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Human Nutrition - Absorption and assimilation

Grade 12IGCSEBiology

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

🔑Concepts

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Absorption is the movement of small food molecules and ions through the wall of the intestine into the blood.

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Assimilation is the movement of digested food molecules into the cells of the body where they are used, becoming part of the cells (e.g., building proteins).

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The small intestine (specifically the ileum) is the primary site of absorption, featuring a large surface area facilitated by villi and microvilli.

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Villi adaptations include: a one-cell thick epithelium to minimize diffusion distance, a rich network of blood capillaries to transport glucose and amino acids, and a lacteal for the absorption of fatty acids and glycerol.

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Glucose is transported to the liver via the Hepatic Portal Vein and converted to glycogen or used in aerobic respiration: C6H12O6+6O2→6CO2+6H2OC_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O.

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Amino acids are used by cells to synthesize proteins; excess amino acids undergo deamination in the liver, where the nitrogen-containing part is removed to form urea: (NH2)2CO(NH_2)_2CO.

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Lipids (fats) are reformed into triglycerides after passing into the lacteals and eventually enter the lymphatic system.

📐Formulae

C6H12O6+6O2→6CO2+6H2O+Energy (ATP)C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O + \text{Energy (ATP)}

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

2NH3+CO2→(NH2)2CO+H2O2NH_3 + CO_2 \rightarrow (NH_2)_2CO + H_2O

Surface Area of a sphere=4πr2\text{Surface Area of a sphere} = 4\pi r^2

💡Examples

Problem 1:

Explain how the structure of the villus facilitates the absorption of glucose against a concentration gradient.

Solution:

Glucose is absorbed through the epithelial cells via active transport.

Explanation:

To move C6H12O6C_6H_{12}O_6 against a concentration gradient, the epithelial cells contain many mitochondria to provide ATPATP. The large surface area provided by microvilli increases the number of carrier proteins available for transport.

Problem 2:

What happens to excess glucose in the liver, and what is the chemical notation of the storage molecule?

Solution:

Excess glucose is converted into glycogen.

Explanation:

Under the influence of the hormone insulin, glucose molecules undergo condensation reactions to form a branched polysaccharide called glycogen, which is represented by the general formula (C6H10O5)n(C_6H_{10}O_5)_n.

Problem 3:

Calculate the theoretical increase in surface area if a flat membrane of 1 cm21 \text{ cm}^2 is replaced by a membrane with 100100 villi, where each villus is a cylinder with radius r=0.01 cmr = 0.01 \text{ cm} and height h=0.1 cmh = 0.1 \text{ cm}.

Solution:

Total Area ≈7.28 cm2\approx 7.28 \text{ cm}^2.

Explanation:

The surface area of one villus cylinder (excluding base) is 2πrh=2×π×0.01×0.1≈0.00628 cm22\pi rh = 2 \times \pi \times 0.01 \times 0.1 \approx 0.00628 \text{ cm}^2. For 100100 villi, the area added is 0.628 cm20.628 \text{ cm}^2 on top of the original base area, significantly increasing the SASA available for absorption.

Absorption and assimilation - Revision Notes & Key Diagrams | IGCSE Grade 12 Biology