Breathing and Exchange of Gases — NCERT Solutions

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):

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

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

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.