### 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

### Related Formula \% mathrmI = frac4 times M_mathrmIM_textThyroxine times 100$\% \mathrm{I} = \frac{4 \times M_{\mathrm{I}}}{M_{\text{Thyroxine}}} \times 100$ ### Core Logic Let's first determine the molecular formula of Thyroxine from its structure shown in below :

- Formula: mathrmC_15mathrmH_11mathrmO_4mathrmNmathrmI_4$\mathrm{C}_{15}\mathrm{H}_{11}\mathrm{O}_4\mathrm{N}\mathrm{I}_4$ Now, compute the molecular mass: - Carbon: 15 times 12 = 180$15 \times 12 = 180$ - Hydrogen: 11 times 1 = 11$11 \times 1 = 11$ - Oxygen: 4 times 16 = 64$4 \times 16 = 64$ - Nitrogen: 1 times 14 = 14$1 \times 14 = 14$ - Iodine: 4 times 127 = 508$4 \times 127 = 508$ textTotal Molecular Mass = 180 + 11 + 64 + 14 + 508 = 777 text g mol^-1$\text{Total Molecular Mass} = 180 + 11 + 64 + 14 + 508 = 777 \text{ g mol}^{-1}$ Now, calculate the percentage of Iodine: \% mathrmI = frac508777 times 100 approx 65.38 \% approx 65 \%$\% \mathrm{I} = \frac{508}{777} \times 100 \approx 65.38 \% \approx 65 \%$ Hence, the percentage of iodine in thyroxine is 65$65$. ### Pattern Recognition Thyroxine molecular mass (777$777$) is composed mostly of Iodine (508$508$), making up approximately 508/777 approx 65\%$508/777 \approx 65\%$ of its entire mass. Always carefully account for the four phenyl-bound iodines. ### Evaluation Rubric / Model Answer Accurate molecular weight summation leading to 65\%$65\%$ nearest integer Iodine composition. ### Chapter Mix Class 12 Chemistry: Biomolecules Class 12 Chemistry: Organic Chemistry - Oxygen containing compounds