GSEB Class 11 Biology Solutions Chapter 19 Excretory Products and their Elimination

Get the most accurate GSEB Solutions for Class 11 Biology Chapter 19 Excretory Products and their Elimination here. Updated for the 2026-27 academic session, these solutions are based on the latest GSEB textbooks for Class 11 Biology. Our expert-created answers for Class 11 Biology are available for free download in PDF format.

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

For Class 11 students, solving GSEB textbook questions is the most effective way to build a strong conceptual foundation. Our Class 11 Biology solutions follow a detailed, step-by-step approach to ensure you understand the logic behind every answer. Practicing these Chapter 19 Excretory Products and their Elimination solutions will improve your exam performance.

Class 11 Biology Chapter 19 Excretory Products and their Elimination GSEB Solutions PDF

Gseb Class 11 Biology Excretory Products And Their Elimination Text Book Questions And Answers

 

Question 1. Define glomerular filtration rate (GFR).
Answer: The amount of filtrate that the kidneys make each minute is known as the glomerular filtration rate (GFR). For a healthy person, GFR is approximately 125 ml per minute, which means about 180 liters per day.
In simple words: GFR measures how much fluid your kidneys filter every minute. A healthy person filters around 125 ml each minute, or 180 liters daily.

Exam Tip: Remember to include both the definition of GFR and its approximate values for a healthy individual, specifying both per minute and per day rates, to score full marks.

 

Question 2. Explain the autoregulatory mechanism of GFR.
Answer: Kidneys possess internal systems that help regulate the glomerular filtration rate. One effective method is carried out by the juxtaglomerular apparatus (JGA). JGA is a specific, sensitive area created by cell changes in the distal convoluted tubule and the afferent arteriole where they meet. If GFR drops, it can prompt the JG cells to release renin, which then boosts glomerular blood flow, bringing GFR back to normal.
In simple words: Your kidneys have a special system called JGA to control how much blood they filter. If filtering slows down, JGA releases a substance called renin to increase blood flow and bring the filtering rate back up.

Exam Tip: When explaining autoregulation of GFR, always mention the juxtaglomerular apparatus (JGA), its composition (distal convoluted tubule and afferent arteriole), and the role of renin in restoring GFR.

 

Question 3. Indicate whether the following statements are true or false:
1. Micturition is carried out by a reflex.
2. ADH helps in water elimination, making the urine hypotonic.
3. The protein-free fluid is filtered from blood plasma into the Bowman's capsule.
4. Henle's loop plays an important role in concentrating the urine.
5. Glucose is actively reabsorbed in the proximal convoluted tubule.
Answer:
1. (T) True. Micturition, or urination, is a reflex action controlled by nerves.
2. (F) False. ADH (Antidiuretic Hormone) actually helps in water reabsorption, making urine more concentrated, not hypotonic.
3. (T) True. The fluid filtered into Bowman's capsule is largely protein-free as proteins are too large to pass through the filtration barrier.
4. (T) True. Henle's loop is crucial for creating a concentration gradient in the renal medulla, which is essential for producing concentrated urine.
5. (T) True. Glucose is actively transported back into the blood from the proximal convoluted tubule, ensuring it's not lost in urine.
In simple words: Micturition is a reflex. ADH helps keep water, not eliminate it. Protein-free fluid is filtered into the Bowman's capsule. Henle's loop helps make urine concentrated. Glucose is reabsorbed in the proximal tubule.

Exam Tip: Carefully read each statement and recall the specific functions of each kidney structure or hormone. Pay attention to terms like "hypotonic" and "actively reabsorbed" as they are key to determining truthfulness.

 

Question 4. Give a brief account of the countercurrent mechanism.
Answer: The Henle's loop and vasa recta play a vital part in making concentrated urine. The liquid flow in the two sides of Henle's loop moves in opposite directions, creating a countercurrent system. Similarly, the blood flow through the two sides of vasa recta also shows a counter-current arrangement.
The close relationship between the Henle's loop and vasa recta, along with their counter-current flow, aids in keeping an increasing osmolarity towards the inner medullary interstitium. This ranges from 300 m Osmol \( L^{-1} \) in the outer part (cortex) to about 1200 m Osmol \( L^{-1} \) in the inner part (medulla).
This gradient mainly occurs because of NaCl and urea. NaCl is moved by the ascending part of Henle's loop, which then gets exchanged with the descending part of the vasa recta. NaCl is sent back to the interstitium by the ascending part of the vasa recta. Likewise, a tiny amount of urea enters the thin section of the ascending part of Henle's loop, which is then moved back to the interstitium by the collecting tubule.
The transport system described above is known as the countercurrent mechanism. It helps keep a concentration gradient in the medullary interstitium. This gradient makes it easy for water to pass out from the collecting tubule, thereby making the filtrate more concentrated.
In simple words: The countercurrent mechanism in kidneys uses Henle's loop and vasa recta to make urine concentrated. Fluids flow in opposite directions in these structures, creating a salt and urea gradient in the kidney. This gradient draws water out of the urine, making it more concentrated.

Exam Tip: When explaining the countercurrent mechanism, focus on the roles of Henle's loop and vasa recta, the opposite flow of fluids, the establishment of the osmolarity gradient (300 to 1200 mOsmol \( L^{-1} \)), and the involvement of NaCl and urea in maintaining this gradient.

 

Question 5. Describe the role of the liver, lungs, and skin in excretion.
Answer: Our lungs get rid of large amounts of CO2 (about 18 liters per day) and also considerable quantities of water daily. The liver, our body's biggest gland, secretes bile that contains substances like bilirubin, biliverdin, cholesterol, broken-down steroid hormones, vitamins, and drugs. Most of these substances are eventually passed out with digestive wastes.
The skin's sweat and sebaceous glands can remove certain substances through their secretions. The sweat made by sweat glands is a watery liquid containing NaCl, a small quantity of urea, lactic acid, and other compounds. While sweat's main job is to cool the body surface, it also assists in removing some of the wastes mentioned above. Sebaceous glands get rid of certain substances like sterols, hydrocarbons, and waxes via sebum. This secretion offers a protective oily layer for the skin.
In simple words: Lungs remove CO2 and water. The liver secretes bile with wastes like bilirubin and old hormones, which leave with digestive waste. Skin's sweat glands remove water, salt, and urea, while sebaceous glands secrete oily substances like sterols and waxes to protect the skin and excrete other wastes.

Exam Tip: For each organ (lungs, liver, skin), clearly state the specific waste products it excretes and briefly explain the mechanism or secretion involved (e.g., CO2 and water vapor from lungs, bile pigments from liver, sweat and sebum from skin).

 

Question 6. Explain micturition.
Answer: Urine, formed by the nephrons, is ultimately moved to the urinary bladder, where it is kept until a conscious signal is given by the central nervous system (CNS). This signal starts when the urinary bladder stretches as it fills with urine. In response, the stretch receptors on the bladder walls send signals to the CNS. The CNS then sends motor messages to start the contraction of the bladder's smooth muscles and the simultaneous relaxing of the urethral sphincter, allowing urine to be released. The process of urine release is called micturition, and the neural mechanisms that cause it are known as the micturition reflex.
In simple words: Micturition is when you urinate. It happens when your bladder fills and stretches, sending a signal to your brain. Your brain then tells the bladder muscles to squeeze and the opening to relax, releasing urine.

Exam Tip: When explaining micturition, emphasize the roles of the urinary bladder's stretching, stretch receptors, the central nervous system (CNS) signal, bladder muscle contraction, and urethral sphincter relaxation.

 

Question 7. Match the items of Column I with those of column II
Column I
(i) Ammonotelism
(ii) Bowman's capsule
(iii) Micturition
(iv) Uricotelism
Column II
(a) Birds
(b) Water reabsorption
(c) Bony fish
(d) Urinary bladder
(e) Renal tubule
Answer:
(i) Ammonotelism - (c) Bony fish (Bony fish often excrete ammonia as their primary nitrogenous waste).
(ii) Bowman's capsule - (e) Renal tubule (Bowman's capsule is the starting part of the renal tubule, where filtration begins).
(iii) Micturition - (d) Urinary bladder (Micturition, or urination, involves the emptying of the urinary bladder).
(iv) Uricotelism - (a) Birds (Birds primarily excrete uric acid as their nitrogenous waste to conserve water).
In simple words: Ammonotelism means bony fish excrete ammonia. Bowman's capsule is the start of the renal tubule. Micturition is emptying the urinary bladder. Uricotelism means birds excrete uric acid.

Exam Tip: To effectively match these terms, recall the primary excretory products of different animal groups and the specific functions of various parts of the nephron and urinary system.

 

Question 8. What is meant by the term osmoregulation?
Answer: The body's metabolic activities create various by-products. Most of these products are useful, but a few are harmful and need to be removed from the body. During cellular respiration, a large amount of CO2 and water is produced from the breakdown of carbohydrates, fats, and proteins. CO2 is very volatile and is removed as a gas from the respiratory surface during breathing.
It is also an essential part of the synthetic and regulatory systems of animals and plants. CO2 helps maintain acid-base balance in the body. Water is also volatile, and some of it is removed from the lungs as vapor with exhaled air. The removal of extra water from the body to keep the water content steady is called osmoregulation.
In simple words: Osmoregulation is how living things control the amount of water and salt in their bodies to keep a steady internal balance. It involves removing extra water and waste to maintain proper fluid levels.

Exam Tip: When defining osmoregulation, highlight that it is the process of maintaining a constant water content and solute concentration within the body, emphasizing the role of removing excess water and other metabolic by-products.

 

Question 9. Terrestrial animals are generally either ureotelic or uricotelic, not ammonotelic, why?
Answer: Many aquatic animals are ammonotelic, meaning they excrete ammonia as a waste product. Ammonia is typically removed by diffusion across body surfaces as ammonium ions. Kidneys do not play a significant part in its removal. Living on land required producing less toxic acid for water conservation. Therefore, terrestrial animals are mostly ureotelic (meaning they excrete urea) and uricotelic (meaning they excrete uric acid).
In simple words: Land animals are mostly ureotelic (excrete urea) or uricotelic (excrete uric acid) instead of ammonotelic (excrete ammonia). This is because ammonia is very toxic and needs a lot of water to be flushed out, which is hard to do on land where water is scarcer. Urea and uric acid are less toxic and require less water for excretion.

Exam Tip: The key to this answer is water conservation. Explain that ammonia is highly toxic and requires large amounts of water for excretion, which is not feasible for terrestrial animals. Contrast this with urea and uric acid, which are less toxic and require less water, making them suitable for land-dwelling organisms.

 

Question 10. What is the significance of the juxtaglomerular apparatus (JGA) in kidney function?
Answer: The kidneys have internal systems for regulating the glomerular filtration rate. One such effective method is performed by the juxtaglomerular apparatus (JGA). JGA is a specific, sensitive area created by cellular changes in the distal convoluted tubule and the afferent arteriole where they meet. If GFR falls, it can prompt the JG cells to release renin, which then boosts glomerular blood flow, bringing GFR back to normal.
JGA, a specialized part of the nephrons, plays an important part in GFR regulation. A drop in glomerular blood flow, glomerular blood pressure, or GFR can activate JG cells to release renin. Renin then converts angiotensinogen in the blood to angiotensin I, and further to angiotensin II. Angiotensin II is a strong vasoconstrictor, which boosts glomerular blood pressure and thus GFR. Angiotensin II also activates the adrenal cortex to release aldosterone. Aldosterone causes the reabsorption of Na+ and water from the distant parts of the tubule. This also results in an increase in blood pressure and GFR. This complex system is generally known as the Renin-Angiotensin mechanism.
An increase in blood flow to the heart's atria can cause the release of Atrial Natriuretic Factor (ANF). ANF can cause vasodilation (blood vessel widening), thereby lowering blood pressure. The ANF mechanism, therefore, serves as a check on the renin-angiotensin mechanism.
In simple words: The JGA is super important for kidney function because it helps control blood filtering. If the filtering rate drops, JGA releases renin, which starts a chain reaction (Renin-Angiotensin mechanism) that narrows blood vessels and makes you retain water, both raising blood pressure and filtering. ANF acts as a counter to prevent blood pressure from getting too high.

Exam Tip: For the significance of JGA, explain its role in detecting changes in GFR, initiating the Renin-Angiotensin-Aldosterone System (RAAS), and how Angiotensin II and Aldosterone work to restore GFR. Also, mention ANF as a counter-regulatory 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.
Answer:
1. Protonephridia (or flatworms, e.g., Planaria)
2. Columns of Bertini
3. Vasa recta
In simple words: Flame cells for excretion are found in protonephridia (like flatworms). Cortical parts between medullary pyramids are Columns of Bertini. Capillaries running parallel to Henle's loop are called vasa recta.

Exam Tip: Be precise with anatomical and zoological terms. Ensure you correctly recall the specific excretory structures for different organisms and the names of kidney parts.

 

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 average) gm of urea/day....................
Answer:
1. Ascending limb of Henle's loop is **impermeable** to water whereas the descending limb is **permeable** to it.
2. Reabsorption of water from distal parts of the tubules is facilitated by hormone **antidiuretic hormone ADH**.
3. Dialysis fluid contains all the constituents as in plasma except **the nitrogenous wastes**.
4. A healthy adult human excretes (on average) **25-30** gm of urea/ day.
In simple words: The ascending loop of Henle doesn't let water through, but the descending loop does. ADH helps water reabsorption in the tubules. Dialysis fluid is like plasma but without nitrogenous waste. A healthy adult passes about 25-30 grams of urea daily.

Exam Tip: For fill-in-the-blanks, recall specific properties and functions. Remember the differential permeability of Henle's loop, the role of ADH in water balance, the composition of dialysis fluid, and the average daily urea excretion.

Free study material for Biology

GSEB Solutions Class 11 Biology Chapter 19 Excretory Products and their Elimination

Students can now access the GSEB Solutions for Chapter 19 Excretory Products and their Elimination prepared by teachers on our website. These solutions cover all questions in exercise in your Class 11 Biology textbook. Each answer is updated based on the current academic session as per the latest GSEB syllabus.

Detailed Explanations for Chapter 19 Excretory Products and their Elimination

Our expert teachers have provided step-by-step explanations for all the difficult questions in the Class 11 Biology chapter. Along with the final answers, we have also explained the concept behind it to help you build stronger understanding of each topic. This will be really helpful for Class 11 students who want to understand both theoretical and practical questions. By studying these GSEB Questions and Answers your basic concepts will improve a lot.

Benefits of using Biology Class 11 Solved Papers

Using our Biology solutions regularly students will be able to improve their logical thinking and problem-solving speed. These Class 11 solutions are a guide for self-study and homework assistance. Along with the chapter-wise solutions, you should also refer to our Revision Notes and Sample Papers for Chapter 19 Excretory Products and their Elimination to get a complete preparation experience.

FAQs

Where can I find the latest GSEB Class 11 Biology Solutions Chapter 19 Excretory Products and their Elimination for the 2026-27 session?

The complete and updated GSEB Class 11 Biology Solutions Chapter 19 Excretory Products and their Elimination is available for free on StudiesToday.com. These solutions for Class 11 Biology are as per latest GSEB curriculum.

Are the Biology GSEB solutions for Class 11 updated for the new 50% competency-based exam pattern?

Yes, our experts have revised the GSEB Class 11 Biology Solutions Chapter 19 Excretory Products and their Elimination as per 2026 exam pattern. All textbook exercises have been solved and have added explanation about how the Biology concepts are applied in case-study and assertion-reasoning questions.

How do these Class 11 GSEB solutions help in scoring 90% plus marks?

Toppers recommend using GSEB language because GSEB marking schemes are strictly based on textbook definitions. Our GSEB Class 11 Biology Solutions Chapter 19 Excretory Products and their Elimination will help students to get full marks in the theory paper.

Do you offer GSEB Class 11 Biology Solutions Chapter 19 Excretory Products and their Elimination in multiple languages like Hindi and English?

Yes, we provide bilingual support for Class 11 Biology. You can access GSEB Class 11 Biology Solutions Chapter 19 Excretory Products and their Elimination in both English and Hindi medium.

Is it possible to download the Biology GSEB solutions for Class 11 as a PDF?

Yes, you can download the entire GSEB Class 11 Biology Solutions Chapter 19 Excretory Products and their Elimination in printable PDF format for offline study on any device.