NCERT Exemplar Solutions for class 11 Biology Practice Questions – Page 3

NCERT Exemplar Solutions for Class 11 Biology Chapter 12 Mineral Nutrition

January 5, 2022September 9, 2018 by Raju

NCERT Exemplar Solutions for Class 11 Biology Chapter 12 Mineral Nutrition

These Solutions are part of NCERT Exemplar Solutions for Class 11 Biology. Here we have given NCERT Exemplar Solutions for Class 11 Biology Chapter 12 Mineral Nutrition.

VERY SHORT ANSWER QUESTIONS

Question 1.
Name a plant, which accumulate silicon.
Solution:
Oryza sativa and Triticum aestivum are the plants that accumulates silicon. These plants absorbs silicon actively and accumulate them in their biomass.

Question 2.
Mycorrhiza is a mutualistic association. How do the organisms involved in this association gain from each other?
Solution:
Mycorrhiza is a mutualistic (symbiotic) association between fungus and roots of plants. The roots provide shelter and food to the fungus and the fungus helps plants in absorption of minerals, water uptake and protection against fungus.

Question 3.
Nitrogen fixation is shown by prokaryotes and not eukaryotes. Comment.
Solution:
Prokaryotes like Rhizobium and Anabaena are capable of nitrogen fixation as they contain enzyme nitrogenase but eukaryotes lack this enzyme.

Question 4.
A farmer adds Azotobacter culture to soil before sowing maize. Which mineral element is being replenished?
Solution:
Azotobacter provides nitrogen fixing bacteria which converts free nitrogen into nitrate and nitrites. It increases soil fertility.

Question 5.
What type of conditions are created by leghaemoglobin in the root nodule of a legume?
Solution:
Leghaemoglobin present in the root nodules of leguminous plants is responsible for creating anaerobic conditions and hence acts as an oxygen scavenger, protecting enzyme nitrogenase to come in contact with oxygen and help in the proper functioning of enzyme, i. e., conversion of atmospheric nitrogen to ammonia (NHj).

Question 6.
Yellowish edges appear in leaves deficient in.
Solution:
Yellowish edges or chlorosis appears in the leaves due to the deficiency of nitrogen. Its deficiency also causes delaying of flowering, interference in protein synthesis and dormancy of lateral buds.

Question 7.
Name the macronutrient which is a component of all organic compounds but it not obtained from soil.
Solution:
Carbon is an essential macronutrient, which is a component of all organic compounds but is not obtained by soil. Plant take it from atmosphere in the form of C0₂. Its concentration in atmosphere is about 0.03%. Plants use C0₂ for photosynthesis (as a source of carbon) to synthesises glucose.

Question 8.
Name one non-symbiotic nitrogen fixing prokaryote.
Solution:
Azotobacter is a non-symbotic nitrogen fixing prokaryote. It flourishs in the rice fields.

Question 9.
Complete the equation for reductive amination …….. .

Solution:
NCERT Exemplar Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 2

Question 10
Excess of Mn in soil leads to deficiency of Ca, Mg and Fe. Justify.
Solution:
When higher amounts of Mn₂₊ is absorbed by plants. The toxicity expressed in the form of brown sports surrounded by chlorotic vein.
It is due to the following reasons
(i) Reduction in uptake of Fe₃₊ and Mn₂₊.
(ii) Inhibition of binding of Mn₂₊ to specific enzymes.
(iii) Inhibition of Ca₂₊ translocation in shoot apex.
Thus, excess of Mn₂₊ causes deficiency of iron, magnesium and calcium.

SHORT ANSWER QUESTIONS

Question 1.
How is sulphur important for plants? Name the amino acids in which it is present.
Solution:
Sulphur is a macronutrient that is important for normal plant growth and development. It is also an integral part of some amino acids, proteins and helps in deciding the secondary structure of proteins as it forms disulphide bonds.

It is absorbed by the plants as SO₄^2- ion. It is present in vitamins (biotin, thiamine), proteins, coenzyme-A, amino acid (cystein and methionine) etc. It is also an essential component of plants like (onion, garlic) and mustard.

Its deficiency causes chlorosis in young leaves, extensive root growth, formation of hard and woody stem. It also causes the reduction in juice content of citrus fruit and tea yellow disease of tea.

Question 2.
How are organisms like Pseudomonas and Thiobacillus of great significance in nitrogen cycle?
Solution:
In biological nitrogen fixation, the atmospheric N₂ gets reduced to NH₃ by the help of enzyme nitrogenase reductase present in some prokaryotes. NH₃ is then oxidised in to N0₂ and NO₃ by some other bacteria (Nitrosomonas and Nitrobacter) following are the various steps involved in nitrogen fixation.
NCERT Exemplar Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 3
Pseudomonas and Thiobacillus are involved in the process of denitrification. They convert nitrate (NO₃^–) and nitrite (NO₂^–) into free nitrogen (N₂), that is released into the atmosphere.

Question 3.
Carefully observe the following figure
NCERT Exemplar Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 4
(a) Name the technique shown in the figure and the scientist who demonstrated this technique for the first time.
(b) Name atleast three plants for which this technique can be employed for their commercial production.
(c) What is the significance of aerating tube and feeding funnel in this setup?
Solution:
(a) Hydroponics,’Julius Von Sachs (1860)
(b) (i) Solanum lycopersicum (tomato)
(ii) Hibiscus asculentus (ladiesfinger)
(iii) Solanum melongena (brinjal)
(c) Aerating tube provides oxygen for the normal growth and development of the roots growing in the liquid solution. Feeding funnel is used to add water and nutrients in the hydroponic system when required.

Question 4.
Name of most crucial enzyme found in root nodules for N2-fixation? Does it require a special pink coloured pigment for its functioning? Elaborate.
Solution:
The most crucial enzyme found in the root nodules for N2-fixation is nitrogenase. It is a Mo – Fe protein that catalyses the conversion of atmospheric nitrogen to ammonia. Pink colour 8.
pigment present is root nodules of leguminous plants is called leghaemoglobin creates * anaerobic conditions for the functioning of nitrogenase enzyme.

Question 5.
Carnivorous plants exhibit nutritional adaption. Citing an example explain this fact.
Solution:
Carnivorous plants fulfill their nutritional requirements by feeding on small animals, like insects or protozoans, e.g. Nepenthes, Venus fly trap, Utricularia etc. Carnivorous plants grow in soil deficient in nitrogen.

In pitcher plant leaves are modified into pitcher which stores the juice to lure an insect. When the insect come to suck this juice, chemicals present in nectar dissolve the skin of the prey and the plant obtains nutrients (mainly nitrogen) from its skin.

Question 6.
A farmer adds/supplied Na, Ca, Mg and Fe regularly to his field and yet he observes that the plants show deficiency of Ca, Mg and Fe. Give a valid reason and suggest a way to help the farmer improve the growth of plants.
Solution:
Plant can tolerate a specific amount of micronutrients. A lesser amount of micro- nutrient can cause deficiency symptoms and higher amount can cause toxicity.

The concentration of mineral ion which reduces the dry weight of the tissues by 10% is called toxic concentration. This concentration is different for different micronutrients as well as for different plants e.g., Mn²⁺ is toxic beyond 600 mgg ^-1; (for soyabean) and (for sunflower) and beyond 5300 μgg^-1.

It has also been observed that the toxicity of one micronutrient causes the deficiency of other nutrients.
To overcome such problems, farmers should use these nutrients in prescribed concentration so that the excess uptake of one element do not reduce the uptake of the element.

Question 7.
We find that Rhizobium forms nodules on the roots of leguminous plants. Also Frankia another microbe forms nitrogen fixing nodules on the roots of non-leguminous plant Alnus.
(a) Can we artificially induce the property of nitrogen-fixation in a plant, leguminous or non leguminous?
(b) What kind of relationship is observed between mycorrhiza and pine trees?
(c) Is it necessary for a microbe to be in close association with a plant to provide mineral nutrition? Explain with the help of one example.
Solution:
(a) Artificial induction in leguminous and non- leguminous plants have been tried by scientists. It’s success rate is very low because expression of gene is highly specific phenomenon. When it desired gene is introduced that may not work because conditions for its expressions are very specific.
(b) Symbiotic mutualistic relationship (mutualism) is observed between the pine roots and mycorrhiza as both are benefitted mutually.
(c) Yes it is necessary for a microbe to be in close association with plant to provide mineral nutrition, to develop a physical relationship for example Rhizobium gets into the root and involve root tissues, then only helps in nitrogen-fixation.

Question 8.
With the help of examples describe the classification of essential elements based on the function they perform.
Solution:
Based on the diverse functions of essential elements, these are categorised into following categories given below:
(i) Constituent of biomolecules: These are the essential component of biomolecules. Hence, known as structural elements of cells, e.g., carbon, hydrogen, oxygen and nitrogen.
(ii) Energy related Chemical compound: Some elements also function in providing energy to the cell e.g. phosphorus is a component of ATP which function as energy currency -of all the living system in which magne¬sium is a component of chlorophyll, which is involved in the conversion of light en¬ergy to chemical energy.
(iii) Enzyme showing catalytic effects: Many of the essential elements are required in the form of cofactors by enzymes. They function as the activator or inhibitor of enzymes, e.g., Mg²⁺ acts as an activator of several enzymes in both photosynthesis e.g., Ribulose bisphosphate(RuBP), Car boxylase , Phosphoenol pymvate carboxy¬lase and respiration (e.g., hexokinase and phosphofructokinase). While Zn²⁺ acts as an activator of alcohol dehydrogenase while Mo of nitrogenase during the course of nitrogen fixation.

Question 9.
Trace the events starting from the coming in contact of Rhizobium to a leguminous root till nodule formation. Add a note on importance of leghaemoglobin.
Solution:
Formation of Root Nodule The coordinated activities of the Rhizobiam bacteria regume depends on the chemical interaction between these symbiotic partners.
In the following diagram the principle stages in the nodule formation are summarised.
NCERT Exemplar Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 5
Leghaemoglobin is an oxygen scavenger, that protects nitrogen enzyme from 02 and also creates anaerobic conditions for the reduction of N2 to NH3 by Rhizobium bacteria.

Question 10.
Give the biochemical events occurring in the root nodule of a pulse plant. What is the end product? What is its fate?
Solution:
Formation of root nodule in pulse plant is the result of infection of roots by Rhizobium. The following figure shows the process of nodule formation.
NCERT Exemplar Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 6
(b) Successful infection of the root hair causes it to curl

(c) Infected thread carries the bacteria to enter the cortex. Bacteria cause cortical and pericycle cells to divide, lead to nodule formation

(d) Mature nodule with vascular tissues continuous with those of the roots.
The chemical reaction is as follows
N₂ + 8e^–+ 8H⁺+ 16 ATP ->2NH₃ + H₂ + 16ADP + P₁i
The reaction takes place in the presence of enzyme nitrogenase that acts in anaerobic conditions, which is created by leghaemoglobin.
Fate of Ammonia
There are two ways by which ammonia is further used.
NCERT Exemplar Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 9
This reaction and transfer of NH2 group take place for amino acid to other amino acid catalysed by enzyme transaminase.

Question 11.
Hydroponics have been shown to be a successful technique for growing of plants. Yet most of the crops are still grown on land. Why?
Solution:
Hydroponics is a soil less culture and successful technique for plants, still many crops are grown on land because
(i) The major concern is its cost. The setting and handling of hydroponics requires much more investment than that of the soil based production.
(ii) Sanitization is extremely important, because especially with indoor hydroponic environments. Water borne disease can spread quickly through some methods of hydroponic production.
(iii) It is relatively a new technique and not used by the traditional farmers due to lack of knowledge.
(iv) Plants are less adaptable to the surrounding atmosphere. However weather and narrow oxygenation may minimise the production and quality of plant yield.

We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 12 Mineral Nutrition, help you. If you have any query regarding NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 12 Mineral Nutrition, drop a comment below and we will get back to you at the earliest.

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

January 5, 2022September 9, 2018 by Raju

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

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

VERY SHORT ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

SHORT ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

LONG ANSWER QUESTIONS

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

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

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

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

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

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

January 5, 2022September 7, 2018 by Raju

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

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

VERY SHORT ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

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

SHORT ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

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

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

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

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

LONG ANSWER QUESTIONS

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

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

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

NCERT Exemplar Solutions for Class 11 Biology Chapter 9 Biomolecules

January 5, 2022September 7, 2018 by Raju

NCERT Exemplar Solutions for Class 11 Biology Chapter 9 Biomolecules

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

VERY SHORT ANSWER QUESTIONS

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

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

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

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

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

SHORT ANSWER QUESTIONS

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

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

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

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

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

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

LONG ANSWER QUESTIONS

Question 1.
What are different classes of enzymes? Explain any two with the type of reactions they catalyse.
Solution:
Enzymes are divided into six classes each with
4-13 sub-classes and named accordingly by a number comparising of four digits.
(i) Oxidoreductases/dehydrogenases : These enzymes take part in oxidation, reduction or transfer of electrons,
(ii) Transferase : These enzymes transfer a functional group (other than hydrogen).
from one molecule to another. The transfer , chemical group does not occur in free state.
(iii) Hydrolases : These enzymes catalyse the hydrolysis of bonds like ester, ether,
peptide, glycosidic C-C, C-halide, P-N etc.

(iv) Lyases cause cleavage, removal of groups without hydrolysis and addition of groups to double bonds or removal of groups producing double bonds.
NCERT Exemplar Solutions for Class 11 Biology Chapter 9 Biomolecules 1.5
(v) Isomerases rearrangement of molecular structure to effect isomeric changes. They are of three types isomerases, epimerases and mutases.

(vi) Ligases catalyse bonding of two chemicals with the help of energy obtained from ATP resulting formation of bonds such as C—O, C—S, C—N and P—O e.g., pyruvate carboxylase
Pyruvric acid + C0₂ + ATP + H₂0
NCERT Exemplar Solutions for Class 11 Biology Chapter 9 Biomolecules 1.7

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

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

January 5, 2022September 7, 2018 by Raju

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

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

VERY SHORT ANSWER QUESTIONS

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

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

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

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

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

SHORT ANSWER QUESTIONS

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

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

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

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

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

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

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

LONG ANSWER QUESTIONS

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

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

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