### Related Formula textDipeptide X xrightarrowtextHydrolysis textAmino Acid y + textAmino Acid z$\text{Dipeptide } X \xrightarrow{\text{Hydrolysis}} \text{Amino Acid } y + \text{Amino Acid } z$ ### Core Logic - Since **y** reacts with nitrous acid (HNO_2$HNO_2$) to give lactic acid (CH_3-CH(OH)-COOH$CH_3-CH(OH)-COOH$), **y** must be alanine (CH_3-CH(NH_2)-COOH$CH_3-CH(NH_2)-COOH$). - When glycine (NH_2-CH_2-COOH$NH_2-CH_2-COOH$) is heated, two molecules undergo intermolecular cyclization to produce a six-membered diketopiperazine ring as shown in the problem diagram. Therefore, **z** is glycine. Hence, combining residue **y** (alanine) and **z** (glycine), the dipeptide X is **alanine-glycine**. ### Step 1: Stepwise Degradation Overview Reaction scheme: 1. Alanine-Glycine linkage rightarrow$\rightarrow$ Alanine + Glycine 2. textAlanine + HNO_2 rightarrow textLactic acid + N_2uparrow + H_2O$\text{Alanine} + HNO_2 \rightarrow \text{Lactic acid} + N_2\uparrow + H_2O$ 3. 2 times textGlycine xrightarrowDelta textCyclic diketopiperazine + 2H_2O$2 \times \text{Glycine} \xrightarrow{\Delta} \text{Cyclic diketopiperazine} + 2H_2O$ ### Pattern Recognition Lactic acid generation from alpha-amino acids via nitrous acid deamination is a definitive chemical fingerprint for alanine. The unsubstituted cyclic diketopiperazine product confirms glycine as the second component. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Biomolecules

### Core Logic A dipeptide sequence is parsed strictly from the **N-terminus** to the **C-terminus**. Therefore, in Gly-Ala: * **Glycine (Gly)** must supply its carbonyl end for coupling. * **Alanine (Ala)** must provide its free amine end. To drive peptide bond formation via the explicit elimination of HCl$HCl$, the carboxyl group of glycine must be pre-activated as an acyl chloride variant: mathrmNH_2-CH_2-COCl$\mathrm{NH_2-CH_2-COCl}$. This reacts smoothly with the unsubstituted amine terminus of alanine, mathrmNH_2-CH(CH_3)-COOH$\mathrm{NH_2-CH(CH_3)-COOH}$, liberating HCl$HCl$ to form the amide bridge linkage: mathrmNH_2-CH_2-CONH-CH(CH_3)-COOH$\mathrm{NH_2-CH_2-CONH-CH(CH_3)-COOH}$. ### Pattern Recognition peptide naming structure sequence convention dictates: textFirst name = textN-terminus acyl donor$\text{First name} = \text{N-terminus acyl donor}$, textSecond name = textC-terminus amine acceptor$\text{Second name} = \text{C-terminus amine acceptor}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Biomolecules

### Related Formula textG-C base pairing implies 3 text Hydrogen bonds$\text{G-C base pairing} \implies 3 \text{ Hydrogen bonds}$ textA-T base pairing implies 2 text Hydrogen bonds$\text{A-T base pairing} \implies 2 \text{ Hydrogen bonds}$ ### Core Logic Let's audit the nucleotide base distribution across the given single strand structure: 5' - G - G - C - A - A - A - T - C - G - G - C - T - A - 3'$5' - G - G - C - A - A - A - T - C - G - G - C - T - A - 3'$ * **Count the Guanine (G) and Cytosine (C) bases:** Bases present: G_1, G_2, C_3, C_8, G_9, G_10, C_11$G_1, G_2, C_3, C_8, G_9, G_{10}, C_{11}$ implies$\implies$ Total of 7 bases. Each G-C interaction forms 3 hydrogen bonds: textBonds_G-C = 7 times 3 = 21$\text{Bonds}_{G-C} = 7 \times 3 = 21$ * **Count the Adenine (A) and Thymine (T) bases:** Bases present: A_4, A_5, A_6, T_7, T_12, A_13$A_4, A_5, A_6, T_7, T_{12}, A_{13}$ implies$\implies$ Total of 6 bases. Each A-T interaction forms 2 hydrogen bonds: textBonds_A-T = 6 times 2 = 12$\text{Bonds}_{A-T} = 6 \times 2 = 12$ Summing them up yields the total hydrogen bonds in the helix: textTotal Htext-bonds = 21 + 12 = 33$\text{Total } H\text{-bonds} = 21 + 12 = 33$ ### Pattern Recognition Quick check optimization: Total Bonds = 3 times (\#G + \#C) + 2 times (\#A + \#T)$3 \times (\#G + \#C) + 2 \times (\#A + \#T)$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Biomolecules

### Core Logic Let's identify the chemical structures of the nitrogenous bases used in nucleic acids: * **(A) Adenine:** A purine derivative featuring a characteristic fused bicyclic ring system with an amino substituent at position 6 (6-aminopurine) ightarrow$ ightarrow$ **(III)**. * **(B) Cytosine:** A pyrimidine monocyclic derivative with an amino group at position 4 and a carbonyl group at position 2 (4-amino-2-oxo-pyrimidine) ightarrow$ ightarrow$ **(IV)**. * **(C) Thymine:** Found in DNA, this pyrimidine derivative features a methyl substituent at position 5 along with carbonyl groups at positions 2 and 4 (5-methyl-2,4-dioxo-pyrimidine) ightarrow$ ightarrow$ **(II)**. * **(D) Uracil:** Found in RNA, this pyrimidine derivative lacks the methyl group found in thymine, featuring just carbonyl groups at positions 2 and 4 (2,4-dioxo-pyrimidine) ightarrow$ ightarrow$ **(I)**. Matching these structures gives the sequence: (A)-(III), (B)-(IV), (C)-(II), (D)-(I). ### Step-by-Step Structural Validation The biochemical structures correspond to the following configurations:

### Pattern Recognition Quick identification keys: - Bicyclic ring = Adenine - Monocyclic ring with a -mathrmCH_3$-\mathrm{CH}_3$ group = Thymine - Monocyclic ring without a -mathrmCH_3$-\mathrm{CH}_3$ group = Uracil - Monocyclic ring with an amino group (-mathrmNH_2$-\mathrm{NH}_2$) = Cytosine ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Biomolecules