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

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

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

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

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

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

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

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

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

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

Question 7.
How is respiration regulated?
Solution:

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

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

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

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

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

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

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

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

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

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

VERY SHORT ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

SHORT ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

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

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

LONG ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Question 11.
Name the following:

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

Solution:

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

Question 12.
Fill in the gaps:

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

Solution:

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

VERY SHORT ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

SHORT ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

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

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

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

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

Long ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

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

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

NCERT Solutions for Class 11 Biology Chapter 4 Animal Kingdom

NCERT Solutions for Class 11 Biology Chapter 4 Animal Kingdom

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

Question 1.
What are the difficulties that you would face in the classification of animals if common fundamental features are not taken into account?
Solution:
There are millions of living organisms with a vast variety of shapes, sizes,s, and forms. They occur in a variety of habitats. It is difficult to identify and describe them at random, so they are classified into various categories. For identification and classification, different types of classification are required. There are fundamental features common to various individuals in relation to the arrangement of cells, body symmetry, nature of coelom, patterns of digestive, circulatory, or reproductive systems, etc. These features are used in the classification of animals.

Question 2.
If you are given a specimen, what are the steps that you would follow to classify it?
Solution:
Classification of specimen deals according to a systematic plan on the basis of their v similarities, differences and relationship.
We will take the following steps
(i) Level or grade of organisation:
There are different types of cell which are organised into functional units of progressively increasing complexity. Such as acellular, cellular, tissue, organ and organ system.
(ii) Pattern in organ system :
There are different organ system in which specific group of organs work together to do a specific function. Such as digestive organs in digestive system. Respiratory organs in respiratory system, etc. in animal body.
(iii) Symmetry:
In animals, 4-types of symmetry are seen. These are spherical, radial, bilateral, Asymmetrical species are classified according to symmetry.
(iv) Diploblastic and triploblastic organisation: According to number of germ layers which differentiate at the time of gastrulation in developing embryo. Species are classified as diploblastic i.e. two germ layer and Triploblastic i.e. three germ layers.
(v) Body cavity or coelom:
According to types of coelom, species are classified into acoelom, pseudocoelom and eucoelom animals.
(vi) Segmentation:
Species are classified according to segmentation. It is of three types i.e. pseudometamerism or false segmentation, internal and external segmentation.
(vii) Notochord:
On the basis of notochord, animals or species are divided into chordates and non-chordates.
We will follow the above steps in classification of animal, then fair idea of that animal can be derived.

Question 3.
How useful is the study of the nature of body cavity and coelom in the classification of animals?
Solution:
Coelom is the gap between gut and body wall. Coelom is the characteristic feature of complex or higher animals while lower animals like Platyhelminthes are acoelomate. The absence of coelom indicates that the animal is yet to develop a functional division of labour to carry out various activities. So the coelom characterises complexity of the animals and represents organic evolution.
There are three types of coelom
(i) Acoelom – It means absence of body cavity which is due to the failure of mesoderm to cavitate during embryogency, so there is no coelom, no peritonium. ex-porifera, colenterata, etc.
(ii) Pseudocoelom — It means presence of coelom that develops from the blastocoel but not lined by mesoderm, ex-nematodes, etc.
(iii) Eucoelom – It means hue coelom, which is lined by mesoderm resulting in tube within -tube design.
ex- higher invertebrates, chordates, etc.

Question 4.
Distinguish between intracellular and extracellular digestion?
Solution:

  1. Intracellular digestion takes place inside the cells by cellular enzymes, which are secreted by the surrounding cytoplasm into the food vacuole and the digestive products are then diffused in cytoplasm.
  2. It mainly occurs in unicellular organism and also is a less efficient method.
  3. Extracellular digestion occurs with the help of digestive enzymes poured into gastrovascular cavity by secretory cells and then the digestive products are diffused across the intestinal wall into various parts of the body.
  4. It mainly occurs in multicellular organisms and is more efficient.

Question 5.
What is the difference between direct and indirect development?
Solution:
In oviparous animals, the newly hatched young may resemble the adult. Such a development is called direct development. In some cases, the young hatched from eggs do not resemble the adult. These young are called larvae, nymphs, or naiad. They lead an independent life for some period and finally undergo important change to become adults. These changes from larvae to adults are called metamorphosis and such a development is called indirect development.

Question 6.
What are the peculiar features that you find in parasitic Platyhelminthes?
Solution:
Members of Platyhelminthes possess the following characteristics.

  • These are mostly endoparasites of animals including human.
  • Hook and suckers are present for attachment to host body.
  • They absorb nutrient from the host directly through their body surface.
  • These are free-living, parasitic forms. Tissue, organ, grade of body organization is seen.
  • The digestive tract is incomplete or absent.
  • Respiration is anaerobic.
  • Reproductive system of parasitic forms is highly developed with enormous power of reproduction.
  • Well defined excretory organs such as flame cells are present.

Question 7.
What are the reasons that you can think of for the arthropods to constitute the largest group of the animal kingdom?
Solution:
The phylum Arthropoda constitutes the largest group of animals which include many economically important insects. Over two-thirds of all named species on earth are arthropods. They have an organ-system level of, body organization. They are bilaterally symmetrical, triploblastic, segmented, and acoelomate animals. The body of arthropods is covered by a chitinous cuticle which forms the exoskeleton. The body segments are fused to form the head, thorax, and abdomen. They have jointed appendages. The appendages are variously modified to form antennae, mouthparts, pincers, or walking legs.

The digestive system is complete. Respiratory organs are gills, book-gills, book lungs, or tracheal systems. The circulatory system is open type. The nervous system is almost similar to that of the annelids. Sensory organs include antennae for perceiving odor, receptors for taste, eyes, statocysts or balance organs, and sound receptors. Excretion takes place through green glands or malpighian tubules. They are mostly dioecious, Reproduction is sexual. Fertilization is usually internal. They are mostly oviparous. Development may be direct or indirect, passing through many larval stages. The process of transformation of a larva into an adult is called metamorphosis.

Question 8.
Water vascular system is the characteristic of which group of the following:
(a) Porifera
(b) Ctenophora
(c) Echinodermata
(d) Chordata
Solution:
Echinodermata – Its vascular system with tube feet helps in locomotion. A perforated plate , madreporite, permits entry of water into ambulacral system which also help in food and gas transport system.

Question 9.
“All vertebrates are chordates but all chordates are not vertebrates”. Justify the statement.
Solution:
The members of subphylum Vertebrata possess notochord during the embryonic period, which is replaced by a cartilaginous or bony vertebral column in the adult. Thus, all vertebrates are chordates but all chordates are not vertebrates. Besides the three basic chordate characters, vertebrates have a ventral muscular heart with two, three, or four chambers, kidneys for excretion and osmoregulation, and paired appendages which may be fins or limbs.

Question 10.
How important is the presence of an air bladder in Pisces?
Solution:
The class Osteichthyes possesses Air bladder. It helps the organism in regulating buoyancy. If the air bladder were absent the animal had to swim constantly to avoid sinking.

Question 11.
What are the modifications that are observed in birds that help them fly?
Solution:
The birds are adopted for flying by reducing the weight and other modifications which are as follows :

  • The forelimb modified into wings to assist in flight.
  • Left ovary absent or reduced
  • Presence of pneumatic or hollow bones for making a lightweight skeleton.
  • The aerodynamic body helps in flying.
  • Excretion of urine and faeces occurs through single opening.

Question 12.
Could the number of eggs or young ones produced by an oviparous and viviparous mother be equal? Why?
Solution:
Normally the number of eggs produced by oviparous is greater than the number of young ones by viviparous. The development of young ones is oviparous takes place outside the mother whereas in the viviparous the development of the embryo takes place inside the uterus of the mother hence a viviparous mother can have very less number of embryos at a time in her uterus. Whereas oviparous mothers have no need to feed their young ones (embryos) until hatched. Hence, largest in number.

Question 13.
13. Segmentation in the body is first observed in which of the following:
(a) Platyhelminthes
(b) Aschelminthes
(c) Annelida
(d) Arthropoda
Solution:
(c) Annelida.

Question 14.
Match the following:
(a) Operculum      (i) Ctenophora
(b) Parapodia       (ii) Mollusca
(c) Scales          (iii) Porifera
(d) Comb plates    (iv) Reptilia
(e) Radula          (v) Annelida
(f) Hair            (vi) Cyclostomata and Chondrichthyes
(g) Choanocytes  (vii) Mammalia
(h) Gill slits      (viii) Osteichthyes
Solution:
(a) Operculum            (viii) Porifera
(b) Parapodia            (v) Annelida
(c) Scales               (iv) Reptilia
(d) Comb Plates    (i) Ctenophora
(e) Radula               (ii) Mollusca
(f) Hair                 (vii) Mammalia
(g) Choanocytes    (iii) Porifera
(h) Gill slits           (v) Cyclostomata and Chondrichthys

Question 15.
Prepare a list of some animals that are found
parasitic on human beings.
Solution:

  • Taenia (Tapeworm)
  • Fasciola (liver fluke)
  • Planaria
  • Echinococcus
  • Pheretima (Earthworm)
  • Hirudinaria (leech)

VERY SHORT ANSWER QUESTIONS

Question 1.
What is common name of Euspongia ?
Solution:
Bath sponge

Question 2.
Name the class which is characterised by pneumatic bones.
Solution:
Aves

Question 3.
Give the characteristic feature of Echinodermata.
Solution:
Water vascular system

Question 4.
Name the phyla which shows metamerism.
Solution:
Annelida, Arthropoda and Chordata.

Question 5.
Name two animal groups with incomplete digestive tract.
Solution:
Cnidariaandplatyhelminthes.

Question 6.
Name a fresh water sponge and coelenterate.
Solution:
Spongilla and Hydra

Question 7.
What is radula ?
Solution:
Rasping organ of molluscs.

Question 8.
Name two characteristic organs of chordates.
Solution:
Notochord and gill slits.

Question 9.
To which phyla Balanoglossus belong ?
Solution:
Hemichordata

Question 10.
Name two flightless birds.
Solution:
Ostrich, Kiwi

Question 11.
Give one example each of ectothermal and endothermal animals.
Solution:
Reptiles, mammals.

Question 12.
Give one example of cephalochordata.
Solution:
Branchiostoma (Amphioxus)

Question 13.
Why do lampreys and snakes lack girdles ?
Solution:
Due to lack of limbs.

Question 14.
What is the difference between epidermis of invertebrates and vertebrates ?
Solution:
Epidermis is a stratified epithelium in vertebrates and a simple epithelium in invertebrates.

Question 15.
Mention modification of coelom in echinoderms.
Solution:
A part of the echinoderm coelom is modified into a water-vascular system for help in locomotion.

Question 16.
What are setae ?
Solution:
Locomotive organs of phylum annelida.

Question 17.
What are endothermic animals ?
Solution:
Endothermic animals (birds and mammals) generate most of their body heat by metabolism.

Question 18.
Why is the Ornithorhynchus considered an 3 exceptional mammal ?
Solution:
Ornithorhynchus, though a mammal, lays egg and has a cloacal aperture.

SHORT ANSWER QUESTIONS

Question 1.
What are pneumatic bones ? Where do you find them?
Solution:
Bones having air spaces in them are called pneumatic bones. Birds have such types of bones to make body light for flying.

Question 2.
How do earthworm and leech differ with regard to their coelom ?
Solution:
Coelom is spacious in earthworm and greatly reduced in leech by formation of botryoidal tissue in it.

Question 3.
Cite unique features of nematodes.
Solution:
Syncytial epidermis, pseudocoel, non-muscular intestine and body wall musculature of longitudinal fibres only.

Question 4.
What is marsupium ?
Solution:
Marsupium is pouch on the females belly for rearing the young one in metatherian mammals. e.g. Kangaroo.

Question 5.
Why is coelentron called gastrovascular cavity ?
Solution:
It is a cavity in which both digestion and circulation occur.

Question 6.
What is the fate of notochord in higher chordate ?
Solution:
Notochord is replaced by vertebral column partly or fully in chordates.

Question 7.
State the two types of fishes. Give their examples.
Solution:
The two types of fishes are
(1) Chrondrichthyes example Trygon, electric ray
(2) Osteichthyes example Rohu, Sea horse

Question 8.
What is metamerism?
Solution:
It is a phenomenen found in organisms like earth worm that they have their bodies segmented and this parttem is called metameric segmentation.

Question 9.
Why are Urochordates called Tunicata?
Solution:
As the adult bodies of Urochordata is covered by a tunic like cover to these are called Tunicata.

Question 10.
Give an example of an organism showing.
(1) Radial symmetry
(2) Bilateral symmetry
Solution:
Radial symmetry: Star fish Bilateral symmetry: Octopus, men.

LONG ANSWER QUESTIONS

Question 1.
(i) What are different symmetry that exists in animal kingdom.
(ii) Distinguish between diploblastic and triploblastic
Solution:
(a) Asymmetrical symmetry
I. In many animals the body can not be divided in two equal halves on any plane.
II. Ex. Amoeba being irregular in shape, some sponges with various branching. Gastropod molluscs (snails, conch) have no symmetry due to torsion in embryonic stage.
(b) Radial symmetry
I. Any plane passing through a central axis of the body divides the organism into two identical halves.
II. Examples are Coelentrates, ctenophores and echinoderms.
(c) Bilateral symmetry
I. In most animals body can be divided into two equal halves on only one plane, hence body has left and right sides front and rear (back) sides; anterior and posterior ends; dorsal and ventral sides.
II. Their advance symmetry and such animals are most mobile.
Diploblastic animals: Animals in which cells are arranged in two embryonic layers, an external ectoderm and internal endoderm are called diploblastic animals.
Triploblastic animals: Animals in which developing embryo has the third germinal layers, mesoderm, in between the ectoderm and endoderm are called triploblastic animals.

Question 2.
Describe economic importance of the largest phylum.
Solution:
The largest phylum is Arhtropoda and its economic importance is as follows:

  • They are major agents for cross pollination. Insects, like butterflies and honey bee facilitate cross pollination.
  • Honey is an important food for human.
  • Many crustaceans, like lobsters and prawns are art of cuisine around the world.
  • The red dye cochineal, produced from a Central American species of insect, was economically important to the Aztecs and Mayans.
  • The blood of horseshoe crabs a clotting agent Limulus Amebocyte Lysate which is now used to test that antibiotics and kidney machines are free of dangerous bacteria, and to detect spinal meningitis and some cancers.
  • Maggots of housefly are used to treat those wounds which take time to heal because of absence of blood supply.
  • The relative simplicity of the arthropods ’ body plan, allowing them to move on a variety of surfaces both on land and in water, have made them useful as models for robotics.
  • They are carriers of many human parasite causing diseases like malaria, filaria and sleeping sickness.
  • Cockroaches are one of the major nuisance as they contaminate food in kitchens.
  • Termites are major causes of playing havoc with wooden furnitures.
  • Scorpions are known for their deadly sting, which can kill human and livestock.

Question 3.
What are the peculiar characters of Aschelminthes?
Solution:
(i) The body of aschelminthes is circular in cross-section, hence named round worms.
(ii) They may be free living, a quatic and terrestrial or parasitic in plants and animals.
(iii) They have organ-system level of body organisation.
(iv) They are bilaterally symmetrical, triploblastic andpseudocoelomate.
(v) Alimentary canal is complete with well developed muscular pharynx.
(vi) An excretory tube removes body waste from the body cavity through the excretory pore.
(vii) Sexes are separate (dioecious), i.e., males and females are distinct. Often females are longer than males.
(viii) Fertilisation is internal and development may be direct or indirect.
(ix) Examples. Ascaris (Round worm), Wuchereria (Filaria worm), Ancylostoma (Hookworm).

We hope the NCERT Solutions for Class 11 Biology at Work Chapter 4 Animal Kingdom, help you. If you have any query regarding NCERT 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 Solutions for Class 11 Biology Chapter 18 Body Fluids and Circulation

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

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

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

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

2. Agranulocytes are cells which lack granules

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

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

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

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

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

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

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

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

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

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

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

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

 

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

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

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

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

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

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

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

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

VERY SHORT ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Question 16.
Name two vital organs affected by high blood
1. pressure or hypertension.

Solution:
Brain, kidney

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

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

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

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

SHORT ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Long ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

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

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

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

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

Question 1.
Briefly describe the structure of the following:
(a) Brain
(b) Eye

(c) Ear
Solution:
Structure of brain:

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

Meninges consist of 3 layers which are as follows :

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

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

  • forebrain
  • midbrain
  • hindbrain

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

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

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

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

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

(b) Structure of eye :

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

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

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

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

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

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

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

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

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

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

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

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

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

Solution:
(a) Neuron
NCERT Solutions for Class 11 Biology Chapter 21 Neural control and co-ordination 6

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

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

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

Question 5.
Write short notes on the following
(a) Neural coordination
(b) Forebrain

(c) Midbrain
(d) Hindbrain

(e) Retina
(f) Ear ossicles

(g) Cochlea
(h) Organ of Corti

(i) Synapse
Solution:
(a) Neural co-ordination:

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

(b) Forebrain :

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

(c) Midbrain:

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

(d) Hindbrain:

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

(e) Retina :

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

(f) Ear ossicles –

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

(g) Cochlea :

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

(h) Organ of corti:

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

(i) Synapse:

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

Electrical synapse:

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

Chemical synapse:

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

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

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

(b) Mechanism of vision

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

Mechanism of hearing

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

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

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

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

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

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

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

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

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

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

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

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

Solution:
(c) Blindspot

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

VERY SHORT ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

SHORT ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

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

Long ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

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

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

Question 1.
What is meant by modification of root? What type of modification of root is found in the:
(a) Banyan tree
(b) Turnip
(c) Mangrove trees.
Solution:
Root in some plants change their shape and structure and become modified to perform functions other than absorption and conduction of water and minerals. This change is called the modification of root.
(a) Banyan Tree has roots called prop roots which are hanging structures that help in support.
(b) Turnip has tap roots which get swollen and store food.
(c) Mangrove trees are found in a marshy area. The roots get modified into pneumatic structures providing extra passage to allow additional oxygen to the plant.

Question 2.
Justify the following statements on the basis of external features:
1. Underground parts of a plant are not always roots.
2. Flower is a modified shoot.
Solution:
1. It is true that roots develop below the ground but there are exceptions. Potato is one such example. Here, in this case, the stem gets modified into a ‘tuber’ like structure for the storage of reserve food material. These tubers develop and grow under the ground.
This can be proved by the following:

  • The potato bears scale leaves (Leaves are found only in the stems)
  • They contain buds in the regions called eyes.
  • They contain nodes. It is justified from the above statements that underground parts of a plant are not always roots.

2. The flower is considered to be a modified shoot (it was suggested by Goethe 1760) because the internodes in a flower are highly condensed and the appendages such as sepals, petals, stamens, and carpels are generally large in number.

Question 3.
How is a pinnately compound leaf different from
a palmately compound leaf?
Solution:
In pinnately compound leaf, a number of leaflets are present on rachis (e.g., neem) whereas in palmately compound leaf, leaflets are attached at a common point i. e., at the tip of petiole e g., silk cotton.
NCERT Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 1

Question 4.
Explain with suitable examples the different types of phyllotaxy.
Solution:

  • In a pinnately compound leaf, a number of leaflets are present on a common axis. Example: Neem leaves.
  • In palmately compound leaf, a number of leaflets are attached at the common point.
    Example: Cotton leaves.

Question 5.
Define the following terms:

  1. Aestivation
  2. Placentation
  3. Actinomorphic
  4. Zygomorphic
  5. Superior ovary
  6. Perigynous flower
  7. Epipetalous stamen.

Solution:

  1. Aestivation: The arrangement of sepals or petals with respect to one another in the floral bud is called ‘aestivation’.
  2. Placentation: It is the arrangement of placentae in the ovary. It may be marginally axile, parietal, basal, free central, and superficial.
  3. When a flower can be divided into t\Vo equal radial halves by any radial plane passing through the center, it is said to be actinomorphic as in mustard, Datura, chili.
  4. When a flower can be divided into two similar halves only in one particular vertical plane, it is zygomorphic, as in pea, Gulmohar, beam, Cassia, etc.
  5. Superior ovary: In the hypogynous flower the gynoecium occupies the highest position while the other parts are situated below it. The ovary in such flowers is said to be superior e.g., mustard, china-rose, and brinjal.
  6. Perigynous flower: If gynoecium is situated in the center and other parts of the flower are located on the rim of the thalamus almost at the same level, it is called a perigynous flower.
  7. Epipetalous stamen: When stamens are attached to the petals, they are epipetalous as in brinjal or epiphyllous when attached to the perianth as in the. flower of the lily.

Question 6.
Differentiate between
(a) Racemose and cymose inflorescence
(b) Fibrous root and adventitious root
(c) Apocarpous and syncarpous ovary
Solution:
The main difference between racemose and cymose inflorescence are as following:
NCERT Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 2
NCERT Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 3

Question 7.
Draw the labelled diagram of the following:
(i) Gram seed
(ii) Y. S. of maize seed
Solution:
NCERT Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 4

Question 8.
Describe modifications of the stem with suitable examples.
Solution:

The stem may not always be typically like what it is expected to be. They are modified to perform different functions, (fig. 5.2)
(1) For storage food: Underground stems of potato, ginger, turmeric, Samarkand, colocasia are modified to store food in them. They also act as organs of presentation to tide over conditions unfavorable for growth.

(2) Support: Stem tendrils, which develop from axillary buds, are slender and spirally coiled and help the plant to climb such as in gourds (cucumber, pumpkins, watermelon) and grapevines.

(3) Protection: Axillary buds of stems may also get modified into woody, straight and pointed thorns. Thoms is found in many plants which protect them from browsing animals such as Citrus, Bougainvillaea.

(4) For Photosynthesis: Some plants of arid regions modify their stems into flattened (Opuntia), or fleshy cylindrical (Euphorbia) structures. They contain chlorophyll and carry out photosynthesis.

(5) Spread Underground: Stems of some plants such as grass and strawberry, etc. spread to new niches, and when older parts die new plants are formed.

(6) Vegetative propagation: In plants like mint and jasmine a slender lateral branch arises from the base of the main axis and after growing aerially for some time arch downwards to touch the ground. A lateral branch with short internodes and each node bearing a rosette of leaves and a tuft of roots are found in aquatic plants like Pistia and Eichhomia. In banana, pineapple, and Chrysanthemum, the lateral branches originate from the basal and underground portion of the main stem, grow horizontally beneath the soil, and then come out obliquely upward giving rise to leafy shoots.NCERT Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 5

Question 9.
Take one flower each of the families Fabaceae and Solanaceae and write its semi-technical description. Also draw their floral diagram after studying them.
Solution:
Fabaceae: This family was earlier called Papilionoideae, a subfamily of family Leguminosae. It is distributed all over the world.
Example: Pisam sativum
Semi technical description of Pisum sativum ^ are as follows :
Vegetative characters :
Habit: An annual herb.
Root: Nodulated tap root.
Stem: Climber, leaflet tendrils
Leaves: Alternate, pinnately compound or
simple; leaf base, pulvinate, stipulate, venation reticulate.
NCERT Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 6
Floral characters:
Inflorescence: Racemose
Flower: Bisexual, zygomorphic, complete,
irregular, hypogynous
Calyx: Sepal 5, gamosepalous, valvate aestivation.
Corolla: Petals 5, polypetalous, papilionaceous consisting of a posterior standard, two lateral wings, two anterior ones forming a keel. Thus, flower becomes zygomorphic, with descending imbricate aestivation i.e. vexillary aestivation.
Androecium: Stamen 10, diadelphous [1 + (9)], anther dithecous, introrse.
Gynoecium: Ovary superior, monocarpellary, unilocular with many ovules, marginal placentation.
Fruit: Legume
Seed: One to many, non-endospermic.
NCERT Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 7
Economic importance
Many plants belonging to the family are sources of pulses (gram, arhar, sem, moong, soyabean, edible oil (soybean, groundnut); fibres (sun hemp); fodder (Sesbania, Trifolium), ornamentals (lupin, sweet pea); medicine (muliathi).
Solanaceae:
It is large family, commonly called as the potato family. It is widely distributed in tropics, subtropics and temperate ones. Example: Datura
Semi technical description of Solatium nigrum.
Vegetative characters:
Habit: Annual herb
Stem: Erect, cylindrical, hairy, slightly fistular. Leaves: Alternate, simple, petiolate, ovate with acute apex, venation reticulate.

Floral characters:
Inflorescence: Solitary, axillary
Flower: Ebracteate, actinomorphic, hypogynous
Calyx: Sepals 5, gamosepalous, persistent, valvate aestivation.
Corolla: Petals 5, gamopetalous, valvate aestivation
Androecium: Stamen 5, epipetalous Gynoecium : Bicarpellary, syncarpous, ovary superior, bilocular but four celled by formation of false septum, placenta swollen with many ovules.
Fruit: Spinous capsule with septifragal dehiscence.
NCERT Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 8
NCERT Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 9

Question 10.
Describe the various types of placentations found in flowering plants.
Solution:
The arrangement of ovules within the ovary is known as placentation. The placentation of different types namely, marginal, axile, parietal, basal, central, and free central. In marginal placentation, the placenta forms a ridge along with the central structure of the ovary and the ovules are borne on this ridge forming two rows, as in pea. When the placenta is axial and the ovules are attached to it in a multilocular ovary, the placentation is said to be axile, as in china-rose, tomato, and lemon.

In parietal placentation, the ovules develop on the inner wall of the ovary or on the peripheral part. The ovary is one-chambered but it becomes two-chambered due to the formation of the false septum e.g. mustard and Argemone. When the ovules are borne on the central axis and septa are absent, as in Dianthus and Primrose; this type of placentation is called free central. In basal placentation, the placenta develops at the base of the ovary and a single ovule is attached to it, as in sunflower, marigold.

Question 11.
What is the flower? Describe the parts of a typical angiosperm flower.
Solution:
The placentation of flowering plants is the distribution of ovule-bearing cushions or placentae inside the ovary. It is of the following types.

  1. Marginal. A monocarpellary unilocular, ovary bears ovules longitudinally along the ventral suture in one or two alternate rows, e.g., Pea.
  2. Parietal. A syncarpous, unilocular ovary bears two or more placentae longitudinally along the wall, e.g., Fumaria, Viola. A false septum occurs between two parietal placentae in the Mustard. It makes the ovary falsely bilocular. In cucurbits, the three parietal placentae grow inwardly, meet in the centre and bend outwardly. The ovary becomes trilocular.
  3. Axile. A syncarpous bilocular to multilocular ovary bears ovules on the central axile column where the septa meet, e.g., China rose, Petunia, Asphodelus.
  4. Free central. Polycarpellary syncarpous but unilocular, ovary bears ovules around a central column which is not connected to the ovary wall.
  5. Basal. Unilocular ovary bears a single ovule from the basal region, e.g., Ranunculus, Sunflower.
  6. Apical. Unilocular ovary bears a single ovule from the apical region, e.g., Cannabis.
  7. Superficial. Ovules are borne along the inner surface of the ovary including the septa if present, e.g., Butomus (unilocular), Nymphaea (multilocular).

Question 12.
How do the various leaf modifications help plants?
Solution:

  • Tendrils: Leaves are converted into tendrils for climbing as in pear or into spines for defense as in cacti.
  • Bulb: The fleshy leaves of onion and garlic store food. In some plants such as Australian acacia, the leaves are small and short-lived.
  • The petioles these plants expand, become green, and synthesize food. Leaves of certain insectivorous plants such as pitcher plants, Venus flytrap are also modified leaves.

Question 13.
Define the term inflorescence. Explain the basis for the different types of inflorescence in flowering plants.
Solution:

  1. The arrangement and distribution of flowers on the floral axis are termed inflorescence.
  2. Depending on whether the apex gets converted into a flower or continues to grow, two major types of inflorescences are defined – racemose and cymose.
  3. In the racemose type of inflorescence, the main axis continues to grow, the flowers are borne laterally in acropetal succession, i. e., older flowers are at the base and younger flowers are at the top.
  4. In cymose type of inflorescence, the main axis terminates in a flower, hence is limited in growth.
  5. The flowers are borne in a basipetal order, i.e., younger flowers are near the base and older
    NCERT Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 10

Question 14.
Writethefloral formula of an actinomorphic, biserial, hypogynous flower with five united sepals, five free petals, five free stamens and two united carples with superior ovary and axile placenta tion.
Solution:
NCERT Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 11

Question 15.
Describe the arrangement of floral members in relation to their insertion on the thalamus. Calyx, corolla, androecium, and gynoecium.
Solution:
The flower is the reproductive unit in the angioSperms. It is meant for sexual reproduction.
A typical flower has four different kinds of whorls. arranged successively on the swollen end of the stalk or pedicel, called thalamus or receptacle. These are calyx, corolla, androecium, and gynoecium. Calyx and corolla are accessory organs, while androecium and gynoecium are reproductive organs. In some flowers like lily, the calyx and corolla are not distinct and are termed as perianth. When a flower has both androecium and gynoecium, it is bisexual. A flower having either only stamens or only carpels is unisexual.

In symmetry, the flower may be actinomorphic (radial) or zygomorphic (bilateral). When a flower can be divided into two equal radial halves in any radial plane passing through the center, it is said to be actinomorphic e.g. mustard, datura, chili. When it can be divided into two similar halves only in one particular vertical plane, it is zygomorphic e.g., pea, Gulmohar, bean, cassia. A flower is asymmetric (irregular) if it cannot be divided into two similar halves by any vertical plane passing through the center as in canna.

A flower may be trimerous, tetramerous, or pentamerous when the floral appendages are in multiples of 3,4 or 5, respectively. Flowers with bracts (reduced leaf found at the base of the medical) are called bracteate and those without bracts, ebracteate. Based on the position of calyx, corolla, and androecium in respect of the ovary and thalamus, the flowers are described as: hypogynous; perigynous, and epigynous. In the hypogynous flower, the gynoecium occupies the highest position while the other parts are situated below it. The ovary in such flowers is said to be superior e.g., mustard, china-rose, and brinjal.

If gynoecium is situated in the center and other parts of the flower are located on the rim of the thalamus almost at the same level, it is called perigynous. The ovary here is said to be half inferior e.g., plum, rose, peach. In epigynous flowers, the margin of the thalamus grows upward enclosing the ovary completely and getting fused with it, the other parts of the flower arise above the ovary. Hence, the ovary is said to be ‘inferior’ as in flowers of guava and cucumber, and the ray florets of sunflower.

VERY SHORT ANSWER QUESTIONS

Question 1.
What does take over the function of photosynthesis in Opuntia?
Solution:
Strm

Question 2.
Which plant part has transformed into the following different modifications
(i) tendril of pumpkin
(ii) thorn of Citrus.
Solution:
Stem(Axillary buds)

Question 3.
Name a cultivated plant in which neither fruits nor seeds are formed.
Solution:
Sugarcane

Question 4.
What term is given to the arrangement of leaves on the stem?
Solution:
Phyllotaxy

Question 5.
Give one example where the epigynous type of flower is present.
Solution:
Sunflower

Question 6.
Which type of placentation is present in Lathyrusl.
Solution:
Marginal

Question 7.
Why are potato and sweet potato called tubers?
Solution:
Potato and sweet potato are called tubers because they are irregularly shaped swollen stem that stores plenty of food.

Question 8.
What is phyllode? Give one example of it.
Solution:
Petiole and rachis modified into leaf-like structures are called phyllodes. e.g., Parkinsonia Australian Acacia.

Question 9.
Distinguish between alternate and whorled phyllotaxy.
Solution:
In alternate phyllotaxy, only one leaf is borne at each node whereas in whorled phyllotaxy, more than two leaves are borne at each node.

Question 10.
What is a tetradynamous condition of stamens?
Solution:
Two out of six stamens are short while the remaining four are long.

Question 11.
Describe the corolla of the family – Fabaceae
Solution:
Corolla of family Fabaceae is papilionaceous i. e., consisting of posterior standard or vexillum, two lateral wings and anterior petals fused along margin to form keel or carina.

Question 12.
What is a floral diagram?
Solution:
Floral diagram is an illustration of the relative and number of parts in each of the sets of organs comprising a flower.

Question 13.
Give any two reasons to justify that the onion bulb is a modified stem.
Solution:
It bears a large number of fibrous adventitious roots at its base. It bears several fleshy sheathing leaf bases and a terminal bud.

Question 14.
What are the main characters of the family Brassicaceae?
Solution:
Tetramerous flowers, six stamens, bicarpellary gynoecium, siliqua type fruit.

Question 15.
Name the food-yielding plants of Liliaceae.
Solution:
Allium cepa, A. sativum and Asparagus racemosus.

Question 16.
What is meant by maturation zone?
Solution:
The part of the root which is most active in water absorption is called the maturation zone.

Question 17.
What type of function is performed by the fleshy leaves of onion and garlic?
Solution:
The function of fleshy leaves of onion and garlic is storage

SHORT ANSWER QUESTIONS

Question 1.
What is a fruit? Describe the parts of a fruit.
Solution:
The fruit is a characteristic feature of flowering plants. It is a mature or ripened ovary, developed after fertilisation.
The fruit consists of a wall or pericarp and seeds. The pericarp may be dry or fleshy. When pericarp is thick and fleshy, it is differentiated into the outer epicarp, the middle mesocarp and the inner endocarp.
NCERT Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 12

Question 2.
Distinguish between prop roots and stilt roots
Solution:
The main differences between prop roots and stilt roots are as following
NCERT Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 13

Question 3.
What is pneumatophore? How do they help the plant? Name an example.
Solution:
Pneumatophores:

  • These are the roots that grow vertically upwards and come above the soil surface; they bear opening called pneumatophores, for the exchange of gases.
  • This feature is an adaptation for plants growing in marshy/swampy areas, where oxygen is deficient in the soil.
  • These roots help the plants to get oxygen from the air for respiration e.g., Rhizophora.

Question 4.
What is the function of the leaf?
Solution:
The leaf is a green, flattened outgrowth of the plant arising from the node of the stem and is specialized to perform the process of photosynthesis. Therefore, the leaf is also known as the kitchen or food factory of the plant.

Question 5.
What is the main function of the root system?
Solution:
The root, system generally grows beneath the ground into the soil, functions of the root system are as follows:

  • It provides great anchorage and support to the plant. Huge trees such as mango, redwood stand erect due to the root.
  • The root hair absorbs nutrients, water, and oxygen from the soil and conducts them to the upper parts of my plants.
  • some of the taproots are specially modified for the storage of carbohydrates and water.

Question 6.
What is the aleurone layer?
Solution:
The major part of the grain is occupied by large endosperm which is rich in starch. The endosperm has one to three-layered peripheral protein layer called the aleurone layer which separates the embryo with endosperm.

LONG ANSWER QUESTIONS

Question 1.
Describe the various parts of an angiosperm plant with a well labelled diagram.
Solution:
Parts of an angiosperm plant: The body of an angiosperm plant consists of the following parts:

  • root
  • stem
  • leaves
  • flowers
  • fruits
  • seeds

(i) Root: It is mostly underground and colourless. It is profusely branched. Its main function is to give support to the plant, fix the plant in the soil, and absorb water and food from the soil.

(ii) Stem: It is the aerial part. It bears fruit, leaves, branches, flowers, etc., and conducts water and minerals from the roots to the various parts of the plant body. The leaves on the stem arise from nodes. The region of the stem between two nodes is called the internode. Leaf axil is the angle formed by the base of the leaf and stem. At each leaf, axil is present a bud, which gives rise to a branch.

(iii) Leaf: These are green in colour. Leaves are termed food factories. The large portion of the leaf is termed as lamina while the stalk is called as petiole.

(iv) Flowers: These are variously formed and attractively coloured structures of the plant. They produce fruits and seeds.

(v) Fruits: The fruit is a characteristic feature of flowering plants. Generally, the fruit consists of a wall or pericarp and seeds. The pericarp may be dry or fleshy. When pericarp is thick and fleshy, it is differentiated into the outer epicarp, the middle mesocarp and the inner endocarp.

(vi) Seeds: The ovules after fertilisation, develop into seeds. A seed is made up of a seed coat and an embryo.
The embryo is made up of a radicle, an embryonal axis, and one (as in wheat, maize) or two cotyledons (as in gram and pea)
NCERT Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 14

Question 2.
Mention the diagnostic characters of the family Fabaceae and write the floral formula
Solution:
Diagnostic characters of family Fabaceae are:

  1. Presence of nodulated roots.
  2. Inflorescence racemose.
  3. Perigynous ovary.
  4. Flower zygomorphic and papilionaceous.
  5. Calyx 5, gamosepalous.
  6. Corolla 5, petals unequal and differentiated into standard, 2 lateral wings and two smallest anterior petals (keel).
  7. Androecium commonly diadelphous (1+9 or 5 + 5) or monoadelphous (10 or 9)
  8. Gynoecium monocarpellary, ovary is unilocular with marginal placentation.
  9. Fruit legume.
    NCERT Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 15

Question 3.
Write five differences between a dicot seed and a monocot seed.
Solution:
The differences between dicot seeds and monocot seeds are:
NCERT Solutions for Class 11 Biology Chapter 5 Morphology of Flowering Plants 16

Question 4.
Describe placentation in flower:
Solution:
The arrangement of the placenta in the ovary of the flower is known as placentation. Its main function is to transfer nutrients from maternal tissue to the growing embryo.

  • Marginal placentation: It is found in the monocarpellary ovary. In this, the ovary is unilocular and ovules are arranged along the margin of the unilocular ovary. Ex- Pea, Clitoria, etc.
  • Axile placentation: It is found in bi or multicarpellary and multilocular ovary. Ovules are arranged along the central axis of placenta and the number of chambers corresponds to the number of carpels. Ex- Lemon, Tomato, Hibiscus, Cotton, etc.
  • Parietal placentation: It is found in bi or multicarpellary ovary but unilocular. Ovules are arranged along periphery or the inner walls of ovary and the number of placenta corresponds to the carpels. Ex – Cucurbita, Argemone, etc.
  • Free central placentation: It is found in multicarpellary syncarpous ovary. Ovules are borne along the central axis. Which is not connected with the ovary wall by the septum.
    Ex- Dianthus Rome primrose, etc.
  • Basal placentation: It is found in monocarpellary but unilocular. In this placentation, the placenta develops at the base of the ovary and a-single ovule is attached to it. Ex – Sunflower, etc.

We hope the NCERT Solutions for Class 11 Biology at Work Chapter 5 Morphology of Flowering Plants, help you. If you have any query regarding NCERT 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 Solutions for Class 11 Biology Chapter 11 Transport in Plants

NCERT Solutions for Class 11 Biology Chapter 11 Transport in Plants

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

Question 1.
What are the factors affecting the rate of diffusion?
Solution:
Diffusion rates are affected by the gradient of concentration, the permeability of the membrane separating them, temperature, and pressure.

Question 2.
What are porins? What role do they play in diffusion?
Solution:
Porins are proteins that form huge pores in outer membranes of plastids, mitochondria and some bacteria, allowing molecules upto the size of small proteins to pass through. Thus porins facilitate diffusion.

Question 3.
Describe the role played by protein pumps during active transport in plants.
Solution:
Active transport uses energy to pump molecules against a concentration gradient. Active transport is carried out by membrane-proteins. Hence different proteins in the membrane play a major role in both active as well as passive transport. Pumps are proteins that use energy to carry substances across the cell membrane.

This pump can transport substances from a low concentration to a high concentration. The transport rate reaches a maximum when all the protein transporters are being used or are saturated. Like enzymes, the carrier protein is very specific in what it carries across the membrane. These proteins are sensitive to inhibitors that react with protein side chains.

Question 4.
Explain why pure water has maximum water potential?
Solution:
Water molecules possess kinetic energy. In liquid and gaseous form they are in random motion that is both rapid and constant. The greater the concentration of water in a system, the greater is its kinetic energy or water potential. Hence, it is obvious that pure water will have the greatest water potential.

Question 5.
Differentiate between the following:
(a) Diffusion and Osmosis
(b) Transpiration and Evaporation
(c) Osmotic Pressure and Osmotic Potential
(d) Imbibition and Diffusion
(e) Apoplast and Symplast pathways of movement of water in plants
(f) Guttation and Transpiration
Solution:
(a) The differences between diffusion and osmosis are as following:
NCERT Solutions for Class 11 Biology Chapter 11 Transport in Plants 1
(b) The differences between transpiration and evaporation are as following:
NCERT Solutions for Class 11 Biology Chapter 11 Transport in Plants 2
(c) The main differences between osmotic pressure and osmotic potential are as following:
NCERT Solutions for Class 11 Biology Chapter 11 Transport in Plants 3
(d) The main differences between imbibition and diffusion are as following:
NCERT Solutions for Class 11 Biology Chapter 11 Transport in Plants 4
(e) The main differences between apoplast and symplast pathways of movement of water in plants are as following :
NCERT Solutions for Class 11 Biology Chapter 11 Transport in Plants 5
(f) The main differences between guttation and transpiration are as following :
NCERT Solutions for Class 11 Biology Chapter 11 Transport in Plants 6

Question 6.
Briefly describe water potential. What are the factors affecting it?
Solution:
Water potential: Free energy per mole of water molecule is called water potential. It is represented by Greek letter ‘φ’ (Psi). Water potential of pure water is zero and addition of solute, decreases its free energy or water “‘potential.
Factors affecting water potential are as following:
(i) Matric potential (φm) causes by adsorbent or colloidal particles (always negative and almost negligible).
(ii) Solute potential (φs) caused by presence of solute particles (decreases the water potential).
(iii) Pressure potential (φp) caused by entry or exit of water (increases the water potential).
φw= φp + φs + φm

Question 7.
What happens when a pressure greater than the atmospheric pressure is applied to pure water or a solution?
Solution:
If a pressure greater than atmospheric pressure is applied to pure water or a solution, its water potential increases. It is equivalent to pumping water from one place to another.

Question 8.
(a) With the help of well-labelled diagrams, describe the process of plasmolysis in plants, giving appropriate examples.
(b) Explain what will happen to a plant cell if it is kept in a solution having higher water potential.
Solution:
(a) Plasmolysis is the shrinkage of the protoplast of a cell from its cell wall under the influence of a hypertonic solution (solution of higher concentration).
NCERT Solutions for Class 11 Biology Chapter 11 Transport in Plants 7
(b) The process of plasmolysis is usually reversible. When the cells are placed in hypotonic solution (solution with low concentration), water diffuses into the cell causing the cytoplasm to swell up. The swelling of shrunken protoplast is called as deplasmolysis.

Question 9.
How is the mycorrhizal association helpful in the absorption of water and mineral in plants?
Solution:
Some plants have additional structures associated with them that help in water (and mineral) absorption. A mycorrhiza is a symbiotic association of a fungus with a root system. The fungal filaments form a network around the young root or they penetrate the root cells. The hyphae have a very large surface area that absorbs mineral ions and water from the soil from a much larger volume of soil that perhaps a root cannot do.

The fungus provides minerals and water to the roots, in turn, the roots provide sugars and N-containing compounds to the mycorrhizae. Some plants have an obligate association with the mycorrhizae. For example, Pinus seeds cannot germinate and establish without the presence of mycorrhizae.

Question 10.
What role does root pressure play in water movement in plants?
Solution:
As various ions from the soil are actively transported into the vascular tissues of the roots, water flows and increases the pressure inside the xylem. This positive pressure is called root pressure and can be responsible for pushing up water to small heights in the stem.

Root pressure can only provide a modest push in the overall process of water transport. The greatest contribution is to re-establish the continuous chains of water molecules in the xylem which often break under the enormous tensions created by transpiration.

Question 11.
Describe the transpiration pull model of water transport in plants. What are the factors influencing transpiration? How is it useful to plants?
Solution:
Transpiration pull theory: The theory was put forward by Dixon and Jolly.
The main features of the theory are :
(i) There is a continuous column of water (present in the tracheary element) from the root through the stem and into the leaves.
(ii) Water molecules attached to one another, by cohesion force, adhesion force (between their walls and water molecules), and surface tension.
(iii) The water in the tracheary element would come under tension, due to transpiration also called transpiration pull.

Factor affecting transpiration: External factors e.g. atmospheric pressure, temperature, humidity, CO2, sunlight, wind velocity, etc. The thickness of cuticle, number, position and closing and opening of stomata, pH, and hormones are internal factors which affect transpiration. Significance of transpiration:

  • Ascent of sap
  • Removal of the excess water
  • Cooling effect
  • Distribution of minerals.
  • Transpiration supplies water for photosynthesis.

Question 12.
Discuss the factors responsible for the ascent of xylem sap in plants.
Solution:
The ascent of xylem sap is supported by the following factors.
(i) Root pressure (positive pressure that develops in the xylem sap of root).
(ii) Cohesion (adhesion of water molecules due to hydrogen bonding).
(iii) Adhesion (force between tracheary wall and water molecule).
(iv) Transpiration pull (tension develops due to transpiration) is the main cause of ascent of xylem sap.

Question 13.
What essential role does the root endodermis play during mineral absorption in plants?
Solution:
Water flowing through apoplast contains minerals useful to plants and also toxins. Endodermal cells have many transport proteins embedded in their plasma membrane; they let some solutes cross the membrane, but not others. Transport proteins of endodermal cells are control points, where a plant adjusts the quantity and types of solutes that reach the xylem.

Question 14.
Explain why xylem transport is unidirectional and phloem transport bidirectional.
Solution:

  1. The plant part which synthesise the food, i.e. leaf known as source and part that needs or stores the food called as sink.
  2. Food, primarily sucrose, is transported by vasculai tissue phloem, from a source to a sink. But depending on season or plants need, the source sink may be reversed.
  3. Sugar stored in roots may be mobilised to become a source of food source in early spring when buds of trees require energy for their growth and development and act as sink.
  4. Since the source – sink relationship is bidirection in phloem while in xylem it is unidirection.

Question 15.
Explain pressure flow hypothesis of translocation of sugars in plants.
Solution:
Mass flow or Pressure Flow Flypothesis : It was put forward by Munch (1927,1930).

  1. According to this hypothesis organic substance move from region of high osmotic pressure to low osmotic pressure due to turgor pressure.
  2. A high osmotic concentration is present in source e.g. mesophyll cells (due to photosynthesis).
  3. Sugar present in them passed into sieve tube therefore high osmotic concentration develops and it absorbs water from xylem and develop a high turgor pressure. It causes flow of sugar towards area of low turgor pressure or sink.
    NCERT Solutions for Class 11 Biology Chapter 11 Transport in Plants 8

Question 16.
What causes the opening and closing of guard cells of stomata during transpiration?
Solution:

  1. The opening and closing of stomata depend upon the turgid or flaccid state of the guard cells.
  2. The inner wall of guard cells (towards pore) is thick and outer wall (towards other epidermal cells) is thin.
  3. When the turgor pressure of the guard cells in increased the outer thinner wall of the guard cell is pushed out (towards the periphery) thus pulling the inner thicker wall leading to the opening of stomatal pore.
  4. When the guard cells are in a flaccid state the outer thinner wall of guard cells returns to original position (moves towards pore) due to which tension on the inner wall is released and stomatal aperture gets closed.
    NCERT Solutions for Class 11 Biology Chapter 11 Transport in Plants 9

VERY SHORT ANSWER QUESTIONS

Question 1.
What fraction of soil water is readily available to plants?
Solution:
Capillary water.

Question 2.
What is a hypertonic solution? (Oct. 87, M.Q.P., Apr. 2000)
Solution:
The external solution which has a concentration higher than that of the cell sap is called a Hypertonic solution.

Question 3.
What is the chemical potential of pure water at normal temperature and pressure?
Solution:
Zero.

Question 4.
What is transmembrane pathway?
Solution:
The movement of water through the cell membrane is called transmembrane pathway or symplast pathway.

Question 5.
What happens to the plant cell when it is plasmolyzed? (Apr. 95)
Solution:
becomes flaccid

Question 6.
Identify a type of molecular movement which is highly selective and requires special membrane proteins, but does not require energy.
Solution:
Facilitated diffusion.

Question 7.
How can you revert a freshly plasmolysed plant cell to its normal state?
Solution:
By placing it in aqueous solution.

Question 8.
What is meant by uniport?
Solution:
A molecule moves across the membrane independent of other molecule called as uniport.

Question 9.
What is the use of imbibition pressure to plants?
Solution:
Responsible for the seedling to emerge out of the soil.

Question 10.
What is a casparian strip?
Solution:
Casparian strip is band of suberin on the tangential and radial walls of endodermal cells.

Question 11.
Name the form in which cabohydrates are transported in plants and the tissue through which it is transported.
Solution:
Sucrose, phloem.

Question 12.
What is the pressure potential of a plasmolysed cell?
Solution:
Pressure potential of a plasmolysed cell is negative.

Question 13.
Give reason:
Excessive use of chemical fertilizers results in wilting of plants. (June 2009)
Solution:
Excessive use of fertilizers increases the solute concentration of the soil solution causing exosmosis resulting in the wilting of plants.

Question 14.
If water enters a cell what is the pressure exerted by its swollen protoplasts?
Solution:
The pressure exerted will be turgor pressure.

SHORT ANSWER QUESTIONS

Question 1.
List any four mechanisms that contribute to the ascent of sap in tall trees.
Solution:
Mechanisms that contribute to the ascent of sap in tall trees are :

  1. Adhesion
  2. Cohesion
  3. Transpirational pull
  4. Root pressure.

Question 2.
Mention two advantage of transpiration.
Solution:
Advantages of transpiration:
(1) It helps in mineral absorption and ascent of sap.
(2) Transpiration, by evaporating of water causes a cooling effect of plant body.

Question 3.
How do rise in temperature and wind velocity affect transpiration? Write any three adaptations shown by plants to reduce transpiration.
Solution:
Rise in temperature, wind velocity increases the rate of transpiration.
Three adaptations shown by plants to reduce transpiration are:
(a) leaves with thick cuticle
(b) sunken stomata
(c) modification of leaves into scales or spines and stem into phylloclades.

Question 4.
What is meant by apoplast pathway? Why does it occur in cortex and not in endodermis?
Solution:
The movement of water exclusively through cell wall without crossing cell membrane, is called apoplast pathway. It occurs in cortex due to the absence of casparian strip and presence of loosely packed cells. It does not occur in endodermis due to the presence of suberin in casparian strip.

Question 5.
What is the significance of osmosis?
Solution:
Significance of osmosis:
(1) Osmosis is important in the absorption of water by plants.
(2) Helps in cell to cell movement of water.
(3) Provides rigidity to the plant organs.
(4) Helps in opening and closing of stomata.

Question 6.
Describe the apoplastic and symplastic movements of water in plants.
Solution:
Apoplastic movement of water occurs exclusively through the non-living cell wall without crossing any membrane. Symplastic movement of water occurs from one cell to the other through plasmodesmata it passes across membrane and i.e. living components.

Question 7.
Sugar crystals do not dissolve easily in ice cold water. Explain.
Solution:
Rate of diffusion is increased when temperature increase. So, sugar crystal do not dissolve easily in ice cold water.etrads of homo logous chromosomes and crossing as the genetic material exchange takes place. Thus this stage is marked with genetic recombination.

Question 8.
What is imbibition pressure? What is the usefulness of imbibition pressure to seed germination?
Solution:
Imbibition pressure is the pressure developed in an adsorbent due to diffusion of water into it.
This pressure makes the seedlings to emerge above the ground during seed germination.

Question 9.
Differentiate between active and passive transport.
Solution:
The main difference between active transport and passive transport are :
NCERT Solutions for Class 11 Biology Chapter 11 Transport in Plants 10

Question 10.
Differentiate between diffusion and facilitated diffusion.
Solution:
The main difference between diffusion and facilitated diffusion are :
NCERT Solutions for Class 11 Biology Chapter 11 Transport in Plants 11

Question 11.
What is meant by source and sink in plants, with regard to translocation?
Solution:
Source : Source is the part of the plant which synthesises the food, they are the leaves.
Sink: Sink is the part of the plant that needs or stores the food like roots, tubers, any storage organ etc.

Question 12.
Differentiate between transportation of water and translocation of Food.
Solution:
NCERT Solutions for Class 11 Biology Chapter 11 Transport in Plants 12

LONG ANSWER QUESTIONS

Question 1.
Describe with the help of labelled diagrams, the closing and opening of stomata in plants.
Solution:

  • Stomata are minute openings formed in the epidermis of leaves, stem and in (some cases) even flowers.
  • Each stomata contains pore surrounded by two guard cells. The guard cells are joined at both ends but separate in the mid-region of their length.
  • Stomata are mostly present on the lower epidermis of the leaf. The open stomata account for diffusion of water vapour through them.
  • During the day, the cell- sap concentration becomes high due to accumulation of sugar as a result of photosynthesis.
  • This results into endosmosis and tl\e water is withdrawn into guard cells from the neighboring cells.
  • This makes the guard cells turgid and stomata open. If the availability of water is reduced, the guard cells lose their turgidity and they become flaccid by exosmosis of water from guard cells.
  • This leads to the closing of stomata and transpiration stops.
    NCERT Solutions for Class 11 Biology Chapter 11 Transport in Plants 13

Question 2.
Describe the root pressure theory and its demerits.
Solution:
Root pressure theory : The theory was put forward by Priestley (1916). Root pressure refers to positive hydrostatic pressure which is develops in xylem sap, when rate of water absorption is more than the rate of transpiration and as a result of which water is pushed up in tracheary element of root.
Demerits of root pressure concept :
(1) Magnitude of root pressure is maximum upto 2 atmosphere which can raise water upto 64ft. only. It can’t drive water in tall trees (300-400 feet) there is need of 20 atmosphere root pressure.
(2) No root pressure in conifers.
(3) Root pressure is low in summer when transpiration is high and high in spring when transpiration is low
(4) Root pressure is not seen in rapidly transpiring plants.

Question 3.
Discuss the factor affecting the water absorption.
Solution:
The factor affecting the water absorption are :
(1) Temperature : Optimum temperature for
absorption is 20-30°C. Absorption decrease at low temperature by decreasing permeability, retarding in growth of root and availability of water and by increasing viscosity of water.
(2) Concentration of soil solution: Absorption decreases with increase in concentration of soil solution, e.g. Halophytes (Rhizophora) are able to grow in swamps with high salt cont-entration because their cell sap concentration is also high.
(3) Soil aeration : Flooding of fields reduces soil air (02). Hence metabolic processes and absorption are also reduced.
(4) Increase in C02 : High C02 concentration lowers the rate of respiration and thus influence metabolic processes. These result low water absorption.
(5) Available soil water : Maximum absorption occurs when the available soil water is in the range of field capacity and permanent wilting coefficient.
(6) Root growth : Deep rooted plant absorb more water. Older roots are suberized and absorb slowly.

Question 4.
Differentiate the following
(a) Simple diffusion and active transport
(b) Turgid cell and flaccid cell
(c) Isotonic solution and hypotonic solution
Solution:
(a) The main difference between simple diffusion and active transport are :
NCERT Solutions for Class 11 Biology Chapter 11 Transport in Plants 14
(b) The main difference between the turgid cell and flaccid cell are :
NCERT Solutions for Class 11 Biology Chapter 11 Transport in Plants 15
(c) The main difference between isotonic solution and hypotonic solution are :
NCERT Solutions for Class 11 Biology Chapter 11 Transport in Plants 16

Question 5.
What are the properties that favours the upward movement water through xylem?
Solution:
The properties are:
(1) Cohesion : mutual attraction between water molecules
(2) Adhesion : attraction of water molecules to polar surfaces
(3)Surface Tension : water molecules are attracted to each other and give water high tensile strength i.e., an ability to resist a pulling force
(4) Capillarity : ability to rise in their tubes. These are also related to the transpiration pull. With this pull water comes to leaves for photosynthesis as well.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

VERY SHORT ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

SHORT ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Functions of ethephon are as follows:

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

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

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

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

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

LONG ANSWER QUESTIONS

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

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

The significance of vernalization is as follows:

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

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

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

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

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

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

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

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

We hope the NCERT Solutions for Class 11 Biology at Work Chapter 15 Plant Growth And Development, help you. If you have any query regarding NCERT Solutions for Class 11 Biology at Work Chapter 15 Plant Growth And Development, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules

NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules

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

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

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

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

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

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

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

Question 4.
Find and write down structures of 10 interesting small molecular weight biomolecules. Find if there is any industry which manufactures the compounds by isolation. Find out who are the buyers?
Solution:
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 2

NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 3
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 4
Fat is being manufactured by many companies in pharmaceuticals business as well as in food business. Vitamins come in many combination and are being used as supplementary medicines. Lactose is made by companies in manufacturing baby food. All of us are buyers of fat, protein and lactose.

Question 5.
Proteins have primary structures. If you are given a method to know which amino acid is at either of two termini (ends) of a protein, can you connect this information to purity or homogeneity of a protein?
Solution:
The primary structure of proteins is described as the type, number, and order of amino acids in the chain. A protein is imagined as a line whose left end represents the first and right end represents the last amino acid. But in fact, this is not so simple. Actually, the number of amino acids in between the two termini determines the purity or homogeneity of a protein.

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

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

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

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

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

Question 11.
Draw the structure of the amino acid, alanine.
Solution:
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 7

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

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

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

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

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

VERY SHORT ANSWER QUESTIONS

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

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

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

Question 4.
Name one heteropolysaccharide.
Solution:
Chitin

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

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

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

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

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

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

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

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

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

SHORT ANSWER QUESTIONS

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

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

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

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

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

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

Question 8.
Differentiate between essential amino acids and non-essential amino acids.
Solution:
NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 9

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

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

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

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

LONG ANSWER QUESTIONS

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

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

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

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

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

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

The relation between temperature and enzyme controlled reaction velocity

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

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

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

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

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

NCERT Solutions for Class 11 Biology Chapter 9 Biomolecules 16

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

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

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

NCERT Solutions for Class 11 Biology Chapter 20 Locomotion and Movement

NCERT Solutions for Class 11 Biology Chapter 20 Locomotion and Movement

These Solutions are part of NCERT Solutions for Class 11 Biology. Here we have given NCERT Solutions for Class 11 Biology Chapter 20 Locomotion and Movement.

Question 1.
Draw the diagram of a sarcomere of skeletal.
Solution:
NCERT Solutions for Class 11 Biology Chapter 20 Locomotion and Movement 1

Question 2.
Define sliding filament theory of muscle contraction.
Solution:
The mechanism of muscle contraction is best explained by the sliding filament theory which states that contraction of a muscle fiber takes place by sliding of the thin filaments over the thick filaments.

Question 3.
Describe the important steps in muscle contraction.
Solution:
Muscle contraction is initiated by a signal sent by the central nervous system (CNS) via a motor neuron. A motor neuron along with the muscle fibers connected to it constitutes a motor unit. The junction between a motor neuron and the sarcolemma of the muscle fiber is called the neuromuscular junction or motor-end plate. A neural signal reaching this junction releases a neurotransmitter (Acetylcholine) which generates an action potential in the sarcolemma. This spreads through the muscle fiber and causes the release of calcium ions into the sarcoplasm. An increase in Ca++ level leads to the binding of calcium with a subunit of troponin on actin filaments and thereby removes the masking of active sites for myosin.

Utilizing the energy from ATP hydrolysis, the myosin head now binds to the exposed active sites on actin to form a cross bridge. This pulls the attached actin filaments towards the center of the A-bonds. The Z-line attached to these actions is also pulled inwards thereby causing a shortening of the sarcomere i.e., contraction. It is clear from the above steps, that during shortening of the muscle i.e., contraction, the ‘I’ bonds are getting reduce whereas the A-bonds are retaining the length. The myosin, releasing the ADP and p1 goes back to its relaxed state.

A new ATP binds and the cross-bridge is broken. This causes the return of Z-lines back to their original position i.e., relaxation. The reaction time of the fibers can vary in different muscles. Repeated activation of the muscles can lead to the accumulation of lactic acid due to the anaerobic breakdown of glycogen in them, causing fatigue. Muscle contains a red-colored oxygen storing pigment called myoglobin. Myoglobin content is high in some of the muscles which gives a reddish appearance. Such muscles are called the Red muscles. These muscles also contain plenty of mitochondria which can utilize a large amount of oxygen stored in them for ATP production.

These muscles, therefore, can also be called aerobic muscles. On the other hand, some of the muscles possess a very little quantity of myoglobin and therefore, appear pale or whitish. These are the white fibers. The number of mitochondria is also few in them, but the amount of sarcoplasmic reticulum is high. They depend on the anaerobic process for energy.

Question 4.
Write true or false. If false, change the statement so that it is true.

  1. Actin is present in the thin filament.
  2. H-zone of striated muscle fibre represents both thick and thin filaments.
  3. The human skeleton has 206 bones.
  4. There are 11 pairs of ribs in man.
  5. The sternum is present on the ventral side of the body.

Solution:

  1. True
  2. False: The H-zone of striated muscle fibre represents only thick filaments.
  3. True
  4. False: There are 12 pairs of ribs in man.
  5. True

Question 5.
Write the differences between.
(a) Actin and Myosin
(b) Red and White muscles
(c) Pectoral and Pelvic girdle
Solution:
(a) Differences between actin and myosin are as following;
NCERT Solutions for Class 11 Biology Chapter 20 Locomotion and Movement 2
(b) The main difference between red muscles and white muscles are as following :
NCERT Solutions for Class 11 Biology Chapter 20 Locomotion and Movement 3
(c) The main difference between the pectoral girdle and pelvic girdle are as following:
NCERT Solutions for Class 11 Biology Chapter 20 Locomotion and Movement 4

Question 6.
Match Column I with Column II
Column I                                            Column II
(a) Smooth muscle                         (i) Myoglobin
(b) Tropomyosin                            (ii) Thin filament
(c) Red muscle                               (iii) Sutures
(d) Skull                                         (iv) Involuntary
Solution:
(a)– (iv)
(b)-(ii)
(c) -(i)
(d) – (iii)

Question 7.
What are the different types of movements exhibited by the cells of the human body?
Solution:
Cells of the human body exhibit three main types of movements, namely, amoeboid, ciliary and muscular. Some specialized cells in our body like macrophages and leucocytes in blood exhibit amoeboid movement. It is affected by pseudopodia formed by the streaming of protoplasm (as in Amoeba). Cytoskeletal elements like microfilaments are also involved in the amoeboid movement.

Ciliary movement occurs in most of our internal tubular organs which are lined by ciliated epithelium. The coordinated movements of cilia in the trachea help us in removing dust particles and some of the foreign substances inhaled along with the atmospheric air. Passage of ova through the female reproductive tract is also facilitated by the ciliary movement. Movement of our limbs, jaws, tongue, etc, requires muscular movement. The contractile property of muscles is effectively used for locomotion and other movements by human beings and the majority of multicellular organisms. Locomotion requires a perfect coordinated activity of muscular, skeletal, and neural systems.

Question 8.
How do you distinguish between a skeletal muscle and a cardiac muscle?
Solution:
The main difference between skeletal muscle and cardiac muscle are as following:
NCERT Solutions for Class 11 Biology Chapter 20 Locomotion and Movement 5

Question 9.
Name the type of joint between the following:
(a) Atlas/Axis
(b) Carpal/metacarpal of the thumb
(c) Between phalanges
(d) Femur/acetabulum
(e) Between cranial bones
(f) Between pubic bones in the pelvic girdle
Solution:
(a) Pivot joint
(b) Saddlejoint
(c) Gliding joint
(d) Ball and socket joint
(e) Fibrous joint
(f) Cartilagenousjoint

Question 10.
Fill in the blank spaces:
(a) All mammals (except a few) have ………….. cervical vertebra.
(b) The number of phalanges in each limb of a human is …………
(c) Thin filament of myofibril contains 2 ‘F’ actins and two other proteins namely ………. and ……………….
(d) In a muscle fibre Ca++ is stored in ……………
(e) …….. and ……….. pairs of ribs are called floating ribs.
(f) The human cranium is made of ……………. bones.
Solution:
(a) Seven
(b) Fourteen
(c) Troponin, tropomyosin
(d) Sarcoplasmic reticulum
(e) 11th, 12th
(f) Eight

VERY SHORT ANSWER QUESTIONS

Question 1.
What causes gouty arthritis in humans?
Solution:
Gouty arthritis (= Gout) is caused either due to excessive formation of uric acid or inability to excrete it.

Question 2.
How many tarsals are there in the ankle?
Solution:
Seven.

Question 3.
What are the bones of the heel called?
Solution:
Metatarsals.

Question 4.
How many types of movement shows by the human body?
Solution:
Three types of movements: amoeboid, ciliary, and muscular movement.

Question 5.
Name the lubricant which is responsible for the movable joint at the shoulder.
Solution:
Synovial fluid.

Question 6.
Give two disorders of skeleton and joints.
Solution:
Arthritis and Osteoporosis.

Question 7.
Mention two sites on all bodies where striated muscles are present.
Solution:
Limbs and tongue.

Question 8.
Name the two filaments which form the cross-bridges during muscle contraction?
Solution:
Actin and myosin.

Question 9.
Name the monomers of myosin.
Solution:
Meromyosins.

Question 10.
How many ribs are present in adult men?
Solution:
Twelve pairs.

Question 11.
Name the single U-shaped bone present at the base of the buccal cavity.
Solution:
Hyoid.

Question 12.
Name the location where Z-line is present in the sarcomere.
Solution:
Centre of I band.

Question 13.
What is the total number of bones present in the left pectoral girdle and the left arm respectively in a normal human?
Solution:
Left pectoral girdle – 2
Left-arm – 30

Question 14.
Name the kind of joint which permits movements in a single plane only.
Solution:
Hinge joint.

Question 15.
What are neuromuscular junctions?
Solution:
The junction between a motor neuron and the sarcolemma of a muscle fiber is known as the neuromuscular junction.

Question 16.
Why are the ribs described as bicephalic?
Solution:
Since each rib has two articulation surfaces on its dorsal end, it is described as bicephalic.

Question 17.
What is acromion?
Solution:
It is a flat expanded process projecting from the spine of the scapula; the clavicle articulates with it.

Question 18.
What is arthritis?
Solution:
Arthritis is painful stiffness and inflammation of joints.

Question 19.
What is sarcomere?
Solution:
A sarcomere is a structural unit within a microfibril bounded by Z lines that contain actin and myosin.

Question 20.
Which muscle protein acts as ATPase?
Solution:
Myosin.

SHORT ANSWER QUESTIONS

Question 1.
What causes osteoporosis?
Solution:
Osteoporosis is a disease in which bone loses minerals and fibres from its matrix. There are more chances of fractures. Decreased level of estrogen is a common cause.

Question 2.
Why a red muscle fibre can work for a prolonged period, while a white muscle fiber suffers from fatigue soon?
Solution:
Red muscle fibres contain myoglobin that stores oxygen in the form of oxymyoglobin.
Since, there is a continuous supply of oxygen; for the oxidation of food materials to release energy, the red muscle fibers retain energy and do not become fatigued and work for long periods whereas white muscle fibres lack myoglobin. At times they carry out anaerobic respiration and become fatigued.

Question 3.
Name the major components of the appendicular skeleton.
Solution:
It is situated at the lateral sides which actually extend outwards from the principal axis. It consists of pectoral and pelvic girdles and bones of arms and legs.

Question 4.
What is the sarcoplasmic reticulum? What is its function?
Solution:
The endoplasmic reticulum of muscle fiber is called the sarcoplasmic reticulum which acts as a storehouse of calcium ions.

Question 5.
Differentiate between A and I bands.
Solution:
The main differences between A-band and I-band are as following :
NCERT Solutions for Class 11 Biology Chapter 20 Locomotion and Movement 6

Question 6.
Draw the labeled diagram of the pectoral girdle and upper arm.
Solution:
NCERT Solutions for Class 11 Biology Chapter 20 Locomotion and Movement 7

Question 7.
Differentiate between bone and cartilage.
Solution:
The main differences between bone and cartilage are as following:
NCERT Solutions for Class 11 Biology Chapter 20 Locomotion and Movement 8

Question 8.
Describe the vertebro-chondral ribs.
Solution:
Vertejno-chondral ribs

  • 8th, 9th and 10th pairs of ribs are called vertebro-chondral (false) ribs.
  • They remain attached dorsally to the respective thoracic vertebrae and vertrally to the sternum through the seventh rib by hyaline cartilage.

Question 9.
How muscular contraction is triggered?
Solution:
It is triggered by nerve releasing a neurotransmitter, which in turn triggeres the sarcoplasmic reticulum to release calcium ions into muscle interior. Where they bind to troponin, thus causing tropomyosin to shift from the face of the actin filament to which myosin heads need to produce a contraction.

Long ANSWER QUESTIONS

Question 1.
Draw a well diagram of human skull.
Solution:

NCERT Solutions for Class 11 Biology Chapter 20 Locomotion and Movement 9

Question 2.
Write short notes on:
(a) Muscular dystrophy
(b) Tetany
(c) Myasthenia gravis
Solution:
Muscular dystrophy
The abnormality of muscles associated with dysfunction and ultimately deterioration is called muscular dystrophy. It is a genetic disorder caused by lack of dystrophin.
Myasthenia gravis: It is an auto-immune disorder that affecting neuromuscular junction and leads to fatigue, weakening, and paralysis of skeletal muscles. ‘
Tetany: The rapid spasm and (wild contractions) is called tetany. In this case, the muscles do not get a chance to relax at all. It is caused due to deficiency of parathyroid hormone and thus lowering Ca++ in blood fluid.

Question 3.
Give differences between movable and immovable joints?
Solution:
Differences between movable and immovable joint are tabulated below:
NCERT Solutions for Class 11 Biology Chapter 20 Locomotion and Movement 10

Question 4.
Describe the structure of the rib cage of a human.
Solution:
Rib Cage: There are 12 pairs of ribs. Each rib is a thin flat bone connected dorsally to the vertebral column and ventrally to the sternum. It has two articulation surfaces on its dorsal end and is hence called bicephalic. First, seven pairs of ribs are called true ribs.

Dorsally, they are attached to the thoracic vertebrae and ventrally connected to the sternum with the help of hyaline cartilage. The 8th, 9th, and 10th pairs of ribs do not articulate directly with the sternum but join the seventh rib with the help of hyaline cartilage. These are called vertebrochondral (false) ribs. The last 2 pairs (11th and 12th) of ribs are not connected ventrally and are, therefore, called floating ribs. Thoracic vertebrae, ribs, and sternum together form the rib cage.

NCERT Solutions for Class 11 Biology Chapter 20 Locomotion and Movement 11

We hope the NCERT Solutions for Class 11 Biology at Work Chapter 20 Locomotion and Movement, help you. If you have any query regarding NCERT Solutions for Class 11 Biology at Work Chapter 20 Locomotion and Movement, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption

NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption

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

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

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

Question 3.
Answer briefly:
(a) Why are villi present in the intestine and not in the stomach?
(b) How does pepsinogen change into its active form?
(c) What are the basic layers of the wall of the alimentary canal?
(d) How does bile help in the digestion of fats
Solution:
(a) Villi increases surface area for absorption and maximum absorption takes place in the intestine.
(b) Coming in contact with hydrochloric acid in stomach proenzyme pepsinogen convert to its active form pepsin, the proteolytic enzyme of the stomach.
(c) There are four basic layers in the wall of alimentary canal i.e. serosa, muscularis, submucosa and mucosa.
(d) Bile helps in the emulsification of fats i.e. breakdown the fats into very small micelles. Bile also activates lipases.

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

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

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

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

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

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

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

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

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

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

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

Question 12.
Discuss the main steps in the digestion of proteins as the food passes through different parts of the alimentary canal.
Solution:
(a) In the stomach the proenzyme pepsinogen, on exposure to HC1 converted into active pepsin that converts proteins into proteases and peptones.
NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 3
(b) Proteins, proteases and peptides in the chyme reaching the intestine are acted upon by proteolytic enzymes of pancreatic juice and converted to dipeptides.
NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 4
(c) The enzymes in succuss entericus act on the end product to form amino acids.
NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 5

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

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

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

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

VERY SHORT ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Question 22.
Name the enzyme involved in the breakdown of nucleotides into sugars and bases.
Solution:
NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 6

SHORT ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

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

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

LONG ANSWER QUESTIONS

Question 1.
What is meant by vernalization? Explain the significance of vernalization.
Solution:
NCERT Solutions for Class 11 Biology Chapter 16 Digestion and Absorption 8

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

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

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

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

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

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

(ii) Constipation

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

(iii) Diarrhoea:

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

(iv) Vomiting:

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

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

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