Biophysics problem - amino acid dipoles

[wfu113]

1. The dipole moment of a peptide bond is 3.7 Debye in water. Assuming that a hydrogen bond is essentially a dipole-dipole interaction, estimate the energy of a hydrogen bond between two peptides in water and in the interior of a protein (neglect the competing interactions with the solvent).

Homework Equations

$$_{}V_d_d = -2\left|\mu\right|^{}^2/D\left|r\right|^{}^3$$



3. The Debye part really confuses me. I tried plugging in 0.5 for the r value and using 1.23*10^-29 for the mu value. The D values are given as $$78.5\kappa\epsilon_{}_0$$ for water and $$3.5\kappa\epsilon_{}_0$$ for the interior of the protein. I feel like I'm almost there, but I don't quite have it. Please help! Thanks!

 

[member 553083]

Answer: The energy of a hydrogen bond between two peptides in water is calculated as follows: E_{}H_B = -2\left|\mu\right|^{}^2/D\left|r\right|^{}^3 = -2(1.23*10^{-29})^2/(78.5 * 8.85 * 10^{-12} * 0.5^3) = -9.32 * 10^{-21} J The energy of a hydrogen bond between two peptides in the interior of a protein is calculated as follows: E_{}H_B = -2\left|\mu\right|^{}^2/D\left|r\right|^{}^3 = -2(1.23*10^{-29})^2/(3.5 * 8.85 * 10^{-12} * 0.5^3) = -3.96 * 10^{-20} J