Match List I with List II:

List I (Bio Polymer)	List II (Monomer)
A. Starch	I. nucleotide
B. Cellulose	II. alpha$\alpha$-glucose
C. Nucleic acid	III. beta$\beta$-glucose
D. Protein	IV. alpha$\alpha$-amino acid

Choose the correct answer from the options given below :

### Core Logic A. Arginine and Tryptophan are indeed essential amino acids (cannot be synthesized by the human body). (Correct) B. Histidine contains an imidazole ring, which is a heterocyclic ring. So statement B is incorrect. C. Proline contains a pyrrolidine ring, which is a five-membered cyclic ring. So statement C is incorrect. D. Glycine is mathrmNH_2-CH_2-COOH$\mathrm{NH_2-CH_2-COOH}$. Because its alpha carbon has two identical hydrogen atoms attached, it lacks a chiral center. (Correct) E. Cysteine's side chain is a thiol group (-mathrmCH_2-SH$-\mathrm{CH_2-SH}$), not a thioether (mathrmMeS-CH_2-CH_2-$\mathrm{MeS-CH_2-CH_2-}$ which belongs to Methionine). So statement E is incorrect. Correct statements: A and D. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Biomolecules

### Related Formula textNumber of unique sequences = n!$$\text{Number of unique sequences} = n!$$ ### Core Logic A **tetrapeptide** is formed by connecting four amino acids through three peptide linkages. Since the problem specifies that complete hydrolysis of the tetrapeptide produces Gly, Ala, Val, and Leu in *equimolar proportions*, the peptide must contain exactly one molecule of each of these four distinct amino acids. ### Step 1: Calculate the Permutations The number of unique peptide sequences corresponds to the number of ways we can arrange these 4 distinct amino acids: textNumber of permutations = 4! = 4 times 3 times 2 times 1 = 24$$\text{Number of permutations} = 4! = 4 \times 3 \times 2 \times 1 = 24$$ ### Pattern Recognition Combinatorics in Chemistry: If we have n$n$ unique, non-repeating amino acids, the number of linear isomeric peptides is n!$n!$. If repetition were permitted, the number of possible peptides would be n^n$n^n$ (which would be 4^4 = 256$4^4 = 256$ in this case). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Biomolecules

### Related Formula Standard molecular weight calculation formula system: textMolar Mass = sum (textAtom Count times textAtomic Weight)$$\text{Molar Mass} = \sum (\text{Atom Count} \times \text{Atomic Weight})$$ textMass = textMoles times textMolar Mass$$\text{Mass} = \text{Moles} \times \text{Molar Mass}$$ ### Core Logic Let's tabulate and sum up the constituent atoms based on the verified structural formula layout : * Counting positions meticulously yields a molecular identity formula of mathrmC_9H_10FIN_2O_4$\mathrm{C_9H_{10}FIN_2O_4}$. * Calculating the total molar mass: textMass = (9 times 12) + (10 times 1) + 19 + 127 + (2 times 14) + (4 times 16)$$\text{Mass} = (9 \times 12) + (10 \times 1) + 19 + 127 + (2 \times 14) + (4 \times 16)$$ textMass = 108 + 10 + 19 + 127 + 28 + 64 = 372mathrm~g/mol$$\text{Mass} = 108 + 10 + 19 + 127 + 28 + 64 = 372\mathrm{~g/mol}$$

### Step 1: Final Mass Scaling Find the mass of 0.1mathrm~mol$0.1\mathrm{~mol}$ of the sample: textWeight = 0.1 times 372 = 37.2mathrm~g = 372 times 10^-1mathrm~g$$\text{Weight} = 0.1 \times 372 = 37.2\mathrm{~g} = 372 \times 10^{-1}\mathrm{~g}$$ Hence, the integer value for the blank box is 372. ### Pattern Recognition When counting atoms in nucleoside structures, remember that each vertex in the ribose sugar ring represents a fully saturated carbon component unless double bonds or heteroatoms are explicitly drawn. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Biomolecules Class 11 Chemistry: Some Basic Concepts of Chemistry

### Related Formula Dimerization of thiol pairs yields a disulfide linkage via an oxidation mechanism: mathrm2R-SH xrightarrow[-H_2] R-S-S-R$$\mathrm{2R-SH \xrightarrow{[-H_2]} R-S-S-R}$$ ### Core Logic Let's review each choice item factually layout-by-row: * Option 1 is inaccurate: Isoleucine and threonine contain two distinct chiral centers. * Option 2 is inaccurate: Glycine lacks a asymmetric center, rendering it optically inactive. * Option 3 is inaccurate: Aspartic acid similarly contains a carboxylic acid functional group along its side chain. * Option 4 is correct: Cysteine contains a highly active free functional -mathrmSH$-\mathrm{SH}$ group that easily dimerizes to form Cystine through a disulfide loop matrix. ### Pattern Recognition Disulfide bridges formed by cysteine play a crucial role in maintaining the tertiary and quaternary structural configurations of complex protein folds. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Biomolecules

### Related Formula Vitamins are classified into two categories based on solubility: - Water-soluble: Vitamin B-complex and Vitamin C. - Fat-soluble: Vitamins A, D, E, and K. ### Core Logic Categorize each given vitamin: - Vitamin mathrmB_1$\mathrm{B}_1$ (Thiamine) rightarrow$\rightarrow$ Water-soluble - Vitamin mathrmC$\mathrm{C}$ (Ascorbic acid) rightarrow$\rightarrow$ Water-soluble - Vitamin mathrmE$\mathrm{E}$ (Tocopherol) rightarrow$\rightarrow$ Fat-soluble - Vitamin mathrmB_12$\mathrm{B}_{12}$ (Cobalamin) rightarrow$\rightarrow$ Water-soluble - Vitamin mathrmK$\mathrm{K}$ (Phylloquinone) rightarrow$\rightarrow$ Fat-soluble ### Step 1: Select correct options Only C (Vitamin mathrmE$\mathrm{E}$) and E (Vitamin mathrmK$\mathrm{K}$) are fat-soluble. Thus, the correct option is (4). ### Pattern Recognition Use the mnemonic 'ADEK' to remember the fat-soluble vitamins. All other vitamins (mainly B-complex and C) are water-soluble. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Biomolecules