NCERT Solutions for Class 11 Biology Chapter 14 Breathing and Exchange of Gases

NCERT
SOLUTIONS
CLASS - 11th

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Class : 11th
Subject : Biology
Chapter : 17
Chapter Name : Breathing And Exchange Of Gases

Q1 De ne vital capacity. What is its signi cance?

Answer. Vital capacity is the maximum volume of air that can be exhaled after a maximum
inspiration. It is about 3.5 — 4.5 litres 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 a normal breathing.

Answer. 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.

Q3 Diffusion of gases occurs in the alveolar region only and not in the other parts of respiratory
system. Why?

Answer. Each alveolus 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 (carbon dioxide-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 O and CO takes place between the blood capillaries surrounding
2 2

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 02 and
higher partial pressure of C02 as compared to alveolar air. Hence, oxygen diffuses into blood.
Simultaneously, carbon dioxide diffuses out of blood and into the alveoli.

Q4 What are the major transport mechanisms for CO ? Explain 2

Answer. Answer Plasma and red blood cells transport carbon dioxide. This is because they are
readily soluble in water.
(I) Through plasma:
About 7% of CO is carried in a dissolved state through plasma. Carbon dioxide combines with
2

water and forms carbonic acid.
CO + H O = H CO
2 2 2 3

Since the process of forming carbonic acid is slow, only a small amount of carbon dioxide is carried
this way.
(2) Through RBCs: About 20 — 25% of CO is transported by the red blood cells as
2

carbaminohemoglobin. Carbon dioxide binds to the amino groups on the polypeptide chains of
haemoglobin and forms a compound known as carbaminohemoglobin.
(3) Through sodium bicarbonate: About 70% of carbon dioxide is transported as sodium
bicarbonate. As CO diffuses into the blood plasma, a large part of it combines with water to form
2

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 H CO and hydrogen ions H .
+
2 3

Carbo nicanhydrase
CO2 + H2 O H2 CO3
⟶
Cantonic − +
H2 CO3 → HCO + H
antly drase 3

Q5 What will be the pO and pCO in the atmospheric air compared to those in the alveolar air ?
2 2

(i)
pO2 lesser, pCO2

(ii)
pO2 higher, pCO2

(iii)
pO2 higher, pCO2

(iv)
pO2 lesser, pCO2

Answer. (ii) pO higher, pCO lesser
2 2

The partial pressure of oxygen in atmospheric air is higher than that of oxygen in alveolar air. In
atmospheric air, pO is about 159 mm Hg. In alveolar air, it is about 104 mm Hg. The partial
2

pressure of carbon dioxide in atmospheric air is lesser than that of carbon dioxide in alveolar air.
In atmospheric air, pCO is about 0.3 mmHg. In alveolar air, it is about 40 mm Hg.
2

Q6 Explain the process of inspiration under normal conditions.

Answer.

Inspiration or inhalation is the process of bringing air from outside the body into the lungs. It is
carried out by creating a pressure gradient between the lungs and the atmosphere. When air
enters the lungs, the diaphragm expands toward the abdominal cavity, thereby increasing the
space in the thoracic cavity for accommodating the inhaled air. The volume of the thoracic
chamber in the anteroposterior axis increases with the simultaneous contraction of the external
intercostal muscles. This causes the ribs and the sternum to move out, thereby increasing the
volume of the thoracic chamber in the dorsoventral axis. The overall increase in the thoracic
volume leads to a similar increase in the pulmonary volume. Now, as a result of this increase, the
intrapulmonary pressure becomes lesser than the atmospheric pressure. This causes the air from
outside the body to move into the lungs.

Q7 How is respiration regulated?

Answer. 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 signalling 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

Q8 What is the effect of pCO on oxygen transport?
2

Answer. pCO plays an important role in the transportation of oxygen. At the alveolus, the low
2

pCO and high pO favours the formation of haemoglobin. At the tissues, the high pCO and low
2 2 2

pO favours the dissociation of oxygen from oxyhaemoglobin. Hence, the af nity of haemoglobin
2

for oxygen is enhanced by the decrease of pCO in blood. Therefore, oxygen is transported in
2

blood as oxyhaemoglobin and oxygen dissociates from it at the tissues.

Q9 What happens to the respiratory process in a man going up a hill?

Answer. 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.

Q10 What is the site of gaseous exchange in an insect?

Answer. 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 C02 from the cells of the
body rst enters the tracheae and then leaves the body through the spiracles.

Q11 De ne oxygen dissociation curve. Can you suggest any reason for its sigmoidal pattern?

Answer. The oxygen dissociation curve is a graph showing the percentage saturation of
oxyhaemoglobin at various partial pressures of oxygen. The curve shows the equilibrium of
oxyhaemoglobin and haemoglobin at various partial pressures. In the lungs, the partial pressure of
oxygen is high. Hence, haemoglobin binds to oxygen and forms oxyhaemoglobin. Tissues have a
low oxygen concentration. Therefore, at the tissues, oxyhaemoglobin releases oxygen to form
haemoglobin. The sigmoid shape of the dissociation curve is because of the binding of oxygen to
haemoglobin. As the rst oxygen molecule binds to haemoglobin, it increases the af nity for the
second molecule of oxygen to bind. Subsequently, haemoglobin attracts more oxygen.

Q12 Have you heard about hypoxia? Try to gather information about it, and discuss with your
friends.

Answer. 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 p02, 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 a
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 de ciency 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.

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

(b) Inspiratory capacity and Expiratory capacity

(c) Vital capacity and Total lung capacity

Q14 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 inspired or expired during normal respiration. 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 = 6000 to 8000 mL/minute Tidal volume in an hour = 6000 to 8000 mL x (60 min)
5 5
= 3.6 × 10 mL to 4.8 × 10 mL

Therefore, the hourly tidal volume for a healthy human is approximately
= 3.6 × 10 mL to 4.8 × 10 mL.
5 5