Biomolecules

Step 1: D and L Configurations.
The D and L notations of carbohydrates are based on the configuration of the asymmetric carbon atom farthest from the carbonyl group (the penultimate carbon).
- If the OH group on the penultimate carbon is on the right → D-configuration.
- If the OH group on the penultimate carbon is on the left → L-configuration.
Step 2: Example.
- D-Glucose and D-Fructose both belong to the D-series, meaning that the OH group on the penultimate carbon is on the right-hand side in their Fischer projections. Step 3: Structures.
D-Glucose (Fischer projection): D-Fructose (Fischer projection): Conclusion:
D and L configurations are based on the arrangement of the OH group on the penultimate carbon atom, and both D-Glucose and D-Fructose belong to the D-series.

Step 1: Primary structure.
- The primary structure of proteins refers to the linear sequence of amino acids in a polypeptide chain, joined by peptide bonds. - This sequence determines the unique characteristics and function of the protein. - Even a single change in the amino acid sequence can alter protein activity (e.g., sickle-cell anemia caused by a single amino acid substitution in hemoglobin). Step 2: Secondary structure.
- The secondary structure refers to the regular folding or coiling of the polypeptide chain due to hydrogen bonding between the C=O and N–H groups of peptide bonds. - Two major forms are: \begin{enumerate} \item
Alpha-helix: Right-handed coiled structure stabilized by intramolecular hydrogen bonds. \item
Beta-pleated sheet: Zig-zag structure where polypeptide chains lie side by side, stabilized by intermolecular hydrogen bonds. \end{enumerate} - These structures provide stability and define the shape of proteins. Conclusion:
The primary structure gives the amino acid sequence, while the secondary structure defines local folding patterns such as alpha-helices and beta-sheets, both crucial for protein function.

Step 1: Recall DNA bases.
DNA contains four nitrogenous bases: Adenine (A), Thymine (T), Guanine (G), and Cytosine (C).

Step 2: RNA bases.
In RNA, uracil (U) replaces thymine and pairs with adenine.

Step 3: Analyzing options.
- (A) Adenine: Present in DNA.
- (B) Thymine: Present in DNA.
- (C) Uracil: Not present in DNA, found in RNA instead.
- (D) Guanine: Present in DNA.

Step 4: Conclusion.
Thus, uracil is not present in DNA.

Final Answer:

Step 1: Definition.
Denaturation of proteins is the process in which proteins lose their native three-dimensional structure without breaking peptide bonds. This leads to the loss of their biological activity.
Step 2: Causes of Denaturation.
Denaturation can be caused by heat, strong acids or bases, heavy metal ions, organic solvents, or radiation.
Step 3: Effects of Denaturation.
- Hydrogen bonds and other non-covalent interactions are disrupted.
- The protein unfolds, losing its specific shape and biological function.
- Example: When egg white (albumin) is boiled, it turns from transparent to opaque due to protein denaturation.
Conclusion:
Denaturation of proteins is an irreversible process in most cases, resulting in the loss of their natural activity and function.

The structural formula of fructose is: Fructose is a monosaccharide and a ketose sugar, meaning it contains a ketone group at the second carbon atom. Glucose, on the other hand, is an aldose sugar, which means it contains an aldehyde group at the first carbon atom. Both sugars are isomers, but they differ in the functional group present and their structural configurations. While glucose forms a six-membered ring in its cyclic form, fructose forms a five-membered ring.

Polypeptides are long chains of amino acids linked together by peptide bonds. Each peptide bond forms between the carboxyl group of one amino acid and the amino group of the next. The sequence of amino acids in a polypeptide determines its three-dimensional structure and its biological function. Polypeptides can vary in length from just a few amino acids to hundreds or thousands.
An example of a polypeptide is insulin, a hormone involved in glucose regulation. Insulin consists of two polypeptide chains: one with 21 amino acids and the other with 30 amino acids. These chains are linked by disulfide bonds, and they fold into a specific structure that allows insulin to bind to its receptor and regulate glucose metabolism in cells.

Step 1: DNA Base Pairs.
DNA contains four nitrogenous bases: adenine (A), thymine (T), guanine (G), and cytosine (C). Uracil (U) is found in RNA, not DNA.
Step 2: Conclusion.
Hence, the correct answer is (D) Uracil.

Step 1: D-Glucose.
It is an aldohexose (6-carbon sugar with an aldehyde group at C-1). Open-chain form: . In cyclic form, it forms a pyranose ring.
Step 2: D-Fructose.
It is a ketohexose (6-carbon sugar with a ketone group at C-2). Open-chain form: . In cyclic form, it forms a furanose ring.
Step 3: Key Difference.
Glucose has an aldehyde (aldohexose), Fructose has a ketone (ketohexose).

(i) Secondary structure of proteins.
Formed by hydrogen bonding between peptide chains. Two main types: -helix (spiral) and -pleated sheet (zig-zag). Gives stability and shape to proteins.
(ii) Peptide bond.
A covalent linkage formed between carboxyl group of one amino acid and amino group of another. Responsible for formation of polypeptides and proteins.
(iii) Monosaccharides.
Simplest carbohydrates (C H O ). Cannot be hydrolysed further. Examples: glucose, fructose, galactose. They are reducing sugars.

Step 1: Bases in RNA.
RNA contains Adenine (A), Guanine (G), Cytosine (C), and Uracil (U).
Step 2: Absence of Thymine.
Thymine is present only in DNA, where it pairs with Adenine. In RNA, Uracil replaces Thymine.
Step 3: Conclusion.
Therefore, the base not present in RNA is
Thymine.

Step 1: Definition.
Electrophoresis is the process of movement of charged particles under the influence of an electric field. It is commonly observed in colloidal solutions where particles carry either positive or negative charges. Step 2: Principle.
When an electric potential is applied across the colloidal solution, the charged colloidal particles migrate towards the electrode of opposite charge: - Positively charged particles ( ) move towards the cathode. - Negatively charged particles ( ) move towards the anode. Step 3: Applications.
Electrophoresis is widely used in: \begin{enumerate} \item
Separation of proteins and nucleic acids: In biochemical and medical laboratories, gel electrophoresis is used to separate DNA, RNA, and proteins. \item
Determining charge on colloids: The direction of movement helps identify whether the colloid is positively or negatively charged. \item
Purification processes: Used to remove impurities from colloidal systems. \end{enumerate} Conclusion:
Electrophoresis is a fundamental technique to study and separate charged particles in a colloidal system under the influence of an electric field.

Step 1: Definition of disaccharide.
A disaccharide is a carbohydrate formed when two monosaccharides are joined by a glycosidic bond.
Step 2: Example.
The most common disaccharide is sucrose, formed from glucose + fructose.
Step 3: Molecular formula.
The molecular formula of sucrose is C H O .
Step 4: Conclusion.
Thus, one disaccharide example is sucrose with formula C H O .

Step 1: Structure of DNA and RNA.
DNA (Deoxyribonucleic Acid) and RNA (Ribonucleic Acid) are nucleic acids but differ in sugar, bases, structure, and function.
Step 2: Key differences.
\begin{tabular}{|c|c|c|} \hline Feature &
DNA &
RNA
\hline Sugar & Deoxyribose & Ribose
\hline Bases & A, T, G, C & A, U, G, C (U replaces T)
\hline Strands & Double-stranded helix & Single-stranded
\hline Function & Genetic material, stores information & Protein synthesis (mRNA, tRNA, rRNA)
\hline Location & Mostly in nucleus & Nucleus and cytoplasm
\hline \end{tabular} Conclusion:
DNA stores and transmits genetic information, while RNA helps in protein synthesis and gene expression.

Step 1: Understanding protein structure.
Proteins have four levels of structure: primary (sequence of amino acids), secondary (α-helix, β-sheet), tertiary (3D folding), and quaternary (multiple chains). Step 2: What is denaturation?
Denaturation is the process in which the secondary, tertiary, and quaternary structures of proteins are destroyed, while the primary structure (amino acid sequence) remains unchanged. Step 3: Causes.
Denaturation occurs due to heat, acids, alkalis, organic solvents, or heavy metals that disrupt hydrogen bonds, ionic bonds, and hydrophobic interactions. Step 4: Example.
The most common example is the coagulation of egg white (albumin) when boiled – it turns from soluble to insoluble solid. Step 5: Conclusion.
Hence, denaturation destroys biological activity of proteins by altering their shape.

Step 1: Nature of amino acids.
Amino acids contain both an acidic group (–COOH) and a basic group (–NH ). Step 2: Behaviour in water.
In aqueous solution, –COOH donates a proton forming –COO , while –NH accepts a proton forming –NH . Step 3: Dipolar form.
Thus, the molecule carries both positive and negative charges simultaneously, existing as a zwitterion. Step 4: Example.
For glycine: Step 5: Conclusion.
At isoelectric point, the amino acid exists mainly in zwitterionic form with no net charge.

Step 1: Structural role.
Proteins provide support and strength to body structures (collagen in connective tissues, keratin in hair, nails, skin). Step 2: Catalytic role.
Proteins act as enzymes that catalyse biochemical reactions (e.g., amylase, protease). Step 3: Transport and storage.
Some proteins carry vital substances – haemoglobin transports oxygen, myoglobin stores oxygen in muscles. Step 4: Defence mechanism.
Antibodies are protein molecules that protect the body against infections. Step 5: Hormonal regulation.
Certain hormones like insulin and glucagon are proteins that regulate metabolism. Step 6: Conclusion.
Proteins are indispensable biomolecules for structural, catalytic, protective, transport, and regulatory functions.

Step 1: Understanding amino acids.
Amino acids contain both an amino group (-NH ) and a carboxyl group (-COOH). They are the building blocks of proteins.

Step 2: Protein structure.
Proteins are long chains of amino acids linked by peptide bonds. Each protein has a unique sequence of amino acids that determines its structure and function.

Step 3: Analyzing the options.
- (A) Carbohydrates: Built from monosaccharides, not amino acids.
- (B) Proteins: Correct. Proteins are polymers of amino acids.
- (C) Lipids: Composed of fatty acids and glycerol, not amino acids.
- (D) Vitamins: Organic micronutrients, not polymers of amino acids.

Step 4: Conclusion.
Therefore, amino acids are structural units of proteins.

Final Answer:

Denaturation disrupts the secondary and tertiary structures of proteins due to heat, acids, or chemicals, causing loss of function.

Step 1: Definition:
Nucleotides in nucleic acids (DNA/RNA) are linked by phosphodiester bonds between sugar and phosphate groups. Step 2: Why not other options:
- Peptide bond: Found in proteins.
- Glycosidic bond: Links sugars in carbohydrates.
- Hydrogen bond: Stabilizes base pairing in DNA.

Fructose is considered the sweetest of all the common sugars. It is a monosaccharide that is sweeter than glucose, sucrose, and lactose. Fructose is found in honey, fruits, and root vegetables, and is often used as a sweetener in foods and beverages.
- Glucose is sweet but not as sweet as fructose.
- Lactose is a disaccharide found in milk and has a less sweet taste than fructose.
- Sucrose (table sugar) is a disaccharide consisting of glucose and fructose but is not as sweet as fructose alone.
Thus, the correct answer is option (D) Fructose.

Tollen's reagent: Ammoniacal silver nitrate containing the complex ion (often represented as ); it oxidises aldehydes to carboxylates giving a silver mirror.
With glucose:
On acidification: (gluconic acid).

Step 1: Recall the composition of inulin.
Inulin is a naturally occurring polysaccharide composed mainly of fructose units, with a terminal glucose sometimes present.

Step 2: Hydrolysis of inulin.
On hydrolysis by dilute acids or enzymes, inulin breaks down to yield fructose as the major product.

Step 3: Analyze options.
- (A) Glucose → Incorrect, since inulin mainly yields fructose, not glucose.
- (B) Fructose → Correct.
- (C) Glucose and fructose → Incorrect, as this is the case for sucrose hydrolysis, not inulin.
- (D) Lactose → Incorrect, lactose is a disaccharide of glucose + galactose.
Step 4: Final Answer.
Hence, the hydrolysis of inulin gives fructose.

Two preparations of glucose:
(1) Hydrolysis of starch

(2) Hydrolysis of sucrose

From glucose to saccharic acid and gluconic acid:
(a) Gluconic acid (oxidise CHO only)

(b) Saccharic (glucaric) acid (oxidise CHO and CH OH)

Tollen's reagent: Freshly prepared ammoniacal silver nitrate containing the complex ion (often written as ). It oxidises aldehydes to carboxylates giving a silver mirror.
With glucose (an aldose):
(on acidification , gluconic acid).
With fructose (a ketose):
In alkaline medium fructose enediol glucose/mannose, hence it also reduces Tollen's reagent:

Proteins are large biomolecules made of amino acids linked by peptide bonds.
The function and properties of proteins depend on the arrangement of amino acids and their folding into higher structures.
Protein structure can be described at four different levels:
1. Primary Structure:
- The primary structure refers to the linear sequence of amino acids in a polypeptide chain.
- This sequence is determined by the genetic code (DNA).
- Even a small change in the sequence (mutation) can drastically alter the protein's function.
2. Secondary Structure:
- The secondary structure is the regular folding or coiling of the polypeptide chain due to hydrogen bonding between the carbonyl oxygen and the amide hydrogen of the peptide backbone.
- The two most common patterns are:
(a) -helix (coiled structure)
(b) -pleated sheet (zig-zag structure).
3. Tertiary Structure:
- The tertiary structure refers to the overall three-dimensional folding of a polypeptide chain.
- This folding occurs due to interactions among side chains (R-groups), such as:
- Hydrogen bonds
- Ionic interactions
- Hydrophobic interactions
- Disulfide bonds (covalent).
- The tertiary structure determines the biological activity of the protein (e.g., enzyme active site).
4. Quaternary Structure:
- The quaternary structure arises when two or more polypeptide chains (subunits) associate together to form a functional protein.
- Examples include hemoglobin (composed of four polypeptide subunits).
- This structure is stabilized by the same interactions as tertiary structure but involves multiple polypeptide chains.
Thus, proteins exhibit a hierarchical structure: primary → secondary → tertiary → quaternary, each level being essential for their stability and biological function.

Property 1: Reduction of Fehling's solution.
Glucose is an aldohexose containing an –CHO group. It reduces Fehling's solution (alkaline Cu ions) to red cuprous oxide (Cu O).
Property 2: Reduction of Tollens' reagent.
Glucose reduces Tollens' reagent (ammoniacal silver nitrate) to metallic silver (silver mirror).
Note:
These reactions confirm that glucose contains a free aldehyde (–CHO) group in its open-chain structure.

Step 1: Recall the nature of glucose.
Glucose is an aldohexose (contains an aldehyde group at C-1). However, in aqueous solution, glucose predominantly exists in cyclic hemiacetal form (pyranose or furanose), in which the aldehyde group is not free but can open to show reducing properties.

Step 2: Tests for aldehyde group.
- Tollen's test: Glucose gives Tollen's test (silver mirror test) because it is a reducing sugar.
- Fehling's test: Glucose also gives Fehling's test, producing a red precipitate of Cu O.
- Schiff's test: This test is generally not given by glucose. Schiff's reagent detects free aldehydes that immediately react to give a magenta colour. Since glucose mainly exists in cyclic form and its aldehyde group is not freely available, it gives only a very weak or negative Schiff's test.

Step 3: Conclusion.
Thus, among the given options, glucose does not give Schiff's test effectively.

(i) Zwitter ion:
A zwitter ion is a dipolar ion that contains both a positive and a negative charge within the same molecule but is overall electrically neutral. Amino acids in aqueous solution commonly exist as zwitter ions, where the amino group ( ) is protonated to and the carboxyl group ( ) is deprotonated to . (ii) Peptide bond:
A peptide bond is a covalent bond formed between the group of one amino acid and the group of another with the elimination of a water molecule (condensation reaction). It has partial double bond character due to resonance, making it rigid and planar. (iii) Primary structure of protein:
The primary structure of a protein is the linear sequence of amino acids in a polypeptide chain, linked together by peptide bonds. This sequence determines the higher levels of protein structure and ultimately governs its biological function. Example: the exact order of amino acids in insulin. (iv) Polysaccharides:
Polysaccharides are long-chain carbohydrates formed by the condensation of numerous monosaccharide units joined by glycosidic bonds. They can be storage polysaccharides (e.g., starch, glycogen) or structural polysaccharides (e.g., cellulose, chitin). They are usually insoluble in water and serve as energy reserves or structural materials in living organisms.

Nucleic acids are complex biomolecules that store and transmit genetic information in living organisms.
They are polymers of nucleotides, which consist of a nitrogenous base, a pentose sugar, and a phosphate group.
The two main types of nucleic acids are:
- DNA (Deoxyribonucleic acid)
- RNA (Ribonucleic acid)
Two important properties:
1. Genetic Information Storage: DNA carries hereditary information that controls the development, functioning, and reproduction of all living organisms.
2. Self-Replication and Protein Synthesis: Nucleic acids can replicate themselves and guide protein synthesis through transcription (DNA → RNA) and translation (RNA → protein).
Thus, nucleic acids are essential molecules of life.

Step 1: Recall what Fehling's test detects.
Fehling's solution test is used to detect reducing sugars. Reducing sugars have a free aldehyde group ( ) or a free ketone group ( ) that can be oxidized. On heating with Fehling's solution, such sugars reduce Cu ions to Cu O, producing a red precipitate.

Step 2: Analyze each option.
- Glucose is an aldohexose and a reducing sugar. It contains a free aldehyde group that can reduce Fehling's solution, giving a positive test (red precipitate of Cu O).
- Sucrose is a non-reducing sugar because the glycosidic bond involves the reducing groups of both glucose and fructose units, so it does not give a positive Fehling's test.
- Fats are esters of glycerol and fatty acids; they do not have free aldehyde or ketone groups, so they do not respond to Fehling's test.
- Proteins contain amino acids linked by peptide bonds; they do not act as reducing sugars, hence no reaction with Fehling's solution.

Step 3: Final Answer.
Only glucose gives a positive Fehling's solution test.

Step 1: Recall glucose structure.
Glucose (an aldohexose) has 6 carbons: one aldehyde group (–CHO) at C1, four secondary –OH groups (on C2, C3, C4, C5), and one terminal –CH OH (primary alcohol) at C6.
Step 2: Identify primary alcohol.
Only carbon 6 has –CH OH group, which is primary alcohol.
Step 3: Conclude.
There is exactly one primary alcoholic group in glucose. Final Answer: % Quciktip

(i) -Hexane from glucose
(Often shown via sorbitol: ).
(ii) Pentaacetyl glucose
(iii) Glucose cyanohydrin
(iv) Gluconic acid

Ordinary ketones are not readily oxidised by Fehling's or Tollen's reagents, hence they are non-reducing.
Fructose (a ketohexose), however, in alkaline medium undergoes enediol formation and the Lobry de Bruyn–van Ekenstein rearrangement to give the corresponding aldoses (glucose/mannose), which are reducing and therefore reduce (Tollen's) and (Fehling's).

DNA (Deoxyribonucleic Acid) and RNA (Ribonucleic Acid) are both nucleic acids but differ in their structural and functional characteristics. Below are the key differences:
Structural Differences:
1. Sugar Molecule:}
- DNA contains deoxyribose sugar, which lacks one oxygen atom (at the 2' carbon) compared to ribose.
- RNA contains ribose sugar, which has a hydroxyl group (-OH) attached to the 2' carbon atom.
2. Strand Type:}
- DNA is typically double-stranded, forming a double helix structure.
- RNA is usually single-stranded and can form various structures like hairpins.
3. Nitrogenous Bases:}
- DNA contains the bases adenine (A), thymine (T), cytosine (C), and guanine (G).
- RNA contains the bases adenine (A), uracil (U), cytosine (C), and guanine (G). Note that RNA uses uracil instead of thymine.
4. Helical Structure:}
- DNA has a right-handed double helix structure.
- RNA is single-stranded and may adopt various secondary structures.
5. Length and Stability:}
- DNA is generally longer and more stable than RNA, designed to store genetic information.
- RNA is usually shorter and less stable, designed for transient functions in the cell.
Functional Differences:
1. Genetic Material:}
- DNA carries the genetic blueprint for the synthesis of proteins and is involved in the storage and transmission of genetic information.
- RNA plays a role in transcribing the genetic code from DNA and is involved in the synthesis of proteins (as mRNA, rRNA, and tRNA).
2. Types of RNA:}
- DNA does not have different types but is used in the formation of mRNA, rRNA, and tRNA in RNA transcription.
- RNA has several types: mRNA (messenger RNA), rRNA (ribosomal RNA), and tRNA (transfer RNA), each playing a role in protein synthesis.
3. Replication and Transcription:}
- DNA is self-replicating, meaning it can duplicate itself during cell division.
- RNA is transcribed from DNA and is not capable of replication itself but can be synthesized through transcription.
4. Function in Protein Synthesis:}
- DNA stores the instructions for protein synthesis, while RNA is involved in the direct synthesis of proteins in the cytoplasm.

Definition:
A peptide bond is a chemical bond formed between two amino acid molecules when the carboxyl group ( ) of one amino acid reacts with the amino group ( ) of another, releasing a molecule of water ( ). It is an amide linkage ( ). Explanation with Example:
Consider the formation of a dipeptide Glycylalanine (Gly-Ala) from Glycine and Alanine.

• Glycine:
• Alanine:

Reaction: Here, the linkage connecting the two amino acid residues is the Peptide Bond.