1. Identify the peptide structure

The tripeptide is Ile–Arg–Pro, i.e. it contains the amino acid residues:

• Isoleucine (Ile)
• Arginine (Arg)
• Proline (Pro)

When peptide bonds form, the side chains remain unchanged, so the number of chiral carbons is found by checking each residue.

2. Count chiral carbons in each amino acid residue

(a) Isoleucine (Ile)

Structure of side chain: $-CH(CH_3)CH_2CH_3$

Isoleucine has:

• the $\alpha$-carbon as chiral
• one more chiral carbon in the side chain, namely the $\beta$-carbon in $-CH(CH_3)CH_2CH_3$

So, Ile contributes: $2 \text{ chiral carbons}$

(b) Arginine (Arg)

Arginine side chain is: $-(CH_2)_3-NH-C(=NH)-NH_2$

Only the $\alpha$-carbon is chiral. The side chain contains only $CH_2$ groups and an sp$^2$ carbon.

So, Arg contributes: $1 \text{ chiral carbon}$

(c) Proline (Pro)

Proline side chain forms a ring with the amino nitrogen: $-(CH_2)_3-$ closing back to N.

Only the $\alpha$-carbon is chiral. The other carbons are $CH_2$ and hence achiral.

So, Pro contributes: $1 \text{ chiral carbon}$

3. Total number of chiral carbons

Adding all contributions: $2 + 1 + 1 = 4$

4. Final answer

The number of chiral carbon(s) in the peptide Ile–Arg–Pro is: $\boxed{4}$

5. Comparison with stored correct answer

Stored correct answer = $4$

My derived answer = $4$

So, they agree.