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CLASS 11th CHAPTER – 12 Mineral Nutrition |Biology | NCERT SOLUTION| EDUGROWN

NCERT Solutions for Class 11 Biology Chapter – 12 Mineral Nutrition includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Biology exams must go through NCERT Solutions for Class 11 Chapter -12 Mineral Nutrition. NCERT Solutions will make you understand the topics in most simple manner and grasp it easily to perform better.

Class 11th Chapter -12 Mineral Nutrition | NCERT BIOLOGY SOLUTION |

Page No: 205

Exercises

1. ‘All elements that are present in a plant need not be essential to its survival’. Comment.

Answer

Plants tend to absorb different kinds of nutrients from soil but all of them are not essential to its survival because they do not directly involved in the composition of their body. A nutrient is inessential for a plant if it is not directly involved in the plant’s physiology and metabolism. However, if the concentration of micronutrients such as Fe, Mn, Cu, Zn, Cl, etc., rise above their critical values, they appear to be toxic for the plant.

2.Why is purification of water and nutrient salts so important in studies involving mineral nutrition using hydroponics?

Answer
As in hydroponics, the plants are grown in a nutrient solution in the absence of soil. Since the plant roots are exposed to a limited amount of the solution, there are chances that the concentrations of oxygen and other minerals in the plant roots would reduce. Therefore, in studies involving mineral nutrition using hydroponics, purification of water and nutrient salts is essential so as to maintain an optimum growth of the plants.

3. Explain with examples: macronutrients, micronutrients, beneficial nutrients, toxic elements and essential elements.

Answer

→ Macronutrients: Elements which are present in large amounts in plant tissues are called macronutrients. They are present in plant tissues in amounts more than 10 mmole kg^–1 of dry matter. Carbon, hydrogen, oxygen, nitrogen, phosphorous, sulphur, potassium, calcium and magnesium are examples.

→ Micronutrients: Elements which are present in small amounts i.e. less than 10 mmole per kg of dry matter are called micronutrients. Iron, manganese, copper, molybdenum, zinc, copper, boron, chlorine and nickel are examples.

→ Beneficial Elements: The nutrients that may not be essential, but are beneficial to plants are called beneficial elements. Sodium, silicon, cobalt and selenium are examples.

→ Toxic elements: Any mineral ion concentration in tissues that reduces the dry weight of tissues by about 10 per cent is considered toxic. For example, when manganese is present in large amounts, it induces deficiencies of iron, magnesium, and calcium by interfering with their metabolism.

→ Essential elements: The elements which are absolutely necessary for plant growth and reproduction are called essential elements. The requirement of these elements is specific and non-replaceable. These elements are further divided into two broad categories based on their quantitative requirements.

4. Name at least five different deficiency symptoms in plants. Describe them and correlate them with the concerned mineral deficiency.

Answer

Five different deficiency symptoms in plants are:

→ Chlorosis is the loss of chlorophyll leads to the yellowing of leaves. It is caused by the deficiencies of nitrogen, potassium, magnesium, sulphur, iron, manganese, zinc, and molybdenum.

→ Necrosis is the death of plant tissues as a result of the deficiencies of calcium, magnesium, copper, and potassium.

→ Inhibition of cell division is caused by the deficiencies of nitrogen, potassium, sulphur, and molybdenum.

→ Delayed flowering is caused by the deficiencies of nitrogen, sulphur, and molybdenum.

→ Stunted plant growth is a result of the deficiencies of copper and sulphur.

5. If a plant shows a symptom which could develop due to deficiency of more than one nutrient, how would you find out experimentally, the real deficient mineral element?

Answer

Every element shows certain characteristic deficiency symptoms in the plants also different plants respond in different ways to the deficiency of the same nutrient. Thus, in order to identify the real deficient mineral element, we need note all the available symptoms in different parts of the plant. Then we have to compare the noted information with the available standard symptom table.

6. Why is that in certain plants deficiency symptoms appear first in younger parts of the plant while in others they do so in mature organs?

Answer
Mobility of an element play very important role in determining whether the deficiency symptoms appear first in younger parts or older parts. Elements such as nitrogen, potassium, and magnesium are highly mobile. These elements move from the mature organs to the younger parts of a plant. Therefore, the symptoms for the deficiencies of these elements first appear in the older parts of the plant. Elements such as calcium and sulphur are relatively immobile. These elements are not transported out of the older parts of a plant. Therefore, the symptoms for the deficiencies of these elements first appear in the younger parts of the plant.

7. How are the minerals absorbed by the plants?

Answer

The absorption of minerals by plants takes by the roots of plants in two main phases: apoplast and symplast.
In the first phase, there is a rapid uptake of nutrients from the soil into the free spaces of plant cells. This process is passive and it usually occurs through trans-membrane proteins and ion-channels.
In the second phase, the ions are taken slowly into the inner spaces of the cells. This pathway generally involves the expenditure of energy in the form of ATP.

8. What are the conditions necessary for fixation of atmospheric nitrogen by Rhizobium. What is their role in N₂ -fixation?

Answer

Rhizobium bacteria need symbiotic association with legume plants to carry out nitrogen fixation. Root nodules contain the necessary enzymes for nitrogen fixation and thus enable rhizobium to fix nitrogen. The enzyme nitrogenase facilitates the conversion of nitrogen into ammonia which is the first stable product of nitrogen fixation. Ammonia is then converted into glutamic acid. Glutamic acid is then utilised by plants to make amino acids which are ultimately utilised to make protein.

9. What are the steps involved in formation of a root nodule?

Answer

Nodule formation involves a sequence of multiple interactions between Rhizobium and roots of the host plant. These steps are:

→ Rhizobium bacteria contact a susceptible root hair and divides near it.

→ After successful infection, the root hairs get curled and are invaded by the bacteria.

→ This invasion is followed by the formation of an infection thread that carries the bacteria into the cortex of the root.

→ The bacteria get modified into rod-shaped bacteroides and cause inner cortical and pericycle cells to divide. Division and growth of cortical and pericycle cells lead to nodule formation.

→ The nodules finally get connected with the vascular tissues of the roots for nutrient exchange.

10. Which of the following statements are true? If false, correct them:
(a) Boron deficiency leads to stout axis.
► True

(b) Every mineral element that is present in a cell is needed by the cell.
► Every mineral element that is present in a cell is not needed by the cell.

(c) Nitrogen as a nutrient element, is highly immobile in the plants.
► Nitrogen as a nutrient element,is highly mobile in plants.

(d) It is very easy to establish the essentiality of micronutrients because they are required only in trace quantities.
► True

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CLASS 11th CHAPTER – 11 Transport in Plants |Biology | NCERT SOLUTION| EDUGROWN

NCERT Solutions for Class 11 Biology Chapter – 11 Transport in Plants includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Biology exams must go through NCERT Solutions for Class 11 Chapter -11 Transport in Plants . NCERT Solutions will make you understand the topics in most simple manner and grasp it easily to perform better.

Class 11th Chapter -11 Transport in Plants | NCERT BIOLOGY SOLUTION |

Exercises

Page No: 193

1. What are the factors affecting the rate of diffusion?

Answer

Factors affecting the rate of diffusion are:

→ Gradient of concentration

→ Permeability of the membrane separating them

→ Temperature and pressure

2. What are porins? What role do they play in diffusion?

Answer

Porins are proteins that form huge pores in the outer membranes of the plastids, mitochondria and some bacteria allowing molecules up to the size of small proteins to pass through. They help in facilitating the passive transport of small-sized protein molecules.

3. Describe the role played by protein pumps during active transport in plants.

Answer

Proteins pumps are made up of specific proteins called trans-membrane proteins which use energy to carry substances across the cell membrane. These pumps can transport substances from a low concentration to a high concentration. The rate of transport reaches the maximum when all the protein transporters are being used or are saturated.

4. Explain why pure water has the maximum water potential.

Answer

Water molecules possess kinetic energy. In liquid and gaseous form, they are in random motion that is both rapid and constant. The greater the concentration of water in a system, the greater is its kinetic energy or water potential. Pure water has the highest concentration of water molecules. Therefore, it has the highest water potential. When some solute is dissolved in water, the water potential of pure water decreases.

5. Differentiate between the following:

(a) Diffusion and Osmosis
(b) Transpiration and Evaporation
(c) Osmotic Pressure and Osmotic Potential
(d) Imbibition and Diffusion
(e) Apoplast and Symplast pathways of movement of water in plants.
(f) Guttation and Transpiration.

Answer

(a) Diffusion and Osmosis

Diffusion	Osmosis
It is a movement of molecules from high concentration to low concentration.	It is a movement of molecules from high concentration to low concentration through semi permeable membrane.
It can occur in solids, liquids, and gases.	It occurs in the liquid medium.
It does not require any driving Force.	It occurs in response to a driving force.

(b) Transpiration and Evaporation

Transpiration	Evaporation
It occurs in plants.	It occurs from any free surface and involves living and non-living surfaces.
It occurs mainly through the stomatal pores on plant leaves.	It occurs through any free surface.
It is controlled by environmental factors as well as physiological factors of plants	It is entirely driven by environmental factors.

(c) Osmotic Pressure and Osmotic Potential

Osmotic Pressure	Osmotic Potential
It is expressed in bars with a positive sign.	It is expressed in bars with a negative sign.
It is a positive pressure.	It is a negative pressure.
Its value increases with an increase in the concentration of solute particles.	Its value decreases with an increase in the concentration of solute particles.

(d) Imbibition and Diffusion

Imbibition	Diffusion
Imbibition is a special type of diffusion when water is absorbed by solids and colloids causing them to enormously increase in volume.	Diffusion is the passive movement of particles, ions, and molecules along the concentration gradient.
It usually involves water.	It involves solids, liquids, and gases.

(e) Apoplast and Symplast pathways of movement of water in plants.

Apoplast pathways	Symplast pathways
It is the system of adjacent cell walls that is continuous throughout the plant, except at the casparian strips of the endodermis in the roots	It is the system of interconnected protoplasts.
Water moves through the intercellular spaces and the walls of cells	Water travels through the cytoplasm of cells and intercellular movement is through plasmodesmata.
It is a faster process of water movement and water moves through mass flow.	It is a slower process of water movement.

(f) Guttation and Transpiration.

Guttation	Transpiration
It occurs through the vein endings of leaves.	It occurs through the stomata.
Water is lost from the leaves in the form of liquid droplets.	Water is lost from the leaves in the form of water vapour.
It occurs usually at night.	It occurs usually during the day.
It is regulated by humidity, temperature and presence of water in soil.	It is regulated by a number of external and internal factors such as relative humidity, temperature opening and closing of stomata, etc.

6. Briefly describe water potential. What are the factors affecting it?

Answer

Water potential is the potential energy of water to move from one part to the other during various cellular processes such as diffusion, osmosis, etc. It is denoted by the Greek letter Psi or Ψ and is expressed in Pascals (Pa). A system with higher concentration of water has a higher kinetic energy or water potential. Pure water has the highest water potential while Solutions have lower water potential than it.

Solute potential and pressure potential are the two factors which affect water potential.

7. What happens when a pressure greater than the atmospheric pressure is applied to pure water or a solution?

Answer

If a pressure greater than atmospheric pressure is applied to pure water or a solution than its ater potential increases because of pressure potential which develops due to increased pressure. It is equivalent to pumping water from one place to another.

8. (a) With the help of well-labelled diagrams, describe the process of plasmolysis in plants, giving appropriate examples.

(b) Explain what will happen to a plant cell if it is kept in a solution having higher water potential.

Answer

(a) Plasmolysis occurs when water moves out of the cell and the cell membrane of a plant cell shrinks away from its cell wall. This happens when the plant cell is placed in a hypertonic solution (a solution having more solute concentration than the cell cytoplasm). This causes the water to move out of the cell and toward the solution. The cytoplasm of the cell shrinks and the cell is said to be plasmolysed. This process can be observed in an onion peel kept in a highly concentrated salt solution.

(b) A hypotonic solution has higher water potential. When a plant cell is placed in a hypotonic solution or a solution having higher water potential, water diffuses into the cell causing the cytoplasm to build up a pressure against the wall, called turgor pressure. This leads to the cell getting swollen. But the rigidity of the cell wall, prevents the cell from bursting.

9. How is the mycorrhizal association helpful in absorption of water and minerals in plants?

Answer

Mycorrhiza is a symbiotic association of a fungus with a root system. This fungal hyphae either form a dense network around the young roots or they penetrate the cells of the roots. The large surface area of the fungal hyphae is helpful in increasing the absorption of water and minerals from the soil. In return, they get sugar and nitrogenous compounds from the host plants.

10. What role does root pressure play in water movement in plants?

Answer

Root pressure is the positive pressure which can be responsible for pushing up water to small heights in the stem by the active absorption of nutrients from the soil. However, root pressure does not play a major role in water movement in tall plants. Root pressure contributes towards reestablishment of continuous chains of water molecules in the xylem which often break under enormous tensions created by transpiration pull.

11. Describe transpiration pull model of water transport in plants. What are the factors influencing transpiration? How is it useful to plants?

Answer

Transpiration creates a suction force inside the xylem. This suction force is called transpiration pull. This is powerful enough to pull the water column from beneath. Adhesion, cohesion and surface tension are the important physical properties of water which further help in the upward movement of water through xylem.

Factors that influence transpiration are temperature, light, humidity, wind speed, number and distribution of stomata, number of stomatal aperture with guard cells open, water status of the plant, canopy structure, etc.

It is useful to plants in following ways:
• creates transpiration pull for absorption and transport of plants
• supplies water for photosynthesis
• transports minerals from the soil to all parts of the plant
• cools leaf surfaces, sometimes 10 to 15 degrees, by evaporative cooling

• maintains the shape and structure of the plants by keeping cells turgid.

12. Discuss the factors responsible for ascent of xylem sap in plants.

Answer

Transpirational pull is responsible for the ascent of water in the xylem. This ascent of water is dependent on the following physical factors:

→ Cohesion – Mutual attraction between water molecules
→ Surface tension – Responsible for the greater attraction between water molecules in liquid phase than in gaseous phase
→ Adhesion – Attraction of water molecules to polar surfaces

→ Capillarity -Ability of water to rise in thin tubes
These physical properties give water high tensile strength, i.e., an ability to resist a pulling force and high capillarity, i.e., the ability to rise in thin tubes. The thin tubes of xylem work like capillary tubes.

13. What essential role does the root endodermis play during mineral absorption in plants?

Answer

In plants, nutrients are absorbed through the active and passive transports. The endodermis of roots have many transport proteins embedded in their plasma membrane. Transport proteins in endodermis cells enable plant cells to adjust the quantity and types of solutes to be absorbed from the soil. It regulates the quantity and type of minerals and ions that reach the xylem tissue of plants.

14. Explain why xylem transport is unidirectional and phloem transport bi-directional.

Answer

Water transported through xylem is utilised in photosynthesis and most of the water is lost through transpiration. Renewed demand for water is once again supplied through the same channel. Hence, transport through xylem is unidirectional.

In phloem, food is transported from source to sink. Leaves are the usual source and storage organs are the usual sink. But the storage organs become source when new buds emerge during early spring. Thus, a reverse flow of food is required. Hence, movement through phloem is bi-directional.

15. Explain pressure flow hypothesis of translocation of sugars in plants.

Answer

According to the pressure flow hypothesis, food is prepared in the plant leaves in the form of glucose. Before moving into the source cells present in the phloem, the prepared food is converted into sucrose. Water moves from the xylem vessels into the adjacent phloem, thereby increasing the hydrostatic pressure in the phloem. Consequently, the sucrose moves through the sieve cells of the phloem. The sucrose already present in the sink region is converted into starch or cellulose, thereby reducing the hydrostatic pressure in the sink cells. Hence, the pressure difference created between the source and the sink cells allows sugars to be translocated from the former to the latter. This starch or cellulose is finally removed from the sink cells through active transport.

16. What causes the opening and closing of guard cells of stomata during transpiration?

Answer

The immediate cause of the opening or closing of the Stomata is a change in the turgidity of the guard cells. The inner walls of the guard cells are thick and elastic, while the outer walls are thin. When turgidity increases within the two guard cells flanking each stomatal aperture or pore, the thin outer walls bulge out and force the inner walls into a crescent shape. The stomatal opening is facilitated by the radial arrangement of the microfibrils. At the time of the closing of the stomata, the guard cells lose their turgidity, the outer and inner walls retain their original shapes, and the microfibrils get arranged longitudinally.

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CLASS 11th CHAPTER -10 Cell Cycle and Cell Division |Biology | NCERT SOLUTION| EDUGROWN

NCERT Solutions for Class 11 Biology Chapter – 10 Cell and Cell Division includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Biology exams must go through NCERT Solutions for Class 11 Chapter -10 Cell and Cell Division . NCERT Solutions will make you understand the topics in most simple manner and grasp it easily to perform better.

Class 11th Chapter -10 Cell and Cell Division| NCERT BIOLOGY SOLUTION |

Page No: 171

Exercises

1. What is the average cell cycle span for a mammalian cell?

Answer
The average cell cycle span for a mammalian cell is approximately 24 hours.

2. Distinguish cytokinesis from karyokinesis.

Answer

Cytokinesis	Karyokinesis
Cytokinesis is the biological process involving the division of a cell’s cytoplasm during mitosis or meiosis.	Karyokinesis is the biological process involving the division of a cell’s nucleus during mitosis or meiosis.
Stages such as prophase, metaphase, anaphase, and telophase are not present in cytokinesis.	It is divided into four stages: prophase, metaphase, anaphase, and telophase

3. Describe the events taking place during interphase.

Answer
The interphase, though called the resting phase, is the time during which the cell is preparing for division by undergoing cell growth and DNA replication. The interphase is-divided into three further phases:

→ G₁ phase – It is the stage during which the cell grows and prepares its DNA for replication. In this phase, the cell is metabolically active.

→ S phase – It is the stage during which DNA synthesis occurs. In this phase, the amount of DNA (per cell) doubles, but the chromosome number remains the same.

→ G₂ phase – In this phase, the cell continues to grow and prepares itself for division. The proteins and RNA required for mitosis are synthesised during this stage.

4. What is G₀(quiescent phase) of cell cycle?

Answer
G₀ or quiescent phase is the stage in which cells are metabolically active but no longer proliferate unless called on to do so depending on the requirement of the organism.

Page No: 172

5. Why is mitosis called equational division?

Answer
During mitosis, the genetic constitution of the daughter cells is identical to that of parent cells. A diploid cell produces four diploid cells. Since, the chromosome number of the daughter cells remains identical to that of the parent cell, it is called as Equational division.

6. Name the stage of cell cycle at which one of the following events occur:
(i) Chromosomes are moved to spindle equator
(ii) Centromere splits and chromatids separate
(iii) Pairing between homologous chromosomes takes place
(iv) Crossing over between homologous chromosomes takes place

Answer
(i) Metaphase
(ii) Anaphase
(iii) Zygotene of meiosis I
(iv) Pachytene of meiosis I

7. Describe the following:
(a) synapsis (b) bivalent (c) chiasmata
Draw a diagram to illustrate your answer.
Answer

(i) The pairing of homologous chromosomes is called synapsis. This occurs during the second stage of prophase I or zygotene.

(ii) Bivalent or tetrad is a pair of synapsed homologous chromosomes. They are formed during the zygotene stage of prophase I of meiosis.

(iii) Chiasmata is the site where two non-sister chromatids of homologous chromosomes have crossed over. It represents the site of cross-over. It is formed during the diplotene stage of prophase I of meiosis.

8. How does cytokinesis in plant cells differ from that in animal cells?

Answer

Cytokinesis	Animal Cells
The division of the cytoplasm takes place by cell plate formation.	The division of the cytoplasm takes place by cleavage.
Cell plate formation starts at the centre of the cell and grows outward, toward the lateral walls.	Cleavage starts at the periphery and then moves inward, dividing the cell into two parts.

9. Find examples where the four daughter cells from meiosis are equal in size and where they are found unequal in size.

Answer

Examples are:

→ During microsporogenesis, four daughter cells formed in flowering plants are equal in size while during megasporogenesis, four daughter cells formed in flowering plants are unequal in size.

→ During Spermatogenesis or the formation of sperms in human beings occurs by the process of meiosis, four daughter cells formed are equal in size while during Oogenesis or the formation of ovum in human beings occurs by the process of meiosis, four daughter cells formed are unequal in size.

10. Distinguish anaphase of mitosis from anaphase I of meiosis.

Answer

Anaphase of mitosis	Anaphase I of meiosis
Anaphase is the stage during which the centromere splits and the chromatids separate. The Chromatids move to opposite poles.	The homologous chromosomes separate, while sisterchromatids remain associated at their centromeres

11. List the main differences between mitosis and meiosis.

Answer

Mitosis	Meiosis
This type of division takes place in somatic cells	This type of division takes place in gametic cells
Two daughter cells are formed	Four daughter cells are formed
Number of chromosomes remains diploid in daughter cells	Number of chromosomes becomes haploid in daughter cells are formed
Mitosis is necessary for growth and repair	Meiosis is necessary for sexual reproduction
Crossing over does not take place	Crossing over takes place

12. What is the significance of meiosis?

Answer
Significance of meiosis are:
→ Conservation of specific chromosome number of each species is achieved across successive generations in sexually reproducing organisms through meiosis.
→ It also increases the genetic variability in the population of organisms from one generation to the next.

13. Discuss with your teacher about
(i) haploid insects and lower plants where cell-division occurs, and
(ii) some haploid cells in higher plants where cell-division does not occur.

Answer
(i) Male bees, wasps and ants are haploid as they are produced from unfertilized eggs.

(ii) Cell division does not happen in synergids and antipodal cells in ovule.

14. Can there be mitosis without DNA replication in S phase?

Answer
No, there cannot be mitosis without DNA replication in S phase because additional DNAs are required for the formation of new cells. DNA duplication is important as it maintains the chromosome number in the daughter cells and hence Mitosis is an equational division. Therefore, the duplication of DNA is an essential step and without it, no mitosis can take place.

15. Can there be DNA replication without cell division?

Answer

DNA replication can take place without cells division as in case of formation of new mitochondria and chloroplasts. During cell division, the parent cell gets divided into two daughter cells. However, if there is a repeated replication of DNA without any cell division, then this DNA will keep accumulating inside the cell. This would increase the volume of the cell nucleus, thereby causing cell expansion.

16. Analyse the events during every stage of cell cycle and notice how the following two parameters change
(i) Number of chromosomes (N) per cell
(ii) Amount of DNA content (C) per cell

Answer
(i) The number of chromosomes (N) remains constant throughout the different stages in mitosis. In meiosis the number of chromosome is reduced to half that is from diploid to haploid. During the anaphase I stage this half number is carried on till the end of meiosis.

(ii) The amount of DNA (C) becomes double during the interphase stage which is followed by mitosis. This doubling occurs due to the replication of the DNA.

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CLASS 11th CHAPTER – 9 Biomolecules |Biology | NCERT SOLUTION| EDUGROWN

NCERT Solutions for Class 11 Biology Chapter – 9 Biomolecules includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Biology exams must go through NCERT Solutions for Class 11 Chapter – 9 Biomolecules. NCERT Solutions will make you understand the topics in most simple manner and grasp it easily to perform better.

Class 11th Chapter -9 Biomolecules | NCERT BIOLOGY SOLUTION |

Page No: 160

Exercises

1. What are macromolecules? Give examples.

Answer
Biomolecules with molecular weights more than one thousand Dalton are called macromolecules. These are found in the acid-insoluble fraction. Examples: Protein, polysaccharides, lipids, etc.

2. Illustrate a glycosidic, peptide and a phospho-diester bond.

Answer

(a) Glycosidic bond is formed normally between carbon atoms, 1 and 4, of neighbouring monosaccharide units.

(b) Peptide bond is a covalent bond that joins the two amino acids by – NH – CO linkage.

(c) Phosphodiester bond is a strong covalent bond between phosphate and two sugar groups. Such bonds form the sugar phosphate backbone of nucleic acids.

3. What is meant by tertiary structure of proteins?

Answer
The helical polypeptide chain undergoes coiling and folding to form a complex three-dimensional shape is called tertiary structure of proteins. These coils and folds are arranged to hide the non-polar amino acid chains and to expose the polar side chains. The tertiary structure is held together by the weak bonds formed between various parts of the polypeptide chain.

4. Find and write down structures of 10 interesting small molecular weight biomolecules. Find if there is any industry which manufactures the compounds by isolation. Find out who are the buyers.

Answer

10.

Fat and many hormones are manufactured by isolation. Pharmaceutical and consumer goods industry can be the major buyers of these products.

5. Proteins have primary structure. If you are given a method to know which amino acid is at either of the two termini (ends) of a protein, can you connect this information to purity or homogeneity of a protein?

Answer

Yes, we can connect the given information to purity or homogeneity of a protein. It is known that an accurate sequence of a certain amino acid is very important for the functioning of a protein. If there is any change in the sequence, it would alter its structure, thereby altering the function. If we are provided with a method to know the sequence of an unknown protein, then using this information, we can determine its structure and compare it with any of the known correct protein sequence. Any change in the sequence can be linked to the purity or homogeneity of a protein.

6. Find out and make a list of proteins used as therapeutic agents. Find other applications of proteins (e.g., cosmetics, etc.)

Answer

List of proteins used as therapeutic agents are Insulin, Oxytocin, Antidiuretic Hormone (ADH), Thrombin, Fibrinogen, Renin, Immunoglobulin, Diastase and Streptokinase.

Proteins are also commonly used in the manufacture of cosmetics, toxins, and as biological buffers.

7. Explain the composition of triglyceride.

Answer

Triglycerides are composed of two types of molecules, i.e., glycerol (3 carbon molecules) and fatty acids which attach to the glycerol at the alcohol unit. It is mainly present in vegetable oils and animal fat. The structure is given:

Page No: 161

8. Can you describe what happens when milk is converted into curd or yoghurt from your understanding of proteins.

Answer

Milk is converted into curd or yogurt by the process of fermentation. Milk consists of globular proteins. During fermentation, the milk sugar (lactose) produces lactic acid. Lactic acid acts on the globular proteins present in the milk and denatures them. This denaturation destroys the tertiary and quaternary structures of proteins and the globular proteins are converted into fibrous proteins thus giving a thick texture to the milk due to coagulation of the proteins.

9. Can you attempt building models of biomolecules using commercially available atomic models (Ball and Stick models).

Answer

Yes. The biomolecules can be represented by the ball and stick model. The bonds which hold the atoms are represented by sticks, whereas the atoms are represented by balls.
Example: In the model of D-glucose, the oxygen atoms are represented by red balls, the hydrogen atoms by blue balls, while the carbon atoms are represented by grey balls.

10. Attempt titrating an amino acid against a weak base and discover the number of dissociating (ionizable) functional groups in the amino acid.

Answer

Titratinga neutral or basic amino acid against a weak base will dissociate only one functional group, whereas titration between acidic amino acid and a weak acid will dissociate two or more functional groups.

11. Draw the structure of the amino acid, alanine.
Answer

12. What are gums made of? Is Fevicol different?

Answer

Gums are natural heteropolysaccharides which are formed of a large number of different monosaccharide units linked together by glycosidic bonds.

Yes, Fevicol is different from gums as it comprises synthetic polymers.

13. Find out a qualitative test for proteins, fats and oils, amino acids and test any fruit juice, saliva, sweat and urine for them.

Answer
Qualitative tests for proteins, amino acids and fats:

• Biuret test: The Biuret test for protein identifies the presence of protein by producing light blue to purple colour of the solution.

• Grease test for oil: Certain oils give a translucent stain on brown paper. This test can be used to show the presence of fat in vegetable oils.

• Ninhydrin test: If Ninhydrin reagent is added to the solution, then the colourless solution changes to pink, blue or purple colour depending on the type of amino acid.

Item		Name of the test	Procedure	Result	Inference
1.	Fruit juice	Biuret’s test	Fruit juice + Biuret’s reagent	Colour changes from light blue to purple	Protein is present.
		Grease test	To a brown paper, add a few drops of fruit juice.	No translucent spot	Fats and oils are absent or are in negligible amounts.
		Ninhydrin test	Fruit juice + Ninhydrin reagent + boil for 5 minutes	Colourless solution changes to pink, blue, or purple colour	Amino acids are present.
2.	Saliva	Biuret’s test	Saliva + Biuret’s reagent	Colour changes from light blue to purple	Proteins are present.
		Grease test	On a brown paper, add a drop of saliva.	No translucent spot	Fats/oils are absent.
		Ninhydrin test	Saliva + Ninhydrin reagent + boil for 5 minutes	Colourless solution changes to pink, blue, or purple colour	Amino acids are present.
3.	Sweat	Biuret’s test	Sweat + Biuret’s reagent	No colour change	Proteins are absent.
		Solubility test	Sweat + Water	Oily appearance	Fats/oil may be present.
		Ninhydrin test	Sweat + Ninhydrin reagent + boil for 5 minutes	No colour change, solution remains colourless	Amino acids are absent.
4.	Urine	Biuret’s test	Few drops of urine + Biuret’s reagent	Colour changes from light blue to purple	Proteins are present.
		Solubility test	Few drops of urine + Water	Little bit of oily appearance	Fats may or may not be present.
		Ninhydrin test	Few drops of urine + Ninhydrin reagent + boil for 5 minutes	Colourless solution changes to pink, blue, or purple colour depending on the type of amino acid	Amino acids are present.

14. Find out how much cellulose is made by all the plants in the biosphere and compare it with how much of paper is manufactured by man and hence what is the consumption of plant material by man annually. What a loss of vegetation!

Answer
Approximately, 100 billion tonnes of cellulose are made per year by all the plants in the biosphere and it takes 17 full grown trees to make one ton of paper. Trees are also used to fulfil the other requirements of man such as for timber, food, medicines, etc. Hence, it is difficult to calculate the annual consumption of plant material by man.

15. Describe the important properties of enzymes.
Answer
Important properties of enzymes are:
→ Enzymes are generally complex macromolecules of globular proteins with high molecular weight.
→ Enzymes do not start a reaction but increase the rate of chemical reaction.
→ Enzyme itself is not used up in the reaction it catalyzes so it can be used over again and again.
→ Enzymes have a high turn over number. It means the number of reactions that an enzyme molecule can catalyze per second.
→ Enzymes are specific in action.
→ The action of enzyme is greatly affected by temperatures. Their activity decreases with increase in temperature.
→ They show maximum activity at an optimum pH of 6 – 8.

→ The velocity of enzyme increases with increase in substrate concentration and then, ultimately reaches maximum velocity.

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CLASS 11th CHAPTER -8 Cell : The Unit of Life |Biology | NCERT SOLUTION| EDUGROWN

NCERT Solutions for Class 11 Biology Chapter – 8 Cell : The Unit of Life includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Biology exams must go through NCERT Solutions for Class 11 Chapter – 8 Cell : The Unit of Life. NCERT Solutions will make you understand the topics in most simple manner and grasp it easily to perform better.

Class 11th Chapter -8 Cell : The Unit of Life | NCERT BIOLOGY SOLUTION |

Page No: 141

Exercises

1. Which of the following is not correct?
(a) Robert Brown discovered the cell.
(b) Schleiden and Schwann formulated the cell theory.

(c) Virchow explained that cells are formed from pre-existing cells.
(d) A unicellular organism carries out its life activities within a single cell.

► (a) Robert Brown discovered the cell.

2. New cells generate from
(a) bacterial fermentation

(b) regeneration of old cells
(c) pre-existing cells

(d) abiotic materials

► (c) pre-existing cells

3. Match the following:

Column I	Column II
(a) Cristae (b) Cisternae (c) Thylakoids	(i) Flat membranous sacs in stroma (ii) Infoldings in mitochondria (iii) Disc-shaped sacs in Golgi apparatus

Answer

Column I	Column II
(a) Cristae (b) Cisternae (c) Thylakoids	(i) Infoldings in mitochondria (ii) Disc-shaped sacs in Golgi apparatus (iii) Flat membranous sacs in stroma

4. Which of the following is correct:
(a) Cells of all living organisms have a nucleus.
(b) Both animal and plant cells have a well defined cell wall.
(c) In prokaryotes, there are no membrane bound organelles.
(d) Cells are formed de novo from abiotic materials.
► (c) In prokaryotes, there are no membrane bound organelles.

5. What is a mesosome in a prokaryotic cell? Mention the functions that it performs.

Answer
Mesosome is a convoluted membranous structure formed in a prokaryotic cell by the invagination of the plasma membrane. Its functions are:
→ These extensions help in cell wall formation, DNA replication and distribution to daughter cells.
→ They also help in respiration, secretion processes, to increase the surface area of the plasma membrane and enzymatic content.

6. How do neutral solutes move across the plasma membrane? Can the polar molecules also move across it in the same way? If not, then how are these transported across the membrane?

Answer
Neutral solutes move across the plasma membrane through osmosis. On the other hand polar molecules cannot pass through the non-polar membrane. They require a carrier protein to help them transverse across the membrane. Such a transport requires energy and is called active transport.

7. Name two cell-organelles that are double membrane bound. What are the characteristics of these two organelles? State their functions and draw labelled diagrams of both.

Answer
Mitochondria and chloroplasts are the two organelles that are double-membrane-bound.

Characteristics of Mitochondria are:
→ The mitochondria are sausage-shaped or cylindrical having a diameter of 0.2-1.0 µm and average 0.5 µm and length 1.0-4.1 µm.
→ Each mitochondrion is a double membrane bound structure with the outer membrane and the inner membrane dividing its lumen distinctly into two aqueous compartments,- outer and inner compartments.

→ The outer membrane forms the continuous limiting boundary of the organelle.
→ The inner membrane forms a number of infoldings called the cristae towards the matrix.
→ The two membranes have their own specific enzymes associated with the mitochondrial function. The matrix of mitochonrdria also possess single circular DMA molecule, a few RNA molecules, ribosomes (70 S) and the components required for the synthesis of proteins.

Functions of Mitochondria are:
→ They are the sites for aerobic respiration.
→ They provide energy in the form of ATP for all vital activities of living cells.
→ They have their own DNA and ribosomes. Hence, they are regarded as semi-autonomous organelles.
→ They have several enzymes, intermediately required for the synthesis of various chemicals such as fatty acids, steroids, and amino acids.

Characteristics of Chloroplast are:

→ The Chloroplasts are also double membrane bound organelles.

→ The space limited by the inner membrane of the chloroplast is called the stroma.

→ A number of organized flattened membranous sacs called the thylakoids are present in the stroma.
→ Thylakoids are arranged in stacks-like the piles of coins called grana.→ In addition, there are flat membranous tubules called the stroma lamellae connecting the thylakoids of the different grana.
→ The membrane of the thylakoids enclose a space called a lumen.

Functions of Chloroplast are:
→ They trap solar energy and utilise it for manufacturing food for plants. Hence, they are involved in the process of photosynthesis.
→ They contain the enzymes required for the synthesis of carbohydrates and proteins.

8. What are the characteristics of prokaryotic cells?

Answer

Prokaryotic cell is a unicellular organism lacking membrane-bound organelles.

The characteristics of prokaryotic cells are as follows:
→ Most of them are unicellular.
→ They are generally small in size. The size of a prokaryotic cell varies from 0.5 – 5 µm.
→ The nuclear region of a prokaryotic cell is poorly-defined because of the absence of a nuclear membrane. Hence, a prokaryotic cell lacks a true nucleus.

→ The genetic materials of prokaryotic cells are naked. They contain single, circular chromosomes. In addition to the genomic DNA, they have a small, circular plasmid DNA.
→ They have specialised membranous structures called mesosomes. Mesosomes are formed by the invagination of the cell membrane. These extensions help in the synthesis of the cell wall, replication of DNA. They also help in the equal distribution of chromosomes into the daughter cells.
→ Membrane-bound cell organelles such as mitochondria, plastids, and endoplasmic reticulum are absent from a prokaryotic cell.
→ Most prokaryotic cells contain a three-layered structure – outermost glycocalyx, middle cell wall, and the innermost plasma membrane. This structure acts as a protective unit.

9. Multicellular organisms have division of labour. Explain.

Answer
The body of a multicellular organism has cell as a basic structural unit. The cells organised to form tissues such as blood, bone, etc. The tissues organised to form organs such as heart, kidney, etc. The organs then organised to form organ systems such as digestive system, reproductive system and respiratory system, etc. The various organ systems of organism get arranged to form a complete individual.

10. Cell is the basic unit of life. Discuss in brief.

Answer
Every animal or plant is made of many systems. Every system is made of many organs. Organs are made of many types of tissues and tissues are made of many cells. A cell is an autonomous structure and is capable of carrying out various functions on its own. A cell can do all an organism can do. In a sense cell is full of life at the minutest possible level. That is why a cell is called the basic unit of life.

11. What are nuclear pores? State their function.
Answer

Nuclear pores are tiny holes present in the nuclear membrane of the nucleus. They are formed by the fusion of two nuclear membranes.
These holes allow specific substances to be transferred into a cell and out from it. They allow molecules such as RNA and proteins to move in both directions, between the nucleus and the cytoplasm.

12. Both lysosomes and vacuoles are endomembrane structures, yet they differ in terms of their functions. Comment.

Answer
Both lysosomes and vacuoles are covered by a single membrane. Both of them perform different types of functions. Lysosomes contain hydrolysing enzymes and can hydrolyse all types of organic substances, except cellulose. They perform phagocytic function. Therefore, they are known as suicidal bags.

The vacuoles are non-cytoplasmic sacs which are also covered by a membrane. The sap vacuoles store sap or water with dissolved organic and inorganic substances. They maintain osmotic pressure or turgidity. Some freshwater invertebrates such as Amoeba, Paramecium occur contractile vacuoles, which perform osmoregulation and excretion. There is another type of vacuoles such as food vacuole which store food and gas vacuoles which store metabolic gases and take part in buoyancy regulation.

13. Describe the structure of the following with the help of labelled diagrams.
(i) Nucleus (ii) Centrosome

Answer
Nucleus controls all the cellular activities of the cell. It is spherical in shape. It is composed of the following structures:

• Nuclear membrane: It is a double membrane separating the contents of the nucleus from the cytoplasm. The narrow space between the two membranes is called the perinuclear space. Nuclear membrane has tiny holes called nuclear pores. These holes allow specific substances to be transferred into a cell and out from it.

• Nucleoplasm/Nuclear matrix: It is a homogenous granular fluid present inside the nucleus. It contains the nucleolus and chromatin. Nucleolus is a spherical structure that is not bound by any membrane. It is rich in protein and RNA molecules, and is the site for ribosome formation. Chromatin is an entangled mass of thread-like structures. It contains DNA and some basic proteins called histones.

Centrosome

Centrosome consists of two cylindrical structures called centrioles. Centrioles lie perpendicular to each other. Each has a cartwheel-like organisation.
A centriole is made up of microtubule triplets that are evenly spaced in a ring. The adjacent triplets are linked together. There is a proteinaceous hub in the central part of a centriole. The hub is connected to the triplets via radial spokes. These centrioles help in organising the spindle fibres and astral rays during cell division. They form the basal body of cilia and flagella.

14. What is a centromere? How does the position of centromere form the basis of classification of chromosomes. Support your answer with a diagram showing the position of centromere on different types of chromosomes.

Answer

Centromere is a narrow non-stainable area, which join two similar threads or chromatids of a late prophase or metaphase chromosome. The two parts of the chromosome on either side of the centromere are known as arm. They may be isobranchial (equal) or helerobranchial (unequal in length). Depending upon the position of the centromere, the chromosomes are classified as:
→ Acrocentric chromosome: Centromere sub-terminal, anaphasic stage J-shaped.
→ Sub-metacentric chromosome: The centromere is sub-median and the anaphasic chromosome appear L-shaped.
→ Metacentric chromosomes: The centromere is in the middle and the chromosome appears V-shaped.
→ Telocentric chromosome: Centromere is terminal, anaphasic stage is I-shaped.
Depending upon the number of centromeres a chromosome possess, it may be monocentric, dicentric (two centromeres), polycentric (many centromeres), acentric chromosome (having no centromere).

Types of chromosomes on number of centromere

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CLASS 11th CHAPTER -7 Structural Organisation in Animals |Biology | NCERT SOLUTION| EDUGROWN

NCERT Solutions for Class 11 Biology Chapter – 7 Structural Organisation in Animals includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Biology exams must go through NCERT Solutions for Class 11 Chapter – 7 Structural Organisation in Animals. NCERT Solutions will make you understand the topics in most simple manner and grasp it easily to perform better.

Class 11th Chapter -7 Structural Organisation in Animals | NCERT BIOLOGY SOLUTION |

Page No: 121

Exercises

1. Answer in one word or one line:
(i) Give the common name of Periplanata americana.
► American cockroach.

(ii) How many spermathecae are found in earthworm?
► Four pairs

(iii) What is the position of ovaries in cockroach?
► Between 12th and 13th abdominal segments

(iv) How many segments are present in the abdomen of cockroach?
► Ten segments

(v) Where do you find Malpighian tubules?
► Malpighian tubules are the main excretory organs of the cockroach.

Page No: 122

2. Answer the following:
(i) What is the function of nephridia?
(ii) How many types of nephridia are found in earthworm based on their location?

Answer

(i) Nephridia regulate the volume and composition of the body fluids.

(ii) There are three types of nephridia found in earthworm:
• Septal nephridia: present on both the sides of intersegmental septa of segment 15 to the last that open into intestine
• Integumentary nephridia: attached to lining of the body wall of segment 3 to the last that open on the body surface
• Pharyngeal nephridia: present as three paired tufts in the 4th , 5th and 6th segments.

3. Draw a labelled diagram of the reproductive organs of an earthworm.

Answer

4. Draw a labelled diagram of alimentary canal of a cockroach.

Answer

5. Distinguish between the following

(a) Prostomium and peristomium
(b) Septal nephridium and pharyngeal nephridium

Answer

(a)

Prostomium	Peristomium
Prostomium is a small fleshy lobe, which overhangs the mouth of an earthworm. It helps the organism push into the soil and is sensory in function.	The first body segment in the earthworm is called the peristomium. It surrounds the mouth opening.

(b)

Septal nephridium	Pharyngeal nephridium
They are present on both sides of inter-segmental septa behind the 15^th segment. They open into the intestines.	They are present as three paired tufts in the fourth, fifth, and sixth segments.

6. What are the cellular components of blood?

Answer

The cellular components of blood are red blood cells (RBC), white blood cells (WBC) and platelets.

7. What are the following and where do you find them in animal body
(a) Chondriocytes
(b) Axons
(c) Ciliated epithelium

Answer

(a) Chondriocytes are cells of cartilages and are present in small cavities within the matrix secreted by them.

(b) Axons are long, slender projections of neurons that help in carrying nerve impulses from the neuron body. Axons aggregate in bundles which make up the nerves.

(c) Ciliated epithelium are columnar or cuboidal cells bear cilia on their free surface. They are mainly present in the inner surface of hollow organs like bronchioles and fallopian tubes.

8. Describe various types of epithelial tissues with the help of labelled diagrams.

Answer

Epithelial tissue lines the surface of a body and forms a protective covering. Epithelium cells are packed tightly together with little intercellular matrix. Epithelial tissue in the body is of two types.
→ Simple epithelium: It consists of a single layer of cells where cells are in direct contact with the basement membrane. It is further sub-divided into the following types:
• Simple squamous epithelium: It consists of a single layer of flat cells with irregular boundaries. It is found in the walls of the blood vessels and in the lining of alveoli.
• Simple cuboidal epithelium: It consists of a single layer of cube-like cells. It is present in regions where secretion and absorption of substances takes place such as the proximal convoluted tubule region of the nephron.
• Simple columnar epithelium: It consists of a single layer of tall, slender cells with their nuclei present at the base of the cells. They may bear micro-villi on the free surfaces. Columnar epithelium forms the lining of the stomach and intestines, and is involved in the function of secretion andabsorption.
• Ciliated epithelium: It consists of columnar or cuboidal cells with cilia on their free surfaces. They are present in bronchioles and oviducts from where they direct mucus and eggs in specific directions.
• Glandular epithelium:It consists of columnar or cuboidal cells involved in the secretion of substances. Glands are of two types, unicellular glands (goblet cells of the alimentary canal) and multicellular glands (salivary glands). They can be classified as exocrine (ductless glands) and endocrine glands (duct glands) by the method through which they release enzymes.

→ Compound epithelium: It consists of many layers of cells. It is involved mainly in the function of providing protection and has a limited role in secretion and absorption. Examples of compound epithelium include the dry surface of the skin or moist inner lining of the buccal cavity, pharynx, pancreatic ducts, and the inner lining of ducts of salivary glands.

9. Distinguish between
(a) Simple epithelium and compound epithelium.

(b) Cardiac muscle and striated muscle
(c) Dense regular and dense irregular connective tissues
(d) Adipose and blood tissue
(e) Simple gland and compound gland

Answer

(a)

Simple epithelium	Compound epithelium
• It is composed of only one layer of cells. • It is mainly involved in the function of absorption and secretion. • It is present in the lining of the stomach, intestine.	• It is composed of many layers of cells. • It is mainly involved in the function of protection and has a limited role in absorption and secretion. • It is present in the lining of the pharynx and buccal cavity.

(b)

Cardiac muscles	Striated muscles
• They are involuntary in function. • They are multi-nucleate and branched. • They are found only in the heart.	• They are voluntary in function. • They are multi-nucleate and unbranched. • They are found only in triceps, biceps, and limbs.

(c)

Dense regular connective tissues	Dense irregular connective tissues
• In dense regular connective tissues, collagen fibres are present in rows between parallel boundless fibres. • They are present in tendons and ligaments.	• In dense irregular connective tissues, fibres are arranged irregularly. • They are present in the skin.

(d)

Adipose tissue	Blood tissue
• It is composed of collagen fibres, elastin fibres, fibroblasts, macrophages, and adipociytes. • It helps in the synthesis, storage, and metabolism of fats. • It is present beneath the skin.	• It is composed of RBCs, WBCs, platelets, and plasma. • It helps in the transportation of food, wastes, gases, and hormones. • It is present in the blood vessels.

(e)

Simple glands	Compound glands
• They contain isolated glandular cells. • They are unicellular. • Examples include goblet cells of the alimentary canal.	• They contain a cluster of secretory cells. • They are multicellular. • Examples include salivary glands.

10. Mark the odd one in each series:

(a) Areolar tissue; blood; neuron; tendon

► Neuron, because it is an example of neural tissue while all other are connective tissue.

(b) RBC; WBC; platelets; cartilage

► Cartilage, as all others are components of blood.

(c) Exocrine; endocrine; salivary gland; ligament

► Neuron, because it is an example of neural tissue while all other are connective tissue.

(d) Maxilla; mandible; labrum; antennae

► Neuron, because it is an example of connective tissue while all other are examples of simple glandular epithelium.

(e) Protonema; mesothorax; metathorax; coxa

► Antennae, as it is not a masticating part of cockroach

11. Match the terms in column I with those in Column II:

Column I	Column II
(a) Compound epithelium (b) Compound eye (c) Septal nephridia (d) Open circulatory system (e) Typhlosole (f) Osteocytes (g) Genitalia	(i) Alimentry canal (ii) Cockroach (iii) Skin (iv) Mosaic vision (v) Earthworm (vi) Phallomere (vii) Bone

Answer

Column I	Column II
(a) Compound epithelium (b) Compound eye (c) Septal nephridia (d) Open circulatory system (e) Typhlosole (f) Osteocytes (g) Genitalia	(iii) Skin (iv) Mosaic vision (v) Earthworm (ii) Cockroach (i) Alimentary canal (vii) Bone (vi) Phallomere

12. Mention briefly about the circulatory system of earthworm
Answer

Earthworm exhibits a closed type of blood vascular system. It consists of blood vessels, capillaries and heart. Due to a closed circulatory system, blood is confined to the heart and blood vessels. Contractions keep blood circulating in one direction. Smaller blood vessels supply the gut, nerve cord and body wall. Blood glands are present on the 4th, 5th and 6th segments. They produce blood cells and haemoglobin which is dissolved in blood plasma. Blood cells are phagocytic in nature.

13. Draw a neat diagram of digestive system of frog.

Answer

14. Mention the function of the following

(a) Ureters in frog
(b) Malpighian tubules
(c) Body wall in earthworm

Answer

(a) Ureters in frogs: A ureter acts as a urinogenital duct, which carries sperms along with urine in male frogs.

(b) Malphigian tubules: Malphigian tubules are excretory organs in cockroaches.

(c) Body wall in earthworms: In earthworms, the body wall consists of muscle layers. It helps in movement and burrowing.

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CLASS 11th CHAPTER -6 Anatomy of Flowering Plants |Biology | NCERT SOLUTION| EDUGROWN

NCERT Solutions for Class 11 Biology Chapter – 6 Anatomy of Flowering Plants includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Biology exams must go through NCERT Solutions for Class 11 Chapter -6 Anatomy of Flowering Plants. NCERT Solutions will make you understand the topics in most simple manner and grasp it easily to perform better.

Class 11th Chapter -6 Anatomy of Flowering Plants | NCERT BIOLOGY SOLUTION |

Page No: 99

Exercises

1. State the location and function of different types of meristem.
Answer
Meristems are specialised regions of plant growth. The meristems mark the regions where active cell division and rapid division of cells take place. Meristems are of three types depending on their location.

→ Apical meristem: These are found at the tips of roots and shoots. The shoot apical meristem is present at the tip of the shoots and its active division results in the elongation of the stem and formation of new leaves. The root apical meristem helps in root elongation.

→ Intercalary meristem: These occurs between the mature tissues present at the bases of the leaves of grasses. It helps in the regeneration of grasses after they have been grazed by herbivores.

Both, apical meristems and intercalary meristems are primary meristems because they appear early in life of a plant and contribute to the formation of the primary plant body.

→ Lateral meristem: These occurs in the mature regions of roots and shoots of many plants and helps them in adding secondary tissues to the plant body and in increasing the girth of plants. It is called the secondary meristem as it appears later in a plant’s life.

2. Cork cambium forms tissues that form the cork. Do you agree with this statement? Explain.

Answer
When secondary growth occurs in the dicot stem and root, the epidermal layer gets broken and need to be replaced to provide new protective cell layers. Hence, sooner or later, another meristematic tissue called cork cambium or 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 cells on the outer side get differentiated into the cork or phellem, while the cells on the inside give rise to the secondary cortex or phelloderm.

3. Explain the process of secondary growth in stems of woody angiosperm with help of schematic diagrams. What is the significance?
Answer
In woody dicots, the strip of cambium present between the primary xylem and phloem is called the interfascicular cambium. The interfascicular cambium is formed from the cells of the medullary rays adjoining the interfascicular cambium. This results in the formation of a continuous cambium ring. The cambium cuts off new cells toward its either sides. The cells present toward the outside differentiate into the secondary phloem, while the cells cut off toward the pith give rise to the secondary xylem. The amount of the secondary xylem produced is more than that of the secondary phloem. The secondary growth in plants increases the girth of plants, increases the amount of water and nutrients to support the growing number of leaves, and also provides support to plants.

4. Draw illustrations to bring out anatomical difference between

(a) Monocot root and dicot root
(b) Monocot stem and dicot stem

Answer

(a)

(b)

5. Cut a transverse section of young stem of a plant from your school garden and observe it under the microscope. How would you ascertain whether it is a monocot stem or dicot stem? Give reasons.

Answer

The transverse section of monocot and dicot stem are:

We can ascertain whether it is a monocot stem or dicot stem through the vascular bundles. In dicot system, they are arranged in ring while in monocot system, they are in scattered arrangement.

6. The transverse section of a plant material shows the following anatomical features, (a) the vascular bundles are conjoint, scattered and surrounded by sclerenchymatous bundle sheaths (b) phloem parenchyma is absent. What will you identify it as?

Answer

The monocot stem is characterised by conjoint, collateral, and closed vascular bundles, scattered in the ground tissue containing the parenchyma. Each vascular bundle is surrounded by sclerenchymatous bundle-sheath cells. Phloem parenchyma and medullary rays are absent in monocot stems.

7. Why are xylem and phloem called complex tissues?

Answer

Xylem and phloem are known as complex tissues as they are made up of more than one type of cells. These cells work in a coordinated manner, as a unit, to perform the various functions of the xylem and phloem.

Xylem helps in conducting water and minerals and composed of four different kinds of elements, namely, tracheids, vessels, xylem fibres and xylem parenchyma.

• Tracheids are elongated, thick-walled dead cells with tapering ends.

• Vessels are long, tubular, and cylindrical structures formed from the vessel members, with each having lignified walls and large central cavities. Both tracheids and vessels lack protoplasm.

• Xylem fibres consist of thick walls with an almost insignificant lumen. They help in providing mechanical support to the plant.

• Xylem parenchyma is made up of thin-walled parenchymatous cells that help in the storage of food materials and in the radial conduction of water.

Phloem transports food materials and composed of Sieve tube elements, Companion cells, Phloem parenchyma and Phloem fibres.
• Sieve tube elements are tube-like elongated structures associated with companion cells. The end walls of sieve tube elements are perforated to form the sieve plate. Sieve tube elements are living cells containing cytoplasm and nucleus.
• Companion cells are parenchymatous in nature. They help in maintaining the pressure gradient in the sieve tube elements.
• Phloem parenchyma helps in the storage of food and is made up of long tapering cells, with a dense cytoplasm.
• Phloem fibres are made up of elongated sclerenchymatous cells with thick cell walls.

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

Answer
Stomata are structures present in the epidermis of leaves. Stomata regulate the process of transpiration and gaseous exchange. Each stoma is composed of two bean shaped cells known as guard cells which enclose stomatal pore. The inner walls of guard cells are thick, while the outer walls are thin. The guard cells are surrounded by subsidiary cells. These are the specialised epidermal cells present around the guard cells. The pores, the guard cells, and the subsidiary cells together constitute the stomatal apparatus.

9. Name the three basic tissue systems in the flowering plants. Give the tissue names under each system.

Answer

Tissue System	Tissues present
Epidermal tissue system	Epidermis, trichomes, hairs, stomata
Ground tissue system	Parenchyma, collenchyma, sclerenchyma, mesophyll
Vascular tissue system	Xylem, phloem, cambium

10. How is the study of plant anatomy useful to us?

Answer
The study of plant anatomy is useful in many ways. It helps us understand the way a plant functions carrying out its routine activities like transpiration, photosynthesis and growth and repair. It also helps botanists and agriculture scientists to understand the disease and cure for plants. The study of plant-structure allows us to predict the strength of wood. This is useful in utilising it to its potential. The study of various plant fibres such as jute, flax, etc., helps in their commercial exploitation.

11. What is periderm? How does periderm formation take place in dicot stem?

Answer
Periderm is composed of the phellogen, phellem, and phelloderm.
During secondary growth, the outer epidermal layer and the cortical layer are broken because of the cambium. To replace them, the cells of the cortex turn meristematic, giving rise to cork cambium or phellogen. It is composed of thin-walled, narrow and rectangular cells.

Phellogen cuts off cells on its either side. The cells cut off toward the outside give rise to the phellem or cork. The suberin deposits in its cell wall make it impervious to water. The inner cells give rise to the secondary cortex or phelloderm. The secondary cortex is parenchymatous.

12. Describe the internal structure of a dorsiventral leaf with the help of labelled diagrams.

Answer
Dorsiventral leaves are found in dicots. The vertical section of a dorsiventral leaf contains three distinct parts:
→ Epidermis: The epidermis is present on both upper surface (adaxial epidermis) and lower surface (abaxial epidermis) of the leaf has a conspicuous cuticle.. It is made up of elongated, compactly arranged cells, which form a continuous layer. The epidermis on the outside is covered with a thick cuticle. Abaxial epidermis bears more stomata than the adaxial epidermis.

→ Mesophyll: It is a tissue of the leaf present between the adaxial and abaxial epidermises. It is differentiated into the palisade parenchyma (composed of tall, compactly-placed cells) and the spongy parenchyma (comprising oval or round, loosely-arranged cells with inter cellular spaces). Mesophyll contains the chloroplasts which perform the function of photosynthesis.

→ Vascular system: Vascular system 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.

TS of dicot leaf

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CLASS 11th CHAPTER -5 Morphology of Flowering Plants |Biology | NCERT SOLUTION| EDUGROWN

NCERT Solutions for Class 11 Biology Chapter – 5 Morphology of Flowering Plants includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Biology exams must go through NCERT Solutions for Class 11 Chapter -5 Morphology of Flowering Plants. NCERT Solutions will make you understand the topics in most simple manner and grasp it easily to perform better.

Class 11th Chapter -5 Morphology of Flowering Plants | NCERT BIOLOGY SOLUTION |

Page No: 82

Exercises

1. What is meant by modification of root? What type of modification of root is found in the
(a) Banyan tree
(b) Turnip

(c) Mangrove trees

Answer
The primary function of the root is absorption of water and minerals and giving mechanical support to the plant. Sometimes roots get modified to carry out other functions like storage of food or additional mechanical support.

(a) Banyan Tree: In banyan trees long roots develop from branches and they go deep down to reach the ground. They are for providing additional mechanical support to the huge banyan tree. This modification is called as prop root.

(b) Turnip: In turnip the root is modified to store extra food.

(c) Mangrove trees: Mangrove trees are found in marshy area. The roots get modified into pneumatic structures providing extra passage to allow additional oxygen to the plant.

2. Justify the following statements on the basis of external features
(i) Underground parts of a plant are not always roots
(ii) Flower is a modified shoot
Answer
(i) In some plants stems remain underground to carry out some additional functions, like food storage and vegetative reproduction. For example, potatoes are modified form of underground stems.

(ii) When the stem takes on the role of sexual reproduction it is modified into a flower. That is why flower id often called a modified shoot.

3. How is pinnately compound leaf different from palmately compound leaf?
Answer
In pinnately compound leaf a number of leaflets are attached on a common axis. Example, neem leaves while in palmately compound leaf number of leaflets are attached at the common point on the leaf stalk. Example, cotton leaves.

4. Explain with suitable examples the different types of phyllotaxy?
Answer
Phyllotaxy is the pattern of arrangement of leaves on the stem or branch. It is of three types: alternate, opposite and whorled.

• Alternate Arrangement of Leaves: In alternate type of phyllotaxy, a single leaf arises at each node in alternate manner, as in china rose, mustard and sun flower plants.

• Opposite Arrangement of Leaves: In opposite type, a pair of leaves arise at each node and lie opposite to each other as in Calotropis and guava plants.

• Whorled Arrangement of Leaves: If more than two leaves arise at a node and form a whorl, it is called whorled, as in Alstonia.

Page No: 83

5. Define the following terms:
(a) Aestivation
(b) Placentation
(c) Actinomorphic
(d) Zygomorphic
(e) Superior ovary
(f) Perigynous flower
(g) Epipetalous Stamen

Answer
(a) The mode of arrangement of sepals or petals in floral bud with respect to the other members of the same whorl is known as aestivation. The main types of aestivation are valvate, twisted, imbricate and vexillary.

(b) The arrangement of ovules within the ovary is known as placentation. The placentation are of different types namely, marginal, axile, parietal, basal, central and free central.

(c) When a flower can be divided into two equal radial halves in any radial plane passing through the centre, it is said to be actinomorphic, Example: mustard and chilli.

(d) When a flower can be divided into two similar halves only in one particular vertical plane, it is zygomorphic. Example: pea, gulmohur, bean and Cassia.

(e) Superior ovary flowers are those flowers in which the gynoecium is present at the highest position, while other floral parts are arranged below it. A flower with this arrangement is described as hypogynous. Example: brinjal and mustard.
(f) In perigynous flowers, the gynoecium is present in the centre and the rest of the floral parts are arranged at the rim of the thalamus at the same level. Example: plum and rose.
(g) When stamens are attached to the petals, they are epipetalous as in brinjal.

6. Differentiate between
(a) Racemose and cymose inflorescence
(b) Fibrous roots and adventitious roots
(c) Apocarpous and syncarpous ovary

Answer
(a)

Racemose inflorescence	Cymose inflorescence
→ Younger flowers are present at the tip while older flowers are arranged at the base of this inflorescence. Such an arrangement is called acropetal succession. → The main axis in racemose inflorescence continues to grow and produce flowers laterally.	→ Younger flowers are present at the base of the inflorescence, while older flowers are present at the top. Such an arrangement is called basipetal succession. → The main axis in cymose inflorescence has limited growth, which later terminates into a flower.

(b)

Fibrous root	Adventitious root
→ In monocots, the primary root which develops from the radicle of the seed is short-lived and is replaced by a large number of roots arising from the base of the stem. → It is found in wheat and other cereals.	→ These roots arise from any part of the plant other than the radicle of seeds. → It is found in banyan, Monstera, and other plants.

(c)

Apocarpous ovary	Syncarpous ovary
→ The flowers with apocarpus ovary have more than one carpel. These carpels are free. → It is found in lotus and rose flowers.	→ The flowers with syncarpous ovary have more than one carpel. However, these carpels are fused. → It is found in the flowers of tomato and mustard.

7. Draw the labelled diagram of the following:
(i) Gram seed
(ii) V.S. of maize seed

Answer

(a)

(b)

8. Describe modifications of stem with suitable examples.

Answer

The modifications of Stem with suitable examples are:
→ Food Storage: Stems are modified to perform different functions. Underground stems of potato, ginger, turmeric, zaminkand, Colocasia are modified to store food in them. They also act as organs of perennation to tide over conditions unfavourable for growth.

Tendrils: Stem tendrils which develop from axillary buds, are slender and spirally coiled and help plants to climb such as in gourds (cucumber, pumpkins, watermelon) and grapevines.

→ Thorns: Axillary buds of stems may also get modified into woody, straight and pointed thorns. Thorns are found in many plants such as Citrus, Bougainvillea. They protect plants from browsing animals. Some plants of arid regions modify their stems into flattened (Opuntia), or fleshycylindrical (Euphorbia) structures. They contain chlorophyll and carry out photosynthesis.

→ Vegetative Reproduction: Underground stems of some plants such as grass and strawberry, etc., spread to new niches and when older parts die new plants are formed. In plants like mint and jasmine a slender lateral branch arises from the base of the main axis and after growing aerially for some time arch downwards to touch the ground. A lateral branch with short internodes and each node bearing a rosette of leaves and a tuft of roots is found in aquatic plants like Pistia and Eichhornia. In banana, pineapple and Chrysanthemum, the lateral branches originate from the basal and underground portion of the main stem, grow horizontally beneath the soil and then come out obliquely upward giving rise to leafy shoots.

9. Take one flower each of families Fabaceae and Solanaceae and write its semi-technical description. Also draw their floral diagrams after studying them.

Answer

→ Family Fabaceae (Pea)

Fabaceae (earlier called Papilionaceae) is a sub-family of the Leguminoseae family.

• Vegetative features:
Habit: Pinnately compound, alternately arranged with leaf tendrils with the pulvinus present at the leaf base along folacious stipules.
Root: Tap root system with root nodules.

• Floral features:
Inflorescence: Racemose, generally axial than terminal
Flower: Zygomorphic and bisexual flowers are found
Calyx: It contains five sepals which are gamosepalous while aestivation is imbricate.
Corolla: It contains five petals (polypetalous) with vexillary aestivation.
Androecium: It consists of ten anthers that are diadelphous with dithecous anthers.
Gynoecium: Monocarpellary superior ovary which is unilocular with marginal placentation.
Fruit: Legume pod with non-endospermic seeds

Floral formula:

Economic importance: Peas are used as vegetables for making various culinary preparations.

→ Flowers of Solanum nigrum Family Solanaceae

• Vegetative features:
Habit: Erect, herbaceous plant
Leaves: Simple, exstipulate leaves with reticulate venation
Stem: Erect stem with numerous branches.

• Floral features:
Inflorescence: Solitary and axillary
Flowers: Actinomorphic, bisexual flowers
Calyx: Calyx is composed of five sepals that are united and persistent. Aestivation is valvate.
Corolla: Corolla consists of five united petals with valvate aestivation.
Androecium: It consists of five epipetalous stamens.
Gynoecium: It consists of bicarpellary syncarpous superior ovary with axile placentation.
Fruits: Berry
Seeds: Numerous, endospermous
Floral formula:

Economic importance: Used for medicinal purposes.

10. Describe the various types of placentations found in flowering plants.

Answer

Placentation: The arrangement of ovules within the ovary is known as placentation. The placentation are of different types namely, marginal, axile, parietal, basal, central and free central.

The various types of placentations found in flowering plants are:

→ Marginal placentation: The ovary in which the placenta forms a ridge along the ventral suture of the ovary and the ovules develop on two separate rows is known to have marginal placentation. This type of placentation is found in peas.

→ Parietal placentation: When the ovules develop on the inner walls of the ovary, the ovary is said to have parietal placentation.

→ Axile placentation: In axile placentation, the placenta is axial and ovules are attached to it. Examples include China rose, lemon, and tomato.

→ Basal placentation: The ovary in which the placenta develops from its base and a single ovule is found attached to the base is said to have basal placentation. It is found in marigold and sunflower.

→ Free central placentation: In free central placentation, the ovules develop on the central axis while the septa are absent. This type of placentation is found in Dianthus and primrose.

11. What is a flower? Describe the parts of a typical angiosperm flower?

Answer
The flower is the reproductive unit in the angiosperms. It is meant for sexual reproduction. A flower has four different parts i.e., the calyx, corolla, androecium, and gynoecium. Androecium and gynoecium represent the male and female reproductive organs of a flower respectively. Parts of flower are:

→ The calyx: The calyx is the outermost whorl of the flower and the members are called sepals. Generally, sepals are green, leaf like and protect the flower in the bud stage. When the sepals of a flower are free, they are called polysepalous, while fused sepals of a flower are called gamosepalous.

→ Corolla: Corolla is composed of petals. Petals are usually brightly coloured to attract insects for pollination. When the petals are free, they are called polypetalous, while fused petals are called gamopetalous. Corolla may be tubular, bell-shaped, funnel-shaped or wheel-shaped.

→ The androecium or the stamen is the male reproductive part of a flower. It consists of two parts, the filament and the bilobed anther. The bilobed anther is the site for meiosis and the generation of pollen grains.

→ Gynoecium represents the female reproductive part of a flower. It consists of an ovary. The ovary is connected by a long tube (called style) to the stigma. The ovary bears numerous ovules attached to the placenta.

12. How do the various leaf modifications help plants?
Answer
Leaves are often modified to perform functions other than photosynthesis. These are:
→ Tendrils: The leaves of a pea plant are modified into tendrils that help the plant in climbing.
→ Spines: The leaves in cactus are modified into sharp spines that act as an organ of defense.
→Phyllode: The leaves of some Australian acacia are short-lived and soon replaced by flattened, green structures called phyllodes that arise from the petiole of the leaves. The petioles in these plants synthesize food.
→ Pitcher: The leaves of the pitcher plant are modified into pitcher-like structures, which contain digestive juices and help in trapping and digesting insects.

13. Define the term inflorescence. Explain the basis for the different types of inflorescence in flowering plants.

Answer
Inflorescence is the manner in which the flowers are arranged on the flowering axis. During the flowering season, the vegetative apex of the stem gets converted into a floral meristem. Depending on

whether the apex gets converted into a flower or continues to grow, two major types of inflorescences are defined – racemose and cymose.

In racemose inflorescence, the floral axis continues to grow and produces flowers laterally while in cymose inflorescence, the main axis terminates into a flower. Hence, it is limited in growth.

14. Write the floral formula of an actinomorphic bisexual, hypogynous flower with five united sepals, five free petals, five free stamens and two united carpals with superior ovary and axile placentation.

Answer
The floral formula of the described flower is

as
Actinomorphic flowers are represented by the symbol

A bisexual flower is indicated by

The calyx contains five united sepals which can be represented as K₍₅₎.
The corolla consists of five free petals and it represented as C₅.
The androecium consists of five free stamens and is represented by A₅.
The gynoecium consists of a superior ovary with two united carpels and axile placentations, which can be represented as

15. Describe the arrangement of floral members in relation to their insertion on thalamus?

Answer
Based on the position of the calyx, corolla, and androecium (with respect to the ovary on the thalamus), the flowers are described as hypognous, perigynous, and epigynous.
In hypogynous flowers, the ovary occupies the highest position on the thalamus while other floral parts are situated below it. In such flowers, the ovary is superior e.g., China rose, mustard etc.
In perigynous flowers, the ovary is situated at the centre and other floral parts are arranged on the rim of the thalamus. The ovary here is said to be half inferior e.g., plum, rose, peach
In epigynous flowers, the thalamus grows around the ovary fusing with its wall. The other floral parts are present above the ovary. Hence, the ovary is said to be inferior e.g., flowers of guava and cucumber.

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CLASS 11th CHAPTER -4 Animal Kingdom |Biology | NCERT SOLUTION| EDUGROWN

NCERT Solutions for Class 11 Biology Chapter -4 Animal Kingdom includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Biology exams must go through NCERT Solutions for Class 11 Chapter -4 Animal Kingdom. NCERT Solutions will make you understand the topics in most simple manner and grasp it easily to perform better.

Class 11th Chapter -4 Animal Kingdom | NCERT BIOLOGY SOLUTION |

Page No: 62

Exercises

1. What are the difficulties that you would face in classification of animals, if common fundamental features are not taken into account?
Answer
If we consider specific characteristics instead of fundamental features then each organism will be placed in a separate group and the entire objective of classification would not be achieved.
Classification of animals is also important in comparing different organisms and judging their individual evolutionary significance. If only a single characteristic is considered, then this objective would not be achieved.

2. If you are given a specimen, what are the steps that you would follow to classify it?Answer

→ First Step: Look for the presence or absence of vertebral column.
→ Second Step: Identify the level of organization.
→ Third Step: Look for symmetry, i.e. radial or bilateral or asymmetrical.
→ Fourth Step: Look for presence or absence of body cavity.

3. How useful is the study of the nature of body cavity and coelom in the classification of animals?
Answer
The presence or absence of the body cavity gives a clue about a particular animal’s position in evolutionary ladder. The absence of coelom indicates that the animal is yet to develop a functional dividion of labour to carry out various activities. This indicates towards a simplicity of activities going on inside the body. The presence of coelom paves the way for evolution of a more complex and more efficient organ system for various activities.

4. Distinguish between intracellular and extracellular digestion?
Answer

Intracellular digestion	Extracellular digestion
The digestion of food occurs within the cell.	The digestion occurs in the cavity of alimentary canal.
Digestive enzymes are secreted by the surrounding cytoplasm into the food vacuole.	Digestive enzymes are secreted by special cells into the cavity of alimentary canal.
Digestive products are diffused into the cytoplasm.	Digestive products diffuse across the intestinal wall into various parts of the body.
It is a less efficient method.	It is a more efficient method of digestion.
It occurs in unicellular organisms.	It occurs in multicellular organisms.

5. What is the difference between direct and indirect development?
Answer

Direct development	Indirect development
It is a type of development in which an embryo develops into a mature individual without involving a larval stage.	It is a type of development that involves a sexually-immature larval stage, having different food requirements than adults.
Metamorphosis is absent.	Metamorphosis involving development of larva to a sexually-mature adult is present.
It occurs in fishes, reptiles, birds, and mammals.	It occurs in most of the invertebrates and amphibians.

6. What are the peculiar features that you find in parasitic platyhelminthes?
Answer

Peculiar features in parasitic platyhelminthes are:
→ They have dorsiventrally flattened body and bear hooks and suckers to get attached inside the body of the host.

→ Their body is covered with thick tegument, which protects them from the action of digestive juices of the host.
→The tegument also helps in absorbing nutrients from the host’s body.

7. What are the reasons that you can think of for the arthropods to constitute the largest group of the animal kingdom?
Answer
Arthropods are the first phylum to have well developed systems to carry out different activities. There is distinct system for respiration, locomotion and reproduction. Their survival capacity is great because of elaborate system. This has helped them survive in diverse conditions. This can be one of the reasons why arthropods are the largest group among the animal kingdom.

8. Water vascular system is the characteristic of which group of the following:

(a) Porifera (b) Ctenophora (c) Echinodermata (d) Chordata
Answer

(c) Phylum, Echinodermata is having water vascular system. This system is used for locomotion, food and waste transportation, and respiration in this group. This system consists of Madreporite, Stone canal, Circular ring canal, Radial canals, Lateral canals and Tube feet.

9. “All vertebrates are chordates but all chordates are not vertebrates”. Justify the statement.
Answer
The characteristic features of the phylum, Chordata, include the presence of a notochord and paired pharyngeal gill slits. In sub-phylum Vertebrata, the notochord present in embryos gets replaced by a cartilaginous or bony vertebral column in adults. Thus, it can be said that all vertebrates are chordates but all chordates are not vertebrates.

10. How important is the presence of air bladder in Pisces?
Answer
Gas bladder or air bladder is a gas filled sac present in fishes. It helps in maintaining buoyancy. Thus, it helps fishes to ascend or descend and stay in the water current.

11. What are the modifications that are observed in birds that help them fly?
Answer

The modifications modifications in birds help them fly:
→ Streamlined body for rapid and smooth movement
→ Covering of feathers for insulation
→ Forelimbs modified into wings and hind limbs used for walking, perching, and swimming
→ Presence of pneumatic bones to reduce weight
→ Presence of additional air sacs to supplement respiration.

12. Could the number of eggs or young ones produced by an oviparous and viviparous mother be equal? Why?
Answer
The numbers of eggs produced by an oviparous mother will be more than the young ones produced by a viviparous mother because in oviparous animals, the development of young ones takes place outside the mother’s body. Their eggs are more prone to environmental conditions and predators. Therefore, to overcome the loss, more eggs are produced by mothers so that even under harsh environmental conditions, some eggs might be able to survive and produce young ones. On the other hand, in viviparous organisms, the development of young ones takes place in safe conditions inside the body of the mother. They are less exposed to environmental conditions and predators. Therefore, there are more chances of their survival and hence, less number of young ones is produced compared to the number of eggs.

13. Segmentation in the body is first observed in which of the following:

(a) Platyhelminthes (b) Aschelminthes (c) Annelida (d) Arthropoda

Answer
The body segmentation first appeared in the phylum, Annelida.

14.Match the following:

(a) Operculum	(i) Ctenophora
(b) Parapodia	(ii) Mollusca
(c) Scales	(iii) Porifera
(d) Comb plates	(iv) Reptilia
(e) Radula	(v) Annelida
(f) Hairs	(vi) Cyclostomata and Chondrichthyes
(g) Choanocytes	(vii) Mammalia
(h) Gill slits	(viii) Osteichthyes

Answer

(a) Operculum	(viii) Osteichthyes
(b) Parapodia	(v) Annelida
(c) Scales	(iv) Reptilia
(d) Comb plates	(i) Ctenophora
(e) Radula	(ii) Mollusca
(f) Hairs	(vii) Mammalia
(g) Choanocytes	(iii) Porifera
(h) Gill slits	(vi) Cyclostomata and Chondrichthyes

15. Prepare a list of some animals that are found parasitic on human beings.

Answer

Ascaris, Wucheraria, Acylostoma, Taenia, Fasciola.

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CLASS 11th CHAPTER -3 Plant Kingdom |Biology | NCERT SOLUTION| EDUGROWN

NCERT Solutions for Class 11 Biology Chapter -3 Plant Kingdom includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 11 Biology exams must go through NCERT Solutions for Class 11 Chapter -3 Plant Kingdom. NCERT Solutions will make you understand the topics in most simple manner and grasp it easily to perform better.

Class 11th Chapter -3 Plant Kingdom | NCERT BIOLOGY SOLUTION |

Page No: 44

Exercises

1. What is the basis of classification of algae?
Answer
The presence or absence of pigments is the main basis of classification of algae.

• Chlorophyceae: Chlorophyll a and b are present in them and impart green colour. Chlorophyceae are also called ‘blue-green algae’.

• Phaeophyceae: Chlorophyll a and c and fuxoxanthin are present. Fucoxanthin imparts brown colour. Phaeophyceae are also called ‘brown algae’.

• Rhodophyceae: Chlorophyll a and d and phycoerythrin are present. Phycoerythrin imparts red colour. Rhodophyceae are also called ‘red algae’.

2. When and where does reduction division take place in the life cycle of a liverwort, a moss, a fern, a gymnosperm and an angiosperm?
Answer
In liverwort, moss and fern, the saprophytic plant produces haploid spores after meiosis while in gymnosperm and angiosperm, meiosis takes place in antheridium and ovary for the formation of pollen grains and ovules.

3. Name three groups of plants that bear archegonia. Briefly describe the life cycle of any one of them.

Answer

Bryophytes, Pteridophytes and Gymnosperms bear distinct archegonia.

Life cycle of gymnosperms

• Reproduction: The gymnosperms are heterosporous; they produce haploid microspores and megaspores. The two kinds of spores are produced within sporangia that are borne on sporophylls which are arranged spirally along an axis to form lax or compact strobili or cones.

• Male gamete: The strobili bearing microsporophylls and microsporangia are called microsporangiate or male strobili. The microspores develop into a male gametophytic generation which is highly reduced and is confined to only a limited number of cells. This reduced gametophyte is called a pollen grain. The development of pollen grains takes place within the microsporangia.

• Female gamete: The cones bearing megasporophylls with ovules or megasporangia are called macrosporangiate or female strobili. The male or female cones or strobili may be borne on the same tree (Pinus) or on different trees (Cycas). The megaspore mother cell is differentiated from one of the cells of the nucellus. The nucellus is protected by envelopes and the composite structure is called an ovule. The ovules are borne on megasporophylls which may be clustered to form the female cones. The megaspore mother cell divides meiotically to form four megaspores. One of the megaspores enclosed within the megasporangium (nucellus) develops into a multicellular female gametophyte that bears two or more archegonia or female sex organs. The multicellular female gametophyte is also retained within megasporangium.

• Fertilization: The pollen grain is released from the microsporangium. They are carried in air currents and come in contact with the opening of the ovules borne on megasporophylls. The pollen tube carrying the male gametes grows towards archegonia in the ovules and discharge their contents near the mouth of the archegonia. Following fertilisation, zygote develops into an embryo and the ovules into seeds.

4. Mention the ploidy of the following: protonemal cell of a moss; primary endosperm nucleus in dicot, leaf cell of a moss; prothallus cell of a ferm; gemma cell in Marchantia; meristem cell of monocot, ovum of a liverwort, and zygote of a fern.
Answer
• Protonemal cell of a moss – Haploid
• Primary endosperm nucleus in a dicot – Triploid
• Leaf cell of a moss – Haploid
• Prothallus of a fern – Haploid
• Gemma cell in Marchantia- Haploid
• Meristem cell of a monocot – Diploid
• Ovum of a liverwort – Haploid
• Zygote of a fern – Diploid

5. Write a note on economic importance of algae and gymnosperms.
Answer
Economic importance of algae: Algae are useful to man in a variety of ways. They perform half of the total carbon dioxide-fixation on earth by photosynthesis, acting as the primary producers in aquatic habitats. Many species of marine algae such as Porphyra, Sargassum, and Laminaria are edible. Chlorella and Spirulina are rich in proteins. Thus, they are used as food supplements. Agar is used in the preparation of jellies and ice-cream. It is obtained from Gelidium and Gracilaria. Carrageenin is used as an emulsifier in chocolates, paints, and toothpastes. It is obtained from the red algae. Many red algae such as Corallina are used in treating worm infections.

Economic importance of gymnosperms: Gymnospermous plants are widely used as ornamentals. Many conifers such as pine, cedar, etc., are sources of the soft wood used in construction and packing. Medicinal uses: An anticancer drug Taxol is obtained from Taxus. Many species of Ephedraproduce ephedrine, which can be used in the treatment of asthma and bronchitis. The seeds of Pinus gerardiana are edible. Resins are used commercially for manufacturing sealing waxes and water-proof paints. A type of resin known as turpentine is obtained from various species of Pinus.

6. Both gymnosperms and angiosperms bear seeds, then why are they classified separately?
Answer
The seeds of gymnosperms are naked, while that of angiosperms are covered by a membrane. Therefore, they are classified separately.

7. What is heterospory? Briefly comment on its significance. Give two examples.
Answer
Heterospory is a phenomenon in which two kinds of spores are borne by the same plant. These spores differ in size. The smaller one are called microspore and the bigger ones are called megaspore. Microspore produce male gametophytes and megaspores produces female gametophyte. Thus it is considered as a crucial step in evolution as it is a precursor to the seed habit.
This ultimately led to seed development in gymnosperms and angiosperms.

Page No: 45

8. Explain briefly the following terms with suitable examples:-
(i) protonema
(ii) antheridium
(iii) archegonium
(iv) diplontic
(v) sporophyll
(vi) isogamy

Answer
(i) Protonema: It is the first stage in the life cycle of a moss, developing directly from the spore. It consists of creeping, green, branched, and often filamentous structures.

(ii) Antheridium: It is the male sex organ present in bryophytes and pteridophytes and is surrounded by a jacket of sterile cells. It encloses the sperm mother cells, which give rise to the male gametes.

(iii) Archegonium: It is the female sex organ present in bryophytes, pteridophytes, and gymnosperms. In bryophytes and pteridophytes, it generally has a swollen venter and a tubular neck, and contains the female gamete called the egg.

(iv) Diplontic: It is the term used for the life cycles of seed-bearing plants (gymnosperms and angiosperms). In these plants, the diploid sporophyte is dominant, photosynthetic, and independent. The gametophyte is represented by a single-celled (or a few-celled) structure.

(v) Sporophyll: In pteridophytes, the sporophytic plant body bears sporangia. These sporangia are subtended by leaf-like appendages known as sporophylls. In gymnosperms, microsporophylls and megasporophylls are found. These bear microspores and megaspores respectively.

(vi) Isogamy: It is a type of sexual reproduction involving the fusion of morphologically-similar gametes. This means that the gametes are of the same size, but perform different functions. This type of reproduction is commonly observed in Spirogyra.

9. Differentiate between the following:-
(i) red algae and brown algae
(ii) liverworts and moss
(iii) homosporous and heterosporous pteridophyte
(iv) syngamy and triple fusion

Answer

(i)

Red algae	Brown algae
Red algae are grouped under the class Rhodophyceae.	Brown algae are grouped under the class Phaeophyceae.
They contain floridean starch as stored food.	They contain mannitol or laminarin as stored food.
They contain the photosynthetic pigments chlorophylls a and d, and phycoerythrin.	They contain the photosynthetic pigments chlorophylls a and c, and fucoxanthin.
Their cell walls are composed of cellulose, pectin, and phycocolloids.	Their cell walls are composed of cellulose and algin.
Flagella are absent	Flagella are present

(ii)

Liverworts	Moss
They have unicellular rhizoids.	They have multicellular rhizoids.
Scales are present very often	Scales are absent
They are generally thalloid, with dichotomous branching.	They are foliage, with lateral branching.
Gemma cups are present	Gemma cups are absent
Sporophyte has very little photosynthetic tissue	Sporophyte has abundant photosynthetic tissue

(iii)

Homosporous pteridophytes	Heterosporous pteridophytes
They bear spores that are of the same type.	They bear two kinds of spores â€“ microspores and megaspores.
They produce bisexual gametophytes.	They produce unisexual gametophytes.

(iv)

Syngamy	Triple fusion
It is the process of fusion of the male gamete with the egg in an angiosperm.	It is the process of fusion of the male gamete with the diploid secondary nucleus in an angiosperm.
A diploid zygote is formed as a result of syngamy.	A triploid primary endosperm is formed as a result of triple fusion.

10. How would you distinguish monocots from dicots?
Answer

Characteristic	Monocot	Dicot
Morphology
Roots	Fibrous roots	Tap roots
Venation	Generally parallel venation	Generally reticulate venation
Flowers	Trimerous flowers	Pentamerous flowers
Cotyledons in seeds	One	Two
Anatomy
No. of vascular bundles in stem	Numerous	Generally 2 – 6
Cambium	Absent	Present
Leaves	Isobilateral	Dorsiventral

11. Match the followings (column I with column II)

	Column I		Column II
(a)	Chlamydomonas	(i)	Moss
(b)	Cycas	(ii)	Pteridophyte
(c)	Selaginella	(iii)	Algae
(d)	Sphagnum	(iv)	Gymnosperm

Answer

	Column I		Column II
(a)	Chlamydomonas	(i)	Algae
(b)	Cycas	(ii)	Gymnosperm
(c)	Selaginella	(iii)	Pteridophyte
(d)	Sphagnum	(iv)	Moss

12. Describe the important characteristics of gymnosperms.

Answer

Important characteristics of gymnosperms are:

→ The seeds of these plants are not enclosed in fruits.

→The plant-body ranges from medium to tall trees and shrubs. The giant redwood tree Sequoia is one of the tallest trees in the world.

→ The root system consists of tap roots. The coralloid roots present in Cycas are associated with nitrogen-fixing cyanobacteria.

→ The stem can be branched (as in Pinus and Cedrus) or un-branched (as in Cycas).

→ The leaves can be simple (as in Pinus)or compound (pinnate in Cycas). The leaves are needle-like, with a thick cuticle and sunken stomata. These help in preventing water loss.

→ Gymnosperms are heterosporous. They bear two kinds of spores – microspores and megaspores.

→ Flowers are absent. The microsporophylls and megasporophylls are arranged to form compact male and female cones.

→ Pollination occurs mostly through wind and pollen grains reach the pollen chamber of the ovule through the micropyle.

→ The male and female gametophytes are dependent on the sporophyte.

→ The seeds contain haploid endosperms and remain uncovered.

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