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

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

Edited By Irshad Anwar | Updated on Sep 08, 2023 09:32 PM IST

Excretory Products and Their Elimination | NCERT Solutions

NCERT Solutions for Class 11 Biology Chapter 19 Excretory Products and Their Elimination: In this chapter of the Human Physiology unit, you will learn the mechanisms of elimination of these substances, with special emphasis on common nitrogenous wastes. You will learn that ammonia, along with urea and uric acid, are the major forms of nitrogenous waste excreted by animals. NCERT Solutions have been updated to the latest CBSE Syllabus. A thorough strategy, backed by extensive and in-depth learning of concepts, is your best bet for acing the board exams. This is why students can rely on NCERT Solutions for Class 11.

NCERT Solutions For Class 11 Biology Excretory Products and their Elimination

CBSE NCERT Solutions for Class 11 Biology Chapter 19 - Excretory Products and Their Elimination carry an impressive explanation of all the questions that are based on the excretory products and how they got eliminated from the body. And to understand that, you need to learn from Chapter 19 Biology Class 11 NCERT Solutions that ammonia is the most toxic form and requires a large amount of water for its elimination. Whereas uric acid, being the least toxic, can be removed with a minimum loss of water. According to Ch 19 Bio Class 11, this process of excreting ammonia from the body is called ammonotelism. It is very interesting to know in Biology Class 11 Chapter 19 that many bony fish, aquatic amphibians, and aquatic insects are ammonotelic in nature. In order to excel in the CBSE exams, it is advised that students spend a few hours per day working through the Class 11 Biology NCERT Solutions.

In CBSE Solutions for NCERT Class 11 Biology Chapter 19 Excretory Products and Their Elimination, you will get questions and answers related to excretory products and their elimination in different organisms (including humans). Students can assess their knowledge by going through the Excretory Products and Their Elimination Class 11 Questions and Answers provided here. If you are looking for an answer from any other chapter, even from any other class, then go with NCERT Solutions, there you will get all the answers to NCERT easily.

Biology Class 11: In Excretory Products and Their Elimination Class 11, you will study that urine formation involves three main processes: filtration, reabsorption, and secretion. The full form of GRF is glomerular filtration rate. As per the Class 11 Biology NCERT Solutions, filtration is a non-selective process performed by the glomerulus using glomerular capillary blood pressure. you have to study in Ch 19 Bio Class 11 to know more about reabsorption and secretion, as related questions are there in Class 11 Excretory Products and Their Elimination.

After going through the Solutions for Excretory Products and Their Elimination Class 11, you must be able to understand all the Class 11 Biology Chapter 19 Question Answer given below:

NCERT Solutions For Class 11 Biology Excretory Products and their Elimination - PDF Download

According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 16.

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NCERT Solutions for Class 11 Biology Chapter 19 - Excretory Products and Their Elimination (Solved Exercise)

The complete exercise and in-text Excretory Products and Their Elimination NCERT Solutions are given below:

Excretory Products and Their Elimination Class 11 Questions and Answers

Q1. Define Glomerular Filtration Rate (GFR)

Answer:

The Glomerular filtration rate- It refers to the amount of glomerular filtrate formed in all the nephrons of both the kidneys per minute. In a healthy individual, it is about 125 ml/minute. The glomerular filtrate contains glucose, amino acids, sodium, potassium, urea, uric acid, ketone bodies, and large amounts of water.

Q2. Explain the autoregulatory mechanism of GFR.

Answer:

The kidneys have built-in mechanisms for the regulation of glomerular filtration rate. One such mechanism is the Juxtaglomerular apparatus. Juxtaglomerular apparatus is a microscopic structure located between the vascular pole of the renal corpuscle and the returning distal convoluted tubule of the same nephron. It plays a role in regulating the renal blood flow and glomerular filtration rate. When there is a fall in the glomerular filtration rate, it activates the juxtaglomerular cells to release renin. This stimulates the glomerular blood flow, thereby bringing the GFR back to normal. Renin brings the GFR back to normal by the activation of the renin-angiotensin mechanism.

Class 11 Biology Chapter 19 Question Answer

Q3. Indicate whether the following statements are true or false:

(a) Micturition is carried out by a reflex.
(b) ADH helps in water elimination, making the urine hypotonic.
(c) Protein-free fluid is filtered from blood plasma into the Bowman’s capsule.
(d) Henle’s loop plays an important role in concentrating the urine.
(e) Glucose is actively reabsorbed in the proximal convoluted tubule.

Answer:

(a) Micturition is carried out by a reflex. True statement
(b) ADH helps in water elimination, making the urine hypotonic. False statement
(c) The protein-free fluid is filtered from blood plasma into the Bowman’s capsule. True statement
(d) Henle’s loop plays an important role in concentrating the urine. True statements
(e) Glucose is actively reabsorbed in the proximal convoluted tubule. True statements

Q4. Give a brief account of the counter current mechanism.

Answer:

The function of the countercurrent mechanism that operates inside the kidney is to conserve the water and make the urine concentrated. The countercurrent mechanism depends upon the loops of Henle, vasa recta, collecting ducts and interstitial fluid. The blood flows in the two limbs of the tube in opposite directions giving rise to the counter-currents. The proximity between the loop of Henle and vasa recta, as well as the countercurrent in them, help in maintaining an in increasing osmolarity towards the inner medullary interstitial fluid i.e. 300 mOsmol/L in the cortex to 1200 mOsmol/L in the inner medulla. This gradient is mainly caused by NaCl and urea. NaCl is transported by the ascending limb of the loop of Henle which is exchanged with the descending capillary of vasa recta. Similarly, small amounts of urea enter the thin segment of the ascending limb of the loop of Henle which is transported back to the medullary interstitial fluid by the collecting duct.

The countercurrent mechanism helps to maintain a concentration gradient in the medullary interstitial fluid which helps in easy absorption of water from the filtrate present in the collecting duct so that the concentration of the filtrate is increased. The overall function of the counter current mechanism is to concentrate sodium chloride in the interstitial fluid and cause water to diffuse out of the collecting ducts and concentrate the urine. This leads to the production of hypertonic urine.

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

Q5. Describe the role of liver, lungs and skin in excretion.

Answer:

Apart from kidney, liver, lungs, and skin also play important roles in the process of excretion.

1. LIver- The liver is the largest gland in vertebrates which helps in the excretion of compounds like cholesterol, steroid hormones, vitamins, drugs, and other waste materials via the production of bile. Urea is formed in the liver by the ornithine cycle whereas ammonia which is a toxic substance is quickly changed into urea in the liver and hence gets eliminated from the body. The liver is also responsible for changing the decomposed haemoglobin pigment into bile pigments in the form of bilirubin and biliverdin.

2. Lungs: The lungs help in removal of waste materials such as carbon dioxide from the body.

3. Skin: Skin possesses many glands that help in excreting waste products through pores in the skin. The two types of glands present in the skin are sweat and sebaceous glands. The sweat glands are highly vascular and tubular glands which separate the waste products from the blood and excrete them in the form of sweat. Sweat also removes excess salt and water from the body. Sebaceous glands are branched glands that secrete an oily secretion called sebum.

Q6. Explain micturition

Answer:

The process by which the urine from the urinary bladder is excreted out is called micturition. As the urine accumulates, the muscular walls of the bladder expands and cause the stimulation of the sensory nerves in the bladder, setting up a reflex action. This reflex stimulates the urge to pass out urine. In order to discharge urine, the urethral sphincter relaxes and the smooth muscles of the bladder contract. This forces the urine out from the bladder. An adult human excretes about 1 — 1.5 litres of urine per day.

Class 11 Bio Chapter 19 NCERT Solutions

Q7. 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

Answer:

The correct matching is a-iii, b-v, c-iv, d-i, e-ii

Column I Column II

(a) Ammonotelism (iii) Bony fish
(b) Bowman’s capsule (v) Renal tubule
(c) Micturition (iv) Urinary bladder
(d) Uricotelism (i) Birds
(e) ADH (ii) Water reabsorption

Q8. What is meant by the term osmoregulation?

Answer:

Osmoregulation- It is a homeostatic mechanism that regulates the optimum temperature of water and salts in the tissues and body fluids. It maintains the internal environment of the body by maintaining the water and ionic concentration.

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

Q9. Terrestrial animals are generally either ureotelic or uricotelic, not ammonotelic, why ?

Answer:

Terrestrial animals are either ureotelic or uricotelic, and not ammonotelic because of the following two reasons:

1. Since ammonia is highly toxic in nature, it needs to be converted into a less toxic form such as urea or uric acid.

2. Terrestrial animals need to conserve water. Ammonia being soluble in water, cannot be eliminated continuously. Hence, it needs to be converted into urea or uric acid which are insoluble in water. This helps terrestrial animals conserve water.

Q10. What is the significance of juxta glomerular apparatus (JGA) in kidney function?

Answer:

Juxtaglomerular apparatus is an apparatus consisting of a few cells of the glomerulus, distal tubule, and afferent and efferent arterioles It is involved in maintaining the glomerular filtration rate in the kidney. JGA is found to be located in a specialised region of a nephron, where the afferent arteriole and the distal convoluted tubule are in direct contact with each other. The juxtaglomerular apparatus consists of certain specialised cells of the afferent arteriole. These cells are known as juxtaglomerular cells and they contain the enzyme renin that can sense blood pressure. When there is a decrease in the glomerular filtration rate, the juxtaglomerular cells get activated and release renin which functions to converts the angiotensinogen in the blood into angiotensin I and further into angiotensin II. Angiotensin II is a powerful vasoconstrictor that can increase the glomerular filtration rate or glomerular blood pressure. Angiotensin II further stimulates the adrenal cortex of the adrenal gland to produce aldosterone which increases the rate of absorption of sodium ions and water from the distal convoluted tubule and the collecting duct. Later on, it leads to an increase in the blood pressure and glomerular filtration rate. This mechanism is known as the renin-angiotensin mechanism, ultimately leads to increased blood pressure. Juxtaglomerular apparatus plays an important role in the renin-angiotensin mechanism.

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Class 11 Biology Chapter 19 NCERT Question Answer

Q11. Name the following:

(a) A chordate animal having flame cells as excretory structures

Answer:

A chordate animal having flame cells as excretory structures is Amphioxus.

Q11. Name the following:

(b) Cortical portions projecting between the medullary pyramids in the human kidney

Answer:

Cortical portions projecting between the medullary pyramids in the human kidney are called columns of Bertini.

Q11. Name the following:

(c) A loop of capillary running parallel to the Henle’s loop.

Answer:

A loop of capillary running parallel to the Henle’s loop is vasa rectum.

Q12. Fill in the gaps :

(a) Ascending limb of Henle’s loop is _______ to water whereas the descending limb is _______ to it.
(b) Reabsorption of water from distal parts of the tubules is facilitated by hormone _______.
(c) Dialysis fluid contain all the constituents as in plasma except _______.
(d) A healthy adult human excretes (on an average) _______ gm of urea/day.

Answer:

(a) Ascending limb of Henle’s loop is impermeable to water whereas the descending limb is permeable to it.
(b) Reabsorption of water from distal parts of the tubules is facilitated by hormone ADH.
(c) Dialysis fluid contains all the constituents as in plasma except nitrogenous wastes
(d) A healthy adult human excretes (on an average) 20-30 gm of urea/day.

NCERT Class 11 Biology Chapter 19 – Excretory Products and Their Elimination

NCERT Solutions for Class 11 Biology Chapter 19 - Excretory Products and Their Elimination are categorised under Unit 5 Human Physiology and are part of the CBSE Syllabus for the 2022–23 academic year. According to historical patterns, this unit 5 carries roughly 18 of the possible points, or 25% of the total weight. NCERT Solutions for Class 11 Biology provides ansewers to exercise questions for all chapters. As a result, students need to pay close attention to every chapter in this unit including Chapter 19 Biology Class 11 NCERT Solutions. Excretion Class 11 explains the Human excretory system, and its corresponding organs performing respective functions.

If you are unable to determine Class 11 Biology Chapter 19 NCERT Question Answer or facing any problem to understand Excretory Products and Their Elimination NCERT Solutions. Then, you need to go through the Excretory Products and Their Elimination NCERT chapter and then try to solve all the given questions in by yourself. You can match your responses from Excretory Products and Their Elimination Class 11 NCERT Solutions. It will help you to prepare not only for the school exam also for other competitive exams like NEET as you can encounter question based on the concepts of Class 11 Bio Chapter 19 NCERT Solutions.

Important Topics of Solutions for NCERT Class 11 Biology Chapter 19 - Excretory Products and Their Elimination

The important topics and subtopics of Excretory Products and Their Elimination NCERT Solutions are given below:

19.1 Human Excretory System

19.2 Urine Formation

19.3 Function of the Tubules

19.4 Mechanism of Concentration of the Filtrate

19.5 Regulation of Kidney Function

19.6 Micturition

19.7 Role of other Organs in Excretion

19.8 Disorders of the Excretory System

Benefits of the Solutions for NCERT Class 11 Biology Chapter 19 Excretory Products and Their Elimination

The benefits of Excretory Products and Their Elimination Class 11 NCERT Solutions PDF are listed below:

  • NCERT is the base of your learning and Chapter 19 Biology Class 11 NCERT Solutions is a part of this.
  • Excretory products and their elimination Class 11 NCERT Solutions will also help you with competitive exams like NEET.
  • You will get all the answers to ch 19 bio class 11 and excretory products and their elimination class 11 will help you to score good marks in the exam.
  • Solutions for excretory system NCERT will also help you in your 12th board exam.
  • Excretory products and their elimination NCERT will also boost your knowledge.
  • To score well in the examination, follow the NCERT syllabus and solve the exercise given in the NCERT Book and Class 11 Bio Chapter 19 NCERT Solutions will assisst you in this. To practice more problems, students must refer to NCERT Exemplar.
  • Solutions given for Biology Class 11 Chapter 19 are created as per latest CBSE syllabus.
  • Excretory Products and Their Elimination Class 11 Questions and Answers are comprehensive and developed by subject experts.
  • Class 11 Biology Chapter 19 Question Answer are provided in the most efficient and effective pattern.

NCERT Solutions for Class 11 Biology - Chapter Wise

NCERT Solutions for Class 11 - Subject wise

Also Check NCERT Books and NCERT Syllabus here :

Frequently Asked Question (FAQs)

1. What are the important topics of excretory products and their elimination class 11 ncert solutions?
  • Human Excretory System  

  • Urine Formation  
  • The function of the Tubules  
  • Mechanism of Concentration of the Filtrate  
  • Regulation of Kidney Function  
  • Micturition  
  • Role of other Organs in Excretion  
  • Disorders of the Excretory System
2. Name the loop of capillary running parallel to the Henle's loop given in the excretion class 11.

 A loop of capillary running parallel to the Henle’s loop is vasa rectum.

3. What are the benefits of the solutions for NCERT class 11 biology chapter 19 excretory products and their elimination?

The benefits of excretion 11th NCERT are listed below:

  • NCERT is the base of your learning.  
  • chapter 19 biology class 11 ncert solutions will also help you with competitive exams like NEET.  
  • You will get all the answers to excretory system class 11 and excretory products and their elimination ncert solutions will help you to score good marks in the exam.  
  • class 11 excretory products and their elimination will also help you in your 12th board exam.  
  • Biology class 11 chapter 19 will also boost your knowledge
  • To score well in the examination, follow the NCERT syllabus and solve the exercise given in the NCERT Book. To practice more problems, students must refer to NCERT Exemplar.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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