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

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

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

VERY SHORT ANSWER QUESTIONS

Question 1.
Name the blood component which is viscous and straw coloured fluid.
Solution:
Blood is a special connective tissue consisting of a fluid matrix, plasma and cells.

Question 2.
Complete the missing word in the statement given below.
(a) Plasma without …….. factors is called serum.
(b)………and monocytes are phagocytic cells.
(c) Eosinophils are associated with …….. reactions. .
(d) ……. ions play a significant role in clotting.
(e) One can determine the heart beat rate by counting the number of ……….. in an ECG.
Solution:
(a) Clotting
(b) Neutrophils
(c) allergic
(d) Calcium
(e) QRS complex

Question 3.
Name the vascular connection that exists between the digestive tract and liver.
Solution:
Hepatic portal system is the vascular connection that exists between the digestive tract and liver.

Question 4.
Given below are the abnormal conditions related to blood circulation. Name the disorders.
(a) Acute chest pain due to failure of 02 supply to heart muscles.
(b) Increased systolic pressure.
Solution:
(a) Angina
(b) High Blood Pressure

Question 5.
Which coronary artery disease is caused due to narrowing of the lumen of arteries?
Solution:
Atherosclerosis is the coronary artery disease caused due to the narrowing of the lumen of arteries due to deposition of calcium, fat, cholesterol and fibrous tissue the arteries become narrow, affecting vessels that supply blood to the heart muscles.

Question 6.
Define the following terms and give their location.
(a) Purkinje fibre
(b) Bundle of His
Solution:
(a) Purkinje Fibres are the fibres that conduct impulse, and the contraction impulses from AV node into the walls of ventricles.
(b) Bundle of His are mass of specialised fibres that originates from the AV node.

Question 7.
State the functions of the following in blood
(a) fibrinogen
(b) globulin
(c) neutrophils
(d) lymphocytes
Solution:
(a) Fibrinogens are the components of blood plasma that are inactive. In the presence of enzyme thrombin, they form a clot or coagulum of a network of threads called fibrin, in which dead and damaged elements of blood are trapped.
(b) Globulins are primarily involved in immunity,
i. e., defence mechanisms of the body.
(c) Neutrophils are phagocytic cells, that destroy foreign organisms entering the body,
(d) Lymphocytes are specialised cells which are responsible for the immune responses in the body. There are two major types of lymphocytes, that are involved in this process are B and T-lymphocytes.

Question 8.
How will you interpret an electocardiagram (ECG) in which time taken in QRS complex is higher?
Solution:
Electrocardiograph (ECG) is a graphical representation of the electrical activity of the heart during a cardiac cycle. A patient is connected to the machine having three electrical leads (one to each wrist and one to the left ankle) that continuously monitor the activity of heart.
Multiple leads are attached to the chest regions, for a detailed evaluation of the heart functions The QRS complex represent the depolarisation of the ventricles, that initiates the ventricular contraction. The contraction starts shortly after Q and marks the beginning of the systole. The time taken in QRS complex is 0.12 second in normal ECG
NCERT Exemplar Solutions for Class 11 Biology Chapter 18 Body Fluids and Circulation 1
The larger Q and R wave indicate a myocardial infarction (heart attack). The S-T segment is elevated in acute myocardial infarction and depressed when the heart muscle receives insufficient oxygen.

SHORT ANSWER QUESTIONS

Question 1.
The walls of ventricles are much thicker than atria. Explain.
Solution:
The walls of ventricles are thicker than the atria. It is due to the greater pressure exerted by pumping out of blood through ventricles of heart in compare to atria. Ventricles need to pump blood further and with much force.

Question 2.
Differentiate between
(a) blood and lymph
(b) basophilsand eosinophils
(c) tricuspid and bicuspid valve
Solution:
(a) Difference between blood and lymph are as follows
NCERT Exemplar Solutions for Class 11 Biology Chapter 18 Body Fluids and Circulation 2
(b) Difference between basophils and eosinophils are as follows
NCERT Exemplar Solutions for Class 11 Biology Chapter 18 Body Fluids and Circulation 3
(c)Difference between tricuspid valve and bicuspid valve are as follows
NCERT Exemplar Solutions for Class 11 Biology Chapter 18 Body Fluids and Circulation 4

Question 3.
Briefly describe the followings
(a) anaemia
(b) angina pectoris
(c) atherosclerosis
(d) hypertension
(e) heart failure
(f) erythroblastosis foetalis
Solution:
(a) Anaemia This is the most common disorder
of the blood, which is caused due to decrease in the number of RBC than the normal amount and also due to less quantity of haemoglobin than the normal value in blood. This is the most common disorder of the blood.
(b) Angina Pectoris When there is blockage in coronary artery, thus insufficient supply of blood reaches to heart muscles. That results in chest pain, fear, anxiety, pale skin, profuse
v sweating and vomiting. The pain usually starts in the centre of the chest spreads down to the left arm which last for only few second.
(c) Atherosclerosis It refers the deposition of cholesterol or fatty substance in the inner lining of arteries called atherosclerotic plaque. Sometimes arteries get completely blocked, this may result in stroke of heart attack.
(d) Hypertension It is sometimes also known as arterial hypertension. The blood pressure in the arteries gets elevated. It could be primary or secondary hypertension which
are caused by various conditions which affect kidneys, arteries heart or endocrine system.
(e) Heart Failure It is the state of heart, causes in when heart does not pump blood effectively enough to meet the requirement of the body.
(f) Erythroblastosis foetalis It is a haemolytic disease causes in new boms, which is an allo-immune condition that develops in foetus, when igG molecules produced by mother pass through placenta and attack RBC. It causes reticulocytosis and anaemia. It develops due to Rh incompatibility between the couples.
In a man with RH+ blood and women with Rlr blood, the second pregnancy foetus may have this problem due to IgG accumulation in women during first child development and delivery.

Question 4.
Explain the functional significance of lymphatic system?
Solution:
Lymphatic system comprises blood vessels that carries a fluid called lymph. It contains white blood cells, which are responsible for fighting against any diseases. It removes and filter the interstitial fluid from tissues, later absorbs and transports fatty acids and fats as chyle from digestive system and also transport cells to immune system.

LONG ANSWER QUESTIONS

Question 1.
Explain Rh-Incompatibility in humans.
Solution:
In nearly 80% of human Rh antigen is observed on the surface of RBCs. Such individuals are called Rh positive (Rh+) and those individuals where this antigen is not present are called Rh negative (Rh–).
Both Rh+ and Rh– individuals are phenotypically normal. The problem in them arises during blood transfusion and pregnancy.
(i) Incompatibility During Blood Transfusion The first blood transfusion of Rh+ blood to the person with Rh” blood causes no harm because the Rh– person develops anti Rh factors or antibodies in his/ her blood, but second transfusion of Rh+ blood to the Rh– person because anti Rh factors are already formed it destroys the red blood corpuscles of the donor. .
(ii) Incompatibility During Pregnancy If father’s blood is Rh+, mother blood is Rh– and the foetus blood is Rh+. It will lead to a serious problem. Rh antigens of the foetus do not get exposed to the Rh’ve blood of the mother in the first pregnancy as the two bloods are well separated by the placenta.
But in the subsequent Rh+ foetus, the anti Rh factors (antibodies) present in causes mother destroys the foetal RBC due to mixing of blood. This causes the Haemolytic Disease of the New Born (HDN), called as erythroblastosis foetalis. In some cases new born may survive but will be anaemic and may also suffer with jaundice.
NCERT Exemplar Solutions for Class 11 Biology Chapter 18 Body Fluids and Circulation 5
This condition can be avoided by administering anti-Rh antibodies to the mother immediately after the delivery of the first child.

Question 2.
Explain different types of blood groups and donor compatibility by making a table.
Solution:
There are more than 30 surface antigens in blood cells which give rise to different blood groups.
ABO Grouping on basis of the presence or absence of two surface antigens on the RBCs namely, A and B. The plasma of different individuals contain two natural antibodies. The distribution of antigen and antibody in divided four groups into. A, AB, B and O.
NCERT Exemplar Solutions for Class 11 Biology Chapter 18 Body Fluids and Circulation 6
Form the given table it is evident that group ‘O’ blood can donate to persons with any other blood group and hence ‘O’ group individuals are called ‘Universal donors’. Person with ‘AB’ blood can accept blood from persons with AB, as well as the other groups of blood. Hence, such persons are called ‘Universal recipients’.

Question 3.
In the diagrammatic presentation of heart given below, mark and label. SAN, AVN, AV bundles, bundle of his and Purkinje fibres.
Solution:
The diagrammatic presentation of heart with labelled SAN, AVN, AV bundles bundle of His and purkinje fibres in heart is shown below.
NCERT Exemplar Solutions for Class 11 Biology Chapter 18 Body Fluids and Circulation 7

We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 18 Body Fluids and Circulation, help you. If you have any query regarding NCERT Exemplar 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 Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants

NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants

These Solutions are part of NCERT Exemplar Solutions for Class 11 Biology. Here we have given NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants.

VERY SHORT ANSWER QUESTIONS

Question 1.
Roots obtain oxygen from air in the soil for respiration. In the absence or deficiency of 02, root growth is restricted or completely stopped. How do, the plants growing in marsh lands or swamps obtain their 02 required for root respiration?
Solution:
The roots of the plants as Rhizophora that grow in marsh/swamp areas become negatively geotropic. They grow vertically upwards in air, above the soil level and respire. They are thus called respiratory roots or pneumatophores.

Question 2.
In Opuntia, the stem is modified into a flattened green structure to perform the function of leaves, (i.e., photosynthesis). Cite some other examples of modifications of plant parts for the purpose of photosytnthesis.
Solution:
In Opuntia a xerophytic plant leaves are modified into spine to reduce the rate of transpiration and they do not perform the photosynthesis at all.
The function of photosynthesis in Opuntia plant is performed by stem which is thick fleshy and flattened structure containing chlorophyll and stores food and known as phylloclade.
In some plants similarly roots become assimilatory e.g., case of Trapa and Tinospom. These roots grow outside the soil, develop chlorophyll in them and perform photosynthesis.

Question 3.
In swampy areas like the sunderbans in West Bengal, plants bear special kind of roots called……
Solution:
Pneumatophores Roots are meant for the absorption of water and minerals from the soil. Cells of roots require 02 to respire. In swampy areas, soil does not have air, so no 07 is available to them.
In such cases, roots come out of the soil showing negative geotropism and breathe after coming in contact with air, e.g., Rhizophora. Such roots are called pneumatophores or respiratory roots.

Question 4.
In aquatic plants like Pistia and Eichhornia, leaves and roots are found near…..
Solution:
In Pistia and Eichhonia, the stem is like a runner where it branches to form leaves at the
v apex and roots below. Both the plants are hydrophytes and thus the roots are found near the surface of water.

Question 5.
Which parts in ginger and onion are edible?
Solution:
The edible part of ginger is rhizome the modified stem which stores food material whreas the edible part in onion is fleshy leaves, where the internode becomes shortened, leaves get condensed to form a tunic and store food material.

Question 6.
In epigynous flower, ovary is situated below the……….. .
Solution:
Ovary is situated below the thalamus (inferior) in epigynous flower while the other whorls of flower like sepals, petals and androecium grows above the ovary (superior), e.g., carrot, guava, Cucurbit a, sunflower, etc.
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.1

Question 7.
Add the missing floral organs of the given floral formula of Fabaceae.
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.14
Solution:
The floral formula of fabaceae family is
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.2
The flower of fabaceae is bisexual, zygomorphic, pentameros, gamosepalous, corolla-petals 5, androecium is ten diadelphous, gynoecium- superior, ovary monocarpellay.

SHORT ANSWER QUESTIONS

Question 1.
Give two examples of roots that develop from different parts of the angiospermic plant other than the radicle.
Solution:
Prop roots are meant for support. Prop roots develop from the lower nodes of stem of banyan tree. They grow downwards and touch the soil.
Stilt roots arise from the lower nodes of stem in sugarcane and enter the soil to provide strength to the plant. These protect the plant against winds.
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.3

Question 2.
The essential functions of roots are anchorage and absorption of water and minerals in the terrestrial plant. What functions are associated with the roots of aquatic plants. How are roots of aquatic plants and terrestrial plants different?
Solution:
Usually the terrestrial roots show a branched network that helps in anchorage and absorption of water and minerals from soil to the plant. While in aquatic plants, roots show modification and deviation from their normal function.
Ex – in plants like Trapu, Tinospora the roots are green and highly branched to increase the photosynthetic area, whereas in plants like Jussiaecci they get inflated due to air project out of water so a to help the plant in floating and exchange of gases.
Difference between roots of aquatic plants and terrestrial plants are as:
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.4

Question 3.
Draw diagrams of a typical monocot and dicot leaves to show their venation pattern.
Solution:
The pattern of distribution of veins and veinlets in the lamina of leaf is called Venation. It’s pattern is different in monocot and dicot leaf.
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.5

Question 4.
A typical angiosperm flower consists of four floral parts. Give the names of the floral parts and their arrangements sequentially.
Solution:
Following are the four floral parts of typical angiospermic flower.

  1. Calyx is the outermost whorl of the flower and comprised of sepals. These are usually green and (in bud stage) are protective in function.
  2. Corolla is composed of petals, usually bright coloured to attract insects for pollination.
  3. Androecium is composed of stamens, the male reproductive organ. Each stamen consists of stalk or filament and anther (containing pollen sac and pollen grains).
  4. Gynoecium is the female reproductive part and comprised of one or more carpels. Each
    NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.6

Question 5.
Reticulate venation is found in dicot leaves while in monocot leaves venation is of parallel type. Biology being a ‘Science of exceptions’, find out any exception to this generalisation.
Solution:
Reticulate venation is a characteristic of dicots and parallel venation is of monocots. But few exceptions are also seen in this generalisation, parallel venation is also found in dicot plants, e.g, Calophyllum, Corymbium, etc and reticulate venation is also found in monocot plants such as Alocasia, smilax, etc.

Question 6.
You have heard about several insectivorous plants that fee on insects. Nepenthes or the pitcher plant is one such example, which usually grows in shallow water or in march lands. What part of the plant is modified into a pitcher? How does this modification help the plant for food even though it can photosynthesise like any other green plant?
Solution:

  1. In insectivorous plant like Nepenthes, the leaf lamin is modified to form a pitcher and anterior part of petiole coils like tendril which keeps the pitcher in a vertical direction.
  2. Posterior part of the petiole remains flattened like a leaf. The apex of lamina forms a lid.
  3. Pitcher contains digestive enzyme for digesting trapped insects.
  4. All these modifications and adaptations are developed to make up for the nitrogen deficiency in the plant because these plants are found in N2 deficient soil, (marshy/swamp soils).
    NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.7

Question 7.
How can you differentiate between free central and axile placentation?
Solution:
The arrangement of ovules on the walls of ovary with the help of special kind of tissue called placenta is placentation. Plants show different types of placentation.
Difference between free central placentation and axile placentation include:
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.8
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.9

Question 8.
Why is maize grain usually called as a fruit and not a seed?
Solution:
The maize grain is usually known as fruit because it is a ripened ovary which contains a ripened ovule, e.g., a single seed. This fruit is known as caryopsis in which the pericarp is fused with the seed coat. The maize grain occurs attached to a thick cob or peduncle.

Question 9.
Tendrils of grapevins are homologous to the tendril of pumpkins, but are analogous to that of pea. Justify the above statement.
Solution:
Homologous organs are organs that have similar origin but they differ functionally. Axillary bud of stem gives rise to tendril of both grapevine and pumpkins so they have same origin, i.e., homologous, whereas analogous organs are organs having different origin, but perform same function. The tendril of pea arises from the leaf and helps the plant to climb.

Question 10.
Rhizome of ginger is like the roots of other plants that grows underground. Despite this fact ginger is a stem and not a root. Justify.
Solution:
Rhizome of Ginger is a type of modified underground stem which grows horizontally underground and bears nodes, intemodes and scaly leaves and buds, which gives rise to aerial shoots.
The adventitious root arises from the lower surface of nodes. It is not a true root because root does not have nodes and intemodes. The rhizome does not perform the function of anchorage and absorption, rather serve as reservoir for food storage. All these characteristics support the fact that ginger is a stem and not a root.
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.10

LONG ANSWER QUESTIONS

Question 1.
Distinguish between families – Fabaceae, solanaceae, Liliaceae on the basis of gynoecium characteristics (with figures). Also write economic importance of any one of the above family.
Solution:
The families in plant kingdom mainly differ from each other in their reproductive structures.
Based on characteristics of gynoecium the difference between the three families include the following:
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.11
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.12

Question 2.
Describe various stem modifications associated with food storage climbing and protection.
Solution:
The aerial part of plant bearing nodes, intemodes, buds, flowers, fruits and seeds is stem. Besides these functions and forms, it gets modified and perform under spfccial conditions.
The various stem modifications include:
NCERT Exemplar Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1.13

We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 5 Morphology of Flowering Plants, help you. If you have any query regarding NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 5 Morphology of Flowering Plants, drop a comment below and we will get back to you at the earliest.

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

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

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

VARY SHORT ANSWER QUESTIONS

Question 1.
Define the following terms?
(a) Tidal volume
(b) Residual volume
(c) Asthma
Solution:
(a) Tidal Volume (TV) is volume of air inspired or expired during a normal respiration. It is approx. 500 ml. i. e., a healthy man can inspire or expire approximately 6000 to 8000 mL of air per minute.
(b) Residual Volume : (RV) is volume of air remaining in the lungs even after a forcible expiration. This averages 1100 mL to 1200 mL.
(c) Asthma is an allergic reaction that causes constriction of the bronchiole muscles, thereby reducing the air passage thus the amount of the air that can get to the alveoli.

Question 2.
A fluid filled double membranous layer surrounds the lungs. Name it and mention its important function.
Solution:
Pleural membrane is a fluid filled double membranous layer surrounds the lung. It protects the lung and provides lubrication to it.

Question 3.
Cigarette smoking causes emphysema. Give reason.
Solution:
Emphysema is a chronic disorder of respiratory system, in which inflation or abnormal distension of alveolar wall occurs. Cigarette smoking and the inhalation of smoke or toxic substances over a time period causes the damaging of septa present between the alveoli, and its elastic tissue is replaced by the connective tissue in lungs.
Hence, decreases the respiratory surface and causes emphysema. It causes shortness of breath, production of sputum, chronic bronchitis, etc.

Question 4.
What is the amount of 02 supplied to tissues through every 100 mL of oxygenated blood under normal physiological conditions?
Solution:
Every 100 mL of oxygenated blood can deliver around 5 mL of 02 to the tissue under normal physiological conditions.

Question 5.
A major percentage (97%) of 02 is transported by RBCs in the blood. How does the remaining percentage (3%) of 02 transported?
Solution:
About 97% of 02 is transported by RBCs in the blood. The remaining 3% of 02 is carried in a dissolved state through the plasma.

Question 6.
Complete the missing terms
(a) Inspiratory Capacity (IC) =…..+ IRV
(b) …. = TV + ERV
(c) Functional Residual Capacity (FRC) = ERV+ …..
Solution:
(a) Inspiratory Capacity (IC) = (TV) + (IRV) Tidal Volume. Inspiratory Reserve Volume
(b) Expiratory Capacity (EC) = (TV + (ERV) Tidal Volume. Expiratory Reserve Volume.
(c) Functional Residual Capacity (FRC) = (ERV) Expiratory + (RV) Reserve Volume. Residual Volume.

Question 7.
Name the organs of respiration in the following organisms.
(a) Flatworm …..
(b) Birds ……
(c) Frog …..
(d) Cockroach …….
Solution:
(a) Flatworm General body surface
(b) Birds Lungs
(c) Frog Lungs and moist skin
(d) Cockroach Tracheal tubes.

SHORT ANSWER QUESTIONS

Question 1.
State the different modes of CO2 transport in blood.
Solution:
The blood carries carbon dioxide in three forms.
(i) In dissolved State About 7% of C02 is carried by physical solution. Under normal temperature and pressure.
(ii) As carbamino Compounds – Carbon dioxide binds directly with Hb to form an unstable compound carbaminocompounds (COzHb). About 23% CO2 is transported in this form. When pC02 is high and p02 is low as in the tissues, more binding of CO2 occurs whereas, when pCO2 is low and pO2 is high as in alveoli as tissue dissociation of CO2 from carbamino-haemoglobin takes place.
NCERT Exemplar Solutions for Class 11 Biology Chapter 17 Breathing and Exchange of Gases 1
(iii) As bicarbonate Ions C02 reacts with water in the presence of carbonic anhydrase to form carbonic acid (H2C03) in RBC H2C03– dissociates into hydrogen and bicarbonate ions (HCO-).
The whole reaction proceeds as follows
NCERT Exemplar Solutions for Class 11 Biology Chapter 17 Breathing and Exchange of Gases 2
The carbonic anhydrase reaction mainly occur in RBC as it contain high concentration of enzyme carbonic anhydrase and minute quantity of it is present in plasma too.

Question 2.
For completion of respiration process, write the given steps in sequential manner.
(a) Diffusion of gases (O2 and CO2) across alveolar membrane.
(b) Transport of gases by blood.
(c) Utilisation of O2 by the cells for catabolic reactions and resultant release of CO2.
(d) Pulmonary ventilation by which atmospheric air is drawn in and CO2 rich alveolar air is released out.
(e) Diffusion of O2 and CO2 between blood and tissues.
Solution:
NCERT Exemplar Solutions for Class 11 Biology Chapter 17 Breathing and Exchange of Gases 3

LONG ANSWER QUESTIONS

Question 1.
Explain the mechanism of breathing with neat labelled sketches.
Solution:
Mechanism of breathing involves two stages:
Inspiration is the process, during which atmospheric air is drawn in expiration is the process by which the alveolar air is released out.
The movement of air into and out ofa the lungs is carried out by creating a pressure gradient between the lungs and the atmosphere, with the help of diaphragm and inter costal muscles.
NCERT Exemplar Solutions for Class 11 Biology Chapter 17 Breathing and Exchange of Gases 4

We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 17 Breathing and Exchange of Gases, help you. If you have any query regarding NCERT Exemplar 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 Exemplar Solutions for Class 11 Biology Chapter 16 Digestion and Absorption

NCERT Exemplar Solutions for Class 11 Biology Chapter 16 Digestion and Absorption

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

VERY SHORT ANSWER QUESTIONS

Question 1.
The food mixes thoroughly with the acidic gastric juice of the stomach by the churning movements of its muscular wall. What do we call the food then?
Solution:
For 4-5 hours, the food is stored in stomach and gets thoroughly mixed with the acidic gastric juice of stomach by the churning movements of its muscular wall. The food at this stage is called as chyme.

Question 2.
Trypsinogen is an inactive enzyme of pancreatic juice. An enzyme, enterokinase, activates it. Which tissue/cells secrete this enzyme?/ How is it activated?
Solution:
Trypsinogen is activated to trypsin in the presence of which enzyme enterkinase is secreted by the intestinal mucosa.

Question 3.
In which part of alimentary, canal does absorption of water, simple sugars and alcohol takes place?
Solution:
The absorption of water, simple sugars, alcohol and some lipid soluble drugs take place by the stomach wall.

Question 4.
Name the enzyme involved in the breakdown of nucleotides into sugars and bases?
Solution:
The enzymes nucleotidases and nucleosidases are involved in the breakdown of nucleotides into sugars and bases.
NCERT Exemplar Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 14

Question 5.
What do we call the type of teeth attachment to jaw bones in which each tooth is embedded in a socket of jaws bones?
Solution:
The type of attachment where teeth are embedded in the socket of jaw bone is called thecodont.

Question 6.
Stomach is located in upper left portion of the abdominal cavity and has three major parts. Name these three parts.
Solution:
The three major of stomach are cardio, fundus and pylorus.
NCERT Exemplar Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 1

Question 7.
Does gall bladder make bile?
Solution:
Gall bladder is involved in the storage of bile and not associated with the bile formation rather, bile is secreted and from the hepatic cells of the liver.

Question 8.
Correct the following statements by deleting one of entries (given in bold).
(a) Goblet cells are located in the intestinal mucosal epithelium and secrete chymot- rypsin/mucus.
(b) Fats are broken down into di-and monog-lycerides with the help of amylase/lipases.
(c) Gastric glands of stomach mucosa have oxyntic cell/chief which secrete HC1.
(d) Saliva contains enzymes that digest starch/ protein.
Solution:
(a) Goblet cells are located in the intestinal
mucosal epithelium and secrete mucus.
(b) Fats are broken down into di and monoglycerides with the help of lipases. Fats —Llp‘’e’ > Diglycerides —> Monoglycerides.
NCERT Exemplar Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 2
(c) Gastric glands of stomach mucosa have oxyntic cells which secrete HCl
(d) Saliva contains enzymes that digest starch
NCERT Exemplar Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 3

SHORT ANSWER QUESTIONS

Question 1.
What is pancreas? Mention the major secretions of pancreas that are helpful in digestion.
Solution:
The pancreas is both exocrine and as well as gland endocrine situated between the limbs of ‘U’ shaped duodenum.
NCERT Exemplar Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 4
Internal structure of pancreas consist of two parts, i.e., the exocrine and endocrine part.
(i) Exocrine part consists of rounded lobules called acini, that secretes alkaline pancreatic juice of pH 8.4 and is mainly involved in the digestion of starch, proteins, fats and nucleic acids.
(ii) Endocrine part secretes hormones like, insulin and glucagon that regulate glucose metabolism.

Question 2.
Name the part of the alimentary canal where major absorption of digested food takes place. What are the absorbed forms of different kinds of food materials?
Solution:
The principle organ for the absorption of nutrients small intestine is the proces’s of digestion complete here and the final products of digestion are absorbed through the mucosa into the blood stream.
The absorbed form of different food materials are
NCERT Exemplar Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 5

Question 3.
List the organs of human alimentary canal and name the major digestive glands with their location.
Solution:
Humn digestive system consists of two main parts: alimentary canal and digestive glands. The organ of human alimentary canal are mouth, pharynx oesophagus, stomach, small intestine, large intestine, rectum and anus. Major digestive glands with their locations are as follows:
NCERT Exemplar Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 6

Question 4.
What are three major types of cells found in the gastric glands? Name their secretions.
Solution:
Following are the three major types of cells found in gastric glands. –
(i) Mucous neck cells (Goblet cells) are involved in the secretin of mucus and are present throughout the epithelium of gastrointestinal tract.
(ii) Peptic of Chief cells (Zymogenic cells) are Involved in the secretion of gastric enzymes such as proenzymes pepsinogen and prorenin and usually basal in location.
(iii) Parietal or oxyntic cells are large and most numerous present on the side walls of the gastric glands. They are involved in the secretion of HC1 and Castlis intrinsci Factor (CIF).

Question 5.
How is the intestinal mucosa protected from the acidic food entering from stomach?
Solution:
The intestinal mucosal epithelium has goblet cell which secrete mucus. The mucus along with bicarbonate present in the gastric juice help in lubrication and protection of mucosal epithelium from the acidic food entering from the stomach.

LONG ANSWER QUESTIONS

Question 1.
A person had roti and dal for his lunch. Trace the changes in those during its passage through the alimentary canal.
NCERT Exemplar Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 7
Solution:
Digestion of Roti (Carbohydrates)
(a) Digestion of Carbohydrates in the Oral Cavity
In oral cavity, the roti get mixed with saliva that contains an enzyme salivary amylase (ptyalin), which converts starch of roti into maltose, isomaltose and small dextrins called a – dextrin. 30% of starch is hydrolysed in the oral cavity.
NCERT Exemplar Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 8
(b) Digestion of Carbohydrates in the Small Intestine
The partially digested roti passes from oral cavity to oesophagus and then reaches to stomach by peristalsis. The stomach stores the food for 4-5 hours. The gastric juice does not contain carbohydrate digesting enzyme. The partially digested food is now called as chyme. In intestine, following action occurs.
(i) Action of Pancreatic Juice – Carbohydrates in the chyme are hydrolysed by pancreatic amylase into disaccharides.
Polysaccharides(starch) —Amylas’: > Disaccharides
NCERT Exemplar Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 9
(ii) Action of Intestinal Juice – Intestinal Juice contain maltase, isomaltase, sucrase (invertase), lactase and a – dextrinase. hi the presence of these enzymes food is converted into simpler compounds like glucose, fructose, galactose, etc.
NCERT Exemplar Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 10
Digestion of Protein
Proteins are made up of amino acids. So proteins are broken down to amino acid during the process of digestion.
Saliva lacks any protein digesting enzyme so, digestion starts further in stomach.
(a) Digestion of Protein in Stomach. The stomach normally stores food for 4-5 hours. The gastric glands of the stomach secrete gastric juice that contains HCI, proenzymes like-pepsinogen and prorennin. Various reactions in stomach are discussed bwlow.
NCERT Exemplar Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 11
(b) Digestion of Protein in Small Intestine (i) Action of Pancreatic Juice – The enzymes , trypsinogen, chymotrypsinogen and procarboxypeptidase in pancreatic juice are all concerned with the protein digestion.
NCERT Exemplar Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 12
(ii) Action of Intestinal Juice – Intestinal juice contain enzymes enterokinase, amino peptidase and dipeptidase
NCERT Exemplar Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 13
The macromolecules are broken down into simpler components are the products of roti and dal (carbohydrates and proteins) which are further absorbed by the villi in small intestine and the rest undigested food is removed in the form of faeces by large intestine.

Question 2.
Discuss mechanisms of absorption.
Solution:

  • Absorption is a process by which the end product of digestion passes through the intestinal mucosa into the blood or lymph.
  • It is carried out by passive, active or facilitated transport mechanism. Small amount of monosaccharide like glucose, amino acids and some electrolytes like chloride ions are absorbed by simple diffusion.
  • Some of the substance like fructose and some amino acids are absorbed with the help of the carrier ions like Na+ are absorbed by the active transport.
  • Fatty acid and glycerol are insoluble, thus they cannot be absorbed by the blood. They are first incorporated into small droplets called micelles which move into the intestinal mucosa.
  • They are reformed into very small protein coated fat globules called chylomicrons which are transported into the lacteals of the villi. The lacteals ultimately release the absorbed substance into the blood stream.
  • The maximum absorption of food takes place in small intestine.

We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 16 Digestion and Absorption, help you. If you have any query regarding NCERT Exemplar 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 Exemplar Solutions for Class 11 Biology Chapter 15 Plant Growth and Development

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

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

VERY SHORT ANSWER QUESTIONS

Question 1.
Fill the places with appropriate word/words.
(a) A phase of growth which is maximum and fastest is ………
(b) Apical dominance as expressed in dicotyledonous plants is due to the presence of more ……. in the apical bud than in the lateral ones.
(c) In addition to auxin, a ……. must be supplied to culture medium obtain a good callus in plant tissue culture.
(d)…….. of a vegetative plants are the sites of photoperiodic perception.
Solution:
(a) A phase of growth which is maximum and latest is exponential phase.
(b) Apical dominance as expressed in dicotyledonous plants is due to the presence or more auxins in the apical bud than in the lateral ones.
(c) In addition to auxin, a cytokinin must be g supplied to culture medium to obtain a good callus in plant tissue culture.
(d) Leaves of vegetative plants are the sites of photoperiodic perception.

Question 2.
Plant Growth Substances (PGS) have innumerable practical applications. Name the PGS you should use to
(a) increase yield of sugarcane
(b) promote lateral shoot growth
(c) cause sprouting of potato tuber
(d) inhibit seed germination
Solution:
(a) Spraying gibberellins
(b) Application of auxins
(c) Ethylene
(d) ABA

Question 3.
A primary root grows from 5 cm to 19 cm in a week. Calculate the growth rate and relative growth rate over the period.
Solution:
Growth depends upon three factors – initial size (WQ), rate of growth (r) and time interval (+) for which the rate of growth is retained.
NCERT Exemplar Solutions for Class 11 Biology Chapter 15 Plant Growth and Development 1

Thus absolute growth rate is 0.1907 while relative growth rate is 3.8 cm.

Question 4.
Gibberellins were first discovered in Japan when rice plants were suffering from bakane (the foolish seedling disease) caused by a fungus Gibberella fujikuroi.
(a) Give two functions of this phytohormone.
(b) Which property of gibberellin caused foolish seedling disease in rice?
Solution:
Following are the two functions of gibberellin:
(i) It produces the phenomenon of bolting i. e., the growth of the intemodal region of stem in rosette plants.
(ii) It induces seed germination and break bud and seed domancy.
(b) The rice seeding/plant show excessive growth in their intemodal region when gets infected by fungus Gibberellafujikuroi. This fungus produces excessive amount of plant hormone GA that makes plants taller in comparison to the normal plant foolishly and many results into death of the plant.

Question 5.
Classify the following plants into Long Day Plants (LDP), Short Day plants (SDP) and Day Neutral Plants (DNP) Xanthium, henbane (Hyoscyamus niger), spinach, rich, strawberry, Bryophyllum, sunflower, tomato, maize.
Solution:
Long Day Plant (LDP) The plants that requires the exposure light for a longer period exceeding a well defined critical duration of light are long day plants.

Among the above given plant LDP are for henbane, Bryophyllum and spinach. Short Day Plants (SDP) The Plants that requires light for a period less than well defined critical duration of light, e.g., Xanthium, rice, strawberry.

Day Natural Plants (DNP) The exposure to light does not affect the flowering in certain plants, e.g, DNP, sunflower, tomato, maize.

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, is a plant growth regulator that has feminizing effect on sex expression. Ethylene promotes formation of female flowers in monoecious plants like cucumber.

Question 7.
Where are the following hormones synthesised in plants?
(a) IAA
(b) Gibberellins
(c) Cytokinins
Solution:
(a) IAA i.e., Indole acetic acid. It is synthesised at the growing apices of the plant, e.g., shoot tip, leaf primordia and developing seeds.
(b) Gibberellins It is synthesised in the apical shoot buds, young leaves, root tips and developing seeds.
(c) Cytokinins are synthesised mainly in routs, but synthes also occurs in the endosperm of seeds, growing embryo etc.

Question 8.
Growth is one of the charactristic of all living organism? Do unicellular organism also grow? If so, what are the parameters?
Solution:
1. Lag phase- In this phase growth is slow.
2. Exponential phase- It shows rapid growth and maintains maximum growth for sometime.
3. Stationary phase- In this phase Growth diminishes and ultimately stops.

Question 9.
In the figure of sigmoid growth curve given below, label segments 1,2 and 3.
NCERT Exemplar Solutions for Class 11 Biology Chapter 15 Plant Growth and Development 2
Solution:
Growth is the main characteristic that distinguish living organisms from non-living. All living organism grow in number and then accumulate biomass and grow in size as well.

Increase in number of cells as well as increase in size and length of each cell, exhibits growth of all living organism.
In unicellular organism, the growth is synchronized with reproduction.

These organism when divide they produce offspring (reproduction) i.g., each cell accumulate (synthesise) protoplasm and increase in size but at a certain a limit and divide to from two cells.

Question 10.
The rice seedlings infected with fungus Gibberellafujikuroi is called foolish seedlings? What was the reason behind it?
Solution:
The rice seedling infected with fungus Gibberella fujikuroi are called foolish seedling because infected plants grow excessively taller than rest of the non infected rice plants in the field fall over and be unharvestable.

SHORT ANSWER QUESTIONS

Question 1.
Nicotiana tobacum, a short day plant, when exposed to more than critical period of light fails to flower. Explain.
Solution:
Short day plants are those plants that flower only when the exposure to duration of light is below critical period. Tobacco, being a short day plant is unable to show flowering when it is exposed to light above than the critical period.

Question 2.
Explain in 2-3 lines each of the following terms with the help of examples taken from different plant tissues.
(a) Differentiation
(b) De-differentiation
(c) Re-differentiation
Solution:
(a) Differentiation is permanent in composition
structure size and function of cells, tissue or organs. For example the meristematic tissues in plants gives rise to new cells which then mature and get differentiated into tissue or an organ of the plant, e.g., cells, distal to root apical meristem form root cap, cell of the periphery form epiblema, followed by cortex, etc.
(b) De-differentiation is the process of regain of differentiated cells so that they again become differentiated and able to divide, e.g., in dicot stem, the cortical cells get de-differentiate and become meristematic to form cambium (interfascicular cambium, and fascicular cambiums).
(c) Re-differentiation The cambium cells thus formed, again re-differentiate to form secondary cortex cells, secondary xylem and phloem elements and phelloderm in case of secondary growth of woody dicot plants.

Question 3.
The role of ethylene and abscissic acid is both positive and negative. Justify the statement.
Solution:
NCERT Exemplar Solutions for Class 11 Biology Chapter 15 Plant Growth and Development 3

Question 4.
In animals, there are special glands secreting
hormones, whereas there are no glands in plants. Where are plant hormones formed? How are the hormones translocated to the site of activity?
Solution:
The plant hormones are synthesised by the plant cells needed. Few hormones are specifically synthesised at a particular part of the plant like auxin synthesised in growing shoot apices and ethylene is secretes by ripened fruits.

Cytokinin is found in dividing cells. Unlike plants animal 1 being more advanced, and organised they have proper hormone secreting glands and organs.

These are transported through the transport system of their body in both plant and animals. In plants, hormone are translocated via xylem and phloem to the site of activity.

Question 5.
In a slide showing different types of cells can you identify which type of the cell may be meristematic and the one which is incapable of dividing and how?
Solution:
On the basis of the following characteristics the meristemtic cells can be identified.
(i) Cell consist thin cellulose wall and dense cytoplasm with large nucleus.
(ii) Among meristematic cells, plasmodesmal connections are more numerous.
(iii) Cell division, i. e., mitosis arid its various stages are distinctly visible.
(iv) Chromosomes of cells replicate and divide into two homologous chromatids.
All these features contribute to open ended growth where structure is in complete in meristematic regions.
Whereas, cells incapable of dividing show features such as
(i) Attains particular shape, size and thickening.
(ii) Undergoes structural and physiological differentiation.
(iii) Different types of cell are formed such as epidermis, cortex, vascular tissues.

Question 6.
A rubber band stretches and reverts back to its original position. Bubble gum stretches, but it would not return to its original position.
Is there any difference between the two processes? Discuss it with respect to plant growth (hint elasticity (reversible) plasticity (irreversible).
Solution:
NCERT Exemplar Solutions for Class 11 Biology Chapter 15 Plant Growth and Development 4

Question 7.
Label the diagram.
A. This is which part of a dicotyledonous plants?
B. If we remove part 1 from the plant, what will happen?
NCERT Exemplar Solutions for Class 11 Biology Chapter 15 Plant Growth and Development 5
Solution:
Representation the labelling of the given diagram is as follows
NCERT Exemplar Solutions for Class 11 Biology Chapter 15 Plant Growth and Development 6

A. the plant part in the given diagram is growing shoot apex.
B. Removal of shoot apex will help to overcome the apical dominance. Thus, the lateral buds grow faster, giving rise to branches and give the plant a bushy appearance.

Question 8.
Both animals and plants grow. Why do we say that growth and differentiation in plants is open and not so in animals? Does this statement hold true for sponges also?
Solution:

  1. Growth in plant is totally different from the animal growth as growth in plant is unlimited and indefinite.
  2. Root and shoot in tips in the plants are open ended i.e. they always grow and form new organ to replace the older and senescent one due to presence of meristem cells, which are capable to grow and divide.
  3. Thus, the plant growth continues throughout the life. On the contrary, animal growth is limited as growth /stops as soon as they mature. Sponges are those animals which show cellular level of organisation.
  4. These animals posses totipotent cells which are capable of giving rise to all other cells in sponges.
  5. A small part detached from a sponge can regenerate into a whole new sponge.
  6. However growth in sponges cannot be called open or indefinite as they cannot grow beyond a certain size.
    Practically, they do not show open ended growth.

Question 9.
Define parthenocarpy. Name the plant hormone used to induce parthenocarpy.
Solution:
Parthenocarpy is the process where fruit develop without fertilisation and so, it lacks contain seed. Seedless fruits are developed in some plants. Certain phytohormone induce development of fruit without fertilisation. This can also be induced artificially by spraying auxin an*i gibberellins in certain plants like, grapes, papaya, etc. ,

Question 10.
While eating watermelons, all of us wish it was seedless. As a plant physiologist can you suggest any method by which this can be achieve.
Solution:
The seedless, fruits can be produced by the process of parthenocarpy. In this fruits are
developed without fertilisation, so, seeds are not formed in the fruit. Artifically parthenocarpy can be induced by spraying auxin and gibberellin to produce seedless watermelons.

Question 11.
On germination a seed first produces shots with leaves, flowers appear later,
A. Why do you think this happens?
B. How is this advantageous to the plant?
Solution:
A. With the germination of seeds the plant
enters into vegetative growth period. This period tabes light stimulus i. e., photoperiod) and synthesise the florigen (a flowering hormone) that flowering.
B. The vegetative growth period prepares the plant to bear reproductive structure like flower, fruits and seeds, and allows it to grow, mature and reproduce.

Question 12.
Fill in the blanks
A. Maximum growth is observed in …… phase.
B. Apical dominance isdue to ………. .
C. …… initiate rooting.
D. pigment involved in phote photoperception in fiowering plants in ……….
Solution:
A. Exponential
B. Auxin
C. Cytokinin
D. Phytochrome.

LONG ANSWER QUESTIONS

Question 1.
Some varieties of wheat are known as spring wheat while others are called winter wheat. Former variety is sown and planted in spring and is harvested by the end of the same season. However, winter varieties, if planted in spring, fail to flower or produce mature grains within a span of a flowering season. Explain, why?
Solution:

  • Some annual plants such as wheat do not flower, unless they experience a low temperature during spring they remain vegetative but after receiving low temperature (in winter) they grow further to bear flowers and fruits.
  • During winter the low temperature prevents precocious reproductive development in autumn, thus enabling the plant to reach vegetative maturity before reproductive phase.
  • Thus, in spring when spring varieties are planted they flower and bear fruits prior to end of growing season.
  • But, if the winter varieties are planted in spring, they fail to flower and produce mature grains before the end of growing season, as they could not perceive low temperature of winters.

Question 2.
Name a hormone which
A. is gaseous in nature
B. Is responsible for phototropism
C. induces femaleness in flowers of cucumber
D. is used for killing weeds (dicots)
E . induces flowering in long day plants.
Solution:
A. Ethylene is a hormone which is gaseous in nature.
B. Auxin (synthetic auxin 2-4D) is responsible for phototropism.
C. Ethylene induces ferminising effect. External supply of very small quantity of ethylene can increase the number of female flowers and hence fruits as in cucumber.
D. Synthetic auxin (2-4D) that kills broad leaved dicot weeds and is used as weedicides.
E. Gibberellins induces flowering in long-day plants.

We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 15 Plant Growth and Development, help you. If you have any query regarding NCERT Exemplar 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 Exemplar Solutions for Class 11 Biology Chapter 4 Animal Kingdom

NCERT Exemplar Solutions for Class 11 Biology Chapter 4 Animal Kingdom

These Solutions are part of NCERT Exemplar Solutions for Class 11 Biology. Here we have given NCERT Exemplar Solutions for Class 11 Biology Chapter 4 Animal Kingdom.

VERY SHORT ANSWER QUESTIONS

Question 1.
What is the importance of pneumatic bones and air sacs in Aves?
Solution:
Birds possess light wfi’ght bones that contain internal spaces filled with air, which are pneumatic bones. They are an adaptation for flight as they help in, reducing the body weight. Aerodynamic lungs with specialized air sacs are an additional feature that aids birds in flying (e.g., bald eagle, pigeon).

Question 2.
What is metagenesis? Mention an example which exhibits this phenomenon.
Solution:
The phenomenon in which one generation of certain plants and animals reproduce asexually, followed by the sexually reproducing generation is metagenesis. Both the forms in metagenesis are diploid hence, it is known as the false alternation of generation. Coelenterates exhibit metagenesis (e.g., Obelia) where in its life cycle polyp form alternates with medusa.

Question 3.
What is the role of feathers?
Solution:
Feathers are the epidermal out growths that form distinctive outer covering or plumage in birds.
A variety of role are played by feathers which includes:
(i) They provide life and help in flight, by creating airfoil shape for wings.
(ii) They help in maintaining body temperature.
(iii) Feathers play a vital role in mating by providing secondary sexual that characters in both the sexes the colour and markings determine the alteractiveness of mate.

Question 4.
Which group of chordates posses sucking and circular mouth without jaws?
Solution:
Class-Cyclostomata is comprised of living jawless fishes. They have a circular mouth and lack jaws, hence they are also called agnathans. The mouth works like a sucker and is surrounded by tentacles (e.g., lampreys and haglish). These also prosses rectroctable teeth that are homy.

Question 5.
Mention two modifications in reptiles required for terrestrial mode of life.
Solution:
Certain characters acquired by reptiles for the terrestrial adaptations include.
(i) Body is covered with dry and comified skin and epidermal scales or scutes.
(ii) Internal fertilisation.

Question 6.
What is the role of radula in molluscs?
Solution:
The radula is a special rasping structure present many molluscs. It is used to scrape and scratch the food and to create depressions in rocks used as habitat.
It bears many rows of tiny teeth that are replaced
as they wear down e.g., Limplet is a marine invertebrate that uses its radula for creating home by boring a shallow hole in the rock.

Question 7.
Name the animal, which exhibits the
phenomenon of bioluminescence. Mention the
phylum to which it belongs.
Solution:
Bioluminescence is the phenomenon of production and emission of light by an organism as a result of chemical reaction during which chemical energy is converted to light energy. The phenomenon of bioluminescence is exhibited by Ctenoplana from phylum- Ctenophora.

Question 8.
Write one example for each of the following in the space providing.
(a) Cold blooded animal
(b) Warm blooded animal
(c) Animal possessing dry and comified skin
(d) Dioecious animal
Solution:
(a) A cold blooded animal is Crocodilus (crocodile)
(b) Elephas maximus (elephant), (mammal) is a warm blooded animal.
(c) Testudo (tortoise) bears dry and comified skin.
(d) Ascaris (roundworm) is a dioecious animal.

Question 9.
Differentiate between a diplobastic and triploblastic animal.
Solution:
Diploblastic animals are animals in which the cells are arranged in two embryonic layers, an external ectoderm and an internal endoderm (e.g., coelentrates). Animals in which the developing embryo has a third germinal layer, i. e., mesoderm lying between the ectoderm and endoderm are calledtriploblastic animals, (e.g., chordates).

Question 10.
Give an example of the following
(a) Roundworm
(b) Fish possessing poison sting
(c) A limbless reptile/amphibian
(d) An oviparous mammal
Solution:
(a) Roundworm – A scans
(b) Fish possessing poison sting – Trygon
(c) A limbless reptile/amphibian – Ichthyophis
(d) An oviparous mammal – Duck billed platypus.

Question 11.
Provide appropriate technical term in the space provided.
(a) Blood-filled cavity in arthropods
(b) Free-floating form of cnidaria
(c) Stinging organ of jelly fishes
(d) Lateral appendages in aquatic annelids
Solution:
(a) The blood-filled cavity in arthropods containing haemolymph is haemocoel.
(b) A form in cnidarians in which the body is shaped like an umbrella which can float freely in sed water is medusa.
(c) Capsules of specialised cells in cnidarians which act as a paraylysing sting are nematocytes.
(d) The paired unjointed lateral outgrowth in annelids bearing chaetae are parapodia.

Question 12.
Match the following.
NCERT Exemplar Solutions for Class 11 Biology Chapter 4 Animal Kingdom 1.1
Solution:
A. —> (iii)
B. —> (i)
C. —> (iv)
D. —> (ii)
(a) Octopus The appendages in invertebrates that are used for grasping food and for locomotion are tentacles.
(b) Crocodile for locomotion, and swimming limbs are used.
(c) Catta Fins are means of locomotion and are used to generation optimum thrust thus controlling the subsequent motion.
(d) Ctenoplana Locomotory organs formed by strong cilia with fused bases are comb plates.

SHORT ANSWER QUESTIONS

Question 1.
Differentiate between
(a) Open circulatory system and closed circulatory system.
(b) Oviparous and viviparous characteristic.
(c) Direct development and indirect development.
Solution:
Differentiation between these are as below
NCERT Exemplar Solutions for Class 11 Biology Chapter 4 Animal Kingdom 1.2

Question 2.
There has been an increase in the number of chambers in heart during evolution of vertebrates. Give the names of the class of vertebrates having two, three or four chambered heart.
Solution:
(a) In organisms like fishes two chambered heart is present. Mixing of oxygenated and deoxygenated blood blood occurs as only one atria and one ventricle is present which are not separated.
(b) After division of auricle into right and left halves three chambered heart develops and in amphibian. In ventiricles mixing of oxygenated and deoxygenated blood occurs.
(c) In reptilies an intermidiary heart is present in which ventricles get partially divided through a septum which is incomplete thus having a false four-chambered heart e.g., Crocodiles.
(d) Both the auricle and ventricle are divided into two halves in four chambered heart and so no mixing of oxygenated and deoxygenated blood occurs, e.g., birds and mammals.

Question 3.
Fill up the blank spaces appropriately
NCERT Exemplar Solutions for Class 11 Biology Chapter 4 Animal Kingdom 1.3
Solution:
Excretion involves the elimination of metabolic waste products from the animal body. In the process of excretion in different animals different organs are involved.
(a) In arthropods excretory products from haemolymph are removed by the malpighian tubules.
(b) The excretory organ occurs as segmentally arranged coiled tubules called nephridia in annelids.
(c) Excretion occurs by paired structures called organ of Bojanus in molluscs also called metanephridia.
(d) Mesonephric kidneys are associated with excretion in amphibians.
The circulation of blood and lymph along with oxygen carbondioxide, hormones, blood cells, etc, within the body system for the nourishment of cells, fighting diseases, and for stabilising body temperature and pH is involved blood circulation.
NCERT Exemplar Solutions for Class 11 Biology Chapter 4 Animal Kingdom 1.4

Respiratory organs are involved in the
exchange of gases from the atmosphere.
Different respiratory organs in various animals.
(a) Lungs and skin in amphibians.
(b) Lung/gills/tracheal system in arthropoda and molluscs.
(c) Skin in annelids.

Question 4.
Match the following
NCERT Exemplar Solutions for Class 11 Biology Chapter 4 Animal Kingdom 1.5

Solution:
A. —> (v)
B. —> (iii)
C. —> (ii)
D.—> (i)
E. —> (vi)
F. —> (iv)
A. Amphibians are found in both aquatic and terrestrial habitat. Their large is completely aquatic while adult lives in terrestrial as well as in aquatic habitat.
B. Mammals produce milk in the mammary glands and feed their young one. The mammary glands are enlarged exocrine modified sweat glands functional in female mammals.
C. Chondrichthyes have (notochord) in the young stage which is gradually replaced by cartilage.
D. Osteichthyes possess air bladder which is a vesicle or sac containing air.
E. Cyclostomes have sucking and circular mouth without jaws which is surrounded by tentacles and the tongue bears teeth, e.g., lamprey and hagfish.
F. Aves comprise of light weighted bones with internal spaces field with air called pneumatic bones and aerodynamic lungs with specialised air sacs. These are the adaptations which enable birds to fly.

Question 5.
Endoparasites are found inside the host body. Mention the special structure, possessed by these and which enables them to survive in those conditions.
Solution:
Endoparasites such as Taenia solium and Fasciola hepatica (liver fluke), etc., are found inside body the host and survive due to the presence of certain characters.
Endoparasites special characters which include:
(i) The is respiration is anaerobic and the gaseous exchange in via general body surface.
(ii) They bear additional organs for the attachment to the host. Taenia solium posses hooks and suckers for the attachment with the host. Fasciola hepatica possesses acetabulum or posterior sucker for the attachment.
(iii) they have well developed reproductive organs. They are generally, harmaphrodite and self fertilisation occurs commonly.
(iv) They have a thick tegument (body covering) which is resistant to the host’s digestive enzymes and antioxins.
(v) Locomotary organs are absent.
(vi) They lack digestive organs because digested and semi digested food of the host is directly absorbed through their body surface.

Question 6.
Mention two similarities between
(a) Aves and mammals
(b) A frog and crocodile
(c) A turtle and Pila
Solution:
(a) Following are the similarities between aves and mammals
(i) Presence of four chambered heart.
(ii) The members of both the groups are homeotherms, i.e., warm blooded. They are able to maintain constant body temperature.
(b) Similarities between frog and crocodile include:
(i) They are cold blooded animals. The members of both the groups are poikilotherms, i.e., they lack the capacity to regulate their body temperature.
(ii) Frogs and crocodiles are oviparous animals.
(c) Similarities between turtle and Pila include
(i) Body is covered with dry and comified skin in both animals. In turtle, the epidermal covering is known as scales whereas in case of Pila, it is known as calcareous shell.
(ii) Both animals are oviparous.

Question 7.
Name
(a) A limbless animal
(b) A cold blooded animal
(c) A warm blooded animal
(d) An animal possessing dry and comified skin
(e) An animal having canal system and spicules
(f) An animal with cnidoblasts
Solution:
(a) Ichthyophis does not possess limbs.
(b) A cold blooded animal scoliodon (dog fish).
(c) warm blooded animal is Columba (pigeon).
(d) Naja naja (snake) possesses dry and cornified skin.
(e) Sycon (sponge) possesses canal system and bear spicules.
(f) Obelia bears cnidoblast.

Question 8.
Excretory organs of different animals are given below. Choose correctly and write in the space provided.
NCERT Exemplar Solutions for Class 11 Biology Chapter 4 Animal Kingdom 1.6

Solution:
Metabolism in body leads to the formation of waste that can affect body’s vital organs so it has to be removed from body. Different classes of organisms possess different types of excretory organs to eliminate the byproduct of metabolism.
A. —> (vi)
B. —> (ii)
C. —> (v)
D. —> (iii)
E. —> (vi)
F. —> (i)
A. Balanogolossus – Proboscis glands. This gland excretes brown granules and is present in front of central sinus.
B. Nephridia in Leech. It helps in osmoregulation and excretion.
C. Malpighian tubules in Locust open into gut and help in excretion.
D. The Flame cells of liver fluke are specialised cells in Platyhelminthes which helps in osmoregulation and excretion. These are also called protonephridia.
E. Sea urchin-absent Specialised excretory organs are absent in sea urchin.
F. It TV/a-Metanephridia is a type of excretory gland or nephridium found in many types of invertebrates such as annelids, arthropods, and molluscs (in molluscus nephridia is also known as Bojanus organ).

LONG ANSWER QUESTIONS

Question 1.
What is the relationship between germinal layers and the formation of body cavity in case of coelomate, acoelomates andpseudocoelomates?
Solution:
Multicellular organisms typically possess a concentric arrangement of tissues in the body. These tissues are derived from the three embrycnio cell, layers called germinal layers.
(i) The outer layer is the ectoderm, the middle layer is the mesoderm and the innermost layer is the endoderm.
(ii) Ectoderm is associated with the formation of CNS, eye lens, ganglia, nerves and glands.
(iii) Mesoderm forms the that in structural components of the body like the skeletal muscles the skeleton, the dermis of the
skin connective tissue, etc.
(iv) Endoderm layer is associated with the formation of the stomach, colon, liver, pancreas urinary bladder and other vital organs is an organism.
(v) Coelom is the body cavity that is lined by mesoderm and the animals possessing coelom are called as ceolomates. e.g., phylum-Annelida, Mollusca, Arthropoda, Echnidermata, Hermichordata and Chordata.
(vi) In some organisms, body cavity is not lined by mesoderm, instead mesoderm is present in the form of scattered pouches in between ectoderm and endoderm, Such body cavity is called pseudocoelom and animals possessing there stusturs are refered to as pseudocoelomates e.g., As car is.
(vii) The animals in which there is complete absence of body cavity are called acoelomates. e.g., Platyhelminthes.
NCERT Exemplar Solutions for Class 11 Biology Chapter 4 Animal Kingdom 1.7

Question 2.
Comment upon the habitats and external features of animals belonging to class- Amphibia and Reptilia.
Solution:
Amphibians
(i) They can dwell in aquatic as well as terrestrial habitats. They are ectothermic or (cold blooded).
(ii) They are tetrapods having (4 limbs) which facilitate movement on land.
(iii) Their limbs have evolved from the pectoral and pelvic fins.
(iv) Skin is thin, covered by mucus and remains mostly moist. It also serves as an accessory source of oxygen.
(v) They breathe through gills and lung gills usually appear in the larval stage, replaced by lungs in the adults stage.
(vi) Their heart is three chambered with two atria and one ventricle.
(vii) Females are oviparous and fertilisation is mostly external.
(viii) Larva is a tadpole, which metamorphose into adult e.g., Rariu frog, Nectureus (mud puppy), Salamandera (salamander).
NCERT Exemplar Solutions for Class 11 Biology Chapter 4 Animal Kingdom 1.8
Reptiles
(i) They are mostly terrestrial animals and their body is covered by dry, and comified skin, epidermal scales or scutes.
(ii) In reptiles the mode of locomotion is creeping and crawling.
(iii) Lungs are well developed and present in all stages of life.
(iv) Claws are present in toes.
(v) s Appendages are well adapted for land movement.
(vi) Heart possesses a partially divided ventricle and 2 atria.
(vii) They lay amniotic eggs which are inclubated on land.
(viii) They are poikilothermic or cold blooded animals. Temperature is regulated mechanically and not metabolically by moving in and out; source of heat is usually the sun.
(ix) Fertilisation is internal. They are oviparous and development of young ones is direct e.g., Chelone (turtle), Naja (cobra), Crocodicus (crocodile).
NCERT Exemplar Solutions for Class 11 Biology Chapter 4 Animal Kingdom 1.9

We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 4 Animal Kingdom, help you. If you have any query regarding NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 4 Animal Kingdom, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Solutions for Class 11 Biology Chapter 14 Respiration in Plants

NCERT Exemplar Solutions for Class 11 Biology Chapter 14 Respiration in Plants

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

VERY SHORT ANSWER QUESTIONS

Question 1.
Energy is released during the oxidation of 3. compounds in respiration. How is this energy stored and released as and when it is needed?
Solution:
The energy currency of every living cell Adenosine Triphosphate (ATP).
Complex organic food molecules such as sugars, fats and proteins are rich sources of energy for cell because much of the energy used to form these molecules is stored within the chemical bonds that hold them together. So, the cells release the stored energy through a series of oxidation reactions.

During oxidation of food, the product of reaction has a lower energy content than the donor molecule. At the same time, electron acceptor molecules capture some of the energy lost during oxidation and store it for later use.

Cells convert the energy from oxidation reactions to energy-rich molecules such as ATP that can be used through the cell for metabolism and construct new cellular components.

Question 2.
How does Respiratory Quotient 5. (RQ) indicate which type of substrate, i.e., carbohydrate, fat or protein is getting oxidised?
R.Q=A/B
What do A and B stand for?
What type of substrates have R.Q. of 1,< 1 or > 1?
Solution:
The ratio of C02 evolved and 02 consumed in respiration is called the Respiratory Quotient (RQ) or respiratory ratio.
NCERT Exemplar Solutions for Class 11 Biology Chapter 14 Respiration in Plants 1
Example:
(i) During aerobic respiration carbohydrates have RQ = 1
(ii) During germination of seeds proteins and fats have RQ of < 1. (iii) Under aerobic conditions substrates like organic acids have RQ of > 1

Question 3.
F0 – F1 particles participate in the synthesis of …… .
Solution:
F0 – F1 particles present in the inner mitochondrial membrane are involved in the Adenosine Triphosphate synthesis. It is known as the energy currency of the cell.
NCERT Exemplar Solutions for Class 11 Biology Chapter 14 Respiration in Plants 2

Question 4.
When does anaerobic respiration occur in man and yeast?
Solution:
In animals anaerobic respiration occurs in the situation of oxygen deficiency during heavy exercise, when pyruvic acid is reduced to lactic acid by the enzyme lactate dehydrogenase.

In yeast, the incomplete oxidation of glucose occurs in anaerobic conditions, during which pyruvic acid is converted to COz and ethanol by the action of enzyme pyruvic acid decarboxylase and alcohol dehydrogenase.

Question 5.
Which of the following will release more energy on oxidation? Arrange them in ascending order.
(a) 1 gm of fat
(b) 1 gm of protein
(c) 1 gm of glucose
(d) 0.5 gm of protein + 0.5 gm glucose
Solution:
The ascending order of substrate that will release more energy on oxidation will be as follows
1 gm protein < 0.5 gm in protein < 1 gm glucose < 1 gm fat + 0.5 gm glucose

SHORT ANSWER QUESTIONS

Question 1.
If a person is feeling dizzy, glucose or fruit juice is given immediately but not a cheese sandwich, which might have more energy. Explain.
Solution:
The glucose is absorbed and reaches blood quickly and gives instant energy. Whereas, cheese sandwich require time for digestion, and absorption. Sick person needs immediate energy supply, so glucose or fruit juices containing glucose are given to them.

Question 2.
Pyruvic acid is the end product of glycolysis. What are the three metabolic fats of pyruvic acid under aerobic and anaerobic conditions? Write in the space provided in the diagram.
NCERT Exemplar Solutions for Class 11 Biology Chapter 14 Respiration in Plants 3

Solution:
The three metabolic products formed under 3. aerobic and anaerobic conditions are Lactic acid, Ethanol and Acetyl Co-A Lactic acid is formed under anaerobic condition in skeletal muscles by the oxidation of pyruvic acid.

Ethanol is formed under anaerobic condition by the oxidation of pyruvic acid in yeast.

Acetyl Co-A is formed by the oxidation of pyruvic acid that take place within the mitochondria under aerobic condition.
NCERT Exemplar Solutions for Class 11 Biology Chapter 14 Respiration in Plants 4
NCERT Exemplar Solutions for Class 11 Biology Chapter 14 Respiration in Plants 5

Question 3.
Oxygen is an essential requirement for aerobic respiration by it enters the respiratory process at the end? Discuss.
Solution:
Aerobic respiration needs oxygen in order to generate ATP. Oxygen acts as final acceptor in respiratory process.

In pulse e (electrons) that energy from the electron transport chain ETC and take up protons from medium to form water.

It plays a vital role in respiration. 02 enters in the respiratory process at the end. It drives the process of aerobic respiration by removing hydrogen from the system. Thus, acting as final hydrogen acceptor.

By the process of oxidative phosphorylation the energy is produced, utilising the energy of oxidation reduction reactions.

Question 4.
The figure given below shows the steps in glycolysis. Fill in the missing steps A, B, C, D and also indicate whether ATP is being used up or released at step E?
NCERT Exemplar Solutions for Class 11 Biology Chapter 14 Respiration in Plants 6

Solution:
Process ofglycolysis is summarised as follow
NCERT Exemplar Solutions for Class 11 Biology Chapter 14 Respiration in Plants 7

Question 5.
Do you know any step in the TCA cycle where there is substrate level phosphorylation.
Which one?
Solution:
In an intermediate reaction TCA cycle,
succinyl Co-A is converted succinic acid and one GTP molecule is synthesised through substrate level
NCERT Exemplar Solutions for Class 11 Biology Chapter 14 Respiration in Plants 8
GTP formed in this reaction gives rise to ATP as follows
GTP+ADP GDP + ATP

Question 6.
In a way green plants and cyanobacteria have synthesised all the food on the earth. Comment.
Solution:
Cyanobacteria are unicellular prokaryotic organisms. Besides, some primitive cellular cell organelles, they have photosynthetic lamellae where photosynthetic pigments like chlorophyll-a c, phycocyanin and phycoerythrin, are present.

These coloured pigments confer typical blue green colour to the bacteria and enable them to manufacture food for themselves and aquatic animals.

Green plants are multicellular are organisms, which is capable of making food by using C02, H20 and light energy in specialized cell organelles called chlorcplast. So bacteria and green plants make food for living organisms on earth.

Question 7.
When a substrate is being metabolised, why does not all the energy that is produced get released in one step. It is released in multiple steps. What is the advantage of step-wise release?
Solution:
This is an enzyme that has dual nature. When C02 concentration is good enough in atmosphere. It acts as carboxylase. But if concentration of O2 increase, its nature changes and it binds with O2 and acts as oxygenase enzyme that forces CO2 to enter in C2 cycle that leads to photorespiration and loss of CO2.

Question 8.
Respiration requires 02. How did the first cells on the earth manage to survive in an atmosphere that lacked O2?
Solution:
Respiration always does not require O2. There are some organisms which respire in anaerobic condition i.e. in the absence of O2.
The first cells of earth i.g., chemosynthetic bacteria, which are the primitive organisms found earlier on earth. They obtain energy by breaking down inorganic molecules like H2S, NO2 – etc.
12H4S + 6 CO2 -> C6H12O6 + 6H2O + 12S

Question 9.
It is known that red muscle fibres in animals can work for longer periods of time continuously. How is this possible?
Solution:
There are basically two kinds of muscle fibres red muscles and white muscles
Red muscles work continuously for a longer time because
(i) These muscle fibres are dark red, due to the presence of red haemoprotein called myoglobin. It binds and stores oxygen as oxymyoglobin in the red fibres. Oxymyoglobin liberates oxygen for utilisation during muscle contraction.
(ii) Mitochondria are more in numbers, hence they work for long periods of time.
(iii) Red muscles possesses less sarcoplasmic reticulum.
(iv) They carry out considerable aerobic oxidation without accumulating much lactic acid. Thus without fatigue red muscle fibres can contract for a longer period.
(v) These muscle fibre have slow rate of contraction for long periods, e.g., extensor muscles of the human back.

Question 10.
RuBP carboxylase, PEPcase, pyruvate dehydrogenase, ATPase, cytochrome oxidase, hexokinase, lactate dehydrogenase, Select/ choose enzymes from the list above which are involved in
(a) Photosynthesis
(b) Respiration
(c) Both in photosynthesis and respiration
Solution:
RuBP Carboxylase is an enzyme that takes part in dark reaction of photosynthesis. It catalyses the fixing of C02 in C3 cycle
PEPcase an enzymes that takes part in photosynthesis of C4 plants. It catalyses the reaction of fixing of C02 to form first stable product oxaloacetate. 4 carbon compound.

Pyruvate dehydrogenase is an enzyme involved in aerobic respiration and catalyses the reaction ( of formation of acetyle Co-A from pyruvic acid.

It requires the participation of NAD and Co- enzyme-A.
NCERT Exemplar Solutions for Class 11 Biology Chapter 14 Respiration in Plants 9
ATPase is a part of both respiration and photosynthesis. Both these processes uses etc, associated proton pump and ATP synthase.
These all play a key part in the process is used by ETC pump hydrogen ions across a membrane.
The protons flows back through ATP synthase, driving the production of ATP.

Cytochrome Oxidase is involved in both respiration and photosynthesis. It acts as electron carrier in the electron transport chain. Hexokinase is an enzymes which is also involved in, respiration. In glycolysis, it catalyses the first reaction, i.e., formation of glucose -6- phosphate from glucose molecule.

It uses one ATP molecule which transfers P04 group to glucose molecules.
NCERT Exemplar Solutions for Class 11 Biology Chapter 14 Respiration in Plants 10
Lactate Dehydrogenase is an enzyme which is involved in anaerobic respiration in bacteria Lactobacillus.

Pyruvic acid formed at the end of glycolysis is converted to lactic acid by the help of homo- fermentative lactic acid bacteria. Hydrogen from NADH molecule is transferred to pyruvate is then transferred to pyruvate molecule lactic acid molecule leading to the formation of acid.
NCERT Exemplar Solutions for Class 11 Biology Chapter 14 Respiration in Plants 11

Question 11.
How does a tree trunk exchange gases with the environment although it lacks stomata?
Solution:
The old tree trunk is covered by dead woody tissue called cork. The epidermal layers of such tree get ruptured and outer cortical cells are loosely arranged. These structures are called as lenticels.
These are the sites of gaseous exchange and transpiration.
NCERT Exemplar Solutions for Class 11 Biology Chapter 14 Respiration in Plants 12

Question 12.
Mention the important series of events of aerobic respiration that occur in the matrix of the mitochondrion as well as one that take place in inner membrane of the mitochondrion.
Solution:
Kreb ’ s cycle occurs in the matrix of mitochondria. It is given in the following series of reactions
NCERT Exemplar Solutions for Class 11 Biology Chapter 14 Respiration in Plants 13
Electron transport chain is carried out in the inner mitochondria membrane
The inner mitochondrial membrane is specific about possessing proton (H+) and electron (e ) acceptors in a particular sequence called electron transport chain. It consists four enzyme complexes.
The electrons either follow the pathway of complexes I, III and IV or II, III and IV that depends that upon the substrates from Kreb’s cycle.
Following are the ways through which the transfer of electrons and hydrogen atoms takes place.
Complex I It consists of flavoproteins of NADH dehydrogenase (FPN), of which FMN is the prosthetic group. It is combined with the flavoprotein is non-heme iron of NADH dehydrogenase. This complex spans inner mitochondrial membrane and is also able to translocate protons across it form matrix side to outer side.
Complex II It consists of flavoprotein of succinate dehydrogenase, of which FAD is the prosthetic group. It is combined with the flavoprotein is non-heme iron of succinate dehydrogenase.
Between complexes II and in the mobile carrier coenzyme-Q (Co-Q) or ubiquinone (UQ) is present
Complex III It consists of cytochrome-/) and cytochrome-c that is associated with cytochrome-h is non-heme iron of complex III. Between complexes III and IV is the mobile carrier cytochrome-c.
Complex IV It consists of cytochrome-a and cytochrome-a3, and bound copper that are required for this complex reaction to occur. This cytochrome also called cytochrome oxidase. It is the only electron carrier in which the heme iron has a free ligand that can react directly with molecular oxygen.
NCERT Exemplar Solutions for Class 11 Biology Chapter 14 Respiration in Plants 14

  1. Thus, hydride ions are transferred from the substance to be oxidised to NAD+. From NAD+ the hydrogen atoms are transferred to FMN of flavor protein 1 (Fp’N). After FMN the hydrogen atom splits into an electron and a proton.
  2. In further stages transfer of e~s occur but there is no longer a transfer of hydrogens. The electron passes to co-enzyme-Q, and from co-enzyme Q to cytochromes- b, cp c, a and ay The proton is released free.
  3. As the hydrogen atom or electron passes down by F0 – F1 particle at the same time oxidation of one coenzyme and reduction occurs at another steps. Oxygen is able to diffuse inside the mitochondria.
  4. It is converted to anionic form 02-, combines with 2H+ and forms metabolic water reduced co-enzyme NADH + H+ that helps in pushing out three pairs of H+ to outer chamber while FADH2 sends two pairs of H+ to outer chamber.
  5. Oxidative phosphorylation is the synthesis of ATP molecules, with the help of energy liberated during oxidation of reduced co-enzyme (NADH2, FADH2) produced in respiration.
  6. The enzyme required for this synthesis is called ATP synthase present in inner mitochondria membrane.

The following figures shows this process
NCERT Exemplar Solutions for Class 11 Biology Chapter 14 Respiration in Plants 15

LONG ANSWER QUESTIONS

Question 1.
Oxygen is critical for aerobic respiration. Explain its role with respect to ETS.
Solution:
The oxidation of glucose starts with glycolysis in cytoplasm which followed by Krebs’ cycle and finally Electron transport Chain (ETC) in inner mitochondrial membrane. The end of ETC 02 is required.
Where, it acts as final hydrogen acceptor. 02 is responsible for removing electrons from the system. In the absence of oxygen, electrons could not be passed through the co-enzymes, intum proton pump will not be established and ATP will not be produced via oxidative phosphorylation. Thus oxygen plays an important role in aerobic respiration in mitochondrial matrix.

Question 2.
Enumerate the assumptions that we undertake in making the respiratory balance sheet. Are these assumptions valid for a living system? Compare fermentation and aerobic respiration in this context.
Solution:
The assumption that we undertake is making the respiratory balance sheet one as follows:
(i) Respiratory substrate is glucose
(ii) There is sequential pathway i.e., glycolysis in cytoplasm, TCA cycle in mitochondrial matrix and ETS in inner mitochondriol membrane.
(iii) NADH synthesised in glycolysis enters into ETC for phosphorylation.
(iv) None of the intermediates in the pathway are utilised to synthesise any other compound.
These assumptions are not valid for a living system because of following reasons:
(i) Glycolysis, TCA and ETC work simultaneously and do not take place one after the other.
(ii) ATP is uutilised when needed.
(iii) Rate of enzyme actions are controlled by multiple means.
Comparison between fermentation and aerobic respiration in this context is as follows:
(i) Fermentation is partial breakdown of glucose whereas aerobic respiration is complete breakdown of glucose.
(ii) Net gain of only 2 ATP in fermentation whereas in aerobic respiration 38 ATP is produced.

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

NCERT Exemplar Solutions for Class 11 Biology Chapter 3 Plant Kingdom

NCERT Exemplar Solutions for Class 11 Biology Chapter 3 Plant Kingdom

These Solutions are part of NCERT Exemplar Solutions for Class 11 Biology. Here we have given NCERT Exemplar Solutions for Class 11 Biology Chapter 3 Plant Kingdom.

VERY SHORT ANSWER QUESTIONS

Question 1.
Food is stored as floridean starch in Rhodophyceae. Mannitol is the reserve food material of which group of algae?
Solution:
Mannitol is a reserve food material of the members of Phaeophyceae (brown algae).

Question 2.
The plant body in higher plants is well differentiated and well developed. Roots are the organs used for the purpose of absorption. What is the equivalent of roots in the less developed lower plants?
Solution:
Root like structure called rhizoids are present instead of roots in less developed lower plants (bryophytes and pteridophytes). The plant tissue system in these is not differentiated into true leaf, stem and roots as it is found in higher plants (gymnosperm and angiosperm).

Question 3.
Most algal genera show haplontic life style. Name an alga which is
(a) Haplo diplontic
(b) Diplontic
Solution:
Haplo diplontic type of life cycle is exhibited by Ectocarpus, Polysiphonia and Kelps. The main plant body is saprophytic in Fucus and it shows diplontic type of life cycle.

Question 4.
In bryophytes male and female sex organs are called …………….. and …………..
Solution:
In bryophytes the male sex organ in antheridium and female sex organ is archegonium.
Antheridium produces flagellate antherozoids which are male gametes.
Archegonia is the female part which bears a single egg cell.

SHORT ANSWER QUESTIONS

Question 1.
Why are bryophytes called the amphibians of the plant kingdom? Amphibians can their in water as well as on terrestrial habitat.
Solution:
Bryophytes are a group of primitive plants having a dominant gametophytic plant body. These plants can live in soil but depend on water for movement of male gametes called antherozoids to reach the archegonium (female organ bearing egg cell) so that fertilisation can occur, so bryophytes are called the amphibians of the plant kingdom.

Question 2.
Heterospory, i.e., formation of two types of spores— microspores and megaspores is a characteristic feature in the life cycle of a few members of pteridophytes and all spermatophytes. Do you think heterospory has some evolutionary significance in plant kingdom?
Solution:

  1. The production of spores of two different sizes and sexes by the sporophytes of land plants is heterospory. Two types ofspores are produced by heterosporic plants.
  2. Small spores are microspores which germinate into the male gametophyte and large spores are macrospores which develop into the female gametophyte.
  3. Pteridophytes are intermediate between bryophytes and gymnosperms in the evolution of plants.
  4. All bryophytes are homosporous and all gymnosperms are heterosporous. This condition is advanced as sexual dimorphism results in cross fertilisation.
  5. Primitive or earlier pteridophytes are homosporous while later pteriodophytes are heterosporous e.g., Dryopteris, Pteris homosporous Selaginella, Sn/vrao-heterosporous.

Question 3.
Each plant group of plants has some phylogenetic significance in relation to evolution Cycas, one of the few living members of gymnosperms is called as the ‘relic of past’. Can you establish a phylogenetic relationship of Cycas with any other group of plants that justifies the above statement?
Solution:
Cycas is an evergreen plant which resembles palm. It has an unbranched stem and large compound leaves. It exhibits phylogenetic relationship with pteridophyte. Its evolutionary characters include thefollowing:
(i) Growth is redundant.
(ii) Shedding of seed while the embryo is still immature.
(iii) Minimal secondary growth and manoxylic wood.
(iv) Megasporophylls are leaf like.
(v) Sperms are flagellate even when pollen tube is present.
(vi) Leaf bases are persistent.
(vii) Ptysix is circinate.
(viii) Arrangement of microsporangia in well defined archegonia.

Question 4.
Comment on the life cycle and nature of fem prothallus.
Solution:
The life cycle of ferm (Dryopteris) clearly depicts the alternation of generation. The gametophytic stage (n) alternates with the sporophytic stage (2n) in the life cycle as shown in the figure.
NCERT Exemplar Solutions for Class 11 Biology Chapter 3 Plant Kingdom 1.1
The prothallus of the fem is a multicellular, free living, thalloid, haploid and autotrophic structure. It develops from the spores produced by sporophyte after reduction division.
These spore germinate within a germtube with an apical cell and forms a filament of 3-6 cells and one or two rhizoids at the base which later develops into gametophytic plant.
NCERT Exemplar Solutions for Class 11 Biology Chapter 3 Plant Kingdom 1.2

Question 5.
How are the male and female gametophytes of pteridophytes and gymnosperms different from each other?
Solution:
The male and female gametophytes of pteridophytes and gymnosperms different from each other as:
NCERT Exemplar Solutions for Class 11 Biology Chapter 3 Plant Kingdom 1.3

Question 6.
How are the male and female gametophytes of pteridophytes and gymnosperms different from each other?
Solution:

  1. Symbiosis is a type of interaction of two living organisms where both the associated partners derive some benefit from each other both co¬exist and flourish well.
  2. Mycorrhiza is a symbiotic association between fungus and the roots of vascular plants. The fungus colonizes the roots of the host either intra or inter cellularly. It helps in the nutrient absorption from soil for the plant.
  3. Mycorrhizal associations are present in conifers, i.e.,Pinus, Cedrus, Abies and Picea.
  4. Coralloid roots develop in Cycas. It is produced in clusters at the base of the stem and protrudes out on the ground.
  5. It is dichotomously branched and greenish in colour. It contains algal zone in cortex.
  6. This algal zone contains blue green algae like Anabaena and Nostoc which grow in symbiotic association with coralloid roots.

LONG ANSWER QUESTIONS

Question 1.
Explain why sexual reproduction in angiosperms is said to take place through double fertilisation and triple fusion. Also draw a labelled diagram of embryo sac to explain the phenomena.
Solution:

  • An angiospermic plants reproduces sexually by the formation of male and female gametes.
  • The male gamete is a pollen which contains two male nuclei and the female gamete is an egg cell produced in ovule (female gametophyte).
  • The pollen grains germinate on the stigma of a flower and the results in growth of pollen through the tissues of stigma and style and reach the egg apparatus.
  • The two male gametes are discharged within the embryo sac. One of the male gamete fuses with the egg cell to form a diploid zygote.
    This’fusion is known as fertilisation or syngamy. The second male gamete fuses with the diploid secondary nucleus and forms the triploid Primary Endosperm Nucleus (PEN). This fusion is known as triple fusion.
  • Because of the involvement of two fusion, this event in angiosperms is termed as double fertilisation. The zygote then develops into embryo and PEN develops into endosperm which provides nourishment to the developing embryo.
    NCERT Exemplar Solutions for Class 11 Biology Chapter 3 Plant Kingdom 1.4

Question 2.
Draw labelled diagrams of
(a) Female and male thallus of a liverwort.
(b) Gametophyte and sporophyte of Funaria.
(c) Alternation of generation in angiosperm.
Solution:
NCERT Exemplar Solutions for Class 11 Biology Chapter 3 Plant Kingdom 1.5
(a) Liverworts
(i) Male thallus of Marchantia polymorpha
(ii) Female thallus of Marchantia polymorpha
(b) Funaria
(gametophyte and sporophyte)
NCERT Exemplar Solutions for Class 11 Biology Chapter 3 Plant Kingdom 1.6

We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 3 Plant Kingdom, help you. If you have any query regarding NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 3 Plant Kingdom, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Solutions for Class 11 Biology Chapter 2 Biological Classification

NCERT Exemplar Solutions for Class 11 Biology Chapter 2 Biological Classification

These Solutions are part of NCERT Exemplar Solutions for Class 11 Biology. Here we have given NCERT Exemplar Solutions for Class 11 Biology Chapter 2 Biological Classification.

VERY SHORT ANSWER QUESTIONS

Question 1.
What is the principle underlying the use of cyanobacteria in agricultural fields for crop improvement?
Solution:
Cyanobacteria are abfs to fix atmospheric nitrogen and make it available to the plants and thus are used in agricultural crop improvement.
This improves crop yield and also reduces the cost of application of nitrogen fertilisers, e.g., Anabena and Nostoc.

Question 2.
How is the five kingdom classification advantageous over the two kingdom classification?
Solution:
The five kingdom classification, proposed by RH whittaker is based upon cell structure, body structure (unicellular, multicellular), nutrition (autotrophic, heterotrophic) reproduction and habitat either aquatic, terrestrial, or aerial and phylogenetic relationship.
It is thus more useful as compared to two kingdom system of classification which does not distinguish between prokaryotes and eukaryotes and no other kingdom except plant and animal are identified.
Polluted water bodies have high growth of algae due to the presence of nutrient. These nutrients increase the rapid growth of water plants, i.e.,

Question 3.
Polluted water bodies have usually very high abundance of plants like Nostoc and Oscillitoria. Give reasons.
Solution:
algae especially Nostoc and Oscillitoria, etc., and result in colonies. These colonies are generally surrounded by a gelatinous sheath and leads to the formation of blooms in water bodies.

Question 4.
Are chemosynthetic bacteria autotrophic or heteroterophic?
Solution:
Chemosynthetic bacteria are capable of oxidising various inorganic substances such as nitrates, nitrites and ammonia and use the released energy for production of ATP and thus they are autotrophs and not heterotrophs.

Question 5.
The common name of a pea is simpler than its botanical (scientific) name Pisum sativum why then is the simpler common name not used instead of the complex scientific/botanical name in biology?
Solution:
The common or vernacular names cause confusion regarding the identification of specific specimen as they change with the change in place whereas the scientific names are in latin and universally accepted and understood. Scientific names are thus preferred over the common vernacular names.

SHORT ANSWER QUESTIONS

Question 1.
Diatoms are also called as ‘pearls of ocean’, why? What is diatomaceous earth?
Solution:

  1. Diatoms and desmids are included under chrysophytes, kingdom-Protista.
  2. These are the main producers in the ocean. They prepare food for themselves and for the other life forms in the ocean as were a siliceous shell known as frustule cores the body of diatoms, this is the reason they are also called as ‘pearls of ocean. ‘
  3. Diatomaceous earth’ is the accumulation of large deposits of diatoms that forms a siliceous covering extending for several 100 metres formed in billions of years.
  4. The material obtained from these deposits is used in polishing and filtration of oils and syrups.

Question 2.
There is a myth that immediately after heavy rains in forest, mushrooms appear in large number and make a very large ring or circle, which may be several metres in diameter. These are called as ‘fairy rings’. Can you explain this myth of fairy rings in biological terms?
Discuss the mycilial structure in Agaricus and its soil borne nature.
Solution:
The fruiting bodies in Agaricus are known as basidiocarps. They form a concentric ring like structure from the mycelium present in the soil. These basidiocarps resemble button in shape and develop to form a ring like structure.
NCERT Exemplar Solutions for Class 11 Biology Chapter 2 Biological Classification 1.1
This fairy ring structure in Agaricus stimulate productivity in plants. This rings are the fruiting bodies of this fungus and the diameter of this fairy ring increases every year due to the spread of mycelium.

Question 3.
Neurospora an ascomycetes fungus has been •arsed as a biological tool to understand the mechanism of plant genetics much in the same way as Drosophila has been used to study animal genetics. What makes Neurospora so important as a genetic tool?
Solution:
Neurospora fungus can be grown easily under laboratory conditions by providing ‘minimal medium’ like inorganic salts, carbohydrates source and vitamin (biotin) and thus was selected to be a very good tool in genetics. The mutations can be also easily introduced in the fungal cells and meiotic division can be easily seen under X-ray treatment.

Question 4.
At a stage of their cycle, ascomycetes fungi produce the fruiting bodies like apothecium, perithecium or cleistothecium. How are these three types of fruiting bodies different from each other?
Discuss the type of fruiting bodies formed by ascomycetes fungus and differentiate accordingly on the basic of there structures.
Solution:
Ascomycetes consist of sporangial sac called ascus. Asci (singular-ascus) may occur freely or in aggregated form with dikaryotic mycelium to form the fruitification bodies called ascocarps.
The fruitification formed by asci include the following :
(i) Apothecium is cup like structure, e.g., Peziza.
NCERT Exemplar Solutions for Class 11 Biology Chapter 2 Biological Classification 1.2
(ii) Perithecium is flask shaped, e.g., Neurospora
NCERT Exemplar Solutions for Class 11 Biology Chapter 2 Biological Classification 1.3
(iii) Cleistothecium is closed with a slit, e.g., Penicilium
NCERT Exemplar Solutions for Class 11 Biology Chapter 2 Biological Classification 1.4

Question 5.
What obsrevable features in Trypanosoma would make you classify it under kingdom- Protista?
Discuss cell structure of Trypanosoma also discuss its different strain brief.
Solution:
Trypanosoma is included under flagellated protozoans on the basis of locomotary organ. It resembles Protisia in the following characters.
NCERT Exemplar Solutions for Class 11 Biology Chapter 2 Biological Classification 1.5
(i) It is unicellular.
(ii) It reproduces asexually i.e., by binary fission.
(iii) Possess centrally located nucieus and also contain an prominent nucleus endosome.
(iv) Reserve food material is in the form of granules.

LONG ANSWER QUESTIONS

Question 1.
Algae are known to reproduce asexually by variety of spores under different environmental conditions. Name these spores and the conditions under which they are produced.
Solution:
Asexual reproduction in algae is very common mean of reproduction. Algae and their spores exhibit significant diversity and vary greatly in their level of specialisation. Asexual reproduction by spores and their types include:
(i) Zoospores are mobile flagellated spores. In this protoplasm of each vegetative cell undergoes repeated longitudinal division either into 2 or 4, rarely 8 or 16 daughter protoplasts. Before the onset of division the parent cell loses its flagella.

  • Each daughter protoplast after the last series of division secretes a cell wall and a neuromotor apparatus that develops two flagella, eyespots and contractile vacuoles.
  • Each of the daughter cell thus formed resembles the parent cell in all aspects except the small size.
  • Under favourable conditions formation of zoospores is very common.
    NCERT Exemplar Solutions for Class 11 Biology Chapter 2 Biological Classification 1.6

(ii) Aplanospores are the non-motile spores. They are formed asexually within a cell, in which protoplast withdraws itself from the parent wall, rounds up and develops into aplanospores which germinate either directly or may divide to produce zoospores.
NCERT Exemplar Solutions for Class 11 Biology Chapter 2 Biological Classification 1.7

(iii) Hypnospores- In this, the protoplasm withdraws from the cell wall, rounds up and develops a thick wall under unfavourable conditions. These resting spores are called as hypnospores. Due to presence of haematochroma they are red in colour e.g., Vancheria, Ulothrix.
(iv) Akinetes are special vegetative thick walled cells present in the filaments which remain under dormant state and resume germination under favourable conditions. They can also withstand unfavourable condition as Spirogyra.
(v) Statospores are thick walled spores ‘ produced in diatoms.
(vi) Neutral spores are the protoplast, of vegetative cells directly functioning as spores (e.g., Ectocarpus).

Question 2.
Apart from chlorophyll, algae have several other pigments in their chloroptast. What pigments are found in blue, green, red and brown algae, that are responsible for their characteristic colours?
Solution:

  • All photosynthetic organisms comprise of one or more organic pigments that are capable of absorbing visible, radiations, which will initiate the photochemical reaction of photosynthesis.
  • The three major classes of pigments found in plants and algae are the chlorophylls, the carotenoids and the phycobilins.
  • Carotenoid and phycobilins are called accessory pigments since, the quanta absorbed by theese pigments can be transferred to chlorophyll.
  • The diversity of light harvesting pigments in alga implies that the common ancestor was primitive and that no close affinity exist between blue, green, red, brown, golden brown and green algae.
    NCERT Exemplar Solutions for Class 11 Biology Chapter 2 Biological Classification 1.8

 

Question 3.
Make a list of algae and fungi that have commercial value as source of food, chemicals, medicines and fodder.
Solution:
Algae
Around 70 species of marine ailgae are used for food, chemical and medicinal purpose.
NCERT Exemplar Solutions for Class 11 Biology Chapter 2 Biological Classification 1.9
Fungi
The role of fungi was established in early history. Since, the beginning of cultivation yeast have been used in making of bread and alcohol. The discovery of penicillin that marked the beginning of a new approach to microbial dis eases in human health.
Products of fungi in medicine, chemical and food include.
Around 70 species of marine aigae are used for food, chemical and medicinal purpose.
NCERT Exemplar Solutions for Class 11 Biology Chapter 2 Biological Classification 1.10

We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 2 Biological Classification, help you. If you have any query regarding NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 2 Biological Classification, drop a comment below and we will get back to you at the earliest.

NCERT Exemplar Solutions for Class 11 Biology Chapter 13 Photosynthesis

NCERT Exemplar Solutions for Class 11 Biology Chapter 13 Photosynthesis

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

VERY SHORT ANSWER QUESTIONS

Question 1.
Examine the figure.
NCERT Exemplar Solutions for Class 11 Biology Chapter 13 Photosynthesis 1
(b) Can these be passed on to the progeny? How?
(c) Name the metabolic process taking place in the places marked (A) and (B).
Solution:
(a) Figure shows the chloroplast, which is green
in colour and performs photosynthesis in plants. The structure is present in plant cell.
(b) Yes, chloroplast has the power of self replication because of presence of extra nuclear DNA. Hence, known as semi- autonomous organelle.
(c) The metabolic processes that occurs in the marked places are as follows.
A-It is the stroma of chloroplast, where dark reaction of photosynthesis takes place.
B-It is the structure of extra nuclear DNA that is responsible for replication of chloroplast, when it is required in the photo synthesising cells.

Question 2.
2H2O—>4H+ + O2 + 4e–
Based on the above equation, answer the following questions
(a) Where does this reaction take place in plants?
(b) What is the significance of this reaction?
Solution:
(a) This reaction takes place in reaction centre
PS II, that is located on the inner surface of thylakoid membrane. It is known as water splitting centre, where electrons are extracted from water and the reaction is catalysed by Mn+ and Cl– ions.
(b) Spliting of water is an important event in photosynthesis are
(i) It liberates molecular oxygen as byproduct of photosynthesis and is the significant source of oxygen in air, or is essential for all living beings on earth.
(ii) Hydrogen ions produced takes part in reducing NADP to NADPH. It is a strong reducing agent.
(iii) The electrons released are transferred from PS II to PS I through a series of electron carriers thus, creating a gradient for the ATP synthesis.

Question 3.
Cyanobacteria and some other photosynthetic bacteria don’t have chloroplasts. How do they conduct photosynthesis?
Solution:

  1. In Cyanobacteria complex lamellar system (thylakoids) are present instead of chloroplast. These thylakoids are functionally analogous to the plastids of eukaryotic cells.
  2. Pigment like chlorophyll-a, C-phycocyanin, C-phycoerythrin embedded in these lamellar system and they trap solar energy and perform photosynthesis. They perform oxygenic photosynthesis.
  3. Photosynthetic bacteria possess related pigments called bacterichlorophyll which are of different types (a,b,c,d,e,f and g). Groups that contain chlorophyll, perform photosynthesis, but do not evolve oxygen.
  4. Bacteriochlorophyll, perform photosynthesis but do not evolve oxygen.
  5. Bacteriochlorophylls are photoreceptors similar to chlorophylls except for the reduction of an additional pyrrole ring and other minor differences that shift their absorption maxima to near infrared, to wavelength as long as 1000 nm.
  6. Thus, they utilize light wavelengths not used by green plants or cyanobacteria.
  7. Bacteriopheophytin is a variant of bacteriochlorophyll that has two protons are present instead of magnesium ion at its centre.

Question 4.
(a) NADP reductase enzyme is located on ……..
(b) Breakdown of proton gradient leads to release of ……….. .
(b) ATP molecules
Solution:
(a) NADP reductase enzyme is located on the outer side of thylakoid membrane.
(b) ATP molecules

Question 5.
Can girdling experiments be done in monocots? If yes, how? If no, why note?
Solution:
The girdling experiment cannot be done in monocots. The monocots vascular bundles are scattered all over the width of stem, so we cannot get the specific band of the phloem tissue which we get in dicot.

Question 6.
3CO2, + 9ATP + 6NADPH + water –>Glyceraldehyde 3-phosphate + 9ADP + 6NADP+ + 8Pi.
Analyse the above reaction and answer the following questions
(a) How many molecules of ATP and NADPH are required to fix one molecule of CO2?
(b) Where in the chloroplast does this process occur?
Solution:
(a) 2 molecules of ATP for phosphorylation and
two molecules ofNADPH for reduction are required to fix one molecule of CO2 (b) The calvin cycle occurs in the stroma of the chloroplast.

Question 7.
Does moonlight support photosynthesis? Find out.
Solution:
Moonlight does not carry enough energy to excite chlorophyll molecules, i.e; reaction centre PSI and PSII, so light dependent reactions are not initiated. Thus, photosynthesis cannot occur in moonlight.

Question 8.
ATPase enzyme consists of two parts. What are those parts? How are they arranged in the thylakoid membrane? Conformational change occur in which part of the enzyme?
Solution:
ATP synthase enzyme consists of two parts:
(a) F1– head piece is a peripheral membrane protein complex and contain the site for synthesis of ATP from ADP + pi (inorganic phosphate).
(b) F0-integral membrane protein complex that form the channel through which proton cross the inner membrane.
The arrangement of F1 and F0 in thylakoid membrane is as follows.
F0– is a portion present within the thylakoid membrane.
F1 is a portion of ATP synthase enzyme present in the stroma of chloroplast.
The conformational change occurs in F1 portion of ATP synthase thus, it facilitates the ATP synthesis.

SHORT ANSWER QUESTIONS

Question 1.
Succulents are known to keep their stomata closed during the day to check transpiration. How do they meet their photosynthetic C02 requirements?
Solution:
Succulent plants grow in dry and xeric conditions so, to prevent water loss through transpiration the stomata remains closed during day time. So that the gaseous exchange does not take place.

Thus plants have developed the mechanism to fix C02 during night in the form of malic acid, which is a 4 carbon compound and are released during the day, inside the photosynthetic cells.

Question 2.
Chlorophyll-‘a’ is the primary pigment for the light reaction. What are accessory pigments? What is their role in photosynthesis?
Solution:
Accessory pigments are also photosynthetic pigments, like chlorophyll-/), xanthophyll and carotenoids which are not directly involved in emission of excited electron, but they help in harvesting solar radiation and pass it on to chlorophyll-o.

This pigment itself absorbs maximum radiation at blue and red region. So the chief pigment of photosynthesis is chlorophyll and others (i.e, chlorophyll-/; xanthophyll and carotenoid) are accessory pigments.

Question 3.
Do reactions of photosynthesis called, as ‘Dark Reaction’ need light? Explain.
Solution:
Dark reactions is a type of independent reactions. Through various biochemical reactions CO2 is reduced to produce C6H12O6 (glucose) which does not need light. But they depend on the products formed during light reactions, i.e., NADPH2 and ATP.

Question 4.
How are photosynthesis and respiration related to each other?
Solution:
Photosynthesis and respiration are related to each other as in both mechanisms, the plants gain energy.
In photosynthesis, plants gain energy from solar radiations whereas, in respiration, they break down glucose molecule to get energy in the form of ATP molecules.

The product of photosynthesis i.e., glucose (food) is utilised in respiration to yield energy in the form of ATP. While doing so, it release many other simple molecules (CO2 + H2O) that are utilised in photosynthesis to produce more sugar.

Question 5.
If a green plant is kept in dark with proper ventilation, can this plant carry out photosynthesis? Can anything be given as supplement to maintain its growth or survival?
Solution:
The plant in given conditions cannot carry out photosynthesis. Light is necessary for any green plant to make its own food. The plant should be watered properly for its survival.

Question 6.
Photosynthetic organisms occur at different depths in the ocean. Do they receive qualitatively and quantitatively the same light? How do they adapt to carry out photosynthesis under these conditions.
Solution:
Mostly algae are present at various depth in ocean. These show great variations in its photosynthetic pigment. These can absorb different wave lengths of light and could perform photosynthesis.
Green algae-chlorophyll-o, (absorbs red) and b(absorbs blue violet).
Brown algae-chlorophyll-o, c and fucoxanthin (absorbs yellow).
Rhodophyceae-chlorophyll-o, d and phyocoerythrin.

Question 7.
What conditions enable RuBisCO to function as an oxygenase? Explain the ensuing process.
(a) Can we artificially induce the property of nitrogen-fixation in a plant, leguminous or non leguminous?
(b) What kind of relationship is observed between mycorrhiza and pine trees?
(c) Is it necessary for a microbe to be in close association with a plant to provide mineral nutrition? Explain with the help of one example.
Solution:
This is an enzyme that has dual nature. When CO2 concentration is good enough in atmosphere. It acts as carboxylase. But if concentration of O2 increase, its nature changes and it binds with O2 and acts as oxygenase enzyme that forces CO2 to enter in C2 cycle that leads to photorespiration and loss of CO2.

Question 8.
Why does the rate of photosynthesis decrease at higher temperatures?
Solution:
Photosynthesis is an enzyme specific process. All enzymes works at an optimum temperature (/. e., 25-35°C). As temperature increases, enzyme gets denatured thus leading to fall in the rate of photosynthesis.

Question 9.
Explain how during light reaction of photosynthesis, ATP Synthesis is a chemiosmotic phenomenon.
Solution:

  1. In light reaction plants solar radiation is trapped by photosynthetic pigments, which converts light energy into chemical energy.
  2. Photophosphorylation is the main event of light reaction i.e., formation of ATP from ADP + Pi by using energy of excited electron movement through electron transport chain, that is present in thylakoid membrane.
  3. The movement of ions across a selectively permeable membrane, down the electrochemical/ proton gradient is known as chemiosmosis.
  4. Chemiosmosis hypothesis of ATP formation was first proposed by Mitchell (1961), according to ATP generated by enzyme ATP synthase via a membrane, proton pump and proton gradient.
  5. ATP synthase allows ions 02 protons to pass through membrane and proton pump.
  6. Which creates a high concentration of protons (H+) in the lumen and hence diffuses across the membrane to activate ATPase, releasing ATP molecules. One molecule of ATP is released for every two (H+) ions passing through ATPase.
    NCERT Exemplar Solutions for Class 11 Biology Chapter 13 Photosynthesis 2

Question 10.
In What kind of plants do you come across ‘Kranz anatomy’? To which conditions are those plants better adapted? How are these plants better adapted than the plants, which lack this anatomy?
Solution:

    1. Kranz anatomy how dimorphism in the chloroplast structure. It is found in C4 plants. The cells of leaves consists two types of chloroplast in them.
    2. Granal Chloroplast is found in the mesophyll cells of leaves.
    3. Chloroplast have well developed grana in them. These chloroplast fixes C02 effectively even if it is present in lower concentrations.
    4. PEP carboxylase fixes CO2 to form oxaloacetic acid (4 carbon compound).
    5. Agranal Chloroplast is found in bundle sheath cells of the leaves. C3 cycle occurs in these cells in the presence of RuBisCo enzyme.
    6. The C4 plants are well adapted to high O2 concentrations and high temperature.
      C4 plants can absorb CO2 even when CO2 concentration in much low thus C4 plants can perform high rate of photosynthesis even the stomata are closed or there is the shortage of water thus, they can conserve water.
    7. Since, PEP-carboxylase is insensitive to O2 thus excess O2 has inhibitory effect in C4 pathway and no photosynthesis occurs in C4 plant.
    8. Thus, C4 plants are better adapted to tropical and desert (hot acid habitats) areas than the plants, that lack kranz anatomy.

Question 11.
Tomatoes, carrots and chilies are red in colour due to the presence of one pigment. Name the pigment. Is it a photosynthetic pigment?
Solution:
The pigments are chromoplasts, these are fat soluble carotinoid pigments like carotenes and xanthophylls. These are called accessory pigments, they absorb light and transfer energy to Chlorophyll a.

Question 12.
Observe the diagram and answer the following.
NCERT Exemplar Solutions for Class 11 Biology Chapter 13 Photosynthesis 3
(a) Which group of plants exhibit these two types of cells?
(b) What is the first product of C4 cycle?
(c) Which enzyme is there in bundle sheath cells and mesophyll cells?
Solution:
(a) Monocot plants that belongs to Graminae/
Poaceae family, e.g., sugarcane, maize etc., possess these two types of cells, i.e., bundle sheath and mesophyll cell (in kranz anatomy).
(b) First product of C4 cycle is 4-carbon compound oxaloacetic acid.
(c) Mesophyll cells consists PEP carboxylase enzyme to fix atmospheric C02 to form 4-carbon compound oxalo acetic acid, whereas bundle sheath cells consists RuBP carboxylase that fixes C02 to form 3-carbon compound 3 PGA (3 phosphoglyceric acid).

LONG ANSWER QUESTIONS

Question 1.
In the figure given below, the back line (upper) indicates action spectrum for photosynthesis and the lighter line (lower) indicates the absorption spectrum of chlorophyll-a, answer the following
NCERT Exemplar Solutions for Class 11 Biology Chapter 13 Photosynthesis 4
(a) What does the action spectrum indicate? How can we plot an action spectrum? Explain with an example.
(b) How can we derive an absorption spectrum for any substances?
(c) If chlorophyll-a is responsible for light reaction of photosynthesis, why do the action spectrum and absorption spectrum not overlap?
Solution:
(a) Action spectrum depicts the relative rates of
photosynthesis at different wavelenghths of light. Action spectrum of photosynthesis can be plotted by measurement of oxygen evolution at different wavelength. Englemann (1882) by using a green algae plotted action spectrum.
(b) Absorption spectrum of a substance can be derived by calculating amount of energy of different wavelength of light absorbed.
(c) Chlorophyll a is responsible for light reaction of photosynthesis, but the action spectrum and absorption spectrum do not overlap because, though chlorophyll a is the main pigment responsible for the absorption of light, other thylakoids pigment like chlorophyll b, xanthophylls, carotenoids,

which are accessory pigments also absorb and transfer the energy to chlorophyll a. Indeed they not only enable a wider range of wavelength of incoming light to be utilized for photosynthesis but also protect chlorophyll from photo-oxidation.

Question 2.
What are the important events and end products of the light reaction?
Solution:
Following are important events of light reaction:
(i) Excitation of chlorophyll molecule to release a pair of electrons and use their energy in the formation of ATP from ADP + Pi. This process is known as photophosphorylation.
(ii) Splitting of water molecule
(a) 2H2O —» 4H+ + 4e– + 0+↑
(b) NADP + 2H+ ->NADPH2
End products of light reaction are NADPH and ATP. Reducing power is produced in the light reaction i.e., ATP and NADPH2 molecules that are used up in dark reaction and O2 is evolved as a by product by the splitting of water.

Question 3.
Why does not photorespiration take place in C4 plants?
Solution:

  • Photorespiration is associated with C3 cycle, where plant lose CO2 fixation due to the increase in concentrate ion of O2 change in the nature of activity of RuBP carboxylase-oxygenase.
  • While C4 plants have evolved a mechanism to avoid loss of CO2.
  • There is not a direct involvement of RuBP carboxylase-oxygenase as C3 cycle operates in bundle sheath cells, where both temperature and oxygen level low.
  • CO2 fixation occurs by another enzyme PEP carboxylase in mesophyll cells andoxaloacetate is formed, that is later converted into malic acid and transported to bundle sheath cells.
  • There, it liberates CO2, which is used in Calvin cycle, operating in bundle sheath cells of C4 plants.
  • We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 13 Photosynthesis, help you.
  • If you have any query regarding NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 13 Photosynthesis, drop a comment below and we will get back to you at the earliest.

 

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NCERT Exemplar Solutions for Class 11 Biology Chapter 1 The Living World

NCERT Exemplar Solutions for Class 11 Biology Chapter 1 The Living World

These Solutions are part of NCERT Exemplar Solutions for Class 11 Biology. Here we have given NCERT Exemplar Solutions for Class 11 Biology Chapter 1 The Living World.

VERY SHORT ANSWER QUESTIONS

Question 1.
Couplet in taxonomic key means……. .
Solution:
Couplet in taxonomic key is a pair of a contrasting characters used as tool foi” identification to aid in identification of a newly discovered organism.

Question 2.
What is a monograph?
Solution:
Monograph is a specialised work of documenting information on a particular taxon, /. e., family or genus or on aspect of subject, usually by a single author.
The main purpose of monograph is to present primary research and original work.

Question 3.
Amoeba multiplies by mitotic cell division. Is this phenomena growth or reproduction? Explain.
Solution:
Amoeba multiplies by simple mitotic cell divisions giving rise to two daughter Amoebae. Growth here is synchronous with reproduction, i.e., increases in number.

Question 4.
Define metabolism.
Solution:
Metabolism is the sum total of all biological reactions occurring in any living cell, which are controlled absolutely by enzymes. These reactions are of two types breaking down reactions (catabolism, e.g., cell ‘ respiration) and synthesing reactions (anabolism,
e.g., photosynthesis).

Question 5.
Which is the largest botanical garden in world? Name a few well known botanical gardens in India.
Solution:
A botanical garden is dedicated to collection, cultivation and display of wide range of plants labelled with their botanical names.
The largest botanical garden in the world is Royal Botanical Garden (in Kew, London). In India the famous well known botanical gardens are
(i) National Botanical Garden (NBG) Lucknow, UP.
(ii) Botanical Garden of FRI, Dehradun (UK).
(iii) Lloyd Botanical Garden, Daijeeling.
(iv) Indian Botanical Garden, Sibpur, Kolkata.

SHORT ANSWER QUESTIONS

Question 1.
A ball of snow when rolled over snow increases in mass, volume and size. Is this comparable to growth as seen in living organisms? Why?
Solution:

  1. Living organisms, grow, have metabolism and respond to external stimuli and reproduce as well. These characteristics are not shown by non-living objects.
  2. In biological terms growth is characteristic feature of all living organisms. It relates to increase in size by accumulation of protoplasm in the cell thus resulting in increase in the size of the cell.
  3. Increase in number of cell by cell division on other hand results in the size of individual organism.
  4. Snow is an inanimate (non-living) object, while rolling over, it gathers more snow on its surface thus, it increases in size by physical phenomenon but not by biological phenomenon.
  5. This growth cannot be thus compared to that seen in living organisms.

Question 2.
In a given habitat we have 20 plant species and 20 animal species. Should we call this as ‘diversity or biodiversity’? Justify your answer.
Solution:
There are existing 20 plant species and 20 animal species in the given habitat. They will exhibit the biodiversity in that given habitat because diversity refers to variation in a broad term and can be applied to any area whereas biodiversity is a degree of variation of life forms within a specified area.

Question 3.
International Code of Botanical Nomenclature (ICBN) has provided a code for classification of plants. Give hierarchy of units of classification, botanists follow while classifying plants and mention different ‘suffixes’ used for the units.
Solution:
ICBN has specified certain rules and principles in order to facilitate the study of plants by botanists. It helps in correct positioning of any organism newly discovered through the pressure of proper identification and nomenclature.
The taxonomic hierarchy, which is used while
classifying any plant given below
Kingdom-Plan tae
Division-phyta
Class-ae
Order-ales
Family-eae/ceae
Genus-First name of organism usually Latin word and written in italics.
Species-Second word of scientific name, also written in italics.

Question 4.
What are hormone receptors? What are the modes of their action ?A plant species shows several morphological variations in response to altitudinal gradient. When grown under similar conditions of growth, the morphological variations disappear and all the variants have common morphology. What are these variants called?
Solution:
These morphological variants are called bio types. It includes group of genetically similar plants showing similarity when grown in same environmental and geographical regions. The same environment provides them the similar abiotic factors like soil, pH, temperature, etc.
When growth in two different geographical regions, they are exposed to different abiotic characters which affects their growth, and development bringing changes in their external morphological features but, their genetic constitution remain same.

Question 5.
What is the difference between flora, fauna and vegetation? Eichhornia crassipes is called as an exotic species, while Rauwoljia serpentina is an endemic species in India. What do these terms exotic and endemic refer to?
Solution:
Following are the difference between flora, fauna and vegetation
NCERT Exemplar Solutions for Class 11 Biology Chapter 1 The Living World 1.1

Question 6.
Brinjal and potato belong to the same genus Solamim, but to two different species. What defines them as seperate species?
Solution:
Genus is a taxonomic rank used in bionorr.’al nomenclature comprising of a group of related species sharing few common characters.
Solanum is the largest genus of flowering plants which includes few economically important plants, e.g., potato, tomato, tobacco and brinjal. All these plants show some common morphological structures related to vegetative and reproductive similarities. So, they are are included in the same common genus Solanum.

Question 7.
The number and kinds of organism is not constant. How do you explain this statement? Change is law of nature.
Solution:
The number and kind of organisms is not constant, because of the following reasons new organism are added due to mechanisms of.
(i) sexual reproduction
(ii) mutation
(iii) evolution
The number of organisms get reduced due to
(i) environmental threats
(ii) loss of habitat
(iii) anthropogenic activities

LONG ANSWER QUESTIONS

Question 1.
Brassica campestris Linn
(a) Give the common name of the plant.
(b) What do the first two parts of the name denote?
(c) Why are they written in italics?
(d) What is the meaning of Linn written at the end of the name?
Solution:
Brassica campestris Linn
(a) The common name of Brassica compestris Linn is mustard.
(b) The first part of the name denotes the genetic name and the second part is the species name of the plant.
(c) According to ICBN, all scientific names are comprised of one genetic name followed by a species name, which require to be always written in italics. It is a rule of bionomial nomenclature.
(d) Linn means Linnaeus. He was the first to discover the plant. He identified, named and classified the plant, so the plant is named after him by adding suffix ‘Linn’, after the scientific name B. campestris.

Question 2.
What are taxonomical aids? Give the importance of herbaria and museums. How are Botanical gardens and Zoological parks useful in conserving biodiversity?
Solution:
The aids which help in identification, classification and naming of a newly discovered organisms (plant or animal) the taxonomic aids.
It could be in the form of a preserved document like herbaria or specimen kept at museums or scientific institutions. Other aids include written document like monography, taxonomic keys, couplets, etc.
A new organism found can be studied while comparing it with living plants and animals living in protected areas like Botanical gardens, Zoological parks, etc. Botanical gardens helps in conservation of plants by
(i) Plant species growing important local and keeping record of them.
(ii) Growing and maintaining species that rare are and endangered.
(iii) Supplying seeds for different aspects of botanical research.
Zoological parks contribute in conserving biodiversity by
(i) Providing natural environment and open space to animals.
(ii) Providing home to different native and exotic wild animals.
(iii) Rescue of endangered species.
(iv) Facilitating breeding animal and releasing them free. Thus, both botanical gardens and zoological parks play an important role in conservation of biodiversity.

We hope the NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 1 The Living World, help you. If you have any query regarding NCERT Exemplar Solutions for Class 11 Biology at Work Chapter 1 The Living World, drop a comment below and we will get back to you at the earliest.