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Determining the Charge on a Peptide. Peptide. Given the peptide: Leu-Lys-Tyr-Glu Problem: Determine the net charge on this peptide at a given pH. First: Look up and assign pKa values to each ionizable group. (You may want to draw out the structure for the peptide.). Structure and pKa values.

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peptide
Peptide

Given the peptide: Leu-Lys-Tyr-Glu

Problem: Determine the net charge on this peptide at a given pH.

First: Look up and assign pKa values to each ionizable group. (You may want to draw out the structure for the peptide.)

example 1 analysis at ph 2 0
Example 1, Analysis at pH 2.0

Assign charge for each group at pH 2.0

+ + 0 0 0 Net = +2

example 2 analysis at ph 6 0
Example 2, Analysis at pH 6.0

Assign charge for each group at pH 6.0

+ + 0 - - Net = 0

example 3 analysis at ph 9 0
Example 3, Analysis at pH 9.0

Assign charge for each group at pH 9.0

0 + 0 - - Net = -1

ion exchange chromatography
Ion Exchange Chromatography

A strong cation exchange resin contains sulfonate, -SO3- Na+, groups attached to benzene.

A positively charged molecule will bind temporarily to this resin by exchanging with sodium ion and thus movement down the column will be slowed.

A negatively charged molecule will be repelled by the resin and will move down the column more rapidly.

electrophoresis
Electrophoresis

A peptide with a net + charge will move toward the negative electrode on electrophoresis.

A peptide with a net - charge will move toward the positive electrode on electrophoresis.

By varying the pH at which electrophoresis is conducted, movement can be altered.

example 4 determine the pi
Example 4, Determine the pI

Start with the fully protonated form.

+ + 0 0 0

determine the pi
Determine the pI

8.0 10.8 10 4.3 3.1

NH3+ NH3+ OH COOH COOH Net

+ + 0 0 0 +2

1st-3.1

+ + 0 0 - +1

2nd-4.3

+ + 0 - - 0

3rd-8.0

0 + 0 - - -1

4th-10

0 + - - - -2

5th-10.8

0 0 - - - -3

determine the pi11
Determine the pI

The zero net charge form is involved in both the second and third ionizations.

So, the two pKas of these ionizations are used in the calculation.

pI = (4.3 + 8.0)/2 = 6.15