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Molecular Biology - Carbohydrates and lipids

Grade 12IBBiology

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

🔑Concepts

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Carbon atoms can form four covalent bonds, allowing for the synthesis of a large variety of stable organic compounds, including C6H12O6C_6H_{12}O_6 (Glucose) and C5H10O5C_5H_{10}O_5 (Ribose).

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Monosaccharides are the basic monomers of carbohydrates. Common examples include glucose, fructose, and galactose. They are linked together by condensation reactions to form disaccharides and polysaccharides.

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Condensation reactions involve the loss of a water molecule (H2OH_2O) to form a covalent bond (glycosidic bond in carbohydrates, ester bond in lipids). The reverse process is hydrolysis.

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Polysaccharides: Cellulose (straight chain of β\beta-glucose, structural in cell walls), Starch (amylose and amylopectin, α\alpha-glucose, energy storage in plants), and Glycogen (highly branched α\alpha-glucose, energy storage in animals).

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Lipids include triglycerides, phospholipids, and steroids. Triglycerides are formed by the condensation of one glycerol molecule and three fatty acids.

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Fatty acids can be saturated (no double bonds), monounsaturated (one double bond), or polyunsaturated (multiple double bonds).

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Unsaturated fatty acids can be ciscis isomers (hydrogen atoms on the same side of the double bond, causing a bend) or transtrans isomers (hydrogens on opposite sides, straight chain).

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Lipids are more suitable for long-term energy storage in humans than carbohydrates because they release approximately 37 kJ/g37\,kJ/g compared to 17 kJ/g17\,kJ/g for carbohydrates, and they are insoluble in water, avoiding osmotic issues.

📐Formulae

BMI=mass in kilograms(height in metres)2BMI = \frac{\text{mass in kilograms}}{(\text{height in metres})^2}

Cn(H2O)nC_n(H_2O)_n

C6H12O6+C6H12O6→C12H22O11+H2OC_6H_{12}O_6 + C_6H_{12}O_6 \rightarrow C_{12}H_{22}O_{11} + H_2O

Glycerol+3×Fatty Acids→Triglyceride+3H2O\text{Glycerol} + 3 \times \text{Fatty Acids} \rightarrow \text{Triglyceride} + 3H_2O

💡Examples

Problem 1:

Calculate the Body Mass Index (BMI) for an adult who weighs 75 kg75\,kg and has a height of 1.8 m1.8\,m. Determine if they fall within the 'Normal' range (18.518.5 to 24.924.9).

Solution:

BMI=751.82=753.24≈23.15BMI = \frac{75}{1.8^2} = \frac{75}{3.24} \approx 23.15

Explanation:

Using the BMI formula, we divide the mass by the square of the height. A result of 23.1523.15 falls within the 18.5−24.918.5 - 24.9 range, categorizing the individual as having a normal weight.

Problem 2:

A molecule of Maltose (C12H22O11C_{12}H_{22}O_{11}) is hydrolyzed into two Glucose molecules. Write the chemical equation and explain the role of H2OH_2O.

Solution:

C12H22O11+H2O→C6H12O6+C6H12O6C_{12}H_{22}O_{11} + H_2O \rightarrow C_6H_{12}O_6 + C_6H_{12}O_6

Explanation:

This is a hydrolysis reaction. A water molecule (H2OH_2O) is consumed to break the glycosidic bond between the two glucose subunits, adding an −OH-OH group to one and a −H-H to the other.

Problem 3:

Identify the type of fatty acid represented by the formula CH3(CH2)nCOOHCH_3(CH_2)_nCOOH where no carbon-carbon double bonds are present.

Solution:

Saturated Fatty Acid

Explanation:

Because there are no C=CC=C double bonds in the hydrocarbon chain, every carbon atom is 'saturated' with the maximum possible number of hydrogen atoms (HH).

Carbohydrates and lipids - Revision Notes & Key Diagrams | IB Grade 12 Biology