How to calculate the hydrogen bonds in alpha helix?
Answers
Answer:
Explanation:
e can describe the arrangement of atoms around the peptide link (the conformation ) by giving the degree and direction in which the Ca-CO and N-Ca bonds are rotated. When a number of successive peptide links have identical rotations the polypeptide chain takes up a particular secondary structure.
There are several types of secondary structure, but we will concentrate on just two: the a-helix and the b-pleated sheet. In both cases you will see how the regular conformation allows the structure to be stabilised by forming many relatively strong hydrogen bonds .
The a-helix
The a-helix is like a narrow-bore tube. The polypeptide backbone is coiled up like a very tight clockwise screw thread or the cord of a telephone. The peptide link plates form the wall of the tube with the Ca atoms projecting a little from the surface. The side chain groups, attached to the Ca atoms, project outwards from the wall of the tube (figure 8).
As you follow the helix around through 36 a-amino acid units you make 10 complete 360¡ turns and travel 5.4 nm in the forward direction (1 nm = 1x10-9m).
The a-helix conformation has a particular stability for two main reasons. Firstly the side chain groups are quite well separated. Secondly, and most importantly, each peptide link is involved in two hydrogen bonds. The C=O is hydrogen bonded to the N_H of the peptide link four units ahead in the primary structure , while it follows that the N_H is hydrogen bonded to the C=O of the peptide link four units behind (figure 7). The atoms involved are arranged linearly (figure 8) so that the hydrogen bonds are nearly at their maximum strength. The hydrogen bonds run down the length of the a-helix tube and lock the conformation in place.
Answer:
In the α-helix, 4 residues at the N-terminus and 4 at the C-terminus make only 1 bond pre residue. This makes the total number of H-bonds 30-2x4=22. When calculating this number, each H-bond was counted twice: one time for the donor residue and one time for the acceptor. The real number of H-bonds is then 22/2=11.