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Human Nutrition - Balanced diet

Grade 12A LevelBiology

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

🔑Concepts

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

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

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

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

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

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

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

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

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

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

C6H12O6+6O2→6CO2+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=853.0625≈27.75 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 20∘C20^{\circ}C to 35∘C35^{\circ}C. Calculate the energy released per gram of food.

Solution:

Energy=20×4.2×(35−20)=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).