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Eddi Din [679]
3 years ago
14

Alpha helices are a type of secondary structure in proteins. What is the length of a 27.0 kDa single‑stranded α‑helical protein

segment? Assume a mean residue mass of 110 Da .
Chemistry
1 answer:
iogann1982 [59]3 years ago
8 0

Answer : The  length of protein will be, 36.8 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 = 27.0 kDa

Mean residue mass = 110 Da

\text{Amino acid residue}=\frac{27.0kDa}{110Da}=\frac{27000Da}{110Da}=245.4

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}=\frac{245.4}{3.6}=68.17

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}=68.17turns\times 0.54\AA=36.8\AA

Thus, the length of protein will be, 36.8 A⁰

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Cesium has a radius of 272 pm and crystallizes in a body-centered cubic structure. What is the edge length of the unit cell
olchik [2.2K]

Answer: Edge length of the unit cell = 628pm

Explanation: For a body centred cubic structured system, the relationship between the edge length of the unit cell and radius of the atoms in the structure is

Edge length of Unit cell (a) = (4R)/(√3)

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a = (4 × (2.72 × (10^-10)))/(√3)

a = (6.28157 × (10^-10))m = 628pm

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</span>The vapor pressure of pure benzene is 322 torr at 40 C.
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<span>
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