The peptide that gives positive ceric ammonium nitrate and carbylamine tests is:

First, let us recall what each qualitative test detects.

We have the ceric ammonium nitrate test. This test is specific for the presence of an alcoholic -OH group, especially a primary or secondary alcohol. When such an -OH group is present, the yellow $$(NH_{4})2[Ce(NO_{3})6]$$ solution turns red-orange. So, for a peptide to give this test, at least one of its amino-acid residues must carry an -OH group in its side chain.

Next, we have the carbylamine (isocyanide) test. The general reaction involved is

$$R-NH_{2} + CHCl_{3} + 3KOH \;\longrightarrow\; R-NC + 3KCl + 3H_{2}O$$

This reaction occurs only when a primary amine ($$-NH_{2}$$) is present. Thus, the peptide must contain an amino-acid residue that still possesses a free primary amino group that has not been converted into an amide bond.

Now we examine the side chains of the amino acids that appear in the options.

$$\begin{aligned} \text{Serine (Ser)} &:& -CH_{2}-OH\quad(\text{contains -OH group})\\\\ \text{Lysine (Lys)} &:& -(CH_{2})4-NH_{2}\quad(\text{contains a free }\varepsilon\text{-NH2})\\\\ \text{Glutamine (Gln)} &:& -CH_{2}-CH_{2}-CONH_{2}\quad(\text{amide, no -OH, no free -NH2})\\\\ \text{Aspartic\;acid (Asp)} &:& -CH_{2}-COOH\quad(\text{no -OH, no free -NH2}) \end{aligned}$$

• Serine clearly satisfies the requirement for the alcoholic -OH group.

• Lysine clearly satisfies the requirement for a free primary amine.

On the other hand,

• Glutamine’s side chain terminates in an -CONH2 group, which is an amide, not a free -NH2.

• Aspartic acid has only a -COOH group; it lacks both an -OH and a free -NH2.

Hence, the only peptide that simultaneously carries an alcoholic -OH (from Ser) and a free primary -NH2 (from Lys) is the dipeptide

$$\text{Ser-Lys}.$$

This corresponds to Option D (Option 4 in the given numbering).

Hence, the correct answer is Option D.