20.110_PSet_5_Solutions_fall2008

# 20.110_PSet_5_Solutions_fall2008 - a 5 amino acid stretch...

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2. Note that the number of conformations depends on the number of peptide bonds. For 100 amino acids, there are 99 peptide bonds connecting the amino acids. However, since this distinction wasn't made clear in the problem statement, we will accept either. For 100 peptide bonds with n conformations , there are n 100 possible conformations ( ). For simplicity, we assume n=2. I f it takes 1 fs to sample each conformation: 2 100 fs(10 -15 s/fs)(1hr/3600s)(1day/24hr)(1yr/365) = 4.02e7 yrs. Change in Entropy: S= k B ln( unfolded ) - k B ln( _folded) = k B ln(2 100 -1/1) k B ln(2 100 )= 69.31k B = 9.565 x 10 -22 J/K. For 99 peptide bonds: time = 2^99 fs = 2.01e7 yrs. DeltaS = k ln(2^99) = 9.47e-22 J/K Alternative solution: Those who have taken Biochemistry may consider that φ and ψ , the torsion angles about the C α bonds, rotate in 3D and with several preferred values (recall Ramachandran plots, Newman projections) . A more realistic value for n might be 9, for the commonly encountered structures (right- and left-hand alpha- helices, parallel and anti-parallel beta sheets, etc.). In this case: 9 100 possible conformations, which will take 8.4 x10 72 years. The change in entropy is: S= k B ln( 9 100 -1/1) k B ln( 9 100 )= 3.03 x10 -21 J/K. 3.)

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4.) The protein below has four binding sites ( α , β , γ , δ ) for the ligand L. We would like to find its equilibrium binding population. For now assume that the association and dissociation constants are equal. L L + L + α γ δ β α γ δ β α γ δ β a) (15 points) Calculate W and the entropy (in units of k) for the situation in which i) 0 ligands are bound (N L = 0) only 1 way to arrange W = 4! / (0! 4!) = 1 S = k ln(1) = 0 ii) 1 ligand is bound (N L = 1)
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20.110_PSet_5_Solutions_fall2008 - a 5 amino acid stretch...

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