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Schach [20]
1 year ago
10

some regions of a polypeptide may coil or fold back on themselves. this is called , and the coils or folds are held in place by

. view available hint(s)for part a some regions of a polypeptide may coil or fold back on themselves. this is called , and the coils or folds are held in place by . tertiary structure ... hydrogen bonds tertiary structure ... covalent bonds primary structure ... covalent bonds secondary structure ... hydrogen bonds secondary structure ... peptide bonds
Chemistry
1 answer:
diamong [38]1 year ago
5 0

some regions of a polypeptide may coil or fold back on themselves. this is called <u>secondary structure</u> , and the coils or folds are held in place by <u>hydrogen bonds</u>

<u></u>

After translation, primary structure is just the arrangement of amino acids. The interactions between the carbonyl, amino, and side groups of the amino acid polymer backbone inside the chain result in the secondary structure of proteins. These interactions are primarily fueled by hydrogen bonds, which result in the formation of alpha helices and beta sheets, which are the primary features of proteins' secondary structures.

To create a useful three-dimensional structure, tertiary structure requires more interactions within the protein chain. Disulfide bonds between cysteines, hydrophobic contacts, ionic bonding, and dipole-dipole interactions are a few of these interactions. To create a useful, three-dimensional protein structure, several protein chains interact in quaternary structure.

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To learn more about secondary structures:

brainly.com/question/15156619

#SPJ4

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Solid sodium iodide is slowly added to a solution that is 0.0050 M Pb 2+ and 0.0050 M Ag +. [K sp (PbI 2) = 1.4 × 10 –8; K sp (A
UkoKoshka [18]

Answer:

[Ag⁺] = 5.0x10⁻¹⁴M

Explanation:

The product solubility constant, Ksp, of the insoluble salts PbI₂ and AgI is defined as follows:

Ksp(PbI₂) = [Pb²⁺] [I⁻]² = 1.4x10⁻⁸

Ksp(AgI) = [Ag⁺] [I⁻] = 8.3x10⁻¹⁷

The PbI₂ <em>just begin to precipitate when the product  [Pb²⁺] [I⁻]² = 1.4x10⁻⁸</em>

<em />

As the initial [Pb²⁺] = 0.0050M:

[Pb²⁺] [I⁻]² = 1.4x10⁻⁸

[0.0050] [I⁻]² = 1.4x10⁻⁸

[I⁻]² = 1.4x10⁻⁸ / 0.0050

[I⁻]² = 2.8x10⁻⁶

<h3>[I⁻] = 1.67x10⁻³</h3><h3 />

So, as the [I⁻] concentration is also in the expression of Ksp of AgI and you know concentration in solution of I⁻ = 1.67x10⁻³M:

[Ag⁺] [I⁻] = 8.3x10⁻¹⁷

[Ag⁺] [1.67x10⁻³] = 8.3x10⁻¹⁷

<h3>[Ag⁺] = 5.0x10⁻¹⁴M</h3>

6 0
3 years ago
A radioactive nuclide is used to detect eye tumors. An atom of this radionuclide contains 15 protons, 15 electrons, and
statuscvo [17]

Answer: ³²P

Explanation:

The radionuclide in question is known as Phosphorus-32. It is an isotope of Phosphorus that is radioactive and has one more neutron than the normal phosphorus does.

Phosphorus does not exist in nature and is created artificially by bombarding the normal stable phosphorus with neutrons. Due to its radioactive nature, it has proven to be useful in the medical industry where it has been used to detect eye tumors as well as in treating some diseases.

4 0
3 years ago
How long is 5 half lives if the half life is 2.6 hours? Remember to round to the correct number of significant figures and use u
Dmitry [639]

Answer:

\Delta t = 13\,h

Explanation:

Half time is period required to desintegrating the half of the initial number of atoms. Then, the total time is:

\Delta t = 5\cdot (2.6\,h)

\Delta t = 13\,h

3 0
3 years ago
Which term defines the following: moles of solute mass of solvent in kg?
kramer
The answer is B. molality
7 0
3 years ago
What is the freezing point of a solution of 465 g of sucrose c12h22o11 dissolved in 575 ml of water?
Amanda [17]

Answer:

The freezing point of the solution is - 4.39 °C.

Explanation:

We can solve this problem using the relation:

<em>ΔTf = (Kf)(m),</em>

where, ΔTf is the depression in the freezing point.

Kf is the molal freezing point depression constant of water = -1.86 °C/m,

density of water = 1 g/mL.

<em>So, the mass of 575 mL is 575 g = 0.575 kg.</em>

m is the molality of the solution (m = moles of solute / kg of solvent = (465 g / 342.3 g/mol)/(0.575 kg) = 2.36 m.

<em>∴ ΔTf = (Kf)(m</em>) = (-1.86 °C/m)(2.36 m) = <em>- 4.39 °C.</em>

<em>∵ The freezing point if water is 0.0 °C and it is depressed by - 4.39 °C.</em>

<em>∴ The freezing point of the solution is - 4.39 °C.</em>

6 0
3 years ago
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