Question

The lipid portion of a typical bilayer is about 30 Å thick. a) Calculate the minimum number of residues in an α-helix required to span this distance. b) Calculate the minimum number of residues in a β-strand required to span this distance. c) Calculate the minimum number of residues in a β-strand required to span this distance if the strand axis is at 45° from the normal to the membrane d) Explain why α-helices are most commonly observed in transmembrane protein sequences when the distance from one side of a membrane to the other can be spanned by significantly fewer amino acids than in a β strand orientation

Answer 1

a) Minimum 9 residues are required for α-helix, b) Minimum 5 residues are required for β-strand, c) Minimum 7 residues are required for β-strand at 45° angle, d) α-helices are more commonly observed in transmembrane protein sequences due to their compact structure.

Step-by-step explanation:

To calculate the minimum number of residues required to span a distance, we need to know the length of one residue. In this case, we are given that a helical turn in an α-helix has 3.6 amino acid residues. Now let's calculate:

a) The thickness of the lipid portion is 30 Å. Since each helical turn has 3.6 residues, we can calculate the number of turns required to span this distance: 30 Å ÷ 3.6 Å per turn = 8.33 turns. Since we can't have a fractional number of turns, we need a minimum of 9 residues to span this distance in an α-helix.

b) The length of a β-strand is approximately 7 Å per residue. Therefore, the minimum number of residues required to span a distance of 30 Å in a β-strand would be: 30 Å ÷ 7 Å per residue = 4.29 residues. Again, we can't have a fractional number of residues, so we would need a minimum of 5 residues to span this distance in a β-strand.

c) If the strand axis is at a 45° angle from the normal to the membrane, we need to adjust the calculation. We can use trigonometry to find the effective distance spanned by the β-strand. The effective distance can be calculated as: 30 Å ÷ cos(45°) = 42.43 Å. With a length of 7 Å per residue, the minimum number of residues required would be: 42.43 Å ÷ 7 Å per residue = 6.06 residues. Again, we would need a minimum of 7 residues to span this distance in a β-strand at the 45° angle.

d) α-helices are most commonly observed in transmembrane protein sequences because they can span the membrane with significantly fewer amino acids compared to β-strands. As we calculated earlier, α-helices require a minimum of 9 residues, while β-strands require a minimum of 5 residues to span the same 30 Å distance. This is because α-helices have a more compact structure, where the amino acid residues are close together, allowing for efficient packing and spanning of the membrane.

Answer 2

a) The minimum number of residues in alfa helix to span 30 Å, is 20 residues

b) The minimum number of residues in a Beta sheet to span 30 Å, is 8.6 residues.

c) Should be better if you show an image to do this question.

d) Look answer below.

Step-by-step explanation:

a) Due to every 5,4 Å there are 3,6 residues. So doing the calculus give us 20 residues.

b) That’s it because every 7 Å there are 2 residues

c) First of all, is important to know that the residues of a protein define the secondary structure due to intermolecular and intramolecular interactions. Taking in account that the membrane bilayer has a lipid interior α-helical domains which are hydrophobic principally are embedded in membranes by hydrophobic interactions with the lipid interior of the bilayer favored by Van der Waals interactions and probably also by ionic interactions with the polar head groups of the phospholipids.