Question

Calculate the pI for the free floating peptide that has the following sequence (show how you got to your answer)

N-V₁A₂I₃S₄F₅R₆L₇V₈E₉R₁₀S₁₁-C

Answer 1

V is valine. The pKa is 2.29, pKb is 9.74. It does not have any pKx, as it does not have any charged functional group.

A is alanine. The pKa is 2.35, pKb is 9.87. It does not have any pKx, as it does not have any charged functional group.

I is isoleucine. The pKa is 2.32, pKb is 9.76. It does not have any pKx, as it does not have any charged functional group.

S is serine. The pKa is 2.21, pKb is 9.15. It does not have any pKx, as it does not have any charged functional group.

F is phenylalanine. The pKa is 2.58, pKb is 9.24. It does not have any pKx, as it does not have any charged functional group.

R is arginine. The pKa is 2.18, pKb is 9.09 and pKx is 13.2.

L is leucine. The pKa is 2.36, pKb is 9.60. It does not have any pKx, as it does not have any charged functional group.

V is valine again. The pKa is 2.29, pKb is 9.74. It does not have any pKx, as it does not have any charged functional group.

E is glutamic acid. The pKa is 2.19, pKb is 9.67 and pKx is 4.25.

R is arginine again. The pKa is 2.18, pKb is 9.09 and pKx is 13.2.

S is serine again. The pKa is 2.21, pKb is 9.15. It does not have any pKx, as it does not have any charged functional group.

pKa = alpha carboxyl group

pKb = alpha ammonium group

pKx = side chain group

Now we will calculate pI of the peptide. Valine is N terminal, so its c-terminal pKa will be ignored, as it will from bond with N-terminal part of alanine. Like this, pK values will be ignored, when N-terminal or C-terminal part of an amino acid is busy to form bond with each other. So, the pKa of the last one, which is serine will be considered again. Other than this, the pKx values will be considered to calculate the pI of the protein. The pK values which will be taken for the pI calculation are in bold. The total will be divided by the number of amino acids participating in pI calculation.

The pI of the peptide is = (9.74 + 13.2 + 4.25 + 13.2 + 2.21) / 5

= 8.5

Ans: The pI for the free floating peptide is 8.5

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