Eagle(1)_lambert4SG_0.001

26. The length of an α-helical section of a polypeptide chain of 20 residues spanning hydrophobic region of membrane would be
(1) 30 A
(2) 54 A
(3) 5.4 A
(4) 3.6 A

Determining the Length of an α-Helical Section in a Membrane-Spanning Polypeptide

Introduction

The α-helix is a fundamental secondary structure in proteins, playing a crucial role in membrane-spanning domains of transmembrane proteins. Understanding its length helps in predicting the structural organization of proteins within biological membranes. This article explains how to calculate the length of an α-helix composed of 20 residues spanning a hydrophobic region of a membrane.

Key Characteristics of an α-Helix

  • Right-handed helical structure stabilized by hydrogen bonding.
  • 3.6 residues per turn of the helix.
  • Each residue contributes 1.5 Å (angstroms) to the helical length.
  • Common in transmembrane proteins, particularly in hydrophobic regions.

Calculation of the α-Helical Length

Given:

  • Number of residues = 20
  • Rise per residue = 1.5 Å

Total length = 20 × 1.5 Å = 30 Å

Correct Answer: (1) 30 Å

Role of α-Helices in Membrane-Spanning Proteins

1. Transmembrane Domains

  • Many integral membrane proteins contain α-helical transmembrane segments.
  • These helices span the hydrophobic lipid bilayer, allowing protein anchoring and function.
  • Typically, 20–25 hydrophobic residues are required to span the 30 Å hydrophobic core of a membrane.

2. Stability and Function

  • Hydrogen bonding within the helix stabilizes its structure.
  • Hydrophobic side chains interact with the lipid bilayer, anchoring the helix.
  • Some transmembrane α-helices form channels or receptors, facilitating molecular transport and signaling.

Experimental Techniques to Study α-Helical Structures

  1. X-ray Crystallography – Provides high-resolution structures of membrane proteins.
  2. Circular Dichroism (CD) Spectroscopy – Determines secondary structures in solution.
  3. Cryo-EM (Cryogenic Electron Microscopy) – Analyzes membrane proteins in their native state.

Conclusion

The length of an α-helical section spanning a hydrophobic membrane region can be calculated based on the rise per residue (1.5 Å). A 20-residue α-helix extends 30 Å, which matches the thickness of the hydrophobic core of a biological membrane. Understanding these structural properties is essential in membrane protein research and drug design.

37 Comments
  • Suman bhakar
    March 27, 2025

    Done sir 👍

  • Beena Meena
    March 28, 2025

    Done sir

  • Arushi
    March 28, 2025

    👍👍✔️

  • Arushi
    March 28, 2025

    👍✔️

  • pallavi gautam
    March 28, 2025

    done

  • Akshay mahawar
    April 1, 2025

    Done 👍

  • Ujjwal
    April 4, 2025

    Done

  • Kabeer Narwal
    April 6, 2025

    👍

  • Priyam choudhary
    April 24, 2025

    Done 👍

  • Priyam choudhary
    April 24, 2025

    Done

  • Mitali Saini
    August 7, 2025

    👍done sir

  • Komal Sharma
    August 11, 2025

    Done ✅

  • Aakansha sharma Sharma
    September 25, 2025

    Correct Answer: (1) 30 Å

  • Meenakshi Choudhary
    September 26, 2025

    Number of residues = 20
    Rise per residue = 1.5 Å
    Total length = 20 × 1.5 Å = 30 Å

  • Kirti Agarwal
    September 26, 2025

    30

  • Dharmpal Swami
    September 26, 2025

    30 A°

  • Pallavi Ghangas
    September 26, 2025

    30 angstrom

  • Parul
    September 26, 2025

    Length of a 20 amino acid long Alpha helix will be 30 Angstrom.

  • Minal Sethi
    September 27, 2025

    length of 1 AA in alpha helix is 1.5 A°
    so for 20 AA = 20*1.5 = 30 A°

  • Neha Yadav
    September 27, 2025

    length of 1 AA in alpha helix = 1.5 A°(rise per residue )
    So , 20 AA = 20*1.5 = 30 A°

  • Kajal
    September 27, 2025

    30 angstrom

  • Khushi Singh
    September 27, 2025

    No. Of residue=20
    Rise per residue=1.5
    Total=20×1.5 = 30

  • Mohd juber Ali
    September 28, 2025

    residue = 20
    Rise per residue = 1.5 Å
    (Total length of aa in alpha helix. = 20 × 1.5 Å = 30 Å)

  • Santosh Saini
    September 28, 2025

    Number of residue = 20 , Rise per residue = 1.5 A° , Total length = 20×1.5 A° = 30A°

  • Sakshi Kanwar
    September 28, 2025

    Pitch per residue of alpha helix is 1.5 Å for 20 residue total length will be 1.5 x 20 = 30 Å

  • Divya rani
    September 28, 2025

    Pitch for alpha helix is 5.4 so 5.4/3.6 = 1.5
    Then 1.5*20 = 30A°

  • Anurag Giri
    September 28, 2025

    Number of residues = 20
    Rise per residue = 1.5 Å
    Total length = 20 × 1.5 Å = 30 Å

  • Soniya Shekhawat
    September 28, 2025

    Pitch for residue of Alpha helix is 1.5 angstrone and for 20 residue the length is 1.5×20=30 angstrone.

  • Anju
    September 29, 2025

    Ans: 30 angstron

  • Heena Mahlawat
    September 29, 2025

    Length of alpha helix= no of amino acids × pitch per residue
    =20× 1.5 = 30 A°

  • Manisha choudhary
    September 29, 2025

    Number of Amino acid =20
    Rise per residue =1.5
    Total length=1.5 ×20 =30 A°

  • Roopal Sharma
    September 29, 2025

    20×1.5 = 30A

  • Devika
    September 30, 2025

    20×1.5A°=30A°

  • Rishita
    September 30, 2025

    30

  • Priti khandal
    September 30, 2025

    1 st is right

  • Arushi Saini
    October 1, 2025

    Total length = 20 × 1.5 Å = 30 Å

  • Muskan Yadav
    October 3, 2025

    Number of residues = 20
    Rise per residue = 1.5 Å
    Total length = 20 × 1.5 Å = 30 Å is the correct answer.

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