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NCERT Solutions For Class 11 Biology Cell The Unit of Life

April 29, 2023

NCERT Solutions For Class 11 Biology Cell The Unit of Life

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

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

NCERT Solutions Class 11 Biology Biology Sample Papers

NCRT TEXTBOOK QUESTIONS SOLVED

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

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

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

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

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

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

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

Functions of mitochondria:

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

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

Functions of chloroplast:

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

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

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

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

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

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

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

13.Describe the structure of the following withthe help of labelled diagrams.
(i) Nucleus (ii) Centrosome
Soln.(i) Nucleus: Nucleus is double membrane bound principle cell organelle which contains all genetic information for controlling cellular metabolism and transmission of genetic information.
Nucleus is differentiated into following four parts:
(a) Nuclear envelope: It is a double membrane bound envelope that surround the nucleus and separates the latter from the cytoplasm.
(b) Nucleoplasm: Itis clear, non-staining, fluid material present in the nucleus, which contains raw materials (nucleotides), enzymes (DNA/RNA polymerases) and metal ions for the synthesis of RNAs and DNA. The nuclear matrix or the nucleoplasm is composed of nucleolus and chromatin.
(c) Nucleolus: It is a naked, round and slightly irregular structure, which is attached to the chromatin at a specific region. It is a site for active ribosomal RNA synthesis.
(d) Chromatin : It has the ability to get stained with certain basic dyes. It is known to be the hereditary DNA protein fibrillar complex. The chromatin fibres are distributed throughout the nucleoplasm.

(ii) Centrosome: Centrosome is an organelle usually containing two cylindrical structures called centrioles. They are surrounded by amorphous pericentriolar materials. Both the centrioles in a centrosome lie perpendicular to each other. They are made up of nine evenly spaced peripheral fibrils of tubulin protein. Each of the peripheral fibril is a triplet. The adjacent triplets are also linked. The hub of centriole is connected with tubules of the peripheral triplets by radial spokes made of protein.

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

NCERT Solutions For Class 11 Biology Photosynthesis in Higher Plants

April 28, 2023

NCERT Solutions For Class 11 Biology Photosynthesis in Higher Plants

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis in Higher Plants:

Section Name	Topic Name
13	Photosynthesis in Higher Plants
13.1	What do we Know?
13.2	Early Experiments
13.3	Where does Photosynthesis take place?
13.4	How many Pigments are involved in Photosynthesis?
13.5	What is Light Reaction?
13.6	The Electron Transport
13.7	Where are the ATP and NADPH Used?
13.8	The C₄ Pathway
13.9	Photorespiration
13.10	Factors affecting Photosynthesis
13.11	Summary

NCRT TEXTBOOK QUESTIONS SOLVED

1. By looking at a plant externally can you tell whether a plant is C₃ or C₄? Why and how?
Solution: It is not possible to distinguish externally between a C₃ and C4 plant, but generally tropical plants are adapted for C₄ cycle.

2. By looking at which internal structure of a plant can you tell whether a plant is C₃ or C₄? Explain.
Solution: C₄ plants live in hot moist or arid and nonsaline or saline habitats. Internally the leaves show kranz anatomy. In kranz anatomy, the mesophyll is undifferentiated and its cells occur in concentric layers around vascular bundles. Vascular bundles are surrounded by large sized bundle sheath cells which are arranged in a wreath-like manner (kranz – wreath). The mesophyll and bundle sheath cells are connected by plasmodesmata or cytoplasmic bridges. The chloroplasts of the mesophyll cells are smaller. They have well developed grana and a peripheral reticulum but no starch. Mesophyll cells are specialised to perform light reaction, evolve 02 and produce assimilatory power (ATP and NADPH). They also possess enzyme PEPcase for initial fixation of CO₂ The chloroplasts of the bundle sheath cells are agranal.

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3. Even though very few cells in a C₄ plant carry out the biosynthetic – Calvin pathway, yet they are highly productive. Can you discuss why?
Solution: Since, through C₄ cycle, a plant can photosynthesise even in presence of very low concentration of CO₂ (upto 10 parts per million), the partial closure of stomata due to xeric conditions would not bring much
effect. Therefore, the plants can adapt to grow at low water content, high temperature and bright light intensities. This cycle is specially suited to such plants which grow in dry climates of tropics and subtropics. Besides, the photosynthetic rate remains higher due to absence of photorespiration in these plants. It can be visualised that both C₄ cycle and photorespiration are the result of evolution or might have been one of the reasons of evolution for the adaptation of plants to different environments. C₄ plants are about twice to efficient as C₃plants in converting solar energy into production of dry matter.

4. RuBisCO is an enzyme that acts both as a carboxylase and oxygenase. Why do you think RuBisCO carries out more carboxylation in C₄ plants?
Solution: RuBisCO is an enzyme which acts both as carboxylase (carboxylation during photosynthesis) and oxygenase (during photorespiration). But RuBisCO carries out more carboxylation in C4 plants. In C₄ plants, initial fixation of carbon dioxide occurs in mesophyll cells. The primary acceptor of C02 is phosphoenol pyruvate or PER It combines with carbon dioxide in the presence of PEP carboxylase or PEPcase to form oxaloacetic acid or oxaloacetate. Malic acid or aspartic acid is translocated to bundle sheath cells through plasmodesmata. Inside the bundle sheath cells they are decarboxylated (and deaminated in case of aspartic acid) to form pyruvate and CO₂ . CO₂ is again fixed inside the bundle sheath Cells through Calvin cycle. RuBP of Calvin cycle is called secondary or final acceptor of CO₂ in C₄ plants. Pyruvate is sent back to mesophyll cells.

5. Suppose there were plants that had a high concentration of chlorophyll b, but lacked chlorophyll a, would it carry out photosynthesis? Then why do plants have chlorophyll b and other accessory pigments?
Solution: Plants that do not possess chlorophyll a will not carry out photosynthesis because it is the primary pigment and act as the reaction centre. It performs the primary reactions of photosynthesis or conversion of light into chemical or electrical energy. Other photosynthetic pigments are called accessory pigments. They absorb light energy of different wavelengths and hence broaden the spectrum of light absorbed by photosynthetic pigments. These pigments hand over the absorbed energy to chlorophylla.

6. Give comparison between the following:
(a) C₃ andC₄ pathways
(b) Cyclic and non-cydic photophosphorylation
(c) Anatomy of leaf in C3 and C₄ plants.
Solution: (a) The differences between C₃ and C₄

(b) The differences between cyclic and non- cyclic photophosphorylation are as follows :

7. Look at leaves of the same plant on the shady side and compare it with the leaves on the sunny side. Or ompare the potted plants kept in the sunlight with those in the shade. Which of them has leaves that are darker green? Why?
Solution: The leaves of the shaded side are darker green than those kept in sunlight due to two reasons:
(i) The chloroplasts occur mostly in the mesophyll cells along their walls for receiving optimum quantity of incident light.
(ii)The chloroplasts align themselves in vertical position along the lateral walls of high light intensity and along tangential wails in moderate light.

8. The given figure shows the effect of light on the rate of photosynthesis. Based on the graph, answer the following questions.

(a) At which point/s (A, B or C) in the curve is light limiting factor?
(b) What could be the limiting factor/s in region A?
(c) What do C and D represent on the curve?
Solution: (a) At regions A and B light is the limiting factor.
(b) In the region A’, light can be a limiting factor.
(c) C is the region where the rate of photosynthesis is not increased when light intensity is increased. D is the point where some other factors become limiting.

9. Why is the colour of a leaf kept in the dark frequently becomes yellow, or pale green? Which pigment do you think is more stable?
Solution: Carotenoid pigments are found in all photosynthetic cells. They are accessory pigments also found in roots, petals etc. These pigments do not breakdown easily thus temporarily reveal their colour due to unmasking, following breakdown of chlorophylls. Thus the colour of leaf kept in dark is yellow or pale green.

NCERT Solutions For Class 11 Biology Digestion and Absorption

April 14, 2023

NCERT Solutions For Class 11 Biology Digestion and Absorption

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption:

Section Name	Topic Name
16	Digestion and Absorption
16.1	Digestive System
16.2	Digestion of Food
16.3	Absorption of Digested Products
16.4	Disorders of Digestive System
16.5	Summary

NCERT TEXTBOOK QUESTIONS FROM SOLVED

1. Choose the correct answer among the following:
(a) Gastric juice contains
(i) pepsin, lipase and rennin
(ii) trypsin, lipase and rennin
(iii) trypsin, pepsin and lipase
(iv) trypsin, pepsin and rennin.
(b) Succus entericus is the name given to
(i) a junction between ileum and large intestine
(ii) intestinal juice
(iii) swelling in the gut
(iv) appendix.
Solution: (a) (i) Pepsin, lipase and rennin
(b) (ii) Intestinal juice

2. Match column I with column II.
Column I Column II
(a) Bilirubin and (i)Parotid biliverdin
(b) Hydrolysis of (ii)Bile starch
(c) Digestion of fat (iii)Lipases
(d) Salivary gland (iv) Amylases
Solution: (a), – (ii),- (b) – (iv), (c) – (iii),- (d) – (i)

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3. Answer briefly:
(a) Why are villi present in the intestine and not in the stomach?
(b) How does pepsinogen change into its active form ?
(c) What are the basic layers of the wall of alimentary canal?
(d) How does bile help in the digestion of fats ?
Solution: (a) The absorptive surface area of small intestine is enormously increased by microvilli and as maximum absorption
of digested food takes place in small intestine as compared to other organs, therefore, villi are present in small intestine and not in stomach. Moreover, stomach is primarily associated with temporary storage of food.
(b) The proenzyme pepsinogen, on exposure to hydrochloric acid, secreted by oxyntic cells of gastric glands gets converted into the active enzyme pepsin, the proteolytic enzyme of the stomach.
(c) The wall of alimentary canal from oesophagus to rectum possesses four layers, namely serosa, muscularis, sub-mucosa and mucosa. Serosa is the outermost layer and is made up of a thin mesothelium with some connective tissues. Muscularis is formed by smooth muscles. The sub-mucosal layer is formed of loose connective tissues containing nerves, blood and lymph vessels. In duodenum, glands are also present in sub-mucosa. The innermost layer lining the lumen of the alimentary canal is the mucosa. This layer forms irregular folds (rugae) in the stomach and small finger¬like foldings called villi in the small intestine.
(d) Bile has no enzymes but contains bile salts, namely, sodium bicarbonate, sodium glycocholate and sodium taurocholate that reduce the surface tension of large fat droplets and break them into many small droplets by a process known as emulsification. These small fat droplets present large surface area for lipase (fat digesting enzyme) to act upon them. Moreover, bile also activates lipases.

4. State the role of pancreatic juice in digestion of proteins.
Solution: The pancreatic juice contains inactive enzymes – trypsinogen, chymotrypsinogen, procarboxypeptidases. Trypsinogen is acti¬vated by an enzyme enterokinase, (secreted by the intestinal mucosa) into active trypsin, which in turn activates the other enzymes of the pancreatic juice. Proteins, proteoses and peptones (partially hydrolysed proteins) in the chyme reaching the intestine are acted upon by these proteolytic enzymes of pancre¬atic juice.

5. Describe the process of digestion of protein in stomach.
Solution: The gastric glands of the stomach secrete gastric juice that contains HCl and proenzymes – pepsinogen and prorennin. The proenzyme pepsinogen, on exposure to HCl gets converted into the active enzyme pepsin, the proteolytic enzyme of stomach. The pepsin converts proteins into proteoses and peptones (peptides). Prorennin is found in gastric juice of infants and is activated by pepsin into active rennin. It helps in digestion of milk protein casein.

6. Give the dental formula of human beings.
Solution: The dental formula of human beings is

It shows arrangement of teeth in each half of
the upper and lower jaw.

7. Bile juice contains no digestive enzymes, yet it is important for digestion. Why ?
Solution: Bile has no enzymes but contains bile salts, namely, sodium bicarbonate, sodium glycocholate and sodium taurocholate that reduce the surface tension of large fat drop¬lets and break them into many small droplets by a process known as emulsification. These small fat droplets present large surface area for lipase (fat digesting enzyme) to act upon them. Moreover, bile also activates lipases.

8. Describe the digestive role of chymotrypsin. Which two other digestive enzymes of the same category are secreted by its source gland ?
Solution: Chymotrypsin is a proteolytic enzyme of pancreatic juice secreted by exocrine part of pancreas. It helps in digestion of proteins. It converts proteins, peptones and proteoses into oligopeptides and dipeptides. Two other proteolytic enzymes present in pancreatic juice are trypsinogen and procarboxypeptidase.

9. How are polysaccharides and disaccharides digested ?
Solution: Digestion of polysaccharides (starch and glycogen) starts from buccal cavity. In buccal cavity, polysaccharides are acted upon by salivary amylase or ptyalin which splits starch and glycogen into disaccharides and small dextrins called ‘a’ dextrin.

The digestion of carbohydrates does not occur in stomach because gastric juice itself has no carbohydrase.
In small intestine, the food mixes with two juices, pancreatic juice and intestinal juice. Pancreatic juice contains a carbohydrase named pancreatic amylase. This enzyme hydrolyses more starch and glycogen.

Intestinal juice contains carbohydrases; maltase, isomaltase, a-dextrinase, sucrase and lactase which act on disaccharides as follows:

fructose and galactose are monomers of carbohydrates. These are absorbed by intestinal mucosa.

10. What would happen if HCl were not secreted in the stomach?
Solution: HCl is secreted by parietal or oxyntic cells of gastric glands. It serves the following functions:

It activates the pepsinogen and prorennin into their active form pepsin and rennin.
It provides the acidic pH (pH 1.8) optimal for pepsin.
It kills the harmful bacteria present in the food.
It stops the action of saliva on food. Pepsin and rennin are the principle proteolytic enzymes of stomach. If these enzymes are not activated by HCl then digestion of protein will not take place in stomach, and also the harmful bacteria can cause various diseases.

11. How does butter in your food get digested and absorbed in the body ?
Solution: Butter is a saturated fat. Fats and oils of the ingested food are triglycerides.
They are digested by lipases. Small intestine is the principal organ for fat digestion.
In the small intestine food meets three secretions, bile, pancreatic juice and intestinal juice, all alkaline in nature.
Bile contains no enzyme but it contains bile salts which reduces the surface tension of large fat droplets and breaks them into smaller ones (emulsification).

Emulsified triglycerides Pancreatic juice contains pancreatic lipase, which is the principal fat digesting enzyme. It is activated by bile.

Fatty acid + Glycerol Intestinal lipase found in intestinal juice hydrolyses some triglycerides, diglycerides and monoglycerides to fatty acids and glycerol like pancreatic lipase.
Fatty acids, glycerol and monoglycerides are the end products of fat digestion and being insoluble in water cannot be directly absorbed from the intestinal contents. So they combine with the bile salts and phospholipids to form micelles (water soluble). From the micelles fatty acids, glycerides, sterols and fat soluble vitamins are absorbed into the intestinal cells by diffusion where they are resynthesised in the ER and are converted into very small protein coated fat molecules (droplets) called chylomicrons. The latter are released from the intestinal cells into the lymph present in the lymphatic capillaries, the lacteals. These lacteals ultimately release the absorbed substances into the blood stream.

12. Discuss the main steps in the digestion of proteins as the food passes through different parts of the alimentary canal.
Solution: Proteins of ingested food are broken down into amino acids by proteases (peptidases). Proteases are secreted in inactive forms called proenzymes which are converted into active forms at site of their action. Protein digestion starts in the stomach and is completed in the small intestine. Saliva contains no protease.
Digestion of proteins in stomach : Chief cells of gastric gland secrete pepsinogen and prorennin, which act as follows:

Digestion of proteins in small intestine: In small intestine, peptones and proteoses are acted upon by enzymes of pancreatic juice and intestinal juice.
Pancreatic juice contains 3 inactive proteases; trypsinogen, chymotrypsinogen and pro-carboxypeptidase. Their action is as follows:

Dipeptides + Amino acids Intestinal juice contains two digestive pro-teases; aminopeptidases and dipeptidases and a nondigestive enterokinase (enteropep- tidase).

Amino acids are the end products of protein digestion which are absorbed by intestinal cells.

13. Explain the term thecodont and diphyodont.
Solution: Thecodont: In human, each tooth is embedded in a socket of jaw bone. Such teeth are described as thecodont.
Diphyodont: Majority of mammals including human beings form two sets of teeth during their life, a set of temporary milk or deciduous teeth replaced by a set of permanent or adult teeth. This type of dentition is called diphyodont.

14. Name different types of teeth and their number in an adult human.
Solution: Adult human has 32 teeth with the

Human has heterodont dentition i.e., having four different types of teeth. The number of different types of teeth in human are as follows:incisors = 8, canines = 4, premolars = 8, molars = 12

15. What are the functions of liver?
Solution: Liver is the largest gland of the body and consists of hepatic cells. Besides being a digestive gland, the liver performs a number of functions for the welfare of body. Its varied functions are as follows

Secretion of bile.
Glycogenesis, gluconeogenesis and glycogenolysis.
Storage of fat, glycogen, vitamins like A, D, E, K and B₁₂, blood, water, etc.
Deamination of amino acids.
Synthesis of urea.
Elimination of excretory substances.
Detoxification of harmful substances.
Formation and breakdown of blood
corpuscles, i.e., in embryos, liver is haemopoietic (produces red blood corpuscles) and in adults its Kupffer cells phagocytise and destroy worn out and dead RBCs.
Secretion of blood proteins, i.e., prothrombin and fibrinogen.
Secretion of anticoagulant heparin.
Production of heat.
Secretion of enzymes.

NCERT Solutions For Class 11 Biology Respiration in Plants

April 4, 2023

NCERT Solutions For Class 11 Biology Respiration in Plants

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants:

Section Name	Topic Name
14	Respiration in Plants
14.1	Do Plants Breathe?
14.2	Glycolysis
14.3	Fermentation
14.4	Aerobic Respiration
14.5	The Respiratory Balance Sheet
14.6	Amphibolic Pathway
14.7	Respiratory Quotient
14.8	Summary

NCERT Solutions Class 11 Biology Biology Sample Papers

NCERT TEXTBOOK QUESTIONS FROM SOLVED

1. Give the schematic representation of an overall view of Krebs’ cycle.
Solution:

2. Differentiate between
(a) Respiration and Combustion
(b) Glycolysis and Krebs’cycle
(c) Aerobic respiration and Fermentation
Solution: (a) Differences between respiration and combustion are as follows :

(b) Differences between glycolysis and Krebs’ cycle are as follows:

(C)Differences between aerobic respiration and fermentation are as follows:

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3. What are respiratory substrates? Name the most common respiratory substrate.
Solution: Respiratory substrates are those organic substances which are oxidised during respiration to liberate energy inside the living cells. The common respiratory substrates are carbohydrates, proteins, fats and organic acids. The most common respiratory substrate is glucose. It is a hexose monosaccharide.

4. Give the schematic representation of glycolysis.
Solution:

5. Explain ETS.
Solution: An electron transport chain or system (ETS) is a series of coenzymes and cytochromes that take part in the passage of electrons from
a chemical to its ultimate acceptor. Reduced coenzymes participate in electron transport chain. Electron transport takes place on cristae of mitochondria [oxysomes ( F₀ -F₁ , particles) found on the inner surface of the membrane of mitochondria]. NADH formed in glycolysis and citric acid cycle are oxidised by NADH dehydrogenase (complex I) and the electrons are transferred to ubiquinone. Ubiquinone also receives reducing equivalents via FADH2 through the activity of succinate dehydrogenase (complex II). The reduced ubiquinone is then oxidised by transfer of electrons of cytochrome c via cytochrome Fc, complex (complex III). Cytochrome c acts as a mobile carrier between complex III and complex IV. Complex IV refers to cytochrome c oxidase complex containing cytochromes a and a₃and two copper centres. When the electrons are shunted over the carriers via complex I to IV in the electron transport chain, they are coupled to ATP synthetase (complex V) for the formation of ATP from ADP and Pi. Oxygen functions as the terminal acceptor of electrons and is reduced to water along with the hydrogen atoms. Reduced coenzymes (coenzyme I, II and FAD) do not combine directly with the molecular O₂. Only their hydrogen or electrons are transferred through various substances and finally reach O₂. The substances useful for the transfer of electron are called electron carriers. Only electrons are transferred through cytochromes (Cyt F₁ Cyt c,,C₂, a, a₃) and finally reach molecular O₂. Both cytochrome a and a3 form a system called cytochrome oxidase. Copper is also present in Cyt a₃ in addition to iron. The molecular oxygen that has accepted electrons now receives the protons that were liberated into the surrounding medium to give rise to a molecule of water. The liberated energy is utilised for the synthesis of ATP from ADP and Pi.

6. What are the main steps in aerobic respiration? Where does it take place?
Solution: Aerobic respiration is an enzymatically controlled release of energy in a stepwise catabolic process of complete oxidation of organic food into carbon dioxide and water with oxygen acting as terminal oxidant. It
occurs by two methods, common pathway and pentose phosphate pathway. Common pathway is known so because its first step, called glycolysis, is common to both aerobic and anaerobic modes of respiration. The common pathway of aerobic respiration consists of three steps – glycolysis, Krebs’ cycle and terminal oxidation. Aerobic respiration takes place within mitochondria. The final product of glycolysis, pyruvate is transported from the cytoplasm into the mitochondria.

7. What are the assumptions made during the calculation of net gain of ATP?
Solution: It is possible to make calculations of the net gain of ATP for every glucose molecule oxidised; but in reality this can remain only a theoretical exercise. These calculations can be made only on certain assumptions that:

There is a sequential, orderly pathway functioning, with one substrate forming the next and with glycolysis, TCA cycle and ETS pathway following one after another.
transferred into the mitochondria and undergoes oxidative phosphorylation.
None of the intermediates in the pathway are utilised to synthesise any other compound.
Only glucose is being respired – no other alternative substrates are entering in the pathway at any of the intermediary stages.

But these kind of assumptions are not really valid in a living system; all pathway work simultaneously and do not take place one after another; substrates enter the pathways and are withdrawn from it as and when necessary; ATP is utilised as and when needed; enzymatic rates are controlled by multiple means. Hence, there can be a net gain of 36 ATP molecules during aerobic respiration of one molecule of glucose.

8. Distinguish between the following:
(a) Aerobic respiration and Anaerobic respira¬tion.
(b) Glycolysis and Fermentation.
(c) Glycolysis and Citric acid cycle.
Solution: (a) Differences between aerobic and anaerobic respiration are as follows:

(b) Differences between glycolysis and fermentation are as follows:

9. Discuss “The respiratory pathway is an amphibolic pathway”.
Solution: Amphibolic pathway is the one which is used for both breakdown (catabolism) and build-up (anabolism) reactions. Respiratory pathway is mainly a catabolic process which serves to run the living system by providing energy. The pathway produces a number of intermediates. Many of them are raw materials for building up both primary and secondary metabolites. Acetyl CoA is helpful not only in Krebs’ cycle but is also raw material for synthesis of fatty acids, steroids, terpenes, aromatic compounds and carotenoids, a-ketoglutarate is organic acid which forms glutamate (an important amino acid) on amination. OAA (Oxaloacetic acid) on amination produces asparate. Both aspartate and glutamate are components of proteins. Pyrimidines and alkaloids are other products. Succinyl CoA forms cytochromes and chlorophyll.
Hence, fatty acids would be broken down to acetyl CoA before entering the respiratory pathway when it is used as a substrate. But when the organism needs to synthesise fatty acids, acetyl CoA would be withdrawn from the respiratory pathway for it. Hence, the respiratory pathway comes into the picture both during breakdown and synthesis of fatty acids. Similarly, during breakdown and synthesis of proteins too, respiratory intermediates form the link. Breaking down processes within the living organism is catabolism, and synthesis is anabolism. Because the respiratory pathway is involved in both anabolism and catabolism, it would hence be better to consider the respiratory pathway as an amphibolic pathway rather than as a catabolic one.

10. Define RQ. What is its value for fats?
Solution: Respiratory quotient (RQ) is the ratio of the volume of carbon dioxide produced to the volume of oxygen consumed in respiration over a period of time. Its value can be one, zero, more than 1 or less than one.

Volume of C02 evolved Volume of 02 consumed
RQ is less than one when the respiratory substrate is either fat or protein.
C₅₇ H₁₀₄O₆ + 80 O₂-» 57 CO₂+ 52H₂O
RQ = 57CO₂/80O₂ = 0.71
RQ is about 0.7 for most of the common fats.

11. What is oxidative phosphorylation?
Solution: Oxidative phosphorylation is the synthesis of energy rich ATP molecules with the help of energy liberated during oxidation of reduced co-enzymes (NADH, FADH₂) produced in respiration. The enzyme required for this synthesis is called ATP synthase. It is considered to be the fifth complex of electron transport chain. ATP synthase is located in FT or head piece of F₀ -F₁ or elementary particles. The particles are present in the inner mitochondrial membrane. ATP synthase becomes active in ATP formation only where there is a proton gradient having higher concentration of H⁺ or protons on the F₀ side as compared to F x side (chemiosmotic hypothesis of Peter Mitchell).
Increased proton concentration is produced in the outer chamber or outer surface of inner mitochondrial membrane by the pushing of proton with the help of energy liberated by passage of electrons from one carrier to another. Transport of the electrons from NADH over ETC helps in pushing three pairs of protons to the outer chamber while two pairs of protons are sent outwardly during electron flow from FADH_2. The flow of protons through the F₀ channel induces F₁ particle to function as ATP-synthase. The energy of the proton gradient is used in attaching a phosphate radical to ADP by high energy bond. This produces ATP. Oxidation of one molecule of NADH₂ produces 3 ATP molecules while a similar oxidation of FADH₂ forms 2 ATP molecules.

12. What is the significance of step-wise release of energy in respiration?
Solution: The utility of step-wise release of energy in respiration are given as follows :
(i) There is a step-wise release of chemical bond energy which is very easily trapped in forming ATP molecules.
(ii) Cellular temperature is not allowed to rise.
(iii) Wastage of energy is reduced.
(iv) There are several intermediates which can be used in production of a number of biochemicals.
(v) Through their metabolic intermediates different substances can undergo respiratory catabolism.
(vi) Each step of respiration is controlled by its own enzyme. The activity of different enzymes can be enhanced or inhibited by specific compounds.
This helps in controlling the rate of respiration and the amount of energy liberated by it.

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NCERT Solutions For Class 11 Biology Neural Control and Coordination

April 4, 2023

NCERT Solutions For Class 11 Biology Neural Control and Coordination

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 21 Neural Control and Coordination:

Section Name	Topic Name
21	Neural Control and Coordination
21.1	Neural System
21.2	Human Neural System
21.3	Neuron as Structural and Functional Unit of Neural System
21.4	Central Neural System
21.5	Reflex Action and Reflex Arc
21.6	Sensory Reception and Processing
21.7	Summary

NCERT Solutions Class 11 Biology Biology Sample Papers

NCERT TEXTBOOK QUESTIONS FROM SOLVED

1. Briefly describe the structure of the following:
(a) Brain (b) Eye (c) Ear
Solution: (a) Brain: The brain acts as control and command system of the body. It is protected by skull and is covered by three meninges. It is divisible into three main regions: forebrain, midbrain and hindbrain.
(i) Forebrain – It consists of three regions:
(a) Olfactory lobes: These are a pair of very small, solid club-shaped bodies which are widely separated from each
other. They are fully covered by cerebral hemispheres.
(b) Cerebrum – It is the largest and most complex of all the parts of human brain. A deep cleft divides the cerebrum into right and left cerebral hemispheres, connected by myelinated fibres, the corpus callosum.
(c) Diencephalon – It encloses a slit-like cavity, the third ventricle. The thin roof of this cavity is known as the epithalamus, the thick right and left sides as the thalami, and floor as the hypothalamus.
(ii) Midbrain – It is located between thalamus/ hypothalamus of forebrain and pons of hindbrain. Its upper surface has two pairs of rounded protrusious called corpora quadrigemina and two bundles of fibres called crura cerebri.
(iii) Hindbrain – It consists of:
(a) Cerebellum – The second largest part of the human brain is the cerebellum. It consists of two lateral cerebellar hemispheres and central worm-shaped part, the vermis. The cerebellum has its grey matter on the outside, comprising three layers of cells and fibres. It also has Golgi cells, basket cells and granule cells.
(b) Pons varolii – An oval mass, called the pons varolii, lies above the medulla oblongata. It consists mainly of nerve fibres which interconnect different regions of the brain.
(c) Medulla oblongata – It extends from the pons varolii above and is continuous with the spinal cord below. The mid brain, pons varolii and medulla oblongata are collectively called brain stem.

(b) Eye: Eye is a hollow spherical structure composed of three coats:
– Outer fibrous coat
– Middle vascular coat
– Inner nervous coat
(i) Fibrous coat: It is thick and protects the eyeball. It has two distinct regions – sclera and cornea. Sclera covers most of the eye ball. The sclera or white of the eye contains many collagen fibres. Cornea is a transparent portion that forms the anterior one – sixth of the eyeball. The cornea is avascular (i.e., lacks blood supply).
(ii)Vascular coat: It comprises of 3 regions : choroid, iris, ciliary body.
(a) Choroid : It lies adjacent to sclera and contains numerous blood vessels and pigmented cells.
(b) Iris: The iris is a circular muscular diaphragm containing the pigment giving eye its colour. It extends from the ciliary body across the eyeball in front of the lens. It 2. has an opening in the centre called the pupil.
It contains two types of smooth muscles, circular muscles (sphincters) and radial muscles (dilators), of ectodermal origin.
(c) Ciliary body: Behind the peripheral margin of the iris, the vascular coat is thickened to form the ciliary body. It is composed of the ciliary muscles and the ciliary processes.
(iii) Nervous coat: It consists of retina which is neural and sensory layer of an eye ball. It consists of three layers; ganglion cells, bipolar cells and photoreceptor cells (rods and cones).
Lens: It is a transparent, biconvex, elastic structure that bends light waves as they pass through its surface. It is composed of epithelial cells that have large amounts of clear cytoplasm in the form of fibres.
Chambers of eyeball: The lens, suspensory ligament and ciliary body divide the eye into an anterior aqueous chamber and a posterior vitreous chamber which are filled with aqueous humour and vitreous humour respectively.

(c) Ear: There are three portions in an ear:
(i) External ear: It further has 2 regions: pinna and external auditory canal or meatus.
(a) Pinna: The pinna is a projecting elastic cartilage covered with skin. Its most prominent outer ridge is called the helix. The lobule is the soft pliable part at its lower end composed of fibrous and adipose tissue richly supplied with blood capillaries. It is sensitive as well as effective in collecting sound waves.
(b) External auditory canal: It is an S-shaped tube leading inward from the pinna. It is a tubular passage supported by cartilage in its exterior part and by bone in its interior part.
(ii) Middle ear: It consists of 3 small bones called ear ossicles – malleus, incus and stapes, which are attached to one another and increase efficiency of transmission of sound waves to inner ear.
(iii) Internal ear: It consists of bony and

2. Compare the following:
(a) Central neural system (CNS) and Peripheral neural system (PNS).
(b) Resting potential and action potential.
(c) Choroid and retina.
Solution: (a) CNS: It lies along the mid-dorsal axis of the body. It is a hollow, dorsally placed structure and comprises of brain and spinal cord. It is a centre of information processing and control.
PNS: Nerves arising from the central nervous system constitute the peripheral nervous system. It carries information to and from the CNS. It includes spinal nerves and cranial nerves.
(b) Resting potential: Outside the plasma membrane of a nerve fibre is the extracellular fluid which is positively charged with respect to the cell contents inside the plasma membrane. A resting nerve fibre shows a potential difference between inside and outside of this plasma membrane. This difference in the electrical charges across the plasma membrane is called the ‘resting potential’. A membrane with resting potential across it, is said to be electrically polarized. Action potential : Action potential is another name of nerve impulse. The contents inside a cell at the excited state becomes positively charged with respect to extracellular fluid outside it. This change in polarity across the plasma membrane is known as an action potential. The membrane with reversed polarity across it is said to be depolarized.
(c) Choroid: Choroid lies adjacent to the sclera and contains numerous blood vessels that supply nutrients and oxygen to the other tissues especially of retina. It contains abundant pigment cells and is dark brown in colour.
Retina: It is the neural and sensory layer of the eye ball. It is a very delicate coat and lines the whole of the vascular coat. Its external surface is in contact with the choroid and its internal surface with vitreous humour. It contains ganglion cells, bipolar cells and photoreceptor cells. membranous labyrinth. Membranous labyrinth consists of three semicircular ducts, utricle, saccule and cochlea.

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3. Explain the following processes:
(a) Polarisation of the membrane of a nerve fibre.
(b) Depolarisation of the membrane of a nerve fibre.
(c) Conduction of a nerve impulse along a nerve fibre.
(d) Transmission of a nerve impulse across a chemical synapse.
Solution: (a) Polarisation of the membrane of a nerve fibre : In the resting (not conducting impulse) nerve fibre the plasma membrane separates two solution of different chemical composition but having approximately the same total number of ions. In the external medium (tissue fluid), sodium ions (Na⁺) and Cl^– ions predominate, whereas within the fibre (intracellular fluid) potassium ions (K⁺) predominate. The differential flow of the positively charged ions and the inability of the negatively charged organic (protein) ions within the nerve fibre to pass out cause an increasing positive charge on the outside of the membrane and negative charge on the inside of the membrane. This makes the membrane of the resting nerve fibre polarized, extracellular fluid outside being electropositive (positively charged) with respect to the cell contents inside it.
(b) Depolarisation of the membrane of a nerve fibre: During depolarisation, the activation gates of Na channels open, and the K channels remain closed. Na⁺ rush into the axon. Entry of sodium ions leads to depolarisation (reversal of polarity) of the nerve membrane, so that the nerve fibre contents become electropositive with respect to the extracellular fluid.
(c) Conduction of a nerve impulse along a nerve fibre: Nervous system transmits information as a series of nerve impulses. A nerve impulse is the movement of an action potential as a wave through a nerve fibre. Action potentials are propagated, that is, self-generated along the axon. The events that set up an action potential at one spot on the nerve fibre also transmit it along the entire length of the nerve fibre. The action potential then moves to the neighbouring region of the nerve fibre till it covers the whole length of the fibre.
(d) Transmission of a nerve impulse across a chemical synapse: 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. When an impulse (action potential) arrives at the axon terminal, it stimulates the movement of the synaptic vesicles towards the membrane where they fuse with the plasma membrane and burst to 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 which can generate a new potential in the post-synaptic neuron. The new potential developed may be either excitatory or inhibitory.

4. Draw labelled diagrams of the following:
(a) Neuron (b) Brain
(c) Eye (d) Ear
Solution: (a)

(b)

(c)

(d)

5. Write short notes on the following:
(a) Neural coordination (b) Forebrain
(c) Midbrain (d) Hindbrain
(e) Retina (f) Ear ossicles
(g) Cochlea (h) Organ of Corti
(i) Synapse
Solution: (a) Neural coordination : When higher animals respond to various stimuli, each response to a specific stimulus generally involves many organs (parts) of their bodies. Therefore, it is necessary that all the concerned organs (parts) of the body should work in a systematic manner to produce the response. The working together of various organs (parts) of the body of multicelullar organism in a proper manner to complement the functions of each other is called coordination. This is achieved by three overlapping processes of nervous system-sensory input, integration and motor output.
(b) Forebrain: It consists of: Olfactory lobes, the paired structures concerned with the sense of smell. Cerebrum which is the largest and most complex of all the parts of the human brain. It is divided by a cleft into left and right cerebral hemispheres which are connected by a large bundle of myelinated fibres the. corpus callosum. The outer cover of cerebral hemisphere is called cerebral cortex. It consists of sensory and motor areas. Hypothalamus region of forebrain contains centres which control body temperature, hunger and also contains group of neurosecretory cells.
(c) Midbrain: The midbrain is located between the thalamus/hypothalamus of the forebrain and pons of the hindbrain. A canal called the cerebral aqueduct passess through the midbrain. The dorsal portion of the midbrain consists mainly of four round swellings (lobes) called corpora quadrigemina. Midbrain and hindbrain form the brain stem.
(d) Hindbrain: The hindbrain comprises pons, cerebellum and medulla. Pons consists of fibre tracts that interconnect different regions of the brain. Cerebellum has very convoluted surface in order to provide the additional space of many more neurons. The medulla of the brain is connected to the spinal cord. The medulla contains centres which control respiration, cardiovascular reflexes and gastric secretions.
(e) Retina: Retina is the inner layer of an eye and it contains three layers of cells-from inside to outside – ganglion cells, bipolar cells and photoreceptor cells. There are two types of photoreceptor cells, namely, rods and cones. These cells contain the light-sensitive proteins called the photopigments. The daylight (photopic) vision and colour vision are functions of cones and the twilight (scotopic) vision is the function of the rods. The rods contain a purplish-red protein called the rhodopsin or visual purple, which contains a derivative of Vitamin A. In the human eye, there are three types of cones which possess their own characteristic photopigments that respond to red, green and blue lights. The sensations of different colours are produced by various combinations of these cones and their photopigments. When these cones are stimulated equally, a sensation of white light is produced.
(f) Ear ossicles : There is a small flexible chain of three small bones called as ear ossicles – the malleus (hammer shaped), the incus (anvil shaped) and the stapes (stirrup shaped) in the middle ear. Malleus is attached to the tympanic membrane on one side and incus on the other side. Incus in turn is connected with the stapes. Malleus is the largest ossicle, however stapes is the smallest ossicle.
(g) Cochlea : It is the main hearing organ which is connected with saccule. It is a spirally coiled tube that resembles a snail shell in appearance. It tapers from a broad base to an almost pointed apex.
(h) Organ of Corti: It is a structure located on the basilar membrane which contains hair cells that act as auditory receptors. The hair cells are present in rows on the internal side of the organ of Corti.
(i) Synapse : It is the junction between the axon of one neuron and the dendrite or cyton of another neuron for transmission of nerve impulse.

6. Give a brief account of
(a) Mechanism of synaptic transmission.
(b) Mechanism of vision.
(c) Mechanism of hearing.
Solution: (a) Mechanism of synaptic transmission: Refer answer 3 (d)
(a) Mechanism of vision: The light rays in visible wavelength focused on the retina through the cornea and lens generate potentials (impulses) in rods and cones. Light induces
dissociation of the retinal from opsin resulting in changes in the structure of the opsin. This causes membrane permeability changes. As a result, potential differences are generated in the photoreceptor cells. This produces a signal that generates action potentials in the ganglion cells through the bipolar cells. These action potentials (impulses) are transmitted by the optic nerves to the visual cortex area of the brain, where the neural impulses are analysed and the image formed on the retina is recognised based on earlier memory and experience.
(b) Mechanism of hearing : The external ear receives sound waves and directs them to the ear drum. The ear drum vibrates in response to the sound waves and these vibrations are transmitted through the ear ossicles (malleus, incus and stapes) to the oval window. The vibrations are passed through the oval window on to the fluid of the cochlea, where they generate waves in the lymphs. The waves in the lymphs induce a ripple in the basilar membrane. These movements of the basilar membrane bend the hair cells, pressing them against the tectorial membrane. As a result, nerve impulses are generated in the associated afferent neurons. These impulses are transmitted by the afferent fibres via auditory nerves to the auditory cortex of the brain, where the impulses are analysed and the sound is recognised.

7. Answer briefly.
(a) How do you perceive the colour of an object?
(b) Which part of our body helps us in maintaining the body balance?
(c) How does the eye regulate the amount of light that falls on the retina?
Solution: (a)In humans, colour vision results from the activity of cone cells, a type of photoreceptor cells. In the human eye, there are three types of cones which possess their own characteristic photopigments that respond to red, green and blue lights. The sensations of different colours are produced by various combinations of these cones and their photopigments. When these cones are stimulated equally, sensation of white light is produced. Yellow light, for instance, stimulates green’and red cones approximately to equal extent, and this is interpreted by the brain as yellow colour.
(b) Ears (cristae and maculae present in internal ears).
(c) The iris contains two sets of smooth muscles – sphincters and dilators. These muscles regulate the amount of light entering the eyeball by varying the size of pupil. Contraction of sphincter muscles makes the pupil smaller in bright light so that less light enters the eye. Contraction of dilator muscles widens the pupil in dim light so that more light goes in eye to fall on retina.

8. Explain the following.
(a) Role of Na⁺ in the generation of action potential.
(b) Mechanism of generation of light-induced impulse in the retina.
(c) Mechanism through which a sound produces a nerve impulse in the inner ear.
Solution: (a) The action potential is largely determined by Na⁺ ions. The action potential results from the following sequential events
(i) Disturbance caused to the membrane of a nerve fibre by a stimulus results in leakage of Na⁺ into the nerve fibre.
(ii) Entry of Na⁺ lowers the trans-membrane potential difference.
(iii) Decrease in potential difference makes the membrane more permeable to Na⁺ than to K⁺ ions so that more Na⁺ enter the fibre than K⁺ leave it.
(iv) Accumulation of Na⁺ in the nerve fibre initiates depolarisation (action potential), making the axonic contents positively charged relative to the extracellular fluid.
(v) With continued addition of Na⁺ the potential reaches zero and then plus 40-50 millivolts. This is the peak of action potential.
(vi) Permeability of a depolarised membrane to Na⁺ then rapidly drops, there are now as many Na⁺ on the inside of the membrane as on the outside.
(b) Refer answer 6 (b)
(c) Refer answer 6 (c)

9. Differentiate between
(a) Myelinated and non-myelinated axons
(b) Dendrites and axons
(c) Rods and cones
(d) Thalamus and Hypothalamus
(e) Cerebrum and Cerebellum
Solution: (a) Differences between myelinated and non-myelinated axons are as follows:

(b) Axon and dendrites can be differentiated as follows:

(c) The differences between rods and cones are as follows:

(d) Thalamus and hypothalamus can be differentiated as follows:

(e) Cerebrum and cerebellum can be differentiated as follows:

10. Answer the following.
(a) Which part of the ear determines the pitch ofa sound?
(b) Which part of the human brain is the most developed?
(c) Which part of our central neural system acts as a master clock?
Solution: (a) The receptor cells in the organ of Corti (Internal ear).
(b) Cerebrum (cerebral hemispheres).
(c) Pineal gland present in diencephalon of forebrain acts as a master clock, which maintains biological rhythm.

11. The region of the vertebrate eye, where the optic nerve passes out of the retina, is called the
(a) fovea (b) iris
(c) blind spot (d) optic chiasma
Solution: (c) blind spot

12. Distinguish between
(a) Afferent neurons and efferent neurons
(b) Impulse conduction in myelinated nerve fibre and unmyelinated nerve fibre
(c) Aqueous humour and vitreous humour
(d) Blind spot and yellow spot
(e) Cranial nerves and spinal nerves
Solution: (a)

(b) Refer answer 9(a)
(c)

(d)

(e)

NCERT Solutions For Class 11 Biology Plant Kingdom

April 3, 2023

NCERT Solutions For Class 11 Biology Plant Kingdom

Topics and Subtopics in NCERT Solutions for Class 11 Biology Chapter 3 Plant Kingdom:

Section Name	Topic Name
3	Plant Kingdom
3.1	Algae
3.2	Bryophytes
3.3	Pteridophytes
3.4	Gymnosperms
3.5	Angiosperms
3.6	Plant Life Cycles and Alternation of Generations
3.7	Summary

NCERT Solutions Class 11 Biology Biology Sample Papers

NCERT TEXTBOOK QUESTIONS SOLVED

1.What is the basis of classification of algae?
soln. Fritsch (1935), has classified algae considering phylogeny, affinities and inter-relationships of various forms. He classified algae mainly on the basis of the characters like structure of plant body, nature of the pigments, reserve food material, number and position of flagella, chemistry of cell wall and methods of reproduction etc. Algae is divided into 11 classes but among them 3 main classes are Chlorophyceae, Phaeophyceae and Rhodophyceae.

2.When and where does reduction division take place in the life cycle of a liverwort, a moss, a fern, a gymnosperm and an angiosperm?
soln.All of these plants show life cycle with one gametophytic (n) generation and one sporophytic (2n) generation. Reduction division or meiosis that produces haploid (n) cells from diploid cells (2n) is necessary in their life cycles to restore gametophyte generation after sporophytic generation. It occurs in different body structures according to the basic body design of these groups. Reduction division in a liverwort and moss takes place at the end of the sporophytic generation, where haploid spores are formed by reduction division of spore mother cell inside capsule. Spores germinate to produce dominant gametophytic generation. Reduction division in fern takes place at the end of the dominant sporophytic generation inside the sporangium from spore mother cell by reduction division. Spores may be of one type (homospory) or of two types (heterospory).
Reduction division in gymnosperms takes place at the end of dominant sporophytic generation. Megaspore and microspores are produced by the reduction division of diploid megaspore mother cell and diploid microspore mother cell respectively, inside megasporangium and microsporangium. Reduction division in angiosperms takes place at the end of dominant sporophytic generation. The haploid pollen grain or microspore and the haploid egg cell are produced by the reduction division of diploid (microspore) mother cell and diploid megaspore mother cell respectively. Microsporic division occurs inside anther and megasporic division occurs inside gynoecium (ovary).

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3.Name three groups of plants that bear archegonia. Briefly describe the life cycle of any one of them.
soln. The three groups of plants that bear archegonia are bryophytes, pteridophytes and gymnosperms.
Life cycle of a bryophyte is as follows : The main plant body of bryophyte is gametophytic (n), which is independent and may be thallose (no differentiation in root, stem, leaves) e.g., Riccia, or may be foliose (having leafy axis) e.g., Funaria. The dominant phase in the life cycle of Funaria is the gametophyte, which occurs in two stages, the protonema stage and the erect, leafy gametophytic plant.
The leafy gametophyte consists of an upright, slender axis (stem-like) that bears spirally arranged leaves and is attached to the substratum by multicellular, branched rhizoids. Vegetative reproduction takes place
by fragmentation; by the buds formed in secondary protonema etc. The sex organs, antheridia and archegonia are produced in dusters at the apices of the leafy shoots. Antheridia produces antherozoids and archegonia produces egg. Antherozoid (male gamete) and egg (female gamete) fuses and form zygote.Zygote develops into a sporophyte; which is differentiated into foot, seta and capsule and spores are produced in the capsule.
Spores on reaching a suitable substratum germinate to produce a filamentous juvenile stage, .called the primary protonema, which later produces secondary protonema that forms erect leafy plants.

4.Mentiontheploidyofthefollowingrprotonemal cell of a moss; primary endosperm nucleus in dicot, leaf cell of a moss; prothallus cell of a fern; gemma cell in Marchantia; meristem cell of monocot, ovum of a liverwort, and zygote of a fern.
soln. Protonemal cell of a moss – haploid. Primary endosperm nucleus in dicot – triploid.
Leaf cell of a moss – haploid.
Prothallus cell of a fern – haploid.
Gemma cell in Marchantia – haploid. Meristem cell of monocot – diploid.
Ovum of a liverwort – haploid.
Zygote of a fern – diploid.

5.Write a note on economic importance of algae and gymnosperms.
soln. Economic importance of algae is as follows:
The group Algae plays both economically beneficial as well as harmful roles.
Beneficial importance :
(i) People of coastal countries have been using sea weeds & certain other algae as source of food, e.g., Porphyra, Ulva, Laminaria, etc.
(ii)Some algae are used as food for marine as well as domestic animals, e.g., Sargassum, Macrocystis.
(iii)Algae are useful source of many commercial products like agar, a jelly like substance (complex polysaccharide) is extracted form species of red algae belonging to the genera Gelidium, Gracilaria etc. Agar is also used as base in culture media. Carrageenin occurs as a cell wall polysaccharide, esterified with sulphate. It is extracted from red alga like Chondrus Crispin, etc. is used in pharmaceutical emulsifier and textile, leather, cosmetic industries. Alginates are salts of alginic acid found in the cell wall of phaeophyceae (brown algae) like Fucus, Laminaria etc.
(iv)Algae are also useful in medicine industry. Antibiotic chlorellin is obtained from Chlorella. Extracts of Cladophora, Lyngbya kill strains of Pseudomonas and Mycobacterium like bacteria. Nitella is used to destroy mosquitoes growth in ponds and hence used in control of malaria.
(v) Some algae are used in agriculture like Nostoc, Anabaena etc. are used to convert atmospheric N2 into nitrogenous compounds which are absorbed by higher plants. Some sea weeds like Fucus, Litlwphyllum, Lycophyllum etc. are rich io K, P, trace elements and growth substances and are used as fertilisers by coastal people.
(vi)Some algae like Chlorella, Chlamydomonas, etc. are used in sewage disposal in ponds. These algae help in bacterial decomposition by providing 02.
(vii)Some algae like Chlorella, Synecoccus, etc are used in space travels. A person inside a spaceship will need a device to get rid of C02 and other body wastes and will require sources of Oz and food. These algae are very useful for this purpose.
(viii)A large amount of iodine (mineral element present in thyroxine hormone of thyroid gland) is extracted from kelps (brown sea weeds or members of phaeophyceae) like Laminaria, Fucus, Ascophyllum etc. Similarly red algae like Rhodomela, Polysiphonia, Rhodymenia are sources of bromine.
Harmful importance:
(i) Some blue green and green algae like Chroococcus, Oscillatoria grow over the surface of water bodies in abundance and cause water bloom. On death and decay these algae give off bad smell. Some algae secrete poisonous or toxic substances.
(ii)Parasitic algae like Cephaleuros virescens causes red rust of tea, coffee etc.
Economic importance of gymnosperms is as follows:
(i)Some species of Cycas like C. revoluta, C. rumphii look like palm tree and are used for decoration purposes as they remain fresh for long period.
(ii)Stem portion of Cycas revoluta is a good source of ‘sago’, a kind of starch used in making bread by poor people. Seeds of some species of Cycas are roasted and taken as food. Young succulent leaves of some species of Cycas are cooked as vegetable.
(iii)Many gymnospersm have medicinal value. The fresh juice extracted from the Cycas circinalis leaves is used as medicine for stomach disorders, blood vomiting and other skin diseases. Pollen grains of some Cycas plants are reported to have some narcotic effect.
(iv)Some gymnosperms like Pinus, Abies, Cedrus are the chief source of various types of woods. The wood of Juniperus is used in making pencils, scales, holders etc.
(v)Some species of Pinus is a good source of turpentine, wood gas, wood alcohol.

6.Both gymnosperms and angiosperms bear seeds, then why are they classified separately?
soln.’Gymnosperms and angiosperms both bear seeds but they are classified separately because gymnosperms are a group of plants in which the ovules are freely exposed on open megasporophylls, whereas in angiosperms the seeds or ovules are enclosed within ovary which later forms the fruit.

7.What is heterospory? Briefly comment on its significance. Give two examples.
soln. The occurrence of two kinds of spores in the same plant is called as heterospory. Among them the smaller spore is called microspore and the larger spore is called megaspore. Heterospory first evolved in pteridophytes. Significance of heterospory
(i) Heterospory is associated with the sexual differentiation of gametophyte /.<?., a microspore develops into a male gametophyte whereas a megaspore develops into a female gametophyte.
(ii)In homosporous pteridophytes spores have to germinate on soil thus face more environmental problems. In heterosporous pteridophytes, spores germinate within the sporangium and the gametophytes are retained inside for variable periods of time. Hence, germinating gametophyte has better chances of survival. This lays the foundation of complete retention of gametophytes within sporophytes in angiosperms and gymnosperms.
(iii)Heterospory is the basis of development of seed habit in higher plants.

8.Explain briefly thefollowing terms with suitable examples.
(i) Protonema (ii) Antheridium
(iii)Archegonium (iv) Diplontic (v) Sporophyll (vi) Isogamy
soln. (i) Protonema : It is the first, usually branched, green and filamentous structure produced by a germinating moss or fern spore. The protonema of mosses bears buds that develop into the gametophyte plant. In fern the protonema becomes the prothallus.
(ii)Antheridium : The male sex organ of cryptogams (algae, fungi, bryophytes and
pteridophytes) is known as antheridium. It produces the male gametes or anthero- zoids. It may consist of a single cell or it may have a wall that is made up of one or several layers forming a sterile jacket around the developing gametes.
(iii)Archegonium : The multicellular flask shaped female sex organ of bryophytes, pteridophytes and many gymnosperms is known as archegonium. Its dialated base called the venter contains the female gamete or egg or oosphere. The cells of the narrow neck of archegonium liquify to allow the male gametes to swim towards the oosphere.
(iv)Diplontic : It is the kind of life cycle in which the diploid sporophyte is dominant and this diploid phase is photosynthetic. The gametophytic phase is represented either by gametes only, that are formed through meiosis or by a highly reduced few celled gametophyte. E.g., all seed-bearing plants (gymnosperms and angiosperms).
(v) Sporophyll : It is a type of leaf bearing sporangia. In ferns, the sporophylls are the normal foliage leaves, but in other plants the sporophylls are modified and arise in specialised structure such as the strobili of club-moss, gymnosperms and the flower of angiosperms. In most plants sporophylls are of two types – microsporophylls and megasporophylls.
(vi)Isogamy: It is a type of sexual reproduction where fusion takes place between two identical gametes. The gametes are similar in size and structure and they show equal motility during sexual reproduction, e.g., Spirogyra (algae).

9.Differentiate between the following:
(i) Red algae and brown algae
(ii)Homosporous and heterosporous pteridophytes
(iii)Liverworts and moss
(iv)Syngamy and triple fusion.
soln.(i) The differences between red algae and brown algae are as follows :

(ii) The differences between homosporous and heterosporous pteridophytes are as follows:

(iii) The differences between liverworts and mosses are as follows :

(iv) The differences between syngamy and triple fusion are as follows :

10.How would you distinguish monocots from dicots?
Soln. Differences between monocots and dicots are as follows :

11.Match the following (Column I with Column II).
Column I Column II
(a)Chlamydomonas (i)Moss
(b)Cycas (ii)Pteridophyte
(c) Selagmella (iii)Algae
(d) Sphagnum (iv)Gymnosperm
Soln.
Chlamydomonas – Algae
Cycas – Gymnosperm
Selayinella – Pteridophyte
Sphagnum – Moss
12.Describe the important characteristics of gymnosperms.
Soln.The term gymnosperm is derived from two Greek words: Gymnos = naked + Sperma = seed, i.e., naked seeded plants. So gymnosperms are a group of plants inwhich the ovules are freely exposed on open megasporophylls. The important characteristics of gymnosperms are :
– Living gymnosperms are perennial and vary from predominantly medium – sized trees (Cycas) to tall trees (Pinus) and shrubs (Ephedra).
– Plants possess tap root system. Some genera possess symbiotic relationship of N2 fixing algae in coralloid roots (Cycas) and fungi in mycorrhizal roots (Pinus).
– The stems are aerial, erect, branched (unbranched in Cycas) and woody.
– The leaves may be simple or compound. They are scaly and foliage also. Leaves are well adapted to withstand extremes of temperature, humidity and wind.
-Roots are characterised by the presence of diarch to polyarch vascular bundles. Xylem is exarch.
-Stems are provided with collateral, endarch and open vascular bundles which are arranged in a ring. Secondary growth is present and annual rings are formed.
-Xylem contains xylem parenchyma and tracheids with bordered pits and vessels are absent (except in Gnetum; Ephedra and Wehmtschia).
-Phloem contains sieve cells and phloem parenchyma and companion cells are absent (except in Gnetum; Ephedra and Weluhtschia).
-Leaves are protected by thick layers of cuticle. Sunken stomata are present. Mesarch xylem and transfusion tissues are found in the leaves. Palisade tissue and spongy parenchyma may be present in mesophyll or it may be undifferentiated.
-The reproductive organs form cones or strobilus except female organs of Cycas.
-The male cone is made of overlapping microsprophylls, that bear micros¬porangia on the abaxial side which produce microspores.
-Female cone is formed by overlapping megasporophylls which bear ovules (megasporangia).
-Ovule is orthotropous, unitegmic with 3 layers i.e. outer fleshy, middle stony and inner fleshy.
– The nucellus of ovule contains single megaspore mother cell which undergoes reduction division to form 4 megaspores, out of which 3 degenerate and only one survives.
– So gymnosperm is heterosporous i.e. producing microspores and megaspores.
– Single megaspore forms haploid female
gametophyte or endosperm before fertilisation. .
– At micropylar end of female gametophyte 2 or more archegonia are produced. Archegonium is with reduced neck (with no neck canal cell).
– Microspores are released from microsporangium and are carried in air currents and come in contact with the micropyle of the ovules.
– Pollen tube carrying the male gametes grows towards archegonia and discharges its contents near the mouth of the archegonia.
– After fertilisation zygote or oospore gives rise to embryo proper and the ovules develop into seeds.
– Polyembryony i.e., development of more than one embryo is an usual feature of gymnosperms but only one of them survives at later stage.
– In embryo 2 or many cotyledons are present.
– The seeds of gymnosperms are uncovered.

NCERT Solutions For Class 11 Biology Transport in Plants

March 18, 2023

NCERT Solutions For Class 11 Biology Transport in Plants

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

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

NCRT TEXTBOOK QUESTIONS SOLVED

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

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

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

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

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

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

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

7. (a) With fhe help of well-labelled diagrams, describe the process of plasmolysis in plants, giving appropriate examples.
(b) Explain what will happen to a plant cell if it is kept in a solution having higher water potential.
Soln.(a) Shrinkage of the protoplast of a cell from its cell wall under the influence of a hypertonic solution is called plasmolysis. Hypertonic solution causes exosmosis or withdrawal of water from cytoplasm and then the central vacuole of cell. The size of cytoplasm, as well as central vacuole and hence protoplast, becomes reduced. The first stage of plasmolysis is called limiting plasmolysis. At limiting plasmolysis, the pressure potential ( ψ_p) is zero and the osmotic concentration of cejl interior is just equivalent to that of external solution (isotonic). The cell is called flaccid. When pressure potential becomes negative, the protoplast withdraws itself from the comers. This stage is known as incipient plasmolysis. At incipient plasmolysis, the cell wall exerts no pressure on the cell contents (i.e. ψ_p is zero). Hence at this stage ψ_w= ψ_s. The hypertonic solution now enters the cell in between the protoplast and the cell wall. Due to continued exosmosis, protoplast shrinks further and withdraws from the cell wall except one or a few points. It is known as evident plasmolysis.

Examples of plasmolysis :
(i) Pickles, meat and fish are preserved by salting. Similarly, jams and jellies are preserved by sweetening with sugars. Salting and sweetening create hypertonic condition in which the fungi and bacteria get killed by plasmolysis.
(ii)Salting kills the weeds of lawns by inducing plasmolysis in their cells.
(iii)Plasmolytic method is applied for the determination of osmotic pressure of a cell in the laboratory.
(b) When the cells are placed in a solution having higher water potential i.e., hypotonic solution (dilute solution as compared to the cytoplasm), water diffuses into the cell causing the cytoplasm to build up a pressure against the wall, that is called turgor pressure. The pressure exerted by the protoplasts due to entry of water against the rigid walls is called pressure potential ψ_p. Because of the rigidity of the cell wall, the cell does not rupture. This turgor pressure is ultimately responsible for – enlargement and extension of cells.

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

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

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

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

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

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

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

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

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

(b) Differences between transpiration and evaporation are as follows:

(c)Differences between osmotic pressure and osmotic potential are as follows:

(d) Differences between imbibition and diffusion are as follows:

(e) Differences between apoplast pathway and symplast pathway are as follows:

(f) Differences between guttation and transpiration are as follows:

NCERT Exemplar Class 11 Biology Chapter 9 Biomolecules

February 27, 2023

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

NCERT Exemplar Class 11 Biology Chapter 9 Biomolecules

Multiple Choice Questions

Q1. It is said that elemental composition of living organisms and that of inanimate objects (like earth’s crust) are similar in the sense that all the major elements are present in both. Then what would be the difference between these two . groups? Choose a correct answer from the following.
(a) Living organisms have more gold in them than inanimate objects
(b) Living organisms have more water in their body than inanimate objects
(c) Living organisms have more carbon, oxygen and hydrogen per unit mass than inanimate objects
(d) Living organisms have more calcium in them than inanimate objects.
Ans: (c)

Element	% Weight of
Element	Earth’s Crust	Human Body
Hydrogen (H)	0.14	0.5
Carbon (C)	0.03	18.5
Oxygen(0)	46.6	65.0
Nitrogen (N)	very little	3.3
Sulphur (S)	0.03	0.3
Sodium (Na)	2.8	0.2
Calcium (Ca)	3.6	1.5
Magnesium (Mg)	2.1	0.1
Silicon (Si)	27.7	negligible

Q2. Many elements are found in living organisms either free or in the form of compounds. One of the following is not found in living organisms.
(a) Silicon (b) Magnesium (c) Iron (d) Sodium
Ans: (a) See Answer 2.
Q3. Aminoacids, as the name suggests, have both an amino group and a carboxyl group in their structure. In addition, all naturally occurring aminoacids (those which are found in proteins) are called L-aminoacids. From this, can you guess from which compound can the simplest aminoacid be made?
(a) Formic acid (b) Methane (c) Phenol acid (d) Glycine
Ans: (d) Glycine is an amino acid (which have both an amino group and a carboxyl group in their structure).

Q4. Many organic substances are negatively charged, e.g., acetic acid, while others are positively charged e.g., ammonium ion. An aminoacid under certain conditions would have both positive and negative charges simultaneously in the same molecule. Such a form of aminoacid is called
(a) Positively charged form (b) Negatively charged form
(c) Neutral form (d) Zwitterionic form
Ans: (d) In aqueous solution, the carboxyl group can lose a proton and amino group can accept a proton, giving rise to a dipolar ion called Zwitter ion. Zwitter ion is neutral but contains both positive and negative charges.

Q5. Sugars are technically called carbohydrates, referring to the fact that their formulae are only multiple of C(H20). Hexoses therefore have six carbons, twelve hydrogens and six oxygen atoms. Glucose is a hexose. Choose from among the following another hexose.
(a) Fructose (b) Erythrose ~(c) Ribulose (d) Ribose
Ans: (a) Sugars are technically called carbohydrates, referring to the fact that their formulae are only multiple of C(H20). Hexoses therefore have six carbons, twelve hydrogens and six oxygen atoms. E.g., glucose and fructose.

Q6. When you take cells or tissue pieces and grind them with an acid in a mortar and pestle, all the small biomolecules dissolve in the acid. Proteins polysaccharides and nucleic acids are insoluble in mineral acid and get precipitated. The acid soluble compounds include amino acids, nucleosides, small sugars etc. When one adds a phosphate group to a nucleoside one gets another acid soluble biomolecule called
(a) Nitrogen base
(b) Adenine
(c) Sugar phosphate
(d) Nucleotide
Ans: (d) Neucliotide = base + sugar + phosphate

Q7. When we homogenise any tissue in an acid, the acid soluble pool represents
(a) Cytoplasm (b) Cell membrane
(c) Nucleus (d) Mitochondria
Ans: (a) When we homogenise any tissue in an acid, the acid soluble pool represents cytoplasm.

Q8. The most abundant chemical in living organisms could be
(a) Protein (b) Water (c) Sugar (d) Nucleic acid
Ans: (b) Most abundant component of cell is water.

Component	% of the total Cellular Mass
Water	70-90
Proteins	10-15
Nucleic acids	5-7
Carbohydrates	3
Lipids	2
Ions	1

Q9. A homopolymer has only one type of building block called monomer repeated V number of times. A heteropolymer has more than one type of monomer. Proteins are heteropolymers usually made of aminoacids. While a nucleic acid like DNA or RNA is made up of only 4 types of nucleotide monomers, proteins are made of
(a) 20 types of monomers (b) 40 types of monomers
(c) 30 types of monomers (d) only one type of monomer
Ans: (a) A homopolymer has only one type of building block called monomer repeated V number of times. A heteropolymer has more than one type of monomer. Proteins are heteropolymers usually made of amino acids. While a nucleic acid like DNA or RNA is made of of only 4 types of nucleotide monomers, proteins are made of 20 types of monomers.

Q10. Proteins perform many physiological functions. For example, some proteins function as enzymes. One of the following represents an additional function that some proteins discharge
(a) Antibiotics
(b) Pigment conferring colour to skin
(c) Pigment making colours of flowers
(d) Hormones
Ans: (d) Proteins perform many physiological functions. For example, some proteins function as enzymes. Hormones represents an additional function that some proteins discharge (like insulin).

Q11. Glycogen is a homopolymer made of
(a) Glucose units (b) Galactose units
(c) Ribose units (d) Amino acids
Ans: (a) Glycogen is a homopolymer made of glucose units.

Q12. The number of ‘ends’ in a glycogen molecule would be
(a) Equal to the number of branches plus one
(b) Equal to the number of branch points
(c) One ‘
(d) Two, one on the left side and another on the right side
Ans: (d) In a polysaccharide chain (say glycogen), the right end is called the reducing end and the left end is called the non-reducing end.

Q13. The primary structure of a protein molecule has
(a) Two ends (b) One end (c) Three ends (d) No ends
Ans: (a) The primary structure of a protein molecule has two ends.
A protein is imagined as a line, the left end represented by the first amino acid and the right end is represented by the last amino acid. The first amino acid is also called as N-terminal amino acid. The last amino acid is called the C-terminal amino acid.

Q14. Enzymes are biocatalysts. They catalyse biochemical reactions. In general they reduce activation energy of reactions. Many physico-chemical processes are enzyme mediated.’ Some examples of enzyme mediated reactions are given below. Tick the wrong entry.
(a) Dissolving C02 in water
(b) Unwinding the two strands of DNA .
(c) Hydrolysis of sucrose
(d) Formation of peptide bond
Ans: (a) Dissolving C02 in water is a physical process.

Very Short Answer Type Questions
Q1. 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 a synthetic chemical.
a. Penicillin
b. Sulfonamide
c. Vitamin C
d. Growth Hormone
Ans: a. Penicillin: Natural product
b. Sulfonamide: Synthetic chemical
c. Vitamin C: Natural product
d. Growth Hormone: Natural product

Q2. Select an appropriate chemical bond among ester bond, glycosidic bond, peptide bond and hydrogen bond and write against each of the following.
a. Polysaccharide
b. Protein
c. Fat
d. Water
Ans: a. Polysaccharide: Glycosidic bond
b. Protein: Peptide bond
c. Fat: Ester bond
d. Water: Hydrogen bond

Q3. Write the name of any one aminoacid, sugar, nucleotide and fatty acid.
Ans: Glycine (amino acid), Ribose (sugar), Cytidylic acid (nucleotide) and
Arachidonic acid (fatty acid).

Q4. Reaction given below is catalysed by oxidoreductase between two substrates A and A’, complete the reaction. A reduced + A‘ oxidised -»
Ans: A reduced + A’ oxidised —»A oxidised + A’ reduced

Q5. How are prosthetic groups different from co-factors?
Ans: Prosthetic groups are organic compounds and are distinguished from other cofactors in that they are tightly bound to the apoenzyme. For example, in peroxidase and catalase, which catalyze the breakdown of hydrogen peroxide to water and oxygen, haem is the prosthetic group and it is a part of the active site of the enzyme.
Cofactor may be organic or inorganic (metal ions).

Q6. Glycine and Alanine are different with respect to one substituent on the a-carbon. What are the other common substituent groups?
Ans: The R-group in these proteinaceous amino acids could be a hydrogen (the amino acid is called glycine), a methyl group (alanine), hydroxy methyl (serine), etc.

Q7. Starch, Cellulose, Glycogen, Chitin are polysaccharides found among the following. Choose the one appropriate and write against each.
Cotton fibre __________
Exoskeleton of Cockroach __________
Liver __________
Peeled potato __________
Ans: Cotton fibre : Cellulose
Exoskeleton of Cockroach : Chitin
Liver: Glycogen
Peeled potato: Starch

Q4. Nucleic acids exhibit secondary structure, justify with example.
Ans: Nucleic acids exhibit a wide variety of secondary structures. For example, one of the secondary structures exhibited by DNA is the famous Watson—Crick model. This model says that DNA exists as a double helix. The two strands of polynucleotides are antiparallel, i.e. run in the opposite direction. The backbone is formed by the sugar-phosphate-sugar chain. The nitrogen bases are projected more or less perpendicular to this backbone but face inside. A and G of one strand compulsorily base pairs with T and C, respectively, on the other strand. There are two hydrogen bonds between A and T and three hydrogen bonds between G and C. Each strand appears like a helical staircase. Each step of ascent is represented by a pair of bases. At each step of ascent, the strand turns 36°. One full turn of the helical strand would involve ten steps or ten base pairs. Attempt drawing a line diagram. The pitch would be 34 A. The rise per base pair would be 3.4 A. This form of DNA with the above mentioned salient features is called B-DNA.

Q5. Comment on the statement “living state is a non-equilibrium steady state to be able to perform work”.
Ans: The most important fact of biological systems is that all living organisms exist in a steady-state characterised by concentrations of each of these biomolecules. These biomolecules are in a metabolic flux. Any chemical or physical process moves spontaneously to equilibrium. The steady state is a non-equilibrium state. One should remember from physics that systems at equilibrium cannot perform work. As living organisms work continuously, they cannot afford to reach equilibrium. Hence the living state is a non-equilibrium steady-state to be able to perform work; living process is a constant effort to prevent falling into equilibrium. This is achieved by energy input. Metabolism provides a mechanism for the production of energy. Hence the living state and metabolism are synonymous. Without metabolism there cannot be a living state.

Long Answer Type Questions
Q1. Formation of enzyme-substrate complex (ES) is the first step in catalysed reactions. Describe the other steps till the formation of product.
Ans: The catalytic cycle of an enzyme action can be described in the following steps:
(1) First, the substrate binds to the active site of the enzyme, fitting into the active site.
(2) The binding of the substrate induces the enzyme to alter its shape, fitting more tightly around the substrate.
(3) The active site of the enzyme, now in close proximity of the substrate breaks the chemical bonds of the substrate and the new enzyme-product complex is formed.
(4) The enzyme releases the products of the reaction and the free enzyme is ready to bind to another molecule of the substrate and run through the catalytic cycle once again.

Q2. What are different classes of enzymes? Explain any two with the type of reaction they catalyse.
Ans: Enzymes are divided into 6 classes each with 4—13 subclasses and named accordingly by a four-digit number.

Q3. Nucleic acids exhibit secondary structure. Describe through Watson-Crick Model.
Ans: Nucleic acids exhibit a wide .variety of secondary structures. For example, one of the secondary structures exhibited by DNA is the famous Watson- Crick model. This model says that DNA exists as a double helix. The two strands of polynucleotides are antiparallel, i.e. run in the opposite direction. The backbone is formed by the sugar—phosphate—sugar chain. The nitrogen bases are projected more or less perpendicular to this backbone but face inside. A and G of one strand compulsorily base pairs with T and C, respectively, on the other strand. There are two hydrogen bonds between A

and T and three hydrogen bonds between G and C. Each strand appears like a helical staircase. Each step of ascent is represented by a pair of bases. At each step of ascent, the strand turns 36°. One full turn of the helical strand would involve ten steps or ten base pairs. Attempt drawing a line diagram. The pitch would be 34 A. The rise per base pair would be 3.4 A. This form of DNA with the above mentioned salient features is called B-DNA.

Q4. What is the difference between a nucleotide and nucleoside? Give two examples of each with their structure.
Ans: Living organisms have a number of carbon compounds in which heterocyclic rings can be found. Some of these are nitrogen bases—adenine, guanine, cytosine, uracil and thymine. When found attached to a sugar, they are called nucleosides. If a phosphate group is also found esterified to the sugar they are called nucleotides. Adenosine, guanosine, thymidine, uridine and cytidine are nucleosides. Adenylic acid, thymidylic acid, guanylic acid, uridylic acid and cytidylic acid are nucleotides.

Q5. Describe various forms of lipid with a few examples.
Ans: Lipids are generally water insoluble. They could be simple fatty acids. A fatty acid has a carboxyl group attached to an R-group. The R-group could be a methyl (-CH3), or ethyl (—C2H5) or higher number of-CH2 groups (1 carbon to 19 carbons). For example, palmitic acid has 16 carbons including carboxyl carbon. Arachidonic acid has 20 carbon atoms including the carboxyl carbon. Fatty acids could be saturated (without double bond) or unsaturated (with one or more C=C double bonds). Another simple lipid is glycerol which is trihydroxy propane.
• Many lipids have both glycerol and fatty acids. Here the fatty acids are found esterified with glycerol. They can be then monoglycerides,
diglycerides and triglycerides. These are also called fats and oils based on melting point. Oils have lower melting point (e.g., gingely oil) and hence remain as oil in winters.
• Some lipids have phosphorous and a phosphorylated organic compound in them. These are phospholipids. They are found in cell membrane. Lecithin is one example. Some tissues especially the neural tissues have lipids with more complex structures.

NCERT Exemplar Class 11 Biology Solutions

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NCERT Exemplar Class 11 Biology Chapter 14 Respiration in Plants

February 27, 2023

NCERT Exemplar Class 11 Biology Chapter 14 Respiration in Plants are part of NCERT Exemplar Class 11 Biology. Here we have given NCERT Exemplar Class 11 Biology Chapter 14 Respiration in Plants.

NCERT Exemplar Class 11 Biology Chapter 14 Respiration in Plants

Multiple Choice Questions

Q1. The ultimate electron acceptor of respiration in an aerobic organism is
(a) Cytochrome (b) Oxygen
(c) Hydrogen (d) Glucose
Ans: (b) The ultimate electron acceptor of respiration in an aerobic organism is oxygen.

Q2. Phosphorylation of glucose during glycolysis is catalyzed by
(a) Phosphoglucomutase (b) Phosphoglucoisomerase
(c) Hexokinase (d) Phosphorylase
Ans: (c) Phosphorylation of glucose during glycolysis is catalyzed by hexokinase.

Q3. Pyruvic acid, the key product of glycolysis can have many metabolic fates. Under aerobic condition it forms
(a) Lactic acid
(b) C02 + H20
(c) Acetyl CoA + C02
(d) Ethanol + CO
Ans: (c) Pyruvic acid, the key product of glycolysis can have many metabolic fates. Under aerobic condition it forms Acetyl CoA + CO2.

Q4. Electron Transport System (ETS) is located in mitochondrial
(a) Outer membrane
(b) Inter membrane space
(c) Inner membrane
(d) Matrix
Ans: (c) Electron Transport System (ETS) is located in mitochondrial inner membrane.

Q5. Which of the following exhibits the highest rate of respiration? .
(a) Growing shoot apex
(b) Germinating seed
(c) Root tip
(d) Leaf bud
Ans: (b) Germinating seed exhibits the highest rate of respiration.

Q6. Mitochondria are called powerhouses of the cell. Which of the following observations support this statement?
(a) Mitochondria synthesise ATP
(b) Mitochondria have a double membrane .
(c) The en2ymes of the Krebs’ cycle and the cytochromes are found in mitochondria
(d) Mitochondria are found in almost all plant and animal cells
Ans: (a) Mitochondria are called powerhouses of the cell because mitochondria synthesise ATP.

Q7. The end product of oxidative phosphorylation is
(a) NADH
(b) Oxygen
(c) ADP
(d) ATP + H20
Ans: (d) The end product of oxidative phosphorylation isATP+H₂o.

Q8. Match the following and choose the correct option from those given below.

Column A		Column B
A.	Molecular oxygen	(i)	a-Ketoglutaric acid
B.	Electron acceptor	(ii)	Hydrogen acceptor
C.	Pyruvate dehydrogenase	(iii)	Cytochrome C
D.	Decarboxylation	(iv)	Acetyl Co A

(a) A—(ii), B—(iii), C—(iv), D—(i)
(b) A—(iii), B—(iv), C—(ii), D—(i)
(c) A—(ii), B—(i), C—(iii), D—(iv)
(d) A—(iv), B—(iii), C—(i), D—(ii)

Ans. (a)

A.	Molecular oxygen	(ii)	Hydrogen acceptor
B.	Electron acceptor	(iii)	Cytochrome C
C.	Pyruvate dehydrogenase	(iv)	Acetyl Co A
D.	Decarboxylation	(i)	a-Ketoglutaric acid

Very Short Answer Type Questions

Q1. Energy is released during the oxidation of compounds in respiration. How is this energy stored and released as and when it is needed?
Ans: Energy released during the oxidation of compounds in respiration is immediately stored in ATP in the form of chemical bonds.
ADP + iP + energy —> ATP
As and when needed, this bond energy is broken and utilised
ATP —> ADP + iP + energy

Q2. Explain the term “Energy Currency”. Which substance acts as energy currency in plants and animals?
Ans: Every function of the cell requires energy. Energy currency stores and releases the energy as and when needed in the cell. ATP is called energy currency in both plants and animals.

Q3. Different substrates get oxidized during respiration. How does Respiratory Quotient (RQ) indicate which type of substrate, i.e., carbohydrate, fat or protein is getting oxidized?

Q5. When does anaerobic respiration occur in man and yeast?
Ans: In alcoholic fermentation (by yeast), incomplete oxidation of glucose occurs under anaerobic conditions by sets of reactions where pyruvic acid is converted to ethanol and C02.
PA —> Ethanol + C02
In animal cells also, like muscles during exercise, when oxygen is inadequate for cellular respiration pyruvic acid is reduced to lactic acid.

Q6. Which of the following will release more energy on oxidation? Arrange them in ascending order?
a. 1 g of fat b. 1 g of protein
c. 1 g of glucose
d. 0.5 g of protein + 0.5 g glucose

Q7. The product of aerobic glycolysis in skeletal muscle and anaerobic fermentation in yeast are respectively ________ and ______.
Ans. The product of aerobic glycolysis in skeletal muscle and anaerobic fermentation in yeast are respectively pyruvic acid and ethanol + CO₂.

Short Answer Type Questions

Q1. If a person is feeling dizzy, glucose or fruit juice is given immediately but not a cheese sandwich, which might have more energy. Explain.
Ans: Glucose or fruit juice is absorbed easily through the alimentary canal. In the cells glucose is oxidised and energy is released immediately. A cheese sandwich provides energy after digestion and absorption which takes long time.

Q2. What is meant by the statement “aerobic respiration is more efficient”?
Ans: Aerobic respiration is more efficient because fermentation accounts for
only a partial breakdown of glucose, whereas in aerobic respiration it is completely degraded to C02 and H20. Also, in fermentation there is a net gain of only two molecules of ATP for each molecule of glucose degraded to pyruvic acid, whereas many more molecules of ATP are generated under aerobic conditions.

Q3. Pyruvic acid is the end product of glycolysis. What are the three metabolic fates of pyruvic acid under aerobic and anaerobic conditions? Write in the space provided in the diagram.

Q4. The energy yield in terms of ATP is higher in aerobic respiration than during anaerobic respiration. Why is there anaerobic respiration even in organisms that live in aerobic condition like human beings and angiosperms?
Ans: Anaerobic respiration even in organisms that live in aerobic condition like human beings and angiosperms due to the oxygen deficiency. In our skeletal muscles during strenuous exercise oxygen deficiency leads to anaerobic respiration. In germinating seeds anaerobic respiration leads to release of energy for emerging the seedling from the soil.

Q5. Oxygen is an essential requirement for aerobic respiration but it enters the respiratory process at the end? Discuss.
Ans: Although the aerobic process of respiration takes place only in the presence of oxygen, the role of oxygen is limited to the terminal stage of the process. Yet, the presence of oxygen is vital, since it drives the whole process by removing hydrogen from the system. Oxygen acts as the final hydrogen acceptor.

Q6. Respiration is an energy releasing and enzymatically controlled catabolic process which involves a step-wise oxidative breakdown of organic substances inside living cells. In this statement about respiration explain the meaning of
1. Step-wise oxidative breakdown and
2. Organic substances (used as substrates).
Ans: 1. Step-wise oxidative breakdown: During oxidation within a cell, all the energy contained in respiratory substrates is not released free into
the cell, or in a single step. It is released in a series of slow step-wise reactions controlled by enzymes, and it is trapped as chemical energy in the form of ATP. Hence, it is important to understand that the energy released by oxidation in respiration is not (or rather cannot be) used directly but is used to synthesise ATP, which is broken down whenever (and wherever) energy needs to be utilised.
2. Organic substances (used as substrates): The compounds that are oxidised during this process are known as respiratory substrates. Usually carbohydrates are oxidised to release energy, but proteins, fats and even organic acids can be used as respiratory substances in some plants, under certain conditions.

Q7. Comment on the statement – Respiration is an energy producing process but ATP is being used in some steps of the process.
Ans: In the respiration pathway, there are some steps where energy is utilised for phosphorylation. For example, conversion of glucose to glucose-6-phosphate consume one ATP. But at the end of the respiratory process many more ATP are produced.

Q8. The figure given below shows the steps in glycolysis. Fill in the missing steps A, B, C, D and also indicate whether ATP is being used up or released at step E?

Q9. Why is respiratory pathway referred to as an amphibolic pathway? Explain.
Ans: During breakdown and synthesis of protein, respiratory intermediates form the link. Breaking down processes within the living organism is catabolism, and synthesis is anabolism. Because the respiratory pathway is involved in both anabolism and catabolism, it would hence be better to consider the respiratory pathway as an amphibolic pathway rather than as a catabolic one.

Q10. We commonly call ATP as the energy currency of the cell. Can you think of some other energy carriers present in a cell? Name any two.
Ans: Yes, some other energy carriers are also present in a cell like GTP (guanosine triphosphate), ADP (adenosine diphosphate) and creatine phosphate.

Q11. ATP produced during glycolysis is a result of substrate level phosphorylation. Explain.
Ans: ATP produced during glycolysis is a result of substrate level phosphorylation because these ATP are produced without the electron transport system (ETS) and chemiosmosis.
During substrate level phosphorylation ATP is directly synthesised from ADP and inorganic phosphate (iP).

Q12. Do you know any step in the TCA cycle where there is substrate level phosphorylation. Which one?
Ans: During TCA cycle there is one step where substrate level phosphorylation takes place. This occurs during conversion of succinyl-CoA to succinic acid.

Q13. In a way green plants and cyanobacteria have synthesised all the food on the earth. Comment.
Ans: All the energy required for ‘life’ processes is obtained by oxidation of some macromolecules that we call ‘food’. Only green plants and cyanobacteria can prepare their own food; by the process of photosynthesis they trap light energy and convert it into chemical energy that is stored in the bonds of carbohydrates like glucose, sucrose and starch.

Q14.When a substrate is being metabolised, why does not all the energy that is produced get released in one step? It is released in multiple steps. What is the advantage of step-wise release?
Ans: The complete combustion of glucose, which produces C0₂ and H₂0 as end products, yields energy most of which is given out as heat.
C₆H₁₂O₆+6O₂ —> 6CO₂+ 6H₂O + Energy
If this energy is to be useful to the cell, it should be able to utilise it to synthesise other molecules that the cell requires. The strategy that the plant cell uses is to catabolise the glucose molecule in such a way that not all the liberated energy goes out as heat. The key is to oxidise glucose not in one step but. in several small steps enabling some steps to be just large enough such that the energy released can be coupled to ATP synthesis.

Q15. Respiration requires 02. How did the first cells on the earth manage to survive in an atmosphere that lacked 02?
Ans: During the process of respiration, oxygen is utilised, and carbon dioxide, water and energy are released as products. The combustion reaction requires oxygen. But some cells live where oxygen may or may not be available. There are sufficient reasons to believe that the first cells on this planet lived in an atmosphere that lacked oxygen. Even among present-day living organisms, we know of several that are adapted to anaerobic conditions.

Q16. It is known that red muscle fibres in animals can work for longer periods of time continuously. How is this possible?
Ans: Muscle contains a red coloured oxygen storing pigment called myoglobin. Myoglobin content is high in some of the muscles which gives a reddish appearance. Such muscles are called the Red fibres. These muscles also contain plenty of mitochondria which can utilise the large amount of oxygen stored in them for ATP production. These muscles, therefore, can also be called aerobic muscles.

Q17. The energy yield in terms of ATP is higher in aerobic respiration than during anaerobic respiration. Explain.
Ans: The energy yield in terms of ATP is higher in aerobic respiration than during anaerobic respiration because fermentation (anaerobic respiration) accounts
for only a partial breakdown erf glucose, whereas in aerobic respiration it is completely degraded to CO2 and H2O. So, in fermentation there is a net gain of only two molecules of ATP for each molecule of glucose degraded to pyruvic acid, whereas many more molecules of ATP are generated under aerobic conditions.

Q18. RuBP carboxylase, PEPcase, Pyruvate dehydrogenase, ATPase, cytochrome oxidase, Hexokinase, Lactate dehydrogenase. Select/choose enzymes from the list above which are involved in
a. Photosynthesis
b. Respiration
c. Both in photosynthesis and respiration
Ans: a. Photosynthesis: RuBP carboxylase, PEPcase, ATPase
b. Respiration: Hexokinase, ATPase, Pyruvate dehydrogenase, Cytochrome oxidase
c. Both in photosynthesis and respiration: ATPase

Q19. How does a tree trunk exchange gases with the environment although it lacks stomata?
Ans: In stems, the ‘living’ cells are organised in thin layers inside and beneath the bark. They also have openings called lenticels. The cells in the interior are dead and provide only mechanical support. Thus, most cells of a plant have at least a part of their surface in contact with air. This is also facilitated by the loose packing of parenchyma cells in leaves, stems and roots, which provide an interconnected network of air spaces.

Q20. Write two energy yielding reactions of glycolysis.
Ans: The conversion of BPGA to 3-phosphoglyceric acid (PGA), is an energy yielding process; this energy is trapped by the formation of ATP. Another ATP is synthesised during the conversion of PEP to pyruvic acid.

Q21. Name the site(s) of pyruvate synthesis. Also, write the chemical reaction wherein pyruvic acid dehydrogenase acts as a catalyst.

Q22.Mention the important series of events of aerobic respiration that occurs in the matrix of the mitochondrion as well as one that takes place in the inner membrane of the mitochondrion.
Ans: For aerobic respiration to take place within the mitochondria, the final product of glycolysis, pyruvate is transported from the cytoplasm into the mitochondria. The crucial events in aerobic respiration are:
The complete oxidation of pyruvate by the stepwise removal of all the hydrogen atoms, leaving three molecules of C0₂.

The passing on of the electrons removed as part of the hydrogen atoms to molecular 0₂ with simultaneous synthesis of ATP. What is interesting to note is that the first process takes place in the matrix of the mitochondria while the second process is located on the inner membrane of the mitochondria.

Q23.Respiratory pathway is believed to be a catabolic pathway. However, nature of TCA cycle is amphibolic. Explain.
Ans: During breakdown and synthesis of protein, respiratory intermediates form the link. Breaking down processes within the living organism is catabolism, and synthesis is anabolism. Because the respiratory pathway is involved in both anabolism and catabolism, it would hence be better to consider the respiratory pathway as an amphibolic pathway rather than as a catabolic one.

Long Answer Type Questions
Q1. In the following flow chart, replace the symbols a,b,c and d with appropriate terms. Briefly explain the process and give any two application of it.

Q2. Given below is a diagram showing ATP synthesis during aerobic respiration, replace the symbols A, B,C,D and E by appropriate terms given in the box.

Q3. Oxygen is critical for aerobic respiration. Explain its role with respect to ETS.
Ans: Although the aerobic process of respiration takes place only in the presence of oxygen, the role of oxygen is limited to the terminal stage of the process. Yet, the presence of oxygen is vital, since it drives the whole process by removing hydrogen from the system. Oxygen acts as the final hydrogen acceptor. Unlike photophosphorylation where it is the light energy that is utilised for the production of proton gradient required for phosphorylation, in respiration it is the energy of oxidation-reduction utilised for the same process. It is for this reason that the process is called oxidative phosphorylation.

Q4. Enumerate the assumptions that we undertake in making the respiratory balance sheet. Are these assumptions valid for a living system? Compare fermentation and aerobic respiration in this context.
Ans: The Respiratory Balance Sheet:
It is possible to make calculations of the net gain of ATP for every glucose molecule oxidised; but in reality this can remain only a theoretical exercise. These calculations can be made only on certain assumptions that:
There is a sequential, orderly pathway functioning, with one substrate forming the next and with glycolysis, TCA cycle and ETS pathway following one after another.
The NADH synthesised in glycolysis is transferred into the mitochondria and undergoes oxidative phosphorylation.
None of the intermediates in the pathway are utilised to synthesise any other compound.
Only glucose is being respired – no other alternative substrates are entering in the pathway at any of the intermediary stages. But this kind of assumptions are not really valid in a living system; all pathways work simultaneously and do not take place one after another; substrates enter the pathways and are withdrawn from it as and when necessary; ATP is utilised as and when needed; enzymatic rates are controlled by multiple means. Yet, it is useful to do this exercise to appreciate the beauty and efficiency of the living system in extraction and storing energy. Hence, there can be a net gain of 36 ATP molecules during aerobic respiration of one molecule of glucose.

Q5. Give an account of Glycolysis. Where does it occur? What are the end products? Trace the fate of these products in both aerobic and anaerobic respiration.
Ans: Glycolysis occurs in the cytoplasm of the cell and is present in all living organisms. In this process, glucose undergoes partial oxidation to form two molecules of pyruvic acid. In plants, this glucose is derived from sucrose, which is the end product of photosynthesis, or from storage carbohydrates. Sucrose is converted into glucose and fructose by the enzyme, invertase, and these two monosaccharides readily enter the glycolytic pathway. Glucose and fructose are phosphorylated to give rise to glucose-6-phosphate by the activity of the enzyme hexokinase. This phosphorylated form of glucose then isomerises to produce fructose-6-phosphate. Subsequent steps of metabolism of glucose and fructose are same. In glycolysis, a chain of ten reactions,
under the control of different enzymes, takes place to produce pyruvate from glucose. Pyruvic acid is then the key product of glycolysis. The metabolic fate of pyruvate depends on the cellular need. There are three major ways in which different cells handle pyruvic acid produced by glycolysis. These are lactic acid fermentation, alcoholic fermentation and aerobic respiration. Fermentation takes place under anaerobic conditions in many prokaryotes and unicellular eukaryotes. For the complete oxidation of glucose to C02 and H20, however, organisms adopt Krebs’ cycle which is also called as aerobic respiration. This requires 0₂ supply.

NCERT Exemplar Class 11 Biology Solutions

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We hope the NCERT Exemplar Class 11 Biology Chapter 14 Respiration in Plants help you. If you have any query regarding NCERT Exemplar Class 11 Biology Chapter 14 Respiration in Plants, drop a comment below and we will get back to you at the earliest.

NCERT Solutions For Class 11 Biology Biomolecules

February 27, 2023

NCERT Solutions For Class 11 Biology Biomolecules

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

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

NCERT Solutions Class 11 Biology Biology Sample Papers

NCRT TEXTBOOK QUESTIONS SOLVED

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

2. Illustrate a glycosidic, peptide, and a phospho- diester bond.
Solution. (i) Glycosidic bond is the type of chemical linkage between the monosaccharide units of disaccharides, oligosaccharides and polysaccharides, which is formed by the removal of a molecule of water.

(ii)Peptide bonds are formed by the reaction between carboxyl (- COOH) of one amino acid and amino (- NH2) group of other amino acid with the elimination of water.

(iii) In a polynucleotide chain, adjacent nucleotides are joined together by a bond called phosphodiester bond. This bond links a phosphate group and sugar group of two adjacent nucleotides by means of an oxygen bridge.

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

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

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

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

7. Explain the composition of triglycerides. jSfflTriacylglycerols (triglycerides) are the esters of glycerol with fatty acids.
Solution: They are insoluble in water and non-polar in character and commonly known as neutral fats. The neutral or depot fats are composed of carbon, hydrogen and oxygen like carbohydrates but have far fewer oxygen atoms than carbon atoms unlike the carbohydrates.

(i) Glycerol – A glycerol molecule has 3
carbons, each bearing a hydroxyl (-OH) group. .
(ii) Fatty acids – A fatty acid molecule is an unbranched chain of carbon atoms with each carbon atom (C) forming four bonds to other atoms. It has a carboxyl group- COOH at one end and hydrogen atom (H) bonded to all or most carbon atoms forming a hydrogen chain. The carbon- hydrogen bonds are non-polar. Therefore, the hydrocarbon chain does not dissolve in water. Because the carboxyl group contains the polar C = O and OFI groups. It tends to dissolve in water even though the rest of fatty acid molecule will not. Triacylglycerols of plants, in general, have higher content of unsaturated fatty acids as compared to that of animals.

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

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

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

11. Draw the structure of the amino acid, alanine.
Solution:

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

13. Find out a qualitative test for proteins, fats and oils, amino acids and test, any fruit juice, saliva, and urine for them.
Solution: Biuret test for protein : The biuret test is a chemical test used for determining the presence of peptide bonds. In a positive test, a copper II ion (Cu2+ ion) is reduced to copper I (Cu+) which forms a complex with the nitrogen and carbon of peptide bonds in an alkaline solution. A violet colour indicates the presence of proteins.
Ninhydrin test for amino acid: Ninhydrin (2,2 Dihydroxy indane-l,3-dione) is a chemical used to detect ammonia or primary and secondary amines. When reacting with these free amines, a deep blue or purple colour known as Ruhemann’s purple is evolved. Amino acid analysis of proteins is also done by ninhydrine. Most of the amino acids (including a-amino acids) are hydrolysed and reacted with ninhydrin except proline (a secondary amine). Amino acid containing a free amino group and a free carboxylic acid group reacts together with ninhydrin to produce coloured product. When the amino group is secondary, the condensation product is yellow.

Solubility test for fats and oils : A positive solubility test for fats is that the fat dissolves in lighter fluid and not in water. In this test, 5 drops of fat or oil are added in two test tubes containing 10 drops of lighter fluid and 10 drops cold water respectively.
Fruit juice contains sugar so it cannot be tested by the above-mentioned tests. Saliva contains proteins, mineral salts, amylase etc., so it can be tested for protein and amino acids. Urine contains proteins, so it can be tested for it.

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

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

NCERT Exemplar Class 11 Biology Chapter 12 Mineral Nutrition

February 27, 2023

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

NCERT Exemplar Class 11 Biology Chapter 12 Mineral Nutrition

Multiple Choice Questions

Q1. Which one of the following roles is not a characteristic of an essential element?
(a) Being a component of biomolecules
(b) Changing the chemistry of soil
(c) Being a structural component of energy-related chemical compounds
(d) – Activation or inhibition of enzymes
Ans: (b)
(i) Essential elements as components of biomolecules and hence structural elements of cells.
(ii) Essential elements that are components of energy-related chemical compounds in plants.
(iii) Essential elements that activate or inhibit enzymes.
(iv) Some essential elements can alter the osmotic potential of a cell.

Q2. Which one of the following statements can best explain the term critical concentration of an essential element?
(a) Essential element concentration below which plant growth is retarded
(b) Essential element concentration below which plant growth becomes enhanced
(c) Essential element concentration below which plant remains in the vegetative phase
(d) None of the above
Ans: (a) The concentration of the essential element below which plant growth is retarded is termed as critical concentration.

Q3. Deficiency symptoms of an element tend to appear first in young leaves. It indicates that the element is relatively immobile. Which one of the following elemental deficiency would show such symptoms?
(a) Sulphur (b) Magnesium (c) Nitrogen (d) Potassium
Ans: (a) The deficiency symptoms tend to appear first in the young tissues whenever the elements are relatively immobile and are not transported out of the mature organs, e.g., S and Ca.

Q4. Which one of the following symptoms is not due to manganese toxicity in plants?
(a) Calcium translocation in shoot apex is inhibited
(b) Deficiency in both Iron and Nitrogen is induced
(c) Appearance of brown spot surrounded by chlorotic veins
(d) None of the above
Ans: (b) Excess of manganese may, in fact, induce deficiencies of iron, magnesium and calcium.

Q5. Reaction carried out by N2 fixing microbes include

Which of the following statements about those equations is not true?
(a) Step (i) is carried out by Nitrosomonas or Nitrococcus
(b) Step (ii) is carried out by Nitrobacter
(c) Both steps (i) and (ii) can be called nitrification
(d) Bacteria carrying out these steps are usually photoautotrophs
Ans:(d) Bacteria carrying out these steps are usually chemoautotrophs.

Q6. With regard to the Biological Nitrogen Fixation by Rhizobium in association with soyabean, which one of the following statement/statements does not hold true?
(a) Nitrogenase may require oxygen to its functioning
(b) Nitrogenase is Mo-Fe protein
(c) Leghaemoglobin is a pink coloured pigment
(d) Nitrogenase helps to convert N2 gas into two molecules of ammonia
Ans:(a) Nitrogenase is highly sensitive to molecular oxygen (02), thus requires
anaerobic conditions. Nodules have adaptations that ensure that the enzyme is protected from 02. To protect nitrogehase, nodule contains an 02-scavenger celled leghaemoglobin.

Q7. Match the element with its associated functions/roles and choose the correct option among the given below.

A.	Boron	(i) V	Splitting of H₂0 to liberate 0₂ during photosynthesis
B.	Manganese	(ii)	Needed for synthesis of auxins
C.	Molybdenum	(iii)	Component of nitrogenase
D.	Zinc	(iv)	Pollen germination
E.	Iron	(v)	Component of ferredoxin

Options:
(a) A—(i), B—(ii), C—(iii). D—(iv), E—(v)
(b) A—(iv), B—(i), C—(iii), D—(ii), E—(v)
(c) . A—(iii), B—(ii), C—(iv), D—(v), E—(i)
(d) A—(ii), B—(iii), C—(v), D—(i), E—(iv)
Ans: (b)

A.	Boron	(iv)	Pollen germination
B.	Manganese	(0	Splitting of H₂0 to liberate 0₂ during photosynthesis
C.	Molybdenum	(iii)	Component of nitrogenase
D.	Zinc	(ii)	Needed for synthesis of auxins
E.	Iron	(v)	Component of ferredoxin

Q8. Plants can be grown in (Tick the incorrect option).
(a) Soil with essential nutrients
(b) Water with essential nutrients
(c) Either water or soil with essential nutrients
(d) Water or soil without essential nutrients
Ans: (d) Plants can be grown in soil with essential nutrients, water with essential nutrients and either water or soil with essential nutrients.

Very Short Answer Type Questions
Q1. Name a plant, which accumulates silicon.
Ans: Rice, sugarcane, etc.

Q2. Mycorrohiza is a mutualistic association. How do the organisms involved in this association gain from each other?
Ans: Mycorrhiza is a symbiotic association between a fungus and the roots of a vascular plant. Through mycorrhization, the plant obtains phosphate and other minerals, such as zinc and copper, from the soil. The fungus obtains nutrients, such as sugars, from the plant root.

Q3. Nitrogen fixation is shown by prokaryotes and not eukaryotes. Comment?
Ans: Very few living organisms can utilise the nitrogen in the form N2, available
abundantly in the air. Only certain prokaryotic species are capable of fixing nitrogen. The enzyme, nitrogenase which is capable of nitrogen reduction is present exclusively in prokaryotes. Such microbes are called N2-fixers.

Q4. Carnivorous plants like Nepenthes and Venus fly trap have nutritional adaptations. Which nutrient do they especially obtain and from where?
Ans: Carnivorous plants grow in nitrogen deficient soil but they utilise their nitrogen by killing the insect by some special structure.

Q5. Think of a plant which lacks chlorophyll. From where will it obtain nutrition? Give an example of such a type of plant.
Ans: Cuscuta, a parasitic plant that is commonly found growing on hedge plants, has lost its chlorophyll and leaves in the course of evolution. It derives its nutrition from the host plant which it parasitises.

Q6. Name an insectivorous angiosperm.
Ans: Nepenthes, Utricularia, Drosera, Dionea, etc.

Q7. A farmer adds Azotobacter culture to soil before sowing maize. Which mineral element is being replenished?
Ans: Nitrogen

Q8. What type of conditions are created by leghaemoglobin in the root nodule of a legume?
Ans: Anaerobic condition

Q9. What is common to Nepenthes, Utricularia and Drosera with regard to mode of nutrition?
Ans: All are carnivorous plant (angiosperms).

Q10. Plants with zinc deficiency show reduced biosynthesis of .
Ans: Auxin

Q11. Yellowish edges appear in leaves deficient in .
Ans: K (potassium)

Q12. Name the macronutrient which is a component of all organic compounds but is not obtained from soil.
Ans: Carbon

Q13. Name one non-symbiotic nitrogen fixing prokaryote.
Ans: (i) Free-living (non-symbiotic) and non photosynthetic aerobic N2-fixing microbes: Azotobacter and Beijernickia.
(ii) Free-living and anaerobic N2-fixing microbes: Rhodospirillum, Bacillus polymyxa and Clostridium.

Q14. Rice fields produce an important green house gas. Name it.
Ans: CH4 (methane) .

Q15. Complete the equation for reductive amination

Q16. Excess of Mn in soil leads to deficiency of Ca, Mg and Fe. Justify.
Ans: Manganese competes with iron and magnesium for uptake and with magnesium for binding with enzymes. Manganese also inhibits calcium translocation in shoot apex. Therefore, excess of manganese may, in fact, induce deficiencies of iron, magnesium and calcium. Thus, what appears as symptoms of manganese toxicity may actually be the deficiency symptoms of iron, magnesium and calcium.

Short Answer Type Questions
Q1. How is sulphur important for plants? Name the amino acids in which it is present.
Ans: Sulphur, besides being present in some amino acids essential for protein synthesis, is also a constituent of several coenzymes, vitamins and ferredoxin which are involved in some biochemical pathway.

Q2. How are organisms like Pseudomonas and Thiobacillus of great significance in nitrogen cycle?
Ans: Pseudomonas and Thiobacillus carry out denitrification process wherein the nitrate present in the soil is reduced to nitrogen thus contributing to the atmospheric nitrogen.

Q3. Carefully observe the following figure:

a. Name the technique shown in the figure and the scientist who demonstrated this technique for the first time.
b. Name at least 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?
Ans: a. Hydroponics, Julius von Sachs.
b. Tomato, seedless cucumber, lettuce.
c. Aerating tube ensures adequate aeration of the root for optimum growth of the plant. The funnel is used to release water and nutrients into the container with nutrient solution. This solution needs to be replaced every day or two for maximum growth.

Q4. Name the most crucial enzyme found in root nodules for N2 fixation. Does it require a special pink coloured pigment for its functioning? Elaborate.
Ans: Nitrogenase. Yes, it does require the presence of a pink-coloured pigment in the nodule called leghaemoglobin for its functioning. This pigment helps in scavenging oxygen as nitrogenase functions under anaerobic condition.

Q5. How are the terms ‘critical concentration’ and ‘deficient’ different from each other in terms of concentration of an essential element in plants? Can you find the values of ‘critical concentration’ and ‘deficient’ for minerals – Fe and Zn?
Ans: The concentration of the essential element below which plant growth is retarded is termed as critical concentration. The element is said to be deficient when present below the critical concentration. Yes. one can find the values of ‘critical concentration’ and ‘deficient’ for minerals – Fe and Zn through the hydroponics technique.

Q6. Carnivorous plants exhibit nutritional adaptation. Citing an example explain this fact.
Ans: Carnivorous plants have green leaves so they are autotrophic but they grow in nitrogen deficient soil. For nitrogen requirement they capture and digest the insects so they are partially heterotrophic nature.

Q7. A farrper adds/supplies 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.
Ans: This is due to the manganese toxicity. Many a times, excess of an element may inhibit the uptake of another element. For example, the prominent symptom of manganese toxicity is the appearance of brown spots surrounded by chlorotic veins. Manganese competes with iron and magnesium for uptake and with magnesium for binding with enzymes. Manganese also inhibits calcium translocation in shoot apex. Therefore, excess of manganese may, in fact, induce deficiencies of iron, magnesium and calcium.
• The farmer should not supplies Mn to his field.

Long Answer Type Questions
Q1. It is observed that deficiency of a particular element showed its symptoms initially in older leaves and then in younger leaves.
a. Does it indicate that the element is actively mobilised or relatively immobile?
b. Name two elements which are highly mobile and two which are relatively immobile.
c. How is the aspect of mobility of elements important to horticulture and agriculture?
Ans: a. It is actively mobilised.
b. Highly mobile—nitrogen, magnesium Relatively immobile—calcium, boron
c. Symptoms of deficiency of mobile elements are more pronounced in older leaves and symptoms of deficiency of relatively immobile element appear first in younger leaves. This information can be utilised by horticulturist and agriculturist to get a broad idea of the deficiency elements in plants.

Q2. 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 mycorrihiza 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.
Ans: a. Yes, one can artificially induce the property of nitrogen fixation in a plant—leguminous or non-leguminous through genetic engineering which involves introduction of specific genes to the host plant that synthesises nitrogenase enzymes.
b. Symbiotic relationship
c. Yes, it is necessary for a microbe to be in close association with a plant to provide mineral nutrition as seen in leguminous plants. Species of rod-shaped Rhizobium has such relationship with the roots of several legumes such as alfalfa, sweet clover, sweet pea, lentils, garden pea, broad bean, clover beans, etc. The most common association on roots is as nodules. The nodule contains all the necessary biochemical components, such as the enzyme nitrogenase and leghaemoglobin. The enzyme nitrogenase is an Mo-Fe protein and catalyses the conversion of atmospheric nitrogen to ammonia.

Q3. What are essential elements for plants? Give the criteria of essentiality. How are minerals classified depending upon the amount in which they are needed by the plants?
Ans: Essential elements: carbon, hydrogen, oxygen, nitrogen, phosphorous, sulphur, potassium, calcium, magnesium, iron, manganese, copper, molybdenum, zinc, boron, chlorine and nickel.
Criteria for Essentiality:
The criteria for essentiality of an element are given below:
(a) The element must be absolutely necessary for supporting normal growth and reproduction. In the absence of the element, the plants do not complete their life cycle or set the seeds.
(b) The requirement of the element must be specific and not replaceable by another element. In other words, deficiency of any one element cannot be met by supplying some other element.
(c) The element must be directly involved in the metabolism of the plant. Based upon the above criteria only a few elements have been found to be absolutely essential for plant growth and metabolism. These elements are further divided into two broad categories based on their quantitative requirements.
(i) Macronutrients and
(ii) Micronutrients
Macronutrients are generally present in plant tissues in large amounts (in excess of 10 mmole kg~’ of dry matter). The macronutrients include carbon, hydrogen, oxygen, nitrogen, phosphorous, sulphur, potassium, calcium and magnesium. Of these, carbon, hydrogen and oxygen are mainly obtained from C02 and H20, while the others are absorbed from the soil as mineral nutrition.
Micronutrients or trace elements are needed in very small amounts (less than 10 mmole kg-1 of dry matter). These include iron, manganese, copper, molybdenum, zinc, boron, chlorine and nickel.

Q4. With the help of examples describe the classification of essential elements based on the function they perform.
Ans: Essential elements can also be grouped into four broad categories on the basis of their diverse functions. These categories are:
(i) Essential elements as components of biomolecules and hence structural elements of cells (e.g., carbon, hydrogen, oxygen and nitrogen).
(ii) Essential elements that are components of energy-related chemical compounds in plants (e.g., magnesium in chlorophyll and phosphorous in ATP).
(iii) Essential elements that activate or inhibit enzymes, for example Mg2+ is an activator for both ribulose bisphosphate carboxylase/oxygenase and phosphoenol pyruvate carboxylase, both of which are critical enzymes in photosynthetic carbon fixation; Zn +is an activator of alcohol dehydrogenase and Mo of nitrogenase during nitrogen metabolism.
(iv) Some essential elements can alter the osmotic potential of a cell. Potassium plays an important role in the opening and closing of stomata. Minerals also play role as solutes in determining the water potential of a cell.

Q5. We know that plants require nutrients. If we supply these in excess, will it be beneficial to the plants? If yes, how/ If no, why?
Ans: No, excess supply of nutrients is not beneficial for the plants. It is toxic to the plants. Any mineral ion concentration in tissues that reduces the dry weight of tissues by about 10% is considered toxic. Such critical concentrations vary widely among different micronutrients. The toxicity symptoms are difficult to identify. Toxicity levels for any element also vary for different plants. Many a times, excess of an element may inhibit the uptake of another element. For example, the prominent symptom of manganese toxicity is the appearance of brown spots surrounded by chlorotic veins. It is important to know that manganese competes with iron and magnesium for uptake and with magnesium for binding with enzymes. Manganese also inhibits calcium translocation in shoot apex. Therefore, excess of manganese may, in fact, induce deficiencies of iron, magnesium and calcium.

Q6. Trace the events starting from the coming in contact of Rhizobiwn to a leguminous root till nodule formation. Add a note on importance of leghaemoglobin.
Ans: Nodule Formation: Nodule formation involves a sequence of multiple interactions between Rhizobium and roots of the host plant. Principal stages in the nodule formation are summarised as follows: Rhizobia multiply and colonise the surroundings of roots and get attached to epidermal and root-hair cells. The root-hairs curl and the bacteria invade the root-hairs. An infection thread is produced carrying the bacteria into the cortex of the root, where they initiate the nodule formation in the cortex of the root. Then the bacteria are released from the thread into the cells which leads to the differentiation of specialised nitrogen fixing cells. The nodule thus formed, establishes a direct vascular connection with the host for exchange of nutrients.
Importance of leghemoglobin: The enzyme nitrogenase is highly sensitive to the molecular oxygen; it requires anaerobic conditions. The nodules have adaptations that ensure that the enzyme is protected from oxygen. To protect these enzymes, the nodule contains an oxygen scavenger called leghaemoglobin. It is interesting to note that these microbes live as aerobes under free-living conditions (where nitrogenase is not operational), but during nitrogen-fixing events, they become anaerobic (thus protecting the nitrogenase enzyme).

Q7. Give the biochemical events occurring in the root nodule of a pulse plant. What is the end product? What is its fate?
Ans: The nodule contains all the necessary biochemical components, such as the enzyme nitrogenase and leghaemoglobin. The enzyme nitrogenase is an Mo-Fe protein and catalyses the conversion of atmospheric nitrogen to ammonia, the first stable product of nitrogen fixation.

• Glutamic acid is the main amino acid from which transfer of NH2 (amino group) takes place and other amino acids are formed through transamination. Enzyme transaminase catalyses all such reactions.
For example, the two most important amides (asparagine and glutamine) found in plants are a structural part of proteins.
• Asparagine formed from aspartic acid and glutamine is formed from glutamic acid by addition of amino group to each. The hydroxyl part of the acid is replaced by another NH2 radicle.

Q8. Hydroponics have been shown to be a successful technique for growing of plants. Yet most of the crops are still grown on land. Why?
Ans: The technique of growing plants in anutrient solution is known as hydroponics. Since, then a number of improvised methods have been employed to try and determine the mineral nutrients essential for plants. The essence of all these methods involves the culture of plants in a soil-free, defined mineral solution. These methods require purified water and mineral nutrient salts. Hydroponics has been successfully employed as a technique for the commercial production of vegetables such as tomato, seedless cucumber and lettuce.
Yet most of the crops are still grown on land because it must be emphasised that the nutrient solutions must be adequately aerated to obtain the optimum growth. On land no such conditions are required.

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