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Gas Exchange in Humans - Breathing and lung function

Grade 11IGCSEBiology

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

🔑Concepts

The pathway of air involves the nasal cavity, larynx, trachea, bronchi, bronchioles, and finally the alveoli where gas exchange occurs.

The trachea and bronchi are lined with ciliated cells and goblet cells. Goblet cells secrete mucus to trap pathogens, while cilia sweep the mucus towards the throat.

Inhalation (Inspiration) occurs when the external intercostal muscles contract and the diaphragm contracts (moves down/flattens). This increases thorax volume and decreases pressure, drawing air in.

Exhalation (Expiration) occurs when the internal intercostal muscles contract (during forced breathing) and the diaphragm relaxes (moves up/dome-shaped). This decreases thorax volume and increases pressure, pushing air out.

Alveoli are adapted for gas exchange by having a very large surface area, a wall only one cell thick (short diffusion distance), a moist lining to dissolve gases, and a dense network of capillaries to maintain a steep concentration gradient for O2O_2 and CO2CO_2.

Gas exchange occurs via diffusion: O2O_2 moves from the alveoli into the blood, while CO2CO_2 moves from the blood into the alveoli.

Inspired air contains approximately 21%21\% O2O_2 and 0.04%0.04\% CO2CO_2, whereas expired air contains approximately 16%16\% O2O_2 and 4%4\% CO2CO_2 due to aerobic respiration.

The limewater test is used to identify CO2CO_2; it turns from clear to cloudy/milky when CO2CO_2 is bubbled through it.

Physical activity increases the rate and depth of breathing because the body requires more energy, leading to increased CO2CO_2 production from aerobic respiration (C6H12O6+6O26CO2+6H2OC_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O), which lowers blood pHpH and triggers the brain to increase ventilation.

📐Formulae

Minute Ventilation=Tidal Volume×Breathing Rate\text{Minute Ventilation} = \text{Tidal Volume} \times \text{Breathing Rate}

C6H12O6+6O26CO2+6H2O+ATPC_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O + \text{ATP}

Rate of DiffusionSurface Area×Concentration GradientDiffusion Distance\text{Rate of Diffusion} \propto \frac{\text{Surface Area} \times \text{Concentration Gradient}}{\text{Diffusion Distance}}

💡Examples

Problem 1:

A student at rest has a tidal volume of 0.5 dm30.5\text{ dm}^3 and a breathing rate of 1212 breaths per minute. During exercise, the tidal volume increases to 2.5 dm32.5\text{ dm}^3 and the breathing rate increases to 2020 breaths per minute. Calculate the increase in minute ventilation.

Solution:

Resting Minute Ventilation=0.5×12=6.0 dm3/min\text{Resting Minute Ventilation} = 0.5 \times 12 = 6.0\text{ dm}^3\text{/min} Exercise Minute Ventilation=2.5×20=50.0 dm3/min\text{Exercise Minute Ventilation} = 2.5 \times 20 = 50.0\text{ dm}^3\text{/min} Increase=50.06.0=44.0 dm3/min\text{Increase} = 50.0 - 6.0 = 44.0\text{ dm}^3\text{/min}

Explanation:

Minute ventilation is the total volume of air inhaled or exhaled per minute. Exercise requires more oxygen for respiration, so both the depth (tidal volume) and frequency (breathing rate) increase.

Problem 2:

Explain why the concentration of CO2CO_2 is higher in expired air compared to inspired air.

Solution:

Inspired air has a CO2CO_2 concentration of approximately 0.04%0.04\%. Inside the body cells, aerobic respiration occurs: C6H12O6+6O26CO2+6H2OC_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O. The CO2CO_2 produced diffuses into the blood and is transported to the lungs, where it diffuses into the alveoli. Consequently, expired air contains about 4%4\% CO2CO_2.

Explanation:

The increase in CO2CO_2 is a direct result of metabolic activity in the mitochondria of cells where glucose is oxidized to release energy.

Breathing and lung function - Revision Notes & Key Diagrams | IGCSE Grade 11 Biology