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
- X-ray Crystallography – Provides high-resolution structures of membrane proteins.
- Circular Dichroism (CD) Spectroscopy – Determines secondary structures in solution.
- 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, 2025Done sir 👍
Beena Meena
March 28, 2025Done sir
Arushi
March 28, 2025👍👍✔️
Arushi
March 28, 2025👍✔️
pallavi gautam
March 28, 2025done
Akshay mahawar
April 1, 2025Done 👍
Ujjwal
April 4, 2025Done
Kabeer Narwal
April 6, 2025👍
Priyam choudhary
April 24, 2025Done 👍
Priyam choudhary
April 24, 2025Done
Mitali Saini
August 7, 2025👍done sir
Komal Sharma
August 11, 2025Done ✅
Aakansha sharma Sharma
September 25, 2025Correct Answer: (1) 30 Å
Meenakshi Choudhary
September 26, 2025Number of residues = 20
Rise per residue = 1.5 Å
Total length = 20 × 1.5 Å = 30 Å
Kirti Agarwal
September 26, 202530
Dharmpal Swami
September 26, 202530 A°
Pallavi Ghangas
September 26, 202530 angstrom
Parul
September 26, 2025Length of a 20 amino acid long Alpha helix will be 30 Angstrom.
Minal Sethi
September 27, 2025length of 1 AA in alpha helix is 1.5 A°
so for 20 AA = 20*1.5 = 30 A°
Neha Yadav
September 27, 2025length of 1 AA in alpha helix = 1.5 A°(rise per residue )
So , 20 AA = 20*1.5 = 30 A°
Kajal
September 27, 202530 angstrom
Khushi Singh
September 27, 2025No. Of residue=20
Rise per residue=1.5
Total=20×1.5 = 30
Mohd juber Ali
September 28, 2025residue = 20
Rise per residue = 1.5 Å
(Total length of aa in alpha helix. = 20 × 1.5 Å = 30 Å)
Santosh Saini
September 28, 2025Number of residue = 20 , Rise per residue = 1.5 A° , Total length = 20×1.5 A° = 30A°
Sakshi Kanwar
September 28, 2025Pitch 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, 2025Pitch for alpha helix is 5.4 so 5.4/3.6 = 1.5
Then 1.5*20 = 30A°
Anurag Giri
September 28, 2025Number of residues = 20
Rise per residue = 1.5 Å
Total length = 20 × 1.5 Å = 30 Å
Soniya Shekhawat
September 28, 2025Pitch for residue of Alpha helix is 1.5 angstrone and for 20 residue the length is 1.5×20=30 angstrone.
Anju
September 29, 2025Ans: 30 angstron
Heena Mahlawat
September 29, 2025Length of alpha helix= no of amino acids × pitch per residue
=20× 1.5 = 30 A°
Manisha choudhary
September 29, 2025Number of Amino acid =20
Rise per residue =1.5
Total length=1.5 ×20 =30 A°
Roopal Sharma
September 29, 202520×1.5 = 30A
Devika
September 30, 202520×1.5A°=30A°
Rishita
September 30, 202530
Priti khandal
September 30, 20251 st is right
Arushi Saini
October 1, 2025Total length = 20 × 1.5 Å = 30 Å
Muskan Yadav
October 3, 2025Number of residues = 20
Rise per residue = 1.5 Å
Total length = 20 × 1.5 Å = 30 Å is the correct answer.