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jeka57 [31]
3 years ago
10

Peptides can be separated according to their charges via an ion-exchange column. in a cation-exchange column at neutral ph, whic

h peptide would elute last?
Chemistry
2 answers:
Firlakuza [10]3 years ago
4 0

A peptide that contains mostly Lys and Arg residues will be eluted last in a cation-exchange column at neutral pH.

A peptide that contains mostly Asp and Glu residues will be eluted last in an anion-exchange column at neutral pH.

Further Explanation:

Chromatography:

It is a procedure that is used for separation and analysis of the mixtures of the chemicals. This technique is based on the differences in the speed with which the different components travel through a stationary phase. Paper chromatography, thin-layer chromatography, gas chromatography, ion-exchange chromatography, and column chromatography are the different types of chromatography.

Ion exchange chromatography:

This is a type of chromatographic technique that is used to separate ionic compounds. It is based on the electrostatic interactions between various components present in the mixture. This chromatography is used for separation of proteins, peptides and charged biopolymers having different molecular sizes and nature. Ion exchange chromatography is further classified as cation-exchange and anion-exchange chromatography.

Cation-exchange chromatography contains negatively charged resins so it has the tendency to bind with positively charged components of the mixture. Since lysine and arginine are the amino acids that are positively charged at neutral pH, so these interact with the negative resins present in the cation column. Therefore a peptide that contains Lys and Arg residues will be eluted last in a cation-exchange column. But the peptides having negatively charged residues show no interaction with the resin so these will be eluted first.

Anion-exchange chromatography contains positively charged resins so it has the tendency to bind with negatively charged components of the mixture. Since aspartic acid and glutamic acid are the amino acids that are negatively charged at neutral pH, so these interact with the positive resins present in the anion column. Therefore a peptide that contains Asp and Glu residues will be eluted last in an anion-exchange column. But the peptides having positively charged residues show no interaction with the resin so these will be eluted first.

Learn More:

  1. Why extraction is a useful method of pre-purification of organic compounds? brainly.com/question/5881840
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Answer Details:

Grade: Senior School

Chapter: Chromatography

Subject: Chemistry

Keywords: chromatography, lysine, arginine, aspartic acid, glutamic acid, anion column, cation column, ion-exchange chromatography.

fgiga [73]3 years ago
3 0
A cation-exchange column is a type of chromatography that separates substances according to their charges. The cation-exchange column in particular has a negatively charged resin that will attract positively charged ions. 

Applying this principle with amino acids, we can say, for example that the negatively charged ones (e.g., aspartate and glutamate) will elute first compared to the positively charged ones (e.g., lysine, arginine, and histidine), which will elute last.
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The NaCl crystal structure consists of alternating Na⁺ and Cl⁻ ions lying next to each other in three dimensions. If the Na⁺ rad
Yuliya22 [10]

Answer : The radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.

Explanation :

As we are given that the Na⁺ radius is 56.4% of the Cl⁻ radius.

Let us assume that the radius of Cl⁻ be, (x) pm

So, the radius of Na⁺ = x\times \frac{56.4}{100}=(0.564x)pm

In the crystal structure of NaCl, 2 Cl⁻ ions present at the corner and 1 Na⁺ ion present at the edge of lattice.

Thus, the edge length is equal to the sum of 2 radius of Cl⁻ ion and 2 radius of Na⁺ ion.

Given:

Distance between Na⁺ nuclei = 566 pm

Thus, the relation will be:

2\times \text{Radius of }Cl^-+2\times \text{Radius of }Na^+=\text{Distance between }Na^+\text{ nuclei}

2\times x+2\times 0.564x=566

2x+1.128x=566

3.128x=566

x=180.9\approx 181pm

The radius of Cl⁻ ion = (x) pm = 181 pm

The radius of Na⁺ ion = (0.564x) pm = (0.564 × 181) pm =102.084 pm ≈ 102 pm

Thus, the radii of the two ions Cl⁻ ion and Na⁺ ion is, 181 and 102 pm respectively.

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