How long should it take the polypeptide backbone of a 6-residue, 10-residue, 15-residue and 20- residue folding nucleus to explore all its possible conformations? Assume that the polypeptide backbone randomly reorients every 10-13 seconds (s). (1) 10-7s, 10-3s, 102 s, 107 s, respectively (2) 10-1Os, 10-6s, 103 s, 1010 s, respectively (3) 10-5s, 10-2s, 10s, 103s, respectively (4) 1s, 10s, 100s, 107s, respectively
  1. How long should it take the polypeptide backbone of a 6-residue, 10-residue, 15-residue and 20- residue folding nucleus to explore all its possible conformations? Assume that the polypeptide backbone randomly reorients every 10-13 seconds (s).
    (1) 10-7s, 10-3s, 102 s, 107 s, respectively
    (2) 10-1Os, 10-6s, 103 s, 1010 s, respectively
    (3) 10-5s, 10-2s, 10s, 103s, respectively
    (4) 1s, 10s, 100s, 107s, respectively

Time Required for a Polypeptide Backbone to Explore All Possible Conformations

Introduction

Protein folding is a complex and highly regulated process that determines the functional conformation of a polypeptide. The time required for a polypeptide backbone to explore all its possible conformations depends on its length and the frequency of random reorientations. This article explains the calculation and significance of folding time estimation.

Understanding Protein Folding Kinetics

The total time for a polypeptide to explore all conformations is influenced by:

  • Residue count: Longer chains have exponentially more possible conformations.
  • Reorientation frequency: The backbone is assumed to reorient every 10⁻¹³ seconds (s).
  • Levinthal’s Paradox: Randomly exploring all conformations is impractical; proteins follow guided folding pathways.

Calculation of Folding Time

Given Data:

  • Random reorientation time per conformation = 10⁻¹³ s
  • Number of residues: 6, 10, 15, 20
  • Possible conformations per residue: ~ (where is the residue count)

Estimated Folding Times:

  • 6 residues: s ≈ 10⁻⁷ s
  • 10 residues: s ≈ 10⁻³ s
  • 15 residues: s ≈ 10² s
  • 20 residues: s ≈ 10⁷ s

Correct Answer:

(1) 10⁻⁷s, 10⁻³s, 10²s, 10⁷s, respectively

Implications for Protein Folding

1. Levinthal’s Paradox

  • If proteins folded by random conformational search, it would take longer than the age of the universe.
  • Instead, proteins fold via directed pathways with specific folding intermediates.

2. Role of Chaperones in Folding

  • Molecular chaperones assist in proper folding by reducing the conformational search space.
  • Example: Hsp70 and GroEL-GroES chaperone systems.

3. Folding Energy Landscape

  • Folding follows an energy funnel model, where proteins move towards the lowest free-energy state.
  • Misfolding can lead to aggregation and diseases like Alzheimer’s and Parkinson’s.

Experimental Techniques to Study Folding Kinetics

  1. Stopped-Flow Fluorescence – Measures rapid folding events.
  2. Circular Dichroism (CD) Spectroscopy – Analyzes secondary structure formation.
  3. NMR Relaxation Studies – Tracks folding intermediates.
  4. Molecular Dynamics Simulations – Models folding pathways computationally.

Conclusion

The time required for a polypeptide to explore all conformations increases exponentially with chain length. For practical folding, proteins rely on guided pathways rather than exhaustive searches. Understanding these principles is crucial for studying protein folding disorders and designing stable synthetic proteins.

 

19 Comments
  • Suman bhakar
    March 27, 2025

    Done

  • Ujjwal
    March 28, 2025

    ✔️👍

  • pallavi gautam
    March 28, 2025

    done sir

    • Khushi Singh
      September 27, 2025

      Done

  • Priyam choudhary
    April 24, 2025

    Done sir 😄

  • Mitali Saini
    August 4, 2025

    (1) 10⁻⁷s, 10⁻³s, 10²s, 10⁷s, respectively

  • Komal Sharma
    August 8, 2025

    The time required for a polypeptide to explore all conformations increases exponentially with chain length. For practical folding, proteins rely on guided pathways rather than exhaustive searches. Understanding these principles is crucial for studying protein folding disorders and designing stable synthetic proteins. So option 2 is correct

  • Meenakshi Choudhary
    September 26, 2025

    10⁻⁷s, 10⁻³s, 10²s, 10⁷s

  • Kirti Agarwal
    September 26, 2025

    Opt 1

  • Dharmpal Swami
    September 26, 2025

    Option 1 correct

  • Pallavi Ghangas
    September 26, 2025

    1

  • Neelam Sharma
    September 27, 2025

    10⁻⁷s, 10⁻³s, 10²s, 10⁷s, respectively

  • Kajal
    September 27, 2025

    Option 1 is correct

  • Mohd juber Ali
    September 28, 2025

    Correct option 1

  • Rishita
    September 28, 2025

    Done

  • Sakshi Kanwar
    September 28, 2025

    Less residue …will reorienta faster the order will be
    10-7s, 10-3s, 102 s, 107 s,

  • Neha Yadav
    September 29, 2025

    Less residue will reorient faster so the order will be –
    10 -7 s, 10 -3 s, 10 2 s, 10 7 s,

  • Heena Mahlawat
    September 29, 2025

    Option 1

  • Muskan Yadav
    October 3, 2025

    10⁻⁷s, 10⁻³s, 10²s, 10⁷s,is the correct answer.

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