I have a 44 angstrom section of alpha-keratin. How many residues would be required to span...
I have a 44 angstrom section of alpha-keratin. How many residues would be required to span this distance? HINT: Number of turns of a helix of keratin and number of residues/turn.
a. 18
b. 23
c. 27
d. 30
Homework Answers
Answer: Option D is correct
Explanation:
One helical turn = 3.6 aa
One amino acid = 1.5 A'
Total length = 44 A'
Number of amino acids required = 44/1.5
= 29.33
= 30
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