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Nesterboy [21]
4 years ago
8

Which of the following is the correct ranking of the three bonds and interactions in order from highest to lowest in terms of th

eir bond strength between two side chains of a protein in their tertiary structure?
I. Disulfide bond between two cystines
II. Hydrophobic interactions between two leucines
III. H-bonding in water
Chemistry
1 answer:
Nesterboy [21]4 years ago
6 0

Answer:

I > III > II

Explanation:

I) A disulfide bond between two cystines is created when a sulfur atom from one cystine forms a strong, single covalent bond with a sulfur atom from a second cystine. When a disulfide bond is created, each cystine loses one hydrogen atom. The atom count is 11 for a cystine in mid-chain, but changes to 10 if the cystine joins with another in a disulfide bond. This lead to a much more stable intermolecular interaction.

III) Hydrogen Bonding in water

These hydrogen bonds are at best an interaction, inducing slight positive and negative charges in the Hydrogen and Oxygen/Nitrogen atoms.

The Hydrophilic amino acids have O & N atoms, which form hydrogen bonds with water. These atoms have an uneven distribution of electrons, creating a polar molecule that can interact and form hydrogen bonds with water.

The hydrogen bonds aren't as strong as the covalent bonds in disulfides.

II) Hydrophobic interactions between two leucines

A hydrophobic interaction is formed between two nonpolar molecules.

It describes the preference of nonpolar molecular surfaces to interact with other nonpolar molecular surfaces, thereby displacing water molecules from the interacting surfaces.

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3 years ago
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Answer:

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Explanation:

Step 1: Data given

Concentration of sucrose = 0.210 M

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Density of solution = 1.02 g/mL

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Step 2: Convert this value into terms of g/L

(0.210 mol/L) * (342.3 g/mol) = 71.9 g/L

Calculate the molality

Step 1: Calculate mass water

Suppose we have a volume of 1.00L

Mass of the solution = 1000 mL * 1.02 g/mL = 1020 g solution

We know that there are 71.9 g of solute in a liter of solution from the first calculation. This means

(1020 grams solution) - (71.9 g solute) = 948.1 g = 0.9481 kg water

Step 2: Calculate molality

Molality = moles sucrose / mass water

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% MAss = (mass solute / mass solution)*100%

(71.9 g) / (1020 g) *100% = 7.05% by mass

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