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Cell: Structure and Function - Biomolecules (Proteins, Carbohydrates, Lipids, Nucleic Acids)

Grade 11ICSEBiology

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

🔑Concepts

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Biomolecules are organic compounds like carbohydrates, proteins, lipids, and nucleic acids that form the building blocks of life and perform various biological functions.

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Carbohydrates: Primarily composed of carbon, hydrogen, and oxygen in the ratio 1:2:11:2:1. They are classified as monosaccharides (e.g., C6H12O6C_6H_{12}O_6), disaccharides (e.g., C12H22O11C_{12}H_{22}O_{11}), and polysaccharides (e.g., cellulose, starch).

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Proteins: Polymers of α\alpha-amino acids linked by peptide bonds. Each amino acid has an amino group (−NH2-NH_2), a carboxyl group (−COOH-COOH), and a variable RR group.

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Lipids: Hydrophobic molecules including fats, oils, and waxes. Triglycerides are formed by the esterification of one glycerol molecule (C3H8O3C_3H_8O_3) with three fatty acids.

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Nucleic Acids: Polymers of nucleotides (DNA and RNA). Each nucleotide consists of a pentose sugar (Ribose C5H10O5C_5H_{10}O_5 or Deoxyribose C5H10O4C_5H_{10}O_4), a nitrogenous base, and a phosphate group (PO43−PO_4^{3-}).

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Bonding: Glycosidic bonds link sugars, peptide bonds (−CO−NH−-CO-NH-) link amino acids, ester bonds link glycerol to fatty acids, and phosphodiester bonds link nucleotides.

📐Formulae

Cn(H2O)n (General formula for monosaccharides)C_n(H_2O)_n \text{ (General formula for monosaccharides)}

R−CH(NH2)−COOH (General structure of an amino acid)R-CH(NH_2)-COOH \text{ (General structure of an amino acid)}

C12H22O11 (General formula for disaccharides like Sucrose)C_{12}H_{22}O_{11} \text{ (General formula for disaccharides like Sucrose)}

Total Amino Acids=n  ⟹  Number of Peptide Bonds=n−1\text{Total Amino Acids} = n \implies \text{Number of Peptide Bonds} = n - 1

A+G=T+C (Chargaff’s Rule for DNA)A + G = T + C \text{ (Chargaff's Rule for DNA)}

💡Examples

Problem 1:

A double-stranded DNA molecule is found to contain 20%20\% Adenine (AA). Calculate the percentage of Cytosine (CC) present in this DNA sample.

Solution:

30%30\%

Explanation:

According to Chargaff's rule, in a double-stranded DNA, A=TA = T and G=CG = C. If A=20%A = 20\%, then T=20%T = 20\%. The sum of A+T=40%A + T = 40\%. The remaining 60%60\% must be G+CG + C. Since G=CG = C, the percentage of Cytosine is 60%2=30%\frac{60\%}{2} = 30\%.

Problem 2:

Calculate the number of water molecules (H2OH_2O) released during the formation of a polypeptide chain consisting of 5050 amino acids.

Solution:

4949 molecules of H2OH_2O

Explanation:

A peptide bond is formed through a dehydration synthesis reaction where one molecule of H2OH_2O is released for every bond formed. For a linear chain of nn monomers, the number of bonds is n−1n-1. Therefore, 50−1=4950 - 1 = 49 peptide bonds are formed, releasing 4949 water molecules.

Problem 3:

Identify the type of bond formed when Glucose (C6H12O6C_6H_{12}O_6) and Fructose (C6H12O6C_6H_{12}O_6) combine to form Sucrose (C12H22O11C_{12}H_{22}O_{11}).

Solution:

1,21,2-glycosidic bond

Explanation:

When two monosaccharides join, a hydroxyl group (−OH-OH) from one sugar reacts with a hydroxyl group from another, releasing H2OH_2O and forming a glycosidic linkage. In sucrose, this occurs between the C1C_1 of α\alpha-glucose and C2C_2 of β\beta-fructose.

Biomolecules (Proteins, Carbohydrates, Lipids, Nucleic Acids) Revision - Class 11 Biology ICSE