Q.33 A newly identified viral protein contains one long α-helix spanning 60 amino acid residues. The number of main chain H-bonds formed in this helix is __________. (Answer in integer)

Q.33 A newly identified viral protein contains one long α–helix spanning 60 amino acid
residues. The number of main chain H–bonds formed in this helix is __________.
(Answer in integer)

The number of main chain H-bonds in a 60-residue α-helix is 56.​

α-Helix Structure Basics

An α-helix forms a right-handed coil stabilized by hydrogen bonds between the carbonyl oxygen (C=O) of residue i and the amide hydrogen (N-H) of residue i+4 in the polypeptide backbone. This i to i+4 pattern creates 3.6 residues per turn with a 5.4 Å pitch. Each internal residue fully participates in this bonding network.​​

Hydrogen Bond Calculation

For n amino acid residues, the first four N-H groups at the N-terminus and the last four C=O groups at the C-terminus lack intra-helix partners, yielding n – 4 hydrogen bonds. With 60 residues, this gives 60 – 4 = 56 H-bonds. The formula holds for long helices like this viral protein segment.​

Common Misconceptions

Some might think every residue forms two bonds (overestimating at ~120), but each bond links one C=O to one N-H. Shorter helices (e.g., 15 residues: 11 bonds) follow the same rule, but edge effects dominate small ones. No options exist here, but errors like n-5 (55) ignore full pairing.​​

In protein secondary structure, understanding main chain H-bonds in 60 amino acid α-helix is crucial for CSIR NET Life Sciences aspirants tackling questions on viral proteins and α-helical stability. This long α-helix spanning 60 residues forms a rigid rod-like structure through precise hydrogen bonding patterns.​

Hydrogen Bonding Pattern

The α-helix features intra-chain H-bonds where the C=O of residue i pairs with N-H of i+4, creating a 13-atom ring (3.6₁₃-helix). All bonds align parallel to the helix axis, with side chains projecting outward. For competitive exams, recognize this i-to-i+4 rule as the cornerstone.​​

Step-by-Step Calculation

  • Total residues (n) = 60

  • Unpaired groups: First 4 N-H (N-terminus) + last 4 C=O (C-terminus) = 8 unsatisfied sites

  • H-bonds formed = n – 4 = 56​

Visualize: Residues 1-4 lose N-H partners; 57-60 lose C=O partners. Internal residues 5-56 fully bond.​

Helix Length (residues) H-Bonds (n-4) Example Context
15 11 Short helix ​
53 49 Viral hemagglutinin ​
60 56 This viral protein ​

Exam Relevance for CSIR NET

CSIR NET questions test this exact formula, often with viral or membrane proteins. Proline distorts helices by breaking 2 H-bonds, but absent here. Practice variations: For 36 residues (10 turns), expect 32 bonds.​​

Master main chain H-bonds in 60 amino acid α-helix for scoring in biomolecule structure units—answer stays 56.​

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