Insulin has two conformational states, and the main difference in the B chain (t
ID: 90148 • Letter: I
Question
Insulin has two conformational states, and the main difference in the B chain (the long chain) between the two states is the length of the central a-helix. In the more relaxed state, the helix extends from residue 3 to residue 19, while in the other state the helix only extends from residue 8 to residue 19. Given the parameters of an a-helix (i.e. 3.6 residues per turn, and S.4 A rise per turn), what is the difference in length of this central helix between the two conformations of insulin? How many more turns of the helix are in the relaxed conformation? Insulin is stored in the pancreas as an inactive hexamer, and released to the blood stream as an active hormone in a monomer form. The conformational states described above are thought to be essential for the transition from hexamer to monomer. The relaxed state appears to be preferred for hexamer formation. As a budding young biochemist, briefly hypothesize why the conformational change might cause a change in quaternary structure.Explanation / Answer
5) Insulin has two conformational states. The main difference between the two states is the length of the central alpha helix in the B chain. In one conformational state, the helix extends from residue 3 to 19 and in the other conformational state, the helix extends from residue 8 to 19.
There are 3.6 residues per turn and 5.4 A rise per turn. The length of central helix in relaxed state is 16 residues and in the other state, it is 11.
3.6 residues constitute one turn. So, 16 residues constitute 4.4 turns and 11 residues constitute 3.05 turns.
One turn produces 5.4 A rise
Therefore, 4.4 turns produce 23.76 A rise
Similarly, 3.05 turns produce 16.47 A rise
The difference between the two is 7.29 A.
a) The difference in length of central helix between two conformational states of insulin is 7.29 A.
b) The relaxed and the other conformational states of insulin has 4.4 and 3.05 turns respectively.
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