A tetrapeptide "x" on complete hydrolysis produced glycine (Gly), alanine (Ala), valine (Val), leucine (Leu) in equimolar proportion each. The number of tetrapeptides (sequences) possible involving each of these amino acids is

### Core Logic To assign D or L configurations to ketohexoses like fructose, locate the configuration at the highest-numbered asymmetric carbon atom (carbon-5). * In D-fructose, the -textOH$-\text{OH}$ group on C-5 is situated on the \right side in the Fischer projection. * In L-fructose, the configuration of every stereocenter is inverted compared to D-fructose, meaning the -textOH$-\text{OH}$ group on C-5 is situated on the \left hand side.

### Pattern Recognition Shortcut: L-sugar means the stereocenter farthest from the carbonyl group (C5 for hexoses) has its -textOH$-\text{OH}$ pointing \left. Ensure it is the perfect non-superimposable mirror image of standard D-fructose. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Biomolecules

### 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

### Related Formula textProtein xrightarrowtextHydrolysis textPeptides xrightarrowtextHydrolysis alphatext-amino acids$$\text{Protein} \xrightarrow{\text{Hydrolysis}} \text{Peptides} \xrightarrow{\text{Hydrolysis}} \alpha\text{-amino acids}$$ ### Core Logic Statement (I) is incorrect because the complete hydrolysis of both oligopeptides and proteins yields alpha$\alpha$-amino acids, not beta$\beta$-amino acids. Statement (II) is incorrect because fibrous proteins are inherently water-insoluble structural materials. Denaturation typically disrupts the tertiary and secondary structures of water-soluble globular proteins, making them insoluble. ### Step 1: Final Conclusion Since both Statement I and Statement II are false, option (3) is the correct choice. ### Pattern Recognition All naturally occurring proteins are polymers of alpha$\alpha$-amino acids, so any statement mentioning beta$\beta$-amino acids as direct translation products can be confidently ruled out. Fibrous proteins (like keratin or collagen) are structural and always insoluble, unlike globular proteins. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Biomolecules