Given below are two statements: Statement I: D-(+)-glucose + D-(+)-fructose -H₂O sucrose sucrose Hydrolysis D-(+)-glucose + D-(+)-fructose$\text{D-(+)-glucose} + \text{D-(+)-fructose} \xrightarrow{-\mathrm{H}_2\mathrm{O}} \text{sucrose} \qquad \text{sucrose} \xrightarrow{\text{Hydrolysis}} \text{D-(+)-glucose} + \text{D-(+)-fructose}$ Statement II: Invert sugar is formed during sucrose hydrolysis. In the light of given statements, choose the correct answer from the options given below:

Core Logic

Nucleic acids (DNA and RNA) are composed of three fundamental building blocks: nitrogenous bases, a pentose sugar, and phosphate groups. The nitrogenous bases are planar and generally achiral. The phosphate groups are also achiral. The chirality in both DNA and RNA originates strictly from their pentose sugar components, which have multiple chiral carbon centers.

Step 1: Identifying the specific sugar

DNA contains β$\beta$-D-2-deoxyribose, and RNA contains β$\beta$-D-ribose. Both of these are chiral D-sugar components.

Pattern Recognition

Any question asking about the source of chirality in nucleic acid backbones always points to the pentose sugar ring (D-sugar component).

Chapter Mix

Class 12 Chemistry: Biomolecules

Core Logic

Since we have 3 distinct amino acids and each must be used exactly once, the number of unique tripeptides corresponds to the number of permutations of these 3 items. Possible sequences:

• Gly - Ala - Val
• Gly - Val - Ala
• Val - Gly - Ala
• Val - Ala - Gly
• Ala - Val - Gly
• Ala - Gly - Val

Total tripeptides = 3! = 6$= 3! = 6$

Step 1: Final Conclusion

Total possible tripeptides without repetition is 6.

Pattern Recognition

For n$n$ distinct amino acids forming a peptide chain of length n$n$ without repetition, the number of distinct isomers is simply n!$n!$.

Chapter Mix

Class 12 Chemistry: Biomolecules Class 11 Maths: Permutations and Combinations

Core Logic

Breaking down the peptide sequence from the N-terminal (left) to C-terminal (right):

• First amino acid side chain is -CH(OH)CH₃$-CH(OH)CH_3$. This matches Threonine (Thr).
• Second amino acid side chain is -CH₂OH$-CH_2OH$. This matches Serine (Ser).
• Third amino acid side chain is -CH₂COOH$-CH_2COOH$. This matches Aspartic Acid (Asp).
• Fourth amino acid side chain is just -H$-H$. This matches Glycine (Gly).
• Fifth amino acid side chain is -CH₃$-CH_3$. This matches Alanine (Ala).

Sequence: Thr - Ser - Asp - Gly - Ala.

Step 1: Identify the Essential Amino Acid

Among the constituents (Thr, Ser, Asp, Gly, Ala), Threonine is an essential amino acid (cannot be synthesized by the human body).

Final Conclusion

Essential Amino acid (Y) = Threonine. Sequence = Thr-Ser-Asp-Gly-Ala.

Pattern Recognition

N-terminal reading is standard left-to-right. Memory aid for essential amino acids: 'PVT TIM HALL' (Threonine is the 'T').

Chapter Mix

Class 12 Chemistry: Biomolecules

Core Logic

For a sugar to be non-reducing, it must NOT possess a free hemiacetal or hemiketal group. The linkage connecting the monosaccharide units must consume both anomeric carbons via an acetal/ketal linkage. Structure (1) represents Sucrose, which has an α, β -1,2$\alpha, \beta -1,2$-glycosidic bond. Both anomeric carbons are locked in the glycosidic linkage

Step 1: Final Conclusion

Structure (1) is sucrose, which lacks a hemiacetal linkage, making it a non-reducing sugar.

Pattern Recognition

Whenever asked for a non-reducing sugar among standard disaccharides, look for Sucrose (linked at C1 of glucose and C2 of fructose). Both anomeric carbons are tied up.

Chapter Mix

Class 12 Chemistry: Biomolecules

Related Formula

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:

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 nⁿ$n^n$ (which would be 4⁴ = 256$4^4 = 256$ in this case).

Chapter Mix

Class 12 Chemistry: Biomolecules