Class 12 Chemistry Ch 10 Biomolecules Formula Sheet PDF Download

Ans. You can download the Biomolecules Class 12 Chemistry Formula Sheet PDF directly from this Collegedunia page. Both the Normal and HD versions are available and free of cost.

Ans. Yes. This page reflects the current 2026-27 syllabus for Class 12 Chemistry. Biomolecules is fully retained in the new edition with no structural cuts; every concept in Sections 10.1 to 10.6 of the NCERT remains examinable.

Ans. The Formula Sheet PDF runs approximately 18 pages and covers the carbohydrate master table, amino-acid and protein structural rules, enzyme activation-energy data, the vitamin-deficiency reference, the DNA-RNA comparison, and the hormone classification.

Ans. Both have the same molecular formula C₆H₁₂O₆ , but they differ in three structural respects. Glucose is an aldohexose: it carries a −CHO group at C1 and four chiral centres (C2, C3, C4, C5). Fructose is a ketohexose: it carries a >C=O at C2, a primary −OH at C1, and three chiral centres. On cyclisation, the −OH at C5 of glucose attacks the C1 aldehyde to give a six-membered pyranose ring (one O + five C); in fructose the −OH at C5 attacks the C2 keto to give a five-membered furanose ring (one O + four C). Glucose is dextrorotatory ([α]_D = +52.5^∘) ; fructose is laevorotatory ([α]_D = -92.4^∘) . The 1 : 1 mixture obtained when sucrose hydrolyses is called invert sugar because the net rotation flips from + to −.

Ans. A reducing sugar must have at least one free hemiacetal (a C bearing both an −OH and an −OR via the ring oxygen) which can open back to a free −CHO and reduce Fehling or Tollens reagent. Sucrose (α-D-glucose C1 – β-D-fructose C2) uses both anomeric −OH groups (C1 of glucose and C2 of fructose) inside the glycosidic linkage, so neither sugar can open up - sucrose is non-reducing. Maltose (two α-D-glucose units linked C1–C4) and lactose (β-D-galactose C1 – β-D-glucose C4) each leave one anomeric C free, so they retain a hemiacetal and reduce Fehling and Tollens. The same rule explains why all monosaccharides reduce both reagents.

Ans. Denaturation is the loss of the secondary and tertiary three-dimensional structure of a protein, triggered by heat, pH change, urea, organic solvents, or heavy-metal salts. The hydrogen bonds that hold the α-helix and the β-pleated sheet, plus the disulphide −S−S− bridges that lock the 3-D fold, are cleaved. The primary structure (the peptide-bond sequence) is preserved because the −CO−NH− bond itself is not broken. The protein loses biological activity (a denatured enzyme cannot catalyse; a denatured hormone cannot signal). Common examples include the coagulation of egg-white on boiling and the curdling of milk by lactic acid from bacteria. In most cases denaturation is irreversible in a laboratory setting; refolding the chain back into its native conformation requires either chaperone proteins or controlled cellular conditions and is rarely complete.

Ans. DNA and RNA differ at three structural levels. Sugar: DNA contains β-D-2-deoxyribose (no −OH at C2′); RNA contains β-D-ribose (−OH at C2′ retained). Bases: DNA uses adenine, guanine, cytosine, and thymine; RNA uses adenine, guanine, cytosine, and uracil in place of thymine. Strand architecture: DNA is a right-handed double helix with two antiparallel strands held by the Watson-Crick base pairs A···T (two H-bonds) and G···C (three H-bonds), giving Chargaff's 1:1 ratio. RNA is single-stranded and exists in three functional forms (mRNA, rRNA, tRNA) that work together during protein synthesis. DNA stores hereditary information and self-replicates; RNA executes the genetic message, and it does not self-replicate inside cells.

Ans. The five disease-vitamin pairs that NCERT Table 10.3 fixes are: Scurvy → deficiency of Vitamin C (ascorbic acid; sourced from citrus and amla); Beri-beri → deficiency of Vitamin B₁ (thiamine); Night blindness / xerophthalmia → deficiency of Vitamin A; Rickets / osteomalacia → deficiency of Vitamin D; Increased blood-clotting time → deficiency of Vitamin K. NCERT exercise 10.19 asks "name the vitamin responsible for the coagulation of blood" and the marking scheme expects only "Vitamin K".

Ans. The boundary between polypeptide and protein is not a sharp count. NCERT records that compounds with more than 10 amino acid residues are usually called polypeptides, and those with more than 100 residues or a molecular mass above 10,000 u are called proteins. Insulin sits below both of those numerical thresholds (51 residues, M about 5,800 u), but it has a clearly defined three-dimensional fold built from two polypeptide chains held together by disulphide −S−S− bridges, and it performs a well-defined biological function (lowering blood glucose by promoting cellular uptake and storage as glycogen). Because of that well-defined 3-D structure and dedicated function, NCERT and IUPAC convention both classify insulin as a protein hormone, even though by residue count alone it would technically be a polypeptide.

Ans. Anomers are stereoisomers of sugars that differ only at the anomeric carbon - C1 in aldoses or C2 in ketoses - which is the new chiral centre created when the open chain cyclises. Alpha- and beta-D-glucopyranose are anomers. Epimers are stereoisomers that differ at one non-anomeric chiral carbon: glucose and galactose are C4 epimers; glucose and mannose are C2 epimers. Mutarotation is the gradual change in specific rotation when a pure anomer of a reducing sugar dissolves in water and equilibrates with the other anomer through the open-chain form. For D-glucose, pure alpha has [α] = +112°, pure beta has [α] = +19°, and the equilibrium mixture has [α] = +52.5°.

Ans. Haemoglobin is built from four polypeptide subunits - two α chains (141 residues each) and two β chains (146 residues each) - each cradling a heme group with a Fe(II) centre. The four subunits associate through non-covalent interactions and salt bridges; total molecular mass is about 64,500 u. Cooperative O₂ binding (positive cooperativity) gives haemoglobin its sigmoidal binding curve, distinguishing it from monomeric myoglobin (a tertiary-only protein). The association of two or more polypeptide chains defines quaternary structure.

Ans. Purines are double-ring nitrogen bases - adenine (A) and guanine (G); mnemonic "PURe As Gold". Pyrimidines are single-ring bases - cytosine (C), thymine (T) and uracil (U). DNA contains A, G, C, T; RNA replaces thymine with uracil and so contains A, G, C, U. The Watson-Crick base pairs in DNA are A ··· T (2 H-bonds) and G ··· C (3 H-bonds), each pairing one purine with one pyrimidine.