NCERT Exemplar Class 11 Biology Solutions

NCERT Exemplar Solutions for Class 11 Biology Chapter 12 Mineral Nutrition

February 16, 2023

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 10 Cell Cycle and Cell Division

February 16, 2023

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 11 Transport in Plants

February 16, 2023

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

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NCERT Exemplar Solutions for Class 11 Biology Chapter 9 Biomolecules

February 16, 2023

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

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NCERT Exemplar Solutions for Class 11 Biology Chapter 22 Chemical Coordination and Integration

February 16, 2023

NCERT Exemplar Solutions for Class 11 Biology Chapter 22 Chemical Coordination and Integration

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Exemplar Solutions for Class 11 Biology Chapter 22 Chemical Coordination and Integration.

VERY SHORT ANSWER QUESTIONS

Question 1.
Which of the two adrenocortial layers, zona glomerulosa and zona reticularis lies outside enveloping the other?
Solution:
Zona glomerulosa envelops zona reticularis from the outside.

Question 2.
What is erythropoiesis? Which hormone stimulates it?
Solution:
The process of formation of RBC is Erythropoiesis. Erythropoietin, a Peptide hormone secreted from the juxtaglomerular cells of kidney stimulates erythropoiesis.

Question 3.
Name the only hormone secreted by pars intermedia of the pituitary gland.
Solution:
The only hormone secreted by Pars intermedia ^ of pituitary gland is Melanocyte Stimulating hormone (MSH). This hormone causes dispersal of pigment granules in the pigment cells, which darken the colour in certain animals like fishes and amphibians.

Question 4.
Name the endocrine gland that produces calcitonin and mention the role played by this ‘ hormone.
Solution:
Calcitonin/thyrocalcitonin linear polypeptide hormone comprising of 32 amino acids that is produced in humans primarily by the parafollicular cells of the thyroid gland. It checks excess Ca2+ and phosphate in plasma by • decreasing mobilization from bones.

Question 5.
Name the hormone that helps in cell-mediated immunity.
Solution:
The hormone thymosin plays a maj or role in the development and differentiation of T-lymphocytes, which provide cell-mediated immunity.

Question 6.
A patient complains of constant thirst, excessive 3. passing of urine and low blood pressure. When the doctor checked the patients’ blood glucose and blood insulin level, the level were normal or 4. slightly low. The doctor diagnosed the condition a diabetes insipid us. But he decided to measure one more hormone in patients blood. Which hormones does the doctor intend to measure?
Solution:
The doctor intends to measure the hyperglycaemia hormone, and its action is opposite to that of insulin Excess of glucose in blood suppresses the secretion of glucose, whereas fall in glucose level enhance glucose production.

Question 7.
Correct the following statements by replacing the term underlined.
(a) Insulin is a steroid hormone.
(b) TSH is secreted from the corpus leteum.
(c) Tetraiodothyronine is an emergency hormone.
(d) the pineal gland is located on the anterior part of the kidney.
Solution:
(a) Insulin is a peptide hormone
(b) TSH is secreted from the pars distalis region ofpitutary.
(c) Adrenaline is an emergency hormone.
(d) The adrenal gland is located on the anterior part of the kidney.

Question 8.
Match the following columns.
Column I Column II
A. Oxytocin 1. Amino acid derivative
B. Epinephrine 2. Steroid
C. Progesterone 3. Protein
D. Growth hormone 4. Peptide
Solution:
The correct matching is
Column I Column II
A. Oxytocin – Peptide
B. Epinephrine – Amino acid derivative
C. Progesterone – Steroid
D. Growth hormone – Protein

SHORT ANSWER QUESTIONS

Question 1.
What is the role-played by luteinising hormones in males and females respectively.
Solution:

LH and FSH stimulate activity of gonads and hence are called gonadotropins.
Luteinising hormone (LH) in males stimulates the synthesis and secretion of hormones called androgens from testis. Androgens along with FSH (Follicle Stimulating Hormone) regulate spermatogenesis.
LH induces ovulation of fully mature follicles in females and maintains the corpus luteum, formed from the remnants of the graafian follicles after ovulation. This secretes progesterone.

Question 2.
George comes on a vacation to India from US. The long journey disturbs his biological system and he suffers from jet lag. What is the cause of his discomfort?
Solution:

The melatonin hormone secreted by the pineal gland is also called as ‘sleep hormone’ as it promotes sleep-wake cycle.
The disruption of the body clock as it is out of synchronisation because of the unfamiliar time zone of the destination causes Jet lag. The body experiences different patterns of light and dark conditions than it is normally used to, this disrupts the natural sleep-wake cycle.
A hormone that plays a key role in body rhythms and causes jet lag is melatonin. Eyes perceive darkness after the sun sets and alert the hypothalamus to begin releasing melatonin, which promotes sleep. Conversely, when the eyes perceive sunlight, they induce the hypothalamus to with hold melatonin prodtiction.
The hypothalamus however cannot readjust its schedule instantly and it may take several days, to overcome this problem.

Question 3.
Inflammatory responses can be controlled by a certain steroid. Name the steroid, its source and also its other important functions.
Solution:

Glucocorticoids cortisol in particular, produce anti-inflammatory reactions and suppress the immune response.
The middle zone, in adrenal cortex which is the widest of three zones called zona fasciculata is the source for glucocorticoids.
The glucocorticoids as the name suggests affect carbohydrate metabolism and metabolism of proteins and fats.
They stimulate gluconeogenesis, lipolysis and proteolysis. They also inhibit utilization of amino acid and cellular uptake. Cortisol is also called stress hormone as it copes with stress.

Question 4.
Old people have weak immune system. What could be the reasons?
Solution:

A major role in the development of the immune system is played by thymus.
The thymus gland is a lobular structure located on the dorsal side of the heart and the aorta. It is derived from the endoderm of the embryo. Thymus secretes a hormone named thymosin which stimulates the development of white blood cells (WBCs), involved in producing immunity.
In old individuals, thymus is degenerated which results in decreased production of thymosin. The immune system as a result becomes weak, in old people.

LONG ANSWER QUESTIONS

Question 1.
Calcium plays a very important role in the formation of bones. Write on the role of endocrine glands and hormones responsible for maintaining calcium homeostasis.
Solution:
The hormones and endocrine glands that are responsible for maintaining calcium
homeostasis, are thyroid and parathyroid glands j and their associated hormones are calcitonin and Parathyroid Hormone (PTH).
(i) Parathyroid glands – These glands developed from the endoderm of the embryo. The cells of parathyroid glands are
of two types – chief cells and oxyphil cells. The chief cells of the parathyroid glands secrete parathyroid hormone (PTH).
This hormone (PTH) is involved in regulation of calcium and phosphate balance between the blood and other tissue. It mobilises the release of calcium into the blood from bones. PTH increases reabsorption of calcium by the body organs like intestine and kidneys.
(ii) Thyroid gland – It is the largest endocrine gland located anterior to the thyroid cartilage of the larynx in the neck.

This gland plays a major role in maintaining calcium homeostasis. It releases thyrocalcitonin hormone produced by the parafollicular cells, also called, ‘C’ cells. This hormone is secreted when the calcium level in blood gets high.

It is a 32 amino acid peptide hormone that lowers the calcium level by suppressing release of calcium ions from the bones. Calcitonin thus has an action opposite to that of the parathyroid hormone in calcium homeostasis.
NCERT Exemplar Solutions for Class 11 Biology Chapter 22 Chemical Coordination and Integration 1

Question 2.
Hypothalamus is a supper master endocrine gland. Elaborate.
Solution:
Hypothalamus is a minor but extremely important part of the diencephalon that is involved in the mediation of endocrine, autonomic and behavioural fiinction.

It consists of several groups of neuro secretory cells called nuclei which produce hormones. Hypothalamus provides anatomical connection between the nervous and endocrine system. It controls the release of major hormones by

hypophysis which include :
(i) Growth Hormone Releasing Hormone stimulates the anterior lobe of the pituitary gland to release growth hormone or somatostatin.
(ii) MSH Releasing Hormone stimulates the intermediate lobe of the pituitary gland to secrete Melanocyte Stimulating Hormone (MSH)
The hormones released from hypothalamus are involved in the processes like temperature regulation, control of water balance in body, sexual behaviour and reproduction, control of daily cycles in physiological state, behaviour and mediation’ of emotional response hypothalamus is thus called as super master endocrine gland of body.
(iii) Prolactin Releasing Hormone (PRH) stimulates the anterior lobe of the pituitary gland to secret prolactin.
(iv) Gonadotropin Releasing Hormone stimulates the anterior lobe of the pituitary gland to release gonadotropic hormones (FSHandlH).
(v) Thyrotropm Releasing Hormone (TRH) stimulates the anterior lobe of pituitary gland to release Thyroid Stimulating Hormone (TSH).
(vi) Adrenocorticotrophic releasing Hormone (ARH) stimulates the anterior lobe of pituitary gland to secrete Adrenocorticotropic Hormone (ACTH). ACTH stimulates the synthesis and secretion of steroid hormones called glucocorticoids by adrenal glands.
The hormones released from hypothalamus are involved in the processes like temperature regulation, control of water balance in body, sexual behaviour and reproduction, control of daily cycles in physiological state, behaviour and mediation of emotional responses. Hypothalamus is thus called as super master endocrine gland of body.

We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 22 Chemical Coordination and Integration, help you. If you have any query regarding NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 22 Chemical Coordination and Integration, 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

February 16, 2023

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.

NCERT Exemplar Solutions for Class 11 Biology Chapter 21 Neural control and co-ordination

February 16, 2023

NCERT Exemplar Solutions for Class 11 Biology Chapter 21 Neural control and co-ordination

These Solutions are part of NCERT Exemplar Solutions for Class 11 Biology. Here we have given NCERT Exemplar Solutions for Class 11 Biology Chapter 21 Neural control and co-ordination.

VERY SHORT ANSWER QUESTIONS

Question 1.
Rearrange the following in the correct order of involvement in electrical impulse movement.
Solution:
The correct order of involvement in electrical impulse movement is as follows:
(i) Dendrites
(ii) Cell body
(iii) Axon
(iv) Axon terminal (vi) Synaptic knob
NCERT Exemplar Solutions for Class 11 Biology Chapter 21 Neural control and co-ordination 1.1

Question 2.
Which cells of the retina enable us to see coloured objects around us?
Solution:
Cone cells present in unable us to see the colours. There are three types of cones which possess their own characteristic photopigments that respond to red, green and blue light.

Question 3.
Arrange the following in the order of reception and transmission of sound wave from the ear drum. Cochlear nerve, external auditory canal, ear drum, stapes, incus, malleus, cochlea.
Solution:
The reception and transmission of sound waves occurs in following order – External Auditory canal —» Eardrum —» Malleus —> Incus —> Stapes —>• Cochlea —> Cochlear nerve

Question 4.
During resting potential, the axonal membrane is polarized, indicate the movement of H-ve and -ve ions leading to polarisation diagrammatically.
Solution:
NCERT Exemplar Solutions for Class 11 Biology Chapter 21 Neural control and co-ordination 4.1

Question 5.
Our reaction like aggressive behaviour, use of abusive words, restlessness etc. are regulated by brain, name the parts involved.
Solution:
Functions as aggressive behaviour, use or abusive words, restlessness, etc. The inner part of cerebral hemispheres and a group of associated deep structures called limbic lobe or limbic system along with hypothalamus are involved.

Question 6.
What do grey and white matter in the brain represent?
Solution:
A major component of CNS is Grey matter consisting of neutronal cell bodies, dendrite, unmyelinated axons, glial cells and capillaries. White matter is also a component of CNS and consists mostly of gilal cell and myelinated axons.

Question 7.
Where is the hunger centre located in human brain?
Solution:
Hypothalamus in human brain contains many centres which control urge for eating and drinking.

Question 8.
Complete the statement by choosing appropriate match among the following.
NCERT Exemplar Solutions for Class 11 Biology Chapter 21 Neural control and co-ordination 8.1

Solution:
A. -> (3), B. -> (4), C. -> (2), D. -> (1)
NCERT Exemplar Solutions for Class 11 Biology Chapter 21 Neural control and co-ordination 8.2

SHORT ANSWER QUESTIONS

Question 1.
The major parts of the human neural system is depicted below. Fill in the empty boxes with appropriate
words.
NCERT Exemplar Solutions for Class 11 Biology Chapter 21 Neural control and co-ordination s1.1

Solution:
The major parts of the human neural system is filled in the boxes with appropriate words
NCERT Exemplar Solutions for Class 11 Biology Chapter 21 Neural control and co-ordination s1.2

Question 2.
Neuron system and computers share certain common features. Comment in five lines.
Solution:
In various organs the sensory neurons is present to sense the environment and extend the message to the brain. So, it is equivalent to input device of computers.
Brain acts as the CPU, or Central Processing Unit. The information gathered by sensory neurons is processed by brain and it gives command to the concerned organ to act accordingly. This message is taken or conveyed by motor neurons which act as output devices.

Question 3.
What is the function described to Eustachian tube?
Solution:
The eustachian tube forms connection between the middle ear cavity with the pharynx. It helps in equalising the pressure on either sides of the ear drum. At the pharyngeal opening of the eustachian tube there is a valve which normally remains closed.
The valve opens during yawning, swallowing and during an abrupt change in altitude, when air enters or leaves the tympanic cavity to v equalise the pressure of air on the two sides of the tympanic membrane.

LONG ANSWER QUESTIONS

Question 1.
Explain the process of the transport and release of neurotransmitter with the help of a labelled diagram showing a complete neuron, axon terminal and synapse.
Solution:
The three main parts of a neuron include the
(i) Cell body
(ii) Axon
(iii) Dendrites
Stimulus or nerve impulse of any kind passes from one neuron to another via axon. This nerve impulse is wave of bioelectric/electrochemical disturbance that passes along the neuron during conduction of an excitation.
NCERT Exemplar Solutions for Class 11 Biology Chapter 21 Neural control and co-ordination l1.1

Within a synapse transport and release of a neuro transmiter occurs.
At a chemical synapse, the membranes of the pre- and post-synaptic neurons are separated by a fluid-filled space called synaptic cleft. Chemicals called neurotransmitters are involved in the transmission of impulses at these synapses.
The axon terminals contain vesicles filled with these neurotransmitters.
Upon arrival of an impulse (action potential) at the axon terminal, it stimulates the movement of the synaptic vesciles towards the membrane, where they fuse with the plasma membrane and release their neurotransmitters in the synaptic cleft.
The released neurotransmitters bind to their specific receptors, present on the post-synaptic membrane. This binding opens ion channels allowing the entry of ions, that can generate a new action potential in the

Question 2.
Explain the structure of middle and internal ear with the help of diagram.
Solution:
Ears are a part of statoacoustic organ meant for balancing and hearing the external ear in most mammals is a heap of tissue also called pinna. It is a part of auditory system.

The human ear consists of three main parts external ear, middle, ear and internal ear.

Structure of Middle Ear

The middle ear consists of three bones or ossicles-the malleus (hammer), incus (anvil and stapes (stir-up).
These bones are attached to one another in a chain-like manner.
The malleus is attached to the tympanic membrane and the stapes is attached to the oval window (a membrane beneath the stapes) of cochlea.
These three ossicles increase the efficiency of transmission of sound waves to the inner ear.
The middle ear also opens into the eustachian tube, which connects with the pharynx and maintains the pressure between the middle ear and the outside atmosphere.

Structure of Internal Ear

Thd inner ear consists of a labyrinth of chambers filled with fluid within temporal bone of the skull. The labyrinth consists of two parts the bony and membranous labyrinth. The bony labyrinth is a series of channels.
Membranous labyrinth lies inside these channels which is surrounded by a fluid called perilymph. The membranous labyrinth is filled with a fluid called endolymph. The coiled portion of the labyrinth is called cochlea.
The cochlea has two large canal separated by a small cochlear duct (scala media). An upper vestibular canal (scala vestibuli) and a lower tympanic canal (scala tympani). The vestibular and tympanic canals contain perilymph and the cochlear duct is filled with endolymph.
The wall of membranous labyrinth comes in contact with the fenestra ovalis at the base of scale vestibuli while the fenestra rotunda.

We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 21 Neural control and co-ordination, help you. If you have any query regarding NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 21 Neural control and co-ordination, drop a comment below and we will get back to you at the earliest.

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

February 16, 2023

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

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

VERY SHORT ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

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

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

SHORT ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

LONG ANSWER QUESTIONS

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

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

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

NCERT Exemplar Solutions for Class 11 Biology Chapter 20 Locomotion and Movement

February 16, 2023

NCERT Exemplar Solutions for Class 11 Biology Chapter 20 Locomotion and Movement

These Solutions are part of NCERT Exemplar Solutions for Class 11 Biology. Here we have given NCERT Exemplar Solutions for Class 11 Biology Chapter 20 Locomotion and Movement.

VERY SHORT ANSWER QUESTIONS

Question 1.
Name the cells/ tissues in human body which
(a) exhibit amoeboid movement
(b) exhibit ciliary movement
Solution:
(a) Macrophages and leucocytes in blood exhibit amoeboid movement. Cytoskeletal elements like microfilaments are also involved in amoeboid movement.
(b) Ciliary Movement occurs mostly in the internal organs, lined by the ciliated epithelium, e.g., cilia in trachea helps in removing dust particle and foreign substances inhaled along with atmospheric air.
Passage of ova through the female reproductive tract is also facilitated by the ciliary movement. This is due to the presence of ciliated epithelium in the Fallopian tube.

Question 2.
Locomotion requires a perfect coordinated activity of muscular …… systems.
Solution:
Locomotion requires a prefect coordinated activity of muscular, skeletal and neural systems.

Question 3.
Sarcolemma, sarcoplasm and sarcoplasmic reticulum refer to particular type of cell in our body. Which is this cell and to what parts of that cell do these names refer to?
Solution:
Muscle fibre is lined by the plasma membrane called sarcolemma. Muscle fibre is a syncitium because sarcoplasm (the cytoplasm) of muscle fibre contains number of nuclei and sarcoplasmic reticulum is the endoplasmic reticulum of the muscle fibre and is the store house of calcium ions.

Question 4.
Label the different components of actin filament in the diagram given below
NCERT Exemplar Solutions for Class 11 Biology Chapter 20 Locomotion and Movement 4.1v
Solution:
Each actin filament is made of two ‘F (filamentous) actins helically wound to each other and each ‘F’ actin is a polymer of monomeric ‘G’ (globular) actins.
The different components of action filament can be represented as
NCERT Exemplar Solutions for Class 11 Biology Chapter 20 Locomotion and Movement 4.2v

Question 5.
What is the difference between the matrix of bones and cartilage?
Solution:
Difference between the matrix of ones and cartilage
NCERT Exemplar Solutions for Class 11 Biology Chapter 20 Locomotion and Movement 5.1v

Question 6.
Which tissue is affected by mysthenia gravis? What is the underlying cause.
Solution:
Myasthenia gravis is autoimmune disorder of skeletal muscle, which affects neuromuscular junction, that leads to fatigue, weakening and paralysis of the skeletal muscle.

Question 7.
How do our bone joints function without grinding noise and pain?
Solution:
The presence of synovial fluid, between articulating surface of the two bones enclosed within synovial cavity of synovial joints to enables out joints to function without grinding noise and pain.

Question 8.
Give the location of a ball and socket joint in a human body.
Solution:
In human body Ball and socket joint are present between humerus and pectoral girdle. These joints allows free movement of bone in all direction. E.g., shoulder joints (humerus bone in socket of pectoral girdle) and

SHORT ANSWER QUESTIONS

Question 1.
With respect to rib cage, explain the following
(a) bicephalic ribs
(b) true ribs
(c) floating ribs
Solution:
There are 12 pairs of ribs. Each rib consist of a thin flat bone dorsally connected to the vertebral column and ventrally to the sternum.
(a) Bicephalic ribs each rib has two articulating surfaces on its dorsal end hence, are called as bicephatic ribs.
(b) The first seven pairs of ribs are true ribs. These ribs are dorsally attached to the thoracic vertebrae and ventrally connected to the sternum with the help of hyaline cartilage.
(c) The last two pair (11th and 12th) of ribs are not connected ventrally to the sternum therefore, called as floating ribs.
NCERT Exemplar Solutions for Class 11 Biology Chapter 20 Locomotion and Movement 1.1s

Question 2.
Exchange of calcium between bone and ^extracellular fluid takes place under the
influence of certain hormones
(a) What will happen if of Ca²⁺ is in extracellular fluid?
(b) What will happen if very less amount of Ca²⁺ is in the extracellular fluid?
Solution:
Parathyroid and thyroid glands, function under the feed back control of blood calcium
(a) More Ca²⁺ concentration in extracellular fluid is associated with hyperparathyroidism. It causes demineralisation, resulting in softening and bending of the bones. This condition leads to osteoprosis.
(b) Very less amount of Ca²⁺ in extracellular fluid is associated with hypoparathyroidism. This increases the excitability or nerves and muscles, causing cramps, sustained contraction of the muscles of larynx, face, hands and feet. This disorder called parathyroid tetany or hypercalcemic tetany.

Question 3.
Rahul exercises regularly by visiting a gymnasium. Of late he is gaining weight. What could be the reasons? Choose the correct answer and elaborate.
(a) Rahul has gained weight due to accumulation of fats in body
(b) Rahul has gained weight due to increased muscle and less of fat
(c) Rahul has gained weight because his muscle shape has improved
(d) Rahul has gained weight because he is accumulating water in the body
Solution:
(b) Rahul has gained weight because the shape of his muscle has changed. Regular exercise increases the body muscle. There is an enlargement of muscles due to increase in the amount of sarcoplasm and mitochondria and the strength he to developed led him to gain the mass and size of body muscle and reduction in fat content.

Question 4.
Radha was running on a treadmill at a great speed for 15 minutes continuously. She stopped the treadmill and abruptly came out. For the next few minutes, she was breathing heavily/fast. Answer the following questions.
(a) What happened to her muscles when she did strenuously exercised?
(b) How did her breathing rate change?
Solution:
(a) Her muscles got fatigues due to continuous exercise because of the accumulation of lactic acid within skeletal muscles. Pain is also often experienced in the fatigued muscles.
(b) Her breathing rate changes from normal to
high as during as her body muscles require thus oxygen for the ATP production, than the normal value, her breathing thus enhances, to take most oxygen from the atmosphere.

Question 5.
Write a few lines about gout.
Solution:
Gout is a disease caused due to improper purine metabolism. It causes accumulation of uric acid and its crystals in the joints. The level of uric acid and crystals of its salts get raised in blood causing their accumulation in the joint to which causes gouty arthritis. The excess of urates in blood can also lead to the formation stones in the kidneys.

Question 6.
What are the points for articulation of pelvic and pectoral girdles?
Solution:

Each half of the pectoral girdle consist of a clavicle and a scapula.
The dorsal flat, triangular body of scapula has a slightly elevated ridge called the spine that, projects flat expanded process called the acromion and the clavicle articulating with it.
There a depression below the acromion is called the glenoid cavity which articulates with the head of the humerous to form the shoulder joint.
Pelvic girdle consist of two coxal bones, each formed by the fusioin of three bones, ilium, ischium and pubis. It articulates with femur through a cavity called acetabulum forming thigh joint.

LONG ANSWER QUESTIONS

Question 1.
How does a muscle shorten during its contracting and return to its original form during relaxation?
Solution:
Muscles contract due to formation of cross-bridge between the actin and myosin filament
(i) An ATP molecule j oins the active site on the head of myosin myofilament. These heads contains an enzyme, myosin ATPase along with Ca²⁺ and Mg²⁺ ions that catalyses the break down of ATP.

(ii) The energy is transferred to myosin head which straightens to join an active site on actin myofilament, forming a across-
NCERT Exemplar Solutions for Class 11 Biology Chapter 20 Locomotion and Movement 1.1l
(iii) The energised cross-bridges move, causing the attached actin filaments to move towards the centre of A-band. The Z-line is also pulled inwards causing shortening of sarcomere, contraction. During contraction A-bands retain the length, while I-bands get reduced.
(iv) The myosin head releases ADP and Pi where relaxes to its low energy state. The head detaches from actin myofilaments when new ATP molecule joins it and cross-bridge are broken.
(v) In the next cycle, the free head cleaves the new ATP. The cycles of cross-bridge formation and breakage is repeated causing further sliding.
NCERT Exemplar Solutions for Class 11 Biology Chapter 20 Locomotion and Movement 1.2l
(vi) After contraction muscle relaxation occurs when the calcium ions are pumped back to the sarcoplasmic cistemae, thus, blocking the sites on actin myofilaments. The Z-line returns to original positions or relaxation.

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NCERT Exemplar Solutions for Class 11 Biology Chapter 19 Excretory Products and their Elimination

February 16, 2023

NCERT Exemplar Solutions for Class 11 Biology Chapter 19 Excretory Products and their Elimination

These Solutions are part of NCERT Exemplar Solutions for Class 11 Biology. Here we have given NCERT Exemplar Solutions for Class 11 Biology Chapter 19 Excretory Products and their Elimination.

VERY SHORT ANSWER QUESTIONS

Question 1.
Where does the selective reabsorption of glomerular filtrate take place?
Solution:
The selective reabsorption of glomerular filtrate takes place in Proximal Convoluted Tubules (PCT) and Distal Convoluted Tubules (DCT).

Question 2.
What is the excretory product from kidneys of reptiles?
Solution:
The excretory product from the kidney of reptile is uric acid.

Question 3.
What is the composition of sweat produced by sweat glands?
Solution:
Sweat produced by sweat glands is a watery fluid that containing NaCl, small amounts of urea, lactic acid, etc. It’s primary function is to facilitate a cooling effect on the body surface and also to helps in removal of water.

Question 4.
Identify the glands that perform the excretory function in prawns.
Solution:
In prawns, the excretory organs are known as antennary glands or green glands. These glands are white pea sized structures and opaque enclosed in the coxa of each 2nd antenna. They mainly excrete ammonia.

Question 5.
What is the excretory structure in Amoeba?
Solution:
Conractile vacuole performs the function of excretion as well as osmoregulation in amoeba.

Question 6.
The following abbreviations are used in the context of excretory functions, what do they stand for?
(a) ANF
(b) ADH
(c) GFR
(d) DCT
Solution:
(a) ANF Stands for Atrial Natriuretic Factor
(b) ADH Stands for Antidiuretic Hormone
(c) GFR Stands for Glomerular Filtration Rate
(d) DCT Stands for Distal Convoluted Tubule

Question 7.
Differentiate glycosuria from ketonuria.
Solution:
Difference between glycosuria and ketonuria is as follows
NCERT Exemplar Solutions for Class 11 Biology Chapter 19 Excretory Products and their Elimination 1

Question 8.
Mention any two metabolic disorders, which can be diagnosed by analysis of urine.
Solution:
Metabolic disorders that can be diagnosed by analysis of urine are
(i) Hematuria- It is a disorder in which blood cells are present in the urine, which could be a sign of kidney stone or a tumor in urinary tract.
(ii) Albuminuria- It is a disorder in which albumin is present in urine and occurs in nephritis i.e., inflammaton of glomeruli. In this condition the size of filtering slits becomes enlarged.

Question 9.
What are the main processes of urine formation?
Solution:
Urine formation includes glomerular filtration (ultra Alteration), selective reabsorption and tubular secretion that occurs in different parts of the nephron.
Glomerular filteration is carried out by glomerulus and is involve the filteration of blood.
Selective reabsorption is the absorption of filtrate through renal tubules either activity or passively.
Tubular secretin involves secretion through tubular cells in urine in order to maintain ionic and acid-base balance of body fluids.

Question 10.
Fill in the blanks appropriately
Organ Excretory wastes
(a) Kidneys ………..
(b) Lungs ………..
(c) Liver ………..
(d) Skin ………..
Solution:
Organ Excretory wastes
(a) Kidneys ——> Urine
(b) Lungs ——> C02
(c) Liver ——> Urea
(d) Skin ——-> Sweat

SHORT ANSWER QUESTIONS

Question 1.
Show the structure of a renal corpuscle with the help of a diagram.
Solution:
The structure of a renal corpuscle is shown below.
NCERT Exemplar Solutions for Class 11 Biology Chapter 19 Excretory Products and their Elimination 2

Question 2.
What is the role played by renin-angiotensin in the regulation of kidney frictions?
Solution:
On activation by fall in the glomerular blood pressure/flow renin is released from the Juxta-Glomerular Apparatus (JGA).

It converts angiotensinogen in blood to angiotensin I and further to angiotensin II. Angiotensin II, being a powerful vasoconstrictor, increases the glomerular blood pressure and thereby Glomerular Filteration Rate (GFR). Angiotensin II also activates the adrenal cortex to release aldosterone.

This Aldosterone causes reabsprption of Na+ and water from the distal parts of the tubule. Which leads to an increase in blood pressure and GFR. This complex mechanism is generally known as Renin Angiotensin Aldosterone System of RAAS.

Question 3.
The composition of glomerular filtrate and urine is not same. Comment.
Solution:
Glomerular filtrate contains all the content of the blood plasma except proteins. About 180 litres of glomerular filtrate like water, glucose, nutrients ions etc. occurs. As a result, now the composition of urine is quite different from that of the glomerular filtrate. Some ions are also added to this fluid by tubules i.e. tubular secretion to maintain ionic and acid base balance of body fluids. Thus the composition of glomerular filtrate andd urine is not same.

Question 4.
What is the procedure advised for the correction of extreme renal failure? Give a brief account of it.
Solution:
The ultimate method for the correction of acute/ extreme renal failure (kidney failure) is, Kidney transplantation it is to minimise chances of rejection by the immune system of the host, functional kidney is used as a transplant from a donor, preferably close relative modem clinical procedures have increased the success rate of such a complicated technique.

Question 5.
Explain, why a haemodialysing unit called artificial kidney?
Solution:

Haemodialysis is a method that become a boon for thousands of uremic (accumulation of urea in blood) patients all over the world. Haemodialysing unit act as artificial kidney by removing urea from patients blood due to kidney failure.
In this process blood is drained from artery and pumped into a dialysing unit after the addition of an anticoagulant named heparin.
The unit contains a coiled cellophane tube which is surrounded by a dialysing fluid having the same composition as that of plasma except nitrogenous waste.
The porous cellophane mfcmbrane of the tube allows the passage of molecules that is based on concentration gradient.
Absence of nitrogenous water in dialysing fluid these substances freely move out thereby clearing the blood.
In the end the cleared blood is pumped back to the body through a vein after the addition of anti¬heparin to it thereby completing the process.

LONG ANSWER QUESTIONS

Question 1.
Explain the mechanism of formation of concentrated urine in mammals.
Solution:
Mammals have the ability to produce concentrated urine. The loop of Henle and vasa recta play important role in it which is discussed as follows:
(i) the proximity between the Henle’s loop and vasa recta as well as the counter current that is formed due to the fiow of filtrate in two limp’s of Henle’s loop in opposite direction and help in opposite direction and help in maintaining an increasing osmolality towards the inner medullary interstitium, i.e., from 300 mOsmoL^-1 in the cortex to about 1200 mOsmol^-1 in the inner medulla.
NCERT Exemplar Solutions for Class 11 Biology Chapter 19 Excretory Products and their Elimination 4

(ii) This gradient is caused mainly due to NaCl and urea ascending limb of Henle’s loop transports NaCl, that is exchanged with the descending limb of vasa recta.
(iii) Through the ascending portion of vasa recta. NaCl is returned to the interstitium.
(iv) Similarly, a small amount of urea enters the thin segment of the ascending limb of Henle’s loop, which is transported back to the intersitium by the collecting tubule.
(v) This special arrangement of Henle’s loop, and vasa recta, is called the counter current mechanism.
(vi) The rate of dissipation is reduced by the counter current exchange. This in turn, reduces the rate at which the current must pump Na⁺ to maintain any given gradient.
(vii) Presence of such interstitial garden helps in an easy passage of water from the collecting tubule thereby concentrating the filtrate (urine).
(viii) Human kidneys produces urine nearly four times concentrated than the initial filtrate formed.

Question 2.
Describe the structure of a human kidney with the help of a labelled diagram.
Solution:

Human kidney are reddish-brown, bean-shaped structures that is situated between the last thoracic and third lumbar vertebra, which is closer to the dorsal inner wall of the abdominal cavity.
Each kidney of an adult human measures 10-12 cm in length, 5-7 cm in width, 2-3 cm in thickness with an average weight of 120-170 gm
The kidney is covered by a fibrous connective tissue i.e., the renal capsula, that protects the kidney.
Internally, it consists of outer dark cortex and an inner light medulla, both containing nephron, nephron is the structural and functional units of kidney.
The median concave border of a kidney contains a notch called hilum, that functions as route entry and exit of blood vessels, nerves and ureter.
The renal cortex is granular in appearance that contains convoluted tubules that malpighian corpuscles. The renal medulla contains loop of henle, collecting ducts and tubules and ducts ofBertini.
Medulla is divided into conical masses, the medullary pyramids that further form papillae.
The papillae form calyces, which join to renal pelvis leading to ureter. Between the medullary pyramids, cortex extends into medulla and forms renal columns which are called as column of Bertini.

We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 19 Excretory Products and their Elimination, help you. If you have any query regarding NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 19 Excretory Products and their Elimination, drop a comment below and we will get back to you at the earliest.

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

February 16, 2023

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

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

VERY SHORT ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

SHORT ANSWER QUESTIONS

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

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

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

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

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

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

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

The concentric growth rings are called annual rings. These rings are formed due to the secondary growth.
Secondary growth occurs due to the activity of cambium which is a meristermatic tissue in dicot trees.
The rate of activity of cambium is more in spring so wood formed has larger wider xylem cells, whereas wood formed in autumn has narrower and smaller xylem elements.
This results in the formation of two rings called growth rings.
By counting these rings, age of the tree can be determined. This branch of science is known as dendrochronology or growth ring analysis.

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

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

Question 10.
Write the precise function of
(a) sieve tube
(b) interfascicular cambium
(c) collenchyma
(d) aerenchyma
Solution:
Sieve tube It’s function is to transport of synthesised food throughout the plant. It is present in the phleom tissue.
NCERT Exemplar Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 1.7
Interfascicular Cambium It is a kind of secondary meristermatic tissue present in between two vascular bundles. It is function is to bring about secondary growth in the dicot stem and root.

Collenchyma cells have angular thickening at corners. There function is to provide mechanical support to young growing herbaceous stem.
NCERT Exemplar Solutions for Class 11 Biology Chapter 6 Anatomy of Flowering Plants 1.9
Aeronchyma is a specilised parenchyma having large air spaces. It provides buyoncy to the hydrophytic plants.

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

LONG ANSWER QUESTIONS

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

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

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

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

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

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