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1. We must consider the distinctive characters and the general nature of palnts ...
This is the opening sentence of
A: Enquiry into plants


2. In fibrous root system, the roots originate from
A: base of the stem


3. Main functions of the root
     A) Absorption of water          B) Absorption of food
     C) Storage of reserve food     D) Anchorage
     E) Synthesis of some hormones
A: A, C, D, E


4. The cells in this region divide repeatedly
A: Proximal to the root cap region


5. Function of root hairs
     A) Absorption of water         B) Absorption of minerals
     C) Anchorage                         D) Absorption of food
A: A only (Absorption of water)        


6. Differentiation and maturation of the cells take palce in
A: region of maturation


7. The following phenomenon can be seen in the region of elongation
     1) Cell maturation           2) Cell elongation
     3) Cell differentiation      4) 1 & 3
A: 2 (Cell elongation)


8. Root hairs are produced from the following
     1) Region of elongation                          2) Region of maturation
     3) Region of meristematic activity         4) Root cap
A: 2 (Region of maturation)


9. Root hair region is region of
A: maturation


10. Match the following regarding food storage

I. Carrot A. Tap root
II. Turnip B. Fibrous roots
III. Sweet Potato C. Tap root
IV. Asparagus D. Adventitious roots

       I      II     III    IV
A:   C     A     D     B


11. Find out mis-match.
       1) Brace roots - Stilt roots
       2) Adventitious roots - Fibrous roots
       3) Prop roots - Pillar roots
       4) Root hair region - Region of maturation
A: 2 (Adventitious roots - Fibrous roots)


12. Roots are green in
A: Taeniophyllum


13. Match the following.

I. Pneumatophores  A. Viscum
II. Velamen roots B. Vanda
III. Nodular roots C. Rhizophore
IV. Haustoria D. Arachis

      I     II    III    IV
A:  C    B    D    A


14. The following plants have green leaves but still they are grouped as parasites
        1) Viscum - Vanda            2) Striga - Rafflesia
        3) Viscum - Striga            4) Viscum - Cuscuta
A: 3 (Viscum - Striga)


15. Epiphytes have the following adv-roots
       1) Velamen roots                 2) Fibrous roots
       3) Photosynthetic roots      4) Pneumatophores
A: 1 (Velamen roots)


16. Roots of the following enter into the xylem and absorb water and minerals.
     1) Cuscuta         2) Rafflesia          3) Vanda      4) Striga
A: 4 (Striga)


17. Function of nodular roots is
A: N2fixation


18. The adventitious roots of Avicennia are
A: Aerial, respiratory


19. Assertion (A): Viscum is a partial parasite.
       Reason (R): Its haustoria enter into phloem of the host
A: A is false. R is correct.


20. Parasitic roots are called
A: Haustoria


21. Plants stealing water and minerals from other plants are called
A: Parasites


22. Pneumatophores are seen in
A: Mangrooves


23. Assertion (A): Rafflesia is a total parasite.
       Reason (R): Its haustoria enter into the xylem and phloem of the host.
A: A, R are correct. R explains A.


24. Prop roots of Banyan and pneumatophores of Avicennia are
A: Hanging and erect


25. The following roots arise from lower nodes of the stem
       1) Prop roots                   2) Brace roots    

    3) Assimilatory roots        4) Velamen roots
A: 2 (Brace roots)


26. The plants which grow in swampy areas are called
A: Mangrooves


27. The following are not adventitious roots
        1) Velamen roots       2) Brace roots        

        3) Prop roots             4) Nodular roots
A: 4 (Nodular roots)


28. The micro organism present in the modular roots
A: Only one species of Rhizobium


29. Stem is formed from
A: Plumule


30. Root may bear the following
        1) nodes          2) leaves      3) flowers        4) None
A: 4 (None)


31. Nodes are separated from each other by
A: inter nodes


32. Assertion (A): Ginger is a stem.
       Reason (R): It stores food.
A: A, R are correct. R do not explains A.


33. Assertion (A): Zaminkand is a underground stem.
       Reason (R): It bears nodes, buds and adventitious roots.
A: A, R are correct. R explains A.


34. Assertion (A): Colocasia is an organ of perennation.
       Reason (R): It is useful in vegetative reproduction.
A: A, R are correct. R do not explains A.


35. Assertion (A): Potato is an organ of perennation.
       Reason (R): It can tide over conditions unfavourable for growth.
A: A, R are correct. R explains A.


36. Match the following.

I. Turmeric A. Eyes
II. Potato B. Corm
III. Onion C. Rhizome
IV. Zaminkand D. Bulb

     I      II    III     IV
A:  C     A     D     B


37. Potato bears the following distinct characters by which it can be differentiated from others under ground stem modifications
        A) Eyes                                               B) Scale leaves
        C) Absense of adventitious roots        D) Presense of adventitious roots
        E) storage of food


38. Fleshy scale leaves are present in
A: Bulb


39. Adventitious roots are absent in
A: Tuber


40. Stem do not stores food in
A: Bulb


41. Scape is present in
Ans: Bulb


42. The distinct characters of Bulb
       1) Tunic           2) Scape      3) Fleshy scale leaves         4) All
Ans: 4 (All)


43. Apical bud and axillary bud are side by side in
Ans: Bulb


44. Food is stored in the following parts of bulb
       1) Stem     2) Scale leaf      3) Branch      4) Root
Ans: Scale leaf


45. One plant shows many under ground stems storing food in
      1) Solanum tuberosum        2) Zingiber    3) Allium cepa      4) Amorphophallus
Ans: Solanum tuberosum


46. The tendril in cucurbita is the modification of
Ans: stem


47. Axillary bud is modified into tendril in
Ans: Cucumber


48. All bud modifications are modifications of
Ans: stem


49. Axillary bud and terminal bud are modified into tendrils respectively in
Ans: Cucumber, Grape vine


50. Aerial stem modifications meant for photosynthesis
        A) Tendrils        B) Phylloclade     C) Thorns     D) Cladode or Cladophyll
Ans: B D


51. Match the following modification

I. Needle like A. Euphorbia
II. Flattened B. Asparagus
III. Cylindrical C. Casurina
IV. Branch of Limitted growth D. Opuntia

            I  II  III  IV
Ans:  C  D   A    B


52. Thorns are present in
       1) Citrus         2) Bougainvillea         3) Opuntia      4) 1 & 2
Ans: 4 (Citrus & Bougainvillea)


53. Phylloclades are seen in the plants growing in-
Ans: Arid regions


54. Match the following parts useful in veg reproduction

I. Agave A. Underground stem
II. Diascorea B. Floral bud
III. Grasses C. Lateral Slender branch
IV. Nerium D. Vegetative bud

         I   II   III  IV
Ans: B  D   A   C


55. Stem is discoid in
         1) Allium         2) Pistia          3) Eichornea        4) All
Ans: 4 (All)


56. Find out correct pair regarding stem modifications
       1) Strawberry - Jasmine       2) Grasses - Nerium
       3) Pistia - Oxalis                     4) Grasses - Strawberry
Ans: 4 (Grasses - Strawberry)


57. Find out incorrect match regarding stem modifications for vegetative reproduction
       1) Oxalis, Grasses, Strawberry           2) Pineapple, Chrysanthemum, Banana
       3) Pistia, Eichornea, Oxalis                  4) Nerium, Jasmine
Ans: 3 (Pistia, Eichornea, Oxalis)


58. Rosette leaves are present in
Ans: Eichornea, Pistia


59. Underground stems are useful in vegetaive reproduction when older parts die
Ans: Grasses, Strawberry


60. Balancing roots are present in
Ans: Offset


61. A lateral branch of one internode length and useful in vegetative reproduction is called
Ans: Offset


62. A lateral branch grows obliquely down words and upwards respectively in the following pair of plants and useful in vegetative reproduction
       1) Strawberry, Nerium            2) Jasmine, Banana
        3) Pineapple, Jasmine             4) Pineapple, Banana
Ans: 2 (Jasmine, Banana)


63. Tap root system and adventitious root system are present in respectively
Ans: Nerium, Chrysanthemum


64. Stem is spongy, petiole is swollen and balancing roots with root pocket are seen in
Ans: Eichornea


65. Axillary bud is modified into bulbil in
Ans: Diascorea


66. Lateral organ of the plant
Ans: leaf


67. Petiole is swollen in
Ans: Eichornea


68. Pulvinus leaf base is seen in
Ans: Legumes


69. The incisions of the lamina are called
        1) Leaf      2) Lobes          3) Leaflets        4) 2, 3
Ans: 4 (Lobes, Leaflets)


70. Common character between simple and compound leaf is the presence of
Ans: Axillary bud


71. Rachis is absent in
       1) Simple leaf                           2) Palmately compound leaf
       3) Pinnately compound leaf   4) 1 & 2
Ans: 4 (Simple leaf & Palmately compound leaf)


72. If the leaflets are attached at one common point the leaf is-
Ans: Palmately Compound


73. If the incisions of the lamina donot touch mid rib, the leaf is
Ans: Simple lobed


74. The type of leaf in Bombax ceiba is
Ans: Palmate Compound


75. Match the following Phyllotaxy with suitable example-

I. Alstonia A. Neither Alternate nor opposite
II. Hibiscus B. as in Guava
III. Calotropis C. Whorled
IV. Nerium D. Alternate

          I   II   III  IV
Ans: C   D  B   A


76. Number of leaves present at each node in Alstonia.
Ans: More than 2


77. A Modified leaf in Nepenthes is concerned with the following functions
       A) Photosynthesis      B) Climbing       C) Fulfils N2 requirement
Ans: ABC


78. Citrus shows
Ans: spines


79. The idea behind the formation of spines in Cacti-
Ans: 1 & 2(Protection, To reduce the rate of respiration)


80. Match the following leaf modifications

I. Photosynthesis A. Bryophyllum
II. Food storage B. Cacti
III. Defence C. Acacia (Australian)
IV. Vegetative reproduction D. Onion

Ans: I   II   III  IV
         C  D   B   A


81. The following plants do not perform protein synthesis if the area in which they grow is free from insects-
      (A) Nepenthes   (B) Venus flytrap   (C) Dionaea
Ans: ABC


82. Tendrils in Pea are the modifications of the following-
      1) Leaf 2) Petiole 3) Terminal leaflets 4) Basal leaflets
Ans: 3 (Terminal leaflets)


83. Phyllode is seen in-
      1) Australian Acacia   2) Nepenthes   3) 1 & 2   4) Opuntia & Nepenthes
Ans: 1 & 2(Australion Acacia, Nepenthes)


84. One of the following is not concerned with Nepenthes-
      1) Lamina-Pitcher                2) Upper petiole-Tendril 
      3) Lower petiole-tendril      4) Lower petiole-phyllode
Ans: 3(Lower petiole-tendril)


85. The leaf of Bryophyllum develop epiphyllous buds-
Ans: from notches of the leaf margin


86. Find the mis-match-
      1) Nepenthes    -            Pitcher plant
      2) Dionaea           -            Venus flytrap
      3) Bombax        -             Silk cotton
      4) Poaceae        -             Fabaceae
Ans: 4( Poaceae - Fabaceae)


87. Inflorescence deals with arrangement of-
Ans: flowers


88. A flower is a modified-
Ans: shoot


89. Racemose inflorescence has-
Ans: Only lateral flowers


90. Assertion (A): The inflorescence in Cassia and Cauliflower is corymb.
      Reason (R): The flowers are brought to the same height in them.
Ans: A & R are correct. R explains A.


91. Match the following:

I. Flowes appear to arise from the same point on peduncle A. Sun hemp
II. Flowers are produced in acropetal succession B. Achyranthes
III. Flowers are brought to the same height C. Carrot
IV. Sessile flowers are produced in acropetal succession D. Cauliflower

         C A D B


92. Find out mismatch-
      1) Simple raceme     -      Fabaceae     2) Spike     -    Poaceae
      3) Head                      -     Asteraceae   4) Umbel   -    Apiaceae
Ans: 4(Umbel  -  Apiaceae)


93. Find out mis-match-
      1) Graminae     -    Poaceae               2) Umbelliferae   -   Apiaceae
      3) Asteraceae   -    Compositae        4) None
Ans: 4(None)


94. Simple Raceme and spike differ in-
Ans: Flowers in the later are sessile and former are pedicellate.


95. Characters of Spike-
      1) Bisexual, acropetal, sessile           2) Unisexual, acropetal, pedicellate
      3) Bisexual, acropetal, pedicellate   4) Bisexual, spathe, acropetal
Ans: 1(Bisexual, acropetal, sessile)


96. Spathe is the modification of-
Ans: Bract


97. Inflorescence with spathe-
Ans: Spadix


98. Neuter flowers are present in-
Ans: Spadix


99. Match the following.

I. Unisexual flowers A. Head
II. Bisexual flowers B. Musa
III. Unisexual and bisexual flowers C. Spadix
IV. Unisexual and neuter flowers D. Carrot

Ans: I   II  III  IV
         B  D  A   C


100. Immature flowers in umbel are present-
Ans: in the centre


101. Find out mismatch-
        1) Compound raceme   -  Panicle                          2) Head                   -   Capitulum
        3) Cyathium                   -  Hypanthodium            4) Simple Cyme    -   Cymule
Ans: 3(Cyathium - Hypanthodium)


102. Minimum number of flowers in Cymule and dichasial Cyme respectively-
Ans: 3, 7


103. Peduncle is condensed in-
Ans: Capitulum


104. Flowers are produced in centripetal succession in-
Ans: Head


105. Head and Umbel are similar in that both have
Ans: Centripetal succession


106. Involucre forms a cup like structure in-
Ans: Cyathium


107. Gall flowers are present only in-
Ans: Hypanthodium


108. Blastophaga lays its eggs in-
Ans: Gall flowers


109. One of the following entire inflorescence becomes fruit-
        1) Head       2) Hypanthodium      3) Verticellaster       4) Cyathium
Ans: 2(Hypanthodium)


110. False whorl of flowers are formed in-
Ans: Verticellaster


111. Flowers are enclosed by fleshy Cup like peduncle in-
Ans: Hypanthodium


112. Arrange Cyathium (A), Hypanthodium (B) and Verticellaster (C) in ascending order basing on the number of types of flowers in them-
Ans: C, A, B(Verticellaster, Cyathium, Hypanthodium)


113. Arrangement of the folowers is irregular in-
Ans: Hypanthodium


114. Position of the verticellaster is-
Ans: at the node


115. Arrange the following in ascending order-
        A) Types of flowers in Hypanthodium
        B) Minimum number of flowers in dichasial cyme
        C) Number of flowers in the inflorescence of Datura
        D) Types of cymose inflorescence in verticellaster
Ans: C, D, A, B


116. The flower is opposite to the bract in-
Ans: Monochasial Cyme


117. Inflorescence without monochasial Cyme is seen in-
A) Hamelia  B) Euphorbiaceae  C) Ficus  D) Lamiaceae  E) Solanum  F) Ipomoea
Ans: C, F(Ficus, Ipomoea)


118. The following characters are seen only in Hypanthodium-
        1) Fleshy deep cup like peduncle       2) Irregular arrangement of the flowers
        3) Presence of Gall flowers                 4) All
Ans: 4(All)


119. Match the following-

I. Irregular arrangement A. Capitulum
II. Acropetal succession B. Main axis terminates into a flower
III. Centripetal succession C. Fleshy deep cup like peduncle
IV. Basipetal succession D. Continuous growth of the main axis

Ans: I  II  III  IV
        C  D  A  B


120. Inflorescence showing Polygamous condition is seen in-
Ans: Mangifera


121. Varied lengths of pedicels are seen in-
Ans: Corymb


122. Match the following.

I. Only one bisexual flowers A. Lamiaceae
II. Only one  Flower B. Spadix
III. Only one type of flowers C. Datura
IV. Neuter flowers D. Cyathium

          I   II  III   IV
Ans: C  D  A    B


123. Sterile female flowers are called-
Ans: Gall flowers


124. Male and female flowers produced not in the same succession-
Ans: Cyathium


125. Number of sporophylls in a Cyathium is 20. The male and female flowers in its are-
Ans: 17, 1


126. Number of stamens in the male flowers of Cyathium and Ficus respectively-
Ans: 1, 3


127. A cyathium has 16 flowers. The number of sporophylls in it are-
Ans: 18


128. Ratio between the microsporophylls of a male flower and megasporophylls of a female flower in cyathium-
Ans: 1 : 3


129. Special inflorescence with sessile flowers-
Ans: Hypanthodium


130. Special inflorescence with bisexual flowers-
Ans: Verticellaster


131. Match the following-

I. Umbel A. Moraceae
II. Hypanthodium B. Apiaceae
III. Verticellaster C. Euphorbiaceae
IV. Cyathium D. Lamiaceae

Ans:  I   II  III  IV
         B   A   D   C


132. Arrange the following whorls from outside to inside-
        1) P G A       2) C K A G      3) K C A G      4) G A K C
Ans: 3(K C A G)


133. Read the following statements. Choose the Correct one-
        1) Largest flower is present in a saprophytic angiosperm
        2) Smallest flower is present in a smallest angiosperm
        3) Accessory organs are always distinct
        4) A unisexual flower has neither androecium nor gynoecium
Ans: 2(Smallest flower is present in a smallest angiosperm)


134. Maximum Number of whorls on the thalamus in an unisexual flower-
Ans: 3


135. The flower of Gulmohur can be cut into 2 similar halves passing through the centre in-
Ans: vertical plane


136. Spathe is present in-
        1) Cocos        2) Colocasia       3) Banana      4) All
Ans: 4(All)


137. Arrangement of floral parts in a hypogynous flower from top to the bottom of      receptacle-
         1) K C A G   2) G A C K  3) A C G K   4) G A K C
Ans: 2(G A C K)


138. Position of gynoecium is not observed in-


139. If gynoecium occupies highest position-
Ans: Ovary is superior


140. If other parts of flower are arranged on rim of thalamus and gynoecium is in the centre in the same level-
Ans: Flower is perigynous


141. Thalamus fuses with ovary completely in-
Ans: Epigynous flower


142. Match the following-

I. Plum A. Ovary occupies lowest position
II. Brinjal B. Ovary ½ superior
III. Rose C. Ovary not protected by thalamus
IV. Sun flower D. Perigynous

         D C  B   A


143. If other parts of the flower arise above the ovary the flower is-

Ans: Epigynous

144. Find out mismatch-
        1) Guava          -    Superior ovary
        2) Peach           -    Ovary ½  inferior
        3) Mustard      -    Superior ovary
        4) Cucumber   -    Epigynous
Ans: 1 (Guava  -  Superior ovary)


145. Find out the unrelated basing on position of ovary-
        1) Plum 2) Rose 3) Rayflorets 4) Peach
Ans: 3(Rayflorets)


146. Ovary is protected by-
Ans: Thalamus


147. Match the following shapes & bundles of stamens

I. Datura A. Bi-lipped
II. Ocimum B. Citrus
III. Diadelphous C. Pea
IV. Polyadelphous D. Funnel shaped

Ans: I   II  III  IV
         D  A   C     B


148. Shape of the bisexual flowers in Sunflower-
Ans: Tubular


149. Wheel shaped flower is seen in-
         1) Datura 2) Ocimum 3) Rayflorets 4) None
Ans: 4(None)


150. Aestivation means the arrangement of
Ans: Perianth in the bud


151. The perianth members in a whorl are free or just touch without overlapping the aestivation is-
Ans: Valvate


152. Aestivation is twisted in-
Ans: Lady’s finger, Cotton, Chinarose


153. If the margins of sepals or petals overlap one another but not in any particular direction, the aestivation is-
Ans: Inbricate


154. The flower, a pentamerous one has 3 types of petals based on size has the following aestivation.
        1) Imbricate 2) Valvate 3) Vexillary 4) Twisted
Ans: 3(Vexillary)


155. Vexillary aestivation is also called-
Ans: Papilionaceous


156. Number of chambers in the mature and immature anther of a stamen-
Ans: 2, 4


157. If the stamens of a flower are free, the condition is-
Ans: Polyandrous


158. Match the following-

I. Epiphyllous A. Solanaceae
II. Polyadelphous B. Fabaceae
III. Epipetalous C. Citrus
IV. Diadelphous D. Liliaceae

Ans: I   II  III  IV
         D  C  A  B


159. Stamens show variation in the length of filments in-
Ans: Salvia, Brassica


160. Rose shows or Lotus shows
Ans: Perigynous, apocarpous


161. Match the following-

I. Parietal A. Placenta along ventral suture
II. Axile B. Inner wall of the ovary or peripheral part
III. Marginal C. Central axis, Unilocular ovary
IV. Free central D. Placenta is axial

         B D A C


162. Ovary is unilocular showing the following placentation-
        1) Marginal        2) Basal        3) Free central        4) All
Ans: 4(All)


163. Ovary without septum is seen in the following aestivation-
        1) Marginal        2) Basal        3) Free central        4) All
Ans: 4(All)


164. Find mismatch-
        1) Dianthus, Primrose   -   Free Central     2) Sun flower, Marigold     -   Basal
        3) Mustard, Lemon       -   Parietal              4) Pea, Bean                        -    Marginal
Ans: 3(Mustard, Lemon - Parietal)


165. Ovules are arranged on
Ans: Cushion like structure


166. One ovule, one locule and no septum are seen in
Ans: Basal


167. Number of nodes and internodes respectively on the thalamus of a bisexual and complete flower.
Ans: 4, 3


168. One of the following pairs have similar placentation-
        1) Pea, Marigold                   2) Mustard, Argemone
        3) Dianthus, Sunflower         4) Citrus, Sunflower
Ans: 2(Mustard, Argemone)


169. Assertion (A): Banana is a parthenocarpic fruit.
        Reason (R): It is seedless.
Ans: A & R are correct. R explains A.


170. If a fruit is formed from ovary and other parts like thalamus and pedicel, it is called-
Ans: False fruit


171. A runner that produces false fruit
Ans: Strawberry


172. Fleshy fruits are differentiated from dry fruits basing on-
Ans: Nature of pericarp


173. Match the following developments

I. Drupe A. Tricarpellary, unilocular inferior ovary
II. Berry B. Monocarpellary superior ovary
III. Pome C. Bi or multicarpellary superior
IV. Pepo D. Bi or Multicarpellary inferior

Ans: I   II  III  IV
         B  C  D  A


174. Match the following-

I. Endocarp stony A. Mango
II. Mesocarp fibrous B. Drupe
III. Mesocarp fleshy C. Cocos
IV. One seeded D. Mango

         B C  A   D


175. Match the following-

I. Pulp A. Hesperidium
II. Endocarp smooth B. Pome
III. Cartilagenous endocarp C. Pepo
IV. Mesocarp papery D. Berry

         D C  B   A


176.  Match the following-

I. Thalamus fleshy A. Hesperidium
II. Epicarp like a rind B. Pome
III. Glandular epicarp C. Citrus
IV. Juicy hairs on endocarp D. Pepo

         B  D  A  C


177. Inferior berry-
Ans: Guava


178. Match the following-

I. Endocarp smooth A. Berry
II. Endocarp Cartilagenous B. Pepo
III. Endocarp hard C. Pome
IV. Seeds hard D. Drupe

         B C  D   A


179. Mesocarp and endocarp are fused to form pulp in-
Ans: Berry


180. Match the following-

I. Pappus A. Nut
II. Mericarps B. Legume
III. Stony Pericarp C. Cypsela
IV. Bursts dorsiventrally D. Acacia

Ans:  I   II  III  IV
         C  D   A   B


181. Fleshy fruit and dry indehiscent fruit are externally seen together in-
Ans: Anacardium


182. The following fruits are all one seeded.
        1) Fleshy  2) Indehiscent  3) Dehiscent  4) Schizocarpic
Ans: 2 (Indehiscent)


183. Pappus is formed from-
Ans: Calyx


184. Find out mismatch.
        1) Rice  -   caryopsis      2) Cashew    -     nut
        3) Pea   -   capsule          4) Tridax      -    cypsela
Ans: 3 (Pea - capsule)


185. The fruit in Cotton and Datura is-
Ans: Capsule


186. Husk is found in-
Ans: Caryopsis


187. Mericarps are released by
Ans: Schizocarpic


188. Entire inflorescence becomes a fruit in-
Ans: Composite


189. Match the following common names-

I. Jack fruit A. Pyrus malus
II. Pine apple B. Artocarpus integrifolia
III. Apple C. Annona Squamosa
IV. Custard apple D. Ananas sativus

         B D  A   C


190. Find out incorrect pair-
        1) Sweet Orange   -   Citrus sinensis    2) Cashew    -     Anacardiaceae
        3) Pine apple         -   Bromeliaceae      4) Ficus         -    Euphorbiaceae
Ans: 4 (Ficus - Euphorbiaceae)


191. Read the following statements-
        A. Pineapple is a sorosis
        B. Jack fruit is a sorosis
        C. Fiscus is a syconus
        D. Ficus and jack fruit belong to the same family
        E. Cashew and Mango belong to the same family
        Find the incorrect statements.
Ans: None


192. Assertion (A): The fruit of Fiscus is a compound one.
        Reason (R): It is formed from entire inflorescence.
Ans: A & R are correct. R explains A.


193. Cotyledons are edible in-
        A. Bean  B. Pea  C. Groundnut  D. Cashew
Ans: A, B, C, D(Bean, Pea, Groundnut, Cashew)


194. Match the following-

I. Pine apple A. Perianth
II. Custard apple B. Peduncle, juicy bracts
III. Jack fruit C. Endosperm
IV. Cocos D. Mesocarp and endocarp of each fruitlet

         B D A C


195. The edible part in cucumber-
        A) Epicarp  B) Mesocarp  C) Rind  D) Endocarp  E) Pericarp  F) Placenta
Ans: B, D, F(Mesocarp,  Endocarp, Placenta)


196. Peduncle is edible in-
Ans: Ficus & Pineapple


197. Read the following synonyms and find out an incorrect one.
        1) Keel                        -  carina                    2) Wings       -  alae
        3) Compound fruit    -  composite fruit      4) Vexillum  -  standard   5) None
Ans: 5(None)


198. One seeded bits of schizocarpic fruits are called-
Ans: Mericarp


199. Function of involucre-
Ans: Protection to inflorescence


200. Under grond stem that grows erect is called-
Ans: Corm


201. Inflorescence with naked flowers-
Ans: Cyathium


202. Function of the bract-
Ans: Protects the flower in bud condition


203. Bracteole arises from-
Ans: pedicel


204. The fruit of Rice is covered by-
Ans: Husk


205. The stem is sub-aerial and useful as a runner-
Ans: Oxalis


206. Inflorescence with single bisexual flower-
Ans: Datura

Posted Date : 30-11-2020