Breathing and Exchange of Gases Class 11 Biology NCERT Solutions

Step-by-step solution

1. 1Vital capacity is defined in the chapter as the maximum volume of air a person can breathe in after a forced expiration, which is the same thing as the maximum volume of air a person can breathe out after a forced inspiration.
2. 2It is not a single measured quantity but a sum. By adding up the respiratory volumes, vital capacity includes the expiratory reserve volume, the tidal volume and the inspiratory reserve volume, so VC = ERV + TV + IRV.
3. 3Its significance is clinical. By adding up a few respiratory volumes one can derive various pulmonary capacities, and these capacities can be used in clinical diagnosis.
4. 4It is also measurable in the clinic. The volume of air involved in breathing movements can be estimated by using a spirometer, which helps in clinical assessment of pulmonary functions, so vital capacity is one of the numbers a spirometer gives.
5. 5It is useful precisely because it is the largest movable volume. Together with the residual volume it gives the total lung capacity, TLC = VC + RV, so knowing vital capacity separates the air a person can move from the air that is trapped in the lungs and can never be expelled.

Step-by-step solution

1. 1Read the question carefully: it asks for the air left after a normal breathing, that is after a normal expiration, not after a forcible one. That quantity is named in the chapter as the functional residual capacity.
2. 2Functional residual capacity is the volume of air that will remain in the lungs after a normal expiration, and it includes the expiratory reserve volume plus the residual volume, so FRC = ERV + RV.
3. 3The chapter gives the two components. Expiratory reserve volume, the air that can be expired by a forcible expiration, averages 1000 mL to 1100 mL. Residual volume, the air remaining in the lungs even after a forcible expiration, averages 1100 mL to 1200 mL.
4. 4Adding them gives FRC = 2100 mL to 2300 mL, so about 2.1 to 2.3 litres of air remains in the lungs after a normal breathing.
5. 5Guard against the common slip. The residual volume on its own, 1100 mL to 1200 mL, is the air left even after the most forcible expiration possible, and that is not what is asked here; the phrase normal breathing excludes the forcible component, so the answer must include the expiratory reserve volume as well.

Step-by-step solution

1. 1The chapter answers this by splitting the respiratory system into two parts. The part starting with the external nostrils up to the terminal bronchioles constitutes the conducting part, whereas the alveoli and their ducts form the respiratory or exchange part.
2. 2The conducting part does not exchange gases. It transports the atmospheric air to the alveoli, clears it from foreign particles, humidifies it and brings the air to body temperature. All of these are preparation for the exchange, not the exchange itself.
3. 3The exchange part is the site of the actual diffusion. The chapter states that exchange of O2 and CO2 between blood and atmospheric air occurs in the exchange part, and the primary sites of exchange of gases are the alveoli.
4. 4Structurally, only the alveoli can support diffusion. Each terminal bronchiole gives rise to very thin, irregular-walled, vascularised bag-like alveoli, and the three-layer diffusion membrane of squamous epithelium, basement substance and capillary endothelium has a total thickness much less than a millimetre, across a very large surface.
5. 5The conducting passages cannot do this. The trachea and the bronchi are supported by incomplete cartilaginous rings, so they are rigid air ducts; they have neither the enormous moist surface nor the dense capillary network that a gradient needs.
6. 6So diffusion is confined to the alveolar region because the conducting region is built for delivering conditioned air and the exchange region is built for the actual exchange, with the thin membrane, the large surface and the rich blood supply that diffusion requires.

Step-by-step solution

1. 1Blood is the medium of transport for O2 and CO2, and carbon dioxide is carried in it in three distinct forms, which together make up nearly the whole of the CO2 load.
2. 2The first mechanism is carbamino-haemoglobin, which accounts for nearly 20 to 25 per cent of the CO2 and is carried by the RBCs. The chapter notes that roughly 20 to 25 per cent of CO2 is transported by RBCs.
3. 3This binding is related to the partial pressure of CO2, and pO2 is a major factor which could affect it. When pCO2 is high and pO2 is low, as in the tissues, more binding of carbon dioxide occurs, whereas when pCO2 is low and pO2 is high, as in the alveoli, dissociation of CO2 from carbamino-haemoglobin takes place and the CO2 bound at the tissues is delivered at the alveoli.
4. 4The second and by far the largest mechanism is bicarbonate, which carries 70 per cent of the CO2. At the tissue site, where the partial pressure of CO2 is high due to catabolism, CO2 diffuses into the blood in RBCs and plasma and forms bicarbonate and H+.
5. 5This conversion is catalysed by carbonic anhydrase. RBCs contain a very high concentration of this enzyme and minute quantities of the same are present in the plasma too, and it facilitates the reaction CO2 + H2O reversibly forming H2CO3, which in turn reversibly forms HCO3- and H+, in both directions.
6. 6At the alveolar site, where pCO2 is low, the reaction proceeds in the opposite direction, leading to the formation of CO2 and H2O, so the CO2 trapped as bicarbonate at the tissue level and transported to the alveoli is released out as CO2.
7. 7The third mechanism is simple physical solution. About 7 per cent of the CO2 is carried in a dissolved state through the plasma, needing no chemical conversion at all.
8. 8The chapter quantifies the delivery. Every 100 ml of deoxygenated blood delivers approximately 4 ml of CO2 to the alveoli.

Step-by-step solution

1. 1This is an objective question to be settled from the table of partial pressures, which is the reliable route rather than any general impression about fresh air.
2. 2The table gives, for atmospheric air, a pO2 of 159 mm Hg and a pCO2 of 0.3 mm Hg.
3. 3It gives, for the alveoli, a pO2 of 104 mm Hg and a pCO2 of 40 mm Hg.
4. 4Comparing the oxygen values, atmospheric pO2 of 159 is higher than alveolar pO2 of 104, so atmospheric air is richer in oxygen.
5. 5Comparing the carbon dioxide values, atmospheric pCO2 of 0.3 is far lower than alveolar pCO2 of 40, so atmospheric air is poorer in carbon dioxide.
6. 6Option (ii) states both of these correctly. Option (i) reverses both figures, option (iii) wrongly claims atmospheric pCO2 is higher, and option (iv) wrongly claims atmospheric pO2 is lower, so only (ii) is consistent with the table.

Step-by-step solution

1. 1The driving force for breathing is a pressure gradient between the lungs and the atmosphere. Inspiration can occur if the pressure within the lungs, the intra-pulmonary pressure, is less than the atmospheric pressure, that is, if there is a negative pressure in the lungs with respect to the atmosphere.
2. 2The muscles that create this gradient are the diaphragm and a specialised set of muscles, the external and internal intercostals between the ribs. Inspiration under normal conditions is produced by two of them acting together.
3. 3Inspiration is initiated by the contraction of the diaphragm, which increases the volume of the thoracic chamber in the antero-posterior axis.
4. 4At the same time, the contraction of the 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.
5. 5The overall increase in the thoracic volume causes a similar increase in pulmonary volume. This is possible because the lungs are housed in the anatomically air-tight thoracic chamber, so any change in the volume of the thoracic cavity is reflected in the lung cavity, and we cannot directly alter the pulmonary volume.
6. 6An 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, and this is inspiration.
7. 7For scale, a healthy human breathes 12 to 16 times per minute on an average, and the volume of air involved in these breathing movements can be estimated with a spirometer.

Step-by-step solution

1. 1Regulation is neural. The chapter states that human beings have a significant ability to maintain and moderate the respiratory rhythm to suit the demands of the body tissues, and that this is done by the neural system.
2. 2The centre that sets the rhythm is the respiratory rhythm centre, a specialised centre present in the medulla region of the brain, which is primarily responsible for this regulation.
3. 3It is moderated by a second centre, the pneumotaxic centre in the pons region of the brain, which can moderate the functions of the respiratory rhythm centre. A neural signal from this centre can reduce the duration of inspiration and thereby alter the respiratory rate.
4. 4The main chemical input comes from a chemosensitive area situated adjacent to the rhythm centre, which is highly sensitive to CO2 and hydrogen ions. An increase in these substances activates this centre.
5. 5Once activated, the chemosensitive area signals the rhythm centre to make the necessary adjustments in the respiratory process, by which these substances can be eliminated, so the control loop closes on itself.
6. 6Peripheral receptors reinforce the same signal. Receptors associated with the aortic arch and carotid artery also recognise changes in CO2 and H+ concentration and send the necessary signals to the rhythm centre for remedial actions.
7. 7The asymmetry to be noted is that the role of oxygen in the regulation of respiratory rhythm is quite insignificant, so the system is essentially a CO2 and H+ controller rather than an O2 controller.

Step-by-step solution

1. 1Set the framework first. Binding of oxygen with haemoglobin is primarily related to the partial pressure of O2, and the partial pressure of CO2, the hydrogen ion concentration and the temperature are the other factors which can interfere with this binding.
2. 2Apply it to the alveoli, where there is high pO2, low pCO2, lesser H+ concentration and lower temperature. Here the factors are all favourable for the formation of oxyhaemoglobin.
3. 3Apply it to the tissues, where low pO2, high pCO2, high H+ concentration and higher temperature exist. Here the conditions are favourable for the dissociation of oxygen from the oxyhaemoglobin.
4. 4So a high pCO2 promotes unloading. Where pCO2 is high, as in the active tissues, haemoglobin gives up its oxygen, which is exactly the behaviour wanted, since that is the tissue consuming the oxygen.
5. 5Expressed on the curve, a raised pCO2 shifts the oxygen dissociation curve to the right, so that at a given pO2 haemoglobin holds less oxygen and more is released to the tissue. The accompanying rise in H+ and temperature at the tissues has the same effect.
6. 6Conversely, the low pCO2 of the alveoli, with lower temperature and lesser H+, favours loading, and O2 gets bound to haemoglobin at the lung surface and gets dissociated at the tissues.

Step-by-step solution

1. 1Start from the demand. Going up a hill is muscular work, so the muscles use more oxygen and produce more carbon dioxide, and the tissues have a greater claim on the oxygen being carried.
2. 2The rise in carbon dioxide is what the body actually detects. A chemosensitive area situated adjacent to the respiratory rhythm centre is highly sensitive to CO2 and hydrogen ions, and an increase in these substances activates this centre.
3. 3The activated area then signals the rhythm centre in the medulla to make the necessary adjustments in the respiratory process, by which these substances can be eliminated, so breathing becomes both faster and deeper.
4. 4The result is that more fresh air is taken in and more CO2 is breathed out, which restores the CO2 and H+ levels and at the same time raises alveolar pO2, so the extra oxygen demanded by the climbing muscles is supplied.
5. 5Note the control design, which explains the sensations of the climb. Because the role of oxygen in the regulation of respiratory rhythm is quite insignificant, and because the chemosensitive area responds to CO2 and H+, the breathlessness and the heaviness of a climb are driven by the accumulating carbon dioxide rather than by any direct sensing of the missing oxygen.
6. 6The pneumotaxic centre in the pons can moderate this response, since a neural signal from it can reduce the duration of inspiration and thereby alter the respiratory rate, so the rate can be trimmed rather than simply running on.

Step-by-step solution

1. 1Begin with the general principle. The chapter states that mechanisms of breathing vary among different groups of animals depending mainly on their habitats and levels of organisation, so the answer must follow the animal group rather than a mammalian template.
2. 2Insects are placed in their own category. Earthworms use their moist cuticle and insects have a network of tubes, the tracheal tubes, to transport atmospheric air within the body.
3. 3So the site of gaseous exchange in an insect is this tracheal tube network, which conveys atmospheric air from outside the body through the branching tubes to the tissues themselves, so the air reaches the cells directly and there is no lung and no blood-borne carriage of the gas.
4. 4Contrast it with the alternatives to place the answer. Special vascularised structures called gills are used by most of the aquatic arthropods and molluscs, whereas vascularised bags called lungs are used by the terrestrial forms, and among vertebrates fishes use gills while amphibians, reptiles, birds and mammals respire through lungs.
5. 5The contrast with the terrestrial vertebrates is the point of the question. An insect, being a terrestrial arthropod, does not use the vascularised lung of pulmonary respiration; it uses tracheal tubes, which is why the site of exchange in an insect is the tracheal system rather than any lung-like sac.

Step-by-step solution

1. 1Define it precisely. A sigmoid curve is obtained when the percentage saturation of haemoglobin with O2 is plotted against the pO2, and this curve is called the oxygen dissociation curve.
2. 2State its use, which is the reason the chapter devotes space to it. The curve is highly useful in studying the effect of factors like pCO2 and H+ concentration on binding of O2 with haemoglobin.
3. 3The reason for the shape lies in the structure of the carrier. Each haemoglobin molecule can carry a maximum of four molecules of O2, so oxygenation is a four-step process rather than a single all-or-nothing event.
4. 4The four steps are not independent. The binding of one O2 molecule makes haemoglobin more willing to bind the next, so the first oxygen is taken up only slowly, then binding accelerates through the steep middle of the curve, and finally haemoglobin approaches full saturation and the curve flattens. That cooperative, staged binding is what produces the sigmoidal, S-shaped, pattern.
5. 5The practical value of the S shape is that the curve is steep in the middle region, which is the physiological range of tissue pO2, so small falls in pO2 there release a large amount of oxygen, whereas the flat upper part protects loading in the alveoli and the flat lower part protects loading again in a low-oxygen environment.
6. 6Its position can shift. In the tissues, low pO2, high pCO2, high H+ concentration and higher temperature shift the curve to the right and favour dissociation of oxygen from the oxyhaemoglobin, whereas the high pO2, low pCO2, lesser H+ and lower temperature of the alveoli favour the formation of oxyhaemoglobin.

Step-by-step solution

1. 1Hypoxia is the condition in which the tissues do not receive enough oxygen to meet their metabolic needs, that is when the oxygen supply falls short of the oxygen demand. It is best understood in this chapter as the failure of the oxygen supply side of the transport story.
2. 2Why it happens follows from the transport mechanism. About 97 per cent of O2 is transported by RBCs, and saturation of haemoglobin depends on the partial pressure of O2, so wherever pO2 is low, haemoglobin cannot be fully saturated and less oxygen reaches the tissues even though the mechanism itself is working normally.
3. 3A familiar setting for it is altitude, and the hill in Q9 is the same case. Atmospheric air at sea level has a pO2 of 159 mm Hg, whereas at height it is lower; going up a hill the alveolar pO2 falls below its sea-level value of 104 mm Hg, the percentage saturation of haemoglobin falls with it, and each 100 ml of blood delivers correspondingly less than the usual around 5 ml of O2 to the tissues.
4. 4The part worth discussing is that the body is not well equipped to notice. The chemosensitive area adjacent to the respiratory rhythm centre responds to CO2 and to hydrogen ions, and the chapter states plainly that the role of oxygen in the regulation of respiratory rhythm is quite insignificant, so there is no strong direct signal from the falling oxygen itself.
5. 5What is actually felt is therefore second-hand. The breathlessness, rapid breathing, headache and fatigue of a climber are driven by the CO2 and H+ that the increased muscular work produces, acting on the chemosensitive area, and by the general metabolic shortfall, rather than by a direct sensation of lack of oxygen, which is why a person can be markedly hypoxic without a proportionate feeling of breathlessness.
6. 6The correction is simple and shows the logic of the whole system. Because the deficiency is in alveolar pO2 and not in the haemoglobin, breathing more air to raise alveolar pO2 restores the loading of haemoglobin; and because the control loop is tuned to CO2 and H+, it is those substances that provide the reliable signal to increase ventilation.
7. 7There is a second context for the same word. In the chapter's own terms hypoxia is the mirror image of what happens in carbon monoxide or in any condition that blocks oxygen binding, because there the haemoglobin is present and the pO2 may be adequate, yet the oxygen cannot be carried, so the same shortage of oxygen at the tissue is produced by a failure of the carrier rather than of the gradient.

Step-by-step solution

1. 1Start with the two reserve volumes, since the first distinction is between them. Inspiratory reserve volume is the extra volume a person can inspire by a forcible inspiration beyond the tidal volume, and it averages 2500 mL to 3000 mL.
2. 2Expiratory reserve volume is the extra volume a person can expire by a forcible expiration beyond the tidal volume, and it averages 1000 mL to 1100 mL. So IRV is what is held back on inspiration and ERV is what can be pushed out on expiration, and IRV is the larger of the two.
3. 3For the second distinction, both capacities begin from a normal breath. Inspiratory capacity is the total volume of air a person can inspire after a normal expiration, and it includes the tidal volume and the inspiratory reserve volume, TV + IRV.
4. 4Expiratory capacity is the total volume of air a person can expire after a normal inspiration, and it includes the tidal volume and the expiratory reserve volume, TV + ERV. So IC and EC are mirror images taken from opposite starting points, and because IRV exceeds ERV, IC is the larger of the two.
5. 5For the third distinction, vital capacity is the maximum volume of air a person can breathe in after a forced expiration, and it includes the expiratory reserve volume, the tidal volume and the inspiratory reserve volume, so VC = ERV + TV + IRV.
6. 6Total lung capacity is the total volume of air accommodated in the lungs at the end of a forced inspiration, and it includes the residual volume, the expiratory reserve volume, the tidal volume and the inspiratory reserve volume, so TLC = RV + ERV + TV + IRV, which is also vital capacity plus residual volume, TLC = VC + RV.
7. 7State the difference that matters. Every component of vital capacity can be breathed in or out, whereas total lung capacity contains the residual volume, which is the air left in the lungs even after a forcible expiration and can never be expelled, so TLC is always the larger and the difference between the two is exactly the residual volume of 1100 to 1200 mL.

Step-by-step solution

1. 1Define tidal volume. Tidal volume, TV, is the volume of air inspired or expired during a normal respiration, and it is approximately 500 mL.
2. 2Get the rate. On an average a healthy human breathes 12 to 16 times per minute, so the number of breaths in one hour is 12 to 16 multiplied by 60, that is 720 to 960 breaths.
3. 3Check the per-minute figure the chapter gives. Multiplying the tidal volume by the rate gives 500 mL multiplied by 12 to 16, that is 6000 to 8000 mL per minute, and this matches the chapter's own statement that a healthy man can inspire or expire approximately 6000 to 8000 mL of air per minute.
4. 4Scale up to the hour. The tidal volume moved in an hour is therefore 6000 to 8000 mL multiplied by 60 minutes, that is 360000 to 480000 mL.
5. 5Convert to litres. Dividing by 1000 gives about 360 to 480 litres of air moved in an hour at rest, so the answer is approximately 360 to 480 litres per hour.