Breathing and Exchange of Gases

Given / Concept: Vital capacity is a pulmonary volume measured using a spirometer.

Definition: Vital Capacity (VC) is the maximum volume of air a person can exhale after a maximum inhalation (or vice versa). It is the sum of:
$$VC = IRV + TV + ERV$$
where IRV = Inspiratory Reserve Volume (~2500 mL), TV = Tidal Volume (~500 mL), ERV = Expiratory Reserve Volume (~1000 mL).

Thus, $VC \approx 2500 + 500 + 1000 = 4000 \text{ mL (approximately 3.5 – 4.5 L)}$.

Significance:
1. It is of great clinical significance as it indicates the functional capacity of the lungs.
2. It reflects the overall health of the respiratory system — a reduced VC indicates restrictive or obstructive lung diseases (e.g., fibrosis, emphysema).
3. Athletes and trained individuals have a higher VC, indicating better respiratory efficiency.
4. It helps physicians assess the extent of lung damage and monitor recovery.

Conclusion: Vital capacity is the maximum usable volume of air in the lungs and serves as an important diagnostic indicator of respiratory health.

Given: We need to find the volume of air remaining in the lungs after a normal expiration (i.e., after normal tidal breathing).

Concept: After a normal expiration, the volume of air that remains in the lungs is called the Functional Residual Capacity (FRC).

$$FRC = ERV + RV$$

where:
- ERV (Expiratory Reserve Volume) $\approx 1000$ mL
- RV (Residual Volume) $\approx 1100$ mL

$$FRC = 1000 + 1100 = 2100 \text{ mL (approximately)}$$

Note: The Residual Volume (RV) alone (~1100 mL) is the air that can never be expelled even after the most forceful expiration. After a normal expiration, FRC (~2100 mL) remains in the lungs.

Conclusion: Approximately 2100 mL of air remains in the lungs after normal (quiet) expiration.

Given: Gas exchange (diffusion of $O_2$ and $CO_2$) occurs only in the alveolar region.

Reason — Structural features of alveoli that facilitate diffusion:

1. Very thin walls: The alveolar walls are extremely thin (single layer of squamous epithelium, ~0.2 µm), minimising the diffusion distance.

2. Rich blood supply: Alveoli are surrounded by a dense network of pulmonary capillaries, maintaining a steep concentration gradient.

3. Large surface area: The approximately 300 million alveoli provide an enormous surface area (~70 m²) for diffusion.

4. Partial pressure gradient: In the alveoli, $pO_2$ is high (~104 mmHg) and $pCO_2$ is low (~40 mmHg) compared to deoxygenated blood ($pO_2 \approx 40$ mmHg, $pCO_2 \approx 45$ mmHg), creating a favourable gradient.

Why NOT in other parts (trachea, bronchi, bronchioles):
- These are conducting airways (dead space) with thick, cartilaginous or muscular walls — diffusion distance is too large.
- They are not surrounded by capillaries in the same way.
- No significant partial pressure gradient exists there.
- Their primary function is to conduct air, not exchange gases.

Conclusion: The unique structural and physiological features of the alveoli — thin walls, large surface area, rich vascularisation, and favourable partial pressure gradients — make them the exclusive site of gaseous exchange.

Given: Carbon dioxide produced in tissues must be transported to the lungs for elimination.

Three major mechanisms for $CO_2$ transport:

1. As Bicarbonate ions ($HCO_3^-$) — ~70%

This is the most important mechanism. $CO_2$ diffuses into RBCs and reacts with water in the presence of the enzyme carbonic anhydrase:
$$CO_2 + H_2O \xrightarrow{\text{carbonic anhydrase}} H_2CO_3 \rightarrow H^+ + HCO_3^-$$
$HCO_3^-$ ions move out of RBCs into plasma (in exchange for $Cl^-$ ions — chloride shift). In the lungs, the reverse reaction occurs and $CO_2$ is released.

2. As Carbamino-haemoglobin — ~20–25%

$CO_2$ binds directly to the amino groups of haemoglobin (and plasma proteins) to form carbamino-haemoglobin:
$$CO_2 + Hb \rightleftharpoons HbCO_2 \text{ (carbamino-haemoglobin)}$$
This binding is favoured at high $pCO_2$ (tissues) and dissociation is favoured at low $pCO_2$ (alveoli).

3. Dissolved in plasma — ~7%

A small amount of $CO_2$ (~7%) is transported in a dissolved state directly in the blood plasma.

Summary Table:
| Mechanism | Percentage |
|---|---|
| As $HCO_3^-$ (bicarbonate) | ~70% |
| As carbamino-haemoglobin | ~20–25% |
| Dissolved in plasma | ~7% |

Conclusion: $CO_2$ is transported primarily as bicarbonate ions (~70%), with significant contributions from carbamino-haemoglobin (~20–25%) and dissolved form (~7%).

Correct Answer: (ii) $pO_2$ higher, $pCO_2$ lesser

Justification:

| Gas | Atmospheric air | Alveolar air |
|---|---|---|
| $pO_2$ | ~159 mmHg | ~104 mmHg |
| $pCO_2$ | ~0.3 mmHg | ~40 mmHg |

- In atmospheric air, $pO_2$ (~159 mmHg) is higher than in alveolar air (~104 mmHg) because $O_2$ is continuously being absorbed by the blood from the alveoli.
- In atmospheric air, $pCO_2$ (~0.3 mmHg) is lesser than in alveolar air (~40 mmHg) because $CO_2$ is continuously being released into the alveoli from the blood.

Hence, compared to alveolar air, atmospheric air has higher $pO_2$ and lesser $pCO_2$.

Given: Inspiration (inhalation) is the process of taking atmospheric air into the lungs.

Concept: Inspiration is an active process that occurs when the atmospheric pressure exceeds the intrapulmonary pressure, creating a pressure gradient.

Step-by-step process of inspiration:

Step 1 — Contraction of the diaphragm:
The dome-shaped diaphragm contracts and flattens, increasing the vertical diameter of the thoracic cavity.

Step 2 — Contraction of external intercostal muscles:
The external intercostal muscles contract, pulling the ribs and sternum upward and outward. This increases the antero-posterior and lateral diameters of the thoracic cavity.

Step 3 — Increase in thoracic volume:
Due to the above muscular contractions, the volume of the thoracic cavity increases.

Step 4 — Decrease in intrapulmonary pressure:
As thoracic volume increases, the lungs expand (due to the negative intrapleural pressure). This causes the intrapulmonary (intra-alveolar) pressure to fall below atmospheric pressure:
P_{\text{intrapulmonary}} < P_{\text{atmospheric}}
(Intrapulmonary pressure drops by ~1–3 mmHg below atmospheric pressure)

Step 5 — Air flows in:
Air moves from the region of higher pressure (atmosphere) to lower pressure (alveoli) — i.e., air rushes into the lungs until pressures equalise.

\text{Atmospheric pressure} > \text{Intrapulmonary pressure} \Rightarrow \text{Air flows IN}

Conclusion: Inspiration is an active, muscle-driven process. Contraction of the diaphragm and external intercostal muscles increases thoracic volume, decreases intrapulmonary pressure, and causes air to flow into the lungs.

Given: Respiration must be precisely regulated to meet the body's metabolic demands.

Regulation of Respiration:

1. Neural Regulation — Respiratory Rhythm Centre (Medulla oblongata):
- A specialised centre in the medulla oblongata (part of the brain stem) called the respiratory rhythm centre is primarily responsible for regulating the rhythm of respiration.
- It generates rhythmic nerve impulses that control the rate and depth of breathing.

2. Pneumotaxic Centre (Pons):
- Located in the pons region of the brain.
- It can moderate the functions of the respiratory rhythm centre.
- It signals to reduce the duration of inspiration, thereby altering the respiratory rate.

3. Chemosensitive Area (Medulla):
- A region adjacent to the rhythm centre is highly sensitive to $CO_2$ and $H^+$ ions.
- An increase in $CO_2$ or $H^+$ concentration in blood activates this area, which in turn signals the rhythm centre to increase the rate and depth of breathing to eliminate excess $CO_2$.
- $O_2$ levels do not directly stimulate this centre significantly under normal conditions.

4. Receptors in Aortic Arch and Carotid Body:
- Peripheral chemoreceptors in the aortic arch and carotid body can also recognise changes in $CO_2$, $H^+$, and $O_2$ concentrations.
- They send signals to the rhythm centre for appropriate adjustments.

5. Stretch Receptors in Lungs (Hering-Breuer Reflex):
- Receptors in the walls of the bronchi and bronchioles are activated when the lungs are over-inflated.
- They send signals to stop further inspiration, preventing over-distension.

Conclusion: Respiration is regulated by the respiratory rhythm centre in the medulla, modulated by the pneumotaxic centre in the pons, and fine-tuned by chemosensitive areas and peripheral receptors that respond to changes in $CO_2$, $H^+$, and $O_2$ levels in the blood.

Given: We need to understand how $pCO_2$ affects the transport of $O_2$ by haemoglobin.

Concept — Bohr Effect:
The effect of $pCO_2$ (and $H^+$) on oxygen transport is explained by the Bohr Effect.

Effect of increased $pCO_2$ (at tissues):
- At the tissues, cellular respiration produces large amounts of $CO_2$, so $pCO_2$ is high.
- High $pCO_2$ leads to increased $H^+$ concentration (acidic pH) as:
$$CO_2 + H_2O \rightarrow H_2CO_3 \rightarrow H^+ + HCO_3^-$$
- High $pCO_2$ and high $H^+$ decrease the affinity of haemoglobin for $O_2$.
- This causes dissociation of oxyhaemoglobin and release of $O_2$ to the tissues:
$$HbO_2 \xrightarrow{\text{high } pCO_2, \text{ low pH}} Hb + O_2$$

Effect of decreased $pCO_2$ (at alveoli):
- In the alveoli, $CO_2$ is expelled, so $pCO_2$ is low and $pO_2$ is high.
- Low $pCO_2$ and low $H^+$ increase the affinity of haemoglobin for $O_2$.
- This promotes loading of $O_2$ onto haemoglobin to form oxyhaemoglobin:
$$Hb + O_2 \xrightarrow{\text{low } pCO_2, \text{ high } pO_2} HbO_2$$

Conclusion: High $pCO_2$ (and low pH) decreases haemoglobin's affinity for $O_2$, promoting $O_2$ release at tissues. Low $pCO_2$ increases haemoglobin's affinity for $O_2$, promoting $O_2$ loading at the alveoli. This is the Bohr Effect and it ensures efficient $O_2$ delivery to metabolically active tissues.

Given: A person is ascending a hill (high altitude).

Changes in respiratory process at high altitude:

1. Decrease in atmospheric pressure and $pO_2$:
As altitude increases, atmospheric pressure decreases. Consequently, the partial pressure of $O_2$ ($pO_2$) in the atmosphere decreases.

2. Reduced $O_2$ availability:
Lower $pO_2$ means less $O_2$ diffuses into the alveoli and subsequently into the blood. The blood carries less $O_2$ to the tissues — a condition called hypoxia.

3. Increased rate and depth of breathing (Hyperventilation):
The chemosensitive area in the medulla detects low $O_2$ and rising $CO_2$ / $H^+$ levels. It signals the respiratory rhythm centre to increase the rate and depth of breathing (hyperventilation) to compensate.

4. Increased heart rate:
The heart rate also increases to pump more blood and deliver adequate $O_2$ to tissues.

5. Long-term acclimatisation:
Over time (days to weeks), the body acclimatises:
- Increased production of RBCs (erythropoiesis stimulated by erythropoietin).
- Increased haemoglobin concentration in blood.
- Increased capillary density in tissues.
- Breathing rate gradually normalises.

Conclusion: Going up a hill causes a decrease in $pO_2$, leading to hypoxia. The body immediately compensates by increasing the rate and depth of breathing (hyperventilation). With prolonged stay, physiological acclimatisation occurs through increased RBC production and haemoglobin levels.

Given: We need to identify the site of gas exchange in insects.

Answer:

In insects, the site of gaseous exchange is the tracheoles (the finest terminal branches of the tracheal system).

Explanation:
- Insects have a well-developed tracheal system — a network of air tubes (tracheae) that open to the outside through small pores called spiracles on the body surface.
- The tracheae branch repeatedly into finer tubes called tracheoles.
- The tracheoles are in direct contact with individual cells and tissues.
- Gaseous exchange ($O_2$ in, $CO_2$ out) occurs directly between the tracheoles and the body cells by diffusion — without the involvement of blood.
- This is a very efficient system as it delivers $O_2$ directly to the respiring cells.

Conclusion: The tracheoles are the actual sites of gaseous exchange in insects. The tracheal system bypasses the circulatory system and delivers $O_2$ directly to the tissues.

Definition:
The oxygen dissociation curve (or oxygen-haemoglobin dissociation curve) is a graph that shows the relationship between the partial pressure of oxygen ($pO_2$) (on the x-axis) and the percentage saturation of haemoglobin with $O_2$ (on the y-axis). It represents the affinity of haemoglobin for oxygen at different $pO_2$ values.

Sigmoidal (S-shaped) Pattern — Reason:

The curve is sigmoidal due to the cooperative binding of $O_2$ to haemoglobin, which is a tetramer with four haem groups (four subunits: $2\alpha$ and $2\beta$).

Explanation of the shape:

1. Lower portion (low $pO_2$, 0–40 mmHg): At low $pO_2$, haemoglobin has a low affinity for $O_2$. The first $O_2$ molecule binds with difficulty (slow initial binding). The curve rises slowly.

2. Middle steep portion (40–70 mmHg): Once the first $O_2$ molecule binds to one haem group, it causes a conformational change in the haemoglobin molecule (T-state → R-state), which increases the affinity of the remaining haem groups for $O_2$. This cooperative binding causes rapid loading of $O_2$ — the curve rises steeply.

3. Upper plateau (above ~70 mmHg): At high $pO_2$ (as in alveoli, ~104 mmHg), haemoglobin becomes nearly fully saturated (~97–98%). The curve flattens as all binding sites are occupied.

Physiological significance of the sigmoidal shape:
- The steep middle portion ensures efficient $O_2$ loading in the lungs.
- The lower plateau ensures efficient $O_2$ unloading at tissues (where $pO_2$ is ~40 mmHg).

Conclusion: The sigmoidal shape of the oxygen dissociation curve is due to the cooperative binding of $O_2$ to the four subunits of haemoglobin — binding of each $O_2$ molecule increases the affinity of the remaining subunits for $O_2$.

Information about Hypoxia:

Definition:
Hypoxia is a condition in which the body or a region of the body is deprived of adequate oxygen supply at the tissue level, despite potentially normal blood flow.

Types of Hypoxia:
1. Hypoxic hypoxia: Due to low $pO_2$ in inspired air (e.g., at high altitudes). Also caused by respiratory diseases.
2. Anaemic hypoxia: Due to reduced haemoglobin content or defective haemoglobin (e.g., anaemia, CO poisoning — CO binds Hb with 200× more affinity than $O_2$).
3. Stagnant (ischaemic) hypoxia: Due to poor blood circulation (e.g., heart failure, shock).
4. Histotoxic hypoxia: Tissues are unable to use $O_2$ even when supplied (e.g., cyanide poisoning blocks cytochrome oxidase).

Causes:
- High altitude (low atmospheric $pO_2$)
- Respiratory diseases (pneumonia, asthma, COPD)
- Anaemia
- Carbon monoxide poisoning
- Heart failure

Symptoms:
- Shortness of breath (dyspnoea)
- Rapid breathing (hyperventilation)
- Increased heart rate
- Headache, dizziness, confusion
- Bluish discolouration of skin/lips (cyanosis) in severe cases
- Loss of consciousness in extreme cases

Body's Response to Hypoxia:
- Increased breathing rate
- Increased heart rate
- Increased production of RBCs (erythropoiesis via erythropoietin)
- Increased 2,3-BPG in RBCs (reduces Hb affinity for $O_2$, promoting $O_2$ release to tissues)

Treatment:
- Supplemental oxygen therapy
- Treatment of underlying cause
- Acclimatisation at high altitudes

Conclusion: Hypoxia is a serious condition of inadequate oxygen supply to tissues. It can be caused by various factors and the body has several compensatory mechanisms to deal with it. Awareness of its symptoms is important for timely medical intervention.

(a) IRV (Inspiratory Reserve Volume) vs ERV (Expiratory Reserve Volume):

| Feature | IRV | ERV |
|---|---|---|
| Definition | Extra volume of air that can be inspired forcefully after a normal inspiration | Extra volume of air that can be expired forcefully after a normal expiration |
| Normal value | ~2500 mL (2.5 L) | ~1000 mL (1 L) |
| Associated with | Inspiration (inhalation) | Expiration (exhalation) |
| Significance | Represents the reserve for deeper inhalation | Represents the reserve for deeper exhalation |

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(b) Inspiratory Capacity (IC) vs Expiratory Capacity (EC):

| Feature | Inspiratory Capacity (IC) | Expiratory Capacity (EC) |
|---|---|---|
| Definition | Total volume of air a person can inspire after a normal expiration | Total volume of air a person can expire after a normal inspiration |
| Formula | $IC = TV + IRV$ | $EC = TV + ERV$ |
| Normal value | $\approx 500 + 2500 = 3000$ mL | $\approx 500 + 1000 = 1500$ mL |
| Starting point | After normal expiration | After normal inspiration |

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(c) Vital Capacity (VC) vs Total Lung Capacity (TLC):

| Feature | Vital Capacity (VC) | Total Lung Capacity (TLC) |
|---|---|---|
| Definition | Maximum volume of air that can be exhaled after a maximum inhalation | Total volume of air in the lungs after a maximum inhalation |
| Formula | $VC = IRV + TV + ERV$ | $TLC = VC + RV$ or $IRV + TV + ERV + RV$ |
| Normal value | $\approx 4000$ mL (3.5–4.5 L) | $\approx 5100$ mL (4.5–6 L) |
| Includes RV? | No | Yes |
| Can be measured by spirometer? | Yes (directly) | Requires special techniques for RV |

Key difference: TLC includes the Residual Volume (RV) (~1100 mL) which cannot be expelled, while VC does not include RV.

Definition of Tidal Volume (TV):
Tidal Volume is the volume of air inspired or expired during a single normal (quiet) breath — i.e., without any extra effort.

$$TV \approx 500 \text{ mL per breath (approximately 6000–8000 mL/min)}$$

Calculation of Tidal Volume in one hour:

Given:
- Tidal Volume per breath $= 500$ mL $= 0.5$ L
- Normal breathing rate $= 12–16$ breaths per minute
- Let us take breathing rate $= 15$ breaths per minute (standard value)

Step 1: Calculate breaths per hour:
$$\text{Breaths per hour} = 15 \times 60 = 900 \text{ breaths/hour}$$

Step 2: Calculate total tidal volume per hour:
$$\text{Total TV per hour} = 500 \text{ mL} \times 900 = 4,50,000 \text{ mL}$$

$$= 450 \text{ L per hour}$$

Conclusion:
- Tidal Volume per breath $\approx$ 500 mL
- Total Tidal Volume in one hour $\approx$ 450 litres (for a healthy adult breathing at ~15 breaths/minute)

This value can vary slightly depending on the individual's breathing rate (12–16 breaths/min), giving a range of approximately 360 L to 480 L per hour.