//M0//QN1//CS//DL0

Carbohydrates are optically active polyhydroxy aldehydes and ketones. They are also called saccharides. All those carbohydrates which reduce Fehling’s solution and Tollen’s reagent are referred to as reducing sugars. Glucose, the most important source of energy for mammals, is obtained by the hydrolysis of starch. Vitamins are necessary food factors required in the diet. Proteins are the polymers of α-amino acids and perform various structural and dynamic functions in the organisms. Deficiency of vitamins leads to many diseases.(a) What are reducing sugars?(b) Why are carbohydrates considered optically active, and what is their structural significance?(c) What are proteins made of?(d) Why is vitamin C not stored in our body?

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(a) Reducing sugars are carbohydrates that reduce Fehling’s solution and Tollen’s reagent.
(b) Carbohydrates are optically active due to the presence of chiral carbon atoms, and their polyhydroxy aldehyde or ketone structure enables energy production and other biological functions
(c) Proteins are polymers of α–amino acids.
(d) Vitamin C cannot be stored in our body because it is water-soluble and is excreted in urine.

//M0//QN2//CS//DL0//EQ

Pentoses and hexoses undergo intramolecular hemiacetal or hemiketal formation due to the combination of the hydroxyl group (-OH) with the carbonyl group. The resulting structure is either a five- or six-membered ring containing an oxygen atom. In their free state, all pentoses and hexoses exist predominantly in the pyranose form (resembling pyran). However, in the combined state, some of them exist as five-membered cyclic structures called furanose. The cyclic structure of glucose is represented by the Haworth projection. α–D–glucose and β–D–glucose differ in the configuration at the anomeric (C–1) carbon atom and are therefore called anomers. The C–1 carbon atom is referred to as the anomeric carbon. The six-membered cyclic structure of glucose is known as the pyranose structure.(a) What causes pentoses and hexoses to form cyclic structures?(b) What is the difference between pyranose and furanose structures?(c) What are anomers, and how do α and β-D-glucose differ?(d) In the structure of α-D-glucose, which carbon is the anomeric carbon?

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(a) Pentoses and hexoses form cyclic structures due to intramolecular hemiacetal or hemiketal formation involving the hydroxyl and carbonyl groups.
(b) Pyranose structures are six-membered rings resembling pyran, while furanose structures are five-membered rings resembling furan.
(c) Anomers are isomers that differ at the anomeric (C–1) carbon α and β–D–glucose have different configurations at this carbon.
(d) The anomeric carbon is the C–1 carbon, which is bonded to the oxygen atom in the ring and determines the α or β configuration.

//M0//QN3//CS//DL0

When a protein in its native state is exposed to physica l factors such as temperature changes or chemical factors like pH alterations, its hydrogen bonds are disrupted. As a result, the protein’s globular structure unfolds, its helix uncoils, and it loses its biological activity. This process is known as denaturation of protein. During denaturation, the secondary and tertiary structures are altered, but the primary structure remains unchanged. Examples of protein denaturation include the coagulation of egg white upon boiling, the curdling of milk, and the formation of cheese when acid is added to milk.(a) What is protein denaturation?(b) What causes denaturation of proteins?(c) Which structures of a protein are affected during denaturation?(d) Give examples of protein denaturation in daily life.

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(a) Protein denaturation is the loss of biological activity due to disruption of hydrogen bonds, altering secondary and tertiary structures but not the primary structure.
(b) Denaturation is caused by physical factors like temperature changes or chemical factors such as pH changes.
(c) Denaturation affects the secondary and tertiary structures, but the primary structure remains unchanged.
(d) Examples include the coagulation of egg white upon boiling, curdling of milk, and cheese formation when acid is added to milk. (Globular proteins are converted into fibrous proteins and Fibrous proteins are converted into globular protein.

//M0//QN4//CS//DL0//EQ

A medical research insitute is studying how carbohydrates, protein and enzymes behave inside human cells. To understand metabolic pathways, Scientist analyze.

Structure of monosaccarides.
Mutarotation
Glycosidic bond formation
Denaturation and struction of protein Enzyme specificity and inhibition.
During experimentation : 1. An aqueous glucose solution rotates plane - polarised light and its rotation changes with time → mutarotation. 2. Hydrolysis of sucrose gives glucose and fructose. 3. Heating egg protein leads to denaturation. 4. Enzyme show heighest activity at optimum pH and temperature. 5. A competitive inhibitor reduces enzyme activity by occupying the active site. Additional data :
Glucose exists mostly in cyclic hemiacetal form.
Proteins contain peptide bonds (–Co–NH–)
Enzymes are hihgly specific to substrates.
(a) Explain why glucose shows mutarotation in aqueous solution.(b) Why is sucrose a non-reducing sugar even through it contains glucose and fructose ?(c) How does denaturation affect the biological activity of proteins ?(d) Explain the difference between competitive and non-competitive. enzyme inhibition.

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(a) In water, glucose interconverts between α - glucose and β - glucose.
Th is happens via open-chain aldehyle form, causing a change in specific rotation until equilibrium is reached.
(b) In sucrose, the anomeric carbon of glucose
(1) and fructose (2) are both involved in glycosidic bonding.
Thus, neither ring can open → no free aldehyde or keto group → cannot reduce Tollen’s or fehling’s reagent.
(c) Denaturation breaks, hydrogen bonds, hydrophobic interactions and salt bridges. So, protein losses tertiary / Secondary structure, Active site destroyed and loss of biological function.
(d)

//M0//QN5//CS//DL0

A biotechnology firm studies nucleic acids, vitamins, and amino acids to design new nutritional supplements : They analyze : Structure of DNA - RNA, Base pairing rules, essential vs non-essential amino acids, vitamin deficiencies and hydrolysis of nucleotides. Key Observations : 1. DNA contain A, T, G, C, where as RNA contain A, V, G, C 2. Adenine paires with thymine by two H-bonds; guanine pairs with cytosine by 3 H-bonas 3. Vit. C deficiency → Seurvy 4. Thiamin deficiency → beriberi 5. Hydrolysis of RNA gives a mixture of ribose, phosphate and nitrogenous bases.

Amino acid studid :
Amino acid pl Type
Glycine 6.0 Neutral
Lysin 9.7 Basic
Glutamic Acid 3.2 Acidic
(a) Why is DNA more stable than RNA ? Explain using base pairing and sugar structure.(b) What products are obtained on complete hydrolysis of RNA(c) Arrang the amino acid in increasing order of isoelectric points (pl)(d) Explain why glutamic acid behaves as an acidic amino acid.

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(a) Reason,
(i) DNA contain deoxyribose, lacking 2’ - OH (less reactive)
(ii) RNA has 2’ - OH (prone to hydrolysis)
(iii) G ≡ C pairing in DNA (stronger)
(iv) DNA is double stranded
(b) Ribose sugar, Nitrognous bases (A, V, G, C) and phosphoric Acid.
(c) Increasing pl order :
Glutamic acid (3.2) < Glycine (6.0)
(d) Glutamic Acid is acidic, because of...
(i) it contain an extra –COOH group
(ii) Losses H+ easily
(iii) forms stable carboxylate ion.
Carbohydrate: Polyhydroxy aldehyde or ketone, or a compound that yields them on hydrolysis.
Monosaccharide: The simplest carbohydrate unit that cannot be hydrolysed further (e.g., glucose, fructose).
Disaccharide: Carbohydrate made of two monosaccharides linked by a glycosidic bond (e.g., sucrose, lactose).
Polysaccharide: Long-chain polymer of monosaccharide units (e.g., starch, glycogen, cellulose).
Reducing Sugar: Sugar capable of reducing Tollens’/Fehling’s reagent due to free aldehyde or hemiacetal group.
Glycosidic Bond: Covalent linkage between the anomeric carbon of one sugar and the –OH of another.
Anomers: Isomers differing at the anomeric carbon (α- and β-forms) in cyclic sugars.
Amino Acid: Organic molecule containing –NH2
and –COOH groups on the same (α) carbon.
Peptide Bond: –CO–NH– linkage formed between two amino acids during protein formation.
Protein Structure: Primary (sequence), secondary (α-helix/β-sheet), tertiary (3D folding), and quaternary (multi-chain assembly).
Denaturation: Loss of protein structure and function due to heat, pH, or chemical action.
Enzyme: Biological catalyst (mostly proteins) that speeds up biochemical reactions with high specificity.
Cofactor / Coenzyme: Non-protein component required for enzyme activity (metal ion or organic molecule).
Vitamin: Essential micronutrient needed in small amounts; classified as water-soluble (B, C) or fat-soluble (A, D, E, K).
Hormone: Chemical messenger produced by endocrine glands that regulates physiological processes.
Nucleoside: Molecule consisting of a nitrogenous base attached to a pentose sugar (no phosphate).
Nucleotide: Phosphate + sugar + nitrogenous base; basic unit of DNA and RNA.
DNA: Double-stranded nucleic acid carrying genetic information; shows A–T and G–C base pairing.
RNA: Single-stranded nucleic acid involved in protein synthesis; contains uracil instead of thymine.