EAPCET Chemistry · Intermediate second year

Biomolecules

Biomolecules covers carbohydrates, proteins, nucleic acids, vitamins and enzymes.

Subject
Chemistry
Taught in
Intermediate second year
Chemistry in the paper
40 of 160 questions

Key formulas and results

  • Glucose is an aldohexose; fructose is a ketohexose
  • Sucrose is non-reducing; maltose and lactose are reducing sugars
  • Amino acids join through peptide bonds
  • DNA contains A, G, C and T; RNA has U in place of T
  • Vitamins A, D, E and K are fat-soluble
  • Isoelectric point of a neutral amino acid: pI=pKa1+pKa22\text{pI} = \frac{\text{p}K_{a_1} + \text{p}K_{a_2}}{2}
  • Number of peptide bonds in a peptide made of nn amino acids =n−1= n - 1
  • Inversion of cane sugar: CX12HX22OX11+HX2O→HX+CX6HX12OX6\ce{C12H22O11 + H2O ->[H+] C6H12O6} (glucose) + CX6HX12OX6+\ \ce{C6H12O6} (fructose)
  • Glucose + HI (heat) → n-hexane, showing all six carbons lie in a straight chain
  • Glucose + Br₂ water → gluconic acid (only –CHO oxidised); glucose + conc. HNO₃ → saccharic acid (both ends oxidised to –COOH)
  • Glucose +5 HIOX4→5 HCOOH+HCHO+ \ce{5HIO4} \to \ce{5HCOOH + HCHO}
  • Glycosidic links: maltose α(1→4) of two α-D-glucose; lactose β(1→4) of β-D-galactose and β-D-glucose; sucrose C1(α)–C2(β) of glucose and fructose; cellulose β(1→4); amylopectin α(1→4) with α(1→6) branches
  • Vitamin deficiencies: A – night blindness, xerophthalmia; B₁ – beri-beri; B₂ – cheilosis; B₆ – convulsions; B₁₂ – pernicious anaemia; C – scurvy; D – rickets; K – delayed blood clotting

What this chapter covers

  • Carbohydrates
  • Proteins and amino acids
  • Structure of proteins
  • Nucleic acids
  • Vitamins
  • Enzymes

How it is usually asked

  • Reducing and non-reducing sugars
  • Vitamin deficiency diseases
  • Base pairing in DNA
  • Levels of protein structure

Understanding Biomolecules

Carbohydrates are polyhydroxy aldehydes or ketones and the compounds that give them on hydrolysis. Glucose is an aldohexose that exists mainly as a six-membered ring; the α and β forms differ only at C1 (anomers) and interconvert in solution (mutarotation). A sugar is reducing if it has a free aldehyde or ketone group or a free anomeric carbon. Sucrose links C1 of glucose with C2 of fructose, so both are locked and it is non-reducing; maltose and lactose are reducing.

Proteins are polymers of α-amino acids joined by peptide (–CO–NH–) bonds. Their primary structure is the amino-acid sequence; the secondary structure is the α-helix or β-pleated sheet held by hydrogen bonds; tertiary and quaternary structures describe overall folding and the association of subunits. Denaturation by heat or pH destroys the secondary and tertiary structure but leaves the primary sequence intact. Enzymes are proteins that act as specific biological catalysts.

Nucleic acids are chains of nucleotides, each made of a base, a pentose sugar and a phosphate. DNA has deoxyribose and the bases A, G, C and T, pairing A with T and G with C in a double helix; RNA has ribose and uracil in place of thymine. Vitamins A, D, E and K are fat-soluble; B-group vitamins and C are water-soluble. Deficiencies are frequently tested: scurvy (C), rickets (D), night blindness (A), beriberi (B₁) and pernicious anaemia (B₁₂).

Worked example

EAPCET-style question

Which of the following is a non-reducing sugar?

  1. (1)Glucose
  2. (2)Maltose
  3. (3)Lactose
  4. (4)Sucrose

Solution

  1. A sugar reduces Tollens’ or Fehling’s reagent only if it has a free aldehyde/ketone group or a free anomeric carbon.
  2. Glucose is a monosaccharide with a free anomeric carbon, so it is reducing.
  3. Maltose and lactose join C1 of one unit to C4 of the next, leaving one anomeric carbon free — both are reducing.
  4. Sucrose joins C1 of α-glucose to C2 of β-fructose, so both anomeric carbons are used in the glycosidic bond.
  5. With no free anomeric carbon, sucrose is non-reducing.

Answer: Sucrose

Shortcuts and results to remember

  • Glucose gives no 2,4-DNP or Schiff’s test, forms no bisulphite adduct, and its pentaacetate does not react with NH₂OH, which shows there is no free –CHO group and supports the cyclic structure.
  • Glucose, fructose and mannose give the same osazone with excess phenylhydrazine because they differ only at C-1 and C-2.
  • Basic amino acids are lysine, arginine and histidine; acidic ones are aspartic and glutamic acid; glycine is the only optically inactive amino acid.
  • Number of tripeptides from three different amino acids, each used once, is 3!=63! = 6.
  • At the isoelectric point the zwitterion is at its maximum, solubility is at its minimum, and the amino acid does not move in electrophoresis.

More practice questions

Practice question 1

For glycine, pKa1=2.34\text{p}K_{a_1} = 2.34 (–COOH) and pKa2=9.60\text{p}K_{a_2} = 9.60 (–NH₃⁺). Its isoelectric point is

  1. (1)3.63
  2. (2)5.97
  3. (3)7.26
  4. (4)11.94

Solution

  1. Glycine is a neutral amino acid, so pI=pKa1+pKa22\text{pI} = \frac{\text{p}K_{a_1} + \text{p}K_{a_2}}{2}
  2. pKa1+pKa2=2.34+9.60=11.94\text{p}K_{a_1} + \text{p}K_{a_2} = 2.34 + 9.60 = 11.94
  3. pI=11.942=5.97\text{pI} = \frac{11.94}{2} = 5.97

Answer: 5.97

Practice question 2

How many different tripeptides can be formed from glycine, alanine and valine if each amino acid is used only once in a tripeptide?

  1. (1)3
  2. (2)6
  3. (3)9
  4. (4)27

Solution

  1. A peptide has a direction (N-terminal to C-terminal), so Gly-Ala-Val and Val-Ala-Gly are different
  2. Each tripeptide is an ordering of three different amino acids
  3. Number of orderings =3!=3×2×1=6= 3! = 3 \times 2 \times 1 = 6

Answer: 6

Common mistakes to avoid

  • Calling sucrose reducing because it contains glucose — its glycosidic bond locks both anomeric carbons.
  • Thinking denaturation breaks peptide bonds — it destroys secondary and tertiary structure, while the primary sequence remains.
  • Pairing A with G or C with T in DNA — pairing is A with T (two hydrogen bonds) and G with C (three hydrogen bonds).
  • Listing vitamin C or B-complex as fat-soluble — only A, D, E and K are fat-soluble.

How to prepare it

Factual recall — revise from short lists and you will rarely lose marks here.

Once the formulas feel familiar, test them the way the exam does: practise Biomolecules questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Chemistry chapters

See every chapter in the EAPCET Chemistry syllabus.

Practise Biomolecules chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Biomolecules part of the EAPCET Chemistry syllabus?
Yes. Biomolecules is taught in Intermediate second year Chemistry, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Biomolecules for EAPCET?
Factual recall — revise from short lists and you will rarely lose marks here.
How many Chemistry questions are there in EAPCET?
40 of the 160 questions in the Engineering paper are Chemistry, spread across 24 chapters.
What mistakes do students make in Biomolecules?
Calling sucrose reducing because it contains glucose — its glycosidic bond locks both anomeric carbons. Thinking denaturation breaks peptide bonds — it destroys secondary and tertiary structure, while the primary sequence remains. Pairing A with G or C with T in DNA — pairing is A with T (two hydrogen bonds) and G with C (three hydrogen bonds). Listing vitamin C or B-complex as fat-soluble — only A, D, E and K are fat-soluble.