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Biology: Human Physiology, Health, and Response - Digestion, Absorption, and Exchange Surfaces

Grade 8IB

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

🔑Concepts

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The human digestive system is a series of organs responsible for the breakdown of food into smaller, soluble molecules through mechanical and chemical digestion.

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Mechanical digestion involves physical breakdown (e.g., teeth grinding, stomach churning), while chemical digestion uses enzymes such as amylase, protease, and lipase.

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Exchange surfaces, such as the Villi in the small intestine, are adapted for maximum efficiency. They have a large surface area-to-volume ratio (SA:VSA:V), are one-cell thick to minimize diffusion distance, and have a rich blood supply (capillary network).

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Absorption is the process where digested food molecules move from the lumen of the gut into the blood or lymph. This occurs via simple diffusion, facilitated diffusion, or active transport.

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Enzymes act as biological catalysts. Their activity is heavily influenced by factors such as temperature and pHpH. Each enzyme has an 'optimal' range; beyond this, the enzyme may denature, changing the shape of its active site.

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The SA:VSA:V ratio is critical: as a cell or structure increases in size, its volume (VV) grows faster than its surface area (SASA), making exchange less efficient unless specialized structures like microvilli are present.

📐Formulae

SA:V=Total Surface AreaTotal VolumeSA:V = \frac{Total\ Surface\ Area}{Total\ Volume}

Surface Area of a Cube=6×s2Surface\ Area\ of\ a\ Cube = 6 \times s^2

Volume of a Cube=s3Volume\ of\ a\ Cube = s^3

Starch+H2O→AmylaseMaltoseStarch + H_{2}O \xrightarrow{Amylase} Maltose

Lipids+H2O→LipaseGlycerol+Fatty AcidsLipids + H_{2}O \xrightarrow{Lipase} Glycerol + Fatty\ Acids

Proteins+H2O→ProteaseAmino AcidsProteins + H_{2}O \xrightarrow{Protease} Amino\ Acids

💡Examples

Problem 1:

A model of an intestinal cell is represented by a cube with a side length s=2 cms = 2\text{ cm}. Calculate the Surface Area to Volume ratio (SA:VSA:V).

Solution:

First, calculate the Surface Area: SA=6×(2 cm)2=6×4=24 cm2SA = 6 \times (2\text{ cm})^2 = 6 \times 4 = 24\text{ cm}^2. Next, calculate the Volume: V=(2 cm)3=8 cm3V = (2\text{ cm})^3 = 8\text{ cm}^3. The ratio is 24:824 : 8, which simplifies to 3:13 : 1.

Explanation:

The SA:VSA:V ratio is 33. This value indicates how much surface area is available for every unit of volume. In biology, higher ratios (often achieved by folding membranes into villi) allow for faster absorption.

Problem 2:

In a lab experiment, a student measures the mass of a food sample before and after it is dried to calculate water content. Initial mass was 15.50 g15.50\text{ g} and final mass was 12.25 g12.25\text{ g}. Calculate the mass of water lost using vertical subtraction.

Solution:

Mass of water lost: 15.50−12.253.25\begin{array}{r} 15.50 \\ - 12.25 \\ \hline 3.25 \end{array} The mass lost is 3.25 g3.25\text{ g}.

Explanation:

Subtracting the final dry mass from the initial mass gives the mass of the water that was evaporated from the sample.

Problem 3:

An enzyme-controlled reaction produced 15 mg15\text{ mg} of product in 5 minutes5\text{ minutes}. Calculate the rate of reaction in mg/min\text{mg/min}.

Solution:

Rate=Amount of productTime=15 mg5 min=3 mg/minRate = \frac{\text{Amount of product}}{\text{Time}} = \frac{15\text{ mg}}{5\text{ min}} = 3\text{ mg/min}

Explanation:

The rate of reaction tells us how quickly the digestive enzymes are breaking down substrates into products.