NCERT Solutions for Class 11 Biology Chapter 14 (2025-2026)

Life depends on the continuous supply of oxygen and the removal of carbon dioxide. Chapter 14 of Class 11 Biology, Breathing and Exchange of Gases, explores the fascinating mechanisms by which organisms obtain oxygen from the environment and eliminate carbon dioxide produced during cellular respiration. The chapter delves into the human respiratory system, the mechanics of breathing, the exchange of gases at different sites, and the transport of oxygen and carbon dioxide in the blood. This chapter is part of the comprehensive NCERT Solutions Class 11 Biology series, which covers all chapters in detail.

The NCERT Solutions for Breathing and Exchange of Gases provided here offer detailed, step-by-step explanations for all textbook questions, helping students strengthen their conceptual understanding, clear doubts effectively, and prepare efficiently for both school exams and competitive tests like NEET.

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Class 11 Chapter 14 Biology Questions & Answers –Breathing and Exchange of Gases

Q1. Define vital capacity. What is its significance?

Solution:  Vital capacity refers to the volume of air a healthy person can inhale after a forced exhalation or the volume of air a person can exhale after a forced inhalation. It is the sum of Expiratory Reserve Volume (ERV), Tidal Volume (TV), and Inspiratory Reserve Volume (IRV). A normal adult person has a vital capacity of 3-5 litres.

Significance: Vital capacity is highest when physiological competence of the body is highest. It goes down as incompetence increases and becomes zero when respiration ceases. It implies that more the vital capacity, healthier is the body. Also, it helps in getting rid of the foul air and aids in supplying fresh air. Thus, the vital capacity enhances the exchange of gases between the body tissues and the environment.

 

Q2. State the volume of air remaining in the lungs after a normal breathing.

Solution:  The volume of air that remains in the lungs after normal breathing is called FRC (functional residual capacity). This includes expiratory reserve volume (ERV, additional volume of air, a person can exhale by a forcible exhalation) and residual volume (RV, the volume of air remaining in the lungs even after a forcible exhalation). In a normal individual, ERV is about 1000 to 1100 mL while RV is about 1100 to 1200 mL.

Since, FRC = ERV + RV

FRC = 1000 (or 1100 mL) + 1100 (or 1200) mL

= 2100 or 2300 mL.

Thus, the volume of air remaining in the lungs after normal breathing is around 2100 mL to 2300 mL.

 

Q3. Diffusion of gases occurs in the alveolar region only and not in the other parts of respiratory system. Why?

Solution: The gaseous exchange (exchange of O2 and CO2) occurs in our body by simple diffusion between the blood capillaries around the alveoli and gases present in the alveoli. Alveoli are made up of thin and highly permeable layers of squamous epithelial cells. The reasons for gaseous diffusion in the alveolar region are as follows:

  1. Concentration and pressure: Diffusion is a concentration and pressure-dependent process. The barrier between the capillaries and alveoli is thin and it facilitates the gaseous diffusion from the region of higher to lower partial pressure. Pressure contributed by an individual gas in gaseous mixture is called its partial pressure. The partial pressure of Ois high in the air present inside the lungs as compared to deoxygenated blood in lung capillaries. It means that there is a concentration gradient for O2 between the lungs and blood. Thus, O2 can easily diffuse from alveolar air into the blood of alveolar capillaries. A reverse concentration gradient is present for CO2, thus diffusion of COtakes place in the opposite direction.
  2. Nature of the membranes at the site of exchange: The region where diffusion takes place is made up three major layers namely; the thin squamous epithelium of alveoli, the endothelium of alveolar capillaries and the basement substance in between them. The total thickness of these three layers is less than a millimetre which makes it ideal to act as a site for the diffusion of gases.

 

Q4. What are the major transport mechanisms for CO2? Explain.

Solution: Blood is the medium for transport of CO2. There are three main ways by which COis transported in the body:

  • By red blood cells (RBCs): The haemoglobin present in the RBCs carry about 20-25% CO2 as carbamino-haemoglobin. Haemoglobin transports CO2 as well as Oand this transport of COdepends upon the partial pressure of these gases. If there is high partial pressure of O2, then CO2 is dissociated and O2 is acquired by haemoglobin (like in lungs where O2 is present in abundance). When oxygenated blood reaches the tissues, due to the high partial pressure of CO2, haemoglobin loses O2 and acquires CO2. This CO2 is transported to lungs for its exchange with O2.
  • By RBCs and plasma as bicarbonate ion: Nearly 70% of CO2 is transported in the form of bicarbonate ions. RBCs contain a very high amount of an enzyme called carbonic anhydrase. This enzyme is also present in minute amounts in plasma. A large part of CO2, as it diffuses into the blood plasma, combines with water to form carbonic acid with the help of enzyme carbonic anhydrase. Thereafter, the carbonic acid dissociates into bicarbonate and hydrogen ions by the action of the same enzyme. Thus, COis transported in the form of bicarbonate ions.
  • Approximately 7% of CO2 is carried in dissolved state through plasma. COcombines with water to form carbonic acid.

 

Q5. What will be the pO2 and pCO2 in the atmospheric air compared to those in the alveolar air?
(i) pO2 lesser, pCO2 higher
(ii) pO2 higher, pCO2 lesser
(iii) pO2 higher, pCO2 higher
(iv) pO2 lesser, pCO2 lesser

Solution:

(ii) pO2 higher, pCO2 lesser

In the atmosphere, O2 is present in large quantity and thus, its partial pressure is higher (159 mm Hg) than pO2 in alveolar air (104 mm Hg).

The pCO2 is lesser in atmosphere (0.3 mm Hg) as compared to pCO2 in alveoli (40 mm Hg).

 

Q6. Explain the process of inspiration under normal conditions.

Solution: The process of taking the air from outside the body into the lungs is called inhalation or inspiration. The process of inspiration involves the generation of the pressure gradient between the atmosphere and the lungs. Inspiration takes place when the pressure within the lungs is less (negative pressure) than the atmospheric pressure. The process of inspiration involves the following steps:

  • A negative pressure is generated within the lungs with the help of diaphragm and intercostal muscles (internal and external intercostal muscles)
  • Contraction of diaphragm increases the volume of the thoracic chamber in the anteroposterior axis.
  • Contraction of external intercostal muscles increases the volume of the thoracic chamber in the dorsoventral axis by lifting up the ribs and the sternum.
  • Increase in thoracic volume results in an increase in the pulmonary volume.
  • Increased pulmonary volume results in a decrease in intra-pulmonary pressure as compared to atmospheric pressure.
  • A decrease in intra-pulmonary pressure (negative pressure as compared to atmospheric pressure) allows the air to move into the lungs, i.e., inspiration or inhalation.

 

Q7. How is respiration regulated?

Solution: Respiration in human beings is a regulated process. It is regulated by the neural system according to the demands of the body tissues. It is mainly controlled by two regulatory regions present in the brain. These are:

(a) Respiratory rhythm centre: It is present in the medulla region of the brain and is mainly responsible for regulating the process of respiration. The areas adjacent to the respiratory rhythm centre are highly chemo-sensitive. They can sense the COand Hlevels, whose increase activates the respiratory rhythm centre. Once activated, this centre adjusts the respiratory process, which helps in the elimination of COand H+ from the body. Receptors associated with the aortic arch and carotid artery also sense the change in COand Hconcentration and send signals to the respiratory rhythm centre to adjust the respiratory process.

(b) Pneumotaxic centre: It is present in the pons region of the brain. This centre can moderate the function of respiratory rhythm centre and reduces the duration of inspiration, which in turn changes the respiratory rate.

 

Q8. What is the effect of pCO2 on oxygen transport?

Solution: Oxygen binds to haemoglobin reversibly to form oxyhaemoglobin, which transports throughout the body to supply oxygen to different tissues. The binding of one molecule of haemoglobin with four molecules of oxygen depends on the partial pressure of oxygen(pO2). However, the partial pressure of CO2 also plays a significant role in the binding of oxygen with the haemoglobin molecule and its transport from lungs to tissues and vice-versa.

In alveoli, the pOis high but pCOis low. Thus, oxygen binds to haemoglobin and oxyhaemoglobin is formed. It transports the oxygen to body tissues. However, in tissues, the pCO2 is higher than pO2. Thus, Odissociates from haemoglobin and is released. Here, CObinds to haemoglobin to form carbamino-haemoglobin. COIs thus transported to the lungs and is released in the lung cavity in exchange of O2. Thus, it can be stated that the affinity of haemoglobin for O2 increases as pCOdecreases.

 

Q9. What happens to the respiratory process in a man going up a hill?

Solution: As a person goes up a hill (that is, at higher altitude), the level of Oin the atmosphere decreases. So, less O2 is available for breathing and O2 level in the blood starts to decline. It results in an increase in the respiratory rate so as to meet the O2 demands of the body. Heart beat also increases to increase the supply of blood to the tissues so that every body tissue gets sufficient O2.

 

Q10. What is the site of gaseous exchange in an insect?

Solution: Insects have specialized structures called spiracles arranged in a series on the sides of its body. These are small openings through which the oxygen-rich air enters into the body. Spiracles are connected to a network of tubes called trachea which diffuses oxygen into the cells of the body. The transport of CO2 occurs in reverse direction, that is, from the trachea to the spiracles. Thus, the gaseous exchange occurs in the insects through the tracheal system.

 

Q11. Define oxygen dissociation curve. Can you suggest any reason for its sigmoidal pattern?

Solution:  Oxygen is transported in the blood by haemoglobin in the form of oxyhaemoglobin. Each haemoglobin molecule can carry a maximum of four O2 molecules. Binding of oxygen to haemoglobin is primarily affected by the partial pressure of O(pO2). However, the partial pressure of carbon dioxide (pCO2), H+ concentration and temperature also affect the binding of oxygen to haemoglobin. When the partial pressure of oxygen is high like in alveoli, haemoglobin binds with oxygen and forms oxyhaemoglobin. However, in tissues where pO2 is low, oxygen dissociates from oxyhaemoglobin. The per cent saturation of haemoglobin with oxygen at various pOcan be studied using the oxygen dissociation curve. The oxygen dissociation curve is a sigmoid curve which is obtained by plotting the percentage saturation of haemoglobin with O2 against pO2.

The oxygen dissociation curve is sigmoid in shape because the binding of oxygen molecules to haemoglobin is co-operative. It means that binding of first O2 molecule to the haemoglobin increases the affinity of haemoglobin for another oxygen molecule. As a result, the haemoglobin attracts more oxygen and thus, the graph showing the percentage saturation of haemoglobin with O2 against pO2 appears sigmoid.

 

Q12. Have you heard about hypoxia? Try to gather information about it, and discuss with your friends.

Solution: Hypoxia is a condition, wherein the tissues are not oxygenated adequately due to decreased supply of oxygen to the lungs. The oxygen deprivation can have severe adverse effects on various body cells and thus, several important biological processes get hampered. In general, hypoxia results in a pathological condition. There are various types of hypoxia depending upon the reason of inadequate oxygen supply to the body:

  1. Hypoxic hypoxia or generalized hypoxia: This results because of an inadequate saturation of blood with oxygen due to a reduced supply of oxygen in the air, decreased lung ventilation or respiratory disease.
  2. Anaemic hypoxia: This is due to decreased haemoglobin concentration.
  3. Stagnant hypoxia: This is due to poor circulation of the blood.
  4. Histotoxic hypoxia: This occurs when the tissues are unable to use oxygen due to cyanide or carbon monoxide poisoning.

 

Q13. Distinguish between

(a) IRV and ERV
(b) Inspiratory capacity and expiratory capacity.
(c) Vital capacity and Total lung capacity.
Solution:
(a) IRV and ERV:
IRV ERV
IRV (Inspiratory Reserve Volume) is the additional volume of air that a person can inhale by a forceful inspiration.
ERV (Expiratory Reserve Volume) is the additional volume of air that a person can exhale by a forceful expiration.
It is approximately 2500-3500 mL. It is approximately 1000-1100 mL.

(b) Inspiratory capacity and expiratory capacity:

Inspiratory capacity Expiratory capacity
Inspiratory capacity is the volume of air that can be inhaled after a normal expiration.
Expiratory capacity is the volume of air that can be exhaled after a normal inspiration.
It includes tidal volume and IRV.
It includes tidal volume and ERV.

(c) Vital capacity and Total lung capacity:

Vital capacity Total lung capacity
Vital capacity refers to the volume of air a healthy person can inhale after a forced exhalation or the volume of air a person can exhale after a forced inhalation.
Total lung capacity is the total volume of air present in the lungs after a forced inhalation.
It includes ERV, IRV, and tidal volume. This includes ERV, IRV, residual volume and tidal volume.
It is about 4000 mL. It is about 5000-6000 mL.

 

Q14. What is Tidal volume? Find out the Tidal volume (approximate value) for a healthy human in an hour.

Solution:  Tidal volume is the volume of air that is inhaled or exhaled during normal respiration. It is about 500 mL. A healthy person breathes about 12-16 times in a minute.

The approximate value of tidal volume for a healthy human per hour can be calculated as follows:

The tidal volume of a healthy human in an hour = Tidal volume per breath X Number of times an adult human being breathes per minute X 60 minutes

= 500 mL X 12-16 times per minute X 60 minutes

= 3,60,000 – 4,80,000 mL

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NCERT Solutions for Class 11 Biology Chapter 14  – FAQs

1. What is the difference between breathing and respiration?

Breathing (or ventilation) is simply the mechanical process of inhaling oxygen-rich air into the lungs and exhaling carbon dioxide-rich air out. It's a physical process. Respiration, on the other hand, is a biochemical process where glucose is broken down in cells to release energy (ATP), using oxygen and producing carbon dioxide as a waste product. Breathing is just one step that supplies oxygen for cellular respiration.

2. How does oxygen and carbon dioxide exchange occur in the alveoli and tissues?

Gas exchange occurs through diffusion based on partial pressure gradients. In the alveoli, oxygen moves from the air (where its partial pressure is high) into the blood (where it's lower), while CO₂ moves from the blood (high partial pressure) to the alveolar air (low partial pressure). At the tissue level, the process reverses - oxygen diffuses from blood into tissues (where cells are using it up), and CO₂ produced by cellular metabolism diffuses from tissues into the blood.

3. How is oxygen transported in the blood?

Oxygen is transported in two ways: About 97% of oxygen binds to hemoglobin in red blood cells, forming oxyhemoglobin. This is the primary transport method. The remaining 3% dissolves directly in the blood plasma. Hemoglobin can carry up to four oxygen molecules per hemoglobin molecule, and this binding is influenced by factors like partial pressure of oxygen, carbon dioxide concentration, pH, and temperature.

Q.1 Differentiate between

(a) Respiration and Combustion

(b) Glycolysis and Krebs’ cycle

(c) Aerobic respiration and Fermentation F

Ans-

(a)

Respiration Combustion
1. It occurs within living cells only. It does not occur inside a living system.
2. It requires enzymes. It does not require enzymes.
3. It occurs in a highly regulated mode under controlled condition. It is non-regulated and uncontrolled.
4. It produces energy equivalents in the form of high energy ATP molecules. It produces energy in terms of heat and light only.

(b)

Glycolysis Krebs’ Cycle
1. It occurs in the cytoplasm. It occurs in the mitochondria.
2. It is a non-cyclic process. It is a cyclic process.
3. It is common in both aerobic and anaerobic respiration. It takes place only in aerobic respiration.
4. Less productive in terms of ATP and NADP generation. Produces 8 ATP molecules from one glucose molecule Produces 15 ATP molecules from one molecule of Acetyl CoA

(c)

Aerobic respiration Fermentation
1. It occurs in the presence of molecular oxygen only. It does not require molecular oxygen.
2. It takes place in both cytoplasm and mitochondria. It takes place only in the cytoplasm.
3. It is highly efficient and produces 38 ATP molecules per molecule of glucose. It is non-economical, produces only 2 molecules of ATP per molecule of glucose
4. The final products formed from one glucose molecule are carbon dioxide and water. The final products formed from one glucose molecule are ethyl alcohol and carbon dioxide

Q.2 What are respiratory substrates? Name the most common respiratory substrate.

Ans-

The complex organic compound that gets oxidized in the cell during respiration to release large amounts of energy is called respiratory substrate. Under normal condition, glucose is the most common respiratory substrate which is a carbohydrate along with six carbon atoms.

Q.3 Give the schematic representation of glycolysis?

Ans-

Schematic representation of glycolysis:

Q.4 What are the main steps in aerobic respiration? Where does it take place?

Ans-

The four main steps of aerobic respiration are as follows:

S.No Steps of aerobic respiration Site of occurrence in the cell
1. Glycolysis Cytoplasm
2. Krebs’ Cycle Mitochondrial matrix
3. Electron transport chain Inner membrane of mitochondria
4. Oxidative phosphorylation F0-F1 particle of cristae present in the inner membrane of mitochondria

Q.5 Give the schematic representation of an overall view of Krebs’ cycle.

Schematic representation of Krebs’ cycle:

Q.6 Explain ETS.

Ans-

The electrons removed from the substrates of glycolysis and the Krebs’ cycle are stored in the reduction equivalents, namely NADH2 and FADH2. This energy is released when NADH2 and FADH2 are oxidized by passing their electrons to a chain of electrons carrier complex called Electron transport system, present in the inner membrane of mitochondria. These complexes transfer the electron through a series of redox reactions with high energy electrons entering the system and low-energy electrons leaving the system. The energy released through this process is utilized to pump out protons which develop a proton gradient (Proton motive force) across the inner membrane. This proton motive force is utilized by ATP synthase to generate high energy ATP molecules at 3 different sites.

Process: The NADH2 produced during the citric acid cycle are oxidized by an NADH dehydrogenase (complex I), and electrons are then transferred to ubiquinone which gets reduced. Ubiquinone also receives reducing equivalents via FADH2 (complex II). The reduced ubiquinone (ubiquinol) is then re-oxidized by transferring its electrons to cytochrome c via cytochrome bc1 complex (complex III). The electron is transferred from complex III to complex IV through cytochrome C which is a mobile carrier present in the inner membrane. Complex IV is called cytochrome C oxidase complex and consists of cytochromes a-a3, and two copper centers.

Q.7 Distinguish between the following:

(a) Aerobic respiration and Anaerobic respiration

(b) Glycolysis and Fermentation

(c) Glycolysis and Citric acid Cycle

Ans-

(a)

Aerobic respiration Anaerobic respiration
1 It occurs only in the presence of molecular oxygen. It occurs in the absence of molecular oxygen.
2. It is highly efficient and produces 38 ATP molecules. It is less efficient and generates only 2 ATP molecules.
3. It takes place in both cytoplasm and mitochondria. It takes place in the cytoplasm only.
4. It is a multistep process having glycolysis, Krebs’ cycle, and ETS. It is not a multistep process.
5. It produces carbon dioxide and water as the by-product. It produces ethyl alcohol and carbon dioxide as a by-product.

(b)

Glycolysis Fermentation
1. It is a common step in both aerobic and anaerobic respiration. It is strictly an anaerobic mode of respiration.
2. It results in the production of pyruvic acid. It produces ethyl alcohol.
3. Net gain is 8 ATP molecules. Net gain is 2 ATP molecules.
4. The product of glycolysis is used as an intermediate in Krebs’ cycle. The product of fermentation (ethyl alcohol) is not used by cells further.

(c)

Glycolysis Citric acid Cycle
1. It occurs in the cytoplasm. It occurs in the mitochondria.
2. It is a non-cyclic process. It is a cyclic process.
3. It is common in both aerobic and anaerobic respiration. It takes place only in aerobic respiration.
4. It produces 8 ATP molecules from one molecule of glucose. It produces 15 ATP molecule from one molecule of Acetyl CoA.

Q.8 What are the assumptions made during the calculation of net gain of ATP?

Ans-

Many assumptions have been made in order to calculate the net gain of ATP from one molecule of glucose. This is required as the cellular system is very complex where numerous biochemical reactions take place simultaneously. The assumptions are as follows:

  1. All the steps of aerobic respiration (glycolysis, TCA cycle, ETS and oxidative phosphorylation) take place in sequential order where the product of the first step enters the subsequent step as a substrate.
  2. The NADH synthesized in glycolysis enters in the mitochondria and produces ATP through oxidative phosphorylation.
  3. The intermediates formed in various steps of aerobic respiration are not utilized in any other metabolic pathways other than subsequent steps of respiration.
  4. The glucose molecule is the only substrate and no other molecule enters as an intermediate substrate in the process.

Q.9 Discuss “The respiratory pathway is an amphibolic pathway.”

Ans-

The process of metabolism involves both anabolic and catabolic reactions. Anabolism is the synthesis of complex macromolecules like lipids and proteins from simple molecules like glycerol and amino acid respectively. On the other hand, catabolism includes the breakdown of macromolecules into simple molecules so that they can enter in the respiratory pathway as a substrate for the release of energy. If fatty acids are used as a respiratory substrate they are broken down to glycerol and acetyl CoA. Glycerol gets converted to 3-phosphoglyceraldehyde (PGAL) and enters in glycolysis while Acetyl CoA directly enters in Krebs’ cycle. However, when an organism needs to synthesize fatty acids, acetyl CoA is withdrawn from the above-said pathway and is made available for catabolic reaction. Similarly, when proteins are used as a substrate, they are first broken down to amino acid, which in turn, depending on their structure, gets converted into different intermediates of Krebs’ cycle. At the time of need, the same molecules are withdrawn to synthesize new proteins. Most of these reactions are reversible and depending on the requirement, the cell uses the respiratory substrate in the process of anabolism or catabolism. Thus, the respiratory pathway is known as the amphibolic pathway rather than only a catabolic pathway.

Q.10 Define RQ. What is its value for fats?

Ans-

Complete oxidation of substrates during aerobic respiration requires oxygen and apart from energy, carbon dioxide is produced as the by-product. The ratio of the volume of CO2 released to the volume of O2 consumed during complete oxidation of one molecule of a substrate in a given period of time at standard temperature and pressure is called the respiratory quotient (RQ).

For example, during aerobic respiration of one molecule of glucose, 6 molecules of CO2 are released and 6 molecules of O2 are consumed. Thus RQ for glucose is 1.

RQ value for fats: Fats need more oxygen molecule than carbohydrate (glucose) for complete oxidation through aerobic respiration, due to which the value of RQ for fat is always less than 1.

For example, when fatty acid tripalmitin is used as a substrate, 145 molecules of O2 are consumed whereas 102 molecules of CO2 are produced, the RQ value is 0.7.

Q.11 What is oxidative phosphorylation?

Ans-

The metabolic pathway that uses the energy released by the oxidation of nutrients to produce adenosine triphosphate (ATP) is called oxidative phosphorylation. Almost all the forms of life on earth use a range of different nutrients to carry out oxidative phosphorylation to produce the molecule that supplies energy to metabolism i.e. ATP. This is a very efficient process of energy generation.

This process requires the presence of oxygen in the system. Oxygen drives the whole process as it removes hydrogen from the system and acts as the final hydrogen acceptor. During oxidative phosphorylation, electrons are transferred from electron donors like NADH2 to electron acceptors such as oxygen. These redox reactions release energy, which is used to form ATP. In eukaryotes, these redox reactions are carried out by a series of protein complexes within mitochondria, whereas, in prokaryotes, these proteins are located in the cells’ inner membranes. These linked sets of proteins are called electron transport chains. It is the energy of the oxidation-reduction process that is used for the production of proton gradient required for phosphorylation and thus, this process is called oxidative phosphorylation.

Q.12 What is the significance of step-wise release of energy in respiration?

Ans-

Carbohydrates, proteins, fats and organic acids are used as respiratory substrates and oxidation of these compounds releases energy in the cell. However, the energy released is not dissipated freely in the cell. In other words, it does not occur in one step. Instead, it is released in a series of slow step-wise reactions controlled by enzymes and is trapped in the form of ATP. This prevents the sudden increase in the temperature and avoids wastage of energy. This holds a lot of significance as ATP which stores the energy can be broken down whenever and wherever it is needed in the various energy-requiring processes of the organisms.

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