NCERT Exemplar Solutions for Class 11 Biology Chapter 11 Transport in Plants

NCERT Exemplar Solutions for Class 11 Biology Chapter 11 Transport in Plants

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

VERY SHORT ANSWER QUESTIONS

Question 1.
Smaller, lipid soluble molecules diffuse faster through cell membrane, but the movement of hydrophilic substances are facilitated by certain transporters which are chemically…….. .
Solution:
The movement of hydrophilic substances are facilitated by transporters which are chemically proteins. These proteins form porins, which are huge pores in the outer membranes of the plastids, mitochondria and some bacteria. These porins allow passage of small molecules through the membrane.

Question 2.
In a passive transport across a membrane. When two protein molecules move in opposite direction and independent of each other, it is called as………… .
NCERT Exemplar Solutions for Class 11 Biology Chapter 11 Transport in Plants 1.1

Solution:
Antiport which facilitates transport of molecules in both the directions across the membrane and their movement is independent of each other.

Question 3.
Osmosis is a special kind of diffusion, in which water diffuses across the cell membrane. The rate and direction of osmosis depends upon both………. .
Solution:
The rate and direction of osmosis is dependent upon the pressure and concentration gradient.

Question 4.
A flowering plant is planted in a earthen pot and irrigated. Urea is added to make the plant grow faster, but after sometime the plant dies. This may be due to ………… .
Solution:
The solution outside the plant is an hypertonic solution, and the plant cells are hypotonic in nature, so there is a gradual movement of water from plant cell to outside urea solution leading to plasmolysis of root cells and plant dies gradually due to exosmosis.

Question 5.
Absorption of water from soil by dry seeds increases the, thus helping seedlings to come out of soil.

Solution:
Imbibition of water by seed materials as starch and protein, pushes the seedlings out of the soil causing the seed to swell and increase of imbibition pressure inside the seed, contributes for germination of seeds.

Question 6.
Water moves up against gravity and even for a tree of 20 m height, the tip receives water within two hours. The most important physiological phenomenon which is responsible for the upward movement of water is………. .
Solution:
Transpiration pull is the physiological phenomenon which is responsible for the upward movement of water in tall trees the water molecules transpire from stomata, which pulls water molecules upward to the leaf from the continuous chain of water molecules carried by xylem

Question 7.
The plant cell cytoplasm is surrounded by both cell wall and cell membrane. The specificity of transport of substances are mostly across the cell membrane, because……… .
Solution:
Transpiration pull is the physiological phenomenon which is responsible for the upward movement of water in tall trees the water molecules transpire from stomata, which pulls water molecules upward to the leaf from the continuous chain of water molecules carried by xylem

Question 8.
The C4 plants are twice as efficient as C3 plants in terms of fixing C02 but lose only…. as much water C3 plants for the same amount of C02 fixed.
Solution:
C4 plants are twice as efficient as C3 plants in terms of fixing carbon in the form of glucose, but lose only half as much water as a C3 plant for the same amount of C2 fixed.

Question 9.
Movement of substances in xylem is unidirectional while in phloem it is bidirectional. Explain
Cell cycle is under genetic control and is a sequential event. Every cell prepares itself before it starts dividing. This preparation takes place in interphase stage of the cell cycle.
Solution:
Xylem is involved in the one way transport of water and minerals from soil to root ’ —> stem —> leaves. Several forces like imbibition, root pressure and finally transpiration pull. Act in this mechanism, It is a undirectional process as there is continuous loss of water at me body surface of plants.

The main function of Phloem is to transport food from source to sink where source is the part of plant responsible for food synthesis and sink are the organs requiring food for their growth and development.

These source and sink parts of a plants may vary in different phases of growth, thus the food needs to travel in both upwards and downward direction. So, phloem shows bidirectional movement of substances.

Question 10
Define water potential and solute potential.
Solution:
Water potential is a measure of free energy associated with water per unit volume (JM -3).
The water potential of pure (φw) at atmosp-heric Pressure is zero.
Additional of solutes reduce water potential (to a negative value). This reduces the of water concentration. Solutions thus have a lower water potential than pure water, the magnitude of this lowering due to dissolution of solute is called
solute potential of φs.

Question 11
An onion peel was taken and
(a) placed in salt solution for five minutes.
(b) after that it was placed in distilled water. When seen under the microscope what would be observed in (a) and (b) ?
Solution:
(a) When placed in salt solution an onion peel shrinks as water from cytoplasm of cell moves out of the cell to wards hypertonic solution.
(b) When again placed back in distilld water, cell regains it’s shape and absorbs water and become turgid.

Question 12
How does most of the water moves within the root?
Solution:
Water mostly flows in the roots via the apoplast pathway as the cortical cells are loosly packed and hence offer no resistance to water movement, through mass flow. This mass flow of water occurs due to adhesive and cohesive properties of water.
Like, symplast pathway is also involved in the movement of water molecules within the root (like, via endodermis to xylem).

Question 13
Transpiration is a necessary evil in plants. Explain.
Solution:
Loss of water in the form of water vapours from the surface of leaves of plant is called transpiration.
Transpiration a necessary evil because the plant continuously lose water in the vapour form from its body surfaces, Which creates a transpiration pull to absorb more and more water from soil through roots.
If water is not available to plants in soil, even then loss through transpiration does not ceasle, so plants sometimes sbfrws wilting.

Question 14
Describe briefly the three physical properties of water which helps in ascent of water in xylem.
Solution:
The following are physical properties of water that helps in ascent up to xylem.
Cohesive properties — Provider mutual attraction between molecules
Adhesive properties — Causes attraction of water molecules to polar surfaces (of tracheids)
Surface tension — Water molecules get attracted to each other more in liquid phase than in gas phase.

Question 15
Identify a type of molecular movement which is highly selective and requires special membrane proteins, but does not require energy.
Solution:
Facilitated diffusion’s is a highly selective passive process. Facilitated diffusion cause net transport of molecules from a low to high concentration. In facilitated diffusion special proteins help in movement of substances across the membrane without expenditure of ATP energy.

Question 16
Correct the statements.
(a) Cells shrink in hypotonic solutions and swell in hypertonic solutions.
(b) Imbibition is special type of diffusion when water is absorbed ‘*y living cells.
(c) Most of the water flow in the roots occurs via the symplast.
Solution:
(a) The cell swellSHORT ANSWER QUESTIONSter is adsorbed by living cells.
(c) Most of the water flow in roots occurs via the apoplast way.

SHORT ANSWER QUESTIONS

Question 1.
Minerals absorbed by the roots travel up the xylem. How do they reach the parts where they are needed most? Do all the parts of the plant get the same amount of the minerals?
Solution:

  1. The sabsorbed mineral are transported through the transpiration steam up the stem, to all parts of plant.
  2. The growing region of the plant, such as the apical and lateral meristems, young leaves, developing flowers, fruits, seeds and the storage organs are the chief sinks for the mineral elements.
  3. Uploading of the mineral ions occurs via fine vein endings through diffusion and active uptake by the cells.
  4. Xylem are involved in transport of inorganic nutrients where phloem transport only organic materials in plants.
  5. Mineral ions are frequently remobilised from older parts of plant like leaves to the younger regions.
  6. Most readily mobilised elements are phosphorus, sulphur, nitrogen, potassium, and some elements like calcium that forms the structural component are not remobilised.

Question 2.
Water is indispensable for life. What properties of water make it useful for all biological process on the earth?
Solution:
Following are the properties of water that make it useful for all biological processes.
(i) Water is the major solvent through which mineral nutrients enter a Plant from the soil solution.
(ii) It is an ideal solvent with neutral pH.
(iii) Water is the major constituent of protoplasm, it constitutes approximately 90% of the protoplasm.
(iv) Water acts as a medium for translocation of nutritive substances. Mineral nutrients are absorbed by the roots. Carbohydrates that are formed during photosynthesis are transported by water from cell to cell, tissue to tissue and organ to organ.
(v) Water is involved in photosynthesis in plants, as it incorporates hydrogen atom into carbohydrate and releases oxygen atoms as O2.
(vi) Water acts as an agent for temperature control. The specific heat of water helps plant in maintaining a relatively stable internal temperature.
(vii) Water is necessary for pollination in some plants in bryophytes and pteridophytes, water are essentially requires for the fertilisation process.

Question 3.
How is it that the intracellular levels of K+ are higher than extracellular levels in animal cells?
Solution:
The excitability of sensory cells, neurons and muscles is dependent on ion channels, signal transducers that provide a regulated path for the movement of inorganic ions such as Na+, K+, Ca2+, and Cl across the plasma membrane in response to various stimuli.
Ion channels are ‘gated’ mplying that they may be open or closed. The Na+, K+, ATPase create a charge imbalance across the plasma membrane by carrying 3Na+ out of the cell for every 2K+ ion carried inside making the inside relatively negative outside.
The membrane is said to be polarised. That is the reason the intracellular levels ofK+ are higher than extracellular levels in animals cells.
NCERT Exemplar Solutions for Class 11 Biology Chapter 11 Transport in Plants 1.2

Question 4.
In a girdled plant, when water is supplied to the leaves above the girdle, leaves may remain green for sometime then wilt and ultimately die. What does it indicate?
Solution:
When water is supplied in a girdle plant to the leaves above the girdle, leaves may remain green for sometime because leaves can synthesise their own carbohydrate food through photosynthesis, they however, gradually wilt due to non-availability of water.
The system of xylem vessels from root to the leaf vein can supply the needed water during girding there is a possible loss of xylem vessels and the water supply is cut off, resulting in death of the plant.

Question 5.
Various types of transport mechanisms are needed to fulfil the mineral requirements of a plant. Why are they not fulfilled by diffusion alone?
Solution:
Ions, minerals and organic compound are transported in plants in various ways which include.
(i) Food substances ways which include v synthesised in leaves are translocated
downward towards root and stem.
(ii) Food is translocated upwards to the developing leaves, buds and fruits.
(iii) Radial transport of food occurs across the stem from the cells of pith, from cortex etc, towards epidermis.
(iv) Ions and minerals are transported upwards through xylem.
Diffusion is a slow process and allows movement of molecules only for short distances, so it cannot carry out the movements of organic and inorganic substances mentioned above. Therefore, a need arises for special long distance transport systems that permits and moves substances at a much faster rate, i.e., mas of bulk flow system through conducting tissues (translocation).

Question 6.
Will the ascent of sap be possible without the cohesion and adhesion of the water molecules? Explain.
Solution:
Ascent of sap is not possible without the cohesive and adhesive properties of water they play an important role in transport of water due to the following reasons
(i) Cohesion forces hold the water molecule together in the conducting channels, so vaccum is not created.
(ii) Adhesive forces acting between the water molecule and cellulose of cell wall make a thin film of water along the channels so that this film is pulled up by transpiration pull drawing more and more water upwards in the conducting channels from the root.

Question 7.
When a freshly collected Spirogyra filament is kept in a 10% potassium nitrate solution, it is observed that the protoplasm shrinks in size
(a) What is this phenomenon called?
(b) What will happen if the filament is replaced in distilled water?
Solution:
(a) The phenomenon, occurring is Spirogyra filament when placed in 10% potassium nitrate solution (hypertonic solution) is Plasmolysis. It occurs as water from the cell is drawn put to extracellular fluid causing the protoplast to shrink away from cell wall.
(b) The Spirogyra upon reabsorption of water, causes the protoplast to regain its original shape. This phenomenon is known as  deplasmolysis.

Question 8.
What are ‘aquaporins’? How does presence of aquaporins affect osmosis?
Solution:
Aquaporins are integral membrane proteins which form pores or channels in the membrane.
The water flows is more rapid through these pores to inside of the cell, as compared to the process of diffusion.
These are plumbing systems of the cells. They selectively conduct water in and out of the cells, while preventing the passage of ions and other solutes.

Question 9.
ABA (Abscicis Acid) is called a stress hormone.
A.How does this hormone overcome stress conditions?
B. From where does this hormone get released in leave?
Solution:
A. Stress hormone ABA (Abscisic Acid) induces closing of stomata, whenever there is scarcity of water available to the plant. This prevents the loss of water through transpiration by leaves. It also increases the tolerance of plants to various kinds of stresses.
B. (ABA) is released or transported from the stem apices to leaves.

Question 10.
How is facilitated diffusion different from diffusion?
Solution:
Difference between cytokinesis in plant cell and animal cell is as follows.
NCERT Exemplar Solutions for Class 11 Biology Chapter 11 Transport in Plants 1.3

Question 11.
Observe the diagram and answer the following.
NCERT Exemplar Solutions for Class 11 Biology Chapter 11 Transport in Plants 1.4

(a) Are these types of guard cells found in monocots or dicots?
(b) Which of these shows a higher water content (i) or (ii)?
(c) Which element plays an important role in the opening and closing of stomata?
Solution:
(a) The guard cells that are bean-shaped are
found in dicot plants.
(b) The guards cells in figure (i) are turgid as, they pull the inner wall of the cell outside thus, they have more water in figure (ii) cells are flaccid, this condition results when cells lose water and close stomatal pore.
(c) The K+ ions move from neighbouring cells to guards cells, lowering their water potential and as a result the water moves inside making them turgid and thus opening stomata.

Question 12.
Define uniport, symport and antiport. Do they require energy?
Solution:

  1. For movement of substances the biological membranes have many mechanism.
  2. Some are active and some are passive. Specific membrane proteins are also involved for special types of transport mechanisms. These mechanisms include:
  3. Uniport is a membrane transport system by an integral membrane protein that is involved in facilitated diffusion.
  4. These channels open in response to a stimulus for free flow of specific molecules in a specific direction. These channels transport molecule with a solute gradient without energy expenditure.
  5. Symport involves the movement of two or more different molecules or ions, across the membrane in the same direction, with no expenditure of energy.
  6. Antiport is called exchanger. This integral membrane protein is involved in secondary active transport of two or more different molecules or ions across the membrane in opposite directions, without affecting the transport of other molecules.
    NCERT Exemplar Solutions for Class 11 Biology Chapter 11 Transport in Plants 1.5

LONG ANSWER QUESTIONS

Question 1.
Minerals are present in the soil in sufficient amounts. Do plants need to adjust the type of solutes that reach the xylem? Which molecules help to adjust this? How do plants regulate the type and quantity of solutes that reach xylem?
Solution:

  • Plants do need to adjust the type and quantity of solutes that reach the xylem.
  • The transport of proteins in endodermal celr help in maintaining and adjusting solute movement.
  • The minerals are present in soil as charged particles with a very low concentration compared to that of roots, and thus cannot be completely transported passively across the cell membranes of roots hairs.
  • Minerals are thus transported both by active and passive processes, to the xylem.
  • Upon reaching xylem, they are further transported, upwards to sinks through transpiration stream.
  • At the sink regions mineral ions are unloaded through diffusion and active uptake by receptor cells. The mineral ions moving frequently through xylem include.

(i) Sulphur and Phosphorus in small amounts are carried in organic forms.
(ii) Njtrogen travels in plants as inorganic ions N02 and N03 but much of the nitrogen moves in the form of amino acids and related organic compounds.
(iii) Mineral ions are frequently remobilised particularly from older senescing parts. Older dying leaves export much of their mineral content to younger leaves. Similarly, before leaf fall in deciduous plants, minerals are removed to other parts.
The most readily mobilised elements are phosphorus, sulphur, nitrogen and potassium. Structural components elements like calcium are not remobilised.

Question 2.
Plants show temporary and permanent wilting. Differentiate between the two. Do any of them indicate the water status of the soil?
Solution:
The loss of turgidity of leaves and other soft aerial parts of a plant causing dropping, folding and rolling of non-woody plants is wilting. It occurs when rate of loss of water is higher than the rate of absorption.
NCERT Exemplar Solutions for Class 11 Biology Chapter 11 Transport in Plants 1.6

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

NCERT Exemplar Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division

NCERT Exemplar Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division

These Solutions are part of NCERT Exemplar Solutions for Class 11 Biology. Here we have given NCERT Exemplar Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division.

VERY SHORT ANSWER QUESTIONS

Question 1.
Between a prokaryote and a eukaryote, which cell has a shorter cell division time?
Solution:
Prokaryotic cell has simple cell structure and cellular organisation. It’s nucleus does not contain nuclear membrane. Prokaryotic cell thus has shorter cell cycle than the eukaryotic cell.

Question 2.
Name a stain commonly used to colour chromosomes.
Solution:
The chromosomes are the thickest and the shortest at metaphase. Acetocarmine and Giemsa stain can be used to stain the chromosomes. They are stained for karyo-typing for further study of chromosomes.

Question 3.
Which tissue of animals and plants exhibits meiosis?
Solution:
Meiosis is also called as reduction division, it is a special kind of cell division which occurs in germ cells or sex cells of male and female reproductive organs of plants and animals. They produce male (($) and female (C^) gametes that take part in sexual reproduction.

Question 4.
Which part of the human body should one use to demonstrate stages in mitosis?
Solution:
All the cells in the human body except germinal cells in the male and female reproductive organs are somatic cells. The somatic cells divide by mitotic cell division for growth and regeneration and can be used to demonstrate mitosis.

Question 5.
The diagram shows a bivalent at prophase-I of meiosis. Which of the four chromatids can cross over?
NCERT Exemplar Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division 1.1
Solution:
In prophase-I of meiosis, the homologous chromosomes lie parallel to each other in leptotene stage. Each chromosome has four chromatids and are bivalent. The non-sister chromatids of homologous chromosomes cross over in pachytene stage of prophase-I.
NCERT Exemplar Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division 1.2

Question 6.
If a tissue has at a given time 1024 cells, how many cycles of mitosis had the original parental single cell undergone?
Solution:
To give 1024 cells the parental cell undergoes 10 divisions of mitotic cycle.

Question 7.
An anther has 1200 pollen grains. How many pollen mother cells must have been there to produce them?
Solution:
The pollen mother cell (2n) undergoes meiotic cell divisions, each such cell produces four daughter cells with haploid (n) number of chromosomes. Three hundred pollen mother cells would have to be there to produce 1200 pollen grains, because one pollen mother cell will produce four pollen grains.

Question 8.
At what stage of cell cycle does DNA synthesis take place?
Solution:
The stage of cell cycle where DNA synthesis or replication takes place is Synthetic phase or S- phase of interphase.

Question 9.
It is said that the one cycle of cell division in human cells (eukaryotic cells) takes 24 hours. Which phase of the cycle, do you think occupies the maximum part of cell cycle?
Cell cycle is under genetic control and is a sequential event. Every cell prepares itself before it starts dividing. This preparation takes place in interphase stage of the cell cycle.
Solution:
If a cell takes 24 hours to divide, it spends 18-20 hours time in interphase stage to prepare itself to undergo cell division.

Question 10
It is observed that heart cells do not exhibit cell division. Such cells do not divide further and exit… phase to enter in inactive stage called…. of cell cycle. Fill in the blanks
Solution:
It is observed that heart cells do not exhibit cell division. Such cells do not divide further and exit G, phase to enter an inactive stage called quiescent stage (GQ) of cell cycle.
Muscle cells when reach a level of maturity, no longer divide and just perform their function all through it life.

SHORT ANSWER QUESTIONS

Question 1.
State the role of centrioles other than spindle formation.
Solution:
The animal cell are present in few membrane less cell organelles. Centrosome is one of them. Two cylindrical structures called centrioles are the part of centrosome.
In the centrosome the two centrioles lie perpendicular to each other. Each has organisation lie a cart wheel. These form the basal body of cilia and flagella of plant/animal cells besides forming spindle fibre in animal cell division. It also helps in the formation of microtubules and sperm tail.

Question 2.
Label the diagram and also determine the stage at which this structure is visible.
NCERT Exemplar Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division 1.3
Solution:
The transition stage between prophase and metaphase stage of mitotic cell division is shown in the diagram.
NCERT Exemplar Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division 1.4

Question 3.
A cell has 32 chromosomes. It undergoes mitotic division. What will be the chromosome number (n) during metaphase? What would be the DNA content (C) during anaphase?
Solution:
Mitosis helps in the growth of organism and its development. It also plays a vital role in a sexually reproducing organisms. The mitotic cell division occurs in somatic cells of an organism.

The chromosome number in the daughter cells remains same as that of the parent (dividing) cell, so even at metaphase or anaphase, the chromosome number does not change.

The DNA content gets doubled at the synthetic phase of interphase and gets divided at anaphase but the chromosome number remains same

Question 4.
While examining the mitotic stage in a tissue, one finds some cells with 16 chromosomes and some with 32 chromosomes. What possible reasons could you assign to this difference in chromosome number. Do you think cells with 16 chromosomes could have arisen from cells with 32 chromosomes orvice-versa?
Solution:
A condition as such, may arise in case of a mosaic, which denotes presence of two or more populations of cells in one individual with varying genotypes.
It can result from variou: mechanisms including non-disjunction, anaphase lagging and end replication. It may also result from a mutation during development, which is propagated to only a subset of the adult cells. In this case, cells with 16 chromosomes could have arisen from cells with 32 chromosomes.

Question 5.
The following events occur during the various phases of the cell cycle. Name the phase against each of the events.
(a) Disintegration of nuclear membrane ………
(b) Appearance of nucleolus ………
(c) Division of centromere ……..
(d) Replication of DNA ……….
Solution:
(a) Prophase
(b) Telophase
(c) Anaphase
(d) S-phase

Question 6.
Mitosis results in producing two cells which are similar to each other. What would be the consequence if each of the following irregularities occur during mitosis?
(a) Nuclear membrane fails to disintegrate
(b) Duplication of DNA does not occur
(c) Centromeres do not divide
(d) Cytokinesis does not occur
Solution:
(a) The spindle fibres would not be able to reach chromosomes if nuclear membrane fails to disintegrate and they would not move towards opposite poles of the cell.
In certain protozoans, such as Amoeba, the ‘ spindle is formed within the nucleus and this is called intra nuclear mitosis or premitosis.
(b) The cell might not be able to surpass S-phase of cell-cycles. If DNA duplication does not occur as no chromosome formation will take place, and cell will not be able to enter M-(mitotic phase) in case it enters mitosis, the cycle will cease.
(c) If the centromeres do not divide as it may result in trisomy, one of the daughter cell will receive a complete pair of chromosomes and other cell would not get any of them.
(d) Multinucleate condition called coenocyte, syncytium is produced. If cytokinesis does not occur as in Rhizopus and Vaucheria, etc

Question 7.
Both unicellular and multicellular organisms undergo mitosis. What are the difference, if any, observed in the process between the two?
Solution:
The type of cell divisions in unicellular organisms is known as amitosis in which somatic cell is directly divided into the parts. Occurs curs. In multicellular organisms.
In both unicellular and multicellular organisms. The difference between mitosis include:
NCERT Exemplar Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division 1.5

Question 8.
Comment on the statement-meiosis enables the conservation of specific chromosome number of each species even through the process per se results in reduction of chromosome number.
Solution:
Meiosis is the mechansim of conservation of specific chromosome number of each species across generations in organisms reproducing sexually. The process results in reduction of chromosome number by half, which is gradually conserved by union of male gamete 9n) and female gamete (n) in next generation. Meiosis also increases the genetic variability in the population of organisms from one generation to the next.

Question 9.
Name a cell that is found arrested in diplotene stage for months and years. Comment.
Solution:

  1. In mammalian occytes, meiotic arrest at diplotene stage usually occurs.
  2. In females, meiosis starts in the embryo and proceeds as for as diplotene, when the chromosomes become diffused and the cells are referred to as being in the dictyate stage. This arest is under hormonal control.
  3. In many amphibian oocyles, birds and insects with a long period of immaturity, the oocyte may be arrested in the dictyate stage for many years and spend a prolonged period in diplotene.
  4. This stage is characterised by formation of lampbrush chromosomes where intense RNA synthesis occurs and most of the genes in the DNA loops are actively transcribed and expressed.

Question 10.
How does cytokinesis in plant cells differ from that in animal cells?
Solution:
Difference between cytokinesis in plant cell and animal cell is as follows.
NCERT Exemplar Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division 1.6

LONG ANSWER QUESTIONS

Question 1.
Comment on the statement- Telophase is reverse of prophase.
Solution:
The following contrasting differences reveals that telophase is reverse of prophase, in cell division.
NCERT Exemplar Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division 1.7

Question 2.
An organisms has two pair of chromosomes (i.e., chromosome number = 4), Diagrammatically represent the chromosomal arrangement during different phases of meiosis-II.
Solution:
Meiosis is reduction division in which chromosome number reduces tc half in daughter cells. The number reduces as half set of chromosomes move to 2 daughter cells in meiosis-I. Thus two cells with half set of chromosomes again re-enter meiosis-II which is similar to mitotic cell division.
NCERT Exemplar Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division 1.8

We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 10 Cell Cycle and Cell Division, help you. If you have any query regarding NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 10 Cell Cycle and Cell Division, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Solutions for Class 11 Biology Chapter 9 Biomolecules

NCERT Exemplar Solutions for Class 11 Biology Chapter 9 Biomolecules

These Solutions are part of NCERT Exemplar Solutions for Class 11 Biology. Here we have given NCERT Exemplar Solutions for Class 11 Biology Chapter 9 Biomolecules.

VERY SHORT ANSWER QUESTIONS

Question 1.
Medicines are either man made (i.e. synthetic) or obtained from living organisms like plants, bacteria, animals, etc., and hence, the latter are called natural products. Sometimes, natural products are chemically altered by man to reduce toxicity or side effects. Write against each of the following whether they were initially obtained as a natural product or as 3 synthetic chemical.
(a) Penicillin
(b) Sulphonamide
(c) Vitamin-C
(d) Growth hormone
Solution:
(a) Penicillin is a group oT antibiotics derived from fungi Penicillium obtained naturally.
(b) Sulphonamide an antimirobial agent is a synthetic chemical.
(c) Vitamin-C or L-ascorbic acid or ascorbate is a natural product and an essential nutrient for humans. It is present in citrus fruits.
(d) Growth hormone also known as somatotropin or somatropin is a peptide hormone occurring naturally in the body it stimulates growth.

Question 2.
Write the name of any one amino acid, sugar, nucleotide and fatty acid.
Solution:
(a) Amino acid — Leucine
(b) Sugar — Lactose
(c) Nucleotide — Adenosine triphosphate
(d) Fatty acid — Palmitic acid

Question 3.
Reaction given below is catalysed by oxidoreductase between two substrates A and
A’, complete the reaction.
A reduced + A’ oxidised —>
Solution:
Oxidoreductase is an enzyme that catalyses oxidation reduction reactions. This enzyme is associated in catalysing the transfer of electron from one molecule (the reduction), also called as electron donor, to another molecule (the oxidant), also called as electron acceptor.
The complete reaction is
NCERT Exemplar Solutions for Class 11 Biology Chapter 9 Biomolecules 1.1

Question 4.
How are prosthetic groups different from co¬factors?
Solution:
Organic compounds that are tightly bound to the apoenzyme, (an enzyme without cofactor) by covalent or non-covalent bonds are prosthetic groups e.g., peroxidase and
catalase catalyse the breakdown of hydrogen peroxide to water and oxygen where haeme is the prosthetic group and it is a part of the active site of the enzyme.
Co-factor is small, heat stable and non-protein part of conjugate enzyme. It may be inorganic or organic in nature. Co-factors when loosely bound to an enzyme is called coenzyme and when tightly bound to apoenzyme is called prosthetic group.

Question 5.
Glycine and alanine are different with respect to one substituent on the a-carbon. What are 4. the other common substituent groups?
Solution:
The common substituted groups in both the amino acids are NH2 COOH and H.
NCERT Exemplar Solutions for Class 11 Biology Chapter 9 Biomolecules 1.2

SHORT ANSWER QUESTIONS

Question 1.
Enzymes are proteins. Proteins are long chains of amino acids linked to each, other by peptide bonds. Amino acids have many functional groups in their structure.
NCERT Exemplar Solutions for Class 11 Biology Chapter 9 Biomolecules 1.3
These functional groups are many of them at least, ionisable. As they are peak acids and bases in chemical nature, this ionisation is influenced by pH of the solution. For many enzymes, activity is influenced by surrounding pH. This is depicted in the curve below, explain briefly.
Solution:
Enzymes, generally function in a narrow range of pH. Most of the enzymes show their highest activity at a particular pH called optimum pH- activity declines below and above this value. Extremely high or low pH values generally results in complete loss of activity for most enzymes. The given graph represents the maximum enzyme activity at the optimum pH.

Question 2.
Explain the association of carbohydrate to the plasma membrane and its significance.
Solution:
Secondary metabolites are chemicals produced by plants which do not play any [role] in growth, photosynthesis reproduction or other primary functions of the plant. Rubber (cis 1,4- polyisopyrene) is a secondary metabolite.
(i) Rubber is extracted from Hevea brasiliensis (rubber tree)
(ii) It is a byproduct of the lactiferous tissue of the vessels that are in the form of latex.
(iii) It contains over 400 isoprene units and thus is the largest of the terpenoids.
(iv) It is elastic, water proof and a good conductor of electricity.

Question 3.
Nucleic acids exhibit secondary structure, justify with example.
Solution:

  1. Nucleic acids are large biological molecules, essential for all known forms of life.
  2. The secondary structure of a nucleic acid molecule refers to the base pairing interactions within a single molecule or set of interacting molecules.
  3. DNA and RNA represent two main nucleic acids, their secondary structures however differ the secondary structure of DNA comprises of two complementary strands of polydeoxyribonucleotide, spirally coiled on a common axis forming a helical structure.
  4. This double helical structure of DNA is stabilized by phosphodiester bonds (between 5’ of sugar of one nucleotide and 3 sugar of another nucleotide), hydrogen bonds (between bases, and ionic interactions.

Question 4.
Comment on the statement ‘living state is a non-equilibrium steady state to be able to perform work’
Solution:

  1. Living organism are not in equilibrium because work cannot be performed by a system at equilibrium.
  2. The living organisms exist in a steady state characterised by concentration of each of the biomoleculSs.
  3. These biomolecules are in a metabolic flux. Any chemical or physical process moves simultaneously to equilibrium.
  4. Living organisms work continuously and they cannot afford to reach equilibrium.
  5. The living state thus is an a non-equilibrium steady-state to be able to perform work. This achieved by energy input provided by metabolism.

LONG ANSWER QUESTIONS

Question 1.
What are different classes of enzymes? Explain any two with the type of reactions they catalyse.
Solution:
Enzymes are divided into six classes each with
4-13 sub-classes and named accordingly by a number comparising of four digits.
(i) Oxidoreductases/dehydrogenases : These enzymes take part in oxidation, reduction or transfer of electrons,
(ii) Transferase : These enzymes transfer a functional group (other than hydrogen).
from one molecule to another. The transfer , chemical group does not occur in free state.
(iii) Hydrolases : These enzymes catalyse the hydrolysis of bonds like ester, ether,
peptide, glycosidic C-C, C-halide, P-N etc.
NCERT Exemplar Solutions for Class 11 Biology Chapter 9 Biomolecules 1.4
(iv) Lyases cause cleavage, removal of groups without hydrolysis and addition of groups to double bonds or removal of groups producing double bonds.
NCERT Exemplar Solutions for Class 11 Biology Chapter 9 Biomolecules 1.5
(v) Isomerases rearrangement of molecular structure to effect isomeric changes. They are of three types isomerases, epimerases and mutases.
NCERT Exemplar Solutions for Class 11 Biology Chapter 9 Biomolecules 1.6
(vi) Ligases catalyse bonding of two chemicals with the help of energy obtained from ATP resulting formation of bonds such as C—O, C—S, C—N and P—O e.g., pyruvate carboxylase
Pyruvric acid + C02 + ATP + H20
NCERT Exemplar Solutions for Class 11 Biology Chapter 9 Biomolecules 1.7

We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 9 Biomolecules, help you. If you have any query regarding NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 9 Biomolecules, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Solutions for Class 11 Biology Chapter 8 Cell: The Unit of Life

NCERT Exemplar Solutions for Class 11 Biology Chapter 8 Cell: The Unit of Life

These Solutions are part of NCERT Exemplar Solutions for Class 11 Biology. Here we have given NCERT Exemplar Solutions for Class 11 Biology Chapter 8 Cell: The Unit of Life.

VERY SHORT ANSWER QUESTIONS

Question 1.
Mention a single membrane bound organelle which is rich in hydrolytic enzymes.
Solution:
The membrane bourld vesicular structures formed by Golgi apparatus are Lysosomes. These vesicles have been found to be rich in all types of hydrolytic enzymes as hydrolase, lipases, proteases and carbohydrases which digest carbohydrates proteins, lipids and nucleic acid at an acidic pH.

Question 2.
What are gas vacuoles? State their functions.
Solution:
Gas vacuoles also known as pseudovacuoles or air vacuoles are the characteristic feature 1 of prokaryotes. They store metabolic gases and take part in regulation of buoyancy.

Question 3.
What is the function of a polysome? (Gk. Poly – many, Soma = body).
Solution:
A polysome consists a cluster of ribosomes that are held simultaneously by a strand of messenger KNA in rosette or helical group. They contain a portion of the genetic code that each ribosome is translating and are used in formation of multiple copies of same polypeptide. They are found in the cyloplasm during the process of active protein synthesis.

Question 4.
What is the feature of a metacentric chromosome?
Solution:
The centromere is median, in metacentric chromo-some. The centromere lies in the middle portion and forms two equal arms of chromosome.
NCERT Exemplar Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 1.1

Question 5.
What is the feature of a metacentric chromosome?
Solution:
Additional constriction or secondary constriction at the chromosomal ends as distal part of the arm formed by chromatin thread are known satellite chromosomees. These constriction gives appearance of an outgrowth or a small fragment.
These are also known as (sat) chromosomes or marker chromosomes. Chromosomes 13,14, 15, 16, 21 and 21 satellite chromosomes.
NCERT Exemplar Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 1.2

SHORT ANSWER QUESTIONS

Question 1.
Discuss briefly the role of nucleous in the cells activity involved in protein synthesis.
Solution:
The round, naked and a slightly irregular structure, which is attached to the chromatin at a specific region called as Nucleolar Organizer Region (NOR). Nucleous was first discovered by Fontana (1781).
NCERT Exemplar Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 1.3
The role of nucleolus can be described as:
(i) Nucleolus is the chief site for the synthesis of ribosomal RNA.
(ii) It is the centre for the formation of ribosome components.
(iii) It is the colloidal complex that fills the nucleus.
(iv) It combines rRNA with proteins to produce ribosomal sub-units. The ribosomes sub-units after their formation pass out and get established in the cytoplasm.
(v) It also receives and stores ribosomal proteins formed in the cytoplasm.
(vi) These ribosomal proteins formed are the sites for protein synthesis in the cell.
(vii) Nucleolus is essential for spindle formation during nuclear division as well.

Question 2.
Explain the association of carbohydrate to the plasma membrane and its significance.
Solution:

  1. The plasma membrane, surrounds th cell. It consists of lipids, proteins and carbohydrates that are imperative in both structure and function of the cell.
  2. Carbohydrates attach either with proteins or lipids usually making up less than 10% of the membrane weight.
  3. They can give rise to a wide variety of structures in relatively short chains. They give distinguishing features to individual cell types and thus they may be involved.
  4. Cell Recognition like ABC surfaces have carbohydrates arranged in branched chains: difference in the arrangement give rise to different blood group antigens (i.e., A, B and O).
  5. Cell surface differences are also responsible for the specificity of action of cells with hormones, drugs, viruses or bacteria. The cause of difference of cell surface is related to characteristic surface due to carbohydrate component.

Question 3.
Briefly describe the cell theory.
Solution:
Schleiden and Schwann formulated the cell theory, in 1938-39 which stated
(i) All living beings are made up of cells and products formed by the cells.
(ii) Cells are the structural and functional units oflife
The cell theory stated by Schleiden and Schwann failed to explain the question of origin of cells.
A major expansion of the cell theory was expressed by Virchow in his statment ‘Omnis cellula e cellula’ (all cells arise from pre-existing cells) in 1855.
This concept, was the actual idea of Nagelli (1846), which later on was elaborated by Virchow, along with considerable evidences in its support. The work of Nagelli and Virchow established cell division as the central pehnomenon in the continuity oflife.
The modem cell theory is thus based on two facts
(i) All living organisms are composed of cells and products of cells.
(ii) Cells are the basic structural and functional units oflife.
(iii) All cells arise from pre-existing cells. Vimses are exception to cell theory as they are .pot composed of cell. They consist of a nucleic acid (DNA or RNA) surrounded by a protein sheet and are incapable of independant existence, self regulation and self reproduction.

Question 4.
Give the biochemical composition of plasma membrane. How are lipid molecules arranged in the membrane?
Solution:
Chemcial composition of plasma membrane includes
Component                      Composition
Lipids                               (20-79%)
Proteins                           (20-70%)
Carbohydrates                (1-5%)
Water                                20%
Lipids form the continuous structural frame of the cell membrane and hence are the major components of the cell membrane. Lipids such as phqspholipics, glycolipids, and steroids are found
NCERT Exemplar Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 1.4
The lipid molecule possess both polar hydrophilic (water loving) and non-polar hydrophobic (water repelling) ends. The hydrophilic region is in the form of a head, while the hydrophobic part contains fatty acid tails. Hydrophobic tail is present towards the centre of the membrane. This structures results is the formation of lipid bilayer known as unit membrane/biological membrane/cell membrane. Proteins are embedded within the lipid bilayer – Carbohydrates are structure upon proteins.

Question 5.
What are plasmids? Describe their role in bacteria.
Solution:

  1. A plasmid is usually a circular (sometimes linear), double stranted DNA that can autonomously replicate.
  2. These are found in the cytoplam of the bacterial cell. Plasmids normally remain separated from the chromosome, but sometimes may temporarily integrate into it and replicate with it incidentally.
  3. Role and Plasmids in Bacteria Plasmids are the extra chromosomal circular, independently replicating unit besides nucleoid in the bacterial cell.
  4. Plasmids are used to transfer information from one cell to another, i.e., transfer of important genes, enabling to metabolise a nutrient, which normally a bacteria is unable to. It also helps in conjugation of bacteria.
  5. These days plasmids are used in a variety of recombination experiments, as cloning vectors.

LONG ANSWER QUESTIONS

Question 1.
Is there a species specific or region specific type of plastids? How does one distinguish one from the other?
Solution:
Plastids are specific to different species and are found in all plant cells and in euglenoids. They bear certain pigments that impart specific colour^ to the part of the plant possesing them. Plastids ar classified into three main types, based on the type of pigments- leucoplasts, chromoplast and chloroplast.
Leucoplasts are colourless plastids which store food material. They are of three types based on their storage products.
(a) Amyloplasts store starch, e.g., tuber of potato, grain of rice, grain of wheat.
(b) Elaioplasts store fats, e.g., rose
(c) Aleuropiasts are protein storing plastids, e.g., castor endosperm.
Chromoplast are non photosynthetic coloured plastids which synthesise and store carotenoid pigments. They appear orange, red or yellow. These mostly occur in ripe fruits (tomato and chilies) carrot roots, etc.
Chloroplasts are green color plastids which help in synthesising food material by photosyntheis. They contain chrophyll and carotenoid pigments which trap light energy.
Each chloroplast is oval or spherical, double membrane bound cell organelle. The space present inside inner membrane is called stroma Anumberrof oiganised flattenedmembranous sacs called thylakoids are present in the stroma. Thylakoids are arranged in stacks called grana.
The thylakoids of different grana are connected by membranous tubules called the stroma lamellae. The stroma of the lamellae contain the enzymes that are required for the synthesis of carbohydrates and proteins.
NCERT Exemplar Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 1.5

Question 2.
Write the functions of the following
(a) Centromere
(b) Cell wall
(c) Smooth ER
(d) Golgi apparatus
(e) Centrioles
Solution:
(a) Centromere is required for proper chromosome segregation. The centromere consists of two sister chromatids. It is also necessary for attachment of chromosomes to the spindle apparatus during mitosis and meiosis.
NCERT Exemplar Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 1.6
(b) Cell wall gives a definite shape to the cell 1 and protects the cell from mechanical injury
and infections. It also aids in cell to cell interaction and acts as a barrier for undesirable macromolecules.
(c) Smooth ER helps in synthesis of lipids, metabolism of carbohydrates, regulation of calcium concentration, drug detoxification and attachment of receptors on cell membrane proteins.
The smooth ER also contains enzymes- glucose 6 phosphatase, which converts glucose 6 phosphate to glucose essential in glucose metabolism.
NCERT Exemplar Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 1.7
(d) Golgi apparatus is an important site for the formation of glycoprotein and glyco lipids also involved in the synthesis of cell wall materials and plays an important role in formation of cell plate during cell divisionas well.
NCERT Exemplar Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 1.8
(e) Centrioles form the base body of cilia and flagella and spindle fibres that gives rise to spindle apparatus during cell division in ‘animal cells. They help in formation of microtubules and sperm tail. They also help in cell division by forming asters, which acts as spindle pole.

We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 8 Cell: The Unit of Life, help you. If you have any query regarding NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 8 Cell: The Unit of Life, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals

NCERT Exemplar Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals

These Solutions are part of NCERT Exemplar Solutions for Class 11 Biology. Here we have given NCERT Exemplar Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals.

VERY SHORT ANSWER QUESTIONS

Question 1.
State the number of segments in earthworm which are covered by a prominent dark band or clitellum.
Solution:
Segments 14-16 bare covered by a prominent dark band of glandular tissue called clitellum in a mature earthworm. It secretes mucus and albumen that help in formation of cocoon.
NCERT Exemplar Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals 1.1

Question 2.
Where are sclerites present in cockroach?
Solution:
In all the body segments of cockroach sclerites are present. They are of two types dorsal sclerites often known as tergites, and ventral sclerites which are referred to as sternites.

Question 3.
How many times do nymphs moult to reach the adult form of cockroach?
Solution:
The nymph grows by moulting about 13 times to reach. In cockroach’, the development is indirect and paurometabous adult form has three stages, i.e., egg, nymph and adult. The nymph resembles adult except/or undeveloped wings and genitalia.

Question 4.
Identify the sex of a frog in which sound producing vocal sacs are present.
Solution:
Sex of frogs can be distinguished on the basis of presence of sound producing vocal sacs. These organs are present in males which make them crock lauder than females, so as to attract females for mating.

Question 5.
A muscle fibre tapers at both ends and does not show striations. Name the muscle fibre.
Solution:
Muscle fibres that taper at both the ends (fusiform) and do not show striations are smooth muscle fibres. They are also called involuntary muscles.

Question 6.
Name the different cell junctions found in tissues.
Solution:
The different cell junctions found in tissue include:
(i) Tight junctions are regions where plasma membrane of adjacent epithelial cells are held close together. They check the movement of material between then.
(ii) Gap junctions are meant for chemical exchange between adjacent cells.
(iii) Adhering junctions function to keep neighbouring cells together.

Question 7.
Give two identifying features of an adult male frog.
Solution:
The two identifying features of an adult male frog include
(a) Nuptial Pad is a copulatory pad present on the first digit of the forelimb of male frog and helps in closing female during amphelexus.
(b) Vocal Sacs are loose skin folds on throat of male frogs for producing louder croak to attract females for mating purposes.

Question 8.
Which mouth part of cockroach is comparable to our tongue?
Solution:
In cockroach, hypopharynx acts as a tongue and lies within cavity enclosed by the mouth parts.

Question 9.
The digestive system of frog is made of the following parts. Arrange them in an order beginning from mouth. Mouth, oesophagus, buccal cavity, stomach, intestine, cloaca, rectum, cloacal aperture.
Solution:
The correct arrangement of the part of digestive system in frog is
Mouth —> Buccal cavity —> Oesophagus —> Stomach —>Intestine —> Rectum —> Cloaca —> Cloacal aperture.

Question 10.
What is the difference between cutaneous and pulmonary respiration?
Solution:
In frog respiration takes place via the skin as well lungs.
Pulmonary respiration and occurs outside the water through lungs. Cutaneous respiration takes place in water as well as land, occurs through highly vascularised moist skin.

Question 11.
Special venous connection between liver and intestine and between kidney and intestine is found in frog, what are the called?
Solution:
In frog, venous connection between liver and intestine is called hepatic portal system and venus connection between the kidney and the lower parts of the frog is called renal portal system.

SHORT ANSWER QUESTIONS

Question 1.
Stratified epithelial cells have limited role in secretion. Justify their role in our skin.
The edible part of the peach or pear pome fruit for the fleshy thalamus.
Solution:

  1. Stratified epithelium consists of epithelial cells in which the innermost layer is made up of columnar or cuboidal cells.
  2. It is a type of compound epithelium and a waterproof protein called keratin is present few outer layers.
  3. These layers of dead cells is called homy layer which is shed at intervals due to frictions and thus has a limited role in secretions and absorption.
  4. The main function of stratified epithelium is to provide protection to the body against mechanical and chemical stresses.

Question 2.
How does a gap junctions facilitate intercellular communication ?
Solution:
Intercellular communication is facilitated by gap Junction allowing small signaling molecules to pass from cell to cell.
These are fine hydrophilic channels, between two adjacent animal cells that are formed with the help of two protein cylinders called connexus.
Each connexus consists of six proteins subunits that surround a hydrophilic channel. Opening or closing of channel is controlled by pH and Ca2+ ion concentration.

Question 3.
Why are blood, bone and cartilage called connective tissue?
Solution:

  1. Connective tissue pt’ovides the structural framework and support to different organs forming tissue.
  2. Blood is a fluid or vascular connective tissue, which connects various organs and transports substances from one place to another.
  3. Bone is a solid, rigid and strong skeletal connective tissue, which supports the body and helps in locomotion.
  4. Cartilage is also a skeletal connective tissue, not as rigid bone but piable and resists compression.
  5. It plays role in support and protection and present in tip of nose, outer ear joints etc.

Question 4.
How do you distinguish between dorsal and ventral surface of the body of earthworm?
Solution:
The body of an earthworm can be distinguished into dorsal and ventral sides due to the presence of certain peculiar feature in it which include the following.
(i) The dorsal surface is darker than ventral surface because it is marked by a dark median mid dorsal line along the longitudinal axis of body. This is due to dorsal blood vessel, seen through integument.
(ii) Genital openings (pores), are present in the ventral surface of both male and female.
(iii) On vental surface genital papilla is located and helps in copulation.
NCERT Exemplar Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals 1.2

Question 5.
Complete the following statement.
(a) In cockroach grinding of food particle is performed by ……..
(b) malpighian tubules help in removal of …….
(c) Hind gut of cockroach is differentiated into……
(d) In cockroach blood vessels open into spaces called ……
Solution:
(a) Gizzard is a muscular and greatly folded structure which marks the end of foregut in cockroach and bears six plates with teeth for crushing and grinding the food.
(b) Malpighian tubules are excretory in ‘ function as they help in the removal of
nitrogenous wastes in arthropods.
(c) Ileum, colon and rectum and rectum opens and through anus.
(d) Haemocoel is the body cavity of cockroach divided into sinuses and contains visceral organs of cockroach floating in haemolymph.

Question 6.
Mention special features of eye in cockroach. Discuss compound eye in arthropods and mention its structural features.
Solution:

  1. In cockroach the eyes are large sessile, paired bean-shaped and present on either side of head.
  2. The are compound in nature. Each compound eye consists of a large number of visual elements called ommatidia.
  3. Each ommatidium is composed of a diopteric region and reticular (receptor) region. It is capable of producing a separate image of a small part of object seen.
  4. Thus, the image of the object viewed consists of several pieces and is known as mosaic image.
  5. Fine nerve fibres arise from the inner end of each ommatidium all of which combine to form one optic nerve connected to the brain.
    NCERT Exemplar Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals 1.3

Question 7.
Frog is a poikilotherm, exhibits camouflage and undergoes aestivation and hibernation, how are all these benficial to it?
Solution:

  1. A trait with a current functional role in the life history of an organism that is maintained and evolved by means of natural selection and evolution and help organism in its survival is an adapture triat.
  2. Frog is a poikilotherm or a (cold blooded animal). It regulates its body temperature according to its environment.
  3. It undergoes winter sleep (hibernation) for withstanding very cold temperatures and sujnmer sleep in hot temperatures (aestivation).
  4. During this period, it lives in a dormant stage with very minimal vital body activities.
  5. Frog is capable of changing its body colour as well, though gradually, with the change in its surrounding and climatic conditions.
  6. This capability in frog is called as camouflage which lets it escape from the predators, an essential survival parameter for living.

Question 8.
Write the functions in brief in Column II, appropriate to the structures given in column I.
NCERT Exemplar Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals 1.4
Solution:
(a) Nictitating Membrane in frog protects the eye from water and any damage by covering the eye ball of frog.
(b) Tympanum is present on each side of the frog head and is involved in the hearing process.
(c) Copulatory Pad present in the limbs of the male frog and helps in copulation by holding the female during its sexual activity.

Question 9.
Using appropriate examples, differentiate between false and true body segmentation.
Solution:

  1. The serial repetition of similar body parts along the length of an animal is segmentation. The body of animals can be truely segmented or pseudo/false segmented.
  2. True segmentation is found in annelids, arthropods and some chordates. In these organisms there is a linear repetition of body parts and each repeated unit is called somite (metamere).
  3. In earthworms, the successive somites are externally and intemaly. ‘
  4. Pseudosegmentation is seen when body is divided into number of false segments which are independent of each other.
  5. Each segment is able to perform all the vital function of body. Growth occurs by the addition of new segments from the anterior end, e.g., tapeworm.

Question 10.
What is special about tissue present in the heart?
Solution:
Special tissue present in heart is called cardiac muscle, which has the following features
(i) Cardiac muscle fibres are supplied with both central and autonomic nervous system and are not under the control of will of animal.
(ii) These muscles show rhythmicity and are immune to fatigue.
(iii) They have a rich supply of blood.
(iv) They are myogenicas. They possess the property of contraction even if completely isolated from the body.
NCERT Exemplar Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals 1.5

LONG ANSWER QUESTIONS

Question 1.
Comment upon the gametic exchange in earthworm during mating. Discuss the physiology in reproduction of earthworm.
Solution:
In earthworm mating is a unique process. Earthworm is a hermaphrodite. In which breeding takes place during rainy season and copulation begins soon after maturation of the sperms.
The gametic exchange and the physiology of reproduction during mating can described in the following manner.
(i) Earthworms are protandrous animal (i.e., maturation of sperm takes place much earlier then that of ova).
(ii) Mating process in earthworm occurs through process of cross-fertilisation.
(iii) The mating process involves exchange of gametic materials between the two worms.
(iv) Two individuals from adjacent burrows emerge half but and lie in contact with each other, and exchange packets of sperms called spermatophores opposite gonadal opening.
NCERT Exemplar Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals 1.6
(v) The skin encircling male pore, elevates a little during the process to form a temporary papilla that fits like a penis into the opposite spermathecal pore to keep it open.
(vi) The copulating worm after filling of spermatheca moves a bit to adjust another pair of spermathecae to face the other male pores. This is accomplished in about an hour’s copulation.
(vii) The sperms mostly remain in their diverticula within the spermathecae and the ampulla is associated with the secretion of nutritive substances for the sperms.
(viii) The sperm and egg are passed into cocoon, secreted by clitellar gland.
(ix) Fertilisation is therefore external.

Question 2.
Explain the digestive system of cockroach with the help of labelled sketch.
Solution:
The alimentary canal of cockroach is divided
into three regions foregut, midgut and hindgut.
(i) Mouthy cavity, pharynx, oesophagus, crop and gizzard are included in foregut.
(ii) Mouth cavity is a small space, surrounded by mouth parts. Food is crushed and acted upon by the salivary secretion in mouth.
(iii) The mouth opens into a short tubular pharynx, leading towards the narrow tubular passage called oesophagus and then into a sac-like structure called crop which acts as a storage organ.
NCERT Exemplar Solutions for Class 11 Biology Chapter 7 Structural Organization in Animals 1.7
(iv) The crop is followed further by gizzard ‘ (proventriculus). Gizzard is composed of thick circular muscles and thick inner cuticle forming six highly chitinous plates called as teeth. It associated with the grinding and crushing of food particles. A thick cuticle lines the entire foregut.
(v) About one-third middle part of alimentry canal comprises of midgut or mesentron. The internal lining of midgut is an endodermal epithelium of columnal cells raised into several small villi like folds.
(vi) Anterior most part of midgut surrounding the stomadaeal valve is called cardia. Finger like blind processes called as enteric or hepatic caeca are present at the junction of foregut and midgut.
(vii) A ring of yellow filamentous structures is formed between the midgut and hindgut that aid in the removal of excretory products from haemolymph.
(viii) The remaining one-third posterior part of alimentary canal is Hindgut. It is relatively thicker than the midgut lined by cuticle and ectodermal epithelium.
(ix) Hindgut is diffrentiated into three parts anterior Ileum, middle colon and posterior rectum. Ileum is short and relatively narrower and its cuticie bears minute spines. Colon is the longest, relatively thicker and a coiled part of hindgut. Rectum is a small and oval chamber that opens out through anus.

We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 7 Structural Organization in Animals, help you. If you have any query regarding NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 7 Structural Organization in Animals, drop a comment below and we will get back to you at the earliest.

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

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

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

VERY SHORT ANSWER QUESTIONS

Question 1.
Product of photosynthesis is transported from the leaves to various parts of the plants and stored in some cell before being utilised. What are the cells/tissues that store them?
Solution:
The first product of photosynthesis is glucose. It is highly reactive molecule and gets converted into a disaccharide-sucrose for storage.
The food gets stored in specialised prarenchymatous cells present either in roots and stems or in their modifications in the form of a polysaccharide called starch.

Question 2.
Protoxylem is the first formed xylem. If the protoxylem lies next to pholem what kind of arrangement of xylem would you call it?
Solution:
The condition of the xylem arrangement if protoxylem lies next to phloem is called as exarch. It is found in roots.

Question 3.
What is the function of phloem parenchyma?
Solution:
The main function of phloem parenchyma is to store food and other substances like resins, latex and mucilage. They help in transport of food as well.

Question 4.
What is present on the surface of the leaves which helps the plant prevent loss of water but is absent in roots?
Solution:
Cuticle is a waxy coating covering the entire surface of the plant body. It is absent in roots, it prevents the loss of water through the surface of the plant.

Question 5.
What is the epidermal cell modification in plants which prevents water loss?
Solution:
Bulliform or motor cells are modified epidermal cells meant for checking water loss present in monocots or grasses. They help in shutting down stomata and thus reduce water loss through transpiration under stressed conditions.

Question 6.
What constitutes the cambial ring?
Solution:
The cambium present in between the xylem and phloem is called fasicular or intrafasicular cambium and the newly formed cambium between the two vascular bundle is known as interfascular cambium. Both type of cambium combine to form the cambial ring.

Question 7.
Give one basic functional difference between phellogen and phelloderm.
Solution:
Phelloderm is a permanent tissue while phellogen is a meristematic tissue. Phellogen (cork cambium) develops from the cortical cells, sometimes from pericycle cells. These cells actively divide and forms phellem on outerside and phelloderm (cortex cells) innerside so phelloderm originates from phellogen.
NCERT Exemplar Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 1.1

Question 8.
Arrange the following in the sequence you would find them in a plant starting from the periphery-phellem, phellogen, phelloderm.
Solution:
The outer most layer is phellem or cork followed by phellogen (cork cambium) which in turn is followed by phelloderm (secondary cortex.

Question 9.
If one debarks a tree, what parts of the plant is being removed?
Solution:
Debarking refers to removal of bark, i.e., all tissues exterior to the vascular cambium, including secondary phloem. Bark includes periderm (phellogen, phellem and phelloderm) and secondary phloem.

SHORT ANSWER QUESTIONS

Question 1.
While eating peach or pear it is usually seen that some stone like structures get entangled in the teeth, what are these stone like structures called?
The edible part of the peach or pear pome fruit for the fleshy thalamus.
Solution:
The stone cells are present in the pulpy part of fruit of peach and pear. These are sclerenchymatous cells and which are dead in nature. They provide mechanical support to the soft tissue.
NCERT Exemplar Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 1.2

Question 2.
What is the commercial source of cork? How frs it formed in the plant?
Solution:
The source of commercial cork is the cork tissue of Quercus suber, which yields bottle cork. Cork is formed by cork cambium or phellogen cell, cells of cork cambium divide periclinally, cutting cells towards the inside and outside. The cells that cut off towards the outside become suberised and dead.
These are compactly packed in radial rows without intercellular spaces and form cork of phellem. Cork is impervious to water due to presence suberin and provides protection to the underlying tissues.

Question 3.
Below is a list of plant fibres. From which part of the plant these are obtained.
(a) Coir
(b) Hemp
(c) Cotton
(d) Jute
Solution:
(a) Coir is a natural fibre obtained from coconut husk. It is the fibrous mesoderm of the fruit of Cocos nucifera (coconut).
(b) Hemp fibre is obtained from the stems of Cannabis sativa. It is the bast fibre (soft or stem fibre) obtained from secondary phloem.
(c) Cotton fibre is the epidermal growth in cotton (Gossypium hirsutum) seed. It is an elongated structure made up of cellulose.
(d) Jute is a natural bast fibre made up of cellulose and lignin obtained from Corchorus capsularis.

Question 4.
Epidermal cells are often modified to perform specialised functions in plants. Name some of them and function they perform.
Solution:
The epidermal tissue system comprises of one cell thick layer of epidermal tissue and forms the outer most covering of the whole plant body. ,
Modification of Epidermal Cells
Following are the modifications of the epidermal tissue (i) root hair
Structure
These unicellular hairs are the extensions of epidermal cell of roots in the root hair zone.
NCERT Exemplar Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 1.3
Function
They increase the surface area for absorption of water and minerals.
(ii)Epidermal Appendages
Structure
These are called trichomes and are epidermal cell modifications. There any be unicellular or multicellular.
NCERT Exemplar Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 1.4
Appendages of epidermis of leaves
A-Stellate hair of a Alyssum
B-Glandular hair of Pelargonium
C-Short glandular hair of Lavandula
D-Floccose hair of Malva
E-Glandularhair of solanum
F-Urtivating hair of Verbascum
Function
They produce some glandular secretions.

Question 5.
The lawn grass (Cyandon dactyl on) needs to be mowed frequently to prevent its overgrowth. Which tissue is responsible for its rapid growth?
Solution:
The rapid growth of mowed lawn grass is due to meristematic tissue. When the apex of grass is cut frequently, it leads to the growth of the lateral branches, that makes it more bushy.

Question 6.
Plants require water for their survival. But when watered excessively, plants die. Discuss.
Solution:
Plants use water for several metabolic process as photosynthesis, transpiration and respiration. Plants when watered in excess die because excess water removes the air trapped between the soil particles.
The plant roots do not get 02 for respiration. Once cells of root die, water and mineral absorption is stopped and this leads to gradual death of a plant.

Question 7.
A transverse section of the trunk of a tree shows concentric rings which are known as growth rings. How are these rings formed? What is the significance of these rings?
Solution:

  1. The concentric growth rings are called annual rings. These rings are formed due to the secondary growth.
  2. Secondary growth occurs due to the activity of cambium which is a meristermatic tissue in dicot trees.
  3. The rate of activity of cambium is more in spring so wood formed has larger wider xylem cells, whereas wood formed in autumn has narrower and smaller xylem elements.
  4. This results in the formation of two rings called growth rings.
  5. By counting these rings, age of the tree can be determined. This branch of science is known as dendrochronology or growth ring analysis.
    NCERT Exemplar Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 1.5

Question 8.
Trunks of some of the aged tree species appear to be composed of several fused trunks. Is it a physiological or anatomical abnormality? Explain in detail.
Solution:
The appearance of several fused trunks is anatomical abnormality. It is an abnormal type of secondary growth where a regular vascular cambium or cork cambium is not formed in its normal position. Anomalous secondary growth produces cortical and medullary vascular bundles in case of old tree trunks. Thus, the additional or accessory vascular bundles given appurtenance of several fused trunks.

Question 9.
What is the difference between lenticels and stomata? The gaseous exchange in all plants. Occurs by means of several openings present in the plant body.
Solution:
NCERT Exemplar Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 1.6

Question 10.
Write the precise function of
(a) sieve tube
(b) interfascicular cambium
(c) collenchyma
(d) aerenchyma
Solution:
Sieve tube It’s function is to transport of synthesised food throughout the plant. It is present in the phleom tissue.
NCERT Exemplar Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 1.7
Interfascicular Cambium It is a kind of  secondary meristermatic tissue present in between two vascular bundles. It is function is to bring about secondary growth in the dicot stem and root.
NCERT Exemplar Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 1.8
Collenchyma cells have angular thickening at corners. There function is to provide mechanical support to young growing herbaceous stem.
NCERT Exemplar Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 1.9
Aeronchyma is a specilised parenchyma having large air spaces. It provides buyoncy to the hydrophytic plants.
NCERT Exemplar Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 1.10

Question 11.
The stomatal pore is guarded by two kidney shaped guard cells. Name the epidermal cells surrounding the guard cells. How does a guard cell differ from an epidermal cell? Use a diagram to illustrate your answer.
Solution:
Stematal apparatus is a special modification of epidermal tissue present over leaf area. The epidermal cells surrounding the guard cells of stomata are called subsidiary cells include.
Differences between guard cells and epidermal cells include:
NCERT Exemplar Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 1.11

LONG ANSWER QUESTIONS

Question 1.
Is Pirns an evergreen Tree? Comment.
Solution:

  • The plants which have persistent leaves in all the four seasons are evergreen. Deciduous plants in contrast completely loose their foliage during winter or dry season.
  • Pinus belonging to gymnosperms is an evergreen tree. Under conditions of extreme cold the flowering plants shed their leaves and become dormant.
  • In Pinus due to the presence of a thick bark thick needle-like leaves and sunken stomata to reduce the rate of transpiration the leaves we not shed.
  • The cold areas are both physiologically and physically dry due to scanty rainfall, precipitation as snow, decreased root absorption at low temperature and exposed habitats.
  • Pinus however is well adapted to such conditions. It continues to manufacture food during this period and grows to domiante other plants. This show that Pinus is an evergreen tree.
  • It does not shed its leaves or needles under any condition.

Question 2.
Assume that a pencil box held in your hand, represents a plant cell. In how many possible planes can it be cut? Indicate these cuts with the help of line drawings.
Solution:
A. If a plant cell is cut in different plane if result, in radial symmetry.
NCERT Exemplar Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 1.12
B. If a plant cell is cut in two equal halves it result in bilateral symmetry.
NCERT Exemplar Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 1.13

Question 3.
Each of thefollowing terms has some anatomical significance. What do these terms mean? Explain with the help of line diagrams.
(a) Plasmodesmata
(b) Middle Lamella
(c) Secondary wall
Solution:
NCERT Exemplar Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 1.14

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

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

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

These Solutions are part of NCERT Exemplar Solutions for Class 11 Biology. Here we have given NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants.

VERY SHORT ANSWER QUESTIONS

Question 1.
Roots obtain oxygen from air in the soil for respiration. In the absence or deficiency of 02, root growth is restricted or completely stopped. How do, the plants growing in marsh lands or swamps obtain their 02 required for root respiration?
Solution:
The roots of the plants as Rhizophora that grow in marsh/swamp areas become negatively geotropic. They grow vertically upwards in air, above the soil level and respire. They are thus called respiratory roots or pneumatophores.

Question 2.
In Opuntia, the stem is modified into a flattened green structure to perform the function of leaves, (i.e., photosynthesis). Cite some other examples of modifications of plant parts for the purpose of photosytnthesis.
Solution:
In Opuntia a xerophytic plant leaves are modified into spine to reduce the rate of transpiration and they do not perform the photosynthesis at all.
The function of photosynthesis in Opuntia plant is performed by stem which is thick fleshy and flattened structure containing chlorophyll and stores food and known as phylloclade.
In some plants similarly roots become assimilatory e.g., case of Trapa and Tinospom. These roots grow outside the soil, develop chlorophyll in them and perform photosynthesis.

Question 3.
In swampy areas like the sunderbans in West Bengal, plants bear special kind of roots called……
Solution:
Pneumatophores Roots are meant for the absorption of water and minerals from the soil. Cells of roots require 02 to respire. In swampy areas, soil does not have air, so no 07 is available to them.
In such cases, roots come out of the soil showing negative geotropism and breathe after coming in contact with air, e.g., Rhizophora. Such roots are called pneumatophores or respiratory roots.

Question 4.
In aquatic plants like Pistia and Eichhornia, leaves and roots are found near…..
Solution:
In Pistia and Eichhonia, the stem is like a runner where it branches to form leaves at the
v apex and roots below. Both the plants are hydrophytes and thus the roots are found near the surface of water.

Question 5.
Which parts in ginger and onion are edible?
Solution:
The edible part of ginger is rhizome the modified stem which stores food material whreas the edible part in onion is fleshy leaves, where the internode becomes shortened, leaves get condensed to form a tunic and store food material.

Question 6.
In epigynous flower, ovary is situated below the……….. .
Solution:
Ovary is situated below the thalamus (inferior) in epigynous flower while the other whorls of flower like sepals, petals and androecium grows above the ovary (superior), e.g., carrot, guava, Cucurbit a, sunflower, etc.
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.1

Question 7.
Add the missing floral organs of the given floral formula of Fabaceae.
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.14
Solution:
The floral formula of fabaceae family is
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.2
The flower of fabaceae is bisexual, zygomorphic, pentameros, gamosepalous, corolla-petals 5, androecium is ten diadelphous, gynoecium- superior, ovary monocarpellay.

SHORT ANSWER QUESTIONS

Question 1.
Give two examples of roots that develop from different parts of the angiospermic plant other than the radicle.
Solution:
Prop roots are meant for support. Prop roots develop from the lower nodes of stem of banyan tree. They grow downwards and touch the soil.
Stilt roots arise from the lower nodes of stem in sugarcane and enter the soil to provide strength to the plant. These protect the plant against winds.
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.3

Question 2.
The essential functions of roots are anchorage and absorption of water and minerals in the terrestrial plant. What functions are associated with the roots of aquatic plants. How are roots of aquatic plants and terrestrial plants different?
Solution:
Usually the terrestrial roots show a branched network that helps in anchorage and absorption of water and minerals from soil to the plant. While in aquatic plants, roots show modification and deviation from their normal function.
Ex – in plants like Trapu, Tinospora the roots are green and highly branched to increase the photosynthetic area, whereas in plants like Jussiaecci they get inflated due to air project out of water so a to help the plant in floating and exchange of gases.
Difference between roots of aquatic plants and terrestrial plants are as:
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.4

Question 3.
Draw diagrams of a typical monocot and dicot leaves to show their venation pattern.
Solution:
The pattern of distribution of veins and veinlets in the lamina of leaf is called Venation. It’s pattern is different in monocot and dicot leaf.
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.5

Question 4.
A typical angiosperm flower consists of four floral parts. Give the names of the floral parts and their arrangements sequentially.
Solution:
Following are the four floral parts of typical angiospermic flower.

  1. Calyx is the outermost whorl of the flower and comprised of sepals. These are usually green and (in bud stage) are protective in function.
  2. Corolla is composed of petals, usually bright coloured to attract insects for pollination.
  3. Androecium is composed of stamens, the male reproductive organ. Each stamen consists of stalk or filament and anther (containing pollen sac and pollen grains).
  4. Gynoecium is the female reproductive part and comprised of one or more carpels. Each
    NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.6

Question 5.
Reticulate venation is found in dicot leaves while in monocot leaves venation is of parallel type. Biology being a ‘Science of exceptions’, find out any exception to this generalisation.
Solution:
Reticulate venation is a characteristic of dicots and parallel venation is of monocots. But few exceptions are also seen in this generalisation, parallel venation is also found in dicot plants, e.g, Calophyllum, Corymbium, etc and reticulate venation is also found in monocot plants such as Alocasia, smilax, etc.

Question 6.
You have heard about several insectivorous plants that fee on insects. Nepenthes or the pitcher plant is one such example, which usually grows in shallow water or in march lands. What part of the plant is modified into a pitcher? How does this modification help the plant for food even though it can photosynthesise like any other green plant?
Solution:

  1. In insectivorous plant like Nepenthes, the leaf lamin is modified to form a pitcher and anterior part of petiole coils like tendril which keeps the pitcher in a vertical direction.
  2. Posterior part of the petiole remains flattened like a leaf. The apex of lamina forms a lid.
  3. Pitcher contains digestive enzyme for digesting trapped insects.
  4. All these modifications and adaptations are developed to make up for the nitrogen deficiency in the plant because these plants are found in N2 deficient soil, (marshy/swamp soils).
    NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.7

Question 7.
How can you differentiate between free central and axile placentation?
Solution:
The arrangement of ovules on the walls of ovary with the help of special kind of tissue called placenta is placentation. Plants show different types of placentation.
Difference between free central placentation and axile placentation include:
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.8
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.9

Question 8.
Why is maize grain usually called as a fruit and not a seed?
Solution:
The maize grain is usually known as fruit because it is a ripened ovary which contains a ripened ovule, e.g., a single seed. This fruit is known as caryopsis in which the pericarp is fused with the seed coat. The maize grain occurs attached to a thick cob or peduncle.

Question 9.
Tendrils of grapevins are homologous to the tendril of pumpkins, but are analogous to that of pea. Justify the above statement.
Solution:
Homologous organs are organs that have similar origin but they differ functionally. Axillary bud of stem gives rise to tendril of both grapevine and pumpkins so they have same origin, i.e., homologous, whereas analogous organs are organs having different origin, but perform same function. The tendril of pea arises from the leaf and helps the plant to climb.

Question 10.
Rhizome of ginger is like the roots of other plants that grows underground. Despite this fact ginger is a stem and not a root. Justify.
Solution:
Rhizome of Ginger is a type of modified underground stem which grows horizontally underground and bears nodes, intemodes and scaly leaves and buds, which gives rise to aerial shoots.
The adventitious root arises from the lower surface of nodes. It is not a true root because root does not have nodes and intemodes. The rhizome does not perform the function of anchorage and absorption, rather serve as reservoir for food storage. All these characteristics support the fact that ginger is a stem and not a root.
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.10

LONG ANSWER QUESTIONS

Question 1.
Distinguish between families – Fabaceae, solanaceae, Liliaceae on the basis of gynoecium characteristics (with figures). Also write economic importance of any one of the above family.
Solution:
The families in plant kingdom mainly differ from each other in their reproductive structures.
Based on characteristics of gynoecium the difference between the three families include the following:
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.11
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.12

Question 2.
Describe various stem modifications associated with food storage climbing and protection.
Solution:
The aerial part of plant bearing nodes, intemodes, buds, flowers, fruits and seeds is stem. Besides these functions and forms, it gets modified and perform under spfccial conditions.
The various stem modifications include:
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.13

We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 5 Morphology of Flowering Plants, help you. If you have any query regarding NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 5 Morphology of Flowering Plants, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Solutions for Class 11 Biology Chapter 4 Animal Kingdom

NCERT Exemplar Solutions for Class 11 Biology Chapter 4 Animal Kingdom

These Solutions are part of NCERT Exemplar Solutions for Class 11 Biology. Here we have given NCERT Exemplar Solutions for Class 11 Biology Chapter 4 Animal Kingdom.

VERY SHORT ANSWER QUESTIONS

Question 1.
What is the importance of pneumatic bones and air sacs in Aves?
Solution:
Birds possess light wfi’ght bones that contain internal spaces filled with air, which are pneumatic bones. They are an adaptation for flight as they help in, reducing the body weight. Aerodynamic lungs with specialized air sacs are an additional feature that aids birds in flying (e.g., bald eagle, pigeon).

Question 2.
What is metagenesis? Mention an example which exhibits this phenomenon.
Solution:
The phenomenon in which one generation of certain plants and animals reproduce asexually, followed by the sexually reproducing generation is metagenesis. Both the forms in metagenesis are diploid hence, it is known as the false alternation of generation. Coelenterates exhibit metagenesis (e.g., Obelia) where in its life cycle polyp form alternates with medusa.

Question 3.
What is the role of feathers?
Solution:
Feathers are the epidermal out growths that form distinctive outer covering or plumage in birds.
A variety of role are played by feathers which includes:
(i) They provide life and help in flight, by creating airfoil shape for wings.
(ii) They help in maintaining body temperature.
(iii) Feathers play a vital role in mating by providing secondary sexual that characters in both the sexes the colour and markings determine the alteractiveness of mate.

Question 4.
Which group of chordates posses sucking and circular mouth without jaws?
Solution:
Class-Cyclostomata is comprised of living jawless fishes. They have a circular mouth and lack jaws, hence they are also called agnathans. The mouth works like a sucker and is surrounded by tentacles (e.g., lampreys and haglish). These also prosses rectroctable teeth that are homy.

Question 5.
Mention two modifications in reptiles required for terrestrial mode of life.
Solution:
Certain characters acquired by reptiles for the terrestrial adaptations include.
(i) Body is covered with dry and comified skin and epidermal scales or scutes.
(ii) Internal fertilisation.

Question 6.
What is the role of radula in molluscs?
Solution:
The radula is a special rasping structure present many molluscs. It is used to scrape and scratch the food and to create depressions in rocks used as habitat.
It bears many rows of tiny teeth that are replaced
as they wear down e.g., Limplet is a marine invertebrate that uses its radula for creating home by boring a shallow hole in the rock.

Question 7.
Name the animal, which exhibits the
phenomenon of bioluminescence. Mention the
phylum to which it belongs.
Solution:
Bioluminescence is the phenomenon of production and emission of light by an organism as a result of chemical reaction during which chemical energy is converted to light energy. The phenomenon of bioluminescence is exhibited by Ctenoplana from phylum- Ctenophora.

Question 8.
Write one example for each of the following in the space providing.
(a) Cold blooded animal
(b) Warm blooded animal
(c) Animal possessing dry and comified skin
(d) Dioecious animal
Solution:
(a) A cold blooded animal is Crocodilus (crocodile)
(b) Elephas maximus (elephant), (mammal) is a warm blooded animal.
(c) Testudo (tortoise) bears dry and comified skin.
(d) Ascaris (roundworm) is a dioecious animal.

Question 9.
Differentiate between a diplobastic and triploblastic animal.
Solution:
Diploblastic animals are animals in which the cells are arranged in two embryonic layers, an external ectoderm and an internal endoderm (e.g., coelentrates). Animals in which the developing embryo has a third germinal layer, i. e., mesoderm lying between the ectoderm and endoderm are calledtriploblastic animals, (e.g., chordates).

Question 10.
Give an example of the following
(a) Roundworm
(b) Fish possessing poison sting
(c) A limbless reptile/amphibian
(d) An oviparous mammal
Solution:
(a) Roundworm – A scans
(b) Fish possessing poison sting – Trygon
(c) A limbless reptile/amphibian – Ichthyophis
(d) An oviparous mammal – Duck billed platypus.

Question 11.
Provide appropriate technical term in the space provided.
(a) Blood-filled cavity in arthropods
(b) Free-floating form of cnidaria
(c) Stinging organ of jelly fishes
(d) Lateral appendages in aquatic annelids
Solution:
(a) The blood-filled cavity in arthropods containing haemolymph is haemocoel.
(b) A form in cnidarians in which the body is shaped like an umbrella which can float freely in sed water is medusa.
(c) Capsules of specialised cells in cnidarians which act as a paraylysing sting are nematocytes.
(d) The paired unjointed lateral outgrowth in annelids bearing chaetae are parapodia.

Question 12.
Match the following.
NCERT Exemplar Solutions for Class 11 Biology Chapter 4 Animal Kingdom 1.1
Solution:
A. —> (iii)
B. —> (i)
C. —> (iv)
D. —> (ii)
(a) Octopus The appendages in invertebrates that are used for grasping food and for locomotion are tentacles.
(b) Crocodile for locomotion, and swimming limbs are used.
(c) Catta Fins are means of locomotion and are used to generation optimum thrust thus controlling the subsequent motion.
(d) Ctenoplana Locomotory organs formed by strong cilia with fused bases are comb plates.

SHORT ANSWER QUESTIONS

Question 1.
Differentiate between
(a) Open circulatory system and closed circulatory system.
(b) Oviparous and viviparous characteristic.
(c) Direct development and indirect development.
Solution:
Differentiation between these are as below
NCERT Exemplar Solutions for Class 11 Biology Chapter 4 Animal Kingdom 1.2

Question 2.
There has been an increase in the number of chambers in heart during evolution of vertebrates. Give the names of the class of vertebrates having two, three or four chambered heart.
Solution:
(a) In organisms like fishes two chambered heart is present. Mixing of oxygenated and deoxygenated blood blood occurs as only one atria and one ventricle is present which are not separated.
(b) After division of auricle into right and left halves three chambered heart develops and in amphibian. In ventiricles mixing of oxygenated and deoxygenated blood occurs.
(c) In reptilies an intermidiary heart is present in which ventricles get partially divided through a septum which is incomplete thus having a false four-chambered heart e.g., Crocodiles.
(d) Both the auricle and ventricle are divided into two halves in four chambered heart and so no mixing of oxygenated and deoxygenated blood occurs, e.g., birds and mammals.

Question 3.
Fill up the blank spaces appropriately
NCERT Exemplar Solutions for Class 11 Biology Chapter 4 Animal Kingdom 1.3
Solution:
Excretion involves the elimination of metabolic waste products from the animal body. In the process of excretion in different animals different organs are involved.
(a) In arthropods excretory products from haemolymph are removed by the malpighian tubules.
(b) The excretory organ occurs as segmentally arranged coiled tubules called nephridia in annelids.
(c) Excretion occurs by paired structures called organ of Bojanus in molluscs also called metanephridia.
(d) Mesonephric kidneys are associated with excretion in amphibians.
The circulation of blood and lymph along with oxygen carbondioxide, hormones, blood cells, etc, within the body system for the nourishment of cells, fighting diseases, and for stabilising body temperature and pH is involved blood circulation.
NCERT Exemplar Solutions for Class 11 Biology Chapter 4 Animal Kingdom 1.4

Respiratory organs are involved in the
exchange of gases from the atmosphere.
Different respiratory organs in various animals.
(a) Lungs and skin in amphibians.
(b) Lung/gills/tracheal system in arthropoda and molluscs.
(c) Skin in annelids.

Question 4.
Match the following
NCERT Exemplar Solutions for Class 11 Biology Chapter 4 Animal Kingdom 1.5

Solution:
A. —> (v)
B. —> (iii)
C. —> (ii)
D.—> (i)
E. —> (vi)
F. —> (iv)
A. Amphibians are found in both aquatic and terrestrial habitat. Their large is completely aquatic while adult lives in terrestrial as well as in aquatic habitat.
B. Mammals produce milk in the mammary glands and feed their young one. The mammary glands are enlarged exocrine modified sweat glands functional in female mammals.
C. Chondrichthyes have (notochord) in the young stage which is gradually replaced by cartilage.
D. Osteichthyes possess air bladder which is a vesicle or sac containing air.
E. Cyclostomes have sucking and circular mouth without jaws which is surrounded by tentacles and the tongue bears teeth, e.g., lamprey and hagfish.
F. Aves comprise of light weighted bones with internal spaces field with air called pneumatic bones and aerodynamic lungs with specialised air sacs. These are the adaptations which enable birds to fly.

Question 5.
Endoparasites are found inside the host body. Mention the special structure, possessed by these and which enables them to survive in those conditions.
Solution:
Endoparasites such as Taenia solium and Fasciola hepatica (liver fluke), etc., are found inside body the host and survive due to the presence of certain characters.
Endoparasites special characters which include:
(i) The is respiration is anaerobic and the gaseous exchange in via general body surface.
(ii) They bear additional organs for the attachment to the host. Taenia solium posses hooks and suckers for the attachment with the host. Fasciola hepatica possesses acetabulum or posterior sucker for the attachment.
(iii) they have well developed reproductive organs. They are generally, harmaphrodite and self fertilisation occurs commonly.
(iv) They have a thick tegument (body covering) which is resistant to the host’s digestive enzymes and antioxins.
(v) Locomotary organs are absent.
(vi) They lack digestive organs because digested and semi digested food of the host is directly absorbed through their body surface.

Question 6.
Mention two similarities between
(a) Aves and mammals
(b) A frog and crocodile
(c) A turtle and Pila
Solution:
(a) Following are the similarities between aves and mammals
(i) Presence of four chambered heart.
(ii) The members of both the groups are homeotherms, i.e., warm blooded. They are able to maintain constant body temperature.
(b) Similarities between frog and crocodile include:
(i) They are cold blooded animals. The members of both the groups are poikilotherms, i.e., they lack the capacity to regulate their body temperature.
(ii) Frogs and crocodiles are oviparous animals.
(c) Similarities between turtle and Pila include
(i) Body is covered with dry and comified skin in both animals. In turtle, the epidermal covering is known as scales whereas in case of Pila, it is known as calcareous shell.
(ii) Both animals are oviparous.

Question 7.
Name
(a) A limbless animal
(b) A cold blooded animal
(c) A warm blooded animal
(d) An animal possessing dry and comified skin
(e) An animal having canal system and spicules
(f) An animal with cnidoblasts
Solution:
(a) Ichthyophis does not possess limbs.
(b) A cold blooded animal scoliodon (dog fish).
(c) warm blooded animal is Columba (pigeon).
(d) Naja naja (snake) possesses dry and cornified skin.
(e) Sycon (sponge) possesses canal system and bear spicules.
(f) Obelia bears cnidoblast.

Question 8.
Excretory organs of different animals are given below. Choose correctly and write in the space provided.
NCERT Exemplar Solutions for Class 11 Biology Chapter 4 Animal Kingdom 1.6

Solution:
Metabolism in body leads to the formation of waste that can affect body’s vital organs so it has to be removed from body. Different classes of organisms possess different types of excretory organs to eliminate the byproduct of metabolism.
A. —> (vi)
B. —> (ii)
C. —> (v)
D. —> (iii)
E. —> (vi)
F. —> (i)
A. Balanogolossus – Proboscis glands. This gland excretes brown granules and is present in front of central sinus.
B. Nephridia in Leech. It helps in osmoregulation and excretion.
C. Malpighian tubules in Locust open into gut and help in excretion.
D. The Flame cells of liver fluke are specialised cells in Platyhelminthes which helps in osmoregulation and excretion. These are also called protonephridia.
E. Sea urchin-absent Specialised excretory organs are absent in sea urchin.
F. It TV/a-Metanephridia is a type of excretory gland or nephridium found in many types of invertebrates such as annelids, arthropods, and molluscs (in molluscus nephridia is also known as Bojanus organ).

LONG ANSWER QUESTIONS

Question 1.
What is the relationship between germinal layers and the formation of body cavity in case of coelomate, acoelomates andpseudocoelomates?
Solution:
Multicellular organisms typically possess a concentric arrangement of tissues in the body. These tissues are derived from the three embrycnio cell, layers called germinal layers.
(i) The outer layer is the ectoderm, the middle layer is the mesoderm and the innermost layer is the endoderm.
(ii) Ectoderm is associated with the formation of CNS, eye lens, ganglia, nerves and glands.
(iii) Mesoderm forms the that in structural components of the body like the skeletal muscles the skeleton, the dermis of the
skin connective tissue, etc.
(iv) Endoderm layer is associated with the formation of the stomach, colon, liver, pancreas urinary bladder and other vital organs is an organism.
(v) Coelom is the body cavity that is lined by mesoderm and the animals possessing coelom are called as ceolomates. e.g., phylum-Annelida, Mollusca, Arthropoda, Echnidermata, Hermichordata and Chordata.
(vi) In some organisms, body cavity is not lined by mesoderm, instead mesoderm is present in the form of scattered pouches in between ectoderm and endoderm, Such body cavity is called pseudocoelom and animals possessing there stusturs are refered to as pseudocoelomates e.g., As car is.
(vii) The animals in which there is complete absence of body cavity are called acoelomates. e.g., Platyhelminthes.
NCERT Exemplar Solutions for Class 11 Biology Chapter 4 Animal Kingdom 1.7

Question 2.
Comment upon the habitats and external features of animals belonging to class- Amphibia and Reptilia.
Solution:
Amphibians
(i) They can dwell in aquatic as well as terrestrial habitats. They are ectothermic or (cold blooded).
(ii) They are tetrapods having (4 limbs) which facilitate movement on land.
(iii) Their limbs have evolved from the pectoral and pelvic fins.
(iv) Skin is thin, covered by mucus and remains mostly moist. It also serves as an accessory source of oxygen.
(v) They breathe through gills and lung gills usually appear in the larval stage, replaced by lungs in the adults stage.
(vi) Their heart is three chambered with two atria and one ventricle.
(vii) Females are oviparous and fertilisation is mostly external.
(viii) Larva is a tadpole, which metamorphose into adult e.g., Rariu frog, Nectureus (mud puppy), Salamandera (salamander).
NCERT Exemplar Solutions for Class 11 Biology Chapter 4 Animal Kingdom 1.8
Reptiles
(i) They are mostly terrestrial animals and their body is covered by dry, and comified skin, epidermal scales or scutes.
(ii) In reptiles the mode of locomotion is creeping and crawling.
(iii) Lungs are well developed and present in all stages of life.
(iv) Claws are present in toes.
(v) s Appendages are well adapted for land movement.
(vi) Heart possesses a partially divided ventricle and 2 atria.
(vii) They lay amniotic eggs which are inclubated on land.
(viii) They are poikilothermic or cold blooded animals. Temperature is regulated mechanically and not metabolically by moving in and out; source of heat is usually the sun.
(ix) Fertilisation is internal. They are oviparous and development of young ones is direct e.g., Chelone (turtle), Naja (cobra), Crocodicus (crocodile).
NCERT Exemplar Solutions for Class 11 Biology Chapter 4 Animal Kingdom 1.9

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NCERT Exemplar Solutions for Class 11 Biology Chapter 3 Plant Kingdom

NCERT Exemplar Solutions for Class 11 Biology Chapter 3 Plant Kingdom

These Solutions are part of NCERT Exemplar Solutions for Class 11 Biology. Here we have given NCERT Exemplar Solutions for Class 11 Biology Chapter 3 Plant Kingdom.

VERY SHORT ANSWER QUESTIONS

Question 1.
Food is stored as floridean starch in Rhodophyceae. Mannitol is the reserve food material of which group of algae?
Solution:
Mannitol is a reserve food material of the members of Phaeophyceae (brown algae).

Question 2.
The plant body in higher plants is well differentiated and well developed. Roots are the organs used for the purpose of absorption. What is the equivalent of roots in the less developed lower plants?
Solution:
Root like structure called rhizoids are present instead of roots in less developed lower plants (bryophytes and pteridophytes). The plant tissue system in these is not differentiated into true leaf, stem and roots as it is found in higher plants (gymnosperm and angiosperm).

Question 3.
Most algal genera show haplontic life style. Name an alga which is
(a) Haplo diplontic
(b) Diplontic
Solution:
Haplo diplontic type of life cycle is exhibited by Ectocarpus, Polysiphonia and Kelps. The main plant body is saprophytic in Fucus and it shows diplontic type of life cycle.

Question 4.
In bryophytes male and female sex organs are called …………….. and …………..
Solution:
In bryophytes the male sex organ in antheridium and female sex organ is archegonium.
Antheridium produces flagellate antherozoids which are male gametes.
Archegonia is the female part which bears a single egg cell.

SHORT ANSWER QUESTIONS

Question 1.
Why are bryophytes called the amphibians of the plant kingdom? Amphibians can their in water as well as on terrestrial habitat.
Solution:
Bryophytes are a group of primitive plants having a dominant gametophytic plant body. These plants can live in soil but depend on water for movement of male gametes called antherozoids to reach the archegonium (female organ bearing egg cell) so that fertilisation can occur, so bryophytes are called the amphibians of the plant kingdom.

Question 2.
Heterospory, i.e., formation of two types of spores— microspores and megaspores is a characteristic feature in the life cycle of a few members of pteridophytes and all spermatophytes. Do you think heterospory has some evolutionary significance in plant kingdom?
Solution:

  1. The production of spores of two different sizes and sexes by the sporophytes of land plants is heterospory. Two types ofspores are produced by heterosporic plants.
  2. Small spores are microspores which germinate into the male gametophyte and large spores are macrospores which develop into the female gametophyte.
  3. Pteridophytes are intermediate between bryophytes and gymnosperms in the evolution of plants.
  4. All bryophytes are homosporous and all gymnosperms are heterosporous. This condition is advanced as sexual dimorphism results in cross fertilisation.
  5. Primitive or earlier pteridophytes are homosporous while later pteriodophytes are heterosporous e.g., Dryopteris, Pteris homosporous Selaginella, Sn/vrao-heterosporous.

Question 3.
Each plant group of plants has some phylogenetic significance in relation to evolution Cycas, one of the few living members of gymnosperms is called as the ‘relic of past’. Can you establish a phylogenetic relationship of Cycas with any other group of plants that justifies the above statement?
Solution:
Cycas is an evergreen plant which resembles palm. It has an unbranched stem and large compound leaves. It exhibits phylogenetic relationship with pteridophyte. Its evolutionary characters include thefollowing:
(i) Growth is redundant.
(ii) Shedding of seed while the embryo is still immature.
(iii) Minimal secondary growth and manoxylic wood.
(iv) Megasporophylls are leaf like.
(v) Sperms are flagellate even when pollen tube is present.
(vi) Leaf bases are persistent.
(vii) Ptysix is circinate.
(viii) Arrangement of microsporangia in well defined archegonia.

Question 4.
Comment on the life cycle and nature of fem prothallus.
Solution:
The life cycle of ferm (Dryopteris) clearly depicts the alternation of generation. The gametophytic stage (n) alternates with the sporophytic stage (2n) in the life cycle as shown in the figure.
NCERT Exemplar Solutions for Class 11 Biology Chapter 3 Plant Kingdom 1.1
The prothallus of the fem is a multicellular, free living, thalloid, haploid and autotrophic structure. It develops from the spores produced by sporophyte after reduction division.
These spore germinate within a germtube with an apical cell and forms a filament of 3-6 cells and one or two rhizoids at the base which later develops into gametophytic plant.
NCERT Exemplar Solutions for Class 11 Biology Chapter 3 Plant Kingdom 1.2

Question 5.
How are the male and female gametophytes of pteridophytes and gymnosperms different from each other?
Solution:
The male and female gametophytes of pteridophytes and gymnosperms different from each other as:
NCERT Exemplar Solutions for Class 11 Biology Chapter 3 Plant Kingdom 1.3

Question 6.
How are the male and female gametophytes of pteridophytes and gymnosperms different from each other?
Solution:

  1. Symbiosis is a type of interaction of two living organisms where both the associated partners derive some benefit from each other both co¬exist and flourish well.
  2. Mycorrhiza is a symbiotic association between fungus and the roots of vascular plants. The fungus colonizes the roots of the host either intra or inter cellularly. It helps in the nutrient absorption from soil for the plant.
  3. Mycorrhizal associations are present in conifers, i.e.,Pinus, Cedrus, Abies and Picea.
  4. Coralloid roots develop in Cycas. It is produced in clusters at the base of the stem and protrudes out on the ground.
  5. It is dichotomously branched and greenish in colour. It contains algal zone in cortex.
  6. This algal zone contains blue green algae like Anabaena and Nostoc which grow in symbiotic association with coralloid roots.

LONG ANSWER QUESTIONS

Question 1.
Explain why sexual reproduction in angiosperms is said to take place through double fertilisation and triple fusion. Also draw a labelled diagram of embryo sac to explain the phenomena.
Solution:

  • An angiospermic plants reproduces sexually by the formation of male and female gametes.
  • The male gamete is a pollen which contains two male nuclei and the female gamete is an egg cell produced in ovule (female gametophyte).
  • The pollen grains germinate on the stigma of a flower and the results in growth of pollen through the tissues of stigma and style and reach the egg apparatus.
  • The two male gametes are discharged within the embryo sac. One of the male gamete fuses with the egg cell to form a diploid zygote.
    This’fusion is known as fertilisation or syngamy. The second male gamete fuses with the diploid secondary nucleus and forms the triploid Primary Endosperm Nucleus (PEN). This fusion is known as triple fusion.
  • Because of the involvement of two fusion, this event in angiosperms is termed as double fertilisation. The zygote then develops into embryo and PEN develops into endosperm which provides nourishment to the developing embryo.
    NCERT Exemplar Solutions for Class 11 Biology Chapter 3 Plant Kingdom 1.4

Question 2.
Draw labelled diagrams of
(a) Female and male thallus of a liverwort.
(b) Gametophyte and sporophyte of Funaria.
(c) Alternation of generation in angiosperm.
Solution:
NCERT Exemplar Solutions for Class 11 Biology Chapter 3 Plant Kingdom 1.5
(a) Liverworts
(i) Male thallus of Marchantia polymorpha
(ii) Female thallus of Marchantia polymorpha
(b) Funaria
(gametophyte and sporophyte)
NCERT Exemplar Solutions for Class 11 Biology Chapter 3 Plant Kingdom 1.6

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NCERT Exemplar Solutions for Class 11 Biology Chapter 2 Biological Classification

NCERT Exemplar Solutions for Class 11 Biology Chapter 2 Biological Classification

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

VERY SHORT ANSWER QUESTIONS

Question 1.
What is the principle underlying the use of cyanobacteria in agricultural fields for crop improvement?
Solution:
Cyanobacteria are abfs to fix atmospheric nitrogen and make it available to the plants and thus are used in agricultural crop improvement.
This improves crop yield and also reduces the cost of application of nitrogen fertilisers, e.g., Anabena and Nostoc.

Question 2.
How is the five kingdom classification advantageous over the two kingdom classification?
Solution:
The five kingdom classification, proposed by RH whittaker is based upon cell structure, body structure (unicellular, multicellular), nutrition (autotrophic, heterotrophic) reproduction and habitat either aquatic, terrestrial, or aerial and phylogenetic relationship.
It is thus more useful as compared to two kingdom system of classification which does not distinguish between prokaryotes and eukaryotes and no other kingdom except plant and animal are identified.
Polluted water bodies have high growth of algae due to the presence of nutrient. These nutrients increase the rapid growth of water plants, i.e.,

Question 3.
Polluted water bodies have usually very high abundance of plants like Nostoc and Oscillitoria. Give reasons.
Solution:
algae especially Nostoc and Oscillitoria, etc., and result in colonies. These colonies are generally surrounded by a gelatinous sheath and leads to the formation of blooms in water bodies.

Question 4.
Are chemosynthetic bacteria autotrophic or heteroterophic?
Solution:
Chemosynthetic bacteria are capable of oxidising various inorganic substances such as nitrates, nitrites and ammonia and use the released energy for production of ATP and thus they are autotrophs and not heterotrophs.

Question 5.
The common name of a pea is simpler than its botanical (scientific) name Pisum sativum why then is the simpler common name not used instead of the complex scientific/botanical name in biology?
Solution:
The common or vernacular names cause confusion regarding the identification of specific specimen as they change with the change in place whereas the scientific names are in latin and universally accepted and understood. Scientific names are thus preferred over the common vernacular names.

SHORT ANSWER QUESTIONS

Question 1.
Diatoms are also called as ‘pearls of ocean’, why? What is diatomaceous earth?
Solution:

  1. Diatoms and desmids are included under chrysophytes, kingdom-Protista.
  2. These are the main producers in the ocean. They prepare food for themselves and for the other life forms in the ocean as were a siliceous shell known as frustule cores the body of diatoms, this is the reason they are also called as ‘pearls of ocean. ‘
  3. Diatomaceous earth’ is the accumulation of large deposits of diatoms that forms a siliceous covering extending for several 100 metres formed in billions of years.
  4. The material obtained from these deposits is used in polishing and filtration of oils and syrups.

Question 2.
There is a myth that immediately after heavy rains in forest, mushrooms appear in large number and make a very large ring or circle, which may be several metres in diameter. These are called as ‘fairy rings’. Can you explain this myth of fairy rings in biological terms?
Discuss the mycilial structure in Agaricus and its soil borne nature.
Solution:
The fruiting bodies in Agaricus are known as basidiocarps. They form a concentric ring like structure from the mycelium present in the soil. These basidiocarps resemble button in shape and develop to form a ring like structure.
NCERT Exemplar Solutions for Class 11 Biology Chapter 2 Biological Classification 1.1
This fairy ring structure in Agaricus stimulate productivity in plants. This rings are the fruiting bodies of this fungus and the diameter of this fairy ring increases every year due to the spread of mycelium.

Question 3.
Neurospora an ascomycetes fungus has been •arsed as a biological tool to understand the mechanism of plant genetics much in the same way as Drosophila has been used to study animal genetics. What makes Neurospora so important as a genetic tool?
Solution:
Neurospora fungus can be grown easily under laboratory conditions by providing ‘minimal medium’ like inorganic salts, carbohydrates source and vitamin (biotin) and thus was selected to be a very good tool in genetics. The mutations can be also easily introduced in the fungal cells and meiotic division can be easily seen under X-ray treatment.

Question 4.
At a stage of their cycle, ascomycetes fungi produce the fruiting bodies like apothecium, perithecium or cleistothecium. How are these three types of fruiting bodies different from each other?
Discuss the type of fruiting bodies formed by ascomycetes fungus and differentiate accordingly on the basic of there structures.
Solution:
Ascomycetes consist of sporangial sac called ascus. Asci (singular-ascus) may occur freely or in aggregated form with dikaryotic mycelium to form the fruitification bodies called ascocarps.
The fruitification formed by asci include the following :
(i) Apothecium is cup like structure, e.g., Peziza.
NCERT Exemplar Solutions for Class 11 Biology Chapter 2 Biological Classification 1.2
(ii) Perithecium is flask shaped, e.g., Neurospora
NCERT Exemplar Solutions for Class 11 Biology Chapter 2 Biological Classification 1.3
(iii) Cleistothecium is closed with a slit, e.g., Penicilium
NCERT Exemplar Solutions for Class 11 Biology Chapter 2 Biological Classification 1.4

Question 5.
What obsrevable features in Trypanosoma would make you classify it under kingdom- Protista?
Discuss cell structure of Trypanosoma also discuss its different strain brief.
Solution:
Trypanosoma is included under flagellated protozoans on the basis of locomotary organ. It resembles Protisia in the following characters.
NCERT Exemplar Solutions for Class 11 Biology Chapter 2 Biological Classification 1.5
(i) It is unicellular.
(ii) It reproduces asexually i.e., by binary fission.
(iii) Possess centrally located nucieus and also contain an prominent nucleus endosome.
(iv) Reserve food material is in the form of granules.

LONG ANSWER QUESTIONS

Question 1.
Algae are known to reproduce asexually by variety of spores under different environmental conditions. Name these spores and the conditions under which they are produced.
Solution:
Asexual reproduction in algae is very common mean of reproduction. Algae and their spores exhibit significant diversity and vary greatly in their level of specialisation. Asexual reproduction by spores and their types include:
(i) Zoospores are mobile flagellated spores. In this protoplasm of each vegetative cell undergoes repeated longitudinal division either into 2 or 4, rarely 8 or 16 daughter protoplasts. Before the onset of division the parent cell loses its flagella.

  • Each daughter protoplast after the last series of division secretes a cell wall and a neuromotor apparatus that develops two flagella, eyespots and contractile vacuoles.
  • Each of the daughter cell thus formed resembles the parent cell in all aspects except the small size.
  • Under favourable conditions formation of zoospores is very common.
    NCERT Exemplar Solutions for Class 11 Biology Chapter 2 Biological Classification 1.6

(ii) Aplanospores are the non-motile spores. They are formed asexually within a cell, in which protoplast withdraws itself from the parent wall, rounds up and develops into aplanospores which germinate either directly or may divide to produce zoospores.
NCERT Exemplar Solutions for Class 11 Biology Chapter 2 Biological Classification 1.7

(iii) Hypnospores- In this, the protoplasm withdraws from the cell wall, rounds up and develops a thick wall under unfavourable conditions. These resting spores are called as hypnospores. Due to presence of haematochroma they are red in colour e.g., Vancheria, Ulothrix.
(iv) Akinetes are special vegetative thick walled cells present in the filaments which remain under dormant state and resume germination under favourable conditions. They can also withstand unfavourable condition as Spirogyra.
(v) Statospores are thick walled spores ‘ produced in diatoms.
(vi) Neutral spores are the protoplast, of vegetative cells directly functioning as spores (e.g., Ectocarpus).

Question 2.
Apart from chlorophyll, algae have several other pigments in their chloroptast. What pigments are found in blue, green, red and brown algae, that are responsible for their characteristic colours?
Solution:

  • All photosynthetic organisms comprise of one or more organic pigments that are capable of absorbing visible, radiations, which will initiate the photochemical reaction of photosynthesis.
  • The three major classes of pigments found in plants and algae are the chlorophylls, the carotenoids and the phycobilins.
  • Carotenoid and phycobilins are called accessory pigments since, the quanta absorbed by theese pigments can be transferred to chlorophyll.
  • The diversity of light harvesting pigments in alga implies that the common ancestor was primitive and that no close affinity exist between blue, green, red, brown, golden brown and green algae.
    NCERT Exemplar Solutions for Class 11 Biology Chapter 2 Biological Classification 1.8

 

Question 3.
Make a list of algae and fungi that have commercial value as source of food, chemicals, medicines and fodder.
Solution:
Algae
Around 70 species of marine ailgae are used for food, chemical and medicinal purpose.
NCERT Exemplar Solutions for Class 11 Biology Chapter 2 Biological Classification 1.9
Fungi
The role of fungi was established in early history. Since, the beginning of cultivation yeast have been used in making of bread and alcohol. The discovery of penicillin that marked the beginning of a new approach to microbial dis eases in human health.
Products of fungi in medicine, chemical and food include.
Around 70 species of marine aigae are used for food, chemical and medicinal purpose.
NCERT Exemplar Solutions for Class 11 Biology Chapter 2 Biological Classification 1.10

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NCERT Exemplar Solutions for Class 11 Biology Chapter 1 The Living World

NCERT Exemplar Solutions for Class 11 Biology Chapter 1 The Living World

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

VERY SHORT ANSWER QUESTIONS

Question 1.
Couplet in taxonomic key means……. .
Solution:
Couplet in taxonomic key is a pair of a contrasting characters used as tool foi” identification to aid in identification of a newly discovered organism.

Question 2.
What is a monograph?
Solution:
Monograph is a specialised work of documenting information on a particular taxon, /. e., family or genus or on aspect of subject, usually by a single author.
The main purpose of monograph is to present primary research and original work.

Question 3.
Amoeba multiplies by mitotic cell division. Is this phenomena growth or reproduction? Explain.
Solution:
Amoeba multiplies by simple mitotic cell divisions giving rise to two daughter Amoebae. Growth here is synchronous with reproduction, i.e., increases in number.

Question 4.
Define metabolism.
Solution:
Metabolism is the sum total of all biological reactions occurring in any living cell, which are controlled absolutely by enzymes. These reactions are of two types breaking down reactions (catabolism, e.g., cell ‘ respiration) and synthesing reactions (anabolism,
e.g., photosynthesis).

Question 5.
Which is the largest botanical garden in world? Name a few well known botanical gardens in India.
Solution:
A botanical garden is dedicated to collection, cultivation and display of wide range of plants labelled with their botanical names.
The largest botanical garden in the world is Royal Botanical Garden (in Kew, London). In India the famous well known botanical gardens are
(i) National Botanical Garden (NBG) Lucknow, UP.
(ii) Botanical Garden of FRI, Dehradun (UK).
(iii) Lloyd Botanical Garden, Daijeeling.
(iv) Indian Botanical Garden, Sibpur, Kolkata.

SHORT ANSWER QUESTIONS

Question 1.
A ball of snow when rolled over snow increases in mass, volume and size. Is this comparable to growth as seen in living organisms? Why?
Solution:

  1. Living organisms, grow, have metabolism and respond to external stimuli and reproduce as well. These characteristics are not shown by non-living objects.
  2. In biological terms growth is characteristic feature of all living organisms. It relates to increase in size by accumulation of protoplasm in the cell thus resulting in increase in the size of the cell.
  3. Increase in number of cell by cell division on other hand results in the size of individual organism.
  4. Snow is an inanimate (non-living) object, while rolling over, it gathers more snow on its surface thus, it increases in size by physical phenomenon but not by biological phenomenon.
  5. This growth cannot be thus compared to that seen in living organisms.

Question 2.
In a given habitat we have 20 plant species and 20 animal species. Should we call this as ‘diversity or biodiversity’? Justify your answer.
Solution:
There are existing 20 plant species and 20 animal species in the given habitat. They will exhibit the biodiversity in that given habitat because diversity refers to variation in a broad term and can be applied to any area whereas biodiversity is a degree of variation of life forms within a specified area.

Question 3.
International Code of Botanical Nomenclature (ICBN) has provided a code for classification of plants. Give hierarchy of units of classification, botanists follow while classifying plants and mention different ‘suffixes’ used for the units.
Solution:
ICBN has specified certain rules and principles in order to facilitate the study of plants by botanists. It helps in correct positioning of any organism newly discovered through the pressure of proper identification and nomenclature.
The taxonomic hierarchy, which is used while
classifying any plant given below
Kingdom-Plan tae
Division-phyta
Class-ae
Order-ales
Family-eae/ceae
Genus-First name of organism usually Latin word and written in italics.
Species-Second word of scientific name, also written in italics.

Question 4.
What are hormone receptors? What are the modes of their action ?A plant species shows several morphological variations in response to altitudinal gradient. When grown under similar conditions of growth, the morphological variations disappear and all the variants have common morphology. What are these variants called?
Solution:
These morphological variants are called bio types. It includes group of genetically similar plants showing similarity when grown in same environmental and geographical regions. The same environment provides them the similar abiotic factors like soil, pH, temperature, etc.
When growth in two different geographical regions, they are exposed to different abiotic characters which affects their growth, and development bringing changes in their external morphological features but, their genetic constitution remain same.

Question 5.
What is the difference between flora, fauna and vegetation? Eichhornia crassipes is called as an exotic species, while Rauwoljia serpentina is an endemic species in India. What do these terms exotic and endemic refer to?
Solution:
Following are the difference between flora, fauna and vegetation
NCERT Exemplar Solutions for Class 11 Biology Chapter 1 The Living World 1.1

Question 6.
Brinjal and potato belong to the same genus Solamim, but to two different species. What defines them as seperate species?
Solution:
Genus is a taxonomic rank used in bionorr.’al nomenclature comprising of a group of related species sharing few common characters.
Solanum is the largest genus of flowering plants which includes few economically important plants, e.g., potato, tomato, tobacco and brinjal. All these plants show some common morphological structures related to vegetative and reproductive similarities. So, they are are included in the same common genus Solanum.

Question 7.
The number and kinds of organism is not constant. How do you explain this statement? Change is law of nature.
Solution:
The number and kind of organisms is not constant, because of the following reasons new organism are added due to mechanisms of.
(i) sexual reproduction
(ii) mutation
(iii) evolution
The number of organisms get reduced due to
(i) environmental threats
(ii) loss of habitat
(iii) anthropogenic activities

LONG ANSWER QUESTIONS

Question 1.
Brassica campestris Linn
(a) Give the common name of the plant.
(b) What do the first two parts of the name denote?
(c) Why are they written in italics?
(d) What is the meaning of Linn written at the end of the name?
Solution:
Brassica campestris Linn
(a) The common name of Brassica compestris Linn is mustard.
(b) The first part of the name denotes the genetic name and the second part is the species name of the plant.
(c) According to ICBN, all scientific names are comprised of one genetic name followed by a species name, which require to be always written in italics. It is a rule of bionomial nomenclature.
(d) Linn means Linnaeus. He was the first to discover the plant. He identified, named and classified the plant, so the plant is named after him by adding suffix ‘Linn’, after the scientific name B. campestris.

Question 2.
What are taxonomical aids? Give the importance of herbaria and museums. How are Botanical gardens and Zoological parks useful in conserving biodiversity?
Solution:
The aids which help in identification, classification and naming of a newly discovered organisms (plant or animal) the taxonomic aids.
It could be in the form of a preserved document like herbaria or specimen kept at museums or scientific institutions. Other aids include written document like monography, taxonomic keys, couplets, etc.
A new organism found can be studied while comparing it with living plants and animals living in protected areas like Botanical gardens, Zoological parks, etc. Botanical gardens helps in conservation of plants by
(i) Plant species growing important local and keeping record of them.
(ii) Growing and maintaining species that rare are and endangered.
(iii) Supplying seeds for different aspects of botanical research.
Zoological parks contribute in conserving biodiversity by
(i) Providing natural environment and open space to animals.
(ii) Providing home to different native and exotic wild animals.
(iii) Rescue of endangered species.
(iv) Facilitating breeding animal and releasing them free. Thus, both botanical gardens and zoological parks play an important role in conservation of biodiversity.

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