(40 POINTS QUESTION) Insulin has two confromational states, and the main differe

Question

(40 POINTS QUESTION) Insulin has two confromational states, and the main difference in the B chain (the long chain) between the two states is the length of the central alpha-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.

1). Given the parameters of an alpha-helix, what is difference in length of this central helix between the two confromations of Insulin?

2). How many more turns of the helix are in the relaxed conformation?

3). Insulin is stored in the pancreas as an inactive hexamer, and released to the blood stream as an active hormone in 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 the quaternary structure.

Explanation / Answer

In relaxed state, the central helix of insulin extends from residue 3 to 19

In other conformational state, the central helix extends from residue 8 to 19

Length between the two residues = 5.4 Angstrom units

Number of residuces per turn of alpha helix = 3.6

1) The length of central helix in the relaxed confirmation

= (19 - 3) x 5.4 Angstrom units

= 86.4 Angstrom units

The length of central helix in the other conformational state

= (19 - 8) x 5.4 Angstrom units

= 59.4 Angstrom units

Therefore, the difference in length of central axis between the two conformational states

= 86.4 - 59.4

= 27 Angstrom units

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