SolveItClass 11 · NCERT

NCERT Solutions · Class 11 Biology Breathing and Exchange of Gases

14 questions · 14 still being checked

Exercises 14.1–14.10 (part 1 of 2)

  1. Exercise 14.1

    Define vital capacity. What is its significance?

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    Vital capacity (VC) is the maximum volume of air a person can breathe in after a forced expiration — equally, the maximum volume that can be breathed out after a forced inspiration.
    It is made up of three volumes: \(\displaystyle \text{VC} = \text{ERV} + \text{TV} + \text{IRV} \).
    Taking the chapter's values (ERV $\displaystyle 1000$–$\displaystyle 1100$ mL, TV $\displaystyle 500$ mL, IRV $\displaystyle 2500$–$\displaystyle 3000$ mL), VC works out to roughly $\displaystyle 4000$–$\displaystyle 4600$ mL.
    It leaves out the residual volume, the $\displaystyle 1100$–$\displaystyle 1200$ mL that stays in the lungs even after a forcible expiration and can never be driven out.
    Significance:
    It is the largest volume of air that can be exchanged in a single breath, so it shows how much fresh air the lungs can take in when the body's demand is high.
    Being a pulmonary capacity measured with a spirometer, it is used in the clinical assessment of pulmonary function and in clinical diagnosis.

  2. Exercise 14.2

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

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    The volume left after a normal expiration is the Functional Residual Capacity (FRC).
    \(\displaystyle \text{FRC} = \text{ERV} + \text{RV} \) — expiratory reserve volume plus residual volume.
    From the chapter's values: ERV $\displaystyle 1000$–$\displaystyle 1100$ mL and RV $\displaystyle 1100$–$\displaystyle 1200$ mL, so FRC is about $\displaystyle 2100$–$\displaystyle 2300$ mL.

  3. Exercise 14.3

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

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    Because only the alveoli are built for diffusion — they are the respiratory or exchange part of the system, while everything from the external nostrils up to the terminal bronchioles is the conducting part.
    The alveoli are very thin, irregular-walled, richly vascularised bag-like structures, so blood and air are brought close together over a huge surface.
    The diffusion membrane at the alveolus has only three layers — the thin squamous epithelium of the alveoli, the endothelium of the alveolar capillaries, and the basement substance in between — and its total thickness is much less than a millimetre. Rate of diffusion depends on the thickness of the membrane, and here it is minimal.
    A steep partial-pressure gradient exists only here: \(\displaystyle pO_{2} \) is $\displaystyle 104$ mm Hg in the alveoli against $\displaystyle 40$ mm Hg in deoxygenated blood, and \(\displaystyle pCO_{2} \) is $\displaystyle 45$ mm Hg in that blood against $\displaystyle 40$ mm Hg in the alveoli.
    The conducting part cannot do this. Its walls are thick, it is supported by incomplete cartilaginous rings, and it is not a vascularised exchange surface. Its jobs are different — it transports atmospheric air to the alveoli, clears it of foreign particles, humidifies it and brings it to body temperature.
    NCERT_Solution_Class11_Biology_Ch14_Q14-3

  4. Exercise 14.4

    What are the major transport mechanisms for CO2\displaystyle CO_{2}? Explain.

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    \(\displaystyle CO_{2} \) is carried by the blood in three ways:(a) As bicarbonate — about $\displaystyle 70$ per cent (the major mechanism)
    RBCs contain a very high concentration of the enzyme carbonic anhydrase, and minute quantities of it are present in the plasma too.
    The enzyme facilitates this reaction in both directions: \(\displaystyle CO_{2} + H_{2}O \rightleftharpoons H_{2}CO_{3} \rightleftharpoons HCO_{3}^{-} + H^{+} \)
    At the tissue site, where \(\displaystyle pCO_{2} \) is high because of catabolism, \(\displaystyle CO_{2} \) diffuses into the blood (RBCs and plasma) and forms \(\displaystyle HCO_{3}^{-} \) and \(\displaystyle H^{+} \).
    At the alveolar site, where \(\displaystyle pCO_{2} \) is low, the reaction runs the other way, giving back \(\displaystyle CO_{2} \) and \(\displaystyle H_{2}O \). The \(\displaystyle CO_{2} \) trapped as bicarbonate at the tissues is thus released out at the alveoli.
    (b) As carbamino-haemoglobin — about $\displaystyle 20$–$\displaystyle 25$ per cent
    \(\displaystyle CO_{2} \) binds haemoglobin in the RBCs to form carbamino-haemoglobin.
    The binding is related to the partial pressure of \(\displaystyle CO_{2} \), and \(\displaystyle pO_{2} \) is a major factor affecting it.
    Where \(\displaystyle pCO_{2} \) is high and \(\displaystyle pO_{2} \) is low, as in the tissues, more binding occurs; where \(\displaystyle pCO_{2} \) is low and \(\displaystyle pO_{2} \) is high, as in the alveoli, dissociation takes place.
    (c) Dissolved in plasma — about $\displaystyle 7$ per cent
    Carried in simple physical solution in the plasma.
    Overall, every $\displaystyle 100$ mL of deoxygenated blood delivers approximately $\displaystyle 4$ mL of \(\displaystyle CO_{2} \) to the alveoli.

  5. Exercise 14.5

    What will be the pO2\displaystyle pO_{2} and pCO2\displaystyle pCO_{2} in the atmospheric air compared to those in the alveolar air ?
    (i)
    pO2\displaystyle pO_{2} lesser, pCO2\displaystyle pCO_{2} higher
    (ii)
    pO2\displaystyle pO_{2} higher, pCO2\displaystyle pCO_{2} lesser
    (iii)
    pO2\displaystyle pO_{2} higher, pCO2\displaystyle pCO_{2} higher
    (iv)
    pO2\displaystyle pO_{2} lesser, pCO2\displaystyle pCO_{2} lesser

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    (ii) \(\displaystyle pO_{2} \) higher, \(\displaystyle pCO_{2} \) lesser.
    From Table $\displaystyle 14.1$, \(\displaystyle pO_{2} \) is $\displaystyle 159$ mm Hg in atmospheric air against only $\displaystyle 104$ mm Hg in the alveoli — higher outside.
    \(\displaystyle pCO_{2} \) is $\displaystyle 0.3$ mm Hg in atmospheric air against $\displaystyle 40$ mm Hg in the alveoli — very much lesser outside.
    These are exactly the gradients needed: \(\displaystyle O_{2} \) moves from the alveoli into the blood, and \(\displaystyle CO_{2} \) moves out of the blood into the alveolar air to be breathed out.

  6. Exercise 14.6

    Explain the process of inspiration under normal conditions.

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    Inspiration happens when the intra-pulmonary pressure is made less than the atmospheric pressure, so outside air is forced into the lungs.
    The steps:
    Inspiration is initiated by the contraction of the diaphragm, which increases the volume of the thoracic chamber in the antero-posterior axis.
    Contraction of the external inter-costal muscles lifts up the ribs and the sternum, increasing the volume of the thoracic chamber in the dorso-ventral axis.
    The lungs sit in an air-tight thoracic chamber, so any increase in thoracic volume causes a similar increase in pulmonary volume.
    The increased pulmonary volume lowers the intra-pulmonary pressure to less than the atmospheric pressure — a negative pressure with respect to the atmosphere.
    Air from outside then moves into the lungs down this pressure gradient. That is inspiration.
    On an average a healthy human breathes $\displaystyle 12$–$\displaystyle 16$ times a minute, taking in about $\displaystyle 500$ mL (tidal volume) with each normal inspiration.
    NCERT_Solution_Class11_Biology_Ch14_Q14-6

  7. Exercise 14.7

    How is respiration regulated?

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    Respiration is regulated by the neural system, which lets us maintain and moderate the respiratory rhythm to suit the demands of the body tissues.
    The centres involved:
    Respiratory rhythm centre — in the medulla region of the brain; primarily responsible for the regulation.
    Pneumotaxic centre — in the pons region; it moderates the functions of the respiratory rhythm centre. Neural signals from it can reduce the duration of inspiration and thereby alter the respiratory rate.
    Chemosensitive area — situated adjacent to the rhythm centre and highly sensitive to \(\displaystyle CO_{2} \) and hydrogen ions. A rise in these activates the area, which signals the rhythm centre to make the adjustments by which these substances can be eliminated.
    Receptors of the aortic arch and carotid artery — they also recognise changes in \(\displaystyle CO_{2} \) and \(\displaystyle H^{+} \) concentration and send the necessary signals to the rhythm centre for remedial action.
    The role of oxygen in the regulation of respiratory rhythm is quite insignificant — the control is exercised through \(\displaystyle CO_{2} \) and \(\displaystyle H^{+} \).

  8. Exercise 14.8

    What is the effect of pCO2\displaystyle pCO_{2} on oxygen transport?

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    A high \(\displaystyle pCO_{2} \) drives \(\displaystyle O_{2} \) off haemoglobin; a low \(\displaystyle pCO_{2} \) favours the binding of \(\displaystyle O_{2} \) to haemoglobin.
    \(\displaystyle pCO_{2} \) is one of the factors — along with hydrogen ion concentration and temperature — that interferes with the binding of \(\displaystyle O_{2} \) to haemoglobin, which is otherwise primarily related to \(\displaystyle pO_{2} \).
    In the alveoli: \(\displaystyle pO_{2} \) is high, \(\displaystyle pCO_{2} \) is low, \(\displaystyle H^{+} \) concentration is lesser and the temperature is lower — all favourable for the formation of oxyhaemoglobin.
    In the tissues: \(\displaystyle pO_{2} \) is low, \(\displaystyle pCO_{2} \) is high, \(\displaystyle H^{+} \) concentration is high and the temperature is higher — all favourable for the dissociation of oxygen from oxyhaemoglobin.
    The effect is therefore useful, not a nuisance: \(\displaystyle O_{2} \) gets bound to haemoglobin at the lung surface and is released exactly where \(\displaystyle CO_{2} \) is being produced, so about $\displaystyle 5$ mL of \(\displaystyle O_{2} \) is delivered to the tissues by every $\displaystyle 100$ mL of oxygenated blood.
    NCERT_Solution_Class11_Biology_Ch14_Q14-8

  9. Exercise 14.9

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

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    What the chapter lets us say:
    Breathing becomes faster and deeper while he climbs. Climbing is muscular work, so catabolism in the muscles produces more \(\displaystyle CO_{2} \), raising \(\displaystyle pCO_{2} \) and \(\displaystyle H^{+} \) concentration in the blood.
    The chemosensitive area next to the respiratory rhythm centre is highly sensitive to \(\displaystyle CO_{2} \) and hydrogen ions; the rise activates it, and it signals the rhythm centre to adjust the respiratory process so that these substances are eliminated. Receptors on the aortic arch and carotid artery send the same message.
    Less \(\displaystyle O_{2} \) is loaded onto haemoglobin as he goes higher. Exchange of gases is by simple diffusion along the partial-pressure gradient, and binding of \(\displaystyle O_{2} \) with haemoglobin is primarily related to \(\displaystyle pO_{2} \). Table $\displaystyle 14.1$ fixes atmospheric \(\displaystyle pO_{2} \) at $\displaystyle 159$ mm Hg against $\displaystyle 104$ mm Hg in the alveoli; if the air outside supplies a lower \(\displaystyle pO_{2} \), that gradient narrows and less oxyhaemoglobin is formed.
    The higher \(\displaystyle pCO_{2} \) and \(\displaystyle H^{+} \) in his working tissues also favour dissociation of \(\displaystyle O_{2} \) from oxyhaemoglobin, so what oxygen the blood does carry is unloaded readily at the muscles.
    Note that on this chapter's account the low oxygen itself does not speed up his breathing — the role of oxygen in the regulation of respiratory rhythm is stated to be quite insignificant.

  10. Exercise 14.10

    What is the site of gaseous exchange in an insect?

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    The tracheal tubes — a network of tubes forming the tracheal system.
    Insects have this network of tubes to transport atmospheric air directly within the body, so the air reaches the tissues without needing a special respiratory surface at the outside of the body.
    This is unlike the other patterns named in the chapter: gills (branchial respiration) in most aquatic arthropods and molluscs, and lungs (pulmonary respiration) in terrestrial vertebrates.