Question

Isoelectric focusing can separate peptides based on their relative contents of acidic and basic residues. When voltage is applied to an electrophoretic gel with a pH gradient, each peptide migrates until it reaches a point where it has no net charge. The pH at this point is the isoelectric point (pI) for that peptide. One can estimate the pI fairly accurately (within 0.1 or 0.2 pH units) using the pK values of all the proton dissociable groups in the peptide. Consider the hexapeptide Leu-His-Glu-Lys-Met-Asp. Amino acid pK Arg 12.5 Asp 3.7 Cys 8.2 Glu 4.3 His 6.0 Lys 10.5 Tyr 10.5 peptide-NH3+ 8.0 peptide-COOH 3.4 * Amino and carboxy terminal values differ from the ? amino and carboxy values of a single amino acid. Step 1: Determine the total positive charge on the hexapeptide when all acidic and basic groups are fully protonated. Enter your answer without the sign. charge: Step 2: Determine the pK value of the group that loses a proton to form the zwitterionic species with a net charge of zero. pK = Step 3: Determine the pK value of the group that gains a proton to form the zwitterionic species with a net charge of zero. pK = Step 4: Estimate the pI value of the hexapeptide. pI =

          Isoelectric focusing can separate peptides based on their relative contents of acidic and basic residues. When voltage is applied to an electrophoretic gel with a pH gradient, each peptide migrates until it reaches a point where it has no net charge. The pH at this point is the isoelectric point (pI) for that peptide. One can estimate the pI fairly accurately (within 0.1 or 0.2 pH units) using the pK values of all the proton dissociable groups in the peptide. Consider the hexapeptide Leu-His-Glu-Lys-Met-Asp.

Amino acid	pK
Arg	12.5
Asp	3.7
Cys	8.2
Glu	4.3
His	6.0
Lys	10.5
Tyr	10.5
*peptide-NH3+	8.0
*peptide-COOH	3.4
* Amino and carboxy terminal values differ from the ? amino and carboxy values of a single amino acid.

Step 1: Determine the total positive charge on the hexapeptide when all acidic and basic groups are fully protonated. Enter your answer without the sign.
charge:

Step 2: Determine the pK value of the group that loses a proton to form the zwitterionic species with a net charge of zero.
pK =

Step 3: Determine the pK value of the group that gains a proton to form the zwitterionic species with a net charge of zero.
pK =

Step 4: Estimate the pI value of the hexapeptide.
pI =

Isoelectric focusing can separate peptides based on their relative contents of acidic and basic residues. When voltage is applied to an electrophoretic gel with a pH gradient, each peptide migrates until it reaches a point where it has no net charge. The pH at this point is the isoelectric point (pI) for that peptide. One can estimate the pI fairly accurately (within 0.1 or 0.2 pH units) using the pK values of all the proton dissociable groups in the peptide. Consider the hexapeptide Leu-His-Glu-Lys-Met-Asp.

Amino acid pK
Arg 12.5
Asp 3.7
Cys 8.2
Glu 4.3
His 6.0
Lys 10.5
Tyr 10.5
*peptide-NH3+ 8.0
*peptide-COOH 3.4
* Amino and carboxy terminal values differ from the ? amino and carboxy values of a single amino acid.

Step 1: Determine the total positive charge on the hexapeptide when all acidic and basic groups are fully protonated. Enter your answer without the sign.
charge:

Step 2: Determine the pK value of the group that loses a proton to form the zwitterionic species with a net charge of zero.
pK =

Step 3: Determine the pK value of the group that gains a proton to form the zwitterionic species with a net charge of zero.
pK =

Step 4: Estimate the pI value of the hexapeptide.
pI =

Added by Felix M.

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