Answer :
pI is the average of N- and C-terminus pKa. Average = (8.00 + 4.00) / 2 = 6.00. Therefore, the answer is a.
To find the isoelectric point (pI) of a dipeptide, we need to consider the two ionizable groups: the N-terminus and the C-terminus. The pI is the pH at which the molecule carries no net electrical charge.
For a dipeptide EA with an N-terminus pKa of 8.00 and a C-terminus pKa of 4.00, we need to determine the pH at which both groups are in their respective protonation states.
At a pH above the pKa of the N-terminus (pKa = 8.00), it will be deprotonated (NH₂). At a pH below the pKa of the C-terminus (pKa = 4.00), it will be protonated (COOH).
To calculate the pI:
1. Find the average of the pKa values of the N-terminus and the C-terminus:
[tex]\[ \text{pI} = \frac{{\text{pKa}_{\text{N-terminus}} + \text{pKa}_{\text{C-terminus}}}}{2} \][/tex]
[tex]\[ \text{pI} = \frac{{8.00 + 4.00}}{2} \][/tex]
[tex]\[ \text{pI} = \frac{12.00}{2} \][/tex]
[tex]\[ \text{pI} = 6.00 \][/tex]
So, the correct answer is:
a) 6.00
