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

### 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

### Core Logic Statement I: Sucrose is formed by condensation of textD-(+)-glucose$\text{D-(+)-glucose}$ and textD-(-)-fructose$\text{D-(-)-fructose}$ (levorotatory fructose, not dextrorotatory as claimed). Hydrolysis of sucrose yields textD-(+)-glucose$\text{D-(+)-glucose}$ and textD-(-)-fructose$\text{D-(-)-fructose}$. Thus, Statement I is false. Statement II: Hydrolysis of dextrorotatory sucrose (+66.5^circ$+66.5^{\circ}$) yields a mixture of dextrorotatory glucose (+52.5^circ$+52.5^{\circ}$) and highly levorotatory fructose (-92.4^circ$-92.4^{\circ}$). Because the overall specific rotation of the mixture becomes levorotatory (-39.9^circ$-39.9^{\circ}$), the hydrolyzed mixture is called **invert sugar**. Thus, Statement II is true. ### Pattern Recognition Natural fructose is always levorotatory, textD-(-)-fructose$\text{D-(-)-fructose}$. Dextrorotatory fructose mentioned in Statement I is an immediate giveaway that the statement is false. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Biomolecules

### Core Logic Amino acids exhibit amphoteric behavior due to the simultaneously present basic amino (-textNH_2$-\text{NH}_2$) and acidic carboxyl (-textCOOH$-\text{COOH}$) functional groups: 1. **At pH = 2 (Highly Acidic Medium)**: The abundant concentration of hydronium ions (H^+$H^+$) protonates the carboxylate ion back into its non-ionized acid state (-textCOO^- rightarrow -textCOOH$-\text{COO}^- \rightarrow -\text{COOH}$), while the amine group remains securely protonated as an ammonium ion (-textNH_3^+$-\text{NH}_3^+$). Hence, product **A** exists purely as a **cation**. 2. **At pH = 10 (Highly Basic Medium)**: The high concentration of hydroxide ions (OH^-$OH^-$) abstracts protons from the system, deprotonating the carboxyl group into a carboxylate anion (-textCOO^-$-\text{COO}^-$) and neutralizing the ammonium group back into a free amine fraction (-textNH_2$-\text{NH}_2$). Hence, product **B** exists purely as an **anion**.

### Pattern Recognition Acidic environments (low pH) force positive overall charges onto amino structures (cation form). Basic environments (high pH) drive a net negative structure (anion form). This simple rule makes picking Option (1) immediate. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Biomolecules

### Core Logic Essential amino acids cannot be synthesized by the body and must be obtained from diet. From the given choices: * Valine (Essential) * Proline (Non-essential) * Lysine (Essential) * Threonine (Essential) * Tyrosine (Non-essential) Hence, (A), (C), and (D) are the essential amino acids. ### Pattern Recognition Mnemonic for essential amino acids: TV TILL PM MALL (Threonine, Valine, Tryptophan, Isoleucine, Leucine, Lysine, Phenylalanine, Methionine, Arginine, Histidine). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Biomolecules

### Core Logic Statement I is false because glucose pentaacetate fixes the cyclic hemiacetal system structure securely into an unreactive ester configuration. As a result, it cannot revert to an open-chain form containing a free aldehyde group, meaning it does not react with carbonyl reagents like 2,4-DNP. Statement II is true because starch, a polysaccharide composed of glucose monomer blocks, undergoes acid-catalyzed hydrolysis to yield glucose when heated under pressure. ### Pattern Recognition Sees: Pentacetate reactivity vs polysaccharide hydrolysis. Shortcut: Acetylation locks the cyclic structure of glucose, preventing reactions that require an open-chain carbonyl group (like 2,4-DNP). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Biomolecules