cl 11 bio ch 14

AHSEC (ASSEB) Class 11 Biology Chapter 14 Solutions – Breathing and Exchange of Gases | Assam Eduverse

Chapter Overview: 

Assam Eduverse provides comprehensive, accurate, and student-friendly solutions for Class 11 Biology (AHSEC / ASSEB)Unit V: Human Physiology, Chapter 14 – Breathing and Exchange of Gases. These well-structured solutions cover all intext questions, exercise questions, and multiple-choice questions (MCQs) with clear explanations, helping students build a strong conceptual understanding and prepare effectively for exams.

Chapter 14 – Breathing and Exchange of Gases explains the mechanism of respiration in humans and other organisms, focusing on breathing, transport of gases, and exchange of oxygen and carbon dioxide. Students learn about the structure of the human respiratory system, respiratory organs, alveoli, and their role in gas exchange. The chapter also covers respiration in plants and aquatic organisms, regulation of breathing, and the factors affecting respiration, helping students understand the physiological processes that sustain life.

Assam Eduverse’s Class 11 Biology Solutions are written in simple, exam-oriented, and easy-to-understand language, ensuring better clarity, quick learning, and strong academic performance. These reliable solutions help students master the process of respiration, understand gas exchange mechanisms, and excel in AHSEC / ASSEB Biology exams with confidence and conceptual clarity.

AHSEC (ASSEB) Class 11 Biology Chapter 14 : Breathing and Exchange of Gases Solutions & Question Answers

EXERCISES

Q1. Define vital capacity. What is its significance?
Answer: Vital Capacity (VC) is the maximum volume of air a person can breathe in after a forced expiration or breathe out after a forced inspiration. It is calculated as VC = ERV + TV + IRV. It shows the maximum amount of air that can move in and out of the lungs and is used to assess lung health. A lower VC may indicate a respiratory problem.

Q2. State the volume of air remaining in the lungs after a normal breathing.
Answer: The volume of air remaining in the lungs after a normal expiration is called the Functional Residual Capacity (FRC). It is calculated as FRC = ERV + RV. This includes the Expiratory Reserve Volume (ERV) and the Residual Volume (RV).

Q3. Diffusion of gases occurs in the alveolar region only and not in the other parts of the respiratory system. Why?
Answer: Diffusion of gases like O₂ and CO₂ occurs only in the alveolar region because alveoli have very thin walls and are surrounded by numerous capillaries, forming an efficient diffusion membrane. Other parts of the respiratory tract mainly transport, filter, and humidify air; they do not have the thin, vascular structure required for gas exchange.

Q4. What are the major transport mechanisms for CO₂? Explain.
Answer: CO₂ is transported in three main ways:

  1. As bicarbonate ions (HCO₃⁻) – about 70% of CO₂ reacts with water to form H₂CO₃, which dissociates into H⁺ and HCO₃⁻ under the action of carbonic anhydrase.
    CO₂ + H₂O ⇌ H₂CO₃ ⇌ H⁺ + HCO₃⁻
  2. As carbamino-haemoglobin – about 20–25% binds reversibly with haemoglobin to form HbCO₂.
  3. In dissolved form – about 7% is directly dissolved in plasma.
    At tissues, CO₂ is converted to bicarbonate for transport, and at alveoli, it is released back as CO₂ for exhalation.

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

Answer: The partial pressure of oxygen (pO₂) is higher, and the partial pressure of carbon dioxide (pCO₂) is lower in the atmospheric air than in the alveolar air. This happens because when air enters the lungs, it mixes with the air already present in the alveoli and also becomes humidified. As a result, oxygen concentration decreases, whereas carbon dioxide concentration increases slightly.

Hence, the correct option is (ii) pO₂ higher, pCO₂ lesser.


Respiratory GasAtmospheric AirAlveolar AirComparison
pO₂159 mm Hg104 mm HgHigher in atmospheric air
pCO₂0.3 mm Hg40 mm HgLower in atmospheric air

Q6. Explain the process of inspiration under normal conditions.
Answer: Inspiration is the process of drawing air into the lungs. It happens when the pressure inside the lungs becomes lower than atmospheric pressure. The diaphragm contracts and flattens, while the external intercostal muscles lift the ribs and sternum. This increases the thoracic volume, which reduces intrapulmonary pressure. As a result, air flows into the lungs.

Q7. How is respiration regulated?
Answer: Respiration is regulated by the brain to match the body’s oxygen needs and CO₂ removal. The medulla has a respiratory rhythm centre that controls basic breathing. The pneumotaxic centre in the pons modifies breathing rate, and a chemosensitive area responds to CO₂ and H⁺ concentration. Receptors in the aorta and carotid artery also sense these changes. Oxygen has a minor role in regulation.

Q8. What is the effect of pCO₂ on oxygen transport?
Answer: A high pCO₂ reduces oxygen binding to haemoglobin, promoting oxygen release (Bohr effect). In tissues with high pCO₂ and H⁺ levels, O₂ dissociates from oxyhaemoglobin, supplying oxygen to cells. In alveoli, where pCO₂ is low and pO₂ is high, haemoglobin readily binds O₂ to form oxyhaemoglobin.

Q9. What happens to the respiratory process in a man going up a hill?
Answer: At high altitudes, the atmospheric pressure and pO₂ decrease, reducing oxygen availability. The body responds by increasing breathing rate and depth through signals from the chemoreceptors and brain’s respiratory centre. Over time, the body adapts by producing more red blood cells to improve oxygen transport.

Q10. What is the site of gaseous exchange in an insect?
Answer: In insects, gaseous exchange occurs through a network of fine tubes called tracheae. These tubes deliver air directly to the body tissues, ensuring direct oxygen supply and CO₂ removal without blood involvement.

Q11. Define oxygen dissociation curve. Can you suggest any reason for its sigmoidal pattern?
Answer: The oxygen dissociation curve is an S-shaped (sigmoidal) graph showing the relationship between haemoglobin saturation and pO₂. Its shape is due to cooperative binding — once one O₂ molecule binds to haemoglobin, the affinity for the next increases, and once one O₂ is released, others are released more easily.

Q12. Have you heard about hypoxia? Try to gather information about it, and discuss with your friends.
Answer: Hypoxia is a condition where body tissues do not get enough oxygen. It can occur due to high altitudes, respiratory disorders, carbon monoxide poisoning, or poor circulation. Symptoms include breathlessness, dizziness, and bluish skin. At high altitudes, the body adapts by increasing breathing rate and red blood cell production.

Q13. Distinguish between(a) IRV and ERV
(b) Inspiratory capacity and Expiratory capacity.
(c) Vital capacity and Total lung capacity
Answer:  
(a) IRV and ERV

FeatureInspiratory Reserve Volume (IRV)Expiratory Reserve Volume (ERV)
DefinitionExtra air inspired by a forcible inspiration.Extra air expired by a forcible expiration.
Average Volume2500–3000 mL1000–1100 mL

(b) Inspiratory capacity and Expiratory capacity
Answer: 

FeatureInspiratory Capacity (IC)Expiratory Capacity (EC)
DefinitionTotal air inspired after a normal expiration.Total air expired after a normal inspiration.
FormulaIC = TV + IRVEC = TV + ERV

(c) Vital capacity and Total lung capacity
Answer: 

FeatureVital Capacity (VC)Total Lung Capacity (TLC)
DefinitionMaximum air exhaled after a forced inspiration.Total air in lungs after a forced inspiration.
FormulaVC = ERV + TV + IRVTLC = VC + RV
Key DifferenceDoes not include Residual Volume (RV).Includes Residual Volume (RV).

Q14. What is Tidal Volume? Find out the Tidal Volume (approximate value) for a healthy human in an hour.
Answer: Tidal Volume (TV) is the amount of air inhaled or exhaled during normal breathing. For a healthy adult, it is about 500 mL per breath. With an average of 14 breaths per minute, the total air exchanged in an hour is approximately 420 L (500 mL × 14 × 60 = 420,000 mL).

🎓 About Assam Eduverse

Assam Eduverse is the best educational platform in Assam, offering
SEBA, AHSEC (ASSEB), SCERT, CBSE, and Assam Board Solutions along with
study materials, notes, and exam preparation guides to help students
learn smarter and score higher.

Our expert-prepared answers and MCQs follow the latest
Assam Board and NCERT syllabus. We make learning simple, accessible, and effective for
all students preparing for board or competitive exams.

📘 Visit
Assam Eduverse
for free Assam Board Solutions, notes, and study materials prepared by experts.



Leave a Comment