Unit- III, IV & V of Second year
  In the last and concluding part of the series of EAMCET model papers, in the second year syllabus Unit-III , Unit-IV and Unit-V is being discussed. All are of the related topics dealing with classical and molecular genetics. Some sub-sections of Molecular genetics are a new addition to the old syllabus. Both long term and regular students feel difficult to understand these chapters. Most of the question will be knowledge based and easy questions. Problems on DNA and genetics is a regular feature in EAMCET exams. One or two problems may be given from these units. Practice is essential for this type of questions to save time on solving the questions. In Biotechnology part of the syllabus not more than one question can be expected whereas in the application part knowledge based questions can be asked.
  
  Model questions:
  1.    Recessive trait in the pea plant
      1) Yellow pod     2) Yellow seed    3) Green pod     4) Violet flowers
  
  2.    Wrong statement regarding monohybrid cross    
  1)    All F1 plants look alike
  2)    In F2 generation 50% are homozygotes
  3)    Alleles are separated in to gametes
  4)    Two different genotypes are observed
  
  3.    The number of F2 classes in dihybrid cross
      1) 9      2) 6     3) 4         4) 1
  
  4.    One of the scientists who rediscovered Mendel's results
      1) Morgan        2) Tschermark    3) Sutton        4) Sturtevant
  
  5.    Assertion(A): Deletion and insertion of base pairs of DNA cause frame shift mutation
      Reason (R): Genetic is triplet codon without overlapping and commas.
  1)    Both A, R are true and R is the correct explanation of A.
  2)    Both A, R are true but R is not the correct explanation of A.
  3)    A is true but R is false
  4)    A is false but R is true
  
  6.    In a population of pea plants the percentage of RrYy is
      1) 40         2) 20      3) 25        4) 33
  
  7.    True statement regarding reasons for selecting Pisum sativum by Mendel for his     hybridization  experiments
  A.    Though bisexual it is easily cross pollinated    
  B.    It can also self fertilized
  C.    It shows seven contrasting characters
  D.    It is perennial and produce many seeds
  1) A, B & C     2) B, C & D
  3) C, A & D     4) A & B
  
  8.    In Antirrhinum majus colour of the flower can be red or white or pink.    True statement regarding this is    
  1)    Colour of the flower is controlled by three different genes
  2)    Colour is multiple allelic. Red dominant over pink and white.    
  3)    Red and white are true breeds. Red is not completely dominant over white
  4)    Colour is biallelic but do not segregate .
  
  9.    In a population of 936 plants percentage of genotype AABb is    1) 117    2) 702        3) 468    4) 434
  
  10.    In a cross between a homozygous dominant and homozygous recessive parents
  A.    All the F1 generation individuals show similar phenotype
  B.    Parental traits reappear in F2 generation
  C.    Equal ratio of reciprocal heterozygotes appear in F2 generation
  D.    Half of the F2 individuals are homozygous
  1)    A & B      2) A, B & C
  3)    A, B & D    4) A, B, C & D
  
  11.    In garden pea Tallness (T) is dominant over dwarf (t). Green pods (G) are dominant over yellow (g). What will be the appearance of  the offspring of the cross TTGg x ttGg
  1)    All tall and green
  2)    All tall plants with green or yellow flowers.
  3)    Some are dwarf green and some are tall yellow
  4)    All tall and yellow
  
  12.    True statement regarding codominance in Lens culinaris for seed coat is
  1)    All F1 hybrids show dominant traits
  2)    Half of the F2 generation shows dominant traits.
  3)    Half of the F2 generation shows both traits.
  4)    One fourth of F2 generation shows both traits
  
  13.    In a population of true breeding plants of Antirrhinum white flowers are 144. The number of red flowers is    
      1) 144   2) 288      3) 432        4) Nil
  
  14.    RNA functions as a structural molecule in
  1)    Some enzymes
  2)    Inner membrane of mitochondria
  3)    Ribosomes        4) All the above
  
  15.    The distance between two base pairs in a nucleatide strand is
      1) 0.34 Å    2) 20 Å        3) 2.7 Å    4) 3.4 Å
  
  16.    Flow of information from RNA to DNA takes place in
      1) Prokaryotes    2) Eukaryotes    3) HIV    4) fX 174
  
  17.    Histone proteins are rich in
  1)    Lysine and Arginine
  2)    Cysteine and methionine
  3)    Aromatic amino acids    
  4)    Valine and Leusine
  
  18.    In the lytic cycle of viral infection, if phage coat is labeled with radioactive sulphur
  1)    All the progeny resulting from the infection will have radioactive sulphur
  2)    Some viral particles will have radioactivity
  3)    None of the progeny will have radioactive sulphur.
  4)    Only one of the progeny will have radioactive sulphur.
  
  19.    hnRNA is transcribed by
      1) RNA polymerase I
      2) RNA polymerase II    
      3) RNA polymerase III    
      4) DNA polymerase.
  
  20.    Assertion (A): Leusine is coded by six codons
      Reason (R): Code is degenerate.
  1)    Both A & R are true and R is the correct explanation of A.
  2)    Both A & R are true but R is not the correct explanation of A.
  3)    A is true, R is false
  4)    A is false, R is true.
  
  21.    Number of Hydrogen bonds in a DNA molecule of 1 kbps length with 200 Adenines is
      1) 1400    2) 2800
      3) 2400    4) 2000
  
  22.    Length of the DNA molecule with 900 nucleotides is
      1) 1530 Å    2) 3060 Å    3) 6120 Å    4) 715 Å
  
  23.    RNA is less stable than DNA due to
  1)    Lack of double strand in RNA    
  2)    1    RNA cannot generate its replica
  3)    Smaller size of RNA
  4)    Presence of 2'OH in the pentose sugar of RNA.
  
  24.    Wrong statement among the following
      1) RNA cannot mutate
      2) RNA viruses are less stable    
      3) RNA can code for protein    4) RNA can code for DNA
  
  25.    Location of Promoter in an operon is
  1)    Immediately upstream to structural genes
  2)    Immediately downstream to operator genes
  3)    Between repressor and operator
  4)    Upstream to reprssor genes
  
  26.    In the absence of inducer in the lac operon is
  1)    Transcription takes place
  2)    Repressor cannot be synthesized
  3)    Transcription do not start
  4)    Translation does not take place.
  
  27.    Assertion(A): In bacteria translation and transcription takes place simultaneously.
      Reason(R): In bacteria nuclear membrane is absent
  1)    Both A and R are true and R is the correct explanation of A.
  2)    Both A and R are true but R is not the correct explanation of A.
  3)    A is true, R is false
  4)    A is false, R is true
  
  28.    True statement regarding Griffith's experiment
  I.    Biochemical nature of genetic material is known.
  II.    Experimental bacteria is Streptococcus pneumoneae.
  III.    Heat killed virulent bacteria is transformed.
  IV.    Mice developed resistance to avirulent bactereia
  1) I & II    2) I, III & IV
  3)     Only II    4) I, II & IV
  
  29.    Roundup ready soybean is
  1)    Herbicide tolerant
  2)    Pods are round and inflated
  3)    Soybean with reduced maturing time
  4)    Soybean with high protein
  
  30.    Bt cotton is resistant to
      1) Viruses    2) Insects
      3) Fungi    4) Herbicides
  
  31.    Golden rice "Taipei" is rich in    
      1) Vitamin  B    2) Vitamin  A    3) Vitamin C    4) Vitamin K
  
  32.    Flavr Savr is
  1)    A gene for flavour
  2)    Bruise resistant tomato variety     
  3)    Technique in genetic engineering
  4)    Fungus resistant potato variety
  
  33.    Bacterium used in producing transgenic plants.
      1) Bacillus thuringienesis        2) Bacillus subtilis
      3) Escherechia coli        4) Agrobacterium tumifaciens.
  
  34.    Transgenic potatoes are resistant to
      1) Phytophthora 2) Pseudomonas  3) Cold and drought    
      4) Bacterial rot
  
  35.    Polymerase chain reaction (PCR) is a technique of        
      1) Gene insertion        2) Gene Isolation        3) Gene multiplication
      4) Gene sequencing
  
  36.    Male sterility is induced by genetic engineering in        1) Carica papaya        2) Brassica napus
      3) Lycopersicon    
      4) Brassica nigra
  
  KEY
  1)  1    2)  4    3)  3    4)  2
  5)  1    6)  3    7)  4    8)  3
  9)  1    10) 4    11) 2    12) 3
  13) 1    14) 3    15) 4    16) 3
  17) 1    18) 3    19) 2    20) 1
  21) 2    22) 1    23) 4    24) 1
  25) 3    26) 3    27) 1    28) 3
  29) 1    30) 2    31) 2    32) 2
  33) 4    34) 1    35) 3    36) 2


  Unit- II & VI of Second Year
 
 In this part Unit-II and Unit -VI of second year is being clubbed for easy preparation as they related in every aspect. Unit-II is Microbiology with Bacteria and Virus dealt separately. The Unit-VI is the last one dealing Plants, Microbes in Human welfare. In this Unit 13 and 14 chapters are included. 14th chapter is more related to Microbiology. The 13th chapter is more related to genetics and biotechnology. Particularly the 13th chapter is important for match the following type of questions. For long term students the last Unit is new one. The student is advised to go through carefully as we can expect nearly 8 questions from this part. As there is some deviation from the last year syllabus long term student need an extra attention on this chapters.
  
  Model Questions
  1.    True statement among the following
  I.    In viral infections protein coat do not take part
  II.    Without a protein coat viroids can cause disease.
  III.    Only protein can cause 'scrapie' disease in sheep
  IV.    Infectious agent of 'mad cow' disease in protein
      1) I & II    2) II & III
      3) III & IV    4) I, II, III & IV
 
  2.    True statement regarding bacteria
  I.    Arrangement of cells in cocci is more complex     
  II.    Shape of bacteria depends on cell walls
  III.    Bdellovibrio is a virus that lives as parasite on bacteria.
  IV.    Photoheterotrophic bacteria utilizes gaseous CO2.
      1) I & II    2) II & III
      3) I & III    4) III & IV
           A           B          C  
 
  3.    NH3  ____ NO2 ____  NO3 ____    N2
  A, B, C in the above reaction
  respectively
  1)    A-Nitrosomonas, B-
      Pseudomonas, C- Nitrobacter    
  2)    A- Nitrococcus, B-
      Nitrobacter, C- Thiobacillus
  3)    A-Nitrosbacter, B- Nitroso- monas, C-Pseudomonas
  4)    A-Pseudomonas , B-
      Thiobacillus, C- Nitrococcus
  4.    'Germ theory of diseases' fails to explain the diseases caused by
      1) Viruses       2) Mycoplasma
      3) Actinimycetes 4) Protozoa
 
  5.    The eclipse period of the viral infection of bacterium refers to
  1)    Period between attachment and penetration
  2)    A period after maturation  and before lysis of the bacterial cells
  3)    A period of time between infection and appearance of mature viruses within the cell.
  4)    A period of finding the bacteria by the virus before attachment.
 
  6.    'Hidden hunger' is    
  1)    Large requirements of food and nutrients due to metabolic disorder developed as a result of lack of  vegetables in normal diet.
  2)    Hunger due to excess amounts of vegetables and low animal fats in the normal diet.
  3)    Micronutrient deficiency or vitamin deficiency developed in consequent to lack of balanced diet due to poverty.
  4)    Craving for food from plant sources developed as a result of consuming genetically modified plants.
 
  7.    'Probiotics'  are
  1)    Science dealing with primitive biological organi -sms and their classification.
  2)    Eating curd as part of regular diet.
  3)    Kinds of foods that encourages the growth of normal intestinal flora.
  4)    Foods containing lot of beneficial bacteria or microbes that form part of normal diet or therapeutic agents.
 
  8.    Assertion(A): LAB plays very beneficial role in checking disease-causing  microbes in our stomach
      Reason(R): LAB produces antibiotics in our stomach.
  1)    Both A, R are true and R is the correct explanation of A.
  2)    Both A, R are true but R is not the correct explanation of A.
  3)    A is true but R is false
  4)    A is false but R is true
 
  9.    To demonstrate bacteria the best household sample is
      1) Bread    2) Curd
      3) Cheese    4) All the above
 
  10.    Microorganism used in 'Swiss cheese' production is
      1) Penicillium
      2) Propionibacterium
      3) Lactobacillus     4) Aspergillus
 
  11.    Composition of biogas during sewage treatment
  1)    Exclusively methane
  2)    Methane, H2S and CO2
  3)    Methane and CO2    
  4)    Natural gas
 
  12.    True statement regarding biogas production is
  I.    Anaerobic bacteria is involved in the biogas production
  II.    Cattle dung is the medium for the biogas production
  III.    The spent slurry is not useful as fertilizer
  IV.    Every time inoculums should be added to the medium
      1) I & II    2) II & III
      3) III & IV    4) IV & I
 
  13.    Organic farming or biological farming envisages
  1)    Biodiversity    
  2)    Pest eradication    
  3)    Genetic engineering methods in farming
  4)    All the above
 
  14.    The BOD values of four water samples  A B C D respectively are 20 mg/L , 8 mg/ L , 300 mg/ L , 150 mg/L.     The correct statement regarding this is    
  1)    Sample C is less polluted and BOD is very high
  2)    All the samples are equally polluted as BOD is not Zero
  3)    Sample B is highly polluted as BOD is inversely proportional to pollution
  4)    Sample B is less polluted as BOD is very low.
 
  15.    Wrong statement regarding microbes
  1)    The bacteria live in rumen of cattle is an autotroph
  2)    Oscillatoria is N2 fixing symbiotic bacteria
  3)    Yeast is multiplied enormously to be used as an SCP.
  4)    Epstien-Barr virus causes cancer in humans.
 
  16.    Assertion(A): Cyanobacteria are considered useful in paddy fields.
      Reason(R): Cyanobacteria are aquatic. Paddy crop needs water stagnation.
  1)    Both A, R are true and R is the correct explanation of A.
  2)    Both A, R are true but R is not the correct explanation of A.
  3)    A is true but R is false
  4)    A is false but R is true
 
  17.    'Super microbes' are
  1)    Microbes which are highly beneficial to the human beings
  2)    Drug resistant pathogens.
  3)    Microbes with many beneficial characters
  4)    Microbes that are used in genetic engineering.
 
  18.    Not a reemerging infectious disease
      1) Cholera     2) SARS
      3) Tuberculosis    4) Dengue fever
 
  19.    Effluent from primary treatment consisting of
  1)    Filtrate and water with microbes
  2)    All solids that settles from primary sludge
  3)    Water sans(without) floating debris and grit
  4)    Primary sludge and water.
 
  20.    Assertion (A): Secondary treatment is referred as Biological Treatment.
      Reason(R): During secondary treatment methane is released.    
  1)    Both A, R are true and R is the correct explanation of A.
  2)    Both A, R are true but R is not the correct explanation of A.
  3)    A is true but R is false
  4)    A is false but R is true
 
  21.    Contamination of drinking water can spread the disease
      1) Plague    2) Diphtheria
      3) Polio    4) SARS
 
  22.     The most desired character in the parents is
  1)    Variability in their alleles
  2)    Few dominant traits in both parents
  3)    One parent is recessive
  4)    Homozygous dominant for as many traits as possible
 
  23.    Most desired character in the new cultivar
  1)    Homozygosity for dominant character
  2)    Heterozygous for all traits
  3)    Homozygous recessive for all characters
  4)    Homozygous recessive for few traits.
 
  24.    Assertion(A): Genetic variability is the root any breeding programme.
      Reason(R):Variability provides scope to select required traits.
  1)    Both A and R are correct and R is the correct explanation of A.
  2)    Both A and R are correct but R is not the correct explanation of A.    
  3)    A is correct, R is false
  4)    A is false, R is true
 
  25.    True statement regarding
  achievement of Plant Breeding in India    
  1)    Wheat resistant to water stress is developed
  2)    Semi-dwarf variety IR-8 developed.
  3)    Paddy variety of 'Jaya' & 'Ratna' were developed.
  4)    A better Sugarcane is introduced into South India.
 
  26.    Bacterial disease on crucifers    1) Brown rust    2) Red rot
      3) Late blight    4) Black rot
 
  27.    Genetic constitution of the plant decides
  1) Resistance to the pathogen
  2)    Yield of the crop
  3)    Tolerance to thew drought
  4)    All the above
 
  28.    'Pusa komal' is resistant to    1) Leaf curl    
      2) Bacterial blight
      3) Aphids    4) Rust
  29.    Resistant to yellow mosaic virus and powdery mildew is induced in            
     1) Chilli    2) Green gram
      3) Flat bean    4) Mustard
 
  30.    Cauliflower variety that resists Black rot and Curl blight is    1) Pusa Swarnam
      2) Pusa shubra        3) Pusa Gourav    
      4) Pusa Sadabahar.
 
  31.    Maize variety resistant to stem borer shows
  1)    High aspartic acid, low nitrogen and low sugar
  2)    Low aspartic acid, high nitrogen and low sugar
  3)    High aspartic acid, high nitrogen and low sugar
  4)    Low aspartic acid, low nitrogen and low sugar
 
  32.    Assertion (A):'Atlas 66' is used as donor for improving cultivation varieties of wheat.
      Reason(R): 'Atlas 66'is a wheat variety with high protein content.
  1)    Both A and R are correct and R is the correct explanation of A.
  2)    Both A and R are correct but R is not the correct explanation of A.    
  3)    A is correct, R is false
  4)    A is false, R is true
 
  33.    King of green vegetable is enriched with    
  1)    Vitamin C and Fe
  2)    Vitamin C and Fe and Ca
  3)    Mineral Fe and Ca
  4)    Vitamin A and Ca
 
  34.    Spirulina is
  1)    Unicellular
  2)    Multicellular filament
  3)    Prokaryote    4) A bacterium
 
  35.    Temparature in autoclave is    1) 120oC    2) 121oC
      3) 100oC    4) 15oC.
  
  KEY
  1)  4    2)  1    3)  2    4)  1
  5)  3    6)  3    7)  4    8)  3
  9)  2    10) 2    11) 2    12) 1
  13) 1    14) 4    15) 2    16) 1
  17) 2    18) 2    19) 3    20) 2
  21) 3    22) 4    23) 1    24) 1
  25) 3    26) 4    27) 4    28) 2
  29) 2    30) 2    31) 1    32) 1
  33) 2    34) 3    35) 2   


  1.    True statement regarding stamens is / are
  I.    In a single flower stamens differ in their  lengths
  II.    Stamens in different flowers show different shapes
  III.    Stamens of different flowers show different attachments
  IV.    Stamens are always bilobed at their distal ends
      1) I & II        2) II, III & IV
      3) I, II & III          4) I, II & IV
 
  2.    Monothecous condition is seen in    
      1) Papaver    2) Hibiscus        3) Annona    4) Michelia
 
  3.    In a mature anther of Datura the number of pollen sacs are
  1)    4, at each corner of the anther    
  2)    2, on each side of the central sterile tissue
  3)    Only one covered by anther wall            
  4)    Only one due to dissolution of sterile tissue
 
  4.    Assertion A: In fully mature anther lobe tapetal cells are not seen.
      Reason R: Tapetal cells serve as food material for growing spores.    
  1)    Both A and R are true and R is the correct explanation of A.
  2)    Both A and R are true but R is not the correct explanation of A.
  3)    A is true, R is false
  4)    A is false, R is true
 
  5.    Endothecium is present between
  1) Middle layers and tapetum
  2)    Tapetum and sporogenous tissue
  3)    Middle layers and epidermis
  4)    Outside epidermis
 
  6.    First cell of the gametophyte is
  1)    Gamete        
  2) Vegetative cell
  3)    Generative cell    
  4) Spore
 
  7.    True statement regarding tapetum is
  A.    It is the only the layer that completely covers the sporogenous tissue
  B.    Cells of tapetum show more than one nucleus
  C.    In a mature anther lobe tapetum cannot be seen.
  D.    It is the inner most layer of the anther wall
      1. A & B    2. B & C
      3. B, C & D    4. A, B, C & D
 
  8.    Characters of vegetative and generative cells of male spore are    
  1)    Vegetative cell shows
      large round nucleus
  2)    Small generative cell floats in the cytoplasm of vegetative cell
  3)    Nucleus of generative cell is spindle shaped
  4)    generative cell feeds on vegetative cell.
 
  9.    Megasporangium is
      1) Nucellus        2) Carpel    3) Ovule        4) Pistil
 
  10.    Ovules without integument is seen in                    1) Helianthus          2) Loranthus
      3) Datura          4) Monocots
  11.    Female gametophyte is
  1) Megaspore mother cell
  2)    Functional megfa spore
  3)    Embryo sac
  4)    Egg apparatus
 
  12.    Ploidy of MMC and nucellus respectively is
  1) Haploid, Diploid
  2)    Diploid , Haploid
  3)    Haploid, Haploid
  4)    Diploid, Diploid
 
  13.    A typical angiospermic embryo sac shows
      1) 8 celled and 7 nucleate        2) 8 celled and 8 nucleate    
      3) 7 celled and 7 nucleate        4) 8 celled and 7 nucleate
 
  14.    Number of cells that do not participate in reproduction        1) One        2) Four        
      3) Five        4) Seven
 
  15.    Monosporic type of embryosac is
  1)    Embryosac developed from single spore
  2)    Only one spore develops into embryosac
  3)    Embryosac developing into one sporophyte
  4)    Embryosac fertilized by one microspore
 
  16.    Assertion A: Geitonogamy & Xenogamy can take place in Cocos                
      Reason R: Cocos is monoecious plant. Pollen may be from same or different plant.
  1) Both A and R are true and R is the correct explanation of A.
  2) Both A and R are true but R is not the correct explanation of A.
  3) A is true, R is false
  4) A is false, R is true
 
  17.    Synchronization of pollen and pistil maturation takes place in
      A. Viola             B. Oxalis        C. Commelina        D. Solanum
      1) A, B, C        2) Only C    3) Only D        4) A, C
 
  18.    True statement regarding anemophilous flowers
  I.    It is the most common method of pollination    
  II.    Stigmas may be feathery
  III.    Single seeded fruits are a character
  IV.    Sticky pollen grains to stick easily to stigma
      1) I & II    2) II & III
      3) III & IV    4) IV & I
 
  19.    Assertion A: In Commelina always self pollination takes place.
      Reason R: Commelina shows cleistogamous flowers.
  1) Both A and R are true and R is the correct explanation of A.
  2) Both A and R are true but R is not the correct explanation of A.
  3) A is true, R is false
  4) A is false, R is true
 
  20.    Enormous amount and light weight pollen grains are produced in
  1) Anemophilous flowers
  2) Entemophilous flowers
  3) Ornithophilous flowers
  4) Hypohydrophilous flowers
 
  21.    Characteristic feature of plants pollinated by flies and beetles is
  1)    Flowers emanate foul smell.
  2)    Aquatic plants
  3)    Flowers are colourless
  4)    Flowers with short styles
 
  22.    At the time of entry of pollen tube into ovule    
  1)    Generative cell disappears
  2)    Generative cell divides into gametes
  3)    Vegetative cell guides the tube
  4)    Pollen tube disintegrates
 
  23.    In artificial hybridization emasculation is
  1)    To prevent unwanted pollination
  2)    To prevent self pollination
  3)    To make flower into unisexual
  4)    To encourage cross pollination
 
  24.    Assertion A: Endosperm
      development precedes embryo development
      Reason R: Endosperm provides assured nutrition to the developing embryo
  1)    Both A and R are true and R is the correct explanation of A.
  2)    Both A and R are true but R is not the correct explanation of A.
  3)    A is true, R is false
  4)    A is false, R is true
 
  25.    Example for endospermic seed is
      1) Castor    2) Pea             3) Bean    4) Ground nut
 
  26.    Animal dispersed seeds among the following
  I.    Coconut        II. Grass
  III.    Martynia    IV. Figs
      1) I & II    2) II & III        3) III & IV    4) II, III & IV
 
  27.    Production seeds without fertilization is
  1)    Parthenocarpy
  2)    Parthenogenesis
  3)    Apomixes     4) Polyembryony
 
  28.    Sequential order of four basic components of taxonomy is
  A.    Classification
  B.    Nomenclature
  C.    Identification
  D.    Characterization
  1) D  B  C  A     2) A  C  B  D
  3) C  B  D  A     4) D  C  B  A
 
  29.     True statement regarding Artif-icial system of classification is
  1)    Anatomy is one of the criteria for the system of  classification
  2)    Number of petals can be the criteria for the classification
  3)    Identification of an unknown plant is difficult
  4)    'Species Plantarum' is a natural system of classification
 
  30.    True statement regarding Bentham & Hooker's classification
  I.    It is a natural system of classification    
  II.    It is published after Darwin's "Origin of Species'.
  III.    It is the classification of only flowering plants    
  IV.    They published the book in the name 'Historia Plantarum'
      1) I & II    2) II & III
      3) II & IV    4) I, II & III
 
  31.    Total number of cohorts in B & H classification            
     1) 15      2) 25      3) 10    4) 21
 
  33.    Cohorts are equal to present day    1) Orders    2) Families        3) Series    4) Sub class
 
  34.    Total number of angiospermic families in B & H classification is    
      1) 202     2) 165       3) 199    4) 34
 
  35.    Which of the following can be represented in a floral formula    
  1)    Superior ovary, axile     placentation
  2)    Zygomorphic flower, Inferior ovary
  3)    Petal 5, twisted aestivation
  4)    Unisexual female flower, perigynous flower
 
  36.    Plants used as fodder
      1) Sesbania, Tephrosia
      2) Glucine , Arachis    
      3) Dalbergia, Pterocarpus
      4) Crotalaria, Phaseolus
 
  37.    Scientific name of Aswagandha ?
  1)    Solanum nigrum    
  2)    Petunia alba        
  3)    Withania somnifera
  4)    Datura metal
 
  38.     Assertion A: Stamen and petals fall off at the same time in Solanaceae                
      Reason R: Stamen are
      epipetalous in Solanaceae
  1)    Both A and R are true and R is the correct explanation of A.
  2)    Both A and R are true but R is not the correct explanation
  of A.
  3)    A is true, R is false    
  4)    A is false, R is true
 
  39.    Reticulate venation in monocot plant is seen in    
  1) Spanish dagger
  2) Sarsaparilla
  3) Medow saffron
  4) Glory lily
 
  40.    True statement regarding Liliaceae family is
  I.    Undistinguished perianth
  II.    Anterior odd tepal
  III.    Zygomorphi c flowers
  IV.    Trimerous flowers
      1) I, II, III    2) I , II
      3) II, III, IV    4) I, II, IV
  
  Key
  1) 3    2) 2    3) 2     4) 1     5) 3
  6) 4    7) 4    8) 2    9) 3    10) 2
  11) 3    12) 4    13) 4    14) 3    15) 2
  16) 1    17) 1    18) 2    19) 4    20) 1
  21) 1    22) 2    23) 2    24) 1    25) 1
  26) 4    27) 3    28) 4    29) 2    30) 4
  31) 2    32) 3    33) 1    34) 3    35) 2
  36) 4    37) 3    38) 1    39) 3    40) 4


 preparation plan
      As First Year IPE exams are fast approaching it is time for the student to shift the mode of preparation to the IPE.
 
      To get good marks student has to maintain a strategy useful only to the IPE preparation.
 
      As the students are writing this kind of exam for the first time after their 10th class exams they have to be cautious not loose marks due to ignorance of the exam pattern. To avoid this, question paper model should be known first. Then approximate weightage for each chapter also should be known clearly.
 
      Student should write the exam in a booklet of 24 pages. No additional papers will be given. So the student has to confine their answers to that effect.
 
      The Botany exam will be of 3 hours duration. Paper is for 60 marks.
 
      The Botany paper consisting of three Sections viz A,B and C,
 
      Section- A is Very Short Answer Type Questions (VSAQ). 10 questions will be given. All the questions must be answered. Each question carries 2 marks. Section -B is Short Answer Type Questions (SAQ). Out of 8 questions 6 must be answered. Each question carries 4 marks. Section -C is Long answer Type Questions (LAQ). Out of 3 questions 2 must be answered. Each question carries 8 marks.
 
      For answering VSAQs student should not write more than 2-3 sentences. Diagrams are not necessary for this section. All the questions must be attempted at one place and in serial order..
 
      Answering this section should not take more than 20 minutes. A well prepared student should attempt this Section first in the exam.
 
  n    Student must be thorough with all most all the question given at the end of each chapter. Only the answer to the questions must be written. Student is adised not to write any unnecessary things here. It will save the time and probably avoid mistakes.
 
      It is always advisable for an average student to attempt SAQs first. Student can get maximum marks easily in this section.
 
      Answers must be limited to 20 lines for each question. Wherever necessary diagrams must be drawn. Answering each question may take not more than 15 minutes.
 
      Selection of 6 questions out of 8 is also important to save time and get good marks.
 
      The answers for these questions should have all the important points. Diagrams must not be left without labeling.
 
      For some questions in this section there is no need of drawing diagrams. Answers to these questions must contain all the important points. Students who are not good at drawing should select these questions.
 
      Students with good drawing skills always select questions with a scope of diagrams. First a neat labeled diagram should be drawn and description is added to it to get maximum marks.
 
      Diagrams always speak of the students' abilities.
 
      An average student can take full 90 minutes in answering this Section. A well prepared student can complete within 70 minutes. If time permits, at the end of answering the entire paper an additional SAQ can be answered.
 
      In writing LAQs in first year, drawing diagrams is invariable. Selection of the question is also important here. Question with time consuming diagrams can be avoided and selection with easy diagrams is essential. Each question needs 35 minutes to answer efficiently.
  
      Student should read the question carefully and answer to it. Certain question can carry break ups . For example.. " Define root. Mention the types of root system. Explain how root is modified to perform different functions?" Here for every part of the question marks will be allotted separately-marks for definition; marks for mentioning the type of root systems and marks for explanation.
 
      Student should practice to complete the answer for each question within 35 minutes.
 
      Question from anatomy takes longer time to answer as the diagrams are drawn carefully. But if a student can draw a diagram within 20 minutes, seeing his own diagram he can explain it easily within 10 minutes.
 
      LAQs can be given from 2nd chapter and 3rd chapter and 6th chapter.
 
      If a student prepares all the questions from these chapters. Question from Internal organization of Plants (6th chapter) is easy for the student with good drawing skills and practical knowledge.
  
  The weightage for the exam chapter wise can be as follows.
  Sl.No    Name of the Unit/Chapter    Weightage of marks
  1    Diversity in the Living World    14
  2    Structural Organization in Plants-Morphology    12
  3    Reproduction in Plants    12
  4    Plant Systematics    06
  5    Cell Structure and Functions    14
  6    Internal Organization of Plants    12
  7    Plant Ecology    06
      TOTAL MARKS    76
  


 UNIT-III
  The third unit consisting of the chapters 6th and 7th. The entire unit deals with the methods and process of reproduction in the plant kingdom. Reproductive methods of Monera and Protista are also dealt here. More elaborately reproduction in Angiosp-erms is discussed in the seventh chapter. In both the chapters the description as usual is very brief with many questions and directions to the students.
 History of the discovery of the sexuality in plants is completely ignored. Except Maheswari's name no other scientist's names are mentioned in the entire unit. Topic on Artificial Hybridization, an applied aspect, is an aberration in the unit. In the sub-topic 'significance of fruit and seed', less is written about the significance.
  *    Reproduction is intriguing in the plants..
  Reproduction is essential for living organisms for continuity of the species. The most interesting part of the plant life is its reproduction. Sexual reproduction is basically similar in all living organisms; both plants and animals. It involves union of gametes. But plants show, besides sexual reproduction, vegetative and asexual reproduction. Lower as well as higher plants adopt different methods of vegetative and asexual reproduction and in each group and individual plant methods are different. So the process of vegetative and asexual reproduction is much varied in the plant kingdom.
 This is where student has to concentrate to answer knowledge based questions. For examples gemmae are vegetative propagules in Bryophytes and sporangiospores in Rhizopus. In evolution the reproductive process underwent changes. These changes are quite evident in the plant kingdom. Prokaryotes like Bacteria undergo binary fission which is a very primitive form of reproduction. In higher plants sexual reproduction is well developed resulting in much variation in the progeny. At the same time most of the higher plants also reproduce vegetatively and this process is also well developed. Many higher plants do not produce flowers regularly and some like Chrysanthemum and Jasminum lost the ability to produce fertile seeds and has to entirely depend on the vegetative propagation. Vegetative propagation does not produce variation in the progeny. In lower plants like fungi sexuality is much reduced in the evolution. Asexual reproduction is a normal feature in the plants. Spore germination is asexual reproduction. Spores never form in the animal kingdom. Another interesting feature in the Plant kingdom is the presence of a haploid stage with varied nutritional status and morphological structure.
 This we cannot see in the animals. Haploid structures or stages, compared to diploid structures do not ha-ve stability in the nature. Peculiarly in Pteridophytes both haploid and diploid stages live independently. With such an enigmatic and wonderful nature, reproduction must be understood very clearly in plant kingdom. A separate unit for this is aptly justified. Even though the student has been studying this process from the lower classes Intermediate is the correct stage to study and understand this difficult process. Studying an example for every group make it easy for the student.
  *    This Unit must be correlated to Chapters 2 & 4.
  The content of this entire unit must be correlated and coordinated with chapters 2 and 4 of unit-I. Once a student read the unit-I after the unit-III he understands easily many aspects written in Unit-I.
  Actually the chapter 6 is a redundancy of the earlier chapters. Information given in the earlier chapters 2 & 4 are presented differently here but without much effect. The definition and explanation given thereof of reproduction is of very low standard. Certain things student will understand only after reading many times. For example statement like "When two parents (opposite sex) participate in the reproductive proce-ss and also involve fusion of male and female gametes, it is called sexual reproduction". It may be true in higher animals. In lower plants same plant produces both the gametes and results in sexual reproduction.
  Regarding asexual reproduction-examples and methods, much is written in the chapter 5. Same things are once again repeated here but not before creating a confusion. For better understanding of asexual reproduction vegetative reproduction must be separated from it. Definition of spore is given in the glossary of chapters 4 & 6, sporophyte in the glossary of chapters 4 & 7 and sporangium in the glossary of chapter 7. A student need to go through all these chapters twice or thrice before understanding something about these structures.
 
  In the chapter 6 the explanation regarding sexual reproduction is confusing. Student feels like reading once again from the beginning. For example we can see a statement here "All organisms have to reach a certain stage of growth and maturity in their life, before they can reproduce sexually and this stage is known as vegetative phase in plants". With this knowledge, understanding of life cycle of an angiosperm plant is very difficult. For competitive exams this kind of confusion will be a drawback.
 
  For IPE examination point of view this chapter does not have much importance. The entire unit carries 12 marks weightage. One Long Answer Question (LAQ) from this unit is invariable. LAQ is not possible from chapter 6. As entire chapter is a repetition of earlier chapters other type of questions can be answered easily. But regarding the EAMCET point of view no chapter can have a certain weightage. So the student has to go through this chapter carefully and thoroughly and should correlate with other chapters for better understanding.