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Biological Molecules - Carbohydrates, lipids and proteins

Grade 12A LevelBiology

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

🔑Concepts

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Carbohydrates are organic compounds consisting of Carbon (CC), Hydrogen (HH), and Oxygen (OO), usually in a 2:12:1 ratio of Hydrogen to Oxygen.

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Monosaccharides like glucose (C6H12O6C_6H_{12}O_6) are the monomers of carbohydrates; they can exist as α\alpha-glucose or β\beta-glucose isomers.

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Disaccharides are formed when two monosaccharides join via a condensation reaction, creating a glycosidic bond and releasing a molecule of water (H2OH_2O).

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Polysaccharides such as starch and glycogen serve as energy storage, while cellulose (made of β\beta-glucose) provides structural support in plant cell walls.

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Lipids (triglycerides) are formed from one glycerol molecule and three fatty acids joined by ester bonds through the process of esterification.

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Fatty acids can be saturated (no double bonds between carbons) or unsaturated (containing one or more C=CC=C double bonds).

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Proteins are polymers of amino acids; each amino acid contains an amino group (−NH2-NH_2), a carboxyl group (−COOH-COOH), and a variable RR group.

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Amino acids are linked by peptide bonds formed between the −NH2-NH_2 of one and the −COOH-COOH of another.

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The 3D shape of a protein is determined by its primary sequence and maintained by hydrogen bonds, ionic bonds, disulfide bridges (−S−S−-S-S-), and hydrophobic interactions.

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The Benedict's test is used for reducing sugars (color change from blue to brick-red), while the Biuret test detects peptide bonds in proteins (blue to purple).

📐Formulae

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

R−CH(NH2)−COOHR-CH(NH_2)-COOH

💡Examples

Problem 1:

Determine the molecular formula of a polysaccharide consisting of 10 glucose (C6H12O6C_6H_{12}O_6) units joined in a linear chain.

Solution:

C60H102O51C_{60}H_{102}O_{51}

Explanation:

To join 10 glucose molecules, 9 condensation reactions must occur. Each condensation reaction removes one water molecule (H2OH_2O). Calculation: (10×C6H12O6)−(9×H2O)=C60H120O60−H18O9=C60H102O51(10 \times C_6H_{12}O_6) - (9 \times H_2O) = C_{60}H_{120}O_{60} - H_{18}O_9 = C_{60}H_{102}O_{51}.

Problem 2:

Explain why unsaturated fats are typically liquid at room temperature compared to saturated fats using molecular structure.

Solution:

Unsaturated fats contain C=CC=C double bonds which create 'kinks' in the fatty acid chains.

Explanation:

The kinks caused by the double bonds prevent the molecules from packing closely together. This reduces the intermolecular forces (Van der Waals forces), resulting in a lower melting point compared to saturated fats, which have straight chains and can pack tightly.

Problem 3:

Predict the effect of extreme pHpH on the tertiary structure of an enzyme.

Solution:

The enzyme will denature as ionic and hydrogen bonds are disrupted.

Explanation:

Tertiary structure is maintained by interactions between RR groups. Changes in H+H^+ concentration interfere with ionic bonds and hydrogen bonds. This causes the protein to unfold, changing the shape of the active site so it is no longer complementary to the substrate.