NCERT Solutions Class 11 Biology Chapter 17

Q.1 Define vital capacity. What is its significance?

Ans-

Vital capacity refers to the volume of air a healthy person can inhale after a forced exhalation or the volume of air a person can exhale after a forced inhalation. It is the sum of Expiratory Reserve Volume (ERV), Tidal Volume (TV), and Inspiratory Reserve Volume (IRV). A normal adult person has a vital capacity of 3-5 litres.

Significance: Vital capacity is highest when physiological competence of the body is highest. It goes down as incompetence increases and becomes zero when respiration ceases. It implies that more the vital capacity, healthier is the body. Also, it helps in getting rid of the foul air and aids in supplying fresh air. Thus, the vital capacity enhances the exchange of gases between the body tissues and the environment.

Q.2 State the volume of air remaining in the lungs after a normal breathing.

Ans-

The volume of air that remains in the lungs after normal breathing is called FRC (functional residual capacity). This includes expiratory reserve volume (ERV, additional volume of air, a person can exhale by a forcible exhalation) and residual volume (RV, the volume of air remaining in the lungs even after a forcible exhalation). In a normal individual, ERV is about 1000 to 1100 mL while RV is about 1100 to 1200 mL.

Since, FRC = ERV + RV

FRC = 1000 (or 1100 mL) + 1100 (or 1200) mL

= 2100 or 2300 mL.

Thus, the volume of air remaining in the lungs after normal breathing is around 2100 mL to 2300 mL.

Q.3 Diffusion of gases occurs in the alveolar region only and not in the other parts of respiratory system. Why?

Ans-

The gaseous exchange (exchange of O₂ and CO₂) occurs in our body by simple diffusion between the blood capillaries around the alveoli and gases present in the alveoli. Alveoli are made up of thin and highly permeable layers of squamous epithelial cells. The reasons for gaseous diffusion in the alveolar region are as follows:

1. Concentration and pressure: Diffusion is a concentration and pressure-dependent process. The barrier between the capillaries and alveoli is thin and it facilitates the gaseous diffusion from the region of higher to lower partial pressure. Pressure contributed by an individual gas in gaseous mixture is called its partial pressure. The partial pressure of O₂ is high in the air present inside the lungs as compared to deoxygenated blood in lung capillaries. It means that there is a concentration gradient for O₂ between the lungs and blood. Thus, O₂ can easily diffuse from alveolar air into the blood of alveolar capillaries. A reverse concentration gradient is present for CO_2, thus diffusion of CO₂ takes place in the opposite direction.
2. Nature of the membranes at the site of exchange: The region where diffusion takes place is made up three major layers namely; the thin squamous epithelium of alveoli, the endothelium of alveolar capillaries and the basement substance in between them. The total thickness of these three layers is less than a millimetre which makes it ideal to act as a site for the diffusion of gases.

Q.4 What are the major transport mechanisms for CO₂? Explain.

Ans-

Blood is the medium for transport of CO₂. There are three main ways by which CO₂ is transported in the body:

• By red blood cells (RBCs): The haemoglobin present in the RBCs carry about 20-25% CO₂ as carbamino-haemoglobin. Haemoglobin transports CO₂ as well as O₂ and this transport of CO₂ depends upon the partial pressure of these gases. If there is high partial pressure of O₂, then CO₂ is dissociated and O₂ is acquired by haemoglobin (like in lungs where O₂ is present in abundance). When oxygenated blood reaches the tissues, due to the high partial pressure of CO₂, haemoglobin loses O₂ and acquires CO₂. This CO₂ is transported to lungs for its exchange with O₂.
• By RBCs and plasma as bicarbonate ion: Nearly 70% of CO₂ is transported in the form of bicarbonate ions. RBCs contain a very high amount of an enzyme called carbonic anhydrase. This enzyme is also present in minute amounts in plasma. A large part of CO₂, as it diffuses into the blood plasma, combines with water to form carbonic acid with the help of enzyme carbonic anhydrase. Thereafter, the carbonic acid dissociates into bicarbonate and hydrogen ions by the action of the same enzyme. Thus, CO₂ is transported in the form of bicarbonate ions.
• Approximately 7% of CO₂ is carried in dissolved state through plasma. CO₂ combines with water to form carbonic acid.

Q.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) pO₂ higher, pCO₂ lesser

In the atmosphere, O₂ is present in large quantity and thus, its partial pressure is higher (159 mm Hg) than pO₂ in alveolar air (104 mm Hg).

The pCO₂ is lesser in atmosphere (0.3 mm Hg) as compared to pCO₂ in alveoli (40 mm Hg).

Q.6 Explain the process of inspiration under normal conditions.

Ans-

The process of taking the air from outside the body into the lungs is called inhalation or inspiration. The process of inspiration involves the generation of the pressure gradient between the atmosphere and the lungs. Inspiration takes place when the pressure within the lungs is less (negative pressure) than the atmospheric pressure. The process of inspiration involves the following steps:

• A negative pressure is generated within the lungs with the help of diaphragm and intercostal muscles (internal and external intercostal muscles)
• Contraction of diaphragm increases the volume of the thoracic chamber in the anteroposterior axis.
• Contraction of external intercostal muscles increases the volume of the thoracic chamber in the dorsoventral axis by lifting up the ribs and the sternum.
• Increase in thoracic volume results in an increase in the pulmonary volume.
• Increased pulmonary volume results in a decrease in intra-pulmonary pressure as compared to atmospheric pressure.
• A decrease in intra-pulmonary pressure (negative pressure as compared to atmospheric pressure) allows the air to move into the lungs, i.e., inspiration or inhalation.

Q.7 How is respiration regulated?

Ans-

Respiration in human beings is a regulated process. It is regulated by the neural system according to the demands of the body tissues. It is mainly controlled by two regulatory regions present in the brain. These are:

(a) Respiratory rhythm centre: It is present in the medulla region of the brain and is mainly responsible for regulating the process of respiration. The areas adjacent to the respiratory rhythm centre are highly chemo-sensitive. They can sense the CO₂ and H⁺ levels, whose increase activates the respiratory rhythm centre. Once activated, this centre adjusts the respiratory process, which helps in the elimination of CO₂ and H⁺ from the body. Receptors associated with the aortic arch and carotid artery also sense the change in CO₂ and H⁺ concentration and send signals to the respiratory rhythm centre to adjust the respiratory process.

(b) Pneumotaxic centre: It is present in the pons region of the brain. This centre can moderate the function of respiratory rhythm centre and reduces the duration of inspiration, which in turn changes the respiratory rate.

Q.8 What is the effect of pCO₂ on oxygen transport?

Ans-

Oxygen binds to haemoglobin reversibly to form oxyhaemoglobin, which transports throughout the body to supply oxygen to different tissues. The binding of one molecule of haemoglobin with four molecules of oxygen depends on the partial pressure of oxygen (pO₂). However, the partial pressure of CO₂ also plays a significant role in the binding of oxygen with the haemoglobin molecule and its transport from lungs to tissues and vice-versa.

In alveoli, the pO₂ is high but pCO₂ is low. Thus, oxygen binds to haemoglobin and oxyhaemoglobin is formed. It transports the oxygen to body tissues. However, in tissues, the pCO₂ is higher than pO₂. Thus, O₂ dissociates from haemoglobin and is released. Here, CO₂ binds to haemoglobin to form carbamino-haemoglobin. CO₂ Is thus transported to the lungs and is released in the lung cavity in exchange of O₂. Thus, it can be stated that the affinity of haemoglobin for O₂ increases as pCO₂ decreases.

Q.9 What happens to the respiratory process in a man going up a hill?

Ans-

As a person goes up a hill (that is, at higher altitude), the level of O₂ in the atmosphere decreases. So, less O₂ is available for breathing and O₂ level in the blood starts to decline. It results in an increase in the respiratory rate so as to meet the O₂ demands of the body. Heart beat also increases to increase the supply of blood to the tissues so that every body tissue gets sufficient O₂.

Q.10 What is the site of gaseous exchange in an insect?

Ans-

Insects have specialized structures called spiracles arranged in a series on the sides of its body. These are small openings through which the oxygen-rich air enters into the body. Spiracles are connected to a network of tubes called trachea which diffuses oxygen into the cells of the body. The transport of CO₂ occurs in reverse direction, that is, from the trachea to the spiracles. Thus, the gaseous exchange occurs in the insects through the tracheal system.

Q.11 Define oxygen dissociation curve. Can you suggest any reason for its sigmoidal pattern?

Ans-

Oxygen is transported in the blood by haemoglobin in the form of oxyhaemoglobin. Each haemoglobin molecule can carry a maximum of four O₂ molecules. Binding of oxygen to haemoglobin is primarily affected by the partial pressure of O₂ (pO₂). However, the partial pressure of carbon dioxide (pCO₂), H⁺ concentration and temperature also affect the binding of oxygen to haemoglobin. When the partial pressure of oxygen is high like in alveoli, haemoglobin binds with oxygen and forms oxyhaemoglobin. However, in tissues where pO₂ is low, oxygen dissociates from oxyhaemoglobin. The per cent saturation of haemoglobin with oxygen at various pO₂ can be studied using the oxygen dissociation curve. The oxygen dissociation curve is a sigmoid curve which is obtained by plotting the percentage saturation of haemoglobin with O₂ against pO₂.

The oxygen dissociation curve is sigmoid in shape because the binding of oxygen molecules to haemoglobin is co-operative. It means that binding of first O₂ molecule to the haemoglobin increases the affinity of haemoglobin for another oxygen molecule. As a result, the haemoglobin attracts more oxygen and thus, the graph showing the percentage saturation of haemoglobin with O₂ against pO₂ appears sigmoid.

Q.12 Have you heard about hypoxia? Try to gather information about it, and discuss with your friends.

Hypoxia is a condition, wherein the tissues are not oxygenated adequately due to decreased supply of oxygen to the lungs. The oxygen deprivation can have severe adverse effects on various body cells and thus, several important biological processes get hampered. In general, hypoxia results in a pathological condition. There are various types of hypoxia depending upon the reason of inadequate oxygen supply to the body:

1. Hypoxic hypoxia or generalized hypoxia: This results because of an inadequate saturation of blood with oxygen due to a reduced supply of oxygen in the air, decreased lung ventilation or respiratory disease.
2. Anaemic hypoxia: This is due to decreased haemoglobin concentration.
3. Stagnant hypoxia: This is due to poor circulation of the blood.
4. Histotoxic hypoxia: This occurs when the tissues are unable to use oxygen due to cyanide or carbon monoxide poisoning.

Q.13 Distinguish between

(a) IRV and ERV

(b) Inspiratory capacity and expiratory capacity.

(c) Vital capacity and Total lung capacity.

Ans-

(a) IRV and ERV:

(b) Inspiratory capacity and expiratory capacity:

(c) Vital capacity and Total lung capacity:

Q.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 that is inhaled or exhaled during normal respiration. It is about 500 mL. A healthy person breathes about 12-16 times in a minute.

The approximate value of tidal volume for a healthy human per hour can be calculated as follows:

The tidal volume of a healthy human in an hour = Tidal volume per breath X Number of times an adult human being breathes per minute X 60 minutes

= 500 mL X 12-16 times per minute X 60 minutes

= 3,60,000 – 4,80,000 mL