Question

Calculate pI value of Arg (pka of COOH is 1.82, pka of NH3+ is 8.99, pKa of R group is 12.48).Indicate species predominate at pH 1,4,8,13,

Answer 1

The isoelectric point or pI of an amino acid is the pH at which an amino acid has a net charge of zero.

At pH = 1.82 concentration of COOH and COO^- would be equal

So, Net charge on Arg at pH (1.82) =  +1(NH₃⁺) + 1 (R gorup) - 0.5 ( as COO^-) = 1.5

Simillarly at pH= 8.99, Concentration of NH₃⁺ and NH₂ would be equal but group which was inially COOH would now be in majority as COO^- .

So, net charge on Arg at pH (8.99) = +0.5 (NH₃⁺ as half of NH₃⁺ is converted to NH₂) + 1 (R group) - 1 (COO^-) = 0.5

Simillarly at pH= 12.48, Concentration of R in protonated form and concentration of R group in deprotonated form would be equal but group which was inially COOH would now be in majority as COO^- and in same way NH₃⁺ would be in NH₂ now.

So, net charge on Arg at pH (8.99) = +0 (as NH₃⁺ is converted to NH₂) + 0.5 (R group) - 1 (COO^-) = -0.5

So ,

pI = 10.735

So, at pH = 1 Arg would be as R⁺CH(COOH)(NH₃⁺)

At pH = 4 Arg would be as R⁺CH(COO^-)(NH₃⁺) as pH is more than pKa of COOH

At pH = 8 Arg would be as R⁺CH(COO^-)(NH₃⁺) as pH is more than pKa of COOH

At pH = 13 Arg would be as RCH(COO^-)(NH₂) as pH is more than pKa of COOH,NH₃⁺ and R group.