NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules

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

protein molecular weight calculator. Molecular mass is the most fundamental characteristics for proteins and peptides.

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 2
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 3
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 4
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:
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 7

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 pH.
  • Activity declines both below and above optimum temperature and pH.
  • 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:
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 9

Question 9.
Differentiate between Structural Proteins and Functional Proteins.
Solution:
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 11

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

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

Question 2.
Distinguish between the primary, secondary, and tertiary structures of proteins.
Solution:
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 13

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

  1. Oxidoreductases: eg Alcohol dehydrogenase, oxidation, Reduction occurs
  2. Transferases: transfer a particular group to another substrate, eg. transavninase
  3. Hydrolases: cleave their substrates by hydrolysis of a covalent bond e.g. Urease, amylase.
  4. Lyases: break the covalent bond eg. Deaminase
  5. Isomerase: by changing the bonds they make isomers. eg: Aldolase.
  6. 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 17 Breathing and Exchange of Gases

NCERT Solutions for Class 11 Biology Chapter 17 Breathing and Exchange of Gases

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 17 Breathing and Exchange of Gases.

Question 1.
Define vital capacity. What is its significance?
Solution:
The maximum volume of air a person can breathe in after a forced expiration it is about 4000mL in a normal adult person. Vital capacity is higher in athletes and singers. Cigarette smokers have a lower vital capacity of the lungs. This includes ERC, TV, and IRV, or the maximum volume of air a person can breathe out after a forced inspiration.

Question 2.
State the volume of air remaining in the lungs after a normal breathing.
Solution:
The volume of air remaining in the lungs after normal respiration is called functional residual capacity. It includes ERV + RV = Expiratory reserve volume + residual volume

Question 3.
volume Diffusion of gases occurs in the alveolar region only and not in the other parts of the respiratory system. Why?
Solution:
Alveoli are the primary sites of the exchange of gases. The alveolar region is having enough pressure gradient to facilitate the diffusion of gases. Other regions of the respiratory system don’t have the required pressure gradient. Additionally, the membrane of alveoli is thin enough to facilitate the exchange of gases in a convenient manner.
NCERT Solutions for Class 11 Biology Chapter 17 Breathing and Exchange of Gases 1

Question 4.
What are the major transport mechanisms for CO2? Explain.
Solution:
Transport of carbon dioxide: About 4 ml of carbon dioxide is transported by every 100 ml of blood.
C02 is transported in three forms in the blood.
(i) In the dissolved form in plasma about 7% of C02 dissolves in the plasma of blood, just as it gets dissolved in water.
(ii) As bicarbonates
Erythrocytes have a high concentration of the enzyme, carbonic anhydrase which catalyzes the following reactions;
NCERT Solutions for Class 11 Biology Chapter 17 Breathing and Exchange of Gases 2
About 70% of C02 is transported as bicarbonates.

(iii) As carbaminohaemoglobin
C02 combines with the globin part of haemoglobin and forms carbamino haemoglobin. About 23% of CO2 transported in this manner.

Question 5.
What will be the pO2 and pCO2 in the atmospheric air compared to those in the alveolar air?
(i) pO2 lesser, pCO2higher
(ii) pO2 higher, pCO2lesser
(iii) pO2 higher, pCO2 higher
(iv) pO2 lesser, pCO2lesser
Solution:
(ii) pO2 higher, pCO2 lesser
pO2 higher will create the pressure gradient to facilitate the movement of O2 from atmosphere to alveoli and pCO2 lesser will create the movement of CO2 from alveoli to atmosphere.

Question 6.
Explain the process of inspiration under normal conditions.
Solution:
The intake of air into the lungs is known as inspiration. Inspiration occurs when the pressure within the lungs is less than the atmospheric pressure. It is initiated by the contraction of the diaphragm which increases the volume of the thoracic chamber in the anteroposterior axis. The contraction of external intercostal muscles lifts up the ribs and the sternum causing an increase in the volume of the thoracic chamber in the dorsoventral axis. The overall increase in the thoracic volume causes a similar increase in pulmonary volume which decreases the intra-pulmonary pressure to less than the atmospheric pressure which forces the air from outside to move into the lungs.

Question 7.
How is respiration regulated?
Solution:

  1. Respiratory rhythm centre, present in medulla region of brain is responsible for respiration regulation.
  2. Its function can be moderate by pneumotaxic centre, present in pons region of brain.
  3. A chemosensitive area present adjacent to rhythm centre, is highly sensitive to C02 and H+.
  4. Chemosenstive centre due to increase in C02 and H+ can signal the rhythm centre to make adjustment to eliminate these substances.
  5. Receptors associated with aortic arch and carotid artery also can recognise changes in CO2 and H+ concentration and send necessary signals to rhythm centre for remedial actions.

Question 8.
What is the effect of pC02 on oxygen transport?
Solution:
At low pC02, blood can carry the maximum amount of oxygen as oxyhemoglobin. At high pCO2, the affinity for oxygen decreases and oxyhaemoglobin dissociates to free oxygen.
NCERT Solutions for Class 11 Biology Chapter 17 Breathing and Exchange of Gases 3
So at high pC02, oxygen transport is inhibited.

Question 9.
What happens to the respiratory process in a man going up a hill?
Solution:
There is a fall of PO2 level at high altitudes. This lowers alveolar PO2 and consequently reduces the diffusion of oxygen from the alveolar air to the blood. So oxygenation of the blood is decreased progressively.

After some time, the affected person gets adjusted to the surroundings due to which the heart rate is accelerated, RBC count in the blood is increased, haemoglobin level and oxygen-carrying capacity are also increased.

Question 10.
What is the site of gaseous exchange in an insect?
Solution:
Insects have a complex system of intercommunicating air tubes called tracheae to enable them to exchange gases between the environment and the body cells (tracheal respiration).

Question 11.
Define oxygen dissociation curve. Can you suggest any reason for its sigmoidal pattern?
Solution:
The curve in which the percentage saturation of hemoglobin with O2 is plotted against the partial pressure of oxygen (PO2) is called the oxygen dissociation curve. At a PO2 of 100 mm Hg, 100 percent saturation of Hb takes place 90% saturation of Hb takes place even at a P02 of 60mm Hg. An I fall of PCX, from 100 to 60mm Hg will cause only 10% decrease in saturation of Hb. Hence the curve takes the shape of a sigmoid.

Question 12.
Have you heard about hypoxia? Try to gather information about it, and discuss it with your friends.
Solution:
Hypoxia is the condition in which there is a deficiency of oxygen at the tissue level.

  • Arterial hypoxia: It’s because of low level of oxygen in the blood. It occurs when the atmosphere does not contain enough oxygen and there is obstruction in the respiratory passage.
  • Anaemic hypoxia: It is due to very low level of haemoglobin in the blood.
  • Stagnant hypoxia: It is due to the inadequate blood flow to deliver oxygen to the tissue.
  • Histoxic hypoxia: It is due to the presence of toxic substances in the oxygen inhaled,  e.g.: Cyanide poisoning.

Question 13.
Distinguish between
(a) IRV and ERV
(b) Inspiratory capacity and Expiratory capacity.
(c) Vital capacity and Total lung capacity.
Solution:
NCERT Solutions for Class 11 Biology Chapter 17 Breathing and Exchange of Gases 4

Question 14.
What is tidal volume? Find out the tidal volume (approximate value) for a healthy human in an hour.
Solution:
The volume of air inspired or expired/breath during normal respiration is approx. 500 ml., i.e., a healthy man can inspire or expire approximately 6000 to 8000 ml of air per minute.

VERY SHORT ANSWER QUESTIONS

Question 1.
Name the enzyme which catalyses the bicarbonate formation in RBCs.
Solution:
Enzyme carbonic anhydrase.

Question 2.
What is carbamino haemoglobin?
Solution:
It is a complex formed by the combination of carbon dioxide with the globin part of haemoglobin.

Question 3.
What is tidal volume?
Solution:
The volume of air inspired or expired with every normal breath during effortless respiration is called tidal volume.

Question 4.
What term is used for the volume of air left in the lungs even after the most powerful expiration?
Solution:
Residual volume.

Question 5.
Name the respiratory organ of
(a) Butterfly
(b) Frog larva.
Solution:
(i) Butterfly – trachea
(ii) Frog larva – gills.

Question 6.
What is the role of oxyhaemoglobin after releasing molecular oxygen in the tissue?
Solution:
Oxyhaemoglobin after releasing oxygen collects carbon dioxide from the tissue and form carbaminohaemoglobin.

Question 7.
What are the two factors that contribute to the dissociation of oxyhaemoglobin in the atrial blood to release’ molecular oxygen in active tissue?
Solution:
The two factors are :
(a) Lowp02
(b) HighpC02

Question 8.
Name the double-walled sac which covers the lungs in mammals.
Solution:
Pleura.

Question 9.
What prevents the collapsing of our trachea during breathing ?
Solution:
C-shaped cartilages at regular intervals.

Question 10.
Define inspiratory reserve volume.
Solution:
The extra volume of air that can be inspired beyond the normal tidal volume, is called inspiratory reserve volume.

Question 11.
Which part(s) of the brain control(s) breathing movements?
Solution:
Medulla and pons.

Question 12.
What is oxyhaemoglobin?
Solution:
Oxyhaemoglobin is a complex formed when oxygen combies with the Fe2+ part of haemoglobin.

Question 13.
How much of oxygen is transported by 100 ml of blood under normal physiological conditions?
Solution:
About 5 mL.

Question 14.
Write the chemical reaction catalysed by enzyme carbonic anhydrase.
Solution:
Carbonic anhydrase catalyses the following reaction:
NCERT Solutions for Class 11 Biology Chapter 17 Breathing and Exchange of Gases 5

Question 15.
How does penumotaxic centre alter the respiratory rate?
Solution:
Pneumotaxic centre can reduce the duration of inspiration and alter the respiratory rate.

Question 16.
What is the percentage of C02 transported as sodium bicarbonate?
Solution:
70% of C02 is transported as sodium bicarbonate.

Question 17.
What will happen if the human blood becomes acidic?
Solution:
Oxygen carrying capacity of haemoglobin will decrease.

SHORT ANSWER QUESTIONS

Question 1.
Write four conditions necessary to facilitates efficient gaseous exchange between human respiratory surface and the environment.
Solution:
Conditions for efficient gas exchange are as follows:
(a) The membrane should be thin.
(b) It should be highly vascularized.
(c) It should be highly permeable to gases.
(d) There should a partial pressure difference on both sides of lung.

Question 2.
Differentiate between pharynx and larynx.
Solution:
The main differences between the pharynx and larynx are as follows:
NCERT Solutions for Class 11 Biology Chapter 17 Breathing and Exchange of Gases 6

Question 3.
Why is hemoglobin called a conjugated protein? What happens to the molecule at a high and low partial pressure of oxygen?
Solution:
Hemoglobin: It is called conjugated protein because it consists of a basic protein globin and a non-protein heme.
The haemoglobin when exposed to the high partial pressure of oxygen combines with oxygen to form oxyhaemoglobin which carries 4 molecules of oxygen loosely bound to the four Fe2+ ions. When this oxyhaemoglobin reaches the tissues where there is low oxygen pressure oxyhaemoglobin dissociates into oxygen and deoxyhemoglobin.

Question 4.
Differentiate between inspiratory capacity and expiratory capacity.
Solution:
The differences between inspiratory and expiratory capacity are :
NCERT Solutions for Class 11 Biology Chapter 17 Breathing and Exchange of Gases 7

Question 5.
Diffusion of gases occurs in the alveolar region only and not in the other parts of the respiratory system. Why?
Solution:
The alveoli have very thin walls consisting of squamous epithelium. The alveolar wall is provided with an extensive network of blood capillaries; due to the intimate contact of the blood capillaries and alveolar wall, there is an exchange of gases taking place easily. In the other part the membrane/wall is not so thin to allow for diffusion.

Question 6.
What percentage of oxygen is transported by erythrocytes in the blood? What happens to the remaining?
Solution:
About 97% of the oxygen is transported by erythrocytes. The remaining 3% is transported in dissolved form in the plasma.

Question 7.
What is asthma? Explain.
Solution:
Asthma: It is the hypersensitivity of bronchioles to any foreign substance, characterized by the spasm of the smooth muscles of the walls of the bronchioles.

Question 8.
What is emphysema? What is its major cause?
Solution:
Emphysema: Emphysema is a chronic disorder is which alveolar walls are damaged and hence the surface area for exchange of. gases is reduced. It is caused mainly by cigarette smoking.

Question 9.
Draw a labelled diagram of a section of an alveolus with a pulmonary capillary.
Solution:
NCERT Solutions for Class 11 Biology Chapter 17 Breathing and Exchange of Gases 8

Question 10.
What will happen if the patient has been inhaling polluted air containing high content of CO?
Solution:
Hemoglobin has much more affinity about 250 times for CO than oxygen. It readily combines with CO to form the most stable compound called carboxyhemoglobin. It may be fatal for the patient.

Question 11.
What is pneumonia? What are its causes?
Solution:
Pneumonia is a respiratory disease in which oxygen has difficulty in diffusing through the flammed alveoli and the blood O2 may be drastically reduced and blood PCO2 remains normal. It is caused by streptococcus pneumonia. Its symptoms are trembling, pain in the chest, fever, cough, etc. It is mostly observed in children and old age.

LONG ANSWER QUESTIONS

Question 1.
What do you mean by occupational lung disease? Enumerate the prevention measure that should be adopted by a person likely to be exposed to substances that cause occupational diseases?
Solution:
Occupational lung disease as the name suggests it is the disease of lung due to the occupation of the human.
Cause: These are caused by harmful substances, such as gas fumes or dust, present in the environment where a person works. Silicosis and asbestoses are common examples, which occur due to chronic exposure of silica and asbestos dust in the mining industry.
Symptoms: It is characterized by the proliferation of fibrous connective tissue (fibrosis) of upper part of lung, causing inflammation.
Prevention: The occupational disease expresses symptoms after chronic exposure (i.e., 10-15 years or even more). Most of the occupational diseases including silicosis and asbestosis is are incurable. Therefore, the person which is exposed to such irritants should adopt preventive measures. These protective measures are as follows:
(i) Minimize the exposure of harmful dust at the workplace.
(ii) Workers should be informed about the harm of exposure to such dust.
(iii) Workers must have protective gear and clothing at the workplace.
(iv) health of the workers should be regularly checked up.

Question 2.
Explain the regulation of respiration by the nervous system.
Solution:
Regulation of respiratory rhythm

  • The ability to maintain and moderate the respiratory rhythm according to the demand of the body tissues is due to neural control.
  • The respiratory rhythm centre located in the medulla of the brain, is primarily responsible for this regulation, f Pneumotaxic centre present in the brain functions as the ‘switch off point for regulation; by altering the duration of inspiration, it can alter the respiratory rate.
  • A chemosensitive area is situated adjacent to the rhythm centre; it is highly sensitive to carbon dioxide and hydrogen ions.
  • An increase in the concentration of these substances activates this centre which in turn sends signals to the rhythm centre to make necessary adjustments in the respiratory process.
  • Receptors associated with aortic arch and carotid artery also are sensitive to carbon di oxide and H+ ions; they too send signals to the respiratory rhythm centre.
  • Oxygen plays only an insignificant role in the regulation of respiratory rhythm.

Question 3.
How does the exchange of respiratory gases take place in the alveoli or lungs?
Solution:
Gaseous exchange in alveoli:

  • The alveolar wall is very thin and contains a rich network of interconnected capillaries.
  • Due to this, the alveolar wall seems to be a sheet of flowing blood and is called the respiratory membrane.
  • It consists mainly of the alveolar epithelium, epithelial basement membrane, a thin interstitial space, capillary basement membrane, and capillary endothermal membrane. All these layers cumulatively form a membrane of 0.2 mm thickness.
  • The respiratory membrane has a limit of gas exchange between alveoli and pulmonary blood. It is called diffusing capacity. It is dependent on the solubility of respiratory gases.
  • The partial pressure of oxygen (p02) in the alveoli is higher (104 mm Hg) than that in the deoxygenated blood in the capillaries of the pulmonary arteries (40 mm Hg). As the gases diffuse from higher to a lower concentration, the movement of oxygen is from the alveoli to the blood. The reverse is the case in relation to carbon dioxide.
  • The partial pressure of carbon dioxide (pC02) is higher in deoxygenated blood (45 mm Hg), than in alveoli (40 mm Hg), therefore, CO2 passes from the blood to the alveoli.

Question 4.
How are inspiration and expiration take place in humans?
Solution:
The inflow (inspiration) and outflow (expiration) of air occur between the atmosphere and lungs by the expansion and contraction of lungs.
Inspiration: It is the process by which fresh air enters the lungs.

  • The external intercostal muscles present between the ribs contract and pull these ribs and sternum upward and outward increasing the volume to the thoracic cavity.
  • The diaphragm becomes flats and gets lowered by the contraction of its muscles thereby increasing the volume of the thoracic cavity.
  • The abdominal muscles relax and allow the compression of the abdominal organ by the diaphragm.
  • As the volume of the thoracic cavity increases and as a result, there is a decrease in air pressure in the lungs. The greater pressure outside the body causes air to flow rapidly into the nostrils through the respiratory tract to the lungs.

Expiration: It is a process by which foul air (CO2) is expelled out from the lungs.

  • Internal intercostal muscles contract so that they pull in ribs downward and inward decreasing the size of the thoracic cavity.
  • The muscle fibres of the diaphragm relax making it convex, decreasing the volume of the thoracic cavity.
  • Contraction of abdominal muscles compresses this abdomen and pushes it towards the diaphragm.
    The overall volume of the thoracic cavity decrease and foul air goes outside from the cavities of alveoli through the respiratory tract.

We hope the NCERT Solutions for Class 11 Biology at Work Chapter 17 Breathing and Exchange of Gases, help you. If you have any query regarding NCERT Solutions for Class 11 Biology at Work Chapter 17 Breathing and Exchange of Gases, drop a comment below and we will get back to you at the earliest.

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

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

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

Question 1.
Define Glomerular Filtration Rate (GFR).
Solution:
The amount of filtrate formed by the kidneys per minute is called glomerular filtration rater (GFR). GFR in a healthy individual is approximately 125ml/minute i.e. 180 lines per day.

Question 2.
Explain the autoregulatory mechanism of GFR.
Solution:
This mechanism in Kidney is present to regulate the glomerular filtrate rate.

  • Juxtaglomerular apparatus (JGA) is a specialized cellular region located where the distal convoluted tubule and afferent arteriole, come in contact with each other.
  • A fall in GFR can activate the JG cells to release renin which acts through a complex series of reactions called renin-angiotensin aldosterone mechanism, which can stimulate blood flow and thereby the GFR back to normal.

Question 3.
Indicate whether the following statements are true or false :
(i) Micturition is carried out by a reflex.
(ii) ADH helps in water elimination, making the urine hypotonic.
(iii) Protein-free fluid is filtered from blood plasma into the Bowman’s capsule.
(iv) Henle’s loop plays an important role in concentrating the urine.
(v) Glucose is actively reabsorbed in the proximal convoluted tubule.
Solution:
(i) True
(ii) False
(iii) True
(iv) True
(v) True.

Question 4.
Give a brief account of the countercurrent mechanism.
Solution:
The loop of Henle and vcisa recta are responsible for concentrating the filtrate. The mechanism is called a counter-current mechanism. The flow of filtrate in limbs of Henle’s loop and vasa recta is in opposite direction so, forms a counter-current system. The proximity, as well as counter-current in them, maintains osmolarity increasing 300 mOsmolL-1 in the cortex and 1200 mOsmolL- 1 in inner medulla.

Question 5.
Describe the role of the liver, lungs and skin in excretion.
Solution:
Liver: This largest gland in our body, secretes bile-containing substances like bilirubin, biliverdin, cholesterol, degraded steroid hormones, vitamins and drugs. Most of these substances ultimately pass out along with digestive wastes.

Lungs: Lungs remove large amounts of CO2, about 18 liters/day, and also significant qualities of water every day.

Skin: The sweat and sebaceous glands in the skin eliminate certain substances through their secretions. Sweat is a watery fluid containing NaCI, small amounts of urea, lactic acid etc. Sebaceous glands eliminate certain substances like sterols, hydrocarbons, and waxes through sebum.

Question 6.
Explain micturition.
Solution:
Urine formed by the nephrons is ultimately carried to the urinary bladder where it is stored till a voluntary signal is given by the central nervous system (CNS). This signal is initiated by the stretching of the urinary bladder as it gets filled with urine. In response, the stretch receptors on the walls of the bladder send signals to the CNS. The CNS passes on motor messages to initiate the contraction of smooth muscles of the bladder and simultaneous relaxation of the urethral sphincter causing the release of urine. The process of release of urine is called micturition and the neural mechanisms causing it is called the micturition reflex.

An adult human excretes on average, 1.5 liters of urine per day. The urine formed is a light yellow colored watery fluid that is slightly acidic (pH-6.0) and has a characteristic odor. On average, 25-30 gm of urea is excreted out per day. Various conditions can affect the characteristics of urine. Analysis of urine helps in the clinical diagnosis of many metabolic disorders as well as malfunctioning of the kidneys. For example, the presence of glucose (Glycosuria) and ketone bodies (ketonuria) in urine are indicative of diabetes mellitus.

Question 7.
Match the items of Column I with those of column II.
Column I                                              Column II
(a) Ammonotelism                            (i) Birds
(b) Bowman’s capsule                       (ii) Water reabsorption
(c) Micturition                                    (iii) Bony fish
(d) Uricotelism                                   (iv) Urinary bladder
(e) ADH                                               (v) Renal tubule
Solution:
(a) – (iii) Bony fish
(b) – (v) Renal tubule
(c) – (iv) Urinary bladder
(d) – (i) Birds
(e) – (ii) Water reabsorption

Question 8.
What is meant by the term osmoregulation?
Solution:
Osmoregulation is the mechanism of maintaining water, blood, body fluid, and ionic (salt) balance in the body.

Question 9.
Terrestrial animals are generally either ureotelic or uricotelic, not ammonotelic, why?
Solution:
Land animals have an integument that is impervious to gas exchange. Ammonia is highly toxic and it has to be eliminated as rapidly as it is formed. It requires a large volume of water for its elimination. They do not have access to such a large volume of water needed for the elimination of ammonia. So, they are either ureotelic or uricotelic.

Question 10.
What is the significance of the juxtaglomerular apparatus (JGA) in kidney function?
Solution:
The JGA plays an important role in the regulation of kidney function. A fall in glomerular blood flow / glomerular blood pressure / GFR can activate the JG cells to release renin which converts angiotensinogen in the blood to angiotensin I and further to angiotensin II. Angiotensin II, being a powerful vasoconstrictor, increases the glomerular blood pressure and thereby GFR. It also activates the adrenal cortex to release Aldosterone. Aldosterone causes reabsorption of Na+ and water from the distal parts of the tubule. This also leads to an increase in blood pressure and GFR. This mechanism is known as the Renin-Angiotensin mechanism.

Question 11.
Name the following:

  1. A chordate animal having flame cells as excretory structures.
  2. Cortical portions projecting between the medullary pyramids in the human kidney.
  3. A loop of capillary running parallel to Henle’s loop.

Solution:

  1. Amphioxus.
  2. Columns of Bertini.
  3. Vasa recta.

Question 12.
Fill in the gaps:

  1. Ascending limb of Henle’s loop is……………… to water whereas the descending limb is………….to it.
  2. Reabsorption of water from distal parts of the tubules is facilitated by hormone………………
  3. Dialysis fluid contains all the constituents as in plasma except…………….
  4. A healthy adult human excretes (on an average)………….. gm of urea/day.

Solution:

  1. impermeable, permeable
  2. ADH (vasopressin)
  3. nitrogenous wastes
  4. 25 to 30 gm

VERY SHORT ANSWER QUESTIONS

Question 1.
Name the nitrogenous waste excreted by larval and adult stage of frog respectively.
Solution:
Larval stage – ammonia, Adult stage – urea.

Question 2.
In which organ ammonia is converted to urea?
Solution:
Liver.

Question 3.
Define ammonotelism.
Solution:
The excretion of ammonia is called ammonotelism.

Question 4.
What is hemodialysis?
Solution:
The process of removal of excess urea from the blood of a patient (normally suffering from uremia) using an artificial kidney is known as hemodialysis

Question 5.
Define ketonuria.
Solution:
The presence of high ketone bodies in the urine is called ketonuria.

Question 6.
What are the columns of Bertini?
Solution:
These are the extension of the renal cortex between the medullary pyramids as renal columns.

Question 7.
In which part of the nephron does filtration take place?
Solution:
Bowman’s Capsule/Renal Corpuscle

Question 8.
What difference is observed in the ascending and descending limbs of Henle’s loop with g reference to permeability to water?
Solution:
Ascending limb is impermeable to water and permeable to solutes.
Descending limb is permeable to water and impermeable to solutes.

Question 9.
Name the body part through which ammonia is eliminated in a bony fish.
Solution:
Gill membranes.

Question 10.
What is vasa recta ?
Solution:
The U-shaped peritubular capillary that runs parallel to the Henle’s loop is called vasa recta.

Question 11.
What is the driving force for glomerular filtration?
Solution:
The driving force for filtration is the blood pressure in the glomerular capillaries.

Question 12.
How are the filtration slits formed?
Solution:
The podocytes are arranged in an intricate manner so as to leave some minute spaces called filtration slits.

Question 13.
Why is glomerular filtration also called as ultrafiltration?
Solution:
Blood is filtered so finely through these membranes, that almost all the constituents of the plasma except the proteins pass onto the lumen of the bowman’s capsule. Therefore, it is considered as a process of ultrafiltration.

Question 14.
Name the mechanism that acts as a check for the Renin-angiotensin mechanism.
Solution:
Arterial Natriuretic Factor (ANF) Mechanism.

Question 15.
What is uremia?
Solution:
Uremia is a condition of excess accumulation of urea in the blood caused by the malfunctioning ofkidneys.

Question 16.
What term is given to the inflammation of glomerulus in nephron?
Solution:
Glomerulonephritis.

Question 17.
Which limb of loop of Henle is impermeable to water?
Solution:
Bowman’s capsule and glomerulus are collectively called malpighian body.

Question 18.
What is afferent arteriole?
Solution:
Blood vessels leading to glomerulus is called afferent arteriole.

Question 19.
Which hormone promotes the reabsorption of water from glomerular filtrate?
Solution:
Vasopressin promotes reabsorption of water from the glomerular filtrate.

SHORT ANSWER QUESTIONS

Question 1.
How does the proximal convoluted tubule of the nephron contribute to homeostasis?
Solution:
All essential nutrients and 70-80 percent of electrolytes and water are reabsorbed by the PCT segment. So it helps to maintain the pH and ionic balance of the body fluids by selective secretion of hydrogen ions, ammonia and potassium ions into the filtrate and by absorption of HCO3-” from it.

Question 2.
What are the functions of nephridia? Name an animal having protonephridia
Solution:
Nephridia help to eliminate nitrogenous wastes and maintain a fluid and ionic balance. Protonephridia are present in Amphioxus, Rotifers, Planaria, etc.

Question 3.
Kidney do not play a major role in excretion in ammonotelic animals. Justify.
Solution:
Ammonia is readily soluble in water and diffuses across the body surface.
In fish, it is excreted as ammonium ions through gill surface. So kidneys do not have any significant role in the elimination of ammonia.

Question 4.
What are the functions of ADH?
Solution:
(i) ADH facilites water absorption from the distal tubule.
(ii) It also affect kidney function by its constrictor effects also on blood vessels.

Question 5.
What is the ultimate method of correcting acute renal failure? Describe.
Solution:
Kidney transplantation is the ultimate method in the correction of acute renal failures (kidney failure). A functioning kidney is used in transplantation from a donor, preferably a close relative, to minimize its chances of rejection by the immune system of the host.

Question 6.
Mention the role of DCT in urine formation.
Solution:
Distal convoluted tubule plays the following roles:

  • Conditional reabsorption of Na+ & water takes place in this segment.
  • It also reabsorbs HCO3-”
  • It helps in the selective secretion of hydrogen and potassium ions to maintain the pH and sodium-potassium balance in the blood.

Question 7.
Why do persons suffering from very low blood pressure pass no urine?
Solution:
The blood passes into the glomerulus under high pressure during glomerular filtration. If blood pressure is less then it results in the failure of the ultrafiltration process in the glomerulus and hence, no urine formation occurs.

Question 8.
Name the passage in sequence through which urine passes from kidneys to the outside in humans. How is urine prevented from flowing back into the ureters?
Solution:
Kidneys → ureters → urinary bladder → urethra
Urine is prevented from flowing back into the ureters because the terminal part of each ureter passes obliquely through the bladder wall.

Question 9.
(a) The two human kidneys do not occur at the same level-explain.
(b) Why are Kidneys called retro-peritoneal?
Solution:
(a) Left kidney is slightly longer, narrower, median, and lies at a level 1.25 cm higher than the right kidney. The right kidney is lower in position due to the presence of the right lobe of the liver.
(b) Retroperitoneal. It is space that lies between the peritoneum and vertebral column, Kidneys occur in this space so that they are lined by peritoneum only on the ventral side.

Question 10.
Differentiate between Cortical Nephron and Juxtamedullary Nephron.
Solution:
NCERT Solutions for Class 11 Biology Chapter 19 Excretory Products and their Elimination 1

Question 11.
What is the chemical composition of human urine?
Solution:
Human urine is transparent, yellowish in colour and variable in chemical composition. It
consists primarily of water (95 %), with organic solutes including urea (2.6%), creatinine, uric acid, and trace amounts of enzymes, carbohydrates, hormones, fatty acids, pigments, and mucins, and inorganic ions such as Na+, K+, Cl– Mg2+, Ca2+, and phosphates.

Question 12.
What is Erythropoietin? What is its function?
Solution:
Erythropoietin is a glycoprotein hormone that controls erythropoiesis, or the formation of red blood cells. It acts as a cytokine (proteinsignaling molecule) for RBC precursor in bone marrow. It is produced by enterstitial fibroblasts in the kidney in close association with peritubular capillary and tubular epithelial tubule.

Long ANSWER QUESTIONS

Question 1.
Describe the structure of the kidney.
Solution:

  • Kidneys are reddish-brown, bean-shaped structures situated between the levels of last thoracic and third lumbar vertebra close to the dorsal inner wall of the abdominal cavity.
  • Each kidney of an adult human measures 10-12 cm in length, 5-7 cm in width, 2-3 cm in thickness with an average weight of 120-170 g.
  • Towards the centre of the inner concave surface of the kidney is a notch called hilum through which ureter, blood vessels and nerves enter.
  • Inner to the hilum is a broad funnel-shaped space called th& renal pelvis with projections called calyces.
  • The outer layer of kidney is a tough capsule. Inside the kidney, there are two zones, an outer cortex and an inner medulla.
  • The medulla is divided into a few conical masses (medullary pyramids) projecting into the calyces (sing.: calyx). The cortex extends in between the medullary pyramids as renal columns called Columns of Bertini.
    NCERT Solutions for Class 11 Biology Chapter 19 Excretory Products and their Elimination 2

Question 2.
Describe the role of organs other than the kidney in the process of excretion in human beings.
Solution:

  • The organs other than kidney involved in the process of excretion are (i) Lungs (ii) Skin (iii) Liver (iv) Intestine (v) Salivary glands.
  • Our lungs remove large amounts of C02 (18 liters/day) and also significant quantities of water every day. The liver, the
  • largest gland in our body, secretes bile-containing substances like bilirubin, biliverdin, cholesterol, degraded steroid hormones, vitamins, and drugs. Most of these substances ultimately pass out alongwith digestive wastes.
  • The sweat and sebaceous glands in the skin can eliminate certain substances through their secretions.
  • The sweat produced by the sweat glands is a watery fluid containing NaCl, small amounts of urea, lactic acid, etc.
  • Though the primary function of sweat is to facilitate a cooling effect of the body surface, it also helps in the removal of some of the wastes mentioned above.
  • Sebaceous glands eliminate certain substances like sterols, hydrocarbons, and waxes through sebum. This secretion provides a protective oily covering for the skin.

Question 3.
Describe the structure of nephron.
Solution:
Structure of nephron
A nephron has two parts-the glomerulus and the renal tubule
(i) Glomerulus: It is a tuft of capillaries formed by the afferent arteriole, which is a fine branch of the renal artery.
(ii) Renal tubule has three-part
(a) proximal convoluted tubule
(b) loop of Henle
(c) distal convoluted tubule
(a) Proximal convoluted tubule –

  1. The renal tubule is closed at the proximal, end; it is expanded and curved inwardly. form a double-walled cup-shaped structure called Bowman’s capsule.
  2. The glomerulus is located in the hollow of the Bowman’s capsule and together they constitute the renal corpuscle.
  3. The lumen of the capsule is continuous with the narrow lumen of the entire tubule.
  4. The tubule continues to form a highly convoluted proximal convoluted tubule (PCT).

(b) Loop of Henle – It arises from the end of the proximal convoluted tubule and ends at the starting of the distal tubule.
It is hairpin-like, with a descending limb (that extends into the medulla) and an ascending limb, that crosses back to the cortex.
(c) Distal convoluted tubule –

  1. The ascending limb, on entering the cortex becomes the distal convoluted tubule.
  2. It then continues as a short straight collecting tubule, that joins the collecting duct.
  3. Each collecting duct receives the collecting tubule of a number of nephrons.
  4. Many collects converge, ran through renal pyramids, and open into the renal pelvis through the openings called renal papillae, at the tip of pyramids.

Question 4.
Describe the process of hemodialysis.
Solution:
Haemodialysis:

  • Blood from the artery of a uremia patient is taken, cooled to 0°C, and mixed with an anti-coagulant like heparin.
  • The unit contains a coiled cellophane tube surrounded by a fluid (dialyzing fluid) having the same composition as that of plasma except for the nitrogenous wastes.
  • The porous cellophane membrane of the tube allows the passage of molecules based on the concentration gradient.
  • As nitrogenous wastes are absent in the dialysing fluid, these substances freely move out, thereby clearing the blood.
  • The cleared blood is pumped back to the body through a vein after adding anti-heparin to it. This method is a boon for thousands of uremic patients all over the world.

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

NCERT Solutions for Class 11 Biology Chapter 18 Body Fluids and Circulation

NCERT Solutions for Class 11 Biology Chapter 18 Body Fluids and Circulation

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 18 Body Fluids and Circulation.

Question 1.
Name the components of formed elements in the blood and mention one major function of each of them.
Solution:
Blood is a mobile connective tissue composed of a fluid, the plasma and formed elements.
Formed elements includes erythrocytes, leucocytes and platelets.
NCERT Solutions for Class 11 Biology Chapter 18 Body Fluids and Circulation 1
Blood corpuscles.
(a) Erythrocytes are the most prevalent corpuscles (approx. 5 to5.5 million/mm3 of blood). They contain carbonic anhydrase enzymes which help in the transportation ofCO2 and haemoglobin pigment which helps in the transportation of O2.
(b) Leucocytes lack any pigment and most active motile constituents of blood (approx. 6000-8000/ mm3 of blood). They may be of two types.
1. Granulocytes are the cells containing granules and a polymorphic nucleus.

  • Neutrophils: responsible for protection against infection.
  • Eosinophils: play important role in an allergic reaction.
  • Basophils: significant in inflammatory reaction.

2. Agranulocytes are cells which lack granules

  • Lymphocytes play a key role in immunological reactions.
  • Monocytes are phagocytic in nature.

(c) Platelets: There are involved in the coagulation or clotting of blood.

Question 2.
What is the importance of plasma proteins?
Solution:
Fibrinogen, globulins and albumins are the major plasma proteins. Fibrinogen are needed for clotting or coagulation of blood. Globulins primarily are involved in defense mechanisms of the body. Albumins help in osmotic balance.

Question 3.
Match Column I with Column II:
Column I                                           Column II
(a) Eosinophils                              (i) Coagulation
(b) RBC                                      (ii) Universal recipient
(c) AB blood Group                          (iii) Resist infection
(d) Platelets                               (iv) Contraction of heart
(e) Systole                                  (v) Gas transport
Solution:
Column I                                                           Column II
(a) Eosinophils                                  (iii) Resist infections
(b) RBC                                          (v) Gas transport
(c) AB blood Group                                (ii) Universal Recipient
(d) Platelets                                      (i) Coagulation
(e) Systole                                        (iv) Contraction of heart

Question 4.
Why do we consider blood as a connective tissue?
Solution:
Blood is a special connective tissue consisting of a fluid matrix, plasma, and the formed elements. It is circulated throughout the body.

Question 5.
What is the difference between lymph and blood?
Solution:
Differences between blood and lymph are as following :
NCERT Solutions for Class 11 Biology Chapter 18 Body Fluids and Circulation 2

Question 6.
What is meant by double circulation? What is its significance?
Solution:
The blood pumped by the right ventricle enters the pulmonary artery whereas the left ventricle pumps blood into the aorta. The deoxygenated blood pumped into the pulmonary artery is passed on to the lungs from where the oxygenated blood is carried by the pulmonary veins into the left atrium. This pathway constitutes pulmonary circulation. The oxygenated blood entering the aorta is carried by a network of arteries, arterioles, and capillaries to the tissues from where the deoxygenated blood is collected by a system of venules, veins, and vena cava and emptied into the right atrium.

This is the systemic circulation (fig 18.1). The systemic circulation provides nutrients, O2, and other essential substances to the tissues and takes CO2, and other harmful substances away for elimination. A unique vascular connection exists between the digestive tract and liver called the hepatic portal system. The hepatic portal vein carries blood from the intestine to the liver before it is delivered to the systemic circulation. A special coronary system of blood vessels is present in our body exclusively for the circulation of blood to and from the cardiac musculature.

Question 7.
Write the differences between
(a) Blood and lymph
(b) OPen and closed system of circulation
(c) Systole and Diastole
(d) P-wave and T-wave
Solution:
(a) Differences between Blood and lymph NCERT Solutions for Class 11 Biology Chapter 18 Body Fluids and Circulation 2
(b) Differences between open and closed systems of circulation are as following :
NCERT Solutions for Class 11 Biology Chapter 18 Body Fluids and Circulation 3

(c) Differences between systole and diastole are as following :
NCERT Solutions for Class 11 Biology Chapter 18 Body Fluids and Circulation 4
(d) Differences between P-wave and T-wave are as following :
NCERT Solutions for Class 11 Biology Chapter 18 Body Fluids and Circulation 5

Question 8.
Describe the evolutionary change in the pattern of heart among the vertebrates.
Solution:

  • As it is clear from the following diagram the heart of fish has two chambers.
  • This means there is no separate circulation for oxygenated and deoxygenated blood.
  • There is separation of two chambers in the atrium of amphibians.
  • This has further evolved to partial separation of ventricle in reptiles.
  • Finally in birds there is complete separation of oxygenated and deoxygenated blood circulation with advent of four chambers in the heart.
  • Mammal heart is the most developed having the most efficient double circulatory system.
    NCERT Solutions for Class 11 Biology Chapter 18 Body Fluids and Circulation 6

 

Question 9.
Why do we call our heart myogenic?
Solution:
Normal activities of the heart are regulated intrinsically i.e., auto regulated by specialised muscles, hence our heart is called myogenic.

Question 10.
The sino-atrial node is called the pacemaker of our heart. Why?
Solution:
The SA node is located in the wall of right auricle slightly below the opening of the superior vena cava. It has a unique property of self-excitation which enables it to act as the pacemaker of the heart. It spontaneously initiates a wave of contraction which spreads over both the auricles more or less simultaneously along the muscle fibres.

Question 11.
What is the significance of the atrioventricular node and atrioventricular bundle in the functioning of the heart?
Solution:
Another mass of model tissue is seen in the lower left comer of the right atrium close to the atrioventricular septum called the atrioventricular node (AVN). A bundle of nodal fibers, atrioventricular bundle (AV bundle) continues from the AVN which passes through the atrioventricular septa to emerge on the top of the interventricular septum and immediately divides into a right and left bundle. These branches give rise to minute fibers throughout the ventricular musculature of the respective sides and are called Purkinje fibers. These fibers along with the right and left bundle are known as the bundle of HIS. The nodal musculature has the ability to generate action potentials without any external stimuli i.e., it is auto excitable. However, the number of action potentials that could be generated in a minute varies at different parts of the nodal system.

Question 12.
Define a cardiac cycle and the cardiac output.
Solution:
Cardiac cycle.   A regular sequence of three events :
(i) auricular systole,
(ii) ventricular systole, and
(iii) joint diastole or complete cardiac diastole (relaxation ofboth auricles and ventricles) during the completion of one heart beat is known as heart cycle or cardiac cycle.
Cardiac output. The amount of blood pumped by heart per minute is called cardiac output or heart output. Heart beats 72 times per minute and pumps out about 70 ml of blood during each beat. Therefore, 72 x 70 or 5040 ml (roughly 5 liters) is the cardiac output.

Question 13.
Explain heart sounds.
Solution:
During each cardiac cycle, two prominent sounds are produced which can be easily heard through a stethoscope. The first heart sound (dub) is associated with the closure of the tricuspid and bicuspid valves whereas the second heart sound (dub) is associated with the closure of the semilunar valves. These sounds are of clinical diagnostic significance.

Question 14.
Draw a standard ECG and explain the different segments in it.
Solution:
The recording of electrical potential generated by the spread of cardiac impulse is called an electrocardiogram (ECG).

ECG is the graphic record of electronic current produced by the excitation of cardiac muscles.

A normal electrocardiogram is composed of P wave, QRS complex and T wave, P wave indicate the depolarisation of the atria. QRS complex expresses ventricular depolarisation. T wave indicates repolarization of ventricles.
NCERT Solutions for Class 11 Biology Chapter 18 Body Fluids and Circulation 7

VERY SHORT ANSWER QUESTIONS

Question 1.
What is the Coagulation of Blood?
Solution:
When blood has been shed, it quickly becomes sticky and soon sets as a red jelly. This jelly or clot contracts or shrinks, and a straw-colored fluid called serum is squeezed out from it.

Question 2.
What is meant by an open circulatory system?
Solution:
When the blood does not remain confined to the blood vessels and flows into open spaces called sinus. It is termed as the open circulatory system.

Question 3.
What is the role of basophils in human body?
Solution:
Basophils are significant in allergic reactions.

Question 4.
Write down the main function of neutrophils.
Solution:
Neutrophils are mainly responsible for protection against infection.

Question 5.
What are the functions of lymphocytes?
Solution:
Lymphocytes play an important role in cell- mediated immunity.

Question 6.
What is tunica externa?
Solution:
Tunica externa is the outermost layer of blood vessels (artery and vein) made up of fibrous connective tissue with collagen fibres.

Question 7.
Mention the function of pericardial fluid.
Solution:
Pericardial fluid keeps the surface of heart moist and prevents the friction between heart wall and surrounding tissues.

Question 8.
What does the QRS complex indicate?
Solution:
QRS complex represent the ventricular depolarizations.

Question 9.
Name the most common disorder of blood circulatory system.
Solution:
Hypertension.

Question 10.
Where is tricuspid valve located? (April 87)
Solution:
The tricuspid valve is located between the right Atrium (auricle) and right ventricle.

Question 11.
Name the type of granulocytes that play an important role in detoxification.
Solution:
Eosinophils.

Question 12.
What transmits the cardiac impulse from the atria to the ventricles?
Solution:
Atrio – ventricular bundle from the atrioventricular node transmits the cardiac impulse from the atria to ventricles.

Question 13.
What is RBCs density in the blood of an adult human?
Solution:
About 5.0 – 5.5 millions/mm3 ofblood.

Question 14.
Name the reptile that has a four-chambered heart
Solution:
Crocodile.

Question 15.
What are Purkinje fibres?
Solution:
The minute branches of the right and left AV-bundles, that are found throughout the ventricular musculature of the respective sides, are called Purkinje fibres.

Question 16.
Name two vital organs affected by high blood
1. pressure or hypertension.
Solution:
Brain, kidney

Question 17.
What is the main symptom of heart failure?
Solution:
Congestion of the lungs.

Question 18.
Which vein carries oxygenated blood?
Solution:
Pulmonary vein.

Question 19.
Which is the valve present in between the left auricle and left ventricle? (April 97, M.Q.P.)
Solution:
The Bicuspid or mitral valve.

Question 20.
Which human organ is known as the “graveyard of RBCs”?
Solution:
Spleen.

SHORT ANSWER QUESTIONS

Question 1.
What are thrombocytes? Where are they produced in the human body?
Solution:
Thrombocytes or blood platelets are colourless formed elements of blood which appear round, or biconvex, or irregular and helps in clotting of blood. They are produced from the megakaryocytes (special cells in the bone marrow).

Question 2.
What is a serum?
Solution:
Serum is straw coloured fluid left after the clotting of blood. It is also called blood serum.

Question 3.
Name the different types of granulocytes. Give the function of the one of which constitutes the maximum percentage of total leucocytes.
Solution:
Granulocytes are of three types neutrophils, eosinophils, basophils.
Neutrophils constitute the maximum percentage of the total leucocyte, mainly responsible for protection against infection. They engulf the foreign substances by phagocytosis.

Question 4.
Why the closed circulatory system is more efficient than an open circulatory system?
Solution:
Advantage of closed circulatory system:
(i) Flow of blood is faster in the closed circulatory system as compared to open circulatory system. Sufficiently high blood pressure can be maintained.

  • Blood does not come in contact with the tissues/organs.
  • The volume of blood flowing to a particular tissue/organ can be regulated according to the need.

(ii) The blood flows under pressure so that all parts of the body receive blood with equal efficiency.
(iii) It transports materials efficiently.
(iv) There are checks for the regulation of the amount and speed of blood passing into an organ according to the requirement of that organs.

Question 5.
Name a portal system present in man. Write its one function.
Solution:
Hepatic portal system.
Significance: The blood which comes from the alimentary canal contains digested food like glucose and amino acids. The excess of glucose is converted into glycogen which is stored in the liver for later use.

Question 6.
Write down the functions of lymph.
Solution:
Functions of lymph :
(i) Lymph acts as a ‘middle man’ which transports oxygen, food materials, hormones etc. to the body cells and brings carbon dioxide and other metabolic wastes from body cells to blood.
(ii) It keeps the body cell moist.
(iii) It maintains the volume of blood.
(iv) It absorbs and transports fat and fat-soluble vitamins from the intestine.

Question 7.
Explain the chemical events taking place during clotting /coagulation of blood.
(Foreign 1997)
Solution:
An injury or a trauma stimulates the plate- 195 lets in the blood to release certain factors which activate the mechanism of coagulation. Clot or coagulam informed mainly of network threads called fibrils in which dead and damaged formed elements of blood are trapped. Fibrins are formed by the conversion of inactive fibrinogens in the plasma by the enzyme thrombin.

Thrombins, in turn are formed from another inactive substance present in the plasma called prothrombin. An enzyme complex, thrombokinase, is required for the above reaction which is formed by a series of linked enzymic reactions involving a number of factors present in the plasma in an inactive state. Calcium ions play a very important role in clotting.

Question 8.
What is an average number of thrombocytes in human blood? What is their function?
Solution:
1,50,000 to 3,50,000 platelets/mm3 of blood
– They release substances that are concerned with the clotting of blood.

Question 9.
How many chambers are present in the heart of a fish? Name them.
Solution:
There are two chambers one atrium and one ventricle. ,

Question 10.
What is meant by single circulation? Give an example.
Solution:
Single circulation

  • Single circulation is the phenomenon in which the heart of an animal receives and pumps blood once for e.g. fish.
  • The heart of fish is two-chambered with an atrium and a ventricle.
  • The heart pumps only deoxygenated blood, to the gills for oxygenation.

Question 11.
Where and from which cells do platelets originate? What is their life span? How do they act when blood vessels get injured?
Solution:
Platelets originate from the megakaryocytes in the bone marrow.

  • They live for about seven days.
  • They release thromboplastins, which help convert prothrombin of the plasma into thrombin and thus they are involved in clotting of blood.

Question 12.
Explain any three disorders of the circulatory system.
Solution:
(i) High Blood Pressure (Hypertension):
Hypertension is the term for blood pressure that is higher than normal (120/80). High blood pressure leads to heart diseases and also affects vital organs like the brain and kidney.

(ii) Coronary Artery Disease (CAD):
Coronary Artery Disease, often referred to as atherosclerosis, affects the vessels that supply blood to the heart muscle. It is caused by deposits of calcium, fat, cholesterol, and fibrous tissues, which makes the lumen of arteries narrower.

(iii) Heart Failure:
Heart failure means the state of the heart when it is not pumping blood effectively enough to meet the needs of the body. The main symptoms are congestion of the lungs.

Question 13.
What is Arteriosclerosis? What are its causes?
Solution:
It is thickening, hardening and loss of elasticity of the wall of arteries. This process progressively
restricts the blood flow to one’s organs and tissues and can lead to severe health risks. It is caused by the build-up of fatty plaque, cholestrol and some other substances in and on the artery wall.

Question 14.
Define Vagus escape.
Solution:
The stimulation of vagus nerve decreases the heart rate but its continuous stimulation shows no further decrease. This is known as vagus escape.

Question 15.
What is erythroblastosis foetalis? How it occurs?
Solution:
It is a type of haemolytic disease of new-borns due to ABO blood type A, B, or O is not compatible with blood group of foetus. It develops in a foetus, when IgG molecules produced by the mother passes through the placenta.

Question 16.
Why capillaries are known as exchange vessels?
Solution:
These have very thin walls which allows the passage of nutrients from blood into body tissues. It also allows the passage of waste product came from body tissues. So, capillaries are known as exchange vessels.

Long ANSWER QUESTIONS

Question 1.
Briefly explain the cardiac cycle.
Solution:
The sequential event in the heart which is cyclically repeated is called the cardiac cycle and it consists of systole and diastole of both the atria and ventricles. Initially, all the four chambers of the heart are in a relaxed state, i.e., joint diastole. Tricuspid and bicuspid valves are open resulting in the inflow of blood into the left and right ventricles from pulmonary veins and vena cava. Semilunar valves are closed at this stage. SAN (sino-atrial node) generates an action poten­tial resulting in simultaneous contraction of both atria – the atrial systole.

The action potential is conducted to the ventricular side by the AVN and AV bundle from where the bundle of His transmits it through the entire ventricular musculature. This causes contrac­tion of ventricles (ventricular systole) and relaxation of atria (diastole). Ventricular Systole increases the ventricular pressure causing the closure of tricuspid and bicuspid valves. As pressure increases, the semilunar valves of the pulmonary artery and aorta are forced open, allowing blood to flow through them into circulatory pathways.

The ventricles now relax (ventricular diastole) and semilunar valves close. As the ventricular pressure declines further, tricuspid and bicuspid valves are pushed open by the pressure in the atria. The blood now once again moves freely to the ventricles. The ventricles and atria are now again in a relaxed (joint diastole) state, as ear­lier and the process continue. The duration of a cardiac cycle is 0.8 sec­onds and during this period each ventricle pumps about 70 ml of blood.

Question 2.
(a) Draw the L. S. of a human heart showing the internal structure. Label the parts of the left side of the heart and the blood vessels that enter and leave the chambers of the same side.
(b) What is the significance of the remnant of sinus venosus in the mammalian heart?
Solution:
(a)
NCERT Solutions for Class 11 Biology Chapter 18 Body Fluids and Circulation 8

(b) It is believed that the remnant of sinus venous is modified into a sino-atrial node (SA Node) in the mammalian heart which is nothing but a mass of neuromuscular tissue laying in the wall of right atrium near the openings of the superior vena cava. It originates impulses for the regulation of the heartbeat. Thus acts as the pacemaker of heart.

Question 3.
Describe briefly the steps involved in the coagulation of blood at the site of injury of a blood vessel.
Solution:

  • Blood exhibits coagulation or clotting in response to an injury or trauma.
  • This is a mechanism to prevent excessive loss of blood from the body. We would have observed a dark reddish-brown scum formed at the site of a cut or an injury over a period of time.
  • It is a clot or coagulum formed mainly of a network of threads called fibrils in which dead and damaged formed elements of blood are trapped.
  • Fibrins are formed by the conversion of inactive fibrinogens in the plasma by the enzyme thrombin.
  • Thrombins, in turn, are formed from another inactive substance present in the plasma called prothrombin.
  • An enzyme complex, thrombokinase, is required for the above reaction.
  • This complex is formed by a series of linked enzymic reactions (cascade process) involving a number of factors present in the plasma in an inactive state.
  • An injury or a trauma stimulates the platelets in the blood to release certain factors which activate the mechanism of coagulation.
  • Certain factors released by the tissues at the site of injury also can initiate coagulation. Calcium ions play a very important role in clotting.

Question 4.
Why is it necessary to check the Rh-factor of the blood of a pregnant woman?
Solution:
Rh-Factor

  • Rh-antigen is present on the surface of erythrocytes in about 80-85% of human beings.
  • The individuals who possess this antigen are called Rh-positive and those who do not have it are called Rh-negative.
  • An Rh-negative person, when exposed to Rh-positive blood, develops anti-Rh- antibodies.
  • If a pregnant woman who is Rh-negative, bears an Rh-positive foetus, will develop anti-Rh-antibodies during the first delivery when the foetal blood comes in contact with her blood.
  • These antibodies linger in the blood for sufficiently long periods.
  • If she carries a second foetus, that is Rh-positive, the anti-Rh-antibodies in her blood enter the foetal circulation and cause damage to the foetal RBCs, and could become fatal.
  • This condition is called erythroblastosis foetalis.

We hope the NCERT Solutions for Class 11 Biology at Work Chapter 18 Body Fluids and Circulation, help you. If you have any query regarding NCERT Solutions for Class 11 Biology at Work Chapter 18 Body Fluids and Circulation, drop a comment below and we will get back to you at the earliest.

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

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

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

Question 1.
Briefly describe the structure of the following:
(a) Brain
(b) Eye
(c) Ear
Solution:
Structure of brain:

  • The brain is the central information processing organ of the body,
  • The human brain is well protected by the skull.
  • Inside the skull, cranial meninges cover the brain. These are tough tissue layers.

Meninges consist of 3 layers which are as follows :

  • The outermost – layer is the dura mater
  • The middle layer is arachnoid
  • The inner layer is pia mater.

The brain can be divided into three major parts which are given below:

  • forebrain
  • midbrain
  • hindbrain

NCERT Solutions for Class 11 Biology Chapter 21 Neural control and co-ordination 1
Forebrain:
1. Cerebrum consists two cerebral hemisphere on the dorsal surface. It is connected by a tract of nerve fibres called corpus callosum. Cerebral hemispheres are covered by the layer of cells called cerebral cortex and are thrown into prominent folds referred as grey matter. Inner part of the cerebral hemisphere is white matter.
2. Diencephlon is the posterior part of fore-brain. It consists of thalamus and hypothalamus.

  • Thalamus is a major co-ordinating centre for sensory and motor signaling. It forms 80 % of diencephalon.
  • Hypothalamus contains a number of centres which control many functions Like – hunger, thirst, sleep, sweating, body temperature and emotions

(ii) Midbrain:
It is located between the thalamus/ hypothalamus of the forebrain and pons of the hindbrain.
It forms the brain stem with the hindbrain. Anterior part of mid-brain contains two cerebral peduncles, which controls the muscle of limbs and Posterior part of mid-brain in four optic lobs called corpora quadri
gemiana i.e. two upper and two lower.

(iii) Hindbrain: Hindbrain consists of pons. Cerebellum and medulla of longata.

  • Pons is present below the midbrain and upper side of medulla oblongata. It possesses pneumotaxic area of respiratory centre.
  • Cerebellum is the 2nd largest part of brain, which lies behind cerebrum and provides the additional space for many neuron and maintains equilibrium or posture of the body.
  • Medulla oblongata lies below cerebellum and continues into the spinal-cord. It contains a respiratory centre for regulating breatheing, Cardiac centre for regulating heartbeat and blood pressure, and also has reflex centre for swallowing, coughing, sneezing, etc.

(b) Structure of eye :

  1. Our paired eyes are located in sockets of the skull called orbits.
  2. The adult human eyeball is nearly a spherical structure.
  3. The wall of the eyeball is composed of three layers which are given below :
  4. .The external layer is composed of a dense connective tissue and is called the sclera. The anterior portion of this layer is called the cornea.
  5. The middle layer is called the choroid contains many blood vessels and looks bluish.
  6. The choroid layer is thin over the posterior two-thirds of the eyeball, but it becomes thick in the anterior part to form the ciliary body.
  7. The ciliary body itself continues forward to form a pigmented and opaque structure called the iris which is the visible coloured portion of the eye.
  8. The eyeball contains a transparent crystalline lens which is held in place by ligaments attached to the ciliary body.
  9.  In front of the lens, the aperture surrounded by the iris is called the pupil. The diameter of the pupil is regulated by the muscle fibres of iris.

The inner layer is the retina and contains three layers of cells- ganglion cells, bipolar cells and photoreceptor cells.
NCERT Solutions for Class 11 Biology Chapter 21 Neural control and co-ordination 2

  • In the human eye, there are three types of cones which possess their own characteristic photopigments that respond to red, green and blue lights.
  • The sensations of different colours are produced by various combinations of these cones and their photopigments. When these cones are stimulated equally, a sensation of white light is produced.
  • The optic nerves leave the eye and the retinal blood vessels enter it at a point medial to and slightly above the posterior pole of the eyeball.
  • Photoreceptor cells are not present in that region and hence it is called the blind spot. At the posterior pole of the eye lateral to the blind sport, there is a yellowish pigmented spot called macula lutea with a central pit called the fovea.
  • The fovea is a thinned-out portion of the retina where only the cones are densely packed.
  • It is the point where the visual acuity (resolution) is the greatest.

(c) Structure of ear:
Anatomically, the ear can be divided into three major sections called the outer ear, the middle ear and the inner ear.
Outer ear. The outer ear consists of the pinna and external auditory meatus (canal). The pinna collects the vibrations in the air which produce sound.

  • The external auditory meatus leads inwards and extends up to the tympanic membrane (the ear dfum).
  • There are very fine hairs and wax-secreting sebaceous glands in the skin of the pinna and the meatus.
  • The tympanic membrane is composed of connective tissues covered with skin outside and with mucus membrane inside.
  • Middle ear : The middle ear contains three ossicles called malleus, incus and stapes which are attached to one another in a chain-like fashion. These ossicles transmit sound waves further inside the ear.
  • An Eustachian tube connects the middle ear cavity with the pharynx. The Eustachian tube helps in equalizing the pressures on either sides of the eardrum. Inner ear the fluid-filled inner ear called labyrinth consists of two parts, the bony and the membranous labyrinths. The bony labyrinth is a series of channels.
  • Inside these channels lies the membranous labyrinth, which is surrounded by a fluid called perilymph.
    The membranous labyrinth is filled with a fluid called endolymph.
  • The coiled portion of the labyrinth is called cochlea. The membranes constituting cochlea, the Meissner’s membrane and basilar membrane, divide the surrounding perilymph filled bony labyrinth into an upper scala vestibuli and a lower scala tympani.
  • The organ of corti is a structure located on the basilar membrane which contains hair cells that act as auditory receptors. The hair cells are present in rows on the internal side of the organ of Corti.
  • The inner ear also contains a complex system called vestibular apparatus, located above the cochlea. The vestibular apparatus is composed of three semi-circular canals and the otolith organ consisting of the saccule and utricle.
  • The base of canals is swollen and is called ampulla, which contains a projecting ridge called crista ampullaris which has hair cells. The saccule and utricle contain a projecting ridge called macula.
  • The crista and macula are the specific receptors of the vestibular apparatus responsible for maintenance of balance of the body and posture.

Question 2.
Compare the following:
(i) Central Neural System (CNS) and Peripheral neural system (PNS)
(ii) Resting potential and action potential
(iii) Choroid and retina
Solution:
Differences between central neural system and peripheral nervous system are as follows :
NCERT Solutions for Class 11 Biology Chapter 21 Neural control and co-ordination 3
Differences between resting potential and action potential are as follows:
NCERT Solutions for Class 11 Biology Chapter 21 Neural control and co-ordination 4
Differences between choroid and retina are as follows :
NCERT Solutions for Class 11 Biology Chapter 21 Neural control and co-ordination 5

Question 3.
Explain the following processes:
(a) Polarization of the membrane of a nerve fibre
(b) Depolarization of the membrane of a nerve fibre
(c) Conduction of a nerve impulse along a nerve fibre
(d) Transmission of a nerve impulse across a chemical synapse
Solution:
(a) When a neuron is not conducting any impulse; i.e resting, the axoplasm inside the axon contains high concentration of K+ and negatively charged proteins and low concentration of Na+. In contrast, the fluid outside the axon contains a low concentration of K+, a high concentration of Na+ and thus forms a concentration gradient. These ionic gradients across the resting membrane are maintained by the active transport of ions by the sodium-potassium pump which transports 3 Na+ outwards for 2K+ into the cell. As a result, the outer surface of the axonal membrane possesses a positive charge while its inner surface becomes negatively charged and therefore is polarised.

(b) When a stimulus is applied at a site, (eg: site A) on the polarised membrane, the membrane at site A becomes freely permeable to Na+. This leads to a rapid influx of Na+ followed by the reversal of the polarity at that site, i.e the outer surface of the membrane becomes negatively charged and the inner side becomes positively charged. The polarity of the membrane at site A is thus reversed and hence depolarised.

(c) Conduction of nerve impulse along a nerve fibre:

  • When a stimulus is applied at a site on the polarised membrane the membrane at the site A becomes freely permeable to Na+.
  • Asa result polarity gets reversed by a rapid inflow of Na+.
  • After the reversal of polarity of the membrane, the membrane becomes depolarised.
  • The electrical potential difference across the plasma membrane at the site A is called the action potential; which is termed as nerve impulse.
  • The axon membrane (site B) has a positive charge on the outer surface and a negative charge on its inner surface. As a result, a current flows on the inner surface from site A to site B.
  • On the outer surface current flows from site B to site A, to complete the circuit of current flow.
  • Hence, the polarity at the site is reversed, and an action potential is generated at site B.
  • Thus, the impulse generated at site A arrives at site B. The sequence is repeated along the length of the axon and consequently impulse is conducted.
  • The rise in the stimulus-induced permeability to Na+ is extremely short-lived.
  • It is quickly followed by a rise in permeability to K+. Within a fraction of a second, K+ diffuses outside the membrane and restores the resting potential of the membrane at the site of excitation and the fibre becomes once more responsive to further stimulation.

(d) Transmission of a nerve impulse across chemical synapse:

  • A nerve impulse is transmitted from one neuron to another through junction called synapses.
  • Electrical current can flow directly from one neuron to the other across these synapses.
  • The membrane of the pre-and post-synaptic neurons are separated by fluid-filled space called synaptic deft.
  • Chemicals called neurotransmitters are involved in the transmission of impulses at these synapses.
  • The axon terminals contain vesicles filled these neurotransmitters.
  • When an impulse (action potential) arrives at the axon terminal, it stimulates the movement of the synaptic vesicles towards the membrane where they fuse with the plasma membrane and release their neurotransmitters in the synaptic cleft. The released neurotransmitters bind to their specific receptors, present on the post- synaptic membrane.
  • This binding opens ion channels allowing the entry of ions which can generate a new potential in the post- synaptic neuron. The new potential developed may be either excitatory or inhibitory.

Question 4.
Draw labelled diagrams of the following :
(a) Neuron
(b) Brain
(c) Eye
(d) Ear
Solution:
(a) Neuron
NCERT Solutions for Class 11 Biology Chapter 21 Neural control and co-ordination 6

(b) Brain
NCERT Solutions for Class 11 Biology Chapter 21 Neural control and co-ordination 7
(c) Eye
NCERT Solutions for Class 11 Biology Chapter 21 Neural control and co-ordination 8

(d)Ear
NCERT Solutions for Class 11 Biology Chapter 21 Neural control and co-ordination 9

Question 5.
Write short notes on the following
(a) Neural coordination
(b) Forebrain
(c) Midbrain
(d) Hindbrain
(e) Retina
(f) Ear ossicles
(g) Cochlea
(h) Organ of Corti
(i) Synapse
Solution:
(a) Neural co-ordination:

  • The functions of the organs/ organ systems in our body must be coordinated to maintain homeostasis. Coordination is the process through which two or more organs interact and complement the functions of one another. For example, when we do physical exercises, the energy demand is increased for maintaining an increased muscular activity.
  • The supply of oxygen is also increased. The increased supply of oxygen necessitates an increase in the rate of respiration, heartbeat and increased blood flow via blood vessels.
  • When physical exercise is stopped, the activities of nerves, lungs, heart and kidney gradually return to their normal conditions.
  • Thus, the functions of muscles, lungs, heart, blood vessels, kidney and other organs are coordinated while performing physical exercises.
  • In our body the neural system and the endocrine system jointly coordinate and integrate all the activities of the organs so that they function in a synchronised fashion.

(b) Forebrain :

  • The forebrain consists of cerebrum, thalamus and hypothalamus. Cerebrum forms the major part of the human brain. A deep cleft divides the cerebrum longitudinally into two halves, which are termed as the left and right cerebral hemispheres.
  • The hemispheres are connected by a tract of nerve fibres called corpus callosum.
    The cerebral cortex contains motor areas, sensory areas and large regions that are neither clearly sensory nor motor in function.
  • These regions called as the association areas are responsible for complex functions like intersensory associations, memory and communication.
  • The cerebrum wraps around a structure called thalamus, which is a major coordinating centre for sensory and motor signaling. Another very important part of the brain called hypothalamus lies at the base of the thalamus.
  • The hypothalamus contains a number of centres which control body temperature, urge for eating and drinking. It also contains several groups of neurosecretory cells, which secrete hormones called hypothalamic hormones. The inner parts of cerebral hemispheres and a group of associated deep structures like amygdala, hippocampus, etc., form a complex structure called the limbic lobe or limbic system.
  • Along with the hypothalamus, it is involved in the regulation of sexual behaviour, expression of emotional reactions (e.g., excitement, pleasure, rage and fear), and motivation.

(c) Midbrain:

  • It is located between the thalamus hypothalamus of the forebrain and pons of the hindbrain.
  • The dorsal portion of the midbrain consists of four small lobes called as corpora quadrigemina.
  • A canal called the cerebral aqueduct passes through the midbrain. Neural control and co-ordination.

(d) Hindbrain:

  • It consists of pons, cerebellum and medulla oblongata.
  • Cerebellum has very convoluted surface to provide the additional space for many more neurons.
  • The medulla is the part that continues as a spinal cord.
  • The medulla contains centres which control respiration, cardiovascular reflexes and gastric secretions.

(e) Retina :

  • The retina contains three layers of cells – from inside to outside ganglion cells, bipolar cells and photoreceptor cells.
  • There are two types of photoreceptor cells, namely, rods and cones. These cells contain the light-sensitive proteins called the photopigments.
  • The daylight (photopic) vision and colour vision are functions of cones and the twilight (scotopic) vision is the function of the rods.
  • The rods contain a purplish-red protein called the rhodopsin or visual purple, which contains a derivative of vitamin A.
  • In the human eye, there are three types of cones that possess their own characteristic photopigments that respond to red, green and blue lights.
  • The sensations of different colours are produced by various combinations of these cones and their photopigments. When these cones are stimulated equally, a sensation of white light is produced.

(f) Ear ossicles –

  • The middle ear contains three ossicles called malleus, incus and stapes which are attached to one another in a chain-like fashion.
  • Malleus is attached to the tympanic membrane and the stapes is attached to the oval window of the cochlea.
  • The ear ossicles increase the efficiency of transmission of sound waves to the inner ear.

(g) Cochlea :

  • The fluid-filled inner ear called labyrinth consists of two parts, the bony and the membranous labyrinths. The coiled portion of the labyrinth is called cochlea.
  • The membranes constituting cochlea, the reissner’s and basilar, divide the surounding perilymph filled bony labyrinth into an upper scala vestibuli and a lower scala tympani.
  • The space within cochlea called scala media is filled with endolymph. At ‘ the base of the cochlea, the scala vestibuli ends at the oval window, while the scala tympani terminates at the round window which opens to the middle ear.

(h) Organ of corti:

  • The organ of corti is a structure located on the basilar membrane which contains hair cells that act as auditory receptors.
  • The hair cells are present in rows on the internal side of the organ of corti.
  • The basal end of the hair cell is in close contact with the afferent nerve fibres.
  • A large number of processes called stereo cilia are projected from the apical part of each hair cell.
  • Above the rows of the hair cells is a thin elastic membrane called tectorial membrane.

(i) Synapse:

  • A nerve impulse is transmitted from one neuron to another through junctions called synapses.
  • A synapse is formed by the membranes of a pre-synaptic neuron and a post-synaptic neuron, which may or may not be separated by a gap called synaptic cleft.
  • There are two types of synapses, namely, electrical synapses and chemical synapses.

Electrical synapse:

  • At electrical synapses, the membranes of pre-and post-synaptic neurons are in very close proximity.
  • Electrical current can flow directly from one neuron into the other across these synapses.
  • Transmission of an impulse across electrical synapses is very similar to impulse conduction along a single axon.
  • Impulse transmission across an electrical synapse is always faster than that across a chemical synapse.
  • Electrical synapses are rare in our system.

Chemical synapse:

  • At a chemical synapse, the membranes of the pre-and post-synaptic neurons are separated by a fluid-filled space called synaptic cleft.
  • Chemicals called neurotransmitters are involved in the transmission of impulses at these synapses.
    The axon terminals contain vesicles filled with these neurotransmitters.
  • When an impulse (action potential) arrives at the axon terminal, it stimulates the movement of the synaptic vesicles towards the membrane where they fuse with the plasma membrane and release their neurotransmitters in the synaptic cleft.
  • The released neurotransmitters bind to their specific receptors, present on the post-synaptic membrane.
  • This binding opens ion channels allowing the entry of ions which can generate a new potential in the post-synaptic neuron. The new potential developed may be either excitatory or inhibitory.

Question 6.
Give a brief account of :
(a) Mechanism of synaptic transmission
(b) Mechanism of vision
(c) Mechanism of hearing
Solution:
(a) Synaptic transmission can be of two types :
(i) Transmission of nerve impulse through electrical synapse.
(ii) Transmission of nerve impulse through chemical synapse.

Electrical synaptic transmission: At electrical synapses, the membranes of pre- and post- synaptic neurons are in very close proximity.
Electrical current can flow directly from one neuron into the.other across these synapses. Transmission of an impulse is very similar to impulse conduction along a single axon.
Chemical synaptic transmission : The membranes of the pre- and post-synaptic neurons are separated by a fluid-filled space called synaptic cleft. Neurotransmitters are involved in the transmission of impulses.
When an impulse (action potential) arrives at the axon terminal, it stimulates the movement of the synaptic vesicles towards the membrane where they fuse with the plasma membrane and release their neurotransmitters in the synaptic cleft.
The released neurotransmitters bind to their specific receptors, present on the post-synaptic membrane.
This binding opens ion-channels allowing the entry of ions which can generate a new potential in the post-synaptic neuron.

(b) Mechanism of vision

  • The light rays in visible wavelength focussed on the retina through the cornea and lens generate potentials (impulses) in rods and cones.
  • The photosensitive compounds (photopigments) in the human eyes is composed of opsin (a protein) and retinal (an aldehyde of vitamin A).
  • Light induces dissociation of the retinal from opsin resulting in changes in the structure of the opsin. This causes membrane permeability changes.
  • As a result, potential differences are generated in the photoreceptor cells. This produces a signal that generates action potentials in the ganglion cells through the bipolar cells.
  • These action potentials (impulses) are transmitted by the optic nerves to the visual cortex area of the brain, where the neural impulses are analysed and the image formed on the retina is recognised based on earlier memory and experience.

Mechanism of hearing

  • The external ear receives sound waves and directs them to the ear drum. The ear drum vibrates in response to the sound waves and these vibrations are transmitted through the ear ossicles (malleus, incus and stapes) to the oval window.
  • The vibrations are passed through the oval window on the fluid of the cochlea, where they generate waves in the lymphs.
  • The waves in the lymphs induce a ripple in the basilar membrane.
  • These movements of the basilar membrane bend the hair cells, pressing them against the tectorial membrane.
  •  As result, nerve impulses are generated in the associated afferent neurons. These impulses are transmitted by the afferent fibres via auditory nerves to the auditory cortex of the brain, where the impulses are analysed and the sound is recognised.

Question 7.
Answer briefly:
(a) How do you perceive the colour of an object?
(b) Which part of our body helps us in maintaining the body balance?
(c) How does the eye regulate the amount of light that falls on the retina?
Solution:
(a) Rods and cones are photoreceptor cells that contain light-sensitive proteins called photopigments. The daylight vision and colour vision are functions of cones. There are three types of cones which respond to red, green and blue lights. The sensations of different colours are produced by various combinations of these cones and their photopigments. When these cones are stimulated equally, a sensation of white light is produced.

(b) The crista and macula are the specific receptors of the vestibular apparatus responsible for the maintenance of the balance of the body.

(c) The pupil in the eye functions as an aperture. This dilates in case of low light and constricts in case of intense light thereby regulating the amount of light falling on the retina.

Question 8.
Explain the following:
(a) Role of Na+ in the generation of the action potential.
(b) Mechanism of generation of light-induced impulse in the retina.
(c) Mechanism through which a sound produces a nerve impulse in the inner ear.
Solution:
(a) When a stimulus is applied at a site, (eg: site A) on the polarised membrane, the membrane at site A becomes freely permeable to Na+. This leads to a rapid influx of Na+ followed by the reversal of the polarity at that site, i.e., the outer surface of the membrane becomes negatively charged and the inner side becomes positively charged. The polarity of the membrane at site A is thus reversed and hence depolarised. Thus action potential is generated across the plasma membrane.

(b) Mechanism of vision: The rays in visible wavelength focussed on the retina through the cornea and lens generate potentials (impulse) in rods and cones. The photo-sensitive compounds (photopigments) in the human eyes is composed of opsin and retinal. Light induces dissociation of the retinal from opsin resulting in changes in the structure of the opsin. This causes membrane permeability to change.

As a result, potential differences are generated in the photoreceptor cells. This produces a signal that generates action potentials in the ganglion cells through the bipolar cells. These action potentials are transmitted by the optic nerves to the visual cortex area of the brain, where the neural impulses are analysed and the image formed on the retina is recognized based on earlier memory and experience.

(c) Mechanism of hearing: The external ear receives sound waves and directs them to the eardrum. The eardrum vibrates in response to the sound waves and these vibrations are transmitted through the ear ossicles to the oval window. The vibrations are passed through the oval window on to the fluid of the cochlea, where they generate waves in the lymphs.

The waves in the lymphs induce a ripple in the basilar membrane. These movements of the basilar membrane bend the hair cells, pressing them against the tectorial membrane. As a result, nerve impulses are generated in the associated afferent neurons. These impulses are transmitted by the afferent fibres via auditory nerves to the auditory cortex of the brain, where the impulses are analysed and the sound is recognised.

Question 9.
Differentiate between:
(a) Myelinated and non-myelinated axons
(b) Dendrites and axons
(c) Rods and cones
(d) Thalamus and Hypothalamus
(e) Cerebrum and Cerebellum
Solution:
Differences:
NCERT Solutions for Class 11 Biology Chapter 21 Neural control and co-ordination 10

Question 10.
Answer the following:
(a) Which part of the ear determines the pitch of a sound ?
(b) Which part of the human brain is the most developed?
(c) Which part of our central neural system acts as a master clock ?
Solution:
(a) Pitch represents the perceived fundamental frequency of a sound. Pitch is a subjective sensation in which a listener assigns perceived tones to relative positions on a musical scale based primarily on the frequency of vibration. As spund is finally perceived by the temporal lobe of the cerebral cortex so it can be said that cerebral cortex perceives the pitch of the sound.
(b) Cerebral cortex (c) Hindbrain

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

Question 12.
Distinguish between
(a) afferent neurons and efferent neurons
(b) impulse conduction in a myelinated nerve fibre and unmyelinated nerve fibre
(c) aqueous humor and vitreous humor
(d) blind spot and yellow spot
(e) cranial nerves and spinal nerves.
Solution:
Differences:
NCERT Solutions for Class 11 Biology Chapter 21 Neural control and co-ordination 11

VERY SHORT ANSWER QUESTIONS

Question 1.
Name the band of nerve fibres that joins the cerebral hemispheres in mammals.
Solution:
The hemispheres are connected by a tract of nerve fibres called corpus callosum.

Question 2.
How many types of nerve fibres do PNS have ? Name them.
Solution:
The nerve fibres of the PNS are of two types :
(a) afferent fibres
(b) efferent fibres

Question 3.
Name the functional unit of nervous system.
Solution:
The functional unit of nervous system is neuron.

Question 4.
How many types of axons are present in CNS? Name them.
Solution:
There are two types of axons, namely, myelinated and non-myelinated.

Question 5.
Name the functional junction between two neurons.
Solution:
A synapse is the functional junction between two neurons.

Question 6.
What does synaptic vesicles contain ?
Solution:
Synaptic vesicles contain chemicals called neurotransmitters.

Question 7.
Name the fluid in which membranous labyrinth of the inner ear floats.
Solution:
The fluid in which membranous labyrinth of the inner ear floats is perilymph.

Question 8.
Why is blind spot devoid of the ability of vision?
Solution:
Blind spot have no photoreceptor cells (rods & cones) and hence it is devoid of vision.

Question 9.
Name the canal which passes through midbrain.
Solution:
Cerebral aqueduct is the canal which passes through midbrain.

Question 10.
Name the type of neuron which carries the signal from CNS to the effector organs.
Solution:
Efferent neuron carries the signal from CNS to the effector/organs.

Question 11.
Name the fluid which fills anterior chamber of eye.
Solution:
An aqueous fluid which fills anterior chamber of eye is aqueous humor.

Question 12.
Name the cells which are responsible for photopic (daylight) and colour vision.
Solution:
Cones are responsible for photopic (day light) and colour vision.

Question 13.
Name the part through which middle ear communicates with the internal ear.
Solution:
Oval window helps the middle ear communicates with the internal ear.

Question 14.
Name the specific receptors of the vestibular apparatus.
Solution:
The crista and macula are the specific receptors of the vestibular apparatus.

Question 15.
Give the technical names of the auditory ossicles in their natural sequence.
Solution:
The technical names of the auditory ossicles are malleus, incus, staps.

Question 16.
Name the membranes constituting the cochlea.
Solution:
Reissner’s membrane and basilar membrane constitute the cochlea.

Question 17.
What is brain stem ?
Solution:
Brain stem is part of brain that lies in continuation of spinal cord, viz., medulla oblongata, pons and mid brain (with or without diencephalon of forebrain)

Question 18.
What is optic chiasma?
Solution:
A cross like structure found on anterior surface of hypothalamus is called optic chiasma.

Question 19.
Name the largest and longest cranial nerve?
Solution:
The largest cranial nerve is Trigeminal nerve and the longest cranial nerve is vagus nerve.

SHORT ANSWER QUESTIONS

Question 1.
What are Nissl’s granules.
Solution:
The cell body contains cytoplasm with typical cell organelles and certain granular bodies called Nissl’s granules.

Question 2.
Describe the location and the role of ciliary body in human eye.
Solution:
Ciliary body is thick vascular, less pigmented ring shaped muscular structure occuring at the junction of choroid and iris. Ciliary body controls the size of pupil and in this way controls the amount of light entering the eye.

Question 3.
Explain the structural and functional significance of fovea in human eye.
Solution:
Fovea
– The fovea is a thinned-out portion of the retina where only the cones are densely packed, ft is the point where the visual acuity (resolution) is the greatest.
– It is a slightly depressed, tiny circular area found in the retina, just above the blind spot.

Question 4.
What is a reflex action?
Solution:
A sudden withdrawal of a body part which comes in contact with objects that are extremely hot, cold pointed or animals that are scary or poisonous. The entire process of response to a peripheral nervous stimulation, that occurs involuntarily, i.e., without conscious effort or thought and requires the involvement of a part of the central nervous system is called a reflex action.

Question 5.
Where are synaptic vesicles found? Name their chemical contents? What is the function of these contents?
Solution:
Synaptic vesicles are found in the bulbous expansion called synaptic knob, at the nerve terminal.

  • Each synaptic vesicle contains as many as 10,000 molecules of a neurotransmitter substance, that is responsible for transmission of nerve impulse across the synapse.
  • When a wave of depolarisation reaches the presynaptic membrane, the voltage-gated calcium channels concentrated at the synapse open and Ca ions diffuse into the terminal from the surrounding fluid.
  • The Ca++ ions stimulate the synaptic vesicles to move to the terminal membrane, fuse with it and then rupture by exocytosis into the cleft.
  • This neurotransmitter diffuses across the synapse and stimulates the membrane of the next neuron.

Question 6.
Write short notes on hindbrain.
Solution:
The hindbrain comprises pons, cerebellum and medulla (also called the medulla oblongata). Pons consists of fibre tracts that interconnect different regions of the brain. Cerebellum has very convoluted surface in order to provide the additional space for tnany more neurons. The medulla of the brain is connected to the spinal cord. The medulla contains centres which control respiration, cardiovascular reflexes and gastric secretions.

Question 7.
Enumerate the functions of hypothalamus.
Solution:
Functions of hypothalamus are as follows:
(i) Hypothalamus maintains homeostasis i. e., internal equilibrium of the body.
(ii) It has centres for regulation of hunger, thirst, emotions.
(iii) It organises behaviour like fighting, feeling etc., related to survival of species.
(iv) It maintains a constant body temperature.
(v) It secretes neurohormones, some of which 2.
control the functioning of pituitary glands called hypothalamic hormones.

Question 8.
Write a brief note on autonomic nervous system.
Solution:
Autonomic nervous is a part of peripheral nervous system. It controls activities occur in our body that are normally in voluntary such as heart beat, gut peristalsis, etc. Most of the actions of this system is controlled within the spinal cord or brain by reflexes known as visceral reflexes. It is maintanied by centre in medulla and hypothalamas. It maintains homeostasis. These are divided into two systems sympathetic and parasympathetic nervous system. The sympathetic nervous system mainly functions in quick responses and parasympathetic nervous system functions in actions which do not require immediate response.

Question 9.
What are nodes of ranvier?
Solution:
These are the periodic gaps or breakes in the 3. myelin sheath these breaks helps in the conduction of electricity in neurons the resistance to current flow between the axoplasm and fluid outside the cell is low these nodes set
up the local circuits to flow current inside neurons as a result, the action potential jump from node to node and passes along the myelineted axon faster compared to the series of small local circuits in a non-myelinated axon.

Long ANSWER QUESTIONS

Question 1.
Explain briefly the structure and functions of middle ear.
Solution:
The middle ear is an air-filled chamber on the inner side of eardrum. Its cavity communicates with an air-filled tube called eustachian tube, which maintains a balanced air pressure on either side of the tympanum.
The small bones called auditory ossicles are present in the middle ear. The malleus is attached to the ear-drum on one side and to the incus on the other side. The incus intum articulates with the stapes. The stapes is attached to the membrane over an oval-window between the middle ear and the internal ear.
Functions:

  • The auditory ossicles transmit the sound- induced vibrations of the ear-drum to the endolymph in the internal ear.
  • The eustachian tube balances and maintains a constant pressure on either side of the ear-drum.

Question 2.
Write a note on the retina.
Solution:
Retina

  • The innermost layer of the wall of eyeball is the retina. It is composed of several layers of cells
  • the photoreceptor layer contains rods and cones, the intermediate layer has bipolar neurons, which synapse with retinal ganglion cells and their axons bundle to form optic nerve.
  • The rod cells contain rhodopsin while the cone cells contain iodopsin.
  • The point in the retina where the optic nerve leaves the eye, is called as blind spot.
  • Lateral to the blind spot, there is a yellowish pigmented spot called macula lutea with a central pit called the fovea.
  • Fovea is the region where cones are densely packed and the vision is the sharpest.

Question 3.
Make a brief note on forebrain.
Solution:
Forebrain consists of cerebrum, thalamus and hypothalamus.

  • Cerebrum occurs as two cerebral hemispheres that are joined together by corpus callosum.
  • Each cerebral hemisphere is divided by other grooves into four lobes – frontal, parietal, temporal and occipital.
  • The convolutions and fissures greatly enlarge the surface area of the cortex.
  • The outer surface of cerebrum (cortex) has grey matter and inner to the cortex is white matter.
  • Thalamus lies just underthe cerebrum, i.e., cerebrum wraps around the thalamus.
  • Thalamus is the major coordinating centre for sensory and motor signals.
  • Hypothalamus lies at the base of the thalamus. .
  • It has centres to control body temperature, hunger, thirst, etc.
  • It contains several groups of neuro secretory cells which secrete hormones.
  • Limbic system is constituted by the inner parts of cerebral hemispheres and a group of structures called amygdala and hippocampus.
  •  Along with the hypothalamus, limbic system is involved in the regulation of sexual behaviour, expression of emotions, motivation, etc.

Question 4.
What is a synapse? How is the nerve impulse transmitted across a synapse?
Solution:
Synapse: The functional/intercommunicating, junction between two neurons, the axon of one neuron and the dendron/dendrite/soma of another neuron, through which impulse is conducted, is called a synapse.
Conduction of nerve impulse across a synapse:

  • When a nerve impulse reaches the pre- synaptic membrane (membrane of synaptic button), the voltage-gated calcium channels, concentrated in the synapse, open.
  • Calcium ions from the fluid in the synapse diffuse into the synaptic button and stimulate the synaptic vesicles to move to the terminal membrane, fuse with it and then rupture (exocytosis) to release the neurotransmitter.
  • The neurotransmitter quickly diffuses across the synaptic cleft in the fluid and stimulates certain specific receptor molecules on the post-synaptic membrane (membrane of the next dendron/dendrite) and causes sparking and electrical current, passing the signal.
    NCERT Solutions for Class 11 Biology Chapter 21 Neural control and co-ordination 12

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NCERT Solutions for Class 11 Biology Chapter 15 Plant Growth and Development

NCERT Solutions for Class 11 Biology Chapter 15 Plant Growth and Development

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 15 Plant Growth and Development.

Question 1.
Define growth, differentiation, development, dedifferentiation, redifferentiation, determinate growth, meristem, and growth rate.
Solution:
(1) Growth: Growth is accompanied by metabolic processes (both anabolic and catabolic), that occur at the expense of energy. For example, the expansion of a leaf is growth.

(2) Differentiation: The cells derived from root apical and shoot apical meristems and cambium differentiate and mature to perform specific functions. This act leading to maturation is termed differentiation. They undergo a few or major structural changes both in their cell walls and protoplasm. For example, to form a tracheary element, the cells would lose their protoplasm.

(3) Development: Development is a term that includes all changes that an organism goes through during its life cycle from germination of the seed to senescence. It is also applicable to tissues/organs.

(4) Dedifferentiation: Plants show another interesting phenomenon. The living differentiated cells, that by now have lost the capacity to divide can regain the capacity of the division under certain conditions. This phenomenon is termed dedifferentiation. For example, interfascicular cambium and cork cambium.

(5) Redifferentiation: While doing so, such meristems/ tissues are able to produce cells that once again lose the capacity to divide but mature to perform specific functions, i.e., get redifferentiated. List some of the tissues in a woody dicotyledonous plant that are the products of redifferentiation.

(6) Determinate growth: The growth in plants is open i.e. it can be indeterminate or determinate. Even differentiation in plants is open because cells tissues arising out of the same meristem have different structures at maturity. The final structure at maturity of a cell/’ tissue is also determined by the location of the cell within. For example, cells positioned away from root apical meristems differentiate as rootcap cells, while those pushed to the periphery mature as the epidermis.

(7) Meristem growth: The constantly dividing cells, both at the root apex and shoot apex, represent the meristematic phase of growth.

(8) Growth rate: The increased growth per unit time is termed as growth rate. Thus, the rate of growth can be expressed mathematically. An organism or a part of the organism can produce more cells in a variety of ways. The growth rate shows an increase that may be arithmetic or geometrical.

Question 2.
Why is not any one parameter good enough to demonstrate growth throughout the life of a flowering plant?
Solution:
In plants, growth is said to have taken place when the number of protoplasm increases. Measuring the growth of protoplasm involves many parameters such as weight of fresh tissue sample, the weight of dry tissue sample, differences in length, area, volume and cell number measured during the growth period. Measuring the growth of plants using only one parameter does not provide enough information and hence, is insufficient for demonstrating growth.

Question 3.
Describe briefly:
(a) Arithmetic growth
(b) Geometric growth
(c) Sigmoid growth curve
(d) Absolute and relative growth rates
Solution:
(i) Arithmetic growth: In this type of growth after mitosis, only one daughter cell continues to divide while the others take part in differentiation and maturation e.g., root elongating at constant rate. Here a ‘linear curve is obtained.
(ii) Geometric growth : In most systems, the initial growth is slow (lag phase), and it increases rapidly thereafter – at an exponential rate (log or exponential phase), Here both the progeny cells following mitotic cell division divide continuously.
(iii) Sigmoid growth curve : Sigmoid or S-shaped growth curve consists of three phases i.e., lag phase, log phase and stationary phase. During lag phase plant i growth is slow (in phase of cell division), but increases at log or exponential phase , (due to cell enlargement). During stationary phase the growth again slows down due f to the limitation of nutrients.
(iv) Absolute and relative growth rates : Measurement and comparison of total growth per unit time is called the absolute growth rate. The growth of the given system per unit time expressed on a common basis e.g., per unit initial parameter is called the relative growth rate,

Question 4.
List flve main groups of natural plant growth regulators. Write a note on discovery, physiological functions and agricultural/ horticultural applications of any one of them.
Solution:
The five main groups of natural growth regulators are
(a) auxins
(b) gibberellins
(c) cytokinins
(d) ethylene
(e) abscisic acid
Gibberellins
Discovery: – They are another kind of promotory PGR. There are more than 100 gibberellins reported from different organisms such as fungi and higher plants. They are denoted as GA1, GA2, GA3. E. Kurosawa reported the symptoms of the disease r in infected rice seedings when they were treated with filtrates of the fungus. Gibberalla fujikuroi caused, ‘bakane’ (foolish seedling) a disease of rice seedlings. The active substances were later identified as gibberellic acid.
Physiological functions
(i) They cause an increase in length of axis is used to increase the length of grapes stalk.
(ii) Gibberellins cause fruit like apple to elongate and improve its shape.
(iii) They also delay senescence. Thus the fruits can be left on the tree longer so as to extend the market period.
Agricultural Applications
(i) Spraying sugarcane crop with gibberellins increases the length of stem. Thus increasing the yield as much as 20 tonnes per acre.
(ii) Spraying juvenile conifers with GAs hastens the maturity period, thus leading to early seed production.
(iii) Gibberellins also promotes bolting (internode elongation just prior to flowering) in beet, cabbages and many plants with rosette habit.

Question 5.
What do you understand by photoperiodism and vernalisation? Describe their significance.
Solution:
Photoperiodism refers to the response of plants with respect to the duration of light. On the basis of its response to the duration of light, a plant is classified as a short-day plant, a long-day plant or a day-neutral plant. Photoperiodism helps in studying the response of flowering in various crop plants with respect to the duration of exposure to light.

Vernalisation is the cold-induced flowering in plants. In some plants, exposure to low temperatures is necessary for flowering to be induced. The winter varieties of rye and wheat are planted in autumn. They remain in the seeding stage during winters and flower during summers. However, when these varieties are sown in spring, they fail to flower.

Question 6.
Why is abscisic acid also known as stress hormone?
Solution:
Abscisic acid is known as the stress hormone because it stimulates the closure of stomata in the epidermis and increases the tolerance of plants to various kinds of stresses.

Question 7.
‘Both growth and differentiation in higher plants are open’ Comment.
Solution:
Growth and development in higher plants are referred to as being open because various meristems, having the capacity for continuously dividing and producing new cells, are present at different locations in these plant bodies.

Question 8.
‘Both a short day plant and a long day plant can flower simultaneously in a given place’. Explain.
Solution:
There are two different plants one is Oat which is a long-day plant and the other one is Xanthium which is a short-day plant. Both have different photoperiods i.e. 9 hrs in Oat and 15.6 hrs in Xanthium. At 9.5 hrs both Oat and Xanthium will be flowering simultaneously.

Question 9.
Which one of the plant growth regulators would you use if you are asked to:
(a) induce rooting in a twig
(b) quickly ripen a fruit
(c) delay leaf senescence
(d) induce growth in axillary buds
(e) ‘bolt a rosette plant’
(f) induce immediate stomatal closure in leaves.
Solution:
(a) Auxin
(b) Ethylene
(c) Cytokinin
(d) Cytokinin
(e) Gibberellin
(f) Abscisic acid

Question 10.
Would a defoliated plant respond to the photoperiodic cycle? Why?
Solution:
The flowering in certain plants depends not only on a combination of light and dark exposures but also on their relative durations. This is a photoperiodic cycle. Because, while shoot apices modify themselves into flowering apices prior to flowering, they (i.e., shoot apices of plants) by themselves cannot perceive photoperiods. The site of perception of light/ dark duration is the leaves. It has been hypothesized that there is a hormonal substance(s) called florigen that is responsible for flowering. Florigen migrates from leaves to shoot apices for inducing flowering only when the plants are exposed to the necessary inductive photoperiod.

Question 11.
What would be expected to happen if:
(a) GA3 is applied to rice seedlings
(b) dividing cells stop differentiating
(c) a rotten fruit gets mixed with unripe fruits
(d) you forget to add cytokinin to the culture medium.
Solution:
(a) It causes elongation of stems and leaf sheaths.
(b) A callus of the undifferentiated cells will be produced.
(c) It stimulates the ripening of unripe fruits.
(d) It inhibits the growth of callus.

VERY SHORT ANSWER QUESTIONS

Question 1.
What is the full form of IAA?
Solution:
Indole acetic acid

Question 2.
Name the apparatus used In determining growth in the plant? (Oct. 85)
Solution:
Auxanometer

Question 3.
Name stress hormone in plants that functions during drought.
Solution:
Abscisic acid

Question 4.
Name the hormone that makes the plant more tolerant to various stresses.
Solution:
Abscisic acid

Question 5.
In a wheat field, some broad-leaved weeds were seen by a farmer. Which plant hormone would you suggest to get rid of them?
Solution:
2,4-dichloro phenoxy acetic acid (2,4-D)

Question 6.
A farmer grows cucumber plants in his field. He wants to increase the number of female flowers in them. Which plant growth regulator can be applied to achieve this?
Solution:
Ethylene

Question 7.
What is the result of the addition of gibberellins to plants? (Oct. 91)
Solution:
Bolting

Question 8.
Define growth rate.
Solution:
The increased growth per unit time is called a growth rate.

Question 9.
Who isolated auxin? Name the plant source.
Solution:
F. W. Went isolated auxin. He isolated it from tips of coleoptiles of oat seedlings.

Question 10.
Define photoperiodism. (Apr. 97)
Solution:
The response of a plant to varying photoperiods of light is called photoperiodism.

Question 11.
Name the causative agent of ‘bakane’ disease in rice seedlings.
Solution:
Gibberella fujikuroi

Question 12.
Define climacteric.
Solution:
Climacteric refers to the increased rate of respiration during the ripening of fruits.

Question 13.
What would happen when a branch from a short day plant after floral induction is grafted on a non-induced long day plant?
Solution:
The long day plant would start flowering because a short day plant is capable to induce flowering in the long-day plant.

Question 14.
What is the most abundant natural cytokinin that was isolated from com kernels and coconut milk?
Solution:
Zeatin is the most abundant natural Cytokinin that was isolated from com Kernels and Coconut milk

Question 15.
Which, plant hormone was first isolated from human urine?
Solution:
Auxin was first isolated from human urine.

SHORT ANSWER QUESTIONS

Question 1.
List some structural modifications which occur during cell differentiation.
Solution:
During differentiation, cells undergo few to major structural changes both in their cell walls and
protoplasm. For example, to form a tracheary element, the cells would lose their protoplasm. They also develop very strong, elastic, lignocellulosic secondary cell walls, to carry water to long distances even under extreme conditions.

Question 2.
How do you induce lateral branching in a plant which normally does not produce them? Give reasons in support of your answer.
Solution:
Apical bud checks the sprouting of lateral buds due to the presence of auxins. When the apical bud is removed, lateral branches are produced. Due to the removal of apical bud effect of auxins is destroyed inducing the lateral buds to grow rapidly.

Question 3.
Define growth regulators.
Solution:
The plant growth regulators (PGRs) are small, simple molecules of diverse chemical
composition. They could be indole compounds (indole-3-acetic acid, IAA); adenine derivatives (kinetin), derivatives of carotenoids (abscisic acid, ABA); terpenes (gibberellic acid, GA3) or gases (ethylene, C2H4). Plant growth regulators are variously described as plant growth substances, plant hormones or phytohormones.

Question 4.
Define the term Growth. Mention the phases of growth. (Oct. 1988, 2000, 2003, July 2006, March 2011)
Solution:
Growth is a permanent irreversible change brought about by an increase in size, weight or volume. The phases of growth are

  • Phase of cell division or formation
  • Phase of cell elongation or enlargement
  • Phase of cell maturation or differentiation.

Question 5.
Define plasticity.
Solution:
Plants follow different pathways in response to the environment or phases of life. It leads to formation of different structures. This ability is called plasticity.

Question 6.
What is growth? How will you measure the rate of growth?
Solution:

  1. Growth is defined as a permanent or irreversible increase in dry weight, size, mass or volume of a cell, organ or organism.
  2. Generally growth is accompanied by metabolic processes (both anabolic and catabolic). At the cellular level, growth is due to increase in amount of protoplasm.
  3. However, it is difficult to measure increase in protoplasm.
  4. Increase in protoplasm leads to increase in cell, cell number and cell size. This fact is used in calculating growth which, therefore, is a quantitive or measurable phenomenon.
  5. The parameters used for measuring growth increase in fresh weight, dry weight, length, area, volume and cell number.

Question 7.
Explain the different phases of growth with the help of a diagram.
Solution:
The period of growth is generally divided into three phases – meristematic, elongation, and maturation.
Meristematic phase: The constantly dividing cells, both at the root apex and the shoot apex, represent the meristematic phase of growth. The cells in this region are rich in protoplasm possess large conspicuous nuclei.

Their cell walls are primary in nature, thin, and cellulosic with abundant plasmodesmatal connections.

Elongation phase: The cells proximal to the meristematic zone represent the phase of elongation.

Increased vacuolation, cell enlargement, and new cell wall deposition are the characteristics of the cells in this phase.

Maturation phase: The cells of this zone, attain their maximal size in terms of wall thickening and protoplasmic modifications.
NCERT Solutions for Class 11 Biology Chapter 15 Plant Growth and Development 1

Question 8.
Define growth and describe the three phases of growth. (Oct. 83, 85)
Solution:
Growth is a permanent irreversible change brought about by an increase in height, weight, or volume. The three phases of growth are;

(a) Phase of cell division or cell formation:
This region is located at the tip of shoot and root. It is represented by the apical meristem capable of rapid cell division. The cells are undifferentiated, with a thin cell wall made of cellulose, with an active protoplasm and prominent nucleus. This region is mainly concerned with cell division.

(b) Phase of cell elongation or cell enlargement: This region lies next to the cell formation zone. The cells enlarge because of their elastic cell walls. Growth takes place during this stage either by apposition or intussusception. Cells are turgid.

(c) Phase of cell differentiation or cell maturation: This represents the last region and differentiation based on functions is seen here. Secondary walls are laid down where some have additional deposits of lignin, Suberin, and others lose their protoplast and become dead.

Question 9.
Where are auxins synthesized in plants? Mention any two of their functions.
Solution:
Auxins are produced in the growing shoot apices and root apices.
Functions of auxins are as follows :
(i) Auxins control apical dominance, i.e., they suppress the growth of lateral buds into branches.
(ii) They help to prevent fruit and leaf drop at early stages but promote the abscission of older mature leaves and fruits.

Question 10.
Where are cytokinins synthesised in plants? Mention any two of their functions.
Solution:
Cytokinins are synthesised in plant parts where rapid cell division occurs, like root apices, shoot buds, young fruits, etc. The functions of cytokines are as follows:
(i) Cytokinins influence cell division (cytokinesis), cell enlargement and differentiation.

Question 11.
Explain apical dominance. Name the hormone that controls it.
Solution:
Apical dominance is the phenomenon in which the apical bud suppresses the growth of lateral buds into branches. Auxin is the hormone that controls it.

Question 12.
How does abscisic acid act antagonistically to auxins and gibberellins?
Solution:
ABA induces the formation of the abscission layer, while auxins prevent the formation of the abscission layer.
ABA induces seed dormancy and bud dormancy, while gibberellins break seed dormancy and bud dormancy.

Question 13.
What is ethephon? How does it function in plants? Give any two of its functions.
Solution:
Ethephon:

  •  It is a compound used as a source of ethylene for plant growth.
  •  It is an aqueous solution that is easily absorbed by the plants and transported within the plant.
  •  It releases ethylene slowly.

Functions of ethephon are as follows:

  • It accelerates abscission in flowers and thinning in cotton, walnut, and cherry, etc.
  • It promotes the development of female flowers in cucumbers thereby increasing the yield.

Question 14.
Discuss the practical applications of auxins in Agriculture and Horticulture. (Oct. 96) OR What are auxins? Explain briefly uses of Auxins. (Oct. 99)
Solution:
Auxins are a group of plant growth substances, acidic in nature and bring about over-all growth.

  • Apical dominance: As long as the apical bud is present growth of lateral buds is pre-vented which is used in the long term storage of potato tubers.
  • Rooting: In low concentrations, auxins stimulate root formation which is used to propagate cuttings. When dipped in a dilute solution of auxins the root formation is initiated.
  • Flower initiation: Low Concentration of 2, 4-D and NAA are used to initiate flowering in a pineapple so that harvesting becomes easy.
  • Abscission: Application of auxins increase the concentration and thereby delays the development of abscission which prevents premature leaf and fruitful. This is used in orchards and prevention of defoliation in cabbages and cauliflower.
  • Parthenocarpy: This is the process of obtaining fruits without fertilization and gives rise to seedless varieties which is successfully used in citrus, dates.
  • Sex expression: Use of auxins on cucurbit plants increases the production of female flower.
  • Weedicide / Herbicide: 2-4-D is widely used as a herbicide on broad-leaved forms because of its non-toxic nature. Widely used in crop plant cultivation or lawns.
  • Tissue culture (organogenesis): In tissue culture where micropropagation is carried out the auxins are used to bring about organogenesis.

Question 15.
Differentiate between phototropism and Geotropism.
Solution:
NCERT Solutions for Class 11 Biology Chapter 15 Plant Growth and Development 2

Question 16.
What are Terpenoids?
Solution:
Terpenoids are derivatives of terpenes, includes abscisic acid and gibberellin and the carotenoid and chlorophyll pigments.

LONG ANSWER QUESTIONS

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

  • Vernalization may be defined as the method of inducing early flowering in plants by pretreatment of their seeds at low temperatures.
  • It is the acquisition or acceleration of the ability to flower by chilling treatment.
  • Some cereals such as wheat, barley, oat and rye have two kinds of varieties: winter and spring varieties.
  • The ‘spring’ variety are normally planted in the spring and come to flower and produce grain before the end of the growing season.
  • Winter varieties, however, if planted in spring would normally fail to flower or produce mature grain within a span of a flowering season.
  • Hence, they are planted in autumn. They germinate, and during winter come out as small seedling, resume growth in the spring, and are harvested usually around mid-summer.
  • Another example of vernalization is seen in biennial plants. Biennials are monocarpic plants that normally flower and die in the second season. Sugarbeet, cabbages, carrots are some of the common biennials.
  • Subj ecting the growing of a biennial plant to a cold treatment stimulates a subsequent photoperiodic flowering response.

The significance of vernalization is as follows:

  • It reduces the vegetative period of the plant.
  • It prepares the plants for flowering.
  • It increases yield, resistance to cold and diseases.
  • Vernalization is beneficial in reducing the period between germination and flowering.

Thus, more than one crop can be obtained during a year.

Question 2.
Discuss the role of auxins in plant growth.
OR
Describe any four physiological effects of auxins. (Oct. 89, 2001)
Solution:

  1. Cell division and Differentiation: Auxins promote cell division and their subsequent differentiation into tissues. They are used in cultures to bring about organogenesis.
  2. Apical dominance: Auxins are more concentrated in the terminal buds rather than lateral buds. Therefore the presence of the terminal buds inhibits the growth of lateral buds which is also true when auxins are applied to the cut surface of the stem. This is used in preventing the sprouting of potato buds (axillary buds).
  3. Root Initiation: Low concentrations of auxins promote rooting which can propagate more plants vegetatively.
  4. Abscission formation: The development of abscission is due to a decrease in auxin concentration resulting in fruit fall and defoliation. In young leaves and fruits, the concentration is high. Hence the external application of auxins helps to prevent premature fruit drop of apple, pear, and defoliation of cabbage.
  5. Parthenocarpy: A normal fruit develops after fertilization, during which the auxin concentration increases. The application of auxins stimulates fruit formation without fertilization and is called parthenocarpy.
  6. Herbicide: Synthetic auxins like 2, 4 – D and 2, 4, 5-T are toxic to broad-leaved plants and because of this used as selective herbicides in crop plants, lawn grass, etc.

Question 3.
What is meant by seed dormancy? Describe the methods to overcome seed dormancy.
Solution:
Seed dormancy

  • There are certain seeds which fail to germinate even when external conditions are favourable. Such seeds are undergoing a period of dormancy which is controlled not by the external environment but are under endogenous control or conditions within the seed itself.
  • Impermeable and hard seed coat; the presence of chemical inhibitors such as abscisic acids, phenolic acids, para-ascorbic acid; and immature embryos are some of the reasons which cause seed dormancy.
  • Seed dormancy, however, can be overcome through natural means and various other means e.g. the seed coat barrier in some seeds can be broken by mechanical abrasions using knives, sandpaper, etc. or vigorous shaking. In nature, these abrasions are caused by microbial action, and passage through the digestive tract of animals.
  • The effect of inhibitory substances can be removed by subjecting the seeds to chilling conditions or by application of certain chemicals like gibberellic acid and nitrates.
  • Changing the environmental conditions, such as light and temperature are other methods to overcome seed dormancy.

Question 4.
Describe the phenomenon of photoperiodism.
Solution:
The effect of photoperiods or day duration of light hours (and dark periods) on the growth and development of plants, especially flowering, is called photoperiodism. On the basis of photoperiodic response to flowering, plants have been divided into the following categories:

  • Short-day plants: They flower when the photoperiod or day length is below a critical period. Most winter flowering plants belong to this category, e.g., Xanthium, Chrysanthemum, rice, sugarcane, etc.
  • Long-day plants: These plants flower when they receive long photoperiods or light hours which are above a critical length, e.g., wheat, oat, sugar beet, spinach, radish, barley, etc.
  • Day-neutral plants: There are many plants, however, where there is no such correlation between exposure to the light duration and induction of flowering response; such plants are called day-neutral plants e.g. tomato, cucumber, etc.

We hope the NCERT Solutions for Class 11 Biology at Work Chapter 15 Plant Growth And Development, help you. If you have any query regarding NCERT Solutions for Class 11 Biology at Work Chapter 15 Plant Growth And Development, 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

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:

  1. Kill bacteria and germs
  2. Loosens fibrous material of food.
  3. Activates proenzyme pepsinogen to its active form pepsin.
  4. Gives an acidic medium for action by pepsin.
  5. 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.
NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 3
(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.
NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 5

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:
NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 6

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:
NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 8

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

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.
NCERT Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 2
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.

  1. 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.
  2. 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.
  3. 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.

  1. 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.
  2. Respiration, nutrition, release of energy for the body are carried out within the cells only.
  3. Even the animals and plants reproduce because the cells reproduce individually.
  4. Growth occurs because cell grow and multiply.
    In Amoeba all the life processes are performed within the boundaries of the single cell.
  5. 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.
  6. 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.
  7. Each tissue is specialized to perform specific functions. Different tissues then get organised into organs which perform certain specific functions.
  8. 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.
  9. These tissues, organs and organ system of the organisms work because the cells work.
  10. 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:
NCERT Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 3
(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:

  1. Under the electron microscope, each centriole is seen to be formed of nine sets of tubular structures arranged in a circular fashion.
  2. 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.
  3. Sometimes delicate strands appear to connect sets of the triplet to each other.
  4. 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.
  5. 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.
    NCERT Solutions for Class 11 Biology Chapter 8 Cell The Unit of Life 8

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

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,

  1. Macronutrients
  2. 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:

  1. 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.
  2. Necrosis: It is the death of tissue. It occurs due to deficiency of Ca, Mg, Cu, K.
  3. Inhibition of cell division: It occurs due to deficiency of N, K, S, Mo.
  4. Stunted plant growth: It occurs due to deficiency of Ca, N, etc.
  5. 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:

  1. 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.
  2. 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.
  3. Zinc is moderately mobile in plants and deficiency symptoms, therefore, appear in middle leaves.
  4. The deficiency symptoms of less mobile elements (S, Fe, Mn, and Cu) appear on new leaves.
  5. Ca and B are immobile in plants, deficiency symptoms appear on terminal buds.
  6. Chlorine deficiency is less common in crops.
    NCERT Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 1

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:
NCERT Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 3

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:
NCERT Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 4a
(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
NCERT Solutions for Class 11 Biology Chapter 12 Mineral Nutrition 7

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

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:
NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis 1
(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
NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis 3
(c) Differences between the anatomy of leaf in C3 plants and anatomy of leaf in C4 plants are
NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis 4

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:
NCERT Solutions for Class 11 Biology Chapter 13 Photosynthesis 5

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

  1. The path traversed by an electron is non-cyclic.
    (a) Path traversed by electron is cyclic.
  2. Both PSI and PSII are active.
    (b) Only PSI is active.
  3. It is accompanied by photolysis.
    (c) No photolysis.
  4. 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 Chla680 or 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.

NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants

NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants

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

Question 1.
Differentiate between
(a) Respiration and Combustion
(b) Glycolysis and Krebs’ cycle
(c) Aerobic respiration and Fermentation
Solution:
(a) Differences between respiration and combustion are as follows :
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 1
(b) Differences between glycolysis and krebs’ cycle are as follows :
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 2
(c) Differences between aerobic respiration and fermentation are as follows :
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 3

Question 2.
What are respiratory substrates? Name the most common respiratory substrate.
Solution:
The compounds that are oxidized during this process are known as respiratory substrates. Usually, carbohydrates are oxidized to release energy, but proteins, fats, and even organic acids can be used as respiratory substances in some plants, under certain conditions.

Question 3.
Give the schematic representation of glycolysis.
Solution:
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 4

Question 4.
What are the main steps in aerobic respiration? Where does it take place?
Solution:
In aerobic respiration which takes place within the mitochondria, the final product of glycolysis, pyruvate is transported from the cytoplasm into the mitochondria.
The crucial events in aerobic respiration are:
• The complete oxidation of pyruvate by the stepwise removal of all the hydrogen atoms, leaving three molecules of CO2.
• The passing on of the electrons removed as part of the hydrogen atoms to molecular O2 with the simultaneous synthesis of ATP.
• The first process takes place in the matrix of the mitochondria while the second process is located on the inner membrane of the mitochondria.
• Pyruvate, which is formed by the glycolytic catabolism of carbohydrates in the cytosol, after it enters mitochondrial matrix undergoes oxidative decarboxylation by a complex set of reactions catalysed by pyruvic dehydrogenase. The reactions catalysed by pyruvic dehydrogenase require the participation of several coenzymes, including NAD+ and Coenzyme A.
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 5
During this process, two molecules of NADH are produced from the metabolism of two molecules of pyruvic acid (produced from one glucose molecule during glycolysis).
The acetyl CoA then enters a cyclic pathway, the tricarboxylic acid cycle, more commonly called as Krebs’ cycle.

Question 5.
Give the schematic representation of an overall view of Krebs’ cycle.
Solution:
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 6

Question 6.
Explain ETS.
Solution:
ETS or electron transport system is located in the inner mitochondrial membrane. It helps in releasing and utilizing the energy stored in NADH + H+ and FADH2 NADH+ H+, which is formed during glycolyis and citric acid cycle, gets oxidized by NADH dehydrogenase. The electrons so generated get transferred to ubiquinone through FMN. In a similar manner, FADH2 generated during citric acid cycle gets transferred to ubiquinone. The electrons from ubiquinone are received by cytochrome bc1, and further get transferred to cytochrome C. The cytochrome C acts as a mobile carrier between complex III and cytochrome C oxidase complex containing cytochrome a and a3, along with copper centres.

NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 7
During the transfer of electrons from each complex, the process is accompanied by the production of ATP from ADP and inorganic phosphate by the action ATP synthase. The amount of ATP produced depends on the molecule, which has been oxidized.

Question 7.
Distinguish between the following:
(a) Aerobic respiration and anaerobic respiration.
(b) Glycolysis and fermentation.
(c) Glycolysis and citric acid cycle.
Solution:
Differences between aerobic respiration and anaerobic respiration are as follows :
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 8
Differences between glycolysis and fermentation are as follows :
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 9
Differences between glycolysis and citric acid cycle are as follows :
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 10

Question 8.
What are the assumptions made during the calculation of the net gain of ATP?
Solution:

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

Question 9.
Discuss “The respiratory pathway is an amphibolic pathway”.
Solution:
Respiration is generally assumed to be a catabolic process because, during respiration, various substrates are broken down for deriving energy. Carbohydrates are broken down into glucose before entering respiratory pathways. Fats get converted into fatty acids and glycerol whereas fatty acids get converted into acetyl CoA before entering respiration. In a similar manner, proteins are converted into amino acids, which enter respiration after deamination.

During the synthesis of fatty acids, acetyl CoA is withdrawn from the respiratory pathway. Also, in the synthesis of proteins, respiratory substances get withdrawn. Thus, respiration is also involved in anabolism. Therefore, respiration can be termed as. amphibolic pathway as it involves both anabolism and catabolism.

Question 10.
Define RQ. What is its value for fats?
Solution:
The ratio of the volume of CO2 evolved to the volume of O2 consumed in respiration is called respiratory quotient (RQ) or respiratory ratio.
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 11

Question 11.
What is oxidative phosphorylation ?
Solution:

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

Question 12.
What is the significance of the step-wise release of energy in respiration?
Solution:
During oxidation within a cell, all the energy contained in respiratory substrates is not released free into the cell, or in a single step. It is released in a series of slow step-wise reactions controlled by enzymes, and it is trapped as chemical energy in the form of ATP.

Hence, it is important to understand that the energy released by oxidation in respiration is not used directly but is used to synthesise ATP, which is broken down whenever (and wherever) energy needs to be utilised. Hence, ATP acts as the energy currency of the cell.

This energy trapped in ATP is utilised in various energy-requiring processes of the organisms, and the carbon skeleton produced during respiration is used as precursors for the biosynthesis of other molecules in the cell.

VERY SHORT ANSWER QUESTIONS

Question 1.
What is anaerobic respiration? (Oct. 83)
Solution:
Incomplete or partial breakdown of fuel molecules into compounds such as ethyl alcohol, lactic acid in the absence of molecular oxygen.

Question 2.
Name the final acceptor of an electron in ETC.
Solution:
Oxygen is the electron acceptor of ETC.
Question 3.
The function of oxygen in aerobic respiration:
(i) It acts as the final electron acceptor.
(ii) It drives the whole process by removing hydrogen from the system.
Solution:
The function of oxygen in aerobic respiration:
(i) It acts as the final electron acceptor.
(ii) It drives the whole process by removing hydrogen from the system.

Question 4.
What is respiration? (Oct. 86)
Solution:
The oxidative process in which chemically bound energy from complex organic fuel molecules such as carbohydrates, proteins, and fats is captured in the form of ATP.

Question 5.
Where does the electron transport system operate in the mitochondria?
Solution:
Phosphofructokinase catalyses the formation of fructose 1, 6 bisphosphates from fructose 6-phosphate.

Question 6.
Give the function of phosphofructokinase in glycolysis.
Solution:
Hexokinase-helps in the phosphorylation of glucose.

Question 7.
Name the enzyme that catalyses the phosphorylation of glucose.
Solution:
The formation of acetyl CoA takes place in the mitochondrial matrix.

Question 8.
Where does the formation of acetyl CoA take place in a cell?
Solution:
The first step in the Krebs cycle is the condensation of an acetyl group (acetyl CoA) with oxaloacetic acid (OAA) to form citric acid and release the Coenzyme A.

Question 9.
What is the first step of reaction in the TCA cycle?
Solution:
Fatty acids may be converted to acetyl CoA before they from the respiratory substrates.

Question 10.
What is alcoholic fermentation?
Solution:
Alcoholic fermentation is the process by which yeast cells breakdown glucose into ethyl alcohol and carbon-dioxide under anaerobic conditions.

Question 11.
Name the oxidative pathway through which intermediate metabolites of glucose, fatty acids, and amino acids are finally oxidised.
Solution:
36 ATP/38 ATP molecules are obtained in the process of respiration and it is related to the aerobic respiration type.

Question 12.
What is lactic acid fermentation? (Oct. 2001)
Solution:
It is the process of fermentation by which lactose found in milk is converted to lactic acid by the action of lactobacillus.

Question 13.
What are the two molecules obtained by the action of aldolase from fructose -1, -6- biphosphate?
Solution:
ATP is produced.

SHORT ANSWER QUESTIONS

Question 1.
How is the proton gradient established?
Solution:
The proton gradient is established by passing proton (H+) from the matrix across the inner mitochondrial membrane into intermembrane space with the energy released during electron transfers in ETC.

Question 2.
Describe the steps in the formation of lactic acid from pyruvic acid.
Solution:
Pyruvic acid is catalysed by the enzyme lactic dehydrogenase. NADH formed in glycolysis is used up for the reduction.
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 12

Question 3.
How is ATP formed by the energy released during the electron transport system in mitochondria?
Solution:
ATP formations require an enzyme called ATP synthase. It has two components F0– F1. ATP- synthase becomes active in ATP formation when the concentration of H+ on the Fo side is higher than the F1 side. Fligher proton concentration in the outer chamber causes the proton to pass the inner chamber. F1 particle induced by the flow of proton through Fo channel. The energy of the proton gradient attaches the phosphate radicle to ADP. This produces ATP.

Question 4.
Give a detailed account of the net gain of ATP at a different stages of respiration.
Solution:
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 13
In most eukaryotic cells 2 molecules of ATP are required for transporting NADH produced in glycolysis to mitochondria for further oxidation. Hence net gain of ATP is 36 molecules.

Question 5.
Enumerate the functions of ATP.
Solution:
Functions of ATP:-
(i) ATP functions as a universal energy carrier of living systems.
(ii) ATP stores small packets of energy in its molecules.
(iii) It is mobile in the cell. Therefore, it reaches all parts of the cell away from the region of ATP synthesis.
(iv) It activates a number of chemicals by functioning as a phosphorylating agent.
(v) ATP provides energy for muscle contraction.
(vi) It is involved in the transport of substances against a concentration gradient.

Question 6.
Where is cytochrome c located? What is its function?
Solution:
Cytochrome c is located on the outer surface of the inner mitochondrial membrane. It acts as a mobile carrier for the transfer of electrons between complex III and complex IV of the electron transport system.

Question 7.
Define respiratory quotient.
Solution:
The respiratory quotient is defined as the ratio of the volume of carbon dioxide evolved to the volume of oxygen consumed in respiration.

Question 8.
What is oxidative phosphorylation?
Solution:
The whole process by which oxygen effectively allows the production of ATP by phosphorylation of ADP is called oxidative phosphorylation.

Question 9.
The energy yield in terms of ATP is higher in aerobic respiration than during anaerobic respiration. Why is there anaerobic respiration even in organisms that live in aerobic conditions like human beings and angiosperms?
Solution:
Aerobic organisms do face situations where oxygen availability is little. For example, overworked muscles do not receive enough oxygen during strenuous exercise. Similarly, deep-seated tissues of angiosperms do not receive enough oxygen through diffusion from outside. In such situations, only anaerobic respiration can help in the survival of the tissue.

Question 10.
Comment on the statement- “Respiration is an energy-producing process but ATP is used in some steps of the process”.
Solution:
ATP is required in all those reactions where phosphorylative activation of the substrate is required. Therefore, despite producing energy (as ATP), respiration requires ATP in certain steps, e.g., glucose – glucose 6-phosphate, fructose 6-phosphate —fructose 1, 6- bisphosphate.

LONG ANSWER QUESTIONS

Question 1.
Explain the major steps in Krebs’ cycle. Why is this cycle also called the citric acid cycle?
Solution:
Krebs cycle: This process occurs in the mitochondrial matrix.
Major steps of Krebs cycle are as follows :

  • Acetyl Co-A, formed by the oxidative decarboxylation of pyruvic acid enters the Krebs’ cycle.
  • It combines with oxalo acetic acid (OAA), a 4C-compound, to form a 6C-compound, citric acid; the reaction is catalysed by citrate synthase.
  • Citrate is then isomerised into isocitrate.
    NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 14
  • Isocitrate is converted into oxalosuccinic acid in the presence of NAD and isocitrate dehydrogenase.
  •  Oxalosuccinic acid is then decarboxylated into a-ketoglutaric acid (KG), in the presence of a decarboxylase enzyme.
  •  a-ketoglutaric acid is converted into succinyl Co-A in the presence of NAD, Co- A, and enzyme a-ketoglutarate dehydrogenase.
  •  When succinyl Co-A is converted into succinic acid, one molecule of GTP is formed and Co-A is released.
  •  In the remaining part of the cycle, succinic acid is converted into OAA, so that the citric acid cycle can continue to operate.
  •  During this cycle, three molecules of NAD and one molecule of FAD are reduced to NADH and FADH respectively.
  •  This cycle is called as a citric acid cycle because the first product is citric acid which is 3-C compound.

Question 2.
Name the end product of glycolysis. Where is it produced in the cell? Discuss oxidative decarboxylation.
Solution:
Glycolysis results in the formation of two molecules of pyruvic acid, NADH, and ATP. It occurs in the cytosol of the cell.
Aerobic oxidation: One of the three carbons of pyruvic acid is oxidised to carbon dioxide in the reaction called oxidative decarboxylation. Pyruvic acid is first decarboxylated and then oxidised by the enzyme pyruvic dehydrogenase. The two-carbon units are readily accepted by coenzyme-A (Co-A) to form acetyl Co-A. The summary of the reaction is given in the following equation :
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 15
Thus, pyruvic acid enters the Krebs cycle as acetyl Co-A. Krebs’ cycle occurs in the mitochondrial matrix.
Acetyl Co-A, formed by the oxidative decarboxylation of pyruvic acid enters the Krebs’ cycle.

Question 3.
Represent schematically the interrelationship among metabolic pathways in a plant, showing respiration mediated breakdown of different organic compounds.
Solution:
Schematic representation among metabolic pathways showing respiration mediated breakdown of different organic molecules to CO2 and H2O:
NCERT Solutions for Class 11 Biology Chapter 14 Respiration in Plants 16
Question 4.
How do plants manage the exchange of gases? Give an overview of respiration in plants.
Solution:
Plants, unlike animals, have no specialized organs for gaseous exchange but they have stomata and lenticels for this purpose. There are several reasons why plants can get along without respiratory organs.

  • Each plant part takes care of its own gas- exchange needs. There is very little transport of gases from one plant part to another.
  • Plants do not present great demands for gas exchange. Roots stem and leave respire at a lower rate than animals do.
  • Only during photosynthesis, large volumes of leases exchanged and, each leaf is well adapted to take care of its own needs during these periods.
  • When cells perform photosynthesis, the availability of O2 is not a problem in these cells since 02 is released
  • The distance that gases must diffuse even in large, bulky plants is not great. Each living cell in a plant is located quite close to the surface of the plant.
  • Even in woody stems, the ‘living’ cells are organised in thin layers inside and beneath the bark. They also have openings called lenticels. The cells in the interior are dead and provide only mechanical support.
  • Thus, most cells of a plant have attested to a part of their surface in contact with air. This is also facilitated by the loose packing of parenchyma cells in leaves, stems, and roots, which provide an interconnected network of air spaces.
  • The complete combustion of glucose, which produces C02 and H20 as end products, yields energy. Most of the energy is given out as heat.
    C6H12O6 + 6O2 → 6C02 + 6H20 + Energy
  • If this energy is to be useful to the cell, it should be able to utilise it to synthesis other molecules that the cell requires.
  • The strategy that the plant cell uses is to catabolize the glucose molecule in such a way that not all the liberated energy goes out as heat.
  • The key is to oxidise glucose not in one step but in several small steps enabling some steps to be just large enough so that the energy released can be coupled to ATP synthesis.

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