krit.club logo

Human Physiology - Gas exchange

Grade 12IBBiology

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Difference between ventilation (movement of air), gas exchange (diffusion at the O2O_2 and CO2CO_2 level), and cell respiration (ATP production).

•

Alveoli adaptations for exchange: single layer of thin cells (Type I pneumocytes), dense capillary network, large surface area, and moist lining.

•

Type II pneumocytes: secrete a surfactant that reduces surface tension, preventing the collapse of the alveoli during expiration.

•

Mechanism of ventilation: requires antagonistic muscle pairs. During inspiration, external intercostal muscles and diaphragm contract. During expiration, internal intercostal muscles and abdominal muscles contract.

•

Transport of CO2CO_2: Carried in the blood in three ways: dissolved in plasma, bound to hemoglobin (carbaminohemoglobin), and primarily as bicarbonate ions (HCO3−HCO_3^-).

•

The Bohr Shift: Increased CO2CO_2 concentration leads to a decrease in blood pHpH, which reduces hemoglobin's affinity for O2O_2, shifting the dissociation curve to the right.

•

Oxygen Dissociation Curves: SS-shaped curves showing the relationship between partial pressure of oxygen (PO2P_{O_2}) and percentage saturation of hemoglobin.

•

Health impacts: Emphysema (destruction of alveolar walls reducing surface area for exchange) and lung cancer (caused by carcinogens like tobacco smoke).

📐Formulae

Ventilation Rate=Tidal Volume×Frequency of BreathsVentilation\ Rate = Tidal\ Volume \times Frequency\ of\ Breaths

CO2+H2O⇌H2CO3⇌H++HCO3−CO_2 + H_2O \rightleftharpoons H_2CO_3 \rightleftharpoons H^+ + HCO_3^-

pH=−log⁡10[H+]pH = -\log_{10}[H^+]

💡Examples

Problem 1:

A patient has a tidal volume of 0.5 L0.5\ L and a ventilation rate of 6.0 L⋅min−16.0\ L \cdot min^{-1}. Calculate their breathing frequency per minute.

Solution:

Frequency=Ventilation RateTidal Volume=6.00.5=12 breaths⋅min−1Frequency = \frac{Ventilation\ Rate}{Tidal\ Volume} = \frac{6.0}{0.5} = 12\ breaths\cdot min^{-1}

Explanation:

Using the ventilation rate formula, we rearrange to solve for frequency (breaths per minute) by dividing total volume per minute by the volume per breath.

Problem 2:

Explain the effect of increased metabolic activity in tissues on the oxygen dissociation curve of hemoglobin.

Solution:

Increased metabolism increases PCO2P_{CO_2} and lowers pHpH. This causes a 'right-shift' in the curve.

Explanation:

As tissues respire, they release CO2CO_2, which reacts with water to form H+H^+ ions via the carbonic anhydrase reaction: CO2+H2O→H++HCO3−CO_2 + H_2O \rightarrow H^+ + HCO_3^-. The resulting acidity (H+H^+) binds to hemoglobin, causing a conformational change that decreases its affinity for O2O_2, allowing more O2O_2 to be released to the demanding tissues (The Bohr Effect).

Gas exchange - Revision Notes & Key Diagrams | IB Grade 12 Biology