The width of the lipid bilayer membrane is 30 Å.
It is permeated by a protein which is a right-handed α-helix.
Q.51
The number of amino acid residues present in the protein is
Options:
(A) 15
(B) 18
(C) 17
(D) 20
Transmembrane α-Helix: Amino Acids to Span 30 Å Lipid Bilayer
The lipid bilayer membrane measures 30 Å in width and is permeated by a right-handed α-helix protein, a common transmembrane structure in biology. This MCQ (Q.51) tests helical geometry knowledge crucial for biotechnology and molecular biology exams. The correct answer is (D) 20, based on standard α-helix rise per residue.
α-Helix Geometry Basics
A right-handed α-helix features 3.6 residues per turn and advances 5.4 Å axially per turn
Each amino acid contributes a 1.5 Å rise along the helix axis, enabling precise length calculations for membrane-spanning segments.
Calculation: 1.5 Å/residue = 5.4 Å/turn ÷ 3.6 residues/turn
Transmembrane helices typically span the ~30 Å hydrophobic core of bilayers, requiring 20-25 hydrophobic residues.
Calculation for 30 Å Span
Number of residues n = 30 Å ÷ 1.5 Å/residue = 20
Thus, 20 amino acids fully permeate the bilayer without mismatch.
Options Explained
| Option | Residues | Length (Å) | Explanation |
|---|---|---|---|
| (A) | 15 | 15 × 1.5 = 22.5 | Too short; fails to span full 30 Å hydrophobic core, causing mismatch. |
| (B) | 18 | 18 × 1.5 = 27 | Nearly spans but leaves ~3 Å gap; insufficient for stable transmembrane insertion. |
| (C) | 17 | 17 × 1.5 = 25.5 | Shorter than bilayer; helix would distort or snorkel, destabilizing protein. |
| (D) | 20 | 20 × 1.5 = 30 | Exact match; ideal for right-handed α-helix in 30 Å bilayer (correct). |
Biological Relevance
Such 20-residue helices anchor integral membrane proteins like ion channels in the bilayer’s core. This principle applies to bioinformatics modeling and biophysics, where hydrophobic mismatch affects folding energetics.
Exam Tip: Remember: 1.5 Å/residue is the gold standard metric for α-helical transmembrane domains.
