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Biology: Life and Systems - Human Body Systems (Digestive, Respiratory, Circulatory)

Grade 8IB

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

🔑Concepts

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The Digestive System performs mechanical and chemical digestion. Chemical digestion involves enzymes that break down polymers into monomers, such as starch into glucose (C6H12O6C_6H_{12}O_6).

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Absorption occurs primarily in the small intestine, where villi and microvilli increase the surface area to volume ratio (SA:VSA:V) for efficient nutrient uptake.

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The Respiratory System facilitates gas exchange. Oxygen (O2O_2) diffuses from the alveoli into the blood, while Carbon Dioxide (CO2CO_2) diffuses from the blood into the alveoli to be exhaled.

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Aerobic Respiration is the chemical process in cells that releases energy: C6H12O6+6O2→6CO2+6H2O+ATP (Energy)C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O + \text{ATP (Energy)}.

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The Circulatory System consists of the heart, blood vessels, and blood. Humans have a double circulatory system: pulmonary (to lungs) and systemic (to body).

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Arteries carry blood away from the heart at high pressure; Veins carry blood towards the heart and contain valves to prevent backflow; Capillaries are one-cell thick to allow for diffusion.

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Red blood cells contain Hemoglobin, a protein that binds to O2O_2 to form oxyhemoglobin for transport.

📐Formulae

C6H12O6+6O2→6CO2+6H2O+EnergyC_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O + \text{Energy}

Cardiac Output (CO)=Heart Rate (HR)×Stroke Volume (SV)\text{Cardiac Output (CO)} = \text{Heart Rate (HR)} \times \text{Stroke Volume (SV)}

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

💡Examples

Problem 1:

During exercise, a student's heart rate increases to 120120 beats per minute (bpm). If their heart pumps 8080 mL of blood per beat (stroke volume), calculate the total Cardiac Output in Liters per minute (L/minL/min).

Solution:

CO=120 bpm×80 mL/beat=9600 mL/minCO = 120 \text{ bpm} \times 80 \text{ mL/beat} = 9600 \text{ mL/min}
To convert to Liters:
96001000=9.6 L/min\frac{9600}{1000} = 9.6 \text{ L/min}

Explanation:

Cardiac Output is the volume of blood pumped by the heart per minute. It is calculated by multiplying the heart rate (frequency) by the stroke volume (volume per pump).

Problem 2:

Identify the reactants and products in the process of aerobic respiration that occurs in the mitochondria.

Solution:

Reactants: Glucose (C6H12O6C_6H_{12}O_6) and Oxygen (O2O_2).
Products: Carbon Dioxide (CO2CO_2), Water (H2OH_2O), and Energy in the form of ATPATP.

Explanation:

Cells require oxygen to break down glucose molecules to release energy. The waste products CO2CO_2 and H2OH_2O are removed by the respiratory and circulatory systems.

Problem 3:

Explain how the structure of the alveoli is adapted for gas exchange using the concept of surface area.

Solution:

The lungs contain millions of alveoli, creating a very large surface area (SASA). Since the rate of diffusion is directly proportional to SASA, this maximizes the amount of O2O_2 entering the blood and CO2CO_2 leaving it.

Explanation:

Efficient gas exchange requires a large area, thin membranes (short diffusion path), and a steep concentration gradient maintained by blood flow.

Human Body Systems (Digestive, Respiratory, Circulatory) Revision - Grade 8 Science IB