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

Solution & Explanation

### Related Formula textNumber 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 ### Pattern Recognition Combinatorics in Chemistry: If we have n unique, non-repeating amino acids, the number of linear isomeric peptides is n!. If repetition were permitted, the number of possible peptides would be n^n (which would be 4^4 = 256 in this case). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Biomolecules

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Q63 jee_main_2024_30_jan_morning Carbohydrates
Sugar which does not give reddish brown precipitate with Fehling's reagent is:
  • A. textSucrose
  • B. textLactose
  • C. textGlucose
  • D. textMaltose

Solution

### Core Logic Fehling's reagent is reduced by reducing sugars to give a reddish-brown precipitate of Cu_2O. Reducing sugars must have a free aldehyde/ketone group or a hemiacetal linkage that can open to form an aldehyde. Sucrose is a non-reducing sugar because its anomeric carbons (C1 of glucose and C2 of fructose) are tied up in a glycosidic linkage, leaving no free hemiacetal group. ### Step 1: Analyzing the options Lactose, glucose, and maltose are all reducing sugars and will give a positive Fehling's test. Sucrose does not. ### Pattern Recognition Sucrose = non-reducing sugar. Maltose, Lactose = reducing sugars. Monosaccharides (glucose, fructose) = always reducing. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Biomolecules
Q89 jee_main_2024_31_jan_evening Vitamins and their Classification
From the vitamins A, B_1, B_6, B_12, C, D, E and K, the number vitamins that can be stored in our body is ________
Numerical Answer. Answer: 5 to 5

Solution

### Core Logic Vitamins are broadly classified into two groups based on solubility: 1) Fat-soluble vitamins: Vitamins A, D, E, and K. These are stored in the liver and adipose (fat-storing) tissues. 2) Water-soluble vitamins: B group vitamins and Vitamin C. These are readily excreted in urine and cannot be stored in the body (with the exception of Vitamin B_12, which can be stored in the liver). ### Step 1: Final List The vitamins that can be stored in the body from the given list are A, D, E, K, and B_12. Total number = 5. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Biomolecules
Q78 jee_main_2024_31_jan_morning Reactions of Glucose
Match List I with List II
LIST-ILIST-II
A. Glucose/NaHCO_3/DeltaI. Gluconic acid
B. Glucose/HNO_3II. No reaction
C. Glucose/HI/DeltaIII. n-hexane
D. Glucose/Bromine waterIV. Saccharic acid
Choose the correct answer from the options given below:
  • A. textA-IV, B-I, C-III, D-II
  • B. textA-II, B-IV, C-III, D-I
  • C. textA-III, B-II, C-I, D-IV
  • D. textA-I, B-IV, C-III, D-II

Solution

### Core Logic Matching the reactions of glucose: (A) Glucose does not react with NaHCO_3, so there is no reaction. (A rightarrow II) (B) Oxidation of glucose with strong oxidizing agents like HNO_3 yields a dicarboxylic acid called saccharic acid. (B rightarrow IV) (C) Prolonged heating of glucose with HI forms n-hexane, indicating a straight chain of six carbon atoms. (C rightarrow III) (D) Oxidation with mild agents like bromine water converts glucose to gluconic acid. (D rightarrow I) ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Biomolecules

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