Number of cyclic tripeptides formed with 2 amino acids A and B is:

To find the number of cyclic tripeptides formed from amino acids A and B, we first note that a linear tripeptide consists of three residues. Therefore, there are AAA, AAB, ABA, ABB, BAA, BAB, BBA, BBB — 8 total ($$2^3$$).

Since cyclic peptides are invariant under rotation, sequences that are cyclic rotations of one another represent the same molecule. Consequently, the sequence AAA is invariant under any rotation, yielding one distinct peptide, and similarly BBB yields one. Next, the three sequences with two A's and one B (AAB, ABA, BAA) are cyclically equivalent, as are the three with one A and two B's (ABB, BAB, BBA), with each group contributing one distinct cyclic peptide. From the above, the total number of distinct cyclic tripeptides is $$1 + 1 + 1 + 1 = 4$$, giving the answer as Option D: 4.