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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 con
Question

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 ​101310^{-13} seconds (s).​​

A.

107s,103s,102s,107s,respectively10^{-7}s, 10^{-3}s, 10^{2}s, 10^{7}s, respectively​​

B.

1010s,106s,103s,1010s,10^{-10}s, 10^{-6}s, 10^{3}s, 10^{10}s, respectivley​

C.

105s,102s,10s,103s,10^{-5}s, 10^{-2}s, 10s, 10^{3}s,respectively​

D.

1s,10s,100s,107s,1s, 10s, 100s, 10^{7}s, respectively​

Correct option is A

To determine the time required for the polypeptide backbone to explore all possible conformations, we use the given reorientation time of 10⁻¹³ seconds per conformation. The number of possible conformations increases with the number of residues (n) due to the degrees of freedom in the backbone. The total time is calculated as the number of conformations multiplied by the reorientation time per conformation.

Thus, the times are 10⁻⁷ s, 10⁻³ s, 10² s, and 10⁷ s, respectively.

Answer: 1. 10⁻⁷ s, 10⁻³ s, 10² s, 10⁷ s, respectively

  • For a 6-residue polypeptide: 10⁶ conformations → 10⁶ × 10⁻¹³ = 10⁻⁷ seconds
  • For a 10-residue polypeptide: 10¹⁰ conformations → 10¹⁰ × 10⁻¹³ = 10⁻³ seconds
  • For a 15-residue polypeptide: 10¹⁵ conformations → 10¹⁵ × 10⁻¹³ = 10² seconds
  • For a 20-residue polypeptide: 10²⁰ conformations → 10²⁰ × 10⁻¹³ = 10⁷ seconds

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