NCERT Biology class 11 chapter 8

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

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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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Question 11.
Name the following:

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

Solution:

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

Question 12.
Fill in the gaps:

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

Solution:

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

VERY SHORT ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

SHORT ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

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

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

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

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

Long ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

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

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

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

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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

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

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

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

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

Meninges consist of 3 layers which are as follows :

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

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

  • forebrain
  • midbrain
  • hindbrain

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

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

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

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

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

(b) Structure of eye :

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

(b) Forebrain :

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

(c) Midbrain:

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

(d) Hindbrain:

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

(e) Retina :

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

(f) Ear ossicles –

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

(g) Cochlea :

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

(h) Organ of corti:

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

(i) Synapse:

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

Electrical synapse:

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

Chemical synapse:

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

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

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

(b) Mechanism of vision

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

Mechanism of hearing

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

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

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

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

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

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

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

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

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

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

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

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

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

VERY SHORT ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

SHORT ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

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

Long ANSWER QUESTIONS

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

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

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

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

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

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

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

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

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

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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.