Q.54

The total length of the 50-amino acid protein stretch is 12.7 nm, calculated from the axial rise of the α-helix (0.15 nm per residue) and β-strand (0.35 nm per residue).

Calculation Details

An α-helix advances 1.5 Å (0.15 nm) along its axis per residue, as 3.6 residues form one helical turn with a 5.4 Å pitch. For 24 residues, the length is $24\times 0.15=3.6$ nm.

A β-strand extends 3.5 Å (0.35 nm) per residue in its fully extended conformation. For 26 residues, the length is $26\times 0.35=9.1$ nm.

The “extended structure” implies these elements connect end-to-end without loops or turns adding significant length, yielding a total of $3.6+9.1=12.7$ nm.

Common Misconceptions

Options implying addition of linker residues (e.g., 2-4 nm extra) fail, as the question specifies a “true” 24-residue helix and 26-residue strand within exactly 50 residues. Using β-sheet inter-strand distance (0.5 nm) instead of strand rise confuses topology with linear extension.

Introduction to Protein Secondary Structure Dimensions

Protein secondary structure length calculations determine globular protein dimensions using standard rise-per-residue values for α-helices and β-strands, essential for CSIR NET Life Sciences. A 50-amino acid residue stretch with 24-residue true α-helix and 26-residue β-strand spans 12.7 nm axially.

Alpha Helix Geometry and Length

The α-helix features 3.6 residues per turn and 5.4 Å pitch, yielding 1.5 Å (0.15 nm) rise per residue.

• 24 residues contribute $24\times 0.15=3.6$ nm.

• Hydrogen bonds (n to n+4) stabilize this right-handed coil.

Beta Strand Extension and Dimensions

β-strands adopt fully extended conformations with 3.5 Å (0.35 nm) per residue, repeating every two residues in pleated sheets.

• 26 residues extend $26\times 0.35=9.1$ nm.

• Distance between Cα atoms (i to i+2) measures ~6 Å, confirming linearity.

Total Length Determination

End-to-end connection in extended structure sums helix (3.6 nm) + strand (9.1 nm) = 12.7 nm, matching exam expectations without turns.