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Biomolecules - Chemical constituents of living cells

Grade 11CBSEBiology

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

🔑Concepts

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Chemical Analysis: Living tissues are analyzed by grinding in trichloroacetic acid (Cl3CCOOHCl_3CCOOH). This separates them into an acid-soluble pool (biomicromolecules with molecular weights 1818 to 800800 Daltons) and an acid-insoluble fraction (biomacromolecules with weights >10,000> 10,000 Daltons).

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Amino Acids: These are substituted methanes containing an amino group (−NH2-NH_2), a carboxyl group (−COOH-COOH), and a variable side chain (RR). At a specific pHpH, they exist as Zwitterions: H3N+−CHR−COO−H_3N^+-CHR-COO^-.

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Proteins: Heteropolymers of amino acids linked by peptide bonds (−CONH−-CONH-). They exhibit four levels of structure: Primary (sequence), Secondary (α\alpha-helix, β\beta-sheets), Tertiary (3D folding/functionality), and Quaternary (assembly of subunits).

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Lipids: Water-insoluble molecules, not strictly polymers. They include fatty acids (R−COOHR-COOH) and glycerol (CH2OH−CHOH−CH2OHCH_2OH-CHOH-CH_2OH). Phospholipids like Lecithin are vital components of the cell membrane.

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Polysaccharides: Long chains of monosaccharides linked by glycosidic bonds. Examples include Cellulose (linear glucose polymer), Starch (helical energy store in plants), and Glycogen (branched energy store in animals).

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Nucleic Acids: Polymers of nucleotides. Each nucleotide consists of a nitrogenous base (Adenine, Guanine, Cytosine, Thymine/Uracil), a pentose sugar (C5H10O5C_5H_{10}O_5), and a phosphate group (−PO43−-PO_4^{3-}).

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Enzymes: Biological catalysts that lower the activation energy (ΔG‡\Delta G^{\ddagger}) of a reaction. They show specificity and are sensitive to temperature, pHpH, and substrate concentration [S][S].

📐Formulae

R−CH(NH2)−COOH (General Amino Acid structure)R-CH(NH_2)-COOH \text{ (General Amino Acid structure)}

Peptide Bond Formation: R1−NH2+HOOC−R2→R1−NH−CO−R2+H2O\text{Peptide Bond Formation: } R_1-NH_2 + HOOC-R_2 \rightarrow R_1-NH-CO-R_2 + H_2O

Michaelis-Menten Equation: V=Vmax[S]Km+[S]\text{Michaelis-Menten Equation: } V = \frac{V_{max}[S]}{K_m + [S]}

Glycosidic Bond: C1−O−C4 (in linear polysaccharides)\text{Glycosidic Bond: } C_1-O-C_4 \text{ (in linear polysaccharides)}

💡Examples

Problem 1:

A polypeptide chain consists of 150150 amino acids. Calculate the number of peptide bonds present and the number of water molecules (H2OH_2O) released during its synthesis.

Solution:

Number of peptide bonds = 149149; Number of H2OH_2O molecules released = 149149.

Explanation:

In a linear polymer, the number of bonds between nn monomers is always n−1n-1. Since one molecule of H2OH_2O is eliminated for every peptide bond formed via dehydration synthesis, 150−1=149150 - 1 = 149 water molecules are released.

Problem 2:

In an enzymatic reaction, what happens to the Michaelis constant (KmK_m) in the presence of a competitive inhibitor?

Solution:

The KmK_m increases, while VmaxV_{max} remains unchanged.

Explanation:

A competitive inhibitor mimics the substrate and binds to the active site. This decreases the affinity of the enzyme for the substrate, thus increasing the KmK_m (the concentration of substrate required to reach 12Vmax\frac{1}{2}V_{max}). VmaxV_{max} remains constant because high [S][S] can eventually displace the inhibitor.

Problem 3:

Identify the primary difference between a Nucleoside and a Nucleotide.

Solution:

Nucleotide = Nucleoside + Phosphate group.

Explanation:

A nucleoside consists of a nitrogenous base and a sugar (Base+SugarBase + Sugar). When a phosphate group is attached to the 5′−OH5'-OH of the sugar via a phosphoester bond, it becomes a Nucleotide (Base+Sugar+PhosphateBase + Sugar + Phosphate).