Solution-Class 12th-Biology-Chapter-1-Reproduction in Lower & Higher Plants-Maharashtra Board

Reproduction in Lower & Higher Plants

Maharashtra Board-Class-12th-Biology-Chapter-1


Q. 1 Multiple choice questions.

(1) Insect pollinated flowers usually posses ................

(a) Sticky pollens with rough surface

(b)  Large quantities of pollens

(c)  Dry pollens with smooth surface

(d)  Light coloured pollens

Answer :

(a) Sticky pollens with rough surface

(2) In ovule, meiosis occurs in ..........

(a)  Integument

(b)  Nucellus

(c)  Megaspore

(d)  Megaspore mother cell

Answer :

(d)  Megaspore mother cell

(3) The ploidy level is NOT the same in ......

(a)  Integuments and nucellus

(b). Root tip and shoot tip

(c)  Secondary nucleus and endosperm

(d)  Antipodals and synergids

Answer :

(c)  Secondary nucleus and endosperm

(4) Which of the following types require pollinator but result is genetically similar to autogamy?

(a) Geitonogamy (b) Xenogamy

(c)  Apogamy (d)  Cleistogamy

Answer :

(a) Geitonogamy

(5) If diploid chromosome number in a flowering plant is 12, then which one of the following will have 6 chromosomes?

(a). Endosperm (b). Leaf cells

(c)  Cotyledons (d)  Synergids

Answer :

(d)  Synergids

(6) In angiosperms, endosperm is formed by/ due to ..........

(a)  Free nuclear divisions of megaspore

(b)  polar nuclei

(c)  polar nuclei and male gamete

(d)  synergids and male gamete

Answer :

(c)  polar nuclei and male gamete

(7) Point out the odd one ..........

(a)  Nucellus (b)  Embryo sac

(c)  Micropyle (d)  Pollen grain

Answer :

(d)  Pollen grain

Q. 2 Very short answer type questions :

(1) Name the part of gynoecium that determines the compatible nature of pollen grain.

Answer :

Stigmatic surface of flower determines the compatible nature of pollen grain. It refuse other wrong type or incompatible pollen grains.

(2) How many haploid cells are present in a mature embryo sac ?

Answer :

6 cells, 2 synergids, 1 egg cell, 3 antipodals are present in a mature embryo sac.

(3) Even though each pollen grain has 2 male gametes, why at least 20 pollen grains are required to fertilize 20 ovules in a particular carpel?

Answer :

Angiosperms have the phenomena of double fertilisation, in which both male gametes are used, one for fusion with the egg cell to produce the zygote and the other for fusing with the secondary nucleus to form the endosperm.

(4) Define megasporogenesis.

Answer :

Megasporogenesis : It is the process of formation of haploid megaspores from diploid megaspore mother cell (MMC).

(5) What is hydrophily ?

Answer :

Transfer of pollen grains in pollination process through agency of water is known as hydrophily.

(6) Name the layer which supplies nourishment to the developing pollen grains.

Answer :

Tapetum is the inner most nutritive layer of anther wall, which supplies nourishment to the developing pollen grains.


(7) Define parthenocarpy.

Answer :

The condition in which fruit is developed without the process of fertilization is called parthenocarpy.

(8) Are pollination and fertilization necessary in apomixis ?

Answer :

Apomixis is formation of embryos through asexual method of reproduction without formation of gametes and the act of fertilization, hence there is no need of pollination and fertilization.

(9) Name the parts of pistil which develop into fruits and seeds.

Answer :

Ovaxy develops into fruit and ovules into seed.

(10) What is the function of filiform apparatus ?

Answer :

Filiform apparatus guides the pollen tube towards egg cell.

Q. 3 Short Answer Questions :

(1) How polyembryony can be commercially exploited ?

Answer :

  • Polyembryony is the development of more than one embryo inside the seed.
  • When such polyembiyonic seed germinate we get multiple seedlings from it.
  • This condition increases the chances of survival of new plants.
  • Nucellar embryos are genetically identical to parent plants hence we get uniform plants.
  • In horticulture we can utilize these as rootstock for grafting, hence they have significant role in fruit breeding programmes e.g. Citrus, Mango.

(2) Pollination and seeds formation are very crucial for the fruit formation. Justify the statement.

Answer :

  • After fertilisation, the ovary is changed into fruit, where the ovarian wall becoming the fruit wall, i.e. the pericarp.
  • After fertilisation, mature ovules are turned into seeds.
  • Fertilisation is the process through which male and female gametes combine to produce a zygote, which develops into an embryo.
  • Pollen grains carrying non-motile male gametes are transported on stigma during pollination.
  • Seeds contain embryos that germinate into new plants, achieving the purpose of reproduction to produce offspring for the following generation.
  • Hence pollination and seeds formation are the crucial events for fruit formation.

(3) Incompatibility is a natural barrier in the fusion of gametes. How will you explain this statement?

Answer :

  • Self incompatibility or self-sterility is a genetic mechanism that prevents germination of pollen on stigma of same flower. This favours cross pollination. E.g. Tobacco.
  • In pollen-pistil interaction, when pollen grain is deposited on stigma, pistil has the ability to recognize and allow germination of right type of pollen.
  • Special type of proteins on stigmatic surface determine compatibility or incompatibility.
  • A physiological mechanism operates to ensure successful germination of compatible pollen.
  • Compatible pollen absorbs water and nutrients from stigmatic surface that are absent in pollen and then pollen tube emerges which grows through style.

(4) Describe three devices by which cross pollination is encouraged in Angiosperms by avoiding self pollination.

Answer :

Unisexuality, dichogamy, prepotency, heteromorphy and herkogamy are the outbreeding devices.

  • Unisexuality : The plants bear either male or female flowers. Due to unisexual nature, self-pollination is avoided. Plants are either dioecious, e.g. Papaya or monoecious, e.g. maize.
  • Heteromorphy : In same plants different types of flowers are produced. In these flowers, stigmas and anthers are situated at different levels. There is heterostyly and heteroanthy. This prevents self-pollination e.g. Primrose.
  • Herkogamy : In bisexual flowers we may come across mechanical device to prevent self-pollination. Natural physical barrier avoids contact of pollens with stigma. E.g. Calotropis where pollinia are situated below the stigma.

Q. 4 Long Answer Questions :

(1) Describe the process of double fertilization.

Answer :

Double fertilization : In angiosperms, one of the two male gametes produced by the male gametophyte unites with the female gamete and the other with the secondary nucleus. Since both the male gametes take part in fertilization and fertilization occurs twice, it is called double fertilization.

  • During double fertilization, the pollen tube on reaching the ovule enters the embryo sac through micropyle and bursts in one of the synergids. Owing to this, the two male gametes contained in the pollen tube, are set free.
  • Out of the two male gametes, one unites with the egg or female gamete and the other unites with the secondary nucleus of the embryo sac, forming a triploid or triple fusion nucleus, called the primary endosperm nucleus. The process involving the fusion of one of the male gametes with the egg nucleus, resulting in the formation of a diploid zygote is called syngamy.
  • The reproductive process in which non-motile male nuclei are carried to the egg cell through a pollen tube is called siphonogamy.
  • After fertilization, zygote develops into an embryo. Certain changes take place in the ovule leading to the development of a seed.

(2) Explain the stages involved in the maturation of microspore into male gametophyte.

Answer :

  • Microspore or pollen grain is first cell of male gametophyte.
  • The protoplast of pollen grain divides mitotically to form two unequal cells — a small thin walled generative cell and a large naked vegetative or tube cell.

  • The generative cell possesses thin cytoplasm and a nucleus. It separates and floats in the cytoplasm of vegetative cell.
  • The vegetative, possesses thick cytoplasm, irregular shaped nucleus and the reserved food.
  • In majority of the angiosperms, the pollen grains are liberated at two-celled stage after the dehiscence of the anther.
  • The generative cell of the pollen grain divides by mitosis to form two male non-motile gametes.

(3) Explain the development of dicot embryo.

Answer :

Development of embryo (dicot) in angio-sperm :

  • The development of embryo from a zygote is called embryogenesis.
  • During fertilisation, the fusing of a male gamete and an egg cell results in the production of a diploid zygote. The zygote forms a wall around itself and is transformed into an oospore.
  • The oospore divides transversely to create a big basal cell near the micropyle and a small apical or terminal cell towards the embryo sac's chalaza.
  • This two-celled structure is known as a proembryo. The basal cell or suspensor first divides transversely to form a multicellular structure known as the suspensor. The suspensor directs the embryo's nutrients to the endosperm.
  • The apical cell or embryonal initial of the proembryo undergoes a transverse division followed by two vertical divisions at right angles to form an octant stage.
  • From octant, the lower four cells form hypocotyl and radicle while four cells of upper side form plumule with two cotyledons.
  • The lowermost cell of suspensor is hypophysis and by its further division forms part of radicle and root cap.
  • The cells from upper side of octant divide repeatedly to form heart shaped which elongated further to form two lateral cotyledons.
  • Enlargement of hypocotyl and cotyledon results into curved embryo which appears horse shoe shaped.

(4) Draw a labelled diagram of the L.S. of anatropous ovule and list the components of embryo sac and mention their fate after fertilization.

Answer :

Components of Embryo sac :

  • Mature embryo sac is 7-celled and 8 nucleate.
  • Egg apparatus at micropylar end -with 2 synergids and egg cell.
  • Central cell with secondary nucleus formed by 2 polar nuclei
  • Antipodal cells at chalazal end - 3 cells.
  • Pollen tube enters the synergids, Synergids guide the growth of pollen tube towards egg.
  • Male gamete fuses with female gamete, i.e. syngamy to form zygote which develops into embryo.
  • One male gamete fuses with secondary nucleus to form primary endosperm nucleus (PEN) which forms endosperm, nutritive tissue for embryo.

Q. 5 Fill in the blanks:

(i) The ................ collects the pollen grains.

Answer :

  • biotic agents

(ii) The male whorl, called the ..................... produces ............... .

Answer :

  • androecium, pollen grains

(iii) The pollen grains represent the ..................

Answer :

  • male

(iv) The ............... contains the egg or ovum.

Answer :

  • embryo sac

(v) ................ takes place when one male gamete and the egg fuse together. The fertilised egg grows into seed from which the new plants can grow.

Answer :

  • Fertilization

(vi) The ............... is the base of the flower to which other floral parts are attached.

Answer :

  • thalamus

(vii) ............. is the transfer of pollen grains from anther of the flower to the stigma of the same or a different flower

Answer :

  • Pollination

(viii) Once the pollen reaches the stigma, pollen tube traverses down the .............. to the ovary where fertilization occurs.

Answer :

  • style

(ix) The ................. are coloured to attract the insects that carry the pollen.

Answer :

  • petals

(x) Some flowers also produce ............... or .............. that attracts insects.

Answer :

  • fragrance. nectar

(xi) The whorl ...................... is green that protects the flower until it opens.

Answer :

  • Calyx.

 Q. 6 Label the parts of seed.

Answer :

Q. 7 Match the column.

Column – I (Structure before seed formation.) Column – II (Structure after seed formation.
A. Funiculus I. Hilum
B. Scar of Ovule II. Tegmen
C. Zygote III. Testa
D. Inner integument IV. Stalk of seed
  V. Embryo

(a) A - V, B - I, C - II, D – IV    

(b) A - III, B - IV, C - I, D - V

(c) A - IV, B - I, C - V, D – II     

(d) A - IV, B - V, C - III, D - II

Answer :

(c) A - IV, B - I, C - V, D - II

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