) GIFT and ZIFTFloridean starch has structure similar to(1) Mannitol and alginIn which of the following techniques, theembryos are transferred to assist thosefemales who cannot conceive?(3) ZIFT and IUTIdentify the wrong statement with referenceto transport of oxygen(1) Higher H conc. in alveoli favours theformation of oxyhaemoglobin(4) Ethidium bromide in UV radiation(2) GIFT and ICSI(4) Glutamic AcidGlutamic acid is an acidic amino acid whileTyrosine is an aromatic amino acid.(3) Acetocarmine in bright blue light(1) ICSI and ZIFT(3) TyrosineValine is a neutral amino acid.(2) Ethidium bromide in infrared radiation9.(2) ValineS o l . Option (1) is the correct answer becauselysine is a basic amino acid.In gel electrophoresis, separated DNAfragments can be visualized with the help of(3) DNA ligase(1) LysineAnswer (1)(1) Acetocarmine in UV radiation8.Identify the basic amino acid from thefollowing.(2) Laminarin and cellulose(3) Starch and cellulose(4) Amylopectin and glycogenAnswer (4)Answer (3)S o l . Floridean starch is stored food material inred algae. It's structure is similar toAmylopectin and Glycogen.S o l . Option (3) is the answer because ART in whichembryos are transferred, include ZIFT and IUTi.e. Zygote Intrafallopian Transfer and IntraUterine Transfer respectively, both areembryo transfer (ET) methods.13.By which method was a new breed ‘Hisardale’of sheep formed by using Bikaneri ewes andMarino rams?(1) Cross breedingOption (1), (2) and (4) are incorrect becausein GIFT (Gamete Intrafallopian Transfer),gamete is transferred into the fallopian tubeof female who cannot produce ova. ICSI isIntra cytoplasmic sperm injection in whichsperm is directly injected into the ovum.(2) Inbreeding(3) Out crossing(4) Mutational breedingAnswer (1)3

NEET (UG)-2020 (Code-F1)16.S o l . Hisardale is a new breed of sheep developedin Punjab by crossing Bikaneri-ewe andMarino rams. In cross-breeding, superiormale of one breed are mated with superiorfemales of another breed.14.(1) Nucleases(3) Ligases- Join themoleculestwo(4) Polymerases - Break thefragmentsColumn-IIDNADNAinto(a) Pituitary gland(i) Grave’s disease(b) Thyroid gland(ii) Diabetes mellitusS o l . Ligases join the two DNA molecules.(c) Adrenal gland(iii) Diabetes insipidus17.(d) Pancreas(iv) Addison’s disease(1) Hypogynous ovary(2) Half inferior ovary(b)(c)(d)(3) Inferior ovary(1) (iii)(i)(iv)(ii)(2) (ii)(i)(iv)(iii)(3) (iv)(iii)(i)(ii)(4) (iii)(ii)(i)(iv)(a)Answer (3)Ray florets have(4) Superior ovaryAnswer (3)Sol. 18.Answer (1)Ray floret have inferior ovary.Epigynous flower are formed in familyAsteraceae (e.g., Sunflower)Match the organismbiotechnology.(a) BacillusS o l . Graves' disease is due to excess secretion ofthyroid hormones (T3 & T4).withitsusein(i) Cloning vectorthuringiensis(b) ThermusDiabetes mellitus is due to hyposecretion ofinsulin from -cells of pancreas.15.- Separate the two strandsof DNA(2) Exonucleases - Make cuts at specificpositions within DNAMatch the following columns and select thecorrect option.Column-IChoose the correct pair from the following(ii) Construction ofaquaticusfirst rDNAmoleculeDiabetes insipidus is due to hyporelease ofADH from posterior pituitary.(c) AgrobacteriumAddison's disease is due to hyposecretion ofhormone from adrenal cortex.(d) SalmonellaSelect the option including all sexuallytransmitted diseases.Select the correct option from the following:(a)(b)(c)(d)(1) AIDS, Malaria, Filaria(1) (iii)(ii)(iv)(i)(2) Cancer, AIDS, Syphilis(2) (iii)(iv)(i)(ii)(3) Gonorrhoea, Syphilis, Genital herpes(3) (ii)(iv)(iii)(i)(4) Gonorrhoea, Malaria, Genital herpes(4) (iv)(iii)(i)(ii)(iii) DNA polymerasetumefaciens(iv) Cry proteinstyphimuriumAnswer (4)Answer (3)S o l . (a) Bacillus thuringiensis is a source of Cryproteins.S o l . Gonorrhoea, Syphilis, Genital herpes aresexually transmitted diseases.(b) Thermus aquaticus is a source ofthermostable DNA polymerase (Taqpolymerase) used in PCR.Gonorrhoea is caused by a bacteriumNeisseria gonorrhoeae.(c) Agrobacterium tumefaciens is a cloningvector.Syphilis is caused by a bacterium Treponemapallidum.(d) The construction of 1st recombinant DNAmolecule was performed using nativeplasmid of Salmonella typhimurium.Genital herpes is caused by a virus Type-IIHerpes simplex virus.4

NEET (UG)-2020 (Code-F1)19.The product(s) of reaction catalyzed bynitrogenase in root nodules of leguminousplants is/areS o l . Platyhelminthes are bilaterally symmetrical,triploblastic and acoelomate animals withorgan level of organisation.(1) Ammonia and oxygen24.(2) Ammonia and hydrogen(1) Effluents of primary treatment(3) Ammonia alone(2) Activated sludge(4) Nitrate alone(3) Primary sludgeAnswer (2)S o l . N2 8e– 8H (4) Floating debrisMg 16ATP 2NH3 H2 Answer (2)S o l . The sediment in settlement tank is calledactivated sludge.16ADP 16PiAmmonia and Hydrogen.20.A small part of the activated sludge ispumped back into aeration tankName the plant growth regulator which uponspraying on sugarcane crop, increases thelength of stem, thus increasing the yield ofsugarcane crop.(1) Ethylene(2) Abscisic acid(3) Cytokinin(4) GibberellinRemaining major part of the sludge is pumpedinto large tank called anaerobic sludgedigesters.25.Answer (4)The body of the ovule is fused within thefunicle at(1) Nucellus(2) Chalaza(3) Hilum(4) MicropyleAnswer (3)S o l . The attachment point of funicle and body ofovule is known as hilum.22.The process of growth is maximum during(1) Senescence(2) Dormancy(3) Log phase(4) Lag phaseColumn-II(a) Floating Ribs(i) Located betweensecond andseventh ribs(b) Acromion(ii) Head of theHumerus(c) Scapula(iii) Clavicle(d) Glenoid cavity(iv) Do not connectwith the sternum(a)(b)(c)(d)(1) (iii)(ii)(iv)(i)(2) (iv)(iii)(i)(ii)(3) (ii)(iv)(i)(iii)(4) (i)(iii)(ii)(iv)Answer (2)Answer (3)S o l . (a) 11 th and 12 th pairs of ribs are notconnected ventrally and are therefore,called floating ribs.S o l . In exponential growth, the initial growth isslow (lag phase) and it increases rapidlythereafter at an exponential rate in log orexponential phase.23.Match the following columns and select thecorrect option.Column-IS o l . Spraying sugarcane crop with gibberellinsincreases the length of the stem, thusincreasing the yield by as much as 20 tonnesper acre.21.Which of the following is put into Anaerobicsludge digester for further sewage treatment?(b) Acromion is a flat expanded process ofspine of scapula. The lateral end ofclavicle articulates with acromionprocess.Bilaterally symmetrical and acoelomateanimals are exemplified by(1) Aschelminthes(c) Scapula is a flat triangular bone in thedorsal part of the thorax between 2nd andthe 7th rib.(2) Annelida(3) Ctenophora(d) Glenoid cavity of scapula articulates withhead of the humerus to form the shoulderjoint.(4) PlatyhelminthesAnswer (4)5

NEET (UG)-2020 (Code-F1)26.Identify the wrong statement with regard toRestriction Enzymes.28.(1) They are useful in genetic engineering.If the head of cockroach is removed, it maylive for few days because(1) the head holds a small proportion of anervous system while the rest is situatedalong the ventral part of its body.(2) Sticky ends can be joined by using DNAligases.(2) the head holds a 1/3 rd of a nervoussystem while the rest is situated along thedorsal part of its body.(3) Each restriction enzyme functions byinspecting the length of a DNA sequence.(4) They cut the strand of DNA at palindromicsites.Answer (2)(3) the supra-oesophageal ganglia of thecockroach are situated in ventral part ofabdomen.S o l . Restriction endonucleases make cuts atspecific positions within the DNA.(4) the cockroach does not have nervoussystem.They function by inspecting the length of aDNA sequence.Answer (1)S o l . The head holds a small proportion of anervous system while the rest is situatedalong the ventral part of its body.Restriction endonuclease bind to the DNA andcut the two strands of double helix at specificpoints in their sugar-phosphate backbones.29.They are used in genetic engineering to formrecombinant molecules of DNA.(1) HimalayasDNA ligases join the DNA fragments.27.(2) Amazon forestsMatch the following columns and select thecorrect option.Column-I(3) Western Ghats of India(4) MadagascarColumn-II(a) Gregarious,(i)Answer (2)AsteriasS o l . The largely tropical Amazonian rain forest inSouth America has the greatest biodiversityon earth.polyphagous pest(b) Adult with radial(ii)Which of the following regions of the globeexhibits highest species diversity?Scorpionsymmetry and larva30.with bilateralsymmetryWhich is the important site of formation ofglycoproteins and glycolipids in eukaryoticcells?(c) Book lungs(iii) Ctenoplana(1) Golgi bodies(d) Bioluminescence(iv) Locusta(2) Polysomes(a)(b)(c)(d)(3) Endoplasmic reticulum(1) (iii)(ii)(i)(iv)(4) Peroxisomes(2) (ii)(i)(iii)(iv)Answer (1)(3) (i)(iii)(ii)(iv)(4) (iv)(i)(ii)(iii)S o l . Golgi bodies are site of formation ofglycoproteins and glycolipids in eukaryoticcells.Answer (4)31.S o l . (a) Locusta is a gregareous pest.(b) In Echinoderms, adults are radiallysymmetrical but larvae are bilaterallysymmetrical.Which of the following pairs is of unicellularalgae?(1) Anabaena and Volvox(2) Chlorella and Spirulina(3) Laminaria and Sargassum(c) Scorpions respire through book lungs.(4) Gelidium and Gracilaria(d) Bioluminescence is well marked inctenophores.Answer (2)6

NEET (UG)-2020 (Code-F1)S o l . Chlorella and Spirulina are unicellular algae.Select the correct option from the followingGelidium, Gracilaria, Laminaria andSargassum are multicellular. Volvox iscolonial.32.(a)Which one of the following is the mostabundant protein in the animals?(1) Lectin(2) Insulin(3) Haemoglobin(4) Collagen(d)(1) (i)(ii)(iv)(iii)(2) (ii)(iv)(iii)(i)(3) (iii)(iv)(i)(ii)(4) (iv)(iii)(ii)(i)S o l . Zygotene SynapsisPachytene Crossing overS o l . Collagen is the most abundant protein inanimal world and RuBisCO is the mostabundant protein in the whole of theBiosphere.Diplotene Chiasmata formationDiakinesis Terminalisation37.Dissolution of the synaptonemal complexoccurs duringWhich of the following statements aboutinclusion bodies is incorrect?(1) Diplotene(2) Leptotene(1) They lie free in the cytoplasm(3) Pachytene(4) Zygotene(2) These represent reserve material incytoplasmAnswer (1)(3) They are not bound by any membraneS o l . Dissolution of the synaptonemal complexoccurs During Diplotene stage of Prophase-Iof Meiosis-I.34.(c)Answer (4)Answer (4)33.(b)(4) These are involved in ingestion of foodparticlesAnswer (4)How many true breeding pea plant varietiesdid Mendel select as pairs, which were similarexcept in one character with contrastingtraits?(1) 14(2) 8(3) 4(4) 2S o l . These are not involved in ingestion of foodparticles38.Which of the following would help inprevention of diuresis?Answer (1)(1) AtrialnatriureticvasoconstrictionS o l . Mendel selected 14 True breeding plantvarieties.(2) Decrease in secretion of renin by JG cells35.(3) More water reabsorptionundersecretion of ADHCuboidal epithelium with brush border ofmicrovilli is found incausesdueto(4) Reabsorption of Na and water from renaltubules due to aldosterone(1) Proximal convoluted tubule of nephron(2) Eustachian tubeAnswer (4)(3) Lining of intestineSol. Adrenal cortex secretes mineralocorticoidslike aldosterone which increase thereabsorption of Na and water from renaltubule that prevent diuresis.39. The transverse section of a plant showsfollowing anatomical features :(4) Ducts of salivary glandAnswer (1)S o l . Cuboidal epithelium with brush border ofmicrovilli is found in proximal convolutedtubule of nephron (PCT).36.factor(a) Large number of scattered vascularbundles surrounded by bundle sheathMatch the following with respect to meiosis(a) Zygotene(i) Terminalization(b) Pachytene(ii) Chiasmata(b) Large conspicuous parenchymatousground tissue(c) Diplotene(iii) Crossing over(c) Vascular bundles conjoint and closed(d) Diakinesis(iv) Synapsis(d) Phloem parenchyma absent7

NEET (UG)-2020 (Code-F1)(c) RNAi (RNA interference) takes place in alleukaryotic organisms as a method ofcellular defense.Identify the category of plant and its part :(1) Dicotyledonous stem(2) Dicotyledonous root(d) PCR is now routinely used to detect HIV insuspected AIDS patients.(3) Monocotyledonous stem(4) Monocotyledonous root42.Answer (3)Flippers of Penguins and Dolphins areexamples ofS o l . All features are related to monocotyledonousstems(1) Industrial melanism40.(3) Adaptive radiation(2) Natural selectionWhich of the following statements iscorrect?(4) Convergent evolution(1) Adenine pairs with thymine through threeH-bondsAnswer (4)S o l . The correct option is (4) because flippers ofPenguins and Dolphins are an example ofanalogous organs. Analogous structures are aresult of convergent evolution.(2) Adenine does not pair with thymine(3) Adenine pairs with thymine through two Hbonds43.(4) Adenine pairs with thymine through one HbondThe oxygenation activity of RuBisCo enzymein photorespiration leads to the formation ofAnswer (3)(1) 1 molecule of 6-C compoundS o l . Adenine pairs with thymine through twoH-bonds i.e., A T(2) 1 molecule of 4-C compound and1 molecule of 2-C compound41.(3) 2 molecules of 3-C compoundMatch the following columns and select thecorrect option.Column-I(4) 1 molecule of 3-C compoundColumn-II(a) Bt cotton(i) Gene therapy(b) Adenosinedeaminasedeficiency(ii) Cellular defence(c) RNAi(iii) Detection of HIVinfection(d) PCR(iv) Bacillusthuringiensis(a)(b)(c)(d)(1) (ii)(iii)(iv)(i)(2) (i)(ii)(iii)(iv)(3) (iv)(i)(ii)(iii)(4) (iii)(ii)(i)(iv)Answer (4)S o l . In photorespiration, O2 binds to RubisCo. As aresult RuBP instead to being converted to2 molecules of PGA bind with O2 to form onemolecule each of phosphoglycerate (3 carboncompound) and phosphoglycolate ( 2 carboncompound).44.The infectious stage of Plasmodium thatenters the human body is(1) Female gametocytes(2) Male gametocytes(3) Trophozoites(4) SporozoitesAnswer (4)S o l . Plasmodium enters the human body assporozoites (Infectious stage) through the biteof Infected Female Anopheles mosquito.Answer (3)S o l . The correct option is (3) because45.(a) In Bt cotton the specific Bt toxin gene wasisolated from Bacillus thuringiensis.Identify the incorrect statement.(1) Sapwood is the innermost secondaryxylem and is lighter in colour(b) The first clinical gene therapy was givenin 1990 to a 4-year old girl with adenosinedeaminase (ADA) deficiency.(2) Due to deposition of tannins, resins, oilsetc., heart wood is dark in colour8

NEET (UG)-2020 (Code-F1)48.(3) Heart wood does not conduct water butgives mechanical support(1) When IA and IB are present together, theyexpress same type of sugar.(4) Sapwood is involved in conduction ofwater and minerals from root to leaf(2) Allele ‘i’ does not produce any sugar.Answer (1)(3) The gene (I) has three alleles.S o l . Incorrect statement: Sapwood is theinnermost secondary xylem and is lighter incolour.(4) A person will have only two of the threealleles.Answer (1)Correct statement: Sapwood is outermostsecondary xylem.46.Identify the wrong statement with referenceto the gene ‘I’ that controls ABO blood groups.S o l . ABO blood groups are controlled by the gene I.The gene I has three alleles IA, IB and i. Thealleles IA and IB produce a slightly differentform of the sugar while allele i does notproduce any sugar. Because humans arediploid organisms, each person can possessat the most any two of the three I gene alleles.Which of the following is correct aboutviroids?(1) They have DNA with protein coat(2) They have free DNA without protein coat(3) They have RNA with protein coat49.(4) They have free RNA without protein coatAccording to Robert May, the global speciesdiversity is aboutAnswer (4)(1) 50 million(2) 7 millionS o l . Viroids have free RNA without protein coat.(3) 1.5 million(4) 20 million47.Answer (2)Match the following diseases with thecausative organism and select the correctoption.Column-ISol. Robert May estimated global speciesdiversity at about 7 million. Although some extreme estimates rangefrom 20 to 50 million.Column-II(a) Typhoid(i) Wuchereria(b) Pneumonia(ii) PlasmodiumWhich of the following is not an attribute of apopulation?(c) Filariasis(iii) Salmonella(1) Mortality(d) Malaria(iv) Haemophilus(2) Species interaction(b)(c)(d)(3) Sex ratio(1) (ii)(i)(iii)(iv)(4) Natality(2) (iv)(i)(ii)(iii)(3) (i)(iii)(ii)(iv)(4) (iii)(iv)(i)(ii)(a)50.Answer (2)S o l . NatalityMortality– Population attribute– Population attributeAnswer (4)Species interaction – PopulationinteractionS o l . Typhoid fever in humans is caused bypathogenic bacterium Salmonella typhi.Sex ratio51.Pneumonia is caused by StreptococcusPneumoniae and Haemophilus influenzae.– Population attributeIn water hyacinth and water lily, pollinationtakes place by :(1) Wind and waterFilariasis or elephantiasis is caused by thefilarial worm, Wuchereria bancrofti andWuchereria malayi.(2) Insects and water(3) Insects or wind(4) Water currents onlyMalaria is caused by different species ofPlasmodium.Answer (3)9

NEET (UG)-2020 (Code-F1)S o l . In majority of aquatic plants, the flowersemerge above the level of water.Succinyl Co-AThese may be pollinated by insects or windSuccinateThiokinaseGDPGTPATPADPSuccinateeg.: Water hyacinth and water lily52.The QRS complex in a standard ECGrepresents55.(1) Depolarisation of ventricles(2) Repolarisation of ventricles(3) Repolarisation of auricles(4) Depolarisation of auriclesMatch the following(a) Inhibitor ofcatalytic activity(i) Ricin(b) Possess peptidebonds(ii) MalonateAnswer (1)(c) Cell wall material (iii) Chitinin fungiS o l . QRS complex represents the depolarisation ofventricles.(d) Secondarymetabolite53.Choose the correct option from the followingSelect the correct match(1) Sickle cell anaemia – Autosomalrecessive trait,chromosome-11(a)(b)(c)(d)(1) (iii)(iv)(i)(ii)(2) (ii)(iii)(i)(iv)(2) Thalassemia– X linked(3) (ii)(iv)(iii)(i)(3) Haemophilia– Y linked(4) (iii)(i)(iv)(ii)(4) Phenylketonuria– Autosomaldominant traitAnswer (1)S o l . PhenylketonuriaAnswer (3)S o l . Option (3) is the correct answer becauseMalonate is the competitive inhibitor ofcatalytic activity of succinic dehydrogenase,so (a) matches with (ii) in column II.– Autosomal recessivedisorderThalassemia– Autosomal recessivedisorderHaemophilia– X linked recessivedisorderCollagen is proteinaceous in nature andpossesses peptide bonds, so (b) matches with(iv) in column II.Chitin is a homopolymer present in the cellwall of fungi and exoskeleton of arthropods,so, (c) matches with (iii) in column II.Sickle cell anaemia – Autosomal recessivetrait, caused due tomutation in genepresent onchromosome no. 1154.(iv) CollagenAbrin and Ricin are toxins, secondarymetabolites, so (d) in column I matches with(i) in column II.Thenumberofsubstratelevelphosphorylations in one turn of citric acidcycle isWhich of the following refer to correctexample(s) of organisms which have evolveddue to changes in environment brought aboutby anthropogenic action?(1) Two(a) Darwin’s Finches of Galapagos islands.(2) Three(b) Herbicide resistant weeds.56.(c) Drug resistant eukaryotes.(3) Zero(d) Man-created breeds of domesticatedanimals like dogs.(4) OneAnswer (4)S o l . One substrate level phosphorylation in oneturn of citric acid cycle as per followingreaction:(1) (b), (c) and (d)(2) only (d)(3) only (a)(4) (a) and (c)Answer (1)10

NEET (UG)-2020 (Code-F1)60.S o l . The correct option is (1) because : 57.Herbicide resistant weeds, drug resistanteukaryotes and man-created breeds ofdomesticated animals like dogs areexamples of evolution by anthropogenicaction.Which of the following statements isnot correct?(1) The functional insulin has A and B chainslinked together by hydrogen bonds.(2) Genetically engineered insulin is producedin E.Coli.Darwin's Finches of Galapagos islands areexample of natural selection, adaptiveradiation and founder's effect.(3) In man insulin is synthesised as aproinsulin(4) The proinsulin has an extra peptide calledC-peptide.Some dividing cells exit the cell cycle andenter vegetative inactive stage. This is calledquiescent stage (G0). This process occurs atthe end ofAnswer (1)(1) S phase(2) G2 phaseS o l . The correct option is (1) because functionalinsulin has A and B chains linked together bydisulphide bridges.(3) M phase(4) G1 phase61.Snow-blindness in Antarctic region is due toAnswer (3)(1) High reflection of light from snowS o l . Some dividing cells exit the cell cycle andenter vegetative inactive stage, calledquiescent stage (G0). This process occurs atthe end of M-phase and beginning of G1 phase.(2) Damage to retina caused by infra-redrays58.(3) Freezing of fluids in the eye by lowtemperature(4) Inflammation of cornea due to high doseof UV-B radiationSecondary metabolites such as nicotine,strychnine and caffeine are produced byplants for theirAnswer (4)S o l . UV-B radiations damage DNA and mutationsmay occur.(1) Defence action(2) Effect on reproductionIn human eye, cornea absorbs UV-Bradiations, and a high dose of UV-B causesinflammation of cornea called snowblindness, cataract, etc.(3) Nutritive value(4) Growth responseAnswer (1)62.Sol. A wide variety of chemical substances that weextract from plants on a commercial scale(nicotine, caffeine, quinine, strychnine, opium,etc) are produced by them (plants) asdefences against grazers and browsers.59. Meiotic division of the secondary oocyte iscompletedStrobili or cones are found in(1) Marchantia(2) Equisetum(3) Salvinia(4) PterisAnswer (2)S o l . Strobili or cones are found in Equisetum.63.(1) After zygote formation(2) At the time of fusion of a sperm with anovumFrom his experiments, S.L. Miller producedamino acids by mixing the following in aclosed flask(1) CH4, H2, NH3 and water vapor at 600 C(2) CH3, H2, NH3 and water vapor at 600 C(3) Prior to ovulation(3) CH4, H2, NH3 and water vapor at 800 C(4) At the time of copulation(4) CH3, H2, NH4 and water vapor at 800 CAnswer (2)Answer (3)S o l . Meiotic division of secondary oocyte iscompleted after the entry of sperm insecondary oocyte which lead to the formationof a large ovum and a tiny IInd polar body.S o l . In 1953, S.L. Miller, an American scientistcreated electric discharge in a closed flaskcontaining CH4, H2, NH3 and water vapor at800 C.11

NEET (UG)-2020 (Code-F1)64.(3) Glucocorticoids stimulate gluconeogenesis.In relation to Gross primary productivity andNet primary productivity of an ecosystem,which one of the following statements iscorrect?(4) Glucagonishypoglycemia.S o l . Glucagon is associated with hyperglycemia.Insulin acts on hepatocytes and adipocytesand is associated with hypoglycemia.Glucocorticoids stimulate gluconeogenesis,so increase blood sugar level.(2) There is no relationship between Grossprimary productivity and Net primaryproductivity(3) Gross primary productivity is always lessthan net primary productivity67.(4) Gross primary productivity is always morethan net primary productivity(b) Contraction of external inter-costalmuscles(c) Pulmonary volume decreasesS o l . Gross primary productivity of an ecosystem isthe rate of production of organic matterduring photosynthesis.(d) Intra pulmonary pressure increasesNet primary productivity is GPP-respirationHence gross primary productivity is alwaysmore than NPP(b) Second trophic level (ii) Vulture(d) Third trophic level(iv) Grass(3) (a) and (b)(4) (c) and (d)The contraction of external intercostalmuscles increase the volume of the thoracicchamber in the dorsoventral axis.68.Select the correct optionThe roots that originate from the base of thestem are(a)(b)(c)(d)(1) Prop roots(1) (iv)(iii)(ii)(i)(2) Lateral roots(2) (i)(ii)(iii)(iv)(3) Fibrous roots(3) (ii)(iii)(iv)(i)(4) Primary roots(4) (iii)(ii)(i)(iv)Answer (3)S o l . The roots that originate from the base of thestem are fibrous roots.Answer (3)S o l . Grassland ecosystem is a terrestrialecosystem. It includes various trophic levels69.First trophic level (T1) – GrassGoblet cells of alimentary canal are modifiedfrom(1) ChondrocytesSecond trophic level (T2) – Rabbit(2) Compound epithelial cellsThird trophic level (T3) – Crow(3) Squamous epithelial cellsFourth trophic level (T4) – Vulture66.(2) only (d)S o l . Inspiration is initiated by the contraction ofdiaphragm, which increases the volume ofthoracic chamber in the anterio-posterioraxis.(i) Crow(iii) Rabbit(1) (a), (b) and (d)Answer (3)Match the trophic levels with their correctspecies examples in grassland ecosystem.(c) First trophic levelSelect the correct events that occur duringinspir

of the linked DNA in the vector, is termed (1) Palindromic sequence (2) Recognition site (3) Selectable marker (4) Ori site Answer (4) Sol. The correct option is (4) because Ori sequence is responsible for controlling the copy number of the linked DNA in the vector. Ori i.e. origin of rep

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