Which of the following pairs is not correctly matched?
C.
Mode of reproduction ExampleThe plant body Sargassum is a diploid sporophyte. It does not multiply asexually by means of spores. Instead it reproduce by vegetative means, i.e. fragmentation which is the only known method of vegetative reproduction in the free floating species of Sargassum.
A flower of tomato plant following the process of sexual reproduction produces 240 viable seeds.
Answer the following questions giving reasons:
(a) What is the minimum number of pollen grains that must have been involved in the pollination of its pistil?
(b) What would have been the minimum number of ovules present in the ovary?
(c) How many megaspore mother cells were involved?
(d) What is the minimum number of microspore mother cells involved in the above case?
(e) How many male gametes were involved in this case?
The number of viable seeds produced by the tomato plant through sexual reproduction = 240
(a) The minimum number of pollen grains that must have been involved in the pollination of its pistil are 240 because each pollen grain contains two male gametes. Out of these two gametes, one fuses with polar nuclei and forms endosperm, while, the other male gamete fuses with the egg cell to form the zygote that eventually give rise to seeds. Therefore, in order to obtain 240 seeds, number of pollen grains needed would be 240.
(b) The minimum number of ovules involved in this process would be 240, as the number of viable seeds is 240. After fertilisation, the ovary turns into fruit and the ovules turn into seeds. Therefore, the number of ovules corresponds to the number of seeds formed.
(c) During the process of gametogenesis, 240 megaspore mother cells are involved as only one megaspore of the tetrad becomes functional and develops further and the rest three megaspores get degenerated.
(d) In the above case, 60 microspore mother cells must have undergone reduction division prior to dehiscence of anther, as each microspore mother cell would give rise to 4 microspores. Since 1 microspore mother cell would produce 4 microspores, therefore, to obtain 240 microspores, there must be 60 microspore mother cells.
(e) The number of male gametes involved in seed formation would be 240 as each male gamete will fuse with one egg nuclei to form zygote, which will further give rise to the seed.
In vitro clonal propagation in plants is characterised by
PCR and RAPD
Northern blotting
Electrophoresis and HPLC
Electrophoresis and HPLC
A.
PCR and RAPD
RAPD stand for Random Amplified Polymorphic DNA. It is a type of PCR reaction, but the segments of DNA that are amplified are random. Often, PCR is used to amplify a known DNA sequence.
Male gametes are flagellated in
Polysiphonia
Anabaena
Ectocarpus
Ectocarpus
C.
Ectocarpus
Male gametes are flagellated in Ectocarpus belonging to Phaeophyceae. The flagella of male and gamete plays an important role in establishing initial sexual contact with the female gamete. However, in Ectocarpus the female gamete too is flagellated but is different in structure.
In Polysiphonia (Rhodophyceae) flagellated gametes are not observed, in Anabaena sexual reproduction through gametes is absent while in Spirogya sexual reproduction takes place by conjugation wherein male gamete passes through a tube to the adjacent filament. The male gametes here are non-flagellated and show amoeboid movement.
Monoecious plant of Chara shows occurrence of
antheridiophore and archengoniophore on the same point
stamen and carpel on the same plant
upper antheridium and lower oogonium on the same plant
upper antheridium and lower oogonium on the same plant
D.
upper antheridium and lower oogonium on the same plant
Monoecious or homothallic, a condition in Chara (green algae) is used to denote upper oogonium and lower antheridium on the same plant. The organisms, which possess both the reproductive organs are bisexual. Unisexual condition represents both male or female organs in the same organism.
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