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.​