1. Understand the problem

We need the number of cyclic tripeptides that can be formed using two amino acids, $A$ and $B$.

A tripeptide has 3 amino acid units. Since it is cyclic, arrangements related by rotation are considered the same.

So we must count distinct 3-membered cyclic sequences made from $A$ and $B$.

2. List all possible 3-letter sequences before cyclic equivalence

Using $A$ and $B$, the possible sequences of length 3 are:

Total linear sequences $=2^3=8$.

3. Identify sequences equivalent under cyclic rotation

For cyclic peptides, sequences differing only by rotation are identical.

(i) All same residues

• $AAA$ gives one cyclic peptide.
• $BBB$ gives one cyclic peptide.

So far: $2$ distinct cyclic peptides.

(ii) Two $A$ and one $B$

The linear forms are:

These are all cyclic rotations of one another:

Hence they represent one cyclic tripeptide.

(iii) Two $B$ and one $A$

The linear forms are:

These are also cyclic rotations of one another:

Hence they represent one cyclic tripeptide.

4. Total number of distinct cyclic tripeptides

Thus the distinct cyclic tripeptides are:

Therefore, total number is

5. Check options

• A: $2$ ❌
• B: $3$ ❌
• C: $5$ ❌
• D: $4$ ✅

6. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

They agree.