Final answer:
The length of a 65.0 kDa alpha-helical protein segment, with an average of 3.6 amino acids per turn and a turn length of 5.4 Å, is approximately 88.636 nm.
Step-by-step explanation:
The alpha-helix structure is a common secondary protein structure stabilized by hydrogen bonds. To calculate the length of a 65.0 kDa alpha helical segment, we consider the average molecular weight of an amino acid to be approximately 110 Da. Since each turn of the alpha helix contains 3.6 amino acids, and knowing one turn measures 5.4 Å (angstroms), we can estimate the total number of amino acids and subsequently determine the helix's length.
First, calculate the number of amino acids in the protein by dividing the protein's molecular weight by the average molecular weight of an amino acid: 65,000 Da / 110 Da/amino acid = 590.91 amino acids approximately. Now, find out how many helical turns are present by dividing this number by 3.6 amino acids per turn: 590.91 amino acids / 3.6 amino acids/turn ≈ 164.14 turns. Finally, by multiplying the number of turns by the length per turn, we get the length of the helix: 164.14 turns * 5.4 Å/turn ≈ 886.36 Å, or 88.636 nm space length.