krit.club logo

Biological Molecules - Carbohydrates, lipids and proteins

Grade 12IGCSEBiology

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

πŸ”‘Concepts

β€’

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

β€’

Monosaccharides like glucose (C6H12O6C_6H_{12}O_6) are the monomers of carbohydrates; they can exist as Ξ±\alpha-glucose or Ξ²\beta-glucose isomers.

β€’

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

β€’

Polysaccharides such as starch and glycogen serve as energy storage, while cellulose (made of Ξ²\beta-glucose) provides structural support in plant cell walls.

β€’

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

β€’

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

β€’

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

β€’

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

β€’

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

β€’

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).

πŸ“Formulae

Cn(H2O)nC_n(H_2O)_n

C6H12O6+C6H12O6β†’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

Rβˆ’CH(NH2)βˆ’COOHR-CH(NH_2)-COOH

πŸ’‘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: (10Γ—C6H12O6)βˆ’(9Γ—H2O)=C60H120O60βˆ’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.

Carbohydrates, lipids and proteins Revision - Grade 12 Biology IGCSE