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Human Nutrition - Balanced diet

Grade 12IGCSEBiology

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

🔑Concepts

A balanced diet provides all essential nutrients in the correct proportions: carbohydrates, fats, proteins, vitamins, minerals, water, and fiber.

Carbohydrates (e.g., glucose C6H12O6C_6H_{12}O_6) provide the primary energy source for cellular respiration.

Proteins are essential for growth and tissue repair; they are polymers of amino acids containing nitrogen, carbon, hydrogen, and oxygen.

Lipids (fats and oils) serve as long-term energy stores, provide thermal insulation, and are components of cell membranes.

Vitamin C (C6H8O6C_6H_8O_6) is necessary for maintaining healthy skin and gums; deficiency leads to Scurvy.

Vitamin D and Calcium (Ca2+Ca^{2+}) are required for strong bones and teeth; deficiency can cause Rickets in children.

Iron (Fe2+Fe^{2+}) is a vital component of hemoglobin in red blood cells; deficiency leads to Anemia, reducing oxygen transport.

Dietary fiber (cellulose) adds bulk to food, providing the muscles of the digestive system something to push against, thereby preventing constipation via peristalsis.

Energy requirements vary based on age, sex, activity levels, and biological states such as pregnancy or lactation.

Malnutrition includes both undernutrition (e.g., Marasmus and Kwashiorkor) and overnutrition (e.g., Obesity, leading to Type 2 diabetes and coronary heart disease).

📐Formulae

BMI=mass (kg)height2 (m2)BMI = \frac{mass\ (kg)}{height^2\ (m^2)}

Energy transferred (J)=mass of water (g)×4.2 (J/gC)×ΔT (C)Energy\ transferred\ (J) = mass\ of\ water\ (g) \times 4.2\ (J/g^{\circ}C) \times \Delta T\ (^{\circ}C)

C6H12O6+6O26CO2+6H2O+EnergyC_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O + Energy

💡Examples

Problem 1:

An individual has a mass of 85 kg85\ kg and a height of 1.75 m1.75\ m. Calculate their Body Mass Index (BMI) and determine their weight category.

Solution:

BMI=851.752=853.062527.75 kg/m2BMI = \frac{85}{1.75^2} = \frac{85}{3.0625} \approx 27.75\ kg/m^2

Explanation:

Using the formula BMI=massheight2BMI = \frac{mass}{height^2}, the result is 27.75\approx 27.75. According to standard health classifications, a BMI between 25.025.0 and 29.929.9 is categorized as 'Overweight'.

Problem 2:

In a calorimetry experiment, 0.5 g0.5\ g of a dried food sample is burned to heat 20 g20\ g of water. The temperature of the water rises from 20C20^{\circ}C to 35C35^{\circ}C. Calculate the energy released per gram of food.

Solution:

Energy=20×4.2×(3520)=20×4.2×15=1260 JEnergy = 20 \times 4.2 \times (35 - 20) = 20 \times 4.2 \times 15 = 1260\ J Energy per gram: 1260 J0.5 g=2520 J/g\frac{1260\ J}{0.5\ g} = 2520\ J/g

Explanation:

First, calculate the total energy absorbed by the water using Q=mcΔTQ = mc\Delta T. Then, divide the total energy by the mass of the food sample to find the energy density (J/gJ/g).

Balanced diet - Revision Notes & Key Diagrams | IGCSE Grade 12 Biology