NCERT Solutions of Class 11th: Ch 17 Breathing and exchange of gases Biology

NCERT Solutions of Class 11th: Ch 17 Breathing and exchange of gases Biology

Exercises

Page No: 277

1. Define vital capacity. What is its significance?

Answer

The maximum volume of air a person can breathe in after a forced expiration is called vital capacity. It helps in finding differentiate causes of lung disease.

2. 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 breathing is called Functional residual capacity (FRC). This includes expiratory reserve volume (ERV) and residual volume (RV). ERV=1000 to 1100 ml
RV = 1100 to 1200 ml
Thus, FRC = 2100 to 2300 ml

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

Answer

Alveoli are the primary sites of exchange of gases. Exchange of gases also occur between blood and tissues. O2 and CO2 are exchanged in these sites by simple diffusion mainly based on pressure/concentration gradient. Alveolar region is having enough pressure gradient to facilitate diffusion of gases while other regions of the respiratory system don’t have the required pressure gradient. Solubility of the gases as well as the thickness of the membranes involved in diffusion are also some important factors that can affect the rate of diffusion.

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

Answer

CO2 is carried by haemoglobin as carbamino-haemoglobin (about 20-25 per cent). This binding is related to the partial pressure of CO2 . pOis a major factor which could affect this binding. When pCO2 is high and pO2 is low as in the tissues, more binding of carbon dioxide occurs whereas, when the pCO2 is low and pO2 is high as in the alveoli, dissociation of CO2 from carbamino-haemoglobin takes place, i.e., CO2 which is bound to haemoglobin from the tissues is delivered at the alveoli. RBCs contain a very high concentration of the enzyme, carbonic anhydrase and minute quantities of the same is present in the plasma too. This enzyme facilitates the following reaction in both directions.
Transport of CO2
At the tissue site where partial pressure of CO2 is high due to catabolism, CO2 diffuses into blood (RBCs and plasma) and forms HCO3 and H+, . At the alveolar site where pCO2 is low, the reaction proceeds in the opposite direction leading to the formation of CO2 and H2O. Thus, CO2 trapped as bicarbonate at the tissue level and transported to the alveoli is released out as CO2 as shown in above figure. Every 100 ml of deoxygenated blood delivers approximately 4 ml of CO2 to the alveoli.

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

Answer

(ii) pO2 higher, pCO2 lesser

6. Explain the process of inspiration under normal conditions.

Answer

Inspiration is initiated by the contraction of diaphragm which increases the volume of thoracic chamber in the antero-posterior axis. The contraction of external inter-costal 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. This pressure gradient forces the air from outside to move into the lungs and inspiration takes place.

7. How is respiration regulated?

Answer

The respiration is regulated bu neural system. A specialised centre present in the medulla region of the brain called respiratory rhythm centre is primarily responsible for this regulation. Another centre present in the pons region of the brain called pneumotaxic centre can moderate the functions of the respiratory rhythm centre. Neural signal from this centre can reduce the duration of inspiration and thereby alter the respiratory rate. A chemosensitive area is situated adjacent to the rhythm centre which is highly sensitive to COand hydrogen ions. Increase in these substances can activate this centre, which in turn can signal the rhythm centre 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 recognise changes in CO2 and H+ concentration and send necessary signals to the rhythm centre for remedial actions. The role of oxygen in the regulation of respiratory rhythm is quite insignificant.

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

Answer

pCO2 plays a major role in transportation of oxygen. In the alveoli, the low pCO2 and high pO2 favours the formation of haemoglobin. In the tissues, the high pCO2 and low pOfavours the dissociation of oxygen from oxyhaemoglobin. Hence, the affinity of haemoglobin for oxygen is enhanced by the decrease of pCO2 in blood. Therefore, oxygen is transported in blood as oxyhaemoglobin and oxygen dissociates from it at the tissues.

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

Answer

When a man going up a hill he has to exert more effort to climb which increases the consumption of oxygen. As a result, the partial pressure of oxygen in haemoglobin decreases which creates more demand for oxygen. Thus, the breathing rate increases to fill this gap.

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

Answer

Insects have a network of tubes known as tracheal tubes to transport atmospheric air within the body. The tracheae open on the lateral surface of the animal through minute pores called spiracles.

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

Answer

When percentage saturation of haemoglobin with O2 is plotted against the pO2 a sigmoid curve is obtained which is called Oxygen dissociation curve.
The dissociation curve is sigmoidal pattern because of the binding of oxygen to haemoglobin. As the first oxygen molecule binds to haemoglobin, it increases the affinity for the second molecule of oxygen to bind. Subsequently, haemoglobin attracts more oxygen.

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

Answer

Hypoxia is defined as a condition of the body in which the tissue have shortage of oxygen. It generally happens because of a mismatch between oxygen demand and supply.

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

Answer

(a) IRV and ERV

Inspiratory reserve volume(IRV)
Expiratory reserve volume (ERV)
It is the maximum volume of air that can be inhaled after a normal inspiration. It is the maximum volume of air that can be exhaled after a normal expiration.
It is about 2500-3500 mL in the human lungs. It is about 1000-1100 mL in the human lungs.

(b) Inspiratory capacity and Expiratory capacity

Inspiratory capacity (IC)
Expiratory capacity (EC)
It is the volume of air that can be inhaled after a normal expiration. It is the volume of air that can be exhaled after a normal inspiration.
It includes tidal volume and inspiratory reserve volume. (IC = TV + IRV) It includes tidal volume and expiratory reserve volume. (EC = TV + ERV)

(c) Vital capacity and Total lung capacity

Vital capacity (VC)
Total lung capacity (TLC)
It is the maximum volume of air that can be exhaled after a maximum inspiration.  It is the volume of air in the lungs after maximum inspiration. It includes IC, ERV, and residual volume.
It includes IC and ERV It includes IC, ERV, and residual volume.
It is about 4000 mL in the human lungs. It is about 5000-6000 mL in the human lungs.

14. 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 approximately 500 ml in a healthy man.
The hourly tidal volume for a healthy human (taking 12 breathes/min) can be calculated as:
= 500 ml × 12 × 60 minute = 360000 ml

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