Answer
a) A Helices have 3.6 residues per turn with a pitch of 5.4 residues. Hence they have a rise per residue of (5.4 A/turn)/(3.6 residues/turn) = 1.50 A. It therefore requires an a helix of 30 A/1.50 A = 20 residues to span the hydrocarbon bilayer core.
b) ß sheets have a length of 3.5 A per residue. The length of a ß sheet required to span the 30-A wide hydrocarbon core of a lipid bilayer when it is inclined by 30° to the membrane normal is 30 A/cos 30º = 34.6 A. Hence the number of residues in this ß strand is 34.6 A/3.5A - 10 residues.
c) Besides penetrating the hydrocarbon core of a bilayer, the transmembrane element of a polypeptide must also pass through the flanking layers of head groups. These layers have relatively few hydrogen bonding groups and hence the polypeptide strand continue their a helical or ß. The additional residues form a helix, which partially satisfies backbone hydrogen bonding requirements, where the lipid head groups do not offer hydrogen bonding partners.
Work Step by Step
a) A Helices have 3.6 residues per turn with a pitch of 5.4 residues. Hence they have a rise per residue of (5.4 A/turn)/(3.6 residues/turn) = 1.50 A. It therefore requires an a helix of 30 A/1.50 A = 20 residues to span the hydrocarbon bilayer core.
b) ß sheets have a length of 3.5 A per residue. The length of a ß sheet required to span the 30-A wide hydrocarbon core of a lipid bilayer when it is inclined by 30° to the membrane normal is 30 A/cos 30º = 34.6 A. Hence the number of residues in this ß strand is 34.6 A/3.5A - 10 residues.
c) Besides penetrating the hydrocarbon core of a bilayer, the transmembrane element of a polypeptide must also pass through the flanking layers of head groups. These layers have relatively few hydrogen bonding groups and hence the polypeptide strand continue their a helical or ß. The additional residues form a helix, which partially satisfies backbone hydrogen bonding requirements, where the lipid head groups do not offer hydrogen bonding partners.
