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GATE Life Sciences Chemistry of Biomolecules Syllabus

Every chapter and topic of Chemistry of Biomolecules examined in GATE Life Sciences — 6 chapters, 7 topics and 4 sub-topics, plus 59 flashcards written against it.

6Chapters
7Topics
4Sub-topics
~6hEst. first pass
11%Of GATE Life Sciences
59Flashcards

Chemistry of Biomolecules syllabus — full chapter and topic list

Expand any chapter to see its topics and sub-topics. This is the whole examinable outline for Chemistry of Biomolecules in GATE Life Sciences, not a summary of it.

  1. Amino acids

    1 topic
    • Proteins
      • Peptide sequencing by chemical methods
      • Peptide sequencing by enzymatic methods
  2. Nucleic acids and nucleotides

    1 topic
    • DNA sequencing
      • DNA sequencing by chemical methods
      • DNA sequencing by enzymatic methods
  3. Carbohydrates

    1 topic
    • Hexoses
  4. Lipids

    1 topic
    • Triglycerides
  5. Principles of biomolecule purification

    2 topics
    • Ion exchange chromatography
    • Gel filtration chromatography
  6. Identification of biomolecules

    1 topic
    • Beer-Lambert’s law

Chemistry of Biomolecules flashcards for GATE Life Sciences

25 of 59 cards from the Chemistry of Biomolecules deck — real questions with worked answers.

  1. What is the basic structural unit of a protein, and what bond links them together?

    The basic unit is the amino acid; adjacent amino acids are joined by a peptide (amide) bond formed between the $\alpha$-carboxyl group of one residue and the $\alpha$-amino group of the next, with loss of water.

  2. Name the four levels of protein structure.

    Primary (amino acid sequence), secondary (local folding: $\alpha$-helix, $\beta$-sheet), tertiary (overall 3D fold of a single chain), and quaternary (assembly of multiple subunits).

  3. Which bonds/interactions stabilize the tertiary structure of proteins?

    Hydrophobic interactions, hydrogen bonds, ionic (salt) bridges, van der Waals forces, and covalent disulfide ($\ce{-S-S-}$) bonds between cysteine residues.

  4. What is the general structure (zwitterion form) of an $\alpha$-amino acid at physiological pH?

    $\ce{H3N+-CHR-COO-}$ — a zwitterion with a protonated amino group, a deprotonated carboxyl group, and a variable side chain R attached to the $\alpha$-carbon.

  5. Define the isoelectric point (pI) of an amino acid or protein.

    The pH at which the molecule carries no net electrical charge (net charge $= 0$); for an amino acid with two ionizable groups, $pI = \frac{pK_{1} + pK_{2}}{2}$.

  6. What are the dihedral angles $\phi$ (phi) and $\psi$ (psi), and what plot displays their allowed values?

    $\phi$ is the rotation about the $\ce{N-C_\alpha}$ bond and $\psi$ about the $\ce{C_\alpha-C}$ bond; their sterically allowed combinations are displayed on a Ramachandran plot.

  7. Compare the $\alpha$-helix and $\beta$-sheet in terms of hydrogen bonding.

    In the $\alpha$-helix, H-bonds form within one chain between residue $i$ and residue $i+4$ (intrachain, parallel to the helix axis). In the $\beta$-sheet, H-bonds form between adjacent strands (interstrand), which may be parallel or antiparallel.

  8. How many residues per turn and what is the rise per residue in a right-handed $\alpha$-helix?

    $3.6$ residues per turn with a rise (translation) of $1.5\ \text{\AA}$ per residue, giving a pitch of $5.4\ \text{\AA}$ per turn.

  9. In Edman degradation, which reagent is used and which terminus of the peptide is sequenced?

    Phenyl isothiocyanate (PITC, Edman's reagent) reacts with the free N-terminal residue, cleaving it as a PTH (phenylthiohydantoin)-amino acid — sequencing proceeds from the N-terminus.

  10. Why is Edman degradation advantageous over older N-terminal methods like Sanger's reagent?

    Edman degradation removes and identifies only one residue at a time without destroying the rest of the peptide, so the shortened peptide can be subjected to repeated cycles to read the sequence residue by residue.

  11. What reagent did Sanger originally use to label the N-terminal residue, and what is its limitation?

    1-Fluoro-2,4-dinitrobenzene (FDNB / Sanger's reagent) forms a DNP-amino acid. Its limitation is that the labeling step requires acid hydrolysis of the whole peptide, so only the N-terminal residue is identified per analysis (the peptide is destroyed).

  12. Which chemical reagent is commonly used to identify the C-terminal residue of a peptide?

    Hydrazine (hydrazinolysis): all residues except the C-terminal one are converted to amino acid hydrazides, leaving the C-terminal residue as a free amino acid that can be identified.

  13. Which reagent cleaves peptide bonds specifically on the C-terminal side of methionine residues?

    Cyanogen bromide ($\ce{CNBr}$) cleaves at the C-terminal side of methionine (Met) residues, converting Met to a C-terminal homoserine lactone.

  14. Why must disulfide bonds be reduced and blocked before sequencing a protein?

    Disulfide bonds cross-link cysteines (within or between chains), so they must be cleaved — e.g. reduced with $\beta$-mercaptoethanol/DTT and alkylated with iodoacetate — so that polypeptide chains can be separated and read individually.

  15. In enzymatic peptide sequencing, where does trypsin cleave?

    Trypsin cleaves peptide bonds on the C-terminal side of the basic residues lysine (Lys) and arginine (Arg).

  16. Where does chymotrypsin cleave peptide bonds?

    Chymotrypsin cleaves on the C-terminal side of aromatic/bulky hydrophobic residues — phenylalanine (Phe), tyrosine (Tyr), and tryptophan (Trp).

  17. What is the function of carboxypeptidase in peptide sequencing, and name the two main types.

    Carboxypeptidases are exopeptidases that sequentially remove residues from the C-terminus. Carboxypeptidase A removes most C-terminal residues except Arg/Lys/Pro; carboxypeptidase B removes only C-terminal basic residues (Arg, Lys).

  18. What is the role of aminopeptidases (e.g. leucine aminopeptidase) in sequencing?

    Aminopeptidases are exopeptidases that sequentially cleave residues from the N-terminus, releasing free amino acids that can be identified in order of release.

  19. Why are overlapping peptides generated (using two or more different proteases) needed to determine a protein's sequence?

    A single protease yields a set of fragments whose order is unknown. Cleaving the same protein with a second enzyme of different specificity produces overlapping fragments, and the overlaps allow the fragments to be ordered into the complete sequence.

  20. State the two complementary base-pairing rules in double-stranded DNA and the number of H-bonds in each pair.

    Adenine pairs with thymine via 2 hydrogen bonds ($\ce{A=T}$); guanine pairs with cytosine via 3 hydrogen bonds ($\ce{G\equiv C}$).

  21. What is Chargaff's rule for DNA base composition?

    In double-stranded DNA, $[\text{A}] = [\text{T}]$ and $[\text{G}] = [\text{C}]$, therefore purines = pyrimidines: $[\text{A}] + [\text{G}] = [\text{C}] + [\text{T}]$.

  22. In which direction is a new DNA strand synthesized, and what is read as the template?

    DNA polymerase synthesizes the new strand in the $5' \to 3'$ direction, reading the template strand in the $3' \to 5'$ direction.

  23. What type of bond joins nucleotides along a DNA strand?

    A 3',5'-phosphodiester bond linking the 3'-OH of one sugar to the 5'-phosphate of the next nucleotide.

  24. Name the two classic DNA sequencing methods and their inventors.

    The Maxam–Gilbert method (chemical degradation) and the Sanger method (chain-termination / dideoxy, enzymatic).

  25. On what principle does the Maxam–Gilbert (chemical) DNA sequencing method rely?

    It relies on base-specific chemical modification followed by cleavage of the phosphodiester backbone at the modified bases, generating labeled fragments of different lengths that are separated by gel electrophoresis.

See more Chemistry of Biomolecules flashcards →

Planning Chemistry of Biomolecules for GATE Life Sciences

Chemistry of Biomolecules is about 11% of the GATE Life Sciences syllabus by topic count — 7 of 64 topics, spread over 6 chapters. At roughly 45 minutes per topic plus 12 minutes per sub-topic, a first pass runs to about 6 hours.

The heaviest chapters are Principles of biomolecule purification (2 topics), Amino acids (1 topics), Nucleic acids and nucleotides (1 topics) . Front-load those while your energy is high; the short chapters are better revision filler later.

Work top-down: read the chapter, then tick topics off individually rather than marking the whole chapter done. Sub-topics are where silent gaps hide.

Chemistry of Biomolecules (GATE Life Sciences) FAQ

What is in the GATE Life Sciences Chemistry of Biomolecules syllabus?

Chemistry of Biomolecules is split into 6 chapters — Amino acids, Nucleic acids and nucleotides, Carbohydrates, Lipids, Principles of biomolecule purification and Identification of biomolecules, containing 7 topics and 4 sub-topics in total.

How many chapters are there in Chemistry of Biomolecules for GATE Life Sciences?

6 chapters. Chemistry of Biomolecules accounts for about 11% of the topics in the whole GATE Life Sciences syllabus (7 of 64).

How long should I spend on Chemistry of Biomolecules for GATE Life Sciences?

Budget around 6 hours for a first pass through Chemistry of Biomolecules — about 45 minutes per topic plus 12 minutes per sub-topic across its 7 topics. Add revision cycles on top.

Are there flashcards for GATE Life Sciences Chemistry of Biomolecules?

Yes — a 59-card Chemistry of Biomolecules deck. Sample cards are printed on this page, and the full deck is free in the Examius app with spaced repetition scheduling.