Biomolecules Class 11 NCERT Biology Study Guide

Chemical Analysis and the Two Pools of Biomolecules

To understand the chemical composition of living tissues, biologists perform a chemical analysis. A living tissue (such as a piece of vegetable or liver) is ground in trichloroacetic acid ($Cl_3CCOOH$) using a mortar and pestle. This creates a thick slurry. When we strain this slurry through a cheesecloth or cotton, we obtain two distinct fractions:

1. The Acid-Soluble Pool (Filtrate): This contains thousands of organic compounds with low molecular weights ranging from 18 to around 800 Daltons. These are referred to as biomicromolecules (e.g., monosaccharides, amino acids, and nucleotides).
2. The Acid-Insoluble Pool (Retentate): This contains organic compounds with molecular weights exceeding 10,000 Daltons. These are biomacromolecules (e.g., proteins, polysaccharides, and nucleic acids).

Essential Terminology

Biomolecules

All the carbon-containing compounds that are found in living tissues.

Zwitterion

A neutral molecule with both positive and negative electrical charges, a state commonly assumed by amino acids at specific physiological pH values.

Glycosidic Bond

The covalent bond formed between two monosaccharide units through a dehydration reaction (loss of a water molecule).

Peptide Bond

The covalent amide link formed between the carboxyl group of one amino acid and the amino group of another.

Steps to Perform Chemical Analysis of Living Tissue

1. Homogenization — Take a living tissue sample (e.g., liver tissue or spinach leaf) and grind it thoroughly in Trichloroacetic acid ($Cl_3CCOOH$) using a clean mortar and pestle to break open cell membranes.
2. Filtration — Strain the resulting thick slurry through a cheesecloth or fine cotton to separate the soluble liquid from the insoluble cellular debris.
3. Fraction Separation — Collect the filtrate (acid-soluble pool containing monomers) and the retentate (acid-insoluble pool containing polymers).
4. Analytical Testing — Subject the fractions to various biochemical assays and chromatography to isolate and identify individual biomolecules.

The Lipid Exception: A Classic Exam Trap

In your CBSE Class 11 biology biomolecules exam, a highly popular question focuses on lipids.

The Trap: Lipids are low-molecular-weight compounds (<800 Daltons) and should structurally belong to the acid-soluble biomicromolecule pool. However, they are always recovered in the acid-insoluble retentate fraction!

The Reason: Lipids are hydrophobic and major components of cell membranes. During homogenization (grinding), cell membranes are broken into tiny fragments that self-assemble into water-insoluble vesicles. These macromolecular vesicles cannot pass through the cheesecloth, meaning they remain in the retentate despite their small individual molecular size.

Practice Questions with Solutions

• Q: What is a peptide bond and how is it formed? Illustrate the reaction steps. A: Step 1: Identify the reacting units. A peptide bond forms between two amino acids. Each amino acid has an amino group ($-NH_2$) at one end and a carboxyl group ($-COOH$) at the other. Step 2: Describe the reaction mechanism. The carboxyl group of the first amino acid reacts with the amino group of the second amino acid. Step 3: State the byproduct. This condensation reaction involves the elimination of a water molecule ($H_2O$), forming a covalent $-CO-NH-$ link. Final answer: A peptide bond is a covalent amide linkage formed via dehydration synthesis between the carboxyl group of one amino acid and the amino group of another.
• Q: Why are nucleic acids classified as macromolecules? Describe their monomeric units. A: Step 1: Define macromolecular criteria. Macromolecules are polymers with a molecular weight exceeding 1,000 Daltons located in the acid-insoluble fraction. Step 2: Identify the polymer. Nucleic acids (DNA and RNA) have molecular weights in the millions of Daltons and consist of long chains. Step 3: Describe the monomers. The repeating monomeric subunits are nucleotides. Each nucleotide consists of three parts: a nitrogenous base, a pentose sugar, and a phosphate group. Final answer: Nucleic acids are biomacromolecules because they are high molecular weight polymers made of repeating nucleotide monomers linked by phosphodiester bonds.
• Q: Explain the concept of a Zwitterion structure of an amino acid. A: Step 1: Understand the functional groups of amino acids. Amino acids contain both an acidic carboxyl group ($-COOH$) and a basic amino group ($-NH_2$) attached to the same alpha-carbon. Step 2: Explain the ionization process. In an aqueous solution of neutral pH, the carboxyl group can lose a proton ($H^+$) to become $-COO^-$, while the amino group can accept a proton to become $-NH_3^+$. Step 3: Describe the state of the molecule. The resulting molecule contains both positive and negative charges, making its net electrical charge zero. Final answer: A zwitterion is the dipolar ionic state of an amino acid where it possesses equal positive and negative charges, occurring at a specific pH called the isoelectric point.
• Q: What is the main difference between primary and secondary metabolites? Give examples of each. A: Step 1: Define primary metabolites. These are compounds directly involved in normal growth, development, and physiological reproduction of the organism. Examples include amino acids, sugars, and lipids. Step 2: Define secondary metabolites. These are compounds produced by plants, fungi, and microbes that are not directly involved in primary survival but have ecological and economic roles. Examples include alkaloids (morphine), rubber, essential oils, and pigments (carotenoids). Final answer: Primary metabolites are essential for basic physiological survival (e.g., glucose, alanine), whereas secondary metabolites are accessory compounds with ecological or defense roles (e.g., rubber, antibiotics, alkaloids).

Frequently Asked Questions