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

1

In the 'Breathing and Exchange of Gases' chapter, the functionality of oxygen and carbon dioxide have been thoroughly explained. As we all know, organisms use oxygen to indirectly break down simple molecules, such as amino acids, fatty acids, glucose, etc. and derive energy that is utilized to carry out various activities. Carbon dioxide, which is considered harmful for, both living organisms and the environment, is also let out when the above catabolic reactions occur.

Hence, in this chapter, it has been made evident that the cells require an uninterrupted supply of oxygen and the carbon dioxide that is produced by the cells needs to be sent out continuously. This process of giving and taking carbon dioxide produced by the cells and oxygen from nature respectively is called breathing, and scientifically respiration. Want to know more about the topics of this chapter, read below.

  • Respiratory Organs in organisms
  • Breathing Mechanics
  • How are Gases Exchanged?
  • How are Gases transported?
  • How is respiration regulated?
  • Problems of Respiratory System

The students get an insight into the mechanism of breathing from this chapter. A human's respiratory system consists of respiratory organs that assist in the breathing process. Breathing, technically termed respiration, has been defined in the chapter as an exchange process between two gases, i.e. inhale of oxygen from the atmosphere and exhale of carbon dioxide produced by the cells, from the body into the atmosphere.

The whole chapter majorly talks about the meaning, definitions, importance, and differentiation of respiration and breathing. A few other relevant topics covered under this chapter are different respiratory volumes and capacities, transport and exchange of gases, major respiratory disorders, and more.

 

Q1. Define vital capacity. What is its significance?
Answer: Vital capacity refers to the maximum volume of air that can be breathed in after forced expiration. It is about 3.5 to 4.5 liters in the human body.
It promotes the act of supplying fresh air and getting rid of foul air, thereby increasing the gaseous exchange between the tissues and the environment.

Q2. State the volume of air remaining in the lungs after normal breathing.
Answer: The volume of air remaining in the lungs after normal breath out is called functional residual capacity (FRC). It combines expiratory reserve volume (ERV) and residual volume (RV). ERV is the maximum volume of air that can be exhaled after a normal expiration and it is about 1000 mL to 1500 mL. RV, on the other hand, refers to the volume of air remaining in the lungs after maximum expiration and is about 1100 mL to 1500 mL.
Thus, FRC = ERV + RV 1500 + 1500 = 3000 mL
Hence, the functional residual capacity of the human lungs is about 2500 - 3000 mL.

 Q3. Diffusion of gases occurs in the alveolar region only and not in the other parts of the respiratory system. Why?
Answer: For efficient exchange of gases, the respiratory surface must have certain characteristics such as:
1. It must be thin, moist, and permeable to respiratory gases.
2. It must have a larger surface area.
3.It must be highly vascular. 
Only the functional alveolar region has these characteristics. Thus, diffusion of gases occurs in this region only.

Q4. What are the major transport mechanisms for CO2? Explain.
Answer: The major transport mechanism for CO2 is transported by sodium bicarbonate as well as red blood cells. See, about 70% of carbon dioxide is transported as sodium bicarbonate. As CO2 diffuses into the blood plasma, a large part of it combines with water to form carbonic acid in the presence of the enzyme carbonic anhydrase. Carbonic anhydrase is a zinc enzyme that speeds up the formation of carbonic acid. This carbonic acid dissociates into bicarbonate (HCO3- ) and hydrogen ions (H+ ). About 20 – 25% of CO2 is transported by the red blood cells as carbaminohaemoglobin. Carbon dioxide binds to the amino groups on the polypeptide chains of hemoglobin and forms a compound known as carbaminohaemoglobin.

Q5. What will be the pO2 and pCO2 in the atmospheric air compared to those in the alveolar air?
(i) p O_2 lesser, p CO_2 higher
(ii)
p O_2 higher, p CO_2 lesser
(iii)
p O_2 higher, p CO_2 higher
(iv)
p O_2 lesser, p CO_2 lesser
Answer: PO
2 higher, pCO2 lesser
As the partial pressure of oxygen in atmospheric air is higher than that of oxygen in alveolar air. So, in atmospheric air, pO2 is about 159 mm Hg and 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, pCO 2 is about 0.3 mmHg and in alveolar air, it is about 40 mm Hg.

Q6. Explain the process of inspiration under normal conditions.
Answer: Inspiration is the process, during which atmospheric air is drawn inside the body. It can occur if the pressure within the lungs (intra-pulmonary pressure) is less than the atmospheric pressure, i.e., there is a negative pressure in the lungs for atmospheric pressure.
It is initiated by the contraction of the diaphragm which increases the volume of the thoracic chamber in the anteroposterior axis. The contraction of external intercostal muscles lifts the ribs and the sternum causing an increase in the volume of the thoracic chamber in the dorso-ventral axis. The overall increase in the thoracic volume causes a similar increase in pulmonary volume. An increase in pulmonary volume decreases the intra- pulmonary pressure to less than the atmospheric pressure which forces the air from outside to move into the lungs.


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Q7. How is respiration regulated?
Answer: Respiration is regulated by the respiratory center present in the brain stem. The pneumatic center and the apneustic center in the pons and medulla rhythmicity center that regulates the respiratory system. The neural signal from this center can reduce the duration of inspiration and thereby alter the respiratory rate.
A chemosensitive area is situated adjacent to the rhythm center which is highly sensitive to CO
2 and hydrogen ions. An increase in these substances can activate this center, which in turn can signal the rhythm center to make necessary adjustments in the respiratory process by which these substances can be eliminated. Receptors associated with the aortic arch and carotid artery also can recognize changes in CO 2 and H+ concentration and send necessary signals to the rhythm center for remedial actions. The role of oxygen in the regulation of respiratory rhythm is quite insignificant.

Q8.  What is the effect of pCO2 on oxygen transport?
Answer: The pCO2 (partial pressure of carbon dioxide) plays an important role in the transportation of oxygen as CO2 shows a higher affinity to hemoglobin than that O2. The low pCO2 and high pO2 favor the formation of oxyhemoglobin in the alveoli because CO2 in atmospheric air is lesser than that of gaseous oxygen (O2). And at the tissues, the high pCO2 and low pO2 favor the dissociation of oxygen from oxyhemoglobin. So, the affinity of hemoglobin for oxygen is enhanced by the decrease of pCO2 in blood. Therefore, oxygen is transported in the blood as oxyhemoglobin and oxygen dissociate from it at the tissues.

Q9. What happens to the respiratory process in a man going up a hill?
Answer:
- When a man goes uphill, he gets less oxygen with each breath, and as we know that when altitude increases, the oxygen level in the atmosphere decreases.
- Because of the decrease in atmospheric O2, the amount of oxygen in the blood starts to decline because of less oxygen content breathe in air.
- The respiratory rate increases in order to respond to the decrease in the oxygen content of the blood.
- Simultaneously, the rate of heartbeat increases to increase the supply of oxygen to the blood.

Q10. What is the site of gaseous exchange in an insect?
Answer: In insects, gaseous exchange takes place through a network of tubes collectively known as the tracheal system. The small openings on the sides of an insect’s body are called spiracles and oxygen-rich air enters through the spiracles. These spiracles are connected to the network of tubes. From the spiracles, oxygen enters the tracheae and from here, oxygen diffuses into the cells of the body. The circulation of carbon dioxide follows the reverse path and the CO2 from the cells of the body first enters the tracheae and then leaves the body through the spiracles.

Q11. Define oxygen dissociation curve. Can you suggest any reason for its sigmoidal pattern?
Answer: The oxygen dissociation curve is a graph that shows the percentage of saturation of oxyhemoglobin at various partial pressures of oxygen. This curve shows the equilibrium of oxyhemoglobin and hemoglobin at various partial pressures. In the case of the lungs, the partial pressure of oxygen is high. So, hemoglobin binds to oxygen and forms oxyhemoglobin.
 

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Q12. Have you heard about hypoxia? Try to gather information about it, and discuss it with your friends.
Answer: Hypoxia is a type of condition characterized by an inadequate or decreased supply of oxygen to the lungs. It is caused by several extrinsic factors such as a reduction in pO2, inadequate oxygen, etc. It can be also classified as either generalized, affecting the whole body, or local, affecting a region of the body.
Different types of hypoxia are: Anemic hypoxia and Hypoxemic hypoxia

Q13. (a) Distinguish between IRV and ERV
Answer:

Inspiratory reserve volume(IRV)

Expiratory reserve volume(ERV)

It is the maximum amount of air that can be inhaled after a normal inhalation.

It is the maximum amount of air that can be exhaled out after a normal exhalation.

About 2500-3500 ml of air can be inhaled after a normal inhalation.

About 1000-1500 ml of air can be exhaled after the normal exhalation.

Q14.  Distinguish between inspiratory capacity and Expiratory capacity.
Answer: Difference between Inspiratory capacity and Expiratory capacity: 
Inspiratory capacity: It is the volume of air that can be inhaled after a normal expiration.
- IC = TV + IRV
Expiratory capacity: It is the volume of air that can be exhaled after a normal inspiration.
- EC = TV + ERV

Q15.  Distinguish between Vital capacity and Total lung capacity.
Answer:

Vital lung capacity

Total lung capacity

Vital capacity (VC) is the maximum amount of air a person can breathe in after a forced breathe out.

Total lung capacity (TLC) is the volume of air in the lungs after a forced inspiration.

It includes IC and ERV.

It includes IC, ERV, and residual volume.

It is about 4000 milliliters in the human lungs.

It is about 5000-6000 milliliters in the human lungs.

Vital lung capacity is calculated by summing tidal volume(VT), inspiratory reserve volume (IRV), and expiratory reserve volume (ERV).

 Therefore, VC = TV+IRV+ERV.

Total lung capacity is calculated by summing tidal volume (TV), inspiratory reserve volume (IRV), expiratory reserve volume (ERV), and residual volume(RV).

Therefore, TLC = TV+IRV+ERV+RV

 

Q16. What is Tidal volume? Find out the Tidal volume (approximate value) for a healthy human in an hour.
Answer: Tidal volume is the volume of air that is transported into and out of the lungs ( inspired or expired ) with each normal respiratory cycle. Tidal volume is approximately 6000 to 8000 mL of air per minute for a healthy human.
We can calculate the hourly tidal volume for a healthy human. If, Tidal volume = 6000 to 8000 mL/minute

So, the Tidal volume in an hour will be = 6000 to 8000 mL × (60 min)
= 3.6 × 10 5 mL to 4.8 × 10 5 mL
Hence, the hourly tidal volume for a healthy human is approximately 360000 ml-480000 ml.
 

 

 

Also See    
NCERT Solutions for Class 11 Biology Chapter 1 - The Living World NCERT Solutions for Class 11 Biology Chapter 2 - Biological Classification NCERT Solutions for Class 11 Biology Chapter 3 - Plant Kingdom
NCERT Solutions for Class 11 Biology Chapter 4 - Animal Kingdom NCERT Solutions for Class 11 Biology Chapter 5 - Morphology of Flowering Plants NCERT Solutions for Class 11 Biology Chapter 6 - Anatomy of Flowering Plants
NCERT Solutions for Class 11 Biology Chapter 7 - Structural Organization in Animals NCERT Solutions for Class 11 Biology Chapter 8 - Cells: The Unit of Life NCERT Solutions for Class 11 Biology Chapter 9 - Biomolecules
NCERT Solutions for Class 11 Biology Chapter 10 - Cell Cycle and Division NCERT Solutions for Class 11 Biology Chapter 11 - Transport in Plants NCERT Solutions for Class 11 Biology Chapter 12 - Mineral Nutrition
NCERT Solutions for Class 11 Biology Chapter 13 - Photosynthesis in Higher Plants NCERT Solutions for Class 11 Biology Chapter 14 - Respiration in Plants NCERT Solutions for Class 11 Biology Chapter 15 - Plant Growth and Development
NCERT Solutions for Class 11 Biology Chapter 16 - Digestion and Absorption NCERT Solutions for Class 11 Biology Chapter 18 - Body Fluids and Circulation NCERT Solutions for Class 11 Biology Chapter 19 - Excretory Products and their Elimination
NCERT Solutions for Class 11 Biology Chapter 20 - Locomotion and Movement NCERT Solutions for Class 11 Biology Chapter 21 - Neural Control and Coordination NCERT Solutions for Class 11 Biology Chapter 22 - Chemical Coordination and Integration


 

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