ncert solutions for class 11 biology

NCERT Solutions for Class 11 Biology Chapter 12 Mineral Nutrition

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 12 Mineral Nutrition:

Section NameTopic Name
12Mineral Nutrition
12.1Methods to Study the Mineral Requirements of Plants
12.2Essential Mineral Elements
12.3Mechanism of Absorption of Elements
12.4Translocation of Solutes
12.5Soil as Reservoir of Essential Elements
12.6Metabolism of Nitrogen
12.7Summary

NCRT TEXTBOOK QUESTIONS SOLVED

1.’All elements that are present in a plant need not be essential to its survival’. Comment.
Soln. Most of the mineral elements present in the soil enter plants through roots but all of these may not be essential for their survival. Some are absorbed and accumulated by plant only because they are present in excess amount. For example plants growing near nuclear test sites take up strontium, even though it is not required by them. Thus, an essential element is that which is necessary for supporting normal growth and reproduction, its requirement must be specific i.e. its deficiency cannot be met by supplying other element and it must be directly involved in the metabolism of plant.

2.Why is purification of water and nutrient salts so important in studies involving mineral nutrition using hydroponics?
Soln.Impure water and salts contain a large number of soluble minerals and impurities. When such water and salts are used as solution culture for growing plants in hydroponics then the impurities will interfere with the experiment and will not give correct result about the essentiality of a mineral element. Therefore, purified water with defined mineral nutrients are used in hydroponics.

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3.Explain with examples:macronutrients, micronutrients, beneficial nutrients, toxic elements and essential elements.
Soln. Macronutrients : Those elements which are generally present in plant tissues in large amounts (in excess of 10 mmole Kg-1 of dry matter) and are involved in the synthesis of organic molecules and development of osmotic potential are called macronutrients or macroelement, e.g. carbon, hydrogen, oxygen, nitrogen, sulphur, potassium, calcium and magnesium etc.
Micronutrients : Those elements which are required by plants in’very small amounts (less than 10 mmole Kg-1 of dry matter) are called micronutrients, e.g. iron, zinc, manganese, boron, copper, molybdenum, chlorine and nickel. These elements are mostly involved in the functioning of enzymes as cofactor or metal activators.
Beneficial nutrients : Those elements which are required by higher plants along with the macro and micronutrients are called beneficial nutrients, e.g. cobalt, silicon, sodium and selenium.
Toxic elements : Any mineral element if supplied to plant tissue in such concentration that it reduces the dry weight of tissues by about 10 percent, is called toxic element. e.g. manganese toxicity leads to the appearance of brown spots surrounded by chlorotic veins. Excess of manganese induces deficiency of iron, magnesium and calcium.
Essential elements : Any element required by living organisms to ensure normal gfbwth, development, maintenance, metabolism and causes deficiency symptoms if not supplied to the plant from external medium is called essential element, e.g. C, H, O, N, P, K, S, Mg, Ca, Mn, Cu, Mo, Zn, B, Cl, etc. Potassium plays an important role in opening and closing of stomata, protein synthesis etc. Magnesium is found in chlorophyll and phosphorus in ATP. Mg2+ is an activator for both ribulose bisphosphate carboxylase-oxygenase and phosphsenol pyruvate carboxylase.Zn2+ is an activator of alcohol dehydrogenase and Mo of nitrogenase during nitrogen metabolism.

4.Name at least five different deficiency symptoms in plants. Describe them and correlate them with the concerned mineral deficiency.
Soln.Five different deficiency symptoms in plants are:
(i)Chlorosis – It is the loss of chlorophyll leading to yellowing of leaves. This is caused due to the deficiency of N, K, Mg, S and Fe etc.
(ii)Necrosis – Killing or death of tissue particularly leaf is called necrosis. This is caused due to the deficiency of Ca, Mg, Cu and K etc.
(iii)Whiptail – Degeneration of lamina but not of petiole and midrib , caused by deficiency of molybdenum.
(iv)Die back – It is the killing of shoot apex i.e. stem tip and young leaves. This is caused due to the deficiency of K and Cu.
(v)Little leaf disease – Small sized leaves, caused by zinc deficiency.

5.If a plant shows a symptom which could develop due to deficiency of more than one nutrient, how would you find out experimentally, the real deficient mineral element?
Soln.Deficiency symptoms are first studied by means of pot and culture experiments. Rapidly growing plants which develop characteristic symptoms are used in culture experiments. They are called test (= indicator) plants. They are then grown in soil under test in small pots. The results are compared to know the deficiency elements. Similar tests are performed with selected crops.

6.Why is it that in certain plants deficiency symptoms appear first in younger parts of the plant while in other they do so in mature organs?
Soln. The parts of the plants that show the deficiency symptoms depend on the mobility of the element in the plant. For elements that are actively mobilised within the plants and exported to young developing tissues, the deficiency symptoms tend to appear first in the older tissues. For example, the deficiency symptoms of nitrogen, potassium and magnesium are visible first in the senescent leaves. In older leaves,biomolecules containing these elements are broken down, making these elements available for mobilising to younger leaves. The deficiency symptoms tend to appear first in the young tissues whenever the elements are relatively immobile and are not transported out of the mature organs, for example, elements like sulphur and calcium are a part of the structural component of the cell and hence are not easily released.

7.How are the minerals absorbed by the plants?
Soln. Plants absorb their mineral salt supply from the soil through the roots from the zones of elongation and root hair. The minerals are absorbed as ions which are accumulated by the plants against their concentration in the soil. Plant shows two phases in mineral absorption – initial and metabolic. In the initial phase there is a rapid uptake of ions into outer or free space of the cells (apoplast) that comprises of intercellular spaces and cell walls. Ions absorbed in free space are freely exchangeable, e.g., replacement of unlabelled K+ ions with labelled K+ ions. In the metabolic phase the ions pass into inner space comprising of cytoplasm and vacuole. In the inner space the ions are not freely exchangeable with those of external medium. Entry of ions into outer space is passive absorption as no energy is required for it. Absorption of ions into inner space requires metabolic energy. It is, therefore, an active absorption. Movement of ions into cells is called influx while movement of ions out of the cells is called efflux.

8.What are the conditions necessary for fixation of atmospheric nitrogen by Rhizobiuml What is their role inN2fixation?
Soln.The conditions necessary for nitrogen fixation by Rhizobium are :
(i) Presence of enzyme nitrogenase.
(ii)A protective mechanism for the enzyme nitrogenase against O2
(iii)A non-heme iron protein-ferrodoxin as an electron carrier.
(iv)The hydrogen donating system (viz, pyruvate, hydrogen, sucrose, glucoseetc).
(v) A constant supply of ATP.
(vi)Presence of thiamine pyrophosphate (TPP), coenzyme-A, inorganic phosphate and Mg++ as co-factors.
(vii)Presence of cobalt and molybdenum,
(viii) A carbon compound for trapping
released ammonia.
In the process of biological nitrogen fixation by free living and symbiotic nitrogen fixers, the dinitrogen molecule is reduced step by step to ammonia (NH3) by the addition of pairs of hydrogen atoms. The pyruvic acid mainly serves as an electron donor but in some cases hydrogen, sucrose, glucose, etc., have also been shown to operate. In leguminous plants, the glucose-6-phosphate molecule probably acts as a substrate for donating hydrogen. The overall process occurs in presence of enzyme nitrogenase, which is active in anaerobic condition. The enzyme nitrogenase consists of two sub-units – a non-heme iron protein (or dinitrogen reductase) and an iron molybdenum protein (Mo-Fe protein or dinitrogenase).
The Fe-protein component reacts with ATP and reduces Mo-Fe protein which then converts N2to ammonia. The ammonia is either directly taken by host or is converted to nitrates with the help of nitrifying bacteria (e.g., Nitrosomonas).

9.What are the steps involved in formation of a root nodule?
Soln. Nodule formation involves a sequence of multiple interactions between Rhizobium and roots of the host plant. Main stages in the nodule formation are:
(i) Rhizobia multiply and colonise the surrounding of roots and get attached to epidermal and root hair cells (Figure a).
(ii)The root hair curl and the bacteria invade the root hair.
(iii)An infection thread is produced carrying the bacteria into the inner cortex of the root (Figure b and c).
(iv)The bacteria get modified into rod-shaped bacteroids and cause inner cortical and pericycle cells to divide. Division and growth of cortical and peri cycle cells lead to nodule formation.
(v) The nodule thus formed, establishes a direct vascular connection with the host for exchange of nutrients (Figure d).
(vi)The nodule contains all the necessary biochemical components, such as the enzyme nitrogenase and leghaemoglobin. The enzyme nitrogenase catalyses the conversion of atmospheric nitrogen to ammonia, the first stable product of nitrogen fixation.
NCERT Solutions For Class 11 Biology Mineral Nutrition Q9

10.Which of the following statements are true?
If false, correct them.
(a) Boron deficiency leads to stout axis.
(b) Every mineral element that is present in a cell is needed by the cell.
(c) Nitrogen as a nutrient element, is highly immobile in plants.
(d) It is very easy to establish the essentiality of micronutrients because they are required only in trace quantities.
Soln. (a) True.
(b) False. Every mineral element that is present in a cell is not needed by the cell.
(c) False. Nitrogen as a nutrient element is highly mobile in plants.
(d) False. It is very difficult to establish the essentiality of micronutrients because they are required only in trace quantities.

NCERT Solutions for Class 11 Biology Chapter 11 Transport in Plants

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 11 Transport in Plants:

Section NameTopic Name
11Transport in Plants
11.1Means of Transport
11.2Plant-Water Relations
11.3Long Distance Transport of Water
11.4Transpiration
11.5Uptake and Transport of Mineral Nutrients
11.6Phloem Transport: Flow from Source to Sink
11.7Summary

NCRT TEXTBOOK QUESTIONS SOLVED

1.What are the factors affecting the rate of diffusion?
Solution. Factors affecting the rate of diffusion are :

  • Density – Rate of diffusion of a substance is inversely proportional to square root of its relative density (Graham’s Law).
  • Permeability of medium – Rate of diffusion decreases with density of the medium.
  • Temperature – A rise in temperature increases the rate of diffusion withQ10 = 1.2 -1.3. Because of it sugar crystals do not dissolve easily in ice cold water while they do so easily in warm water.
  • Diffusion pressure gradient – Rate of diffusion is directly proportional to the difference of diffusion pressure at the two ends of a system and inversely proportional to the distance between the two.

2.What are porins? What role do they play in diffusion?
Solution: The porins are proteins that form huge pores in the outer membranes of the plastids, mitochondria and some bacteria allowing molecules up to the size of small proteins to pass through. Thus they play an important ‘ role in facilitated diffusion.

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3.Describe the role played by protein pumps during active transport in plants.
Solution: Active transport uses energy to pump molecules against a concentration gradient. Active transport is carried out by membrane play a major role in both active as well as passive transport. Pumps are proteins that use energy to carry substances across the cell membrane. These pumps can transport substances from a low concentration to a high concentration (‘uphill’ transport). E.g.,H+ pump,K+ pump, Cl- pump, Na+-K pump.The pumps operate with the help of ATP.K+-H+ exchange pump occurs in guard cells. Na+-K+ exchange pump operates across many animal membranes. Transport rate reaches a maximum when all the protein transporters or pumps are being used or are saturated.Like enzymes these carrier proteins are very specific in what they carry across the membrane. These proteins are sensitive to inhibitors that react with protein side chains.

4. Explain why pure water has the maximum water potential.
Solution: Water molecules possess kinetic energy. In liquid and gaseous form they are in random motion that is both rapid and constant. The greater the concentration of water in a system, the greater is its kinetic energy or ‘water potential’. Hence, it is obvious that pure water will have the greatest water potential. Water potential is denoted by the Greek symbol Psi or ψ and is expressed in pressure units such as pascals (Pa). By convention, the water potential of pure water at standard temperatures, which is not under any pressure, is taken to be zero. If some solute is dissolved in pure water, the solution has less free water and the concentration of water’decreases, reducing its water potential. Hence, all solutions have a lower water potential than pure water.

5.Briefly describe water potential. What are the factors affecting it?
Soln. The term water potential was first used by Slatyer and Taylor (1960). The free energy per mole of any particular chemical species in a multicomponent system is defined as the chemical potential of that species. The chemical potential of water is referred to as the water potential (ψw). Since the ψ of pure water is zero (0), the presence of solute particles reduces the free energy of water, thus decreases the water potential (negative value). Therefore,ψ of solution is always less than zero or its highest value is zero.
For solutions water potential is determined by three internal factors, i.e.,ψw = ψm +  ψs + ψp (where ψm is matric potential which is used for the surface such as soil particles or cell wall to which water molecules are absorbed, ψsis solute potential, also called osmotic potential, the amount by which water potential is reduced and ψp is pressure potential such as TP and WP). Since in plant system ψm is disregarded the equation may be simplified as :
ψw = ψs + ψp       

6. What happens when a pressure greater than the atmospheric pressure is applied to pure water or a solution?
Solution: If a pressure greater than atmospheric pressure is applied to pure water or a solution, its water potential increases. It is equivalent to pumping water from one place to another. Pressure can build up in a plant system when water enters a plant cell due to diffusion causing a pressure built up against the cell wall, it makes the cell turgid.

7. (a) With fhe help of well-labelled diagrams, describe the process of plasmolysis in plants, giving appropriate examples.
(b) Explain what will happen to a plant cell if it is kept in a solution having higher water potential.
Soln.(a) Shrinkage of the protoplast of a cell from its cell wall under the influence of a hypertonic solution is called plasmolysis. Hypertonic solution causes exosmosis or withdrawal of water from cytoplasm and then the central vacuole of cell. The size of cytoplasm, as well as central vacuole and hence protoplast, becomes reduced. The first stage of plasmolysis is called limiting plasmolysis. At limiting plasmolysis, the pressure potential ( ψp ) is zero and the osmotic concentration of cejl interior is just equivalent to that of external solution (isotonic). The cell is called flaccid. When pressure potential becomes negative, the protoplast withdraws itself from the comers. This stage is known as incipient plasmolysis. At incipient plasmolysis, the cell wall exerts no pressure on the cell contents (i.e. ψp is zero). Hence at this stage ψw = ψs. The hypertonic solution now enters the cell in between the protoplast and the cell wall. Due to continued exosmosis, protoplast shrinks further and withdraws from the cell wall except one or a few points. It is known as evident plasmolysis.
NCERT Solutions For Class 11 Biology Transport in Plants Q7
Examples of plasmolysis :
(i) Pickles, meat and fish are preserved by salting. Similarly, jams and jellies are preserved by sweetening with sugars. Salting and sweetening create hypertonic condition in which the fungi and bacteria get killed by plasmolysis.
(ii)Salting kills the weeds of lawns by inducing plasmolysis in their cells.
(iii)Plasmolytic method is applied for the determination of osmotic pressure of a cell in the laboratory.
(b) When the cells are placed in a solution having higher water potential i.e., hypotonic solution (dilute solution as compared to the cytoplasm), water diffuses into the cell causing the cytoplasm to build up a pressure against the wall, that is called turgor pressure. The pressure exerted by the protoplasts due to entry of water against the rigid walls is called pressure potential ψp . Because of the rigidity of the cell wall, the cell does not rupture. This turgor pressure is ultimately responsible for – enlargement and extension of cells.

8.How is the mycorrhizal association helpful in absorption of water and minerals in plants?
Soln. Some plants have additional structures associated with them that help in water (and mineral) absorption. A mycorrhiza is a symbiotic association of a,fungus with a root system. The fungal filaments form a network around the young root or they penetrate the root cells. The hyphae have a very large surface area that absorb mineral ions and water from the soil from a much larger volume of soil that perhaps a root cannot do. The fungus provides minerals and water to the roots, in turn the roots provide sugars and N-containing compounds to the mycorrhizae. Some plants have an obligate association with the mycorrhizae. For example Pinus seeds cannot germinate and establish without the presence of mycorrhizae.

9.What role does root pressure play in water movement in plants?
Soln. As various ions from the soil are actively transported into the vascular tissues of the roots, water follows (its potential gradient) and increases the pressure inside the xylem. This positive pressure is called root pressure, and can be responsible for pushing up water to small heights in the stem. Root pressure can, at. best, only provide a modest push in the overall process of water transport. They obviously do not play a major role in water movement up tall trees. The greatest contribution of root pressure may be to re-establish the continuous chains of water molecules in the xylem which often break under the enormous tensions created by transpiration.

10.Describe transpiration pull model of water transport in plants. What are the factors influencing transpiration? How is it useful to plants?
Soln.Transpiration pull or cohesion-tension theory was originally proposed by Dixon and Joly in 1894 and further improved by Dixon in 1914. According to this theory, a continuous
column of water is present in the xylem channels of plant. The continuity of water column is maintained in the plant because of cohesive force of water molecules. There is another force of adhesion which holds water tp the walls of xylem vessels. During transpiration in plants, water is lost, in form of water vapour, from the mesophyll cells to exterior, through stomata. As a result, the turgor pressure of these cells decreases and the diffusion pressure deficit (DPD) increases. Now these cells take water from adjoining cells and the turgor of those adjoining cells decreases. This process is repeated and ultimately water is absorbed from nearest xylem vessels of leaf. As there is a continuous water column inside the xylem elements, a tension or pull is transmitted down and finally transmitted to root, resulting in the upward movement of water.
Factors affecting transpiration include both environmental and internal factors. Environmental factors:
(i) Relative humidity – The rate of transpiration is inversely proportional to the relative humidity, i.e., the rate of transpiration is higher when the relative humidity is lower and lower when the relative humidity is higher.
(ii)Atmospheric temperature – A high temperature opens stomata even in darkness. Besides producing a heating effect, it lowers the relative humidity of the air and increases vapour pressure inside transpiring organ. Consequently, rate of transpiration increases.
(iii)Light – Because most of the transpiration occurs through stomata, the rate of transpiration is quite high is light. It falls down appreciably in the darkness.
(iv)Air movements – Transpiration is lower in the still air because water vapours accumulate around the transpiring organs and reduce the DPD of the air. The movement of the air increases the rate of transpiration by removing the saturated air around the leaves.
(v) Atmospheric pressure – Low atmospheric pressure enhances evaporation, produces air currents and increases the rate of transpiration.
(vi)Availability of water – The rate of transpiration depends upon the rate of absorption of soil water by roots. This is further influenced by a number of soil factors like soil water, soil particles, soil temperature, soil air, etc.
Internal or plant factors :
(i) Leaf area (transpiring area) – A plant with large leaf area will show more transpiration than another plant with less leaf area.
(ii)Leaf structure – Leaf structure affects transpiration in following ways:
(a) Cuticular transpiration decreases with the thickness of cuticle and cutinisation of epidermal walls.
(b) Because most of the transpiration takes place through the stomata, their number and position influences the rate of transpiration.
(c) The sunken stomata are device to reduce the rate of transpiration by providing an area where little air movement occurs.
(iii)Root/shoot ratio – A low root/shoot ratio decreases the rate of transpiration while a high ratio increases the rate of transpiration.
(iv)Mucilage and solutes – They decrease the rate of transpiration by holding water tenaciously.
Transpiration is useful to plants in the following ways:
(i) Removal of excess water – It has been held that plants absorb far more amount of water than is actually required by them. Transpiration, therefore, removes the excess of water.
(ii)Root system – Transpiration helps in better development of root system which is required for support and absorption of mineral salts.
(iii)Quality of fruits – The ash and sugar content of the fruit increases with the increase in transpiration.
(iv)Temperature maintenance – Transpiration prevents overheating of leaves. However, plants growing in areas where transpiration is meagre do not show over¬heating. Some succulents can endure a temperature of 60°C without any apparent damage.
(v)Pole in ascent of sap and turgidity – Ascent of sap mostly occurs due to transpiration pull exerted by transpiration of water. This pull is important in the absorption of water. Further, transpiration maintains the shape and structure of plant parts by keeping cells turgid.
(vi)Distribution of mineral salts- Mineral are mostly distributed by rising column of sap.
(vii)Photosynthesis – Transpiration supplies water for photosynthesis.

11.Discuss the factors responsible for ascent of xylem sap in plants.
Soln. Xylem sap ascends mainly due to forces generating in the foliage of plants as a result of active transpiration. Thus, the factors which enhance the rate of transpiration are also the factors responsible for ascent of xylem sap in plants.
Various factors responsible for ascent of xylem sap in plants are as follows:
(i) Capillarity: There is limited rise of water in narrow tubes or capillaries due to forces of cohesion amongst molecules of water and their property of adhesion to other substance.
(ii)Root pressure: It is positive pressure that pushes sap from below due to active absorption by root.
(iii)Transpiration pull: Transpiration in aerial parts brings the xylem sap under negative pressure or tension due to continuous withdrawal of water by them. Water column does not break due to its high tensile strength related to high force of cohesion and adhesion.

12.What essential role does the root endodermis play during mineral absorption in plants?
Soln. Like all cells, the endodermal cells have many transport proteins embedded in their plasma membrane; they let some solutes cross the membrane, but not others. Transport proteins of endodermal cells are control points, where a plant adjusts the quantity and types of solutes that reach the xylem. Because of the layer of suberin, the root endodermis has the ability to actively transport ions in one direction only.

13.Explain why xylem transport is unidirectional and phloem transport bidirectional.
Soln. Transport over longer distances proceeds through the vascular system (the xylem and the phloem) and is called translocation. In rooted plants, transport in xylem (to water and minerals) is essentially unidirectional, from roots to the stems. Organic and mineral nutrients however, undergo multidirectional transport. Food, primarily sucrose, is transported by the vascular tissue, phloem, from a source to a sink. Usually the source is part of the plant which synthesises the food,
i.e., the leaf, and sink, the part that needs or stores the food. But, the source and sink may be reversed depending on the season, or the plant’s needs. Since the source-sink relationship is variable, the direction of movement in the phloem can be upward or downward, i.e., bi-directional. Hence, unlike one-way flow of water in xylem, food in phloem tissues can be transported in any required direction.

14.Explain pressure flow hypothesis of translocation of sugars in plants.
Soln. The accepted mechanism used for the translocation of sugars from source to sink is called the pressure flow hypothesis. As glucose is prepared at the source i.e., in leaves, (by photosynthesis) it is converted to sucrose (a dissacharide). The sugar is then moved in the form of sucrose into adjacent companion cells and then into the living phloem i.e., in sieve tube cells by active transport. This process of loading at the source produces a hypertonic conditions in the phloem. Water in the adjacent xylem moves into the phloem by osmosis. As osmotic pressure builds up, the phloem sap will move to areas of lower pressure. At the sink, osmotic pressure must be reduced. Again active transport is necessary to move the sucrose out of the
phloem sap and into the cells which will use the sugar converting it into energy, starch, or cellulose. As sugars are removed, the osmotic pressure of the phloem decreases and water moves out of the phloem
NCERT Solutions For Class 11 Biology Transport in Plants Q14

15.What causes the opening and closing of guard cells of stomata during transpiration?
Soln.Transpiration is the evaporative loss of water by plants. It occurs mainly through the stomata in the leaves. The immediate cause of the opening or closing of the stomata is change in the turgidity of the guard cells. The inner wall of each of the guard cells, towards the pore or stomatal aperture, is thick and elastic. When turgidity increases within the two guard cells flanking each stomatal aperture or pore, the thin outer walls bulge out and force the inner walls into a crescent shape and thus the stomata opens. The opening of the stoma is also aided due to the orientation of the microfibrils in the cell walls of the guard cells. Cellulose microfibrils are oriented radially rather than longitudinally making it easier for the stoma to open. When the guard cells lose turgor, due to water loss (or water stress) the elastic inner walls regain their original shape, the guard cells become flaccid and the stoma closes.

16.Differentiate between the following:
(a) Diffusion and Osmosis
(b) Transpiration and Evaporation
(c) Osmotic Pressure and Osmotic Potential
(d) Imbibition and Diffusion
(e) Apoplast and Symplast pathway of movement of water in plants
(f) Gutta’tion and Transpiration
Soln.
(a) Differences between diffusion and osmosis are as follows :
NCERT Solutions For Class 11 Biology Transport in Plants Q16

NCERT Solutions For Class 11 Biology Transport in Plants Q16.1

NCERT Solutions For Class 11 Biology Transport in Plants Q16.2
(b) Differences between transpiration and evaporation are as follows:
NCERT Solutions For Class 11 Biology Transport in Plants Q16.3

NCERT Solutions For Class 11 Biology Transport in Plants Q16.4

NCERT Solutions For Class 11 Biology Transport in Plants Q16.5
(c)Differences between osmotic pressure and osmotic potential are as follows:
NCERT Solutions For Class 11 Biology Transport in Plants Q16.6
(d) Differences between imbibition and diffusion are as follows:
NCERT Solutions For Class 11 Biology Transport in Plants Q16.7
(e) Differences between apoplast pathway and symplast pathway are as follows:
NCERT Solutions For Class 11 Biology Transport in Plants Q16.8
(f) Differences between guttation and transpiration are as follows:
NCERT Solutions For Class 11 Biology Transport in Plants Q16.9

NCERT Solutions For Class 11 Biology Transport in Plants Q16.10
NCERT Solutions For Class 11 Biology Transport in Plants Q16.11

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

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 10 Cell Cycle and Cell Division:

Section NameTopic Name
10Cell Cycle and Cell Division
10.1Cell Cycle
10.2M Phase
10.3Significance of Mitosis
10.4Meiosis
10.5Significance of Meiosis
10.6Summary

NCERT Solutions Class 11 BiologyBiology Sample Papers

NCRT TEXTBOOK QUESTIONS SOLVED

1. What is the average cell cycle span for a mammalian cell?
Solution: 24 hours.

2. Distinguish cytokinesis from karyokinesis.
Solution: Differences between cytokinesis and karyokinesis are:
NCERT Solutions For Class 11 Biology Cell Cycle and Cell Division Q2

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3. Describe the events taking place during the interphase.
Solution: The interphase, though called the resting phase, is metabolically quite active. It is the time during which the cell prepares itself for division by undergoing both cell growth and DNA replication in an orderly manner. The interphase is further divided into three phases:
• G1 (Gap 1) phase
• S (Synthesis) phase
• G2 (Gap 2) phase
G1 phase corresponds to the interval between mitosis of previous cell cycle and initiation of DNA replication. During G1 phase the cell is metabolically active and grows continuously but does not replicate its DNA S or synthesis phase marks the period during which DNA synthesis or replication takes place. During this time the amount of DNA doubles per cell. In animal cells, during the S phase, DNA replication occurs in the nucleus, and the centriole duplicates in the cytoplasm. During the G2 phase synthesis of DNA stops while cell growth continues with synthesis of protein and RNA in preparation for mitosis.

4. What is G0 (quiescent phase) of cell cycle?
Solution: G0 phase is the phase of inactivation of cell cycle due to non-availability of mitogens and energy rich compounds. Cells in this stage remain metabolically active but no longer proliferate i.e., do not grow or differentiate unless called on to do so depending on the requirement of the organism. E.g., Nerve and heart cells of chordates are in permanent G0 phase.

5. Why is mitosis called equational division?
Solution: Mitosis is a type of cell division in which chromosomes replicate and become equally distributed in two daughter nuclei so that the daughter cells come to have the same number and type of chromosomes as present in parent cell. So mitosis is called as equational division.

6. Name the stage of cell cycle at which each one of the following events occur:
(i) Chromosomes are moved to spindle equator.
(ii) Centromere splits and chromatids separate.
(iii) Pairing between homologous chromosomes takes place.
(iv) Crossing over between homologous chromosomes takes place.
Solution: 
(i) Metaphase
(ii) Anaphase
(iii) Zygotene of prophase I of meiosis 1
(iv) Pachytene of prophase I of meiosis I

7. Describe the following:
(a) Synapsis
(b) Bivalent
(c) Chiasmata
Draw a diagram to illustrate your answer.
Solution: 
(a) Synapsis: During zygotene of prophase I stage homologou s chromosomes start pairing together and this process of association is called synapsis. Electron micrographs of this stage indicate that chromosome synapsis is accompanied by the formation of complex structure called synaptonemal complex.
(b) Bivalent: The complex formed by a pair of synapsed homologous chromosomes is called a bivalent or a tetrad i.e., 4 chromatids or a pair of chromosomes.
NCERT Solutions For Class 11 Biology Cell Cycle and Cell Division Q7
(c) Chiasmata: The beginning of diplotene is recognized by the dissolution of the synaptonemal complex and the tendency of the synapsed homologous chromosomes of the bivalents to separate from each other except at the sites of crossovers. These points of attachment (X-shaped structures) between the homologous chromosomes are called chiasmata.

8. How does cytokinesis in plant cells differ from that in animal cells?
Solution: Plant cytokinesis and animal cytokinesis differ in following respects:
NCERT Solutions For Class 11 Biology Cell Cycle and Cell Division Q8

9. Find examples where the four daughter cells from meiosis are equal in size and where they are found unequal in size.
Solution: During formation of male gametes (i.e., spermatozoa) in a typical mammal (i.e., human being), the four daughter cells formed from meiosis are equal in size. On the other hand, during formation of female gamete (i.e., ovum) in a typical mammal (i.e., human being), the four daughter cells are unequal in size.

10. Can there be DNA replication without cell division?
Solution: Yes. Endomitosis is the multiplication of chromosomes present in a set in nucleus without karyokinesis and cytokinesis result-ing in numerous copies within each cell. It is of 2 types.
Polyteny: Here chromosomes divide and redivide without separation of chromatids so that such chromosomes become multistranded with many copies of DNA. Such polytene (many stranded) chromosomes remain in permanent prophase stage and do not undergo cell cycle e.g., polytene (salivary glands) chromosome of Drosophila has 512- 1024 chromatids. Here number of sets of chromosomes does not change.
Polyploidy (endoduplication) : Here all chromosomes in a set divide and its chromatids separate but nucleus does not divide. This results in an increase in number of sets of chromosomes in the nucleus (4x, 8x….). This increase in sets of chromosomes is called polyploidy. It can be induced by colchicine and granosan. These chromosomes are normal and undergo cell cycle.

11. List the main differences between mitosis and meiosis.
Solution: 
NCERT Solutions For Class 11 Biology Cell Cycle and Cell Division Q11

12. Distinguish anaphase of mitosis from anaphase I of meiosis.
Solution: Anaphase of mitosis : It is the phase of shortest duration. APC (anaphase promoting complex) develops. It degenerates proteins -binding the two chromatids in the region of centromere. As a result, the centromere of each chromosome divides. This converts the two chromatids into daughter chromosomes each being attached to the spindle pole of its side by independent chromosomal fibre. The chromosomes move towards the spindle poles with the centromeres projecting towards the poles and the limbs trailing behind. There is corresponding shortening of chromosome fibres. The two pole-ward moving chromosomes of each type remain attached to each other by interzonal fibres. Ultimately, two groups of chromosomes come to lie at the spindle poles.
NCERT Solutions For Class 11 Biology Cell Cycle and Cell Division Q12
Anaphase I of meiosis : Chiasmata disappear completely and the homologous chromosomes separate. The process is called disjunction. The separated chromosomes (univalents) show divergent chromatids and are called dyads. They move towards the spindle poles and ultimately form two groups of haploid chromosomes.
NCERT Solutions For Class 11 Biology Cell Cycle and Cell Division Q12.1

13. What is the significance of meiosis?
Solution: The significance of meiosis is given below:
(i) Formation of gametes – Meiosis forms gametes that are essential for sexual reproduction.
(ii) Genetic information – It switches on the genetic information for the development of gametes or gametophytes and switches off the sporophytic information. ‘
(iii) Maintenance of chromosome number – Meiosis maintains the fixed number of chromosomes in sexually reproducing organisms by halving the same. It is essential since the chromosome number becomes double after fertilisation.
(iv) Assortment of chromosomes – In meiosis paternal and maternal chromosomes assort independently. It causes reshuffling of chromosomes and the traits controlled by them. The variations help the breeders in improving the races of useful plants and animals.
(v) Crossing over – It introduces new combination of traits or variations.
(vi) Mutations – Chromosomal and genomic mutations can take place by irregularities of meiotic divisions. Some of these mutations are useful to the organism and are perpetuated by natural selection.
(vii) Evidence of basic relationship of organisms – Details of meiosis are essentially similar in the majority of organisms showing their basic similarity and relationship.

14. Discuss with your teacher about
(i) haploid insects and lower plants where cell division occurs, and
(ii)some haploid cells in higher plants where cell division does not occur.
Solution: 
(i) Cell division occurs in haploid insect, such as drones of honey bee and lower plant like gametophyte of algae, bryophytes, and pteridophytes.
(ii) Synergids and antipodals in embryo sac of ovule are haploid cells where cell division does not occur.

15. Can there be mitosis without DNA replication in S-phase?
Solution: No there cannot be any mitotic division without-DNA replication in ‘S’ phase.

16. Analyse the events during every stage of ceil cycle and notice how the following two parameters change.
(i) number of chromosomes (N) per cell
(ii) amount of DNA content (C) per cell
Solution: Number of chromosomes and amount of DNA change during S-phase and anaphase of cell cycle. S or synthesis phase marks the period during which DNA synthesis or replication takes place. During this time the amount of DNA per cell doubles. If the initial amount of DNA is denoted as 2C then it increases to 4C. However, there is no increase in the chromosome number; if the cell had diploid or 2N number of chromosomes at G„ even after S phase the number of chromosomes remains the same, i.e., 2N.
In mitotic anaphase, number of chromosomes remains the same. It is only sister chromatids which move towards their respective poles. DNA content remains unchanged. In anaphase I of meiosis, number of chromosomes are reduced to half, i.e., from 2N to IN and also DNA content decrease to one half i.e., from 4C to 2C. In anaphase II of meiosis II DNA content decreases to one half from 2C to 1C but chromosome number remain same.

NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules:

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules:

Section NameTopic Name
9Biomolecules
9.1How to Analyse Chemical Composition?
9.2Primary and Secondary Metabolites
9.3Biomacromolecules
9.4Proteins
9.5Polysaccharides
9.6Nucleic Acids
9.7Structure of Proteins
9.8Nature of Bond Linking Monomers in a Polymer
9.9Dynamic State of Body Constituents – Concept of Metabolism
9.10Metabolic Basis for Living
9.11The Living State
9.12Enzymes
9.13Summary

NCERT Solutions Class 11 BiologyBiology Sample Papers

NCRT TEXTBOOK QUESTIONS SOLVED

1. What are macromolecules? Give examples.
Solution: Macromolecules are large high molecular weight substances with complex molecular structure and occur in colloidal state (being insoluble) in intracellular fluid. These are formed by polymerization of large number of micromolecules. Polysaccharides, proteins and nucleic acids are few examples.

2. Illustrate a glycosidic, peptide, and a phospho- diester bond.
Solution. (i) Glycosidic bond is the type of chemical linkage between the monosaccharide units of disaccharides, oligosaccharides and polysaccharides, which is formed by the removal of a molecule of water.
NCERT Solutions For Class 11 Biology Biomolecules Q2
(ii)Peptide bonds are formed by the reaction between carboxyl (- COOH) of one amino acid and amino (- NH2) group of other amino acid with the elimination of water.
NCERT Solutions For Class 11 Biology Biomolecules Q2.1
(iii) In a polynucleotide chain, adjacent nucleotides are joined together by a bond called phosphodiester bond. This bond links a phosphate group and sugar group of two adjacent nucleotides by means of an oxygen bridge.
NCERT Solutions For Class 11 Biology Biomolecules Q2.2

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3. What is meant by tertiary structure of proteins?
Solution: The helical polypeptide molecule may fold on itself and assume a complex but specific form-spherical, rod-like or any form in between these. These geometrical shapes,are known as tertiary (3°) structure of protein molecules. The coils and folds of the polypeptide molecules are so arranged as to hide the non-polar amino acid chains inside and to expose the polar side chains. The tertiary structure of a protein brings distant amino acid side chains nearer to form active sites of enzymatic proteins. The tertiary structure is maintained by weak bonds such as hydrogen, ionic, disulphide and hydrophilic – hydrophobic bonds, formed between one part of a polypeptide and another. This structure is easily disrupted by pH, temperature and chemicals stopping the function of proteins.

4. Find and write down structures ©f 10 interesting small molecular weight biomolecules.
Solution: Interesfing small molecular weight biomolecules are minerals (like sodium, potassium, calcium, zinc, iodine etc), gases (like Oz, N2, C02, NH3) sugars – (ribose, deoxyribose, glucose, fructose), lipids, amino acids, nucleotides (pyrimidines & purine). Structures of 10 interesting small molecular weight biomolecules are as follows:
NCERT Solutions For Class 11 Biology Biomolecules Q4

NCERT Solutions For Class 11 Biology Biomolecules Q4.1
NCERT Solutions For Class 11 Biology Biomolecules Q4.2
NCERT Solutions For Class 11 Biology Biomolecules Q4.3

5. Proteins have primary structure. If you are given a method to know which amino acid is at either of two termini (ends) of a protein, can you connect this information to purity or homogeneity of a protein?
Solution: There are several methods provided by several scientists to find out the sequence of amino acids. Frederick Sanger proposed Sanger’s reagent to know the amino acid sequence in a polypeptide chain.
Sanger used 1-fluoro 2, 4 dinitrobenzene (FD NB) to determine insulin structure. FDNB specifically binds with N-terminal amino acid to form a dinitrophenyl (DNP) derivative of peptide. This DNP- derivative peptide can be identified by chromatography. The identified sequence of amino acids shows the homogeneity of a protein molecule.

6. Find out and make a list of proteins used as therapeutic agents. Find other applications of proteins.
Solution: Proteins used as therapeutic agents are: thrombin, fibrinogen, enkephalins, antigens, antibodies, streptokinase, protein tyrosine kinase, diastase, renin, insulin, oxytocin, vasopressin etc. Proteins are also used in cosmetics, dairy industries, textile industries, research techniques, biological buffers etc.

7. Explain the composition of triglycerides. jSfflTriacylglycerols (triglycerides) are the esters of glycerol with fatty acids.
Solution: They are insoluble in water and non-polar in character and commonly known as neutral fats. The neutral or depot fats are composed of carbon, hydrogen and oxygen like carbohydrates but have far fewer oxygen atoms than carbon atoms unlike the carbohydrates.
NCERT Solutions For Class 11 Biology Biomolecules Q7
(i) Glycerol – A glycerol molecule has 3
carbons, each bearing a hydroxyl (-OH) group. .
(ii) Fatty acids – A fatty acid molecule is an unbranched chain of carbon atoms with each carbon atom (C) forming four bonds to other atoms. It has a carboxyl group- COOH at one end and hydrogen atom (H) bonded to all or most carbon atoms forming a hydrogen chain. The carbon- hydrogen bonds are non-polar. Therefore, the hydrocarbon chain does not dissolve in water. Because the carboxyl group contains the polar C = O and OFI groups. It tends to dissolve in water even though the rest of fatty acid molecule will not. Triacylglycerols of plants, in general, have higher content of unsaturated fatty acids as compared to that of animals.

8. Can you describe what happens when milk is converted into curd or yoghurt, from your understanding of proteins.
Solution: Milk is converted into curd or yoghurt due to denaturation of proteins. In denaturation, disruption of bonds that maintains secondary and tertiary structure leads to the conversion of globular proteins into fibrous proteins. This involves a change in physical, chemical and biological properties of protein molecules.

9. Can you attempt building models of biomolecules using commercially available atomic models (Ball and stick models).
Solution: Yes, models of biomolecules can be prepared using commercially available atomic models.
Ball and stick models and space filling models are 3D or spatial molecular models which serve to display the structure of chemical products and substances or biomolecules. With ball and stick models, the centers of the atoms are connected by straight lines which represent the covalent bonds. Double and triple bonds are often represented by springs which form curved connections between the balls. The bond angles and bond lengths reflect the actual relationships, while the space occupied by the atoms is either not represented at all or only denoted essentially by the relative sizes of the spheres.

10. Attempt titrating an amino acid against a weak base and discover the number of dissociating (ionizable) functional groups in the amino acid.
Solution: The existence of different ionic forms of amino acids can be easily understood by the titration curves. The number of dissociating functional group is one in case of neutral and basic amino acids and two in case of acidic amino acids.

11. Draw the structure of the amino acid, alanine.
Solution:
NCERT Solutions For Class 11 Biology Biomolecules Q11

12. What are gums made of ? Is fevicol different ?
Solution: Gums are hetero-polysaccharides (poly-mers) of large number of different monosac-charide units. Yes, fevicol is a different kind of polymer. It is a synthetic sticky substance called resin which is manufactured by esteri-fication of organic compounds.

13. Find out a qualitative test for proteins, fats and oils, amino acids and test, any fruit juice, saliva, and urine for them.
Solution: Biuret test for protein : The biuret test is a chemical test used for determining the presence of peptide bonds. In a positive test, a copper II ion (Cu2+ ion) is reduced to copper I (Cu+) which forms a complex with the nitrogen and carbon of peptide bonds in an alkaline solution. A violet colour indicates the presence of proteins.
Ninhydrin test for amino acid: Ninhydrin (2,2 Dihydroxy indane-l,3-dione) is a chemical used to detect ammonia or primary and secondary amines. When reacting with these free amines, a deep blue or purple colour known as Ruhemann’s purple is evolved. Amino acid analysis of proteins is also done by ninhydrine. Most of the amino acids (including a-amino acids) are hydrolysed and reacted with ninhydrin except proline (a secondary amine). Amino acid containing a free amino group and a free carboxylic acid group reacts together with ninhydrin to produce coloured product. When the amino group is secondary, the condensation product is yellow.
NCERT Solutions For Class 11 Biology Biomolecules Q13
Solubility test for fats and oils : A positive solubility test for fats is that the fat dissolves in lighter fluid and not in water. In this test, 5 drops of fat or oil are added in two test tubes containing 10 drops of lighter fluid and 10 drops cold water respectively.
Fruit juice contains sugar so it cannot be tested by the above-mentioned tests. Saliva contains proteins, mineral salts, amylase etc., so it can be tested for protein and amino acids. Urine contains proteins, so it can be tested for it.

14. Find out how much cellulose is made by all the plants in the biosphere.
Solution: About 100 billion tonnes of cellulose is prepared per year by the plants of the world.

15. Describe the important properties of enzymes.
Solution: The important properties of enzymes are as follows:
(i) The enzymes are generally proteins which are high molecular weight complex globular proteins. They can associate with non-protein substance for their activity.
(ii) The enzymes do not start a chemical reaction but only accelerate it. They combine temporarily with the substrate molecules and are not consumed or changed permanently in the reaction which they catalyse.
(iii) The enzyme controlled reactions are reversible.
(iv) The enzymes are specific in action. An enzyme catalyses only a particular kind of reaction or acts on a particular substrate only.
(v) The enzymes are thermolabile i.e., heat sensitive and can function best at an optimum temperature. Similarly, enzymes show maximum activity at optimum pH.
(vi) The enzymes are inactivated by poisons and radiation.

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

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life:

Section NameTopic Name
8Cell The Unit of Life
8.1What is a Cell?
8.2Cell Theory
8.3An Overview of Cell
8.4Prokaryotic Cells
8.5Eukaryotic Cells
8.6Summary

NCERT Solutions Class 11 BiologyBiology Sample Papers

NCRT TEXTBOOK QUESTIONS SOLVED

1.Which of the following is not correct?
(a) Robert Brown discovered the cell.
(b) Schleiden and Schwann formulated the cell theory.
(c) Virchow explained that cells are formed from pre-existing cells.
(d) A unicellular organism carries out its life activities within a single cell.
Soln.(a) Robert Hooke discovered the celland Robert Brown discovered nucleus in the cell.

2.New cells generate from
(a) bacterial fermentation
(b) regeneration of old cells
(c) pre-existing cells
(d) abiotic materials.
Soln.(c)

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3.Match the following.
Column I                                           Column II
(a) Cristae                                 (i) Flat membranous sacs in stroma
(b) Cisternae                           (ii) Infoldings in mitochondria
(c) Thylakoids                        (iii) Disc-shaped sacs in Golgi apparatus
Soln.a – (ii); b – (iii); c – (i).

4.Which of the following is correct?
(a) Cells of all living organisms have a nucleus.
(b) Both animal and plant cells have a well defined cell wall.
(c) In prokaryotes, there are no membrane bound organelles.
(d) Cells are formed de novo from abiotic
Soln. (c) Mature mammalian erythrocytes and sieve tube cells of vascular plants lack nucleus. Animals lack cell wall and only cell membrane is present. Prokaryotes are unicellular organisms which lack nucleus and other membrane bound organelles. All cells arise from pre-existing cells.

5.What is a mesosome in a prokaryotic cell? Mention the functions that it performs.
Soln. Mesosome is a membranous structure in prokaryotic cell, which is formed by the extensions of the plasma membrane into the cell in form of vesicles, tubules and lamellae. Mesosomes are equal to mitochondria in eukaryotes, as they perform aerobic cellular respiration in prokaryotes. It helps in DNA replication and distribution of genetic material to daughter cells. Mesosomes also help in respiration, increase the surface area of the plasma membrane and enzymatic content and cell wall formation.

6.How do neutral solutes move across the plasma membrane? Can the polar molecules also move across it in the same way? If not, then how are these transported across the membrane?
Soln. Neutral solutes move across the membrane by the process of simple diffusion along the concentration gradient i.e., from higher concentration to the lower concentration. Polar molecules cannot pass through the nonpolar lipid bilayer, they require carrier proteip of the membrane to facilitate their transport across the membrane. In facilitated diffusion, molecules are transported along concentration gradient by help of ion channels and permeases and it does not involve energy expenditure (passive transport).

7. Name two cell-organelles that are double membrane bound. What are the characteristics of these two organelles? State their functions and draw labelled diagrams of both?
Solution: Mitochondria and chloroplast are double membrane bound organelles. Mitochondria: Mitochondria are cylindrical or sausage shaped cell organelles and contains two membranes, outer and inner. The inner compartment is called the matrix containing DNA, RNA, ribosomes, enzymes of Krebs cycle etc and outer membrane forms the continuous limiting boundary of the organelle. Inner membrane forms number of infoldings called the cristae which increases the surface area. Oxysomes are present on inner mitochondrial membrane. Mitochondria are semiautonomous organelles, i.e., have their own DNA and ribosomes.
NCERT Solutions For Class 11 Biology Cell The Unit of Life Q7
Functions of mitochondria:

  • Mitochondria are essential for aerobic respiration.
  • Mitochondria provide intermediates for synthesis of important biomolecules such as chlorophyll, cytochrome, steroids etc.
  • Mitochondria regulate the calcium ion concentration in the cell.
  • Mitochondrial matrix contains enzymes for the synthesis of fatty acids.
  • Synthesis of many amino acids takes place here

Chloroplast: They are green coloured plastids which are disc shaped. The space limited by inner membrane of chloroplast is called as stroma. Stroma has organised flattened membranous sacs called the thylakoids. Thylakoids are arranged in stacks called grana. Matrix of a chloroplast contains DNA, RNA, ribosomes and enzymes. Chloroplast is also a semiautonomous organelle.
NCERT Solutions For Class 11 Biology Cell The Unit of Life Q7.1
Functions of chloroplast:

  1. Photosynthesis is performed by chloroplast.
  2. Chloroplast stores starch grains.
  3. Maintains balance of C02 concentration in the air.
  4. Keeps oxygen balance constant in atmosphere by liberating 02 into the atmosphere, used during respiration and combustion.

8.What are the characteristics of prokaryotic cells?
Solution: Characteristics of prokaryotic cells are as follows:

  • The prokaryotic cell is essentially a single – envelope system.
  • Prokaryotes lack membrane bound cell organelles.
  • Prokaryotes have 70S ribosomes.
  • DNA is naked and lies coiled in cytoplasm. It is not covered by nuclear membrane and is termed as nucleoid.
  • Nuclear components, like, nuclear envelope, nucleolus, nucleoplasm are absent.
  • Cell wall is present in bacteria and cyanobacteria, but absent in mycoplasma.
  • Multiplies by asexual reproduction.
  • Transcription and translation takes place in cytoplasm.

9.Multicellular organisms have division of labour. Explain.
Soln. Division of labour is differentiation of certain components or parts to perform different functions for increased efficiency and higher survival. Multicellular organisms often possess millions of cells. Various cells are grouped together to form specific tissue, organ or organ system, with each specialised to perform particular function. Every cell of a multicellular organism cannot obtain food from outside. The organism requires a system for obtaining food, its digestion and distribution. Therefore, a digestive system and system of transport are also required. Certain cells of the body take over the function of reproduction. Others take part in repair and replacement of worn out or injured portions. For optimum functioning of cells, a multicellular organism also requires an internal favourable environment. Therefore, multicellular organisms come to have division of labour.

10.Cell is the basic unit of life. Discuss in brief.
Soln. Cell is fundamental, structural and functional unit of life, as no living organism can have life without being cellular. All life begins as a single cell. An organism is either made of single cell (unicellular) or many cells (multicellular). In unicellular organism, single cell is capable of independent existence and perform all essential functions of life, while in multicellular organism, each group of cells is specialised for specific function. Life passes from one generation to the next in form of cells, and new cell always arise from division of pre-existing cells. Cells are totipotent, i.e., single cell has ability to form whole organism. The activities of an organism are sum total of activities of its cells, therefore, cell is the basic unit of life.

11.What are nuclear pores? State their function.
Soln. Nuclear envelope bounds the nucleus from outside and separates it from cytoplasm. It consists of two membranes, with outer membrane continuous with endoplasmic reticulum. The nuclear envelope is interrupted by minute nuclear pores, at a number of places, which are produced by the fusion of its two membranes. These
nuclear pores are the passages through which movement of RNA and protein molecules takes place in both directions between the nucleus and the cytoplasm.

12.Both lysosomes and vacuoles are endomem-brane ‘structures, yet they differ in terms of their functions. Comment.
Soln. Organelles of endomembrane system such as lysosome and vacuoles function in close coordination with one another but are specialised to perform different functions. Lysosomes breakdown the ageing and dead cells, they help in digestion of food as they contain hydrolytic digestive enzymes. They are involved in cell division also. Vacuoles on other hand, help in excretion and osmoregulation in Amoeba (contractile vacuole) or provides buoyancy, mechanical strength in prokaryotes (air vacuoles).

13.Describe the structure of the following withthe help of labelled diagrams.
(i) Nucleus (ii) Centrosome
Soln.(i) Nucleus: Nucleus is double membrane bound principle cell organelle which contains all genetic information for controlling cellular metabolism and transmission of genetic information.
Nucleus is differentiated into following four parts:
(a) Nuclear envelope: It is a double membrane bound envelope that surround the nucleus and separates the latter from the cytoplasm.
(b) Nucleoplasm: Itis clear, non-staining, fluid material present in the nucleus, which contains raw materials (nucleotides), enzymes (DNA/RNA polymerases) and metal ions for the synthesis of RNAs and DNA. The nuclear matrix or the nucleoplasm is composed of nucleolus and chromatin.
(c) Nucleolus: It is a naked, round and slightly irregular structure, which is attached to the chromatin at a specific region. It is a site for active ribosomal RNA synthesis.
(d) Chromatin : It has the ability to get stained with certain basic dyes. It is known to be the hereditary DNA protein fibrillar complex. The chromatin fibres are distributed throughout the nucleoplasm.
NCERT Solutions For Class 11 Biology Cell The Unit of Life Q13
(ii) Centrosome: Centrosome is an organelle usually containing two cylindrical structures called centrioles. They are surrounded by amorphous pericentriolar materials. Both the centrioles in a centrosome lie perpendicular to each other. They are made up of nine evenly spaced peripheral fibrils of tubulin protein. Each of the peripheral fibril is a triplet. The adjacent triplets are also linked. The hub of centriole is connected with tubules of the peripheral triplets by radial spokes made of protein.
NCERT Solutions For Class 11 Biology Cell The Unit of Life Q13.1

14.What is a centromere? How does the position of centromere form the basis of classification of chromosomes. Support your answer with a diagram showing the position of centromere on different types of chromosomes.
Soln. A chromosome consists of two identical halves, the chromatids held together at one point called the centromere. The centromere is also called as primary constriction. On its side a disc shaped structure called kinetochore is present. Chromosomes are classified into four types according to position of centromere on the chromosome.
(i) Metacentric chromosome: In this chromosome, centromere is in the middle and the two arms are almost equal in length.
(ii)Submetacentric chromosome: The centromere is slightly away from middle point so one arm is slightly shorter than the other.
(iii)Acrocentric chromosome: The centromere is near the end and one arm is extremely short and other arm is extremely long.
(iv)Telocentric chromosome: Centromere is at the tip of chromosome. These chromosomes are not present in humans.

NCERT Solutions For Class 11 Biology Structural Organisation in Animals

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 7 Structural Organisation in Animals:

Section NameTopic Name
7Structural Organisation in Animals
7.1Animal Tissues
7.2Organ and Organ System
7.3Earthworm
7.4Cockroach
7.5Frogs
7.6Summary

NCRT TEXTBOOK QUESTIONS SOLVED

1. Answer in one word or one line.
(i) Give the common name of Periplaneta americana.
(ii) How many spermathecae are found in earthworm?
(iii) What is the position of ovaries in cockroach?
(iv) How many segments are present in the abdomen of cockroach?
(v) Where do you find Malpighian tubules?
Solution: (i) Cockroach.
(ii) Four pairs.
(iii) In cockroach two large ovaries, lie laterally in the 2nd – 6th abdominal segments’.
(iv) Abdomen of cockroach consists of 10 segments.
(v) Malpighian tubules are present at the junction of midgut and hindgut in cockroach.

2. What are the following and where do you find them in animal body?
(a) Chondrocytes
(b) Axons.
(c) Ciliated epithelium
Solution: (a) Chondrocytes – Chondrocytes are the only cells found in cartilage. They are present in spaces called lacunae and they produce and maintain the matrix of cartilage. Bending ability of cartilage is due to chondrocytes. Cartilage is present at tip of nose, pinna of ear, epiglottis etc.
(b) Axon – Axon is one of the processes of neuron, which is the structural and functional unit of nervous system. The part of cyton – n’here axon arises is axon hillock and axon ends in group of branches called terminal arborizations. It conducts impulses away from the cyton. Neurons (nerve cells)
are present in brain and spinal cord.
(c) Ciliated epithelium – If the columnar or cuboidal cells bear cilia on their free surface they are called ciliated epithelium. Their function is to move particles or mucus in a specific direction over the epithelium. They are mainly present in the inner surface of hollow organs like bronchioles and Fallopian tube.

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3. Draw a labelled diagram of the reproductive organs of an earthworm.
Solution: 
NCERT Solutions For Class 11 Biology Structural Organisation in Animals Q3

4. Answer the following.
(i) What is the function of nephridia?
(ii) How many types of nephridia are found in earthworm based on their location?
Solution: (i) Nephridia are excretory organs of earthworm, which perform the function of excretion and osmoregulation. Nephridia regulate the volume and composition of the body fluids. A nephridium is a coiled tubular and microscopic structure which starts out as a funnel that collects excess fluid from coelomic chamber. The funnel connects with a tubular wastes through a pore to the surface in the body wall or into the digestive tube.
(ii) In earthworm, nephridia are present in all segments except the first two. There are three types of nephridia on the basis of their location:
(a) Septal nephridia, present on both the sides of intersegmental septa from segment 15 to the last that open into intestine.
(b) Integumentary nephridia, attached to lining of the body wall of segment 3 to the last that open on the body surface and
(c) Pharyngeal nephridia, present as three paired tufts in the 4th, 5th and 6th segments.
NCERT Solutions For Class 11 Biology Structural Organisation in Animals Q4

5. Draw a labelled diagram of alimentary canal of a cockroach.
Solution: 
NCERT Solutions For Class 11 Biology Structural Organisation in Animals Q5

6. What are the cellular components of blood?
Solution: Blood is a fluid connective tissue. It is composed of plasma (fluid) and blood cells (corpuscles). Cellular components of blood (blood corpuscles) constitute about 45% of blood volume.
Three types of blood cells are:
(i) Erythrocytes or red blood cells: They are most abundant blood cells. Normal RBC count is 5-5.5 million/mm3 in males and 4.5-5 million/mm3 in females) RBCs help in transport of gases and maintain blood pH.
(ii) Leucocytes or white blood cells: The normal WBC count is 5000-6000/mm3 of blood. They are involved in immune response of body and act as soldiers and scavangers.
(iii) Thrombocytes or blood platelets: There are about 2,50,000 platelets/mm3 of blood. They are involved in blood clotting.

7. Distinguish between the following:
(a) Prostomium and peristomium
(b) Septal nephridium and pharyngeal
Solution: (a) Differences between prostomium and peristomium are
NCERT Solutions For Class 11 Biology Structural Organisation in Animals Q7

NCERT Solutions For Class 11 Biology Structural Organisation in Animals Q7.1
(b) Differences between septal and pharyngeal nephridia are:
NCERT Solutions For Class 11 Biology Structural Organisation in Animals Q8

8. Mark the odd one in each series.
(a) Areolar tissue; blood; neuron; tendon
(b) RBC; WBC; platelets; cartilage
(c) Exocrine; endocrine; salivary gland; ligament
(d) Maxilla; mandible; labrum; antennae
(e) Protonema; mesothorax; metathorax; coxa.
Solution: 
(a) Neuron: Areolar tissue, blood and tendon are connective tissues while neuron is a part a nervous tissue.
(b) Cartilage: RBC, WBC and platelets are parts of vascular connective tissue while cartilage is skeletal connective tissue.
(c) Ligament: Ligament is a connective tissue.
(d) Antennae: Maxilla, mandible and labrum are mouth parts of cockroach while antennae are sense organs.
(e) Protonema: Protonema is a filamentous juvenile stage in life cycle of Bryophytes, while mesothorax, metathorax and coxa are appendages of cockroach.

9. Match the terms in column I with those in column II.

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

Solution:  (a) – (iii), (b) – (iv), (c) – (v), (d) – (ii), (e) – (i), (f) – (vii), (g) – (vi)

10. Mention briefly about the circulatory system of earthworm.
Solution: Earthworm possesses a closed type of blood vascular system, as the blood flows through closed blood vessels. Blood is red in colour due to respiratory pigment haemoglobin. Prominent blood vessels in earthworm includes dorsal, ventral, sub- neural, lateral oesophageal and supra- oesophageal blood vessels. There are four pairs of tubular hearts, provided with valves. The anterior two pairs of hearts, known as lateral hearts lie in the 7th and 9th segments and connect the dorsal blood vessel with the ventral blood vessel. They receive blood from the dorsal blood vessel and convey it to the ventral blood vessel. The posterior two pairs of hearts are called latero-oesophageal hearts and are situated in the 12th and 13th segments. The latero-oesophageal hearts apart from connecting the dorsal and ventral blood vessels are also joined with the supra oesophageal blood vessel. Latero-oesophageal hearts carry blood from the dorsal vessel and the supra oesophageal vessel to the ventral blood vessel.Contractions keep blood circulating in one direction. Blood glands are present in the 4th, 5th and 6lh segments which produce blood cells and haemoglobin which is dissolved in blood plasma. Blood cells are phagocytic in nature.

11. Describe various types of epithelial tissues with the help of labelled diagrams.
Solution:  Epithelial tissue is a tissue made of one or more layers of compactly arranged cells that covers external surface and internal free surface of body organs and which is underlined by a basement membrane. The various types of epithelial tissue along with the diagram are given below:
(i) Simple epithelium : It is composed of single layer of cells which rest on basement membrane. Simple epithelium generally occurs over secretory and absorptive surfaces and forms lining of body cavities, ducts and tubes. Simple epithelium is of several types.
(a) Squamous epithelium: It consists of single layer of flat cells, tightly linked together and have centrally located oval or spherical nucleus. It is also called pavement epithelium. It is found in walls of blood vessels, air sacs of lungs, and lining of eye lens.
(b) Cuboidal epithelium: Cells of cuboidal epithelium are as tall as wide, with centrally placed nucleus. Its main functions are secretion and absorption. It lines sweat gland, thyroid follicles, salivary glands. Brush bordered cuboidal epithelium, i.e., cells having microvilli on their free surface lines proximal part of uriniferous tubule, pancreatic duct, testis and ovary.
(c) Columnar epithelium: Cells are with basally located nucleus. It helps in secretion and absorption. It occurs in lining of intestine, stomach, gall bladder.
(d) Ciliated epithelium: Free surface of columnar and cuboidal cells are covered with cilia. Cilia help in moving fluids, particles, mucus, etc. in a specific direction. It occurs in the inner surface of Fallopian tubules, nasal passage, bronchioles.
NCERT Solutions For Class 11 Biology Structural Organisation in Animals Q11
(e) Pseudostratified epithelium: It consists of single layer of cells but some cells are shorter than others. Due to difference in size of cells, the epithelium appears 2-3 layered. Pseudostratified columnar epithelium occurs in urethra and parotid salivary gland. Pseudostratified columnar ciliated epithelium (only larger cells ciliated) occurs in lining layer of nasal’ chambers, trachea and large bronchi. It helps in moving mucus and foreign particles.
NCERT Solutions For Class 11 Biology Structural Organisation in Animals Q11.1
(ii) Compoundepithelium/stratifiedepithelium: It is multilayered epithelium where cells of only the lowermost or basal layer are in contact with basement membrane. It provides protection against mechanical and chemical stresses and has limited role in secretion and absorption. It covers dry surface of skin, moist surface of buccal cavity, pharynx, etc. Different types of compound epithelium are:
(a) Stratified squamous epithelium: The cells of outer layer are flattened and squamous while the inner layers are cuboidal cells. It is of two types: Non- keratinised lining oesophagus, pharynx, buccal cavity, cornea, vagina and anal canal and keratinised (comified): forming epidermis of skin, hair, horn and nail.
NCERT Solutions For Class 11 Biology Structural Organisation in Animals Q11.2
(b) Stratified cuboidal epithelium: The outer layer of cuboidal cells and basal layer of columnar cells. It lines ducts of sweat glands, large salivary and pancreatic ducts.
(c) Stratified columnar epithelium: Both upper and basal layers are made of columnar cells, e.g., epiglottis covering, part of urethra.
(d) Stratified ciliated columnar epithelium: Outer layer consists of ciliated columnar cells and basal layer of columnar cells, e.g., larynx.
(iii) Transitional Epithelium: This is stratified epithelium which contains cuboidal or columnar shaped cells, which are thin and stretchable. No basement membrane is present as it would impede stretchability. It lines the inner surface of renal calyces, urinary bladder, ureter. Because of its t distribution, it is also called urothelium.
NCERT Solutions For Class 11 Biology Structural Organisation in Animals Q11.3
(iv) Glandular epithelium: It consists of specialised epithelial cells which synthesise intracellular macromolecules (protein in pancreas, lipids in adrenal glands, glycoprotein in salivary glands and all the three in mammary glands) and pour out the same in the form of a useful fluid secretion which is different from blood or any other extracellular fluid. Glands can be unicellular or multicellular on the basis of number of cells.
(a) Unicellular glands: Single-celled, e.g., goblet (mucous) cells of respiratory tract and alimentary canal.
(b) Multicellular glands: Consist of cluster of cells, e.g., Salivary glands.
NCERT Solutions For Class 11 Biology Structural Organisation in Animals Q11.4
On the basis of presence or absence of duct glands can be:
(a) Exocrine glands : These glands pour their secretion through a duct. They secrete milk, saliva, mucus, earwax. e.g., goblet cells, salivary glands, tear glands, gastric glands, intestinal glands.
(b) Endocrine glands: They are ductless glands, which pour their secretions into blood or lymph for reaching the target region. Their secretion is called hormone e.g., pituitary gland, thyroid gland, parathyroid glands, adrenal glands.
(c) Heterocrine glands: Both exocrine and endocrine, e.g., pancreas.
On basis of mode of secretion glands can be:
(a) Merocrine: Secretion is discharged
through diffusion, e g., goblet cells, sweat glands.
(b) Apocrine glands: Glandular secretion accumulates in the terminal part of the cell which is pinched off, e.g., mammary glands.
(c) Holocrine glands : The cell filled with secretory product disintegrates during discharge of the product, e.g., sebaceous gland.
(v) Modified epithelium : It is of following types:
(a) Germinal epithelium (generally cuboidal, produces gametes), (b) Glandular epithelium (columnar or cuboidal secretes chemicals and mucus), (c) Sensory epithelium or neuroepithelium. Epithelial cells having sensory hair on free surface and connected with nerve fibres on the other surface (generally columnar, receives and conveys stimuli), e.g, nasal epithelium, taste buds, retina, sensory spots of internal ear. (d) Pigmented epithelium – The cells possess melanin granules, e.g, retinal layer in contact with choroid of eye.

12. Distinguish between
(a) Simple epithelium and compound epithelium.
(b) Cardiac muscle and striated muscle.
(c) Dense regular and dense irregular connective tissues.
(d) Adipose and blood tissue.
(e) Simple gland and compound gland.
Solution: 
(a) Differences between simple and compound epithelium are as follows:
NCERT Solutions For Class 11 Biology Structural Organisation in Animals Q12
(b) Differences between cardiac and striated muscles are as follows:
NCERT Solutions For Class 11 Biology Structural Organisation in Animals Q12.1

NCERT Solutions For Class 11 Biology Structural Organisation in Animals Q12.2
(c) Differences between dense regular and dense irregular connective tissues are as follows:
NCERT Solutions For Class 11 Biology Structural Organisation in Animals Q12.3
(d) Differences between adipose tissue and blood tissue are as follows:
NCERT Solutions For Class 11 Biology Structural Organisation in Animals Q12.4
(e) Differences between simple gland and compound gland are as follows:
NCERT Solutions For Class 11 Biology Structural Organisation in Animals Q12.5

13. Draw a neat diagram of digestive system of frog.
Solution: 
NCERT Solutions For Class 11 Biology Structural Organisation in Animals Q13

14. Mention the function of the following:
(a) Ureters in frog
(b) Malpighian tubules
(c) Body wall in earthworm.
Solution: 
(a) Ureters in frog: Ureter is a transparent duct which arise from outer portion of kidney. In the “male frogs, ureter acts as urinogenital duct which runs backwards from kidneys and opens into the cloaca. It carries both urine and spermatozoa from kidney to the cloaca. In female, ureter conducts only urine from kidneys to the cloaca.
(b) Malpighian tubules: Malpighian tubules are excretory organs present in cockroach. These are present at junction of mid gut and hindgut. These are fine, long, unbranched, yellowish and blind tubules and are 100-150 in number. They help in the removal of excretory products from haemolymph.
(c) Body wall in earthworm: It consists of cuticle, epidermis, muscular layer and parietal peritoneum.
(i) It maintains the characteristic shape of’ the body.
(ii) It protects the internal organs.
(iii) The cuticle prevents excessive evaporation.
(iv) It serves as an ideal respiratory organ.
(v) The receptor cells play a vital sensory function.
(vi) The albumen helps in the formation of cocoon. It also serves as a food for the developing earthworm inside the cocoon.
(vii) Setae and muscles are responsible for locomotion.
(viii) Excretory matter is passed out through nephridiopores.

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

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants:

Section NameTopic Name
6Anatomy of Flowering Plants
6.1The Tissues
6.2The Tissue System
6.3Anatomy of Dicotyledonous and Monocotyledonous Plants
6.4Secondary Growth
6.5Summary

NCRT TEXTBOOK QUESTIONS SOLVED

1.State the location and function of different types of meristems.
Soln. Meristems are of three types on the basis of their location in plant body:
(i) Apical meristem: It is present at the apices of root and shoot and is responsible for increase in length.
(ii)Intercalary meristem: It is present at the bases of leaves above the nodes or below the nodes and is responsible for elongation of the organs.
(iii)Lateral meristem : It is present on lateral side and is responsible for increase in girth or diameter.

2.Cork cambium forms tissues that form the cork. Do you agree with this statement? Explain.
Soln. Yes, I agree with this statement. Cork cambium cuts off cells both on its outer side and inner side. The cells cut off on outer side form cork and cells cut off on inner side form secondary cortex. The cells of cork are dead whereas those of secondary cortex are living.

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3.Explain the process of secondary growth in the stems of woody angiosperms with the help of schematic diagrams. What is its significance?
Soln. Secondary growth is the formation of secondary tissues from lateral meristems. It is found in dicots only. It increases the diameter of the stem. Secondary tissues are formed by two types of lateral meristems, vascular cambium and cork cambium. Vascular cambium produces secondary vascular tissues while cork cambium forms periderm.The vascular bundles in dicot stem are conjoint, collateral, open and are arranged in a ring. The cambium present between xylem and phloem in vascular bundles is called fascicular or intrafascicular cambium. Besides this, some cells of medullary rays also become meristematic and this is called interfascicular cambium. Both these cambia collectively constitute complete ring of vascular cambium. This ring of vascular cambium divides periclinally to cut off cells both on inner side and outer side. The cells cuts off on outer side are secondary phloem and inner side are secondary xylem. Amount of secondary xylem cut off is more than secondary phloem and thus with the formation of secondary tissue, increase in girth or diameter occurs. The structure of secondary xylem and secondary phloem is similar to that of primary xylem and primary phloem. With the increase in secondary tissue, the primary xylem and primary phloem get crushed. The ray initials of vascular cambium ring divide by tangential divisions and add new cells. These new cells produced on both the sides of ray initials remain meristematic for sometime and then differentiate into parenchymatous cells of rays. The rays, produced by vascular cambium in between the secondary xylem and secondary phloem, are called secondary medullary rays. They are usually one to few layers in thickness and one to several layers in height. The medullary rays form the radial systejn responsible for radial conduction of solutes. They maintain connection between pith and cortex There is a marked difference in activity of cambium with change in season. In spring, the activity of cambium is more and hence the wood elements are larger in size with wide lumen. The activity of cambium is less during autumn and the wood elements are smaller in size with narrow lumen. Spring wood and autumn wood of a year constitute annual ring.
In order to increase in girth and prevent harm on the rupturing of the outer ground tissues due to the formation of secondary vascular tissues, dicot stems produce a cork cambium or phellogen in the outer cortical cells. Phellogen cells divide on both the outer side as well as the inner side to form secondary tissues. The secondary tissue formed on the inner side is called secondary cortex while the tissue formed on outer side is called cork.
NCERT Solutions For Class 11 Biology Anatomy of Flowering Plants Q3
Significance of secondary growth is as
follows:
(i) It adds to the girth of the plant thus provides support to increasing weight of aerial parts due to growth.
(ii)It’ produces a corky bark around the tree trunk that protects the interior from abrasion, heat, cold and infection.
(iii)It adds new vascular tissues for replacing old non-functioning one as well as for meeting increased demand for long distance transport of sap and organic nutrients.

4.Draw illustrations to bring out the anatomical difference between
(a) Monocot root and dicot root
(b) Monocot stem and dicot stem
Soln.(a) Differences between monocot root and dicot root are illustrated in the following figure and table.
NCERT Solutions For Class 11 Biology Anatomy of Flowering Plants Q4
NCERT Solutions For Class 11 Biology Anatomy of Flowering Plants Q4.1
(b) Differences between monocot and dicot stems are illustrated in the following figure and table.
NCERT Solutions For Class 11 Biology Anatomy of Flowering Plants Q4.2

NCERT Solutions For Class 11 Biology Anatomy of Flowering Plants Q4.3

5.Cut a transverse section of young stem of a plant from your school garden and observe it under the microscope. How would you ascertain whether it is a monocot stem or a dicot stem ? Give reasons.
Soln. Vascular bundles in dicot stem are arranged in a ring whereas in monocot stem vascular bundles are scattered throughout the ground tissue. On the basis of arrangement of vascular bundles it can be ascertained
whether the young stem is dicot or monocot. Besides undifferentiated ground tissue, sclerenchymatous hypodermis, oval or circular vascular bundles with Y shaped xylem are other differentiating features of monocot stem.

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

7.Why are xylem and phloem called complex tissues?
Soln. A group of different types of cells which perform common function is called complex tissue. Xylem and phloem are called complex tissues as all cells that work as a unit for a common function have different structural organisation. Xylem has four types of cells-tracheids, vessels, xylem parenchyma and xylem fibres. Phloem consists of sieve tube elements, companion cells, phloem parenchyma and phloem fibres. Xylem is associated with conduction of water and minerals from roots to top of plants and phloem is responsible for transport of organic food.

8.What is stomatal apparatus? Explain the structure of stomata with a labelled diagram.
Soln.Stomata are structures present in the epidermis of leaves. Stomata regulate the process of transpiration and gaseous exchange. Each stoma is composed*of two bean shaped cells known as guard cells which enclose stomatal pore. The outer walls of guard cells (away from the stomatal pore) are thin and the inner walls (towards the stomatal pore) are highly thickened. The guard cells possess chloroplasts and regulate the opening and closing of stomata. Sometimes, a few epidermal cells, in the vicinity of the guard cells become specialised in their shape and size and are known as subsidiary cells. The stomatal aperture, guard cells and the surrounding subsidiary cells are together called stomatal apparatus.
NCERT Solutions For Class 11 Biology Anatomy of Flowering Plants Q8
9.Name the three basic tissue systems in the flowering plants. Give the tissue names under each system.
Soln. The three basic tissue systems in flowering plants are epidermal tissue system, ground tissue system and vascular tissue system.
Epidermal tissue system comprises epidermal cells, stomata, trichomes and hairs.
Ground tissue system consists of cortex, endodermis, pericycle, pith and medullary rays, in the primary roots and stems. In¬leaves, the ground tissue consists of thin walled chloroplast containing cells and is called mesophyll.
The vascular tissue system consists of complex tissues, the phloem and the xylem.

10.How is the study of plant anatomy useful to us?
Soln. Study of internal structures of plants is called plant anatomy. Study of plant anatomy is useful:
-for solving taxonomic problems.
-for knowing homology and analogy of various plant groups.
-to differentiate the superior and inferior, standard and substandard or specified and unspecified woods.
-in establishing purity and correct identity of plant parts in pharmacognosy (science connected with sources, characteristics and possible medicinal uses).
-in knowing the structural peculiarities of different groups of plants.

11 .What is periderm? How does periderm formation take place in the dicot stems?
Soln. phelloderm, phellogen and phellem together constitute the periderm. Periderm is protective in function.Dicot stems produce cork cambium or phellogen in the outer cortical cells. Phellogen cells divide on both the outer side as well as the inner side to form secondary tissues. The secondary tissue produced on the inner side of the phellogen is called secondary cortex or phelloderm. On the outer side phellogen produces cork or phellem.

12.Describe the internal structure of a dorsiventral leaf with the help of labelled diagram.
Soln.
NCERT Solutions For Class 11 Biology Anatomy of Flowering Plants Q12
Dorsiventral leaves are found in dicots. The important anatomical features of dorsiventral leaves are discussed below:
(a) Upper epidermis : This is generally outermost single layer made of parenchymatous cells. The epidermal cells have sometimes outgrowths called papillae, e.g., in Gladiolus. The epidermal cells are devoid of chloroplast and stomata are absent on upper epidermis.
(b) Lower epidermis : It is just like upper epidermis but here stomata are present. Chloroplasts are absent in lower epidermis also, except the guard cells of stomata.
(c)Mesophyll: In between upper and lower epidermis mesophyll tissue is present which can be divided into two regions:
(i)Palisade parenchyma : These are elongated columnar cells without intercellular spaces. These have chloroplast in them and are generally arranged in two layers.
(ii)Spongy parenchyma : It is found below palisade parenchyma and are spherical or oval with intercellular spaces. They also have chloroplasts but number of chloroplasts is more in palisade parenchyma than spongy parenchyma.
(d)Vascular bundles : Vascular bundles are. generally found at the boundary between the palisade and the spongy regions. The vascular bundle in midrib region is largest. Vascular bundles are conjoint, collateral and closed. Each vascular bundle is surrounded by a bundle sheath of parenchymatous cells. In the vascular bundle, xylem is present towards upper epidermis and phloem towards lower epidermis. Further in xylem, protoxylem is towards upper epidermis.

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

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants:

Section NameTopic Name
5Morphology of Flowering Plants
5.1The Root
5.2The Stem
5.3The Leaf
5.4The Inflorescence
5.5The Flower
5.6The Fruit
5.7The Seed
5.8Semi-technical Description of a Typical Flowering Plant
5.9Description of Some Important Families
5.10Summary

NCRT TEXTBOOK QUESTIONS SOLVED

1.What is meant by modification of root? What type of modification of root is found in the:
(a) Banyan tree
(b) Turnip
(c) Mangrove trees
Soln.Roots of some plants change their shape and structure and become modified to perform certain functions other than absorption and conduction of water and minerals. It is called modification of roots. Roots are modified for support, storage of food and respiration, etc.
(a) Root modification in banyan tree : In banyan tree, the root modifies to form prop roots. Prop roots arise from branches and enter the soil. Thus, they provide mechanical support to densely branched, huge trees.
(b) Root modification in turnip : The
modification of root found in turnip is napiform for food storage. The upper portion of these fleshy roots is inflated or swollen which tapers towards the lower end.
(c) Root modification in mangrove trees : In mangrove plants, i.e., plants growing in saline marshes, the branches of tap root come out of the ground and grow vertically upwards showing negative geotropism. These roots are called pneumatophores. They help to get oxygen for respiration.

2.Justify the following statements on the basis of external features:
(i) Underground parts of a plant are not always roots.
(ii) Flower is a modified shoot.
Soln. (i) Underground parts of plant are not always roots because sometimes the stem also becomes underground and gets modified into various forms to perform different functions of storage, vegetative propagation, perennation, etc. Underground modifications of stems are tuber, rhizome, corm and bulb. The underground stems can be distinguished from roots externally by the presence of nodes and internodes, axillary buds, scale leaves etc. and by absence of root cap and root hairs.
(ii) Flower is the reproductive part of the angiospermic plant and it is defined as the modified shoot because (a) like shoot, flower develops from an axillary or rarely terminal bud. (b) flowers may get modified into fleshy buds or bulbils, (c) A transition from foliage leaves to floral leaves is found in Paeonia. (d) Nymphaea shows transition from sepals to petals and petals to stamens, (e) In Passiflora and Cleome long intemodes occur below gynoecium and stamens.

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3.How is a pinnately compound leaf different from a palmately compound leaf?
Soln.The compound leaves may be of two types, pinnately compound leaf and palmately compound leaf. In pinnately compound leaf, a number of leaflets are present on a common axis, the rachis, which represents the midrib of the leaf as in neem. Pinnately compound leaf may be of different types as unipinnate, bipinna te, tripinna te and decompound. In palmately compound leaf, the leaflets are attached at a common point, i.e., at the tip of petiole, as in silk cotton. Palmately compound leaf may be of different types as unifoliate, bifoliate, trifoliate, quadrifoliate and multifoliate.

4.Explain with suitable examples the different types of phyllotaxy.
Soln.Phyllotaxy is the pattern of arrangement of leaves on the stem or branch. It is usually of three types – alternate, opposite and whorled. In alternate type of phyllotaxy, a single leaf arises at each node in alternate manner, as in china rose, mustard and sunflower plants. In opposite type, a pair of leaves arises at each node and lie opposite to each other as in Calotropis and guava plants.
If more than two leaves arise at a node and form a whorl it is called whorled phyllotaxy as in Alstonia.

5.Define the following terms:
(a) aestivation (b) placentation
(c) actinomorphic (d) zygomorphic
(e) superior ovary (f) perigynous flower (g) epipetalous stamen.
Soln. (a) Aestivation : The mode of arrange¬ment of accessory floral organs (sepals and petals) in relation to one another in floral bud is known as aestivation. The main type of aestivation are valvate, twisted, imbricate, and vexillary.
(b) Placentation : The arrangement of ovules within the ovary is known as placentation. The placentation are of different types namely, marginal, axile, parietal, basal, and free central.
(c) Actinomorphic : When flower can be divided into equal radial halves in any radial plane passing through the centre, it is said to be actinomorphic, e.g., mustard, Datura etc.
(d) Zygomorphic : When a flower can be divided into two similar halves only in one particular vertical plane, it is said to be zygomorphic, e.g., pea, gulmohar, bean, Cassia.
(e) Superior ovary : In hypogynous flower, the gynoecium occupies the highest position while the other parts are situated below it. The ovary in such flowers is said to be superior, e.g., mustard, brinjal.
(f) Perigynous flower: If gynoecium is situated in the centre and other parts of the flower are
located on the rim of the thalamus almost at the same level, it is called perigynous. Here ovary is half superior, e.g., peach, plum.
(g) Epipetalous stamen : When stamens are attached to the petals, they are called epipetalous stamens e.g., brinjal.

6.Differentiate between
(a) Racemose and cymose inflorescence
(b) Fibrous root and adventitious root
(c) Apocarpous and syncarpous ovary
Soln.
(a) Differences between racemose and cymose inflorescence are as follows:
NCERT Solutions For Class 11 Biology Morphology of Flowering Plants Q6
(b) Differences between fibrous and adventitious roots are as follows :
NCERT Solutions For Class 11 Biology Morphology of Flowering Plants Q6.1

NCERT Solutions For Class 11 Biology Morphology of Flowering Plants Q6.2
(c) Differences between apocarpous and syncarpous ovary are as follows :
NCERT Solutions For Class 11 Biology Morphology of Flowering Plants Q6.3

7.Draw the labelled diagram of the following:
(i) Gram seed (ii) V. S. of maize seed.
Soln.
(i) Gram seed.
NCERT Solutions For Class 11 Biology Morphology of Flowering Plants Q7
(ii) V.S. of maize seed.
NCERT Solutions For Class 11 Biology Morphology of Flowering Plants Q7.1

8.Describe modifications of stem with suitable examples.
Soln. Stems are modified to perform different functions. Underground stems of some plants are modified to store food in them. They also act as organs of perennation to tide over conditions unfavourable for growth. Different modifications of stem are :
(i) Underground modifications
(ii)Sub-aerial modifications
(iii)Aerial modifications
(i)Underground modifications of stem are discussed as follows:
(a)Tuber: It is the branch of main stem which accumulates or stores food in it and swells up, e.g., Solarium tuberosum (potato).
(b) Rhizome: It is a branched, prostrate horizontally growing stem having nodes and internodes. On the nodes sessile scale leaves are formed, e.g., Carina, Zingiber officinale (ginger), Curcuma domestica (turmeric) etc.
(c) Corm: This is a spherical,
branched, vertically growing thick underground stem with more diameter than length, e.g., Crocus sativus (saffron), Gladiolus, Colocasia esculenta (arvi) etc.
(d) Bulb: In bulb the stem is highly reduced and can be seen only as a disc-like structure bearing numerous fleshy scaly leaves, e.g., Allium cepa (onion), Allium sativum (garlic) etc.
(ii) Subaerial modifications : Subaerial part of stem grows horizontally on the ground while some part remains underground. Vegetative propagation takes place by means of these. They may be of following kinds.
(a) Runner: It grows prostrate on the surface of soil. It develops at the base of erect shoot called crown. A number of runners arise from one erect shoot which spread in different directions. Each runner has one or more nodes which bear scale leaves and axillary buds, e.g., Cynodon (doob grass).
(b) Stolon: The nodes of horizontally growing underground stem give rise to branches which come out of the soil, e.g., Fragaria (strawberry).
(c) Sucker: Suckers are formed from the node of underground stem. Sucker comes up obliquely in the form of leafy shoot, e.g., Mentha (mint).
(d) Offset: Stem consists of thick and short intemodes. The branches are formed from the main stem and upper portion of each branch bears a group of leaves while the lower portion bears the roots. Each branch is capable of growing as an independent plant after separating from the parent plant, e.g., Eichhornia (water hyacinth), Pistia, etc.
(iii)Aerial modifications : The aerial portion of stem is modified to perform different functions, e.g., climbing, protection, food manufacturing, etc. It may show following types of modifications:
(a) Twinners : The stem is long, flexible and sensitive which can coil around an upright support like a rope, e.g., Ipomoea, Convolvulus.
(b) Climbers : The stem is weak and flexible but is unable to coil around an upright support by itself. It requires the help of clasping or clinging structures. Accordingly, climbers are of four types : root climbers, e.g., Betel; tendril climber, e.g., Passiflora; scramblers, e.g., Bougainvillea and lianas, e.g., Bauhinia.
(c) Phylloclade: The stem performs the function of photosynthesis. The stem modifies into green fleshy leaf-like
structure having distinct nodes and intemodes. Leaves of such plants are reduced into spines in order to prevent loss of water, e.g., Opantia (prickly pear), Euphorbia.
(d) Cladode: It is similar to phylloclade with only one internode, e.g., Asparagus.
(e) Thorn: Stem is modified into stiff, pointed unbranched or branched structures which have lost their growing point and become hard, called as thorns, e.g., Bougainvillea,Pomegranate, Citrus, etc. They perform defensive function.
(f) Tendrils : These are thread like sensitive structures which can coil around a support and help the plant
in climbing, e.g., Cucurbita.
(g) Bulbils: In some plants vegetative buds or floral buds modify into a swollen structure called bulbil. It separates from the parent plant and on approach of favourable condition gives rise to a new plant, i.e., it is an organ of vegetative reproduction, e.g., Agave, Oxalis.

9.Take one flower each of the families Fabaceae and Solanaceae and write their semi-technical description. Also draw their floral diagram after studying them.
Soln.Family Fabaceae (e.g., Pisum sativum) Systematic position:
Class – Dicotyledoneae
Subclass- Polypetalae
Series – Calyciflorae
Order – Rosales
Family – Fabaceae
Vegetative characters:
Habit: herb. Root: tap, branched, with root nodules.
Stem: herbaceous, climbing.
Leaves : pinnately compound, leaf base pulvinate, stipulate, venation reticulate.
Floral characters:
Inflorescence: racemose.
Flower : bisexual, zygomorphic, irregular, hermaphrodite, white or pink, complete, hypogynous to perigynous.
Calyx : sepals five, gamosepalous, ascending, imbricate aestivation, campanulate calyx tube.
Corolla : petals five, polypetalous, vexillary aestivation, papilionaceous, consisting of a posterior standard or vexillum two lateral wings or alae, two anterior ones forming a keel.
Androecium : 10 stamens in two bundles (diadelphous) of (9) + 1, anthers dithecous (bilobed), basifixed, introrse.
Gynoecium : ovary superior, monocarpellary, unilocular with many ovules, marginal placentation, style bent and long, stigma simple and-hairy.
Fruit : legume; seeds one to many, non- endospermic.
Floral formula :  NCERT Solutions For Class 11 Biology Morphology of Flowering Plants Q9.2
NCERT Solutions For Class 11 Biology Morphology of Flowering Plants Q9
Family Solanaceae (e.g., Solanum nigrum) Systematic position:
Class Subclass Series Order Family
Vegetative characters:
Habit: herbs Stem : herbaceous, aerial, erect, cylindrical, branched.
Leaves: alternate, simple, exstipulate, venation reticulate.
Floral characters:
Inflorescence: cymose.
Flower : ebracteate, ebracteolate, bisexual, actinomorphic, white, hypogynous.
Calyx : sepals five, gamosepalous, persistent, valvate aestivation.
Corolla : petals five, gamopetalous, valvate. aestivation.
Androecium : stamens five, epipetalous, polyandrous, anthers large, bithecous and basifixed.
Gynoecium : bicarpellary, syncarpous,
ovary, obliquely placed carpels in the flower, bilocular, axile placentation, placenta swollen with many ovules.
Fruits : berry with persistent calyx.
Floral formula :NCERT Solutions For Class 11 Biology Morphology of Flowering Plants Q9.3
NCERT Solutions For Class 11 Biology Morphology of Flowering Plants Q9.1

10.Describe the various types of placentations found in flowering plants.
Soln.Placenta is a parenchymatous cushion present inside the ovary where ovules are borne. The number, position, arrangement or distribution of placentae inside an ovary is called placentation. The placentation are of different types namely, marginal, axile, parietal, basal and free central.
(i)Marginal placentation : The placenta forms a ridge along the ventral suture of the ovary and the ovules are borne on this ridge forming two rows, e.g., pea.
(ii)Axile placentation : When the placenta is axial and the ovules are attached to it in a multilocular ovary, the placentation is said to be axile, e.g., china rose, tomato and lemon.
(iii)Parietal placentation : The ovules develop on the inner wall of the ovary or on peripheral part. Ovary is one-chambered but it becomes two-chambered due to the formation of the false septum, e.g., mustard and Argemone.
(iv)Free central placentation : When the ovules are borne on central axis and septa are absent, as in Dianthus and primrose the placentation is called free central.
(v)Basal placentation: The placenta develops at the base of ovary and a single ovule is attached to it, as in sunflower, marigold.

11.What is a flower? Describe the parts of a typical angiosperm flower.
Soln.Flower is the reproductive unit in the angiosperms. It is meant for sexual reproduction. A typical flower has four different kinds of whorls arranged successively on the swollen end of the stalk or pedicel, called thalamus or receptacle. These are calyx, corolla, androecium and gynoecium.
Calyx and corolla are accessory organs, while androecium and gynoecium are reproductive organs. In some flowers like lily, the calyx and corolla are not distinct and are termed as perianth. Some flowers have both androecium and gynoecium and are termed hermaphrodite flowers while some flowers have only one of these two whorls.
Calyx : The calyx is the outermost whorl of the flower and its units are called sepals. Generally, sepals are green, leaf like and protect the flower in the bud stage. The calyx may be gamosepalous (sepals united) or polysepalous (sepals free).
Corolla : Corolla is composed of petals. Petals • are usually brightly coloured to attract insects for pollination. Like calyx, corolla may also be free (polypetalous) or united (gamopetalous). The shape and colour of corolla vary greatly in plants. Corolla may be tubular, bell-shaped, funnel-shaped or wheel-shaped.
Androecium : Androecium is the male reproductive part of the flower. It is composed of stamens. Each stamen which represents the male reproductive organ consists of a stalk or a filament and an anther. Each anther is usually bilobed and each lobe has two chambers, the pollen-sacs. The pollen grains are produced in pollen-sacs. A sterile stamen is called staminode.
Gynoecium : Gynoecium is the female reproductive part of the flower and is made up of one or more carpels. A carpel consists of three parts namely stigma, style and ovary. Ovary is the enlarged basal part, on which lies the elongated tube, the style. The style connects the ovary to the stigma. The stigma is usually at the tip of the style and is’ the receptive surface for pollen grains. Each ovary bears one or more ovules attached to a flattened, cushion-like placenta. When more than one carpel is present, they may be free (as in lotus and rose) and are called apocarpous. They are termed syncarpous when carpels are fused, as in mustard and tomato. After fertilisation, the ovules develop into seeds and the ovary matures into a fruit.

12. How do the various leaf modifications help plants?
Soln.Leaves perform various functions besides photosynthesis and thus they are modified into different forms such as –
(i)Leaf tendrils: The different parts of a leaf are modified into tendrils which help the plant in climbing up. Parts of leaf modified into tendrils include stipules e.g., Smiiax ; petiole e.g., Clematis ; leaf apex e.g., Gloriosa ; leaflets e.g., Pisum; whole leaf e.g., Lathyrus.
(ii)Leaf spines: Either for the protection of plant or to lessen the rate of transpiration in xerophytic plants, the leaves modify into sharp, pointed spines. Parts of leaf modified into leaf spines include stipules e.g., Zizyphus; leaf margins e.g., Argemone; leaf apex e.g.r Yucca; entire leaf e.g., Berberis.
(iii)Phyllode: Petioles modify into leaf¬like green, photosynthesising structure e.g., Parkinsonia, Acacia auriculiformis.
(iv)Scale or protective leaves : The leaves modify into hard scaly leaves which protect the vegetative bud by covering them, e.g., Ficus, Artocarpus, Casuarina, etc.
(v) Leaf hooks : They help in climbing e.g., Bignonia.
(vi)Leaf roots : A leaf transforms into roots for balancing on water e.g., Salvinia.
(vii)Leaf pitchers : Leaf is modified into pitcher e.g., Nepenthes (insectivorous), Dischidia (non-insectivorous).
(viii)Leaf bladder: The leaves modify to form bladder like structure which trap insects and then it is closed by a valve present on the mouth of bladder e.g., Utricularia (bladderwort).
(ix) Leaf tentacles: The leaf of sundew plant, Drosera bear minute hairs which have shinning, sticky substance at their tips (tentacles). When any insect sits on the leaf, it is covered by these hairs.

13. Define the term inflorescence. Explain the basis for the different types of inflorescence in flowering plants.
Soln. The arrangement of flowers on the floral axis is termed as inflorescence. A flower is a modified shoot wherein internodes do not elongate and the axis gets condensed. The apex produces different kinds of floral appendages laterally at successive nodes instead of leaves. When a shoot tip transforms into a flower, it is always solitary. Depending on whether the apex gets converted into a flower or continues to grow, two major types of inflorescence are defined – racemose and cymose. In racemose type of inflorescence the main axis continues to grow, the flowers are borne laterally in acropetal succession. In cymose type of inflorescence the main axis terminates in a flower, hence is limited in growth. The flowers are borne in a basipeta! order.

14. Write the floral formula of an actinomorphic, bisexual, hypogynous flower with five united sepals, five free petals, five free stamens and two united carples with superior ovary and axile placentation.
Soln. The floral formula for actinomorphic, bisexual, hypogynous flower with five united sepals, five free petals, five free stamens and two united carples with superior ovary and
axile placentation is:

15.Describe the arrangement of floral members in relation to their insertion on thalamus.
Soln. In a typical flower, the floral members like calyx, corolla, androecium and gynOecium are arranged over the thalamus! Based on the position of calyx, corolla and androecium in respect to ovary on thalamus, the flowers are described as hypogynous, perigynous and epigynous ones. In the hypogynous flower the gynoecium occupies the highest position while the other parts are situated below it. The ovary in such flowers is said to be superior, e.g., mustard, china rose and brinjal. If gynoecium is situated in the centre and other parts of the flower are located on the rim of the thalamus almost at the same level, it is called perigynous. The ovary here is said to be half inferior or sub superior, e.g., plum, rose, peach. In epigynous flowers, the margin of thalamus grows upward enclosing the ovary completely and gets fused with it; the other parts of flower arise above the ovary. Hence, the ovary is said to be inferior as in flowers of guava and cucumber, and the ray florets of sunflower.

NCERT Solutions for Class 11 Biology Chapter 4 Animal Kingdom:

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 4 Animal Kingdom:

Animal Kingdom Phylum List

Section NameTopic Name
4Animal Kingdom
4.1Basis of Classification
4.2Classification of Animals
4.3Summary

NCERT Solutions Class 11 BiologyBiology Sample Papers

NCRT TEXTBOOK QUESTIONS SOLVED

1.What are the difficulties that you would face in classification of animals, if common fundamental features are not taken into account?
soln. The common fundamental features used for classifying animals include body symmetry, arrangement of cells, nature of coelom, level of organisation. Animal classification would be very confusing if fundamental features are not considered.
(i)Animals having different levels of organisation would have been placed in same group. E.g., Sponges and Cnidarians having cellular and tissue level of organisation respectively.
(ii)Animals showing varied types of germinal layers would have been placed together, as diploblastic cnidarians and triploblastic platyhelminthes.
(iii)Animals having different body symmetry would have been placed together, as coelenterates with radial symmetry and platyhelminthes with bilateral symmetry.
(iv)There would have been no classification of animals based on with or without body cavity..
(v)Placing of oviparous and viviparous animals together.

2.If you are given a specimen, what are the steps that you would follow to classify it?
soln. Various steps considered to classify a specimen are:
(i)Mode of nutrition – It can be autotrophic, holozoic, saprophytic or parasitic.
(ii)Complexity of body structure – Whether the specimen is unicellular or multicellular.
(iii)Presence or absence of membrane bound organelles.
(iv)Body symmetry, i.e., the plane by which organism can be divided into two equal halves.
(v)Presence or absence of coelom, it can be acoelomates, pseudocoelomates, eucoelo- mates.
(vi)Phylogenetic relationship.

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3.How useful is the study of the nature of body cavity and coelom in the classification of animals?
soln. Organisms can be classified according to presence or absence of the coelom. The body cavity, which is lined by mesoderm is called coelom. Animals possessing coelom are called coelomates, e.g., annelids, molluscs, arthropods, echinoderms, hemichordates and chordates. In some animals, the body cavity is not lined by mesoderm, instead, the mesoderm is present as scattered pouches in between the ectoderm and endoderm. Such a body cavity is called pseudocoelom and the animals possessing them are called pseudocoelomates, e.g., aschelminthes. In pseudocoelomates, body cavity is derived from blastocoel of the embryo. The animals in which the body cavity is absent are called acoelomates, e.g., platyhelminthes.

4.Distinguish between intracellular and extra-cellular digestion.
soln. Differences between intracellular and extracellular digestion are:
NCERT Solutions For Class 11 Biology Animal Kingdom Q4

NCERT Solutions For Class 11 Biology Animal Kingdom Q4.1

5.What is the difference between direct and indirect development?
soln. Differences between direct development and indirect development are :
NCERT Solutions For Class 11 Biology Animal Kingdom Q5

6.What are the peculiar features that you find in parasitic platyhelminthes?
soln.Following are the peculiar features of parasitic platyhelminthes:
(i) The thick tegument (body covering) resistant to the host’s digestive enzymes and anti-toxins.
(ii)Adhesive organs like suckers in flukes and the hooks and suckers in tapeworms for a firm grip on or in the host’s body.
(iii)Loss of locomotory organs.
(iv)Digestive organs are absent in tapeworms because digested and semidigested food of the host is directly absorbed’ through the body surface.
(v) Reproductive system is best developed in parasitic flatworms.
(vi)Parasitic flatworms, such as liver fluke and tapeworms perform anaerobic respiration.
(vii)They possess a considerable osmotic adaptability, as they can successfully live in different media.

7.What are the reasons that you can think of for the arthropods to constitute the largest group of the animal kingdom?
soln. Arthropods are most successful animals and constitute the largest group of the animal kingdom. They have conquered land, sea and air and make up over three fourth of currently known living and fossil organisms. They range in distribution from deep sea to mountain peaks. Thick, tough, non-living chitinous cuticle forms the exoskeleton which protects the organism from predators, help to withstand temperature upto 100°C or more and prevents water loss. They have ability to reproduce very fast and less time is needed for young ones to hatch from their eggs. Due to metamorphosis, there is less competition among larval and adult forms for food. Cockroaches can even survive nuclear radiations and poisoned earth. All these factors made arthropods the largest phylum among animals.
8.Water vascular system is the characteristic of which group among the following ?
(a) Porifera
(b) Ctenophora
(c) Echinodermata
(d) Chordata
soln. (c) Echinodermata

9.”All vertebrates are chordates but all chordates are not vertebrates”. Justify the statement.
soln. Chordates are the animals that possess notochord (a stiff, supporting rod like structure present on the dorsal side) at some stage of their lives. Phylum Chordata is divided into three Subphyla: Urochordata or tunicata, Cephalochordata and Vertebrata. Subphyla Urochordata and Cephalochordata are often referred to as protochordates and are exclusively marine. In urochordata, notochord is present only in tail of larva and disappears in adults, while in cephalochordata, it extends from head to tail region and persists throughout the life.
The members of Subphylum Vertebrata a possess notochord during the embryonic period and is replaced by a cartilaginous or bony vertebral column in the adult. Thus all vertebrates are chordates but all chordates are not vertebrates.

10.How important is the presence of air bladder in
Pisces?
soln. Bony fishes have a sac-like outgrowth, the swim bladder also called air bladder, that arises as an outgrowth from the dorsal wall of oesophagus. It is hydrostatic in function. It regulates buoyancy and helps them to swim up and down, thus preventing them from sinking. In some species air bladder also helps in respiration. It also serves as resonating chamber to produce or receive sound.

11.What are the modifications that are observed
in birds that help them fly?
soln. Birds have adapted to aerial mode of life through the following modifications:
(i) Body is streamlined and spindle shaped which minimise resistance to the wind.
(ii)Body is covered with feathers. It reduces the friction, prevent loss of heat and help to maintain constant temperature.
(iii)Forelimbs are modified into wings, which help during flight.
(iv)Flight muscles are greatly developed
(v) Most of the bones are pneumatic, hollow and filled vvith air which makes the body lighter and helps in flight.
(vi)Birds are warm-blooded. They maintain a high body temperature (40° – 46°C). This is necessary for flight.
(vii)Heart is four-chambered and functions efficiently with double circulation.
(viiiJAir sacs are present which act as reservoir of air and helps in temperature regulation
(ix)Urinary bladder is absent (except in Rhea) and only one ovary is present which reduces the weight, which is essential for flight.

12. Could the number of eggs or young ones produced by an oviparous and viviparous mother be equal? Why?
soln. No, the number of eggs or young ones produced by an oviparous and viviparous mother respectively cannot be equal. Oviparous mother lays large number of eggs, as the eggs are laid outside the body, so they are not protected from predators and harsh environmental conditions, and therefore destroyed. However in viviparous mother, eggs are not laid outside, but the embryos develop inside the mother and thus are protected from the outside harsh environment, thus, the number of eggs produced are less. Therefore, the number of eggs or young ones produced by an oviparous and viviparous mother respectively cannot be equal.

13.Segmentation in the body is first observed in which of the following?
(a) Platyhelminthes
(b) Aschelminthes
(c) Annelida
(d) Arthropoda
soln. (c) Annelida

14.Match the following:
(a) Operculum (i) Ctenophora
(b) Parapodia (ii)Mollusca
(c) Scales (iii)Porifera
(d) Comb plates (iv)Reptilia
(e) Radula (v) Annelida
(f) Hair (vi)Cyclostomata and
Chondrichthyes
(g) Choanocytes (vii)Mammalia
(h) Gill slits (viii Osteichthyes
soln.(a) – (viii), (b) – (v), (c) – (iv), (d) – (i),
(e) – (ii), (f) – (vii), (g) – (iii), (h) – (vi).

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

NCERT Solutions for Class 11 Biology Chapter 2 Biological Classification

Section NameTopic Name
2Biological Classification
2.1Kingdom Monera
2.2Kingdom Protista
2.3Kingdom Fungi
2.4Kingdom Plantae
2.5Kingdom Animalia
2.6Viruses, Viroids and Lichens
2.7Summary

NCERT Solutions Class 11 BiologyBiology Sample Papers

NCRT TEXT BOOK QUESTIONS SOLVED

1.Discuss how classification systems have undergone several changes over a period of time?
Soln. Biological classification is the scientific procedure of arranging organisms in a hierarchical series of groups and sub-groups on the basis of their similarities and dissimilarities. Scientists have proposed different systems of classification which have undergone several changes from time to time.
Earlier Aristotle proposed artificial system of classification, which divided animals and plants on basis of habitat. E.g., Aquatic (fish, whale), terrestrial (e.g., reptiles, cattle) and aerial (e.g., bat, birds). Then, natural system of classification was based on morphology^ anatomy, physiology, reproduction, ontogeny, cytochemistry, etc. After natural system, organisms were classified on basis of evolutionary relationships called phyloge¬netic system. It is based on cytotaxonomy, chemotaxOnomy, numerical taxonomy and cladistic taxonomy.

2.State two economically important uses of:
(a) heterotrophic bacteria
(b) archaebacteria
Soln. (a) Heterotrophic bacteria: They include saprotrophic, symbiotic and parasitic bacteria. They act as natural scavengers as they dispose off the dead bodies, organic wastes, release raw materials for reutilisation. They also help in sewage disposal, manure production etc. Symbiotic bacteria help in nitrogen fixation. Some bacteria arq employed in the production of a number of industrial products like lactic
acid, curd, cheese, butter, vinegar etc. Some bacteria are used in preparation of serum, vaccines, vitamins, enzymes, antibiotics etc. e.g., Pseudomonas, Xanthomonas, etc.
(b) Archaebacteria : Archaebacteria are employed in the production of gobar gas from dung and sewage and in ruminants, they cause fermentation of cellulose.

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3.What is the nature of cell-wall in diatoms?
Soln. The cell walls of diatoms are called frustules. The cell wall is chiefly composed of cellulose impregnated with glass-like silica. It is composed of two overlapping halves (or theca) that fit together like two parts of a soap box or petri dish. The upper half (lid) is called epitheca and the lower half (case) is called hypotheca. The outer covering possesses very fine markings, pits, pores and ridges. The siliceous frustules of diatoms do not decay easily. They pile up at the bottom of water reservoirs and form big heaps called diatomite or diatomaceous earth. It may extend for several hundred metres in certain areas from where the same can be mined.

4.Find out what do the terms ‘algal bloom’ and ‘red tides’ signify.
Soln.The rapid increase in populations of algae and other phytoplanktons, in particular cyanobacteria, in water bodies rich in organic matter is called algal bloom. The density of the organisms may be such that it may prevent light from passing to lower depths in the water body. Algal blooms are caused by an increase in levels of nitrate, a mineral ion essential for algal and bacterial growth.
The source of increased nitrate may be from agricultural fertilizers, which are leached – into water systems from the land, or sewage effluent.
Red tides are caused by a sudden, often toxic proliferation of marine phytoplankton, notably dinoflagellates, that colour the sea red, brown, or yellowish due to the high concentration of the photosynthetic accessory pigments. Some dinoflagellates, such as Gonyaulax, produce potent toxins, which may kill fish and invertebrates outright or accumulate in the food chain, posing a hazard to humans eating shellfish and other seafood. These phytoplanktonic blooms may be related to nutrient-rich inputs from the land, or upwelling oceanic waters, and are initiated by the activation of cyst-like forms lying on the sea bed.

5.How are viroids different from viruses?
Soln. Viroids are the smallest known agent of infectious diseases that contain small single-stranded RNA molecule. They lack capsid and have no proteins associated with them. Viroids infect only plants. Whereas, viruses have genetic material surrounded by a protective coat of protein or lipoprotein. The genetic material of viruses are of 4 types – double-stranded DNA, double-stranded RNA, single-stranded DNA, single-stranded RNA. They infect both plants and animals.

6.Describe briefly the four major groups of protozoa.
Soln. All protozoans are heterotrophs and live as predators or. parasites. They are be-lieved to be primitive relatives of animals. They are classified into four groups on the basis of locomotory organelles.
(i) Amoeboid protozoans : These organisms live in fresh water, sea water or moist soil. They move and capture their prey by developing pseudopodia (false feet) as in Amoeba. Some of them such as Entamoeba are parasites.
(ii)Flagellated protozoans : The members of this group are either free-living or parasitic. They have flagella for locomotion. The parasitic forms cause diseases such as sleeping sickness e.g., Trypanosoma.
(iii)Ciliated protozoans : These are aquatic, actively moving organisms because of the presence of thousands of cilia. They have
a cavity (gullet) that opens to the outside ‘
of the cell surface. The coordinated movement of rows of cilia causes the water laden with food to be steered into the gullet e.g., Paramecium. ~
(iv)Sporozoans: This includes diverse parasitic organisms that have an infectious spore¬like stage in their life cycle. Locomotory organs are absent. The most notorious N . is Plasmodium (malarial parasite) which
causes malaria which has a staggering effect on human population.

7.Plants are autotrophic. Can you think of some
plants that are partially heterotrophic?
Soln. Some insectivorous plants like Drosera,
Nepenthes, Utricularia are partially heterotrophic
plant. These plants are deficient in nitrogen
content but are otherwise autotrophic. They, trap various insects to obtain nitrogen from them. Rest, the food i.e., carbohydrate is
manufactured by themselves.

8.What do the terms phycobiont and mycobiont signify?
Soln. A lichen is structurally organised
entity consisting of the permanent association
of a fungus and an alga. The fungal component of a lichen is called mycobiont and the algal component is called phycobiont. Both mycobiont and phycobiont are associated
in symbiotic union in which the fungus is predominant and alga is subordinate partner. – ; Fungus provides the structural covering that protects alga from unfavourable conditions,
i.e., drought, heat, etc. It also traps moisture from the atmosphere and anchors the
lichen to a rock, tree bark, leaves and other similar supports. The alga prepares organic food by the process of photosynthesis from carbon dioxide. If the algal component is cyanobacteria (blue-green alga), they fix atmospheric nitrogen in addition to preparation of food.

9.Organise a discussion in your class on the topic – ‘Are viruses living or non-living’?
Soln. Viruses are regarded as intermediate between non-living entities and living
organisms. It is very difficult to ascertain whether they are living or non-living. Some . characters of viruses suggest their non-living nature whereas many other characters suggest their living nature.
They resemble non-living objects in –
(i) Lacking protoplast.
(ii)Ability to get crystallised.
(iii)Inability to live independent of living cell.
(iv)High specific gravity which is found
.only in non-living objects.
(v)Absence of respiration.
(vi)Absence of energy storing system.
(vii)Absence of growth and division. Instead different parts are synthesized separately.
Viruses resemble living beings in –
(i)Being formed of organic macromolecules which occur only in living beings.
(ii)Presence of genetic material.
(iii)Ability to multiply or reproduce although only inside living cell.
(iv)Occurrence of mutations.
(v) Occurrence of enzyme transcriptase in most viruses.
(vi)Some viruses like Pox virus contains vitamins like riboflavin and biotin.
(vii)Infectivity and host specificity.
(viii)Viruses are ‘killed’ by autoclaving and ultraviolet rays.
(ix)They breed true to their type. Even variations are inheritable.
(x) They take over biosynthetic machinery of the host cell and produce chemicals required for their multiplication.
(xi)Viruses are responsible for a number of infectious’ diseases like common cold, epidemic influenza, chicken pox.

10.What are the characteristic features of Euglenoids?
Soln. The euglenoid flagellates are the most interesting organisms having a mixture of animal and plant characteristics. The characteristic features are:
(i) They are unicellular flagellates.
(ii)These protists lack a definite cellulose cell wall. Instead the cells are covered by
a thin membrane known as pellicle. The pellicle is composed of protein, lipid and carbohydrates.
(iii)One or two flagella which help these protists in active swimming are present. If two flagella are present, then one is long and other is short. They are tinsel – shaped i.e., with two longitudinal rows of fine hairs. Each flagellum has its own basal granule. The two flagella join with each other at a swelling, called paraflagellar body and finally only one long flagellum emerges out through the cytostome.
(iv)Cell at the anterior end possesses an eccentric mouth or cytostome which leads into a flask-shaped cavity viz. gullet or cytopharynx. Gullet opens into a large basal reservoir.
(v) At one end of the reservoir, the cytoplasm contains an orange red stigma (eye spot). The eye spot is a curved plate with orange-red granules and contains red pigment astaxanthin. Both paraflagellar body and eye spot act as photoreceptors.
(vi)Just below the reservoir is found a contractile vacuole having many feeding canals. The contractile vacuole takes part in osmoregulation. It expands and pumps its fluid contents in the reservoir.
(vii)The mode of nutrition in euglenoids is holophytic or photoautotrophic. Some euglenoids show mixotrophic nutrition (both holophytic as well as saprobic mode).
(viii)Cytoplasm is differentiated into ectoplasm and endoplasm. Nucleus is large and occurs roughly in middle. The envelope and nucleolus persist during cell division.
(ix)Each chloroplast is composed of a granular matrix traversed by 10-45 dense bands and is covered by 3-membraned envelope. They contain the photo¬synthetic pigments-chlorophyll – n, b. They store carbohydrates as paramylon bodies, scattered throughout the cytoplasm.
(x) Asexual reproduction occurs by longitudinal binary fission. The flagellum is duplicated before cell division.
(xi)Under unfavourable condition the euglenoids form cysts to perennate the dry period.
(xii)Sexual reproduction is not observed.

11.Give a brief account of viruses with respect to their structure and nature of genetic material. Also name four common viral diseases.
Soln. Virus (L. poisonous fluid) is a group of ultramicroscopic, non-cellular, highly infectious agents that multiply only intracellularly- inside the living host cells without involving growth and division. Outside the host cells, they are inert particles. They are nucleoproteins having one or more nucleic acid molecule, either DNA or RNA, encased in a protective coat of protein or lipoprotein. A virus consist of two parts – nucleoid (genome) and capsid. An envelope and few enzymes are present in some cases,
(i) Nucleoid : The nucleic acid present in the virus is called nucleoid and it represents viral chromosome. It is made up of a single molecule of nucleic acid. It may be linear
or circular and nucleic acid can be DNA or RNA. It is the infective part of virus which utilizes the metabolic machinery of the host cell for synthesis and assembly of viral components.
(ii)Capsid : It is a protein covering around genetic material. Capsid have protein subunits called capsomeres. Capsid protects nucleoid from damage from physical and chemical agents. ,
(iii)Envelope : It is the outer loose covering present in certain viruses like animal viruses (e.g., HIV) but rarely present in plant and bacterial viruses and made of protein of viral origin and, lipid and carbohydrate of host. Outgrowths called spikes may be present. Envelope proteins have subunits called peplomers. A virus without envelope is naked virus.
(iv)Enzymes : Rarely, lysozymes are found in bacteriophages. Reverse transcriptase enzyme (catalyses RNA to DNA synthesis) is found in some RNA viruses like HIV. Some common viral diseases are – influenza, polio, measles, chickenpox, hepatitis, AIDS, bird flu, SARS (Severe Acute Respiratory Syndrome) etc.

12.Give a comparative account of the classes of Kingdom Fungi under the following:
(i) mode of nutrition (ii) mode of reproduction
Soln.
NCERT Solutions For Class 11 Biology Biological Classification Q12