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mote1985 [20]
3 years ago
10

Alpha helices are a type of secondary structure in proteins. What is the length of a 33.0 kDa single‑stranded α‑helical protein

segment? Assume a mean residue mass of 110 Da . length: Å
Chemistry
1 answer:
WARRIOR [948]3 years ago
7 0

Answer:

450 Å

Explanation:

The amino acids in an α-helix are arranged in a right-handed helical structure where each amino acid residue corresponds to a 100° turn in the helix, and a translation of 1.5 Å (0.15 nm) along the helical axis

Mass of protein= 33.0 kDa

33.0 kDa ×(1000Da/kDa) × ( 1 residue/ 110 Da)

=33.0 ×1000 × 0.0091

= 300 residues

300 residues × (1.5Å /residue)

=450 Å

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One liter of oxygen gas at standard temperature and pressure has a mass of 1.43 g. The same volume of hydrogen gas under these c
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Indeed, the two samples should contain about the same number of gas particles. However, the molar mass of \rm O_2\; (g) is larger than that of \rm H_2\; (g) (by a factor of about 16.) Therefore, the mass of the \rm O_2\; (g) sample is significantly larger than that of the \rm H_2\; (g) sample.

Explanation:

The \rm O_2\; (g) and the \rm H_2\; (g) sample here are under the same pressure and temperature, and have the same volume. Indeed, if both gases are ideal, then by Avogadro's Law, the two samples would contain the same number of gas particles (\rm O_2\; (g) and \rm H_2\; (g) molecules, respectively.) That is:

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Note that the mass of a gas m is different from the number of gas particles n in it. In particular, if all particles in this gas have a molar mass of M, then:

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In other words,

  • m(\mathrm{O_2}) = n(\mathrm{O_2}) \cdot M(\mathrm{O_2}).
  • m(\mathrm{H_2}) = n(\mathrm{H_2}) \cdot M(\mathrm{H_2}).

The ratio between the mass of the \rm O_2\; (g) and that of the \rm H_2\; (g) sample would be:

\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})}\end{aligned}.

Since n(\mathrm{O_2}) = n(\mathrm{H}_2) by Avogadro's Law:

\begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})} = \frac{n(\mathrm{O_2})\cdot M(\mathrm{O_2})}{n(\mathrm{H_2})\cdot M(\mathrm{H_2})} = \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}.

Look up relative atomic mass data on a modern periodic table:

  • \rm O: 15.999.
  • \rm H: 1.008.

Therefore:

  • M(\mathrm{O_2}) = 2 \times 15.999 \approx 31.998\; \rm g \cdot mol^{-1}.
  • M(\mathrm{H_2}) = 2 \times 1.008 \approx 2.016\; \rm g \cdot mol^{-1}.

Verify whether \begin{aligned}& \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})}= \frac{M(\mathrm{O_2})}{M(\mathrm{H_2})}\end{aligned}:

  • Left-hand side: \displaystyle \frac{m(\mathrm{O_2})}{m(\mathrm{H_2})}= \frac{1.43\; \rm g}{0.089\; \rm g} \approx 16.1.
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Note that the mass of the \rm H_2\; (g) sample comes with only two significant figures. The two sides of this equations would indeed be equal if both values are rounded to two significant figures.

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