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

Breathing and Exchange of Gases Class 11 Questions and Answers

Q1. Define vital capacity. What is its significance?

Answer:

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

Answer:

The volume of air remaining in the lungs after normal breathing 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, functional residual capacity of the human lungs is about 2500 - 3000 mL.

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

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 be very large

3. It must be highly vascular

Only alveolar region has these characteristics. Thus, diffusion of gases occurs in this region only.

Q4. What are the major transport mechanisms for CO_2? Explain.

Answer:

The major transport mechanisms 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 CO ₂ 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 (HCO ₃ ^– ) and hydrogen ions (H ⁺ ). About 20 – 25% of CO ₂ 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.

Chapter 17 Biology Class 11 Exercise Solutions

Q5. What will be the pO_2 and pCO_2in the atmospheric air compared to those in the alveolar air?

(i) lesser, higher
(ii) higher, lesser
(iii) higher, higher
(iv) lesser, lesser

Answer:

PO ₂ higher, pCO ₂ lesser

As the partial pressure of oxygen in atmospheric air is higher than that of oxygen in alveolar air. So, in atmospheric air, pO ₂ 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 with respect to atmospheric pressure.

It is initiated by the contraction of diaphragm which increases the volume of the thoracic chamber in the anteroposterior axis. The contraction of external intercostal muscles lifts up 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,

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

Q7. How is respiration regulated?

Answer:

A center present in the medulla region of the brain called respiratory rhythm center and it is primarily responsible for respiration regulation. And another center which is present in the pons region of the brain called pneumotaxic center. It can moderate the functions of the respiratory rhythm center. 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. 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 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 pCO_2 on oxygen transport?

Answer:

Breathing and exchange of gases chapter will tell you that pCO_2 plays an important role in the transportation of oxygen. The low pCO_2 and high pO_2 favors the formation of oxyhaemoglobin take place at alveolus. And at the tissues, the high pCO_2 and low pO_2 favor the dissociation of oxygen from oxyhaemoglobin. So, the affinity of haemoglobin for oxygen is enhanced by the decrease of pCO_2 in blood. Therefore, oxygen is transported in the blood as oxyhaemoglobin and oxygen dissociate from it at the tissues.

Chapter 17 Biology Class 11 NCERT Solutions

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 this, the amount of oxygen in the blood start to decline. The respiratory rate increases in the response 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 movement of carbon dioxide follows the reverse path and the CO ₂ from the cells of the body first enters the tracheae and then leaves the body through the spiracles.

Class 11 Biology Chapter 17 NCERT Question Answer

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

Answer:

The oxygen dissociation curve is a graph which shows the percentage of saturation of oxyhaemoglobin at various partial pressures of oxygen. This curve shows the equilibrium of oxyhaemoglobin and haemoglobin at various partial pressures. In the case of the lungs, the partial pressure of oxygen is high. So, haemoglobin binds to oxygen and forms oxyhaemoglobin.

Q12. Have you heard about hypoxia? Try to gather information about it, and discuss 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 pO_2, 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:

• Anaemic hypoxia
• Hypoxemic hypoxia

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

Q13. (a) Distinguish between IRV and ERV

Answer:

Difference between the IRV and ERV are given below:

IRV :

• Total lung capacity minus the volume of air in the lung at the end of a normal inspiration. This means that we have a reserve volume that we can tap into as tidal volume increases with exercise or activity.
• IRV is about 2500 – 3500 mL in the human lungs.

ERV :

• Difference between the volume of air left in the lung at the conclusion of normal expiration versus at the conclusion of maximal expiration. That means that we have a "reserve" volume which we can tap into when our tidal volume increases with exercise or activity.
• ERV is about 1000 – 1500 mL in the human lungs.

Q13. (b) Distinguish betweenInspiratory 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

Q13 (c).Distinguish between Vital capacity and Total lung capacity.

Answer:

Vital capacity

• It is the maximum volume of air that can be exhaled after a maximum inspiration. It includes IC and ERV.
• It is about 4000 mL in the human lungs.

Total lung capacity

• It is the volume of air in the lungs after a maximum inspiration. It includes IC, ERV, and residual volume.
• It is about 5000-6000 mL in the human lungs.

Exchange of Gases Class 11 NCERT Solutions

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 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.