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NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis

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NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis.

Question 1.
By looking at a plant externally can you tell whether a plant is C3 or C4 ? Why and how?
Solution:
Plants that are adapted to dry tropical regions have the C4 pathway. They have a special type of leaf anatomy, they tolerate higher temperatures, they show a response to highlight intensities. Study vertical sections of leaves, one of a C3 plant and the other of a C4 plant.

Question 2.
By looking at which internal structure of a plant can you tell whether a plant is C3 or C4? Explain.
Solution:
In C4 plant internal structure of the leaf possess a special type of anatomy called ‘Kranz’ anatomy. ‘Kranz’ means ‘wreath’ and is a reflection of the arrangement of cells.

The bundle sheath cells may form several layers around the vascular bundles; they are characterised by having large number of chloroplasts, thick walls impervious to gaseous exchange and no intercellular spaces.

While in C3 plants, there is no special type of leaf anatomy. There is only a single type of chloroplast inC3 i.e. granal, while in C4 chloroplasts are dimorphic, i.e, granite in the mesophyll cells and agranal in the bundle sheath cells.

Question 3.
Even though very few cells in a C4 plant carry out the biosynthetic-Calvin pathway, yet they are highly productive, can you discuss why?
Solution:
Though these plants have the C4 oxalacetic acid as the first CO2 fixation product they use the C3 pathway or the Calvin cycle as the main biosynthetic pathway.
In C4 plants photorespiration does not occur. This is because they have a mechanism that increases the concentration of CO2 at the enzyme site.
This takes place when the C4 acid from the mesophyll is broken down in the bundle cells to release CO2 this results in increasing the intracellular concentration of CO2 In turn, this ensures that the Rubisco functions as a carboxylase minimizing the oxygenase activity.

Now that you know that the C4 plants lack photorespiration, you probably can understand why productivity and yields are better in these plants. In addition, these plants show tolerance to higher temperatures.

Question 4.
RuBisCO is an enzyme that acts both as carboxylase and oxygenase. Why do you think RuBisCO carries out more carboxylation in C4 plants.
Solution:
RuBisCO or Ribulose bisphosphate carboxylase – oxygenase enzyme can bind to both C02 and O2. This binding is competitive. The relative concentration of C02 and 02 determines which one of the two will bind to the enzyme.

In C4 plants photorespiration does not occur. This is because they have a mechanism that increases the concentration of C02 at the enzyme site.

This takes place when oxaloacetic acid is broken down in the bundle sheath cells to release C02.

It results in increased intracellular concentration of C02. This ensures that the RuBisCO functions as a carboxylase and minimising the oxygenase activity.

Question 5.
Suppose there were plants that had a high concentration of chlorophyll b, but lacked chlorophyll a, would it carry out photosynthesis? Then why do plants have chlorophyll b and other accessory pigments?
Solution:
Though chlorophyll is the major pigment responsible for trapping light, other thylakoid pigments like chlorophyll b, xanthophylls, and carotenoids, which are called accessory pigments, also absorb light and transfer the energy to ‘chlorophyll a’.

Indeed, they not only enable a wider range of wavelengths of incoming light to be utilized for photosynthesis but also protect ‘chlorophyll a’ from photo-oxidation. Reaction centre chlorophyll-protein complexes are capable of directly absorbing light and performing charge separation events without other chlorophyll pigments but the absorption cross-section is small.

Question 6.
Why is the colour of a leaf kept in the dark frequently yellow, or pale green? Which pigment do you think is more stable?
Solution:
Chlorophyll is unable to absorb energy in the absence of light and loses its stability, giving the leaf a yellowish colour. This shows that xanthophyll is more stable.

Question 7.
Look at leaves of the same plant on the shady side and compare it with the leaves on the sunny side. Or, compare the potted plants kept in the sunlight with those in the shade. Which of them has leaves that are darker green? Why?
Solution:
Light is a limiting factor for photosynthesis Leaves get lesser light for photosynthesis when they are in shade. Therefore, the leaves or plants in shade perform lesser photosynthesis as compared to the leaves or plants kept in sunlight. In order to increase the rate of photosynthesis, the leaves present in shade have more chlorophyll pigments.

This increase in chlorophyll content increases the amount of light absorbed by the leaves, which in turn increases the rate of photosynthesis. Therefore, the leaves or plants in shade are greener than the leaves or plants kept in the sun.

Question 8.
The figure shows the effect of light on the rate of photosynthesis. Based on the graph, answer the following questions.
(a) At which point/s (A, B, or C) in the curve is light a limiting factor?
(b) What could be the Jimiting factor/s in region A?
(c) What do C and D represent on the curve?
Solution:
NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis 1
(a) In the region ‘A’ and half of ‘BTight is limiting factor because rate of photosynthesis is increasing with the intensity of light.
(b) All the other factors except light.
(c) C represents a region where a factor other than light is limiting, e.g., CO2. D represents the light intensity at which rate of photosynthesis is maximum under existing conditions (e.g., CO2).

Question 9.
Give a comparison between the following:
(a) C3 and C4 pathways
(b) Cyclic and non-cyclic photophosphorylation
(c) Anatomy of leaf in C3 and C4.
Solution:
(a) Differences between C3 and C4 pathway
NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis 2
(b) Differences between cyclic and non-cyclic photophosphorylation are
NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis 3
(c) Differences between the anatomy of leaf in C3 plants and anatomy of leaf in C4 plants are
NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis 4

VERY SHORT ANSWER QUESTIONS

Question 1.
Write one anatomical feature of C4 plants.
Solution:
Kranz anatomy in leaf.

Question 2.
Which of the following is not a useful function of the light reaction in photosynthesis?
(a) splitting water
(b) synthesis of NADPH
(c) converting light energy into chemical energy
(d) releasing oxygen for photorespiration
Solution:
(d) Releasing oxygen for photorespiration.

Question 3.
What is the starting substance in the CO2 fixation cycle? (Apr. 91)
Solution:
RuMP.

Question 4.
Where is PS II located in a chloroplast?
Solution:
PS II is located in the appressed regions of grana thylakoid

Question 5.
Name the reaction centre of PS I and PS II.
Solution:
P700 & P680‘

Question 6.
What type of light causes maximum photo-synthesis? (Oct. 1995)
Solution:
Red light

Question 7.
How many molecules of ATP and how many molecules of NADPH are spent to fix three molecules of CO2 in the Calvin cycle?
Solution:
9 ATP and 6 NADPH

Question 8.
Why do the stomata of CAM plants open during the night?
Solution:
As these plants grow in dry areas, they keep stomata close during the day to conserve water and open their stomata during the night for the diffusion of gases.

Question 9.
Mention one useful role of photorespiration in plants.
Solution:
It protects the plants from photooxidative damage.

Question 10.
Cyanobacteria and some other photosynthetic bacteria don’t have chloroplasts. How do they conduct photosynthesis?
Solution:
Cyanobacteria have bluish pigment phycocyanin, which they use to capture light for photosynthesis. Some green bacteria (cyanobacteria) are red or pink due to pigment phycoerythrin. Whatever the colour of cyanobacteria, they are photosynthetic and so can manufacture food.

Question 11.
What is phosphorylation? (M.Q.P.)
Solution:
Synthesis of ATP either with the help of light (during photosynthesis) or in presence of oxygen (during respiration) is called phosphorylation.

Question 12.
Name the organism Englemann used in his experiment.
Solution:
Cladophora.

Question 13.
Write the currently accepted equation of photosynthesis in plants.
Solution:
NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis 5

Question 14.
What is a pigment?
Solution:
A pigment is a substance that absorbs light of certain wavelength(s).

Question 15.
Write the full form of NADP
Solution:
NADP – Nicotinamide adenine dinucleotide Phosphate.

Question 16.
Expand RuBP
Solution:
Ribulose 1, 5 bisphosphates.

Question 17.
Give a reason for the following:
Some bacteria exhibit photosynthesis but they do not produce oxygen. (July 2006)
Solution:
Some photosynthetic bacteria do not use water as their source of hydrogen, hence do not liberate oxygen.

Question 18.
Mention two conditions where light can become a limiting factor.
Solution:
Conditions in which light can become a limiting factor:
(i) Plants in the shade.
(ii) Plants growing under the canopy in a dense forest.

Question 19.
What are antenna molecules?
Solution:
Antenna molecules are light-harvesting pigment molecules that occur on the outer side of a photosynthetic unit.

Question 20.
What is a quantasome? Where is it present?
Solution:
Quantasome means photosynthetic units. It is equivalent is 230 chlorophyll molecules. These are present in the grana lamellae.

SHORT ANSWER QUESTIONS

Question 1.
Specify how C4 photosynthetic pathway increases carbon dioxide concentration in bundle sheath cells of sugarcane?
Solution:
In C4 pathway of sugarcane, C02 from atmosphere enters through the stomata in the mesophyll cell and combines with phosphoenol pyruvate to form a 4-C compound oxaloacetic acid. The OAA is then transported to the bundle sheath where it is decarboxylatedto release C02 in bundle sheath.

Question 2.
Differentiate between absorption spectrum and action spectrum.
Solution:
The main differences between absorption spectrum and action spectrum are as follows.
NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis 6

Question 3.
What are quantasomes? (Oct. 94)
Solution:
Quantasome is a functional unit (Photo-synthetic unit) made of a group of pigment molecules required for carrying out a photochemical reaction. The Pigment molecules are embedded in the grana and differentiated as pigment system I (with chi 670, chi 680, P 700) and pigment system ll(with chi 670, chi 680, P 680, and Xanthophylls)

Question 4.
Distinguish between cyclic and non-cyclic photophosphorylation. (M.Q.P., March 2011)
Solution:
Non cyclic photophosphorylation (a) Cyclic photophosphorylation

  1. The path traversed by an electron is non-cyclic.
    (a) Path traversed by electron is cyclic.
  2. Both PSI and PSII are active.
    (b) Only PSI is active.
  3. It is accompanied by photolysis.
    (c) No photolysis.
  4. The major pathway that takes place.
    (d) Secondary pathway when additional ATP is needed.

Question 5.
What is Blackman’s law of limiting factors?
Solution:
F.F. Blackman (1905) extended a law to formulate the principle of limiting factors. “When a process is conditioned as to its rapidity by a number of separate factors, the rate of the process is limited by the pace of slowest factors.”

Question 6.
In the condition of water stress why the rate of photosynthesis declines?
Solution:
Due to water stress, stomata remain closed and so there is a decrease in CO2concentration and the leaf water potential is also reduced, decline the rate of photosynthesis.

Question 7.
What is a reaction centre? Give the reaction centres of PSI and PSII.
Solution:
Reaction centre is a chlorophyll component of the photosystem and it absorbs as well as accepts energy from other pigments and ejects an electron. The reaction centre of PSII is Chla680 or P680 and PSI is Chla700 or P700

Question 8.
Why is photorespiration considered a wasteful process?
Solution:
Photorespiration considered a wasteful process because
(i) 25% of photosynthetically fixed carbon is lost in the form of C02.
(ii) There is no energy-rich useful compound produced during this process.

Question 9.
Give two reasons as to why photosynthesis is important for sustaining life on earth.
Solution:
Photosynthesis is the most important process because;
(i) it is the only natural process by which oxygen is liberated into the atmosphere.
(ii) it is the process by which food is manufactured for all living organisms.

Question 10.
Why does the rate of photosynthesis decrease at higher light intensities? What plays a protective role in such situations?
Solution:
Rate of photosynthesis decreases for two reasons :
(i) Other factors required for photosynthesis become limiting.
(ii) Destruction of chlorophyll by photo-oxidation.
Carotenoids play a protective role by:
(i) absorbing the excess light and
(ii) acting as an antioxidant to detoxify the effect of activated oxygen species.

Question 11.
What is C4 -pathway? Give an example. (March 2008)
Solution:
CA -pathway is an alternative photosynthetic pathway seen in plants like sugarcane/sorghum/ maize in which the stable compound is oxaloacetate a 4-C compound. It is called the Hatch-slack pathway.

Question 12.
What is kranz anatomy in plants?
Solution:
In Kranz Anatomy vascular bundles are surrounded by a layer of bundle sheath that contains a large number of chloroplasts in mesophyll cells and it is present in C4 plants e.g, Maize, Sugarcane, etc.

LONG ANSWER QUESTIONS

Question 1.
How is photosystem I different from photosystem II?
Solution:
The main differences between photosystem I and photosystem II are
NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis 7

Question 2.
Describe the factors that influence the rate of Photosynthesis. (Oct. 1989)
Solution:
The factors that affect photosynthesis may be both internal & External.

Internal factors:

  • Chlorophyll: it is the light-absorbing pigment and only portions of the plant having chlorophyll can help in photosynthesis.
  • Protoplasmic factor: young seedlings when transferred from darkness to light show the presence of some factors which is believed to be enzymatic initiates photosynthesis and is called the protoplasmic factor.

External factors:

  • Light: It is one of the most important factors which affects the process in 3 ways i.e. quantity, quality, and intensity. Quantity of light is otherwise duration and depends upon the photoperiod that is required by the plant quality refers to the wavelength, maximum photosynthesis occurs in red and blue light while minimum in green light. Intensity favours the process and low intensity decreases the rate of photosynthesis. Very high intensity brings about photooxidation of pigments which is called solarization.
  • CO2: An increase in CO2 concentration favours the process provided other factors are not limiting but very high concentrations are toxic and inhibit photosynthesis.
  • Temperature: Increase in temperature favour photosynthesis but above the optimum range the process decreases due to the denaturation of enzymes.

Question 3.
Explain the process of the biosynthetic phase of photosynthesis occurring in the chloroplasts.
Solution:
The biosynthetic phase of photosynthesis :

  • It occurs in the stroma of chloroplasts.
  • These reactions reduce the carbon dioxide into carbohydrates, making use of the ATP and NADPH2 produced in the photochemical reactions.
  • The reactions are also called as Calvin cycle.
  • The three phases of the Calvin cycle are as follows:

(i) Carboxylation
Six molecules of Ribulose 1,5 bisphosphate
react with six molecules of carbon dioxide to form six molecules of a short-lived 6C- compound.
The reaction is catalysed by RuBP carboxylase (RuBisCo).
The six molecules of the 6C-intermediate break into 12 molecules of 3- phosphoglyceric acid (3-PGA), an SC- compound.
It is through this step that carbon dioxide is fixed in the plant.
NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis 8
(ii) Reduction
12 molecules of 3-phosphoglyceric acid are converted into 12 molecules of 1, 3 diphosphate-glyceric acid, utilising 12 molecules of ATP and then reduced to 3- phosphoglyceraldehyde making use of 12 molecules of NADPH. Two molecules of phosphoglyceraldehyde react to form one molecule of glucose. It is in this step that there is an actual reduction of carbon dioxide leading to sugar formation.

(iii) Regeneration of RuBP
10 molecules of phosphoglyceraldehyde, by a series of complex enzyme-catalyzed reactions, are converted into six molecules of ribulose 1,5-bisphosphate; six molecules of ATP are needed for this step. This step of ‘ regeneration of RuBP is important for the cycle to continue

We hope the NCERT Solutions for Class 11 Biology at Work Chapter 13 Photosynthesis, helps you. If you have any query regarding NCERT Solutions for Class 11 Biology at Work Chapter 13 Photosynthesis, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants

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NCERT Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants.

Question 1.
State the location and function of different types ofmeristems.
Solution:
NCERT Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 1

Question 2.
Cork cambium forms tissues that form the cork. Do you agree with this statement? Explain.
Solution:
Sooner or later, another meristematic tissue called cork cambium or phellogen develops, usually in the cortex region. Cork cambium is a couple of layers thick. It is made of narrow, thin-walled, and nearly rectangular cells. Cork cambium cuts off cells on both sides. The outer cells differentiate into cork or phellem while the inner cells differentiate into secondary cortex or phelloderm. The cork is impervious to water due to suberin deposition in the cell wall. The cells of the secondary cortex are parenchymatous. Phellogen, phellem, and phelloderm are collectively known as periderm.

Question 3.
Explain the process of secondary growth in the stems of woody angiosperms with the help of schematic diagrams. What is its significance?
Solution:
NCERT Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 2
showing secondary growth.
Secondary growth in dicot stem:

  1. It is a “permanent increase in thickness due to the activity of vascular cambium and cork cambium in stellar and extrasolar regions”. In dicot stem intra fascicular cambium is present.
  2. The cells of the medullary ray become meristematic and form interfascicular cambium.
  3. These two cambiums unite and make a complete cambial ring.
  4. The cells of it divide and produce new cells both on its outer and inner sides.
  5. The cells formed on the outer side differentiate into secondary phloem while the cells of the inner side form secondary xylem.
  6. The epidermis is replaced by a secondary protective tissue by an increase in the growth of the stem of the plant. It is made of phellogen (cork cambium).
  7. It arises from the peripheral cells of the cortex. The phellogen forms new cells on the outer side which make phellem (cork) and phelloderm on its inner side also.
  8. Significance: Secondary growth increases the girth or thickness of the plant.
  9. Annual rings of woody angiosperms are very distinct and thus helps in determining the age of the plant.

Question 4.
Draw illustrations to bring out the anatomical difference between
(a) Monocot root and dicot root
(b) Monocot stem and dicot stem
Solution:
NCERT Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 3
NCERT Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 4

Question 5.
Cut a transverse section of the young stem of a plant from your school garden and observe it under a microscope. How would you ascertain whether it is a monocot stem or a dicot stem? Give reasons.
Solution:
After observing the transverse section of the stem we can differentiate that stem is monocot or dicot on the basis of the following characters:
NCERT Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 5
Question 6.
The transverse section of a plant material shows the following anatomical features –

  1. the vascular bundles are conjoint, scattered, and surrounded by a sclerenchymatous bundle sheath,
  2. phloem parenchyma is absent What will you identify it as?

Solution:
The transverse section of a typical young monocotyledonous stem shows that

  1. The vascular bundles are conjoint, scattered, and surrounded by sclerenchymatous bundle sheaths
  2. Phloem parenchyma is absent, and water containing cavities are present within the vascular bundles.

Question 7.
Why xylem and phloem are called complex tissues?
Solution:
Xylem and phloem a composed of several types of cells and they work as a unit. Hence they are called complex tissues.

Question 8.
What is the stomatal apparatus? Explain the structure of stomata with a labelled diagram.
Solution:

  1. Several minute openings or stomata are found on the epidermis of all the green aerial parts of plants but are abundant on the lower surface on the leaves as they regulate the process of transpiration.
  2. A large number of stomata occur on the upper surface of leaves of aquatic plants.
  3. Each stomata is surrounded by two cells known as the guard cells. In the dicotyledons plants these are bean-shaped, but in sedges and grasses these are dumb-bell-shaped.
  4. The guard cell is living. Their outer walls are thin where as the inner ones surrounding the aperture are highly thickened.
  5. Due to this variation in the thickening, the guard cell may become turgid and flaccid, depending upon the supply of water in them, which makes the opening and closing of stomata possible.
  6. Some times a few neighbouring epidermal cells in the vicinity of guard cells become specialized in their shape and size and contents. These are known as subsidiary cells.
  7. The stomatal aperture, guard cells and the surrounding subsidiary cell are together called stomatal apparatus.

Question 9.
Name the three basic tissue systems in the flowering plants. Give the tissue names under each system.
Solution:
On the basis of their structure and location, there are three types of tissue systems. These are the

  1. Epidermal tissue system,
  2. The ground or fundamental tissue system and
  3. The vascular or conducting tissue system.

1. Epidermal tissue system The epidermal tissue system forms the outer-most covering of the whole plant body and comprises epidermal cells, stomata, and the epidermal appendages the trichomes, and hairs.
2. All tissues except epidermis and vascular bundles constitute the ground tissue. It consists of simple tissues such as parenchyma, collenchyma, and sclerenchyma.
3. The vascular system consists of complex tissues, the phloem, and the xylem. The xylem and phloem together constitute vascular bundles.

Question 10.
How is the plant anatomy useful to us?
Solution:
‘The study of plant anatomy is useful in many ways. First of all the study helps us understand the way a plant functions carrying out its routine activities like transpiration, photosynthesis, and growth and repair. Second, it helps botanists and agriculture scientists to understand the disease and cure for plants. Plants are important to maintain the ecological balance of the earth, so understanding plant anatomy is a way to understand the large system of the ecology on this planet.

Question 11.
What is periderm? How does periderm formation take place in the dicot stems?
Solution:
Phellogen, phellem, and phelloderm are collectively known as periderm. Phellogen develops, usually in the cortex region. Phellogen is a couple of layers thick. It is made of narrow, thin-walled, and nearly rectangular cells. Phellogen cuts off cells on both sides. The outer cells differentiate into cork or phellem while the inner cells differentiate into secondary cortex or phelloderm. The cork is impervious to water due to suberin deposition in the cell wall. The cells of the secondary cortex are parenchymatous.

Question 12.
Describe the internal structure of a dorsiventral leaf with the help of labelled diagrams.
Solution:
Dorsiventral (dicotyledonous) leaf: The vertical section of a dorsiventral leaf through the lamina shows three main parts, namely, epidermis, mesophyll, and vascular system.
Epidermis: The epidermis which covers both the upper surface (adaxial epidermis) and lower surface (abaxial epidermis) of the leaf has a conspicuous cuticle. The abaxial epidermis generally bears more stomata than the adaxial epidermis. The latter may even lack stomata.
Mesophyll:

  1. The tissue between the upper and the lower epidermis is called the mesophyll.
  2. It possesses chloroplasts and carries out photosynthesis, is made up of parenchyma.
  3. It has two types of cells – the palisade parenchyma and the spongy parenchyma.
  4. The adaxially placed palisade parenchyma is made up of elongated cells, which are arranged vertically and parallel to each other.
  5. The oval or round and loosely arranged spongy parenchyma is situated below the palisade cells and extends to the lower epidermis.
    NCERT Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 11
    6. There are numerous large spaces and air cavities between these cells.

Vascular system:

  • This includes vascular bundles, which can be seen in the veins and the midrib.
  • The size of the vascular bundles is dependent on the size of the veins.
  • The veins vary in thickness in the reticulate venation of the dicot leaves. The vascular bundles are surrounded by a layer of thick-walled bundle sheath cells.

VERY SHORT ANSWER QUESTIONS

Question 1.
Vascular bundles having cambium are known as.
Solution:
Open, Vascular bundle

Question 2.
Name the two types of sclerenchyma.
Solution:
Sclerenchyma fibers and stone cells.

Question 3.
From where do the secondary meristems originate?
Solution:
Permanent tissue.

Question 4.
What does make the root apical meristem subterminal?
Solution:
The presence of the root cap makes the root apical meristem subterminal.

Question 5.
Where are companion cells located in flowering plants? What are their functions?
Solution:
Companion cells are located in phloem cells of vascular tissues, they support the sieve tubes in water conduction.

Question 6.
What is the advantage of lignocellulose in the wall of the xylem?
Solution:
It provides rigidity, thickness, and resistance

Question 7.
A cross-section of a plant material shows the following features under the microscope: vascular bundles are radially arranged. These are found xylem strands showing exarch condition. What type of plant part of is this?
Solution:
Dicot root.

Question 8.
Based on position, classify various types of meristems
Solution:
Apical, intercalary and lateral meristems.

Question 9.
Name the various component cells of xylem. Which of them does not have a nucleus?
Solution:
Tracheids, vessels, xylem parenchyma andxylem fibres. Only xylem parenchyma have nucleus and living.

Question 10.
Give an example of a secondary meristem.
Solution:
Examples of secondary meristem are cork cambium and interfascicular cambium.

Question 11.
Name the tissue involved in linear and lateral growth in plants.
Solution:
Linear growth is caused by apical meristem and lateral growth is caused by lateral meristem.

Question 12.
Heartwood is more durable than springwood. Why?
Solution:
Heartwood is more durable than spring wood due to its little susceptibility to the attack of pathogens and insects.

Question 13.
Where these present:

  1. Hypodermis layer
  2. Mesophyll tissue
  3. Stomata
  4. Cambium

Solution:

  1. Hypodermis layer – is found in stems
  2. Mesophyll tissue – in leaves
  3. Stomata – lower epidermis in leaves
  4. CambiumIn vascular bundles which are open

SHORT ANSWER QUESTIONS

Question 1.
What are the differences between root hairs and stem hairs?
Solution:
The main difference between stem hairs and root hairs are :
.NCERT Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 9

Question 2.
Draw well labelled diagrams of the T.S. of dicotyledonous leaf.
Solution:
NCERT Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 10

Question 3.
Why is cambium considered to be a lateral meristem?
Solution:
Cambium is responsible for the increase in the thickness of stems and roots as a result of the addition of secondary tissues (secondary cortex, secondary phloem and secondary xylem). They are located at the lateral position so-known as lateral meristems.

Question 4.
Name the plant part in which the endodermis is absent. Give one basic difference between the endodermis and epidermis.
Solution:
The endodermis is absent in leaves. Cells of endodermis possess Casparian strips or bands in their radial and transverse walls which are not found in the epidermis.

Question 5.
What are Casparian strips?
Solution:
These are thickenings of lignin and suberin formed around the lateral walls of the endodermis to prevent plasmolysis.

Question 6.
Which tissue is most abundantly found in plants? Where all is it present in plants?
Solution:
The tissue most abundantly found in plants is parenchyma. It is found in pith, cortex, and in entire mesophyll of the leaves.

Question 7.
What is present in the phloem of leaves besides sieve elements and is it living or dead? How are these functional & used?
Solution:
Besides sieve elements, in phloem parenchyma, living cells are present. These store food other cells are phloem fibres that are dead and provide mechanical strength. These are also used in making ropes and coarse textiles.

LONG ANSWER QUESTIONS

Question 1.
Describe the structure and functions of xylem tissues in angiosperm plants.
Solution:

  • Xylem is a complex tissue. It forms a part of the vascular bundle.
  • It is mainly concerned with the conduction of water and minerals. It also provides mechanical support to the plant.
  • As a conducting strand, xylem forms a continuous channel through the roots, stem, leaves and other aerial parts.
  • It consists of four different types of cells—xylem vessels, trachieds, xylem fibres and xylem parenchyma.
  • Xylem vessels and tracheids are concerned with the conduction of water and minerals from roots to aerial parts of the plant.
  • Xylem fibres provide mechanical strength to the plant body. Xylem parenchyma are the only living components of xylem.
  • These are concerned with the storage of food and other vital functions.

Question 2.
What is collenchyma? Explain its structure and function in the plant body of a herbaceous angiosperm.
Solution:

  • The cells of collenchyma are somewhat elongated with cellulose thickening, found as longitudinal strips.
  • These are usually confined to the comers of the cells.
  • Collenchyma cells appear circular, oval or angular in the transverse section. Internally, each cell possesses a large 4 central vacuole, peripheral cytoplasm and a nucleus.
  • Collenchyma is usually found beneath the epidermis in stem, petiole and leaves of herbaceous dicot plants. It is usually absent in monocot stems and monocot roots.

Functions:

  • It provides tensile strength and rigidity to the plants due to thickening.
  • Chloroplasts containing collenchyma cells are responsible for photosynthesis.
  • Collenchyma also provides elasticity to the plant organs.
  • Collenchyma are alive and also stores food.

Question 3.
Explain sclerenchyma with a well labelled diagram.
Solution:
Sclerenchyma is a simple permanent tissue. It consists of two types of cells. They are sclerenchyma fibres and sclereids.
(a) Sclerenchyma fibres –

  • These are much elongated fibers with tapering ends.
  • On ipaturity, they lose their protoplasm and become dead. Their cell wall is made up of cellulose or lignin, or both.
  • Central cavity of the cell is greatly reduced due to the formation of secondary thickening. Sclerenchyma provides mechanical strength to the plants.
  • They help in conduction when present in the secondary xylem.

(b) Sclereids –

  • They develop from ordinary parenchyma cells by the deposition of lignin.
  • These cells are thick-walled and highly lignified and become dead on maturity.
  • They are broader as compared to fibers and their cell lumen is veiy narrow.
  • Sclereids protect the plant from environmental forces like a strong wind.
  • They provide mechanical strength and rigidity to the plant.
    NCERT Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 7

Question 4.
Describe the structure of a monocotyledonous leaf.
Solution:
Anatomy of Monocot/isobilateral leaf: The upper and lower surfaces are covered by a single-layered epidermis.

  • The upper epidermis has some cells larger than the others; such large cells are known as bulliform/motor cells.
  • Stomata are found on both upper and lower epidermal layers. The mesophyll is not differentiated into palisade and spongy parenchyma.
  • Mesophyll cells are isodiametric and are arranged compactly; they contain a number of chloroplasts. Since monocot leaves have parallel veins, a number of vascular bundles can be seen in a row in the section.
  • Each vascular bundle has sclerenchyma cells (caps) on its upper and lower edges.
  • The xylem is on the upper side and the phloem on the lower side. There is a parenchymatous bundle sheath, which often contains chloroplasts and performs the function of photosynthesis.
    NCERT Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 8

Question 5.
Give two examples & salient features of
(1) Simple Tissue
(2) Complex Tissue
Solution:
(1) Simple Tissue:
(i) Parenchyma
(ii) Collenchyma
(2) Complex Tissue:
(i) Xylem
(ii) Phloem
(1) Simple Tissue:
(i) Parenchyma: These are living, thin-walled cells. It is used for storage of food, induction of substances, provides turgidity to softer parts of the plants
(ii) Collenchyma: These are longer than parenchyma. These are living mechanical tissue, it provides mechanical strength to organs and is present in peripheral position in plants to resist bending my the mind.
(2) Complex Tissue:
(i) Xylem: This is also called Hadrome, which is a water-conducting tissue. It is made up of cells like tracheids, xylem fibers, and xylem parenchyma only xylem parenchyma is living and all others are dead.
(ii) Phloem: This is also called Bast, which is a conducting tissue of food from leaves to all parts of the body. The parts of phloem are sieve elements, companion cells, phloem fibres, and phloem parenchyma. Phloem fibres are dead while parenchyma is living. Together these perform their function.

We hope the NCERT Solutions for Class 11 Biology at Work Chapter 6 Anatomy of Flowering Plants, help you. If you have any query regarding NCERT Solutions for Class 11 Biology at Work Chapter 6 Anatomy of Flowering Plants, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants

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NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants.

Question 1.
Differentiate between
(a) Respiration and Combustion
(b) Glycolysis and Krebs’ cycle
(c) Aerobic respiration and Fermentation
Solution:
(a) Differences between respiration and combustion are as follows :
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 1
(b) Differences between glycolysis and krebs’ cycle are as follows :
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 2
(c) Differences between aerobic respiration and fermentation are as follows :
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 3

Question 2.
What are respiratory substrates? Name the most common respiratory substrate.
Solution:
The compounds that are oxidized during this process are known as respiratory substrates. Usually, carbohydrates are oxidized to release energy, but proteins, fats, and even organic acids can be used as respiratory substances in some plants, under certain conditions.

Question 3.
Give the schematic representation of glycolysis.
Solution:
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 4

Question 4.
What are the main steps in aerobic respiration? Where does it take place?
Solution:
In aerobic respiration which takes place within the mitochondria, the final product of glycolysis, pyruvate is transported from the cytoplasm into the mitochondria.
The crucial events in aerobic respiration are:
• The complete oxidation of pyruvate by the stepwise removal of all the hydrogen atoms, leaving three molecules of CO2.
• The passing on of the electrons removed as part of the hydrogen atoms to molecular O2 with the simultaneous synthesis of ATP.
• The first process takes place in the matrix of the mitochondria while the second process is located on the inner membrane of the mitochondria.
• Pyruvate, which is formed by the glycolytic catabolism of carbohydrates in the cytosol, after it enters mitochondrial matrix undergoes oxidative decarboxylation by a complex set of reactions catalysed by pyruvic dehydrogenase. The reactions catalysed by pyruvic dehydrogenase require the participation of several coenzymes, including NAD+ and Coenzyme A.
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 5
During this process, two molecules of NADH are produced from the metabolism of two molecules of pyruvic acid (produced from one glucose molecule during glycolysis).
The acetyl CoA then enters a cyclic pathway, the tricarboxylic acid cycle, more commonly called as Krebs’ cycle.

Question 5.
Give the schematic representation of an overall view of Krebs’ cycle.
Solution:
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 6

Question 6.
Explain ETS.
Solution:
ETS or electron transport system is located in the inner mitochondrial membrane. It helps in releasing and utilizing the energy stored in NADH + H+ and FADH2 NADH+ H+, which is formed during glycolyis and citric acid cycle, gets oxidized by NADH dehydrogenase. The electrons so generated get transferred to ubiquinone through FMN. In a similar manner, FADH2 generated during citric acid cycle gets transferred to ubiquinone. The electrons from ubiquinone are received by cytochrome bc1, and further get transferred to cytochrome C. The cytochrome C acts as a mobile carrier between complex III and cytochrome C oxidase complex containing cytochrome a and a3, along with copper centres.

NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 7
During the transfer of electrons from each complex, the process is accompanied by the production of ATP from ADP and inorganic phosphate by the action ATP synthase. The amount of ATP produced depends on the molecule, which has been oxidized.

Question 7.
Distinguish between the following:
(a) Aerobic respiration and anaerobic respiration.
(b) Glycolysis and fermentation.
(c) Glycolysis and citric acid cycle.
Solution:
Differences between aerobic respiration and anaerobic respiration are as follows :
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 8
Differences between glycolysis and fermentation are as follows :
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 9
Differences between glycolysis and citric acid cycle are as follows :
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 10

Question 8.
What are the assumptions made during the calculation of the net gain of ATP?
Solution:

  • There is a sequential, orderly pathway functioning, with one substrate forming the next and with glycolysis, TCA cycle, and ETS pathway following one after another.
  • The NADH synthesized in glycolysis is transferred into the mitochondria and undergoes oxidative phosphorylation.
  • None of the intermediates in the pathway are utilized to synthesize any other compound.
  • Only glucose is being respired -no other alternative substrates are entering the pathway at any of the intermediary stages.

Question 9.
Discuss “The respiratory pathway is an amphibolic pathway”.
Solution:
Respiration is generally assumed to be a catabolic process because, during respiration, various substrates are broken down for deriving energy. Carbohydrates are broken down into glucose before entering respiratory pathways. Fats get converted into fatty acids and glycerol whereas fatty acids get converted into acetyl CoA before entering respiration. In a similar manner, proteins are converted into amino acids, which enter respiration after deamination.

During the synthesis of fatty acids, acetyl CoA is withdrawn from the respiratory pathway. Also, in the synthesis of proteins, respiratory substances get withdrawn. Thus, respiration is also involved in anabolism. Therefore, respiration can be termed as. amphibolic pathway as it involves both anabolism and catabolism.

Question 10.
Define RQ. What is its value for fats?
Solution:
The ratio of the volume of CO2 evolved to the volume of O2 consumed in respiration is called respiratory quotient (RQ) or respiratory ratio.
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 11

Question 11.
What is oxidative phosphorylation ?
Solution:

Although the aerobic process of respiration takes place only in the presence of oxygen, the role of oxygen is limited to the terminal stage of the process. Yet, the presence of oxygen is vital, since it drives the whole process by removing hydrogen from the system. Oxygen acts as the final hydrogen acceptor. Unlike photophosphorylation where it is the light energy that is utilized for the production of proton gradient required for phosphorylation, in respiration, it is the energy of oxidation-reduction utilized for the same process. It is for this reason that the process is called oxidative phosphorylation.

Question 12.
What is the significance of the step-wise release of energy in respiration?
Solution:
During oxidation within a cell, all the energy contained in respiratory substrates is not released free into the cell, or in a single step. It is released in a series of slow step-wise reactions controlled by enzymes, and it is trapped as chemical energy in the form of ATP.

Hence, it is important to understand that the energy released by oxidation in respiration is not used directly but is used to synthesise ATP, which is broken down whenever (and wherever) energy needs to be utilised. Hence, ATP acts as the energy currency of the cell.

This energy trapped in ATP is utilised in various energy-requiring processes of the organisms, and the carbon skeleton produced during respiration is used as precursors for the biosynthesis of other molecules in the cell.

VERY SHORT ANSWER QUESTIONS

Question 1.
What is anaerobic respiration? (Oct. 83)
Solution:
Incomplete or partial breakdown of fuel molecules into compounds such as ethyl alcohol, lactic acid in the absence of molecular oxygen.

Question 2.
Name the final acceptor of an electron in ETC.
Solution:
Oxygen is the electron acceptor of ETC.
Question 3.
The function of oxygen in aerobic respiration:
(i) It acts as the final electron acceptor.
(ii) It drives the whole process by removing hydrogen from the system.
Solution:
The function of oxygen in aerobic respiration:
(i) It acts as the final electron acceptor.
(ii) It drives the whole process by removing hydrogen from the system.

Question 4.
What is respiration? (Oct. 86)
Solution:
The oxidative process in which chemically bound energy from complex organic fuel molecules such as carbohydrates, proteins, and fats is captured in the form of ATP.

Question 5.
Where does the electron transport system operate in the mitochondria?
Solution:
Phosphofructokinase catalyses the formation of fructose 1, 6 bisphosphates from fructose 6-phosphate.

Question 6.
Give the function of phosphofructokinase in glycolysis.
Solution:
Hexokinase-helps in the phosphorylation of glucose.

Question 7.
Name the enzyme that catalyses the phosphorylation of glucose.
Solution:
The formation of acetyl CoA takes place in the mitochondrial matrix.

Question 8.
Where does the formation of acetyl CoA take place in a cell?
Solution:
The first step in the Krebs cycle is the condensation of an acetyl group (acetyl CoA) with oxaloacetic acid (OAA) to form citric acid and release the Coenzyme A.

Question 9.
What is the first step of reaction in the TCA cycle?
Solution:
Fatty acids may be converted to acetyl CoA before they from the respiratory substrates.

Question 10.
What is alcoholic fermentation?
Solution:
Alcoholic fermentation is the process by which yeast cells breakdown glucose into ethyl alcohol and carbon-dioxide under anaerobic conditions.

Question 11.
Name the oxidative pathway through which intermediate metabolites of glucose, fatty acids, and amino acids are finally oxidised.
Solution:
36 ATP/38 ATP molecules are obtained in the process of respiration and it is related to the aerobic respiration type.

Question 12.
What is lactic acid fermentation? (Oct. 2001)
Solution:
It is the process of fermentation by which lactose found in milk is converted to lactic acid by the action of lactobacillus.

Question 13.
What are the two molecules obtained by the action of aldolase from fructose -1, -6- biphosphate?
Solution:
ATP is produced.

SHORT ANSWER QUESTIONS

Question 1.
How is the proton gradient established?
Solution:
The proton gradient is established by passing proton (H+) from the matrix across the inner mitochondrial membrane into intermembrane space with the energy released during electron transfers in ETC.

Question 2.
Describe the steps in the formation of lactic acid from pyruvic acid.
Solution:
Pyruvic acid is catalysed by the enzyme lactic dehydrogenase. NADH formed in glycolysis is used up for the reduction.
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 12

Question 3.
How is ATP formed by the energy released during the electron transport system in mitochondria?
Solution:
ATP formations require an enzyme called ATP synthase. It has two components F0– F1. ATP- synthase becomes active in ATP formation when the concentration of H+ on the Fo side is higher than the F1 side. Fligher proton concentration in the outer chamber causes the proton to pass the inner chamber. F1 particle induced by the flow of proton through Fo channel. The energy of the proton gradient attaches the phosphate radicle to ADP. This produces ATP.

Question 4.
Give a detailed account of the net gain of ATP at a different stages of respiration.
Solution:
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 13
In most eukaryotic cells 2 molecules of ATP are required for transporting NADH produced in glycolysis to mitochondria for further oxidation. Hence net gain of ATP is 36 molecules.

Question 5.
Enumerate the functions of ATP.
Solution:
Functions of ATP:-
(i) ATP functions as a universal energy carrier of living systems.
(ii) ATP stores small packets of energy in its molecules.
(iii) It is mobile in the cell. Therefore, it reaches all parts of the cell away from the region of ATP synthesis.
(iv) It activates a number of chemicals by functioning as a phosphorylating agent.
(v) ATP provides energy for muscle contraction.
(vi) It is involved in the transport of substances against a concentration gradient.

Question 6.
Where is cytochrome c located? What is its function?
Solution:
Cytochrome c is located on the outer surface of the inner mitochondrial membrane. It acts as a mobile carrier for the transfer of electrons between complex III and complex IV of the electron transport system.

Question 7.
Define respiratory quotient.
Solution:
The respiratory quotient is defined as the ratio of the volume of carbon dioxide evolved to the volume of oxygen consumed in respiration.

Question 8.
What is oxidative phosphorylation?
Solution:
The whole process by which oxygen effectively allows the production of ATP by phosphorylation of ADP is called oxidative phosphorylation.

Question 9.
The energy yield in terms of ATP is higher in aerobic respiration than during anaerobic respiration. Why is there anaerobic respiration even in organisms that live in aerobic conditions like human beings and angiosperms?
Solution:
Aerobic organisms do face situations where oxygen availability is little. For example, overworked muscles do not receive enough oxygen during strenuous exercise. Similarly, deep-seated tissues of angiosperms do not receive enough oxygen through diffusion from outside. In such situations, only anaerobic respiration can help in the survival of the tissue.

Question 10.
Comment on the statement- “Respiration is an energy-producing process but ATP is used in some steps of the process”.
Solution:
ATP is required in all those reactions where phosphorylative activation of the substrate is required. Therefore, despite producing energy (as ATP), respiration requires ATP in certain steps, e.g., glucose – glucose 6-phosphate, fructose 6-phosphate —fructose 1, 6- bisphosphate.

LONG ANSWER QUESTIONS

Question 1.
Explain the major steps in Krebs’ cycle. Why is this cycle also called the citric acid cycle?
Solution:
Krebs cycle: This process occurs in the mitochondrial matrix.
Major steps of Krebs cycle are as follows :

  • Acetyl Co-A, formed by the oxidative decarboxylation of pyruvic acid enters the Krebs’ cycle.
  • It combines with oxalo acetic acid (OAA), a 4C-compound, to form a 6C-compound, citric acid; the reaction is catalysed by citrate synthase.
  • Citrate is then isomerised into isocitrate.
    NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 14
  • Isocitrate is converted into oxalosuccinic acid in the presence of NAD and isocitrate dehydrogenase.
  •  Oxalosuccinic acid is then decarboxylated into a-ketoglutaric acid (KG), in the presence of a decarboxylase enzyme.
  •  a-ketoglutaric acid is converted into succinyl Co-A in the presence of NAD, Co- A, and enzyme a-ketoglutarate dehydrogenase.
  •  When succinyl Co-A is converted into succinic acid, one molecule of GTP is formed and Co-A is released.
  •  In the remaining part of the cycle, succinic acid is converted into OAA, so that the citric acid cycle can continue to operate.
  •  During this cycle, three molecules of NAD and one molecule of FAD are reduced to NADH and FADH respectively.
  •  This cycle is called as a citric acid cycle because the first product is citric acid which is 3-C compound.

Question 2.
Name the end product of glycolysis. Where is it produced in the cell? Discuss oxidative decarboxylation.
Solution:
Glycolysis results in the formation of two molecules of pyruvic acid, NADH, and ATP. It occurs in the cytosol of the cell.
Aerobic oxidation: One of the three carbons of pyruvic acid is oxidised to carbon dioxide in the reaction called oxidative decarboxylation. Pyruvic acid is first decarboxylated and then oxidised by the enzyme pyruvic dehydrogenase. The two-carbon units are readily accepted by coenzyme-A (Co-A) to form acetyl Co-A. The summary of the reaction is given in the following equation :
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 15
Thus, pyruvic acid enters the Krebs cycle as acetyl Co-A. Krebs’ cycle occurs in the mitochondrial matrix.
Acetyl Co-A, formed by the oxidative decarboxylation of pyruvic acid enters the Krebs’ cycle.

Question 3.
Represent schematically the interrelationship among metabolic pathways in a plant, showing respiration mediated breakdown of different organic compounds.
Solution:
Schematic representation among metabolic pathways showing respiration mediated breakdown of different organic molecules to CO2 and H2O:
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 16
Question 4.
How do plants manage the exchange of gases? Give an overview of respiration in plants.
Solution:
Plants, unlike animals, have no specialized organs for gaseous exchange but they have stomata and lenticels for this purpose. There are several reasons why plants can get along without respiratory organs.

  • Each plant part takes care of its own gas- exchange needs. There is very little transport of gases from one plant part to another.
  • Plants do not present great demands for gas exchange. Roots stem and leave respire at a lower rate than animals do.
  • Only during photosynthesis, large volumes of leases exchanged and, each leaf is well adapted to take care of its own needs during these periods.
  • When cells perform photosynthesis, the availability of O2 is not a problem in these cells since 02 is released
  • The distance that gases must diffuse even in large, bulky plants is not great. Each living cell in a plant is located quite close to the surface of the plant.
  • Even in woody stems, the ‘living’ cells are organised in thin layers inside and beneath the bark. They also have openings called lenticels. The cells in the interior are dead and provide only mechanical support.
  • Thus, most cells of a plant have attested to a part of their surface in contact with air. This is also facilitated by the loose packing of parenchyma cells in leaves, stems, and roots, which provide an interconnected network of air spaces.
  • The complete combustion of glucose, which produces C02 and H20 as end products, yields energy. Most of the energy is given out as heat.
    C6H12O6 + 6O2 → 6C02 + 6H20 + Energy
  • If this energy is to be useful to the cell, it should be able to utilise it to synthesis other molecules that the cell requires.
  • The strategy that the plant cell uses is to catabolize the glucose molecule in such a way that not all the liberated energy goes out as heat.
  • The key is to oxidise glucose not in one step but in several small steps enabling some steps to be just large enough so that the energy released can be coupled to ATP synthesis.

We hope the NCERT Solutions for Class 11 Biology at Work Chapter 14 Respiration in Plants, helps you. If you have any query regarding NCERT Solutions for Class 11 Biology at Work Chapter 14 Respiration in Plants, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Biology Chapter 2 Biological Classification

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NCERT Solutions for Class 11 Biology Chapter 2 Biological Classification

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 2 Biological Classification.

Question 1.
Discuss how classification systems have undergone several changes over a period of time?
Solution:
The classification was born instinctively out of a need to all organisms for our own use since the dawn of civilization. Aristotle was the earliest to attempt a more scientific basis for classification. He used simple morphological characteristics to classify plants into trees, herbs and shrubs. He also divided animals into Animals with red blood and those who do not have red blood. Linnaeus proposed a two-kingdom system of classification with Plantae and Animalia including plants and animals respectively.

The above system did not distinguish eukaryotes and prokaryotes, unicellular and multicellular, photosynthetic and non-photosynthetic organisms. A need was felt for including besides gross morphology, other characteristics like cell structure, nature of cell wall, mode of nutrition, habitat, method of reproduction, evolutionary relationships etc., Recently R.H. Whittaker proposed a five-kingdom classification to answer above the Five kingdoms are

  • Monera
  • Protista
  • Fungi
  • Plantae
  • Animalia.

Question 2.
State two economically important uses of
(a) heterotrophic bacteria
(b) archaebacteria
Solution:
(a) Heterotropic bacteria : These bacteria are natural scavengers. The souring of milk into lactic acid and alcohol to vinegar is brought about by some saprophytic bacteria, e.g., Lactic acid bacteria and acetic acid bacteria respectively.
A number of antibiotic are extracted from actinomycetes especially from the genus Streptomyces e.g. Streptomycin, Chloramphenicol, Oilorotetracycline, Erythromycin, Terramycin etc.
(b) Archaebacteria live as symbionts in the rumen of herbivorous animals.
Methanogens are present in the guts of several ruminant animals such as cows and buffaloes and they are responsible for the production of methane (biogas) from the dung of these animals.

Question 3.
What is the nature of the cell wall in diatoms?
Solution:
In diatoms, ail walls form two thin overlapping shells which fit together as in a soapbox. The walls are embedded with silica and thus the walls are indestructible. Thus diatoms have left behind a large amount of cell wall deposits in their habitat.

Question 4.
Find out what do the terms ‘algal bloom’ and ‘red tides’ signify?
Solution:
Algal bloom : When colour of water changes due to profuse growth of coloured phytoplankton, it is called algal bloom.
Red tides : Redness of the red sea is due to luxurient growth of Trichodesrium erythrium, a member of cynobacteria (blue green alage).

Question 5.
How are viroids different from viruses?
Solution:
Viroids are simpler than viruses, consisting of
a single RNA molecule that is not covered by protein capsid. The genetic material of viruses are surrounded by protein coat.

Question 6.
Describe briefly the four major groups of protozoa.
Solution:
The four major group of protozoa are flagellated protozoan, amoeboid protozoan, sporozoan, ciliated protozoan. The main characters of these group are as follows:
NCERT Solutions for Class 11 Biology Chapter 2 Biological Classification 1

Question 7.
Plants are autotrophic. Can you think of some plants that are partly hetrotrophic?
Solution:
Bladderwort and venus fly trap are examples of insectivorous plants and Cuscuta is a parasite. These are plants which are partially heterotrophic.

Question 8.
What do the terms phycobiont and mycobiont signify?
Solution:
Lichens shows symbiotic association between algae and fungi. The fungal component of lichen is called mycobiont and the algal component is called phycobiont.

Question 9.
Give a comparative account of the classes of kingdom fungi under the following:
(a) Mode of nutrition
(b) Mode of reproduction
Solution:
Kingdom fungi has four classes, these are Phycomycetes, ascomycetes, basidiomycetes and Deuteromycetes. The comparison between these classes are as follows :
NCERT Solutions for Class 11 Biology Chapter 2 Biological Classification 2

Question 10.
What are the characteristic features of Euglenoids?
Solution:
Euglenoids show the following characteristic features:

  • They store carbohydrates in the form of paramylon.
  • Since euglenoids are green and holophytic like other plants.
  • Few are non-green and saprophytic, some are holotropic.
  • They bear a red-pigmented eyespot and a gullet near the base of flagellum.
  • All the euglenoids have one or two flagella which help in swimming.
  • Absence of cell-wall but contain flexible pellicle made up of protein.
  • Freshwater, free-living found in ponds and ditches.

Question 11.
Give a brief account of viruses with respect to their structure and the nature of genetic material. Also, name four common viral diseases.
Solution:
Viruses have the following characteristics:
(i) All plant viruses have single-stranded RNA and all animal viruses have either single or double-stranded RNA or double-stranded DNA.
(ii) Protein vims also contain genetic material RNA or DNA. A vims is a nucleoprotein and the genetic material is infectious, These are obligate parasites, self replicating, non-cellular organisms.
(iii) Vimses are smaller than bacteria and their genetic material is surrounded by protein I coat called capsid. Capsid is made up of small subunits called capsomeres.
Four common viral diseases are :
(a) Cough and cold
(b) Mumps
(c) Influenza
(d) Smallpox

Question 12.
Organise a discussion is your class on the topic are viruses living or non-living?
Solution:
Vimses are link between living and non-living. They possess some living characters and some non-living characters. Crystallization is a non-living character but it can reproduce inside living body.
Actually vimses are metabolically inert when outside the host-cell. They reproduce using the metabolic machinery of the host cell.

VERY SHORT ANSWER QUESTIONS

Question 1.
Who wrote the books ‘Species Plantarum’ and ‘Systema Naturae?
Solution:
Carolus Linnaeus.

Question 2.
Name the two kingdoms of the living world proposed by Linnaeus.
Solution:
Plantae and Animalia

Question 3.
What are protists?
Solution:
Protists are unicellular, eukaryotic organisms.

Question 4.
Which organism was earlier placed in the plant as well as animal kingdoms and why?
Solution:
Euglena because it has locomotory organelle, flexible pellicle, contractile vacuole and reproduce by binary fission like animals and chloroplasts and pyrenoids like plants.

Question 5.
Name the 5 kingdoms of organisms in the order of their supposed evolution.
Solution:
Monera, Protista, Fungi, Animalia and Plantae.

Question 6.
Mention 2 traits in which fungi resemble animalia.
Solution:
Heterotrophy and glycogen as reserve food.

Question 7.
Define
(a) Plasmogamy
(b) Karyogamy
Solution:
(a) Plasmogamy – Fusion of protoplasms between two motile or non-motile gametes.
(b) Karyogamy – Fusion of two nuclei.

Question 8.
What is a retrovirus? Give an example
Solution:
Retrovirus is organisms that have RNA s as genetic material. For example HTV

Question 9.
Give two salient features of slime moulds.
Solution:
The two salient features of slime moulds are:

  1. These do not have a cell wall
  2. These have pseudopodia for movement

Question 10.
What is called the jokers of microbiology and why?
Solution:
Jokers of microbiology are mycoplasma as they have no cell wall and no definite shape.

Question 11.
Give the names of two diseases caused by Protozoans
Solution:
Two diseases caused by protozoans are
(1) Amoebiasis
(2) Malaria

SHORT ANSWER QUESTIONS

Question 1.
Cyanobacteria play a major role in our ecology. Discuss.
Solution:
Cyanobacteria, also known as ‘blue-green algae’ help in carbon fixation in a major way on the ocean surface.

They are helpful in nitrogen fixation in paddy fields leading to a better harvest. About 80% of photosynthesis on ocean surface is done by cyanobacteria. So, it can be said that they play a major role in our ecology.

Question 2.
What is the role of methanogens?
Solution:
Methanogens are type of bacteria which live in the gut of ruminating animals.

They assist those animals in digestion and the byproduct of that digestive process is methane.

More number of livestock population results in increased methane level in the environment leading to global warming. So, indirectly methanogens can be responsible for global warming.

Question 3.
What are lichens? What are the roles of lichen in water pollution ?
Solution:
Lichens are symbiotic associations i.e. mutually useful associations, between algae and fungi.

The algal component is known as phycobiont and fungal component as mycobiont, which are autotrophic and heterotrophic, respectively.

Algae prepare food for fungi and fungi provide shelter and absorb mineral nutrients and water for its partner. Lichens are very good pollution indicators as they do not grow in polluted areas.

Question 4.
On what factors is the 5 kingdom classification of Whittaker based?
Solution:
The five kingdom classification is based upon the following factors :
(i) Complexity of cell structure – Prokaryotes or Eukaryotes
(ii) Complexity of organisms body – Unicellular or Multicellular
(iii) Mode of obtaining nutrition – Autotrophs or Heterotrophs
(iv) Phylogenetic relationships

Question 5.
Give the technical terms used for the following:
(a) Remains of an organism of a former geological age.
(b) Science of classification of organisms.
(c) Evolutionary history of a group of organisms.
(d) Organisms which synthesize their own food, using chemical energy.
Solution:
(a) Fossils
(b) Taxonomy
(c) Evolution
(d) Autotrophs

Question 6.
What are the kinds (shapewise) bacteria found in nature. Name the pathogen with the disease caused
Solution:
NCERT Solutions for Class 11 Biology Chapter 2 Biological Classification 3

 

Question 7.
Why is
(i) Basidiomycetes called club fungi?
(ii) Ascomycetes called sac fungi?
Solution:
(i) After sexual reproduction basidium is formed which form the shape a club and this chin these fungi are called Club Fungi.
(ii) In sexual reproduction ascospores are formed in a sac like asci and thus this fungi is called sac fungi.

LONG ANSWER QUESTIONS

Question 1.
Give an account of early work in taxonomy.
Solution:

  • Since the dawn of civilisation, there have been many attempts to classify living organisms.
  • It was done instinctively not using criteria that were scientific but borne out of a need to use organisms for our own use – for food, shelter and clothing.
  • Aristotle was the earliest to attempt a more scientific basis for classification. He used simple morphological characters to classify plants into trees, shrubs and herbs.
  • He also divided animals into two groups, those which had red blood and those that did not.
  • In Linnaeus’ time a Two Kingdom system of classification with Plantae and Animalia kingdoms was developed that included all plants and animals respectively.
  • Classification of organisms into plants and animals was easily done and was easy to understand, inspite, a large number of organisms did not fall into either category.
  • R.H. Whittaker (1969) proposed a Five Kingdom Classification.
  • The kingdoms defined by him were named Monera, Protista, Fungi, Plantae and Animalia.
  • The main criteria for classification used by him include cell structure, thallus organisation, mode of nutrition, reproduction and phylogenetic relationships.

Question 2.
Differentiate briefly characteristics of kingdom Plantae and Animalia.
Solution:

  • Kingdom Plantae includes all eukaryotic chlorophyll-containing organisms commonly called plants.
  • A few members are partially heterotrophic such as the insectivorous plants or parasites.
  • Bladderwort and Venus fly trap are examples of insectivorous plants and Cuscuta is a parasite.
  • The plant cells have an eukaryotic structure with prominent chloroplasts and cell wall mainly made of cellulose.
  • Plantae includes algae, bryophytes, pteridophytes, gymnosperms and angiosperms.
  • The animal kingdom is characterised by v heterotrophic eukaryotic organisms that are multicellular and their cells lack cell walls.
  • They directly or indirectly depend on plants for food. They digest their food in an internal cavity and store food reserves as glycogen or fat.
  • Their mode of nutrition is holozoic – by ingestion of food. They follow a definite growth pattern and grow into adults that have a definite shape and size.
  • Higher forms show elaborate sensory and neuromotor mechanism. Most of them are capable of locomotion.

Question 3.
Give the economic importance of diatoms. Diatoms are used
Solution:
(1) as a cleaning agent in tooth pastes and metal polishes.
(2) Adding to make sound proof rooms.
(3) In Alteration of sugar, alcohol and antibiotics
(4) as put in paints to ad the paint visibility at night
(5) as an insulating material in Refrigerators, fumances etc.

Question 4.
What are the distinguishing characters of kingdom fungi?
Solution:
The distinguishing characters of kingdom fungiare as follows :
(i) Fungi are non-vascular, non-seeded, non-flowering, eukaryotic achlorophyllous (nongreen), heterophic (heterophytic) spore bearing, thalloid, multicellular decomposers and mineralisers of organic wastes and help in recycling of matter in the biosphere.
(ii) In true fungi the plant body is thallus. It may be non-mycelial or mycelial.
a. Non mycelial: The non-mycelial forms are unicellular; however they may form a pseudomycelium by budding,
b. Mycelial: In mycelial form plant body is made up of thread like structures called hyphae. Hyphae are usually branched tube like structure bounded by a cell-wall of chitin. The hyphae may be septate (higher fungi) or aseptate (lower fungi).
Septate hyphae may be of 3 kinds, uninucleate (monokaryotic hyphae), with binucleate cells (dikaryotic hyphae) ormultinucleate. Some fungi are aseptate and known as coenocytic fungi, with hundreds of nuclei in continuous cytoplasmic mass.
(iii) The cell shows eukaryotic organization but lack chloroplast and Golgi bodies. The genetic material is DNA and mitosis is intracellular (karyochorisis).
(iv) Fungi lack chlorophyll, hence, they do not prepare food by photosynthesis. Thus they can grow everywhere, where organic material is available.
(v) Fungi are heterotrophs that acquire their nutrient by absorption. They store their food in the form of glycogen.
(vi) The primitive fungi have oogamous type of sexual reproduction where as most advanced ones do not have sexual reproduction.

Question 5.
Compare the main features of Monera with Protista.
Solution:
NCERT Solutions for Class 11 Biology Chapter 2 Biological Classification 4

 

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NCERT Solutions for Class 11 Biology Chapter 1 The Living World

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NCERT Solutions for Class 11 Biology Chapter 1 The Living World

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 1 The Living World.

Question 1.
Why are living organisms classified?
Solution:
Classification of living organisms grouped them in special categories, which is based on observable characters. It makes their study easy and convenient. For example, Mammals are those who possess mammary glands, the hair on the body, external pinnae, etc.

Question 2.
Why are classification systems changing every now and then?
Solution:
The classification system changes when more information becomes available about the organisms. Additional information are updated from time to time about different organisms at this stage there is a need arises to make changes in the classification system.

Question 3.
What different criteria would you choose to classify people that you meet often?
Solution:
Classification means the arrangement of organisms into groups on the basis of their affinities or relationships. The branch of biology that deals with the study of principles and procedures of biological classification are called taxonomy. Some fundamental elements of taxonomy are discussed below.

Nomenclature: It is the science of providing distinct and proper names to organisms. It is the determination of the correct name as per established universal practices and rules.

Classification: It deals with the mode of arranging organisms or group^ of organisms into categories according to a systematic plan or ah order. The categories used in the classification of animals are Class, Order, Family, Genus, and Species. Each category is a unit and is also called a taxon (PI. Taxa).

Identification: It is the determination of the correct name and place of an organism in a system of classification. It determines that the particular organism is similar to some other organism of known identity. This implies assigning an organism to a particular taxonomic group. Suppose there are three plants say x, y, z. AH represent different species. Another plant w resembles y. The recognition of the plant was identical to the already known plant y is its identification.
One of the important features of systematics is the naming of living organisms. The organisms have been given two types of names i.e

  • common or vernacular names
  • Scientific names.

Question 4.
What do we learn from the identification of individuals and populations?
Solution:
Identification of individuals and populations determines their exact place or position in the set plan of classification.

Question 5.
Given below is the scientific name of mango. Identify the correctly written name.
(a) Mangifera Indica
(b) Mangifera indica
Solution:
(b) Mangifera indica

Question 6.
Define a taxon. Give some examples of taxa at different hierarchical levels.
Solution:
“Taxon is a unit of classification or a rank or a level of hierarchy in system of classification. The following chart gives taxonomical categories showing a hierarchical arrangement in ascending order.

Kingdom
↑
phylum or Division
Class
↑
Order
↑
Family
↑
Genus
↑
Species

Question 7.
Can you identify the correct sequence of taxonomical categories?
(a) Species → Order → Phylum → Kingdom
(b) Genus → Species → Order → Kingdom
(c) Species → Genus → Order → Phylum
Solution:
(c) Species → Genus → Order → Phylum

Question 8.
Try to collect all the currently accepted meanings for the word ‘species’. Discuss with your teacher the meaning of species in the case of higher plants and animals on one hand and bacteria on the other hand.
Solution:

  1. Species is one of the basic units of biological classification. A species is often defined as a
    group of organisms capable of interbreeding aid in producing fertile offspring.
  2. Sometimes more precise or differing measures such as similarity of DNA, morphology o,^ecological niche are used to define the basis of species.
  3. In case of animals, the name of species is defined by the specific name or the specific epithet. For example, gray wolves belong to the species Canis lupus, golden Jackals to Cam’s aureus etc.
  4. Both of them belong to same genus Canis, but species name varies. But species name of plant is only called species epithet.
  5. The ‘specific name’ in botany is always the combination of genus name and species epithet such as saccharum in Acer saccharum (Sugar maple).
  6. But bacteria are grouped under four categories based on their shape – spherical, rod-shaped, comma and spiral shaped and species of bacteria is according to their shapes. Thus the meaning of species in higher organism and bacteria are different.

Question 9.
Define and understand the following terms:
(i) Phylum (ii) Class (iii) Family (iv) Order (v) Genus
Solution:
(i) Phylum: A phylum is a group of related classes having some common features, e.g., protozoa.
(ii) Class: A class is a group of related orders, for e.g., order Rodentia, Lagomorpha and Carnivora all having hair and milk glands are placed in class Mammalia.
(iii) Family: A family is a group of related genera. The genus Felis of cats and the genus Panthera of lion, tiger and leopard are placed in the family Felidal.
(iv) Order: An order is a group of related families. The family Felidae of cats and the »family Coridal of dogs are assigned to the order Carnivora. Cats and dogs have large canine teeth and are flesh-eaters.
(v) Genus: A genus is a group of species alike in the broad features of their organization but different in detail. As per the rules of binomial nomenclatures, a species can not be named without assigning it to a genus.
NCERT Solutions for Class 11 Biology Chapter 1 The Living World 1

Question 10.
How is key helpful in the identification and classification of organisms?
Solution:
Keys are contrasting pairs of characters (couplet), it represents the choice made between two opposite options. This results in acceptance of only one and rejection of the other. Each statement in the key is called a lead. Separate taxonomic keys are required for each taxonomic category such as family, genus, and species for identification purposes.

Question 11.
Illustrate the taxonomical hierarchy with suitable examples of a plant and an animal.
Solution: 
NCERT Solutions for Class 11 Biology Chapter 1 The Living World 2

VERY SHORT ANSWER QUESTIONS

Question 1.
Name the basic Unit of classification.
Solution:
Species.

Question 2.
Who introduced the hierarchy in taxonomy?
Solution:
Linnaeus

Question 3.
Who is the father of taxonomy?
Solution:
Carolus Linnaeus.

Question 4.
What is meant by cytotaxonomy?
Solution:
Classification based on chromosome number.

Question 5.
Who devised the binomial nomenclature?
Solution:
Carolus Linnaeus

Question 6.
What is a type specimen?
Solution:
Establishment of the name of the new species on the basis of the original specimen is called type specimen

Question 7.
In which language binomial nomenclature is written?
Solution:
Latin

Question 8.
What term is used to describe organisms without a well-developed nucleus?
Solution:
Prokaryote

Question 9.
Is inter-specific breeding possible?
Solution:
Yes, both.

Question 10.
What are DNA viruses / RNA viruses?
Solution:
Viruses that possess DNA as the genetic material are called DNA viruses.

Question 11.
What is speciation?
Solution:
Formation of a new species from an existing one by the appearance of mutation.

Question 12.
What are correlated characters?
Solution:
The common features the species have to qualify for inclusion in a genus are called correlated characters

Question 13.
Why classification of plants and animals is necessary?
Solution:
Classification divides millions of plant and animal species into convenient groups that make their study easier

Question 14.
What is cohort or order?
Solution:
The cohort is a unit of classification higher than the 6. family

Question 15.
Give an example of symbiotic bacteria.
Solution:
Rhizobium leguminosarum

Question 16.
Give botanical and zoological names of the following:
(1) Pea
(2) Wheat
(3) Man
(4) Potato
Solution:
(1) Pea → Pisumsatinum
(2) Wheat → Triticumaextivum
(3) Man → Homo sapiens
(4) Potato → Solanum tuberosum

SHORT ANSWER QUESTIONS

Question 1.
Write a note on bacteriophages. (Dharwar. 2004, Belgaum. 04,2005)
Solution:
The viruses that infect bacteria are called bacteriophages. They were discovered by Twort. They are Tadpole shaped. They have DNA as their genetic material. They are distinguished into T – odd phages as well as T – even phages.

Question 2.
What is a taxonomic aid?
Solution:
A taxonomic aid is storage of record of either live or dead specimens of flora or fauna, which helps scientists in taking reference to study classification

Question 3.
Give the classification of man.
Solution:
Common Name – Human
Scientific Name – Homo sapiens
Genera – Homo
Families – Hominidae
Orders – Primata
Classes – Mammalia
Phyla/Division – Chordate

Question 4.
What is a museum? How many kinds of museums are found?
Solution:
Museum in an institution where artistic and educational materials are exhibited to the public. The material available for observation and study is called a collection.
Kinds of Museums:

  • Art Museum
  • History Museum
  • Applied Science Museum
  • Natural Science Museum

Question 5.
Give a reason for the following.
Bacteria are the Natural Scavengers ‘ (D.Kannada 2006)
Solution:
because they bring about the decomposition of organic debris and clean the earth’s surface.

Question 6.
What is the role of characteristics of living beings in classification?
Solution:
A group of common features of living beings are placed under a common category of classification and when uncommon under a different category. It means more systematic a process for further study, research, protection and recording.

Question 7.
What is the significance of a HERBARIUM?
Solution:
HERBARIUM:- A book, case, or room containing an orderly collection of dried plants is called Herbarium. It develops interest in Nature for the activists in it. It can be used to gain knowledge and be updated about plants and their scientific names and even compare various samples. It is a small scale it can be proactive to do. One can make projects too from it for schools, colleges and research institutions.

Question 8.
Explain the role of blue-green algae in soil fertility.
Solution:
Blue-green algae like Nostoc, Anabaena fix atmospheric nitrogen. Heterocyst contains nitrogens enzyme that helps in nitrogen fixation. Nitrogen-fixing blue-green algae are inoculated in the rice field to increase soil fertility.

LONG ANSWER QUESTIONS

Question 1.
Write a short note on Binomial Nomencia? ture and guidelines for Binomial nomenclature.
Solution:
Binomial Nomenclature was introduced by Carolus Linnaeus. In this method every organism is given a scientific name, which has two parts, the first is the name of the genus (generic name) and the second is the name of the species (specific epithet) e.g.: Homo sapiens In the above examples, Homo is a generic name, while sapiens is the name of the species belonging to Homo.

Guidelines:

  • scientific names are generally in Latin or derived from Latin irrespective of their origin
  • The scientific names are written in italics or underlined (when handwritten)
  • The first word denotes the name of the genus and the second word denotes the specific epithet
  • The generic name starts with a capital letter, while the specific name starts with a small letter (If a specific name starts with a capital letter it denotes the name of a person or place)
  • The name of the author is written in an abbreviated form after the specific name. e.g.: Homo sapiens Linn.

Question 2.
What is the difference between living and nonliving?
Solution:
Question 3.
Explain the binomial system of nomenclature.
Solution:
Binomial nomenclature system was developed by Linnaeus. Binomial nomenclature is the system of providing organisms with appropriate and distinct names consisting of two words, first generic and second specific. The first or 4.

  • generic word is also called genus. It is like a noun and its first letter is written in capital form.The second word or specific epithet represents the species.
  • It is like an adjective. Its first letter is written in small form except occasionally when it denotes a person or place. The two word name is appended with the name of the taxonomist who discovered the organism and provided with its scientific name, e.g., Ficus bengalensis L., Mangifera indica Linn, The name of taxonomist can be written in full or in abbreviated form.
  • There are several technical names which have three words, e.g., Homo sapien sapiens, Acacia nilotica indica, Gerilla gorilla. Here the first word is generic, the second specific while the third word represents variety (mostly in botanical literature) or subspecies (mostly in zoological literature).
  • If the same scientific name is to be written time and again, the name of the genus can be abbreviated, e.g., F. bengalensis.

Question 4.
What is the role of zoological parks in wildlife conservation?
Solution:

  • In the early stages, the zoological parks were considered as places of relaxation and enjoyment for public, however, there has been a change in the objective of the purposefulness of these parks.
  • The establishment of zoological parks help in providing knowledge about different native and exotic wild mammals, birds, reptiles, fish and flora to the public in general and school children in particular.
  • Since the key to wildlife conservation lies in the education of the masses and the involvement of voluntary organisations, zoological parks are very useful in spreading knowledge on the wildlife wealth of the country.
  • These are also important centres for organising seminars, training and researches on the management of wildlife species and for study of their social behaviour, breeding and ecological species.

We hope the NCERT Solutions for Class 11 Biology at Work Chapter 1 The Living World, help you. If you have any query regarding NCERT Solutions for Class 11 Biology at Work Chapter 1 The Living World, drop a comment below and we will get back to you at the earliest.