NCERT Solutions for Class 11 Biology Chapter 17: Breathing and Exchange of Gases
1. Define vital capacity. What is its significance?
Ans. Vital capacity (VC) is the maximum volume of air a person can breathe in after a forced expiration. It is about 4000 ml–4600 ml for human body. Significance: It promotes the act of supplying fresh air and getting rid of foul air. In this way it increases the gaseous exchange between the tissues and the environment.
2. State the volume of air remaining in the lungs after a normal breathing.
Ans. The volume of air remaining in the lungs after a normal expiration is known as functional residual capacity (FRC). It includes expiratory reserve volume (ERV) and residual volume (RV). ERV is the maximum volume of air that can be exhaled after a normal expiration. It is about 1000 mL to 1500 mL. RV is the volume of air remaining in the lungs after maximum expiration. It is about 1100 mL to 1500 mL.
∴ FRC = ERV + RV
≅ 1500 + 1500
≅ 3000 mL
Functional residual capacity of the human lungs is about 2500 – 3000 mL.
3. Diffusion of gases occurs in the alveolar region only and not in the other parts of respiratory system. Why?
Ans. The alveolar region consists of following features that makes it suitable for diffusion of gases which other parts of respiratory system do not have.
(a) The diffusion membrane of alveoli is extremely thin.
(b) It consists of thin squamous epithelium of alveoli, the endothelium of alveolar capillaries and the basement substance in between them which makes it highly vascular and extensive for gas exchange.
(c) All these layers have total thickness of much less than a millimeter.
Due to these features, only the alveolar region is suitable for the gas exchange.
4. What are the major transport mechanisms for CO2? Explain.
Ans. CO2 is carried by haemoglobin in RBCs as carbamino-haemoglobin (about 20-25 percent). About 70 percent of it is carried as bicarbonate at the tissue level and transported to the alveoli and is released out as CO2. The remaining 7 percent of CO2 is carried in a dissolved state through plasma.
5. What will be the pO2 and pCO2 in the atmospheric air compared to those in the
alveolar air ?
(a) pO2 lesser, pCO2 higher
(b) pO2 higher, pCO2 lesser
(c) pO2 higher, pCO2 higher
(d) pO2 lesser, pCO2 lesser
Ans. (b) pO2 higher, pCO2 lesser
The partial pressure of oxygen (pO2) in atmospheric air is higher than that of oxygen in alveolar air. In atmospheric air, pO2 is about 159 mm Hg while in alveolar air, it is about 104 mm Hg. The partial pressure of carbon dioxide (pCO2) in atmospheric air is lesser than that of carbon dioxide in alveolar air. In atmospheric air, pCO2 is about 0.3 mm Hg while, in alveolar air, it is about 40 mm Hg.
6. Explain the process of inspiration under normal conditions.
Ans. The respiratory process by which fresh air enters the lungs is called inspiration. The following events occur during this process.
(a) Diaphragm flattens, extending the superior/ inferior dimension of the thoracic cavity.
(b) External intercostal muscles elevate the ribs and sternum, extending the anterior/ posterior dimension of the thoracic cavity.
(c) The increase in the thoracic volume results in increase in pulmonary volume.
(d) The increase in pulmonary volume decreases the intra-pulmonary pressure to less than the atmospheric pressure.
(e) This pressure gradient forces the air from outside to move into the lungs and inspiration takes place.
7. How is respiration regulated?
Ans. Respiration is regulated by the following mechanisms:
(a) A specialised respiratory rhythm centre is present in the medulla region of the brain. It is primarily responsible for the regulation of respiration.
(b) The pneumotaxic centre present in the pons region of the brain can moderate the functions of the respiratory rhythm centre. Neural signal from this centre can reduce the duration of inspiration and thereby alter the respiratory rate.
(c) A chemosensitive area is situated adjacent to the rhythm centre which is highly sensitive to CO2 and hydrogen ions. The increase in CO2 and hydrogen ions can activate this centre, which in turn can signal the rhythm centre to make necessary adjustments in the respiratory process by which these substances can be eliminated.
(d) The receptors associated with aortic arch and carotid artery recognise changes in CO2 and H+ concentration and send necessary signals to the rhythm centre for remedial actions.
(e) The role of oxygen in the regulation of respiratory rhythm is quite insignificant.
9. What happens to the respiratory process in a man going up a hill?
Ans. The concentration of atmospheric oxygen is lower at the higher altitude so the partial pressure of oxygen decreases. The deficiency of oxygen in this condition needs more oxygen to fulfill the body demand. As a result, the man begins to breathe rapidly to compensate the intake of oxygen to the blood. This leads to increased heart rate to meet the demand of oxygen supply.
10. What is the site of gaseous exchange in an insect?
Ans. Tracheal system in insects is responsible for the gaseous exchange. It is the network of tubes in which the small openings called spiracles are located. The oxygen rich air enters through these spiracles. The oxygen then 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 leaves the body through the spiracles.
11. Define oxygen dissociation curve. Can you suggest any reason for its sigmoidal pattern?
Ans. The oxygen dissociation curve is a graph which shows the percent saturation of oxyhaemoglobin at various partial pressures of oxygen. The sigmoid curve shows the equilibrium of oxyhaemoglobin and haemoglobin at various partial pressures. The oxygen content of hemoglobin increases as PO2 increases until the maximum capacity is reached. As this limit is approached, very little additional binding occurs. Due to this, the curve levels out as the hemoglobin becomes saturated with oxygen. This makes the curve sigmoid or S-shaped.
12. Have you heard about hypoxia? Try to gather information about it, and discuss with your friends.
Ans. The condition in which there is an insufficient supply of oxygen to the tissues to maintain normal cellular function.
Symptoms: The symptoms of hypoxia are headache, difficulty in breathing, coughing, wheezing, confusion and bluish colour in skin, fingernails and lips (cyanosis).
Causes: The following are the causes of hypoxia:
(a) Low oxygen pressure at higher altitudes. It is very common in mountain climbers and is known as altitude sickness.
(b) Paralysis of respiratory muscles as in the case of poliomyelitis, motor neuron diseases.
(c) Damage to the respiratory centers of brain due to brain tumors or due to the consumption of toxic drugs.
13. Distinguish between:
(a) IRV and ERV
(b) Inspiratory capacity and Expiratory capacity.
(c) Vital capacity and Total lung capacity.
Ans. (a) Differences between Inspiratory Reserve Volume (IRV) and Expiratory Reserve Volume (ERV):
(b) Differences between Inspiratory Capacity (IC) and Expiratory Capacity (EC):
|Inspiratory Reserve Volume (IRV)||Expiratory Reserve Volume (ERV)|
||ERV is the additional volume of air that can be expired by a forcible expiration.|
||It is about 1000 – 1100 mL in the human lungs.|
|Inspiratory Capacity (IC)||Expiratory Capacity (EC)|
||Expiratory capacity is the volume of air that can be exhaled after a normal inspiration.|
||It includes tidal volume and expiratory reserve volume.|
||EC = TV + ERV|
(c) Differences between Vital Lung Capacity (VC) and Total Lung Capacity (TLC):
|Vital Lung Capacity (VC)||Total Lung Capacity (TLC)|
||It is the volume of air in the lungs after maximum inspiration.|
||It includes IC, ERV, and residual volume.|
||It is about 5100 - 5800 mL in the human lungs.|
14. What is Tidal volume? Find out the Tidal volume (approximate value) for a healthy human in an hour.
Ans. Tidal volume is the volume of air inspired or expired during a normal respiration.
In a healthy man, tidal volume is about 500 mL. It means a healthy man can inspire or expire approximately 6000 to 8000 mL of air per minute (because a healthy human breathes 12-16 times per minute). Thus, Tidal volume (approx. value) for a healthy human in one hour will be = 360000 ml – 480000 ml.
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