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

NCERT Solved Exercise Questions – Class 11 Biology Chapter 17 Breathing and Exchange of Gases

17.1. Define vital capacity. What is its significance?

Ans – After performing a forced expiration, a normal adult can breathe in a maximum of around 4000 mL of air. Singers and sportsmen have larger vital capacities. The lung vital capacity is lower in cigarette smokers. This includes the greatest amount of air a person may exhale following a forceful inhalation, also known as ERC, TV, and IRV.

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

Ans – The amount of air in the lung following a normal expiration is known as functional residual capacity when someone breathes normally. It is the result of adding the expiratory reserve volume and the residual volume (FRC = RV + ERV). There are about 2100 – 2300 mL of air.

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

Ans – Respiratory surfaces must have specific qualities in order to effectively exchange gases, such as 

(i) I being thin, permeable to respiratory gases, and being meist.

(ii) It must be very vascular; 

(iii) it must have a vast surface area. These qualities are unique to the alveolar area. Gases therefore only diffuse in this area.

Exchange of gases in alveoli

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

Ans – For every 100 ml of blood, approximately 4 ml of carbon dioxide is carried.

The blood carries CO2 in three different forms.

(i) About 7% of carbon dioxide dissolves in blood plasma in the dissolved state, exactly like it does in water.

(ii) As erythrocytes contain a significant amount of carbonic anhydrase, an enzyme that catalyses the following processes;

mechanisms for CO2

(iii) Carbaminohaemoglobin is created when carbon dioxide (C02) reacts with the globin portion of haemoglobin. This method is used to transfer about 23% of CO2.

17.5. 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

Ans – (ii)Alveolar air is that which has passed through the bronchioles and into the alveoli. CO2 and 02 partial pressures are the same as those found in ambient air. Then, gaseous exchange takes place between the alveoli and nearby blood capillaries. O2 diffuses from the alveolar air into the blood, while CO2 diffuses from the blood into the alveolar air. As a result, compared to atmospheric air, new alveolar air has higher pCO2and lower pO2.

17.6. Explain the process of inspiration under normal conditions.

Ans – The process through which new air enters the lungs is called inspiration. An essential role is played by the diaphragm, intercostal muscles, and abdominal muscles. The main muscles of inspiration are the diaphragm and external intercostal muscles. Through the contraction of the diaphragm and external intercostal muscles, the thoracic cavity’s volume increases. The relaxation of the abdominal muscles during inspiration enables the diaphragm to compress the abdominal organs. As a result, the thoracic cavity’s overall volume rises, which lowers the air pressure in the lungs. Air now enters the lungs quickly due to the increased pressure outside the body. The order of the air flow is.

17.7. How is respiration regulated?

Ans – Both the neurological system and the body’s chemistry control respiration.
Neuronal clusters in the medulla oblongata and pons varolii make up the brain’s respiratory centre. The respiratory centre controls how quickly and deeply we breathe.
The dorsal region of the medulla oblongata houses the dorsal respiratory group of neurons. Inspiration is mostly produced by this cluster of neurons.
The ventrolateral region of the medulla oblongata houses the ventral group of neurons. Both inspiration and expiry may result from these.

The dorsal portion of the pons varolii is home to the pneumotaxic centre. It communicates with all of the dorsal respiratory group’s neurons and only the ventral respiratory group’s inspiratory neurons. Its primary duty is to restrict inspiration. Chemoreceptors, which are found in great numbers in the carotid and aortic bodies, control respiration chemically. The respiratory centre of the brain is primarily stimulated by excess carbon dioxide or hydrogen ions, which enhances the inspiratory and expiratory impulses to the breathing muscles. Acidosis results from lowered pH caused by increased CO2. The modulation of respiratory rhythm is mostly independent of oxygen.

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

Ans – Blood pressure increases cause a rightward shift in the oxygen dissociation curve of haemoglobin, which lowers the molecule’s affinity for oxygen. It is known as the Bohr effect. In the release of oxygen into the tissues, it is crucial.

 pCO2 on oxygen transport

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

Ans – High heights cause the PO2 level to decline. As a result, the transport of oxygen from the alveolar air to the blood is reduced, lowering the alveolar PO2. As a result, the blood’s oxygenation is gradually reduced.

After some time, the affected person adjusts to their circumstances, which causes an increase in their heart rate, blood RBC count, haemoglobin level, and oxygen-carrying capacity.

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

Ans – In order to exchange gases between their surroundings and their bodily cells, insects develop a sophisticated network of air tubes called tracheae (tracheal respiration).

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

Ans – A curve known as the oxygen haemoglobin dissociation curve serves as a visual representation of the relationship between the partial pressure of oxygen (pO2) and % saturation of the haemoglobin with oxygen (O2) (also called oxygen dissociation curve).

Two characteristics that are crucial to the transport of oxygen lead to the sigmoidal shape of the oxygen haemoglobin dissociation curve. These two attributes are:

(i) Significant variations in oxygen tension occur above p02 of 70–80 mm Hg despite minimal oxygen loss from haemoglobin above this. This is represented by the curve’s somewhat flat part.

(ii) The oxygen released from haemoglobin is disproportionately increased as p02 continues to fall below 40 mm Hg. It causes the curve and the steeper part of the curve. to be sigmoid.

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

Ans – When there is a lack of oxygen in the tissues, it is known as hypoxia.

1. Arterial hypoxia: This occurs when there is not enough oxygen in the blood. It happens when there is an obstruction in the respiratory route and there is not enough oxygen in the atmosphere.

2. Anaemic hypoxia: This condition results from extremely low blood haemoglobin levels.

Stagnant hypoxia: This condition occurs when there is insufficient blood flow to provide oxygen to the tissue.

3. Histoxic hypoxia: This condition results from the inhalation of toxic compounds, such as cyanide poisoning, that are present in the oxygen.

17.13. Distinguish between
(a) IRV and ERV
(b) Inspiratory capacity and Expiratory capacity.
(c) Vital capacity and Total lung capacity.
Ans – (a)



 Inspiratory capacity and Expiratory capacity.


Vital capacity and Total lung capacity

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

Ans – The amount of air that is inhaled or exhaled with each regular breath is known as tidal volume. In an adult, this equates to around 500 mL. It consists of around 150 mL of dead space and 350 mL of alveolar volume. The air that enters the alveoli’s respiratory surfaces and conducts gas exchange makes up the alveolar volume. The air that does not reach the respiratory surfaces makes up the dead space volume.

An average guy can breathe in or out between 6000 and 8000 mL of air every minute. A healthy person’s hourly tidal volume is therefore between 360 and 480 mL of air.

Class 11 Biology Chapter 17 Breathing and Exchange of Gases Solved Exercise Questions Free PDF Download

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