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Chapter 6: Q4P. (page 177)

Calculate the length in angstroms of a 100-residue segment of the 伪 keratin coiled coil.

Short Answer

Expert verified

The length of a 100-residue segment of the 伪 keratin coiled coil is 145.7脜.

Step by step solution

01

α keratin coiled coil.

伪 keratin is a fibrous protein, made up of a repeating sequence of amino acids. It is the major component of vertebrate nails, hair, claw, epidermis of skin, etc.

The structure of 伪 keratin resembles the structure of 伪 helix. It contains 3.5 residues per turn and a length of 5.1 脜 for each turn.

02

Calculation for the number of turns.

One turn in 伪 keratin contains 3.5 residues. Thus,the number of turns that would contain 100 residues will be:

1 x 100/3.5

= 28.57 turns.

03

Calculation for the length of the α keratin.

In 伪 keratin, one turn exhibits a length of 5.1脜. Thus, the total length exhibited by 28.57 turns will be:

28.57 x 5.1

= 145.707 脜.

Hence, the length of the 伪 keratin will be 145.707 脜 = ~145.7 脜.

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