NCERT Biology class 11 chapter 8 – Page 2

NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules

January 5, 2022June 18, 2021 by Raju

NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules

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

Question 1.
What are macromolecules? Give examples.
Solution:
Biomolecules i.e. chemical compounds found in living organisms are of two types. One, those which have molecular weights less than one thousand and are usually referred to as macromolecules or simply as biomolecules while those which are found in the acid-insoluble fraction are called macromolecules or as biomacromolecules.

The molecules in the insoluble fraction with the exception of lipids are polymeric substances. Then why do lipids, whose molecular weights do not exceed 800, come under acid-insoluble fractions i.e., macromolecular fractions?

Question 2.
Illustrate a glycosidic, peptide and a phospho-diester bond.
Solution:
(a) Glycosidic bond: It is a bond formed between two monosaccharide molecules in a polysaccharide. This bond is formed between two carbon atoms of two adjacent monosaccharides.

(b) Peptide bond: Amino acids are linked by a peptide bond which is between the carboxyl (- COOH) group of one amino acid and the amino (- NH₂) group of the next amino acid which is formed by the dehydration process.

(c) Phosphodiester bond: This is the bond present between the phosphate and hydroxyl group of sugar which is called an ester bond. As this ester bond is present on either side, it is called a phosphodiester bond.

Question 3.
What is meant by the tertiary structure of proteins?
Solution:
Tertiary structure of protein : When the individual peptide chains of secondary structure of protein are further extensively coiled and folded into sphere-like shapes with the hydrogen bonds between the amino and carboxyl group and various other kinds of bonds cross-linking on-chain to another they form tertiary structure. The ability of proteins to carry out specific reactions is the result of their primary, secondary and tertiary structure.
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 1

Question 4.
Find and write down structures of 10 interesting small molecular weight biomolecules. Find if there is any industry which manufactures the compounds by isolation. Find out who are the buyers?
Solution:

NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 3

Fat is being manufactured by many companies in pharmaceuticals business as well as in food business. Vitamins come in many combination and are being used as supplementary medicines. Lactose is made by companies in manufacturing baby food. All of us are buyers of fat, protein and lactose.

Question 5.
Proteins have primary structures. 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:
The primary structure of proteins is described as the type, number, and order of amino acids in the chain. A protein is imagined as a line whose left end represents the first and right end represents the last amino acid. But in fact, this is not so simple. Actually, the number of amino acids in between the two termini determines the purity or homogeneity of a protein.

Question 6.
Find out and make a list of proteins used as therapeutic agents. Find other applications of proteins (e.g., Cosmetics, etc.)
Solution:
Haemoglobin, Insulin, thyroxine, growth hormone, other hormones of the adenohypophysis, serum albumen, serum globulin, fibrinogen, etc. are used as the therapeutic agents. Proteins are also used for the synthesis of food supplements, film, paint, plastic, etc.

Question 7.
Explain the composition of triglyceride.
Solution:
Triglycerides are esters of three molecules of fatty acids and one molecule of glycerol.
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 5

Question 8.
Can you describe what happens when milk is converted into curd or yoghurt, from your understanding of proteins.
Solution:
Conversion of milk into curd is the digestion of milk protein casein. Semi digested milk is the curd. In the stomach, renin converts milk protein into paracasein which then reacts with Ca⁺⁺ ion to form calcium paracaseinate which is called the curd or yoghurt.

Question 9.
Can you attempt building models of biomolecules using commercially available atomic models (Ball and Stick model)?
Solution:
Yes, the Three-dimensional structure of cellulose can be made using balls and sticks. Similarly, models of other bimolecular can be made
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 6

Question 10.
Attempt titrating an amino acid against a weak base and discover the number of dissociating (ionizable) functional groups in the amino acid.
Solution:
When an amino acid is titrated with weak base then its-COOH group also acts as weak acid. So it forms a salt with weak base then the pH of the resulting solution is near 7, so there is no sudden change. Number of dissociating functional groups are two, one is amino group (NH2) and another is carboxylic group ( – COOH). In the titration, amino acid acts as an indicator. Amino acids in solution acts as basic or acidic as situation demands. So these are also called amphipathic molecules.

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

Question 12.
What are gums made of? Is fevicol different?
Solution:
Gums are categorized into secondary metabolites or biomolecules. Thousands of compounds one present in plant-fungal and microbial cells. They are derived from these things. But is different. Fevicol has not derived from paper written cells.

Question 13.
Find out a qualitative test for proteins, fats and oils, amino acid and test any fruit juice, saliva, sweat and urine for them.
Solution:
Qualitative Tests for proteins, amino acids, and fats:
Biuret Test: Biuret test for protein identifies the presence of protein by producing violet colour of solution. Biuret H2NCONHCONH2 reacts with copper ion in a basic solution and gives violet colour.
Liebermann-Burchard Test for cholesterol:
This is a mixture of acidic anhydride and sulphuric acid. This gives a green colour when mixed with cholesterol.
Grease Test for oil: Certain oils give a translucent stain on clothes. This tesi can be used to show presence of fat in vegetable oils. These tests can be performed to check presence of proteins and amino acids and fats in any of the fluid mentioned in the question.

Question 14.
Find out how much cellulose is made by all the plants in the biosphere and compare it with how much of paper is manufactured by man and hence what is the consumption of plant material by man annually. What a loss of vegetation?
Solution:
According to a 2006 report from the UN, forests store about 312 billion tons of carbon in their biomass alone. If you add to that the carbon in deadwood, litter, and forest soil, the figure increases to about 1.1 trillion tons! The UN assessment also shows that the destruction of forests adds almost 2.2 billion tons of carbon to the atmosphere each year, the equivalent of what the U.S. emits annually. Many climate experts believe that the preservation and restoration of forests offers one of the least expensive and best ways to fight against climate change.
Although it is difficult to get exact data about the quantum of cellulose produced by plants, but above information can give some idea. About 10% of cellulose is used in paper making. The percentage is less but wrong practice of cutting wood and re-plantation makes the problem complicated. Usually older trees are cut for large quantity of cellulose and re-plantation is limited to selected species of plants. Selected species disturb the biodiversity as it leads to monoculture.
Add to this the problem of effluents coming out of a paper factory and the problem further aggravates.

Question 15.
Describe the important properties of enzymes.
Solution:
Properties of enzymes

Enzyme catalysis hydrolysis of ester, ether, peptide, c-c, c-halids, or P-N bonds.
Enzymes catalysis removal of the group from the substrate by mechanisms other than hydrolysis of leaving double bonds.
Enzymes generally function in a narrow range of temperature and p^H.
Activity declines both below and above optimum temperature and p^H.
The higher the affinity of the enzyme for its substrate the greater is its catalytic activity.
The activity of an enzyme is also sensitive to the presence of specific chemicals that bind to the enzyme.
For eg: Inhibitors that shuts off enzyme activity and Co-factors that facilitate catalytic activity.
Enzymes retain their identity at the end of the reaction.

VERY SHORT ANSWER QUESTIONS

Question 1.
Which organic compound is commonly called animal starch?
Solution:
Glycogen

Question 2.
Name the biomolecules of life.
Solution:
Carbohydrates, Lipids, Proteins, Enzymes, and nucleic acids.

Question 3.
Name one basic amino acid.
Solution:
Lysine.

Question 4.
Name one heteropolysaccharide.
Solution:
Chitin

Question 5.
Name the biomolecules present in the acid-insoluble fraction.
Solution:
Protein, polysaccharide, nucleic acid, and lipids.

Question 6.
Name the bond formed between sugar molecules.
Solution:
Glycosidic bond.

Question 7.
Name three pyrimidines.
Solution:
Thymine, cytosine, and uracil

Question 8.
Which enzyme does catalyse covalent bonding between two molecules to form a large molecule?
Solution:
Ligases.

Question 9.
On reaction with iodine, starch turns blue-black, why?
Solution:
The appearance of blue colour with the addition of iodine is due to its reaction with amylose fraction of starch.

Question 10.
Which type of bonds are found in proteins and polysaccharides?
Solution:
Peptides bond in protein and glycosidic bonds in polysaccharides.

Question 11.
Name one neutral amino acid.
Solution:
Valine.

Question 12.
Where does histone occur?
Solution:
Chromosomes.

Question 13.
Name two different kinds of metabolism.
Solution:
Anabolism and catabolism.

SHORT ANSWER QUESTIONS

Question 1.
Which type of bonds are found in nucleic acids?
Solution:
Phosphodiester bond.

Question 2.
What are the monosaccharides present in DNA and RNA? (Chikmagalur 2004)
Solution:
Deoxyribose in DNA and Ribose in RNA.

Question 3.
What are fatty acids? Give two examples.
Solution:
Fatty acids are compounds which have a carboxyl group attached to an R-group, which could be a methyl (CH3), or ethyl (C2H5) group or a higher number of CH2 groups e.g., Linoleic acid, Palmitic acid.

Question 4.
What are co-enzymes? Give two examples.
Solution:
Coenzymes are the non-protein organic ^compounds bound to the apoenzyme in a conjugate enzyme, their association with the apoenzyme is only transient, e.g., Nicotinamide adenine dinucleotide (NAD). Flavin adenine dinucleotide (FAD), Nicotinamide adenine dinucleotide phosphate (NADP).

Question 5.
(i) What is meant by complementary base pairing?
(ii) What is the distance between two successive bases in a strand of DNA?
(iii) How many base pairs are present in one turn of the helix of a DNA strand?
Solution:
(i) Complementary base pairing is the type of
pairing in DNA, where a purine always pairs with a pyrimidine, i.e., adenine pairs with thymine (A=T) and guanine pairs with cytosine (G=C).
(ii) 0.34 nm or 34 A is the distance between two successive bases in the strand of DNA
(iii) 10 base pairs

Question 6.
Differentiate between DNA and RNA.
Solution:
The main differences between DNA add RNA are as following
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 8
Question 7.
What la a prosthetic group? Give an example.
Solution:
The non-protein part of a conjugated protein is called a prosthetic group. For example in a nucleoprotein (nucleic acid is the prosthetic group).

Question 8.
Differentiate between essential amino acids and non-essential amino acids.
Solution:

Question 9.
Differentiate between Structural Proteins and Functional Proteins.
Solution:

Question 10.
What is activation energy?
Solution:
Activation Energy: An energy barrier is required for the reactant molecules for their activation. So this energy with enzyme-substrate reaction is called Activation energy.
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 12

The activation energy is low for reactions with catalysts [enzymes] than those with Non enzymatic reactions.

Question 11.
What are the components of enzymes?
Solution:
Enzymes are made up of protein as well as non – protein parts. The protein part is called an apoenzyme and the non-protein part is a coenzyme. These two together are called a holoenzyme.

LONG ANSWER QUESTIONS

Question 1.
How many classes are enzymes divided into? Name all the classes.
Solution:
Enzymes are divided into 6 classes. Namely

Oxidoreductases/dehydrogenases: Enzymes which catalyze oxidoreduction between two substrates
Transferases: Enzymes catalyzing a transfer of group between a pair of substrates.
Hydrolases: Enzymes catalyzing the hydrolysis of ester, ether, peptide, glycosidic, C-C-C-halide or P.N bonds.
Lyases: Enzymes catalyze the removal of groups from – substrates by mechanisms other than hydrolysis leaving double bonds.
Lyases: Enzymes catalyzing the interconversion of optical geometric or positional isomers.
Ligases: Enzymes catalyzing the linking together of 2 compounds.

Question 2.
Distinguish between the primary, secondary, and tertiary structures of proteins.
Solution:

Question 3.
Explain the effect of the following factors on enzyme activity:
(i) Temperature
(ii) pH.
Solution:
Temperature: An enzyme is active within a narrow range of temperature. The temperature at which an enzyme shows its highest activity is called optimum temperature.

It generally corresponds to the body temperature of warm blood animals e.g., 37°C in human beings. Enzyme activity decreases above and below this temperature. Enzyme becomes inactive below minimum temperature and beyond maximum temperature.

Low temperature present inside cold storage prevents spoilage of food. High temperature destroys enzymes by causing their denaturation.
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 15

The relation between temperature and enzyme controlled reaction velocity

pH – Every enzyme has an optimum pH when it is most effective.

A rise or fall in pH reduces enzyme activity by changing the degree of ionisation of its side chains. A change in pH may also reverse the reaction.

Most of the intracellular enzymes function near-neutral pH with the exception of several digestive enzymes which work either in acidic range of pH or alkaline range of pH. pH for trypsin is 8.5.

Question 4.
Discuss the B-DNA helical structure with the help of a diagram.
Solution:

Watson & Crick suggested the double-helical structure of DNA in 1953.
The backbone of the DNA molecule is made up of deoxyribonucleotide units joined by a phosphodiester bond.
The DNA molecule consists of two chains wrapped around each other.
The two helical strands are bound to each other by Hydrogen Bonds.
Purines bind with pyrimidines A = T, C = G
The pairing is specific and the two chains are complementary.
One strand has the orientation 5’ → 3’ and other has 3’ → 5’.
Both polynucleotides strands remain separated with a 20A° distance.
The coiling is right-handed.

NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 16

Question 5.
What are different kinds of enzymes? Mention with enzyme examples.
Solution:
Enzymes with substrate bonds are broken and changed to different kinds as

Oxidoreductases: eg Alcohol dehydrogenase, oxidation, Reduction occurs
Transferases: transfer a particular group to another substrate, eg. transavninase
Hydrolases: cleave their substrates by hydrolysis of a covalent bond e.g. Urease, amylase.
Lyases: break the covalent bond eg. Deaminase
Isomerase: by changing the bonds they make isomers. eg: Aldolase.
Ligase: These bind two substrate molecules eg: DNA ligase, RNA ligase

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

NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption

January 5, 2022June 18, 2021 by Raju

NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption.

Question 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) Succuss enterics is the name given to:
(i) a junction between the ileum and large intestine
(ii) intestinal juice
(iii) swelling in the gut
Solution:
(a) (i) Pepsin, lipase, and rennin
(b) (ii) Intestinal juice

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

Question 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 the alimentary canal?
(d) How does bile help in the digestion of fats
Solution:
(a) Villi increases surface area for absorption and maximum absorption takes place in the intestine.
(b) Coming in contact with hydrochloric acid in stomach proenzyme pepsinogen convert to its active form pepsin, the proteolytic enzyme of the stomach.
(c) There are four basic layers in the wall of alimentary canal i.e. serosa, muscularis, submucosa and mucosa.
(d) Bile helps in the emulsification of fats i.e. breakdown the fats into very small micelles. Bile also activates lipases.

Question 4.
State the role of pancreatic juice in the digestion of proteins.
Solution:
Pancreatic juice contains inactive enzymes like trypsinogen, chymotrypsinogen, procarboxypeptidases. Trypsinogen is activated by an enzyme enterokinase secreted by the intestinal mucosa into active trypsin which in turn activates other enzymes in the pancreatic juice. Proteins, proteases, and peptones in the chyme are digested by the proteolytic enzymes of pancreatic juice.

Question 5.
Describe the process of digestion of protein in the stomach.
Solution:
The mucosa of the stomach has gastric glands that secrete mucus, proenzyme pepsinogen, HCl, and intrinsic factor. Intrinsic factor is essential for the absorption of vitamin B12. The food mixes thoroughly with acidic gastric juice of the stomach and called the chyme. The proenzyme pepsinogen on exposure to HC1 gets converted to active enzyme pepsin. Pepsin converts proteins into proteases and peptones. Renin found in the gastric juice of infants also helps in the digestion of milk protein.

Question 6.
Give the dental formula of human beings.
Solution:
Dental formula of human beings is 2123/2123.

Question 7.
Bile juice contains no digestive enzymes, yet it is important for digestion. Why?
Solution:
The bile juice released into the duodenum contains bile pigments (bilirubin and biliverdin), bile salts, cholesterol, and phospholipids but no enzymes. Bile helps in the emulsification of fats i.e. break down the fats into very small micelles. Bile also activates lipases. Bile is alkaline (pH about 8). Bile pigments are excreted in faeces. In absence of HCl, the above-mentioned functions will not occur and digestion of food will be affected.

HCl of the gastric juice may cause gastric or duodenal ulcers and damage, the underlying blood vessels. These cause hemorrhage inside. Strong HCl is produced in stress conditions also. Parental or oxyntic cells which secrete HCl and intrinsic factor (Intrinsic factor is essential for the absorption of vitamin B₂)

Question 8.
Describe the digestive role of chymotrypsin. Which two other digestive enzymes of the same category are secreted by their source gland?
Solution:
Chymotrypsin changes proteins into peptides and also milk protein, changes into paracasein (curd). The pancreatic juice contains inactive enzymes, trypsinogen, chymotrypsinogen, procarboxypeptidases, amylases, lipases and nucleases. Trypsinogen is activated by an enzyme, enterokinase, secreted by the intestinal mucosa into active trypsin, which in turn activates the other enzymes in the pancreatic juice.

Trypsin changes proteases and peptones into peptides and amino acids. Procarboxypetidases change peptides into small peptides and amino acids. Amylases change starch into maltose. Lipases change emulsified fat into fatty acids and glycerol. Nucleases changes nucleotides into phosphate, sugar and nitrogen bases.

Question 9.
How are polysaccharides and disaccharides digested?
Solution:
(a) Digestion of carbohydrates starts in the mouth cavity with the help of enzymes salivary amylase.
NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 1

Question 10.
What would happen if HCl were not secreted in the stomach?
Solution:
HCl is secreted by oxyntic cells in the stomach wall and performs five functions:

Kill bacteria and germs
Loosens fibrous material of food.
Activates proenzyme pepsinogen to its active form pepsin.
Gives an acidic medium for action by pepsin.
Curdles milk. Pepsin changes proteins into proteases and peptones.

Question 11.
How does butter in your food get digested and absorbed in the body?
Solution:
Butter is a kind of fat. Fats are broken down by lipases with the help of bile into di- and monoglycerides:
NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 2

Question 12.
Discuss the main steps in the digestion of proteins as the food passes through different parts of the alimentary canal.
Solution:
(a) In the stomach the proenzyme pepsinogen, on exposure to HC1 converted into active pepsin that converts proteins into proteases and peptones.

(b) Proteins, proteases and peptides in the chyme reaching the intestine are acted upon by proteolytic enzymes of pancreatic juice and converted to dipeptides.
NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 4
(c) The enzymes in succuss entericus act on the end product to form amino acids.

Question 13.
Explain the terms thecodont and diphyodont.
Solution:
Thecodont: In human beings, teeth are
embedded in pits, the sockets of the jawbones. Such teeth are called the thecodont.
Diphyodont: The teeth that appear in two sets, i. e., milk-teeth which are later replaced by permanent teeth. This condition is called diphyodont.

Question 14.
Name different types of teeth and their number in an adult human.
Solution:
Incisors – 8
Canines – 4
Premolars – 8
Molars – 12

Question 15.
What are the functions of the liver?
Solution:

Secretion of bile
Synthesis of blood clotting factors
Regulating carbohydrate, protein, and lipid metabolism.
Synthesis of amino acids, plasma proteins, cholesterol, etc.
Stores glycogen, fats, fat-soluble vitamins, etc.
Detoxifies toxic substances
Elimination of foreign bodies
Produces heat through metabolism
Produces anticoagulant heparin
Synthesis of urea from ammonia.

VERY SHORT ANSWER QUESTIONS

Question 1.
What is digestion?
Solution:
Digestion is the mechanical, enzymatic, and biochemical transformation of complex (polymers) food molecules into simple molecules (monomers) which are suitable for absorption.

Question 2.
Which is the food constituent that bile helps to digest and absorb?
Solution:
Fats.
Question 3.
What is the function of enterokinase?
Solution:
Enterokinase of intestinal juice activates the inactive trypsinogen into trypsin which digests protein in the duodenum.

Question 4.
Which is a nondigestive activating enzyme?
Solution:
Enterokinase.

Question 5.
Mention the role of bile salt in the digestion of fats.
Solution:
Bile salts emulsify fat particles and reduce the surface tension of fat droplets to increase the action of enzyme lipase.5. Bile salts emulsify fat particles and reduce the surface tension of fat droplets to increase the action of the enzyme lipase.

Question 6.
What is the role of HCl in protein digestion?
Solution:
Role of HCl:

It activates pepsinogen into active pepsin.
It provides a suitable acidic medium for the action of proteases in the stomach.

Question 7.
Name the hardest substance in the body.
Solution:
Enamel.

Question 8.
What is a cystic duct?
Solution:
Duct of gall bladder.

Question 9.
Mention two functions of mucus.
Solution:
Role of mucus:
(a) Acts as a lubricant.
(b) Protects the epithelial surface of the stomach from the corrosive effect of hydrochloric acid and digestion by pepsin.

Question 10.
Name the secretion of goblet cells in the human stomach.
Solution:
Goblet cells secrete mucus.

Question 11.
Where the taste buds located?
Solution:
Taste buds are located in the papillae on the upper surface of the tongue.

Question 12.
What is the ‘Pancreatic Enzyme’ which acts on starch? (Oct. 83, 01)
Solution:
The pancreatic amylase(Amylopsin).

Question 13.
What is the function of epiglottis?
Solution:
Epiglottis prevents the entry of food into the trachea, by closing its opening called the glottis.

Question 14.
Which part of the stomach continues into the duodenum?
Solution:
Pyloric region

Question 15.
What name is given to the major lymph vessel present in the intestinal villi?
Solution:
Lacteal

Question 16.
Where are the crypts of Leiberkuhn located?
Solution:
Crypts of Lieberkuhn are located in between the bases of the villi in the intestine.

Question 17.
Name the structural and functional unit of the liver.
Solution:
Hepatic lobules

Question 18.
Mention the secretion of Goblet cells.
Solution:
Mucin (Mucus) (April 93)

Question 19.
What is a bolus?
Solution:
When the thoroughly masticated food mixes with the saliva, the food particles become adhered together by the mucus known as bolus.

Question 20.
What is chyme?
Solution:
After partial digestion in the stomach, the form of food is called chyme.

Question 21.
What is the meaning of deglutition?
Solution:
The act of swallowing is called deglutition.

Question 22.
Name the enzyme involved in the breakdown of nucleotides into sugars and bases.
Solution:

SHORT ANSWER QUESTIONS

Question 1.
Mention the juice secreted by the liver. State its function indigestion. (April 1985)
Solution:
The liver cells secrete a juice called the ‘Bile juice’. ‘Bile salts’ are one of the components of Bile juice. These are very important in digesting lipids or fats. Specifically ‘Bile salts’ aid in digestion than bringing about digestion of fat by emulsifying fats and making the fat molecules accessible to the action of the pancreatic Lipase (or Lipid digesting) enzyme. Apart from this the bile juice also creates an alkaline medium in the intestine so that food particles can be actively acted upon by the pancreatic enzymes.

Question 2.
Name the organs which secret carboxypeptidases and aminopeptidases respectively. Give the function performed by these enzymes
Solution:
Carboxypeptidases are secreted by the pancreas. Aminopeptidases are secreted by the intestine. Both the enzymes act on the terminal peptide bonds and release the terminal/last amino acids of the peptide chain.

Question 3.
State the Role of HCI indigestion. (Oct. 88, April 93, 98)
Solution:
The main role of HCI is to convert the inactive enzyme pepsinogen to pepsin which in turn helps in the digestion of proteins. It also creates an acidic medium in the stomach for the activity of pepsin.

Question 4.
What is succus entericus? Mention any two carbohydrate digesting enzymes present in it? (April 2006)
Solution:
Succus entericus intestinal juice is the secretion of intestinal glands in the ileum of the small intestine. The carbohydrate digesting enzymes are maltase, lactase, and sucrase.

Question 5.
Differentiate between micelles and chylomicrons.
Solution:
The differences between micelles and chylomicrons are
NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 7

Question 6.
How is our gut lining protected from its own secretion of proteases?
Solution:
(i) Protease is secreted in an inactive form and poses no threat to the gut lining.
(ii) The mucus provides protection to the epithelial lining.

Question 7.
Name the organs which secrete carboxypeptidases and aminopeptidases respectively. Give the functions performed by these enzymes.
Solution:
Carboxypeptidases are secreted by the exocrine part of the pancreas. Aminopeptidases are secreted by the intestinal mucosa. These enzymes act on the terminal peptide bonds and release the last amino acid from the polypeptide chain, thereby progressively shortening the peptide chain.

Question 8.
Where is the ileocaecal valve present? What is its function?
Solution:
The ileo-caecal valve is present at the junction of the ileum of the small intestine and the caecum of the large intestine.
It prevents the backflow of the matter from the caecum into the ileum.

Question 9.
How does the nervous system control the activities of the gastro-intestinal tract?
Solution:
The sight, smell and presence of food in the oral cavity can stimulate the secretion of saliva. Gastric and intestinal secretions are also stimulated by similar neural signals. Muscular activities of the alimentary canal are coordinated by both local and CNS neural mechanisms. Hormonal control of secretion of digestive enzymes is carried out by local hormones.

Question 10.
Name the final products of the digestion of proteins. How and where are they absorbed from the alimentary canal?
Solution:
The final products of digestion of proteins are amino acids They are absorbed by an active process utilising energy against the concentration gradient in the ileum.

Question 11.
What is the pancreas? Mention the major secretions of the pancreas that are helpful in digestion.
Solution:
The pancreas is a carrot-shaped soft greyish pink gland that lies transversely below the stomach between the duodenum and spleen, which secretes digestive enzymes from its exocrine parts and hormones from its endocrine parts.
The pancreas secretes three enzymes in inactive proenzyme or zymogen state and three in active enzyme state.
Proenzymes. Trypsinogen, chymotrypsinogen, and procarboxypeptidases.
Active Enzymes. Amylase, lipase, and nucleases.

LONG ANSWER QUESTIONS

Question 1.
What is meant by vernalization? Explain the significance of vernalization.
Solution:

Question 2.
Write short notes on
(a) Liver
(b) Layers of the alimentary canal
Solution:
(a) Liver: The liver is the largest gland of the body. It is situated in the abdominal cavity, just below the diaphragm and has two lobes. The hepatic lobules are the structural and functional units of the liver containing hepatic cells arranged in the form of cords. Each lobule is covered by a thin connective tissue sheath called the Glisson’s capsule. The bile secreted by the hepatic cells passes through the hepatic ducts and is stored and concentrated in a thin muscular sac called the gall bladder. The duct of gall bladder (Cystic duct) along with the peptic duct from the liver forms the common bile duct. The bile contains pigments like bilirubin and biliverdin, bile salts, cholesterol and phospholipids but no enzymes. Bile helps in the emulsification of fats and activates lipases.

(b) Layers of the alimentary canal:
The wall of the alimentary canal possesses four layers namely serosa, muscularis, sub-mucosa, and mucosa. The serosa is the outermost layer and is made up of a thin mesothelium with some connective tissues. Muscularis is formed by smooth muscles usually arranged into an inner circular and an outer longitudinal layer. An oblique muscle layer may be present in some regions.

The sub-mucosal layer is formed of loose connective tissues containing nerves, blood, and lymph vessels. In the duodenum, glands are also present in the sub-mucosa. The innermost layer lining the lumen of the alimentary canal is the mucosa. This layer forms irregular folds in the stomach and small finger-like foldings called villi in the small intestine. The cells lining the villi produce numerous microscopic projections called microvilli giving a brush border appearance.

Villi are supplied with a network of capillaries and a large lymph vessel called the lacteal. The mucosal epithelium has goblet cells which secrete mucus that help in lubrication. Mucosa also forms glands in the stomach and crypts in between the bases of villi in the intestine (crypts of lieberkuhn).

Question 3.
Describe the major disorders of the human digestive system.
Solution:
Disorders of the digestive system:
(i) Indigestion:

It is the condition in which the food is not properly digested leading to a feeling of fullness.
It is caused by inadequate secretion of digestive enzymes, food poisoning, overeating or spicy food.

(ii) Constipation

It refers to the condition where the faeces are retained in the rectum for longer periods as the bowel movements occur irregularly.

(iii) Diarrhoea:

It refers to the abnormal frequency of bowel movement and increased liquidity of the faecal discharge; absorption of food is impaired.

(iv) Vomiting:

It is the ejection of stomach contents through the mouth; this reflex action is controlled by the vomit centre in the medulla.
It is due to viral infection, where liver is affected and digestion of fats is impaired.
The eyes and skin turn yellow due to the deposit of bile pigments.

Question 4.
How is the DNA content in our diet digested in the body?
Solution:
DNA content is digested in the intestinal part of our alimentary canal by the enzymes present in pancreatic juice and succus entericus.
NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 9
DNAase is found in pancreatic juice while nucleotidase and nucleosidase occur in succus entericus and hydrolyse the DNA content in our diet.

We hope the NCERT Solutions for Class 11 Biology at Work Chapter 16 Digestion and Absorption, help you. If you have any query regarding NCERT Solutions for Class 11 Biology at Work Chapter 16 Digestion and Absorption, drop a comment below and we will get back to you at the earliest.

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

January 5, 2022June 18, 2021 by Raju

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

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

Question 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 from pre¬existing cells.
(d) A unicellular organism carries out its life activities within single cell.
Solution:
(a) Robert brown discovered the cell.

Question 2.
New cells generate from
(a) bacterial fermentation
(b) regeneration of old cells
(c) pre-existing cells
(d) abiotic material
Solution:
(c) pre-existing cells

Question 3.
Match the following
Column A Column B
(a) Cristae (i) Flat membranous sac in stroma
(b) Cisternae (ii) Infoldings in mitochondria
(c) Thylakoids (iii) Disc-shaped sacs in Golgi apparatus
Solution:
(a) Cristae (ii) Infoldings in mitochondria
(b) Cisternae (iii) Disc-shaped sacs in Golgi apparatus
(c) Thylakoids (i) Flat membranous sac in stroma

Question 4.
Which of the following is correct
(a) Cells of living organisms have a nucleus.
(b) Both animals 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 materials
Solution:
(c) In prokaryotes, there are no membrane-bound organelles.

Question 5.
What are mesosomes in a prokaryotic cells? Mention the function that it performs.
Solution:
A special membranous structure is a mesosome that is formed by the extensions of the plasma membrane into the cell. These extensions are in the form of vesicles, tubules, and lamellae. They help in cell wall formation, DNA replication, and distribution to daughter cells. They also help in respiration, in the secretion process increase plasma membrane surface and enzymatic content.

Question 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?
Solution:
Neutral solutes may move across the membrane by the process of simple diffusion along the concentration gradient i.e. from higher concentration to the lower. Water may also move across this membrane from higher to lower concentrations. The movement of water by diffusion is called osmosis. As the polar molecules cannot pass through the non-polar lipid bilayer, they require a carrier protein of the membrane to facilitate their transport across the membrane.

A few ions or molecules are transported across the membrane against their concentration gradient i.e. from lower to the higher concentration. Such transport is an energy-dependent process, in which ATP is utilized and is called active transport. e,g., Na’/K* Pump.

Question 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:
Two double membrane-bound cell organelles:
(a) Mitochondria: It has finger-like folds in the inner membrane called cristae. Mitochondria is the place for aerobic respiration.
NCERT Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 1
(b) Chloroplast: Chloroplast is responsible for converting light energy into chemical energy. Chloroplast contains stacked thylakoid in its matrix.

Functions of mitochondria: The double membrane mitochondria are actively associated with aerobic respiration and the release of energy for cellular activity. The biological oxidation of the fats and carbohydrate release much amount of energy which is utilised by mitochondria for ATP synthesis. The required energy is released from ATP molecules for various cell processes in cells so they are termed as “the powerhouse of the cell”
Functions of chloroplast :
(i) Their main function of the chloroplast is to trap the sun’s energy and to convert it into chemical energy of food by photosynthesis.
(ii) Storage of starch.
(iii) Chloroplast in fruits and flowers changes into chromoplasts.

Question 8.
What are the characteristics of prokaryotic cells?
Solution:
The prokaryotic cells are represented by bacteria, blue-green algae, mycoplasma, and PPLO (Pleuro Pneumonia-like organisms). They are generally smaller and multiply more rapidly than the eukaryotic cells. They may vary greatly in shape and size. The four basic shapes of bacteria are bacillus (rod-like), coccus (spherical), vibrio (comma-shaped), and spirillum (spiral).

The organization of the prokaryotic cell is fundamentally similar even though prokaryotes exhibit a wide variety of shapes and functions. All prokaryotes have a cell wall surrounding the cell membrane. The fluid matrix filling the cell is the cytoplasm. There is no well-defined nucleus. The genetic material is basically naked being not enveloped by a nuclear membrane.

In addition to the genomic DNA (the single chromosome/ DNA circle), many bacteria have small DNA circles outside the genomic DNA. These smaller DNA circles are called plasmids, The plasmid DNA confers certain unique phenotypic to antibiotics. Prokaryotes have something unique in the form of inclusions.

A specialised differentiated form of the cell membrane called mesosome is the characteristic of prokaryotes. They are essentially infoldings of the cell membrane.

Question 9.
Multicellular organisms have a division of labour. Explain.
Solution:
In unicellular organisms, there is no division of labour.

The single cell of the organisms is capable of performing all the vital activities of life i.e., respiration, movement, digestion and reproduction, etc. Respiration, nutrition, and excretion in most of these unicellular organisms takes place through the general body surface.
No special organs for these are present in them because they are too small to need them. Most of these unicellular organisms reproduce by simple binary division, to maintain their continuity.
However, in some, sexual reproduction has also been observed.

Question 10.
Cell is the basic unit of life. Discuss in brief.
Solution:
Cell: The Basic Unit of life: All living organisms are composed of small, tiny structures or compartments called cells. These cells are called the ‘building blocks’ of life.

The cells in true sense are considered as the basic unit of life because all the life processes i.e., metabolism, responsiveness, reproduction are carried out by the cells.
Respiration, nutrition, release of energy for the body are carried out within the cells only.
Even the animals and plants reproduce because the cells reproduce individually.
Growth occurs because cell grow and multiply.
In Amoeba all the life processes are performed within the boundaries of the single cell.
This is true of all other multicellular organisms. The only difference in the multicellular organisms is that the body of these organisms is made up of many cells.
In these organisms, the cell do not behave independently but get organized into tissues. Each tissue is specialized to perform specific functions. Different tissues then get organised into tissues.
Each tissue is specialized to perform specific functions. Different tissues then get organised into organs which perform certain specific functions.
Different organs are finally organised to form organ systems. Now it must be very clear that the basic structure to tissues, organs and organ system are the cells only.
These tissues, organs and organ system of the organisms work because the cells work.
Thus “the cells are structural and functional unit of the living beings” hence it is the basic unit of life.

Question 11.
What are nuclear pores? State their function.
Solution:
At a number of places, the nuclear envelope is intercepted by minute pores which are called nuclear pores. These are formed by the fusion of two nuclear membranes. These nuclear pores are the passages through which the movement of RNA and protein molecules takes place in both directions between the nucleus and the cytoplasm.

Question 12.
Both lysosomes and vacuoles are * endomembrane structures, yet they differ in
terms of their functions. Comment.
Solution:
Lysosomes are filled with hydrolytic enzymes that are capable of digesting carbohydrates, proteins, lipids and nucleic acids whereas vacuoles contain water, sap, excretory product and other materials not useful for cell.

Question 13.
Describe the structure of the following with the help of labelled diagrams.
(i) Nucleus
(ii) Centrosome
Solution:

(i) Nucleus:

1. The nucleus is a large organelle controlling all the activities of the eukaryotic cells. Some cells have more than one nucleus.
2. Binucleate cells have 2 nuclei per cell eg. Paramoecium. Multinucleate cells have many nuclei e.g. Ascaris.
3. Some cells lack nucleus (anucleate) at maturity. Examples: mammalian RBCs and sieve tube cells in vascular plants.
4. The nucleus is bounded by two membranes, which make the nuclear envelope.
5. The outer and inner membranes are separated by a narrow space, perinuclear space.
6. The outer membrane remains in continuation with endoplasmic reticulum (ER) and the inner one surrounds the nuclear contents.
7. At some points, the nuclear evelope is interrupted by the presence of small structures called nuclear pores.
8. These pores help in exchange of materials between nucleoplasm and cytoplasm. Nuclear membrane dissappears during cell division. It reappears during nuclear reorganization in stage.
9. The nucleoplasm contains chromatin and nucleolus. The nucleolus is a rounded structure. It is not separated from the rest of the nucleoplasm by membrane.
10. It is associated with a specific nucleolar organizing region (NOR) of some chromosomes. Nucleolus is the “site for ribosomal RNA synthesis”.
11. The cells which remain engaged in protein synthesis have larger and more numerous nuclei in their nucleoplasm.

NCERT Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 4

(ii) Centrosome:

Under the electron microscope, each centriole is seen to be formed of nine sets of tubular structures arranged in a circular fashion.
Each of these sets is a triplet composed of three microtubules. Each microtubule has a diameter of about 250A. The triplets are found in the matrix.
Sometimes delicate strands appear to connect sets of the triplet to each other.
Also can be seen radiating from the central core of the cylinder, delicate strands which connect sets of the triplets to each other giving a cartwheel appearance.
Basal bodies are structures similar to the centrioles. They produce cilia and flagella.

Question 14.
What is a centromere? How does the position of the centromere form the basis of the classification of chromosomes? Support your answer with a diagram showing the position of the centromere on different types of chromosomes.
Solution:
Eukaryotic chromosomes: The chromosomes are uncoiled in a loose, indistinct network called the chromatin that contains DNA, RNA and protein in interphase.

The types of proteins present and associated with DNA are histone and non-histone proteins.

Chromosomes are thread-like structures. They become visible (under light microscope) during cell division.

In higher organisms, the well-developed nucleus contains a definite number of chromosomes of definite size and shape.
NCERT Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 5
The shape of a chromosome is usually observable at metaphase and anaphase when the position of primary constriction {centromere) is clearly seen. Based on the position of the centromere, chromosomes are of 3 types:

telocentric – with terminal centromere,
the acrocentric – terminal centromere is capped by a telomere
submetacentric – the centromere is subterminal in position
metacentric – these have median centromere.

VERY SHORT ANSWER QUESTIONS

Question 1.
Who discovered the Golgi body?
Solution:
Camillo Golgi (1898).

Question 2.
Give the location of 70S ribosomes.
Solution:
Prokaryotic cells, plastids, and mitochondria

Question 3.
Name the cell organelle rich in acid hydrolases.
Solution:
Lysosomes

Question 4.
Who proposed the cell theory?
Solution:
Schleiden and Schwann.

Question 5.
Expand PPLO.
Solution:
PPLO (Pleuro Pneumonia Like Organisms)

Question 6.
Name the organelle responsible for protein synthesis in a cell.
Solution:
Ribosome

Question 7.
Give the full form of SER and RER.
Solution:

SER – Smooth endoplasmic reticulum.
RER – Rough endoplasmic reticulum.

Question 8.
Name the membrane which surrounds the vacuole in the cell.
Solution:
Tonoplast

Question 9.
Name two types of constituents of the plasma membrane.
Solution:
Proteins and lipids.

Question 10.
Name two processes of passive transport.
Solution:

Osmosis
Diffusion

Question 11.
What is plasmodesmata? What is its function?
Solution:
Plasmodesmata: Adjoining the cells and the linking gap of cytoplasmic protoplasmic presence is called Plasmodesmata. It links the neighbouring cells together.

SHORT ANSWER QUESTIONS

Question 1.
Why fluid-mosaic model is more accepted than other models of the plasma membrane?
Solution:
The fluid mosaic model explains
(i) quasifluid state of the plasma membrane,
(ii) It differentiates two 6. types of proteins
(iii) It explains functional specificity and variability in two surfaces of PM.

Question 2.
Name three types of elements in the Golgi body. List two major functions of the Golgi body.
Solution:
Three types of elements in golgi body are cistemae, vesicles and vacuoles. The main function of golgi bodies are cellular secretion and acrosome formation.

Question 3.
What is peculiar about mitochondrial DNA?
Solution:
Mitochondrial DNA is circular double-stranded and not associated with histone proteins.

Question 4.
How does cytokinesis take place in plant and animal cells?
Solution:
In plant cell cytokinesis take place by cell plate formation and in animal cells it occurs by constriction

Question 5.
Differentiate between prokaryotic and eukaryotic cells.
Solution:
The main differences between prokaryotic cell and eukaryotic cell are
NCERT Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 6

Question 6.
Differentiate between active and passive transport across the membrane.
Solution:
The main differences between active transport and passive transport are
NCERT Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 7

Question 7.
Describe the different methods of transport of nutrients in cell.
Solution:
Different methods of transport of nutrients in the cell are:
(i) Simple diffusion : It is the movement of ions/ molecules of any substance from a region of higher concentration to ‘the region of lower concentration, until equilibrium is reached. Many neutral solutes move by diffusion.
(ii) Osmosis : It is the movement of solvent molecules across a semipermeable membrane, from the region of higher concentration to the region of lower concentration, until equilibrium is reached. Water moves by osmosis from one cell to the other.
(iii) Facilitated diffusion : It refers to the movement of ions/molecules across the membrane with the help of transmembrane proteins.

Question 8.
Why are ribosomes of prokaryotes different from eukaryotes?
Solution:
The type of ribosomes of prokaryotes is different from eukaryotes because the prokaryotes were primitive, simpler and have remained intact during evolution while at the base level eukaryotes have adapted with the environment and are retaining their kind of entities for the complex structure. Prokaryotes have 70S ribosomes with 30S and 50S subunit and eukaryotes have 80S ribosome with 40S subunit and 60S sub unit.

Question 9.
What is the function of
(1) Nuclear Pores
(2) Slimy Capsule in Bacteria
(3) Golgi Bodies:
(4) Centrosome with Centrioles
Solution:
(1) Nuclear Pores: There is exchange of RNA and proteins through the nuclear pores
(2) Slimy Capsule in Bacteria : A slimy capsule is the outer covering of cell wall of bacteria and is an additional protection for the bacteria.
(3) Golgi Bodies :
(i) It takes part in packaging materials delivered either to the intra-cellular targets or secteted outside the cell.
(ii) It is also a important site of formation of glycoproteins and glycolipids.
(4) Centrosome with Centrioles
(i) Centrioles help in organising the spindle fibres and astral rays during cell division.
(ii) It also provides basal bodies which give rise to cilia and flagella.

LONG ANSWER QUESTIONS

Question 1.
Describe the structure of cell wall.
Solution:

A non-living rigid structure called the cell wall forms an outer covering for the plasma membrane of fungi and plants.
Cell wall does not only give shape to the cell and but protect the cell from mechanical damage and infection.
It also helps in cell-to-cell interaction and provides barrier to undesirable macromolecules.
Algae have cell wall, made of cellulose, galactans, mannans and minerals like calcium carbonate, while in other plants it consists of cehulose, hemicellulose, pectins and proteins.
The cell wall of a young plant cell, the primary wall is capable of growth, which gradually diminishes as the cell matures and the secondary wall is formed on the inner side of the cell.
The middle lamella is a layer mainly of calcium pectate which holds or glues the different neighbouring cells together.
The cell wall and middle lamellae may be transferred by plasmodesmata which connect the cytoplasm of neighbouring cells.

Question 2.
Give an account of prokaryotic cells.
Solution:

The prokaryotic cells are represented by bacteria, blue green algae, mycoplasma and PPLO (Pleuro Pneumonia Like Organisms). They are generally smaller and multiply more rapidly than the eukaryotic cells.
They may vary greatly in shape and size. The four basic shapes of bacteria are bacillus (rod like), coccus (spherical), vibrio (comma shaped) and spirillum (Spiral).
The organisation of the prokaryotic cell is fundamentally similar even though prokaryotes exhibit a wide variety of shapes and functions. All prokaryotes have a cell wall surrounding the cell membrane.
The fluid matrix filling the cell is the cytoplasm. There is no well defined nucleus.
The genetic material is basically naked, not enveloped by a nuclear membrane.
In addition to the genomic DNA (the single chromosome/circular DNA), many bacteria have small circular DNA outside the genomic DNA. These smaller DNA is called plasmids.
The plasmid DNA confers certain unique phenotypic characters to such bacteria. One such character is resistance to antibiotics. Nuclear membrane is found in eukaryotes.
No organelles, like the ones in eukaryotes, are found in prokaryotic cells except ribosomes.
Prokaryotes have something unique in the form of inclusions. A specialised differentiated form of cell membrane called mesosome is the characteristic of prokaryotes which helps in respiration process.
They are essentially infoldings of cell membrane.

Question 3.
Give an ultrastructure of mitochondria.
Solution:

Mitochondria, unless specifically stained, are not easily visible under the microscope.
The number of mitochondria per cell is variable depending on the physiological activity of the cells.
In terms of shape and size also, considerable degree of variability is observed.
Typically it is sausage shaped or cylindrical having a diameter of 0.2-1.0 ft m (average 0.5 film) and length (1.0 -4.1 ft).
Each mitochondrion is a double membrane-bound structure with the outer membrane and the inner membrane dividing its lumen distinctly into two aqueous compartments, i.e. the outer compartment and the inner compartment.
The inner compartment is called the matrix. The outer membrane forms the continuous limiting boundary of the organehe.
The inner membrane forms a number of infoldings called the cristae. The cristae increase the surface area.
The two membranes have their own specific enzymes associated with the mitochondrial function. Mitochondria are the sites of aerobic respiration.
They produce cellular energy in the form of ATP, hence they are called. “Power houses” of the cell.
The matrix also possesses single circular DNA molecule, a few RNA molecules, ribosomes (70s) and the components required for the synthesis of proteins. The Mitochondria divide by fission.

Question 4.
Describe the fluid mosaic model of membrane.
Solution:
The characteristic features of fluid mosaic model
• This model was proposed by Singer and Nicholson.
• According to this model, there is a central bilipid layer (of phospholipids) with their polar head group toward the outside and the non-polar tails pointing inwards.
• Some proteins which are embedded in the lipid layer are called integral proteins and they cannot be separated from the membrane easily.
• Some large globular integral proteins which project beyond the lipid layer on both the sides are believed to have channels through which water soluble materials can pass across.
• Those proteins which are superficially attached are called peripheral (extrinsic) proteins and they can be easily removed.
• Some membrane lipids and integral proteins remain bound to oligosaccharides; such oligosaccharides project into the extracellular fluid and they influence the manner in which cells interact with the other cell.
• There are also certain specific proteins called membrane receptors, which mediate the flow of materials and information into the cell.

Question 5.
What are plastids? How are they classified on the basis of the type of pigments? Name them and their pigments and mention their functions.
Solution:
Plastids are double-membrane bound organelles
of different shapes, that are found only in plant
cells and contain pigments and storage products.
They are of three types :
(i) Leucoplasts
These are the oval, spherical, rod-like or filamentous colourless plastids which are found in storage organs. Their main function is to store reserve materials like starch (amyloplasts), proteins (aleuroplasts) and fats (elaioplasts).
(ii) Chromoplasts
• These are coloured plastids containing mainly the yellow, red and orange pigments (carotene and xanthophyll).
• These are found in petals of flowers and skin of fruits.
• They attracts agents for pollination and dispersal of fruits/seeds.
(iii) Chloroplasts
• These are the green plastids containing mainly chlorophylls and very little carotene and xanthophyll.
• Their main function is photosynthesis and the formation of starch.

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

NCERT Solutions for Class 11 Biology Chapter 12 Mineral Nutrition

January 5, 2022June 18, 2021 by Raju

NCERT Solutions for Class 11 Biology Chapter 11 Transport in Plants

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

Question 1.
“All elements that are present in a plant need not be essential to its survival”. Comment.
Solution:
Plants obtain their inorganic nutrients from the air, water, and soil. Plants absorb a wide variety of mineral elements. Not all the mineral elements that they absorb are required by plants. Out of the more than 105 elements discovered so far, less than 21 are essential and beneficial for normal plant growth and development. The elements required in large quantities are called macronutrients. While those required in fewer quantities or in the trace are termed micronutrients. These elements are either essential constituents of proteins, carbohydrates, fats, nucleic acid, etc. and/or take part in various metabolic processes.

Question 2.
Why is the purification of water and nutrient salts so important in studies involving mineral nutrition using hydroponics?
Solution:
The technique of growing plants in a nutrient solution is known as hydroponics. Since a number of improvised methods have been employed to try and determine the mineral nutrients essential for plants. The essence on all these methods involves the culture of plants in a soil-free, defined mineral solution. These method require purified water and mineral nutrients salts. Purification of water and nutrient salt is important to find out other influencing factors

Question 3.
Explain with examples:
Macronutrients, micronutrients, beneficial nutrients, toxic elements, and essential elements.
Solution:
Based upon the 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,

Macronutrients
Micronutrients

Macronutrients must generally be present in plant tissues in the concentration of 1 to 10 mg/L of dry matter. The macronutrients include carbon, hydrogen, oxygen, nitrogen, phosphorous, sulfur, potassium, calcium, and magnesium. Of these, carbon, hydrogen, and oxygen are mainly obtained from CO₂, and H₂0, while the others are absorbed from the soil as mineral nutrition.

Micronutrients or trace elements are needed in very small amounts (equal to or less than 0.1 mg/L of dry matter). These include iron, manganese, copper, molybdenum, zinc, boron, chlorine, and nickel. In addition to the 17 essential elements named above, there are some beneficial elements such as sodium, silicon, cobalt, and selenium. They are required by higher plants.

Essential elements can also be grouped into four broad categories on the basis of their diverse functions. These categories are:
(1) Essentia] elements as components of biomolecules and hence structural elements of cells, (eg: carbon, hydrogen, oxygen, and nitrogen).

(2) Essential elements that are components of energy-related chemical compounds in plants, for example, magnesium in chlorophyll and phosphorous in ATP.

(3) Essential elements that activate or inhibit enzymes, for example, Mg²⁺ is an activator for both ribulose bisphosphate carboxylase-oxygenase and phosphoenolpyruvate carboxylase, both of which are critical enzymes in photosynthetic carbon fixation; Zn²⁺ is an activator of alcohol dehydrogenase and Mo of nitrogenase during nitrogen metabolism.

(4) Some essential elements can alter the osmotic potential of a cell. Potassium plays an important role in the opening and closing of stomata. Any mineral ion concentration in tissues that reduces the dry weight of tissues by about 10 percent is considered toxic. Such critical concentrations vary widely among different micronutrients. The toxicity symptom! are difficult to identify.

Toxicity levels for any element also vary for different plants. Many times excess of an element may inhibit the uptake of another element. For example, the prominent symptoms of manganese toxicity are the appearance of brown spots surrounded by chlorotic veins.

Question 4.
Name at least five different deficiency symptoms in plants. Describe them and correlate them with the concerned mineral deficiency.
Solution:

Chlorosis: Chlorosis is the loss of chlorophyll leading to yellowing in leaves. It is caused by a deficiency of N, K, Mg, S, Fe, Mn, Zn, and Mo.
Necrosis: It is the death of tissue. It occurs due to deficiency of Ca, Mg, Cu, K.
Inhibition of cell division: It occurs due to deficiency of N, K, S, Mo.
Stunted plant growth: It occurs due to deficiency of Ca, N, etc.
Premature fall of leaf and buds: It occurs due to deficiency of calcium, magnesium.

Question 5.
If a plant shows a symptom which could develop due to deficiency of more than one nutrient, how would you find out experimentally, the real deficient mineral element?
Solution:

The deficiency symptoms can be distinguished on the basis pf the region of occurrence, presence or absence of dead spots, and chlorosis of entire leaf or interveinal chlorosis.
The region of the appearance of deficiency symptoms depends on the mobility of nutrients in plants. The nutrient deficiency symptoms of N, P, K, Mg, and Mo appear in lower leaves.
Zinc is moderately mobile in plants and deficiency symptoms, therefore, appear in middle leaves.
The deficiency symptoms of less mobile elements (S, Fe, Mn, and Cu) appear on new leaves.
Ca and B are immobile in plants, deficiency symptoms appear on terminal buds.
Chlorine deficiency is less common in crops.

Question 6.
Why is it that in certain plants deficiency symptoms appear first in younger parts of the plant while in other they do so in mature organs?
Solution:
The deficiency symptoms tend to appear first in the young tissues whenever the elements are relatively immobile and are not transported out of the mature organs, for example, elements like sulphur and calcium are a part of the structural component of the cell and hence are not easily released.

Question 7.
How are the minerals absorbed by the plants?
Solution:
Uptake of mineral ions, by plants, occurs through
two main phases.
Passive Absorption: It is the process of absorption of minerals through it’s outer space(Intercellular space and cell wall) by physical process. Direct expenditure of metabolic energy is not involved. A substance moves from a region of higher chemical potential to lower chemical potential. It occurs through ion channels (transmembrane protein). The theories to explain the movement of ions:
(a) Ion exchange: Both cation and anion gets absorbed on the surface of cell wall. The absorbed ions are exchanged with ions present in soil solution.
(b) Mass flow hypothesis: According to this hypothesis mass flow of ions occur along with absorption of water as a result of transpirational pull.
Active Absorption: It is the process of movement of ions against a concentration gradient, by utilizing ATP as energy. Both influx and efflux of ions are carried out by carrier mechanism. The activated ions combine with carrier proteins and form ion carrier complex. This complex moves
across all the membrane and reaches inner surface, where it breaks and releases ions into the cytoplasm.

Question 8.
What are the conditions necessary for the fixation of atmospheric nitrogen by Rhizobium? What is their role in nitrogen fixation?
Solution:
Rhizobia are unique because they live in a symbiotic relationship with legumes. Necessary conditions:

Requires a strong reducing agent and energy in the form of ATP.
The enzyme nitrogenase which is very sensitive to oxygen is required.
The processes take place in an anaerobic environment
The energy is provided by the respiration of host cells.

The reduction of nitrogen to ammonia by living organisms is called biological nitrogen fixation. The enzyme, nitrogenase which is capable of nitrogen reduction is present exclusively in prokaryotes. Several types of symbiotic biological nitrogen-fixing associations are known. The most common association on roots is nodules. Their role in N₂– fixation is to supply the plants with nitrogenase that converts nitrogen to amino acids.

Question 9.
What are the steps involved in the formation of root nodule?
Solution:
Nodule formation involves a sequence of multiple interactions between Rhizobium and the roots of the host plant. Stages in the nodule formation are summarised as follows:
Steps in the development of root nodules:
(a) When a root hair of a leguminous plant comes in contact with Rhizobium, it is deformed due to the secretion from the bacterium.
(b) At the site of curling Rhizobia invades the root tissue and proliferate within root hairs.
(c) Some bacteria enlarge to form membrane-bound structures, bacteroids which cannot divide.
(d) The plants form the infection thread, made up of plasma membrane that grows inward, separating the infected tissue from the rest of the plant.
(e) Cell division is stimulated in the infected tissue and more bacteria invade the newly formed tissues.
(f) It is believed that a combination of cytokinin produced by invading bacteria and auxins produced by plant cells, promotes cell division and extension, leading to nodule formation.
NCERT Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 2

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

VERY SHORT ANSWER QUESTIONS

Question 1.
Name the enzymes that reduce nitrogen in the root nodules of a bean plant.
Solution:
Nitrogenase.

Question 2.
Name the enzymes used in biologically nitrogen fixation. What are the mineral elements needed for the activity of the enzyme?
Solution:
Nitrogenase enzyme.

Question 3.
Name the enzyme that can reduce nitrogen to ammonia.
Solution:
Nitrogenase.

Question 4.
What is the importance of phosphorus for plants?
Solution:
Phosphorus is a constituent of cell membranes, certain proteins, all nucleic acids, and nucleotides, and is required for all phosphorylation reactions.

Question 5.
Name two crops that are commonly produced by hydroponics.
Solution:
Tomato, Lettuce

Question 6.
Name the group of enzymes activated by zinc.
Solution:
Carboxylases.

Question 7.
Define critical concentration of elements with reference to plant nutrition.
Solution:
Critical concentration refers to the concentration of the essential element, below which the plant growth is retarded.

Question 8.
Name the element which is a limiting nutrient for both natural and agricultural ecosystems.
Solution:
Nitrogen.

Question 9.
Name two bacteria that oxidise ammonia into nitrite.
Solution:
Nitrosomonas, Nitrococcus

Question 10.
Name two symbiotic nitrogen-fixing bacteria.
Solution:
Rhizobium, Frankia.

Question 11.
What is hydroponics (Tank farming)?
Solution:
It is plant growth in the liquid culture medium.

Question 12.
What are the framework elements of a plant?
Solution:
Carbon, hydrogen, and oxygen are called framework elements.

Question 13.
What type of condition is created by leghaemoglobin in the root nodules of legumes?
Solution:
Anaerobic condition.

Question 14.
What is meant by active absorption?
Solution:
Active absorption: The uptake of mineral ions against the concentration gradient is called active absorption.

Question 15.
How are amides transported In plants?
Solution:
Amides are transported along with water through the xylem.

Question 16.
What is the function of the enzyme nitrite reductase?
Solution:
It reduces nitrate ions to ammonia.

Question 17.
Name two free-living micro-organisms which can fix nitrogen.
Solution:
Azotobacter, Beijemickia.

SHORT ANSWER QUESTIONS

Question 1.
What is nitrification?
Solution:
Questions It is the conversion of ammonium ion to nitrite and then to nitrate. Nitrosomonas converts ammonium into nitrites, Nitrobacter converts nitrites into nitrates.

Question 2.
Prior to sowing rice, a legume crop was cultivated and ploughed back in this field. Why? Explain.
Solution:
Leguminous plants possess root nodules in which the symbiotic bacteria Rhizobium fixes nitrogen. The fixed nitrogen makes the soil rich in nitrogen fertilizer where the leguminous plant is ploughed back into the field.

Question 3.
Name the organism that fixes nitrogen in symbiotic association with a legume. Where does it live in such plants?
Solution:
Rhizobium. It lives in the root nodules of leguminous plants.

Question 4.
Bring out at similarity and difference between leghaemoglobin and hemoglobin.
Solution:
Both leghaemoglobin and hemoglobin are iron-containing molecules but leghaemoglobin is present in the root nodules in the plants belonging to the family Fabaceae while hemoglobin in human blood pigment.

Question 5.
Bring out at similarity and difference between leghaemoglobin and hemoglobin.
Solution:

Question 6.
In what form is boron absorbed by plants from the soil? Mention its two uses in plants and give two deficiency symptoms of boron in them.
Solution:
Boron is absorbed as \({ H }_{ 2 }{ Po }_{ 4- }\quad and\quad { HPo }_{ 4 }^{ 2- }\)

Translocation of carbohydrates
Pollen germination

Deficiency symptoms

Death of root and shoot tips
Abscission of flowers

Question 7.
List the macronutrients and mention three major function, (any three)
Solution:
NCERT Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 4

Question 8.
In what form is boron absorbed by plants from the soil? Mention its two uses in the plants and give two deficiency symptoms of boron in them.
Solution:

(i) Translocation of carbohydrates.
(ii) Pollen germination.
(iii) Absorption and utilisation of calcium.
(iv) Cell elongation and differentiation.

Question 9.
Distinguish between micronutrients and macronutrients.
Solution:
Differences between micronutrients and macronutrients are as following:
NCERT Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 5

LONG ANSWER QUESTIONS

Question 1.
With the help of a suitable diagram describe the nitrogen cycle.
Solution:
Nitrogen cycle: Plants compete with microbes for the limited nitrogen that is available in the soil. Thus, nitrogen is a limiting nutrient for both natural and agricultural ecosystems.

In nature, lightning and ultraviolet radiation provide enough energy to convert nitrogen to nitrogen oxides (NO, NO2, N2O).
Industrial combustions, forest fires, automobile exhausts, and power-generating stations are also sources of atmospheric nitrogen oxides.
The decomposition of organic nitrogen of dead plants and animals into ammonia is called ammonification. Some of this ammonia

volatilises and re-enters the atmosphere but most of it is converted into nitrate by soil bacteria in the following steps
NCERT Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 6

Question 2.
Name at least five different deficiency symptoms in plants. Describe them and correlate them with concerned mineral deficiency.
Solution:

Chlorosis – yellowing of leaves due to loss of chlorophyll caused by the deficiency of N, S, mg, Fe.
Necrosis – Death of tissues, especially in leaves caused by the deficiency of Ca, Mg, Ca & K.
Delay in flowering caused by the deficiency of molybdenum, nitrogen, and sulphur.
Dieback of roots caused by a deficiency of copper.
Inhibition of cell division caused by a deficiency of potassium, calcium and nitrogen.

Question 3.
Write notes on :
(a) Reductive animation
(b) Transamination
Solution:
(a) Reduction animation – In this process ammonia (formed by nitrogen assimilation) reacts with a ketoglutaric acid to form the amino acid – glutamic acid. Here ∝ – ketoglutaric acid comes from Kreb’s cycle and hydrogen is donated by co-enzyme NADH or NADPH.
NCERT Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 8

(b) Transamination – Once the glutamic acid is synthesized by reductive amination, other amino acids are synthesized by the transfer of an amino groups to other carbon skeletons. Glutamic acid is the starting material from which 17 other amino acids are formed by the transfer of the amino group of an amino donor compound to the carboxyl position of an amino acceptor compound. Transaminase is the enzyme responsible for such a reaction.

Question 4.
Write an account of the role of mineral elements in a plant.
Solution:
Role of the mineral elements: Plants require mineral elements for various metabolic activities of their body. The following are some of the important functions which the mineral elements perform:

(i) Constituents of the plant body: Elements form the constitution of the plant body. For example carbon, hydrogen and oxygen are essential for the production of carbohydrates hence they are termed framework elements. Nitrogen, sulphur, and phosphorus are required for the synthesis of proteins. Magnesium is an important part of the chlorophyll molecule.

(ii) Influence on the pH of the cell sap: They also influence the pH of the cell sap.

(iii) Maintenance of osmotic pressure in the plant cells: The mineral salts and organic compounds of the cell sap produce necessary osmotic pressure.

(iv) They influence the permeability of cytoplasmic membrane: Different minerals decrease or increase the permeability of plasma membrane.

(v) They have balancing function reactions: Some of the minerals balance the effects of the other.

(vi) They take part in enzymatic reactions: Some elements work as activators while others work as inhibitors in various enzymatic reactions.

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

NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis

January 5, 2022June 18, 2021 by Raju

NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis.

Question 1.
By looking at a plant externally can you tell whether a plant is C₃ or C₄ ? Why and how?
Solution:
Plants that are adapted to dry tropical regions have the C₄ pathway. They have a special type of leaf anatomy, they tolerate higher temperatures, they show a response to highlight intensities. Study vertical sections of leaves, one of a C₃ plant and the other of a C₄ plant.

Question 2.
By looking at which internal structure of a plant can you tell whether a plant is C₃ or C₄? Explain.
Solution:
In C₄ plant internal structure of the leaf possess a special type of anatomy called ‘Kranz’ anatomy. ‘Kranz’ means ‘wreath’ and is a reflection of the arrangement of cells.

The bundle sheath cells may form several layers around the vascular bundles; they are characterised by having large number of chloroplasts, thick walls impervious to gaseous exchange and no intercellular spaces.

While in C₃ plants, there is no special type of leaf anatomy. There is only a single type of chloroplast inC₃ i.e. granal, while in C₄ chloroplasts are dimorphic, i.e, granite in the mesophyll cells and agranal in the bundle sheath cells.

Question 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:
Though these plants have the C₄ oxalacetic acid as the first CO₂ fixation product they use the C₃ pathway or the Calvin cycle as the main biosynthetic pathway.
In C₄ plants photorespiration does not occur. This is because they have a mechanism that increases the concentration of CO₂ at the enzyme site.
This takes place when the C₄ acid from the mesophyll is broken down in the bundle cells to release CO₂ this results in increasing the intracellular concentration of CO₂ In turn, this ensures that the Rubisco functions as a carboxylase minimizing the oxygenase activity.

Now that you know that the C₄ plants lack photorespiration, you probably can understand why productivity and yields are better in these plants. In addition, these plants show tolerance to higher temperatures.

Question 4.
RuBisCO is an enzyme that acts both as carboxylase and oxygenase. Why do you think RuBisCO carries out more carboxylation in C₄ plants.
Solution:
RuBisCO or Ribulose bisphosphate carboxylase – oxygenase enzyme can bind to both C0₂ and O₂. This binding is competitive. The relative concentration of C0₂ and 0₂ determines which one of the two will bind to the enzyme.

In C₄ plants photorespiration does not occur. This is because they have a mechanism that increases the concentration of C0₂ at the enzyme site.

This takes place when oxaloacetic acid is broken down in the bundle sheath cells to release C0₂.

It results in increased intracellular concentration of C0₂. This ensures that the RuBisCO functions as a carboxylase and minimising the oxygenase activity.

Question 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:
Though chlorophyll is the major pigment responsible for trapping light, other thylakoid pigments like chlorophyll b, xanthophylls, and carotenoids, which are called accessory pigments, also absorb light and transfer the energy to ‘chlorophyll a’.

Indeed, they not only enable a wider range of wavelengths of incoming light to be utilized for photosynthesis but also protect ‘chlorophyll a’ from photo-oxidation. Reaction centre chlorophyll-protein complexes are capable of directly absorbing light and performing charge separation events without other chlorophyll pigments but the absorption cross-section is small.

Question 6.
Why is the colour of a leaf kept in the dark frequently yellow, or pale green? Which pigment do you think is more stable?
Solution:
Chlorophyll is unable to absorb energy in the absence of light and loses its stability, giving the leaf a yellowish colour. This shows that xanthophyll is more stable.

Question 7.
Look at leaves of the same plant on the shady side and compare it with the leaves on the sunny side. Or, compare the potted plants kept in the sunlight with those in the shade. Which of them has leaves that are darker green? Why?
Solution:
Light is a limiting factor for photosynthesis Leaves get lesser light for photosynthesis when they are in shade. Therefore, the leaves or plants in shade perform lesser photosynthesis as compared to the leaves or plants kept in sunlight. In order to increase the rate of photosynthesis, the leaves present in shade have more chlorophyll pigments.

This increase in chlorophyll content increases the amount of light absorbed by the leaves, which in turn increases the rate of photosynthesis. Therefore, the leaves or plants in shade are greener than the leaves or plants kept in the sun.

Question 8.
The 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 a limiting factor?
(b) What could be the Jimiting factor/s in region A?
(c) What do C and D represent on the curve?
Solution:

(a) In the region ‘A’ and half of ‘BTight is limiting factor because rate of photosynthesis is increasing with the intensity of light.
(b) All the other factors except light.
(c) C represents a region where a factor other than light is limiting, e.g., CO₂. D represents the light intensity at which rate of photosynthesis is maximum under existing conditions (e.g., CO₂).

Question 9.
Give a comparison between the following:
(a) C₃ and C₄ pathways
(b) Cyclic and non-cyclic photophosphorylation
(c) Anatomy of leaf in C₃ and C₄.
Solution:
(a) Differences between C₃ and C₄ pathway
NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis 2
(b) Differences between cyclic and non-cyclic photophosphorylation are

(c) Differences between the anatomy of leaf in C₃ plants and anatomy of leaf in C₄ plants are

VERY SHORT ANSWER QUESTIONS

Question 1.
Write one anatomical feature of C₄ plants.
Solution:
Kranz anatomy in leaf.

Question 2.
Which of the following is not a useful function of the light reaction in photosynthesis?
(a) splitting water
(b) synthesis of NADPH
(c) converting light energy into chemical energy
(d) releasing oxygen for photorespiration
Solution:
(d) Releasing oxygen for photorespiration.

Question 3.
What is the starting substance in the CO₂ fixation cycle? (Apr. 91)
Solution:
RuMP.

Question 4.
Where is PS II located in a chloroplast?
Solution:
PS II is located in the appressed regions of grana thylakoid

Question 5.
Name the reaction centre of PS I and PS II.
Solution:
P₇₀₀ & P₆₈₀‘

Question 6.
What type of light causes maximum photo-synthesis? (Oct. 1995)
Solution:
Red light

Question 7.
How many molecules of ATP and how many molecules of NADPH are spent to fix three molecules of CO₂ in the Calvin cycle?
Solution:
9 ATP and 6 NADPH

Question 8.
Why do the stomata of CAM plants open during the night?
Solution:
As these plants grow in dry areas, they keep stomata close during the day to conserve water and open their stomata during the night for the diffusion of gases.

Question 9.
Mention one useful role of photorespiration in plants.
Solution:
It protects the plants from photooxidative damage.

Question 10.
Cyanobacteria and some other photosynthetic bacteria don’t have chloroplasts. How do they conduct photosynthesis?
Solution:
Cyanobacteria have bluish pigment phycocyanin, which they use to capture light for photosynthesis. Some green bacteria (cyanobacteria) are red or pink due to pigment phycoerythrin. Whatever the colour of cyanobacteria, they are photosynthetic and so can manufacture food.

Question 11.
What is phosphorylation? (M.Q.P.)
Solution:
Synthesis of ATP either with the help of light (during photosynthesis) or in presence of oxygen (during respiration) is called phosphorylation.

Question 12.
Name the organism Englemann used in his experiment.
Solution:
Cladophora.

Question 13.
Write the currently accepted equation of photosynthesis in plants.
Solution:

Question 14.
What is a pigment?
Solution:
A pigment is a substance that absorbs light of certain wavelength(s).

Question 15.
Write the full form of NADP
Solution:
NADP – Nicotinamide adenine dinucleotide Phosphate.

Question 16.
Expand RuBP
Solution:
Ribulose 1, 5 bisphosphates.

Question 17.
Give a reason for the following:
Some bacteria exhibit photosynthesis but they do not produce oxygen. (July 2006)
Solution:
Some photosynthetic bacteria do not use water as their source of hydrogen, hence do not liberate oxygen.

Question 18.
Mention two conditions where light can become a limiting factor.
Solution:
Conditions in which light can become a limiting factor:
(i) Plants in the shade.
(ii) Plants growing under the canopy in a dense forest.

Question 19.
What are antenna molecules?
Solution:
Antenna molecules are light-harvesting pigment molecules that occur on the outer side of a photosynthetic unit.

Question 20.
What is a quantasome? Where is it present?
Solution:
Quantasome means photosynthetic units. It is equivalent is 230 chlorophyll molecules. These are present in the grana lamellae.

SHORT ANSWER QUESTIONS

Question 1.
Specify how C₄ photosynthetic pathway increases carbon dioxide concentration in bundle sheath cells of sugarcane?
Solution:
In C₄ pathway of sugarcane, C0₂ from atmosphere enters through the stomata in the mesophyll cell and combines with phosphoenol pyruvate to form a 4-C compound oxaloacetic acid. The OAA is then transported to the bundle sheath where it is decarboxylatedto release C0₂ in bundle sheath.

Question 2.
Differentiate between absorption spectrum and action spectrum.
Solution:
The main differences between absorption spectrum and action spectrum are as follows.
NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis 6

Question 3.
What are quantasomes? (Oct. 94)
Solution:
Quantasome is a functional unit (Photo-synthetic unit) made of a group of pigment molecules required for carrying out a photochemical reaction. The Pigment molecules are embedded in the grana and differentiated as pigment system I (with chi 670, chi 680, P 700) and pigment system ll(with chi 670, chi 680, P 680, and Xanthophylls)

Question 4.
Distinguish between cyclic and non-cyclic photophosphorylation. (M.Q.P., March 2011)
Solution:
Non cyclic photophosphorylation (a) Cyclic photophosphorylation

The path traversed by an electron is non-cyclic.
(a) Path traversed by electron is cyclic.
Both PSI and PSII are active.
(b) Only PSI is active.
It is accompanied by photolysis.
(c) No photolysis.
The major pathway that takes place.
(d) Secondary pathway when additional ATP is needed.

Question 5.
What is Blackman’s law of limiting factors?
Solution:
F.F. Blackman (1905) extended a law to formulate the principle of limiting factors. “When a process is conditioned as to its rapidity by a number of separate factors, the rate of the process is limited by the pace of slowest factors.”

Question 6.
In the condition of water stress why the rate of photosynthesis declines?
Solution:
Due to water stress, stomata remain closed and so there is a decrease in CO₂concentration and the leaf water potential is also reduced, decline the rate of photosynthesis.

Question 7.
What is a reaction centre? Give the reaction centres of PSI and PSII.
Solution:
Reaction centre is a chlorophyll component of the photosystem and it absorbs as well as accepts energy from other pigments and ejects an electron. The reaction centre of PSII is Chla₆₈₀or P₆₈₀ and PSI is Chla₇₀₀ or P₇₀₀

Question 8.
Why is photorespiration considered a wasteful process?
Solution:
Photorespiration considered a wasteful process because
(i) 25% of photosynthetically fixed carbon is lost in the form of C0₂.
(ii) There is no energy-rich useful compound produced during this process.

Question 9.
Give two reasons as to why photosynthesis is important for sustaining life on earth.
Solution:
Photosynthesis is the most important process because;
(i) it is the only natural process by which oxygen is liberated into the atmosphere.
(ii) it is the process by which food is manufactured for all living organisms.

Question 10.
Why does the rate of photosynthesis decrease at higher light intensities? What plays a protective role in such situations?
Solution:
Rate of photosynthesis decreases for two reasons :
(i) Other factors required for photosynthesis become limiting.
(ii) Destruction of chlorophyll by photo-oxidation.
Carotenoids play a protective role by:
(i) absorbing the excess light and
(ii) acting as an antioxidant to detoxify the effect of activated oxygen species.

Question 11.
What is C₄ -pathway? Give an example. (March 2008)
Solution:
CA -pathway is an alternative photosynthetic pathway seen in plants like sugarcane/sorghum/ maize in which the stable compound is oxaloacetate a 4-C compound. It is called the Hatch-slack pathway.

Question 12.
What is kranz anatomy in plants?
Solution:
In Kranz Anatomy vascular bundles are surrounded by a layer of bundle sheath that contains a large number of chloroplasts in mesophyll cells and it is present in C₄ plants e.g, Maize, Sugarcane, etc.

LONG ANSWER QUESTIONS

Question 1.
How is photosystem I different from photosystem II?
Solution:
The main differences between photosystem I and photosystem II are
NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis 7

Question 2.
Describe the factors that influence the rate of Photosynthesis. (Oct. 1989)
Solution:
The factors that affect photosynthesis may be both internal & External.

Internal factors:

Chlorophyll: it is the light-absorbing pigment and only portions of the plant having chlorophyll can help in photosynthesis.
Protoplasmic factor: young seedlings when transferred from darkness to light show the presence of some factors which is believed to be enzymatic initiates photosynthesis and is called the protoplasmic factor.

External factors:

Light: It is one of the most important factors which affects the process in 3 ways i.e. quantity, quality, and intensity. Quantity of light is otherwise duration and depends upon the photoperiod that is required by the plant quality refers to the wavelength, maximum photosynthesis occurs in red and blue light while minimum in green light. Intensity favours the process and low intensity decreases the rate of photosynthesis. Very high intensity brings about photooxidation of pigments which is called solarization.
CO₂: An increase in CO₂ concentration favours the process provided other factors are not limiting but very high concentrations are toxic and inhibit photosynthesis.
Temperature: Increase in temperature favour photosynthesis but above the optimum range the process decreases due to the denaturation of enzymes.

Question 3.
Explain the process of the biosynthetic phase of photosynthesis occurring in the chloroplasts.
Solution:
The biosynthetic phase of photosynthesis :

It occurs in the stroma of chloroplasts.
These reactions reduce the carbon dioxide into carbohydrates, making use of the ATP and NADPH₂ produced in the photochemical reactions.
The reactions are also called as Calvin cycle.
The three phases of the Calvin cycle are as follows:

(i) Carboxylation
Six molecules of Ribulose 1,5 bisphosphate
react with six molecules of carbon dioxide to form six molecules of a short-lived 6C- compound.
The reaction is catalysed by RuBP carboxylase (RuBisCo).
The six molecules of the 6C-intermediate break into 12 molecules of 3- phosphoglyceric acid (3-PGA), an SC- compound.
It is through this step that carbon dioxide is fixed in the plant.
NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis 8
(ii) Reduction
12 molecules of 3-phosphoglyceric acid are converted into 12 molecules of 1, 3 diphosphate-glyceric acid, utilising 12 molecules of ATP and then reduced to 3- phosphoglyceraldehyde making use of 12 molecules of NADPH. Two molecules of phosphoglyceraldehyde react to form one molecule of glucose. It is in this step that there is an actual reduction of carbon dioxide leading to sugar formation.

(iii) Regeneration of RuBP
10 molecules of phosphoglyceraldehyde, by a series of complex enzyme-catalyzed reactions, are converted into six molecules of ribulose 1,5-bisphosphate; six molecules of ATP are needed for this step. This step of ‘ regeneration of RuBP is important for the cycle to continue

We hope the NCERT Solutions for Class 11 Biology at Work Chapter 13 Photosynthesis, helps you. If you have any query regarding NCERT Solutions for Class 11 Biology at Work Chapter 13 Photosynthesis, drop a comment below and we will get back to you at the earliest.