Question

Alpha helices are a type of secondary structure in proteins. What is the length of a 69.0 69.0 kDa single‑stranded α‑helical protein segment? Assume a mean residue mass of 110 Da 110 Da .

Answer 1

Answer : The  length of protein will be, 94.1 A⁰

Explanation :

First we have to calculate the amino acid residue.

$\text{Amino acid residue}=\frac{\text{Length of single strand of protein}}{\text{Mean residue mass}}$

Given:

Length of single strand of protein = 69.0 kDa

Mean residue mass = 110 Da

$\text{Amino acid residue}=\frac{69.0kDa}{110Da}=\frac{69000Da}{110Da}=627.3$

Now we have to calculate the number of turns in protein.

$\text{Number of turns in protein}=\frac{\text{Amino acid residue}}{3.6\text{amino acid per turn of the alpha-helix}}$

As there are 3.6 amino acid per turn of the alpha-helix.

$\text{Number of turns in protein}=\frac{627.3}{3.6}=174.25$

Now we have to calculate the length of the protein.

$\text{Length of protein}=\text{Number of turns in protein}\times \text{Length of each turn}$

As, we know that the length of each turn = 0.54 A⁰

$\text{Length of protein}=174.25turns\times \0.54\AA=94.1\AA$

Thus, the length of protein will be, 94.1 A⁰