CBSE class 11 Biology Solutions

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

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

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

VERY SHORT ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

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

SHORT ANSWER QUESTIONS

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

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

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

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

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

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

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

  1. Haemodialysis is a method that become a boon for thousands of uremic (accumulation of urea in blood) patients all over the world. Haemodialysing unit act as artificial kidney by removing urea from patients blood due to kidney failure.
  2. In this process blood is drained from artery and pumped into a dialysing unit after the addition of an anticoagulant named heparin.
  3. The unit contains a coiled cellophane tube which is surrounded by a dialysing fluid having the same composition as that of plasma except nitrogenous waste.
  4. The porous cellophane mfcmbrane of the tube allows the passage of molecules that is based on concentration gradient.
  5. Absence of nitrogenous water in dialysing fluid these substances freely move out thereby clearing the blood.
  6. In the end the cleared blood is pumped back to the body through a vein after the addition of anti¬heparin to it thereby completing the process.
    NCERT Exemplar Solutions for Class 11 Biology Chapter 19 Excretory Products and their Elimination 3

LONG ANSWER QUESTIONS

Question 1.
Explain the mechanism of formation of concentrated urine in mammals.
Solution:
Mammals have the ability to produce concentrated urine. The loop of Henle and vasa recta play important role in it which is discussed as follows:
(i) the proximity between the Henle’s loop and vasa recta as well as the counter current that is formed due to the fiow of filtrate in two limp’s of Henle’s loop in opposite direction and help in opposite direction and help in maintaining an increasing osmolality towards the inner medullary interstitium, i.e., from 300 mOsmoL-1 in the cortex to about 1200 mOsmol-1 in the inner medulla.
NCERT Exemplar Solutions for Class 11 Biology Chapter 19 Excretory Products and their Elimination 4
NCERT Exemplar Solutions for Class 11 Biology Chapter 19 Excretory Products and their Elimination 6
(ii) This gradient is caused mainly due to NaCl and urea ascending limb of Henle’s loop transports NaCl, that is exchanged with the descending limb of vasa recta.
(iii) Through the ascending portion of vasa recta. NaCl is returned to the interstitium.
(iv) Similarly, a small amount of urea enters the thin segment of the ascending limb of Henle’s loop, which is transported back to the intersitium by the collecting tubule.
(v) This special arrangement of Henle’s loop, and vasa recta, is called the counter current mechanism.
(vi) The rate of dissipation is reduced by the counter current exchange. This in turn, reduces the rate at which the current must pump Na+ to maintain any given gradient.
(vii) Presence of such interstitial garden helps in an easy passage of water from the collecting tubule thereby concentrating the filtrate (urine).
(viii) Human kidneys produces urine nearly four times concentrated than the initial filtrate formed.

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

  • Human kidney are reddish-brown, bean-shaped structures that is situated between the last thoracic and third lumbar vertebra, which is closer to the dorsal inner wall of the abdominal cavity.
  • Each kidney of an adult human measures 10-12 cm in length, 5-7 cm in width, 2-3 cm in thickness with an average weight of 120-170 gm
  • The kidney is covered by a fibrous connective tissue i.e., the renal capsula, that protects the kidney.
  • Internally, it consists of outer dark cortex and an inner light medulla, both containing nephron, nephron is the structural and functional units of kidney.
  • The median concave border of a kidney contains a notch called hilum, that functions as route entry and exit of blood vessels, nerves and ureter.
  • The renal cortex is granular in appearance that contains convoluted tubules that malpighian corpuscles. The renal medulla contains loop of henle, collecting ducts and tubules and ducts ofBertini.
  • Medulla is divided into conical masses, the medullary pyramids that further form papillae.
  • The papillae form calyces, which join to renal pelvis leading to ureter. Between the medullary pyramids, cortex extends into medulla and forms renal columns which are called as column of Bertini.
    NCERT Exemplar Solutions for Class 11 Biology Chapter 19 Excretory Products and their Elimination 5

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

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

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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 17 Breathing and Exchange of Gases

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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

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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

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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.