Class 11 · Biology · Chapter 14

Breathing and Exchange of Gases

The chapter follows one breath of air from the nostrils to the alveoli, across the diffusion membrane into the blood, through the tissues and back out again: the respiratory organs of animals, the human respiratory system, the pressure-gradient mechanism of breathing, lung volumes and capacities, partial pressures, transport of O2 by haemoglobin and of CO2 as bicarbonate, neural regulation from the medulla and pons, and four disorders. It is a short chapter that examiners lean on heavily because almost every line of it can be turned into a question, and it is the foundation for Class 12 boards and NEET.

11 sections14 NCERT questions31 board questions18-question quiz

Before you startWhat the board asks from this chapter

Class 11 Biology is a 70-mark theory paper set by the school on the CBSE pattern. Breathing and Exchange of Gases belongs to Unit V, Human Physiology, which is the largest unit and carries about 18 marks shared among Breathing and Exchange of Gases, Body Fluids and Circulation, Excretory Products, Locomotion and Movement, Neural Control and Chemical Coordination. In a typical school final this chapter alone contributes 4 to 7 marks: one or two 1-mark items (an MCQ on lung volumes, partial pressures, the percentage carried in each form, or the site of the rhythm centre, and an Assertion-Reason on the sigmoid curve or carbonic anhydrase), a 2-mark or 3-mark question on the mechanism of inspiration, the dissociation curve or CO2 transport, and every second year a 5-mark question that pairs a labelled diagram of the respiratory system with the steps of respiration or the transport of gases. In NEET it reliably gives one or two questions, nearly always lifted from the NCERT numbers and figures.

Question typeMarksUsually asked
MCQ1Tidal volume (500 mL); residual volume (1100 to 1200 mL); which capacity equals ERV + RV (functional residual capacity); pO2 in alveoli (104 mm Hg) and in deoxygenated blood (40 mm Hg); pCO2 in tissues (45 mm Hg); percentage of CO2 carried as bicarbonate (about 70 per cent) and of O2 carried by RBCs (about 97 per cent); the enzyme carbonic anhydrase; the trachea dividing at the fifth thoracic vertebra; the pneumotaxic centre in the pons; incomplete cartilaginous rings; which animal uses which organ (earthworm moist cuticle, insect tracheal tubes, fish gills, frog skin and lungs); mL of O2 delivered per 100 mL blood (about 5 mL); emphysema caused by smoking.
Assertion–Reason1The oxygen dissociation curve is sigmoid because of cooperative binding of four O2 to one haemoglobin; inspiration needs a negative intra-pulmonary pressure because air moves down a pressure gradient; the conducting part does not exchange gases because it is not thin-walled and vascularised; CO2 diffuses faster per unit gradient because its solubility is 20 to 25 times that of O2; the pleural fluid reduces friction; oxygen plays an insignificant role in regulating breathing.
Very short answer2Define tidal volume, vital capacity or residual volume with values; distinguish IRV from ERV; name the three layers of the diffusion membrane; state the function of the epiglottis, the pleura or the conducting part; give the equation catalysed by carbonic anhydrase; name the site of gas exchange in an insect and an earthworm; say why oxygen has little role in regulation.
Short answer3Explain inspiration or expiration step by step with muscles, volumes and pressures; describe the oxygen dissociation curve and the conditions that shift it; explain the three modes of CO2 transport with percentages; list the five steps of respiration; compare partial pressures in alveoli, blood and tissues from Table 14.1; describe the regulation of respiration; explain three respiratory disorders.
Long answer5Draw and label the human respiratory system and describe the conducting and exchange parts; explain the mechanism of breathing with a diagram of the thoracic cavity during inspiration and expiration; explain the transport of O2 and CO2 with the equations and the reason for the sigmoid curve; explain lung volumes and capacities with a spirogram and compute the capacities from given volumes.
Case-based4A spirogram or a table of lung volumes with numerical sub-questions; Table 14.1 partial pressures with questions on direction of diffusion; a passage on a smoker with emphysema or a stone-crusher with fibrosis; a passage on a person climbing a hill (hypoxia, increased breathing rate); a passage on the dissociation curve with a shift to the right.

Hot topics, in order

  1. Mechanism of breathing: inspiration and expiration, diaphragm and intercostals, intra-pulmonary pressure
  2. Respiratory volumes and capacities with values; IC, EC, FRC, VC, TLC as sums
  3. Transport of CO2: carbamino-haemoglobin, bicarbonate via carbonic anhydrase, dissolved; the 20 to 25, 70 and 7 per cent split
  4. Oxygen dissociation curve: sigmoid shape, factors favouring loading in alveoli and unloading in tissues
  5. Partial pressures of O2 and CO2 in atmosphere, alveoli, blood and tissues (Table 14.1) and the three-layered diffusion membrane
  6. Labelled diagram of the human respiratory system; conducting versus exchange part
  7. Regulation of respiration: rhythm centre in medulla, pneumotaxic centre in pons, chemosensitive area, aortic and carotid receptors
  8. Respiratory organs across animals and disorders (asthma, emphysema, occupational fibrosis)

The whole chapter, in orderNotes

01 Introduction: why we breathe

Every cell in the body oxidises food to release energy. In that process the cell uses oxygen to break down glucose, amino acids and fatty acids, and it gives out carbon dioxide, which is harmful if it accumulates. So the body needs a continuous supply of O2 to the cells and a continuous removal of CO2 from them. The process of exchange of O2 from the atmosphere with CO2 produced by the cells is called breathing, commonly known as respiration. NCERT uses that exact sentence, and a 1-mark question can ask you to reproduce it.

Keep two words apart from the start, because the first question in many school papers is exactly this distinction. Breathing is the physical, mechanical process of taking air into the lungs and pushing it out; no energy is released, no enzymes act, and it happens at the level of the organism. Cellular respiration is the biochemical, enzyme-driven oxidation of food inside the cell that actually releases energy as ATP. Breathing supplies the O2 that cellular respiration uses and removes the CO2 that it produces; the two are linked but not the same thing.

The chapter then moves in NCERT order: the organs that different animals use to breathe, the human respiratory system in detail, how the air is moved in and out, how much air moves, how the gases cross into the blood, how the blood carries them, how the brain sets the rhythm, and what goes wrong in disease. Read it as one journey of a molecule of O2 from the nostril to a mitochondrion, and one molecule of CO2 making the reverse trip.

Exam tip If a question says 'define breathing', write the NCERT sentence with the phrase 'exchange of O2 from the atmosphere with CO2 produced by the cells'. Do not write 'breathing is respiration'; the examiner is testing whether you know they differ.

02 14.1 Respiratory Organs

The mechanism of breathing varies among animals depending on their habitat and their level of organisation. The examiner expects you to match each animal group with its organ, so learn this list as a table.

  • Lower invertebrates such as sponges, coelenterates and flatworms exchange O2 and CO2 by simple diffusion over the entire body surface. They are small or thin enough that no cell is far from the outside.
  • Earthworms use their moist cuticle. Gases dissolve in the film of moisture and diffuse into the blood vessels just beneath the skin.
  • Insects have a network of tubes, the tracheal tubes, opening to the outside through spiracles and carrying atmospheric air directly to the tissues. Blood plays no part in gas transport in insects.
  • Aquatic arthropods and molluscs use special vascularised structures called gills; this is branchial respiration.
  • Terrestrial forms use vascularised bags called lungs; this is pulmonary respiration.

Among vertebrates, fishes use gills, while amphibians, reptiles, birds and mammals respire through lungs. Amphibians like frogs can also respire through their moist skin, which is cutaneous respiration. Note the pattern: as animals get larger and more complex, the exchange surface moves from the general body surface to a specialised, folded, richly vascularised organ, because diffusion alone cannot serve a large body.

Two sentences to keep for 1-mark questions: the site of gaseous exchange in an insect is the tracheal tube system (the exchange itself happens where the fine tracheoles meet the tissues), and a frog uses both lungs and skin. Mammals, including humans, use lungs alone.

Exam tip For 'name the respiratory organ of' questions write the organ the NCERT names: earthworm moist cuticle (not 'skin'), insect tracheal tubes (not 'spiracles'; spiracles are only the openings), frog moist skin and lungs.

03 14.1.1 Human Respiratory System

Trace the path of air in order, because the diagram and the sequence are what the 5-mark question rewards. Air enters through a pair of external nostrils into the nasal chamber, then the nasal passage, which opens into the pharynx. The pharynx is a common passage for food and air. From the pharynx air enters the larynx, a cartilaginous box that helps in sound production and is therefore called the sound box. During swallowing the epiglottis, a thin elastic cartilaginous flap, closes the glottis so that food does not enter the larynx. Below the larynx is the trachea, a straight tube that runs down to the mid-thoracic cavity and at the level of the fifth thoracic vertebra divides into a right and a left primary bronchus. Each primary bronchus divides repeatedly into secondary and tertiary bronchi and then into bronchioles, ending in very fine terminal bronchioles. The trachea, primary, secondary and tertiary bronchi and the initial bronchioles are supported by incomplete cartilaginous rings, which keep the airway open without preventing it from flexing. Each terminal bronchiole gives rise to a number of very thin, irregular-walled and vascularised bag-like structures called alveoli. The branching network of bronchi, bronchioles and alveoli makes up the lungs.

We have two lungs covered by a double-layered pleura, with pleural fluid between the two layers. The fluid reduces friction on the lung surface as it expands and relaxes. The outer pleural membrane lies in close contact with the thoracic lining and the inner one with the lung surface.

NCERT divides the system into two functional parts, and this division is a very common 2-mark question. The conducting part runs from the external nostrils up to the terminal bronchioles. It transports atmospheric air to the alveoli, clears it of foreign particles, humidifies it and brings it to body temperature. The exchange part is the alveoli and their ducts, the site of actual diffusion of O2 and CO2 between blood and atmospheric air. No exchange happens in the conducting part because its walls are neither thin enough nor vascularised.

The lungs sit in the thoracic chamber, which is anatomically an airtight chamber formed dorsally by the vertebral column, ventrally by the sternum, laterally by the ribs and on the lower side by the dome-shaped diaphragm. This setup is important: any change in the volume of the thoracic cavity is reflected in the lung (pulmonary) cavity. We cannot directly alter the volume of the lungs, so we alter the volume of the chest and the lungs follow.

Finally, NCERT lists the five steps that together make up respiration. Write them as a numbered list whenever the question says 'steps of respiration'. (i) Breathing or pulmonary ventilation, by which atmospheric air is drawn in and CO2-rich alveolar air is released out. (ii) Diffusion of O2 and CO2 across the alveolar membrane. (iii) Transport of gases by the blood. (iv) Diffusion of O2 and CO2 between blood and tissues. (v) Utilisation of O2 by the cells for catabolic reactions with the resultant release of CO2, that is, cellular respiration.

Exam tip Two numbers examiners love: the trachea divides at the fifth thoracic vertebra, and the cartilaginous rings are incomplete (C-shaped), not complete. Also remember that the pharynx is common to food and air; the larynx is not.

04 14.2 Mechanism of Breathing

Breathing has two stages: inspiration, during which atmospheric air is drawn in, and expiration, by which alveolar air is released out. Air moves only because of a pressure gradient between the lungs and the atmosphere. Inspiration occurs when the pressure within the lungs, the intra-pulmonary pressure, is less than atmospheric pressure, that is, there is a negative pressure in the lungs with respect to the atmosphere. Expiration occurs when the intra-pulmonary pressure is higher than atmospheric pressure. The gradient is created by the diaphragm and a specialised set of muscles, the external and internal intercostal muscles that lie between the ribs.

Inspiration is initiated by contraction of the diaphragm. The dome flattens and moves downward, which increases the volume of the thoracic chamber along the antero-posterior axis. At the same time contraction of the external intercostal muscles lifts the ribs and sternum up and outward, increasing the volume of the thoracic chamber along the dorso-ventral axis. The overall increase in thoracic volume produces a similar increase in pulmonary volume, because the lungs follow the chest wall. A larger volume means a lower pressure, so the intra-pulmonary pressure falls below atmospheric pressure and air is forced from outside into the lungs. That is inspiration: an active process, driven by muscle contraction.

Expiration under normal conditions is passive. The diaphragm and the external intercostal muscles relax, the diaphragm returns to its dome shape and the ribs and sternum return to their normal positions. Thoracic volume decreases, pulmonary volume decreases, intra-pulmonary pressure rises slightly above atmospheric pressure, and air is expelled. We can increase the strength of both inspiration and expiration with the help of additional muscles in the abdomen; forced expiration uses the internal intercostals and abdominal muscles.

Two facts finish this section. On average a healthy human breathes 12 to 16 times per minute. The volume of air involved in breathing movements can be measured with a spirometer, which helps in the clinical assessment of pulmonary functions.

For a 3-mark answer on inspiration the marking scheme wants, in this order: contraction of diaphragm, contraction of external intercostals lifting ribs and sternum, increase in thoracic volume in the two axes, increase in pulmonary volume, fall in intra-pulmonary pressure below atmospheric, air rushes in. Miss the pressure line and you lose a mark; the pressure gradient is the whole point.

Inspiration: diaphragm + external intercostals contract → thoracic volume ↑ → pulmonary volume ↑ → intra-pulmonary pressure below atmospheric → air in
intra-pulmonary pressurepressure of air within the lungsantero-posterior axisincreased by the diaphragm flatteningdorso-ventral axisincreased by the ribs and sternum lifting
Expiration: diaphragm + external intercostals relax → thoracic volume ↓ → pulmonary volume ↓ → intra-pulmonary pressure above atmospheric → air out
normal expirationpassive, no muscle contractionforced expirationinternal intercostals and abdominal muscles contract
Exam tip Do not write that the lungs 'pull air in' or 'push air out'. Lungs have no muscle. Write that the volume changes and the pressure gradient moves the air.

05 14.2.1 Respiratory Volumes and Capacities

NCERT defines four volumes and then builds five capacities by adding volumes together. Memorise the four volumes with their values; the capacities you can then derive in the exam rather than memorise.

Tidal Volume (TV) is the volume of air inspired or expired during a normal respiration, approximately 500 mL. A healthy man can inspire or expire roughly 6000 to 8000 mL of air per minute (500 mL × 12 to 16 breaths). Inspiratory Reserve Volume (IRV) is the additional volume of air a person can inspire by a forcible inspiration, on average 2500 to 3000 mL. Expiratory Reserve Volume (ERV) is the additional volume of air a person can expire by a forcible expiration, on average 1000 to 1100 mL. Residual Volume (RV) is the volume of air remaining in the lungs even after a forcible expiration, on average 1100 to 1200 mL. The residual volume can never be breathed out; it keeps the alveoli from collapsing and is the one volume a spirometer cannot measure directly.

The capacities are sums. Inspiratory Capacity (IC) is the total volume of air a person can inspire after a normal expiration: IC = TV + IRV. Expiratory Capacity (EC) is the total volume a person can expire after a normal inspiration: EC = TV + ERV. Functional Residual Capacity (FRC) is the volume of air that remains in the lungs after a normal expiration: FRC = ERV + RV. Vital Capacity (VC) is the maximum volume of air a person can breathe in after a forced expiration, or equivalently the maximum volume a person can breathe out after a forced inspiration: VC = ERV + TV + IRV. Total Lung Capacity (TLC) is the total volume of air accommodated in the lungs at the end of a forced inspiration: TLC = RV + ERV + TV + IRV, which is the same as VC + RV.

Put in typical numbers so the sums feel real. Taking TV = 500, IRV = 3000, ERV = 1100 and RV = 1200 mL: IC = 3500 mL, EC = 1600 mL, FRC = 2300 mL, VC = 4600 mL and TLC = 5800 mL. Any capacity that includes RV (FRC and TLC) cannot be measured by a spirometer; the rest can.

Vital capacity is the one most often asked, together with its significance: it tells us the maximum amount of air that can be exchanged in one breath, so it is a measure of the strength of the respiratory muscles and the elasticity of the lungs. It is higher in athletes, mountain dwellers and young adults, and lower in smokers, the elderly and patients with emphysema or fibrosis. Doctors use it to assess lung function.

IC = TV + IRV ≈ 3000 to 3500 mL
TVtidal volume, about 500 mLIRVinspiratory reserve volume, 2500 to 3000 mL
EC = TV + ERV ≈ 1500 to 1600 mL
ERVexpiratory reserve volume, 1000 to 1100 mL
FRC = ERV + RV ≈ 2100 to 2300 mL
RVresidual volume, 1100 to 1200 mL
VC = ERV + TV + IRV ≈ 4000 to 4600 mL
VCvital capacity, maximum air breathed out after a forced inspiration
TLC = RV + ERV + TV + IRV = VC + RV ≈ 5100 to 5800 mL
TLCtotal lung capacity, air in the lungs after a forced inspiration
Exam tip Read the question stem carefully: 'after a normal expiration' points to FRC (ERV + RV), 'after a forced expiration' points to RV alone, 'after a normal inspiration' with 'expire' points to EC. Students often answer RV when FRC is asked.

06 14.3 Exchange of Gases

Alveoli are the primary sites of exchange of gases, and exchange also occurs between blood and tissues. In both places O2 and CO2 move by simple diffusion, mainly down a pressure or concentration gradient. Two other factors affect the rate of diffusion: the solubility of the gas and the thickness of the membranes involved.

The pressure contributed by an individual gas in a mixture of gases is its partial pressure, written pO2 for oxygen and pCO2 for carbon dioxide. NCERT Table 14.1 gives the values in mm Hg and you must know them. Atmospheric air: pO2 159, pCO2 0.3. Alveoli: pO2 104, pCO2 40. Deoxygenated blood: pO2 40, pCO2 45. Oxygenated blood: pO2 95, pCO2 40. Tissues: pO2 40, pCO2 45. Read the gradients off the table. For O2 the pressure falls from alveoli (104) to deoxygenated blood (40), so O2 diffuses into the blood; it falls again from oxygenated blood (95) to tissues (40), so O2 diffuses into the tissues. For CO2 the gradient runs the other way: tissues (45) to blood (40 in arterial blood arriving), and deoxygenated blood (45) to alveoli (40), so CO2 diffuses out of the tissues into blood and out of blood into the alveoli.

Notice that the CO2 gradients are small, only about 5 mm Hg, while the O2 gradient at the alveolus is about 64 mm Hg. Yet roughly as much CO2 leaves as O2 enters. The reason is solubility: the solubility of CO2 is 20 to 25 times higher than that of O2, so the amount of CO2 that can diffuse across the membrane per unit difference in partial pressure is much higher than that of O2. This is a favourite Assertion-Reason item.

The diffusion membrane is made of three major layers: the thin squamous epithelium of the alveoli, the endothelium of the alveolar capillaries, and the basement substance (basement membrane) in between them. Its total thickness is much less than a millimetre. Thin membrane, large gradient for O2, high solubility for CO2: all the factors are favourable for the diffusion of O2 from alveoli to tissues and of CO2 from tissues to alveoli.

Why does alveolar air have a lower pO2 (104) than atmospheric air (159) and a higher pCO2 (40 against 0.3)? Because fresh air mixes with the residual air already in the lungs, which has given up O2 to the blood and received CO2 from it, and because the air is humidified so water vapour takes part of the total pressure. NCERT exercise question 5 tests exactly this comparison.

Table 14.1 (mm Hg): atmosphere pO2 159 / pCO2 0.3; alveoli 104 / 40; deoxygenated blood 40 / 45; oxygenated blood 95 / 40; tissues 40 / 45
pO2partial pressure of oxygenpCO2partial pressure of carbon dioxide
Rate of diffusion ∝ partial-pressure gradient × solubility / membrane thickness
solubility of CO220 to 25 times that of O2membrane thicknessthree layers, much less than 1 mm
Exam tip Write the direction of diffusion with the two numbers that justify it, for example 'O2 diffuses from alveoli (104 mm Hg) into blood (40 mm Hg)'. Numbers earn the mark; 'from high to low' alone does not.

07 14.4 Transport of Gases

Blood is the medium of transport for both gases, and NCERT gives a clear percentage split that examiners ask as MCQs and 2-mark questions.

About 97 per cent of O2 is transported by the RBCs in the blood, bound to haemoglobin, and the remaining 3 per cent is carried in a dissolved state through the plasma. O2 is poorly soluble in water, so a carrier pigment is essential.

CO2 is carried in three ways. Nearly 20 to 25 per cent is carried by RBCs bound to haemoglobin as carbamino-haemoglobin, about 70 per cent is carried as bicarbonate ions (HCO3−), and about 7 per cent is carried in a dissolved state through the plasma. CO2 is far more soluble than O2, which is why a larger fraction can travel dissolved and why it can be converted to bicarbonate in the blood.

Write these numbers as a small table in your notes and re-read it the night before the exam; a mismatch such as writing '70 per cent as carbamino-haemoglobin' is one of the most common ways students lose a mark in this chapter. The two sub-sections that follow explain the chemistry behind each figure.

O2: 97% as oxyhaemoglobin in RBCs, 3% dissolved in plasma
oxyhaemoglobinHb·4O2, reversible
CO2: 20 to 25% as carbamino-haemoglobin in RBCs, 70% as HCO3−, 7% dissolved in plasma
carbamino-haemoglobinCO2 bound to the amino groups of haemoglobinHCO3−bicarbonate formed by carbonic anhydrase in RBCs
Exam tip Three percentages for CO2 must add to about 100: 20 to 25 + 70 + 7. If your numbers do not add up, you have swapped something.

08 14.4.1 Transport of Oxygen

Haemoglobin is a red-coloured, iron-containing pigment present in the RBCs. O2 binds with haemoglobin in a reversible manner to form oxyhaemoglobin. Each haemoglobin molecule can carry a maximum of four molecules of O2, one on each of its four haem groups. Binding of O2 with haemoglobin is primarily related to the partial pressure of O2. Three other factors also interfere with the binding: the partial pressure of CO2, the hydrogen ion concentration and the temperature.

When the percentage saturation of haemoglobin with O2 is plotted against pO2, a sigmoid (S-shaped) curve is obtained. This is the oxygen dissociation curve, NCERT Fig. 14.5. The curve is sigmoid because of cooperative binding: the binding of the first O2 molecule changes the shape of haemoglobin so that the second, third and fourth bind more easily. At low pO2 saturation rises slowly, then steeply over the middle range, and finally flattens as the molecule approaches full saturation. The shape is very useful physiologically: the flat top means haemoglobin stays almost fully loaded even if alveolar pO2 drops a little, and the steep middle means a small fall in pO2 in active tissues unloads a lot of O2.

The curve is highly useful in studying the effect of pO2, pCO2 and related factors on binding. In the alveoli, where there is high pO2, low pCO2, lesser H+ concentration and lower temperature, the factors all favour the formation of oxyhaemoglobin. In the tissues, where there is low pO2, high pCO2, high H+ concentration and higher temperature, the conditions favour the dissociation of O2 from oxyhaemoglobin. On the graph, high pCO2, high H+ (low pH) and high temperature shift the curve to the right, meaning lower affinity and easier unloading, which is exactly what working muscle needs. This shift with pCO2 and pH is the Bohr effect, a name that is not in the rationalised NCERT but that examiners still use.

Every 100 mL of oxygenated blood can deliver around 5 mL of O2 to the tissues under normal physiological conditions. Learn this number alongside the 4 mL of CO2 delivered per 100 mL of deoxygenated blood to the alveoli.

A person going up a hill meets lower atmospheric pressure and therefore lower pO2. Less O2 loads onto haemoglobin, the tissues receive less oxygen (hypoxia), and the body responds with a faster, deeper breathing rate and a faster heartbeat; over days it makes more RBCs. This is NCERT exercise question 9.

Hb + 4O2 ⇌ Hb(O2)4 (oxyhaemoglobin)
forward directionfavoured in alveoli: high pO2, low pCO2, low H+, low temperaturereverse directionfavoured in tissues: low pO2, high pCO2, high H+, high temperature
Oxygen dissociation curve: % saturation of Hb (y-axis) against pO2 in mm Hg (x-axis), sigmoid
right shifthigh pCO2, high H+, high temperature: O2 unloadedleft shiftlow pCO2, low H+, low temperature: O2 loaded
Exam tip When asked the effect of pCO2 on oxygen transport, do not stop at 'it decreases'. Say: high pCO2 in tissues lowers the affinity of haemoglobin for O2, shifts the curve to the right and favours dissociation; low pCO2 in alveoli favours binding.

09 14.4.2 Transport of Carbon dioxide

CO2 is carried by haemoglobin as carbamino-haemoglobin (about 20 to 25 per cent). This binding is related to the partial pressure of CO2, and pO2 is a major factor that affects it. When pCO2 is high and pO2 is low, as in the tissues, more binding of CO2 occurs; when pCO2 is low and pO2 is high, as in the alveoli, CO2 dissociates from carbamino-haemoglobin. So haemoglobin picks up CO2 where it drops O2, and drops CO2 where it picks up O2.

The bulk of CO2, about 70 per cent, travels as bicarbonate, and the chemistry depends on one enzyme. RBCs contain a very high concentration of carbonic anhydrase, and minute quantities are present in the plasma too. This enzyme catalyses the following reaction in both directions:

CO2 + H2O ⇌ H2CO3 ⇌ HCO3− + H+

At the tissue site, where pCO2 is high because of catabolism, CO2 diffuses into the blood (into the RBCs and the plasma) and forms HCO3− and H+. At the alveolar site, where pCO2 is low, the reaction proceeds in the opposite direction, forming CO2 and H2O. Thus CO2 trapped as bicarbonate at the tissue level is transported to the alveoli and released out as CO2. Every 100 mL of deoxygenated blood delivers approximately 4 mL of CO2 to the alveoli.

One detail that older NCERT editions and most school papers still ask: the bicarbonate formed inside the RBC diffuses out into the plasma, and to keep the RBC electrically neutral chloride ions move from the plasma into the RBC. This exchange is the chloride shift (Hamburger phenomenon). At the alveoli the movement reverses. The H+ released stays inside the RBC and is buffered by haemoglobin, which is also what lowers haemoglobin's affinity for O2 in the tissues.

Read the two sub-sections together and the logic becomes one story. At the tissue: O2 unloads, CO2 loads as carbamino-Hb and bicarbonate, H+ rises. At the alveolus: CO2 unloads from both forms and leaves, H+ falls, O2 loads. The same haemoglobin molecule, the same enzyme, and the direction is set purely by the partial pressures on each side.

CO2 + H2O ⇌ H2CO3 ⇌ HCO3− + H+ (carbonic anhydrase)
forward directionat the tissues, high pCO2reverse directionat the alveoli, low pCO2carbonic anhydrasevery high concentration in RBCs, minute in plasma
Hb + CO2 ⇌ carbamino-haemoglobin
favouredhigh pCO2, low pO2 (tissues)reversedlow pCO2, high pO2 (alveoli)
100 mL deoxygenated blood → about 4 mL CO2 to alveoli; 100 mL oxygenated blood → about 5 mL O2 to tissues
undernormal physiological conditions
Exam tip Write the reversible arrows (⇌) and state where each direction runs. An answer with a one-way arrow and no location earns half.

10 14.5 Regulation of Respiration

Human beings have a significant ability to maintain and moderate the respiratory rhythm to suit the demands of the body tissues. This is done by the neural system; there is no hormonal regulation to write about in this chapter.

A specialised centre in the medulla region of the brain, the respiratory rhythm centre, is primarily responsible for the regulation. It sets the basic rhythm of inspiration and expiration. Another centre in the pons region, the pneumotaxic centre, can moderate the functions of the rhythm centre: a neural signal from the pneumotaxic centre can reduce the duration of inspiration and thereby alter the respiratory rate. So the medulla produces the rhythm and the pons fine-tunes it.

The chemical inputs come next. A chemosensitive area situated adjacent to the rhythm centre is highly sensitive to CO2 and hydrogen ions. An increase in these substances activates this centre, which in turn signals the rhythm centre to make the necessary adjustments in the respiratory process so that the excess CO2 and H+ are eliminated. In addition, receptors associated with the aortic arch and the carotid artery also recognise changes in CO2 and 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. That last sentence is asked almost every year in some form, either directly or as an Assertion-Reason.

Why CO2 and not O2? Because CO2 in blood becomes H2CO3 and then H+, so a small rise in CO2 changes blood pH quickly and dangerously; the body guards pH first. O2 stores on haemoglobin are large and the flat top of the dissociation curve means saturation barely changes with moderate falls in pO2, so O2 is a poor signal. Only a severe fall in pO2 stimulates breathing through the peripheral receptors.

For a 3-mark answer write four points: rhythm centre in medulla; pneumotaxic centre in pons reduces inspiration duration; chemosensitive area near the rhythm centre senses CO2 and H+; aortic and carotid receptors also sense CO2 and H+; oxygen insignificant. A simple flow diagram earns the same marks faster.

Medulla (respiratory rhythm centre) sets rhythm; pons (pneumotaxic centre) shortens inspiration → alters rate
chemosensitive areaadjacent to rhythm centre, senses CO2 and H+peripheral receptorsaortic arch and carotid artery, sense CO2 and H+O2insignificant role in regulation
Exam tip Students write 'the pneumotaxic centre is in the medulla'. It is in the pons. Medulla: rhythm centre and chemosensitive area. Pons: pneumotaxic centre.

11 14.6 Disorders of Respiratory System

NCERT lists three disorders and you should be able to define each in one sentence with its cause, because a 3-mark question of the form 'explain any three respiratory disorders' is standard.

Asthma is a difficulty in breathing that causes wheezing, due to inflammation of the bronchi and bronchioles. The inflamed, narrowed airways with excess mucus resist airflow, especially during expiration, which produces the whistling sound. It is usually an allergic response to dust, pollen or smoke and is treated with bronchodilators.

Emphysema is a chronic disorder in which the alveolar walls are damaged, due to which the respiratory surface is decreased. Neighbouring alveoli merge into larger sacs with much less surface area, the lungs lose elasticity, and the patient cannot exchange enough gas even though the airways are open. One of the major causes is cigarette smoking. Because the loss of alveolar wall is permanent, emphysema does not reverse.

Occupational respiratory disorders arise in certain industries, especially those involving grinding or stone-breaking, where so much dust is produced that the defence mechanism of the body cannot fully cope with the situation. Long exposure can give rise to inflammation leading to fibrosis, the proliferation of fibrous tissues, and thus cause serious lung damage. The stiff fibrous lung cannot expand, and vital capacity falls. Workers in such industries should wear protective masks. Silicosis and asbestosis are the named examples used in school papers.

Beyond NCERT, two terms appear in questions on this chapter. Hypoxia is a shortage of oxygen at the tissue level; it can be caused by low atmospheric pO2 at high altitude, by anaemia, by carbon monoxide poisoning (CO binds haemoglobin more strongly than O2) or by lung disease, and its symptoms are breathlessness, headache, fatigue and, in severe cases, confusion. Pneumonia is an infection (usually bacterial, Streptococcus pneumoniae) in which the alveoli fill with fluid and gas exchange falls; you will meet it again in Class 12 Human Health and Disease.

Exam tip For emphysema, the phrase the examiner looks for is 'alveolar walls damaged, respiratory surface decreased, caused by cigarette smoking'. For asthma, 'inflammation of bronchi and bronchioles, wheezing'. For fibrosis, 'proliferation of fibrous tissue after long exposure to dust'.

VocabularyKey terms

BreathingThe process of exchange of O2 from the atmosphere with CO2 produced by the cells; a physical process with no energy release, also called pulmonary ventilation.
Cellular respirationThe enzyme-catalysed oxidation of glucose, amino acids and fatty acids inside the cell using O2, releasing energy as ATP and giving out CO2; the fifth step of respiration.
Tracheal tubesThe network of air-filled tubes in insects that opens through spiracles and carries atmospheric air directly to the tissues without using blood.
EpiglottisA thin, elastic cartilaginous flap that closes the opening of the larynx (glottis) during swallowing so that food does not enter the trachea.
AlveoliVery thin, irregular-walled, richly vascularised bag-like structures at the end of terminal bronchioles; the primary site of gas exchange.
PleuraThe double-layered membrane covering each lung, with pleural fluid between the layers to reduce friction on the lung surface; the outer layer touches the thoracic lining and the inner layer the lung.
Conducting partThe airway from the external nostrils to the terminal bronchioles; it transports, filters, humidifies and warms the air but does not exchange gases.
Exchange partThe alveoli and their ducts, where actual diffusion of O2 and CO2 between blood and alveolar air takes place.
Intra-pulmonary pressureThe pressure of the air within the lungs; inspiration occurs when it is below atmospheric pressure and expiration when it is above.
Tidal volume (TV)The volume of air inspired or expired during a normal breath, about 500 mL in a healthy adult.
Inspiratory reserve volume (IRV)The additional volume of air a person can inspire by forcible inspiration beyond the tidal volume, 2500 to 3000 mL.
Expiratory reserve volume (ERV)The additional volume of air a person can expire by forcible expiration beyond the tidal volume, 1000 to 1100 mL.
Residual volume (RV)The volume of air remaining in the lungs even after a forcible expiration, 1100 to 1200 mL; it cannot be measured by a spirometer.
Vital capacity (VC)The maximum volume of air a person can breathe out after a forced inspiration (or in after a forced expiration); VC = ERV + TV + IRV.
Functional residual capacity (FRC)The volume of air that remains in the lungs after a normal expiration; FRC = ERV + RV.
Total lung capacity (TLC)The total volume of air in the lungs at the end of a forced inspiration; TLC = RV + ERV + TV + IRV = VC + RV.
Partial pressureThe pressure contributed by an individual gas in a mixture of gases, written pO2 and pCO2; the gradient of partial pressure drives diffusion.
Diffusion membraneThe barrier between alveolar air and blood, made of the squamous epithelium of the alveolus, the basement substance and the endothelium of the capillary; much less than 1 mm thick.
OxyhaemoglobinThe reversible compound formed when O2 binds to the iron-containing pigment haemoglobin in RBCs; each haemoglobin can carry four O2 molecules.
Oxygen dissociation curveThe sigmoid curve obtained by plotting percentage saturation of haemoglobin with O2 against pO2; used to study the effects of pCO2, H+ and temperature on binding.
Carbamino-haemoglobinCO2 bound to haemoglobin; carries about 20 to 25 per cent of blood CO2, favoured by high pCO2 and low pO2.
Carbonic anhydraseThe enzyme present in very high concentration in RBCs (and minutely in plasma) that catalyses CO2 + H2O ⇌ H2CO3 ⇌ HCO3− + H+ in both directions.
Chloride shiftThe movement of Cl− from plasma into RBCs at the tissues to balance the HCO3− ions leaving the RBCs, keeping the cell electrically neutral; reversed at the alveoli.
Respiratory rhythm centreThe centre in the medulla oblongata primarily responsible for regulating the rhythm of breathing.
Pneumotaxic centreThe centre in the pons that moderates the rhythm centre; its signals reduce the duration of inspiration and thereby alter the respiratory rate.
Chemosensitive areaAn area adjacent to the rhythm centre, highly sensitive to CO2 and H+, which signals the rhythm centre to adjust breathing and eliminate them.
EmphysemaA chronic disorder in which alveolar walls are damaged and the respiratory surface is decreased; a major cause is cigarette smoking.
FibrosisProliferation of fibrous tissue in the lungs after long exposure to dust, as in grinding and stone-breaking industries, causing serious lung damage.
HypoxiaA condition of insufficient oxygen at the tissue level, caused by high altitude, anaemia, carbon monoxide poisoning or lung disease.

Keep this on one pageFacts and values to remember

NameExpressionNotes
Respiratory organs of animalsSponges, coelenterates, flatworms: whole body surface; earthworm: moist cuticle; insects: tracheal tubes; aquatic arthropods and molluscs, fishes: gills; amphibians, reptiles, birds, mammals: lungs; frog also moist skinMatch-the-following and 1-mark favourite; write the exact organ name.
Path of airExternal nostrils → nasal chamber → nasal passage → pharynx → larynx → trachea → primary bronchi (divide at 5th thoracic vertebra) → secondary → tertiary bronchi → bronchioles → terminal bronchioles → alveoliConducting part = nostrils to terminal bronchioles; exchange part = alveoli and ducts.
Five steps of respiration(i) Breathing / pulmonary ventilation (ii) Diffusion across alveolar membrane (iii) Transport by blood (iv) Diffusion between blood and tissues (v) Cellular respiration (utilisation of O2, release of CO2)Write as a numbered list; 3-mark question.
Muscles of breathingInspiration: diaphragm contracts (antero-posterior volume ↑) + external intercostals contract, ribs and sternum lift (dorso-ventral volume ↑); normal expiration: both relax; forced breathing: abdominal muscles and internal intercostalsInspiration active, normal expiration passive; 12 to 16 breaths per minute; spirometer measures volumes.
Lung volumesTV ≈ 500 mL; IRV = 2500 to 3000 mL; ERV = 1000 to 1100 mL; RV = 1100 to 1200 mL; 6000 to 8000 mL breathed per minuteRV cannot be exhaled or measured by spirometer.
Lung capacitiesIC = TV + IRV; EC = TV + ERV; FRC = ERV + RV; VC = ERV + TV + IRV; TLC = RV + ERV + TV + IRV = VC + RVTypical: IC 3500, EC 1600, FRC 2300, VC 4600, TLC 5800 mL.
Partial pressures (Table 14.1, mm Hg)Atmosphere pO2 159 / pCO2 0.3; alveoli 104 / 40; deoxygenated blood 40 / 45; oxygenated blood 95 / 40; tissues 40 / 45O2 gradient at alveolus about 64 mm Hg; CO2 gradient about 5 mm Hg but solubility 20 to 25 times higher.
Diffusion membraneSquamous epithelium of alveolus + basement substance + endothelium of capillary; total thickness much less than 1 mmThree layers, in that order, from air to blood.
Transport of O<sub>2</sub>97% as oxyhaemoglobin in RBCs, 3% dissolved in plasma; Hb + 4O2 ⇌ oxyhaemoglobin; 5 mL O2 delivered per 100 mL bloodBinding depends on pO2, also pCO2, H+, temperature; sigmoid curve.
Oxygen dissociation curve conditionsAlveoli: high pO2, low pCO2, low H+, low temperature → loading; tissues: low pO2, high pCO2, high H+, high temperature → unloading (right shift)Sigmoid because of cooperative binding of four O2.
Transport of CO<sub>2</sub>20 to 25% carbamino-haemoglobin; 70% bicarbonate; 7% dissolved in plasma; CO2 + H2O ⇌ H2CO3 ⇌ HCO3− + H+ (carbonic anhydrase, high in RBCs); 4 mL CO2 delivered per 100 mL bloodForward at tissues (high pCO2), reverse at alveoli (low pCO2); chloride shift keeps RBC neutral.
Regulation of respirationRespiratory rhythm centre: medulla; pneumotaxic centre: pons (reduces inspiration duration); chemosensitive area next to rhythm centre: senses CO2 and H+; aortic arch and carotid receptors: sense CO2 and H+; O2 insignificantNeural only; no hormone in this chapter.
DisordersAsthma: inflammation of bronchi and bronchioles, wheezing; emphysema: chronic, alveolar walls damaged, surface decreased, cigarette smoking; occupational: dust from grinding or stone-breaking → inflammation → fibrosis; protective masksEach definition with its cause is a 1-mark line.

Practise with a pencilDiagrams that earn marks

Human respiratory system (NCERT Fig. 14.1a)

Draw the head in outline with the nasal chamber opening through the external nostrils, the pharynx behind the mouth, a small box for the larynx with the epiglottis as a flap above it, and the trachea as a straight tube with C-shaped rings drawn as short horizontal bars. Divide it at chest level into right and left primary bronchi entering two lung outlines (right lung with three lobes, left with two), each bronchus branching into bronchioles ending in small clusters of alveoli. Draw the dome-shaped diaphragm under the lungs and the ribs as curved lines at the sides. Labels the marking scheme looks for: external nostrils, nasal chamber, pharynx, larynx, epiglottis, trachea, cartilaginous rings, bronchus, bronchioles, alveoli, lungs, pleural membranes, diaphragm, ribs.

Alveoli with blood capillaries (NCERT Fig. 14.1b inset)

Draw a terminal bronchiole ending in a cluster of grape-like sacs. Around one sac draw a fine capillary network entering from a branch of the pulmonary artery (deoxygenated, draw with a darker line) and leaving into a branch of the pulmonary vein (oxygenated). Show the alveolar wall as a single thin line. Label: terminal bronchiole, alveolus, alveolar wall (squamous epithelium), capillary network, pulmonary arteriole, pulmonary venule. Add two small arrows: O2 from alveolus into capillary, CO2 from capillary into alveolus.

Mechanism of breathing: inspiration and expiration (NCERT Fig. 14.2)

Draw two side-by-side outlines of the thoracic cavity. Left panel, inspiration: rib cage expanded outward and upward, diaphragm flattened and lower, lungs larger, arrow into the trachea, note 'intra-pulmonary pressure below atmospheric'. Right panel, expiration: rib cage back in, diaphragm domed upward, lungs smaller, arrow out of the trachea, note 'intra-pulmonary pressure above atmospheric'. Label in each: sternum, ribs, external intercostal muscles, diaphragm, lung, trachea, and use arrows to show the direction of rib and diaphragm movement. Under each panel write the muscle state: contracted / relaxed.

Spirogram of lung volumes and capacities (NCERT Fig. 14.3)

Draw a horizontal time axis and a vertical volume axis marked 0 to 6000 mL. Draw a wavy line of small regular waves (tidal breathing, about 500 mL peak to trough), then one tall peak (forced inspiration), then one deep trough (forced expiration), then normal waves again. Mark with vertical brackets: TV on a small wave, IRV from the top of a normal wave to the tall peak, ERV from the bottom of a normal wave to the deep trough, RV from the deep trough down to zero. Then draw longer brackets for IC (TV + IRV), EC (TV + ERV), FRC (ERV + RV), VC (from deep trough to tall peak) and TLC (from zero to tall peak). Write the typical value beside each bracket.

Diffusion membrane between alveolus and capillary (NCERT Fig. 14.4)

Draw a section showing, from top to bottom, alveolar air space, a thin layer labelled 'squamous epithelium of alveolus', a thin layer labelled 'basement substance (basement membrane)', a thin layer labelled 'endothelium of capillary', and then the capillary lumen with one or two RBCs. Draw an arrow for O2 crossing downward from air into an RBC and an arrow for CO2 crossing upward from blood into air. Write beside it: total thickness much less than 1 mm; pO2 104 in alveolus, 40 in blood; pCO2 40 in alveolus, 45 in blood.

Oxygen dissociation curve (NCERT Fig. 14.5)

Draw x-axis 'pO2 (mm Hg)' from 0 to 100 and y-axis 'percentage saturation of haemoglobin with O2' from 0 to 100. Draw an S-shaped curve: slow rise at low pO2, steep between about 20 and 60 mm Hg, flattening near 100 per cent above 80 mm Hg. Mark the alveolar point (pO2 about 95 to 104, saturation about 97 per cent) and the tissue point (pO2 about 40, saturation about 75 per cent). If asked about the effect of pCO2, draw a second curve shifted to the right and label it 'high pCO2, high H+, high temperature (tissues)'. Title it 'oxygen dissociation curve, sigmoid'.

Transport of CO<sub>2</sub> in an RBC at the tissue (bicarbonate pathway)

Draw a tissue cell on the left, a capillary in the middle with one RBC drawn large, and plasma around it. Arrow CO2 from tissue into plasma and into the RBC. Inside the RBC write CO2 + H2O → H2CO3 → HCO3− + H+ with 'carbonic anhydrase' over the first arrow. Arrow HCO3− out of the RBC into plasma and Cl− into the RBC, labelled 'chloride shift'. Show H+ binding to haemoglobin and O2 leaving haemoglobin toward the tissue. A second small arrow shows some CO2 binding directly to Hb as carbamino-haemoglobin. Label percentages: 70 per cent bicarbonate, 20 to 25 per cent carbamino-Hb, 7 per cent dissolved.

Regulation of respiration flow chart

Draw a brain outline with the pons above the medulla. In the medulla draw a box 'respiratory rhythm centre' and next to it a smaller box 'chemosensitive area (CO2, H+)' with an arrow into the rhythm centre. In the pons draw a box 'pneumotaxic centre' with an arrow down to the rhythm centre labelled 'reduces duration of inspiration'. Below, draw the aortic arch and carotid artery with small circles for receptors and arrows to the rhythm centre labelled 'CO2, H+'. From the rhythm centre draw arrows to the diaphragm and intercostal muscles. Write in a corner: 'role of O2 insignificant'.

TextbookNCERT questions the board keeps asking

1 Define vital capacity. What is its significance?2 m

Vital capacity (VC) is the maximum volume of air a person can breathe in after a forced expiration, or the maximum volume a person can breathe out after a forced inspiration. It is the sum of the tidal volume, the inspiratory reserve volume and the expiratory reserve volume: VC = TV + IRV + ERV, about 4000 to 4600 mL in a healthy adult.

Significance: it represents the largest amount of air that can be exchanged in a single breath and so measures the strength of the respiratory muscles and the elasticity of the lungs. It is higher in athletes and people living at high altitude, and lower in smokers, the elderly and patients with emphysema or fibrosis. It is measured with a spirometer and used clinically to assess lung function.

2 State the volume of air remaining in the lungs after a normal breathing.2 m

The air left in the lungs after a normal expiration is the functional residual capacity (FRC). It is the sum of the expiratory reserve volume and the residual volume: FRC = ERV + RV = (1000 to 1100 mL) + (1100 to 1200 mL) ≈ 2100 to 2300 mL. Note that this is not the residual volume alone; the residual volume (1100 to 1200 mL) is what remains after a forced expiration.

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

The part of the respiratory system from the external nostrils to the terminal bronchioles is the conducting part. It is built for transporting, filtering, humidifying and warming air, not for exchange: its walls are thick, supported by cartilage and lined by thick epithelium, and they are not richly supplied with blood capillaries. The alveoli, on the other hand, are very thin-walled (a single layer of squamous epithelium), highly vascularised with a dense capillary network, and extremely numerous, giving an enormous surface area. The diffusion membrane between alveolar air and blood is made of only three thin layers (alveolar epithelium, basement substance, capillary endothelium) and is much less than a millimetre thick. Only here does a steep partial-pressure gradient meet a thin, moist, vascularised membrane, so gaseous exchange occurs in the alveolar region alone.

4 What are the major transport mechanisms for CO2? Explain.3 m

CO2 is transported by blood in three forms.

  1. As carbamino-haemoglobin (20 to 25 per cent): CO2 binds to the amino groups of haemoglobin in RBCs. High pCO2 and low pO2 in the tissues favour binding; low pCO2 and high pO2 in the alveoli favour its release.
  2. As bicarbonate ions (about 70 per cent): RBCs contain a very high concentration of the enzyme carbonic anhydrase (minute amounts in plasma), which catalyses CO2 + H2O ⇌ H2CO3 ⇌ HCO3− + H+. At the tissues, where pCO2 is high, CO2 diffuses into blood and is converted to HCO3− and H+. At the alveoli, where pCO2 is low, the reaction runs in reverse to give CO2 and H2O, and the CO2 is released into the alveolar air.
  3. Dissolved in plasma (about 7 per cent): CO2 is fairly soluble in water, so a small fraction travels in simple solution.

Every 100 mL of deoxygenated blood delivers about 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 lesser1 m

(ii) pO2 higher, pCO2 lesser. Atmospheric air has pO2 of 159 mm Hg and pCO2 of 0.3 mm Hg, while alveolar air has pO2 of 104 mm Hg and pCO2 of 40 mm Hg. Alveolar air is poorer in O2 and richer in CO2 because fresh air mixes with residual air that has already given O2 to the blood and received CO2 from it, and because the air is saturated with water vapour.

6 Explain the process of inspiration under normal conditions.3 m

Inspiration is the active phase of breathing in which atmospheric air is drawn into the lungs. It happens when the intra-pulmonary pressure becomes less than atmospheric pressure. The steps are:

  1. The diaphragm contracts and flattens, increasing the volume of the thoracic chamber along the antero-posterior axis.
  2. The external intercostal muscles contract, lifting the ribs and the sternum upward and outward, increasing the volume of the thoracic chamber along the dorso-ventral axis.
  3. The overall increase in thoracic volume causes a similar increase in pulmonary volume, because the lungs, enclosed in the airtight thoracic cavity by the pleura, follow the chest wall.
  4. The increase in pulmonary volume decreases the intra-pulmonary pressure to below atmospheric pressure.
  5. Air from outside moves down the pressure gradient into the lungs until the pressures equalise.

About 500 mL of air (the tidal volume) enters in one normal inspiration.

7 How is respiration regulated?3 m

Respiration is regulated by the neural system.

  • A respiratory rhythm centre in the medulla region of the brain is primarily responsible; it generates the basic rhythm of inspiration and expiration.
  • A pneumotaxic centre in the pons moderates the rhythm centre. Its neural signals reduce the duration of inspiration and thereby alter the respiratory rate.
  • A chemosensitive area adjacent to the rhythm centre is highly sensitive to CO2 and H+. A rise in these activates the area, which signals the rhythm centre to adjust breathing so that the excess CO2 and H+ are eliminated.
  • Receptors in the aortic arch and carotid artery also detect changes in CO2 and H+ concentration and send signals to the rhythm centre for remedial action.
  • The role of oxygen in regulating the respiratory rhythm is quite insignificant.
8 What is the effect of pCO2 on oxygen transport?2 m

pCO2 affects the binding of O2 to haemoglobin. In the alveoli, pCO2 is low (40 mm Hg) and pO2 is high (104 mm Hg); low pCO2 together with low H+ and lower temperature increases the affinity of haemoglobin for O2, so oxyhaemoglobin is formed. In the tissues, pCO2 is high (45 mm Hg) and pO2 is low (40 mm Hg); high pCO2 (and the H+ it generates through carbonic acid) lowers the affinity of haemoglobin for O2, shifting the oxygen dissociation curve to the right, so oxyhaemoglobin dissociates and O2 is released to the cells. Thus a high pCO2 favours unloading and a low pCO2 favours loading of O2; in this way CO2 ensures that O2 is delivered exactly where metabolism is producing CO2.

9 What happens to the respiratory process in a man going up a hill?2 m

As a person climbs, atmospheric pressure falls and with it the partial pressure of oxygen in the air. The pO2 in the alveoli therefore falls, the gradient for diffusion into the blood becomes smaller, and less haemoglobin gets saturated with O2. The tissues receive less oxygen, a condition called hypoxia, producing breathlessness, fatigue, headache and dizziness (altitude sickness). The body compensates: the rate and depth of breathing increase (hyperventilation) to bring more air to the alveoli, the heart rate rises to circulate blood faster, and over days the kidneys stimulate the bone marrow to produce more RBCs and haemoglobin, so that the blood can carry more O2. This is why people living at high altitude have a higher RBC count and a higher vital capacity.

10 What is the site of gaseous exchange in an insect?1 m

Insects have a network of air-filled tracheal tubes (tracheae) that open to the outside through pores called spiracles. The tracheae branch into finer and finer tubes, the tracheoles, which reach every tissue of the body. Gaseous exchange takes place by diffusion at the ends of these tracheoles, directly between the air in the tubes and the tissue cells; the blood (haemolymph) of insects plays no role in transporting respiratory gases.

11 Define oxygen dissociation curve. Can you suggest any reason for its sigmoidal pattern?3 m

The oxygen dissociation curve is the sigmoid (S-shaped) curve obtained when the percentage saturation of haemoglobin with O2 is plotted against the partial pressure of O2 (pO2). It is used to study the effect of pO2, pCO2, H+ concentration and temperature on the binding of O2 to haemoglobin.

Reason for the sigmoid shape: each haemoglobin molecule has four haem groups and can bind four O2 molecules, and the binding is cooperative. The first O2 binds with difficulty, but its binding changes the shape (conformation) of the haemoglobin molecule so that the affinity for the second, third and fourth O2 increases. Hence at low pO2 the saturation rises slowly, in the middle range it rises steeply, and near full saturation it flattens because few sites remain. This shape means haemoglobin loads fully at the high pO2 of the alveoli and unloads a large amount for a small drop of pO2 in the tissues.

12 Have you heard about hypoxia? Try to gather information about it, and discuss with your friends.2 m

Hypoxia is a condition in which the tissues do not receive enough oxygen to meet their metabolic needs. Its main causes are: (i) low pO2 in the inspired air, as at high altitude, so that haemoglobin is not fully saturated; (ii) anaemia, in which there is too little haemoglobin to carry O2; (iii) carbon monoxide poisoning, because CO binds haemoglobin far more strongly than O2 and blocks the sites; (iv) lung diseases such as emphysema, fibrosis or pneumonia that reduce the diffusion of O2 into the blood; (v) poor circulation to a tissue. Symptoms include breathlessness, rapid breathing, headache, fatigue, dizziness, bluish discolouration of lips and nails (cyanosis) and, if severe, confusion and unconsciousness. The body responds by increasing the breathing rate and heart rate and, in chronic cases, by producing more RBCs. Treatment is to supply oxygen and remove the cause.

13 Distinguish between (a) IRV and ERV (b) Inspiratory capacity and Expiratory capacity (c) Vital capacity and Total lung capacity.3 m

(a) Inspiratory reserve volume (IRV) is the additional volume of air a person can inspire by a forcible inspiration after a normal inspiration; it is 2500 to 3000 mL. Expiratory reserve volume (ERV) is the additional volume of air a person can expire by a forcible expiration after a normal expiration; it is 1000 to 1100 mL.

(b) Inspiratory capacity (IC) is the total volume of air a person can inspire after a normal expiration; IC = TV + IRV, about 3000 to 3500 mL. Expiratory capacity (EC) is the total volume of air a person can expire after a normal inspiration; EC = TV + ERV, about 1500 to 1600 mL.

(c) Vital capacity (VC) is the maximum volume of air a person can breathe out after a forced inspiration; VC = ERV + TV + IRV, about 4000 to 4600 mL; it excludes the residual volume and can be measured with a spirometer. Total lung capacity (TLC) is the total volume of air in the lungs at the end of a forced inspiration; TLC = RV + ERV + TV + IRV = VC + RV, about 5100 to 5800 mL; it includes the residual volume and cannot be measured directly by a spirometer.

14 What is Tidal volume? Find out the Tidal volume (approximate value) for a healthy human in an hour.2 m

Tidal volume (TV) is the volume of air inspired or expired during a normal, quiet respiration. It is approximately 500 mL in a healthy adult.

A healthy human breathes 12 to 16 times per minute. Air breathed per minute = 500 mL × 12 to 16 = 6000 to 8000 mL. In one hour = 6000 to 8000 mL × 60 = 3,60,000 to 4,80,000 mL, that is, about 360 to 480 litres of air per hour. Check: 500 × 12 × 60 = 3,60,000 and 500 × 16 × 60 = 4,80,000.

Previous years and frequently askedBoard questions with model answers

MCQ The volume of air remaining in the lungs after a forcible expiration is called
  1. Tidal volume
  2. Expiratory reserve volume
  3. Residual volume
  4. Functional residual capacity
Frequently asked1 m
Answer: (c)
Residual volume (1100 to 1200 mL) is what stays in the lungs even after forcible expiration. Functional residual capacity is the air left after a normal expiration (ERV + RV).
MCQ Which of the following capacities is equal to ERV + RV?
  1. Inspiratory capacity
  2. Expiratory capacity
  3. Functional residual capacity
  4. Vital capacity
Frequently asked1 m
Answer: (c)
FRC = ERV + RV, the air remaining after a normal expiration. IC = TV + IRV, EC = TV + ERV, VC = ERV + TV + IRV.
MCQ The partial pressure of oxygen in the alveolar air is about
  1. 159 mm Hg
  2. 104 mm Hg
  3. 95 mm Hg
  4. 40 mm Hg
Frequently asked1 m
Answer: (b)
From NCERT Table 14.1: atmosphere 159, alveoli 104, oxygenated blood 95, deoxygenated blood and tissues 40 mm Hg.
MCQ The largest fraction of carbon dioxide in the blood is transported as
  1. Carbamino-haemoglobin
  2. Bicarbonate ions
  3. Dissolved CO2 in plasma
  4. Carbonic acid
Frequently asked1 m
Answer: (b)
About 70 per cent travels as bicarbonate, 20 to 25 per cent as carbamino-haemoglobin and about 7 per cent dissolved in plasma.
MCQ The trachea divides into the two primary bronchi at the level of the
  1. Third thoracic vertebra
  2. Fifth thoracic vertebra
  3. Seventh cervical vertebra
  4. First lumbar vertebra
Frequently asked1 m
Answer: (b)
NCERT: the trachea extends up to the mid-thoracic cavity and divides at the level of the fifth thoracic vertebra.
MCQ The pneumotaxic centre that can moderate the respiratory rhythm centre is located in the
  1. Medulla oblongata
  2. Cerebrum
  3. Pons
  4. Cerebellum
Frequently asked1 m
Answer: (c)
The rhythm centre is in the medulla; the pneumotaxic centre is in the pons and its signals reduce the duration of inspiration.
MCQ The enzyme carbonic anhydrase is present in very high concentration in
  1. Blood plasma
  2. Red blood cells
  3. Alveolar epithelium
  4. Liver cells
Frequently asked1 m
Answer: (b)
RBCs contain a very high concentration of carbonic anhydrase; only minute quantities occur in plasma.
MCQ Which of the following animals respires through a moist cuticle?
  1. Cockroach
  2. Earthworm
  3. Frog
  4. Hydra
Frequently asked1 m
Answer: (b)
Earthworms exchange gases through their moist cuticle. Cockroach uses tracheal tubes, frog uses skin and lungs, Hydra uses its entire body surface.
MCQ Every 100 mL of oxygenated blood delivers approximately how much O2 to the tissues under normal physiological conditions?
  1. 4 mL
  2. 5 mL
  3. 20 mL
  4. 50 mL
Frequently asked1 m
Answer: (b)
NCERT: about 5 mL of O2 per 100 mL of oxygenated blood; 4 mL is the CO2 delivered to the alveoli by 100 mL of deoxygenated blood.
Assertion–Reason Assertion (A): The oxygen dissociation curve of haemoglobin is sigmoid in shape. Reason (R): Binding of one O2 molecule to haemoglobin increases the affinity of the remaining haem groups for O2.
  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true but R is not the correct explanation of A
  3. A is true but R is false
  4. A is false but R is true
Frequently asked1 m
Answer: (a)
Cooperative binding of the four O2 molecules is exactly what produces the slow-steep-flat S shape.
Assertion–Reason Assertion (A): Inspiration occurs when the intra-pulmonary pressure falls below atmospheric pressure. Reason (R): Contraction of the diaphragm and external intercostal muscles increases the volume of the thoracic cavity and hence of the lungs.
  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true but R is not the correct explanation of A
  3. A is true but R is false
  4. A is false but R is true
Frequently asked1 m
Answer: (a)
The muscle contraction raises thoracic and pulmonary volume, which lowers intra-pulmonary pressure below atmospheric; that gradient draws air in.
Assertion–Reason Assertion (A): Oxygen plays a very important role in the regulation of respiratory rhythm. Reason (R): The chemosensitive area adjacent to the rhythm centre is highly sensitive to CO2 and H+.
  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true but R is not the correct explanation of A
  3. A is true but R is false
  4. A is false but R is true
Frequently asked1 m
Answer: (d)
NCERT states that the role of oxygen in regulating respiratory rhythm is quite insignificant, so A is false. The chemosensitive area does sense CO2 and H+, so R is true.
Assertion–Reason Assertion (A): CO2 diffuses across the alveolar membrane in adequate amounts even though its partial-pressure gradient is only about 5 mm Hg. Reason (R): The solubility of CO2 is 20 to 25 times higher than that of O2.
  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true but R is not the correct explanation of A
  3. A is true but R is false
  4. A is false but R is true
Frequently asked1 m
Answer: (a)
High solubility means much more CO2 crosses per unit difference in partial pressure, which compensates for the small gradient.
Very short answer Differentiate between breathing and cellular respiration.Frequently asked2 m

Breathing is the physical process of exchange of O2 from the atmosphere with CO2 produced by the cells; it takes place in the lungs, involves muscles, needs no enzymes and releases no energy. Cellular respiration is the biochemical oxidation of food (glucose, amino acids, fatty acids) inside the cells using O2; it is enzyme-catalysed, occurs in the cytoplasm and mitochondria, releases energy as ATP and produces CO2. Breathing supplies the O2 for cellular respiration and removes the CO2 it produces.

Very short answer Name the three layers of the diffusion membrane in the lungs. What is its total thickness?Frequently asked2 m

The diffusion membrane consists of (i) the thin squamous epithelium of the alveoli, (ii) the basement substance (basement membrane) in between, and (iii) the endothelium of the alveolar capillaries. Its total thickness is much less than a millimetre, which allows rapid diffusion of O2 and CO2.

Very short answer What is the function of the epiglottis and of the pleural fluid?Frequently asked2 m

The epiglottis is a thin elastic cartilaginous flap that closes the glottis (the opening of the larynx) during swallowing, preventing food from entering the larynx and trachea. The pleural fluid lies between the two layers of the pleura that cover each lung; it reduces friction on the lung surface as the lungs expand and relax during breathing and keeps the lungs attached to the thoracic wall so that changes in thoracic volume are transmitted to the lungs.

Very short answer Write the reaction catalysed by carbonic anhydrase and state where the forward and the reverse reactions occur.Frequently asked2 m

CO2 + H2O ⇌ H2CO3 ⇌ HCO3− + H+

The forward reaction (formation of bicarbonate and H+) occurs at the tissue site, where pCO2 is high, inside the RBCs which contain a very high concentration of the enzyme. The reverse reaction (release of CO2 and H2O) occurs at the alveolar site, where pCO2 is low, so that the CO2 can diffuse into the alveoli and be breathed out.

Very short answer Name the respiratory organs of the following: sponge, earthworm, cockroach, fish, frog.Frequently asked2 m

Sponge: entire body surface by simple diffusion. Earthworm: moist cuticle. Cockroach: tracheal tubes (tracheal system). Fish: gills. Frog: lungs and moist skin (cutaneous respiration).

Very short answer A person's tidal volume is 500 mL, inspiratory reserve volume 2500 mL, expiratory reserve volume 1000 mL and residual volume 1200 mL. Calculate the vital capacity and the total lung capacity.Frequently asked2 m

Vital capacity = ERV + TV + IRV = 1000 + 500 + 2500 = 4000 mL.

Total lung capacity = VC + RV = 4000 + 1200 = 5200 mL (check: RV + ERV + TV + IRV = 1200 + 1000 + 500 + 2500 = 5200 mL).

Short answer Explain the mechanism of expiration under normal conditions. How can expiration be made forceful?Frequently asked3 m

Normal expiration is a passive process.

  1. The diaphragm relaxes and returns to its dome shape, moving upward, and the external intercostal muscles relax, so the ribs and sternum return to their normal positions.
  2. This reduces the thoracic volume in both the antero-posterior and dorso-ventral axes, and consequently the pulmonary volume decreases.
  3. The decrease in lung volume raises the intra-pulmonary pressure to slightly above atmospheric pressure.
  4. Air moves down the gradient from the lungs to the atmosphere: this is expiration, and about 500 mL (tidal volume) leaves.

Forceful expiration: we can increase the strength of expiration with additional muscles. The internal intercostal muscles contract to pull the ribs down and in, and the abdominal muscles contract, pushing the abdominal organs against the diaphragm so that it rises higher. This expels the expiratory reserve volume (1000 to 1100 mL) in addition to the tidal volume.

Short answer Describe the oxygen dissociation curve. Explain the conditions in the alveoli and in the tissues that affect the binding of O2 to haemoglobin.Frequently asked3 m

The oxygen dissociation curve is the sigmoid curve obtained when the percentage saturation of haemoglobin with O2 is plotted against pO2. It is S-shaped because each haemoglobin binds four O2 molecules cooperatively: the first binds slowly, and each binding increases the affinity for the next, so saturation rises slowly, then steeply, then flattens near 100 per cent.

Binding is primarily related to pO2, but pCO2, H+ concentration and temperature also interfere. In the alveoli, there is high pO2 (104 mm Hg), low pCO2 (40 mm Hg), lesser H+ concentration and lower temperature; all these favour the formation of oxyhaemoglobin (the curve sits to the left). In the tissues, there is low pO2 (40 mm Hg), high pCO2 (45 mm Hg), high H+ concentration and higher temperature; these favour the dissociation of O2 from oxyhaemoglobin (the curve shifts to the right). As a result every 100 mL of oxygenated blood delivers about 5 mL of O2 to the tissues.

Short answer Distinguish between the conducting part and the exchange part of the human respiratory system, giving their extent and functions.Frequently asked3 m

Conducting part: extends from the external nostrils up to the terminal bronchioles (nasal chamber, pharynx, larynx, trachea, bronchi, bronchioles). Its walls are relatively thick, supported by cartilage, and not adapted for diffusion. Functions: it transports atmospheric air to the alveoli, clears it of foreign particles (mucus and cilia), humidifies it and brings it to body temperature. No exchange of gases occurs here.

Exchange (respiratory) part: consists of the alveoli and their ducts. The alveoli are very thin-walled (squamous epithelium), irregular, bag-like and richly vascularised. Function: it is the site of actual diffusion of O2 and CO2 between blood and atmospheric air across the three-layered diffusion membrane.

Together, the branching network of bronchi, bronchioles and alveoli constitutes the lungs.

Short answer Write short notes on asthma, emphysema and occupational respiratory disorders.Frequently asked3 m

Asthma: a difficulty in breathing causing wheezing, due to inflammation of the bronchi and bronchioles. The narrowed airways with excess mucus resist airflow; it is often an allergic response to dust, pollen or smoke and is treated with bronchodilators.

Emphysema: a chronic disorder in which the alveolar walls are damaged, so that adjacent alveoli merge and the respiratory surface area is decreased. Gas exchange falls and the lungs lose their elasticity. One of the major causes is cigarette smoking; the damage is permanent.

Occupational respiratory disorders: in industries involving grinding or stone-breaking so much dust is produced that the body's defence mechanism cannot cope. Long exposure causes inflammation leading to fibrosis (proliferation of fibrous tissue), which stiffens the lung and reduces vital capacity, causing serious lung damage (for example silicosis, asbestosis). Workers should wear protective masks.

Short answer Using the partial pressures of O2 and CO2 in the alveoli, blood and tissues, explain the direction of diffusion of the two gases at the alveolar and tissue sites.Frequently asked3 m

Values in mm Hg (NCERT Table 14.1): alveoli pO2 104, pCO2 40; deoxygenated blood pO2 40, pCO2 45; oxygenated blood pO2 95, pCO2 40; tissues pO2 40, pCO2 45.

At the alveoli: pO2 is 104 in alveolar air and 40 in the deoxygenated blood arriving from the tissues, so O2 diffuses from alveoli into blood. pCO2 is 45 in that blood and 40 in the alveoli, so CO2 diffuses from blood into the alveoli.

At the tissues: pO2 is 95 in the oxygenated blood and 40 in the tissues, so O2 diffuses from blood into the tissues. pCO2 is 45 in the tissues and 40 in arterial blood, so CO2 diffuses from the tissues into the blood.

Although the CO2 gradient is only about 5 mm Hg compared with about 64 mm Hg for O2 at the alveolus, the solubility of CO2 is 20 to 25 times higher, so enough CO2 diffuses per unit gradient. The thin (less than 1 mm) three-layered diffusion membrane makes all these exchanges rapid.

Long answer Draw a labelled diagram of the human respiratory system. Describe the path of air from the nostrils to the alveoli and list the steps involved in respiration.Frequently asked5 m

Diagram: draw the nasal chamber, pharynx, larynx with epiglottis, trachea with C-shaped cartilaginous rings, the two primary bronchi entering the lungs, bronchioles ending in clusters of alveoli, the pleura around each lung, ribs at the sides and the dome-shaped diaphragm below (NCERT Fig. 14.1). Label at least: external nostrils, nasal chamber, pharynx, larynx, epiglottis, trachea, bronchus, bronchiole, alveoli, lung, pleural membranes, diaphragm.

Path of air: air enters through the external nostrils into the nasal chamber and nasal passage, which open into the pharynx, a common passage for food and air. From the pharynx it enters the larynx (cartilaginous sound box), the epiglottis preventing entry of food during swallowing. It then passes down the trachea, a straight tube that divides at the level of the fifth thoracic vertebra into right and left primary bronchi, which divide into secondary and tertiary bronchi and then bronchioles, ending in terminal bronchioles. Trachea, bronchi and initial bronchioles are supported by incomplete cartilaginous rings. Each terminal bronchiole opens into many thin-walled, vascularised alveoli, where gas exchange occurs. Nostrils to terminal bronchioles form the conducting part (transport, filtering, humidifying, warming); alveoli and ducts form the exchange part.

Steps of respiration: (i) breathing or pulmonary ventilation, drawing in atmospheric air and releasing CO2-rich alveolar air; (ii) diffusion of O2 and CO2 across the alveolar membrane; (iii) transport of gases by the blood; (iv) diffusion of O2 and CO2 between blood and tissues; (v) utilisation of O2 by cells for catabolic reactions with release of CO2 (cellular respiration).

Long answer Explain the mechanism of breathing in humans with the help of a diagram. Name the muscles involved and the pressure changes that take place.Frequently asked5 m

Breathing consists of inspiration (air drawn in) and expiration (air released out), and is driven by a pressure gradient between the lungs and the atmosphere created by the diaphragm and the intercostal muscles. The lungs lie in the airtight thoracic chamber (vertebral column dorsally, sternum ventrally, ribs laterally, diaphragm below), so any change in thoracic volume is reflected in pulmonary volume; we cannot change lung volume directly.

Diagram: two panels of the thorax (NCERT Fig. 14.2). Inspiration: ribs and sternum raised, diaphragm flattened, lungs expanded, arrow into trachea. Expiration: ribs and sternum lowered, diaphragm domed, lungs smaller, arrow out. Label sternum, ribs, external intercostals, diaphragm, lungs.

Inspiration (active): (1) the diaphragm contracts and flattens, increasing thoracic volume in the antero-posterior axis; (2) the external intercostal muscles contract, lifting the ribs and sternum, increasing thoracic volume in the dorso-ventral axis; (3) pulmonary volume increases likewise; (4) intra-pulmonary pressure falls below atmospheric pressure (negative pressure); (5) air moves in from outside.

Expiration (passive under normal conditions): (1) the diaphragm and external intercostals relax, restoring the diaphragm dome and the rib position; (2) thoracic and pulmonary volumes decrease; (3) intra-pulmonary pressure rises slightly above atmospheric; (4) air is expelled.

Forceful breathing recruits additional abdominal muscles and the internal intercostals. A healthy adult breathes 12 to 16 times per minute, moving about 500 mL (tidal volume) each time; the volumes can be measured with a spirometer for clinical assessment of lung function.

Long answer Describe how oxygen and carbon dioxide are transported in the blood. Give the percentages carried in each form and the relevant equations.Frequently asked5 m

Transport of oxygen. About 97 per cent of O2 is carried by RBCs bound to haemoglobin and 3 per cent dissolved in plasma. Haemoglobin, a red iron-containing pigment, binds O2 reversibly: Hb + 4O2 ⇌ oxyhaemoglobin. Each molecule carries a maximum of four O2. Binding depends primarily on pO2, and also on pCO2, H+ and temperature; the plot of percentage saturation against pO2 is the sigmoid oxygen dissociation curve. In the alveoli (high pO2, low pCO2, low H+, low temperature) oxyhaemoglobin forms; in the tissues (low pO2, high pCO2, high H+, high temperature) it dissociates. Every 100 mL of oxygenated blood delivers about 5 mL of O2 to the tissues.

Transport of carbon dioxide. (i) About 20 to 25 per cent binds to haemoglobin as carbamino-haemoglobin, favoured by high pCO2 and low pO2 in tissues and released in the alveoli where pCO2 is low and pO2 high. (ii) About 70 per cent travels as bicarbonate. RBCs contain a very high concentration of carbonic anhydrase (minute amounts in plasma), which catalyses CO2 + H2O ⇌ H2CO3 ⇌ HCO3− + H+. At the tissues, high pCO2 drives the reaction forward, trapping CO2 as HCO3−; at the alveoli, low pCO2 drives it in reverse, releasing CO2 and H2O. (Bicarbonate leaves the RBC in exchange for Cl−, the chloride shift.) (iii) About 7 per cent is carried dissolved in plasma, because CO2 is far more soluble than O2. Every 100 mL of deoxygenated blood delivers about 4 mL of CO2 to the alveoli.

Long answer Define the four respiratory volumes with their values and derive the five pulmonary capacities from them. Which of these cannot be measured with a spirometer and why?Frequently asked5 m

Volumes. Tidal volume (TV): air inspired or expired in a normal breath, about 500 mL. Inspiratory reserve volume (IRV): additional air that can be inspired by forcible inspiration, 2500 to 3000 mL. Expiratory reserve volume (ERV): additional air that can be expired by forcible expiration, 1000 to 1100 mL. Residual volume (RV): air remaining in the lungs after forcible expiration, 1100 to 1200 mL.

Capacities. Inspiratory capacity (IC) = TV + IRV, total air inspired after a normal expiration, about 3000 to 3500 mL. Expiratory capacity (EC) = TV + ERV, total air expired after a normal inspiration, about 1500 to 1600 mL. Functional residual capacity (FRC) = ERV + RV, air left after a normal expiration, about 2100 to 2300 mL. Vital capacity (VC) = ERV + TV + IRV, maximum air breathed out after a forced inspiration, about 4000 to 4600 mL. Total lung capacity (TLC) = RV + ERV + TV + IRV = VC + RV, air in the lungs after a forced inspiration, about 5100 to 5800 mL.

Spirometer limitation. A spirometer records only air that actually passes through the mouthpiece. The residual volume never leaves the lungs, so RV and every capacity that contains it, namely FRC and TLC, cannot be measured directly with a spirometer; they are found by indirect methods such as gas dilution. TV, IRV, ERV, IC, EC and VC can be measured directly.

Case-based A student recorded the following spirometer readings for a healthy adult: tidal volume 500 mL, inspiratory reserve volume 3000 mL, expiratory reserve volume 1100 mL. The residual volume was estimated separately as 1200 mL. The subject breathed 15 times per minute at rest.
(i) Calculate the vital capacity.
(ii) Calculate the functional residual capacity and the total lung capacity.
(iii) How much air does the subject breathe in one minute at rest?
(iv) Why could the spirometer not record the residual volume directly?
Frequently asked4 m

(i) VC = ERV + TV + IRV = 1100 + 500 + 3000 = 4600 mL.

(ii) FRC = ERV + RV = 1100 + 1200 = 2300 mL. TLC = VC + RV = 4600 + 1200 = 5800 mL (check: 1200 + 1100 + 500 + 3000 = 5800).

(iii) Minute volume = TV × breathing rate = 500 × 15 = 7500 mL per minute, within the NCERT range of 6000 to 8000 mL.

(iv) The residual volume stays in the lungs even after the most forcible expiration; it never passes through the spirometer mouthpiece, so the instrument cannot record it. It has to be estimated indirectly.

Case-based Read the passage and answer the questions. Ravi, who has smoked for twenty years, complains of breathlessness even on walking a short distance. His doctor finds that many of his alveolar walls have broken down so that neighbouring alveoli have merged into large sacs, and his vital capacity is much reduced. Ravi's neighbour Suresh works in a stone-crushing unit without a mask and has developed stiff, scarred lungs.
(i) Name Ravi's disorder and state its major cause.
(ii) Explain why the merging of alveoli causes breathlessness.
(iii) Name Suresh's condition and the process that produced it.
(iv) Suggest how Suresh's condition could have been prevented.
Frequently asked4 m

(i) Ravi has emphysema, a chronic disorder in which the alveolar walls are damaged. Its major cause is cigarette smoking.

(ii) The alveoli are the exchange part of the respiratory system. When their walls break down and they merge into fewer, larger sacs, the total respiratory surface area decreases sharply, so less O2 can diffuse into the blood and less CO2 out in each breath. The lungs also lose elasticity, so expiration becomes inefficient. The tissues are starved of O2 even on mild exertion, producing breathlessness.

(iii) Suresh has an occupational respiratory disorder (silicosis-type fibrosis). Long exposure to stone dust that the body's defence mechanism could not clear caused chronic inflammation leading to fibrosis, the proliferation of fibrous tissue in the lungs, which makes them stiff and reduces their ability to expand.

(iv) Wearing a protective mask at work to filter out the dust, along with dust-suppression measures at the site and regular lung check-ups, would have prevented the disorder.

Case-based The table gives partial pressures (mm Hg) of respiratory gases at different sites: atmospheric air pO2 159, pCO2 0.3; alveoli pO2 104, pCO2 40; deoxygenated blood pO2 40, pCO2 45; oxygenated blood pO2 95, pCO2 40; tissues pO2 40, pCO2 45.
(i) In which direction will O2 move at the alveoli, and what is the size of the gradient?
(ii) The CO2 gradient at the alveoli is small. Why does enough CO2 still leave the blood?
(iii) Why is alveolar pO2 lower than atmospheric pO2?
(iv) Name the layers through which O2 passes from alveolar air to the blood.
Frequently asked4 m

(i) O2 moves from the alveoli (104 mm Hg) into the deoxygenated blood (40 mm Hg); the gradient is 104 − 40 = 64 mm Hg.

(ii) The gradient is only 45 − 40 = 5 mm Hg, but the solubility of CO2 is 20 to 25 times higher than that of O2, so the amount of CO2 diffusing per unit difference in partial pressure is much greater. This compensates for the small gradient.

(iii) Fresh atmospheric air mixes in the lungs with residual air that has already given up O2 to the blood and received CO2; the air is also saturated with water vapour, whose pressure takes part of the total. So alveolar pO2 is 104 rather than 159 and pCO2 is 40 rather than 0.3.

(iv) The squamous epithelium of the alveolus, the basement substance (basement membrane) and the endothelium of the alveolar capillary; total thickness much less than 1 mm.

Test yourselfPractice quiz

Marks lost every yearCommon mistakes

  • Confusing residual volume with functional residual capacity: RV (1100 to 1200 mL) is the air left after a forced expiration; FRC (ERV + RV, about 2300 mL) is the air left after a normal expiration. NCERT exercise 2 asks for FRC, and most students write RV.
  • Swapping the CO2 transport percentages, writing 70 per cent for carbamino-haemoglobin. The correct split is 20 to 25 per cent carbamino-haemoglobin, 70 per cent bicarbonate, 7 per cent dissolved; the examiner deducts for each wrong figure.
  • Writing that the lungs 'pull in' or 'push out' air. The lungs have no muscle; marks go to the sequence muscle contraction, thoracic volume change, pulmonary volume change, pressure change, air movement. Leaving out the intra-pulmonary pressure step loses a mark every time.
  • Placing the pneumotaxic centre in the medulla. The rhythm centre and the chemosensitive area are in the medulla; the pneumotaxic centre is in the pons.
  • Saying oxygen regulates breathing. NCERT is explicit that the role of oxygen in regulating respiratory rhythm is quite insignificant; the stimuli are CO2 and H+. This turns up as an Assertion-Reason with the answer 'A false, R true'.
  • Misreading Table 14.1: alveolar pO2 is 104, oxygenated blood 95, deoxygenated blood and tissues 40; pCO2 is 40 in alveoli and oxygenated blood, 45 in deoxygenated blood and tissues. Mixing 95 and 104 or 40 and 45 costs the mark in a direction-of-diffusion question.
  • Naming the wrong organ: 'skin' for earthworm (NCERT says moist cuticle), 'spiracles' for insects (the organ is the tracheal tubes; spiracles are only openings), 'lungs' alone for frog (frog also uses moist skin).
  • Describing complete cartilaginous rings in the trachea, or saying the pharynx is the sound box. The rings are incomplete (C-shaped) and the larynx is the sound box; the pharynx is the common passage for food and air.
  • Drawing the oxygen dissociation curve as a straight line or a hyperbola and labelling the axes the wrong way round. It is sigmoid, with pO2 on the x-axis and percentage saturation of haemoglobin on the y-axis; a right shift for high pCO2 must be drawn to the right, not the left.

How to prepareStrategy for this chapter

This is a short chapter with a very high question-to-page ratio, so prepare it in a single focused sitting of two to three hours and then revise it from a one-page sheet. Start by reading NCERT once from 14.1 to 14.6, underlining every number: 500, 2500 to 3000, 1000 to 1100, 1100 to 1200, 12 to 16, 6000 to 8000, 159, 104, 95, 40, 45, 0.3, 97 and 3, 20 to 25, 70 and 7, 5 mL and 4 mL, 20 to 25 times, fifth thoracic vertebra. Copy these onto a single card; a large share of the MCQ, Assertion-Reason and case questions are built from them.

Next, practise the three diagrams that carry marks: the respiratory system (Fig. 14.1), the two-panel breathing mechanism (Fig. 14.2) and the spirogram (Fig. 14.3). Draw each twice from memory with all labels; in a 5-mark question the diagram is often two of the five marks. Add the dissociation curve with its axes and the right shift.

Then write out, as numbered lists, the four sequences the examiner wants verbatim: the path of air, the five steps of respiration, the steps of inspiration and expiration with pressures, and the regulation flow from medulla and pons to the muscles. Learn the CO2 equation with reversible arrows and the two sites where each direction runs. Finally work the NCERT exercise questions 1 to 14 in writing; every one of them has appeared in school papers, and question 14 (tidal volume per hour) is the only calculation in the chapter. For NEET, add the NCERT exemplar MCQs on the same numbers. In the exam, answer the 1-mark items from the card, and for the 3- and 5-mark questions write points with numbers rather than paragraphs.

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