NCERT Exemplar Class 11 Biology Solutions Chapter 5 Morphology of Flowering Plants

NCERT Exemplar Class 11 Biology Solutions Chapter 5 Morphology of Flowering Plants

Edited By Priyanka kumari | Updated on Aug 16, 2022 05:12 PM IST

In NCERT Exemplar Class 11 Biology chapter 5 solutions we will learn about the various classifications of plants. In this chapter, we shall look into the morphology of plants, i.e. the structure of plants. The pioneer in this field was Katherine Esau, a Ukrainian woman who made a tremendous contribution in this field of study. Each plant has certain features that perform certain functions which enable the plant to thrive in its environment. In NCERT Exemplar Class 11 Biology solutions chapter 5, we shall study each feature of a flowering plant in detail, starting from its roots and up to the flower itself. We shall also study the formation of seeds and fruits to a great extent in the NCERT Exemplar Class 11 Biology chapter 5 solutions.

By utilising the NCERT Exemplar Class 11 Biology chapter 5 solutions PDF Download function students can make learning even more convenient as they will get access to quality study material effectively constructed by experts for the best learning experience.

NCERT Exemplar Class 11 Biology Chapter 5 Solutions:

Multiple Choice Questions:

Question:1

Rearrange the following zones as seen in the root in vertical section and
choose the correct option.
A. Root hair zone
B. Zone of meristems
C. Rootcap zone
D. Zone of maturation
E. Zone of elongation
Options:
(a) C, B, E, A, D
(b) A, B, C, D, E
(c) D, E, A, C, B
(d) E, D, C, B, A

Answer:

Ans. The answer is the option (a) C, B, E, A, D
Explanation: Roots cap is at the tip; followed by a zone of meristems. This is followed by the zone of elongation and then the root hair zone. Zone of maturation comes after all of them.


Question:2

In an inflorescence where flowers are borne laterally in an acropetal succession, the position of the youngest floral bud shall be
(a) Proximal
(b) Intercalary
(c) Distal
(d) Any where

Answer:

Ans. The answer is the option (c) Distal
Explanation: Acropetal succession is an arrangement of flowers when the youngest flower is at the top of the floral axis.


Question:3

The mature seeds of plants such as gram and peas, possess no endosperm,because
(a) Endosperm gets used up by the developing embryo during seed development
(b) There is no double fertilization in them
(c) Endosperm is not formed in them
(d)These plants are not angiosperms

Answer:

Ans. The answer is the option (a) Endosperm gets used up by the developing embryo during seed development

Explanation: During or after germination the developing embryo derives its nourishment from the endosperm, that is why plants do not possess endosperm.


Question:4

Roots developed from parts of the plant other than radicle are called
(a) Taproots
(b) Adventitious roots
(c) Fibrous roots
(d) Nodular roots

Answer:

Ans. The answer is the option (b) Adventitious roots
Explanation: Taproots and fibrous roots come out from the radicles; nodular roots emerge from root nodules. Whereas the adventitious roots generally emerge from the stems and sometimes from the leaves.


Question:5

Venation is a term used to describe the pattern of arrangement of
(a) Veins and veinlets in a lamina
(b) Flower in inflorescence
(c) Floral organs
(d) All of them

Answer:

Ans. The answer is the option (a) Veins and veinlets in a lamina

Explanation: The veins of the leaves which carry the food and water are designed in a complex net-like structure. This complex arrangement of veins in a leaf is called venation. Venation is of two types based on the arrangement of veins and veinlets.


Question:6

Endosperm, a product of double fertilization in angiosperms is absent in the seeds of
(a) Coconut
(b) Maize
(c) Orchids
(d) Castor

Answer:

Ans. The answer is the option (c) Orchids
Explanation: The embryo in the orchid plant absorbs the endosperm during growth.


Question:7

Many pulses of daily use belong to one of the families below (tick the correct answer)
(a) Solanaceae
(b) Poceae
(c) Liliaceae
(d) Fabaceae

Answer:

Ans. The answer is the option (d) Fabaceae
Explanation: Leguminosae and pulses are referred as Fabaceae and are also known as legumes.


Question:8

The placenta is attached to the developing seed near the
(a) Hilum
(b) Testa
(c) Micropyle
(d) Chalaza

Answer:

Ans. The answer is the option (a) Hilum
Explanation: The scar on the seed coat through which the seed attaches it to the fruit is known as Hilum. Chalaza is the base of the ovule, Testa is the outer covering of the seed, and lastly, micropyle is a tiny opening in the seed coat through which water gets absorbed in the seed.


Question:9

Which of the following plants is used to extract the blue dye?
(a) Trifolium
(b) Lupin
(c) Indigofera
(d) Cassia

Answer:

Ans. The answer is the option (b) Indigofera
Explanation: ‘Indigo’ is a blue colour. And Indigofera is used for blue dye.


Question:10

Match the following and choose the correct option

Column IColumn II
A. Aleurone layeri. without fertilization
B. Parthenocarpic fruitii. Nutrition
C. Ovuleiii. Double fertilization
D. Endospermiv. Seed


Options:

a. A-i, B-ii, C-iii, D-iv

b. A-ii, B-i, C-iv, D-iii

c. A-iv, B-ii, C-i, D-iii

d. A-ii, B-iv, C-i, D-iii

Very Short Answer Type Questions:

Question:1

Roots obtain oxygen from air in the soil for respiration. In the absence or deficiency of O2, root growth is restricted or completely stopped. How do the plants growing in marshlands or swamps obtain their O2 required for root respiration?

Answer:

The plants which are growing in marshlands or swamps have vertical growths from their roots. These roots are hollow from within, and thus they make way for air to enter through these roots. These are known as pneumatophores. In consequence, pneumatophores assist the plants in swamps and marshland areas to acquire oxygen for root respiration.

Question:3

In Opuntia the stem is modified into a flattened green structure to perform the function of leaves (i.e., photosynthesis). Cite some other examples of modifications of plant parts for the purpose of photosynthesis.

Answer:

Ans.
Euphorbia is a large genus of flowering plants, also known as spurge. The stem of Euphorbia is fleshy and cylindrical in structure. The stem in this plant helps to carry out photosynthesis because of the presence of chlorophyll in the stem.


Question:4

In swampy areas like the Sunderbans in West Bengal, plants bear a special kind of roots called _____.

Answer:

Ans. Pneumatophores, which helps for root respiration.


Question:5

In aquatic plants like Pistia and Eichhornia, leaves and roots are found near _____.

Answer:

Ans. Leaves and roots are found near the surface of the water.



Question:7

Which parts in ginger and onion are edible?

Answer:

The underground stem in ginger is good for consumption, whereas the altered leaves are edible in the case of onion.


Question:10

Name the body part modified for food storage in the following

(a) Carrot

(b) Colocasia

(c) Sweet potato

(d) Asparagus

(e) Radish

(f) Potato

(g) Dahlia

(h) Turmeric

(i) Gladiolus

(j) Ginger

(k) Portulaca

Answer:

Ans.
(a) Tap Root
(b) Stem
(c) Adventitious Roots
(d) Root
(e) Tap Root
(f) Stem
(g) Adventitious Roots
(h) Stem
(i) Stem
(j) Stem
(k) Adventitious Roots


Short Answers Type Questions:

Question:1

Give two examples of roots that develop from different parts of the angiospermic plant other than the radicle.Answer:

Ans.
Adventitious roots are the term used when the roots develop from different plant parts in angiosperm. Two examples of adventitious roots are: -
  • Prop Roots in Banyan: The roots of a banyan tree is an example for prop roots. These prop roots deeply penetrate the soil and help the tree to maintain the balance and support itself.
  • Stilt Roots in Maize: The roots in maize plants are an example for stilt roots.These roots enter the ground and are developed from the nodes just above the ground; they provide extra support to the maize plant.


Question:2

The essential functions of roots are anchorage and absorption of water and minerals in the terrestrial plant. What functions are associated with the roots of aquatic plants? How are the roots of aquatic plants and terrestrial plants different?

Answer:

Ans. As known by all, the basic function of roots is that of anchorage and absorption of water and minerals. Availability of water is not an issue for aquatic plants. That is why the roots are not properly grown in most of the aquatic plants. In these free-floating plants, the roots are very few in number and are in a structure of fine hairs. In the plants that are submerged, roots play the important role of anchorage.

Question:3

Draw diagrams of a typical monocot and dicot leave to show their venation pattern.

Answer:

Ans. Reticular Venation

Parallel Venation


Question:4

A typical angiosperm flower consists of four floral parts. Give the names of the floral parts and their arrangements sequentially.

Answer:

Ans. The arrangement of four floral parts is as follows; starting from the periphery of the flower: Calyx, Corolla, Androecium And Gynoecium.
Calyx is the part of the flower which is made up of green leaf-like sepals. Corolla is the part of the flower which is composed of colourful petals. Androecium is the part of the flower which consists of stamens (filament and anther). Gynoecium is the part of the flower which is composed of carpel (stima, style and ovary).

Question:6

Reticulate venation is found in dicot leaves while in monocot leaves venation is of parallel type. Biology being a ‘Science of exceptions’, find out any exception to this generalization.

Answer:

Ans. These are some examples of exception to this generalisation:

  • Parallel venation in dicot: Calophyllum, Corymbium
  • Reticulate venation in monocot: Alocasia, Smilax

Question:7

You have heard about several insectivorous plants that feed on insects. Nepenthes or the pitcher plant is one such example, which usually grows in shallow water or in marshlands. What part of the plant is modified into a 'pitcher'? How does this modification help the plant for food even though it can photosynthesize like any other green plant?

Answer:

Ans. The leaves of the pitcher plant are modified into a pitcher. The pitcher is facilitated with a lid which is developed from the tip of the lamina. The anterior part of the petiole, which is coiled like a little tendril. It helps to keep the pitcher in a vertical position.

Question:8

Mango and coconut are ‘drupe’ type of fruits. In mango fleshy mesocarp is edible. What is the edible part of coconut? What does milk of tender coconut represent?

Answer:

The endosperm is the edible part of a coconut. The oily endosperm in the liquid form constitutes the milk of tender coconut. After some time, this milk becomes flesh as it gets deposited along the walls of the endocarp.


Question:9

How can you differentiate between free central and axile placentation?

Answer:

Placentation: It is regarded as free central when the septa is absent, and the ovules are born on the central axis.

Placentation: It is regarded as axile when the placenta is axial, and the ovules are attached to the placenta in a multilocular ovary.

Question:10

Tendrils are found in the following plants. Identify whether they are stem tendrils or leaf tendrils.

a. Cucumber

b. Peas

c. Pumpkins

d. Grapevine

e. Watermelons

Answer:

a. Cucumber – Stem tendrils

b. Peas – Leaf tendrils

c. Pumpkins – stem tendrils

d. Grapevines – stem tendrils

e. Watermelons – stem tendrils

Question:11

Why is maize grain usually called as a fruit and not a seed?

Answer:

The maize grain is generally known as a fruit and not seed as it is a ripened ovary that contains the ripened ovule.

Question:12

Tendrils of grapevines are homologous to the tendril of pumpkins but are analogous to that of a pea. Justify the above statement.

Answer:

Tendrils of pumpkins are homologous to the tendrils of grapevines because both originate from the same part of the plan i.e., the stem despite having different functions. But the tendrils of grapevines and tendrils of pumpkins are analogous to that of a pea because the function of tendrils on pumpkin is to creep while in grapevines it is to climb.

Question:13

The rhizome of ginger is like the roots of other plants that grow underground. Despite this fact, ginger is a stem and not a root. Justify.

Answer:

The rhizome of ginger is like the roots of other plants that grow underground. Despite this fact, ginger is a stem and not a root. Because ginger is not a root but steam as it has internodes and nodes that roots do not possess.


Question:14

Differentiate between

a. Bract and Bracteole

b. Pulvinus and petiole

c. Pedicel and peduncle

d. Spike and spadix

e. Stamen and staminoid

f. Pollen and pollenium

Answer:

a. The bracteolate is found between the bract and flower while a bract is found towards the base of the pedicle.

b. Petiole is a subcylindrical stalk that connects the lamina with the leaf base while Pulvinus is a swollen leaf base which is found in leguminous plants.

c. The Peduncle is a stalk of the whole inflorescence while a pedicle is the stalk of a flower.

d. Spadix is covered by several large bracts known as spates while in spike sessile flowers are attached on an elongated peduncle.

e. Every stamen represents a male reproductive organ and androecium is composed of stamens. The staminode is a sterile stamen.

f. Polonium is defined as the group of pollen grains while a pollen is a male gametophyte of angiosperms.

Long Answer Type Questions:

Question:1

Distinguish between families Fabaceae, Solanaceae, Liliaceae on the basis of gynoecium characteristics (with figures), Also write economic importance of any one of the above family

Answer:

Gynoecium CharacteristicsSolanaceaeLiliaceaeFabaceae
OvulesMany in each locularTrilocularMargin in two alternate rows
PlacentationAxile2-many ovules in each loculesMarginal
OvarySuperior bilocular(2-4locular in tomato)SuperiorSuperior unilocular
CarpelsBicarpellarySyncarpousTricarpellarysyncarpousMonocarpellary free, single
StyleSimpleSimple but may be united or separateBent, single








The economic importance of Fabaceae can be specified as follows: -
  • Source of pulses
  • Medicine
  • Edible oil
  • Dye
  • Fibres
  • Fodder
  • Ornamental
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Question:2

Describe various stem modifications associated with food storage,climbingand protection.

Answer:

The various stem modifications that are associated with food storage climbing, and protection are as follows: -
  • For storage of food, underground stems of ginger, turmeric, potato, and colocasia are modified. These stems also perform the function of organs of perennation that helps get over unfavourable growth conditions.
  • Developed from axillary buds, stem tendrils are dare spirally coiled and are slender which helps the plants to climb. Some examples include grapevines, pumpkins, cucumber, and watermelon etc.
It is also possible for axillary buds of the stems to be modified into straight, woody and pointed thorns. Some of the famous examples being Citrus, Bougainvillea. Thorns help to protect the plants from animals.


Question:3

Stolon, offset and rhizome are different forms of stem modifications. How can these modified forms of stem be distinguished from each other?

Answer:

OffsetStolon

Rhizome
It is a lateral branch that has short intermodes. Each node bears a tuft of roots and rosette of leaves which are often found in aquatic plants.
For eg: - Eichhornia and Pistia
This form of stem is a creeper. Which is in the form of thicker internodes and a long distance runner with horizontal branches arising from the internodes,
For e.g.: - Strawberry
It is an underground stem which grows parallel to the surface of the soil.
For e.g.: - Ginger and Turmeric.


Question:5

The arrangements of ovules within the ovary is known as placentation. What does the term placenta refer to? Name and draw various types of placentations in the flower as seen in T.S. or V.S

Answer:

The placenta is flattened, soft pillow-like tissue on which the ovules get attached. The placenta supplies the developing embryo with nutrients.


Question:6

Sunflower is not a flower. Explain.

Answer:

Ans. Sunflower is actually not a flower, but a bunch of flowers, altogether. Inflorescence is the term used for a bunch of exceedingly small flowers, for what it appears to be a single flower. These flowers are put on a receptacle in a special arrangement. This type of inflorescence of flowers is called capitulum.

These are some characteristics of capitulum: -

  • There are many tiny, sessile florets present on the receptacle.
  • The youngest florets are present at the center, while the oldest florets are at the periphery.
  • The receptacle is flattened.
  • There are two kinds of florets in a sunflower, viz. ray florets and disc florets.
  • Ray florets are present on the rim of the receptacle. These are of yellow colour.
  • Disc florets are present in the center. They are bisexual in nature and are actinomorphic.

Question:7

How do you distinguish between hypogeal germination and epigeal germination? What is the role of cotyledon (s) and the endosperm in the germination of seeds?

Answer:

Epigeal GerminationHypogeal Germination

Epigeal germination is described as the type of germination when the hypocotyl grows initially. It then pushes the cotyledons and other parts of the seed out from the soil.Hypogeal germination is described as the type of germination when the epicotyl grows first and only the plumule is pushed out from the soil. But the other parts including the cotyledons remain under the soil.

This type of germination takes place in Onion, Castor, Mustard etc.This type of germination takes place in Monocots and some dicots.


Question:8

Seeds of some plants germinate immediately after shedding from the plants while in other plants they require a period of rest before germination. The latter phenomenon is called dormancy. Give the reasons for seed dormancy and some methods to break it.

Answer:

Reasons for seed dormancy:

The seed coat is hard, and the impermeable layers of the seeds prevent uptake of water, this inhibits growth and delays the process of germination. In some cases, the seeds have undeveloped embryos which is the reason for the delay in germination.

Methods to break seed dormancy:
Seed dormancy can be broken by changes in temperature of the environment and other conditions. Physical dormancy can be also broken when the seed goes through the gut of an animal who has consumed it. Mechanical dormancy is broken with the help of some external factors. Example: An animal might break the hard coat of the seed.

Main Subtopics of NCERT Exemplar Class 11 Biology Chapter 5 Solutions Morphology of Flowering Plants:

  • 5.1 The Root
  • 5.1.1 Regions of the Root
  • 5.1.2 Modifications of Root
  • 5.2 The Stem
  • 5.2.1 Modifications of Stem
  • 5.3 The Leaf
  • 5.3.1 Venation
  • 5.3.2 Types of Leaves
  • 5.3.3 Phyllotaxy
  • 5.3.4 Modifications of Leaves
  • 5.4 The Inflorescence
  • 5.5 The Flower
  • 5.5.1 Parts of a Flower
  • 5.5.1.1 Calyx
  • 5.5.1.2 Corolla
  • 5.5.1.3 Androecium
  • 5.5.1.4 Gynoecium
  • 5.6 The Fruit
  • 5.7 The Seed
  • 5.7.1 Structure of a Dicotyledonous Seed
  • 5.7.2 Structure of Monocotyledonous Seed
  • 5.8 Semi-technical Description of a Typical Flowering Plant
  • 5.9 Description of Some Important Families
  • 5.9.1 Fabaceae
  • 5.9.2 Solanaceae
  • 5.9.3 Liliaceae

What will the students Learn using the NCERT Exemplar Solutions for Class 11 Biology Chapter 5?

  • NCERT Exemplar Class 11 Biology solutions chapter 5 will teach the reader about various shapes and sizes of the various features of a flowering plant in detail using a diagram. They will learn about the variations of such plants in terms of size, structure, method of gaining nutrition and habitat. There are two types of roots which are tap and fibrous.
  • Further, they will also learn which types of plants possess which kind of roots. Further, the morphological features of the stem-like nodes and the internodes will be taught to them.
  • Using the NCERT Exemplar Class 11 Biology solutions chapter 5 students will learn why the leaf possesses the colour it has and what function it performs. Moreover, they will understand how different leaves get modified into different features during its lifecycle.

NCERT Exemplar Class 11 Biology Solutions Chapter Wise:

Important Topics in NCERT Exemplar Class 11 Biology Chapter 5 Solutions Morphology of Flowering Plants

· Class 11 Biology NCERT Exemplar solutions chapter 5 has detailed that digestive system, digestive glands, absorption of digested products(nutrients), digestive tract and disorders of the digestive system are important topics which students should pay extra attention to.

· The students will learn that the flower is a modified shoot that helps in sexual reproduction. These flowers come in a variety of shapes and colour and all of them serve a unique purpose which the student will understand after studying NCERT Exemplar solutions for Class 11 Biology chapter 5.

· They will also learn of the process of formation of fruits and how seeds get fertilized for this purpose. This chapter will enable the student to build a strong foundation in the field of Botany which will help him as an aspiring scientist in this area.

NCERT Exemplar Class 11 Solutions

Check Chapter-Wise Solutions of Questions Given in Book:

Also, Read NCERT Solution subject wise

Must Read NCERT Notes subject wise

Also Check NCERT Books and NCERT Syllabus here

Frequently Asked Questions (FAQs)

1. What makes this chapter important?

Those students who want to have a career in the field of botany or pharmacological science should make sure that they understand every topic with utmost clarity.

2. Are questions solved as per the CBSE pattern?

Yes, these NCERT Exemplar Class 11 Biology Chapter 5 Solutions are designed keeping in mind the pattern and marking scheme of CBSE.

3. How were the solutions for this chapter prepared?

These NCERT Exemplar Class 11 Biology Chapter 5 Solutions are made by the highly experienced biology teachers of ours, who have CBSE teaching experience.

4. Are these solutions helpful from the board exam point of view?

Yes, these solutions are prepared in a manner in which the student should answer in the board exams, and also help them in clarifying any doubts they might have.

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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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