(I) At pH = 2
(highly acidic), the structure of the peptide will be as shown below.
Net charge +2
|Z1| = 2 at pH = 2
(ii) At pH = 6 (neutral solution)
In neutral medium, the given tripeptide exists as Zwitter ion.
net charge = 0
|Z2| = 0 at pH = 6
(iii)At pH = 11 (basic medium)
In basic medium the given tripeptide exists in anionic form.
Net charge = β3
So, |Z3| = |-3|
|Z1| + |Z2| + |Z3| = 2 + 0 + 3 = 5
Given below are two statements: