NCERT Exemplar Class 11 Biology Solutions Breathing and Exchange of Gases
Multiple Choke Questions
1.Respiration in insects is called direct because
(a) The cells exchange O2 /CO2 directly with the air in the tubes
(b) The tissues exchange O2 /CO2 directly with coelomic fluid
(c) The tissues exchange O2 /CO2 directly with the air outside through body surface
(d) Tracheal tubes exchange O2 /CO2 directly with the haemocoel which then exchange with tissues.
Solution.(d):Direct respiration is the exchange of gases, without any special respiratory organ and blood. In insects, the tracheal tubes exchange O2 and CO2 directly with the ‘ haemocoel which then exchange them with tissues.
2. A person suffers punctures in his chest cavity in an accident, without any damage to the lungs, its effect could be
(a) Reduced breathing rate
(b) Rapid increase in breathing rate
(c) No change in respiration
(d) Cessation of breathing.
Solution.(d): The movement of air into and out of lungs is due to development of pressure gradient between the lungs and the atmosphere. Punctures in a person’s chest may lead to loss of pressure gradient, and thus resulting in cessation of breathing.
3.It is known that exposure to carbon monoxide is harmful to animals because
(a) It reduces CO2 transport
(b) It reduces O2 transport
(c) It increases CO2 transport
(d) It increases O2 transport.
Solution. (b): Haemoglobin has about 250 times more affinity for carbon monoxide than for oxygen. In the presence of carbon monoxide, it readily combines to form a stable compound called carbonmonoxyhaemoglobin (HbCO). The oxygen combining power decreases and as a result tissues suffer from oxygen starvation which leads to asphyxiation and in extreme cases to death.
4.Mark the true statement among the following with reference to normal breathing.
(a) Inspiration is an passive process whereas expiration is active
(b) Inspiration is an active process whereas expiration is passive
(c) Inspiration and expiration are active processes
(d) Inspiration and expiration are passive processes.
Solution. (b): Inspiration is the process by which fresh atmospheric air enters into the alveoli of the lungs. It is an active process and is brought about by activity of inspiratory muscles. Expiration is the process by which foul air is expelled out of the lungs. Expiration is a normal passive process that involves relaxation of inspiratory muscles.
5.Mark the incorrect statement in context to O2 binding to Hb.
(a) Higher pH (b) Lower temperature
(c) Lower pCO2
(d) Higher pO2
Solution.(None of the options is correct) :
Oxygen dissociation curve is highly useful in studying the effect of factors like pCO2 , H+concentration, etc., on binding of O2 with haemoglobin. In the alveoli, high pO2, low pCO2, lesser H+ concentration i.e. higher pH and lower temperature, all these factors are favourable for the formation of oxyhaemoglobin.
6.Mark the correct pair of muscles involved in the normal breathing in humans.
(a) External and internal intercostal muscles
(b) Diaphragm and abdominal muscles
(c) Diaphragm and external intercostal muscles
(d) Diaphragm and intercostal muscles
Solution.(d): Breathing involves two stages : inspiration during which atmospheric air is drawn in and expiration by which the alveolar air is released out. Inspiration 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. Similarly, expiration takes place when the intra-pulmonary pressure is higher than the atmospheric pressure. The diaphragm and a specialised set of muscles-extemal and internal intercostals between the ribs, help in generation of such gradients.
7.Incidence of emphysema-a respiratory disor-der is high in cigarette smokers. In such cases
(a) The bronchioles are found damaged
(b) The alveolar walls are found damaged
(c) The plasma membrane is found damaged
(d) The respiratory muscles are found dam-aged.
Solution. (b) : Emphysema is a condition of short breath due to breakdown of alveolar walls and reduction of respiratory area mainly due to smoking.
8.Respiratory process is regulated by certain specialized centres in the brain. One of the following listed centres can reduce the inspiratory duration upon stimulation.
(a) Medullary inspiratory centre
(b) Pneumotaxic centre
(c) Apneustic centre
(d) Chemosensitive centre
Solution. (b) : Pneumotaxic centre located in the dorsal part of pons varolii, regulates rate of respiration by reducing the duration of inspiration.
9. CO2 dissociates from carbaminohaemoglobin when
(a) pCO2 is high and pO2 is low
(b) pO2 is high and pCO2 is low
(c) pCO2 and pO2 are equal
(d) None of the above.
Solution.(b) : High pO2 and low pCO2 in the lung alveoli causes dissociation of CO2 from carbaminohaemoglobin.
10.In breathing movements, air volume can be estimated by
(a) Stethoscope (b) Hygrometer
Solution. (d) :Spirometry is the process of recording the changes in the volume and movement of air in and out of the lungs and the instrument used for this purpose is called spirometer or respirometer, which measures the volume of air inhaled and exhaled by lungs.
11. From the following relationships between respiratory volumes and capacities, mark the correct option.
(i) Inspiratory capacity (1C) = Tidal Volume + Residual Volume
(ii)Vital Capacity (VC) = Tidal Volume (TV) + Inspiratory Reserve Volume (IRV)+Expiratory Reserve Volume (ERV)
(iii)Residual Volume (RV) = Vital Capacity (VC) – Inspiratory Reserve Volume (IRV)
(iv)Tidal Volume (TV! = Inspiratory Capacity (1C) – Inspiratory Reserve Volume (IRV)
(a)(i) Incorrect, (ii) Incorrect, (iii) Incorrect, (iv) Correct
(b) (i) Incorrect, (ii) Correct, (iii) Incorrect, (iv) Correct
(c) (i) Correct, (ii) Correct, (iii) Incorrect, (iv) Correct
(d) (i) Correct, (ii) Incorrect, (iii) Correct, (iv) Incorrect
12.The oxygen – haemoglobin dissociation curve will show a right shift in case of ,
(a) High pCO2 (b) High pO2
(c) Low pCO2 (d) Less H+ concentration.
Solution.(a) : The oxygen haemoglobin dissocia-tion curve is shifted either to right or left by various factors. The shift of curve to right indicates dissociation of oxygen from haemoglobin. It occurs in the following conditions :
(i) Decrease in partial pressure of oxygen.
(ii)Increase in partial pressure of carbon dioxide (Bohr effect).
(iii)Increase in hydrogen ion concentration and decrease in pH (acidity).
(iv)Increased body temperature.
(v) Excess of 2, 3-diphosphoglycerate (DPG).
13.Match the following and mark the correct options
Animal Respiratory Organ
A.Earthworm (i)Moist cuticle
B.Aquatic Arthropods (ii)Gills
(a) A-(ii), B-(i), C-(iv), D-(iii)
(b) A-(i), B-(iv), C-(ii), D-(iii)
(c) A-(i), B-(iii), C-(ii), D-(iv)
(d) A-(i), B-(ii), C-(iv), D-(iii)
Solution.(None of the options is correct) : Both fishes as well as aquatic arthropods (like crustaceans) have gills as respiratory organs. Earthworms have moist cuticle, while birds/ reptiles have lungs as respiratory organs.
Short Answer Type Questions
1.Define the following terms.
(a) Tidal volume
(b) Residual volume
Solution.(a) Tidal volume is the volume of air inspired or expired during a normal breath. Its value is approximately 500 mL.
(b) Volume of air remaining in the lungs even after a forcible expiration is called residual volume. This average 1100 mL to 1200 mL.
(c) Asthma is a respiratory disorder which causes difficulty in breathing, wheezing, coughing due to inflammation of bronchi and bronchioles.
2.A fluid filled double membranous layer surrounds the lungs. Name it and mention its important function.
Solution.The fluid (pleural fluid) filled double membranous layer surrounding the lungs is called pleura. The pleural fluid lubricates the pleurae so that the may slide over each other without friction during breathing.
3.Name the primary site of exchange of gases in our body?
Solution. Alveoli are the primary site of exchange of gases in our body.
4.Cigarette smoking causes emphysema. Give reason.
Solution.Cigarette smoke contains various harmful chemicals like tar, nicotine, hydrogen cyanide and different metals. They damage alveolar walls due to which respiratory surface in decreased and it causes emphysema. Cigarette smoking is one of the major causes of emphysema. It is a chronic disorder.
5.What is the amount of O2 supplied to tissues through every 100 mL of oxygenated blood under normal physiological conditions?
Solution. 5 mL of O2 is the amount of O2 supplied to tissues through every 100 mL of oxygenated blood under normal physiological conditions.
6.A major percentage (97%) of O2 is transported by RBCs in the blood. How is the remaining percentage (3%) of O2 transported?
Solution. The remaining percentage (3%) of O2 is transported in a dissolved state through the plasma.
7.Arrange the following terms based on their volumes in an ascending order:
(a) Tidal Volume (TV)
(b) Residual Volume (RV)
(c) Inspiratory Reserve Volume (1RV)
(d) Expiratory Capacity (EC).
Solution. TV (500 mL) < RV (1100-1200 mL) < EC (1600/mL) < IRV (2000 – 3000 mL)
8.Complete the missing terms
(a) Inspiratory Capacity (1C)=_____+IRV
(b)_____= TV + ERV
(c) Functional Residual Capacity (FRC) = ERV +_____
Solution.(a) Inspiratory Capacity (IC) = Tidal volume (TV) + IRV
(b) Expiratory capacity (EC) = TV + ERV
(c) Functional Residual Capacity (FRC) = ERV + Residual volume (RV).
9.Name the organs of respiration in the following organisms:
(a) Flatworm ____________
(b) Birds ____________
Solution. (a) Body surface
(c) Moist skin and lungs
(d) Tracheal tubes
10. Name the important parts involved in creating a pressure gradient between lungs and the atmosphere during normal respiration.
Solution.Diaphragm and external and internal intercostal muscles of ribs are the important parts involved in creating a pressure gradient between lungs and the atmosphere during normal respiration.
Short Answer Type Questions
1.State the different modes of CO2 transport in blood.
Solution.CO2 is transported in the blood by the following modes:
(i) Dissolved in plasma (about 7 %).
(ii)As bicarbonate ions in plasma (about 70%).
(iii)As carbaminohaemoglobin (about 20 – 25%).
2.Compared to O2 , diffusion rate of CO2 through the diffusion membrane, unit difference in partial pressure is much higher. Explain.
Solution.This is due to the solubility of CO2 which is 20 – 25 times higher than that of O2 .
3.For completion of respiration process, write the given steps in sequential manner.
(a) Diffusion of gases (O2 and CO2) across alveolar membrane.
(b) Transport of gases by blood.
(c) Utilisation of O2 by the cells for catabolic reactions and resultant release of CO2 .
(d) Pulmonary ventilation by which atmos-pheric air is drawn in and C02 rich alveolar air is released out.
(e) Diffusion of 02 and CO2 between blood and tissues.
Solution. The correct sequence is d â€”> a â€”> b â€”> e â€”> c i.e. Pulmonary ventilation by which atmospheric air is drawn in and CO2 rich alveolar air is released out â€”> Diffusion of gases (02 and CO2 ) across alveolar membraneâ€”> Transport of gases by blood â€”> Diffusion of Oz and C02 between blood and tissues â€”> Utilisation of O2 by the cells for catabolic reactions and resultant release of CO2 .
(a) Inspiratory and expiratory reserve volume
(b) Vital capacity and total lung capacity
(c) Emphysema and occupational respiratory disorder.
Solution.(a) The differences between inspiratory reserve volume and expiratory reserve volume are as follows:
(b) The differences between vital capacity and total lung capacity are as follows:
(c) The differences between emphysema and occupational respiratory disorder are as follows:
Long Answer Type Questions
1.Explain the transport of O2 and CO2 between alveoli and tissue with diagram.
Solution. Transport of O2 : Blood carries oxygen from the alveoli to various body tissues. About 3% of 02 is carried in a dissolved state through the plasma. About 97% of O2 is transported in combination with haemoglobin of the RBCs as oxyhaemoglobin. Partial pressure of O2 is high in the alveoli as compared to pulmonary
blood capillaries, therefore 02 diffuses from alveoli into the pulmonary capillaries and combines with Hb to form oxyhaemoglobin. When this oxygenated blood reaches the different tissues having low partial pressure of 02, the bonds holding O2 to Hb become unstable. As a result, O2 is released from the blood capillaries into the tissues.
Transport of CO2 : Blood carries CO2 from various body tissues to the alveoli. About 7% of CO2 gets dissolved in the blood plasma and is carried in solution. About 70% of CO2 is transported by plasma as bicarbonate ions. From the tissues (which have high pCO2), C02 diffuses into the blood capillaries (which have low pCO2). In the RBCs, CO2 combines with water, to form carbonic acid (H2 CO3 ).H2CO3 is unstable and quickly dissociates into hydrogen ions and bicarbonate ions.
The above reaction is thousand times faster in RBCs as compared to plasma because RBCs contain carbonic anhydrase enzyme that reversibly catalyses the conversion of CO2 and water to H2CO3. About 20-35% CO2 is carried by Hb as carbaminohaemoglobin.
The blood carries CO2 in these three different forms towards the alveoli. CO2 is less soluble in arterial blood than in venous blood. Therefore, some CO2 diffuses from the plasma of the pulmonary capillaries into the alveoli. For the release of CO2 from the bicarbonate, a series of reverse reactions takes place and CO2 is released into the alveoli of the lungs. High pO2 in the pulmonary capillaries due to oxygenation of Hb favours separation of CO2 from carbaminohaemoglobin.
2.Explain the mechanism of breathing with neat labelled sketches.
Solution.Breathing involves two stages : inspiration during which atmospheric air is drawn in and expiration by which the alveolar air is released out. The movement of air into and out of the lungs is carried out by creating a pressure gradient between the lungs and the atmosphere. The diaphragm and a specialised set of muscles – external and internal intercostal muscles between the ribs help in the mechanism of breathing. The given figure shows thoracic cavity Inspiration:
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 sternum causing an increase in the volume of the thoracic chamber in the dorsal-ventral axis. This increase in the thoracic volume leads to 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, i.e., inspiration. Relaxation of the diaphragm and intercostal muscles cause the diaphragm and sternijm to go back to their normal positions and thus reduce the thoracic volume and thereby the pulmonary volume. This results in an increase in intra-pulmonary pressure to slightly above the atmospheric pressure causing the expulsion of air from the lungs i.e., expiration.
The given figure shows thoracic cavity during expiration:
3.Explain the role of neural system in regulation of respiration.
Solution. Neural system plays a significant role in maintaining and moderating the respiratory rhythm. Medulla oblongata has a specialised centre called respiratory rhythm centre, that regulates the respiration. The functions of the respiratory rhythm centre are controlled by another centre present in the pons varolii, called pheumotaxic centre. Neural signals from this centre can reduce the duration of inspiration and thereby alter the respiratory rate. Adjacent to the rhythm centre is situated a chemosensitive area which is highly sensitive to CO2 and H+ 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 present on aortic arch and carotid artery also can recognize changes in CO2 and H+ concentration and send necessary signals to the rhythm centre for remedial actions.