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

Rajasthan Board RBSE Solutions for Class 11 Biology Chapter 17 Breathing and Exchange of Gases Textbook Exercise Questions and Answers.

RBSE Class 11 Biology Solutions Chapter 17 Breathing and Exchange of Gases

RBSE Class 11 Biology Breathing and Exchange of Gases Textbook Questions and Answers

Question 1. 
Define vital capacity. What is its significance? 
Answer:
Vital capacity is the maximum volume of air that can be exhaled after a maximum inspiration. It is sum total of inspiratory reserve volume, tidal volume and expiratory reserve volume.
VC = IRV + TV + ERV 
or = 3000 + 500 + 1000 
= 4500 mL
It promotes the act of supplying fresh air and getting rid of foul air, thereby increasing the gaseous exchange between the tissue and the environment.

Question 2. 
State the volume of air remaining in the lungs after a normal breathing.
Answer:
The volume of air in the lungs after a normal expiration is known as functional residual capacity (FRC). It equals to the sum of total expiratory reserve volume and residual volume. It is about 2200 (500 + 3000) mL and 2400 (500 + 1900) mL in men and women respectively.

Question 3. 
Diffusion of gases occurs in the alveolar region only and not in the other parts of respiratory system. Why?
Answer:
Each alveoli is made up of highly permeable and thin layers of squamous epithelial cells. Similarly, the blood capillaries have layers of squamous epithelial cells. Oxygen rich air enters the body through the nose and reaches the alveoli. The deoxygenated (CO₂ - rich) blood from the body is brought to the heart by the veins. The heart pumps it to the lungs for oxygenation. The exchange of O₂ and CO₂ takes place between the blood capillaries surrounding the alveoli and the gases present in the alveoli.

Thus, the alveoli are the sites for gaseous exchange. The exchange of gases takes place by simple diffusion because of pressure or concentration differences. The barrier between the alveoli and the capillaries is thin and the diffusion of gases takes place from higher partial pressure to lower partial pressure. The venous blood that reaches the alveoli has lower partial pressure of O₂ and higher partial pressure of CO₂ as compared to alveolar air. Hence, oxygen diffuses to blood. Simultaneously, CO₂ diffuses out of blood and into alveoli.

Question 4. 
What are the major transport machanisms for CO₂? Explain.
Answer:
The CO₂ is produced during oxidation of food substances stored in tissues. It moves into blood capillaries by diffusion. The transport of CO₂ in blood occurs in the following ways:
1. In the form of carbonic acid: CO₂ is highly soluble in water. About 10% of CO₂ combines with water of plasma to form carbonic acid (H₂CO₃).
CO₂ +H₂O → H₂CO₃
About 10% part of total CO₂ remains as H₂CO₃ in blood and rest part breaks down rapidly into hydrogen and bicarbonate ions.
H₂CO₃ → HCO₃^- + H^-

2. In the form of Bicarbonate: About 70% of the CO2 is transported through the plasma in the form of bicarbonates of sodium and potassium.
CO₂ + H₂O⇌ H₂CO₃ ⇌ HCO_3^-  + H⁺
HCO₃ ^- + Na⁺ → NaHCO₃ (Sodium bicarbonate) 
HCO₃ ^- + K⁺ → KHCO₃ (Potassium bicarbonate)

3. In the form of Carboxy haemoglobin : Some CO₂ combines with haemoglobin of RBCs to form an unstable compound carboxyhaemoglobin.
Hb + 4CO₂ → Hb(CO₂)₄

4. In the form of carbamino compound: About 7% CO₂ combines with protein of plasma to form carbamino compound.
Plasma protein + CO₂ → Carbamino compound (unstable)
The blood with carbonic acid, sodium and potassium bicarbonates, carboxyhaemoglobin and carbamino compound is impure blood. This blood goes from capillaries to heart through veins and from here to lungs through pulmonary arteries. From lungs CO₂ diffuses to alveoli.

5. Release of CO₂ from unstable substances: The amount of carboxyhaemoglobin is more in blood capillaries near the lungs. This is more acidic. Due to having acidic nature of carboxyhaemoglobin, all the unstable compounds break down from it and release CO₂.
2NHCO₃ → Na₂CO₃ + H₂O + CO₂
2KHCO₃ → K₂CO₃ + H₂O + CO₂
H₂CO₃ → H₂O + CO₂
Hb(CO₂ )₄ → Hb + 4CO₂
Such released CO₂ reaches to lung alveoli by diffusion through walls of thin capillaries. From where it is expelled out by expiration.

Question 5. 
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:
pO₂ higher, pCO₂ lesser: The partial pressure of oxygen in atmospheric air is higher than that of oxygen in alveolor air. In atmospheric air, pO₂ is about 159 mm Hg. In alveolar air it is about 104 mm Hg. The partial pressure of carbon dioxide in atmospheric air is lesser than that of carbon dioxide in alveolar air. In atmospheric air, pCO₂ is about 0.3 mm Hg. In alveolar air, it is about 40 mm Hg.

Question 6. 
Explain the process of inspiration under normal conditions.
Answer:
 Inspiration: The entry of atmospheric oxygen rich air into the lungs is called inspiration. This occurs as a result of increase in the volume of thoracic cavity due to contraction of external intercostal muscles and diaphragm. As a result of contraction in the muscle fibres of diaphragm it becomes flattened and comes downwards thereby displacing the abdominal organs laterally and anteriorly which finally increases the volume of thoracic cavity. With contraction of the diaphragm, external intercostal muscles also get contracted which raises every rib upward and the sternum is also moved upward and forward resulting into increase in the volume of thoracic cavity and lungs get expended. In this way, due to increase in the volume of lungs, the air pressure inside the alveoli becomes 1 to 3 mm Hg lesser than the atmospheric air pressure. As a result of this atmospheric air enters through external nares and after passing through nasal passage, internal nares, pharynx, larynx, trachea, primary, secondary and tertiary bronchi, bronchioles, respiratory bronchioles, alveolar ducts and sacs, finally gets filled in the lung alveoli.

 
Question 7. 
How is respiration regulated?
Answer:
Regulation of respiration: The respiratory rhythm centre present in the medulla region of the brain is primarily responsible for the regulation of respiration. The pneumotaxic centre can alter the function performed by the respiratory rhythm centre by signaling to reduce the inspiration rate. The chemosensitive region present near the respiratory centre is sensitive to carbon dioxide and hydrogen ions. This region then signals to change the rate of expiration for eliminating the compounds.
The receptors present in the carotid artery and aorta detect the levels of carbon dioxide and hydrogen ions in blood. As the level of carbon dioxide increases, the respiratory centre sends nerve impulses for the necessary changes.

Question 8. 
What is the effect of pCO₂ on oxygen transport?
Answer:
Effect of pCO₂ on transport of Oxygen: pCO₂ plays, an important role in the transportation of oxygen. At the alveolus, the low pCO₂ and high pO₂ favours the formation of oxyhaemoglobin. At the tissues, the high pCO₂ and low pO₂ favours the dissociation of oxygen from oxyhaemoglobin. Hence, the affinity of haemoglobin for oxygen is enhanced by the decrease of pCO₂ in blood. Therefore, oxygen is transported in blood as oxyhaemoglobin and oxygen dissociates from it at the tissues.

Question 9. 
What happens to the respiratory process in a man going up a hill?
Answser:
As altitude increases, the oxygen level in the atmosphere decreases. Therefore, as a man goes uphill, he gets less oxygen with each breath. This causes the amount of oxygen in the blood to decline. The respiratory rate increases in response to the decrease in the oxygen content of blood. Simultaneously, the rate of heart beat increases to increase the supply of oxygen to blood. 

Question 10. 
What is the site of gaseous exchange in an insect?
Answer:
Breathing Mechanism in insect: In insects, gaseous exchange occurs through a network of tubes collectively known as the tracheal system. The small openings on the sides of an insect’s body are known as spiracles. Oxygen rich air enters through the spiracles. The spiracles are connected to the network of tubes. From the spiracles, oxygen enters the tracheae. From here, oxygen diffuses into the cells of the body. The movement of carbon dioxide follows the reverse path. The CO₂ from the cells of the body first enters the tracheae and then levaes the body through the spiracles.

Question 11. 
Define oxygen dissociation curve. Can you suggest any reason for its sigmoidal pattern?
Answer:
Usually 98.5% of the total oxygen is transportated by the haemoglobin and rest 1.5% oxygen is transported by the blood plasma. The haemoglobin combines with oxygen to gives rise oxyhaemoglobin at high oxygen tension and get dissociated to release this oxygen at low oxygen tension.

It is clear that the association of oxygen and haemoglobin is reversible process which depends upon the oxygen tension. Since there are four ferrous ions per haemoglobin molecule, hence a haemoglobin molecule can be associated with upto four oxygen molecules.
Hb + 4O₂ ⇌ Hb(O₂)₄
This sigmoid shape of the dissociation curve is because of the binding of oxygen to haemoglobin. As the first oxygen molecule binds to haemoglobin, it increases the affinity for the second molecule of oxygen to bind. Subsequently, haemoglobin attracts more oxygen.

Question 12. 
Have you heared about hypoxia? Try to gather information about it, and discuss with your friends.
Answer:
Hypoxia: Hypoxia is a condition characterised by an inadequate or decreased supply of oxygen to the lungs. It is caused by several extrinsic factors such as reduction in pO₂ . Inadequate oxygen etc. The different types of hypoxia are discussed below:

Hypoxemic hypoxia: In this condition, there is a reduction in the oxygen content of blood as result of the low partial pressure of oxygen in the arterial blood. 
Anaemic hypoxia: In this condition there is a reduction in the concentration of haemoglobin.

Stagnant or ischemic hypoxia: In this condition, there is a deficiency in the oxygen content of blood because of poor blood circulation. It occurs when a person is exposed to cold temperature for a prolonged period of time.

Histotoxic hypoxia: In this condition, tissues are unable to use oxygen. This occurs during carbon monoxide or cyanide poisoning.

Question 13. 
Distinguish between:
(a) IRV and ERV
(b) Inspiratory capacity and expiratory capacity
(c) Vital capacity and total lung capacity.
Answer:
(a) Differences between IRV and ERV:

(b) Differences between Inspiratory Capacity and Expiratory Capacity:

(c) Differences between Vital Capacity and Total Lung Capacity:

Question 14. 
What is Tidal volume? Find out the Tidal volume (approximate value) for a healthy human in an hour.
Answer:
Tidal volume: Tidal volume is the volume of air inspired or expired during normal breathing. It is about 6000 to 8000 mL of air per minute.
The hourly tidal volume for a healthy human can be calculated as:
Tidal volume in an hour
= 6000 to 8000 mL x (60 min)
= 3.6 x 10⁵ to 48 x 10⁵ mL
Therefore, the hourly tidal volume for a healthy human is approximately 3.6 x 10⁵ to 48 x 10⁵ mL.