CBSE Class 12 Chemistry Biomolecules Notes and Questions

Bio molecule

Gist Of The Chapter

1.Carbohydrates- These are optically active polyhydroxy  aldehydes or ketones due to presence of chiral `C’ or the compounds which produce these on hydrolysis except dihydroxy acetone is not optically active.

2. Classification-

(i) Monosaccharide’s – Those carbohydrates which cannot get hydrolysed e.g. glucose ,fructose, galactose etc.

(ii) Oligosaccharides- Those carbohydrates which give to or more monosaccharide’s on hydrolysis e.g. sucrose on hydrolysis gives glucose and fructose. Raffinose on hydrolysis gives glucose, fructose and galactose.

(iii) Polysaccharides- Those carbohydrates which on hydrolysis give large number of monosaccharide’s hydrolysis.eg  starch, cellulose, glycogen.

3. Sugar-

(i) Reducing Sugars- Those which reduce Fehling’s or Tollen’s reagent. They have free aldehydic groups, eg , glucose, fructose , galactose

(ii) Non Reducing Sugars- Those which do not reduce Fehling’s or Tollen’s reagent. They do not have free functional group ,e.g.,  sucrose

4. Glucose- It is a monosaccharide’s with molecular formula C₆H₁₂O₆

5. Preparation

(i) From Sucrose

     C₁₂H₂₂O₁₁ + H₂O ------->  C₆H₁₂O₆ + C₆H₁₂O₆ ( Only from sucrose)                                        

(ii)From Starch

(C₆H₁₀O₅)_n + nH₂O -------> C₁₂H₂₂0₁₁ + H₂0 ------> 2C₆H₁₂O₆

6.Structure

(i)Fischer structure –

CHO — [CHOH] — CH2OH

[+] Glucose has `D’ configuration as shown

`D’ means —OHgroup on first chiral `C’ from the bottom is on right hand and + means it is dextro rotator, i.e, it rotates plane polarized light towards right.

(ii) Cyclic Structure OF Glucose: the straight chain is unable to explain the following reactions.

(a) It does not give the 2, 4‐DNP test, Schiff’s Test and does not form the hydrogensulphide product with NaHSO3 .

(b) The pentacetate of glucose does not react with NH2OH, indicating the absence of free aldehydic group.

(iii) Glucose exist in 2 different crystalline forms α and β forms. These are called anomers. They differ in optical rotation, they also differ in melting point.

Anomers are isomers which have a different configuration across C‐1 (first chiral ‘C’ atom).

7. Glycosidic Linkage:The linkage between two monosaccharide units through oxygen is called the glycosidic linkage.

Biomolecules

VSA Type Questions - (1 Mark)

Q1. Which functional groups are present in monosaccharides?
Answer: Monosaccharides contain hydroxyl (\(-\text{OH}\)) groups along with either an aldehydic (\(-\text{CHO}\)) group or a ketonic (\(-\text{CO}-\)) group.
In simple words: They contain alcohol groups combined with either an aldehyde or a ketone group.

Exam Tip: Remember that aldoses contain \(-\text{CHO}\) while ketoses contain \(\ge \text{C}=\text{O}\).

Q2. Name an aldopentone, aldohexone and ketohexone.
Answer: Aldopentose: Ribose
Aldohexose: Glucose
Ketohexose: Fructose
In simple words: Ribose has 5 carbons with an aldehyde, glucose has 6 carbons with an aldehyde, and fructose has 6 carbons with a ketone.

Exam Tip: Pay attention to the prefix 'aldo-' (aldehyde group) and 'keto-' (ketone group), and the suffix '-pentose' (5 carbons) and '-hexose' (6 carbons).

Q3. What is animal starch?
Answer: Glycogen is known as animal starch because its structure is similar to amylopectin and it is stored in animal tissues.
In simple words: Glycogen is the stored carbohydrate in animals, much like starch in plants.

Q4. Which types of bonds are present in a protein molecule?
Answer: Protein molecules contain peptide bonds, hydrogen bonds, disulphide linkages, ionic bonds, and van der Waals forces.
In simple words: Proteins are held together by strong peptide bonds in their backbone, along with weaker bonds that maintain their 3D shape.

Exam Tip: Peptide bonds link amino acids in the primary structure, while hydrogen bonds stabilize the secondary and tertiary structures.

Q5. Which \(\alpha\)-helix or \(\beta\)-helix is more stable?
Answer: The \(\alpha\)-helix is right-handed and is more stable due to intramolecular hydrogen bonding between the carbonyl oxygen of the first amino acid and the amino hydrogen of the fourth amino acid.
In simple words: The \(\alpha\)-helix forms a stable coiled structure locked by internal hydrogen bonds.

Exam Tip: Specify that \(\alpha\)-helix relies on internal hydrogen bonding within the same chain.

Q6. The sequence of bases in one strand of DNA is TACGGACA. What is the sequence of bases of complementary strand of DNA.
Answer: According to Chargaff's base pairing rules (\(\text{A}\) pairs with \(\text{T}\), and \(\text{G}\) pairs with \(\text{C}\)), the complementary sequence is ATGCCTGT.
In simple words: Swap every A with T, T with A, G with C, and C with G along the strand.

Exam Tip: Always remember the base pairing rule: Adenine pairs with Thymine, and Guanine pairs with Cytosine.

Q7. Name the vitamin whose deficiency causes rickets?
Answer: Vitamin D.
In simple words: A lack of Vitamin D leads to weakened bones, known as rickets in children.

Exam Tip: Vitamin D is a fat-soluble vitamin crucial for calcium absorption.

Q8. Name the purines present in DNA.
Answer: Adenine and guanine.
In simple words: The two purine bases in DNA are adenine and guanine.

Exam Tip: Purines are double-ring nitrogenous bases, whereas pyrimidines (thymine and cytosine in DNA) are single-ringed.

Q9. Give an example of (a) water soluble (b) fat soluble is
Answer: (a) Water soluble: Vitamin C
(b) Fat soluble: Vitamin D
In simple words: Vitamin C dissolves in water, while Vitamin D dissolves in fats.

Exam Tip: B-complex and Vitamin C are water-soluble; Vitamins A, D, E, and K are fat-soluble.

Q10. Name a protein which is insoluble in water.
Answer: Keratin (or Fibroin).
In simple words: Fibrous proteins like keratin found in hair and nails do not dissolve in water.

Exam Tip: Distinguish between fibrous (insoluble) and globular (soluble) proteins.

SA I Type Questions

Q1. Name polysaccharides that make up starch and what is the difference between them.
Answer: Starch is made of amylose and amylopectin.
Amylose is a linear polymer of \(\alpha\)-D-glucose units linked by \(\alpha\text{-}(1 \to 4)\)-glycosidic bonds and is water-soluble.
Amylopectin is a branched polymer of \(\alpha\)-glucose with \(\alpha\text{-}(1 \to 4)\)-glycosidic linkages in the chain and \(\alpha\text{-}(1 \to 6)\)-glycosidic linkages at branching points, and it is water-insoluble.
In simple words: Amylose is straight-chain and soluble, while amylopectin is branched and insoluble.

Exam Tip: Mention both solubility and branching differences to secure full marks.

Q2. What are anomers?
Answer: Monosaccharides that differ in configuration only at the \(\text{C}-1\) carbon atom (the hemiacetal carbon) are called anomers, e.g., \(\alpha\)-glucose and \(\beta\)-glucose.
In simple words: Anomers are sugar isomers that differ only in the orientation of the hydroxyl group at the first carbon.

Exam Tip: Always associate anomers specifically with the configuration at \(\text{C}-1\) for aldoses.

Q3. Where does the water present in the egg go after boiling the egg?
Answer: On boiling, protein undergoes denaturation. The globular proteins unfold and coagulate, and free water molecules get adsorbed or trapped within the denatured protein network.
In simple words: Boiling denatures egg proteins, causing them to trap the water inside a solid gel matrix.

Exam Tip: Link this question directly to the process of protein denaturation.

Q4. Write two main functions of carbohydrates in plants.
Answer: (i) Structural material: Cellulose forms the plant cell wall.
(ii) Reserved food material: Starch serves as energy storage in plants.
In simple words: Carbohydrates build plant cell walls and store energy.

Q5. What do you understand by glycosidic linkage?
Answer: The oxide linkage formed by the loss of a water molecule when two monosaccharide units condense through their hydroxyl groups is called a glycosidic linkage (\(-\text{O}-\)).
In simple words: It is the oxygen bridge joining two sugar units together.

Exam Tip: Mention the elimination of water during condensation.

Q6. What are essential and non essential amino acid? Give two examples of each type.
Answer: Essential amino acids: Amino acids that cannot be synthesized in the human body and must be obtained through diet. Example: Valine, Leucine.
Non-essential amino acids: Amino acids that can be synthesized by the human body. Example: Glycine, Alanine.
In simple words: Essential ones must come from food; non-essential ones are made by our body.

Exam Tip: Clearly distinguish based on whether the body synthesizes them or not.

Q7. How do you explain the amphoteric behavior of amino acids?
Answer: Amino acids contain both an acidic carboxyl group (\(-\text{COOH}\)) and a basic amino group (\(-\text{NH}_2\)). In aqueous solution, the carboxyl group can lose a proton and the amino group can accept a proton, forming a zwitterion. Thus, they react with both acids and bases.
In simple words: Because they have both acid and base parts in the same molecule, they can react with either acids or bases.

Exam Tip: Mention the zwitterionic form to explain amphoteric nature clearly.

Q8. What is the difference between a nucleoside and a nucleotide?
Answer: Nucleoside = Pentose Sugar + Nitrogenous Base
Nucleotide = Pentose Sugar + Nitrogenous Base + Phosphoric Acid Group
In simple words: A nucleoside lacks the phosphate group that a nucleotide has.

Exam Tip: Equation-based answers work best for this definition.

Q9. Define (a) Enzymes (b) Antibody
Answer: (a) Enzymes: They are biological catalysts (usually globular proteins) that speed up biochemical reactions in living systems. For example, invertase catalyzes the hydrolysis of sucrose into glucose and fructose:
\(\text{C}_{12}\text{H}_{22}\text{O}_{11} + \text{H}_2\text{O} \xrightarrow{\text{Invertase}} \text{C}_6\text{H}_{12}\text{O}_6 + \text{C}_6\text{H}_{12}\text{O}_6\)
(b) Antibody: They are specialized protein molecules (immunoglobulins) produced in the body in response to antigens to destroy invading pathogens or foreign substances.
In simple words: Enzymes speed up body reactions, while antibodies fight off infections.

Exam Tip: Include the chemical reaction equation for invertase when defining enzymes.

Q10. What is invert sugar?
Answer: An equimolar aqueous mixture of D-(+)-glucose and D-(-)-fructose obtained by the hydrolysis of sucrose is called invert sugar.
In simple words: A 1:1 mixture of glucose and fructose formed from sucrose hydrolysis.

Exam Tip: Mention that the optical rotation inverts from dextrorotatory to laevorotatory during this process.

SA II Type Questions

Q1. Give three differences between DNA and RNA.
Answer:

In simple words: DNA stores genetic data with deoxyribose and thymine, while RNA helps make proteins using ribose and uracil.

Exam Tip: Tabular comparisons are preferred for clarity and full credit.

Q2. Difference between globular protein and fibrous protein.
Answer:

In simple words: Globular proteins are ball-shaped and soluble; fibrous proteins are thread-like and insoluble.

Exam Tip: Mention shape and water solubility as primary differentiating points.

Q3. Give reactions with support cyclic structure of glucose.
Answer: (a) Glucose does not give the 2,4-DNP test, Schiff's test, or form a hydrogensulphite addition compound with \(\text{NaHSO}_3\), proving the absence of a free aldehydic group in its open-chain form.
(b) The pentaacetate of glucose does not react with hydroxylamine (\(\text{NH}_2\text{OH}\)), confirming that the free \(-\text{CHO}\) group is absent.
(c) Glucose exists in two distinct crystalline anomeric forms (\(\alpha\) and \(\beta\)), which can only be explained by a cyclic hemiacetal structure.
In simple words: Certain typical aldehyde tests fail for glucose because its aldehyde group is tied up in a ring structure.

Exam Tip: Mention the failure of Schiff's test and 2,4-DNP test as key evidence.

Q4. Define with example (a) Isoelectric point (b) Mutarotation (c) Transcription
Answer: (a) Isoelectric point: The specific \(\text{pH}\) at which amino acids exist as zwitterions and show no net migration toward any electrode during electrolysis. Example: Most amino acids have an isoelectric point around \(\text{pH} = 5.5 - 6.3\).
(b) Mutarotation: The spontaneous change in the specific optical rotation of an optically active compound in aqueous solution over time until equilibrium is reached. Example: \(\alpha\)-D-glucose dissolved in water changes its optical rotation from \(+111^\circ\) to \(+52.5^\circ\).
(c) Transcription: The biochemical process by which genetic information is copied from one strand of DNA into messenger RNA (\(\text{m-RNA}\)). Example: If the DNA template has sequence ATACA, the transcribed \(\text{m-RNA}\) has sequence UAUGU (or complementary rules applied).
In simple words: Isoelectric point is the pH of zero charge; mutarotation is the shifting of optical rotation; transcription is copying DNA to RNA.

Exam Tip: Provide clear examples for each definition to score full marks.

Q5. What happens when glucose reacts with (a) \(\text{HI}\) (b) \(\text{HNO}_3\) (c) \(\text{Br}_2\) water
Answer:
(a) With concentrated \(\text{HI}\) on prolonged heating, glucose reduces to n-hexane:
\(\text{C}_6\text{H}_{12}\text{O}_6 + \text{HI} \xrightarrow{\text{heat}} \text{CH}_3-(\text{CH}_2)_4-\text{CH}_3\ (\text{n-hexane})\)
(b) With nitric acid (\(\text{HNO}_3\)), glucose is oxidized to saccharic acid (glucaric acid):
\(\text{C}_6\text{H}_{12}\text{O}_6 + \text{HNO}_3 \longrightarrow \text{COOH}-(\text{CHOH})_4-\text{COOH}\ (\text{saccharic acid})\)
(c) With bromine water (\(\text{Br}_2\) water), glucose is selectively oxidized to gluconic acid:
\(\text{C}_6\text{H}_{12}\text{O}_6 + \text{Br}_2\text{ water} \longrightarrow \text{COOH}-(\text{CHOH})_4-\text{CH}_2\text{OH}\ (\text{gluconic acid})\)
In simple words: HI gives straight-chain hexane, nitric acid oxidizes both ends to dicarboxylic acid, and bromine water oxidizes only the aldehyde end.

Exam Tip: Remember that \(\text{Br}_2\) water is a mild oxidizing agent that converts only the aldehyde group of glucose into a carboxylic acid group.

Q6. Differentiate primary, secondary and tertiary structure of protein.
Answer:
- Primary structure refers to the specific sequence of amino acids linked together by covalent peptide bonds in a polypeptide chain.
- Secondary structure refers to the local folded shape assumed by the polypeptide chain (such as \(\alpha\)-helix and \(\beta\)-pleated sheets) stabilized by hydrogen bonding between peptide bonds.
- Tertiary structure represents the overall 3D folding of the entire polypeptide chain, giving rise to fibrous or globular molecular shapes stabilized by hydrogen bonds, disulphide linkages, and hydrophobic interactions.
In simple words: Primary is the amino acid sequence, secondary is local coiling, and tertiary is the overall 3D fold.

Exam Tip: Emphasize the type of bonding and spatial arrangement for each structural level.

Q7. Discuss the specificity and mechanism of enzyme action.
Answer: Enzymes are highly specific catalysts. According to the Lock and Key Model, the enzyme has an active site configured to bind to a specific substrate. The mechanism involves the following steps:
Step (i): Binding of substrate (\(\text{S}\)) to enzyme (\(\text{E}\)) to form an enzyme-substrate complex:
\(\text{E} + \text{S} \longrightarrow [\text{ES}]\) (\text{Enzyme-substrate complex})\)
Step (ii): Conversion of substrate into product within the complex:
\([\text{ES}] \longrightarrow [\text{EP}]\) (\text{Enzyme-product complex})\)
Step (iii): Dissociation of the complex to release the product and regenerate the unchanged enzyme:
\([\text{EP}] \longrightarrow \text{E} + \text{P}\)
In simple words: Enzymes act like locks and keys; they temporarily bind to specific substrates, convert them into products, and release them unchanged.

Q8. Mention structural differences between amylopectin and cellulose.
Answer:

In simple words: Amylopectin is branched \(\alpha\)-glucose, while cellulose is linear \(\beta\)-glucose.

Exam Tip: Highlight the difference between \(\alpha\)-glucose and \(\beta\)-glucose units.

Q9. What deficiency diseases are caused due to lack of vitamins \(\text{B}_1\), \(\text{B}_6\) and \(\text{K}\) in human diet.
Answer:

In simple words: Vitamin B1 causes beriberi, B6 causes convulsions, and K causes bleeding disorders.

Exam Tip: Memorize vitamin deficiency disorders as they frequently appear in exams.

Q10. Glucose or Sucrose are soluble in water but cyclohexane and benzene are insoluble in water. Explain.
Answer: Glucose contains five \(-\text{OH}\) groups and sucrose contains eight \(-\text{OH}\) groups. These hydroxyl groups form extensive intermolecular hydrogen bonding with polar water molecules, making them highly soluble. In contrast, benzene and cyclohexane are non-polar hydrocarbons that lack \(-\text{OH}\) groups and cannot form hydrogen bonds with water, making them insoluble.
In simple words: Like dissolves like; sugar molecules form hydrogen bonds with water, whereas benzene and cyclohexane are non-polar and hydrophobic.

Exam Tip: Always mention hydrogen bonding with water molecules as the key factor for solubility.

HOTS Questions

VSA (1 Mark)

Q1. How many atoms are present in the ring of pyranose structure of glucose?
Answer: 6 atoms (5 Carbon atoms and 1 Oxygen atom).
In simple words: The pyranose ring is a six-membered ring containing five carbons and one oxygen.

Exam Tip: Do not confuse pyranose (6-membered ring) with furanose (5-membered ring).

Q2. Write the formula of Zwitter ion for Glycine.
Answer: \(\text{H}_3\text{N}^+-\text{CH}_2-\text{COO}^-\)
In simple words: Glycine in zwitterionic form has a positive charge on the amino group and a negative charge on the carboxyl group.

Exam Tip: For glycine, \(\text{R} = -\text{H}\) in the general amino acid formula.

Q3. Which proteins possess \(\alpha\)-Helix structure?
Answer: Keratin and myosin.
In simple words: Hair and muscle proteins commonly contain the \(\alpha\)-helix structure.

Exam Tip: Remember that fibrous structural proteins like keratin and hair proteins are classic examples.

Q4. What is the native state of protein?
Answer: The energetically most stable 3D conformation of a protein under physiological \(\text{pH}\) and temperature is called its native state.
In simple words: The natural, active, folded shape of a protein in the body is its native state.

Exam Tip: Denaturation destroys this native state.

Q5. Fresh tomatoes are a better source of Vitamin C that which have been stored for some time. Explain.
Answer: Vitamin C (ascorbic acid) is readily oxidized by atmospheric oxygen over time, leading to its destruction. Therefore, stored tomatoes have lower Vitamin C content than fresh ones.
In simple words: Vitamin C gets destroyed by air oxidation when stored.

Exam Tip: Mention oxidation by air as the reason for vitamin degradation.

Q6. Why are carbohydrates generally optically active?
Answer: It is due to the presence of one or more chiral carbon (asymmetric carbon) atoms in their molecular structures.
In simple words: They have asymmetric carbon atoms that rotate plane-polarized light.

Exam Tip: Chiral carbon atoms are the fundamental reason for optical activity.

Q7. What type of linkages hold together monomers in DNA?
Answer: Phosphodiester linkages.
In simple words: Nucleotides in a DNA strand are connected by phosphodiester bonds.

Q8. Why is cellulose not digested in human body?
Answer: It is due to the fact that human digestive tracts lack the specific enzyme (\(\beta\)-glucosidase) required to hydrolyze the \(\beta\text{-}(1 \to 4)\)-glycosidic linkages present in cellulose.
In simple words: Humans lack the enzyme needed to break down cellulose.

Exam Tip: Highlight the absence of the specific enzyme for \(\beta\)-linkages.

Q9. Name the enzyme that is used to dissolve blood clots?
Answer: Streptokinase.
In simple words: Streptokinase is a medical enzyme used to clear blood clots.

Q10. Name two diseases caused due to deficiency of enzymes.
Answer: Albinism and phenylketonuria.
In simple words: Genetic enzyme deficiencies can lead to metabolic disorders like albinism and phenylketonuria.

Exam Tip: These are classic inborn errors of metabolism caused by missing enzymes.

SA Type I (2 Marks)

Q1. Give reasons for the following - (i) On electrolysis in acidic solution amino acids migrate towards cathode, while in alkaline solution these migrate towards anode. (ii) The monoamino monocarboxylic acids have two \(\text{p}K_a\) values.
Answer: (i) In acidic solution, the carboxylate ion accepts a proton to form a cationic species (\(\text{H}_3\text{N}^+-\text{CHR}-\text{COOH}\)), which migrates towards the negatively charged cathode. In alkaline solution, the ammonium ion loses a proton to form an anionic species (\(\text{H}_2\text{N}-\text{CHR}-\text{COO}^-\)), which migrates towards the anode.
(ii) Monoamino monocarboxylic acids have two dissociable functional groups: a carboxyl group (\(-\text{COOH}\)) and an amino group (\(-\text{NH}_3^+\)). Hence, they show two distinct \(\text{p}K_a\) values corresponding to the stepwise dissociation of protons.
In simple words: Amino acids carry positive charge in acid (moving to cathode) and negative charge in base (moving to anode); their two functional groups give two pKa values.

Exam Tip: Explain the ionic forms in acidic and basic mediums clearly.

Q2. Which forces are responsible for the stability of \(\alpha\)-helix? Why is it named as \(3.6_{13}\) helix?
Answer: Hydrogen bonds formed between the \(-\text{NH}\) and \(\text{C}=\text{O}\) groups of peptide bonds give stability to the \(\alpha\)-helix structure. It is named \(3.6_{13}\) helix because each turn of the helix contains approximately 3.6 amino acid residues, and a 13-membered ring is formed by hydrogen bonding.
In simple words: Internal hydrogen bonds stabilize the coil, and 3.6 amino acids make up one full turn containing a 13-atom ring loop.

Exam Tip: Break down both parts of the name: 3.6 residues per turn and 13 atoms in the hydrogen-bonded ring.

Q3. Write about the following protein synthesis - (i) Name the location where the protein synthesis occurs? (ii) How do 64 codones code for only 20 amino acids?
Answer: (i) Protein synthesis occurs at the ribosomes located in the cytoplasm.
(ii) The genetic code is degenerate; more than one codon can code for the same amino acid. For example, leucine and proline are each encoded by multiple codons (e.g., proline is encoded by CCU, CCA, CCG, and CCC).
In simple words: Proteins are made at the ribosomes, and multiple codons can specify the same amino acid.

Exam Tip: Mention ribosome location and the degeneracy of the genetic code.

Q4. Describe the mechanism of replication of DNA.
Answer: DNA replication is the process by which a DNA molecule makes identical copies of itself. The two strands of the double helix partially unwind, and each strand acts as a template for synthesizing a new complementary strand following base-pairing rules (A with T, G with C). This is a semi-conservative process where each daughter molecule retains one original parent strand and one newly synthesized strand. Replication proceeds exclusively in the \(5' \to 3'\) direction.
In simple words: DNA strands unzip, and each acts as a template to build a new matching strand, resulting in two identical DNA copies.

Exam Tip: Emphasize semi-conservative nature and \(5' \to 3'\) synthesis direction.

Q5. Answer the following queries about proteins - (i) How are proteins related to amino acids? (ii) When is protein said to be denatured?
Answer: (i) Proteins are biopolymers made up of a large number of amino acids linked together by peptide bonds in a specific 3D spatial arrangement.
(ii) A protein is said to be denatured when its native 3D structure is disrupted by physical changes (like heat, \(\text{pH}\) changes) causing uncoiling, loss of biological activity, and precipitation while retaining its primary structure.
In simple words: Proteins are polymers of amino acids, and denaturation is the uncoiling of their 3D shape by heat or acid.

Exam Tip: Clarify that denaturation disrupts secondary and tertiary structures while leaving the primary structure intact.

SA(II) - 3 Marks

Q6. Draw simple Fischer projections of D and L- glucose. Are these enantiomers?
Answer:
D-Glucose Fischer Projection:
\(\text{CHO} - (\text{CHOH})_4 - \text{CH}_2\text{OH}\) (with \(-\text{OH}\) at \(\text{C}-5\) on the right).
L-Glucose Fischer Projection:
\(\text{CHO} - (\text{CHOH})_4 - \text{CH}_2\text{OH}\) (with \(-\text{OH}\) at \(\text{C}-5\) on the left).
Yes, these two Fischer projections are non-superimposable mirror images of each other, and therefore they are enantiomers.
In simple words: D and L-glucose are mirror-image isomers called enantiomers.

Exam Tip: Ensure all chiral centers are inverted when drawing D and L pairs.

Q7. A tripeptide on complete hydrolysis gives glycine, alanine and phenylalanine using three letter symbols write down the possible sequence of tripeptide.
Answer: Since all three amino acids are distinct, the number of possible linear sequences is \(3! = 6\):
(i) Gly-Ala-Phe
(ii) Gly-Phe-Ala
(iii) Ala-Gly-Phe
(iv) Ala-Phe-Gly
(v) Phe-Ala-Gly
(vi) Phe-Gly-Ala
In simple words: With 3 different amino acids, there are 6 possible arrangements for the tripeptide.

Exam Tip: Use mathematical permutation (\(n!\)) to verify you have listed all possible sequences.

Q8. Glycine exists as a Zwitter ion but o- and p-amino benzoic acids do not. Explain.
Answer: In o- and p-aminobenzoic acids, the lone pair of electrons on the nitrogen atom of the amino group is involved in resonance with the benzene ring, making it less available to accept a proton from the carboxylic group. Therefore, they do not form zwitterions easily, whereas glycine lacks a benzene ring, allowing proton transfer between groups to form the stable zwitterion.
In simple words: Resonance in aminobenzoic acids pulls electron density away from the amino group, preventing internal proton transfer.

Q9. Write short notes on - (i) Co-enzymes (ii) Prosthetic groups
Answer: (i) Co-enzymes: These are small, organic non-protein molecules (often derived from vitamins like thiamine, riboflavin, and niacin) that are loosely bound to the enzyme protein and can be easily separated by dialysis.
(ii) Prosthetic groups: These are also non-protein organic groups derived from vitamins (like biotin) that are tightly and permanently bound to the protein molecule by covalent bonds, separable only by careful hydrolysis.
In simple words: Co-enzymes are loosely attached helper molecules, while prosthetic groups are tightly and permanently bound.

Exam Tip: Highlight the strength of binding (loose vs. tight/covalent) to differentiate them.

Q10. The melting points and solubility in water of amino acids are generally higher than that of the corresponding halo acids. Explain.
Answer: Amino acids exist as dipolar zwitterions (\(\text{H}_3\text{N}^+-\text{CHR}-\text{COO}^-\)), creating strong electrostatic and dipole-dipole attractions between molecules, which results in high melting points and strong hydration in water. Halo acids lack this zwitterionic structure and intermolecular salt-like bonding, resulting in lower melting points and weaker water interactions.
In simple words: Amino acids have strong ionic-like zwitterion attractions, giving them higher melting points and greater water solubility than halo acids.

Exam Tip: Use the zwitterionic structure and electrostatic attraction to explain both properties.