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Human Physiology - Gas Exchange

Grade 11IBBiology

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

🔑Concepts

Ventilation is the process of bringing fresh air into the lungs and removing stale air to maintain a concentration gradient of O2O_2 and CO2CO_2.

Gas exchange is the passive process of O2O_2 diffusing from the alveoli into the blood and CO2CO_2 diffusing from the blood into the alveoli across the respiratory membrane.

Cell respiration is the metabolic process within the mitochondria where O2O_2 is consumed to produce ATPATP, with CO2CO_2 as a byproduct.

The alveoli are adapted for gas exchange: they have a large surface area, a thin epithelial layer (one cell thick), and are surrounded by a dense network of capillaries.

Type I pneumocytes are extremely thin alveolar cells adapted for gas exchange, while Type II pneumocytes secrete a surfactant (a phospholipid) that reduces surface tension and prevents the alveoli from collapsing.

Inspiration (inhalation) occurs when the external intercostal muscles and diaphragm contract, increasing thoracic volume and decreasing pressure (Pinternal<PatmosphericP_{internal} < P_{atmospheric}).

Expiration (exhalation) occurs when the internal intercostal muscles and abdominal muscles contract (during forced breathing), or simply when the diaphragm relaxes, decreasing volume and increasing pressure (Pinternal>PatmosphericP_{internal} > P_{atmospheric}).

Emphysema is a chronic lung disease where the alveolar walls are destroyed, reducing the surface area for gas exchange and decreasing the elasticity of the lungs.

📐Formulae

Ventilation Rate=Tidal Volume×Respiratory Rate\text{Ventilation Rate} = \text{Tidal Volume} \times \text{Respiratory Rate}

P1VP \propto \frac{1}{V}

CO2+H2OH2CO3H++HCO3CO_2 + H_2O \rightleftharpoons H_2CO_3 \rightleftharpoons H^+ + HCO_3^-

%Change=FinalInitialInitial×100\% \text{Change} = \frac{\text{Final} - \text{Initial}}{\text{Initial}} \times 100

💡Examples

Problem 1:

A student at rest has a tidal volume of 0.5 L0.5\text{ L} and breathes 1212 times per minute. After exercise, the tidal volume increases to 2.5 L2.5\text{ L} and the respiratory rate increases to 3030 breaths per minute. Calculate the ventilation rate for both states.

Solution:

At rest: 0.5 L×12 min1=6.0 L min10.5\text{ L} \times 12\text{ min}^{-1} = 6.0\text{ L min}^{-1}. After exercise: 2.5 L×30 min1=75.0 L min12.5\text{ L} \times 30\text{ min}^{-1} = 75.0\text{ L min}^{-1}.

Explanation:

The ventilation rate is the total volume of air inhaled or exhaled per minute. Exercise increases both the depth of breathing (tidal volume) and the frequency to meet the increased demand for O2O_2 and the removal of CO2CO_2.

Problem 2:

Explain the role of CO2CO_2 in blood pHpH and how the body responds to high levels of CO2CO_2 during exercise.

Solution:

High CO2CO_2 levels lead to the reaction CO2+H2OH2CO3H++HCO3CO_2 + H_2O \rightarrow H_2CO_3 \rightarrow H^+ + HCO_3^-, which increases the concentration of H+H^+ ions and lowers the pHpH.

Explanation:

Chemoreceptors in the medulla oblongata and the carotid/aortic bodies detect the drop in pHpH. This triggers the respiratory center to increase the ventilation rate, thereby expelling more CO2CO_2 and restoring the blood pHpH to its homeostatic range of approximately 7.47.4.

Gas Exchange - Revision Notes & Key Diagrams | IB Grade 11 Biology