Important Questions and Answers — Class 11 Biology Chapter 14 Notes

How to Use This Section

This chapter is short, but it is dense with things that must be got exactly right. Almost every question on it is "give the exact definition", "give the exact number", "which way does this gradient run", or "which of these two words was in the question". The marks sit in exact wording, exact figures and exact directions, and they are lost by one swapped word - normal written where the source says forcible, medulla written where the answer is pons, 104 written where the answer is 95.

Three habits will carry you through it.

First, learn every capacity as a formula, never as a total. The printed volumes are ranges, so any total you memorise will disagree with the next book you open. IC = TV + IRV, EC = TV + ERV, FRC = ERV + RV, VC = ERV + TV + IRV, TLC = RV + ERV + TV + IRV = VC + RV. When a paper hands you numbers, write the formula down first and put the given numbers into it. That is why a whole run of numerical items sits in Tier 2 below.

Second, read the adjective before the noun. "Remaining after a NORMAL expiration" is FRC. "Remaining after a FORCIBLE expiration" is RV. "Additional, by a forcible inspiration" is IRV. One adjective is the entire difference between a right and a wrong answer, and the chapter-end exercises are built on exactly that.

Third, keep the two number lists apart. Oxygen: 97 per cent on RBCs, 3 per cent dissolved, 4 molecules per haemoglobin, 5 mL delivered per 100 mL of oxygenated blood. Carbon dioxide: 20-25 per cent as carbamino-haemoglobin, 70 per cent as bicarbonate, about 7 per cent dissolved, 4 mL delivered per 100 mL of deoxygenated blood. Every distractor in this chapter is a real figure borrowed from the other list.

The items below run in three tiers.

Tier 1 - short recall. The definitions, the numbers, the names, the orders. Answer these aloud until none of them needs thinking about.
Tier 2 - applied reasoning. A situation or a set of numbers is given and you have to work out what follows. Five of these are numerical items on the respiratory capacities, because that is the commonest applied question set on this chapter.
Tier 3 - longer written answers. Full accounts written the way a written paper wants them, with the marking-scheme words in place.

The chapter has fourteen exercises at the end, which come to sixteen questions once the three parts of exercise 13 are counted separately. All sixteen of them are already answered in full inside the ten teaching sections of this chapter, so this section adds no exercise item of its own. That is not an omission - it means the exercise set is completely covered elsewhere. The last block of this section is a table telling you exactly where each of the sixteen sits, so you can attempt the whole set on paper and then check every answer against a complete one.

Tier 1 - Short Recall

Question 1

Q. Fill in the respiratory surface and the name of the respiration for each group: earthworm, insect, aquatic arthropod, terrestrial vertebrate.

Answer. Mechanisms of breathing vary among different groups of animals depending mainly on their habitats and levels of organisation. The four names all end in "-al" and are asked as a matching set.

Group	Respiratory surface	Name of the respiration
Earthworm	moist cuticle	cutaneous
Insect	a network of tracheal tubes	tracheal
Aquatic arthropod or mollusc	gills, special vascularised structures	branchial
Terrestrial vertebrate	lungs, vascularised bags	pulmonary

Below all four sit the lower invertebrates - sponges, coelenterates and flatworms - which exchange gases by simple diffusion over the entire body surface, and that has no "-al" name of its own.

Question 2

Q. Write the route of air from outside the body to the exchange surface, and mark on it the exact point at which the conducting part ends.

Answer. External nostrils -> nasal chamber (through the nasal passage) -> pharynx -> larynx -> trachea -> primary bronchi -> secondary bronchi -> tertiary bronchi -> bronchioles -> terminal bronchioles || alveoli.

The double line is the boundary. The conducting part runs from the external nostrils up to the terminal bronchioles. The respiratory or exchange part is the alveoli and their ducts. The terminal bronchiole is the last station that only conducts; the alveolus is the first that exchanges, and no diffusion of gases takes place anywhere in the conducting part.

Question 3

Q. Which structure is the sound box, which one is the lid, and at what level does the trachea divide?

Answer. Three separate facts that are constantly swapped for one another.

The larynx is a cartilaginous box which helps in sound production, and is therefore called the sound box.
The epiglottis is a thin elastic cartilaginous flap that covers the glottis during swallowing, to prevent the entry of food into the larynx. It is the lid, not the box.
The trachea divides at the level of the 5th thoracic vertebra into a right and a left primary bronchus.

Add one adjective while you are here: the trachea, the primary, secondary and tertiary bronchi and the initial bronchioles are supported by incomplete cartilaginous rings. Incomplete is the examinable word.

Question 4

Q. Name the muscle that contracts in each axis during inspiration, and say what each one moves.

Answer. Two muscles, two axes, and the pairing is the question.

Muscle that contracts	What it moves	Axis in which thoracic volume increases
Diaphragm	the floor of the thoracic chamber	antero-posterior axis
External inter-costal muscles	the ribs and the sternum, which are lifted up	dorso-ventral axis

Diaphragm goes with antero-posterior; external inter-costals go with dorso-ventral. Swapping the two axes is the commonest error in this part of the chapter. Note also that it is the external inter-costals that act during inspiration, and that they lift the ribs and the sternum, not the ribs alone.

Question 5

Q. State the pressure condition for inspiration and for expiration, and say which of the two is a passive process.

Answer. The movement of air into and out of the lungs is carried out by creating a pressure gradient between the lungs and the atmosphere.

Inspiration occurs if the intra-pulmonary pressure is less than the atmospheric pressure, that is, a negative pressure in the lungs with respect to atmospheric pressure.
Expiration takes place when the intra-pulmonary pressure is higher than the atmospheric pressure.

Normal expiration is the passive one. It is brought about by the relaxation of the diaphragm and the inter-costal muscles, which returns the diaphragm and sternum to their normal positions - there is no contraction of its own behind it. Inspiration is the active process.

Question 6

Q. Name the four respiratory volumes with their average values, and then arrange them in increasing order of value.

Answer. All four, as printed.

Volume	Definition	Average value
Tidal Volume (TV)	volume of air inspired or expired during a normal respiration	approximately 500 mL
Inspiratory Reserve Volume (IRV)	additional volume of air a person can inspire by a forcible inspiration	2500 mL to 3000 mL
Expiratory Reserve Volume (ERV)	additional volume of air a person can expire by a forcible expiration	1000 mL to 1100 mL
Residual Volume (RV)	volume of air remaining in the lungs even after a forcible expiration	1100 mL to 1200 mL

In increasing order of average value: TV, then ERV, then RV, then IRV. ERV and RV are the close pair - 1000 mL to 1100 mL against 1100 mL to 1200 mL - and that is where the ordering question is usually lost.

Question 7

Q. Write the five respiratory capacities as formulae, without any numbers at all.

Answer. Every capacity is a sum of volumes, and the formula is the answer.

Inspiratory Capacity: IC = TV + IRV.
Expiratory Capacity: EC = TV + ERV.
Functional Residual Capacity: FRC = ERV + RV.
Vital Capacity: VC = ERV + TV + IRV.
Total Lung Capacity: TLC = RV + ERV + TV + IRV, that is TLC = VC + RV.

Only two of the five contain the residual volume - FRC and TLC. IC, EC and VC do not. That is why total lung capacity always exceeds vital capacity by exactly RV, and why neither FRC nor TLC can be measured directly with a spirometer.

Question 8

Q. Reproduce the partial pressure table of the chapter, in mm Hg, for both respiratory gases.

Answer. Partial pressure is the pressure contributed by an individual gas in a mixture of gases, written $p\mathrm{O_2}$ and $p\mathrm{CO_2}$ .

Respiratory gas	Atmospheric air	Alveoli	Deoxygenated blood	Oxygenated blood	Tissues
Oxygen	159	104	40	95	40
Carbon dioxide	0.3	40	45	40	45

All values are partial pressures in mm Hg. Two repeats inside the table are used as distractors: oxygen reads 40 both in deoxygenated blood and in the tissues, and carbon dioxide reads 40 both in the alveoli and in oxygenated blood. And atmospheric $p\mathrm{CO_2}$ is 0.3, not zero.

Question 9

Q. Name the three factors that affect the rate of diffusion of the respiratory gases, and the three layers of the diffusion membrane.

Answer. Two lists of three, both asked as complete sets.

Factors affecting the rate of diffusion:

The pressure or concentration gradient - the main basis of the exchange.
The solubility of the gases.
The thickness of the membranes involved in diffusion.

Layers of the diffusion membrane:

The thin squamous epithelium of the alveoli.
The endothelium of the alveolar capillaries.
The basement substance in between them.

Three major layers, not two and not four - the basement substance is the one students drop - and the total thickness of the membrane is much less than a millimetre.

Question 10

Q. Give the oxygen transport figures: how it is carried, on what, and how much each molecule can hold.

Answer. Blood is the medium of transport for oxygen and carbon dioxide.

About 97 per cent of oxygen is transported by RBCs in the blood.
The remaining 3 per cent of oxygen is carried in a dissolved state through the plasma.
Haemoglobin is a red coloured, iron containing pigment present in the RBCs.
Oxygen binds with haemoglobin in a reversible manner to form oxyhaemoglobin. Reversible is the load-bearing word - a permanent bond would make the carrier useless.
Each haemoglobin molecule can carry a maximum of four molecules of oxygen.

Question 11

Q. Name the three routes by which carbon dioxide travels in the blood, with the share of each.

Answer. Three routes, and the three figures are the most asked numbers in gas transport.

Route	Form	Share
Bound to haemoglobin in the RBCs	carbamino-haemoglobin	nearly 20-25 per cent
In the blood after the carbonic anhydrase reaction	bicarbonate	70 per cent
Simply in solution in the plasma	dissolved carbon dioxide	about 7 per cent

Quote them exactly as printed even though 20-25 plus 70 plus 7 does not come to exactly 100 - one figure is a range and another is prefixed with "about". Do not swap 7 with 70. The bulk of oxygen rides on haemoglobin; the bulk of carbon dioxide rides as bicarbonate.

Question 12

Q. State the two delivery figures of the chapter and tie each one to the right blood.

Answer. They are offered as each other's distractors, so learn them as a pair.

Every 100 mL of oxygenated blood delivers around 5 mL of oxygen to the tissues under normal physiological conditions.
Every 100 mL of deoxygenated blood delivers approximately 4 mL of carbon dioxide to the alveoli.

Oxygen 5, carbon dioxide 4 - and the blood matches the gas it is named for: oxygenated blood hands over oxygen, deoxygenated blood hands over carbon dioxide.

Question 13

Q. Name the four structures involved in the regulation of respiration, say where each sits, and say what each does.

Answer. Human beings have a significant ability to maintain and moderate the respiratory rhythm to suit the demands of the body tissues, and this is done by the neural system.

Structure	Where it sits	What it does
Respiratory rhythm centre	the medulla region of the brain	primarily responsible for maintaining and moderating the respiratory rhythm
Pneumotaxic centre	the pons region of the brain	moderates the rhythm centre; its signals reduce the duration of inspiration and thereby alter the respiratory rate
Chemosensitive area	adjacent to the rhythm centre	highly sensitive to carbon dioxide and hydrogen ions; on a rise it signals the rhythm centre so that these substances are eliminated
Receptors of the aortic arch and the carotid artery	on those two blood vessels, outside the brain	recognise changes in carbon dioxide and hydrogen ion concentration and send signals to the rhythm centre for remedial actions

The role of oxygen in the regulation of respiratory rhythm is quite insignificant.

Question 14

Q. Define asthma, emphysema and hypoxia in one line each, and name the industries that cause occupational respiratory disorders.

Answer. Four one-line answers, each worth a mark on its own.

Asthma is a difficulty in breathing causing wheezing, due to inflammation of the bronchi and bronchioles.
Emphysema is a chronic disorder in which the alveolar walls are damaged, due to which the respiratory surface is decreased. One of the major causes is cigarette smoking.
Hypoxia is a condition in which the tissues receive an inadequate supply of oxygen.
Occupational respiratory disorders arise in industries 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 gives rise to inflammation leading to fibrosis - proliferation of fibrous tissues - and thus serious lung damage. Workers in such industries should wear protective masks.

Tier 2 - Applied Reasoning

Question 15

Q. A spirometry report on a healthy young man reads TV = 500 mL, IRV = 2500 mL, ERV = 1000 mL and RV = 1100 mL. Work out all five respiratory capacities.

Answer. Write the formula first, then put the given numbers into it. Never substitute values you have memorised.

Capacity	Formula	Arithmetic	Value
Inspiratory Capacity	IC = TV + IRV	500 + 2500	3000 mL
Expiratory Capacity	EC = TV + ERV	500 + 1000	1500 mL
Functional Residual Capacity	FRC = ERV + RV	1000 + 1100	2100 mL
Vital Capacity	VC = ERV + TV + IRV	1000 + 500 + 2500	4000 mL
Total Lung Capacity	TLC = RV + ERV + TV + IRV	1100 + 1000 + 500 + 2500	5100 mL

Always check the last row the second way: TLC = VC + RV = 4000 + 1100 = 5100 mL. The two routes agree, as they must, and that check catches an arithmetic slip in seconds.

Question 16

Q. A report gives IC = 3500 mL, TV = 500 mL, ERV = 1100 mL and RV = 1200 mL. Find the inspiratory reserve volume, and then the vital capacity and the total lung capacity.

Answer. The capacity is given and a volume is missing, so run the formula backwards.

IC = TV + IRV, so IRV = IC - TV = 3500 - 500 = 2500 mL. That sits at the bottom of the printed range of 2500 mL to 3000 mL, so the figure is a reasonable one.
VC = ERV + TV + IRV = 1100 + 500 + 2500 = 4100 mL.
TLC = VC + RV = 4100 + 1200 = 5300 mL.

Check TLC the long way as well: RV + ERV + TV + IRV = 1200 + 1100 + 500 + 2500 = 5300 mL. The two agree.

The point of an item like this is that the five formulae work in both directions. Any one unknown - a volume or a capacity - can be found as long as the rest of the sum is given.

Question 17

Q. For a man whose TV = 500 mL, IRV = 3000 mL, ERV = 1000 mL and RV = 1200 mL, how much air is left in his lungs at the end of (i) a normal expiration, (ii) a forcible expiration, and (iii) a normal inspiration?

Answer. Read the adjective before the word "expiration" every single time - that is the whole question.

(i) After a NORMAL expiration only the tidal volume has been let out, so the expiratory reserve volume is still sitting on top of the residual volume. The answer is the Functional Residual Capacity. FRC = ERV + RV = 1000 + 1200 = 2200 mL.
(ii) After a FORCIBLE expiration the expiratory reserve has been pushed out as well, so only the Residual Volume is left. RV = 1200 mL.
(iii) After a NORMAL inspiration the man is holding the FRC plus the breath he has just drawn in, that is FRC + TV = 2200 + 500 = 2700 mL.

One word - normal against forcible - changes the answer from a capacity of 2200 mL to a volume of 1200 mL, which is roughly half as much. A chapter-end exercise turns on exactly this distinction.

Question 18

Q. A patient's total lung capacity is 5000 mL and his residual volume is 1200 mL. His tidal volume is 500 mL and his inspiratory reserve volume is 2400 mL. Find his vital capacity, his expiratory reserve volume, his inspiratory capacity, his expiratory capacity and his functional residual capacity. Comment on one of the figures.

Answer. Start from the relation that links the two given capacities.

TLC = VC + RV, so VC = TLC - RV = 5000 - 1200 = 3800 mL.
VC = ERV + TV + IRV, so ERV = VC - TV - IRV = 3800 - 500 - 2400 = 900 mL.
IC = TV + IRV = 500 + 2400 = 2900 mL.
EC = TV + ERV = 500 + 900 = 1400 mL.
FRC = ERV + RV = 900 + 1200 = 2100 mL.

The figure worth commenting on is the expiratory reserve volume of 900 mL, which is below the printed range of 1000 mL to 1100 mL. The vital capacity of 3800 mL is also low. A reduced vital capacity is the standard clinical signal that the lungs are not expanding or emptying as they should, and that is precisely why the capacities are used in clinical diagnosis.

Question 19

Q. A man breathes 14 times a minute with a tidal volume of 500 mL. How much air does he move in a minute, and how much in an hour? Check your minute figure against the chapter's printed range.

Answer. Tidal volume multiplied by breathing rate gives the air moved per unit time.

In one minute: 500 mL x 14 = 7000 mL.
In one hour: 7000 mL x 60 = 4,20,000 mL, that is 420 litres.

The check. On an average a healthy human breathes 12-16 times per minute, and from that a healthy man can inspire or expire approximately 6000 to 8000 mL of air per minute. 7000 mL sits neatly inside that range, as it must, since 14 breaths a minute is the middle of 12 to 16.

Watch the unit of time before you read the number. 500 mL is per breath; 6000 to 8000 mL is per minute. Attaching "per minute" to 500 mL is the standard planted error.

Question 20

Q. A long-term cigarette smoker is breathless on climbing one flight of stairs, but his airways are open and he does not wheeze. Name the disorder, say what has been damaged, and say which respiratory capacity you would expect to be reduced.

Answer. The disorder is emphysema. It is a chronic disorder in which the alveolar walls are damaged, due to which the respiratory surface is decreased, and cigarette smoking is one of the major causes.

The absence of wheezing rules out asthma. Asthma is a difficulty in breathing causing wheezing, due to inflammation of the bronchi and bronchioles - it narrows the tubes. Emphysema leaves the tubes open and destroys the exchange surface instead.

The capacity that suffers is the vital capacity. With less alveolar surface, less air is usefully moved and less oxygen can diffuse across per breath, so the tissues receive an inadequate supply of oxygen - hypoxia - and the man is breathless on effort. Note where the failure sits in the five steps of respiration: it is step 2, the diffusion of gases across the alveolar membrane, not step 1 and not step 3.

Question 21

Q. In one patient the diaphragm is paralysed. In another, a tight plaster cast around the chest holds the ribs so that they cannot be lifted. Explain what is lost in each case.

Answer. Each patient has lost one of the two axes in which the thoracic volume is increased.

The paralysed diaphragm means the thoracic volume can no longer be increased in the antero-posterior axis, because inspiration is initiated by the contraction of the diaphragm, which increases the volume in that axis.
The rigid cast means the ribs and the sternum cannot be lifted up by the external inter-costal muscles, so the thoracic volume cannot be increased in the dorso-ventral axis.

In both cases the consequence runs down the same chain. A smaller increase in thoracic volume means a smaller increase in pulmonary volume, because the thoracic chamber is air-tight and any change in its volume is reflected in the pulmonary cavity. The intra-pulmonary pressure therefore does not fall as far below atmospheric, so less air is forced in and the tidal volume drops.

Neither patient can compensate from inside the lung, because there is no muscle in the lung and we cannot directly alter the pulmonary volume. The one thing they can add is the additional muscles in the abdomen, which increase the strength of inspiration and expiration.

Question 22

Q. The carbon dioxide gradient between the tissues and the blood is only 5 mm Hg, while the oxygen gradient between the alveoli and the blood is 64 mm Hg. Why does the small gradient not leave carbon dioxide stranded in the tissues?

Answer. Because the rate of diffusion depends on solubility as well as on the gradient, and carbon dioxide wins overwhelmingly on solubility.

Read the two gradients off the table first. Carbon dioxide runs 45 in the tissues against 40 in oxygenated blood, a difference of 5 mm Hg. Oxygen runs 104 in the alveoli against 40 in deoxygenated blood, a difference of 64 mm Hg.

Now the solubility. The solubility of $\mathrm{CO_2}$ is 20-25 times higher than that of $\mathrm{O_2}$ . Therefore the amount of carbon dioxide that can diffuse through the diffusion membrane per unit difference in partial pressure is much higher compared to that of oxygen.

So a small gradient is enough for carbon dioxide, while oxygen needs a large one. That single fact is why the two rows of the partial pressure table look so different in scale, and it is a favourite one-mark reason question.

Question 23

Q. A man rescued from a smoky room has a normal breathing rate, a normal $p\mathrm{O_2}$ in his alveoli and undamaged lungs, yet his tissues are starved of oxygen. What has happened, and what is the condition called?

Answer. The condition is hypoxia - a condition in which the tissues receive an inadequate supply of oxygen - and the cause here is carbon monoxide poisoning.

The failure is in the carrier, not in the exchange. About 97 per cent of oxygen is transported by RBCs, bound to haemoglobin, a red coloured, iron containing pigment. Carbon monoxide blocks haemoglobin, so it cannot pick oxygen up. The air is fine, the alveoli are fine, the diffusion membrane is fine - but the vehicle that should carry 97 per cent of the oxygen is out of service, so only the 3 per cent that travels dissolved in plasma is left, and that is nowhere near enough.

Two of the four causes of hypoxia work like this. Anaemia and carbon monoxide poisoning cause hypoxia with perfectly normal lungs, because the failure is in the carrier; high altitude and lung disease such as emphysema break the chain earlier, at the air and at the exchange surface respectively.

Question 24

Q. A muscle in heavy exercise becomes warmer, more acidic and richer in carbon dioxide. Say what those three changes do to oxyhaemoglobin, and what they do at the same moment to carbamino-haemoglobin.

Answer. The three changes all push in the same useful direction - oxygen off, carbon dioxide on - which is exactly what a working muscle needs.

For oxygen. In the tissues, where low $p\mathrm{O_2}$ , high $p\mathrm{CO_2}$ , high hydrogen ion concentration and higher temperature exist, the conditions are favourable for dissociation of oxygen from the oxyhaemoglobin. A hard-working muscle exaggerates every one of those four, so oxyhaemoglobin gives up its oxygen even more readily there, and more oxygen is unloaded exactly where it is being spent.

For carbon dioxide. The binding of carbon dioxide to haemoglobin is related to $p\mathrm{CO_2}$ , and $p\mathrm{O_2}$ is a major factor which could affect this binding. When $p\mathrm{CO_2}$ is high and $p\mathrm{O_2}$ is low, as in the tissues, more binding of carbon dioxide occurs, so more carbamino-haemoglobin is formed.

One molecule, two cargoes, two opposite stations. At the tissues haemoglobin drops oxygen and picks up carbon dioxide; at the alveoli it drops carbon dioxide and picks up oxygen.

Question 25

Q. Two blockages are compared: in patient A a terminal bronchiole is plugged, in patient B the alveoli beyond an open bronchiole have had their walls destroyed. In which patient has gas exchange itself been stopped, and in which has only the delivery of air been stopped?

Answer. The answer follows from the boundary between the two parts of the respiratory system.

Patient A has lost part of the conducting part. The conducting part runs from the external nostrils up to the terminal bronchioles, and it transports the atmospheric air to the alveoli, clears it of foreign particles, humidifies it and brings it to body temperature. No diffusion of gases takes place anywhere in it. So the plug does not stop diffusion directly - it stops air reaching the alveoli where diffusion would have happened.
Patient B has lost part of the exchange part. The respiratory or exchange part is the alveoli and their ducts, the site of the actual diffusion of oxygen and carbon dioxide between blood and atmospheric air. Here the exchange surface itself is gone, which is emphysema.

Both patients end up short of oxygen, but the step that failed is different - step 1, breathing or pulmonary ventilation, in patient A; step 2, diffusion of gases across the alveolar membrane, in patient B.

Question 26

Q. A swimmer takes several deep, rapid breaths before diving so that he can stay under longer. Explain why the trick works, and why it is dangerous.

Answer. It works because it removes the stimulus that would have forced him to breathe, and it is dangerous for exactly the same reason.

Start from what actually drives the urge to breathe. A chemosensitive area is situated adjacent to the rhythm centre and is highly sensitive to carbon dioxide and hydrogen ions. An increase in these substances activates this centre, which signals the rhythm centre to make the necessary adjustments so that these substances can be eliminated. The role of oxygen in the regulation of respiratory rhythm is quite insignificant.

Breathing hard beforehand blows off carbon dioxide, so the swimmer starts the dive with an unusually low $p\mathrm{CO_2}$ . The chemosensitive area therefore stays quiet for much longer than usual, and the urge to breathe is delayed.

The danger is that the alarm he has silenced was never the oxygen alarm. His oxygen is being used up all the same, and since oxygen has no significant regulatory role, nothing warns him. He can reach a level of hypoxia that causes him to lose consciousness under water before any signal to breathe arrives. The body regulates breathing by watching what has to be thrown out, not what is running low, and this is the case where that design fails.

Question 27

Q. Two students of the same build, one who has lived at sea level all his life and one who has spent a month at a high-altitude camp, are tested at the same $p\mathrm{O_2}$ . Whose blood carries more oxygen, and what has changed in him?

Answer. The student who has spent a month at altitude carries more oxygen at the same $p\mathrm{O_2}$ .

What changed is the amount of the carrier, not the physics of the binding. Over days of acclimatisation the body produces more red blood cells and therefore more haemoglobin. With more haemoglobin, the blood can carry more oxygen even at the lower partial pressure, because about 97 per cent of oxygen is transported by RBCs and each haemoglobin molecule can carry a maximum of four molecules of oxygen - more molecules of the carrier means more oxygen carried for the same saturation.

Why he needed the change in the first place. At altitude the percentage of oxygen in the air is unchanged; it is the atmospheric pressure and therefore $p\mathrm{O_2}$ that falls. Less oxygen diffuses into the blood at the alveoli, the percentage saturation of haemoglobin falls, and the tissues receive an inadequate supply of oxygen - hypoxia. This is why climbers spend days at intermediate camps instead of going straight to the top.

Question 28

Q. A sharp object punctures the chest wall so that air enters the space between the two pleural membranes on one side. Explain why that lung stops inflating, even though the diaphragm and inter-costal muscles on that side are working normally.

Answer. Because the mechanism of breathing depends entirely on the thoracic chamber being air-tight.

We have two lungs, covered by a double layered pleura, with pleural fluid between them; the outer pleural membrane is in close contact with the thoracic lining and the inner pleural membrane is in contact with the lung surface. The lungs are situated in the thoracic chamber, which is anatomically an air-tight chamber.

The consequence of that arrangement is the whole mechanism. Any change in the volume of the thoracic cavity will be reflected in the lung cavity, and such an arrangement is essential for breathing, as we cannot directly alter the pulmonary volume.

Let air in between the two pleural membranes and the coupling is broken. The chest wall can still expand, but the lung no longer follows it, because the fall in pressure is now taken up by the air in the pleural space instead of being passed on to the lung. The intra-pulmonary pressure of that lung does not fall below atmospheric, so no air is forced into it, and it collapses.

Say the key line plainly: there is no muscle in the lung. The lung is moved by the box around it, and a hole in the box stops the lung.

Tier 3 - Longer Written Answers

Question 29

Q. Describe the mechanism of breathing in full - both stages, both axes, and the pressure changes - and say which stage is active and which passive.

Answer. Breathing involves two stages: inspiration, during which atmospheric air is drawn in, and expiration, by which the alveolar air is released out. The movement of air into and out of the lungs is carried out by creating a pressure gradient between the lungs and the atmosphere. The diaphragm and a specialised set of muscles - the external and internal intercostals between the ribs - help in the generation of such gradients.

Inspiration, step by step.

Inspiration is initiated by the contraction of the diaphragm, which increases the volume of the thoracic chamber in the antero-posterior axis.
The contraction of the external inter-costal muscles lifts up the ribs and the sternum, causing an increase in the volume of the thoracic chamber in the dorso-ventral axis.
The overall increase in the thoracic volume causes a similar increase in the pulmonary volume.
An increase in pulmonary volume decreases the intra-pulmonary pressure to less than the atmospheric pressure, that is, a negative pressure in the lungs with respect to atmospheric pressure.
This forces the air from outside to move into the lungs - that is inspiration.

Expiration, step by step.

Relaxation of the diaphragm and the inter-costal muscles returns the diaphragm and sternum to their normal positions.
This reduces the thoracic volume and thereby the pulmonary volume.
The fall in pulmonary volume leads to an increase in intra-pulmonary pressure to slightly above the atmospheric pressure.
This causes the expulsion of air from the lungs - that is expiration.

Inspiration is an active process; normal expiration is a passive process, driven by relaxation and the elastic recoil that follows, not by a contraction of its own. We can, however, increase the strength of inspiration and expiration with the help of additional muscles in the abdomen.

Two facts complete the account. On an average, a healthy human breathes 12-16 times per minute, and the volume of air involved in breathing movements can be estimated by using a spirometer, which helps in the clinical assessment of pulmonary functions.

Nothing pumps air in. The lungs are made low-pressure and the atmosphere pushes air in, and that is possible only because the thoracic chamber is air-tight and we cannot directly alter the pulmonary volume.

Question 30

Q. Give a complete account of the respiratory volumes and capacities, and work every capacity out for a man with TV = 500 mL, IRV = 2800 mL, ERV = 1050 mL and RV = 1150 mL.

Answer. There are four volumes, which are measured, and five capacities, which are sums of those volumes. By adding up a few respiratory volumes one can derive various pulmonary capacities, which can be used in clinical diagnosis.

The four volumes, with the printed ranges.

Volume	Definition	Printed value
Tidal Volume (TV)	volume of air inspired or expired during a normal respiration	approximately 500 mL
Inspiratory Reserve Volume (IRV)	additional volume of air a person can inspire by a forcible inspiration	2500 mL to 3000 mL
Expiratory Reserve Volume (ERV)	additional volume of air a person can expire by a forcible expiration	1000 mL to 1100 mL
Residual Volume (RV)	volume of air remaining in the lungs even after a forcible expiration	1100 mL to 1200 mL

The five capacities, with the given numbers put into each formula.

Capacity	Definition	Formula	Arithmetic	Value
Inspiratory Capacity (IC)	total volume of air a person can inspire after a normal expiration	IC = TV + IRV	500 + 2800	3300 mL
Expiratory Capacity (EC)	total volume of air a person can expire after a normal inspiration	EC = TV + ERV	500 + 1050	1550 mL
Functional Residual Capacity (FRC)	volume of air that will remain in the lungs after a normal expiration	FRC = ERV + RV	1050 + 1150	2200 mL
Vital Capacity (VC)	maximum volume of air a person can breathe in after a forced expiration, or breathe out after a forced inspiration	VC = ERV + TV + IRV	1050 + 500 + 2800	4350 mL
Total Lung Capacity (TLC)	total volume of air accommodated in the lungs at the end of a forced inspiration	TLC = RV + ERV + TV + IRV	1150 + 1050 + 500 + 2800	5500 mL

Check the last row the short way: TLC = VC + RV = 4350 + 1150 = 5500 mL.

Three things to say once the table is written.

Only FRC and TLC contain the residual volume; IC, EC and VC do not. TLC therefore exceeds VC by exactly RV.
Residual volume cannot be measured by a spirometer, because a spirometer records only the air that actually moves in and out of the lungs and residual volume never leaves them. It is determined indirectly, and every capacity that contains it - FRC and TLC - inherits the same limitation.
The totals depend entirely on which end of each printed range is chosen, so the formula is the answer, not a memorised total. If a numerical question gives you volumes, use them.

Question 31

Q. Explain the exchange of gases at the alveoli and at the tissues in full: the mechanism, the factors, the gradients and the membrane.

Answer. Alveoli are the primary sites of exchange of gases, and exchange of gases also occurs between blood and tissues - two sites, one at each end of the circulation, running on the same physics.

The mechanism. Oxygen and carbon dioxide are exchanged at these sites by simple diffusion, mainly based on pressure or concentration gradient. There is no pump, no carrier protein and no energy spent anywhere in gas exchange.

The three factors that affect the rate of diffusion are the pressure or concentration gradient, the solubility of the gases, and the thickness of the membranes involved in diffusion.

The gradients, read off the table. Partial pressure is the pressure contributed by an individual gas in a mixture of gases, written $p\mathrm{O_2}$ and $p\mathrm{CO_2}$ , in mm Hg.

Respiratory gas	Atmospheric air	Alveoli	Deoxygenated blood	Oxygenated blood	Tissues
Oxygen	159	104	40	95	40
Carbon dioxide	0.3	40	45	40	45

There is a concentration gradient for oxygen from alveoli to blood and from blood to tissues - 104 against 40 at the alveoli, then 95 against 40 at the tissues.
A gradient is present for carbon dioxide in the opposite direction, that is, from tissues to blood and from blood to alveoli - 45 against 40 at the tissues, then 45 against 40 at the alveoli.

Solubility. The solubility of $\mathrm{CO_2}$ is 20-25 times higher than that of $\mathrm{O_2}$ , so the amount of carbon dioxide that can diffuse through the diffusion membrane per unit difference in partial pressure is much higher compared to that of oxygen. That is why carbon dioxide manages on a 5 mm Hg gradient while oxygen needs a large one.

The membrane. The diffusion membrane is made up of three major layers: the thin squamous epithelium of the alveoli, the endothelium of the alveolar capillaries, and the basement substance in between them. Its total thickness is much less than a millimetre.

Therefore, all the factors in our body are favourable for the diffusion of oxygen from alveoli to tissues and of carbon dioxide from tissues to alveoli - the gradients run the right way, carbon dioxide is highly soluble, and the membrane is extremely thin.

Question 32

Q. Give a full account of the transport of oxygen, including the dissociation curve and the conditions at the two ends of the circulation.

Answer. Blood is the medium of transport for oxygen and carbon dioxide.

How much travels how. About 97 per cent of oxygen is transported by RBCs in the blood; the remaining 3 per cent of oxygen is carried in a dissolved state through the plasma.

The carrier. Haemoglobin is a red coloured, iron containing pigment present in the RBCs. Oxygen can bind with haemoglobin in a reversible manner to form oxyhaemoglobin, and each haemoglobin molecule can carry a maximum of four molecules of oxygen.

What governs the binding. Binding of oxygen with haemoglobin is primarily related to the partial pressure of oxygen, $p\mathrm{O_2}$ . Partial pressure of carbon dioxide, hydrogen ion concentration and temperature are the other factors which can interfere with this binding. One primary factor and three modifiers - an option that calls $p\mathrm{CO_2}$ the primary factor is wrong.

The curve. A sigmoid curve is obtained when percentage saturation of haemoglobin with oxygen is plotted against $p\mathrm{O_2}$ , and this curve is called the oxygen dissociation curve. Percentage saturation goes up the vertical axis and $p\mathrm{O_2}$ in mm Hg runs along the horizontal axis. The curve is highly useful in studying the effect of factors like $p\mathrm{CO_2}$ and hydrogen ion concentration on the binding of oxygen with haemoglobin.

The two opposed sets of conditions.

Condition	In the alveoli	In the tissues
$p\mathrm{O_2}$	high	low
$p\mathrm{CO_2}$	low	high
Hydrogen ion concentration	lesser	high
Temperature	lower	higher
What is favoured	the formation of oxyhaemoglobin	the dissociation of oxygen from oxyhaemoglobin

Learn one column and flip every entry to get the other. This clearly indicates that oxygen gets bound to haemoglobin at the lung surface and gets dissociated at the tissues - loading where the air is, unloading where the cells are.

And the amount delivered. Every 100 mL of oxygenated blood can deliver around 5 mL of oxygen to the tissues under normal physiological conditions.

Question 33

Q. Give a full account of the transport of carbon dioxide, including the reaction that carries most of it.

Answer. Carbon dioxide leaves the tissues by three routes, and the three figures are printed as follows.

Route	Form	Share
Transported by RBCs	carbamino-haemoglobin	nearly 20-25 per cent
Carried in the blood	bicarbonate	70 per cent
In a dissolved state through plasma	dissolved carbon dioxide	about 7 per cent

Reproduce these three figures as printed even though they do not total exactly 100.

Route one - carbamino-haemoglobin. Carbon dioxide is carried by haemoglobin as carbamino-haemoglobin, about 20-25 per cent of the total. This binding is related to the partial pressure of carbon dioxide, and the partial pressure of oxygen is a major factor which could affect this binding. When $p\mathrm{CO_2}$ is high and $p\mathrm{O_2}$ is low, as in the tissues, more binding of carbon dioxide occurs; whereas when $p\mathrm{CO_2}$ is low and $p\mathrm{O_2}$ is high, as in the alveoli, dissociation of carbon dioxide from carbamino-haemoglobin takes place. That is, carbon dioxide which is bound to haemoglobin from the tissues is delivered at the alveoli.

Route two - bicarbonate, and the enzyme behind it. RBCs contain a very high concentration of the enzyme carbonic anhydrase, and minute quantities of the same are present in the plasma too. This enzyme facilitates the following reaction in both directions.

$\mathrm{CO_2 + H_2O} \xrightleftharpoons{\text{carbonic anhydrase}} \mathrm{H_2CO_3} \xrightleftharpoons{\text{carbonic anhydrase}} \mathrm{HCO_3^- + H^+}$

At the tissue site, where the partial pressure of carbon dioxide is high due to catabolism, carbon dioxide diffuses into blood (RBCs and plasma) and forms $\mathrm{HCO_3^-}$ and $\mathrm{H^+}$ .
At the alveolar site, where $p\mathrm{CO_2}$ is low, the reaction proceeds in the opposite direction, leading to the formation of carbon dioxide and water.

Thus, carbon dioxide trapped as bicarbonate at the tissue level and transported to the alveoli is released out as carbon dioxide. The gas is disguised for the journey and undisguised on arrival, and the direction the reaction runs is decided entirely by the local $p\mathrm{CO_2}$ .

Route three - dissolved. About 7 per cent is simply carried in a dissolved state through the plasma.

The amount delivered. Every 100 mL of deoxygenated blood delivers approximately 4 mL of carbon dioxide to the alveoli.

Question 34

Q. How is the respiratory rhythm regulated, and why is a fall in oxygen not the signal that makes you breathe?

Answer. Human beings have a significant ability to maintain and moderate the respiratory rhythm to suit the demands of the body tissues, and this is done by the neural system. Four structures do the work.

The respiratory rhythm centre is a specialised centre present in the medulla region of the brain, and it is primarily responsible for this regulation.
The pneumotaxic centre is present in the pons region of the brain and can moderate the functions of the respiratory rhythm centre. Neural signals from this centre can reduce the duration of inspiration and thereby alter the respiratory rate. Shortening inspiration is its only stated action - it does not start breathing and it does not stop it.
A chemosensitive area is situated adjacent to the rhythm centre, and it is highly sensitive to carbon dioxide and hydrogen ions. An increase in these substances activates this centre, which signals the rhythm centre to make the necessary adjustments so that these substances can be eliminated.
Receptors associated with the aortic arch and the carotid artery also recognise changes in carbon dioxide and hydrogen ion concentration and send necessary signals to the rhythm centre for remedial actions.

Read the four as a direction of traffic. The chemosensitive area and the aortic and carotid receptors send information in to the rhythm centre. The pneumotaxic centre modifies the rhythm centre. Only the rhythm centre actually drives the breathing.

Now the part that is most often got backwards. The role of oxygen in the regulation of respiratory rhythm is quite insignificant. Neither the chemosensitive area nor the aortic and carotid receptors is described as an oxygen sensor - both watch carbon dioxide and hydrogen ions. The body regulates breathing as a waste alarm, not as a fuel gauge: it watches what has to be thrown out, not what is running low. Learn the sentence in those exact words, because the standing distractor is an option claiming that a fall in oxygen is the main stimulus for breathing.

Question 35

Q. Write an account of the disorders of the respiratory system and of hypoxia, including its causes.

Answer. Three disorders, each defined by one damaged structure and one giveaway word.

Disorder	What is damaged	Cause	Giveaway word
Asthma	the bronchi and bronchioles are inflamed	inflammation of the airways	wheezing
Emphysema	the alveolar walls are damaged, so the respiratory surface is decreased	cigarette smoking is one of the major causes	decreased respiratory surface
Occupational respiratory disorders	serious lung damage after long dust exposure	dust in industries involving grinding or stone-breaking	fibrosis

In full sentences. Asthma is a difficulty in breathing causing wheezing, due to inflammation of the bronchi and bronchioles. Emphysema is a chronic disorder in which the alveolar walls are damaged, due to which the respiratory surface is decreased. In certain industries, especially those involving grinding or stone-breaking, 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 - proliferation of fibrous tissues - and thus causing serious lung damage. Workers in such industries should wear protective masks.

Two of the three hit the airway and one hits the exchange surface. Asthma narrows the tubes, which is why it wheezes; emphysema destroys the alveolar walls, so the tubes are open but there is less surface left to diffuse across. A chronic smoker's breathlessness is emphysema, not asthma.

Hypoxia. Hypoxia is a condition in which the tissues receive an inadequate supply of oxygen. The definition is about the tissues, not about the lungs and not about the air, and anything that breaks the chain between the atmosphere and the cell can cause it.

Cause	Where the chain breaks
Going to a high altitude	the partial pressure of oxygen in the air is low, so less oxygen enters the blood
Lung disease such as emphysema	the respiratory surface is reduced, so less oxygen diffuses across
Anaemia	there is too little haemoglobin to carry the oxygen
Carbon monoxide poisoning	haemoglobin is blocked, so it cannot pick oxygen up

Only one of the four is a lung problem. Anaemia and carbon monoxide poisoning cause hypoxia with perfectly normal lungs.

Question 36

Q. Compare the respiratory organs used by different animal groups, and then explain how the human respiratory system is divided by function.

Answer. Mechanisms of breathing vary among different groups of animals depending mainly on their habitats and levels of organisation.

Group	Respiratory surface	Name
Sponges, coelenterates, flatworms	simple diffusion over the entire body surface	-
Earthworms	moist cuticle	cutaneous
Insects	a network of tubes - tracheal tubes - carrying atmospheric air within the body	tracheal
Most aquatic arthropods and molluscs	gills, special vascularised structures	branchial
Terrestrial forms	lungs, vascularised bags	pulmonary

Among the vertebrates: fishes use gills; amphibians, reptiles, birds and mammals respire through lungs; and amphibians like frogs can respire through their moist skin as well - cutaneous respiration. A frog therefore has two respiratory routes, not one.

The insect is the odd one out. Air enters through openings in the body wall called spiracles, travels along the tracheae, which branch into finer and finer tubes, and the finest branches, the tracheoles, end in direct contact with the tissues - so gas exchange happens right there, and the blood plays no part in carrying oxygen.

Now the human system, divided by function.

Part	Where it runs	What it does
Conducting part	from the external nostrils up to the terminal bronchioles	transports the atmospheric air to the alveoli, clears it of foreign particles, humidifies it, and brings the air to body temperature
Respiratory or exchange part	the alveoli and their ducts	the site of the actual diffusion of oxygen and carbon dioxide between blood and atmospheric air

The boundary is the terminal bronchiole, and no diffusion of gases takes place anywhere in the conducting part. Notice the common thread across every group in the table: gills are "special vascularised structures", lungs are "vascularised bags", and alveoli are "very thin, irregular-walled and vascularised bag-like structures". A respiratory surface is always thin and always vascularised - that is what makes it one.

Question 37

Q. Follow one molecule of oxygen from the atmosphere to a muscle cell, and one molecule of carbon dioxide from that cell back to the air, naming the process at each stage.

Answer. The chapter lists respiration as five steps, and the journey is those five steps run in order and then in reverse. The five are breathing or pulmonary ventilation; diffusion of gases across the alveolar membrane; transport of gases by the blood; diffusion of oxygen and carbon dioxide between blood and tissues; and utilisation of oxygen by the cells for catabolic reactions and the resultant release of carbon dioxide.

The oxygen molecule, outward.

It is drawn in by inspiration, because the intra-pulmonary pressure has been made less than atmospheric. It travels the conducting part - external nostrils to terminal bronchioles - and is cleaned, humidified and warmed on the way, arriving at an alveolus.
Its partial pressure has fallen from 159 mm Hg in atmospheric air to 104 mm Hg in the alveoli.
It crosses the diffusion membrane by simple diffusion down the gradient of 104 in the alveoli against 40 in deoxygenated blood, passing the squamous epithelium of the alveolus, the basement substance and the capillary endothelium - in all, much less than a millimetre.
In the alveolar conditions of high $p\mathrm{O_2}$ , low $p\mathrm{CO_2}$ , lesser hydrogen ion concentration and lower temperature, it binds haemoglobin reversibly to form oxyhaemoglobin, one of a maximum of four molecules on that haemoglobin, and rides inside an RBC as part of the 97 per cent carried that way.
At the muscle, where $p\mathrm{O_2}$ is 40 mm Hg against 95 in the arriving oxygenated blood, and where high $p\mathrm{CO_2}$ , high hydrogen ion concentration and higher temperature all favour dissociation, it leaves the oxyhaemoglobin and diffuses into the cell - part of the 5 mL delivered by every 100 mL of oxygenated blood.
Inside the cell it is used for catabolic reactions - that step alone is cellular respiration.

The carbon dioxide molecule, homeward.

It is released by catabolism in the muscle, so $p\mathrm{CO_2}$ there is 45 mm Hg against 40 mm Hg in the arriving blood, and it diffuses into the blood down that small gradient - small, but enough, because the solubility of $\mathrm{CO_2}$ is 20-25 times higher than that of $\mathrm{O_2}$ .
Most likely it is converted to bicarbonate inside an RBC by carbonic anhydrase, joining the 70 per cent that travels that way; it might instead join the 20-25 per cent bound as carbamino-haemoglobin, favoured here because at the tissues $p\mathrm{CO_2}$ is high and $p\mathrm{O_2}$ is low, or the roughly 7 per cent dissolved in plasma.
At the alveolus, where $p\mathrm{CO_2}$ is low, the carbonic anhydrase reaction runs the other way and carbon dioxide and water are re-formed, and carbamino-haemoglobin dissociates.
It diffuses out down the gradient of 45 in deoxygenated blood against 40 in the alveoli - part of the 4 mL given up by every 100 mL of deoxygenated blood.
It leaves the body in expiration, when relaxation of the diaphragm and inter-costal muscles raises the intra-pulmonary pressure slightly above atmospheric.

One sentence holds the whole journey together: oxygen always runs alveoli to blood to tissues, and carbon dioxide always runs tissues to blood to alveoli.

Question 38

Q. Distinguish between the members of each of these five pairs: TV and RV; FRC and RV; VC and TLC; oxyhaemoglobin and carbamino-haemoglobin; asthma and emphysema.

Answer. Five distinctions, each turning on one word.

Pair	The first one	The second one	The word that separates them
TV and RV	Tidal Volume - the volume of air inspired or expired during a normal respiration, approximately 500 mL	Residual Volume - the volume of air remaining in the lungs even after a forcible expiration, 1100 mL to 1200 mL	moved against never moved; RV cannot be measured by a spirometer
FRC and RV	Functional Residual Capacity - the volume of air that will remain in the lungs after a NORMAL expiration, FRC = ERV + RV	Residual Volume - what remains after a FORCIBLE expiration	normal against forcible; FRC is a capacity and is larger by exactly ERV
VC and TLC	Vital Capacity - the maximum volume a person can breathe in after a forced expiration, or breathe out after a forced inspiration, VC = ERV + TV + IRV	Total Lung Capacity - the total volume of air accommodated in the lungs at the end of a forced inspiration, TLC = VC + RV	TLC includes the residual volume and VC does not; the gap between them is air that can never be breathed out
Oxyhaemoglobin and carbamino-haemoglobin	haemoglobin carrying oxygen, formed in the alveoli where $p\mathrm{O_2}$ is high; carries about 97 per cent of the oxygen	haemoglobin carrying carbon dioxide, formed in the tissues where $p\mathrm{CO_2}$ is high and $p\mathrm{O_2}$ is low; carries nearly 20-25 per cent of the carbon dioxide	which gas is bound, and at which end of the circulation it is loaded
Asthma and emphysema	difficulty in breathing causing wheezing, due to inflammation of the bronchi and bronchioles	a chronic disorder in which the alveolar walls are damaged, so the respiratory surface is decreased; cigarette smoking is one of the major causes	airway against exchange surface; wheezing belongs to asthma only

Two of the five - FRC against RV, and VC against TLC - are separated by the residual volume or by the single adjective in front of "expiration". Those two are the most examined distinctions in the chapter.

Where Every Chapter-End Exercise Is Answered

This chapter has fourteen exercises at the end. Counted properly they come to sixteen questions, because exercise 13 has three parts, a, b and c, while every other exercise is a single part.

All sixteen of them are already answered in full inside the ten teaching sections of this chapter. This section therefore adds no exercise item of its own - there was nothing left over to add. That is a complete outcome, not an omission: it means every set question has a full worked answer somewhere in the chapter, and the table below is the map to them.

Attempt each exercise on paper first and then turn to the section named. The answer there is written out in full, with the wording a marking scheme is looking for.

Exercise	The question, in short	Section	Answered as
1	Define vital capacity and give its significance	Section 5 - Respiratory Capacities	Question 1
2	The volume of air remaining in the lungs after a normal expiration	Section 5 - Respiratory Capacities	Question 2
3	Why diffusion of gases occurs in the alveolar region only and not elsewhere	Section 6 - Exchange of Gases - Partial Pressures and the Diffusion Membrane	Question 13
4	The major transport mechanisms for carbon dioxide	Section 8 - Transport of Carbon Dioxide	Question 2
5	The multiple-choice item on $p\mathrm{O_2}$ and $p\mathrm{CO_2}$ in atmospheric air compared with alveolar air	Section 6 - Exchange of Gases - Partial Pressures and the Diffusion Membrane	Question 7
6	Explain the process of inspiration under normal conditions	Section 3 - The Mechanism of Breathing	Question 6
7	How is respiration regulated	Section 9 - Regulation of Respiration	Question 9
8	The effect of $p\mathrm{CO_2}$ on oxygen transport	Section 7 - Transport of Oxygen and the Oxygen Dissociation Curve	Question 9
9	What happens to the respiratory process in a man going up a hill	Section 10 - Disorders of the Respiratory System, and Hypoxia	Question 9
10	The site of gaseous exchange in an insect	Section 1 - Respiratory Organs Across the Animal Kingdom	Question 6
11	Define the oxygen dissociation curve, and account for its sigmoid pattern	Section 7 - Transport of Oxygen and the Oxygen Dissociation Curve	Question 6
12	Hypoxia - what it is and what causes it	Section 10 - Disorders of the Respiratory System, and Hypoxia	Question 8
13 (a)	Distinguish between Inspiratory Reserve Volume and Expiratory Reserve Volume	Section 4 - Respiratory Volumes	Question 2
13 (b)	Distinguish between Inspiratory capacity and Expiratory capacity	Section 5 - Respiratory Capacities	Question 3
13 (c)	Distinguish between Vital capacity and Total lung capacity	Section 5 - Respiratory Capacities	Question 4
14	What is Tidal Volume, and its approximate value for a healthy human in an hour	Section 4 - Respiratory Volumes	Question 3

Read the table as a revision plan, because it tells you where the marks are.

Six of the sixteen parts - exercises 1, 2, 13 (b), 13 (c), 13 (a) and 14 - are about the volumes and capacities alone. That is well over a third of the whole exercise set sitting in two short sections. Learn the four volumes with their printed values and the five capacities as formulae, and you have answered six of the sixteen before you start.

Four more - exercises 3, 4, 5 and 8 - come out of the partial pressure table and the transport of the two gases. Learn the oxygen row as 159, 104, 40, 95, 40 and the carbon dioxide row as 0.3, 40, 45, 40, 45, keep the three carbon dioxide routes and the two oxygen figures apart in your head, and those four follow.

The remaining six are one-topic accounts - the insect tracheal system, inspiration, the dissociation curve, the regulation of respiration, hypoxia, and the man going up a hill. Two of those, hypoxia and the hill, are set as activities rather than as questions, and both are answered out in full in Section 10 - Disorders of the Respiratory System, and Hypoxia, arithmetic and all, so that you never have to go looking for them.