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What is the length of a 71.0 kDa single‑stranded α‑helical protein segment? Assume a mean residue mass of 110 Da .

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Answer : The length of protein will be, 96.8 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 = 71.0 kDa

Mean residue mass = 110 Da


\text{Amino acid residue}=(71.0kDa)/(110Da)=(71000Da)/(110Da)=645.4

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}=(645.4)/(3.6)=179.28

Now we have to calculate the length of the protein.


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

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


\text{Length of protein}=179.28turns* \0.54\AA=96.8\AA

Thus, the length of protein will be, 96.8 A⁰

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