NCERT Exemplar Class 11 Biology Solutions Chapter 3 Plant Kingdom

NCERT Exemplar Class 11 Biology Solutions Chapter 3 Plant Kingdom

Edited By Priyanka kumari | Updated on Aug 22, 2022 10:56 AM IST

NCERT exemplar Class 11 Biology solutions chapter 3 explains both flora and fauna. In this chapter, we shall study the Plant kingdom which is a specialised branch of study. It is termed as botany. In NCERT Solutions for Class 11 Biology, we will learn the basic concepts of this particular branch in detail. For the study, scientists have classified plants into different categories to simplify this process. NCERT Exemplar solutions for Class 11 Biology chapter 3 starts with the most basic plant organism which is algae and depending on the type of pigment it possesses, it has further been classified into other subcategories. Everything is explained in detail in the NCERT exemplar Class 11 Biology solutions chapter 3.

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This Story also Contains
  1. Multiple Choice Questions:
  2. Very Short Answer Type Questions:
  3. Short Answer Type Questions:
  4. Long Answer Type Questions:
  5. NCERT Exemplar Solutions for Class 11 Biology Chapter 3 Main Subtopics:
  6. What the students will learn in NCERT Exemplar Solutions for Class 11 Biology Chapter 3?
  7. NCERT Exemplar Class 11 Biology Solutions Chapter Wise
  8. NCERT Exemplar Class 11 Biology Solutions Chapter 3 Important Topics

Also, check NCERT solutions for Class 11 other subjects.

Multiple Choice Questions:

Question:1

What are Cyanobacteria are classified under?
(a) Plantae
(b) Protista
(c) Algae
(d) Monera

Answer:

The answer is the option (d) Monera.

Question:2

What is the term for the fusion of two motile gametes which are dissimilar in size?
(a) Anisogamy
(b) Isogamy
(c) Oogamy
(d) Zoogamy

Answer:

The answer is the option (a) Anisogamy.
Explanation: The size of gametes is dissimilar in accordance to each other and hence is termed as ‘Anisogamy’.

Question:3

Holdfast, stipe and frond constitutes the plant body in case of
(a) Phaeophyceae
(b) All of the above
(c) Rhodophyceae
(d) Chlorophyceae

Answer:

The answer is the option (a) Phaeophyceae.

Question:4

A plant shows thallus level of organisation. It shows rhizoids and is haploid. It needs water to complete its life cycle because the male gametes are motile. Identify the group to which it belongs to
(a) Pteridophytes
(b) Bryophytes
(c) Monocots
(d) Gymnosperms

Answer:

The answer is the option (b) Bryophytes.
Explanation: Plants from monocots and gymnosperms do not require any water for the process of fertilisation. Bryophytes and pteridophytes require water for fertilisation. But pteridophytes have actual roots. Hence, ‘d’ is the correct choice.

Question:5

A Prothallus is
(a) A structure in pteridophytes formed before the thallus develops
(b) A sporophytic free living structure formed in pteridophytes
(c) A primitive structure formed after fertilisation in pteridophytes
(d) A gametophyte free living structure formed in pteridophytes

Answer:

The answer is the option (d) A gametophyte free-living structure formed in pteridophytes

Question:6

Plants of this group are diploid and well adapted to extreme conditions. They grow bearing sporophylls in compact structures called cones. The group in reference is
(a) Monocots
(b) Gymnosperms
(c) Pteridophytes
(d) Dicots

Answer:

The answer is the option (b) Gymnosperms.
Explanation: Cones are a primary feature of gymnosperms. Furthermore, gymnosperms are admirably adapted to extreme conditions.

Question:7

The embryo sac of an Angiosperm is made up of
(a) 8 cells
(b) 7 cells and 7 nuclei
(c) 8 nuclei
(d) 7 cells and 8 nuclei

Answer:

The answer is the option (d) 7 cells and 8 nuclei.
Explanation: The seven cells are one egg cell, two synergids, three antipodal cells and one secondary nucleus. The secondary nucleus consists of two nuclei, while the other cells consist of one nucleus each.

Question:8

If the diploid number of a flowering plant is 36. What would be the chromosome number in its endosperm?
(a) 36
(b) 54
(c) 18
(d) 72

Answer:

The answer is the option (c) 54.
Explanation: Endosperm is formed through the fusion of a haploid male gamete and diploid secondary nucleus. Consequently, the endosperm has a triploid nucleus, which will have three times the number of chromosomes as in a haploid nucleus.

Question:9

Protonema is
(a) Haploid and is found in mosses
(b) Haploid and is found in pteridophytes
(c) Diploid and is found in pteridophytes
(d) Diploid and is found in liverworts

Answer:

The answer is the option (b) Haploid and is found in mosses.

Question:10

The giant Redwood tree (Sequoia sempervirens) is a/an
(a) Gymnosperm
(b) Free fern
(c) Pteridophyte
(d) Angiosperm

Answer:

Ans. The answer is the option (a) Gymnosperm.
Explanation: Redwood trees are a Gymnosperm which have earned many labels; known as the tallest tree, oldest tree, largest by trunk volume, etc.

Very Short Answer Type Questions:

Question:1

Food is stored as Floridean starch in Rhodophyceae. Mannitol is the reserve food material of which group of algae?

Answer:

In Phaeophyceae or Brown Algae, mannitol is the reserve food material used by them.

Question:2

Give an example of plants with
(a) Haplontic life cycle
(b) Diplontic life cycle
(c) Haplo- diplontic life cycle

Answer:

(a) Volvox, Spirogyra and some types of Chlamydomonas
(b) Alga; Fucus
(c) Kelps, Polysiphonia, ectocarpus etc.

Question:4

Most algal genera show haplontic lifestyle. Name an alga which is

(a) Haplo-diplontic, (b) Diplontic.

Answer:

(a) Ectocarpus and Polysiphonia are Haplo-diplontic
(b) Fucus is Diplontic.

Question:5

In Bryophytes, male and female sex organs are called ____ and ____.

Answer:

Antheridium (male sex organ), archegonium (female sex organ).

Short Answer Type Questions:

Question:1

Why are bryophytes called the amphibians of the plant kingdom?

Answer:

Bryophytes require water for one stage of their life cycle, i.e. in the process of reproduction. Similarly, amphibians, too, require water to lay their eggs and their tadpoles grow and develop in water. Thus, bryophytes are called the amphibians of the plant kingdom.

Question:2

Heterospory, i.e., the formation of two types of spores – microspores and megaspores are a characteristic feature in the life cycle of a few members of pteridophytes and all spermatophytes. Do you think heterospory has some evolutionary significance in the plant kingdom?

Answer:

Yes, heterospory has evolutionary significance in plant kingdom because heterospory is the formation of two types of spores differentiating in the size and the sex, microspores, and megaspores. Microspores are the male who is smaller in size, and megaspores are the female who is larger in size. In Pteridophytes, the microspores and megaspore sprout together to give rise to male and female gametophytes. In the end, the zygote matures in the female gametophyte. Seeds make it possible for gymnosperms and angiosperms to survive long periods of hostile conditions. Seeds have also made it possible for them to spread to a different geographical area by seed dispersal.

Question:3

How far does Selaginella, one of the few living members of lycopodiales (pteridophytes) fall short of seed habit?

Answer:

Ans. In Selaginella, the development of the zygote into the embryo takes place inside of the female gametophyte. The female gametophyte holds on to the parent plant for fluctuating periods. This is considered as a precursor, or the parent of seed habit as an improved version of this occurrence is observed in higher plants. Although, unlike seeds in higher plants; the embryo in Selaginella requires it to mature quickly to start the new generation.

Question:4

Each plant or group of plants has some phylogenetic significance in relation to evolution: Cycas, one of the few living members of gymnosperms, is called the ‘relic of past’. Can you establish a phylogenetic relationship of Cycas with any other group of plants that justifies the above statement?

Answer:

Answer: Cycas is the only living species in the family Cycadaceae. All other species of the Cycadaceae family are extinct, as of now. Cycas plant is often considered as the living fossil and the plant of the Old World; because it is found only at those places where some of the oldest rocks in the world are found. Because of these reasons, Cycas is also known as the ‘relic of the past’.
Besides that, Cycas resembles more with Pteridophytes than any Gymnosperms. This proves a phylogenetic relationship between Pteridophytes and Gymnosperms.
Some features of Cycas which prove this are as follows: -
  • In Cycas, the seeds are shed when the embryo is still developing or mature. In gymnosperms, seeds are shed solely when the embryo is mature.
  • The megasporophylls look like a leaf.
  • Even though the pollen tubes are present, the male gametes are flagellate.
  • Leaf base is persistent in nature.
  • Circinate ptyxis; which is similar as in Pteridophytes.

Question:7

Comment on the lifecycle and nature of a fern prothallus.

Answer:

Fern, which is a type of pteridophytes; reveals a haplo-diplontic state. In this situation, both the stages of gametophytic and sporophytic are multicellular. The diploid sporophyte is represented by a prepotent and independent, photosynthetic, vascular plant body. The haploid gametophyte, which is also independent in nature, but the duration of its life is shorter than that of the sporophytes.



Question:8

How are the male and female gametophytes of pteridophytes and gymnosperms different from each other

Answer:

PteridophytesGymnosperms
(a)Flagellated male gamete(a)Non-flagellated male gamete
(b)Archegonia with neck canal cells(b)Neck canal cells are absent
(c)Pollen tubes are not formed(c)Pollen tubes are formed
(d)Water is essential for fertilisation(d)Water is not essential

The size of the gametophyte of pteridophytes quite large compares to gymnosperm's gametophytes. Gametophyte of pteridophytes are photosynthetic in nature while gametophyte of gymnosperms are not photosynthetic in nature

Question:9

In which plant will you look for mycorrhiza and coralloid roots? Also, explain what these terms mean.

Answer:

Mycorrhiza, which is also known as ‘fungus-root’, is basically a mutually beneficial relationship between that of plant root and a fungus. In most cases, the fungus grows within the roots of the plants and helps the plant to absorb water and nutrients, and in return, the plant provides the fungus with food. E.g., Pinus. In some cases, the fungus could be harmful to the plants.

Coralloid Roots: These roots are associated with nitrogen-fixing cyanobacteria. In coralloid roots, the nodules are formed in large numbers, in the roots giving them coralloid appearance. E.g. Cycas and plants of Leguminosae.

Long Answer Type Questions:

Question:1

Gametophyte is a dominant phase in the life cycle of a bryophyte. Explain.

Answer:

The dominant phase in bryophytes is represented by gametophytic plants. The gametophyte is self-reliant, photosynthetic, and erect. Antheridia and archegonia are developed and found on the gametophyte. These male and female organs are multicellular in nature. The Antheridia and Archegonia carry the male and female gametes. The antheridium produces biflagellate antherozoids, while the archegonium produces a single egg. Water helps to carry the antherozoids to the archegonium. The antherozoids flagellate, which gives them the feature to be motile. After the process of fertilisation, the zygote is formed. The zygote develops into a multicellular figure called a sporophyte.

The sporophyte in bryophytes is smaller comparably and is not self-reliant. The sporophyte is attached to the gametophyte and gets it nutrition from the gametophyte. The gametophyte has control over the life cycle of a bryophyte in every phase. The gametophyte is self-independent and photosynthetic. The size of the gametophyte is larger in size as compared to the sporophyte. Also, the sporophyte is not independent and holds on to the gametophyte for nutrition.

Question:2

With the help of a schematic diagram, describe the haplo-diplontic life cycle pattern of a plant group.

Answer:

Ans. Pteridophyte and Bryophyte both reveal a haplo-diplontic life cycle. The diploid phase is not independent in bryophytes; however, it is independent in the case of pteridophytes. Furthermore, the sporophyte is more clearly visible in the case of pteridophytes than in bryophytes.



Question:3

Lichen is usually cited as an example of ‘symbiosis’ in plants where an algal and a fungal species live together for their mutual benefit. Which of the following will happen if algal and fungal partners are separated from each other?
(a) Both will survive and grow normally and independent of each other.
(b) Algal components will survive while the fungal component will die.
(c) Both will die
(d) The Fungal component will survive while the algal partner will die.
Based on the answer, how do you justify this association as symbiosis?

Answer:

Option(c) is the correct, i.e. both will die.
Explanation: The process of ‘symbiosis’ explains that the plant and the fungus which are growing together depend on each other for nutrients and if they are bound to get separated, they will both die because they are codependent.

Question:4

Explain why sexual reproduction in angiosperms is said to take place through double fertilisation and triple fusion. Also, draw a labelled diagram of embryo sac to explain the phenomena.

Answer:

In the process of double fertilisation in case of angiosperms; one male gamete fuses with the female gamete, and the second male gamete fuses with the secondary nucleus. This fusion of the male and female gametes results in the outcome of the zygote. And the fusion between the second male gamete and the secondary nucleus results in the outcome of the primary endosperm nucleus. Since two occurrences of fusion of nuclei take place, thus it is called double fertilisation. These following figures show double fertilisation:



Question:5

Draw labelled diagrams of the following:
(a) Female and male thallus of a liverwort.
(b) Gametophyte and sporophyte of Funaria.
(c) Alternation of generation in Angiosperm.

Answer:

(a) Female and male thallus of a liverwort:

(b) Funaria, gametophyte and sporophyte: -

(c) Alternation of generation in Angiosperm:


NCERT Exemplar Class 11 Biology chapter 3 solutions will introduce the reader to several concepts which form the foundation of Botany, a subcategory of biology. Students looking for NCERT exemplar Class 11 Biology solutions chapter 3 pdf download must know that this function is to be available soon. Till then, they can refer to this webpage or use an online webpage to pdf tool to save this page as a pdf.

NCERT Exemplar Solutions for Class 11 Biology Chapter 3 Main Subtopics:

  • ALGAE
  • Chlorophyceae
  • Phaeophyceae
  • Rhodophyceae
  • BRYOPHYTES
  • Liverworts
  • Mosses
  • PTERIDOPHYTES
  • GYMNOSPERMS
  • ANGIOSPERMS
  • Plant Life Cycles and Alternation of Generations

What the students will learn in NCERT Exemplar Solutions for Class 11 Biology Chapter 3?

A detailed study of NCERT Exemplar Class 11 Biology solutions chapter 3 will enable the students to grasp the core requirement in the study of plant. They will learn the different modes of reproduction in plants- asexual, sexual and vegetative reproduction. They will learn Bryophytes which live on land but depend on water to reproduce sexually.

Further in class 11 Biology NCERT exemplar solutions chapter 3, they will learn the male and female sex organs called antheridia and archegonia, respectively. They will also learn sporophytes that produce spores. They will also be introduced to the functions of a flower which acts as a reproductive organ in angiosperms.

The angiosperms are divided into monocotyledons and dicotyledons which is dealt in great detail in the coming chapter. The students will also learn the different life cycles of plants and the various factors influencing these lifecycles, either of a single plant or a group of plants.

NCERT Exemplar Class 11 Biology Solutions Chapter Wise

NCERT Exemplar Class 11 Biology Solutions Chapter 3 Important Topics

Class 12 Chemistry NCERT exemplar solutions chapter 3 has detailed that Algae and its types, Bryophytes and its types, Gymnosperms, Angiosperm, methods of reproduction and Plant life-cycles and alternation of generations are important topics which students should pay extra attention to.

· A thorough understanding of class 11 Biology NCERT exemplar solutions chapter 3 will enable students to develop a genuine interest in this subject and fuel their spirit of research and inquiry.

NCERT Exemplar Class 11 Solutions

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Frequently Asked Questions (FAQs)

1. What will the students learn from this chapter?

Students will learn in detail about the plant kingdom and the phylums included in it.  This chapter includes all phyla like bryophytes, algae, pteridophytes etc.

2. How are these questions solved?

These questions are solved in the most detailed manner which includes, pointers, features, diagrams and theory wherever needed.

3. Can one download these solutions?

Yes, one can easily download these NCERT exemplar Class 11 Biology solutions chapter 3 pdf download to their computers or devices so the PDF can be used offline as well.

4. How many questions are solved in this chapter?

All the 12 questions mentioned in the main exercise of the chapter are solved in complete detail in NCERT Exemplar Class 11 Biology chapter 3 solutions.

5. What pattern is used in the NCERT Exemplar Solutions?

Latest  NCERT and CBSE pattern is used for solving the questions.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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